Geomet for Colege Sdents
I. Martin Isaacs
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saacs, Martin, (date) Geomety for college students Martin saacs p cm ncludes index SBN 0534351794 (text) 1. Geometry Title 00056461 QA445 63 2000 516dc21
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Preface
his book is primariy intended for coege mathematics students who enjoyed high schoo geometry and who wish to ea more about the amaing properties of ines circles triange triangess and other geometric gures I n pparticuar articuar I hope that those who are are preparing preparing to become highschoo high schoo mathematics mathematics teachers teachers wi nd nd inspiration in spiration here here to hep them share their enthusiasm enthusia sm and enjoyment of geometry geometry with their own future future students utt why u why shoud sho ud anyone study geometry? geo metry? One reas reason on of course cou rse is that geometry geo metry and its descendant desce ndant trigo trigonometr nometry y are essentia ess entia toos in engineering architecture navigation and other other discipines hese practica appications however however surey sure y do not expain why it is that for for centurie centuriess geometry has been taught to amost every every student and why at east recenty a person who knew no geometry geometry was not considered to be propery edu until recenty cated I think there there are at east ea st two reaso reasons ns more important important than usefunes us efunesss that expain why geometry has been and shoud continue to be a part of the schoo schoo curricuum Since Eucid some 300 years ago geometry geometry has been taught as a deductive deductiv e science with theorems and proofs s a consequence, generations of geometry students have eaed how to draw vaid concusions from hypotheses and how to detect and avoid invaid invaid reasoning reasoning In other words by studying geometry geometry students students can c an e how to think are other subjects subjects that coud aso be used to teach deductive reasoning reasoning Of course there are but geometry is especiay eective because it seems to have the perfect baance of depth and concretenes s Many of the theorems that we prove prove in geometry are dee in the sense that that they they assert as sert something nonobvious nonobvious and sometimes even surprising hey are are concrete because students can easiy e asiy draw draw the appropr appropriat iatee trianges circes circes or whatever, whatever, and they they can see that what is aeged to happen actuay does appear appear to happen happe n Geometry is aso beautifu and some of its theorems are so amaing as to seem amost mracuou mracuouss In fact fact much the same s ame comment comment coud be made about most areas of mathematics but geometry is unique in that its miraces are visua so they can readiy be appreciated even by the uninitiated Surey one does not need a great dea of mathematica sophi s ophisti stication cation to marve marve at the fact fact that if a ine is drawn through each vertex vertex of any triange triange and if each of these these ines is i s perpendicuar to the opposite oppos ite side si de of the triange triange then these three ines a go through a common point One coud cou d argue that that the fundamenta fundamenta theorem theorem of cacuus cacuu s for exampe is equay equay amaing and beautifu beautifu but unfort unfortunatey unatey it is not possibe poss ibe to appreciate appreciate it without without rst studying cacuus cacu us ut ut the aesthetic vaue of geometry geo metry extends beyond beyon d the strikng statements of its theorems
PREACE
Many Many of the the proofs have their own subte and eegant beauty which whic h unfortunatey is a itte harder harder to perceive. Nevetheess Nevetheess we expect that that most readers of this this book wi ea to enoy the beauty of o f the the proofs proofs as a s we as that of the theorems theorems.. In shot we shoud continue to study and teach geometry because it is a highy attracti attractive ve subect and because b ecause we can c an ea from from it something about deductive reasoning and the nature of mathematica mathem atica proof. It seems se ems cear ce ar therefoe therefoe that as in the past students shoud shou d continue to see theorems theorems being proved in their their geometry geometry cass cas s they should sh ould be taught how to understand proofs and perhaps even more impotant they shoud shou d lea how to invent and write proofs. I have seected se ected for this book b ook some so me of the the more spectacuar theorems in pane ge ometry ometry and I have presented ustications usti cations of thes thesee facts facts using u sing a vaiety of diere dierent nt techniques of proof. Mosty ignored however are the kind of unsurprising theorem which whie required by mode standards of mathematica rigor can seem rather pointess to students students It requires considerabe sophistication to appreciate why why anyone woud want to prove a fact fact that that seems competey competey obvious obviou s . Even professiona professi ona mathe matcians, most of whom do understand the the signicance signi cance of these resuts often nd their their forma forma axiomatic proofs proo fs somewhat du earning proofs proofs shoud not be an unewarding chore instead we expect that students wi demand proofs proofs because becaus e the assertions being estabished are otherwis otherwisee so incredibe. is book was written as a text for oege Geometry which is a course that I have taught severa times at the niversit of Wisconsin Madison. The courses principa audience consi c onsists sts of o f sophomore and unior unior undergradua undergraduate te math maors who are speciaiing in secondary s econdary education. For them the course is required but there there is aso a so a substantia sub stantia minority who take the course eectivey eectivey Some S ome of these students s tudents simpy want to earn geometry whie others take oege Geometry because they nd it to be a more gente gente and accessibe acces sibe introduction to to mathematica proof proof than than a course cours e in abstract abstract agebra or advanced cacuus. cacuus . It is my beief bei ef that for some students s tudents the study of geometry geometry is i s an exceent preparation for these these more difcut difcut abstract courses . I have been dissatised with the avaiabe texts that might be used for suh a course. course . Many Many incude in cude a umber umber of interesting topics topics within a arge arge samping s amping of assorted geometric materia. ut they do not put the focus where I think it beongs: on the reay pretty theorems and their proofs proofs that in my opinion opini on shoud shou d be at the heat heat of a geometry course for coege students. so they often devote much more space than I think think is appopriate appopriate to fomaism fomaism and axiomatics Most Mo st students probaby do not nd this especiay esp eciay exciting excitin g and I share that opinion. here aso exist exi st some s ome wonderfu wonderfu books that that are ed with spectacuar s pectacuar theorems theorems and eegant proofs but none of these seems quite suitabe as a text for this geometry course either I have found that that most mos t students who register for for oege Geometry Ge ometry claim to remember remember very itte from from their one previous previous exposure to geometry in high schoo. schoo . For the sake of this maority some review and accimatiation are necessary. text is needed therefore that starts from a point cose to the beginning and introduces the notion of proof genty and then then gets to the the good stu" stu " as quicky as pos sibe. sibe . ecause ecause I coud not no t nd a book b ook that covered the right materia materia at the right pace pac e and an d that stated stated from the right pace pac e I taught the course cou rse many times without a text.
PREACE
Most of my students seemed to enjoy the course and many of them became very excited about about geometry. In this this book which is an expansion expansion of my course lecture notes I have have tried tried to reproduce as closely clos ely as pos sible the experience experience of the classro cla ssroom om and so so I hope that my readers readers will wi ll also als o nd that geometr geometryy is an enjoyable enjoyable and an d exciting sub s ubject ject Most of all I hope that those of my students and readers readers who wh o are or will be teaching highschoo high schooll mathematics will wil l convey some of that excitement to their their own students students I am grateful to the follo following wing reviewers for their their helpful comments comme nts:: Fred Flene, Notheaste llinois Univesity Aa Hoe Uivesity of Caifoia Ivne Kathyn Lenz, Uivesity of Minesota, Duuth David Poole, Tent Univesity Ron Soomo, Ohio State Univesity Lay Sowde, San Diego State Univesity Alex Turull Univesity of Florida, Gansville Jeane Wald Michigan State Uivesity
I Martin Isaacs
Contents
CHPER 1
1 1 E 1F G 1
Introduction and poogy 1 ongruent rianges rianges 5 nges and and arae arae ines 1 1 araeograms 14 rea 18 irces and rcs rcs 23 oygons in irces irces 34 Simiarit Simi arityy 39
CHPER 2
2 2 2 2 2E 2F 2G 2
5
he ircumcirce 50 he entroid entroid 5 he h e Euer ine, Orthocenter Orthocenter and and Nineoint irce irce 0 omputations 7 he Incirce Incirce 7 3 Exscribed irces irces 8 0 Moreys Moreys heore heorem m 82 Optimization Optimization in rianges rianges 85
CHPER 3
3 3 3 3
4
Simson ines 94 he uttery uttery heorem 105 ross ross atios 10 he adica xis 119
CONTENTS
CHPER 4
v' v 4 4 4 4
eva evass heorem heorem 1 25 Interior Interior and and Exterio Exteriorr evians evians 1 3 1 eva evass heorem heorem and nges 1 3 Meneaus Meneaus heore heorem m 14
CHPER 5
56
V M f f 5 5 5 5 5E 5F
ectors 15 ectors and Geometry 158 ot roducts roducts 1 3 heckerboards 1 it it of o f rigonometr rigonometryy 170 inear Operators 172
CHPER 6
E F
ues of the Game 1 82 econstructing econs tructing rianges rianges 1 87 angents 191 hree ard ard robems 1 9 onstructibe ons tructibe Numbers 203 hanging the ues 208
6
5
GEOMETRY or f
College Students College
CHPER ONE
Te Basics
A
ntroduction ntroduction and Apology
Elemen ts, it has been Since the appearance about about 2300 years years ago of Eucids book b ook The Elements, tradtiona to study geometry as a deductive discipine, using the socaed axiomatic method. Ideay this means that the geometer (or geometry student) starts with a few assumed as sumed facts facts caed ca ed axioms or postuates postu ates and then systematicay system aticay and and carefu carefuyy derives the entire entire content of the subject subject using nothing but pure ogic. ogic . he he axiomatic method requires requires that this content, which for geometry is a coection c oection of o f facts facts about trianges circes circes and other gures shoud be presented as a sequence sequen ce of ever ever deeper deeper theorems theorems each rigorousy rigorou sy proved using usi ng the axioms axio ms and earier earier theorems . Geomet Geometry ry has continued to deveop in the centuries centuries since si nce Eucid. Eucid . Mathematicans Mathematicans incuding many taented amateurs amateurs have discovered a weath of beautifu, beautifu, surprising and even spectacuar properties properties of geometric gures gures and these these together together with the geomet geo metry ry Element s, constitute what is today of The Elements, today caed cae d Eucidean geometry. geometry. mazing as many of these asserted facts may be we can be condent that they are correct because they are proved they are not merey tested by experiment and conrmed by measurement Furthermore ore we do not have to rey on the authority authority of o f Eucid or his and observation. Furtherm success succe ssors, ors, because becaus e with a itte practice and prepar preparation, ation, we can read and undertand undertand these proofs ourseves. With uck we might nd simper and more eegant proofs for some som e of the the known theorems of o f Eucidean Euc idean geometry geo metry.. It is even even possib pos sibe e that we might discover disc over new geometric facts that have never been seen s een before and invent our own proofs for them. geometry has resuted not ony in the discover disco veryy of o f amazing he deveopment of geometry facts, facts, but aso in the invention of new and powerfu powerfu techniques of proof proof and these too are considered consi dered part of Eucidean geometry. geometry. escartes escartes invention invention of coordinate geometry is an exampe of this. third thread thread in the history of geometry, geometry, especiay e speciay in mode times has been an investigation into the foundations foundations of the subject subject.. It has been asked for instance whether or not a of Eucids Eucid s axioms and postuates are reay vaid and whether or not we reay need to assume them a. ut we say amost nothing about these foundationa foundationa issues is sues in this this book. bo ok. Instead, I nstead, we concentrate concentrate on the two more cassica cass ica themes of geometry: facts and proofs. We oer the reader a seection of some of the
2
CHAPTER
THE BASCS
more striking facts of Eucidean geometr ge ometry y and we present enough proof proo f machinery machinery to estabish estabish them. In the centuries since Eucid the accepted standards standards of mathematica mathematica rigor have have su fciency ci ency of proofs proofs in the stye of Eucid has been chalenged. One changed changed and the suf objection for example is that Eucid reied too heaviy on diagrams and that he and the other cassica geometers did not aways prove facts they considered to be obvious from the gures. In this book we foow the tradition of Eucid and of most of his succes sors and so s o we are wiing to t o use some of the information information contained in caref carefuy uy drawn drawn gures g ures.. ut ut we must mus t not rey on diagrams for for certain certain other types of infoation, infoation, and unfort unfortunatey unatey it is i s difcult difcult to be precise about exacty what we are wiling to read read off off from a diagram and what requires a proof. We are are condent cond ent that readers wil catch on to this fairly fairly quicky quick y however how ever and we apoogie for the ambiguity. erhaps an example wi hep carify carify the situation. si tuation. In Figure Figure 1 . 1 ange ange bisectors A B , and C Z have been drawn in ABC hree facts that seem cear c ear in the gure are are : A
Figure 11
1 B < C In other words words line segment B is shorter shorter than ine segment C 2 . he point , where where the bisector of L A meets ine BC, ies on o n the ine segment BC 3 he three ange bisectors are concurrent. efore we discuss these observations et us be clear about the meanings of the technica words and notation. We assume that the nouns point, line, angle, triangle, and bisector are famiiar to readers of this book. wo ines, uness they happen to be parae parae aways meet meet at a point but if three or more ines a go through a common point, then something unexpected is happening and we say that that the ines are concurrnt he remaining undened technica word in the precedng ist of facts is segment line mnt is that part of a ine that ies between two given points on the ine. Observe that that in assertion assertion ( 1 ) we are are using the notation notation B in two dierent dierent ways: ways : In the inequality, B represents represents the length of the ine segment, se gment, which is a number and then ater B is used us ed to name the segment itsef. In addition the notation B is often used us ed to represent represent the entire ine containing the points B and , and not just ju st the segment s egment they determine. determine. ut we sha always provide enough infoation infoation so that it is cear c ear from from context which of the three three possible pos sible meanings meaning s is intended. For exampe in inequaities or equations it is obvious obvio usy y the numerica numeric a interpretation interpretation that is wanted. Each of ( 1 ) (2) ( 2) and (3) is true but we need to distinguish distingui sh among three di diere erent nt types of truth here. First note that the fact in (1) that B < C is an accident. his
A NTRODUCTON AND APOLOGY
inequality happens to be true in this gure, gure, but it is not an instance of some s ome genera generall or universal truth Even Eve n in a particular particular case, this sort of information can be be unreliable unrelia ble since sinc e it depends on the accuracy of the the diagram and the care with which measurements are made It is never considered con sidered valid to read read o an equality equalit y from from a diagram and it is rare that that one is justi ju stied ed in reading of o ff an inequality Fact (2) that point lies between between points B and on line B , is not not accidenta accidentall Indeed, it seems completely obvious that the bisector of each angle of an arbirary triangle must always intersect the opposite side of the triangle In other words, the bisector intersects the line segment determined by the other two vertices of the triangle triangle Note that that although althoug h this is true about angle bisectors bi sectors,, it can fail fail for other important lines associated asso ciated with a triangle triangle For example example the attud drawn from from A, which is the line through A perpendicular to line B , may not meet the segment B lthough this failure ailure (accidentally) (accidentally) does not happen happen for for AB A B of Figure 1 1 , it certainly certainly can happe happenn for other triangles triangles he obvious" obvious " fact that angle bisectors of triangles triangles meet the opposite sides is the sort of information that Euclid was, and we are willing to read off from diagrams lthough lthoug h it is a fact fact that for every triangle triangle,, the three angle bisec bi sectors tors are concurrent, co ncurrent, as in assertion ass ertion (3) we w e follow follow Euclid in that that we do not consider cons ider this this to be obvious he experimental evidence evid ence of o f even an accurate computerdrawn diagram (or of many many such diagrams) is not sufcient for us to accept this as a universal universal truth we require a proof proof eaders will probably have seen se en a proof of this theorem in highscho highs chool ol geometry, ge ometry, and it is also als o proved here here in hapter hapter 2 hus there is an a n inherent ambiguity ambiguity about which informa information tion can ca n be reliably estab lished from from diagrams and which cannot ecause ec ause of this ambiguity, modern standards of mathematical mathematical rigor require require that there there must be no reliance whatsoever on gures gure s Geome try without without diagrams diagrams was made possible by avid avid ilbert ilbert (1 821 82 1 943) who, building on the work work of his predecessors predeces sors constructed co nstructed an an appropriate appropriate set of precisely precise ly stated axioms from which he was able to prove everythi everything ng formall formallyy and without diagrams Euclid Eu clid s ideal that geometry geometry should shoul d be a purely deductive enter enterprise prise was thus nally nally realized by ilbert at the turn of o f the 20th century In particular ilberts ilbert s axioms allowed him to prove our observation (2), that an angle bisector bis ector of a triangle always meets mee ts the opposite oppos ite side sid e of the triangl trianglee ut ut it is far far from a triviality to prove rigorously, rigorously , and without relying on a diagram, that the bisector bise ctor of B A meets liline B at a point that lies between between B and Further Furthermore, more, there there would be no hope hop e of proving such a thing thing without having a precise denition of the word between, which whic h ilbert was was able to provide provid e researchers into the foundations foundations of geometry geometry he achievements of ilbert and other researchers were substantial and signic sig nicant, ant, but it is my opinion that, that, by far, far, the most interesting theorems of geometry are those that provide surprises We feel, for example, that the fact fact that the three angle bisectors of a triangle are always concurrent is much more exciing than is the fact fact that the the angle bisectors bis ectors meet the opposite sides s ides of the the triangle For his reason we have chosen chose n to present in this book as many unexpected facts facts and surprisin surprisingg theorems theorems as space allows allows Since Sinc e we want to get to these these quickly quickly we must omit ilbert ilbertss careful and rigorous treatment of the foundations foundations of geometry, geometry, and instead, we will follow tradition and rely on diagrams without specif spe cifying ying exactly exactl y the extent of our
4
CHAPTER
THE BASCS
reiance. so since the reader reader of this book wi have studied some geometry g eometry in high schoo schoo we wi not start our presentation presentation at the very very beginning of the the subject. We have aready apoogize apo ogizedd for the ambiguity abou a boutt how much mu ch information we are aowed to obtain obtain from from diagrams. Some apoogy is aso as o appropriate appropriate concerning the the issue is sue of how much highschoo geometry we are assuming. Students may fairy compain when they are doing homework exercises exercises that it is uncear uncea r which facts facts they must prove and which they can c an merey quote as remembered from from schoo s choo . We do not attempt to give a compete ist of assumed resuts resu ts but we sha s ha show by exampe the eve of proof that we expect and we sha devote much of the rest of this chapter to a review of some of the essentia facts denitions and theorems from from highschoo highschoo geometry. efore proceeding with our review of highschoo geometry we discuss briey some of the issues in the foundations of geometry to which we referred earier We mentioned that Eucid caed his unproved assumptions axioms and postuates. he distinction, which is not considered signicant today today is that Eucids Eucid s axioms ax ioms conceed conc eed genera ogica reasoning, whie his postuates were more specicay geometric. For exampe one of Eucid s axioms is things equa equa to the same thing are are equa to each other" other" whereas his parae postuate ess entiay asserts that given a ine and a point not on that ine there exists exis ts one and ony one ine through the given point parae to the given ine." ctuay ctuay Eucid s parae postuate postuate is not stated in precisey precisey this way it appears appears in a somewhat so mewhat more compicated but ogicay equivaent formuation. formuation. Over the centuries, Euclids Euclid s parae postuate has engendered a great dea of interest and controversy. controversy. S omehow the existence and uniqueness of a ine ine parae to a given ine through through a given point seemed see med ess obvious than the facts facts asserted as serted by the the other postuates. Geometers fet uncomfortabe unc omfortabe assung assu ng the parae postuate pos tuate and they attempted instead to prove it they tried to deduce it from from the rest of Eucids Eucid s axioms and postuates efore efore we discuss discu ss these attempts we w e shoud shou d stress that the the parae parae postuate makes two separ s eparate ate assert ass ertions ions each of which woud have to be proved. proved. It woud be neces nec essary sary to show that a parae ine (through the given point) exists, and one woud aso as o need to prove that it is unique ow might a proof of the parae postuate proceed? proceed? One coud ass assume ume that it is fase fase and a nd then try try to derive derive some s ome contradictory contradictory concusions concusion s ssume s sume for exampe exampe that there there is some s ome ine A B and some point P not on AB, and there there is no ine parae to A B through P If by by means of this assumption a ssumption one coud deduce the existence o f a triange triange �Z for which Z and aso < Z, then this contradiction woud prove at east the existence part of the parae parae postuate. What actuay happened when this was tried was that apparent contradictions contradictions were derived derived and the existence of gures that seem impossibe impos sibe was proved. For exampe the assumption that no ine parae parae to A B goes goe s through through point P yieds a triange triange �Z that has three right anges. anges . Surey Surey this is impossibe" imposs ibe" we might migh t say but how can we prove prove this imposs impo ssibiity ibiity?? We We know from from high schoo that the three three anges of a triange triange sum to 1 80 and an easy experiment conrms conrms this fact (ear o the three coers of a paper triange and ine them up to see that the three three anges tota tota a straight ange.) ange. ) It may seem that by proving the existence of a triange with with three rght anges, we have have the desired contradiction contradiction but this is wrong. he highschoo highs choo proof that that L + L + L Z 1 80 utimatey reies reies on the parae parae postuate which we are temporariy temporariy refusin refusingg to assume. as sume. so, so , the experiment with paper trianges trianges is certainy not a mathematica proof.
B CONGRUENT TRANGLES
fter fter repeated attempts to obtain obtain contradiction con tradictionss from from the denia den ia of o f either either the ex istence or the uniqueness part of the parae postuate it was reaized by J oyai N. obachevski and Gauss in the 19th century that athough such denias yied seemingy ridicuous ridicu ous situations situations such suc h as trianges trianges with three right anges anges no proof of a contradiction contradiction was possibe pos sibe.. In fact fact it was proved that no such such proof is possib pos sibe.e. In other words one coud c oud buid a perfecty perfecty consistent consi stent deductive deductive geometry geometry by repacing Eucid s parae parae postuate po stuate with wi th either one one of o f two ateative ateative new new postuates. po stuates. One of o f these denies o f a parae to some s ome ine through through some point poi nt and the other asserts the the existence of existence of at east two such suc h paraes. paraes . Each of the two two types of geometry that arise in this way is said to be nonEucidean and each has its own set s et of proved theorems. theorems. he geometry where where no parale exists exis ts is i s caed eiptic geometry, geometry, and when more than one parae to a ine goes goe s through a point we have hyperboic hyperboic geometry. he deductions of each of the two types of nonEucidean geometry contradict contradict each other and they aso contradict contradict the theorems of cassica cass ica Eucidean geometry geometry but each of the three three avo avors rs of geometry geometry appears to be inteay consistent consi stent and from from a mathematician mathematician s point p oint of view view they are equay vaid. ctuay it is known that if Eucidean geometry is internay consistent then the two nonEucidean geometries are aso consistent but no forma proof proof of the the consi co nsistency stency of Eucidean geometry has been found. he questio que stionn of which if any any of the three three avors of geometry that we have been discussing describes the rea word is interesting, but it is not reay a mathematica question question it beongs beon gs in i n the ream ream of physics. physics . he itte experiment experiment with the paper triange certainy suggests that we ive in a Eucidean universe but this has not been rmy estabished on a arge arge scae. sc ae. Indeed modern theories of the structure structure of the cosmos cos mos incuding Einstein Eins teinss theory of genera reativit reativity y suggest sugge st that none of the three three geometries provides an entirey accurate description of the universe in which we actuay ive. Nevert Neverthees heesss Eucidean Euci dean geometry provides a very good approximation to reaity on a human human scae sc ae and so it is usefu u sefu for for practica purposes such s uch as engineering e ngineering navigation, and architecture.
IB
Congruent Con gruent iangles
We devote the rest of this chapter to some s ome of the basic facts facts and a nd techniques of Eucidean geometry geometry much of which wi be a review for for most readers. eca that two geometri g eometricc gures are conrunt if informay speaking they have the same size and shape. S omewhat more preci precisey sey two gures gures are congruent congruent if one can be subected to a rigid rigid motion so s o as to make it coincide with the other. other. y a rigid motion motion we mean a transation transation or shift a rotation in the pane p ane or a reection in a ine . he atter can be viewed informa informay y ifting the gure gure from from the pane ipping it over and pacing it back in the pane pan e as ifting In Figure 1.2, for exampe the three trianges are congruent athough to make Z coinc co incide ide with either of the other two triang trianges es it is neces nec essary sary to ree reect ct it or ip ip it over. We write ABC �RST to report that that the rst two trianges in Figure 1 are congruent congruen t but note that there is more to thi thiss notation than may at rst be apparent. he ony way that these two trianges can be made to coincide is i s for point R to coincide with point A , for S to coincide with B , and for T to coincide with C We say that A and R B and S, and C and T are corrpondin point of these these two congruent trianges trianges.. he ony
6
CHAPTER
THE BASCS z R s
T
Figure 1.2
correc correctt ways to report report the congruence is to ist corresponding points in corresponding positions posi tions It is correct therefore therefore to write ABC RST or BAC SRT but it is wrong to write ABC S RT his atter atter assertion is not true because there is no way that that these two trianges trianges in Figure 1 2 can be made to coincide with A and S, B, and R , and C and T being corresponding points points Since RST Z, it is cear cear that corresponding corresponding sides of thes thesee trianges have equa ength and that corresponding correspond ing anges have equa measure (contain (con tain equa numbers numbers of degrees or radians) radians ) We can thus write for exampe R S and LSRT L Z Note that in this context, co ntext, the notation L S RT refers refers to the measure o f the ange in some convenient units units such as degrees or radians radians In other other situations situations however we may write write S R T to refer refer to the the ange ang e itsef its ef his hi s is entirey entirey anaogous anaog ous to t o the fact fact that R S can ca n refer either to a ine segment or to its ength in centimeters inches mies or whatever In some geometry books books the notation mLSRT is used us ed to denote the the measure measu re of LSRT Whenever Whenever we know that two trianges trianges are congruent congruent we can deduce six equaities three three of engths and three three of measures of anges anges It is aso as o reasonaby ovious that given given two trianges trianges,, if a six equaities hod, then the the trianges trianges can be made to t" one on top of the other other and they are congruent con gruent s readers readers of this book boo k are surey aware it is not necessary neces sary to know a six equaities to concude that two trianges trianges ae a e congruent If we ow, for exampe that the three sides side s of one triange equa respectivey, the three corresponding sides of the other triange we can safey deduce that the trianges are congruent cong ruent If we know, know, for exampe that A B RS, AC RT, and B C ST, we can concude that ABC RST In a proof proof where where each assertion ass ertion must be justied ju stied we say that these two trianges are congruent by SS S S S " he abbre abbreviation viation SS S S S which stands for sideside side sideside side"" refers refers to the theorem that says that if the the three sides of a triange triange are equa in ength to the three corresponding corresponding sides of o f some other triange triange then the two trianges trianges must be congruent co ngruent Other vaid criteria that can be used to prove that two trianges are congruent are SS S and S hese of course are abbreviations for sideangeside" angesideange" and sideangeange" respectivey In the expectation that these are entirey famiiar to readers of this book we iustrate ony one of them with an exampe In Figure 12 if we somehow know that S T Z and that L S L and L R L, we can write in a proof We concude by S that SRT Z. (Notice (Notic e that we we wrote as a short form form for for LRST, which is the fu fu name of this this ange ange his is acceptabe accepta be when it cannot resut in ambiguity) ambiguity ) What What is the ogica status of the four congruence criteria criteria S S S SS S S S, S , and S? For each of these these the fact that the criterion criterion is sufcient sufcient to guarant gu arantee ee the congruence of of
B CONGRUENT TRANGLES
7
two triagles triagles is actually a theorem, proved proved by Euclid from from his postlat pos tlates es These The se four four theorems are amog amo g the the basic bas ic results resul ts that we are are acceptig as kow ko w to be valid ad that we are willig to use without providig proofs proofs I fact, however, however, it is ot hard to prove prove the sufciecy of some s ome of these criteria if we are willig to accept some of o f the others s a example, example , ad as the rst proof proof that we we actually preset i this book, bo ok, let us deduce the sufciecy sufciecy of the SSS S SS criterio, criterio, with the uderstadig uderstadig that we may freely freely use ay of the the other three triaglecogruece triaglecogruece coditios ssume that i ABC ad R, we kow that AB R, A C R , a ad BC rov rovee tha thatt A B C R without without usig the the SSS SS S cogruece critero critero
(1.1) PROBLEM.
To do this, this , we w e shall shal l appeal to aother theorem that that readers readers surely s urely remember from from thei previous study of geometry: The base agles of a isosceles triagle are equal ecall that a triagle U V W is ioc if two of its sides have equal legths I Figure 1.3 for example, example, the triagle triagle is isosceles isos celes because U V UW. The third side V W occ urs at the bottom of the the is called the ba of the triagle, whether or ot it actually occurs diagram The ba an of a a isosc is osceles eles triagle are the the two agles at the eds of the the base, base , ad the theorem theorem asserts that they they are ecessarily equal equal I Figure 1.3 therefore, we have U V W U W V. We metio that this baseagles theorem for isosceles triagles triagles is used so s o ofte ofte that it is give a ame: the pon ainorum This ati phrase meas bridge of asses " pparetly, pparetly, the theorem theorem has acquired this this ame partly because the diagram i The Elements used i its proof vaguely resembles a bridge bridge
Figre 13
y reamig the poits, poits , if ecessar ecess ary, y, we ca ass assume ume that that AC is the logest side of ABC, ad cosequetly, R is the logest side of R Sice Si ce we we are are give give that that AC R , we ca mov movee R , ippig it over over,, if ecessary, so that poits A ad R coicide, as do poits C ad , ad so that lie A C poits B ad lie o opposite sides of lie Now draw lie segmet segm et B What must result is a situatio si tuatio resembli res emblig g the left left diagram i Figure 14. It is ot possibl pos siblee for B to fail to meet A C , as i the ight F igure 1.4 or for B to go through oe of the poits A or C because diagram of Figure that would would require require oe of BC or BA to be loger tha tha A C (We (We are are sham s hamelessly elessly relyig o the diagram to see this ) We ca thus assume as sume that we are i the situatio of the left left gure g ure Sice A B R , we see that A B is isos is oscel celes es with base B , ad hece hec e by the pos asiorum, asi orum, we deduce that y , where, as idicated idic ated i the diagram, diagram, we are writig ad y to represet represet the measures measures of A A B ad R B , respectivel respectivelyy Simlarly, Simlarly, by a secod s ecod applicatio of the the pos asiorum, as iorum, we obtai u v ad it
oution to Probm 1.1.
8
CHAPTER
THE BASCS
A
C
R
T
T
s
s
Figre 1.4
follows by additio that that x + u y + v I other words, LABC R Sice Sice we already kow that A B R ad B C , we ca coclude coclude by SS that ABC R , as requ require ired d • Observe Ob serve that we have writte writte this proof i a coversatioal coversa tioal sty e, usi u sig g complete comp lete Eglish seteces s eteces grouped ito logical paragraphs paragraphs We eded by establishig establis hig what we set out to prove, ad we clearly marked the the ed of the the proof (It has become customar cu stomaryy for a box box to replace the oldfashioed oldfash ioed E as a edofproof marker) marker) This style s tyle,, with mior variatios, has become the accepted model for what a proof should look like throughout throughout most mos t areas of mathemat mathematics ics,, ad we follow it cosistetly cosi stetly i this book We expect studets to do the homework exercises with proofs writte i the same style, usig usi g complete seteces s eteces We do ot recommed the the twocolum proof format ormat that is ofte ofte required i highschool high school geometry geometry classes classe s Note that the secod setece of the last paragraph of the proof begis with the word similarly This Thi s word ca be a powerful tool for for simplif simpli fyig yi g ad shortei sh orteigg proofs proofs ad makig them them more itelligible itelligibl e ike most powerful powerful tools, however, however, this oe ca be dagerous dagero us if use improperly improp erly We ecourage ecour age studets to use the word similarly to avoid uecess uece ssar aryy repetitio i their proofs, proofs, but to use it carefully carefully lthough we expect that most readers of this this book remember how to prove the pos asiom, as iom, we preset more tha oe proof here to illustrate a few poits poits ad to provide provide furt further her models of proofwri proofwritig tig style styl e Our rst proof also als o yields some additioal iformatio iformatio about is osceles osc eles triagles triagles We remid the reader reader that a mdian of a triagl triaglee is the lie segmet j oiig a vertex to the the mdpoit of the opposite side s ide THEOREM et A B C be isosc isoscele eless with base BC Then L B L C Also Also the median from ve v e rtex A the bisec bis ector tor of of L A and the altitud a ltitudee from vertex v ertex A a re all a ll the same line
I Figure 1.5 we have draw the bisect bis ector or A of LA, ad thus LBA LCA y hypothesi hypothesis,s, we kow that that AB AC ad of course, A A Thus BA CA by SS S S It follows follows that L B L C sice these are correspodig agles i the coget triagles triagles We also eed to show that the agle bisector A is a media medi a ad that it is a altitude too To see s ee that it is a media, it sufces sufces to check that is the midpoit
Proof.
B CONGRUENT TRANGLES
of segmet B C, ad this is true because B X X C sice sic e these are correspod ig sides of our cogruet triagles Fi ally, to prove prove that AX is also a altitude, altitude, we must show that AX is perpedicular to BC I other words, we eed to estab lish li sh that B X A 90° . From the cogruet triagles, we kow that that the the correspodig agles agl es B X A ad C X A are equal, equal , ad thus BXA BXC 90° sice the the straight agle B X C 180°. •
9
A
B
C
Figre 1.5
To prove that the agle ag le bisector, media, ad a d altitude fro from m vertex vertex A are all the same, s ame, we started started by drawig oe of these lies (the bisector) ad showig that it was aso as o a media ad a altitude Sice there is oly oe media ad oe altitude from A, we ko ko that the bisector bi sector is the media media ad the altitude What would have happeed happee d if we had started by by drawig the media from A i istead stead of the bisector bis ector of A A ? This is, is , after after all, the same lie I this situatio, we could deduce that that B AX C AX by S SS SS It would would the follow ollow that that B AX C AX AX We We would deduce that AX is the agle bisector, ad everythig would proceed as before This appoach would have bee less s atisfacto atisfactory, ry, because it makes the pos asiorum deped o the SSS SS S cogruece co gruece criterio criterio I roblem 1 . 1 , however, however, we proved proved the the validity of of the S S S criterio usig the pos asiorum, ad this this would be a example of ivalid circular reasoig What if we had started by drawig alti A tude AX ? We We would the the kow kow that that B XA C X A sice si ce both of o f these are right agles agle s We also kow kow that that AB AC ad AX AX t this poit, we mght be tempted tempted to coclude that BAX C A X by SS, SS , but we we would would resist B D that temp temptat tatio, io, of course, course, because SS is ot a valid cogruece co gruece criterio To see why, cosider co sider Figre 1.6 Figure Figure 1 .. . . I this gure, the base base B C of isosce les ABC has bee exteded to a arbitrary poit D beyod C The two triagles ADC ad AD B are clearly clearly ot cogue coguett because because DB D C, ad yet the triagles triagles agree agree sidesideaglee sice AB A B AC, AD AD, ad ad D D i sidesideagl There is oe case where SS is a valid cogruece criterio: whe the agle is hypoteu sear arm" m" criterio, criterio, abbreviated abbreviated eca ecallll tha the a right agle This is the hypoteuse logest loges t side of a right right triagle triagle,, the side opposite oppo site the right right agle, is i s called the hypotnu of the triag triagle le The other two sides si des of the triagle are ofte ofte called its arm
two right triangles tri angles have equal eq ual hypoten hypotenuse usess and an arm a rm of of one of the triangles equals equ als an arm of of the othe othe then the n the triangles tri angles are congrue co ngruent nt
( 1 . 3) THEOREM.
10
CHAPTER
THE BASCS
B Figre 1.7
We are give triagles ABC ad D with right agles at C ad We kowthatAB = DadAC = D,adwewatto showthatABC D Move D , ippig it over over ifif ecessary eces sary,, so that poits A ad D ad poits C ad coic co icide ide ad the diagram diagram res resembles embles Figure 17 I Figure 17 we have B C = B C A + D = 90° + 90° = 180° ad thus B C is a lie segmet, which we ca ow call B Sice Sic e A B = D , we see that A B is isosceles isos celes with base base B Thus altitu altitude de AC is a media by Theor Theorem em 1 ad hece BC B C = The desired cogece cogece ow o w follows follows by S SS •
Proof.
We ed this sectio s ectio with with yet aother proof proof of the pos asiorum This Thi s oe o e is amusig ad very short, but it is somewhat tricky tricky ad should sho uld be read carefully carefully We are give give isosceles iso sceles A B C with with base bas e B C , ad we wat wat Sice also also A = A, to show show that that B = C We have AB = AC ad AC = AB Sice we ca coclude that ABC A CB by SS It follo follows ws that that B C sic sicee these are corres correspodig podig agles of the cogruet triagles triagles •
Proof of pon ainorum.
xercses B
rove the coverse of the pos asiorum Show, i other other words, that if B = C rove i A B C, th the AB = AC If the altitude from from vertex vertex A i A B C is also the bisector bise ctor of A , show s how that that A B = AC altitude from from ver verte texx A i A A B C is i s also a media, show that AB = AC If the altitude This fact fact will be used later, later, s o please do this this problem I Figure 18 medias B Y ad CZ have bee bee draw draw i isosceles isos celes A B C with with basee B C Show bas S how that that B Y = CZ assume me ow that BY B Y ad C Z are agle bisectors bis ectors of sig Figure 18 agai, assu isosceles isosceles A B C wit with base base B C Show Show tha thatt B Y = CZ gai i Figure 18 assume that that A B C is isosceles isos celes with with base BC, B C, but this this time, assume tha B Y ad C Z are alti altitudes tudes Show that that B Y = C Z s ssume sume for for this problem that, that, as i the gure, gure, the two altitudes actually actually lie iside isi de the triagle triagle but observe that this does ot always happe happe
ANGLES AND PARALLEL LNES
11
A
Figure 1.8
Now i Figure 1. assume that BY ad CZ are equal altitudes Show that A B AC eferrig eferrig agai aga i to Figure 1. let P b e the the poit where where B Y ad C Z meet ssume that that B Y CZ a ad P Y PZ Show that that A B AC Oce more i Figure Figure 1. let P b e the the itersectio itersectio poit o f BY ad C Z ssume s sume that A B A C ad B Y C Z We ask if it is ecessarily eces sarily te that P Y P Z ou are asked, i other words, either to prove prove that P Y P Z or else to show show how to draw a couterexample couterexample diagram diagram where all of the hypotheses hypothes es hold but where P Y ad P Z clearly c learly have differet differet legths legths We drew drew Figure 1 so that P Y ad P Z actually are equal, ad so if a couterexample couterexample exists, exists , it would ecessarily eces sarily have to look somewhat so mewhat dieret from from the gure
C
Angles Angl es and Parallel Lines
tranvra is a lie that cuts across two give give lies sually, sually, the two give lies are are parallel whe we use the word transversal but we do ot ot absolutely isis i sistt o this I Figure 1.9 for for example, lie li e t is a trasversal trasversal to lies a ad b There There is some omeclature that that is usefu us efull for for discussig discus sig the eight agles a gles that we ave l abeled with 1 through i Figure 1.9 gles that are are o the same side of the trasversal trasversal ad o corresp correspodig odig sides of the two lies a ad b are said to be corrpondin ales I Figure 19 therefore, L 1 ad L 5 are correspodig agles, as are L ad L ad also L3 ad L 7 ad, of course, L 4 ad L. It is a theorem theorem that that correspodig agles agl es are are equal whe a trasversal cuts a pair of parallel lies, ad coversely, coversely, if i Figure 19 ay ay oe oe
a b
Figure 1.9
12
CHAPTER
THE BASCS
of the equalities L 1 L5 L L L3 L7 or L4 L 8 is kow to hold, the it is a theorem theorem that lies a ad b are parallel, parallel, ad thus thus the other three three equalities also als o hold airs of agles such as L4 ad L or L 3 ad L5 that lie o opposite oppos ite sides of the trasversal ad betwee the two give lies are called atrnat intrior agles ad pairs such as L 1 ad L 7 or L ad L 8 that lie o opposite sides s ides of o f the the trasversal ad outside of the space betwee the two parallel lies are atat xtrior agles It is a theorem that alteate iterior agles are equal ad that alterate exterior agles are equal whe a trasversal cuts two parallel lies It is also als o true that, coversely c oversely,, if i Figure 19 ay oe of the equalit equalities ies L 1 L7 L L8 L3 L5 or L4 L is kow to hold, the lies a ad b must be parallel, ad thus the other three three equalities also hold, as do the four four equalities eq ualities metioed me tioed i the previous paragraph paragraph We recall also that whe two lies cross, as do lies a ad t i Figure 19 the agles , of o f course, L 1 ad L 3 are said to be vrtica an, as are L ad L4 ertical agles, are always equal While reviewig reviewig omeclatur omecl aturee for for agles, agles , we metio that two agles whose measures sum to 180° are said to be uppmntary, ad if the sum is i s 90° the an, ad a agle ages age s are are compmntary Of course, a agle of 180° is a traiht an, of 90° is a riht an I Figure 19 we see se e that L 1 ad L4 are supplemetary, ad so if a ad b are parallel, parallel, the L 1 L5 ad it follow followss that L4 ad L 5 are supplemetary, suppl emetary, as are L 3 ad L We ca apply some s ome of this to the agles ag les of a triagle Give Give ABC exted side B C LAC D is said to poit D as show i Figure 110 I this situatio, LACD said to be a xtrior agle of the triagle at vertex C, ad the two agles L A ad L B are the rmot iterior agles with respect to this exterior agle THEOREM. An exterior angle of a triangle equals the sum of the two remote interior inte rior angles Also Also the sum of of all three interior interio r angles angles of of a triangle is 1 8 0° 0°
Figre 1.10
We eed to show i Figure 110 that LACD LA + LB. raw a lie CP through C ad parallel to AB as show show Now L A LACP sice sic e these are alteate alteate iterior agles for parallel parallel lies li es A B ad P C with respect to the trasversal A C lso, L B L PC D sice these thes e are correspodig correspodig agles It follows follows that LACD LAC D required ed LACP LACP + L P C D LA + B , as requir LAC D+ LAC B L B CD C D 180° The substitutio We see se e i Figure 110 that LACD+ of L A + L B for L AC D i this equatio yields the desired coclusio coclu sio that the sum of the the three iterior agles of ABC is 180° •
Proof.
Observe that our proof that the iterior agles of a triagle total a straight agle relies o the fact fact that it is pos sible to draw a lie through C parallel to A B I fact,
I ANGLES AND PALLEL LNES LNES
1
there does ot exist a proof of this result that does ot, omehow, deped o paallel lies This is because i oEuclidea geometries, it is ot true that the agles of a triagle must total 1 80 80 , ad yet the oly fudametal fudametal diff differece betwee betwee Euclidea Euclid ea ad ad oEuclidea oEucli dea geometries is i the parallel postulate triagle, of course, is a polygo with three sides We digress to cosider the questio of how to d the sum of the iterior iterior agles agles of a polygo with w ith sides, sides , where where may be larger tha thee Fid a formula formula for the sum of the iterior iterior agles of a go go osider, o sider, for for example, the case 6 I Figur Figuree 1 1 1 , we see two two 6gos, which which are usually usu ally called hexagos hexago s I the left left hexago, we have draw the three diagoals diagoals from vertex A ( diaona of a polygo is a lie segmet joiig two of its oadj oadj acet vert vertices ices ) I geeral, a go has exactly exactly  3 diagoals diagoal s termiatig termiatig at each of its vertices vertices,, ad this gives a total total of (  3 /2 diagoals i all all (o you see s ee why we had to divide divid e by 2 here?) polygo is convx if all of its diagoals lie etirely i the iterior The iterior agles of a polygo are the agles as see from iside, ad for a covex covex polygo such as the left left hexago i Figure Figure 1 1 1 , these agles are all less tha tha 1 80
A
D
Figre 1.11
I the right hexago o f Figure Figure 1 1 1 , which is ot covex, covex, we see s ee that that two two of its six iterior agles exceed 180 ( agle with measure lager tha 180 is sai to be a rx agle ) I geeral, a polygo is covex covex if ad oly if its iterior iterior agles all measure meas ure lss l ss tha 1 80 If som itrior agle agle of a go g o is exactly exactly equal w e do ot cosider it to be a covex go, go, but i this situatio, si tuatio, there are are to 1 80, we s ides that together together form form a sigle lie segmet, ad the go go ca c a be two adjacet sides cosider cosidered ed to to be a a ( 1) go iewed iewed as a ( 1 )go, the give polygo polygo may may be covex Suppos Suppo s w have have a cov covxx go go such as hxago AB A B CD F of Figu Figurre 1 1 1 Fix some som e particular vertx vertx A ad draw the 3 diagoals from from A A This divides the origal polyg ito it o exctly 2 triagles triagles ad it should be clear that the sum of all th trior agls of all of ths triagls is xactly the sum of all iterior agles of th origial polygo po lygo It follows follows that th sum of the iterior iterior agles of a covex covex polygo is xactly xactly 1 80( 80 ( 2) dgrs dgrs W have ot yt yt fully fully solve so lve roble roblem m 1 5 , of course, because becaus e we have have oly cosiderd cosi derd covex covex polygos polygo s ctually, it is ot quite ecessary for for the polygo poly go to b covx to mak th prvious argumt work What is really required is that
14
CHAPTER CHAPTER
THE BAS CS
there should exist exi st at least oe vertex vertex from from which all of the diagoals lie iside i side the polygo polygo ecall that the deitio of a covex covex polygo requires requires that all diagoals from all vertices should be iterior It is coceivable that every polygo has at least oe vertex from which all diagoals are iterior, but ufortuately, that is ot true true car c aref eful ul ispectio of the right right hexago hexago i Figure Figure 1 1 1 shows that it is a couterexample; at least oe diagoal diago al from from every every oe of its vertices fails to be iterior iterior It is true, true, but ot easy to prove, that every every polygo has at least oe iterior diagoal, diago al, ad it is possibl pos siblee to use this hard hard theorem theorem to costruct co struct a proof that that for for ever everyy go go, , the the sum of the the iterior agles is 1 80( 80 ( 2) degrees There is aother way to thik about roblem 15 that may give additoal isight isi ght Imagie walkig clockwise clockwi se aroud a covex covex polygo, startig from from some poit other tha a vertex o oe of the sides Each time you reach a vertex, you must tu tu right by a certai umber of degrees degrees If the iterior agle at the th vertex vertex is fk the it is easy e asy to see that the the right tu at that vertex vertex is a tu through precisely preci sely 1 80  fk degrees Whe you retu to your startig poit, you will be facig i the same s ame directio as whe whe you started, started, ad a d it should be clear that you have tured clockwise clockwi se through a total total of exactly exactly 360 360 • I other words, words,
L
(1 80  f L k=
360
L
Sice Si ce the quatity 1 80 is added times times i this sum ad each each quatity f is subtracted oce, we see that that 1 80  fk 3 60, ad hece hece fk 1 80 80  360 180(  2) This provides provid es a seco s ecodd proof of the formula for the sum of o f the iterior iterio r agles of a covex covex go ow importat is covexity for this secod s ecod argumet? argumet ? If whe walkig wal kig clock wise aroud the polygo pol ygo you yo u reach the th vertex vertex ad see a reex iterior agle there, you actually tu left, ad ot right I this case, your left left tu is easily see to be through exactly exactl y f  1 80 degr degrees ees,, where, where, as before before,, f is the iterior iter ior agle at the vertex If we we view vie w a left tu as beig bei g a right tu through some so me egative e gative umber of degrees, degrees , we see s ee that at at the th vertex vertex we are are turig right by 1 80  fk degrees, ad this is i s true regardless regardless of whether whether fk < 1 80 as i the covex covex case or f 1 80 at a reexagle vertex vertex It is also als o clearly true true at a straightagle straightagle vertex, where fk 180 The pre previous vious calculatio thus works works i all cases, ad it shows shows that 1 80( 80 (  2) is the sum of o f the iterior agles for every every polygo, polygo , covex covex or ot ot
ID
Parallelograms
mog polygos, perhaps parallelograms are secod i importace after triagles ecall that, that, by deitio, dei tio, a paraoram is a quadril quadrilate ateral ral AB C D for for which which A B CD ad A D B C I other words, the opposite oppos ite sides sid es of the the quadrilateral quadrilateral are parallel It is also true that the opposite oppos ite sides si des of a parallelogra parallelogram m are equal, but this is i s a cos c osequece equece of te te assumptio assumptio that the the opposite oppos ite sides are parallel; it is ot part of the deitio deitio (1.6)
pposi te sides s ides of of a parallelogram pa rallelogram are a re equal equ al Opposite
D PARALLELOGRAMS
15
I Figure 1 1 2, we are give that that A B C D ad AD A D B C, ad our task task is to show thatAB CDadAD BC raw diagoal BD ad ote ote that A BD CDB sice these these are are alterat alteratee iterior iterior agles for for the paralle parallell lies A B ad C D , ad simila similarly, rly, D BC A DB Sic Sicee BD B D, we see see tha thatt D AB B CD by by • S, ad it follows follows that that AB A B CD ad ad AD B C, as require required d
Proof.
B
C
D
Figre 1.12
Figre 1.13
There are are two useful coverses for Theorem Theorem 1 6 In quadril quadrilate aterral A B C D sup suppose that that A B CDandAD BC ABC D is a parallelogram. Then ABCD In quadrilateal quadrilate al ABCD suppose suppose that A B C D and ABCD Then ABCD ABC D is a parallelogram.
We leave the proofs of these thes e two theorems to the exercises exercis es A quadrilateral is a parallelogram and only its diagonals bisect each each othe
s show i Figur Figuree 1 1 3 , we let X be be the the poi poitt where where diagoa diagoalsls AC ad B D of quadrilat quadrilatera erall A B C D cross Suppose rst that that X i s the commo commo midpoit of of lie lie segmet segmetss A C ad BD The The AX XC ad BX XD ad ad aalso lso A XB CXD because these are are vertica verticall agles It follows that that AX B C X D by b y SS S S , ad thus AB CD Similarly, AD BC, ad hece ABCD ABCD is a parallelogrm parallelogrm by Theorem 17 oversely oversely ow, ow, we assume that that AB A B C D is a parallelogram, ad we show sh ow that that X is the midpoit of each of the diagoals A C ad B D We have have B A X X C D ad ABX AB X CDX CD X because bec ause i each each case, cas e, these are pairs pairs of alteate alteate itrior itrior agles agles for the the parallel parallel lies lies A B ad CD C D lso, A B C D by Theorem 1 6, 6 , ad thus ABX C DX by S We deduce deduce that that AX CX ad BX DX, as • required
Proof.
Observe that that by Theorem Theorem 1 7, 7 , a quadrilateral quadrilateral i which all four four sides are are equal must be a parallelogram parallelogram Such a gure g ure is called a rhombu I the case of o f a rhombus the diagoals ot ot oly bisect bise ct each other, but they they are als alsoo perpedicular perpedicular lthough lthou gh this fact fact is ot difcult to prove directly, we prefer prefer to derive it as a coseque cos equece ce of the follo followig wig more geeral result
16
CHAPTER
THE BASCS
iven a line segment BC BC the locus of of all points po ints equidistant eq uidistant from B and C is the perpendi perpendicular cular bisector bise ctor of of the segment
(1.10)
We eed to show that every poit o the perpedicular bisector of BC is equidistat equidistat from from B ad C ad we must also show that every every poit that is equidistat from rom B ad C lies o the perpedicular bisetor of B C ssume ss ume that A X is the perpedic perpedicular ular bisector bisector of o f B C i Figur Figuree 1 14 14 This This meas meas that X is the midpoit of B C ad AX is perpedicular to to BC I other words, we are assumig that A X is simultaeously s imultaeously a media ad a altitude altitude i ABC ad we wat to deduce that A B AC This is precisely precisely Exercise Exercise 1 3 , ad we will will ot give the proof here ssu s sume me ow that A is equidistat from B ad C i Figure Figure 1 1 4 ad draw draw meda meda A X of ABC Sice A B AC this triagl triaglee is isosceles, isosc eles, ad thus by Theorem Theorem 1 2, 2 , media A X is also a lso a a altitude altitude I I other other words, words, A X is the perpedicular bisector of B C ad of course, A lies o this lie lie •
Proof.
A
B
X
C
Figre 1.14
(1.11)
of rhombus A B C D are per pe rpendicu pe ndicula la The diagonals of
Sice A B AD we kow by Theorem Theorem 1 1 0 that that A lies o the perpedicular bisector bis ector of diagoal B D ad similarly, C lis o the perpedicular bisectr of B D ut A C is the oly li that cotais the two poits A ad C ad thus A C is the perpedicular bisector of B D. I particular, diagoal A C is prpeicular to diagoal B D •
Proof.
We close clo se this sectio sectio by metioig aother spcial typ typ of paralllogram paralllogram a rct rct agle y deitio, a rctan is a quadrilateral quadrilateral all of whos whos agls agl s r right agles agles It is easy it see that th opposite sies s ies of a rctagl rctagl parallel parallel d so a rctagle is automatically automatically a parallelogram W also kow (by roblm 1 , fr fr xampl) tha th sum of th four four agls agl s of a a arbitrary arbitrary quarilatral is 0 0 hus if i f w hav ay quadri later la teral al with all four agls qual each agl must b 90 ad th gur is a rctagl rctagl W rmark also that adjact vrtics of a parllelogram hav supplmt itrior agles agles It follows easily that if o gle of a paralllogram paralllogram is a right agl ag l th th parallogram must mu st b a rctagl rctagl Fially w mtio that a quadrilatrl that is both a rectagle ad a rhombus rhombus is by itio a ar
ID PARALLELOGRAMS
xercses
17
rove rove Theorem Theorem 1 7 by showig sho wig that that quadrila quadrilatera terall A B C D is a parallelogram if A B CD ad ad AD BC rove Theorem 1 8 by showig that that quadrilat quadrilateral eral AB A B C D is i s a parallelogram if A B CD a ad AB CD rove that the diagoal diag oalss of a rectagle are equal equal rove that a parallelogram paralle logram havig perpedicular diagoals diagoal s is a rhombus rhombus rove that a parallelogram parallelogram with equal diagoals is a rectagle rectagle I Figur iguree 1 15, weare weare giv give that AB CD ad that A D ad B C are ot parallel paralle l Show Sh ow that D C if ad oly if A D BC HINT raw a lie through B parallel Figure 1.15 toAD NOTE: ecall ec all that a quadrilateral is said to be a trapzoid if it has exactly oe o e pair of parallel sides si des If the oparallel pair pair of sides are equal, the trapezoid is said to be ioc Show that opposite agles of a parallelogram parallelogram are equal equal I quadrila quadrilater teral al AB A B C D, suppose that that AB C D ad B D Show that that AB C D is a parallelogram I quadr quadrililat ater eral al AB CD , supp suppose ose that A C a d B D (We are are refer eferrrig to the iter iterior ior agles, agles, of course) course ) Show Sh ow that that AB C D is a arallelogram HINT: Show that A ad C are supplemeta supplemetary ry that the diagoals of a iso sceles scele s trapezoid trapezoid are equal Show that I Figure Figure 1 6,we are are give give that that �AX �A X B ad �A Y B are are isosceles isosc eles ad share share base S how that poits X, Y, ad Z are are colliear (lie o a commo lie) if ad ad A B Show oly if �A ZB is isosceles wit withh base base AB AB x
Figure 1.16
18
CHAPTER
THE BASCS
I Figure 117, poit is equidis tat from the the vertice verticess of �A B C Thi Thi s poit is reected i each side of the triagle, yeldig poits X, Y, ad Z Z This meas that X is perpedicu lar to B C ad that the perpedicular perpedic ular C distace dis tace from X to B C isi s equal equ al to that from to BC Similar asser x tios hold for Y ad Z rove that �ABC �XYZ ad that corre Figure 1.17 spodig sides side s of these two triagles triagles are parallel HINT: Show that quadrilaterals ZBA ad CAY are rhombuses educe that B Z ad C Y are are equal ad parallel NOTE Give ay triagle, there always exi sts a poit equidistat equidi stat from from the three three vertices vertices This poit, called the circumcntr of the triagle, triagle, does ot always lie iside isid e the triagle The assertio as sertio of this problem is still s till valid if lies outside of �A B C or eve if it lies lie s o oe of the sides side s Each pair of parallel lies i Figure 118 represets a railroad track with with two parallel rails The distaces betwee the rails rail s i each of the tracks tracks are equal rove rove that the parallelogram formed formed where where the tracks tracks cross cros s is is a rhombus rhombus Figure 1.18 NOTE The distace betwee the rails is, of course, measured perpedicular perpedicul arly ly It shou s hould ld be clear that for for each track, the perpedicular perpedic ular distace dista ce from from a poit o oe rail to the other rail is costat, cos tat, idepedet idepe det of the poit dis tace from from a poit to a lie is measured perpedicularly Give ecall that the distace A B C, show that the locus locu s of all all poits iside i side the agle ad equidistat equidis tat from from the two lies B A ad a d B C is i s the bisector bis ector of the agle
IE
Area
We assume ass ume as a s kow the t he fact fact that the area of a rectagle is i s the product of its legth ad width The area of a geometric gure g ure is ofte deoted K , ad so the formula for the area of a rectagle is K = bh where b is the legth of oe of the sides si des of o f the rectagle ad h is the legth of the perpedicular sides side si de of legth legth b is referred referred to as the base b ase of the rectagle, rectagle, ad the height h is the legth of the sides perpedicular to the base Of course, the base of a rectagle rectagle does ot have to be at the the bottom; it ca be ay side si de Now suppose that we have a parallelogram that is ot ecessarily a rectagle ga we desigate de sigate oe o e side si de of the the parallelogram parallelogram as the base ad write b to deote its legth ut ut this time, the height h is the perpedicular distace betwee the two parallel
E AREA AREA
x
I I hi I b
19
b
x
b Figure 1.19
b
sides of legth Of course, if the parallelogram parallelogram happes to be a rectagle rectagle,, the this perpedicular distace is the legth of the the other other two sides side s O the left of Figure 119, we see a rectagle with base ad height h . O the right, we have draw a parallelogram that also has base ad height h ad we claim areas To see why this is true with that the parallelogram ad the rectagle have equal areas a very iformal argumet, drop perpediculars from oe ed of each of the sides of legth of the the parallelogram parallelogram to the extesio extesios s of the the opposite sides s ides,, as a s show sho w i the gure What results result s is a rectagle with base base + x ad height h where, as show, x represets represets the amout that the the bas basee of the parallelogram parallelogram had to be b e exteded to meet the perpe perpedicu dicular lar The area of this rectagle rect agle is + x )h ad to obtai obtai the area o the origial parallelo para llelogram, gram, we eed to subtract from from this the area of the two right triagles learly, learly, the two tw o right triagles triagles pasted together would form form a rectagle with bas e x ad height h ad so the total area of the two triagles triagle s is x h The area of the parallelogram parallelo gram is therefore + x)h  xh h , ad so the formula K h works to d areas of arbir arbirary ary parallelograms ad ot just of rectagles rectagles other way to see why the the rectagle ad ad parallelogram parallelogram i Figure 1 1 9 have equal area is from the poit of view of calculus Imagie slicig each of the areas ito iitesimally iites imally thi horizotal strips of legth We see se e that each of the the two areas is is give by the same s ame formula formula
b b
b
b
(b
(b
b
b
b. K lhbdY,
equal , as claimed cl aimed I fact, fact, if we carry carry out the itegratio, itegra tio, we we ad thus the two areas are equal, expect get the th e formula formula K h , as we expect Next, we cosider co sider areas of triagle triagless I the left left diagram diagram of Figure 1 20, w have draw draw a triagle with base ba se ad height h where ay ay oe of the three three sides ca c a be viewed as the base ad the legth le gth of the altitude draw to that that side is the correspo correspodig dig heght heg ht emember emember that this this altitude altitude may lie l ie outside of the triagle triagle
b
b
B
Figure 1.20
X
C
20
CHAPTER
THE BAS CS
To compute comput e the area K of the the give triagle, we have costructed co structed a parallelogram parallelogram by drawig lies parallel to our base ad to oe of the other sides of the the triagle, as as show i the gure This parallelogram has base b ad height h , ad so its area is bh The parallelogram parallel ogram is divided divid ed ito two cogruet triagles by a diagoal diago al,, ad so the area of each of these thes e tria triagle gless is exactly exac tly half the area of the parallelogram (The two tria triagles gles are easily easily see to be cogr co gruet uet by by SSS S SS)) Sice the origial origial triagle triagle is i s oe of the two halves of the the parallelogram, we see that K bh ad this is i s the basic formula formula for for the area of a triagl triaglee s is idicated idi cated with A A B C o the right of Figure 1 .20, .20 , ay oe of the three sides coud have bee desigated de sigated as the base, bas e, ad for for each, there is a correspo correspodig dig altitude (It i ot a coicidece coi cidece that the the three three altitudes are cocurret; cocurret; we shall sh all see i hapt h apter er 2 that thi thiss is guarateed to happe) happe ) We thus thus hav h avee three dieret formula formulass for the the area area KAB of A B C We have
KAB
AXBC 2
BY·AC 2
CZ·AB 2
ad we ca deduce ded uce usefu u sefull iformatio iformatio relatig the legths of the sides side s ad altitudes of a triagle triagle For example, i Exercis Exercisee 1 .6, .6 , you were asked asked to show that that i a isosceles iso sceles A B C with base B C, the t he two altitude altitudess B Y ad ad CZ C Z are are equal, ad you were were allowed allowed to asume that the altitudes were iside is ide the triagle triagle y the area foula foula we have just ju st derived, derived, we kow that B Y · A C C Z A B sice each of these these quatities quatitie s equals equal s twice the area of the triagle triagle We ca cacel the equal equal quatities AC A C ad AB A B to obtai B Y C Z, as desired This proof, furt furthermore, hermore, works idepedetly of whether or ot the altitudes lie iside i side the triagle We remark remark als alsoo that the coverse coverse is true: If we are give that that altitude altitudess B Y ad C Z are equal, equal, it follows follows fro from m the the foula foula BY B Y · A C CZ ·A B that that A B A C (This coverse coverse appea appeared red as Exer Exercise cise 1 .7.) .7 .) We ca get some additioal ifoa ifoatio tio about areas of triagles by b y usig usi g a ittle elemetary elemetary trigo trigoometr ometry y I Figure Figure 1 .2 1 , we have have followed followed custom ad used the symbols a , b ad to deote the the legths of the sides of A B C opposite vertices vertices A, B, ad C, respectively respectively lso, lso , w e have draw draw the altitude of legth h from A (I the gure, this altitude happes happes to lie outside the triagle, triagle, but this is irrelevat irrelevat to our calculatio) calcul atio) We have, of course, K ah, where, as usual, K KA B represets the area of the triagle triagle We see that that si (C) h j b , ad hece h b si(C) si( C) , ad if we substitute substitute this this ito the area formula K ah, we obtai K ab si(C) si( C) Similar Si milarly, ly, we also also have have ad K b si(A) K a si(B) ad I t is ow clear c lear that for for ay ay triagle, the equatios ab si(C) b si(A) a si(B)
mus always hold I f w e divide tough tough by the quatity quatity abc ad take reciprocals, reciprocals, we obtai obtai the socalled soc alled law of sies : si(C)
a si(A)
b si(B)
IE AREA
21
A A
h
B B
Figure 121
C
a Figure 122
simpl e proof of a more powerf powerful formula formula called the exteded law of o f sies that There is a simple applic atio of the sie formula for the area of a triagl triaglee we prove i hapter 2 other applicatio is the followi followig g
et AX be b e the bisector of LA in 6A 6 A B C. Then BX AB XC AC In oher oh er words words X divides B C into int o pieces pi eces propo proportiona rtiona to the engths e ngths of of the near nea rer sides si des of of the trian t riange. ge. THEOREM.
I the otatio otatio of Figure 1 22, we must show that that u / = /, where where ad c the legths le gths of B X ad XC as show show To see see wh thi thiss are as usual ad u ad are the equatio holds, let h be the height of 6ABC with respect respect to the base BC. The h is also als o the height of each each of 6A B X ad 6A C X with with respect to bases base s B X ad XC respectivel We have h cx si() uh b x si() ad = KABX = = KA x = 2 2 2 2
Proof.
where x = A X ad we have writte = LA. ivisio of the rst of these • equatios equatios b the th e secod seco d ields the th e desired propo proportio rtio s a applicatio, applicatio , we have have the followig followig,, which whic h should shoul d be compared with Exer Exer cises cis es 1 2 ad 1 1 3 like those exerci exercises ses,, it it difcult difcult to see a direct ad ad elemetar elemetar proof via cogruet co gruet triagles for for this fact PROBLEM. Suppose Suppos e that that i 6ABC the media me dia from vertex A ad the bisec tor of L A are the same lie Show that A B = AC. I the otatio of Figure 1 22, we have have u = sice the agle bisector bi sector A X is assumed assu med to be a media It follows follows from from Theorem Theorem 1 1 that that a = , as required required •
oution.
22
CHAPTER CHAPTER
THE BASC S
We have ow give two t wo dieret types of formula formula for the area of a triagle: triagle : oe usig oe side ad a altitud altitudee ad the other other usig two sides ad a agle Sice the SS S cogruece criterio tells u s that a triagle is determed determed by its three sides, sides , w wee might expect that there there should be a ice way to compute the area of a triagl triaglee i terms of the the legths of its sides There is I hapter 2, we prove prove the followig followig formula, formula, which is attribut attributed ed to ero of lexadria (c 50 AD : KA B C
a ) (s  b ) ( s
=

c)
,
where s (a + b + c) / 2 is called call ed the miprimtr of the triagle triagle Of course course,, a, b, ad c retai their usual usu al meaigs meaigs We close clo se this sectio with with what what seems to be a amazig fact fact s show i Figure Figure 1 23 , poits P Q, ad a d R lie o the sides of �A B C oit P lies li es oe third of the way from from B to C poit poi t Q lies lie s oe third of of the l ies oe third third of the the way from from A to B ie segmets way from from C to A ad poit R lies A P B Q, ad ad C R subdivide the iterior of the triagle triagle ito three quadrilaterals quadrilaterals ad four triagles triagles,, as sh show, ow, ad we see that exactly oe of the the four four small triagles has o vertex vertex i commo com mo with ABC rove that the the area of this this triagle is i s exactly oe seveth of the area of the orgial triag triagle le
(1.14)
We eed to compute compu te the area Kxyz i Figu Figure re 1 23 y choosig uits appro appro priately, priately, we ca c a assume ass ume that KAB 3 ad we write k to deote KByP We draw lie segmet Y C ad start computig areas Sice �CYP has the same s ame height height as 6BYP but its base P C is twice as log, we deduce that K yp 2k Similarly Si milarly,, sice A Q 2 Q C , we see s ee that KABQ 2 K B Q , ad thus K B Q KA B 1 This eables eables us to compute that that K YQ 1 KBY 1 3. y the usual reasoig, reasoig, KA YQ 2 KYQ 2  6k, ad we kow that KABQ KAB 2 It follows that KA B y 2  KAYQ 2  (2 6k) 6k) 6k owever, KABP KA B 1, ad thus KByp 1  6k ut we ow that KB yP k, ad hece 1  6k k ad k 1/7 Simlar reasoig shows that KARX 1/7 K Q z ad sice KA R area of quadrilatera quadrilaterall A X Z Q is 1 2/7 5/7 K AB 1 , we deduce that the area Fially, we recall that KA YQ 2  6k 8/7, 8/ 7, ad it follows follows that Kxyz 8/7 5 /7 3/ 7 This is exactly oe seveth of the area area of the origial tragle, tragle, as desired desired •
oution.
A
B
P
Figure 123
C
I CRCLES AND ARCS
23
xercses
raw raw two of the the medias of a triagle triagle This subdivides the iterior of the triagle triagle ito four pieces : three triagles triagle s ad a quadrilatera quadrilaterall Show S how that two of the three three small triagles triagl es have equal area ad that the area area of the third is equal eq ual to that of te quadr lateral arbitrary poit P is chose chose o side B C of AB C ad perpediculars P U ad P V ae draw from P to the other two sides of the trigle (It may be that or ies o a extesio of AB or AC ad ot o the th e actual side of the triagle triagle This ca c a happe, happe, for for istace, if L A is obtuse obtus e ad poit P is very ear B or C.) Show Sh ow that that the sum P U + P V of the the legths leg ths of the two perpediculars is costat as P moves alog B C. I other words, this thi s quatity is idepedet of the choice of P Sice Si ce a triagle i s determied determied by aglesideagle, aglesi deagle, there there should shou ld be a foula foula for for formula KA B C expressed terms of a ad L B ad LC. erive such a formula et P be a arbitrar arbitraryy poit i the iterior iterior of a covex quadrilateral quadrilateral raw the li segmets joiig joii g P to the midpoits of each of the four four sides, side s, thereby thereby subdividig the iteror ito four four quadrilaterals quadrilaterals Now choose choos e two of the small quadrilaterals quadrilaterals ot havig a side i i commo ad show that the the areas of thes thesee two total exactly half the area of the the origial gure
IF
Circes an Arcs
s all readers of this this book bo ok surely ow, a circ is the locus lo cus of all poits equidistat equidistat from some give poit called the cntr The commo distace r from the ceter to the poits of the circle is the radiu, ad the word radius is also used to deote deote ay ay oe of of the lie segmets s egmets joiig jo iig the ceter to a poit of the circle chord is ay lie segmet j oiig two poits of a circle, circle , ad a diamtr is a chord that goes through te ceter The legth d of ay diameter is give by d 2r ad this this is the maximum of the legths chords Fially, we metio that ay ay two circles with equal radii are cogruet, of al chords o e of two two cogruet circles ca c a be made to to correspod to ay poit ad ay poit o oe o the other other circle circle Just as ay a y two poit deermie a uique le, le , it is lso true tat ay three poits, uless ul ess they happe to lie o a lie, lie o a uique circle e e is exacty one circe c irce through through any three three given noncoinear Tee points
all the poits A B ad C. Sice by hypothesis, there there is o lie though though these poits poit s we ca be sure that we are are dealg with three three distict poits poi ts,, ad we s egmets A B ad A C. et et ad ad c be the perpedicular bisectors of these draw lie segmets segets ad obs obser erve ve that that lies ad c caot be parallel parallel because lies A B ad A C are either parallel or are are they the same lie lie (Si ( Sice ce A B ad A C have poit A i
Proof.
24
CHAPTER CHAPTER
THE BAS CS
commo, commo , they surely sure ly are ot parallel parallel;; they caot be the same lie because be cause we w e are are assumig ass umig that o lie cotais all all three poits A B , ad ad C.) et P be the poit wher wheree les ad meet Sice P lies o the perpedic perpedicular ular bisector of A B we kow kow by by heorem heorem 1 1 0 that that P is equidistat equid istat from from A ad B. I other words, words, P A P B Similarl Si milarly, y, sice P lies o lie we deduce that P A PC If we let r deote the commo c ommo legth of the three three segmets P A P B, a a PC we see that the circle of radius r cetered at P goes through the three three give poit o see that our three three poits caot c aot lie o ay other circle, circle , we could c ould appeal appeal to the obvious" fact that two dieret circles meet i at most two poits, but it is almost as easy to give a real proof proof If a circle circle cetere ceteredd at some poit Q, say, goes through A B , ad C the Q is i s certaily equidistat from from A ad B , ad hece by heorem heorem 1 1 0, it lies o the perpedic perpedicular ular bisector of segmet A S imilarly, imilarly, we see that Q lies o lie ad thus Q P because P is the oly poit poi t commo to the two two lies lies Sice S ice the distace P A r , it follows follows that t hat the oly circle c ircle through A B, ad ad C is the circle of o f radius radius r cetered at P . • Give � A B C the uique circle that goes through the three vertices is called the circumcirc of the triagle triagle,, ad the triagle is s aid to be incribd i the circle More More geerally, ay polygo all of whose vertices lie o some give circle is referred to as beig incribd i that circle, ad the the circle is circumcribd about the polygo lthough every every triagle is iscribed i some s ome circle, the same caot be said for for gos go s whe 3 he statemet statemet of heor heorem em 1 1 5 would b e eate eaterr if w e did ot have have to deal with the exceptioal except ioal case cas e where the three give poits poi ts are colliear collie ar If we were willig will ig to say that a lie is a certai kid of circle (which, (which , of course, it is ot), we could the s ay that that every every choice of three distict distic t poits determies a circle It is sometimes coveiet to preted preted that a lie is a circle with iite" radius, radiu s, but of course, this should s hould ot be take take too literally We retu retu ow to our study of geuie circles wo poits A ad B o a circle divide the circle ito ito two pieces, pieces , each each of which is called a arc We write A B to deote oe of these two arcs arcs,, usually usu ally the smaller smaller s this i ambiguous, ambiguous , a threepo threepoit it desigatio for a a arc is ofte prefera preferable ble I Figure 1 24, for example, the smaller of the two arcs determed by poits poi ts A ad B would w ould be desigated ad the larger is he ambiguity i the otatio A B is related to a similar ambiguity ambiguity i the otatio for agles For example, example, if we write L A O B we geerally mea the agle that i Figure 124 icludes poit X i its iterior; we do ot mea the reex agle, with Y its iterior ow big is i s a arc arc ? he most commo commo way way to discus disc usss the size of a arc arc is i terms of the the fractio fractio of the circle it is, is , where the whole circle is take to be 360 36 0 or 2 2 radias rad ias arc extedig over a quarter quarter of the the circle, therefor therefore,e, is i s referre referredd to as a s a 90 9 0 arc arc , ad /2 radias i this case 90 or c ase Of course, this size we would write descriptio for a arc is meaigful oly relative relative to the circle of which it is a part If we are told, for exampl, exampl, that we have two 90 arcs arcs,, we w e caot say s ay that they are cogruet or that that they have equal legth ules u lesss we kow that these are two arcs of the same s ame crcle or of two circles circle s havig havi g equal radii radii o o remid us that the the umber of degrees degre es (or ( or radias) that we assig assig to a arc gives oly relative iforma iformatio, tio, we use the symbol , which
A
A
CRCLES AND ARCS
25
A B p
y
Figure 1.24
Figure 1.25
we ead as equal n degees (o radans) radans),"," and we avod te te use of n ts context I nfoally, nfoally, oweve, we do speak of equal acs acs , but t t s pobably best to avod ts t s pase except wenwe know tat tat te two tw o acs acs ae n te same o equal ccles Gven an ac A B on a ccle centeed at pont , we say tat L A O B s te cntra an corespondng to te ac Snce S nce a full full ccle s 3 60 of ac ac and one full otaton otaton s 360 360 of angle angle,, t sould sou ld be clea tat te number number of degees n te measure meas ure of cental cental angle LAO B s equal to te numbe of degees n 90 angle at te cental pont fo example, cuts o a quate quate ccle, wc s a 90 ac ac Te standa s tandadd j agon fo fo te pase pase cuts o" n te pevous pevous sentence s ubtnd In geneal, we w e can wte wte cental angle, teef teefoe, oe, s equal n degees to te arc t subtends subtends We LAOB can also s ay tat te ac ac s s maurd by te cental angle In a gven ccle, ccle , te angle foed by two cods tat sae an endpont s c alled an incribd angle S ome of te te nscbed angles n Fgue 1 25 2 5 , fo fo example, example, ae LAP B , LAQB and LARB
An inscribe in scribedd ange in a circe is equa equ a in degrees to to one ha its subtended sub tended arc. arc. Equivaen Equiv aenttyy the arc subtended by by an inscribed inscri bed ange ang e is measured measu red by twice the ange.
(1.16)
In Fgue 125, fo example, LA P B and also L A Q B In partcula, we can deduce tat LA P B LAQB and n geneal, any two nscbed equal s we sall s all see, ts povdes a angles angles tat subtend te same ac n a ccle ae equal useful tecnque fo povng equalty of angles angles ow consde co nsde LAR B n Fgue 125 ke L A P B and L A Q B ts too s an nscbed angle foed by cords troug A but we cannot c annot conclude tat tat LARB s equal to te ote two angles because t and B , but subtends te ote ac detemned by ponts A and B B In fact, fact, L A R B P Snce follows tat and togete consttute te wole ccle, t follows 1 1 LARB + LAPB LAP B (AP B + A R B ) 360 1 80 2 2 and tus L A R B and LA Q B ae supplementay Ts poves te followng coollay to tat a polygon s nscbed n n a ccle f all of ts ts vertces vertces le on Teoem 1 1 6 ecall tat te ccle ccle
6
CHAPTER CHAPTER
THE BAS CS
(1.17) COROLLARY.
menta
Opposite anges of an inscribed quadriatera are suppe • B
p
p
Q
B
Q
p
Figure 1.26
Gven Gven APB nscbed n a ccle c cle centeed at pont te tree cases we need to consde ae ae llustated llustated n Fgue 1 .26 .2 6 It may be tat pont falls falls on one of o f te te sdes sd es of APB, APB, as n te left left dagam ltenatvely, ltenatvely, mgt le n te nteo of te angle, as n te mddle dagam dagam of Fgue 1 . 26, o t mgt be exteo exteo to te angle, angle , as n te gt dagam Suppose st tat tat les on a sde (say (s ay,, P B ) . aw adus adus A and obseve tat P s sosceles so sceles wt base A P , and so by te te pons pons asnoum, A P ental ental AB s an exteo angle of � A P , and ence t s equal to te sum of te two emote nteo angles by Teoem Teoem 1 .4. Tus ABA+P2P, and ence P ° as equed ow ass assume ume tat te cente of te ccle les n te nteo of te angle and daw damete P Q , as n te mddle dagam of Fgue 1 26 26 y te part of te teoem tat we ave aleady poved, we know tat
Proof of Thorm 1.16.
Ao
APQ
and
ddng tese tese equaltes gves gves A P B ° as equ equed ed Fnally, we can as sume te stuaton of te gt dagam n n Fgue 1 . 26. gan we daw damete P Q and we get te te same two equaltes as n te pevous cas e Ts tme, subtacton yelds te desed esult. • Imagne Imagn e te foll follown owngg expement Mak a lage ccle ccl e on te gound and eect two vertc vertcal al poles pole s at ponts A and B on te te ccle ow stand somewee else on te ccle and obseve ow fa apart te two poles appea to be, as seen fom fom you pespectve Te appaent sepaaton of te poles s detemned by te angle fom fom one pole pol e to you eye and and back to te te ote pole. pole . Snce S nce ts angle s nscbed n te te ccle, Teoem 1 1 6 guaantees tat, as we walk along te ccle, ccle , te appaent appaent sepaaton of te poles pole s emans uncanged, as long as we stay on one of te two acs detemned by A and B Te angula angula sepaaton we see fom anywee on te ote ac detemned by A and B, oweve, oweve, s te supplement of ts ts angle
I CRCLES AND ARCS
27
ppo se now tat te lne toug te two poles pole s uns due not and sout so ut We know S uppose tat fom all ponts on te ccle and east of te poles, te sepaaton between te poles appea app eass consta cons tant. nt. In Fgue 1 .27 , n ote wods, wods, L A P B s ndependent ndependent of of te te coce of te pont P on te easte ac of te ccle
v
Figure 127
outs de of te ccle, to Wat appens f we eman east of te poles but move outsde pont po nt X, fo example? ommon ommon sense tells tel ls us tat snce X s s fate fate fom te poles tan s P, te angula sepaaton sould decease We can quantfy quantfy ts ts because be cause,, as a s we see n gue,e, LA P B s an exteo exteo angle of of �A P X. Tus L A X B L A P B  L XA X A P , an andd te gu te appaent decease decea se n angula sepaaton sep aaton as we move fom P to X s pecsel pec selyy equal notce ce tat tat LXA LX A P L A P we see tat to L X A P . lso, f we not 1 � LAXB o (AB  P) . lne segment seg ment (suc (s uc as A X and B X) tat extends a cod beyond a ccle s called calle d a cant, and so we ave tus poved te followng followng esult.
U
The angle between two secants drawn to a circle from an exterior point po int is equal in degrees degrees to ha the derence ofthe two subtended su btended ar a rcs. •
(1.18) COROLLARY.
etunng now to ou two poles, pole s, we see n Fgue Fgue 1 . 27 tat f we move fom fom pot P appaentt sepaat s epaaton on of te poles to a pont nsde te ccle and east of te poles, te appaen nc nceas eases es f fom L A P B to L A B. Snce Snce L AB A B s an exteo exteo angle angle of � A P, t follows follows tat tat te te amou amount nt of of nc nceas easee equals equals L A B  L A P B L P A Tus 1 � + PV). LAB LA B o (AB 2 We ave poved te t e foll follown owngg
The angle betwee be tweenn two chords chords that intersect inter sect in the interio inte riorr of • a cir ci rcle is equal eq ual in degrees degrees to ha the sum of the two subtended su btended arcs.
(1.1) COROLLARY.
We ave ave seen se en tat te pat of te plane east of ou two poles pole s s subdvded subd vded nto tee sets. sets . Eveywee on te ccle, te angula sepaaton sepaaton of te poles s Insde Ins de te ccle, te angula sepaaton sepaaton s always always geate tan tan ts quantty, quantty, and outsde outs de te ccle, sepaaton s always smalle We We close ts dscus d scusson son wt a lttle execse execse te angula sepaaton
28
CHAER CHAER
THE BAS CS
Given two points and some angle sketc te locus of all points on te plane from wic te angular separation s eparation of te two given points is equal eq ual to onsider ons ider separa s eparatel telyy tree cases < 90°, 90°, and 90°. te tree 6foottall rectangular painting is ung ig on a wall, wi its bottom edge 7 feet above te oor n lover wose eyes are 5 feet above te oor wants as good a view as possibl pos sible,e, and so se wants to maximize te angular angular separation sep aration from from er eye to te top and te bottom of te painting painting ow ow far from from te wall wall sould se stand?
(1.0)
b
M
e p
Figure 1.28
In Figure 128, orizontal line e represents te possible positions of te viewer's eye, 5 feet above above te oor ine B is te wall on wic te picture picture is ung, and and B represent, represent, respectively, respectively, te top and bottom of te picture. We seek a point P on line e tat maximizes maximizes L B P ltoug tis can be done by calculus calculu s tecniques, tecnique s, we present present an an easier metod We kow tat point B is 2 feet above line e and tat is 6 feet iger, or 8 feet above abov e e . Te mdpoint M of B is tus at te te average average eigt eigt (2 + 8) /2 5 feet feet above e raw te perpendicular bisector b of B so tat b is orizontal orizo ntal and 5 feet above e and coose point on b so tat B 5 raw te circle of radius radius 5 centered at and note tat tis tis circle c ircle is tangent to e at some point P Te circle, circle, in oter words, words , touces line e at P , but it does not extend below e . We argue tat tis tis point po int P solves sol ves our problem Every oter point poin t on line e lies outside te circle and tus sees" sees " te picture B wit a smaller angle tan does P In oter words, L B P i s te maximum we seek seek To answer te question que stion tat was asked, we need to kow ow fa far point P is from from te wall Tis distance is equal to M, and so we examine te rigt triangle triangle � M B We kow tat tat ypotenuse B 5 because B is a radius radius of te circle lso lso,, M B 3, and ence by te te ytagorean teorem, we see tat M 4 Te answer ans wer to te problem, proble m, terefore, is tat te art lover sould s ould stand 4 feet feet from from te wall wall •
oution.
S ince we used use d te ytagorea ytagoreann teorem teorem in te solution to roblem roblem 1 20, 2 0, peraps tis tis s a good place to digress to give an elegant noncomputational proof proof
I CRCLES AND ARCS
9
(Pythaora) Ifa right r ight triangle trian gle has arms a rms of of lengths a an andd b and its hy hypotenu pote nuse se has length c then t hen a + b c
(121
b
b
b
t b
b
b
b
Figure 1.29
s we sall explain, te wole proof proof isis visible in Figure 1.29 On te left, we see our ou r given triangle triangle.. Te middle diagram sows a s quare of side a +b decomposed into two squares sq uares and an d four four rigt triangle triangless . We observe, obs erve, furtermore, furtermore, tat eac of te rigt triang triangles les as as arms of lengt lengt a and b and so eac is congruent to te original triangle by SS S S . Te two smaller squares in te middle diagram diagram ave side lengts a and b and so te area remaining in te big square s quare of side a + b wen four copies of our given triangle are removed is exactly a + b On te rigt, we see anoter square of side a + b decomposed tis time into one square s quare and four rigt triangles triangles.. gain, g ain, eac of te rigt triangles triangles is congruent si de lengt lengt of te smaller square is exactly exactly c It to te given one by SS , and so te side follows tat te area remaining wen four copies copi es of our tri tri angle are removed from from a square of side a + b is c We just ju st saw s aw,, owever, tat tis area is equal equa l to a + b and tus a + b c as required. Tere is one crucial detail we ave omitted. It is clear tat te square" of side c in te gure on te rigt is a rombus, but wy wy are its angles equal to 90°? ook, for example, exampl e, at te top of te diagram on te rig rigt.t. We see tat te anle of te rombus togeter wit L 1 and L 2 make a straigt angle of 180° In te original triangle, triangle , owever, we know tat L 1 , L 2, and a rigt angle sum to 180°, and it follows • o f te rombus is 90°, as required tat te angle of
Proof.
We mention an alternative argument tat tat could be used us ed to see s ee wy te rombus wit side lengt c in te rigt rig t diagram of Fiure 129 as to be a square. Observe Obs erve tat tere is a rotational symmetry in te gure In oter words, if we we rotate rotate te entire large square by a quarter turn, wat results resu lts is identical to te gure wit wic we started. It follo follows ws from tis tat all four four angles of te rombus must be equal equal Since we kow tat te sum of te angles of any quadrilateral is 30°, we deduce tat eac corner of te rombus is 90°, and tus tus te rombus is a square. s quare. We return return now to our study of angles in circles . Tere is a special speci al case of Teo rem 11 tat is used us ed so s o often often tat it deserves special spe cial mention
30
CHAPTER CHAPTER
THE BAS CS
Given ABC the angle at vert ve rtex ex C is a right angle and onl on ly side A B is a diameter diamete r of of the crcumcircle c rcumcircle
(1.2) COROLLARY.
�
Proof. We ow tat te cicumcicle exists by Teoem 15 15 In tis cicle, AB en B to denot written n ot den otee the ac not couse, w e have writt Hee, of couse, i s measued by 2 L C. He ly when determine. ine. hord A B i s a diameter precise precisely se points determ these containing C that the of. proof. • completes the pro 90 ° . This comp when L C 90° equivalently, wh B ° 1 80°, or equ =
eaps we sould comment on te pase if and only if," wic appeas in te statement of oollay ool lay 1 22 Te if pat of te statement statement assets ass ets tat A B is a diamete, then L C is a igt angle In ote wods, tis pat of te coollay says tat tat weneve we know tat AB is a diamete, diamete, we can conclude tat tat L C is a igt igt angle Te only only if on ly wa wayy tat L C can be a igt angle is pat of te te coollary coo llary tells us tat te only i s fo A B to be a diamete In ote wods, L C is a igt igt angle, then A B must be a diamete diamete e only if pat pat of te assetion, ass etion, teefoe, teefoe, is pecisely peci sely te convese of te if pat Te if and only if fom fom of matematical matematic al statements statem ents is so common comm on tat tese fou wods ae often often combined combin ed into te single sing le abbeviation i," altoug we do not use us e te abbeviation in in tis book Geneally, Geneally, wen we ae asked to pove an assetion tat says someting in te fo fo " abc if and only if xz, we ae expected to povide two poofs We pove te if pat by assuming xz and someow deducing abc We pove pove te only if pat by stating stat ing all ove and assuming abc and deducing Of couse, couse, it doesn doesn t eally te wete wete we do if o only if st, as long as bot get done In some exceptional exceptional cases cas es,, oweve, oweve, it is possible to pove bot te if and te only if pats of an if and only if assetion simultaneously simultaneous ly Te poof we just ju st gave fo fo oollay oollay 1 22 2 2 is an example of t t ee is an amusing amusing pactical" application fo fo oollay 1 22 22 Suppos S upposee a cicle is printed printed on a piece of pape, pape, and we want to nd its exact cente If te cicle ad been dawn wit a compass, we could old te pape up to te ligt to nd te tiny ole tat would mak te cente, but fo a pinted cicle, we need a dieent metod Take anote piece of pape, wit a 90° coe, and place it down ove te printed cicle so tat it coves coves te appoximate appoximate location loc ation of te cente and so tat its coe lies exactly exactly on te cicle cicle ow mak te point on te cicle wee eac of te two sides si des of te te igt angled coe cosses coss es te cicle cicle emove te coveing coveing pape and and use a staigtedge staigtedge to join tese two maks maks y oolla oollayy 1 22, te line segment tat esults is i s a diamete diamete of te cicle epeat tis pocess poce ss to daw a second diamete and mak mak te point wee te two diametes coss Tis is clealy te cente of te cicle Obseve Obs eve tat tat given any line segment A B , tee is a unique uni que cicle aving A B as a diamete Tis, of couse, is te cicle centeed at at te midpoint of A B and aving adius A B . note way to state oollay 1 22, 2 2, teefoe, teefoe, is tat L C is i s a igt angle in A B C if and and only if point C lies on te unique cicle aving side A B as a diamete diamete ote tat L C < 90° if C lies outside of tis cicle and L C 90° if C is in te inteio ecall tat angles smalle tan 90° ae said to be acut, and angles between 90° and 180° ae obtu ecall tat a line is tannt to a cicle if it meets meets te cicle in exactly e xactly one point Troug evey evey point P on a cicle, cicle, tee is a unique tangent line, wic is i s necessarily neces sarily
I CRCLES AND ARCS
31
perpendicular to the radius terminating terminating at P One wa to see s ee wh the tangent tangent must be perpendicular to this radius is from the point of view of calculus calculus In Figure 130we have drawn tangent line P T and we want to compute LPT, where P is the diameter diam eter that extends radius P. hoose hoos e a point Q on the circle near draw the secant seca nt line P Q If we move Q closer and closer to P , we see that LA P T P and draw is the limit of L P Q ° utt as Q approaches P , we observe u obs erve tha Q 180° since Q P is a semicircle semi circle It follows follows that LP Q approaches 90° and so P is perpendicular to the tangent tangent as claimed nother fact that that we can see in Figure 1 30 is that L Q P T between chord cho rd P Q and tangent P T is the complement of LP Q Thus Thus 1 1 1 L A P Q ° 90° 90 °  A Q  ( 1 8 0° 0° A Q ) °  P Q . L QPT 90 ° 2 2 2 In oher words words we have the the foll following owing
=

=

=
A
Figure 130
=

s
Figure 1.31
The age betwee betwe e a chor cho rd ad a d the taget at oe o e of its ed e dpoits po its • is equa eq ua i degrees to ha the subteded sub teded arc arc
(1.) THEOREM.
We leave leave as an exercis exercisee the following following consequence c onsequence The age betwee be twee a secat ad a taget meetig meet ig at a poit po it outside a cir ci rce is equa equa i degrees degrees to t o ha the derece erec e of of the sub s ubteded teded arc
(1.4) COROLLARY.
Two circles are said to be mutuay tannt at a point P if P lies on both circles circles This can happen exteall when and the same line through P is tangent to both circles tangent line or inteally inteally when the circles te two circles are on oppos ite sides of te tangent are on the same side si de of the tangent and one circle is inside the other Given two exteall mutuall tangent circles with common point P , draw two common secants D and and B C through P , as in Figure 131 Show that B and C D are parallel parallel
(1.5) PROBLEM.
Since Si nce B C is a transversal to the two lines B and and CD it suf sufces ces to show show L B and ST and L C are equal raw the common tangen that the alteate interior angles Theorem 1 23 lso of course L C ° P D, and and note that L D P T ° P D b Theorem
oution.
32
CHAPTER CHAPTER
THE BAS CS
so itit follows ollows that that L C L D P Similar Similarly ly L B L AP ut ut L D P T and and LAP are vert vertical ical angles angl es so s o they they are are equal and it follows follows that L C L B as desired d esired • Exe rc is e s IF
Suppose Suppo se that A B and CD are are chords on on two circles circles with equal radii Show that that A B C D if an a nd on ff A B C D NOTE: The assertion as sertion of this problem i s really really pretty pretty obvious obviou s but nevertheless nevertheless here you should provide a proof here F.2 et A B C and D be placed consecutively on a circle et , X, , and Z Show of A be the the mdponts mdponts B B C C D and DA respect respectvely vely S how tha thatt chords chords and X Z are peendicular peendi cular F. et P be a point exterior to a circle centered at point poin t and draw the two tangents tangent s to the the circle circle from from P et and T be the two points of tangency S how that P bise bisect ctss L P an and P P 1 80 F.4 In the situation of the 80 L P the previous previous exercis exercisee show that F.S In A B C prove that that L A is a right angle if and only if the length of the median from A to B C is i s exactly exactly half the length length of side B C F.6 In quadrila quadrilater teral al AB C D assume that L A 90 L C raw diagonals A C and BD and show show that that L D A C LDBC F.7 In the situation of the the previous previous exercis exercisee ass assume ume that that diagonal A C bisects bis ects diagonal B D rove that he quadrilatera quadrilaterall is a rectangle rectangle F.8 In Figure 1.32, two circles meet at points P and Q and diameters diameters P A and P B are drawn drawn Show S how that that line line A B goes throug throughh point Q F.
p
Figure 132
Figure 133
I CRCLES AND ARCS
33
p
Q
Figure 134
Figure 135
In Figure 133, point is the center of the circumcircle of �ABC and the bisector of L A is extended to meet the the circle at P rove that radius P is perpendicul perpendicular ar to B B F. Given �ABC inscribed in a circle draw the bisector b of the exterior angle at A Suppose that line b is not tangent to the circle and let P be the point other than A where b meets the the circle circle Show that that P is equidistant from B and C In circle show that A is equidistant e quidistant the exceptional case where b is tangent to the circle from rom B and and C F. In Figure 134, w e have drawn drawn three circles of equal radius that g o through a common point P and we have have designated b A B and C the thee thee other other points where these these circles cros s Show S how that that the unique circle through A B and C has the same radius as the original three circles circles H se s e Exercis Exerc isee 112 to show that �ABC is congruent to the triangle foed b the centers of the thee thee given circles circles se the fact that that circumcircles of congruent triangles triangles have equal radii F.1 In Figure 135, we have have selected sele cted two points poi nts P and Q outsid ou tsidee of a circle circl e and we have drawn drawn two secants se cants through each point in such a wa that these four secants intersect the circle in the the four four points A B C and D D as shown shown Show S how that that the the angle bis b isectors ectors P X and Q of L P and L Q are peendicular peendicu lar to each other H et U and V be the points where bisector bis ector Q meets secants P D and PC respec respectiv tively ely Show that that �P U V is isosceles b b provi proving ng tha thatt L P U V LPVU F.1 rove orollar 124 Show that if point P is extea to a circle and tangent P and secant P A B are drawn drawn wher wheree A and B lie on the circle circle then o LPA ( B  A ) F
34
CHAPTER CHAPTER
THE BAS CS
Figure 136 F.14
In Figure Figure 1 3 6 segment segment P Q is a chord common common to two circles circles and it bisects L R P where where R and and lie on the circles circles as shown Each of the the chords chords P R and P iiss cut b the other other circle at points and U U rove rove that that R U
Polygons in Circles
polgon is said to be ruar if all of its sides are are equal and also als o all al l of its angles are equal n equilateral equ ilateral triangle triangle for example certainl has equal sides si des b denition and b two applications applications of the pons asinorum it is eas to see that all three three angles must be equal too n equilateral triangle triangle theref therefore is a regular regular 3gon 3go n n equilateral equilateral 4gon is a rhomb rhombus us but of course course a rhombus rhombus need not have equal angles angles and so it is not necessaril neces saril a regular polgon polg on square however however is a regular regular 4gon In general general for 3 a regular gon can be drawn drawn b markng markng equall e quall spaced points around a circle and then drawing the chords connecting consecutive points ( equall equall spaced sp aced we mean that that the arcs formed formed b pairs pairs of conse con secutive cutive points point s are all equal in degrees degrees ) It foll follows ows b Exercise Exerci se F 1 that that the chords are equal in length To see that the polgon formed ormed b the equall spaced sp aced points is regular regular we must mu st also establish that the angles are are all equal Each of the arcs is clearl equal in degrees to 3 60/ degrees degrees and each angle of the polgon subtends subtends an arc arc consisting cons isting of  2 of these small arcs It follows b Theorem 116 that each of these angles is equal to (  2)(360/) 2)(360/) 1 80( 80 (  2)/ 2) / degrees degrees and and thus thus the the polgon is regula regularr as as desired desired Note that that the sum of all the angles of our regular regular gon is times 1 80( 80 (  2) / This is 1 80( 80 (  2) degrees degrees which fort fortunat unatel el agrees agrees with what we know to be the sum of the the angles of an arbitrar arbitrar gon gon Now draw the radii radii joining the center of the the circle to the equall equ all spaced points points This Thi s subdivides the interior of the gon into into isos i sosceles celes triangles triangles with equal bases of length length s , the common side s ide length of the polgon These triangles triangles are all congruent congruent b S S S and thus thus the lengths of the the altitudes altitudes of these triangle triangless (drawn (drawn from from the center of the circle) are all equal n one of these altitudes altitudes is said s aid to be an apothm of the regular regular polgon polgo n and we write a to denote their their common length Since Sin ce the area of each of the the isosceles iso sceles triangles triangles is sa, we see s ee that the area area of the entire regular gon gon is sa pa, where we have written p s, the perimeter perimeter of the the polgon pol gon Fix an integer 3 Given a circle how should shou ld points on this circle be chosen so as to maximze the area area of the corresponding corresponding g gon? on?
(1.6)
I G POLYGONS POLYGONS N CRCLES
35
sol ution to an optimization (max It is often the case that there is smmetr in the solution min") problem and so it seems natural to guess that the areamaximizing inscribed polgon should be regular In other words the points should sh ould be equall spaced s paced around the circle circle s s we shall see this is correct R In Figure 1.37, for instance we A 5. In the illustrate illu strate the case where whe re left eft dia diagr gram am AB BC C D D and and S B are all equal equal and thus thus each is 72°. v T The circle in the right diagram has an equal radius radius and we have placed place d ve u points around it in such a wa that not all of the arcs are equal equal We need Figure 137 to show that the area of the regular penta pentagon gon A B C D is i s strictl s trictl greater ST U V than that of pentagon R ST We begin with a discu d iscussi ssion on of general general strategies strategies for solving geometric g eometric optimization problems as illustrated il lustrated in this this case ca se There are are at least two was we might mi ght proceed The straightfor straightforwar wardd approach would be to show s how directl that the the area of pentagon A B C D exceeds exceed s that of an other pentagon (such (s uch as RSTUV) inscribed ins cribed in the same or an equal circe and which does not have have all arcs equal lteativel lteativel we could show that given an pentagon pentagon (such (s uch as R ST U V) in which the arcs arcs are not all equal it is possible pos sible to nd a pentagon larger in area inscribed in the same circle This would wou ld show that no pentagon other than one with equal arcs could c ould maximize the area If we somehow knew that an areamaximizing areamaximi zing pentagon pentago n necessaril neces saril exists exis ts it would follo follow w that that all all of its its arcs must mu st be equal and the problem problem would be solved For roblem 1.2, the direct approach approach seems seem s difcult difcult but it is not hard to show that given an gon go n inscribed ins cribed in a circle circle and having arcs that that are are not all equal equal there exists another gon gon with w ith larger area inscribed in the same circle To see this observ obs ervee that since the arcs are are not all equal we can surel nd two consecutive cons ecutive unequal arcs ar cs and hence we can nd three consecutive vertices R, S, and T of our our gon gon where R S and are unequal We show that it is possible pos sible to move point S, leaving all of the remaining 1 points xed xed so s o that the area area is increased The area of the whole polgon pol gon can be viewed as the t he area of �RST plus the t he area of pol gon that lies on the other side of chord R T. We are assumin ass umingg that S that part of the polgon is not the midpoint of R ST, and we label the midpoint S', as in Figure 1.38.
Sf
Figure 138
36
CHAPTER CHAPTER
THE BAS CS
It is clear from from the diagram that the perpendicular perpendicular distance h from S to line R T is less than the distance h' from S' to R T. s h is the height of � R S T with respect to the base R T and h' is the height hei ght of �RS'T with respect to the same base it follows follows that the area of �RST is les l esss than that of �RS'T. If we move point S to S', the eect is to increase increas e the area of the triangle without withou t aecting aecting the rest of the polgo pol gonn and the eect eect on the area of the whole polgon pol gon is thus an increase increase.. This shows that an gon gon inscribed ins cribed in a circle where the arcs are unequal cannot have have the maximum maximum pos sible area. area. To complete the solution of roblem 12, it sufces now to show that among all possible gons inscribed in a given circle there is one for which the area is a maximum. This This maximzing polgon is necessaril nece ssaril regular b the foregoing foregoing argument. The existence of an areamaximizing gon gon inscribed in a circle follows from a general principle called compactness co mpactness which we explain somewhat informall informall and without proof. proof. erhaps readers will recall from their stud of calculus that if (x is a function function of a real variable x dened for a : x : b, and (x is continuous c ontinuous in this interval then the function necess nece ssar aril il takes on a maximum value at at sme point p oint in the interval. interval . There are two crucial hypothes hypo theses es here here that is continuous continu ous and that that x runs runs over a closed clos ed interval interval . (The interval interval cons con s isting is ting of those thos e real numbers numbers x such that a x b is said s aid to be be cod because it includes includ es both endpoints.) endpoints .) More generall generall it is true that that a continuous function even of several variables variables takes on a maximum value if each variable runs over a closed clos ed and bounded bounde d set. (We (We shall dene these terms presentl.) presentl.) In I n our geometr problem we can think of the area of an gon gon inscribed ins cribed in a circle as a continuous c ontinuous function of variables the points which we do not require to be all dierent. That this function is continuous means esse e ssentiall ntiall that a small pertur perturbatio bationn in the locations of the the points results in at most a small change in the area. area. Each point is required to to lie on o n our circle which as we shall sha ll see se e is closed clo sed and bounded. It follows that that for for some choice of points on the circle the area functi function on takes on a maximum value value.. set of points poi nts is boundd if itit is contained in the interior interior of some circle pos p ossibl sibl a ver large circle. Some So me examples of bounded sets are a line segment a circle and the interior of of a circle; unbounded sets s ets include a line the exterior of a circle and the interior interior of an an angle. angle . efore we we can explain exp lain what it means for an an arbitrar arbitrar set of points poi nts to be close cl osedd we we need nee d another another denition. denition . Given a set of points in the plane we sa s a that a point P is adjacnt to if ever ever circle circl e centered at P, no matter matter how small contains at least one point of in its interior. It is obvious that if P is actuall a member of then P is adjace adjacent nt to but it is also poss po ssible ible for for a point to to be adjacent adjacent to a set s et without actuall being in the set. se t. n endpoint of a line segment for for example is not no t in the interior of the segment but it is adj adj acent to the interior. plane set is cod if ever point adjacent adjacent to the set is actuall actual l a member membe r of the set. For example example the interior of of a circle is not a closed set because becau se the points of the circle are adjacent adjacent and et are not in the the set which consists onl of interior interior points. The disk formed b a circle together with its interior is a closed set s et however as is i s the circle itself. line is a closed set and so s o is a line segment provided that that we include the endpoins of the segment. line segment s egment without its endpoints however is not closed clos ed nor is a circle with one point deleted.
IG POLYGONS N CRCLES
37
set that is both closed clo sed and bounded is said s aid to be compact If we are willing to believe the theorem theorem that a realvalued continuous function of several several variables variables,, each of which runs over a compact set, attains a maximum and also a minimum value, then roblem 1.2 is solved s olved This is because bec ause the domain domain of choice for eah point is a circle, which is a compact set It is vital vita l that that the domain domain of choice for for each point be the entire entire circe, circe , with no restriction restrictio n We must not insist, insi st, therefore, that our poins poi ns all be diffe different rent Observe that the maxmin existence exi stence theorem also guarantees that a minimum area can be obtained b a suitable s uitable choice of points on a circle circle The minimum area of zero zero is attained, attained, for example, example, when whe n all points are the the same same There are, of course, course , also other congura con gurations tions that ield this minimum area Our discussion of circles would not be complete without some mention of the amazing number 31415 denition, is the ratio ratio of the perimeter, perimeter, also als o called the circumference, circumference, of a circle to its diameter diameter It is not amazing that this ratio is the same for for all circles, circles , independent of size If we change the scale scal e and multipl the diameter diameter b some so me constant cons tant k, it is clear cl ear that the perimeter perimeter is also multiplied b k and the ratio ratio remains unchanged unch anged What does seem remarkable is that this this same number is also involved in the formula formula K giving the area of a circle in terms of its radius It seems that this calls for for an an explanation Fix a circle of radius and let Kn denote the area area of a regular regular gon gon inscribed ins cribed in the circle It should sho uld be reasonabl clear, and we will not give a formal formal proof, that that the area K of the circle is the limit of the polgon areas Kn as 0 We have seen that Kn Pnn, where Pn and n are, respectivel, respective l, the perimeter and apothegm of a regular gon gon inscribed ins cribed in our given given circle Obs erve that as gets gets lar l arge, ge, Pn approaches of the the circle and n approaches the radius Thus the circumference
K nlim Kn (nl im Pn)(nlim n) n n
as we expected We mention that the formulas for the surface area and volume of a sphere sph ere in terms terms of its radius also involve the seemingl ubiquitous number Without giving proofs, proofs , we w e remind the reader that these these formulas formulas are S and V We cannot resist resis t mang a few few more comments comments about about the sometimes misunderstood misun derstood number J We wrote earlier that tha t J 3.1415 where where the dots represent represent an innite number of omitted decimal places There is certainl nothing msterious or unusual in the fact that decimal expansion of J does does not terminat terminatee ; the same s ame can be said of wellundersto od numbers numbers as 1/3 or / 7 These are are rationa numbers, which whi ch means means such su ch wellunderstood that the are quotients of integers It is well known that the decimal expansions of rational numbers either terminate or eventuall repeat, but the number J is irrational, irrational, and its decimal expansion never repeats The same can be said of numbers such as 1.4142 but in a certain certain sen sense, se, J is even more unlike most of the numbers we meet ever da than is To understand the true nature of J we must distinguish between algebraic and transcendental transcendental numbers numbers ecall that a poynomia is an expres expression sion of o f the the form (x) n n the constants cons tants i are are called c alled the cofcint of nx n x + n_Ix  1 + + l X + o where the ass ume that the coefcient n of the the highest highes t power of x is the polnomial (x) and we assume
J
38
CHAPTER CHAPTER
THE AS CS
nonzero number r is said to be a root of the polnomial polnomial () if we get 0 when we plug in r in place of Thu the number number r = is a root of the polnomial () = 2 Note that the coefcients of this polnomial pol nomial are a = 1 , a = 0, and a = 2; in particular, all the coefcients are intgers number r is said s aid to to be abraic if it is a root root of some polnomi po lnomial al with integer coefcients coefcients Thus Thus is algebraic and so too is ever rational number For example, example , 2/7 2 /7 is a root root of the polnomial poln omial () ( ) = 7 2 number is trancndnta if it is not no t algebraic, algebraic, which means that it is not a root of an an polnomial with integer coefcients coefcients In fact, fact, 1 is transcendental, and this this ma be what makes makes ths number seem so msterious m sterious and unusual unusual (ctuall, transcendental numbers are not reall unusual; there is a sense in which it is true to sa that most numbers numbers are are transcendental What is unusual un usual is i s to have in hand a particular number such as 1 or e = 2 7 1 82 8 2 that is actuall known to be transcendental) transcendental ) It is i s the fact that that 1 is transcendental transcendental that that explains wh it isis impossible imposs ible to square a circle We shall explain what that means in hapter The fact fact that there is i s no n o polnomial poln omial equation for 1 does not, of course, mean mean that there is anthing haz or ambiguous ambiguo us about abou t this number There are, in fact, fact, man formulas formulas that can be proved to give 1 exactl To To mention just three of these, these , we have
1 =
4
1 +
xercses G Gl
G
G.
1
In Figure 138, show that as moves along the distance h from to R varies in direct proportion proportion to the the product R H se s e areas Suppose Suppo se that A, B , C, C , and D are four four consecutive vertices vertices of a regular polgon polg on but do not assume as sume that this this polgon pol gon is inscribed i nscribed in a circle circle We know that that there there is ssome ome unique circle throu through gh points points A B C rove rove that that this this circle also goes throug throughh D D H et be the center of the circle so that A = B = C se congruent triangles triangles to show that C = D It follo follows ws easil e asil from from the result of o f this exercis exerc isee that ever ever regular polgon polgo n can be inscribed in a circle In Figur Figuree 1 39, 3 9, equilater equilateral al A A B C is inscribed inscribed in a circle and and point point P is chosen arbitraril arbitraril on Show that that A P + P C = P B H Extend chord PC P C to point Q , as shown, so that PQ P Q = P B and and then then draw draw B Q
H SMLARTY
39
A
Figure 1.39 G4
et K and a denote respectivel the area and and the apothegm apothegm of a regular regular gon g on inscribed in a circl of radius r Show that r
1 G5
a) /2 where as in the previous In a circle of radius radius 1 , show that a = ( 1 + a) previous exercise a is the apothegm apothegm of a reguar reguar gon gon ins cribed in the circle educe that the followin followingg simple s imple iterative iterative algorithm can be used to o mpute an approximation to 1 Start with numbers u = 2 and = 1 / t each step replace u b u / S how that that the limiting limiti ng value for u is 1 and replace b ( 1 + /2 Show It is not ver hard hard to prove that that atat each step of this algorit hm the error" 1 u is actuall less le ss than half half of wat it was was at the previous previous step and so it does not take ver man iterations to get a fairl fairl decent approximatio app roximationn to 1 Tr it on programmable calculator c alculator a computer or programmable
IH
imilarity
Informall Informall speaking w sa s a that two geometric geometric gures are similar if the have the same shape shape If the gures are actuall congruent the have have both the sam shape s hape and the same size size and so similarit simi larit is a weaker weaker condition than than congruence though we have not dened shape recisel let us agree that the shape of a trangle s determined b its angles angles This motivates motivates our ofcial" denition of similarit for for triangles triangles Two triangles ang les of one are equal to the three angles angl es of the other If for are imiar if the three angles example we are given ABC and XYZ and w know that LA = LX, LB = L Y, and and the two two triangles are are similar and we write write A A B C r XYZ XYZ s was the L C = L Z , then the case for for congruent triangles triangles the order in which we write vertices vertices is sgnicant sgni cant here; it is the corresponding angles an gles that are equal other than than triangle triangless the angles do not necessaril necess aril determine determine the shape For gures other Two rectangles for for example example agree in all four four angles angle s and et one ma be a square sq uare and sim ilarit s we shall s hall see it is possible pos sible to give an altea alteative tive denition of similarit the other not s geometric gures besides bes ides triangles triangles but in general, one must that applies to all sorts of geometric
40
CHAPTER CHAPTER
THE BAS CS
know more than the equalit of corresponding corresponding angles to establish establis h that two gures are simlar si mlar For triangles triangle s , on the other other hand, it is not even necess nece ssar ar to check all three three pairs of angles angles If two angles of one one triangle are equal to the two correspo corresponding nding angles of anoher triangle triangle then we automaticall have equalit of the other pair of angles angles Gien ABC and XYZ suose L A L X and L B LY Th en L C LZ and so ABC XYZ
(17)
Proof.
Since Sinc e the the sum su m of the angles angles of a triangle triangle is 1 80 we have L C 1 8 0 (LA + LB)
1 80 80 (LX + LY)
as required
LZ •
When we use Theorem 1 27 to prove prove that two triang triangles les are similar, sim ilar, we sa that the trangles trangles are are silar si lar b Of course, is an abbreviation abbreviation for for angleangle" angle angle" ut what is the use of knowing that two triangles are simlar? What can we deduce from similarit? ABC XYZ then the t he lengths of of the co rresondi rresonding ng sides of these two t wo trian t riangles gles are a re roo roortion rtional al
(18)
efore we prove Theorem Theorem 1 28 , which is of o f fundamental importance, let us be sure that we understand understand precisel what it is telling us The assertion ass ertion that that the corresponding corresponding sides are proportional proportional tells us that there is some positive po sitive number , , called c alled a ca factor, such su ch that A B XY X Y B C · YZ , and A C XZ For example, if i f the scale scal e factor factor is equal to 3, this would sa that that each each s de of A B C s three three tmes as long as the correspondng side of XYZ XYZ and f ) 1/2 this would sa sa that the sdes of A B C are half as long as the corresponding corresponding sides of XYZ XYZ If the sides of ABC are 3 km, 4 km, and 5 k and the corresponding sides of XYZ are 3 cm, 4 cm, and 5 cm, then then the sids s ids of these two two triangles are proportional with a scale factor actor 1 00, 00 , 000 Observe Obs erve that proportionalit proportionalit of side lengths is i s smmetric s mmetric If I f the sides of A A B C are proportional propo rtional to the sides s ides of XYZ XYZ with scale s cale factor , then the sides s ides of XYZ XYZ are proportional to those thos e of A B C with scale sc ale factor factor 1 / There is another wa to think think about proportionalit Suppo se we know know that A B C XYZ XYZ Theorem 1 28 2 8 tells us that the side lengths of A B C are proportional to to those of
IH SMLARTY
41
�XZ, but it does not mention an particular sc ale factor factor We We can determine A however, from an an one of the the equations AB A XY BC A or YZ AC A . XZ The signi sign icance cance of o f proportionalit proportionalit is that we get the same value for A from each e ach of these equations. equations . In other words, words , the corresponding sides are proportional proportional if and and onl if AB BC AC XY YZ XZ The ke t o the proof of Theorem Theorem 28 i s the following following lemma, which relates paral lelism lelis m to proportionalit. proportionalit. et and V be oints on sides AB and AC of �ABC. Then V B C and only only A AV AB AC
(1) LEMMA
Write a BA B A B and VC AC. Then A A B B ( )AB, and similarl, simi larl, A V ( )AC. Thus A AV a and · AB AC It follows that the ratios A AB and A V A C are are equal if and onl if a . It sufces now no w to show that a and are equal if and onl if V B C . raw C , a s in Figure 40 and compare the area of �B C with that of �ABC. iewing A B and B as bases, we see that these triangles have equal heights, heights , and thus KB U C B = a AB KABC and KB U C aKABC Similarl, KBVC fKABC and it follows that the quantities a and are equal if and onl if �B C and �B V C have equal areas areas We therefore therefore this equalit e qualit of areas happens if and onl onl if V B C . need to show that this Since � B C and �BVC share base BC, their areas areas are equal if i f and onl if therefore, is i s to show s how that E V F the have equal heights E and V F. Our task, therefore, if and and onl if V and E F are parallel. parallel . Observe Obs erve that E and V F are parallel since si nce each of these these lines is perpendicular to B C. If E V F therefore, it follows follows b Theorem 8 that V F E is a parallelogram, parallelogram, and thus V E F as required, as in
Proof.
42
CHAPTER
THE BAS CS A
E
F
Figure 1.40
Figure 1 40 4 0 onversel, onversel, if U V E F then U V E F is a parallelogram b denition, and so U E V F b Theorem 1.. This completes the t he proof • ecall that that w are are given given that that A A B C XYZ, XYZ, and we will prove prove that XY A B X Z A C Smilar rasoning rasoning would also show that XY A B Y Z B C , and thus all three ratios are equal and the sides sid es are proportional If XY AB , th then A BC XY XYZ b S In this this case, c ase, X Z AC AC,, and and so XYAB XYAB 1 XZ XZ A C , as desired desi red We We can thus thus suppose suppo se that XY and AB are are unequal unequa l , and it is no loss los s to assume ass ume that that X Y is the shorter of these thes e two segments hoose pont pont U on side side AB A B of ABC A BC so tha thatt AU A U X Y and draw draw U V parallel parallel to BC, B C, where where V lies lies on side side AC Sinc S incee U V B C, we have have LA U V L B L Y and and LA VU L C L Z ut ut AU X Y, and and so we we deduc deducee that that A U V XYZ b S In particular, particular, we have A V X Z Since Since U V B C, we ow ow b em emma ma 1 29 that that AU AB A VAC Sinc Sincee A U X Y and A V XZ, XZ , t follows follows that XY A B XZA XZ A C , and the the proof proof is complete •
Proof Proof of of Thorm 1 8
What What is wrong with with the following eas proof" of Theore Theorem m 1 28 ? In A B C , the the law of snes gives sin(A) sin( A) sin(B) sin( B) , and and thus thus sin(A) sin(B) sin( B) In an an tria triangle ngle,, therefor therefore,e, the ratio of two two ssides ides is equal to the ratio of the sines si nes of o f the opposite oppos ite angles angles In the the situation of Theor Theorm m 1 28 , we have have BC sin(A) sin(X) Y Z AC sin(B) sin(Y) X Z and a bit of algebra ields XZAC YZBC The equalit XZAC XYAB fol folows ows simil si milarl arl and thus the sides side s of the trangles are are proportional as desired des ired and the proof is complete To see wh this argument argument is not valid, valid , recall that in proving the law of sines sin es,, we used the fact that if ABC is a right triangle with LC 90° and LA , then then sin si n () B C A B nderling this oppositeoverhpotenuse" formula for for the sine of an angle is i s the assumption a ssumption that the ratio B CA C A B has a value that is independent independent of the particular particular triangle in question que stion In other words words if XYZ XYZ is another triangle triangle and L Z 90° and and L L , we expect that it is also true that that sin () ( ) Y Z X Y This is correc cor rectt since A B C XYZ XYZ b , , and hence b Theorem Theorem 1 28 , corresponding sides are proportional proportional and we have Y Z BC and XY A B for some scale sc ale fact factor or
IH SMLARTY
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It foll follows ows that Y Z X Y B C A B , as expcted We need need Theore Theorem m 1 28 to justif the the oppositeoverhpotenuse oppositeoverhpo tenuse formula for for the the sine, sine , and hence it would be circular reasoning reasoning to use sines to prove Theorem Theorem 1 28 28 Of course, now that that we have proved Theorem 1 28, 28 , other trigonometric trigonometric functions functions to stud stud triangles we can safel use sines and other s a demonstration of the the power of similar triangles triangles to prove prove interesting theorems, we present the the followi following ng In 6BC, draw U V parallel to BC as in Figure Figure 1 4 1 Show tha thatt the intersection of B V and UC lies on the median of 6BC from point A
(1 0) PROBLEM
A
E
n
Figure 1.41
s shown shown in in Figu Figure re 1 41 4 1 , let P be the intersection point of B V and U C. raw line A P and let D and E b e the the points points where this this line crosss cros ss U V and B C For convenience, w e denote the lengths U D, D V, B E, E C, D P, and P E b r, s, m, n, x, and y , respectivel, as indicated in i n the the gure Our task is to prove that E is the midpoint midpoint of AC , and so we need to to show that m n First, note that 6U D 6BE b because U D BE, and and so LA U D LBE and LADU LEB We conclude that rm ADAE, AD AE, and similar si milarl, l, sn AD AE This gives rm sn, and hence nm sr Now consider 6DU P and 6EC P cause U D EC, we have have L DU P L PC E and L U D P L C E P. Thus 6DU P 6EC P, and hence r n x Similar reasoning with with 6D V P and 6EBP ields sm xy, and we conclude that rn sm, and thus mn sr Sinc we prviousl had nm sr, we • see that mn nm, and it follows follows that tha t n m , as rquired
oution
There is a case of emma 129 that occurs sufcientl oftn to deserve separate mention et U and V be the mido midoints ints of of sides B and C resec resec tivel tivelyy in 6 B C. Then U V B C and U V BC
(11) COROLLARY
That UVBC is immediate from emma 129 since UAB 12 , and thus b Theorem 1 28, 28 , we V A C In this situation, 6UV 6BC b , know that U V B C A UA B 1 /2 /2 It follows follows that U V B C, as required required •
44
CHAPTER CHAPTER
THE BAS CS
There are two useful similarity criteria other than These are SSS and SS which should sho uld not be conf co nfused used with with the congruence criteria criteria having the same names The SSS SS S similarity criterion criterion asserts as serts that two two triangles must be similar simil ar if we know that the the three sides of one of the triangles are proportional to the three corresponding sides of the other triangle triangle We state this formally formally as follows Suose that the sides of of 6 A B C are a re roo roortional rtional to the corre THEOREM Su sonding sides of of 6XYZ 6XYZ Then 6 A B C 6XYZ
y hypothesi hypothesiss each each of XY, XZ, and Y Z is equal rspectively to some scale factor times AB, AC, and BC If it happens that th at AB XY, then 1 and an d all three sides of 6ABC are equal to to the corresponding corresponding sides of 6XYZ In this this case case the triangles triangles are are congruent by SS S and so they are automatically automatically similar s imilar We We can thus assume ass ume that A B and X Y are are unequal and it is no loss lo ss to assume ass ume that X Y is the shorter shorter of these these two segments hoose point U on segment A B such that A U = X Y and draw U V parallel to B C with V on AC It follows that 6AUV 6ABC by and hence the corresponding sides of these these two triangles are proportional proportional The scale s cale factor factor for for this proportionality is A U / A B X Y / A B , and thus A V AC X Z and U V BC YZ It follows that 6AUV 6XYZ by SSS, and hence L A U V L Y and LA VU LZ Thus L B L Y and L C LZ, and hence 6ABC 6XYZ by •
Proof
One way to appreciate the signicance of the SSS similarity criterion in Theo rem 1 . 32 is to imagine a photocopy machine with a control control that that enables the user us er to make make a reducedsize image im age If the reducing reducing control is set to 75 %, for example then the copy copy will be only three quarters quarters as large as the orginal but in other respec respectsts it resembles the orgnal document Suppose Suppo se that the the original happens happens to be a geometry geometry diagram such su ch as those in this book bo ok (Of course course owners of this book would not really really allow unauthorized reproduction in violation of copyright rules but this is just a hypothetical example) engths of line segments se gments in the image are are shorter than the the corresponding lengths in the orignal gure g ure and they are proportional with a scale factor factor of 7 5 ut how do do the angles in the image compare with the correspondng correspondng angles in the original? original ? We know of course that the image has the same shape (whatever that means) means) as the original original and so w expect every angle in the image image to equal its corresponding correspon ding angle angl e in the orgnal orgnal Theorem Theorem 1 32 justies this this expectati expectation on as follows follows Suppose the original gure contains some angle (say LABC) and suppose that the points in the reducedsize reduceds ize copy that correspond to to A, B, and C, are X, Y, and Z , respectively respectiv ely To To see why LABC = LXYZ, we can suppose that egment A C was drawn in the original so that segment X Z appears in the the copy For simplicity we shall assume ass ume that that neither LABC nor LXYZ is a straight angle so that we have actual triangles triangles 6ABC in the orignal and 6XYZ in the photocopy photocop y We ow that the the sides sid es of 6XYZ are proportional to the sides of ABC, with scale s cale factor factor 75, 7 5, and thus 6ABC 6XYZ by SSS In particular LABC LXYZ, as expected
IH SMLARTY
45
We are are now read to to dene simi s imilarit larit for gures that ar not nees saril saril triang triangles les Two arbitrar gures gu res ar imiar if for for each each point in one o fthem, there is a corresponding point in the other so that all distances in the rst gure are proportional, with some particular scal factor A to the corresponding distances in the second gure gure aution: aution : This is not how how we dened dene d similari simil arit t for for triangls triangls,, but we know know b Theorem 1 28 that similar triangles do satisf this condition, and conversel, b Theorem 132, we see that triangles triangles that are similar simil ar in this proportionalit proportionalit sense se nse are actuall similar s imilar triangles triangles,, according to our earlier denition denition Finall, we coe to to the S SSS similarit sim ilarit criterion criterion,, which guarantees that two two trian triangles gles are similar simil ar if two pairs of corresponding sides s ides are proportional proportional and the included inc luded anges are qual THEOREM Given �ABC and �XYZ assume that LX XYjAB = X Z j AC The Thenn �A BC �XYZ
LA and that
If X Y = AB, then X Z A C and the the triangles triangles are congruent congruent b SS S S,, and hence the the are automaticall similar s in the proof of Theorem 32, 32 , therefore, therefore, we can assume assum e that X Y < AB, and we choose point U on side A B of �ABC so that previous proof, we draw U V parallel to B C with V on sid A U = XY s in the previous obs erve that that � A U V I �ABC b Then A V j A C = A U AB AC, and we observe X Y j A B X Z j AC, and hence A V XZ We conclude that �AUV �XYZ b SS , and hus LXYZ = L A U V LABC It now foll follows ows b b that �ABC • �XYZ, as required required
Proof
strikng strikng and nontrivial nontrivial application application of Theor Theorm m 1 3 3 is the followi following ng PROBLEM In Figur Figuree 1 42, 42 , point P lies lie s outside of paralllogram ABC D and LPAB = LPCB LPC B Show that LAPD LCPB lthough we mentioned that the ke ke to this problem is the SS similarit simi larit criterion, criterion, there seem to be few similar triangles in Figure 142 If we believe the assertion of the problem, however, then � PC B � P AX , where X is the unlabeled point of the diagram where P D crosses A B If we could somehow prove that these these triangles triangles actuall are similar, then the equalit of LA P D and LC P B would follow follow ut it is not clear how pos sibl l use the SS criterion to prove that � PC B � P AX . we can possib p
Figure 1.42
46
CHAPTER CHAPTER
THE BAS CS
Th secret to slving s lving this problem pro blem is to draw a number number of additional additiona l lines. lines . Extend sids C D and A D he given parallelogram so s o that they meet lines P A and P C at E ad F , respectivey, and draw segments A C and EF, as shown in Figure 1 43 We We now now hav an embarrassnent of richesthere are so many pairs of similar triangls in this gure that that it is still not quite obvious how we should proceed proceed.. Observe that LDEA L B A P because becau se these are corre sponding angle for the parallel parallel lines E D and A B . lso, we have have L D F C L B C P by similar reaoning Since B A P LBC P by hypothesis, we conude at LDEA LDFC LDFC lso, LADE = LCDF since these are vertical angles, and so ADE CDF by y Theorm 128, we conclude that ADCD E D F D , and hence by lementary algebra, we have A DE D E D C D F D Since we know that ADC = LEDF, it follows via SS that ADC EDF In particular, particular, we w e av L CAD = L FED Observe w w that LFEP LF D + LDEA = LCAD + LBA P LACB + LBC P = LACP where the penltimate equality holds bcause L B A P = LBC P by hypothesis Since LE P F LC P A, we can now conclude con clude that E P F C P A by , and so we have P C P E = AC FE W also have AC F E AD ED because becaus e we know that AC EDF ombining these these equalities , we obtain PC AC AD C B P E FE ED E D where we have se sedd the fact fact that A D C B to obtain obtain the last equality We can no no argue that PC B P ED ertainly, L P E D L PC B since each of these s qual to L P AB , and so the similarity follows by SS sinc we
ouio o Probm 14
p
F
Figure 1.43
H SMLART
have have just seen that P C j P E required
=
47
C B j ED. We conclude that L C P B LEPD, as •
What What follows is a much easier easi er result that that can be proved proved using simil s imilar ar triangles Given a cir c ircle cle and apoint P not n ot on the cr c rcle choose an a n arbitra line through through P meeting the circle circle at points X and Y. Then the quanti P X P Y dep dep ends only on ly on the poin po intt P and is independ independent ent of of the choice cho ice of of the line li ne thr th rough oug h P .
(1.5) THEOREM.
V
X
X
U
U
Figure 1.44
raw a second secon d line through through P that also meets the circle in two points U and V , as shown in Figure Figure 1 44 We must show that that P X P Y = P U P V There are two cases that ve ve must conside con sider:r: where P lies inside i nside the circle and where P lies outside In either case draw draw line segments U X and V Y, as shown in Figure 1 44, and observe that L U = L Y since these inscribed inscribed angles subtend the same arc arc ext observe obs erve that L X P U = L V P Y since these are are vertical vertical angles when wh en P is inside ins ide the circle and the are in fact fact the same angle when P is outside the circle In either case cas e therefore therefore we have � P X U " � P V Y b , and thus P X j P V P U j P Y. • Elementar algebra algebra now ields P X P Y = P U P V, as required required
Proof.
a pplication tion of simil si milarit arit in this chapter we offer offer another proof of the s our nal applica thagorean theorem It is based b ased on the following following eas e as lemma Sup Suppose � A B C is a right triangle triangle with hypotenuse hypotenuse A B and let C P be the altitude alti tude drawn drawn to to the hypoten hypotenuse use Then � A C P " �ABC " �C B P.
(1.6) LEMMA.
The rst similar s imilarit it follows follows b b since L A = L A and LAPC = 90° = LAC B. • The proof of the the second s econd similarit s imilarit is similar
Proof. Pro of.
In the situation of emma 136 write as usual a = BC, b = AC, and c = AB Since �ACP �ABC, we have A P j A C = A C j AB, and it follo follows ws that A P = b j c Similarl B P = a j c , ad since c = A B = A P + P B, we conclude concl ude that c = a j c + b j c Multiplication b • c ields the equation c = a + b , as desired
Arnaiv proof of Pyhagora' horm.
48
CHAPTER CHAPTER
THE BAS CS
We close clos e this section on similarit with a brief discuss discu ssion ion of the areas of similar gures Suppose we have two two similar polgons / and where the the distances in are obtained from from those thos e in / b mltiplication b some xed scale s cale factor factor If it happens that / and are squares with side lengths p and q respectivel then q = p and hence K = q = p = K In other other words words to nd the area area of we mltipl mltipl the the area of / b the square of the scale factor that that was used us ed for for distances The rule multipl mult ipl b the square s quare of the scale scal e factor" factor" works for arbitrar arbitrar polgon pol gonss and not just ju st for squares To see wh imagine im agine that / is subdivided into a ver arge number of ver small squares s quares with a little left left over at the the edges If is subdivided into corresponding correspo nding squares s quares then each square in has area equal to to times the area of the correspondng square sq uare n / Thus the sum of the the areas of the little squar squ ares es that almos almostt comprise is times the sum of the little squares in / and it follows that with vanishi vanishingl ngl small s mall error error we have K = K We do not n ot pretend that the discus disc ussion sion of o f the previous previous paragraph paragraph provides a rigorous proof of the theorem that the areas of similar gures are related to each other b the square of the scale sc ale factor factor that that relates corresponding distances Our argument argument does explain explain wh this fact fact should shou ld be true true however however and it can be transform transformed ed into a correct proof proof using limits and other ideas from from calculus
xercses 1H.1
1 H .
1H. 1H.4
1H5
1H6
Given �ABC, let X, Y, and Z be the midpoints of sides BC, AC, and AB respectivel and draw �XYZ. Show that this divides the original triangle into four congruent triangles triangles Show that the the three medians of a triangle triangle go through a common point po int HINT: se roblem roblem 1 30 3 0 poin t is called the cnroid of the triangle triangl e We shall s hall have more to NOTE: This point sa about abo ut it in hapter 2 et ABC D be a n arbitr arbitrar ar quadrilatera quadrilaterall Show that the midpoints m idpoints o f the the four four sides are the vertices vertices of a parallelogram parallelogram ABC D , draw the two diagonals A C and B D, and Given a convex quadrilateral ABCD assume assu me that LADB = LACB. Show that LABD = LACD. NOTE: One wa to prove this i s to show that point C must lie on the circumcircle of �AB D. There is also a wa to prove this without circles using similar triangles triangles Tr to to nd such su ch a proof Given two points X and Y on a circle a point P is chosen on line X Y, outside of the circle circle and tangent P T is drawn where T lies on the circle Show that P X PY = (PT) In Figure Figure 1 45 45 line segments segments P A, P B and and P C join point P to the three vertice verticess of �ABC We have chosen cho sen point Y on P B and drawn Y X and Y Z parallel to B A and B C, respectivel where X lies on P A and Z lies on PC. rove that XZAC
IH SLARTY
A
49
X
c
B
c
Figure 1.45 1H7
1H8
1H
1H10
11 1H11 1H
Figure 1.46
In � A B C draw line segments A Q and B P where P and Q lie on sides A C and B C respectivel. Now draw draw P Y parallel to A Q and Q X parallel to B P lie on AC and B C as shown shown in Figur Figuree 1 .46 Show that that X Y A B . where X and Y lie et X be a point on on side B C of �ABC �A BC and and ddra raw w AX A X . et U and V b e points on side AB and AC AC respec respectiv tivel el chosen chosen so that that U V B C and let Y be the the ppoint oint where where AX AX cuts U V . Show that that UY Y V = BX XC. NOTE: I n parti particula cularr this exercise exercise tells tells us u s that that if AX i s a median o f �A B C then then AY is a medi median an of of �AU �A U V . In Figure 1 .47 .47 we started started with with a circle having parallel chords chords A B and CD C D and we chose points P and Q as shown. s hown. et X be the the intersection point of chords P B and Q C and let let Y be the intersecti intersection on point of of chords chords Q A and P D . Show S how that XY CD. HINT: et Z b e the the intersection intersection point point of P D and Q C and show that that � P ZX � � Q ZY . se this this to to show show tha thatt ZX ZC = ZYZD. Given a circle centered cente red at point a and an arbi arbitra trar r point P consi c onsider der the locus of all points Y that occur as midpoints of segments segmen ts P X where wher e X lies li es on the given circle. Show that this locus loc us is a circle with radius radius half that of the the original circle. ocate ocate the center of the locus. locu s. tangent P A and a secant P B are are drawn drawn to a circle from from an outside outsi de point P and the the circle goes through the the midpoint midpoint of P B as shown in Figure 1 .48 If compute the the leng length th AB AB. the length A = 1 compute p
A
B
Figure 1.47
Figure 1.48
CH PER O
Trianges
2A
The Th e Circumcircle Circumcircle
s we saw in Theorem Theorem 1 1 5 there there is a unique circle circle thro through ugh an three three noncollinear points Each Each triangle therefore therefore is inscribed in scribed in exactl one circle called c alled its circumcirc The center and the length of the radius of this circumscribed circle are referred to as the circumcnr and the circumradiu of the the triangle triangle and the the usual usu al notation is for the t he circumcenter and R for the circuadius Given ABC, we know that its circumcenter is equidistant from vertices A and B , and so it lies on the perpendicular perpendicular bisector of side A B Similarl lies on the perpendicular bisectors bi sectors of sides B C and AC and we have the following following The three perpendcular bsectors of the sdes of a trangle are concurrent conc urrent at the cr c rcumcente cumc enterr of of the trangle. •
(2.1) THEOM.
In Figure 2 1 we see that that there there are are at least two possibilities possibili ties In the diagram on the the left the circumcenter , where the three perpendicular bisectors meet is an interior point of ABC, while on the right lies outside ou tside of the the triangl trianglee The explanation explanati on for for this thi s dierence is that on the left left L A is acute and on the right it isis obtuse obtu se In general gene ral given an arbitrar ABC, we see s ee that if L A < 90° , then the arc arc of the circumcircle subtended b L A is less than a semicircle In this case cas e the center of the circle lies on the same side of line B C as does A . If all all three angles angles are acute similar si milar reasoning shows that lies on the inside" ins ide" side s ide of all three three sides si des of the triangle triangle and so it is interior to ABC. If the angle at some vertex (s a A) exceeds 90° however however then the angle subtended is i s more A
A
Fgure 2.1
2A THE CRCUMCRC LE
51
than a semicircle and an d the center lies on o n the side of line B C opposite opp osite from from A In other words wor ds the circumcenter circumcenter lies outside of the triangle triangle in this case cas e Finall Finall if L A 90° then then is a semicircle In this case case the circumcente circumcenterr lies on the hpotenuse B C and thus B C is a diameter of the circumcircle circumcirc le For a right right triangle triangle therefore therefore the circumcenter circumc enter is the midpoint of the hpotenuse and an d the circuadius circuadius R B C t which points do the perpendicular perpendicular bisectors of the sides si des of of a triangle meet the circumcircle?
(22) PROBLEM
y
Fgure 2.2
In Figure Figure 22 2 2 line XY X Y is the perpe perpendic ndicular ular bisector of side B C of �A B C and X and Y lie on the circumcircle circumcircl e We are (rather ambiguous ambigu ousl) l) asked to describe or somehow characterize characterize the points X and Y Y Since Y lies on the perpendicular bisector bis ector of B C we know that Y i s equidistant and C C and thus thus chords chords B Y and C Y are are equal equal It follows follows tha thatt B� Y o C� Y, from from B and and thus thus L B A Y L C AY AY This shows that Y i s the point point where where the the bisector o f LA LA in the original triangle meets the circumcircle It is not quite so s o obvious how ho w to character characterize ize the point X but a stud stu d of the the diagram and the smmetr s mmetr of the the situation leads one to gue ss that X is where where the bisector of the exteror angle at A meets the circumcircle circumci rcle To To prove this this we exend A B to D D as shown and we dra draw w AX A X Our goal isis to show show that that LD L D AX LCAX Note that the circumcenter of �ABC lies on the perpendicular bisecto XY thu s X Y i s a diameter of the circle It follows that L X A Y 90°, o f B C and thus and thus 80 °  (90° (90° + LBAY) 90° 90 ° LBA Y LDAX 1 80 ° LBAX 1 80° Since we know know that that L B A Y L C A Y this ields LDAX 90° 90 ° LCAY L XA X A Y  L C AY A Y LC AX • as required
ouion
ecall that in hapter hapter 1 we used area principles principles to derive derive the law of sines : a b sin(A) sin(B) sin(C)
52
CHAPTER 2
TRANGLES
Thi Thi formula tell u that there i a certain mteriou quantit that can be computed in three three dierent dierent wa wa from the perpective perpective of each of the the three three vertice vertice of A B C The fact fact that that we w e get the ame ame anwer b computing computing each of o f a j in(A) b j in(B) an andd c j in(C) in( C) ugget that thi quantit hould mean omething but our earlier proo did not make clear what thi meaning might be In fact the common value of the three fraction raction in the law of ine tu out to be exactl the diameter of the circumcircle circumcircle Gven ABC wth crcumradus R wrte as usual usu al a b and c to denote the lengths of of the sdes sd es opposte opposte vertces ve rtces A B and C respectvely. Then a b c = =2R. in(A) in(B) in(C)
(2.) THEOREM (Exndd Law of in).
A
A
Figure 2.3
We hall how h ow that a j in(A) = 2R; the other equalitie are proved imilarl imil arl raw the circumcircle of A B C and let B P be the diameter through through point B We have drawn drawn two of the poible poi ble cae cae in Figure 2.3; there there i alo the poibilit that P C which occur oc cur if L A i i a right angle We will need n eed to treat thee thee three ituation eparatel eparatel Firt as asume ume that that LA L A < 90° a in the left diagram diagram of Figure 2.3. Since B P i a diameter we ee ee that P B C i a right triangle with hpotenu hpot enue e B P of length 2R, and thu thu in( P) = B C j B P = a j 2R. ut LA = L P becau bec auee the ubtend the ame arc; arc; hence in(A in (A)) = a j 2R, and the deired equalit follow follow If LA 90° then A B C i i a right triangle triangle and the hpotenue hp otenue B C i i a diameter of the the circumcircle Thu Thu a = BC 2R, and and ince in(A in (A)) = 1 in thi cae ca e we have a j in(A) = a 2R, a required Finall we refer to the diagram on the right of Figure 2.3, wher wheree L A 90° in the rt cae ca e we ee that that here here too P B C i a right triangle and and that that in( in ( P) a j 2R. In thi cae A and P are oppoite vertice of an incribed quadrilateral and and hence hence L A = 1 80° L P we remember from from trigonometr trigonometr it follo follow w that in(A) = in( P) and and thu thu in(A) in(A) aj2R in thi thi cae too too •
Proof.
ere i a nice nic e little application of the the extended law law of ine We leave to the exercie exercie a direct nontrigonometric proof of thi fact Give Givenn iocele A A BC chooe a point point P on the bae B C Show Show that that A B P and AC AC P have have equal equal circuadii circuadii
(2.4) PROBLEM.
2A THE CRCUMCRCLE
53
ppl the extended law law of sines at vertex vertex B in A B P to deduce dedu ce that the diameter diameter of of the the circumcircle circumcircle of �A B P is A P / sin( B) Similarl Similarl b appling appling o f sines at vertex vertex C in �A C P , we see s ee that the diameter of the the extended law of circumc circumcirc ircle le of �AC � AC P is A P / sin(C) ut since �ABC is isosceles the pons asinorum asinorum tells us that that LB L B = L C, and thus thus sin(B) sin( B) = sin (C) and the two dime dimeter terss • are equal
ouio.
nother application of the extended law of sines is the followin followin formula relating relating the circumradius circumradius area and side lenths of a triangle triangle COROLLARY. et R and K denote the cr c rcum radus and area of �ABC respec respectve tvely ly and let let a b and c denote denote the sde lengths as usual usu al Then 4 K R = abc We know kno w from from hapter 1 that tha t 2 K = ab sin(C) sin( C) and b the extende extendedd law of Multiplication ation of these these two equations ields the sin si nes we have 2R = c / sin C) Multiplic • desired result We begin now to explore the connection between altitudes altitudes and circumcircles circumcircles PROBLEM. Given acute angled ABC draw altitude AD and note that point D must lie on line segment BC, as in Figure 24 Extend AD beond D circumcircle cle Observe that that AD A D > D X and thus thus there there is a point point H to point X on the circumcir on segment A D for for which H D = D X Show that that line line B H is perpendicular perpendicular to AC Since b assumption L B and L C are acute acute it should should b e clear clear that point D D which is the foot foot of altitude AD lies between B and C and so it lies on line segment B C as asserted ass erted The fact that AD > DX is is perhaps perhaps not quite as obvious and so we sketch an argument that explains wh D must lie below the midpoint of AX in Figure 24 onsider B C and the perpendicular bisector bis ector of AX These Thes e are parallel lines that cut cu t AX at point point D and at the the midpoint midpoint of AX respective respectivel l and so it is enough to show that that B C is below the perpendicular bisector ut u t the perpendicular perpendicular bisector of AX goes goe s through the center of the circle circle and thus it sufces to observe that B C is i s below the center This is true because L A is i s acute and an d hence hen ce it subtends an arc arc that that is less les s than a semicicle semicicle We have now justied ju stied the ass assert ertion ion that the the point H actuall lies inside insi de the tangle as we have ha ve drawn drawn it in Figure 2.4 A
x
Figure 24
54
CHAPTER
TRANGLES TRANG LES
et E and Y be the points where B H meets side AC and where where it meets the circle circle respectivel Since S ince D is the midpoint of H X, we see s ee that B is the perpendicular bisector of H X, and thus B H B X Since B D s an altitude altitude of isosc is osceles eles � H B X, it is also an angle bisector and so L Y B C LX B C. It follo follows ws from this that Y C 0 XC two applications applications of orollar orollar 1 1 9 we w e have have 1 � 0 1 � LAE B + XC) XC) (AB + 90 0 LADB 0 (AB 2 2 • and thus LA E B is a rght angle angle as desired desired
ouion o Probm 2.6.
In the situation si tuation of roblem roblem 2., we know that the altitude of �ABC from from vertex B crosses altitude A D at the specied point H similar reasoning, we can deduce that the altitude from vertex C also crosses A D at this same point H , and thus the three altiudes are concurrent lthough lthou gh it is true for ever triang triangle le that the altitudes are concurrent con current at a point poi nt called call ed the orhocnr, we have have so far proved this strikng theorem onl in the case cas e of acute angled triangles triangle s We give a diff different and better proof that works works in all all cases in Section 2 roblem roblem 2, we know that the three points that result resu lt when the orthocenter orthoc enter of an acute angled triangle is reected in the sides of the triangle triangle all lie on the circumcircle (ecall that a point Q is the rcion of a point P in a given line if the line is the perpendicular bisector of segment P Q ) It is natural to ask if there is an point po int other than the orthocenter orthocenter whose reections reection s in the sides side s of the the triangle triangle all lie on the circumcircle To see that the the answer answ er is no consider cons ider Figure 25. Start with acute angled �ABC and reect each of o f the the vertices A, B, and C in the opposite opposi te side of the triangle triangle to obtain points U, V and W , respectivel Next draw the circumcircles circumcircles of o f �BCU, �CAV, and � A B W and observ obs ervee that these appear to to go through a common point To understand what is going on here note that � U B C is the reection of �ABC in line BC and thus the circumcircle of �U BC is just the reection of the circumcircle of �ABC in this line line It follows that the circumcirce of � U B C is the locus locu s of all points whose who se reection reection in line B C lies on the circumcirce of the original triangle triangle In particular particular b roblem 2, we know that the orthocenter H of �ABC lies on this circle v
""
""
""
""
Fgure 2.5
A THE CRCUMCRCLE
55
two circles in similar reasoning, we know that H lies on each of the other two Figure 25 2 5 and hence the three three circles circles do indeed have a point in common commo n The ortho center H lie on all three circles circles and since these circles clearl cannot have more than reectio ns in the sides side s one common point we see that H is the onl point all of whose reections of ABC lie on the circumcircle of this this triangle We can draw a furthe furtherr conclusion conclu sion in this situation The three three circles circle s through point H in Figure Figure 2 2 5 are clearl the circumcircles of H BC, AHB AH B and ABH Since each of thes thesee circles is a reection of the circumcircle circumcircle of ABC, all four circumcircles have equal radii Ths Th s fact appears with a diere dierent nt proof as orollar 2 1 5 in S ection 2 2
xercses xercs es 2Al
2A.2
2A.
2A.4
ABC D can be inscrbed in a circle Show Sh ow that quadrilateral ABCD circle if and and onl if L B and L D are upplementar sh ow that D lies on the unique circle through A B, and C HINT: To prove if show NOTE: quadrilateral cycic quadrilateral inscribed in i n a circle circle is i s said to be cycic Given ABC and XYZ, assume A B = X Y and that that LC L C and L Z are supple mentar mentar rove without using the extended extended law of sines that the two tangles have equal circumra circumradii dii HINT: Move XYZ so that A B and X Y coincide alteative solution to roblem 2 4 NOTE: This proides an alteative Given acute anged ABC extend the altitudes from A, B and C to meet the circumcircle at points X, Y, and Z, respecti respectivel vel Show S how that lines lines A X BY BY and C Z bisect bis ect the three three angles of X Y In Figure Figure 26 2 6 altitude altitude A D of ABC has been extended to meet the circumcircle at point X oint H was chosen chos en on on line A D so that H D = DX Show that B H is perpendicular to A C HINT: et P be the point where B H crosses the circle and show that 0 PAC = NOTE: similar reasoning C H is perpendicular to AB, and thus all three altitudes of ABC are concurrent at H Observe that the reason H lies above H
B
C
Figure 26
56
CHAPTER
2A.5
2B
TRANGLES TRANG LES
A on line A D in Figure 2 is that L A is obtuse This exercise together with roblem 2, shows that in all cases cases the altitues altitues of a trianle trianle are concuent In the situation si tuation of Figure 2, show that the reection reection of o f H in line A C lies on the circle H et P be as in the hint for for the previous exercis exerc isee and show that P is the reection of H in A C b proving that H A P is isosceles iso sceles Observe that that L C an AP B L H P A are both supplementr to APB NOTE We now see that even when ABC is not acute angled angled the three three reec reec tions of the orthocenter H in the sies of the the trianle trianle all lie li e on the circumcircle
The Centroid
In ever ever trianle the three medians are concuent This Th is fact appeare appeare as Exercise Exercis e 12, but we present a better proof of a stroner assertion as Theorem 27 The point of concurrence of the meians is i s called cal led the cnroid of the the triangle triangle an it is i s customaril cu stomaril denote G The three medans of an arbtra trangle are concurrent at a pont that les two thrds of the way wa y along each medan from the vertex of the trangle towar towa rd the md m dpo pon ntt of of the oppos opposte te sde s de
(2.7) HEOREM.
et G be the point po int where meian AX crosses cross es meian meian BY, as in Figure 27 We rst show that G lies lie s two thirds of the wa from from A to X along line segment AX We need to show, s how, in other words words that A G = 2G X raw X Y and observe that that this segment joins jo ins the mipoints o f two sides sid es of �ABC orollar 1 3 1 we conclue that X Y must be parallel to the third sie AB, an that X Y = !AB Since XYAB, we have equalit of alteate interior interior angles angles an thus BAG = LYXG an LABG = LXYG It follows that �BAG � �YXG b , and hence A G j G X = A B j X Y = 2, as required Similarl meian C Z crosses AX at a point that lies two thirs of the wa from A to X learl le arl however, G is the onl point on AX that has this thi s propert and thus C Z goes through G The three medians medians are thus concuent conc uent and we know that the point of concurrence lies two thirs of the wa along meian AX similar reasoning reasoning we euce that the point of concurrence concurrence lies two thirs of the wa aong each meian •
Proof. Proof.
A
B
X
Figue 2.7
C
B THE CENTROD
57
s an application, we have the following following converse to the easy obs ervation that that the the median medianss to the two equal sides sides of an isosceles triangle triangle are are equal equal (See (S ee Exercise Exercise 4) 4 ) Suppose that in ABC, medians B Y and C have equal lengths lengths rove that A B = AC
(2.8) PROBLEM.
Medians B Y and C Z intersect intersec t at the centroid G, and we have 2 2 B G = B Y = CZ = C G 3 3 by Theorem 27 Thus B G C is isoscele is osceless and we have have L G B C = L G C B by the pons asinorum We are given that B Y = CZ, and of course, BC = BC Since we now know that LYBC = LZCB, it follows that BYC CZB by SS and thus thus YC ZB We conclude conc lude that AB = 2ZB = 2YC = AC, as required required •
ouio.
y roblem 2 8 , we know that that a triangle triangle having having two equal medians must be isosceles iso sceles , and in Exercise Exercise 7, 7 , we saw that that a triangle having two two equal equal altitudes is isosc iso sceles eles (This was proved more fully fully using area principl principles es in Section S ection E) E ) These two results suggest sugges t the possibility poss ibility that the equality of two angle bisectors of a triangle triangle also al so guarantees that the triangle triangle is isosc is oscele eless Tis is true but notoriously dicult to establish proof appears appears in Section 2 The centroid of a triangle triangle has a signicance signi cance that can be explained by a little physics experiment which whic h we encourage encou rage readers to try try ut out a trian triangle gle from from a uniform piece of cardboard cardboard and then use us e one thumbtack to attach the triangle triangle to a wall w all or to a bulletin board bo ard in such suc h a way that the triangle can swing s wing freely freely in a vertical plane plane To overcome friction riction it may be necess nece ssar aryy to use a somewhat longer tack tack so that the cardboard does not rub against the wall w all The interesting fact fact is that that the cardboard cardboard triangle triangle will come to rest with its centroid directly directly below the point of suspensio suspen sion n and this is true regardless regardless of wher wheree we choose the t he point of suspension to be, although although in practice, practice, this works works best when the suspe su spensi nsion on point is relatively relatively far far from from the centroid centroid To carry out the experiment, determine the centroid by carefully drawin two medians and clearly marking their intersection point To help see that the centroid comes to rest res t directly below the point of suspension, suspensio n, it is usef us eful ul to tie a thread thread to the tack triangle le If I f a weight is i s attached to the free free end of o f the thread, the t he in front of the cardboard triang ve rtically, and an d so it should sh ould pass pas s directly in front front of the marked centroid thread will hang vertically, of the triangle Why does this work? What we are are really saying here is that the center of mass mas s of a triangular sheet she et is located at the centroid centroid of the triangle triangle We expect that most uniform uniform triangular readers readers of this book bo ok will know about centers of mass from from their study of o f calculus or or particular, readers may remember remember that every every rigid rigid physical phys ical body behaves behave s in physics phy sics In particular, static balancing experiments as though all of its mass were concentrated at a single sing le point called the cr of ma of the body To prove that the center of mass mas s M of a uniform u niform cardboard triangle coincides with the centroid, it sufces to show that M lies on each median
58
CHAPTER
TRANGLES
B
c
Figure 2.8
A B C lies To see that the center of mass mas s of the region region bounded bounde d by A lie s on median AX, for for example imagine slicing slic ing the triangular region into a huge number of very very thin strips each parallel to B C. In Figure 2.8, line segment U V represents represents one of these strips Observe that the point Y , where stri U V meets median AX, i s actually the midpoint of U V This fact, proved using similar triangles is contained in Exercise 8 and the note foll following owing it learly, the center of mass mas s of strip U V is at its midpoint Y , and thus the strip behaves as though all of its mass were concentrated concentrated at Y Since Y lies on median AX, we see s ee that each of the pieces into wich our triangle has been decomposed decompos ed appears to have its mass concentrated on this median We can pretend therefore that the entire mass of the triangle is distribut di stributed ed along AX, and it foll follows ows that the center of mass M lies on this median Similarl Simi larly, y, of course, M also lies on the other two medians, and thus thus it coincides with wit h the centro centroid id as claimed cl aimed ecause ec ause of its appeal to physical physic al intuition the foregoing foregoing argument argument should sh ould not, of course, course , be cons c onsidered idered a rigorous mathematical mathematical proof proof It is i s interesting to note, howev ho wever, er, that this physical reasoning demonstrates that there is a particular point (the center of mass) mas s) that lies on each of the the medians and thus thus provides provide s an alteative argumen argumentt to show that the the medians of a triangle must be concurrent concurrent We shall see other examples where physical reasoning can be used to prove" geometry geometry theorems theorems There There is a simple physics phy sics argument, argument, for example, example, that demonstrates demonstrates that the centroid must lie two thirds of the way along each median median This time, ti me, imagine i magine that our ABC is made from from some massless massl ess rigid material material and that a unit point mas mas s is attached at each each vertex There are are two two equal masses mas ses at the the ends of side sid e B C, and so that side behaves as a s though it s mass were concentrated concentrated at the midpoint X . We can thus pretend pretend that the three units of o f mass of the the entire triangle triangle are positioned posi tioned with one unit at A and two units at X . The center of mass M of the the triangle, triangle, theref therefore, lies on AX, closer to X than than to A . More precisely, we see s ee that that since the mass mas s at X is twice that at A , the distance M X = M A In other words words M lies lies on AX two thirds o f the way from A to X Since M also als o lies on the other other medians by similar reasoning reasoning,, it follows that M is the centroid and that it lies two thirds of the way along each e ach median median We have seen that the center of mass of a uniform cardboard triangle is at the centroid and also that the center center of mass of a massless massl ess triangle with with eequal qual point masses masse s at the vertices vertices is aatt the centroid centroid nother interesting interesting case cas e is that of a triangle made of uniform uniform wire We assume as sume in other words words , that the mass is uniformly uniformly distr dis tributed ibuted aong A B C and that the sides of A that the interior interior is massles mass lesss In this case, ca se, we can assume assu me that that the total total mass mas s of side B C is a (the length of BC), and we pretend that it is concentrated co ncentrated at the midpoint X of BC Similarly Si milarly a mass of b units is at the the midpoint Y of AC, and
p
B THE CENTROD
59
a mass of units is at the midpoint Z of AB. We now need to consider only XYZ with with point masses , , and at vert vertices ices X Y, and Z . We mention that XYZ, the trian triangle gle formed by the midpoints of the sides of A B C is called the mdia riag / 2 and of ABC. y orollary 1 3 1 we see s ee that X Y = /2, X Z = /2, and Y Z = /2 thus ABC � XYZ by SSS. SS S. We can now replace the two point masses mass es aatt Y and Z b y a single mass mas s at the center of mass P of side Y Z. Since the masses at Y and Z are and ,, and these are not necessarily equal, point P need not be the midpoint of Y Z. To locate P precisely let Y P = and Z P = . Then = , and thus thus / / = / = X Y / XZ. ence we see that point P divides side Y Z of X Y Z into two pieces whose whos e lengths are in the same ratio as the lengths of the the nearer nearer sides of the triangle triangle.. omparison of o f this this with Theorem 1.12 shows that the bisector of L X meets Y Z at P , and thus thus the center of mass of X Y Z, with with the appropriate appropriate point masses mass es at the vertices vertices lies lie s on the angle bisector bise ctor X P Similarl Sim ilarly y this center of mass lso lies on the other other two angle bisectors bisec tors of X Y Z It follows follows that the center of mass mass oour original uniform uniform wire triangle triangle lies l ies at the point of concurrence of the angle bisectors bis ectors of the the medial triangle. triangle. We shall see in i n SSection ection 2E that the angle bisector bi sectorss of an arbitrar arbitraryy triangle are concurrent at at a point called call ed the icr In general, iti t is not true that the the incenter of the medial triangle is the centroid of the original triangle, triangle, and thu the center of mas mas s of a uniform uniform wire triangle is not always at the same s ame location as the center of mass of the the corresponding uniform uniform cardboard triangle triangle
xercses B 2Bl
2B2
2B
2B4
Show Sho w that the centroid of the medial triangle of o f ABC is the centroid of ABC. Show using us ing centers of mass that the angle bisectors of A B C are concurrent con current at a point I , where I lies on bisector A P at a position such s uch that AI AP
where as usu u sual al and denote the lengths lengths of the sides of o f the triangle triangle.. HINT ut point masses of , and units at points A, B, and C . Show that that there there can be b e no point P in the interior of ABC such that every every line through P subdivides the triangle into into two pieces of equal area. area. HINT: If there is a point P that has this property property show sh ow that the medians would have to go through P lie s at the NOTE The fact that the center of mass mas s of o f a unif uni form triangular sheet lies centroid certainly does not imply that every line through this point divides the area into into two equal parts. parts . ABC D is cut from a uniform piece of Suppose that a convex quadrilateral ABCD cardboard. Show that the center of mass of the cardboard quadrilateral lies on diagonal A C if and only if AC bisects diagonal B D centerss o f mass o f AB D and C B D HINT onsider the line joining the center
60
CHAPTER
2B5 2B 5
2C
TRANGLES TRANGLE S
hoose a point P on side AC of ABC and write p = AP/ AP/ A C et AX/ AX / A , where where X is the point wher wheree B P crosses cross es median median A A , and observe that that = 2/3 2/ 3 when p = 1 /2. /2 . Find a general general form formula ula for in terms of p HINT: ut masses o f one unit each each at points points B and C and nd an appropr appropriat iatee mass to put at point A so that the center center of mass will be at X X
The Euler Line, Orthocenter, and NinePoi Nine Point nt Circle
So far, far, we have discus disc ussed sed two important important points associated assoc iated with a triangle triangle the circum center and the centroid If the given triangle is equilateral, then each median is the perpendicula perpendicularr bisector b isector of the the opposite oppos ite side, side , and it follows that the circumcenter circumc enter and and the centroid centroid are are actually the same point. In all other cases, cases , however, these points are distinct distinct the crcumcen c rcumcenter ter and the centr cen troo d of of a trangle trang le conc c oncde de then the trangle trangle must be equlater equ lateral al
(2) LEMMA
Suppose that the the centroi centroidd of A A B C is also the circumcent circumcenter er et G be the centroid and let X be b e the midpoint of side B C Then X, G , and A are distinct points on the median from from A , and since sinc e these points are collinear, collinear, we see that A lies on line G X. ut G is also the circumcenter, and so it lies on the perpendicular bisector of side B C . Of course, course , the midpoint X of side B C also als o lies on the perpendicular perpendicular bisector bis ector of B C , and it follows follows that the line G X is the perpendicular perpendicular bisector bis ector of B C . Since A lies on G X, we have shown that A lies on the perpendicular bisector of BC , and thus thus A is equidist equidistant ant from B and and C. We now now have have AB = AC AC,, and similar reasoning shows that B A B C . It follows that all three three sides of A B C are equal, and thus thus the triangle is equilate equil ateral ral •
Proof
Given Given any nonequilateral A A B C , we now know know that the circumcenter and the centroid G are two distinct points. These two points determine a unique line that is called the Eur i of the triangle (eonhard Euler was a Swiss astronomer and mathematician mathematici an who lived from from 1707 to 1783 is name is pronounced oiler") oiler") We We stress that for for equilateral triangles there there is no Euler line dened dened We have have already mentione m entionedd that the three three altitudes of a triangle are always al ways concurrent c oncurrent at a point called the orhocr of the triangle. emarkably, the Euler line also goes through the orthocenter In some sense, it is a double miracle when four previously determined determined lines are concurrent. For an equilateral triangle, triangle, the altitudes are the medians, and we know that they are concurrent concurrent at the centroid. Since there there is no Euler line for an equilateral triangle, triangle, we have nothing further further to prove in this this case. For nonequilateral nonequilateral triangles triangles , we w e want to show that all three altitudes meet mee t at some point of the Euler line. line . In fact, fact, we are able to specif sp ecifyy this point in advance.
C THE EULER EULER L L ORTHOCENTER ORTHOCENTER AND NNEPONT CRCLE
61
Assume that A B C s not equlateral equlateral and let G and be ts centrod and crcumcente respectvely et H be the pont on the Euler Euler lne G O that les on the opposte sde of G from and such that H G = 2G O Then all three alttudes of of A B C pass through through H
(2 10) THEOREM
A
A
M
M
H
Figure 2.9
We shall s hall show sho w that the altitude from from A passes through throu gh H; the proofs proofs for for the other two altitudes are similar, similar, of course If H coincides with A, there is nothing to prove, and so we can c an assume that H and A are differ different ent points ut ut note that it really can happen that H and A coincide; this occurs when L A = 90° It sufces now to show that line A H is i s perpendicular to B C et M be the midpoint of side B C and observe that and M are are distinct points points Otherwise, the median G M is the Euler line, and since sinc e we know that that A G = 2G M, it would follow that A is H, which we are are assuming assuming is not the case ca se Since S ince both both and M lie on the perpendicular bisector of B C, line M is perpendicular to B C , and we will be done if we can show sh ow that M is parallel to A H Figure 29 2 9 shows s hows two of the the several possible pos sible congura con gurations tions for this this problem, i dentical in all all cases case s We We will establish that AH M by proving the but the proof is identical equality of the alteate interior angles, L H and L O We know that A G j M G = 2 by Theorem 27 and that H G O G = 2 by the construction of the point H Since and LA H = LMGO, we see that AGH � MGO by A G j MG = H G j O G and • SS SS It follows follows that that L H = L , and the the proof proof is complete complete
Proof
Given ABC, let H be its orthoce orthocenter nter If we start with a right triangle, triangle , then clearly B , C, and H are easily seen H coincides with the right angle, angle, and in all other other cases , A, B, to be four four distinct points po ints Observe that the line determined determined by any two of thes thesee four four is points points is perpendicular to the line determined by the other two For example, A H is perpendicular to BC because becaus e line AH is the altitude from from A in ABC It is amusing amusing to observ obs ervee that each of the four four points A, B, C, and H is i s the orthocenter of the triangle formed by the other three That A is the orthocenter of H B C, for example, example, is i s just anot anothe herr way of saying saying that that AH, A H, A B , and AC are perpendicular to BC, HC, and H B , respectively, respectively, which i s a fact fact that we have already already noted Given Given any four points with the property that each is the orthocenter of the triangle quadrup formed formed by the other three, we w e say s ay that the given set of four four points is i s an orhic quadrup arbitrary set s et of three three points A, B, and C , there there almost almos t always We now know that given an arbitrary exists exist s a fourth fourth point H such suc h that the set {A , B C H} is an orthic quadruple quadruple (Just take H
62
CHAPTER
TRANGLES
to be the orthocenter of A B C ) The only exceptions exceptio ns are when when A B C is a right triangle or when ABC does not exist exi st because becaus e the given three three points are collinear collin ear Given ABC let D, E, and F be the the points point s where the altitudes from A B and C meet lines B C A C and A B respectively respectively These points are called the f of the altitudes, altitudes, and we note that they may not actually lie on o n the line segm segmen ents ts BC AC and AB If A B C is not no t a right right triangle, triangl e, it is not hard to see that the feet D, E, and F are distinct and form a triangl trian glee We refer refer to DE F as the pda riag of ABC although some authors call it the orhic riag In Figure 2 1 0, we have have drawn drawn two two of the situations situations that can occur I each gam, gam, the original triangle is drawn with heavy lines line s In the cong con guration uration on the right, we had to extend two of the sides of the the triangle triangle to meet the altitudes, and these extensions are shown with dashes In both diagrams, the pedal triangle and the altitudes are drawn drawn with solid lighter li ghter lines lines It should sho uld be clear from from the gure that the feet feet of the altitudes altitudes lie on the sides of o f the triangle for acute angled triangles, triangle s, but bu t two of the feet feet lie uts utside ide of the triangle if there is an obtuse obtus e angle
Figure 2.10
Figure Figure 2 1 0 shows s hows something something else of interest interest The triangle triangle on the left left and the one on the right happen to be two of the four four triangles triangles that t hat can be formed using three three of the four points of an an orthic quadruple We see that in both cases , we get g et exactly the same pedal triangle triangle The pedal ped al trangles tran gles of each eac h of th four ou r tangl t angles es determn dete rmned ed by an orthc quadru qu adruple ple ar a re all the same
(2 11) THEOREM
We have seen previously that for for each each choice of two of our four given points, the line determined by those thos e two is perpendicular to the line determned determned by the other other two Since Sinc e there are are exactly three three ways to pair o four four obje objects cts into two sets of wo this gives three points that occur as the intersections of pairs of perpendicular lines determined determined by our orthic quadruple quadruple It is easy to see that these three points must mu st be the vertices of the pedal triangle of each of the the four triangles triangles •
Proof
The crcumcircle of the the pedal triangle of A B C tus out to be be an amazing objec objectt In addition to the feet of the three altitudes, this circle aso contai the mdpoints of the three sides; hence it is also the circumcircle of the medial triangle of ABC emarkab emarkably, ly, this circle has a furth further er unexpected property property It bisects bise cts each of o f the line segments AH B H and CH where H is the orthocent orthocent of o f ABC We refer to midpoint X of segment A H as the Eur poi of ABC opposite to side BC and
C THE EULER LNE RTHOCENTER AND NNEPON NNE PONT T CRCL E
63
similarly, the midpoints Y and Z of B H and C H are the Euler points of o f 6A B C opposite to sdes AC and AB, respectively The common circumcircle circumcircle of the pedal and medial co ntains the three Euler points and even more more is true We can state the full full triangles triangles contains result as as follows Gven any rangle all of of the followng ollown g ponts po nts le l e on o n a common c ommon cr c rcle cle the three feet ee t of of the alttudes alt tudes the three th ree md m dpo ponts nts of of the sdes sde s and the th e three Euler ponts. po nts. Furthermore each of of the lne segments onng o nng an Euler E uler pont po nt to the mdpo mdpont nt of of the oppo opposste te sde s a dameter d ameter of ths crcle. c rcle.
(212) THEOREM
The remarkable remarkable circle whose existence is ass asserte ertedd in Theorem 212 is called c alled the the the triangle We should sh ould point out, however, however, that the nine points ipoi circ of the referre referredd to in the statement s tatement of the theorem theore m are not always distinct dist inct In Figure 2.11 we have draw drawnn one possible pos sible conguratio con gurationn for this theorem The proof will will be the same s ame for all all cases, case s, although the diagram can look loo k a little diff different for for dierent dierent starting triangle triangless In the gure, g ure, point p ointss D, E and F are the feet of the the altitudes of 6ABC oints oints P, Q, and R are the midpoints midpoi nts of the sides, and X, Y, and Z are the Euler points We need to show that all nine of these points lie on a common circle and that X P , Y Q , and and Z R are diameters diameters of of this circle s usual, u sual, we have called the ort orthocen hocenter ter H H raw line segment Y Q and consider co nsider the unique circle that has Y Q a s a diameter diameter We will rst show that points P and R lie on this this circle circl e To see that P lies on the circle with diameter Y Q , it sufces to show that that L Y P Q 90° We We will do this this by proving that Y P C F and P Q AB Since C F and A B are perpendicular, perpendicular, it folows easily that that Y P and P Q are perpendicular, perpendicular, and a nd thus L Y P Q 90° , as required required That P Q is i s parallel to A B is clear cl ear by orollary 1 . 3 1 since P and Q are the midpoints of two two sides side s of 6ABC and AB is the third side Similarly Sim ilarly,, to prove that Y P is i s parallel to C F we work in 6 B H C Of course, P is i s the midpoint of side B C of this triangle, and the the Euler point Y is the midpoint of side B H It follows follows that that Y P is parallel to the third third side of this triangle, which is C H, and thus thus Y P C F as desired des ired We have now shown that L Y P Q 90° and thus point P lies li es on the crcle with diameter Y Q , as desired desired
Proof of Thorm Thorm 2 12
,,, D
, P
Figure 2.11
64
CHAPTER
TRANGLES TRANGLE S
Next, we use us e similar si milar reasoning to show that that R lies on this circle It sufces sufces to prove that L Y R Q = 90°, and we accomplish this by showing that Y R AD and altitude A D is perpendicular to side BC, it will follow that Q R QRBC. Since altitude is perpendicular to Y R, and thus L Y R Q = 90° as claimed That Q R B C is a straightforward application of o f orollary orol lary 1 3 1 in ABC, and to prove that Y R AD , we apply orollary orollary 1 3 1 in ABH. We have now shown that points P, Q , R, and Y all lie on the same circle and that Y Q is a diameter of this circle Since this circle contains P, Q, and R , i t is, o f course, the circumcirc circumcircle le of the the medial medial triangle triangle of ABC. We have now proved that given gi ven an arbitrary arbitrary ABC, the circumcircle of its medial triangle has line segment Q Y as a diameter, where Q is the midpoint of side A C and Y is the opposite oppos ite Euler point It follows similarly that the line segments P X and R Z are also also diameters of the medial circumcircle, and in particular, particular, the other two Euler points , X and Z, Z , lie on this thi s circle circle We have now now shown that six of the required nine points lie on the medial circumcircle and that each of X P, Y Q, and Z R is a diameter of this circle circle Next, we w e observe obs erve that that E lies lie s on the circle with diameter Y Q since L Y E Q = 90° This shows that the altitude foot E lies on the medial circumcircle of an arbitrary ABC . It follows similarly that the medial circumcircle contains the other two altitude feet, D and F , and this concludes the proof • (21) PROBLEM
of ABC?
What is the radius and where where is the t he center of the ninepoint circle
erhaps the easiest easi est way to analyze this situation is via the technique of transform transforma a tional geometry, which whic h we introduce informally informally as we proceed proce ed Since the ninepoint ninepoint circle is, among other things, things, the circumcircle of o f the medial triangle of ABC, we can solve so lve this problem by focusing focusing attention on P Q R, where P, Q, and R are are the mdpoints of sides side s B C, A C , and A B , as in Figure 212 We know that QPAR and RPAQ thus A Q P R is a parallelogram, and diagonal A P bisects diagonal diagonal R Q by Theo Theorem rem 1 9 In other other words, words, median A P of ABC bisects side R Q of the medial triangle PQR, and hence it contains the median from P in triangle P Q R . Similarly, the other two two medians of ABC
ouio o Probm 21
A
P
Figure 2.12
C THE EULER EULER LNE LNE ORTHOCENTER ORTHOCENTER AND NNEPONT CRCLE
65
contain the other medians of PQR, and it follows that the centroid G of ABC i s also the centro centroid id of P Q R. Now imagine imagin e the following twostep two step transforma transformation tion of the plane o f Figure Figure 2.12 First, shrink the plane with a scale factor factor A 5 in such a way that point poi nt G remains xed and every other point moves toward G . In other words, an arbitrary arbitrary point X of the plane is moved to the midpoint of segment G X Next, rotate the the plane 1 8 0 with a center of o f rotation rotation at G . It does not n ot matter matter,, of course, course , if i f this rotation rotation is i s clockwi c lockwise se or counterclockwise counterclockwis e Write Write T to denote the net eect eect of these thes e two operations operation s and view function Observe that (A) P since AG 2G P and LA G P 1 8 0 • T as a function Similarly, (B) Q and (C) R . We now argue without withou t a formal proof that that the transformation T carries ines to lines, triangles to triangles, and circles to circles Furthermore, given a circle centered at some point , the image of that circle under under T is a circle centered at the that of the original circle Of course, all of this this can c an point T ) and having radius half that formalized ed and made more precise preci se,, and rigorous proofs proo fs can be constructed cons tructed We be formaliz trust trust,, however, that the truth truth of our assert ass ertions ions is reasonably clear, and we proceed without witho ut further further proof Now consider the circumcircle of ABC, which is centered at the circum center and has radius R . The transformation T carries ABC to P Q R , and hence it carries carries the circumcircle of ABC to the circumcircle of P Q R, which is the ninepoint circle of ABC. It follows that the center of the the ninepoint circle, which we call N, is exactly the point T ) and the radius radius of the ninepoint ninep oint circle is R /2. lso, since X P, Y Q, and Z R are diameters of the ninepoint circle, it follows that each each of these segments has length R . We record this this information information here • for future future use us e et R be the crcumradus of of A B C . Then the dstance dstan ce from each Euler Eul er pont po nt of of A B C to t o the md m dpo pont nt of of the oppos opposte te sde s R and the radus • of the nnepo nn epont nt crcle of of A B C s R/2 R /2..
(2 14) THEOREM
Sup Suppose A B C s not a rght trangle and let let H be ts ortho cente cente Then A A B C H BC BC AH C and A B H have have equal crcumr crcumrad ad
(215) COROLLARY
We know by Theorem 2 1 1 that these four four triangles triangles share a common commo n pedal triangle triangle Since Sin ce the ninepoint circle of any triangle triangle is the circumcircle of its pedal triangle triangle t follows follows that the the four four triangles triangles share a common ninepoint circl e We have have just ju st seen, see n, however, however, that for an arbit arbitrar raryy triangle, the circumradius i s exactly twice the follows that our four four triangles triangles have equal circumradii c ircumradii • ninepoint ninepoin t radius, and it follows
Proof
Our goal now is to give give a convenient conveni ent description descri ption of the location of the ninepoint nine point center N of ABC. We know, know, of course, that N must be the midpoint of each of the segments P X, Q Y, and R Z but we are are looking characterization In the case where the given trian triangle gle is equilateral, equilateral, for a more useful characterization it is clear c lear that N coincides with G and and H . We assume, therefore, that the given triangle is not equilateral, and thus it has an Euler line GO. We know that
ouio o Probm 2 1 coiud
66
CHAPTER
TRANGLES TRANGLE S
N = ( ), and it follows from from the denition of that N lies on o n the line through G and , which is the Euler line Furthermore, N lies on the opposite side of G from , and we have N G = GO. ecall now that the orthocenter H also lies on the Euler line on the opposite side of G from and that H G = 2G O. It follows that tha t N lies on the segment G H and HN = H G  N G = GO. lso, N = N G + G O = GO, and we deduce that H N = N O . In other words, the ninepoint ninepoi nt center N is the midpoint midp oint of the segment H O . Note that this this is true, in some sense, se nse, even if A B C is equilate equilatera rall since in that case N , H, and are all the same point This completes our solution s olution to roblem roblem 2.13. •
Suppose ABC s not a rght trangle and let H be ts ortho cente cen te Then the Euler Eule r lnes of of AB C H BC AHC and ABH are concur rent. any of these trangles s equlateral then the Euler lnes of the remanng trangles are concurrent.
(216) COROLLARY
We have have seen that these four triangles share s hare a ninepoint ninepo int circle circl e The center N of this circle lies on all of the Euler Euler lines, lines , which are therefor thereforee concurrent •
Proof
xercses xercses
Suppose that the Euler line of ABC is perpendicular to BC Show that AB = AC 2C2 Show that the circumcenter of A B C i s the orthocenter of the medial triangle 2C Given ABC, draw line W V through A parallel to BC, line U W through B parallel to t o A C , and line l ine U V through through C parallel pa rallel to A B Show S how that the orthocenter of A B C is the circum circumcent center er of UVW 2C4 Show that the ninepoint circle of A B C is the locus of all midpoints of seg ments U H, where H is the orthoc orthocent enter er of A B C and U is an arbitrar arbitrar point of the circumcircle circumcircle H Exercis Exerc isee R0, we alread know that that this locus locu s is a circle 2C5 Given Given A B C, let X be the Euler Euler point opposite side BC B C Show that that length length AX is equal to the t he perpendicular distance from from the circumcenter of A B C to side side B C educe that that (AH) (A H) = 4R 4 R  a , where as usual, H is i s the orthocenter orthocenter,, R is the circuadius, and a = BC H Show Sh ow that the diagonals diagona ls of quadrilateral X H P O bisect bise ct each other, where P is the mdpoint of side B C NOTE Imagine holding points B and C xed and letti letting ng point A move on some xed circle through B and C C Then H is a variable variable point, point, but this problem problem shows that the length of the variable variable line segment A H is constant, c onstant, independent of the choice of A Note also that AH will alwas be perpen perpendicul dicular ar to B C 2Cl
D COMPUTATONS
2C6
2D
67
Given four four points on a circle c ircle join jo in each of them them to the orthocenter of the the triangle formed by the other three Show Sh ow that the four four line segments that result have a common midpoint M and so in particular particul ar these thes e four four lines line s are concurrent con current at M . Show also that the the ninepoint ninepo int circles o f the four four triangles triangles formed by each e ach three of the four points point s all go through throu gh M . ons ider just jus t two of the the line segments and show that they are the diago HINT onsider nals of o f a parallelogram se Exercise 2 4 4 for for the second part
Computtions
lengths of the sides of �A B C are are the three three numbers numbers a Suppose Su ppose we are told that the lengths b , and c Without being given any additional data and without even being allowed to see the triangle we are asked to determine its area In theory at least we can see that this should sh ould be pos sible because b ecause we could c ould draw our our own triangle triangle having the same s ame three three lengths and then we could measure measure its area Since our triangle triangle is congruent co ngruent to to the sid si de lengths original original one by S S S iti t follows follows that the the area area we measured is equal to the area of the unseen �A B C Similarl Simi larly y it is in principle poss po ssible ible to determine the circumradius circumradius of �A B C or to determine any other numerical quantity assoc as sociated iated with the triangle triangle when the only data we have available available are the lengths lengths of o f the sides side s In this section sectio n we will show how to compute the area area of a triangle triangle given the lengths of its its sides side s More generally our goal is to develop formul formulas as relating various numerical quantities assoc as sociated iated with a triangle triangle The law l aw of sines is an example of the sort of thing thing we want; it relates the side si de lengths and the angles of a triangle and in its i ts extended form form it relates relates these to the circumradius circumradius nother example example proved proved in orollar oroll aryy 2 2 5 is the equation 4 K R = abc relating the area area circumradius and side lengths lengt hs of a triangl trianglee following question Give Givenn the side lengths lengths of �A � A B C how can We begin with the following the angles be determin determined? ed? erhaps erhaps the easiest way is via the law of cosines cosines Gven �ABC let a b and c denote as a s usual the lengths length s of sdes B C AC and AB respectvely Then c = a + b 2a b cos(C cos(C))
(217) (2 17) THEOREM (Law (Law of Coin)
Given a b, and c the equation of the law of cosines cosin es can c an easily be solved to obtain cos ( C) = (a + b c ) 2a b and of course similar for formula mulass yield yield cos (A) (A ) and cos (B ) in terms terms of a b, and c Instead Instead ofproceeding of proceeding immediately to give a proof o fTheore Theorem m 2 2 17 we w e have have decided decided to devote a few few sentences to a discuss disc ussion ion of how one might nd a proof It I t is often often useful to think think about special speci al cases to help try to understand what is really going on in some formula formula onsider for example what happens in the law of cosines if L C = 90° Then cos (C) (C ) = 0 and in this this situaion situaion the formula formula tells us that c = a + b , which we recognize as the ythagorean theorem The law of cosines thus includes incl udes the ythagorean ythagorean theorem as a special spe cial case c ase and it follows follows that we should try to use the ythagorean theorem theorem the proof of Theore Theorem m 2 2 1 7 Otherwise Otherwise we will nd ourselves reproving reproving this theorem theorem in the
68
CHAPTER 2
TRANGLES
within the proof proof of the the law of o f cosines co sines . To use the ythagorean ythagorean theorem however however we need a right triangle triangle and so it would seem that a good rst rs t step is to draw an altitude altitude of our given triangle. A
A
:h
C
P igure
I I I
2.13
Since Si nce a triangle can have have at most one angle that fails fails to be acute we can be sure that at least one of L A or L B is an acute angle angle and it is therefore therefore no loss lo ss to assum as sumee that L B < 90° . raw altitude altitu de A P from A to B C and write h = A P . Note that there there are are three three possibilities poss ibilities : Either L C < 90° 90° and and P lies on segment s egment B C as in the left left diagram diagram of Figur Figuree 2.13 or LC = 90° and point P coincides coincide s with with point C or L C 90° and P lies on an extension of side B C as in the right diagram of Figure 2.13. Suppose rst that LC < 90° and write = PC Since Since �A PC is a rig right ht triangle we see that cos(C) = PC/AC = / and so = cos(C) cos (C) . lso, lso, two two applications of the ythagorean ythagorean theorem yield = + h and c = h + (a ) . This gives c = (  ) + (a ) =  + a  2ax + = a +  2ax . Since Sin ce = cos ( C) we obtan the desired formula. formula. We have already already seen that the law of cosines cos ines holds hol ds when whe n L C = 90° and so we can now ass assume ume that L C 90° and we w e refe referr to the diagram on the right of Figure 2.13. ere too we write = P C but in this case we have have cos(C) cos( C) = / and =  cos (C) (C ) . Two Two applicatio applications ns of the ytha ythagore gorean an theore theorem m yield + h and c = h + (a + ) and we have c = (  ) + (a + ) =  + a + 2ax + = a + + 2ax Since = cos co s (C) (C ) in this case case the proof is complete •
Proof of Thorm 217
s an application of the the law of cosines cos ines we can derive derive erons eron s formula for the area area of a triangle given the lengths of the sides si des.. To state the formula formula cleanly cle anly we w e introduce a new quantity s = (a + + c) called call ed the the miprimr (218) THEOREM (Hron' Formua)
equaton
The area K of �ABC s gven the
 a )(s  ) (s c ) K= wher whe re a and c are the lengths of of the sdes and s s the sem s emperme pe rmete te
2D COMPUTATONS
69
It may be of some value to try out erons formula in a couple of cases where we already know how to compute the area Suppose for example that �ABC is a right triangle with arms of length 3 and 4 and hypotenus hypotenusee of length 5 . We know that K = bh = (3)(4) = 6 in this case On the other hand we have = (3 + 4 + = 6  3)(6  4) (6 5 ) = 5) = 6 and erons formula gives K = as expected Next consider an equilateral triangle each of whose sides has length 2 It is easy to see that K = h = (2)() = and erons formula gives
 2 ) (3  2)(3 2)(3  2) = , again as expe cted. exp ected.
K=
Proof Proof of Thorm 2 1
We know that K = sin(C) sin(C) and so
4 K 2 = a 2 b 2 s i n2 ( C ) = a 2 b 2 ( 1  c o s 2 ( C ) ) .
The The law of cosines gives cos (C) = (   ) 2 2 and we can substitut subs titutee this this into the previous previou s equation to obtain (  ) ( ) 4K = 1 = 4 4 Thus 16K = 4  (   ) and the right side of this equation equati on factors as a dierence dierence of squares sq uares to yield
2a b 1 6 K = [ 2ab
+
( c  a  b ) ] [ ( 2ab 2a b  (c  a  b ) ]
= [ (  (  ) [ ( + )  Each of the factors factors on the right ri ght of the the previous equatio e quationn factors factors as a dierence of squares squares and we obtain 1 6 K = [ + (  ) ] [ (  (  ) [ ( + ) + ) ] [ ( + )  ] Observe Obs erve that that +  = ( ( + + )  2 = 2( 2 (  ) ) Similarly Similarly the the second factor actor in our formula formula for 1 6K equals 2 (  ) and the third and fourth fourth fact factors ors are 2 and 2(  ) respectiv respectively ely It follows that that K = (  ) (  )(  ) an and erons • formula follow followss (21) PROBLEM
the sides sides ouio
Expres Expresss the circumradius R of �A B C in i n terms of the lengths of
This Thi s is easy e asy We know that 4K R = and hence hence R 4K  )( )(  )( )(  )
•
another her applicati application on of the law of cosines cosines In I n Figure 2 1 4 point P Next we oer anot chos en arbitrarily arbitrarily on side sid e A B of �A B C dividing the side side which is of length length was chosen following result provides a formula formula that into pieces with lengths and y , as shown The following allows allo ws one to compute the length t = C P in terms of the given data: y , = AC A C and and = B C Of course course we also have have = A B = + y
70
CHAPTER 2
TRANGLES TRANG LES c
C igure
(220) THEOREM (wr)
2.14
In the t he stu s tuat atn n f Fgure we have y b xy c = xa + yb ct + xyc
the law of cosines cos ines we have t = a + y  2ay cos(B) and b = a + c 2ac cos(B) We can can eliminate cos (B ) if we multipl the the rst rst equation b c and the second b y and then subtract subtrac t sing sing the fact that that c y = x, we obtain
Proof
cy (y (y  c ) = x a 2 x y c , ct 2  y b 2 = (c  y ) a 2 + cy 2  y c2 = x a 2 + cy
and the desired equation follows follows
•
Show how to compute the lengths of the angle bisectors of a triangle triangle given given the lengths of o f the sides
(221) PROBLEM
To use us e Stewart s theorem to to nd the length t of angle bisector C P of �ABC in term termss of a b and c w e need to express x and y in terms of the given data data We recall from Theorem 2 that bisector C P divides A B into pieces piece s proportional proportional to the lengths of the nearer sides We have in other words xy = ba Since x + y = c a bit of algebra ields x = bc/ (a + b) and y = ac/ (a + b) (To (To check the algebra, observe o bserve that the sum of these these two fractions is c and that the rst divided b the second is equal to b a ) It follow followss that a b c and xyc = (a + b) and we can substitute sub stitute into the Stewts theore theorem m equation to get a bc ct + = abc (a + b) little more algebra now ields c t = ab (a + b) and the length t of the angle bisector bisec tor can be found found b takng takng the square s quare root •
ouion
]
D OMPUTATONS
71
s a check c heck of this this rather rather unpleas unpleasant ant form formula ula we can try it in the cas e of an isosc is osceles eles triangle triangle where a = b and the angle bisector C P is an altitude altitude The thagorean theorem theorem yields = a c /4 and a bit of algebra shows that this agrees with our previous calculation PROBLEM et A X and B Y be angle bisectors in �ABC and suppose suppos e that AX = BY S how that AC = BC We saw in Exercise Exerci se 1 7 that if two altitudes of a triangle are eual, eu al, then the triangle is isosceles iso sceles and in roblem roblem 28 we proved the corresponding corresponding asertio for medans roblem 222 wich is the analog of these facts facts for angle bisectos seems by far the hardest of these three similarsounding results One unusua aspct of the following arguent arguent is that we prove an equality by contradiction hanging the roles of a b and c appropriately appropriately we can apply the formu formula la derived in roblem 221 to compute that a (AX) = b 1 and (b + ) b (B) = a 1 (a + ) and by hypothesis hypothesis we know that that these these two quantities are are equal ivision ivisi on by c thus yelds
ouio o Probm
[ [
] ]
and we get ba (b + c) We need to show that a = b and so we try to derive a contradiction by that b > a and so the supposing su pposing that a and b are unequal It is no loss to ass ume that eft side of the the previous equation equation is i s positive pos itive The right side si de must therefore therefore also be positive pos itive and it follo follows ws that a ( b + )
ut b
b ( a + c )
> a , and so
b a a > > ( a + c ) (a + ) ( b + ) This contradicts the previous inequality and thereby thereby proves the result
•
72
CHAPTER 2
TRANGLES TRANGL ES
igure
215
We close clo se this section s ection by using usin g Stewarts theorem to deduce one on e more more computatonal result. et a b c and d be the lengths lengt hs f f cnsecu cn secutve tve sdes f a quadrlateral quad rlateral nscrbed n a cr c rcle and sup suppse x and y are the lengths f f the dagnals dagnals Then ac a c + bd = xy
(22) THEOREM (Pomy)
et r s u and be the lengths of the four partial diagonals as shown in Figu Figure re 2 1 5 wher wheree r + s = x and u + = y It is easy eas y to see from the two pairs of similar simil ar triangles triangles that b u s a u r and d r c s uc , br = ud and u = rs It follows and thus as = uc, follows that th at s a + rb r b = uac ua c + ubd = u (ac + bd) bd) and x u + xrs = x u + xu = xu (u + ) = x u y y Stewts S tewts theorem, the quantities on the left left sides of these equations equ ations are equal equal and hence the right sides must also be equal We conclude by canceling u that • ac + bd = xy as required
Proof
xercses 2Dl
2D2
2D
rove that the sum of the squares of the lengths of the medians media ns of o f a triangle is three fou fourths rths the sum of the squares of the lengths of o f the sides si des quadri quadrilate lateral ral with side side lengths lengths 1 1 1 and d is inscribed ins cribed in a circle circle Find a formula for the radius R of the circle in terms of d heck your formula by computing R directly when d = 1 HINT: rove that the diagonals agonals of the the quadrilateral quadrilateral must m ust be equal equal Given Given a regular pentagon with side length 1 compute the lengths of its diagonals HINT: y Exercis Exercisee G2 G 2 the polygon can be inscribed in a circle circle Note that that all ve diagonals must m ust have the same length length x and use tolemys tolemy s theorem to to nd a quadratic quadratic equation sat s atisised ed by x
E THE NCRCLE
2E
73
The ncircle
inte rior to the triangle and all circle is said to be incribd in a triangle if its center is interior three sides of o f the triang triangle le are tangent to the circle Given an arbitrary arbitrary triang triangle le we shall sh all see that there must exist a unique inscribed circle This circle ci rcle is called call ed the incirc; its center I is the incnr, and the length r of its radius radius is the inradiu of the triang triangle le We can see informally why the the incircle must exist Start with with a small circle placed ins ide the triangle triangle and let it grow continuously keeping iti t inside the triangle triangle by letting its center move freely freely as the circle circ le grows grows Eventually Eventuall y the circle will reach a maximum size siz e after after which there is no room ro om for for further further growth t that that point the circle circ le will wil l be touching touch ing (that is is tangent to) all three three sides unsatis factory and so we present pres ent a more formal formal This This argument is rather hazy and unsatisf proof proof We begin with a lemma lemm a LEMMA The b sect se ctrr f f LABC s the lcus fpn fpnts ts P n the th e nter nt er rr f f the angle that are a re equ eq u dstant dstan t frm the sdes f f the angle an gle A
c igure
216
ecall ecal l that the distance distance from from a point to a line is i s measured perpendicularly If P is in the interior interior of LA L A B C as in Figure Figure 21 drop perpendiculars perpendicu lars P X and P Y to lines A B and C B S aying that that P is equidistant from from the the sides of the the angle angl e therefore therefore is the same s ame as saying s aying that P X P Y and our task is to show that that this happens if and only if P lies li es on the angle bisector Suppose Supp ose rst that P X P Y Since B P B P we can c an conclude conc lude that that P X B r P Y B by because becau se these thes e are right right triangles with right angles angle s at X and Y It follows follows that that LX L X B P L Y B P and thus line B P is the bisector of L A B C other words P lies li es on the angle angle bisector bis ector as required required In other onversely now suppose that P lies on the angle bisector Since B P is the only line through B that contains P it follows follows that B P is i s the angle bisector bise ctor and and so LX B P L Y B P Sinc Sincee LBX P = 90° L B Y P and and B P B P we see that that • S and we conclude co nclude that P X P Y as desired P X B r P Y B by S
Proof
THEOREM The three angle bsectrs bse ctrs f fa trangle trangle ar a re cncu c ncurr rrent ent at a pnt I equ eq udsta dstant nt frm the t he sdes s des f f the trangle tran gle If we dente dent e by r the dstan d stance ce frm I t each f the sdes then the crcle f radus r centered at I s the unque crcle nscrbed n the gven trangle
74
CHAPTER
TRANGLES
et ABC be the given triangle and let I be the point where the bisectors of L B and L C meet From I drop perpendiculars I , I V and I W from I to sides B C AC and AB respectively respec tively and note that by emma 2.24 we have I I W since I lies on the bisector of LB L B Simila S imilarlrly y since I also lies on the bisector of L C we see that I I V. We conclude concl ude that I W I V, and thus by emma 224 point I must lie on the bisector of LA en ence ce all three three angle angle bisectors go through this point as claimed We can now write I I V I W and we see that I is equidistant from sides B C A C and A B oint lies on the circle of radius centered at I and since B C i s a line through through perpendicular to radius I , i t follows follows that B C is tangent tangent to this circle c ircle at Similarly A C is tangent to to this circle at a t point poin t V and A B is tangent at W . We have now shown that the circle of radius centered at I is inscribed in ABC and what remains is to prove that this this is i s the only circle inscribed ins cribed in this triangle triangle Suppose Supp ose that we are are given some inscribed circle To prove that that it is the circle we have already already found it suf su fces to show that its center is I and that its radius is et P be the center of our our unknown inscribed circle and let X Y and Z be the points points of tangency of this circle with sides si des B C A C and A B, respectively Then radii P X and P Y are are perpen perpendicular dicular to sides B C and A C and since P X P Y we see by emma 2.24 that P must lie on the bisector bisector of LC L C Similarly Similarly P lies on the bisector of L B, and we conclude that P is the point I , as expected It follows follows thateach of P X and is a perpendicular perpendicular drawn drawn from from this point po int to BC and hence these are the same line s egment The radius radius P X of the unknown unknown circle cir cle is thus thu s equal to I r, and the proof is complete •
Proof
In the spirit of the previo previous us ssectio ection n we ask how the inradius can be computed and how the points of tangency tangency , V and W can be located In the following we refer to Figure 217 where we have drawn the incircle of ABC and the three radii I IV and I W of length joining the center I to the three points of tangency of the circle circle with the sides These The se radii are thus thus perpendicular to the correspo corresponding nding sides s ides We have also drawn segments I A, I B and I C which by Theorem 2.25 we know bisect bisec t the angles of the the triangle triangle (.6) PROBLEM
prove that s triangle
Given a triangle with area K semiperimeter s and inradius inradius , expresss in terms of the lengths length s of o f the sides si des of the K se this to expres A
U igure
2.17
E THE NCRCLE
75
onsider �B I C in Figure 217 Since I is perpendicular to BC, we see s ee that the area K1 B C (I)(BC) ra Similarly KIA KI A B rc and K1 c A rb dding dding these w e get K K1 B C + KI A B + K1 C A 1 ra + + c) rs 2 as required. Since we know by erons formula (Theorem 218) that K a)(s a)( s b )(s c) c) we deduce that
ouion
r
= Ks =
(s a)(s b)(s b)(s c) s
which is the desired form formula ula.. •
Next we attempt attempt to pin down the precise precis e locations loc ations of the the points of tangency V and along sides BC, AC and AB (We (We continue cont inue to refer refer to Figure 217) Exercise Exercise F.3 asserte ass ertedd that the lengths of the two two tangents to t o a circle from from an exterior point are equal and so A V A . In fact fact it is easy eas y to see this this directly because �A V I r � A I by . We write x AV A W and similarly y B B and z C CV We have y + z BC a and similarly x + z and x + c It i s easy to solve these three equations equati ons for the three three unknowns x y and z by writing z a y from the rst equation and substituting subs tituting this into the second equation to get x + a y Thus third equation equation we deduce d educe that x a and since x + y c from the third (c + a ) (a + + c) 2a sa 2 2 and similarly y s and z s c that we w e just I t i s interesting to note that in the calculation of the the distances dis tances x y and z that completed we did not fully fully use u se the assumption ass umption that that points V and are are the points points of tangency of the incircle. incircle . ll l l that was actually used us ed were the three three equations A V A information ion to determine the distances dis tances x B B and C C V; that was enough informat the precise locations o f the three three points poi nts , V and y and z and thereby to determine the reference erence iti t is worth stating this observ obs ervation ation as a lemma which we w e have just jus t For future ref proved. he re s exactly exactly ne way t chse pnts p nts V and n sdes sdes BC B C LEMMA Ther respectve pe ctvelly f � A B C s that A V A B B W and and CU A C and A B res C V The nly nly pnts p nts that sats sats these equatns equ atns ar a re the pnts pn ts where the sdes f f the trangle are tangent t the ncrcle Furthermre Furthe rmre the dstances A V B and C • are are equal t s a s and s c respe respectve ctvely ly n the usual ntatn ntatn
PROBLEM Given three three pairwise mutually externally tangent circles show three common commo n tangent lines are concurrent that the three
76
CHAPTER 2
TRANGLES
igure
218
Since robem 228 appears to have have nothing to do with inscribed in scribed circes or even with trianges, trianges, the reader reader may wonder wonder why we have presented it here s we sha see, the ey to the soutio s outionn is the fact fact that the point of concurrence of the three three tangent ines turns out ou t to be the incenter of an appropri appropriate ate triange triange s shown sho wn in Figure Figure 28, we et et U, V and be the three points where two of the circes touch, and we write A, B, and C to denote the centers of the three given circes Observe that radius B U is perpendicuar to the common tangent through through U and that that the same is true of radius C U It foo foows ws that L B U C = 80° and thus thus U ies on the ine ine segment B C S imiary, V ies on A C and and ies ies on on AB We see now that U, V and are are ppoin ointsts on the the sides sides of A B C, and we have A V = A since these thes e two segments are are radii of the same s ame circe Simiar Simi ary, y, B U = B and and C U = C V and we concude c oncude from from emma 227 that that points U, U, are the points points of tange tangency ncy of the the incirc incirce e of A B C with with the sides of V and are the triange triange It foows foows that radius radius I U of the incirce is perpendicuar to side si de B C at U We ow, however, that the common tangent ine for the circes circe s centered ce ntered at B and C is perpendicua perpe ndicuarr to B C at U, U , and thus thus this common com mon tangent must mus t be ine I U Simiay, Simi ay, the other two two common co mmon tangent ines aso as o go g o through through the incenter I of A B C , and hence the three three common tangents tangents are concurrent at I . •
ouion o Probm 28
We return return now to the situation situ ation of Figure 27 from which whic h we can extract sti sti ore infomation Given the engths a b and c of the the sides of A B C, we now now that that we can compute the anges of this triange using the aw of cosines cos ines If we want L C, for exape, we start with the equation c = a + b 2ab cos (C) (C ) From From this, we deduce deduce that that cos(C) cos (C) = ( a + b c )2ab and this this determines determines C since s ince there is ony one on e ange in the range 0° to 80° that has has any given cosine The foowi foowing ng aw of tangents is an aternativ aternativee method method method for for nding C (22) THEOREM (Law of Tangn)
have have ther the r usual meanngs Then
In ABC let the the quanttes a b c and s
(s  a)(s b) s(s c)
2E THE NCRCLE N CRCLE
77
onsider right CU in Figure 17 We now that U/UC is the tan gent of L U C and since C bisects L C of the original original triangle this gives tan(C/) U/UC r/(s r/(s  c) We saw previously however that r •  a)(s a)(s  b )(s )/ and the resul resultt fol follows lows
Proof.
Of course cours e with w ith the obvious modications modi cations the law law of tangents tangents wors to give formulas formulas for for tan( tan(AA /) and and tan(B tan( B /) also s a comparison comparison of the laws of osines and and tangents tangents we use both methods to approximate approximate L C in a triangle for which a = 5 b 6 and c 7 First First using the law of cosines cosines w e have have cos(C) cos (C) (5 + 36 49)/60 49)/60 1/60 1/5 sing the inverse cosin c osinee button on a calclato c alclatorr we w e nd that L C 7 8 4 46 3 To use the law of tangents we rst compute that 9 Thus tan( tan(CC /) = and to to double the result result we we sing a calculator c alculator to compute the inverse tangent o f obtain the same answer for L C as a s before before This calculation gives u s the exact formula: formula: arccos(1/5) which might be a bit bit tedious to try to prove prove directly directl y It is amusing that we can use the law of tangents to produce a proof of the ythagorean theorem If L C 90 then tan( tan ( C /) 1 and the law law of tangents tangents yields (s  a)(s  b) s(s  c) From this and the fact that 2s a + b + c, some algebraic manipulation manipulation which we leave to the reader yields that c a + b , as required In fact fact this proof" of the ythagorean theorem theorem is invalid because it is i s circular: The thagorean theorem underlies underli es our proof of the law of tangents We leave it to the reader to trace through the the arguments and verify verify this assertion as sertion It is easy to see that in the case of an equilater equilateral al triangle triangle all of the special points we have been discus dis cussingthe singthe circumcenter circumcenter the centroid the th e orthocenter orthocenter and the incenter incenter coincide It is not too hard to see that conversely if any two of these points coincide then the triangle must be equilateral equilateral S uppose that that the inc enter enter and the circumcenter of A A B C are are the same point point Show S how that the the triangle triangle must be equilateral equilateral
(.0 PROBLEM.
Join to vertices A and B and note note that A B is isosc is osceles eles since we now that A B y the the pons asinorum asinorum we have have L A B L B A Since we are assuming that is also als o the inc enter enter we now that A and B bisect angles A and B and we conclude that LA = L AB an and LB = L BA It follo follows ws that that the pons asinorum we see that C A = C B L A = L B and so by the converse of the Since Si nce we could co uld have started started with any two of the three three vertic vertices es it foll follows ows that all of • the sides are equal as required
ouio.
We now now that except when the triangle is equilateral points and never coincide co incide It seems natural therefore to as just how far apart these two points are in general general n answer ans wer is given by the follo following wing theorem of Euler Eul er et d the dstance frm the crcumc c rcumcent enter er t the ncenter nc enter f f an arbtra a rbtra trangle Then d R(R  r ) where as usual R and r are the crcumradus c rcumradus and nr n radus f the gven gve n trangle tran gle
(.1 THEOREM (Eur.
78
CHAPTER 2
TANGLES
We need an intrsting preliminary result rst Extend the bisectr f ne f the angles f a triangle t meet the circumcircle at int in t P Then the distancefrm P t each f the ther ther tw vertices vertic es f the triangle is equal t the distance P where is the incenter f the given triangle
(. LEMMA.
A
A
B
C p igure
219
igure
220
In Figu Figure re 1 9 we have have draw drawnn line AP A P bisecting bisecting L A of AB A B C Of cours coursee lies on this ine in e aand nd our task is to show that P = C P y the converse of the the pons asinorum asin orum it suf su fces ce s to show that L C P = L C P and so we star s tartt computing angles Since Sin ce L B C P subtends the same arc as L B A P w e see that L B C P = B A P = lso C bisects bi sects L C of the the original triangle triangle and thus L C B = L C and we L A lso conclude conclude that C P = i (LA + LC) To compue L C P w e observe obs erve rst that L P = L B since sinc e these thes e subtend the same arc It I t follow followss that L C P + LCP = 1 8 0  L P = 1 8 0  L B = L A + C and since w e have seen that L C P i s exactly exactly half of this quant qu antity ity w e euce that that L C P is the her half Thus C P = L C P and the proof is complete complete •
Proof.
We are are now rey to prove prove Euler s theorem s shown in Figure 0 we extend line segment to a diameter X Y f the circumcircle Since = d, we see that X = R  d and Y = R + d, nd thus X · Y = R d y Theore Theorem m 1 3 5 however however we know that that the product o the the lengths of o f the two pieces of each chord through through is a constant independent indepe ndent o the particular chord and it follows oll ows that R  d = X Y = A P
Proof of Thorm .1.
E TE NCRCLE
79
Next we try to compute the length P which wh ich by emma 3 3 we know is equal to to P C To To accomplish this this we use the extend extended ed law law of sines sines in � A P C Since PAC = L A we have have PC =2R sin(A2) and thus thus sin(A)) R d = A P = A PC = A R sin(A Finally Finally we compute compute A A · sin(A sin( A 2) by working working in right right �A � A F , where F is the point of tangency of the incircle with side A C Since F = r, we see s ee that and thus thus A sin(A sin(A ) = r If we substitute this into our sin si n (A2) = r A and • previou previouss formula formula we w e get R d = R and Eulers Eulers formula formula follows follows Fr any trangle R r and equal equ al hld h ldss and nl the trange s equ eq ulateral lateral
(. COROLLARY.
Since d R (R ) we see that that R can never never be negativ negativee Furthe Furthermo rmore re words R r if and and only if points and are R = 2 i and only ifif d = 0 in other words identical identi cal We already know however that that and coincide for equilatera equ ilaterall triangles triangles • and by roblem roblem 30 30 only for for equilateral equilateral tria triangle ngless
Proof.
xercses E.1 E. E. E.4
E.5
Suppose Suppo se that the orthocente orthocenterr and incenter incenter of �A B C are the same point rove rove that the the triangle is equilateral Suppose Suppo se that the centroid and incenter incenter of �A B C are are the same s ame point rove that that the triangle triangle is equilateral Show Sh ow that in a right triangle triangle the inradius circumradius circumradius and semiperimeter are are related by the formula formula s = r + R et CD C D be b e an altitude altitude o f �A B C and assume that C = 90 et r and r be the the �C A D and �C � C B D respectiv respectively ely and show that that r + r + r CD inradii of �CA where where as usual usu al r is the inad inadius ius of �A B C Extend Extend the the bisectors bisectors of LA B and LC L C of �A B C to meet meet the the circumcir circumcircle cle at at points X, Y, and Z Show Sh ow that that is the orthocenter of � X Y Z H S how that tha t X Y is the perpendicular bisector o f C
80
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2F
TRANGLES
xcribed ircle
The incircle of o f a triangle triangle is i s sometimes referred referred to as a riag circle because it is tangent to all three three sides side s of the triangle triangle We can see in igure igure 1 that if we are willing to consider circles tangent to etensins of the sides a triangle also has three other tritangent tritangent circles circles Each of these is tangent tangent to one side and extensions extensio ns of the the other two sides
igure
221
The three tritangent tritangent circles whose centers centers are are exterior exterior to the given triangle triangle are called c alled the xcribd circ or the xcirc of the triangle triangle lthough lth ough we w e shall sh all not give a formal formal proof proof of the existence and uniqueness uniquenes s of the three three exscribed circles it should shou ld be clear cl ear by analogy with the inscribed circle how to construct construct such a proof It should shou ld also be clear that as was the case c ase for the inc enter enter the center of each of the exscribed circles c ircles lies at the intersectio intersectionn of thee thee angle angle bisecto bis ectors rs In igure igure 1 for example example the center center of the excircle opposite vertex A has been denoted I; it lies at the point of concurrence of the bisector bisector of L A and the bisectors bis ectors of the eterir angles angles at points points B and and C The centers centers of o f the excircles excircles o f �A B C are are the xcr of the trian triangle gle and it is is I , and I, as indicated i n the gure customary to label these points , I gure The corresponding radii are denoted r, r, and r , and they are referred referred to as the xradii f the triangle The three exradii exradii together with the inadius inadius r, are are collectively col lectively known as the tritangent tritangent radii radii and as it tus out there is a pretty pretty relationship relationship among these four four quantities quantities (.4) THEOREM.
Given an arbitra arbitra �ABC we have 1 1 1 + + r = r r r
Just as it was useful u seful to locate the points of tangency tangency of the the incircle along alon g the sides si des of the triangle triangle so too too to prove prove Theorem Theorem 3 34 4 will wi ll we need to nd the points of tangency tangency of the three three excircles excircle s We We are esp especia ecially lly interested in the exteal points poi nts of tangency The length leng th f the tangent tangen t frm a verte ve rte f f a triangle tria ngle t the ppsite ppsit e escribed circle is equal t the semiperimeter s
(.5) LEMMA.
2 EXSCRBED CRCLES
81
Q igure
222
s in Figure . 2 let Y be the point of tangency of side A C with the excircle opposite B and let P and Q be the points of tangency of this circle with the extensions extensions of A B and B C . ince the two tangents tangents to a circle from from an exterior exterior point are equal we know that that B P = B Q and we need to show that this common length is s Since A P = AY we have B P = BA + A P = BA + AY and similarly dding these equations equations we obtain B Q B C + C Y . dding B P + B Q BA + AY + BC + CY = BA + BC + AC = s ut B P and B Q are equal equal and since sin ce their sum is 2 it foll follows ows that each of them • is equal to s .
Proof.
=
=
We can use emma .35 to express the exradii of �ABC in terms of the side lengths a b, and c Observe Obs erve that ex center I lies on the bisector of B. If as in write P to denote the point of tangency tangency of the excircle opposite oppos ite B with Figure 2 2 we write the exte extension nsion of side A B we see from from emm emmaa 35 35 that that the the � B P I is a right triangle triangle with B P of length s Since Sinc e arm IP has length r, it follows follows that tan( tan( B) B ) = rs y the law of tangents tange nts we obtain (s a)(s a)(s  c s (s b) Proof of Thorm .4 .
=
a)(s
sb
c)
.
We have
sa sb 1 1 1 sc ++= + . r r r s(s b ) (s s(s s( s a)(s a)(s c + s(s a)(s b) (s  c ombining ombin ing thes e over a common commo n denominator we see that the above sum is equal to s 1 (s a) + (s b) + (s c s K r a)(s a)(s b )(s )( s c a)(s a)( s c) • by robl roblem em 6. 6.
xercses F Fl
In Figu Figure re 2.2 2. 2 show that that A Y = s c and and C Y = s a
82
CHAPTER 2 TRANGLES
B igure
2F2
2.23
In Figure Figure 33 three common tangent tangent lines have have been drawn to to two circles circles Show that that A V BU HINT: iew the two circles as the incircle and one of o f the excircles excircl es of an appro appro priate triangle
Morley's Theorem
We devote this section to an amazing theorem discovered by Fran Morley at the beginning of the 0th century lthough this result is somewhat awward to state in words the diagram of Figure Figure 24 2 4 maes the assertio ass ertionn striingly clear cl ear A
B
C igure
2.24
Draw the sx angle trsectrs trsect rsf r an arbtra arb tra trangle and f r each sde s de f f the trangle t rangle mark the nter n tersect sectn n pnt p nt f f the tw t w trsec tr sectrs trs near nea rest that sde Then the tranglef trangle frmed by by the three marked pnts p nts s equ eq ulateral lateral
(2.6 THEOREM (Mory.
In the the gure gure lines lines A Y and AZ divide divide A into thr three ee equal equal parts parts Similar S imilarly ly B X and BZ trisect trisect L B and CX and CY C Y trisect trisect C C Morleys Morleys theorem theorem says that that �XY Z must be equilateral equilateral We need a preliminary result resul t about incenters et A X be the bsec b sectr tr f f A n �ABC where X les n sde sde BC Then the ncenter f �ABC s the unque pnt P n segment A X such that B PC 90 + i LBAC
(2.7 LEMMA.
2G MORLEY' S THEOM THEOM
83
First First ass assume ume that that P is the incenter. incenter. Then Then P lies on all three angle bisectors bis ectors and and we compute compute in in � B P C that that B C BPC = PBC LPCB = o = LA) = + L A
Proof.
1 0 1 10 0
0 90 1
and thus the incenter has the property that was claimed. claime d. To see that the incenter is the only point on AX with the stated property consider what happens happens to L B PC as point P moves moves along segment A X from from point A from a diagram diagram that that B PC is monotonically monotonic ally strictly increasing to point X It is clear from from a minimum equal to A when whe n P is at A to a maximum of when P is at X there can be just one point P where B P C taes o n any particular I t follows that there • value and thus the incenter is the only point where where L B P C = + A
10 90
We shall use us e a rather rather unusual unusu al method to prov provee Morleys Morley s theorem. S o before before we begin begi n it might be a good idea to explain the strategy. We We shall describe des cribe the diagram of Figure 4 as the Mory cograio associated asso ciated with with �A B C . y y this we mean that that the six six lines lines AY AZ A Z B X B Z C X and CY are are the the trisect trisector orss of A B an andd L C . Our Our goal of course is to show that that for every Morley Morley conguration � X Y Z is i s equilateral, eq uilateral, and our proof will proceed by constructing a large variety variety of Morley Morley congurations con gurations for which �X Y Z actually is equilateral. (We (We shall sh all say s ay that a Morley conguration is ccf if suc cessfful congurations congu rations prove � X Y Z is equilateral) ut how does the construction of success the theorem theorem The ey here is to observe ob serve that if we we start with a pair pair of similar simil ar triangles triangle s then each diers from the other by expansion or contraction via an appropriate scale factor. It follows that the the entire Morley congurations ass a ssociated ociated with these two triangles triangles dier dier expansio n or contraction via the same scale sc ale factor factor and thus if one from one another by expansion is successful then so is the other. Given an arbitrary triangle therefore it sufces to produce a success succe ssfful Morley triangle triangle associ as sociated ated with with some triangle similar to the given triangle. To guarantee guarantee that the Morley Morley conguration co nguration we produce is successful we wor bacward. We start with equilatera equilaterall �X Y Z as shown in Figure 2.5 2. 5 and we wor wor to construct construct a larg larger er �A BC which together with �X �XY Z
Proof of Thorm .6.
A
B
igure
c
225
84
CHAPTER 2
TRANGLES
will form a Morley Morley conguration con guration We will be done if we can do this in i n such suc h a way way that A B C will be similar to the the given unshown U V W et et and be U L V and and W respectively respecti vely and note that + + 1 80° 60° 60° uild isosceles isosceles Z P Y on side side YZ Y Z of XY Z as shown shown in in Figu Figure re 25, 25 , and and do this so that P Z Y P Y Z + To see that this this is pos sible it sufces to observe that + < 90° Similarly construct points Q and R so that X Q Z is iso is o sceles with with Q ZX Q XZ and and X R Y is isosceles wit withh RX Y R Y X + · Next extend segment Q Z in the direction direction of Z and extend RY R Y in the direction of Y We claim that these extensions meet at some point A in the vicinity of the label A in Figure 5 To see why this is so it sufces to consider the sum L Q ZY + R Y Z Observe that if 180° that would imply that QZ RY Otherwise lines Q Z and R Y meet and we observe that if i f < 180° then the intersection inters ection point po int would woul d lie below line Y Z in the gure and if > 1 80° then then the the intersection inters ection point p oint is above Y Z as we have claime c laimedd In fact fact Q Z Y Q Z X + X ZY + + 60° 60° an and simil similaarly R YZ + + 60° 60° Sinc Sincee + 60 60 it follows that
%
%
%
%
+ + + 10° 1 0° + 180° 1 80° This exceeds exc eeds 1 80° and we conclude that the intersection intersection point A of lines Q Z and R Y is positioned as indicated indicated above above line Y Z lso lso woring in A Z Y we see that that LA 1 80° 80°  AZ Y  A YZ and and thus A 180° 180 °  ( 1 80 8 0°  L Q Z Y) Y )  ( 1 80 8 0°  LRYZ)  180° 180° Similar Simil arly ly lines lin es P Z and R X meet at B and lines Q X and P Y meet at C C were points B and C are positioned as as indicated indicated in in Figure Figure 5 lso ls o we have have B and LC We now wor to show that Z is i s the inc enter enter of A B R Observe rst rs t that Z Y R Z X R by SSS and thus thus L Y RZ XR X R Z and Z lies on the bisector bis ector of A RB If we can can show show that that A ZB 90° + A R B it will will follo follow w by emma emma 23 7 that Z is the inc enter of A R B as desired We have
AZB L P Z Q PZY + Y ZX + XZ Q ( + ) + 60° + ( + ) 120 120° + since si nce + + 60° Y RX that that 60° lso we see in YRX 2(60°  ) R 180° 180 °  RYX  RXY 180° 180°  ( ( + ) 1 80 °  2(60° 60 ° + and thus thus
as required required It follows follows that Z is the incent incenter er of A R B as claimed c laimed and thus AZ bisects bisects B A R and and we have have B AZ L ZA Y Similarl Similarly y Y is the incent incenter er of
2 OPTMZATON N TRANGLES TRAN GLES
85
�A QC and and we we dedu deduce ce that that Y Y A C = a . Thus LB AC = 3a, and and A Y and and A Z are are the the trise trisector ctorss of LA in �A B C Similar Si milar reasoning shows that L B = 3 and B Z and B X are the trisectors trise ctors o this angle in � A B C lso L C = 3 and C Y and an d C X are are angle angl e trise trisectors ctors and thus we have the the Morley Morley conguration conguration associated with �A � A B C In fact fact this is a successul conguration sinc s incee we started started with �X Y Z being bei ng equilate equil ateral ral Furthermore Furthermore �A B C • is simil si milar ar to our given given � U V W by and the proof proof is complete comp lete
y
xercses G 2G.l
Join Join each each vertex of �A B C to the points of trisection of the opposite oppos ite side si de and the points of intersection of these sidetrisectors side trisectors as shown s hown in let X Y and Z be the 6. Show that that the sides si des of �X Y Z are parallel parallel to corresponding sides Figure 6. of �A B C and and tha thatt �X YZ �ABC A
igure
226
Optiization in iangle
Our goal in this this section is i s to solve certain geometric geometric optimization or maxmin maxmin problems by geometric metods metods s an example of what we mean mean we begin with something s omething easy Given points points P and Q on the same side of a given line nd the poit X on the line that minimizes the total distance P X X Q Show that at this
(2.8 PROBLEM.
Q
F
R
igure
227
86
CHAPTER 2
TRANGLES
point poi nt the bisec bi sector tor of L P X Q is perpendicular perpendi cular to the line and that P X and Q X ae equal angles with with the line First reect one of the given given points points (say Q) in the given given line to get R as in Figure 7 7 ecall that this eans that that we drop drop a perpendicular Q fro fro Q to to the line and then choose R on the line Q on the opposite side of the given given line from Q so s o that that Q = R Next draw draw P R and let X be the point poin t where where P R eets the given line line We claim cl aim that this point X is the th e unique solution to our optiization problem To establish e stablish this this we ust us t prove that P X + X Q < P Y + Y Q for every pos sible choice cho ice of a point po int Y on the line with wit h Y dierent fro fro X To prove the inequality we observe ob serve that X Q = X R and and Y Q = Y R since sin ce line X Y is the t he perpendicular bisector of segent segent Q R Thus
ouio.
P X + X Q = PX + XR < P Y + Y R
=
PY + YQ
as required where the inequality holds since P X + X R = P R is i s the straightline distance fro fro P to R and so is shorter than the broenline broenline distance P Y + Y R fro P to R Note that that we could also have have found found a solution by reecting P rather rather than Q This would wo uld necessarily neces sarily have have yielded the sae point X because we have seen that X is the unique solution to the proble proble We ust show that the bisector bis ector X B of P X Q is perpendicular to the th e given line line To see this thi s note that t hat L P X Q is an exterior angle of Q X R and thus P X Q = R + L Q L R where where the last equality equality follows follows via the the pons asinorum asi norum fro fro the the fact act that that X Q = XR Thu Thuss L PXB = L P X Q = L R and it follows ollows tha thatt XB R Q Since Q R is perpendicular to the the given line so too is X B Finally to see that P X and Q X ae equal angles angl es with with the line observe that that these two angles ang les are just ju st the coplements coplements of L P X B and Q X B which are are equal • Next we we cons co nsider ider a harder harder and more interesting proble proble Give Givenn A B C nd points points X X Y and and Z on sides BC AC and and A B respectively such that the perieter perieter of X Y Z i s iniized iniiz ed
(. PROBLEM.
First First let us consider whether whether or not rob roble le 39 39 necessarily has a solution If we wish wis h to prove the existence of a miniu using the compactness technique that was discusse discu ssedd in Section Sec tion G we need to to be sure that that the doain of choice for each each of the variables variables X Y and Z is close cl osedd This will wi ll be the the case if we interp interpret ret the requirement requirement in in the problem problem that the points are are chosen on the sides of A B C to allow the possibility that one or ore of the points are actually at a vertex If we do that however however we can no longer be sure that X Y and Z are distinct points and in that case we ust agree about the eaning of the perieter perieter of the triangle fored fored by these thes e points Our interpreta interpretation tion is that that the perimeter perimeter of X Y Z is the su of o f the distances X Y Y Z and Z X X even if one one or more more of these distances di stances is zero With this interpretation interpretation and allowing X Y or Z to be at a vertex vertex roble 39 39 always has a solution
H OPTMZATON OPTMZATON N TRANGLES
87
We shall see that if one of the the angles (say (s ay LA) of the original triangle is not acute �XY Z occurs when both Y and Z are at A It is clear then the minimum minimu m perimeter for �XYZ in this case that X must be at the foot of the altitude from A and that the minimum perimeter is 2AX If we had decided to interpret the problem strictly and we did not allow Y and Z to coincide with A there would be no no solution in this this case. c ase. This is because wherever wherever we place Y and Z we can always always reduce the perimeter perimeter of �XYZ by moving these points closer clo ser to to A and adjusting X appropriately. The ey to the solution sol ution of roblem 239 is the following following lemma. le mma. Given Given � A BC with with acute acute anles anles at B and and C and and L A = x a poi po int P on side B C and let Q and R be the reections reections of of P in sides A B and AC A C A B and AC and consider the respectively et U and V be variable points on sides AB perimeter p of � a. If < 90 then the mnimum value Pm of P occurs when and only when U and V coinc co incide ide ith the points po ints Z and Y where line Q R meets sides sides AB A B and and AC AC Furthermore Pm = Q R = 2A· sin«) min imum value val ue of of p occurs occ urs when and an d only only when U and V b If 90 the the minimum both coincde with point A A
(2.40) LEMMA.
Suppose rst that that < 90 and refer to Figure 228 sing congruent triangles eas y to see that L QA B = LB AP and L C AR = L PAC It foll follows ows that it is easy L Q A R = LQAB+LBAP+LPAC+LCAR = 2(LBAP+LPAC) = 2 < 1 8 0 and thus thus segment s egment Q R really does intersect intersect A B and AC as shown in Figure 228 that segment QR Q R passes pas ses above point A Note that if i f > 90 90 this calculation shows that pass es through through A and that if ( = 90 then Q R passes Since A B i s the perpendicular bisector of segment segment Q P w e see that that U P = U Q and similarly sim ilarly V P = V R . It follows that p = P U + U V + V P = Q U + U V + V R is a broenline broenlin e distance dista nce from from Q to R unless unless U coincides coincides with with Z and V coincides i s the straightline distance from from Q to R It follows follows with wi th Y in ich case p = Q R is that p is minimized when U and V are at Z and Y and that every every other possible poss ible choice of U and V yields p > Q R To compute the length QR consider �AQR We now that A Q = P = AR and so this triangle is isosceles If we let M be the midpoint of Q R then
Proof.
Q p
igure
228
88
APTER 2
TRANGLES
A is perpendicuar perpendi cuar to Q R and A bisects bi sects Q A R Thus Q A = and sin«) = Q A Q It fo foows ows that that Q R = 2Q = 2A Q sin«) sin«) = 2A sin« sin«) ) This competes co mpetes the proof of part part (a) ssume now now tha thatt B A C 90 90 s we hav havee seen ine Q R passes passe s above above or or through point A in this case and it is cear from a picture that the shortest path from Q to U to V to R with U on AB and V on AC occurs when U and V are both at A • If the given triange is not acute anged we can assume that A 90 Wherever Wherever we pace point X on side B C we now from from emma 240 2 40 that that the optimum position posi tion for Y and Z so as to minimize the perimeter perimeter p is to have them them both coincide with A A In this situation p = 2AX and so we can minimize p by minimizing minimizi ng A X We We accompish accompis h this by pacing X at the the foot foot of the atitude atitude from A Much more interesting interesting is the case where where a anges of A B C are are acute and we state the resut in that case as a theorem •
ouio o Probm 2..
The edal trianle ofa iven i ven acute a cute anled anle d trianle has a smaller sm aller perimeter pe rimeter than any other othe r trianle whose vertices v ertices lie li e on the three thre e ides of of the iven iv en trianle
(2.41) THEOREM.
For any choice of point X on side B C we now by emma 240 that we can choose Y on A C and Z on AB so that that the the perimet perimeter er of X Y Z is equa to 2AX 2A X sin( A) and this this is i s the smaest possibe perimete perimeterr for the given given point X and for any choice of Y and Z To nd the overa minimum perimeter therefore we must choose X so as to minimize minimize AX since sin( A) is i s constant The smaest smaest possibe perimeter occurs therefore when X is the foot of the atitude from A and Y and Z are as specied spec ied by emma 2 40 N o other other choice of the point X aows for a pperimeter erimeter as as sma s ma as this this no matter how points Y and Z are seected We have hav e seen se en that there is a minimum poss po ssibe ibe perimeter perim eter for X Y Z and a nd that it can occur ony when when X is at the foot foot of the atitude atitude from from A A S imiar reasoning shows show s that for X Y Z to have the minimum minim um perimeter Y and Z must be b e the feet of the atitudes from from B and from from C It foows that among among a pos p ossibii sibiities ties for X Y Z the peda triange triange and ony the peda triange triange has h as the minimum perimeter •
Proof.
There is more information information avaiabe from our anaysis anaysi s Given acute anged A B C with atitude A X w e now from from emma 2 40 that that to minimize m inimize the perieter of X Y Z with with Y on AC A C and Z on AB we must tae tae points Y and Z on ine Q R where where Q and R are the reections of point X in sides AB and AC On the other hand Theorem 241 tes te s us that if we we tae Y and Z to be the other other two atitude feet feet the this thi s wi wi minimize the perimeter of X Y Z ombining these facts facts we deduce that atitude feet Y and Z ie on on Q R This proves the the foowing oowing
2H OPTMZATON N TRANGLES
89
In an acute anled tranle the reecton of the foot of an alttude n ether of the sdes not contann t s collnear wth the other two • alttude altt ude fee feet.t.
(2.42) COROLLARY.
We can get still stil l more information from from Theorem 2.41. Fix altitude feet Y and Z on sides A C and A B of acute acute angled �ABC and consider consi der the point X on line B C that minimizes the distace ZX + XY. We know from robem 2.38 that there is such a point. It is clear c lear from from a diagr diagram am that X must lie lie between B and C . lso lso because we we are holding Y and Z xed we see that minimizing ZX X Y is exactly equivalent to the problem problem of choosing choo sing X to minimize the perimeter perimeter of of XY Z. Since Y and Z are two of the vertices of o f the the pedal triangle triangle it folows by Theorem 241 that the minimum perimeter will occur when X coincides c oincides with the third third vertex vertex of the pedal triangle. We have now shown show n that if X is chosen cho sen on side sid e B C so as to minimize Y X + X Z then A X is an altitude of the given acute angled A BC . We know from roblem roble m 2.38 however, however, that for for this this point X X the bisector bi sector of LYXZ is perpendic perpendicular ular to BC. ut A X is an altitude o f A BC and and so iti t is perpendicular perpendicular to B C at X and we conclude that that altitud altitudee A X bisects L Y X Z . We have proved the following followi ng The alttudes alt tudes of of an acute acu te anled anle d tranle are the an a nle le bsec bs ec tors of of ts pedal pe dal tran tr anle le •
(2.4) COROLLARY.
We consider consi der next a problem proposed propos ed by ierre Fermat Fermat (10115) ocate the point F i n the interior of triangle ABC that minimizes FA + F B + FC.
(2.44) PROBLEM (Frma).
Since the interior of a triangle is not a compact set we have no guarantee that roblem 2.44 has a solution If we reinterpret the problem however and allow the possibiity pos sibiity that that F lies either inside insid e the triangle triangle or on one of its sides sid es or vertices vertices then a solution must mus t exist exis t It tus tus out that that if one one of the angles (say ( say LA) of the given triangle is 120° or larger larger then the solution so lution to the modied problem is to take F to be A . f we triangle there is no solution in this case. s we realy insist that F be interior to the triangle shall see sha see there is a very very pretty pretty solution solu tion to Fermats Fermats problem when whe n all three three angles of of are less than 120°. In that that case case it tus out ou t that that there there is a unique uniqu e point F in the �ABC are for which the quantity qu antity F A + F B + F C interior of the triangle (called the Frma poi for s ome additional amazing amazing properties properties:: It is is minimized. We sha see that this point has some the point of concurrence of three three interesting ines and three three interesting interesting circles . Buld Bu ld an outward outwardpontn pontn equ lateral lateral A B R on sde sde A B of A B C po nt X n the plane we have X A + X B + XC R C. Equa Equall can can a. For eve pont occur onl on ly when X les on lne sement sement RC. R C. b. all anles of of A B C are less than 1 20° let F be the pont other othe r than than R where lne RC meets the crcumcrcle crcumcrcle of of �AB �A B R. Then F les nsde A B C and FA + FB + F C = RC.
(2.45) LEMMA.
90
HAPTER 2
TRANGLES
A R
C
B
igure
2.29
We begin by constucting c onstucting equite X B T s shown in igue 2.29. To be moe pecise, pecis e, we dw ine segment B X we otte it though though n nge of 0° with point B s the cente of ottion ottion to obtin segment B T nd then we dw X T to compete the equite eq uite tinge uthe uthemoe moe we specify tht the diection of the the s me s the diection diection though though which wh ich one woud wou d hve to to 0° ottion of B X is the sme otte B A by 0° to get B R. Obseve tht tht since ech possible poss ible position posi tion fo X yieds n unmbiguous unmbiguo us detemin detemintion tion of T we cn thin of T s function of the vibe point X . In pticu if X is t A then T is t R nd if X is t B then T is so t B nd in this this cse c se X B T degenetes degenetes to singe sin ge point We cim tht AX B r RTB tht RT B . The esiest wy to see this is to obseve tht 0° ottion bout point B cies point A to point R nd the sme ottion cies point X to point T Since this ottion tnsfomtion cey cies point B to itsef the net effect of the tnsfomtion is to move AX B so s to me it coincide with with RT B nd thus the tinges e conguent moe tdition gument using us ing SS is i s so vilbe since B A = B R B X = B T nd it is esy to see tht LABX = LRBT We now hve X A = RT We so s o now tht X B = XT nd thus XA + XB + thiss pth fom R to C cnnot be ess hn X C = R T + T X + X C . The ength of thi the stightline distnce R C. In fct fct it cnnot even be equ to R C uness point X (nd so point T ) ies on segment RC This competes comp etes the poof of pt (). () . It is not hd to see in igue 2.30 tht if L A < 120° nd L B < 120° hen ine R C cosses line line A B between points A nd B . Note tht if L A = 120° hen point R woud ie on n extension of side AC nd R C woud cut A B t A simiy if L B = 120° then R C woud cut A B t B . ontinuing to ssume tht L A nd L B e ess thn 120° we dw the cicumcice of equite ARB nd dene F to be the point othe thn R
Proof.
A R
igure
2.30
2H OPTMZATON N TRANGLES TRANGLE S
91
where ine R C cuts the circe. s ssuming suming aso that C < 120° we caim that point F ies inside � A B C as iustrat iustrated ed in Figu Figure re 2.30. 2.3 0. To see this observe that that AB ° = (240) = 120 120°. Sin Since C < A B it foows foows that that point point C ies outside outs ide the circe as shown and thus F ies inside ins ide the the triange triange.. In this case where where a thr three ee anges anges of �A B C are are ess than 1 20° 20° we tae tae the point X of the rst part of the proof to be F and we argue that the corresponding correspondi ng point T ies on segment RF as shown in Figure 2 30. 30 . To To see that this this is i s true we = 60° rst observe that R F B ° 60° and hence T ies somewhere so mewhere on ine R C to the eft of point F in the diagram. To see that T ies between F and R it sufces sufces to show that T B F < R B F . ut this is cear since T B F = 60° and R B F > LRBA = 60°. We ow from the rst rst part of the proof that F A+ FB+ FC = RT +T F + FC When a thee thee anges of o f the origina triange triange are are ess than 1 20° 20° w wee have seen that that the broen broen ine" RTF C is actuay actua y straight and the atter sum is thus exacty exact y equa • to R C . This competes c ompetes the proof proof of part (b). (b) .
R
s promised we can now sove Fermats Fermats probe probem m (robem 2.44) 2. 44) in the case where none of of the anges anges of �A B C is as arge arge as 1 20° 20° . Some S ome of the amazing properties of the Femat Femat point are visibe in Figure 23 2 3 1 but Theorem 2.46 2. 46 tes us that even more more is true true.. Given a trianle trian le with all three three anles less than 120° construct outwar outwa rdfacin ac in equilateral equ ilateral trianles on each eac h of of the sides. Then he n the line sements semen ts oinin the vertices of the iven trianle with the far vertices of the equilateral trianles on the op opposite sides have equal lenths len ths and they are concu rrent rrent at some point F inside the oriinal trianle Also Also the quanti FA + FB + FC F C is equal to the common c ommon lenth of the three sements and is smaller than X A + X B + X C for any any other oth er point po int X in the plane In addition addition the point po int F lies on the circumc c ircumcircles ircles of of all three equilateral equ ilateral trianles trianles
(2.46) THEOREM.
p igure
2.31
92
HAPTER
TRANGLES
We abe points as in Figure 23 1 so that it is ine segments P A QB an R C that w e must prove are concurrent concurrent an of equa equa ength ene F as in emma emma 25(b) 25 (b) to be the point poi nt where R C meets the circumcirce of AB R in the interior of ABC . y the emma w e know know that F A + F B + FC = R C an that this quantity is the minimum poss po ssibe ibe vaue that X A + X B + X C can have as X runs over a points in the pane It foows foows simiary that Q B is aso the minimum minimum pos sibe vaue of thus since sinc e there is at at most one minimum possible pos sible vaue we X A + X B + X C an thus must have Q B = RC. Simiary A P RC and so the three ine segments segment s have equa equa engths ength s as a s require require We now know that F A + F B + F C Q B and so by emma emma 245( 2 45(a) a) appie to Q in pace of R it foows that F must ie on ine segment Q B Simiary F ies on A P an the three segments segmen ts are concurrent concu rrent at F This argument as asoo shows that if X is any point for which the quantity X A + X B + X C is minimize then X must aso ie on a three segments an so X is the point F In other wors wors F is the unique point po int where the the minimum is attaine attaine This This reasoning shows that the the point of concurrence concurrence of P A Q B an R C ies on the circumcirce of AB R It foows foows simiary that this this point ies ie s on the other two circumcirces an the proof is compete co mpete •
Proof.
=
=
In the situati situ ation on of of Theorem he orem the six anles an les fo rmed by the three concurrent conc urrent line sements se ments are all all equal to 60°
(247) COROLLARY.
LAF R ° = 60° where F is the point of Observe in Figure Figure 23 1 that that LAFR concurrence The coroary foows foows by appying simiar si miar reaso reasoning ning to the remaiing ve anges anges •
Prof.
=
=
If F is the Fermat point of A BC which has has a anges smaer than than 1 20° it folows folows from oroary 247 that LA F B L B F C L C F A = 1 20° 20 ° There is aso a physics proof" of this fact Imagine that the pane of ABC is the t he horizonta surface surface of a tabe an that tiny hoes ho es have been bore bore through the tabetop at points point s A B an C Strings are threae threae through the three three hoes an tie together together above the tabe Equa weights are attache to the ower ens of the tee tee strings an the system is i s then aowe to come into equiibrium equiib rium (The tabe is high enough so that none of the the weights touches the oor) The ot which whi ch we assume is just a singe point, c omes to rest at some point on the pane an we attempt attempt to ocate this this point poin t It is perhaps not obvious that ther theree is just ju st one such equiibrium point but we sha see that that in fact fact this is the case cas e The potentia energy of the the system s ystem is proportiona proportiona to the sum su m of the heights above the groun of the three three weights weights an the system is in stabe equiibrium precisey when this this energy is at a oca oc a minimum minimum or equivaenty when the tota ength of string beow the tabetop tabetop is at a oca maximum In other wors wors the knot is in stabe equiibrium at some point X precisey precise y when the quantity X A + X B + X C which is i s the tota ength of string above the tabe is i s at a oca oc a minimum In particu particuar ar the Fermat point F is one possibe pos sibe resting pace for the the ot When the knot comes to rest rest there is no net force force acting on it an so the three force vectors corresponing to the tension in the three strings strings must sum to zero The magnitues magnitues of these vectors are equa, however since the
H OPTMZATON OPTMZATON N TRANGLES
9
thre weights are equa and from from this it easiy easi y folo folows ws that the three anges ange s between betwee n the strings are each 120° as we wanted to show In Figure 231 we se e that the the ocus of points X inside ABC such that LAXB see 120° is exacty arc A B of the circumcirce of equiatera AB R It foows in our physics expeiment that that every every possibe pos sibe equiibrium position pos ition for the knot knot ies i es on this arc arc and simiary, simiary, a equiibrium positions must aso as o ie on the other other two two circumcirces It foows from this that the Fermat point which is where the three circes meet is the unique unique equiibrium equ iibrium point and thus the function function X A + X B + X C has no oca minimum other than than the goba gob a minimum that occurs when whe n X is at the Fermat Fermat point F . In hapter 5 w e estabish estabis h yet another another remarkab remarkabe e fact fact about the con co nguration guration of Figure 231 The centers of the three circes form an equiatera triange a theorem sometimes attribute attributedd to Napoeon onaparte We cose by mentioning a remarkabe generaization of part of Theorem 24 Suppose that ABC is competey arbitrary with no restriction on its anges and buid outwardpointing outwardpointing isoscees is oscees AB R BC P and and C A Q having bases AB BC and CA. We replace replac e the assumption ass umption of Theorem 24 that these thes e triang trianges es are equiatera with with the weaker condition that a of the the base anges are are equa equa but not necess nece ssariy ariy equa to 0° as in Theorem 24 The amazing fact is that even in this generaity it is stil true that ines A P B Q and and C R are concurrent con current proof of of this is presented in hapter h apter 4
xercse xercsess 2H.l
rectange has side engths a and b and as shown in Figure 232 points P and Q are seected on the the sides side s of ength a so that P Q is parae p arae to the sides si des of ength b Show Sh ow that that there there exist uniquey uniq uey determined points X and Y on the sides of ength b such that P X + X Y + Y Q is i s minimized and show that the the minimum possibe pos sibe vaue for this this quantity quantity is 4a + b b
a
a b igure
232
CH PER HRE E
Cir Ci rc es an Lines
3A
imson Lines
The next severa topics reay have itte in common except that as with much of geometry geo metry they invove circes circe s and ines ines Our rst resut resu t which is a theorem attributed attributed to obert obert Simson ( 1 6871 68 71 768) 76 8) estabishes the existence existence of certa certain in ines associated as sociated with with the circumcirce of a triange triange s we sha sh a see there there is aso as o a connection between these these Simson ines i nes and a nd the ninepoint circe circe of the triange triange and an d so this section s ection provides provides a continuation continu ation of some so me of the materia materia from the previou previouss chapter ch apter Our concern conc ern here is the conguration conguration iustrated iustrated in Figure Figure 3 1 z
c igure
3.1
Given �ABC choose choo se an a n arbitra arbitrary ry point P on its circumcirce and drop perpen perpen dicuars P X P Y and P Z to sides BC AC and AB as shown It is amost aways necessary to extend extend at east one one of the the sides of the the triange triange to do this In Figure Figure 3 1 for for exampe exampe we had to extend side s ide A B to meet the perpendicuar from P Sisons theorem asserts as serts the amazing fact that the feet feet X Y and Z of the three perpendicuars perpendicuars from point poin t P are aways coin co inear ear ppropriatey the ine through X Y and Z is caed the imo i of � ABC with respect to point P and P is referred referred to as the po for this Simson ine efore we prove Simsons theorem et us consider the degenerate" case where the poe P is chosen to be one of the vertices of the triange Suppose P is at A for exampe Then X is ceary the foot of the atitude from A in �ABC but where are Y 94
A SM SON LNES
95
and Z in this case ca se The ine perpendicuar perpendicuar to to AC through P which is A eets AC and thus Y i s the poin A and siiar siiary y Z i s aso at A In this situation points X at A and Y and Z are reay just ju st two points and so they are are certainy certainy coinear. co inear. The Sison S ison ine of a triange triange with w ith respect to a poe at one of its vertices theref therefore ore is i s just the atitude of the triange fro fro that vertex and in particuar particuar the Siso Si sonn ine goes go es through throu gh its own poe in that case case It is not no t hard hard to show that in fact fact the ony way that a Sison Sis on ine can go through its poe po e is if the poe is at one of the vertice vertice of o f the triang triangee This T his is aso a so the ony way that two of the points X Y and Z can coincide and as asoo i is ony in this case that that it ight not be necess nece ssar aryy to extend one of the sides of the triange to construct con struct thes thesee points The foowing foowing resut res ut is the ey to our proof proof of Sison s theore but it aso yeds soe usefu u sefu additiona inforation inforation po int P on o n the circumcircle of of � A B C and let Q be the THEOREM. Choose a point other point where wh ere the per pe rpendicular to to B C throuh throuh P meets m eets the circumcircle et X be the point where this perpendicular perpendicular meets line B C and let Z be the point wher whe re the perpendicular perpendicular to A B throuh P meets A B Q is derentfrom derentfrom A then Z lies on the line parallel to to Q A throuh throuh X X
Figure 32 shows two of the the severa possibe possib e congurations that can occur in The ore 3 . 1 In the diagra on the eft we chose point P so that the foot X of the perpendicuar fro P to side B C fas on an extension of that side On the right we started with the sae triange but here P was chosen so s o that that X actuay ies on se ment B C In each of these these diagras diagras point Z fas on side A B but it can aso happen that A B has to be extended to construct Z To see an exape of this this cons c onsider ider the right diagra diagra but exchange the abes abes B and C so s o that that A B is i s now the neary neary vertica side of the the triange triange and the new Z is not no t on this this ine in e segent Note that X and Q are not affected by the reabeing reabeing of B and CC and so the theore tes us that the new point Z aso ies on the ine through through X parae to A Q . Since the origina Z aso ies on this ine the new Z the origina Z and X are coinear coinear s we sha see this is how S isons ison s theore theore is proved efore efore we proceed with the proof of Theore Theore 3 1 we shoud mention a possibe pos sibe degeneracy degeneracy though thou gh Q was dened de ned as the other" point where the perpendicuar to B C p
igure
3.2
Q
96
APTER
RLES AND LNES
through P meets the circe circe it can c an happen happen that this this perpendicuar is tangent to the crce and in that case cas e there is no second sec ond point of intersection intersection The theorem is sti s ti true in this situation if we tae Q to be P and the the proof wi go through through with ony sight change change We eave the verication of this to the exercises If X and Z happen to be the same point point there is reay nothing to prove So we can assume that X and Z are distinct and our goa is to prove that X Z Q A If P is at the point B then X and Z are aso at B and since we are assuing that X and Z are different this does not happen Thus P and B are are dierent dierent and we can consider co nsider the unique circe having diameter P B Since L P X B = 90 = L P Z B it foows that points X and Z ie on this circe and thus L P X Z = L P B Z since these anges anges subtend subtend the same arc in that that circe ut ut assuming that P is different from Q we see that L P B Z = L P Q A because these anges subtend s ubtend the same arc arc in the origina circe circe Thus T hus LPXZ = L P Q A and it foows that t hat X Z and Q A are parae parae as desired •
Proof Proof of Thorm Thorm . 1.
In the statement of Theorem 3 1 it was necessary to assume that Q is different fro A since otherwise the line Q A woud not be dened and the theorem theorem would mae no sense sens e To see how the assuption assu ption that Q and A are diff different coud possiby pos siby fai fai suppose suppos e that that these these points actuay do coincide Then AX is the ine QX which is perpendicuar to BC and thus thus Q X is the atitde atitd e from from A in �ABC ut P lies on QX and thus P ies o n the atitude from from A In other words words if P i s not on o n the atitude fro fro A then the assuption i n Theorem Theorem 31 that Q and A are are distinct is guarante guaranteed ed to t o hold To prove Simsons theorem we need to consider the reationship between the orthocenter and the circumcirce circum circe of a triange triange If a of the anges ange s of �ABC are acute then then we now that the orthocenter ies inside the triange and hence it ies ie s inside i nside the circumcirce ut if one of the anges of the triange is obtuse then it is easy to see fro a diagra that the orthocenter orthocenter ies outside of the circumcirce circumcirce Finay for a right triange triange the orthocenter orthocenter coincides with a verte vertex x and so it ies on the circumcirce et P be any point po int on the t he circumci c ircumcirrcle of of � A B C and let X Y and Z be the feet ee t of of the perpen perpendiculars diculars drop dropped ped from P to lines B C A C and an d A B res respectivel pective ly Then points X Y and Z are are collinea collinea
(.2) THEOREM (imo' Thorm).
Suppose rst that that P does not ie on the atitude from from A in �ABC and et Q be as in Theorem 3 1 Since P is not on the atitude from from A we have seen s een tha the hypothesis hypothesi s in Theorem 31 that Q and A are are distinct points is guaranteed guaranteed to hod y Theore 3 1 therefore therefore we now that Z ies on the ine through through X parae to s imiar reasoning shows s hows that Y aso ies on the ine through through X parale A Q Exacty simiar to A Q and thus thus X Y and Z ie on a comon ine as required We have now shown that X Y and Z are coinear if P does not ie on the atitude from A y siiar s iiar reasoning we get the sae concusion concusi on if P fais to ie on on the atitude from B or if it fais fais to ie i e on the atitude from C The ony case in which we have not yet proved the theorem there therefor foree is when P ies on al three atitudes atitudes in which case P s the orthocenter of of �ABC ut P ies i es on the circumcirce of o f this
Proof
A SMSO N LNES
97
triang triange,e, and we have seen s een that it is ony for a right right triange triange that the orthocenter can ie on the circumcirce We can now assume as sume that �A B C is a right right triange triange and that P is its orthocenter orthocenter We can suppo s uppose se that L B is the right ange, ange , and it foo foows ws that that P is at B and thus X and Z are as asoo at B . Since X and Z are are the same same point in this case, ca se, the points X Y and Z are are certainy coinear, co inear, and the proof is compete co mpete Where on the circumcirce of �ABC shoud we tae the the poe P so that the corresponding Simson ine wi be parae to one of the sides of the triange
() PROBLEM
It may not be obvious that it is possibe to nd such a poe, but et us anayze anayze the situation For For deniteness deniteness,, suppose that that we want a Simson ine parae parae to side BC. y denition, every Simson Sim son ine contains at east east one point on ine BC and and so s o if a Simson ine is parae parae to BC this Simson Sims on ine must actuay be BC. We as, therefore, therefore, if it is possibe pos sibe to nd a poe P for which the the corres correspondi ponding ng Simso Si msonn ine is the ine B C. If such a point exists and we draw the perpendicuar from P to side A C then the foot Y of this perpendicuar perpendicuar must ie on the Sims S imson on ine B C Since vertex vertex C is the ony point common to A C and BC we see that Y must be at C and hence P C is perpendicuar to A C If it exists , therefore, therefore, the poe P must ie somewhere on the ine perpendicuar to to A C at C Simiary, P must ie on the perpendicuar to A B at B and thus P can ony be the point where thes thesee two ines intersect
ouio
igure
3.3
ut there is another requirement requirement The poe must aso ie on the given circe, circe , and so we need to as whether or not the intersection point P of the two perpendicu ars P B and P C actuay ies on the circe, as a s it appears to to do in Figure 33. uting the question another way, way, we as if we can nd a point P on the circe such that obvious Just choose choo se P so that L P B A 90° LPCA. The answer shoud now be obvious A P is a diameter. We We have shown, therefore, therefore, that if we tae as a poe a point on the circumcirce of a triange diametricay diametricay opposite a vertex, then the corresponding Simson Si mson ine is the side opposite oppos ite that that verte vertexx In robem 3.3 we were given a xed triange and we were ased to ocate o cate a poe for for which the corresponding corresponding Simson S imson ine was parae to a specic given ine: ine : a side of te origina triange We coud as the same question more generay, as foows Suppose Suppo se that we are given some arbitra arbitrary ry ine s the poe P moves around the circe circe,,
98
CHAER
CRCLES AND LNES
does doe s the Simson Sim son ine ine move in such a way way that it can be made parae to the given ine ine The answer is yes In fact, the direction of the Simson Sims on ine varies in a uniform uniform way way as its poe moves : The Sims S imson on ine tus at exacty haf haf the anguar rate at which the poe P moves around the circe In particuar 80 as P particuar the Simson ine tus though though a fu fu 1 80 traves 30 around around the circe, and hence it sopes in every every possibe pos sibe direction direction We can state this this a itte more precisey as foows foows et U and V b e points o n the cir cumcircle of �AB Then the anle between the Simson lines havin these points poin ts as poles is equal eq ual in derees de rees to ha of of
(4) THEOREM.
There is some ambiguity here since when two ines cross, cross , they determine determine two anges, and this this corresponds to the fact fact that the points U and V deteine two two arcs arcs What is meant by Theorem 3.4 is that the smaer of the two anges ange s is equa in degrees to haf of the smaer of the two arcs We wi not give an absoutey rigorous proof that covers a cases cas es of Theorem 34 but we hope the argument that foows oow s is convincing co nvincing
A
U
R gure
3.4
In Figure 34 we have extended the perpendicuars from U and V to B to meet meet the circe at at R and y Theorem Theorem 3 1 we now that the Simson Sim son ine having poe U is parae to AR and the Simson ine with poe V is parae to A The ange ange betwee betweenn these these two Simson ines is thus thus equa to RA U R and V S are parae chords however we see that Snce UV R, and the resut foows foows
Proof of Thorm .4
s a specia speci a case, case , we have the the foowin foowingg Two Simson line l iness for a iven iven trianle are per pe rpendicular pen dicular and only their the ir poles po les are at opposi opposite te ends of of a diamete diame te
(5) COROLLARY.
s we move the poe P around the circumcirce of �AB, we now that the moving Simson Si mson ine in e tus tus so as to sope so pe in ever everyy possibe pos sibe direction We stres stress,s, however however that the the Simson Sims on ine does doe s not simpy rotate about about a singe point its motion is much more more compicated There can be no one point common to to a of the Simson Sims on ines of �AB �A B because we have seen s een that each each of o f the sides of o f the triang triange e is i s a Simson ine Simson Sims on's's theorem tes us that ifif we choose any any point on the circumcirce of a ttriange riange and drop perpendicu perpendi cuars ars from from it to the sides of o f the triang triange e extended if necess neces s ary, ary, hen the feet feet of these these perpendicuars are coinear coinear This suggests sugges ts the question q uestion of whether whether or or not the same thing thin g can happen if we start with a point not on the circumcirce circumci rce
A SMSON SM SON LNES
Given �ABC suppose suppose that thefeet thefeet of the per pe rpendi pe ndicul culars arsfrom som s omee poin po intt Q to the three sides of of the trianle tri anle ar a re colline colli nea a Then Q must lie on the circu c ircumc mcircle ircle of of �ABC
(6) THEOREM
In other other words, the converse of Sison Si son's's theore theore is true true To To prove it, we need a fairy fairy easy ea, whose state ent, unfortunate unfortunatey, y, is soewhat copicated Figure 35 shoud hep untange the hypotheses hypotheses
99
m
n igure
3.5
Sup Suppose that tha t lines m and n are parallel parallel lines b and c meet at a point A line line m meets b and c at points V and respectively and line n meets b Perpendiculars ulars to to b and c are erected at V and and c at Y and Z respectively respectively Perpendic and these these meet me et at a point poin t Q Similarl Simi larlyy the per pe rpendiculars pen diculars to b and c at Y and Z meet mee t at P P Then points poin ts P A and Q are collinea collinea
(.7) LEMMA.
First, First, we dispose of soe unintere uninterestin stingg degene degenerat rate" e" cases case s If A happens to be one of the points P or Q there is reay reay nothing to to prove So we can assue as sue that A is neither P nor Q , and it foows foows that that neither neither of the the ines m or n passes passe s thro throug ughh A A Next, we consider cons ider what what happens happens if i f point Q ies on ine b It is not n ot very difcut to see se e that that in this this case, ca se, ine m ust be perpendicuar to c and it fo foows ows that n is aso perpendicuar to c and therefore therefore P ust aso ie on b In this case, P A, and Q are coinear, as desired We can suppose therefore that Q does not ie on b and siiar sii ary, y, we can assue that Q is not on c and that P is not on either b or c In particuar Q is dierent fro V and , and P is different fro Y and Z Next we observe ob serve that V is the foot foot of the perpendicuar fro Q to b Since A ies on b fro V it foo foows ws that A Q is not perpendicuar to b Thus A Q is and A is dierent fro sii ary A Q is i s not parae parae to Z P not parae to Y P and siiary Figure 35 shows one possibe conguration: where A ies between ines m and n It is aso poss po ssibe, ibe, of course course that A is above both ines or beow both ines The proof in a cases cas es is i s identica, however however Our goa is to show that P ies o A Q, as is i s indicated by the dashed ine in the gure Since Y P is not parae to to A Q we consider the point R where Y P eets eets A Q, and siiary, we et S be the point where Z P eets eets A Q Of course, course, we expect expect that that R and S are actuay the point P but we do not yet now that P ies on A Q We propos proposee to to show that in fact R and S are are the sae point Since P is the ony point comon to ines Y P and Z P , it wi wi foow that P R, and S are a the sae point, and in particu particuar ar,, this wi show that P ies on A Q, as desired Now Y R V Q since sinc e both bo th of these ines are are perpendicuar to b It foows A R/ A Q = A Y / A V and siiary si iary,, we get A / A Q = using us ing siiar trianges that AR/ AZ/ AZ / A ut since Y Z V , we see see that that AY/ AY / A V = A Z/ A and hence we have foows ows that that AR A R = AS and hus hus A R / AQ = A Y / AV = AZ/ A = A / A Q It fo R and S are are the sae point, poi nt, as desired This copetes the proof
Proof.
100
CAER
CRCLES AND LNES
et et U, V, and and W be the feet of the perpendicuars perpendic uars fro Q to B A C and A B respectivey, respectivey, and suppose that these three points a ie on soe ine m We have seen that a Sison ine can be found parae to any given ine, and so we can choose a poe P (on the circucirce, of course) for which the corresponding corresponding Sison S ison ine n is parae to m y denition, n runs through the points X Y and Z which are the feet of the perpendicuars fro P to ines B We see s ee now that if w e dene b to be the ine A C and c A C and A B respectvey We to be the ine A B we are preci precise seyy in the situation of o f ea ea 37. We concude fro ea 3.7 that points P A and Q are coinear Exacty simiar reasoning shows show s that points P B and Q are aso coinear and that P C and Q are coinear too too If P and Q are are not the the sae point, i t fo foows ows that ine P Q runs through a three vertice verticess A B and C of �ABC ut since s ince these points are are the vertices vertices of a triange, triange, they cannot be coinear, and hence we have have a contradiction Our assuption that P and Q are dierent ust be wrong, therefore, and so we concude that P and Q are the sae point Since P ies on the circucirce of �A B it foows that Q ies on the circucirce, as required
Proof of Thorm 6
There There is a reay pretty pretty consequence of Sison' Siso n' s theore (oroary 3.2) and its converse (Theore 3). This resut conces four ines in gra poiio, which eans that no two of the ines are parae and no three of the are concurrent ote that four ines in i n genera posi po sition tion deterine four four triang trianges es by tang the th e ines three thre e at a tie, and there are are six points of o f intersection intersection of the ines ines As illustated illustated in Fiure the circu c ircumc mcircles ircles of of the the our ou r trian les detemined by by any four throuh a common ou r lines line s in eneal e neal position always always o throuh point
(8) COROLLARY
raw two of the circucirces circuci rces and observe that one of their points poi nts of intersec tion, which we ca P ies on o n none of the given ines rop perpendicuars perpendicuars fro fro P to each of the the four ines , thereby deterini deterining ng four four feet, feet, one on each ine ine y Siso S ison' n'ss theore appied in one of the circes, three three of the four four feet feet are coinear coine ar,, and by a second appication of Sison Si son'' s theore, in the other other circe, another thre threee of the feet feet are coinear coi near It foows foow s that a four of the feet feet of the perpendicuars perpendicu ars are coinear, and thus thus by Theore 3 the point P ust ie on a four four circuci circucirces rces
Proof
igure
3.6
A SMS ON LNES LNES
101
There There are some intimate connections between the the Simson Si mson ines ine s assoc as sociated iated with with a triange and and the ninepoint ninepo int circe of that that triange triange erhaps the the most mos t striing of these is the foowing fact, reated to oroary 35 y that resu resut,t, we now that two Simson Sims on ines with poes at the ends of a diameter diameter are perpendicuar perpendicuar,, and we as now where tese tes e two perpendicuar perpendicuar Simson ines meet Fix �A � A B C and let let U V be an diameter diamete r of of its circumc c ircumcircle ircle Then the Simson lines of of � A B C havin havin poles U and V meet on the t he nine n inepoin pointt circle of of the tianle
(.) THEOREM.
In fact fact if we aow ao w diameter U V to rotate through 180 the intersection point of the two two Simson S imson ines traces out the the whoe ninepoint circe In preparation for proving Theorem 39 we reca a usefu fact Given �ABC having orthocenter H consider cons ider the ocus of a midpoints of ine segments s egments H P joining Exercise 110 this ocus is a circe, circe, H to points P on the circumcirce of �ABC y Exercise and since the ocus ceary contains the three Euer points of the triange, it must be the unique circe though though these points po ints In other words, words, the ocus of the midpoints of o f the segments H P is the ninepoint circe of �ABC The foowing foowing resut, resu t, which wh ich we sha sh a use to prove Theorem 39 tes u s that every every point of the ninepoint circe of �ABC ies on an appropriate appropriate Simson Simso n ine et H be the orthocenter of �A B C and let P be any any point on the cicum ci cumcircle circle of of this trian tr ianle le Then the mid m idpo poin intt of of H P lies on the Simson Sim son line lin e of � A B C with wi th pole P
(.10) THEOREM
Figure 37 is somewhat compicated, and so s o we begin with a carefu carefu description of how it was drawn We started with the given �ABC shown in heavy in and a point P on its circumcirce we drew atitude A and mared the orthocenter H on this ine, and we drew ine segment PH Next, we drew the ine throug P perpendicuar to side B and meeting B at X ; we extended P X to meet the circe circ e again at Q and then we w e drew A Q
Proof
R
B igure
37
10
CAER
CRCLES AND LNES
The Simson ine with poe P certainy goes through X, and we now rom Theorem 31 that it is parael to A Q, and so we have drawn (with dashes) dash es) the ine throug throughh X parae to to A Q Q This is i s the Simson ine with poe P , and it meets meets P H at point M Our tas is to show that M is actua actuay y the midpoint midpoint o P H The diagram o Figure Figure 3.7 is competed as oows oows : Extend atitude A to meet the circe circe at ; ; draw P meeting side B at D and nay, draw draw H D and extend it to meet the the extension extensi on o P Q at R We We propo propose se to show that M X is parae para e to the base H R o � P H R and that X is the midpoint o side P R o this triange It wi oow that M s the midpoint o o side P H , as required required Since is i s the oot oot o an atit atitude ude o �A B , it ies on the niepoint circe, circe, and hence it is the midpoint o H since s ince the ninepoint circe i s the ocus ocus o a midpoints o ne segments joining jo ining H to points on the circumcirce circumcirce Thus D is the perpendicuar bisector o H H , and we concude that D H D We now have have PQA P A RH P RH where where the rst equaity hods since si nce both anges anges subtend P A , the second sec ond hods by the pons asinorum in iso i sosce scees es � D H , and the the third third hods because becau se these thes e are ateat ateatee interior anges ang es or parae ines i nes P R and A (Note that P R and A are parae because each is perpendicuar to to B ) Since P Q A P R H , it oows that H R is parae to A Q , and thus thus H R is parae to the the Sims S imson on ine M X This is one o the two acts acts we need What remains is to prove that X is the midpoint o P R, which we wil es tabish by sho\ing that tha t � P D r �X R D We certainy certainy have h ave X D X D and PXD 90° RX D, and so by S, S, it suces suces to to show that that X X P D XRD ut P R A , and so these these ange angess are are equa equa to to D H and and D H , respect respectiv ive ey, y, and these are equa to each other because �DH �DH is isoscees The proo is thus compete The proo o Theorem 3.9 is now airy easy et X denote the intersection point o the the Simson Si mson ines having poes U and V and et H be the orthocenter o the given triang triangee y Theorem 3.10 the Simson Sims on ines go throug throughh the midpoints M and P o H U and H V , respectivey, and so the two two Simson Si mson ines are M X and P X, drawn with heavy in in in Figure 38 We have drawn the ninepoint circe o �A � A B , athough the origina triange and its circumcirce are not shown Midpoints M and P ie on the inepoint inepoi nt circe,
Proof of Thorm
x
igure
3.8
A SMSON LNES
103
and since UV is a diameter of the circumcirce, we now b oroary 3.5 that L M X P = 90°. To prove that X ies on the ninepoint circe heref herefore, ore, it sufce sufcess to show that chord M P is actuay a diameter diameter of this circe circe Since M and P are are the midpoints of two two sides s ides of o f � U H V t foows that M P U We reca reca,, however, that for for any any triang triange, e, the diameter of he ninepo nin epoint int circe is exacty haf the diameter of of the circumcirce circumcirce It foows foows h cord M P of the ninepoint ninepoi nt circe has ength equa to that of a dimeter dimeter o thi rce, rce , and thus M P must be a diameter, as required This competes the proof
There is a remarabe rest that we woud ie to mention efore cosing this secton secton Given Given fou points on a circe, we can get four four Simson Sim son ine by considering cons idering each point in turn turn as a poe and constructing the corresponding corresponding Simso Sims o ine with respect to the riange formed formed by the remaining remaining three points erhaps erhaps the rader can gues gu esss what happens The four ou r Simson Sims on lines lin es deter d etermine minedd by an f r disti d istinct nct points poin ts of of a cicle ae concur concu rent ent The point poi nt of of concurrence conc urrence moeove mo eove les on the ninepo nin epoint int circle ci rcle of each ea ch of the fo ur tianles tian les fo rmed by taki taki thee of t iven ive n points po ints
(11) THEOREM
We aready now that the four four ninepoint circes go through a common c ommon point; poi nt; that fact was part of Exerci Exercise se 2 2 66 In fact, fact, that exercise exercise actuay actuay proves Theorem Theorem 3.11. It asserts that the four ine segments obtained by joining each of th given points to the orthoceter orthoceter of o f the trian triange ge formed by the other three a share sh are a OrO midpoint midpo int We We now that that midpoint midpoi nt ies on the four ninepoi nine point nt circes, circes , and by by Theorem 3.9 it aso ies on the fou fourr Simson Sims on ines We now present a proof of the the ey step in the soution o f Execise 2 6 et P P Q B and C be four distinct points on a circle and let let H and K be the orthocenter orthocen terss of of � P B C and � Q B C res respectivel pective ly Then sements P K and Q H have a common commo n midpoi midpoint nt
(12) LEMMA
efer rst to the eft diagram of Figure 3.9 and observe tht P K and Q are diagonas diagona s of quadriatera quadriatera P Q K H We need need to show that that these dagonas d agonas bisect bis ect each
Proof
c
igure
c
3.9
104
CAER
CRCLES AND LNES
other, and so it sufces to show sh ow that the quadriatera is a paraeogra parae ogra ertainy, P H and Q K are parae since these ines are atitudes of � P B C and �QB and hence they are each perpendicuar to B C. It sufces, therefore, to show that PH = QK et be the center of the given circe and drop a perpendicuar perpendic uar M fro to BC. (efer to the right diagra in Figure 39 for this) this ) We wi wi prove prove that P H = 2M. It foows siiary that Q K = 2M and this shows that P H = QK as required requir ed To prove that P H = 2M et X be the idpoint of P H and consider quadriatera X M H . We wi show that this this is a paraeogra, paraeogra, and it wi foow foow that M = X H = PH as required It sufce sufcess to show s how that diagonas H and X M bisect bis ect each other The idpoint of H we now, is the center of the ninepoint circe of � P B C. To ocate the midpoint of X M we observe obs erve that that since is the center center of the given circe, M is the idpoint of chord B C so, X is the Euer point opposite M in � P B C and so by Theore 212 segent X M is a diameter of the ninepoint circe Its idpont is thus aso the ninepoint center, center, and so X M and H have the the same s ame idpoint The proof is now compete
We ention that there there is an easy atern aternati ative ve arguent arguent based on Exercise 2 2 y y that exercise, PH = QK. QK . so, PH and QK are both perpendicuar to BC and so they are parae parae Therefore, PHQK is a paraeogra and and the resut foows foows
xercses 3 Al A2 A A4
A5
A6
Suppose Suppo se that a Sison Siso n ine of a triang triangee goes through its own poe po e Show that the poe ust ust be one of the vertices vertices of the triange triange Suppose that a Sison Sis on ine of a triange triange is perpendicuar perpendicuar to one o ne of the sides of the triange triange Show Sh ow that the the poe ust be one of o f the vertices vertices of the triange triange Suppose that a Sison ine of a triange goes through the orthocenter of the triange triange Show Sh ow that the the poe must be one of the the vertices vertices of the the triange Show that for an acute anged triange, every Simson ine eets the interior of the triang triange e onversey, o nversey, show that if every Siso Si sonn ine of a triang triange e contans a point of the interior, interior, then the anges ust a be acute rove Theore 3 . 1 in the case ca se where where the ine though though P perpendicuar to B C is tangent to the circucirce of �ABC In this this case P and Q coincide, and a nd you ust show s how that P A is parae to X Z Given a point P on the circucirce circuc irce of �A B C reect i n each of the sides of the triange Show that the thee points thus obtained a ie on the ine through the orthocenter of �ABC parae to the Simson Simso n ine with poe P .
B TE BUTTERF BUTTERFLY LY TEOREM
3
105
The uttery Theorem
The socaed uttery theore is the foowing pretty resut, which is notoriousy difcut difcut to prove if you don't don 't now how Supposee that t hat chor cho rds P Q and R S of of a iven iven ir i rcle meet me et at the THEOREM. Suppos midpoint midpoint M of of chord A B. X and Y are are the points poin ts where where P S and Q R meet AB respectively respectively then X M Y M.
We have have drawn two possibe pos sibe congurations c ongurations for this theore in Figure 3.10 using the sae" chords PQ R, and AB in the two diagras The ony diff difference is that on the eft, points point s P and R ie on the sae side s ide of ine ine A B whie whie on the right, we have intercha interchanged nged the the abes R and S so that P and R ie on opposite opposite sides si des of A B . Note that in the atter atter situation, situation , chord A B had to be extended to eet ines P and Q R The nae uttery" u ttery" refers to the sefintersecting sefinters ecting quadriatera P S R Q which, which , with soe so e eort, eort, can be iagined to res resebe ebe the eponyous insect ins ect p
p
Q
Q igure
310
We present rst an eementary, but soewhat copicated, proof using siiar trianges trianges of the case of Theore Theore 3.13 where the points X and Y ie inside the circe, as o f Figure 310. though it is possibe pos sibe to construct c onstruct a siiar s iiar proof for for in the eft diagra of other case, we sha s ha not do so so instead, ins tead, we present in in the next section a uch ore ore the other powerfu technique tech nique that wi aow us to prove both cases cas es of Theorem 3 . 1 3 easiy We begin by exracing he h e hardes par of the arguent as a separate ea, which is a u nice appication of siiar trianges trianges LEMMA In Fiure Fiu re sem se men entsts of of lenth lenth x y u u v s and t are marked. the anles anle s marked with with dots are equal equ al then x u st
igure
311
106
CHAER
CRCLES AND LNES
A
u X
B
D igure
3.12
abe the points poi nts in the origina diagram as shown sh own in Figure 312 and draw X R and Y U parae parae to B C and X and Y T parae to AD �YU M and X SM r Y T M by , we concude Since �X RM r �YUM co ncude that that RM SM UM T M and this this yieds yi eds RMSM TMUM LAM B = L C M D we must aso Now L A = L C b y hypothesis, hypothesis, and since LAMB have L B = LD so, LAX R = L B L D = LCYT and it foows by that �AX R r �CYT We concude that AX XR SM CY YT UM where where the ast equaity foows because becaus e X RM and Y T M U are paraeograms Simiary, Si miary, = R M T M and we concude from the equaity of the previous paragrap paragraphh that that = ( ) (), ( ), as req requi uire red d
Proof
=
We prove the case of the theorem that appears in the eft diagram diagram of Figure 310 Since L P = LR we can appy emma 314 to concude conc ude that
Proof of Thorm 1
PX·XS RYYQ Now PXXS = AXXB by Theorem 135 and simiary R Y Y Q BYYA Writing = XM = Y M and A M = = BM we now see se e that AXXB (  )() ( + ) BY·YA (  ) ( ( + ) Since 0 eementary agebra now yieds yie ds that , and we w e concude that = , as caimed caimed
=
C CROSS RATOS RATOS
17
xercses 3B Bl
In Figure 3.13 line segments of length length and y have been drawn from vertex A of A B C to base B C and these these mae equal equal angles angl es with the sides, side s, as indicated by the dots These Thes e lines divide B C into segments segmen ts of length U v and w as shown Find a formula formu la for the ratio / y in terms of the the quantities quantiti es u , v and w A
B
v igure
3C
w
C
3.13
Cross Ratios Ratios
We now begin beg in a discu di scussi ssion on of the theory theory of cross ratios ratios,, which (after (after all of the preliminary preliminary wor has has been done) can c an be used use d to give give an easy proof of the uttery theorem th eorem s we shall explain presently, the cross ratio ratio of four four distinct collinear colli near points is a certain number uniquely uniqu ely determined by these points To To motivate the idea, we consid con sider er rst the case cas e of two distinct points A and B . These Thes e determine the number A B which is the length of the line segment seg ment they determine or the distance between them them seful s eful and important though it is, the dis distance tance function function has certain certain deciencies deciencie s Most obvious obvious among these is the fact fact that the number A B is not unambiguously unambiguou sly determined determined by the the two points; it depends also on the unit of measurement Three distinct collinear points A B and C determine a number that is independent indepe ndent of the unit of measurement We We can dene de ne the associated ass ociated quantity r (A B C) to be the ratio (unitless ) number that that conveys conveys some interesting informaion informaion A B / BC This is a pure (unitless) For example, rA B C) 1 if and only if B is the midpoint of AC. lthough lthou gh the ratio r (A B C) is independent of the unit of measurement, it suers from a more subtle deciency that is also shared by the distance dis tance function function of two points points To explain this, are three point sources of light attached attached to a rigid straight rod imagine that A B and C are If we observe from from afar, afar, we see three apparently apparently collinear lights , but bu t we clearly cannot determine the distances A B and B C without nowing how far away away the rod is and how it is oriented ith respect to our line of sight It is also al so imposs impo ssible ible to determine the ati atioo r(A B C) from afar, afar, although althoug h in this case, case , the orientation of the rod is more important than the the distance dist ance X Y Z , and P Q R Figure 3.14 shows three three sets of collinear points A B C , X that would loo identical to an eye located at point E . In this situation, w e say that these these thee sets are in prpciv from E. Since we have drawn A C P R it is an easy exercise exercise using u sing similar triangles triangles to show that rA B C) rP Q R) but r X Y Z) is denitely de nitely unequal unequ al to the other two ratios ratios In the diagram, we have taen taen Y to be the midpoint of segment X Z , and hence hence rX Y Z) 1 but it should be clear that B is not no t the the midpoint of AC and so rA B C) 1 . iewing iewin g from E in other words, we
18
CHAER
CRCLES AND LNES
c igure
3.14
cannot deteine the ratio r because we cannot distinguish { } rom {X Y Z } and yet ye t the associated ass ociated ratios ratios or or these two sets are dierent dierent emarably emarably,, i we start with our collinear points and it is possible pos sible to dene a unitless quantity, denoted cr that is invarian invariantt under perspective In other words, i { } and {W X, Y Z } are sets o distinct collinear colli near points in cr cr W X, Y Z). This means that, in perspective perspecti ve rom rom a point E then cr theory, it is possi po ssible ble to see the number cr even i we now neither the distances dist ances to the our points poi nts nor the orientation in space o the the line containi co ntaining ng them We are assuming that the the eye E is not collinear with and otherwise, it would be impossible impos sible even to see that there there are are our our points s we shall s hall see, it is also al so necessary neces sary or or the the eye to be able to deteine which point poi nt is which We can can imagine imagi ne that the the points have our our dierent dierent colors , or instance The quantity cr is called c alled the cro raio o the our our distinct di stinct collinear points and and it is dened by the oula cr
·
For example, example, suppose suppos e that the the points and are equally spa spaced ced and that they are are arrayed arrayed in that order along a line line Then cr 4/ 3 and cr / 3 /4 The reader reader should chec these calculations calculation s We were somewhat sloppy when we discussed two sets o points as being in perspective perspect ive rom rom some so me given point poi nt we should sho uld have ree reerred rred to two o rdered lists instead The reason is that the cross ratio o o our our points depends on the order in which the points occur For example, we have seen that it is not generally true true that cr cr I we we expect that, as advertised, advertis ed, the cross cros s rato will be invariant invariant under under perspective perspectiv e rom rom a point, poi nt, then clearly, the notion noti on o perspective must mus t eep trac o the order order in which the points occur occ ur precise precise denition is as ollows ollow s Suppose Suppo se and are lines and is a point not on either o them Then points point s and o line are in prpciv from with points W X , Y and Z o line respectively, i W X , Y and Z lie on lines A and respectively Figure 3 1 5 shows show s two examples o a perspective rom rom In both cases cas es,, we started started with the same our collinear points and and the same point but we varied the second line containing W X , Y and Z Observe that may lie between two corresponding points, as it lies between and W in the right diagram, or it may not, as with points Z and in both diagrams diagrams It is also poss po ssible ible or or corresponding points actually to be identical, identic al, although we have not drawn drawn a diagram where that occurs
C CROSS RATOS RATOS
p x
n igure
19
y
3.15
Suppose that A B C and D are distinct collinear collinea r points in perspective from some point P with distinct distinct collinear co llinear points X Y and Z ( , X, Y, Y, Z) respectively respectively Then (A, B , C D) = (,
(15) THEOREM
315 Observe n examination examinat ion of angles angle s is the key to the proof of Theorem 315 Ob serve that in the left diagram of Figure 315 we have LAPC = L P Y LBPC = LXPY LAPD LPZ LBPD = LXPZ, but that not all four of these thes e relations hold ho ld in the right right diagram In fact, fact, the two equations equatio ns on the rst line are ar e valid in the right diagram diagram because of the equality of vertical vertical angles , but the two equations equati ons on the second seco nd line do not hold in the right right diagram they must be replaced by L A P D + L P Z = 1 8 0 and L B P D + LXPZ = 1 8 0 experimentation will show that, that, in general, no matter matter how the points and lines litle experimentation are are ar arranged, nged, each each of the the angle angless LA PC , L B P D, L A P D, and L B P C is eithe eitherr equal equal to or supplementary to the corresponding angle among L P Y , L X P Z, L P Z, and L X P Y Since S ince supplementar suppl ementaryy angles have equal equal sines, however, however, we always have sin(LAPC) sin(L sin(L PY) PY ) sin( L B P D) D) = sin( sin(LX LX P Z) sin(LA sin(LA P D) = sin(L P Z) sin(LBPC) = sin(LXPY) whenever collinear points A, B, C, and D are in perspective from P with collinear points , X, Y, and Z, respectively To To prove Theorem 315 therefore, therefore, it suf su fces c es to show that for any any point P not on the line of the collinear points A, B , C, and D, the cros crosss ratio ratio expressed in in term termss of the four quantities quantities sin( L A P C) , sin( sin ( L B D) , (A (A B C, D) can be expressed sin(L sin( L A P D) , and sin(L sin( L B P C) The theo theore rem m is thus thus an immed immediat iatee onsequence onsequence of of the the following lemma
110
CHAPTER
RCLES AND LNES
et A B C and D be distict distict collinear collinea r points poin ts and suppose suppose P is any point poin t not o the line throuh them. Then sin( sin(LA LA PC) sin(L B P D) (A (A B , C D ) = sin(LAPD) sin(LB PC)
(16) LEMMA
et h be the perpendicular distance from P to the line containing A, B, C, and D and noe that we can compute the area area KA P c of A P C in two ways We have A C KA P c ( P A . P C) sin(L sin( L A P C) , and there there are are similar foula foulass for KB P D K KAA P D , and KB P c . Thus sin(LA PC) h · A C (P A· PC) sin(LA · B D (P B· P D) D) sin(L sin(L B P D) h A D (P A·P D) sin(LA sin(LA P D) ·BC (P B· PC) sin(L sin(L B PC) and we see ta t a if we divide the product of the rst two of these thes e eqations eq ations by the product product of the the second two, two , the lengths lengths P A, P B , P C , P D, and h all cancel and we get AC ·B D sin(L sin(L AP C) sin(L sin(L B P D) , B C D ) AD ·B C sin(L sin(L AP D) sin(LB PC) as required required • The foll followin owingg shows show s how Theorem Theorem 3 5 can be used in i n certain certain types of problems. (17) PROBLEM
Given that th at A B 3 and B C in Figure 3 nd nd CD
There hardly hardly seems seem s to be b e enough informati information on here I n fact, fact, the location of point P is completely arbitr arbitrary ary and cannot be determined determined from from the given data but nevertheless nevertheless,, we shall see that it is possible poss ible to deteine deteine CD C D , as required. ef efore ore we explain ho to do this, however, we would lie to describe the procedure that was use u sedd to draw the diagram of Figure Figure 3 We started started with with collinear points A, B , and C with diances dian ces as given, and we chose point P not o n line A C but otherwise otherwise completely arbrrily. arbrrily. ines P A, P B , and and PC were drawn drawn and point Q different
ouio
A
B
C
igure
316
C CRSS RATOS RATOS
111
from A and P was chosen cho sen arbitrar arbitrariy iy on A P Next, C Q was drawn, meeting P B at a point abeed X and then A X was drawn, meeting P C at R . Finay, point po int D was determined as the point where Q R meets the origina ine AC earaby, the information contained in this paragraph is suf su fcient ci ent to compute the cross cros s ratio cr B C D) even without specifying the distances A B and BC We state thi • resut as a separate emma (18) LEMMA cr A B have cr
In the conuration of points and lines shown in Fiure we C D) D) 2.
Since coinear points points Q Y R and D are in perspective from point P with coinear points A B C and D we now from from Theorem Theore m 315 that c r Q Y R D = cons ider the perspecive from from X instead (A B C C D) On the other hand, if we consider of from P we see that Q Y R and D ae in perspective from C B A and D respectivey, and thus c r Q Y R D) = (C B A D) I t fow fows,s, therefor therefore,e, that ACBD CABD = cr ( A , B C, D = cr ( C , B A D ) = CD·BA ADBC Since the numeratrs numeratrs of thes thesee fr fractions are are equa, the denonators must mu st be equa s o if we write = AB y = BC and = D we have ( + y + y = too, too, and so AD·BC = CDBA = Thus y (y + ) =  , and we can recomute the numerator AC·BD = ( + y) (y + ) = y + ) + y(y y( y + ) = y + ) + (  y) = 2 C D = 2, as caimed • Since the denominator is equa to , we have A B C
Proof
We have A B = 3 and B C = 1 and we write C D = , an unnown y emma 318 we have 2 A B C D) • 4 1 + )/(4 + ), and soving this, this, we get get = 2
ouio o Probm 17 coiud
More generay, if we foow the notation of the proof of emma 1 8 and write A B = , B C = y and C D = in the situation of Figure 31 we can sove so ve fr in terms of and a nd y to get = y ( + y) / ( y) (hec that the substitution sub stitution of 3 and saw before before)) y = 1 yieds = 2, as we saw Now imagine distorting Figure 31 eeping a ines straight, so that B moves toward A decreasing and increasing y whie hoding + y constant con stant It is cear c ear from from both the diagram and the formua formua that increases increase s as B moves It bows up" to innity when = y which occurs when the moving point reaches the midpoint of A C What does it mean for point poi nt D to move innitey in nitey fa" to to the right What is this thi s trying to te us us eary, the concusion concusi on w shoud s houd draw draw here is that Q R is para to A C when A B is i s indeed inde ed a theorem, athough the argument of this this paragraph a median of A PC This is qe stionabe abe as a proof In fact, fact, there is a way way to interpret parae ines in es as intersecti i ntersecting ng is qestion ratios The branch of at innity" and thereby to dea with innite quantities in cross ratios geometry geometry where this is i s done is i s caed ca ed projective projective geometry, geometry, but we wi wi not pursue that any further further here here
112
CHAPTER
CRCLES AND LNES
We mention one oter case were te formula z y(x + y)/ (x y) gives a nonsensical result If y > , ten according to te formula, z D is negative. Obviously, Obvious ly, te lengt of a line segment cannot be negative, so wat is going go ing on ere? Te answer is tat if y > A ten te diagram of Figure 3 1 is not correct In tat tat case, te point D , were Q R meets lies to te left of and not to te rigt of as sown sown Interestingly, te distance D as given by our formula formula isi s as correct co rrect as it can be Te actual distance di stance is te absolute absolut e value of te computed negative quantity, and we can interpret te negative sign si gn as meaning tat D lies tat many units unit s on te wrong side of te side sid e opposite oppo site from from tat indicated in te picture. picture . To apply te teory of cross cros s ratios to prove te u uttery ttery teorem, we need to dene te cross ratio of four distinct points on a circle. Ju st as points on a line are said to be collinear, we sall s all say s ay tat points on a circle are cocircuar Note tat if i f tree or more distnct points are cocircular, cocircular, ten tere is a unique circle circle containing all of tem. Suppose and D are are any four four distinct cocircular points. We dene te cro raio of tese tese points to be te quant qu antity ity
sin � B cr D) sin D sin sin
were te arcs, of course, are on te common circle troug te four given points. Tere are several several potential pot ential ambiguiti ambi guities es tat we need to to resolve reso lve before we can say tat tis really makes sense sen se.. First, observe tat we can measure te arcs in eiter degrees degrees or radians, radians, as we please, plea se, provided tat tat we interpret interpret te te sine sin e function function accordingly. Te stadard function f () sin () assumes assume s tat is measured in radians, but if we wis, we can measure angles and arcs in degrees and use te function function () sin(). Observe for example, tat f( /2) 1 (90) Next, note tat it is i s not actually true tat two points X and Y of a given circle dene a unique arc; tey actually determine two arcs wose wo se angular measurs sum to 30 Neverteless Neverteless,, te quantity quantity sin( sin ( is unambiguously unambiguou sly dened because becaus e te two possible pos sible meanings of correspond corresp ond to two two angles tat tat sum to 1 80 , and tus tey tey ave ave equal equal sines. sines . y te discussion of te previous paragrap te quantity cr D) is un ambiguously ambiguousl y de dened ned for any any four distinct cocircular points A and D. For ex ample, te reader can ceck tat if is a square, ten cr D) 2 and cr D) 1/2. Tere remains a potential ambiguity in our notation, ow ever: Te denition of cr D) is different if te points and and D are collinear coll inear rater rater tan cocircular. cocircu lar. We We are are safe, owever, because becau se four distinct distin ct points can never be bot colline co llinear ar and cocircular, and ence at most one of te two two denitio de nitios s of cr D) applies. applies . Tere sould tus be no danger danger of confusi confusion. on. We are using te t e same name cross cros s ratio and te same notation "cr , ) for for four collinear points and for four cocircular points because becau se tere is an intimate intimate connection beteen tese two concepts, concepts , as te following following teorem demonstrates.
�
X
befour distinct distinct cocir coci rcular points and sup suppose THEOREM. et A and D befour P is a point poin t on the same cir ci rcle derentf eren tfrrom all of of them Given Giv en a line lin e not no t throu throuh h P let W X Y and Z be the four ou r necessarily necessarily distinct distinct and collinear points where
C CROSS RATOS RATOS
113
D)
P A P B P and P D respectivel meet the iven line Then r (A (A B r ( X, Y Z)
Figure 3.17 sows one o ne of te te many possible congurations co ngurations for tis tis teorem We know from emma e mma 3.1 tat sin ( P Y) sin ( X P Z) r( r ( W X Y Z ) sin(PZ)sin(XPY) and by denition, denitio n,
Prof.
=
( ! A sin ( ! ) (A (A B, D ) . sin ( 4 A D n ( 4 B sin
It sufces, terefore, to sow tat te four sines in te rst formula are equal, respectively, to te four four sines sine s in te second s econd formula formula Tus sin( PY) In Figure 317 we see tat tat PY APC were in tis equation of sines, we need not specify wic of te two sin poss ible arcs arcs we mean mean Similarl Similarly, y, X P Y B PC were we w e refer refer ere to te te arc arc between B and tat tat excludes excludes point P Tus sin ( X P Y) were again, because sines of supplementary supplementary angles are equal, we can sin( aord to be sloppy sl oppy in our designation des ignation of te arc arc In te conguration sown in Figure 3.17 te oter two equalities of sines for te angles tat involve point Z old for a sligtly dierent reason We see tat tat X P Z is i s supplement supplementar aryy to B P D B" C D and tus XP XP Z B D ave sin( XP Z) sin( Similarly, P P Z and tus and we ave sin( P Z) Z) sin( as required Tis proves te equality of te two cross ratios ratios in te diagram of Figure 3.17. In general, it is i s easy to see tat eac angle angle is always equal in degrees to alf of one of te te two possible pos sible arcs corresponding to it Te precise rule is tat we need to use te arc arc tat contains P wen P lies li es between exactly one of te two pais of correspond corresponding ing points, and oterwise, oterwise, we use te arc excluding P In Figure 3.17 not lie between between A and , B , fr instance, P lies between D and Z, but it does not and X, or and Y Tis is i s te reason we ad to to coose coos e te arcs arcs containing P fo fo te te angles involving Z in te previous previous paragrap paragrap Te four four sines of angles, teref terefore, ore, • e qual to te four four sines sine s of arc arc s, and te proof is complete comp lete are always equal
4
A
4
4
A y
D igure
3.17
=
114
CHAER
CRCLES CRCL ES AND LNES
projcio cio poi Tis In te te situation of Teorem 3 1 9, we sall refer refer to P as te proj result is very very powerful powerful and an d it is especially esp ecially usef us eful ul in te restricted restricted case were te line Z cuts te circle at two of te original four points To demonstrate tis, we consider te following problem, wic wic appears appears to be very difcult difcult by oter metods.
point on te circumcircle of a square is joined to t o te two mos mostt distant vertices vertices,, tereby tereby cutting te nearest side of te square into tree pieces . Find te lengt of te middle segment if te oter two pieces ave lengts len gts 3 and 1 0.
(20) PROBLEM
igure
3.18
In Figure 3 . 1 8 , denote te intersection points of P and PC wit wit by respectively, and write = y Teo Teore rem m 3 1 9, wit wit pro projection jection poit P, P, w e ave + 3 10) = c r = cr = + 13) were were te rst equality olds because is a square Elementary algebra now yields te equation + 13  30 = 0, and tis quadratic equation as roots = and = 1. Since we are looking for a positive pos itive number, number, we rej rej ect  1 and conclude tat te te lengt of te te middle segment is •
ouio and
We now now give give te promised promise d quick proof of te te utter u ttery y teorem using cross raios raios.. We rep repea eatt Figure 3.3 . 1 0 ere ere as Figure 3.3 . 1 9 and we recall recall tat tat is te midpoint of and at our task is to prove tat X = Y in eac of o f te two diagrams. In eiter diagram, we see tat by using P as te projection point in Teorem Teorem 3 1 9, we get cr X = cr Q Wit as te Y, = cr Q and so we projection point, owever, we get cr X cr Y deduce tat cr X Now write = X y = Y, and and = = Working in te left diagram of Figure 3. 1 9 , we see tat tat X + 1 cr X = = + X
Proof of Thorm 1
C CROSS RATOS RATOS p
115
p
Q
Q igure
3.19
and
AYB ( ( + ) 1 ABY 2 2 2 Since Sin ce we know tat tese are equal, we deduce tat , as required required In te rigt diagram, a similar calculation yields yield s tat (  ) 1  (A , X , B) 2 2 2 and (  ) 1 (A Y B) 2  2 2 and we deduce dedu ce tat in tis case too. too. (A Y B)
•
Tis proof of te te uttery uttery teore teorem m yields yi elds some additional information, information, wic we discuss somewat info informa rmally lly.. Suppose P Q, RS and AB A B are any tree tree cords of a circle tat are are concurrent concurrent at a point tat is not necessarily neces sarily te midpoint of A B . et X ad Y , respectively, be te intersections of A B wit P S and wit Q R as in Figur Figuree 3 1 9 We now ave ve distinct collinear points A X, , Y, and B, and te rst part of our proof of te uttery teorem teorem sows sows tat (A , X B ) (A, (A , Y B) B ) . We sress sress tat tis tis part of te argument argument did not rely on te assumption as sumption tat A B i te tteryy teorem. We We sall say tat tat ve distinct collinear points points A , X, X , , Y , and B ave utter u te ury propry wenever (A X B) (A , Y, B) , and we observ observee tat tis property is invariant under perspective since te cross ratios are invariant by Teo Teore rem m 3. 1 . We claim now tat if we ad started wit tree concurrent cords of an ellipse instead of a circle, te t e ve points A, X, Y, and B would woul d nnever everteles telesss continue to to enjo enjoyy te buttery property. property. To see wy wy tis is so, s o, we use te te fact fact tat an an ellipse ellip se is a conic section. section . Witout going into detail ere, we can exploit tis by imagining te folloin folloingg situation. We are in a room wit a point poi nt source of ligt in te ceiling ceil ing and a clean wite w ite oor. We ave a planar seet of glass on wic an ellipse as been drawn wit opaque ink, and we old it so s o tat a sadow of e ellipse is cast onto te oor. oo r. Te fact tat te
116
CHAER
CRCLES AND LNES
ellipse is a conic section tells us tat itit is possible pos sible to position and tilt te te glass in suc a way tat te sadow is a circle. In fact, fact, tis can be done so tat te center of te circle is directly below te ligt, but we sall not need tis additional addition al information information.. Wile our assistant ass istant olds te glas s steady, wit te ellipse casting a circular sadow, we draw tre treee concurrent cords on te ellipse, and we construct and mark te points , X M Y and , as we discus dis cussed sed previously. Te boldface boldface font font indicates tat we used opaque , and it allows us to distinguis tese tese ve points points from from teir teir sadows sadows A , X, , Y, and B , wic are points on te oor. Since sadows of lines are lines, we see tat te sadows of te tree concurrent cords of te ellipse are tree concurrent cords of its circular sadow. Furtermore, te sadow of te buttery pattern pattern on te ellipse ellips e is a buttery pattrn pattrn on te circle, circle , as in Figure Figure 3. 1 9 , and we we deduce deduce tat te ve collinear sadow sadow points A, A , X, , Y , and B satisfy sati sfy te te buttery property. Te key key observation ere is tat te ve sadow s adow points A , X , , Y , and B are are in perspective perspective from from te te ligt wit te ve points , X M Y and marked on te glas s. It follows follows tat tat te latter latter ve points also al so enjoy enjoy te butter buttery y property, as claimed. Suppose Suppo se now tat te te point of concurrence M of te te tree cords of te ellipse ellips e appens to be te midpoint of cord . (aution: It does not follow tat is te mdpoint mdpoint of A B .) Since we know tat tat (, X M ) (, M Y ) , we can ow apply te second part of te crossratio proof of te uttery teorem to deduce at XM MY Tis part of te argument did not rely on te assumption tat we started wit a circle. Te preceding disc discus ussion sion sows tat Teorem Teorem 3. 3 . 1 3 , te uttery teorem, olds for ellipses ellips es as well as for circles circles .
xrcss 3 3C.l
3C2
Suppose tat tat A, B , X, and and Y are are colline collinear ar and and ddistin istinct. ct. If AX/ A Y BX/ BY, sow tat exactly one of te points A and B lies on line segment X Y . Suppose Supp ose tat A, A , B , and C are distinct distinct and collinear coll inear and tat tat X and Y lie on te line toug toug tem. tem .
a. If (A , B, C X) (A ( A , B B C Y ) , sow tat tat eiter eiter X and Y are are te te same same point, point , or else exactly one of A and B lies on segment X Y . b. If(A, If(A, B, C X) (A B C, Y) and and also also (C B A, X) (C B A, Y), sow tat X and Y must be te same point.
3Cl
NOTE Tis Ti s provides a tool tool tat can be used to prove tat tat points X and Y on te line t toug oug A , B , and C are actually actually identical. identical. oints and X are are cosen on side side AB A B of A B C and points Y and Z are are cosen on side AC . Suppose tat tat (A , , X, B) (A Y Y Z C) C) and and tat tat Y XZ . rov rovee ta tatt XZ BC . HINT et T b e te point were te te parallel parallel to X Z troug B meets line A C . Note tat tat neiter neiter A nor Y can lie on segment C and use us e Exercis Exercisee 3.2 3 .2(a) (a) to sow tat tat T is C.
C CROS S RATOS RATOS
117
A
Fiue 320
In igure igure 3 0 points P and Q were chosen chos en in in the interior interior of A B C and coinear with A ines in es B Q and B P meet side A C at Y and Z and ines C Q and C P meet meet side A B at and and X Show tha thatt (A (A , X X B ) (A Y Z C) 3C.3 In the the situation situation of igu igure re 30 3 0 suppose that that ines Y and and X Z meet at a point R Show that R ies on ine B C HINT et T be the point point wher wheree ine ine R B crosses crosses ine AC se xercise xercise 3 (b) to show sho w that C and T are the same point 3C.4 In igure igure 33 1 we hae hae chosen a point X on the the extension extension of diagona diagona B D of square square A B C D and we dropped a perpendicuar perpendicuar C P from C to A X If C P meets DB at Y and and BY B Y X D show show that that BY B Y BA HINT oints A B C D and P are are cocircua cocircuarr 3C. igure igure 33 shows a sefintersect sefintersecting ing hexagon hexagon AB A B C D inscribed in a circe Sides A B and and D intersect intersect at U sides B C and intersect at V and sides side s C D and A intersect interse ct at Show that points U V and are are coinear coinear HINT et X and Y be the points poin ts where A and CD C D respectiey respe ctiey meet mee t U V roe that X and Y are the same point by showing that ( P T Q X) (P (P U Q V) (P T Q Y) where P and Q are the points where U V meet the circe as shown shown and T is the unabee unabeedd point where where A D meets UV Since points P and Q can be interchanged the same reasoning yieds that (Q T P X) (Q T P Y) NOT en en if hexagon hexagon A B C D inscribed in a circe is not no t sefintersecting or if diff different pairs pairs of sides intersect we can amost amo st aways dene thee points U intersectionn points of the the three three pairs pairs of ines A B and D B C V and a s the intersectio 3C.
E A
C
Fiue 321
B
Fiue 322
118
CAPTER
3C.6
CRCLES AND LNES
and and C D and A We can can run run into difcuty difcuty because becau se the ines in one or more of these pairs might might be parae parae so so we note that the sides side s of the the hexagon may need to be extended to construct U V and It I t is a theorem of asca asca ( 1 6231 662) that that points points U V and are are aways coinear coi near In fact this works works for hexagons inscribed in arbitra arbitrary ry conic sections sec tions and not just ju st in circes et A B C and D be b e distinct distinct cocircu cocircuar ar points and x a ine ine m not throu through gh A A et Y and Z be the the points where where AB A B A C and AD AD respectiey respectiey meet meet ine ine m and et be the intersection point of m with the tangent at A to the circ circe e through A B C C an andd D Sho S how w that that (A B, B , C D) = (W Y Z) NOTE: This resut can be iewed iewed as th imiting imiting case of Theorem Theorem 3 . 1 9 as the point of that theorem theorem approaches approaches A
Fiue 3.23
3C7
3C8 3C9
et AB C D be a quadriate quadriatera ra inscribed in a circe circe as shown in igure igure 33 23 . If U is the intersection of its diagonas and V is the the intersectio intersectionn of sides A B and CD CD show sho w that the tangents to the circe at A and at D meet ine in e U V at the same point these tangents is parae to U V. ssume s sume that neither of these HINT: et and Y be the points where the two tangents meet U V and show that cr U Q , ) = cr V Q R) = cr U Q Y) where and Q are the points where ine U V crosses the circe and R is the intersection of side A D with U V as in the diagram This cacuation cacu ation requires four four appications of Exerc Exercise ise 3. 3 . 8 . NOT Simiar reasoning shows that the tangents at B and D aso as o intersect at a point on ine U V et AB A B be a diameter diameter of a circe and and suppose that U V is a perpendic perpendicuar uar chord chord Show that cr A U B V) = 2. In the situation si tuation of the the preious probem prob em x x an arbitrar arbitraryy point on ine A B and and construct cons truct the point Q on A B as foows raw ine ine U and et be the point other than U where U meets meets the circe circe If PU P U is tangent to the the circe at at U ake to be U Now et Q be the point where V X meets A B Show that point point Q is independe i ndependent nt of th choice choic e of the the chord U V perpendic perpendicuar uar to A B igu igure re 3.24 3 .24 shows one possibe pos sibe conguration con guration for this probem
D TE RADCAL AXS
1 19
x
ure
3C.0
3D
3.24
ure
3.25
In Figur Figur 3.2 3. 2 w ar ar givn givn � X P wit wit angl angl bisctor XB , wr X and and B cntr lis on lin P . Sow tat if Y is any point p oint of t li on a circl wos cntr circl circl otr tan B tn Y B biscts bis cts Q Y P .
The Radical Ais
onsidr tr circls ac xtrnally tangnt to t otr two. In roblm 2.28 we sowd tat in tis tis situation t tr common tangnt lins ar concurrnt. T pras common tangnt lin" lin" is i s somwat impr imprcis. cis. r and in wat follows follows w intnd tis to rf rfrr to t common co mmon tangnt at t point of tangncy tangncy of t two circls . ctually as indicatd in Figur 3.2 3 .266 somting muc mor mor gnral gnral is going on r. First as a s te diagram on t lft dmonstrats dmonstrats t rquirmnt rquirmnt tat t t circls b tangnt in roblm 2.28 2. 28 is not no t rall rallyy ncssary. ncssary. In fact fact tanency is rally externally tangnt ilvant too. To xplain tis, w obsrv tat tr is an intrsting lin naturally associatd assoc iatd wit any pair of circls aving at last on point in common. comm on. For tangnt tangnt circls circls tis line is t common co mmon tangnt tangnt for nontangnt nontangnt intrscti intrscting ng circls circl s t lin to wic w rfr is t lin toug t two points of intrsction wic w somwat imprcisly call t common scant. (Not tat t common tangnt is rally just a limiting cas of tis common co mmon scant. sc ant. If w gradually gradually incras t sparation btwn t centers of t two circl circls,s, t two intrsction points coalsc co alsc to a singl point po int of tangncy tangncy and te common scant s cant bcoms te common tangnt. tangnt.)) If w start wit any any tr circles ac aving a point in common wit wi t t otr otr two two w gt tr intrsting lins on from ac pair of circls. W will prov t rmarkabl rmarkabl fact fact tat as sown in Figur Figur 3. 3 . 26, ts ts tr tr lines ar always concunt.
iure
3.26
10
CAER
CRCLES AND LNES
ctually, thr is an xcption. xcptio n. If th thr thr cntrs ar collin col linar ar thn th th thr lins will wil l b paralll Given three circles with noncollinear centers assume that eve two of them have a point in common If for each pair of circles we draw as appropriate propriate either eithe r the common secant se cant or the common tanent tanent then the three t hree lines thus constructed are concurrent
(3.1) THEOREM.
onsidr what happns if w hold two of th circls xd and vary th third Of cours th lin dtrmind by th two two xd circls rmains rmains xd and th othr othr two lins will mov. mov . W obsrv howvr that th th (moving) point of intrsction of th two moving lins is constra cons traind ind by Thorm Thorm 33 2 1 to travl along along th third third (xd) lin To s that thr is still s till mor going goin g on hr consid con sidrr what happns if i f w w try th sam xprimnt in a situation whr th two xd circls hav no point in common but th th varying varying circl mts both of thm. n xampl of this appars in Figur 3 27 whr w show four positions pos itions and sizs siz s for th th variabl variabl circl To mak th diagram appar appar lss cluttrd w chos chos on of th th two xd circls insid ins id th othr othr but this aangmnt aangmnt is not ssntia s sntial.l. Th only on ly rquirm rquirmnt nt is that that the two xd circls should not no t b concntric concntric W again hav two moving lins and again w obsrv th bhavior of thir intrsction point markably markably its trajctory trajctory still appars appars to b a straight lin lin In th gur P Q R and S ar collin col linar ar What What is th th locus lo cus of th th intrsction point of th two moving commn c ommn scants? It is clarly som objct dtrmind by th two xd circls and in th cas whr ths circls hav a point in common th locus is thir common scant or tangnt tangnt lin W will prov that th locus is always a line line vn whn th two two givn circl circ lss hav no point in common . This lin lin which is dashd in in Figur 3 27 is calld th radical radical axis axis of th to circls but bfor w can dn this trm ofcially w nd to introduc an auxiliary ida Givn a point P and a circl of radius radius r cntrd cntrd at som point w say s ay that th por of P with rspct to th givn circl is th quantity quantity p d r whr d P O is th distanc from from th point poi nt to th cntr of th circl Th points xtrior to th circl clarly hav positiv pos itiv powr Th intrior points point s hav ngativ powr and th points of th circl itslf it slf hav powr p o
Q
_ _ _ _p _ _ _ _ _ _ _ _ R _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ s
Fiue 3.27
D TE RADCAL S
121
To mak t dnition d nition of t powr of a point wit rspct to a circl sm a littl arbitrary ary w rcall Torm 1 . 3 . In tat rsult w sowd tat if P is i s any point ls s arbitr not on a givn circl and w slct any lin troug troug P mting t circl at two points X coic of t t lin lin and Y tn t quantity P X P Y is a constant indpndnt of t coic T following lmma tlls us tat tis mystrious constant is itr t powr of t point P or its ngativ W sall not actually us mot of tis lmma lmma i$ purpos er is to rlat t t powr of a point to idas tat w av av sn s n prvious prviously ly Fix Fix a cir ci rcle and an d a point poi nt P and let p be the power pow er of of P with respect to the iven circle ou tside the circle and a line throu throuhh P cuts the circle at X and Y then a If P lies outside P X P Y p b If P is inside insi de the circle on chord XY then P X P Y p circle at point T then P T p c If P lies on the line tanent to the circle
(3.2) (3 .2) LEMMA.
ssum rst tat P lis outsid t circl y Torm 1.3 t quantity P X P Y will b t sam s am for for all lins toug P tat mt t circl in two points It is no loss los s trf trfor or to assum as sum tat t lin lin X Y troug P actually gos troug t cntr of t t circl wic w dnot Tus X Y is a diamtr diamtr and w can assum tat X is t narr narr of ts two points to P Writing P d and X r radius w s s tat P X P  X d  and and P Y P + Y d + r t radius (d  r) (d + r) d r = p as rquird Tus P X P Y (d Suppos now tat P lis insid t circl y Torm 13 t quantity P X P Y is a constant cons tant indpndnt indpndnt of t t particular particular cord X Y toug toug P To prov (b) trfor w can assum tat cord XY is a diamtr and w can furtr tat P lis on t sgmnt X W s tat P X X  P r  d assum as sum tat and P Y Y + P r + d Tus P X P Y = (r  d) (r + d) r  d p a s rquird Finally if P is i s tangnt to to t circl at at T w can us (a) and a limit argumnt argumnt to sow tat (PT) p ltativly w obsrv tat T P is a rigt triangl wit sid T r and ypotnus ypotnus P T d It follo follows ws by t ytagoran ytagora n torm • tat (PT) d  r p as rquird
Prf.
Fix Fix two circles centered cent ered at distinct points poi nts A and B B Then there exist points po ints whose powers po wers with respec respectt to to the two iven circles c ircles are are equal eq ual The locus lo cus of all such points poin ts is a line lin e perpen perpendicl diclar ar to AB A B
(3.3) (3 .3) THEOREM.
It is convnint c onvnint to us coordinat gomtry gomtry for tis proof proof W can suppos su ppos tat points A and B li on t x axis so tat A is t point (a 0) and B is (b 0) wr a b If P is i s an arbitrary arbitrary point wit coordinats (x y) tn tn ( P A ) y + (x  a) and and (P B ) y + (x  b) Writi Writing ng and and s to dnot t radii of t givn circls cntrd at A and B rspctivly, w s s tat t t powrs of P wit rsp rspct ct to t t two circls ar ar qual if and only if
Prf
122
CHAPTER
CRCLES CRC LES AND LNES
Th y trms cancl and when w too . w xpand th parnthss parnths s,, th x trms cancl too. Th condition condit ion that th two powrs of P ar qual qual thus rducs rduc s to th linar quation a  2ax  b  2bx  s Sinc b  a is nonzro this is quivalnt quivalnt to  s + b  a x 2(b  a) whr whr w obsrv ob srv that that th right right sid is som consta cons tant. nt. Sinc Sinc this is th quation of a lin prpndicular to th x axis th proof proof is complt. • Finally w can den th objec objectt in th hading of this sction s ction.. Given Given two circles having dirnt cntrs thir radica axi is th lin consis con sisting ting of all points that av av qual qual powrs with with rspect rspec t to th th two circls. circls . f o cicles inesec a o poins A and B hen hei adical axis is hei common secan AB o ci ci cles ae angen angen a a poin po in T hen hei adical adical axis is hei common angen a T
(3.4) COROARY.
point common to two circles has powr 0 with rspct to ach of thm thm and and thus its two powrs ar qual qual and th point point lies on th radical axis. axis . If A and B ar ar two diffrnt points common to two circls thn A and B both li on th radical axis axis which w ow is a line. It follows that that th radical axis is th lin A B In th cas cas whr two circls ar tangnt at T thn sinc T is on both circles circles it lis on th radical axis. axis . To s that th radical radical axis is tangnt to ach circl at T i t sufcs sufcs to show that T i s th only point whr whr this this lin mts ither circl. This is clar cl ar howvr bcaus b caus if a point P of th radical radical axis lis l is on on of th circls circls its it s power with rspect to that circl and hnc with rspct rspct to th othr circl too i s O It follows follows that P lis on o n both circls circls and hnc P is th uniqu point common to th two two circls namly, T •
Proof
It is now clar that th th following asy rsult includs Thorm 2 Given Give n hee cicles wih noncolli non collinea nea cenes cen es he hee adi adi cal axes of of he cicles aken in pais ae ae disinc concuen conc uen lines lines
(3.) COROARY.
Sinc Sinc th radical axis of a pair of circls is i s pndicular pndicula r to to th lin of cntrs cntrs of t t circls, circls , it follows from th noncollinar noncolli narity ity of th thr cntrs that th tr tr radical axs ar distinct dis tinct and nonparalll. nonparall l. Every two of thm thrfor hav a point of intrsction intrsction.. For any point P lt us writ P P and P to dnot th powrs of P with rspct to th thr thr givn circls. circls . For points points on on radical axis w hav P P and on anothr, w hav P P t th point P whr ths two radical axs mt w hav P P P and thus P P and P also lis on th third radical axis.
Proof.
D THE RADCAL AXS AXS
igue
123
3.28
We mention tat in te situation of orollary oroll ary 3 2 2 te unique point common co mmon to te tree tree radical radical axes is called te radica cr of te te tree tree circles. circles . Given two nonconcentric circles printed printed on a pice o f paper paper ow can c an w actually nd and draw draw tir radical axis? axis ? If te te two circles intersect at two points tis ti s is i s an asy task: Just draw t line troug t two points points If te two circles are are tangent, te radical axis is te common tangent line line and tis is only a bit more more difcult difcult to draw. ut ow can c an w draw te radical axis if i f te two circles circles ave no points po ints in i n common? If te circls are xtal xtal to ac otr, otr, w can draw a line lin e tangent to bot and w let le t S and T b te two two points of tangency tangency as sown in Figure Figure 3.2 3 .2 8. I f M is any any point on line ST ten ten by emma 3.22(c) 3. 22(c) we w e know know tat distances distances M S and M T are te powers of point M wit wit respect to te two circles. circle s. It follow followss tat tat if we take M to be te midpoint of segment ST as in Figure Figure 326 3 26 ten ten MS MT and so te two powers powers ar qual. Te midpoint M tus lies on te radical radical axis, and since we know tat tat te radical axis is i s perpendicular to te te line of centers centers it sufces to draw te te perpendicular to tis line troug M to complete te construction lternatively, lternatively, we can draw one of te oter oter tree lines tangent to bot circles and we let U and V be its two points of tangency. Te mdpoint N of segmnt U V must also lie on t radica radicall axis wic can tus be constructd co nstructd by drawing drawing line M N We can avoid te possibility tat M and N ar t same sam e point if w coos co osee te second tangent appropriately. If one of te two circles is inside i nside t oter tere are are no lines tangnt tangnt to bot bo t circles and te metod of te previous paragrap cannot be used. ltoug it may not be obvious ow o w to proced proced it is really not ard to construct te radical radical axis in tis situation too Te metod metod we ar about about to describe is suggested sugge sted by Figure Figure 3. 27 27 but it actually works in all cases cas es.. In fact, fact, iti t is probably even easier to carry carry out tan tan te construction cons truction of te previous paragrap. raw an auxiliary circle meeting eac of te two given circles circles in two points and draw te line troug eac of tes pairs of points. Tese two lines are te te radical axes of te auxili auxiliar aryy circle wit eac of te two original original circles circles and we know know by orollar oroll aryy 3 25 tat te point P were tese lines meet mus mustt lie on te radical axis tat we seek. Now coose coos e a second auxiliary auxil iary circle and perf perform a similar similar construction to to obtain a point Q . Since bot P and Q are are known to lie on te radical radical axis of te two given circles we can complete complet e our construction con struction by drawing drawing line P Q . Given any two xed nonconcentric circles we mentioned previously tat teir radical radical axis is te locus locu s of all points P tat can be obtained using an auxiliary circle, as i te previous paragrap We ave already already argued tat every point constructed cons tructed tis way must lie on te radical radical axis but to prove prove tat tat tis is actually actually te locus, locus , we must
=
14
CAER
CRCLES AND LES
also als o sow tat vry point on tis lin can b obtaind from from som som coic coi c of an an auxiliary auxil iary circl. Givn Givn any point P on t t radical axis draw draw two two lins troug P , on o n for for ac ac of t t givn circl circls s,, intrscting it in two points. points . Tis givs four points points two on ac circl and w s s tat a circl troug ts four four points would srv as an auxiliary circl tat yilds P . Our task tn tn is i s to sow tat ts four four points actually actually ar cocircular cocircular Given poins poi ns A and and B on one on e cicle and C and D on anohe le P be he he ineseci i nesecion on of of lines A B and and C D Then Then P lies li es on he h e adical adica l axis of of he o given cicles and only he fou fou poins poin s A B C and D ae ae cocicula cocicula
(3.6) COROARY.
If t four four points all li on som circle tn by orollary 3. 24 lins lin s A B and C D ar ar t radical axs of tis tis circl wit ac of t givn circls . y y orollary 3.2 3 .2 tir tir intrsction P lis on t radical axis axis of t two two givn givn circls. circls . onvrsly suppos suppo s P lis on o n t radical axis axis of o f t two tw o givn circls and nam nam ts circls X and Y wr A and B li on X and C and D li on Y onsidr t circl Z troug A B and C and obsrv obsrv tat tat A B is t radical radical axis of X and Z Sinc P lis l is on tis radical axis and als alsoo on t radical axis of t original original two circls X and Y it follows follows by orollary 3. 2 tat P also lis li s on t radical axis of Y and Z ut C also lis li s on tis radical axis axis and w conclud tat t t radical axis axis of circl circlss Y and Z must b t lin lin c Tis lin gos troug D owvr and tus D as qual powrs powrs wit rspct to circls circls Y and Z Sinc D lis on Y w conclud tat it also also lis on Z and t proof is complt. •
Prf.
xercses 3 3.1
3.
onsidr onsi dr t t tr circls circls wos diamt diamtrs rs ar t t sids sid s of a givn givn triangl triangl.. Sow S ow tat t radical cntr of ts ts circls is t ortocentr ortocentr of t triangl triangl.. In Figur Figur 3. 29 t common common cord cord P of two two circls circls biscts lin sgmnt A B wr A and B li on t circls as sown. s own. If X and Y ar ar t otr otr points wr A B mts t two two circl circlss sow tat B X AY.
iure
3.29
C H P E R F OU R
Ceva 's Teo Teorrem an Its Rea Re atives tives
4A
Ceva's Ceva' s Theorem Theorem
We know several theorems that have the form form the three three somethings s omethings of a triangle are are concurrent. In every triangle for example te three medians are concurrent and so too are te three angle bisectors and the three altitudes Notice that in each of these situations situations the th e three three concurrent concurrent somethings are are three three lines lines each ea ch of them passing pas sing through one vertex of the triangle Of course, tere are many many other ways to dene de ne three lines passing through the vertices vertices of a triangle, triangle, and it seems amazing how often there there is a theorem that guarantees guarantees that these three lines must mus t go trough a common point ut not every concurrence theorem associated with a triangle has the form we are discussing here. We know for example that the perpendicular bisectors of the sides of a triangle are always concurrent but this is i s not a result of the type we are are considering consi dering because becau se the perpendicular bisectors bis ectors generally do not go through the the vertices. concurrence fact fact that actually is an example of the phenomenon we have in mind owever owever is the followi followigg PROBEM. et P and R be the points of tangency tangency of the incircle of ABC with sides BC CA and AB respective respectively ly as in Figure Figure 4. 1 . Show that lines A P B and and C R are concurrent. co ncurrent. Te point of concurrency of the three three lines in roblem roblem 4. 1 is sometimes s ometimes called c alled the Gro poi of the the triangle triangle s s apparent apparent in Figure 4 1 the Gergonne point need A
P
gure . 125
16
CAPTER
CEVA CEVA S TEOM AND TS RELATVES RELATVES
not be the center of the circle and thus in general the tee lines are not the angle bisectors . We shall present the solution of roblem 4 as an application appli cation of the theorem of Giovanni eva (4877) that is the main resul resultt of this section. section . et AP A P B and C R be b e thee lnes onng the vetces po nts P and R on the the op opposte sdes Then these thee lnes ae ae of A B C to ponts concuent con cuent and only ARBPC . RB PC A
(4.2) THEOREM (Cva).
We shall refer to any line lin e going goi ng though exactly one vertex of a triang triangle le as a Cvian of the triangle. triangle. y Theorem 42 we see s ee that if we wish wi sh to prove that three evians are are concuent (as in roblem 4 for for example) all we need to do is compute compu te the the product of the the tee fractions fractions in evas eva s theorem and show that the resulting quantity is equal to In general general we w e will refer to this quantity as the Cvian prodct ass associated ociated wit the thee thee given evians . Of course we can c an also think of this this product of the theee fractions fractions as the quotient obtained by dividing the product product of thre threee segment lengths by the product product of another another three three segment s egment lengths. One On e way to remember the denition of the evian evian product is to imagine traveling traveling around aroun d the triangle from A through R to B then from B through P to C and then along the third side from C through and back to A For each side of the triangle, triangle, we get one fraction raction:: rst A R / R B then B P / PC and nally C / A The evian product product is just jus t the product of these three fractions. fractions. ecall that in Figure 4 we know that A R A B P RB and C PC (ctually by emma 227 we know know more more The quantities AR B P and C are equal to s a s b and s c respectively in the usual notation. We shall s hall not n ot need that additional infor information mation however.) however.) Observe that A R occurs in the numerator of the evian product and that A occurs in the denominator. denominator. These two equal equa l quantities thus cancel when we compute the product product and similarly sim ilarly,, B P cancels with R B and C cancels with PC ll six quantities quanti ties cancel therefore therefore and the value of the evian product is It follows by eva e va s theorem that the three evians are concuent as required.
otion otion to Probm Probm 4. 1.
Notice that eva s theorem is an if and and only if statement Thus Thu s whenever we have three concuent line segments joining the vertices of a triangle with points on the opposite sides the evian product must be trivial. y trivial we mean of course that the product is equal to Suppose for example that our three evians are the medians of ABC so that P and R are the midpoints of BC CA and AB respectively, and thus A R RB BP PC and C A Since the medians are concuent we know that the evian product has to be trivial and indeed it is because here too the six quantities cancel. Next consider the case where the evians A P B and C R are the the angle bisectors. bi sectors. gain the evian product must be trivial trivial
A CEVA CEVA S TOM
17
since te angle bisectors bise ctors are are always concurrent, concurrent, and we can conrm tat tat by recalling Teorem 2 wic tells us tat an angle bisector of a triangle triangle divides te opposite side into pieces wose lengts are proportional to te nearer sides of te triangle. sing te usual notation, were we write a b and c to denote te lengts of te sides opposite vertices A B and C we ave ARRB ba BPPC cb and C Q Q QA ac. Te evian product (AR RB) (B P PC)(C Q Q A) is tus equal to (ba)(cb)(ac) and again everyting cancels and te evian product is trivial, as expected We need ne ed a simple s imple lemma for te proof of Teorem 42 and we state it somewat so mewat more generally tan is really neces sary EMMA. Gven dstnct ponts A and B and a postve numbe � thee s exactly exactly one pont po nt X on the lne lne segment AB A B such that that AX A X X B � A lso thee s at most one othe pont pon t on the lne lne A B o whch ths equaton eq uaton hold ho ldss iew X as a variable point and let (X) be te function wose value at X is te quantity A X X B Tus (X) is a nonnegative real number, and it is de ned everywere except wen X B s X moves from A toward B along segment AB we see tat AX increases and X B decreases, and tus (X) is monotonically increasing from 0 wen X is at A and it approaces innity as X approaces B Tere is tus exactly one point X between A and B were (X) � If X is on line A B outside of segment AB tere are just two possiblities: Eiter B is between X and A or else A is between X and B In te rst case, AX XB BA and (X) AXXB (BAXB) > Oterwise, AX X B  B A and (X) AXXB  (BAXB) < For any given value (X) � teref terefore, ore, at most one of tese tese two situations can occur occ ur depending on weter � > or � < If B is between X and A te functio functionn (X) + ( B A X B) is monotonically monotonically (BAXB) is decreasing as X moves farter from B Oterwise, (X)  (BAXB) monotonically increasing as X gets ge ts farter from B In eiter case, we see s ee tat tere tere can be at most one point X suc tat (X) = �
Proof.
We sall also need an elementary algebraic fact about ratios of real numbers. If (s ay, ab c d) ten we automatically get two more ratios equal two ratios are equal (say, to tese two, namely, (a c)/(b d) and (a  c )/(b  d) (Of course, course, we need to assume tat b d is nonzero for te rst of tese and tat b  d is nonzero for te second seco nd)) To see wy tis works, works , write ab cd so tat a b and c d Ten a c (b+d) and a  c (b  d) and tus (a + c)/(b + d) (a  c)/(b  d) as we wanted We will refer to tese as te addition and subtraction principles for ratios We are now ready to present te surprisin surprisingly gly easy proof of evas eva s powerful teorem
18
CAPTER
CEVAS TEOREM AND TS RELATVES
ssume s sume rst tat te te tree evians are concurrent at some point T as in Figure 42 We compute com pute te evian product by considering cons idering te areas of certain triangles triangles.. iew iew B P and P C as te bases base s of 6AB P and 6APC respectively, and observe tat tese triangles ave equal eigts. It follows tat B P / P C is te ratio of te areas of tese two triangles. Since segments B P and P C can also als o be viewed as te bases of 6 T B P and 6 T PC and tese two tri angles also als o ave equal equal eigts, eigt s, we conclude tat
Proof of Thorm 4.
A
P
igure
4.2
KA B P KT B P P C K P C APC T were as usual, usu al, we w e are are writing K z to denote te t e area of an arbitrary 6 X Y Z y te subtraction subtrac tion principle for ratios, ratios , we deduce tat
KABP  K T B P P C KAPC  K T P c Exactly similar simi lar reasoning yields AR RB
KA B T KAB KC A T
C Q KB C T KC A T and Q A KA B T KB C T and tus tus we can compute comp ute te evian product. produc t. We We ave

,
AR BP C Q KC A T K A B T K B C T 1 KABB T R B P C Q A K B C T KC A T KA
as we wanted wanted To prove te converse, converse , we assume as sume tat te evian product is trivial. trivial . We We prove tat A P B Q and C R are concurrent by dening T to be te intersection of A P and B Q in Figure 42 and sowing sow ing tat line C R must also als o pass troug troug T Tis Tis is equivalent, of course, to sowing tat line CT goes goe s troug troug R, and so we we let R be te point were line C T actually does meet side A B. Ten C R is a evian evian tat tat is concurrent concurrent wit wit A P and B Q and so by te rst part of te te proof, te correspo corre sponding nding vian product is trivial trivial We now ave AR BP CQ A R B P CQ 1 R B PC QA R B PC QA were were te second seco nd equality is by assumption. assumption . ancellation yields y ields tat A R / R B A R / R B , and we write write � to denote te te common value of tese two equal eq ual ratios ratios y emma 4 tere tere can only be one point on line segment AB for for wic A X/ X B � ut R and R bot lie lie on segment AB , and for eac eac of tese points, points , we get get te
A CEVA CEVA S TEOREM
19
same ratio � Tus Tus R and and R' must actually be te te same point, and ence C T goes troug R as required. We can also als o tink about eva s teorem from from te point of view of centers of mass mas s in pysics. Imagine tat 6ABC is manufactured from weigtless rods, and we place nonzero masses m m and m at vertices A B and C respectiely. Te center of mass of side B C will be at tat point P on side B C were te moments m B P and m P C are are equal, equal , and tus B P / P C = m / m We know tat in every balancing bal ancing experiment, experiment, side s ide B C beaves as toug its total mass were concentrated at te point P and tus we can c an pretend tat te te entire mass m + m of side B C i s placed at point P Te center of mass T of te entire triangle, terefore, must lie on te evian A P y exactly similar simil ar reasoning, reasonin g, we w e know tat tat T also lies on o n evians evians B Q and C R , wee Q and R are te te centers of mass masses es of sides C A and AB respectively. Te tree evians determined determined by te masses masse s m m are tus guaranteed to be concurrent at te m and m are center of mass T of te triangl trianglee . s a ceck, we compute comput e te evian product. produ ct. We saw tat B P / PC = m /m and similar reasoning sows tat C Q/ QA m / m and AR / RB = m / m We terefore ave A R B P C Q m m m = , RB PC QA m m m and te te evian product is trivial, trivial, a s expected. Now, suppose tat AP B Q and C R are evians of some triangle 6ABC and assume tat te corresponding evian product is trivial trivial.. We will sow so w tat it is always pos po s sible to coose nonzero masses m m and m so tat if we place tese masses masse s at vertces A B and C respectively, te centers of mass of te tree sides will be at P Q and R. R . We We are continuing to assume, assume , of course, tat te actual actual sides sid es are weigtless and tat all of te mass is concentrated at te vertices. ssuming tat we can nd suc masse mas ses,s, tis will w ill provide an alteative alteative proof tat tat te tree tree given evians must be concurrent. concurrent. Tis is because bec ause eac evian will ave to pass troug te center of mass of te triangle. triangle. m and m we view tese quantities as To nd appropriate nonzero masses m m unknowns and we attempt attempt to solve te tree simultaneous equations m B P = m P C sol ution for for tis tis system of equations equations mC Q = m m QA and m AR = m m RB One solution would be to set eac eac of te unknowns to zero, but tis is of no value value to us since centers of mass of massles mass lesss objects are undened. We We can coose our units of o f mass mas s so tat m = and tis forces m = QA / C Q and m = AR/ RB from te second sec ond and tird equations. To verify tat te rst equation also olds for tese values of m m and m we need to ceck tat te quantities ARBP QAPC and RB CQ are equal. ut ARBPCQ = QA·RB·PC since we are assuming tat te evian product is equal to and tus m BP = mPC as desired. Tis completes te pysics pysic s proof tat tree evians must mu st be concurrent concurrent wen te corresponding evian product is trivial.
130 13 0
CAPTER
CEVA CEVA S TEOREM AND TS RELATVES RELATVES
s a lagniappe, tis metod of mass points enables us to compute te te position pos ition along eac evian of te point of concurrence of tree cocurrent evians We know, for example, example , tat te point poi nt of concurrence con currence of te te tree medians of a triang triangle le lies li es two tirds of te way along eac median, and we would like to ave analogous analogo us information information for for any any tree concurrent evians evians We ave seen s een tat te center of mass T of 6ABC wit masses m m and m at te vertices vertices lies on evian AP Furterm Furtermore, ore, we can assume ass ume tat te wole mass c oncentrated ed at P It follows tat m A T m + m )T P m + m of side B C is concentrat sing sing te fact tat A P AT + T P some elemetary elemetary algebra yields m +m AT A P m + m +m We can now plug pl ug in m m ARRB and m QACQ and wit a little algebra, we deduce deduc e te foll following owing formula, formula, wic w ic we state as a teorem. teore m. et AP A P B Q and C R be Cevans Cevans n 6A 6 A B C whee P Q and and R le on sdes B C C A and AB A B espe espect ctvely vely f f these Ceva C evans ns ae conc co ncuent uent at a pont T then AT AR·C Q + QA·RB AP AR· CQ + QA·R B + RB· C Q and smla fomulas holdfo holdfo B T B Q and C T CR
(4.4) THEOREM.
Teorem 44 tells us were to look along evian A P to nd te point T of con currence. More specically, it tells us wat fraction of te route from A to P we must traverse to nd T We can ceck c eck tis somewat upleasant formula formula in te case were were we are are dealing wit tree medians and we already already know tat te intersection, intersec tion, te centroid of 6ABC lies two tirds of te way way along eac median. In tis situation, si tuation, we can write write A R R B and C Q q QA lugging lug ging tis into te formula formula of Teorem 44 we get A T AP q 3q) /3 as expected.
xercses 4A.l
4A.2
4A3
et T be te Gergonne point poi nt of 6ABC ecall tat ti tiss is te point po int of concurrence of te te evians evi ans in te situation situa tion of roblem roblem 4 Sow So w tat if T coincides coinci des wit te incenter incente r or te circumcenter or te ortocenter or te centroid centro id of 6A B C en te triangle triangle must be equilate equil ateral. ral. et U and V be points on sides A B and A C respectively, of 6 ABC and suppose tat U V is parallel to B C Sow tat te intersectio of U C and V B lies on median AM NOTE Tis is essentially ess entially roblem roblem 130 but but you sou s ould ld now be able to nd a muc easier eas ier proof tan tan we gave in apter apter Sow tat te lines joining te vertices vertices of a triangle triangle to te points of tangecy tangecy of te opposite oppos ite exscribed circles are concurrent. concurrent.
B NTEROR ND EXTEROR EXTEROR CEVNS CEV NS
4A.4 4A.5
4A.6
4B
131
In Figure Figure 4.2, if AR/AB 2 / 3 and BP/BC 2 / 3 , compute CQ/CA and AT/AP Given tree tree concurrent concu rrent evians evian s in a triangle, sow tat te tree tree lines lin es obtaied obta ied by joining joi ning te midpoints of te te evians to te midpoints of te te correspondig sides are concurrent. concurrent. ons ider te te medial triangle. HNT onsider Suppose evians evians A P B Q and C R are are concurrent conc urrent at poit poi t T as in Figure Figure 4.2, so tat 6ABC is decomposed decomposed into six small triangles. triangles. If te te areas of 6ART 6B P T and 6C QT are are equal, sow tat in fact fact all six small s mall triangles ave equal areas. make te algebra a little little neater, assume assum e tat units ave been cosen co sen HNT To make so tat te areas of 6ART 6 BP T and 6CQT are eac equal to
nterior and Eterior Cevians
In te statement and proof of Teorem 42 4 2 and in all of te examples we ave cons c onsidered idered so far, evians AP B Q and C R begin at vertices of 6ABC tey cut across te interior of te triang triangle le,, and tey terminate at points poin ts lying on o n te opposite opposi te sides sid es of te triangle triang le.. We We refer refer to suc suc lines as intrior evians. Interestingly, Interestingly, evas teorem almost almost remains valid even if we expand our denition and allow xtrior evians. Tese are lines line s tat join joi n a vertex of a triangle to a point on an extenson of te opposite side, si de, and wic wic tus do not cut across te interior of te triangle. triangle. We sall insist, insi st, owever, tat a evian must go g o troug just one vertex of te triangle. triangle. Oterwise, te evian product would would be te te meanigless expression expression 0/0 For example, A P is an exterior evian if point P lies on line B C but it does not lie on te line segment B C and in particular, particul ar, P is not one of te points point s B or C We sall see tat if evians A P B Q and C R are concurrent, ten te evian product is guaranteed to equal even if not all of te given evians are interior. If it is suitably interpreted and modied, te converse statement also remains true wen exterior evians are allowed. If we know tat te te evian product is trivial, w sall see at under appropriate appropriate additional additional conditions , we can conclude conc lude tat te evians are are concurrent concurre nt To understand unders tand te additional complicati comp lications ons in te stateent and proof of tis part of te teorem, consider cons ider Figure Figure 43 wic sows te tree tree possible possi ble congurat cong urations ions tat we need to consider. con sider. In eac of tese tree diagrams , A P B Q and C R are are evians tat are are concurrent concurrent at a point T On te left, all are interior evians, evian s, but in te oter two diagrams diag rams,, A P is interior wile bot B Q and C R are are exterior. little experimentation sows sow s tat tese are are indeed te te only pos p ossibilitie sibilities,s, except for for a possible pos sible renaming renaming of points, points , and and we observe obs erve tat if tee evians are concurrent, ten te number number of interior evians evian s among tem is necess nece ssarily arily eiter one or or tree. In sort, given tree concurrent con current evians evi ans,, te number of of interior interior evians among tem must be odd ltoug it is pos sible for te evian product to be trivial trivial wen te number of interior interior evians is even, we see tat in tis case, cas e, te evias evias cannot be concurrent. It follows tat some additional assumption must be made
13
CAPTER
CEVAS TEOREM AND TS LATVES T
Q
A
B
p
C
B
R A
C
p
B R
igure
II I
Q
4.3
if we wis wis to conclude conc lude tat tree tree evias are concurrent wen te corresponding correspond ing evian product is trivial. Te appropriat appropriatee additional additional assumption as sumption is preci precisely sely tat te number of interior interior evians sould so uld be odd. Tere is yet anoter complication. onsider, for example, te middle diagram in Figure 43 and imagine point T moving upward along line A P farter and farter from A s T moves, we adjust Q so tat it remains at te intersection of lines B T and AC and similarly, we keep R at te intersection of lines C T and A B t every stage, A P B Q, and and C R are are concurrent conc urrent evians wit a evian product equal equ al to I te limiting situation, situ ation, terefor tereforee , te evian product will still be but te tree tree evians will be parallel, and not concurrent. It follows tat in te general form form of eva evass teorem, paallel te word concuent must be replaced by te prase prase concuent o paallel S ometimes it is i s elpf el pful ul to pretend tat tat parallel parallel lines meet at some imaginar i maginaryy point tat is innitely far far away. Wit tat tat ction, we could say tat tree tree parallel evians are concurrent, concurrent, and we would woul d avoid aving aving to consider an exceptional case in eva s teorem. To To be a little more precise, we imagine tat eac line in te plane contains one extra ideal point (or point poi nt at inity) i nity) in addition to its real points . We We want eve two lines, lines , parallel or not, to ave exactly one point in common, and so we declare tat parallel lines sare s are a common ideal point, but nonparallel nonparallel lines ave distinct ideal points . lso, lso , since we want w ant every every two points to t o determine determine a lie, we create one ideal line consi co nsisting sting of al al te ideal points . ll of tis can c an be be made precis precisee and rigorous rigorous,, and wat results is a extension of te Euclidean plane in wic every two two lines ave a common point and every two points lie on a common line. Tis extended plane, wic is called te projctiv pan, is often useful for for avoiding avoiding annoying annoying special speci al cases and exceptios exceptios.. We can expand our o ur notion of evians a bit furter furter.. y y our denitio deni tion, n, almost almos t every every line troug troug a vertex vertex of a triangle is a evian of tat triangle. triangle . In fact, fact, tere are are exactly exac tly tree exceptions except ions for eac vertex. vertex . Troug vertex A of 6AB C for example, exampl e, te exceptions exceptio ns are lines A B and A C and te line troug A parallel to BC We ave ave explicitly excluded A B and A C from consideration, con sideration, and te te tird exclusi exclusion on is a consequence of our requirement requirement tat a evian troug A must be a line of te form A P were P is some poit of line B C oter tan B or C Of course, te line parallel to B C troug A contains no suc point P and so by our current denition, tis line would not qualif qu alifyy as a evian. evian.
B NTEROR AND EXTEROR EXTEROR CEVANS CEVA NS
133
If we work in te projective plane, ten parallel lines meet at some ideal point at innity In tis situation, te line troug A parallel to B C is a evian, and te corresponding point P is te ideal point on line B C We adopt tis point of view now and declare tat te line troug A parallel to B C is a evian ut tere is a problem problem wen we try to to extend evas teorem in tis situation. situation . If P is te ideal point on BC ow sould we compute te factor B P / P C in te evian product? Te answer is easy: We simply dene B P / PC = in tis case so tat te evian product product becomes reaso nable since te limit of B P / P C is as P moves move s off A R / R B) B) C C Q / Q A) Tis is reasonable to innity innity We can c an now state te fully fully general g eneral form form of evas teorem teorem Cevans of 6A B C whee ponts P Q THEOREM. et A P B Q and C R be Cevans and R le on lnes B C C A and AB A B espectve espectvelly. Then these Cevans Cev ans ae ae ethe concuent o o paallel paallel and and onl o nlyy an odd numbe numbe of of them ae a e nte nteo o and AR BP CQ RB PC QA
=
.
We will only sketc a proof ere, omitting many of te details If te tree evians evi ans are concurrent con current or parallel, parallel , we observe o bserve from from an appropriate diagram tat a odd number of tem must be interior, and in particular, at least one is interior, ad so we can assume as sume tat A P is interior If te tree tree lines line s are actually conc c oncurrent urrent ad not parallel, we can c an assume ass ume tat we are are in te situation situati on of one of te tree diagrams of Figure 43 and we can c an compute te evian product using usin g areas of o f appropri appropriate ate triangle triangless , just as we did in te proof of Teorem Teorem 4 In all cases, cases , we ave ave
Proof.
AR RB KR T B BP KB T P P C K p A C Kp T C C R KC B Q  KC T Q R A KQBA KQ T A
Next, we apply te te addition and subtraction principles principle s for ratios, ratios , as we did in te proof of Teorem 4 (Te appropriate appropriate principle in eac case depends depen ds on wic of te tee diagrams diagr ams of Figure 43 is under under consideation) consideation) Tis yields AR RB
KC A T KB C T '
BP PC

KA B T KC A T
,
and
and tus AR BP CQ
RB PC QA
CR = KB c T RA
KA B T
134
CAPTER
CEVAS TEOREM AND TS RELATVES
as required. In te case were te tree evians are parallel, we get te same conclusion conclu sion eiter by taking limits or by a direct direct argument argument using usi ng similar s imilar triangles triangles (See (S ee roblem roblem 4) To prove te converse, ass assume ume tat te number o f interior evians evi ans is i s odd and tat te evian product is trivial. Tere is noting to prove if te tree lines are parallel, and so we can assume tat two of tem (say, A and B Q) meet at some point T and we use essen es sentially tially te same sam e argument as in te proof of Teorem 4 to sow tat line C R must mu st also als o pass troug T We propose to t o sow s ow tat line C T goes toug R, and so we let R be te point were C T meets meets line A B , and we work to sow tat R and R are are te same point. Now C R is a evian tat tat is concurrent wit A and B Q, and so we know tat te corresponding evian product is trivial trivial.. easoning easoning exactly as in te proof of Teorem 4 we deduce tat AR/RB AR/RB. Next, w e observe tat eiter bot of te evians C R and C R are are interior or else els e neiter neiter is . Tis is because bec ause te total total number of inter interior ior evians in eac of te sets { A B Q , CR } and {A B Q CR } is odd. Te Te rs rs set set con conta tains ins an odd odd nnum umbe berr of interior evians by ypotesis, ypotesi s, and te second set s et contains an odd number since tese tree evians are known known to be concurr co ncurrent ent by construction. construction . It follows eiter tat bot bot of te points R and R lie on te line segment A B or else neiter of of tem does Te fact fact tat R and R must be te te same point is now a consequence con sequence of emma 43 se evas eva s teorem to nd an alteative proof of te fact tat tat te altitudes of o f a triangle are concurrent. co ncurrent. (Se ( Seee Teorem 10)
(4.6) PROBEM.
Observ Obs ervee tat tere tere are are exactly tree tree possibilities possib ilities for our given A B C . Eiter all of te te angles ang les of te te triangle are acute, as in e left diagram diagram of Figure 44 one of te angles (say, A) is obtuse, as in te rigt diagram of te gure, or te triangle as a rigt angle In te latter situation, were we ave a rigt triangle, eac altitude clearly goe s troug te te rigt angle angle,, and so s o tere is really noting to prove in tis c ase. We can c an tus assume ass ume tat one on e of te t e two diagrams of Figure Figure 44 applies, and tus tus te altitudes A , B Q, and C R are are evians tat we want to sow are concurrent We see tat in eiter case, cas e, te number numb er of altitudes altitude s tat are interior is odd, and ence it sufces by Teorem 45 to sow tat te evian product (AR/ (A R/ R B) (B / C) (C Q/ Q/ QA) QA ) is triv trivia ial.l.
otion.
A
H
p
igure 4.4
B NTEROR NTEROR AND AND EXTEROR CEVANS
135
In te left diagram of Figure 44 we ave A R = cos(A) B = cos(B) C Q = a cos(C)
RB = a cos(B) C = cos(C) Q A = cos( cos(A) A)
were, as usual u sual,, we ave written written a , and to denote denote te lengts of te te sides si des opposite opp osite A , B , and C, C , respectivel respectivelyy Eac of of a , and tus occurs once in te numerator an once in te denominator denominator of te evian product, product, and similarly si milarly,, eac of cos(A cos (A)) , cos(B), and cos(C) occurs once in te numerator and once in te denominator Everyting cancel canc els,s, terefore, and te evian evian product is trivial, as required require d In te rig rigtt diagram, were were B A C 90° exactly te same equations eq uations old for for te te lengts le ngts of te te six si x line segments se gments provided pro vided tat we we interpret A as referring referring to te exteo angle of te te triangle triangle at A so tat cos (A) (A ) will be positive po sitive It follows follows tat in tis case c ase too te evian product is trivial trivial
xerc xercses B 4B.l
4B.2
In Figure 45 compute co mpute eac of te te ratios X/Y and BQ/QC in terms of te lengts = AX, s = XB, t = AY, and u = Y C Sow tat te only way tese ratios can be equal is if bot ratios ratios equal 1 and sow tat in tat case, ca se, X Y is parallel paralle l to B C ee co HNT: Tere are two ways to ssee current evians in Figure 45 Tere are tree tree concurrent c oncurrent interior evians evians for for A B C , but tere tere are are also tree concur rent rent evians for for AX A X Y Sow using u sing similar triangles triangles tat if e vians A , B Q , and C R are are parallel, ten te evian product is trivial (See Figure 4) HNT et a , and denote te lengts of te sides of A B C and write write x = B and y = C Express Express eac eac of te six lengts tat appear appear in te evian product in terms of te ve quantities qu antities a , , x and y
A
igure
Q
A
4.5
p
igure
4.6
136
CAPTER
4B.3
4C
CEVAS TEOREM AND TS RELATVES
onsider 6ABC in Figure 47 e vians C R and B Q are concurrent wit line AT wic is parallel to BC If we view A T as a evian of 6ABC in te generalized sense, ten we know tat te corresponding evian product is (AR/RB)(CQ/QA). Sow witout reference reference to eva s teorem tat tis quantity is equal to 1 .
R
igure
4.7
Ceva's Ceva 's Theorem and Anles
Suppose tat tat A is a evian evian in 6ABC were were as usual, usua l, lies on line BC Tis evian determines determines two distances B and C , neiter of wic is zero since is i s not allowed to be one of te points B or C Given tree evians, one troug eac vetex of 6ABC ten, o f course, six distances are determ determined. ined. If we know tese six distances, ten ten by evas eva s teorem, teorem , we can decide weter or not te tree tree given evians evian s are concurrent In addition to determining te two distances B and C, te evian A also determines determines two angles: angles : B A and A C, neiter neiter of wic is zero Given tree tree evians, evians, teref terefore, ore, one o ne troug eac vetex of 6ABC tere are are six angles angle s determined determined For some problems, problems , it would wou ld be usefu us efull to be able to determine determine from from a knowledge of tese six angles weter or or not te tree evians are concurrent concurrent It sould be clear tat it is imposs imp ossible ible to determine te six distances appearing appearing in te evian product from from a knowledge of tese six angles, angles , but nevet neveteles eless,s, and peraps peraps surprisingly, surprisingly, if we know know te six angles, angle s, it is not ard ard to compute te value of te evian product product Te six angles can tus be used to detemine weter weter or not te te tree evians evians are concurent, concurent, and as we sall see, tis is exactly wat we need to s olve a number of interesting interesting problems In fact, te evian product is always equal to te quantity we call te anar Cvan prodct, wic is te product of te tee fractions in te statement of te following result
Suppose that AP B Q and C R ae ae Cevans Cevans n 6A 6 A B C Then the the coe co esspondng pondn g Cevan C evan poduct poduct s equal eq ual to
(4.7 THEOREM.
sin(ACR) sin(BA) sin(CBQ) sin( sin( RCB) sin( sin( AC) sin( sin( QBA) n patcula the thee Cevans ae concuent o paallel and only onl y an odd numbe numb e of of them ae nteo nteo and ths ths angula Cevan poduct poduct s equal eq ual to 1
C CEVA' CEVA'S S THEOREM AND ANGLES A
B
13
A
P
C
igure
C
P
4.8
Te tree factors factors of te evian product are, of course, A R / R B , B P / P C , and C Q / Q A , and we will use te law of sines to express eac of tese in ters of angles. angles . (To review te law of sines, sines , see te discuss disc ussion ion in Section E or refer to Teore .3) We work rst wit te ratio B P / P C , and we begin b egin wit w it te case ca se were te evian A P is interior, interior, as in te t e left left diagram diagram of Figure 4.8 Worki orking ng i n �A B P, we ave ave AP BP , sin( B A P) sin( sin( B) and siilarly, siilarly, in � A C P , we ave ave AP PC sin( P AC) sin(C) If we we solve s olve tese tes e equations for B P and P C and ten divide and cancel A P , we obtain B P sin( sin( BA P) sin(C) sin(C) PC sin( sin( P AC) sin( sin( B) I n te case were A P i s an exterior evian, iti t turns turns out tat we get exactly te sae formula. To see ti s, consider con sider te rigt rigt diagra of Figure 48 and notice tat tere tere is no cange can ge at all in te forula forula tat we obtained from from te law of sines in �AB P. In �P A C, oweve owever,r, we observe tat te angle angle opposite side A P is not te original C ACB AC B , but instead, ins tead, it is te corresponding corresponding exterior angle o te te 80  C . ut triangle, triangle, namely, namely, A C P 1 80 ut te te sine of any angle is equal to te sine sine of its supplement, suppl ement, and so wen we apply te law law of sines , we get te sae s ae forula forula we ad previously. In all cases cas es,, terefore, terefore, we ave B P sin( sin( BA P) sin(C) sin(C) , PC sin(PAC) sin(P AC) sin(B) and silarly, we get sin( C B Q) sin( sin( A) C Q sin( AR sin( in( ACR) sin( sin( B) and RB sin( sin( RCB) sin(A) sin(A) QA sin( sin( QB A) sin(C) sin(C)
Proof.
138
CHAPTER
CEVA'S THEOREM AND TS RELATVES
Wen we multiply te tree ratios ratios to compute te evian product, we get AR BP C Q (LACR) sin si n (LBAP) (LCBQ) si n ((LQBA) RB PC QA (LRCB) (LPAC) sin LQBA) as required
If te tree evians appen to be te angle bisectors of te triangle, it is obvious tat te angular evian product is trivial, trivial, and a nd tis is i s consi co nsistent stent wit te fact tat tat te angle bisectors are always concurrent It is a little more more difcult, difcult, owever, to see tat te ordinary ordinary evian product product is trivial trivial in tis case, cas e, altoug altoug we ave done tat calculation On t oter and, if te evians are te tree medians medians,, it is immediate tat te ordinary ordinary evian product is trivial, but tis is not so obvious for te angular evian product ltoug te two forms of te te evian product always ave equal values, value s, we see s ee tat tey are often often not equally equall y easy to evaluate evaluate For any any particular problem, problem , terefore, one or te oter of tese tese products migt be te more appropriate appropriate tool n application of o f te te angular evian product is te following following surpris surprising ing result result Gven an abta abta �A � A B C buld the theee outwa ou twad dpo pontng ntng tan tan gles BC B C U C A and A B W each shang shang a sde wth the ognal ognal tang tangle le as llustated n Fgu Fgue Assume that
(4.8) THEOREM.
L BA B A W L CCAA V LC BU LAB W and LAC V LB CU Assume futhe that lnes A U B V and CW cut acoss acoss the nteo of of � A B C as n thegue thegue Then lnes AU AU B V and C W ae ae concuent concuent
A
v
u
igure
4.9
C CEVA'S CEVA'S THEOREM AND ANGLES
139
ines AU, BV, and C are intrior evians, and so by Teorem 4.7, it sufces to sow s ow tat te angular evian product product is trivial trivial.. In oter words, word s, we w e must establis establi s tat sin(AC) sin(BAU) sin(CBV) sin( sin( CB) sin( sin( U AC) sin( V BA) For clarity of notation, we write B A C A V , and similarly, we let and denote te measures of te oter two pairs of equal angles, as indicated in Figure 4.9 To compute compute sin( sin( A C ) , we use us e te te law law of sines in �A C to deduce deduce tat tat A C sin(A C) C ) sin( AC) Since Since A C A , we ave sn(AC) A sin ( A ) C
Proof.
and similarly, intercanging te te roles of A and B , we see tat B sin(B ) sin(CB) C and we obtain sin(AC) A sin( A ) sin( sin( CB) B sin( sin( B ) Te ratio A / B can be b e computed using te law o f sines again, tis time in �A B . We ave ave B A sin() sin() and tus A sin() B sin() Substitution Subs titution of ti tiss into our previous formula ormula yields yield s sin() sin(A ) sin(AC) sin() sin() sin( sin( B ) sin( sin( C B) sin() Similar Simi lar reasoning yields sin(BAU) sin() sin( B ) sin(CBV) sin() sin( C ) and sin() sin( sin( C ) sin(UAC) sin(U AC) sin() sin( sin( V BA) sin() sin( A ) and so wen we multiply multip ly tese tes e tree ratios ratios of sines to compute te angular angu lar evian product, everyting cancels and te result result is 1 , as desired
140
CHAPTER
CEVA'S THEOREM AND TS RELATVES
We mention tat tat te ypotesis in Teorem 4.8 tat te tree tree lines all cut across te interior interior of �A B C is not automatically automatically satised, s atised, and it can fail fail Neverteles Neverteless,s, Teo rem 4.8 remains true witout tis assu a ssumption, mption, except tat it may turn out tat te point of concurrence concurrence is at innity, innity,"" wic means tat lines A U , B V , and C are are parallel, parallel, and tey tey are not not really concurrent concurre nt lso l so,, te teorem continues contin ues to old ol d if we build te inwardpointing triangles on te sides of �A B C Only Only minor minor canges in our arguent arguent are needed to prove tese te se and oter even more general forms of Teorem 4.8, but we sall not pursue tese variations variations any an y furt furter er n interesting interestin g case ca se of Teorem Teorem 48 is wen a in te notation notation of Figur Figuree 4 9 Wen tat tat appens appens,, ten of course, te tree triangles triangles attaced attaced to te sides o f � A B C are similar isoscele iso sceless triangl triangles es wose wos e bases are are te sides of te te original triangle triangle Te proof proof of Teorem 48 would be a bit bit sort s orter er if we assumed tis equality of o f angle angless because bec ause ten it would woul d not be necessary neces sary to use u se te t e law of sines sin es in �A B to evaluate te ratio A / B since si nce te ratio would wou ld automatically be 1 in tis tis case cas e lso, lso , even witout te assumption as sumption ta ta te tee tee lines A U , B V , and C cut across te interior interior of of te te original original triangle triangle it is not possible for tese lines to t o be parallel in tis isoscel iso sceles es situation; situation ; tey are necessaril neces sarilyy concurrent Tese lines lin es can be parallel parallel,, owever, owever, if we allow inwardpointing isoscel is osceles es triangles et us consider co nsider te limiting cases ca ses of Teo Teorem rem 48 under te assumption tat a If we let tese angles approac 0, ten points U, V, and approac te midpoi midpoints nts of of sides sides B C, CA CA,, and AB , respec respectitive vely ly It follo follows ws tat tat AU , B V , and C approac te te medians of o f �A B C , and te point of concurrence approaces approaces te centroid centroid of te te triangle Te fact tat tat te medians of a triangle are concurrent can tus be b e viewed as a limitng case of Teorem 4.8 t te oter extreme, we can let te equal angles approac 90• s recedes, recedes , te tree tree lines A , B , and a nd C approac parallelis parallelism, m, wit A and B approacing perpendiculari perpendicularity ty to A C Te limit of line C is i s tus an altitude of �A B C , and te concurrence point approaces approaces te ortocenter of te triangle triangle Te fact tat te altitudes of a triangle are concurrent can tus also be viewed as a limiting case ca se of Teorem Teorem 4.8. Tere Tere is i s anoter anoter case o f Teor Teorem em 4.8 tat w e knew knew previousl previouslyy : wen all tree of a , and and equal 0 and te tree attaced triang triangles les are actually equilateral equi lateral In tat case, cas e, if eac angle angle of �A B C is less tan tan 10 so tat tat we are in te situation s ituation of Figure 4.9, te concurrence concurrence point is te Fermat point point of te triangle, triangle, te unique uni que point were te su of te distances to te tree vertices vertices is i s a minimum (See (S ee Teorem Teorem .4.) s anoter anoter application applic ation of Teorem 4.7, we offer offer te follo following wing
PROBEM. In Figur Figuree 4 1 0, te pedal triangle triangle of acut acutee �A � A B C is � D , and perpendiculars A , B V , and C are dropp dropped ed from from te vertices vertices of te te original triangle triangle to te te sides of te te pedal pedal triangle triangle Sow tat tat lines A U , B V , and C are are concurrent and detene te point of concurrence concurrence y evas teorem for for angles, angles , it is enoug to ceck tat te angular angular evian product is trivial, trivial , and so we want to sow tat sin(AC) sin(BAU) sin(CBV) 1 . sin(CB) sn(UAC) sn(VBA)
otion.
.
.
=
C CEVA' CEVA' S THEOREM THEOREM AND AND ANGES
141
A
D
igure
410
� C is a rigt rigt triang triangle, le, we see tat tat sin( sin( AC ) cos( C) , and and S ince � ly,, sin( sin( U AC AC)) cos ( U A) A ) We recall, recall, oweve owever,r, tat tat since �D � D F is te similar similarly peda pedall trian triangle gle of � A B C, we av avee U A C, wic is immediate rom orollary 43 It follows tat tat sin( A C ) sin( U A C) , and similarly, similarly, te oter oter factors in te angular evian product all cancel, canc el, and te lines line s are concurrent at some point point X Since CA X and A C X are are comple complemen menta tarry to te te equal equal angles angles A U and and C , we ave ave CAX A CX, and and tu tuss � �AX AX C is isosceles isosceles aand nd AX C X Similarly, Si milarly, B X C X , and tus X must be te circumcenter of of �A � A BC A
P
B
Q
X
R
igure
C
s
4.11
Given any vian in � A B C , tere is a natural natural way to construct con struct a new evian from from it Simply reect ree ct it in te bisector bi sector of te te appropriat appropriatee angle of te triangle. triangle. Te resuting line is called te ioona of te te original evian wit respect to te given triange In Figure 411 line A X is i s te bisector of A A , and lines A Q and A R are te images of eac oter upon reection reection in te te bisector bisector AX . In oter oter words words Q A X RAX and A are are reect reections ions of eac ot oter er in in o r equivalently, QAB RA C lso, AP and and P A B AC. In tis AX, an and so so P A X AX an tis situation, A Q and AR A R are isogonals is ogonals of eac oter, oter, as are are A P and A . Note also tat te isogonal isogo nal of any interor interor evian is again aga in an interior evian, evian , and te isogon is ogonal al of an exterior evian is an exterior evian Te following following string result is an immediate immediate consequence cons equence of Teorem Teorem 47 f Cevns A P B Q nd C R of �A B C e concuent co ncuent o pllel then the sogonl Cevns e e lso concuent o pllel
(4.10) COROARY.
142
CHAPTER
CEVA' CEVA'S S THEOREM AND TS RELATVES RELATVES
Observe tat te angular evian evian product of te te isogon is ogonals als of o f tree tree evians is exactly te reciprocal of te te angular evian evian product of te original evians . lso, lso , tang isogonal is ogonalss does do es not no t cange te te number of interior interior evians among te t e tree tree I f A P, B Q, and C R are are concurr con current ent or parallel, ten an odd o dd number of tem are are interior, and teir angular angula r evian product is trivial Te same is terefore true true about • te isogonals isogo nals of tese tese tree evians, evians, and and te result follows follows
Proof.
For example, te isogonals isog onals of o f te medians of � A B are called called te te ymmdian of te triangle, triangle, and tey are are necessarily neces sarily concurrent concurrent since te medians are concurrent. (Te symmedians cannot c annot be paralll paralll because te medians are interior interior evians, and tus te symmedians are interior too. ) Te point of concurrence concurrence of te te symmedians symmedi ans of a triangle triangle is called c alled te moin point Given �ABC coose any point X tat is not on one of te lines AB, BC or CA Te tree lines AX, BX and C X are evians of �ABC, and tey are obviously concurrent at te point X y orollary 49, te isogonals of tese tese evians are eiter eiter concurrnt concurrnt or parallel, and if tey tey are concurrent concurrent at some so me point Y , ten Y is called te ioona conjat of X (In tis situation, it sould be clear tat X is also te isogonal cougate cou gate of Y) For example, example, te centroid centroid and te emoine point are isogonal is ogonal cougates co ugates of eac oter. If X is in te interior of te triangle, ten te tree evians AX B X, and C X are interior, interior, and tus tir isogonal is ogonal evians are interior ence tey are necessarily concurrent since tey cannot be parallel. It follows tat every point in te interior interior of te triangle as an isogonal co ugate, and tat cougate cou gate is also in te interior interior of te te triangl triangle.e. It is easy to see tat te incenter incenter of te triangle triangle is i s te only point in te interior tat is its own isogonal iso gonal cougate co ugate s we ave observed, a point X may may not ave an isogonal isog onal cougate co ugate because becaus e t isogonals iso gonals of te te tree tree evians A X, B X, and C X may be parallel rater tan concurrent. In fact, fact, tis appens appen s wen X lies on te circumcircle circumcircle of �ABC et X be a pont othe than A B o C o n the ccumccle ccumccle of of � A B C Then the sogonl Cevans of A X B X and CX ae paallel
(411) THEOREM.
It is no loss los s to assume tat X lies between points A and B on te circumcircle circumcircle,, LCAY AY = LBAX and as sown s own in Figure Figure 4.1. We ave drawn A Y and BZ so tat tat LC for convenience, we ave labeled tese angles a and and Tus LABX LAB X L C B Z , and for evian A Y is te isogonal of evian evian AX, and evian B Z is te isogonal isog onal of evian
Proof.
x
igure
4.12
C CEVA'S CEVA'S THEOREM AND ANGLES
14
B X , and we will wi ll sow tat A Y and B Z are parallel parallel.. Note tat by orollar oroll aryy 410 we know tat if A Y and B Z are parallel, parallel , ten te isogonal is ogonal of evian C X is guaranteed gu aranteed to be parallel to tese two lines l ines,, and so tere tere is no need to consider con sider it any furt furter. er. Since Si nce te sum of te angle angless in �A X B is 1 8 0 , we see tat weree te second equali equality ty olds because quadrila quadrilater teral al AX AXBC 180 X C , wer i s inscribed in a circle. It follows follows tat Y A B ABZ CAB ABC + C CAB ABC 1 8 0 Since Si nce Y A B and A B Z are supplementary, supplementary, it follows follows tat A Y and B Z are parallel, parallel, as required. In fact, fact, it is not ard to see tat it is only for points poi nts X on te circumcircl circum circlee tat te te tree tree evians isogo i sogonal nal to AX A X , B X , and C X can be parallel parallel.. It follows tat an isogonal isog onal cougate co ugate point is i s dened dene d for every every point X in te plane tat tat is neiter on one of te lines A B , B C, or C A nor on te te circumcir circumcircle cle s we see in Figure Figure 413, if we exclude tese tree lines and circle, te remainder remainder of te plane is divided into ten regions tat we ave labeled a, a', , , c, c', w y , and z .
a
c ' I
a x Fiue 413
little experimentation experimentation sows tat te isogonal is ogonal cougate co ugate of any any point poin t in one of te regions w y, or z is back in te same s ame region region.. (We already already knew tis, tis , of couse, in te case of region region w wic is te interior of of te te triangle triangle ) Te points of regions regions a and a', owever, owever, are intercanged intercanged by te proce process ss of isogonal isogon al cougation, cou gation, as are te points regions and and of of regions regions c and c of regions ltoug muc more is known about symmedians and te emoine point and more generally, about isogonal isog onal cougates co ugates,, we sall give give only one furter furter result in tis direction. a ny point in the t he inteio of of �ABC and let Y be the THEOREM. et X be any cicumcente of the tiangle whose vetices ae the eetions of X in the sides of t he isogona iso gonall con co nugate ugat e of of X with wi th espec espectt to �ABC. �ABC. Then Y is the
It sufces to sow tat AY is te isogonal evian of AX. Similar reasoning would ten ten sow s ow tat B Y and C Y are are te isogonals of B X and and C X , and it would
Proof.
144
CHATER
CEVA'S THEOREM AND TS RELATVES
follow tat is te isogonal isog onal cougate of X Wat we must prove, terefore, is tat LB AX LCA A
B
Figue 4.14
et et and and V be te reections reections of X in A B and AC respectively, as sown in Figure 414 Ten is te circumcenter of a triangle triangle aving aving W V as a side, and it follows tat te perpendicular bisector of W V passes troug troug lso, A B is te perpendicular bisector bis ector of X W and A C is te perpendicular bisector of X V So it follows tat te point A were A B and A C meet, i s te circumcent circumcenter er of X W V and tus A lies on te perpendicular perpendicular bisector of W V Since also lies on te perpendicular bisector of W V it foll follows ows tat A is tis perpendicular bisector bis ector and it meets W V at its midpoint D Next, note tat A WE AXE since A B i s te perpendicuar perpendicuar bisector bisector of WX Tus LE A W LE AX and we see tat LBAX is alf of L W AX We now consider cons ider arcs of te circumcircle of X W V centered at A Writing R to denote te point were A C mets WV we compute tat
W0
L BA B A X LWAX ° LRAD CA as desired
L V 9 0 L VR V R F 9 0  A R D •
Tis section began beg an wit wit a transi transition tion fro from m lengts to angles Star S tarting ting from from evas teorem, teorem, wic is a concurrence concurrence criterio criterionn based o n lengts, we derive derivedd an a n analogous criterion criterion based on angles angles We end wit a transition transition from from angles back b ack to lengts lengts sing s ing te angular evian product we w e derive anoter anoter lengtbased lengtbased concurrence criterion criterion Tis time, owever, owever, we w e consider cons ider te te diagonals of an inscribed exagon et ABC D E F be b e hexag h exagon on nscbed nscb ed n c c cle cle Then the THEOREM. et dagonals dagonals A D B E and C F e concuent nd only only AB CD EF 1 . BC DE FA
C CEV CEV S THEOREM THEOREM AND GLES
145
F
E
Figue 4.15
raw raw lines A C, C , and A, as a s sown in Figur Figuree 415, and note tat we can ca n view te tree tree given diagonals of te exagon as evians of A C y Teorem Teorem 47, tese diagonals dia gonals are concurrent concurrent if and only if sin(AB) sin(CAD) sin(C) sin(BC) sin(DA) sin(CA) It sufces, terefore, to sow tat tis angular evian product is equal to te exagonal exago nal evian ev ian product in te statement of te teorem y te extended law of sines (Teorem 3) in AB , we ave ave AB sin(AB) were R is te radius of te given given circle, and ence is te circumradius circumradius of A B Tus AB sin(AB sin(A B ) R and and similarly, similarly, u sing te extended law of sines i n B C, we get BC . sn(BC) R Tus sin(AB) AB sin(BC) BC and similarly, simi larly, te oter two ratio ratioss of sines tat appear in te angular evian product two ratios of side lengts of te exagon in te exagonal are equal to te oter two evian product. Te angular evian product is terefore terefore equal to te exagonal • evian product, product, and te proof proof is complete
Proof.
xercses 4C.l
Give Givenn an acute angled angled A B C , sow s ow tat tat te te lines j oining A, A , B , and C to te te midpoints midpoi nts of te nearer sides of te pedal triangle are concurrent
146
CHAPTER
4C.2
4C.3 4C.4 4C.5
4D
CEVA' CEVA' S THEOREM AND TS RELATVES RELATVES
Sow tat te only point in te interior of �A B C tat is its own isogonal cougate co ugate is i s te incenter and nd all oter points in te plane tat are teir own isogonal cougates I f �ABC is not no t a rigt triangl triangle,e, sow tat its circumcenter circumcent er and ortocenter are are isogonal isogonal cougat c ougates es I f L C 90° i n �ABC, sow tat te te emoine point o f te triangle triangle lies lie s on te altitude from vertex C If te te emoine point o f �ABC lies o n te altitude from from vertex C , sow tat 90 ° eiter A C B C or L C 90°
Menel Menelaus' aus' Theor Theorem em
Given �ABC let P , Q , and R be arbitra arbitrary ry points on lines line s BC C A, and AB, respectively, and assume assu me tat none of tese points is i s a vertex of te triangle triangle If te points P , Q , and R are someow cosen cose n at random, random, it is possible, pos sible, but igly unlikely, tat te tree tree evians co ncurrentt or parallel parallel Te purpos purposee of evas teorem is to AP B Q, and C will be concurren tell us wen tis miracle miracle occurs occurs Te evians will be concurrent concurrent or parallel precisely wen te evian evian product is trivial trivial and te number of interior evians among t tree tree is odd odd Note tat te condition on te interior interior evians can easily be express expre ssed ed in terms of te te tree tree random points points Tis oddness condition is exactly exactly equivalent to s aying tat tat te number of points in te set {P Q , R } tat lie lie on actual actual sides sid es of te te triangle triangle is odd If we coose coos e P , Q , and R at random, as in te previous previou s paragrap, tere is anoter miracle tat tat could appen appen Te tee points migt be collinear coll inear Note tat tis tis cannot appen if eac of P , Q , and R lies on an actual side of te triangle rater tan on an extension of te side, sid e, and in fact, fact, some experimentation experimentation sows tat for for P , Q , and R to to be collinear, coll inear, it is nece n ecess ssar aryy tat eiter exactly two of tem or none of tem lie on sides sid es of te triangle triangle Tese Tes e two possibili po ssibilities ties are illustrat illus trated, ed, respectively, by te left left and rigt rigt diagrams of Figure 41 were we ave drawn dased lines to indicate te extensions of te sides sid es In oter words, te t e number of member memberss of o f te set {P, Q , R } tat tat lie lie on sides of te triangle triangle must mus t be even for for tese points to be collinear Tis is, of course, te exact exact opposite oppo site of te situation in eva' s teorem classic clas sical al teorem of Menelaus, proved proved about 100 AD, tells us wen te miracle of collinearity appens appens Given tat tat te number of P , Q, and R lying on actual sides Q
"
"
Q "
"
" A
C
P
Figure 4.16
P
D MENELAUS THEOREM
147
of te triangle is even, te condition for for collinearity is tat te evian product product is i s equal to 1 . In oter words, and rater rater amazingly, exactly te same numerical condition co ndition tat is equivalent equivale nt to te te concurre co ncurrence nce of te te evians is also equivalent to te collineariy co llineariy of te points. Te only dierence dierence in te conditions for eva's eva' s and Menelaus' Menelaus ' teorems teorems is tat for te former, former, an odd number of members members of te set { P Q R} lie on sides si des of te triangle, triangle , wile for te latter, an even number do. do . Gven �ABC let ponts P Q and R le on lnes BC CA and AB espectvely and assume that none of these ponts a vete ve texx of of the tang t angle le Then he n P Q and R ae collnea and only only an even ev en numbe n umbe of them le on segments segments B C C A and AB and AR BP CQ = 1 RB PC QA
(4.14) THEOREM (Mna).

We sould soul d mention tat tat i n some geometry geometry books, books , eva's eva' s and Menelaus' Menelaus ' teorems are stated somewat dierently from te way we ave presented tem ere In tose works, te ratios tat occur in te evian product, suc as A R / R B are sometimes considered cons idered to be negative negative numbers numbers . Specically, Speci cally, AR/R AR/ R B is negative if R does not lie between A and B and it is positive oterwise. Wit tat convention, we see tat in te situation of eva' s teorem, were an odd number of te evians are interior, an even number of te ratios AR/ RB B P / PC and C Q/ Q/ QA are negative. In te case of Menelaus Menelaus'' teorem, on te oter and, an odd number of tese ratios are negative. sing tis sceme, sceme , te concurrence condition of e eva va is tat te evian product product sould s ould equal + 1 wile te collinearit coll inearityy condition of Menelaus is tat it sould be 1 ut we prefer prefer to consider cons ider all lengts and ratios of lengts to be positive, and so s o our stateme st atements nts of eva's and Menelaus' Menelaus ' teorems refer to oddness and evenness evenness and not to positvity and negativity. First, First, assume tat tat P, P , Q and R are collinear. We can tus assume tat we are in one of te two situations depicted in Figure 4.1 and in particular, particular, tis forces forces te number of members of te set { P Q R} tat are are on sides of te te triangle to be even. We need to sow tat te evian product is trivial raw A P and C R as sown sown i n Figur Figuree 4.17 and work wit bot diagrams
Proof of Thorm 4.14.
"
B
"
"
"
" A
p
C
C
Figure 4.17
p
148
CHAPTER
CEVA'S THEOREM AND TS RELATVES
simultaneously. simultaneous ly. Observe tat B P and P C can c an be viewed as te bases b ases of B B PR and C C P R , wic ave equal equal eigts . Tis yields yiel ds B P KBPR P C KCPR and and similarl similarly, y, using A P R and and B P R, wit wit bases AR A R and and R B , we w e get A R KAPR R B KB P R We compute co mpute te ratio C Q / Q A twice, twic e, usin u singg two pairs of o f triangles triangle s aving tese te se segments segments as as bases, namel namely, y, C Q P and A Q P , and also C Q R and A QR Q R . Tus Tus C Q  Kc Q P  KC Q R Q A KA Q P KA Q R and we can apply appl y te subtraction subtracti on principle for ratios to get ge t C Q Kc Q P  KC Q R Q A K A Q P  KA Q R Ever Everyt yti ing ng now now cancel cancelss wen wen we we com compu pute te (AR /R B) (B P/ P C) (C Q/ QA), QA ), and tus tis evian product is equal to 1 as desired. We now assume tat te evian product is trivial and tat an even umber of P, Q and R lie on sides of te triangle. We sketc a proof tat tese points are collinear, omitting a number of details details . little experimentation experimentation wit diagra di agrams ms sows tat te only way it c an appen appen tat tat P Q A B , Q R B C , and R P C A i s for for all tree of P , Q , and R to lie li e on sides of te te triangle, triangle , and we know tat is not te case. cas e. We can assume, assu me, terefore, terefore, tat tat P Q is not parallel parallel to A B , and we let R be te point were were P Q meets A B . Our goal now now is to sow tat R and R are actually actually te te same point Te proof proceeds proceeds almost almo st exactly as for for evas teorem. It is easy to see tat R is neiter A nor B and tat it lies between A ad B if and only if R lies between betwee n A and B . Te latter latter assertion ass ertion olds because even numbers numbers of points in eac of te te sets { P Q R } and and { P , Q R R } lie on sides of te te tria triang ngles, les, and and so te numb numbers ers fo for te te two sets set s cannot c annot dier dier by exactly 1 y ypotesis, ypotesis , te evian evian product for points P , Q, and R is trivial, and also, since P, Q, and R are collinear, te new evian product tat tat res results ults wen R is replaced by by R must also be trivial. (We (We are are using te rst rs t part of te proof for tis, tis , of course.) course .) It follows tat A R / R B A R / R B , and tus by emma 43 we conclude conclude ta tatt R and R must be te te same point, point, as desired. • 
,
s an application of Menelaus teorem, teorem, we oer te following following . PROBEM. In Figure 4.18 we ave drawn drawn te tangent lines to te circum circle of A B C at te vertices vertices of te triangle. triangle. Te tangent at A meets line B C at point P , and similarly sim ilarly,, te tangents tangents at B and C meet lines lines C A and A B at poins Q and R, respectively. S ow tat tat points P , Q , and R are are collinear.
D MENELAUS' THEOM
149
p
R
Figure 4.18
Since points P and R li outsid of t circl, non of tm lis on a side of t triangl, and tus Menlaus torm applis It sufcs, trfor, to comput te tre ratios A R / R B B P j PC and C Q j Q A and sow tat teir product product is qual to 1 s we sall s all se, it is esy to xprss ts tr rtios rtios in ts of te lengts b and c of t sids of t original triangl triangl LCB wr t scond qulity follows b W av L B A Q Torem 13 Since L B Q A LCQB w se se tt BAQ CB by It follows tat CQ B CB BQ AQ BA c
ouion
(c j )B Tis yilds C j A and tus C Q (jc)BQ and A (cj A (j ( j c) j (c j ) j c Similrl, A R j R B b j nd B P j P C c j b It follows tat t t vian product wic is t product product of ts tr quntitis i s qual to 1 nd t points r collinr collin r s s ruird
T lin troug P nd R in roblm 41 is somtims clld c lld t oin axi of t tringl It s t proprty tt it is pndiculr to t lin troug t circumcntr nd t moin point ltoug ltoug w w will not prsnt proof of tt fact r r Note tat on or mor of t points point s P Q nd R of roblm 1 cn fil fil to xist It my ppn for instnc tt t tngnt tngnt lin t A is prlll prlll to B C so tt tt P is undnd undnd In tis cs cs it is s to prov prov tt A B A C nd so t tngl tngl is isoscls If t triangl is not ctull quiltrl tn nd R r dnd nd it follows tt R is parlll to B C at is ppning in otr words is tt if P dos not xist tn t tangnt lin t A and lins B C nd R ar parlll, parlll, wil w il if P dos xist ts tr tr lins ar a r concunt at P gin gi n s in v s torm w s tt w tr tr lins ar parlll parll l w v nd of limiting limiti ng cs cs of t situation wr wr t tr lins lin s r concurrnt s anoter application of Mnlus Mnlus torm w prsnt a torm torm of ppus wo lived in te fourt cntury s w will xplain ppus torm is diffrnt in avor from almost evryti evryting ng els in tis book boo k
10
CHAPTER
CEVA'S THEOREM AND TS RELATVES
x
z
Figure 4.19
Suppose Suppose that that points A B and C lie on some so me line and that points X Y and Z lie on line lin e m where the six points are distinct and the two lines are also also distinct Assume that lines B Z and C Y meet me et at P lines A Z and C X meet at at Q and and lines lines A Y and B X meet at R R Then points P Q and and R are are collinea collinea
(4. 16) THEOREM (Papp) (Papp)..
Note tat Figure 419 illustrat illus trates es just one of te many possible poss ible congurations con gurations for appus app us teorem We deliberately deliberately drew drew lines and m to be nearly nearly parallel, parallel , and we placed place d te points A B and C on and X Y and Z on m in te orders sown because becaus e tat arrangem arrangement ent forces forces te tree allegedly collinear co llinear points to remain nearby, and tis keeps te entire diagram diagram from from becoming unmanageably large We stress, stres s, owever, owever, tat te two lines and te six points po ints need nee d not be arranged arranged in tis tis way; tey t ey are completely arbitra arbitrary ry except for te conditions stated i te teorem Note, owever, tat if one of te six points appens appens to be te intersection point of te two lines, lines , ten points P Q and R will not be distinct. Te teorem is triviall true in in tat case because becau se two points are automatic all y collin col linear. ear. appus appus teorem is true true somewat more more generally generally tan we ave ave stated it. Suppose S uppose,, for example, tat lines B Z and C Y appen to be parallel so tat point P does not exis exis (Of course, tis cannot appen if te points are arranged arranged as in Figure 418 but B Z and C Y certainly can be parallel in oter congurations congurations ) We can ten work in te extended projective plane and take P to be te te imaginary ideal point po int at innity, were te parallel lines B Z and C Y meet. To say tat P is collinear wit Q and R in tis situation means tat line Q R contains te ideal point P and tis tells us tat Q R is parallel to B Z and CY We will wil l not prove te versions of appus appus teorem involving ideal points and parallel lies, lies , but it is a fact fact tat tat wit suitable interpretation, interpretation, all suc generalizations are true. true. appus teorem as a dierent dierent avor from from any of te oter results result s we ave studied because it is entirely nonmetric. To understand tis, suppose tat in some particular situation, we w e want to ceck tat te te ypoteses of appus appus teorem old, or we want want to conrm tat its its conclus c onclusion ion is valid ll tat tat we need to do is ceck tat certai certainn points lie on certain lines. line s. Wat is neded for for appus appus teorem are oly te notions notio ns of point, line, and incidence. (We (We ssay ay tat tat a point and line are are incidnt if te te point lies on te line, or equivalently, equivalently, te lie goes troug te point.) Te lengts of line segments and te sizes of angles are completely irreleva irrelevant nt for ap appus pus teorem. teorem. Tere is absolutely noting to be measured, me asured, and so we w e refe referr to appus appus teorem and oter resu results lts of tis type
D MENELA MENELAUS US ' THEOREM THEOREM
as belonging belongi ng to te area of nonmtric omtry Note tat no result involving circles coud be called nonmetric because a circle is dened as te locus of points of some xed distance from from a given point, and distance dis tance is, of course, a metric metric concept. Since Si nce appus appus'' teorem teorem is about points, points, lines, incidence, incidence, and absolutely noting noting proctiv omtry To explain tis else, we can also say tat tis result belongs to proctiv term, we imagine imag ine tat Figure 419 or some oter diagram diagram illustrating appus ' teorem is drawn drawn wit opaque ink on a seet of glass and tat a point source of ligt causes te gure to cast a sadow onto a planar screen Since tis projection from a poit carries points to points and lins to lines, lines , and it preserves preserves incidence, we see tat te sadow of a diagram diagram for for appus' appus ' teorem is again a diagram for appus appus ' teorem In a very roug sense se nse,, proj proj ective geometry geo metry isis tat part part of ordinary (Euclidean) (Eucli dean) geometr ge ometr were te te sadows of diagrams illustrate te relevant information in te original diagrams Tus appus app us'' teorem belongs to t o pro projective jec tive geometry, geometry, but te pons asinorum, asi norum, for example does not because te sadow (projection) (projection) of an isosceles triangle need need not be isoscele is osceless lso, lso , eva's eva' s and Menelau Menelau s' teorems sould probably probably not be considered as belonging to projec projective tive geometry geom etry even toug tou g concurrence of evias evias and colli co llinearity nearity of points poi nts are preserved by proje projections ctions Te nonprojective nonprojective aspect of tese tese results is tat tey conce ratios suc as B P / PC wic are not preserved Note, owever tat cross ratios are preserved preserved ; tis is te ess e ssential ential content of Teorem Teorem 315 ltoug we ave ave just ass a sset eted ed tat tat appus appus ' teorem is unlike anyting tat we ave seen before, before, tere is, is , in fact, fact, a closely relate relatedd result tat appeared appeared in in Exercise 3.7 3 .7.. In tat teorem teorem of ascal, we also als o ave six si x points tat are are j oined by tree pairs of ines intersecting in tee points, and tere too te conclusion conclusi on is tat te tree intersection points are necessarily collinear In ascal's teorem, owever, te six points lie on a circle, wereas in appus' teorem, tey lie on two lines. Note tat, as we stated it, ascal's teorem is neiter nonmetric nor projective because it involves a circle rojec rojections tions of circles are ellipses ellips es and oter oter conic sections sectio ns,, owever, owever, and so if we restate ascal's ascal 's teorem wit an arbit arbitrar raryy conic co nic section instead i nstead of a circle, circle , we get a version of te result tat really is a teorem teorem of projective projective geometry. lso, ls o, it sould be noted tat te trut trut of tis more general teorem follows follows from te special spe cial case ca se for circles by proje projecting cting te appropriate diagram. Te proof of ascal's ascal' s teorem tat tat was suggested s uggested in te int for Exercise 3.7 3 .7 used cross ratios, and in fact, fact, an entirely entirely analogous proof proof is available for for appus appus'' teorem We prefer, owever, to deduce te appus result from Menelaus' teorem Te proof tat follows is not quite complete, owever because it assumes tat certain lines are not parallel parallel One could cou ld prove prove te full full result resu lt from from te case tat we consider con sider by using usi ng limi arguments, but we will omit tis renement better better argument argument tat avoids te consideration cons ideration of suc special cases cas es is also available, available, using u sing tecniques of linear algebra, algebra, but we sall sa ll not pursue tat furter furter ere ene poit to be te intersection of XC wit AY as sown in Figure 40 and an d similarly, similarly, let le t be te intersection of A Y wit B Z ad let be te intersection of B Z wit XC Of course, we are assuming assum ing tat none of te pairs pairs of lines deing dei ng points and is parallel so tat tese tree points are actually actually dened
Proof of Thorm 4.16.
12
CHAPTER
CEVA CEVA S THEOREM AND TS RELATVES RELATVES
igur 4.20
Not tt points P nd R li on lins MN N nd M rspctivel propos to prov tt tt ts tr tr points points r r collinr by appling Mnelaus' Mnelaus ' M N drwn in Figur 0 wit vy ink sall torm to MN sal l comput te vi prouct ( R / R M) (M P / P N) (N Q / ) ad sow tat it is qul to To complt t proo w soul lso ls o cck o cours tat t numbr o membrs membrs M N is vn Tis is crtinly o t st { { , R tt li on ctual sids o MN tru in Figur 0 wr t numbr is but w sll omit t vric ation o tis in otr congurtions o t points, nd w procd dirctly to t computtion o t vin prouct prouct Obsrv tt points A B n C r r collinr n li on lins M MN n N rspctivl nd so w duc by Mnlus Mnlus torm tt tt A M NC NC AM N C
n simi si milrl lrl rom rom t ct ct tt Y nd Z r collinr col linr w gt Y MZ NX YM ZN
v svrl mor tripls o collinr points, n so w gt itionl vin proucts ul to For xmpl R n X r collinr n tus R M NX _ RM N
Similrl rom t collinarit o A and Z nd o C nd Y w get, rspctivl A MZ N AM ZN
nd
Y MP NC YM PN C
D MENELA MENELAUS US ' THEOREM THEOREM
15
Multiplying te last tree equatios equatios and te te rst two, we get A MB N C Y M Z NX R MB NX A MZ NQ Y MP NC = 1 =  AM BN C YM ZN X RM BN X AM ZN ZN Q YM PN PN C Now te six fractions raction s o te rigt cancel wit six of te nine fractio fractions ns o te left, and wat wat results result s is te equation equati on R MP NQ =1 RM PN Q • as we wanted.
xercses 4.l
4.2
4.3
We say tat points X and Y are ymmtric point wit res respect pect to a point M if M is te te midpoint midpoint of segment segment XY X Y . Given Given �ABC suppose tat is a line not parallel to any of its sides, and let P Q and R be te poits of intersection of wit lines BC C A and AB respectively. Suppose Suppo se tat points P and X are symmetric wit respect to te midpoint of segment BC. Similarly Simil arly,, assume a ssume tat symmetric wit wit respect to to te midpoint of C A and and ta tatt R and and Z Q and Y are symmetric are symmetric wit respect to te midpoit of A B. Sow tat tat X , Y , and Z are collinear. I te case tat pass passes es troug a vertex vertex of te triangle, triangle, wic line passes troug troug X, X , Y, and and Z ? et A P and B Q be te bisectors bisectors of L A and L B i n �ABC and suppose suppose tat tat CR CR i s perpendicula perpendicularr to te bisector o f L C were were R lies o n an extension extension of side A B . Sow tat points P Q and R are collinear. col linear. HINT ompute te relevant relevant evian evian product product using angles, angles , as in Teorem 47 Tree concurrent evians AP B Q and CR are drawn drawn in �ABC as sown in Figure 41 and R Q is extended to meet meet te extension of B C at S. pply bot eva's eva' s and Menelaus' Menelaus ' teorem in �ABC to prove tat xz = y(x + y + z) were we ave written B P = x P C = y and C S = as indicated. NOTE Exercise 43 provides an alteative solutio, independent of cross ratios, ratios, to roblem 31 A
p
C
Figure 4.21
154
CHAPTER
CEVAS THEOREM AND TS RELATVES A
Figure 4.23
Figure 4.22 4.4
4.S
46
Figure 42 sows tree tree concurrent interior interior evians in a triangle triangle.. One of tese is divided into tree tree parts by te point of concurrence concurrence and by te line joining te ends of te te oter two evians. evians . If te tree tree pieces ave av e lengts x y and as sown, prove tat xz y(x + y + z) HINT s in Exercise 43 tis can be done by applying bot eva's and Menelaus Menelau s ' teorems to an appropri appropriate ate triangle. triangle. It is not neces nec essary sary to draw draw any extra lines line s or to extend any of te line segmets segm ets in i n te diagram. diagram . s sow in Figure 423 tree concurrent interior evians are drawn in ABC and te intersectio inters ectionn of te evian from from A wit te line joining jo ining te eds o te te oter two evians is denoted T . ine segment S U is drawn troug T parallel to BC as sow, were S lies on side A B and U lies on te evian from B . rove tat S T T U Figure 4 sows tree circles of dierent dierent sizes, sizes , none of o f wic i s inside any any of te oters. oters . For eac pair of tese circles, circles , te two common commo n exterior tangents tangents are drawn, and tese tree pairs of tangents are extended to meet at points P Q , and R rove tat tes tesee tree points point s are collinear. collin ear.
p
Figure 4.24 47
I Figure 45 we see tat ABC and X Y Z are in prpctiv from point p oint S wic means tat lines AX BY and C Z all go troug point S . Te corre sponding side s idess of tese triangle triangles,s, wen extended, extended, meet at points P Q , and and R as sown. rove rove tat tese tese tree tree points are collinear, as indicated ind icated by te dased line. line . HINT pply Menelaus ' teorem teorem a total total of four times times.. In SBC te evian product for te collinear points P Z and Y is trivial. trivial . Two more trivial evian
D MENELAUS THEOREM
155
R
s y
Figure 425
�SC A and SAB. Multiply te tree evian products can be obtained from �SCA products tus obtained NOTE Te result of tis problem is a teorem of Girard esargues (193 1 662). Observe tat like appus' appus' teorem, teorem, esargues' esargues' teorem teorem is entirely entirely non metric and belongs to te area of proje projective ctive geometry
CHPER F E
Vector Metos of Proof
5A
Vectors
In tis capter, we sow ow vectors can be used to prove some striking geometry teorems tat are difcult to prove prove using us ing more conventional tecniq ues. ues . We begin wit a bref review of te deition of vectors ad of some of teir basic properties properties.. First, we w e introduce our notation. To distinguis distinguis vectors from from oter types of o f objects objects,, suc as numbers or points, tey ar generally written like tis: v or or wit a little little arrow over te symbol symb ol or symbols symbol s . ut wat is a vector? plane vector v is simply an ordered pair of real real numbers, wic are called its coordinat We ca tus write v = a b) were te coordinates a and b are are just ju st numbers. ltoug i coordinate coordinate geometry, suc pairs of real numbers are used us ed to denote points in te plane , we prefer prefer not to tink of vectors as points p oints;; vectors are are just ju st pairs of numbers. If one one is worng in tree tree dimensional space and is not conned to a plane, plane, it is convenient to use space vectors, wic are dened to be ordered ordered triples triples a b c) of real numbers. In fact, fact, it is often often useful useful to consider co nsider dimension dimensional al vectors vectors tat look like lik e a l ' a2 a3 . . an , were tere are are coordinat c oordinates es instead in stead of just ju st two or tree and can be ay positive pos itive integer. Our purpose purpos e in introducing vectors is i s to use tem to prove facts facts in plane pl ane geometry geom etry,, owever, and so we will lit ourselves ourselve s encefort encefort to vector vectorss aving aving just ju st two coordinates. coordinates . ectors can be added or subtracted by adding or subtracting te corresponding coordinates. Tus if v = a b) and = c d) we ave v = a c b + d) ad v = a c b d ) . lso, lso , we can mltiply vectors vectors by scalars sc alars simply by multiplying eac coordinate by tat scalar. sc alar. ( caar is otig but an ordinary ordinary real number.) number.) I z is scalar and v = a , b) is a vector, terefore, we write zv to deote te vector za z b ) . Many of te usual rules of aritmetic also old for vectors. For example, te commutative commutative and ass associative ociative laws are valid for vector vector addition, addition, and two distribu d istributive tive laws old ol d for addition and scalar sc alar multiplication. multiplication. (Te two distributive distributive laws are are tese: tese : (y z ) v y v + zv and y ( v + ) y v y w , are vectors. vectors.)) lso, lso , te vecto vectorr (0 0) wic were y and z re scalars nd v and are is called te zro vctor, beaves very muc like te number 0 in ordinary aritme aritmetic tic:: If v i s any vector and z i s any scalar, ten v + = v and z =
156
A VECTORS
17
To relate vectors to geometry, we represent eac vector as an arrow in te plane To be specic, speci c, let us assume as sume tat our our plane comes equipped wit wit a coordinate system sys tem so tat tat eac point P can be described as an ordered ordered pair pair ( y ) . Of course, course, ( ( y ) looks just ju st like a vector, but we refus refusee to tink of it as a vector it is just ju st a way of naming te point P . Suppose Suppo se now no w tat we are are given a vector v a b) et P be any point in te plane and suppose tat its coordinates are ( y) If we let Q be te point wose coordinates are ( + a y + b) ten we can c an tink of te te vector v as instructions about ow to get from point po int P to point Q: Go a units rigt and b units up. Of course, if a is negative, we actually move left, and if b is nnegative, egative, we move down. If we draw an arrow from P to Q wit tail at P and ead at Q , ten we tink of tis arrow as being bei ng a picture of te te Often we tink of te arrow arrow from from P to Q as actually vector v and we write P V Often essential ential to remember remember tat tat being te vector v, but tis can be dangerous because it is ess te point P was c osen arbitrar arbitrarily ily it was not in any sense sens e determined determined by te vector V unambiguously determin determined. ed. Te vector vector v is utt of course, once P i s cosen, ten Q i s unambiguously u represented represented by innitely many diff different arrow arrowss in te plane : one for eac coice c oice of te te tail point P We can tink of any any one of tese arrows arrows as being being a picture or representation representation of v, and we sall see tat all of te te arrows tat represent represent v are parallel and ave equal lengts. Eac of tem can be obtained from from any of te oters by a translation, translation , wic is a motion in te plane pl ane witout rotation rotation.. We ave a sligt problem if v is te zero vector (0 0) since in tat case, te points P and Q are identical, and we cannot draw an actual arrow from P to Q. Neverteless does give instructions ins tructions for ow ow to get from from P to P and we can at east imagine a corresponding correspo nding arrow wit zero zero lengt and no particular directio direction.n. Given two points P and Q and an d an arro arrow w wit tail at P and ead at Q, we can can reconstruct rec onstruct te vector v P by subtracting te corresponding corresponding co ordinates of P (Xl YI ) and Q (X Y) Tus P (X Xl Y YI) , and we see tat t at every arrow we can draw repres represents ents some vector Note tat we need te arrow from from P to Q and not just ju st te te line segment P Q , because becau se we need need to know wic point is te ead ead and wic is te tail so tat we can subtract te tail coordinates coordinates from from te ead coordinates, and not vice versa. In fact, we see tat Q P Wat is te geometric meaning of vector addition? Given vectors v and W we represent v as an arrow from P to Q, were P is arbitr arbitrary ary ltoug ltoug we can c an represent W as an arrow wit any starting point (tail) tat we like, we coose coos e to draw W starting from Q, and we write W Q We ave tus placed te arrows representin g v and w wit te ead of v at te tail of W It is easy ea sy to see tat ta t te arrow arrow from from P to R represents v + W In oter words, we ave te vector equation P + Q We can also tink of tis tis in te following way way Te instructio ins tructions ns for going goi ng from from P to R are are rst to go from P to Q and ten to go from from Q to R We ave already remarked remarked tat any given vector can b e represented represented by b y innitely many different different arrows arrows . Given Gi ven te four points point s P Q, R and S suppose suppos e it appens tat We entioned previous previously ly tat in in tis case, cas e, line segments P Q and R S must P be equal and parallel. To see wy tis is true, consider Figure were we ave drawn rigt rigt triangles P Q X and RSY wit orizontal and vertical arms and wit our given equal vectors vectors as ypotenuses ypotenuse s . (We really really sould say, of course, course , tat te arrows arrows
=
18
CAPTER
VECTOR METODS O PROO s
I Ib I
R
a y
p
Figure 51
representing representing te vector are te ypotenuses, ypotenuses , but it is convenient to talk about te aows aows as toug tey actually actually were te vectors. vectors.)) (a b) = If we write write we see se e tat P X a R Y and X Q b YS = (a and tus te two rigt triangl triangl es are congruent by SS S S . (For simplicit simpl icity,y, we are working in te case were te coordinates a and b are positive, but tis tis is not no t really really essential essential ) I t follows follows tat te lengts P Q and R S are equal. I n fact, fact, by b y te ytagorean teorem, teorem, we see se e tat te lengts of and Q are bot equal to a + b Tis sows s ows tat two two arrows arrows represen ting te same vector must ave equal lengt s . ut u t more more is true. true . We We ave ave = + P+ and if we subtract te equal vectors from bot sides, we deduce tat Q It follows from wat we just proved tat te corresponding arrows ave equal lengts, and so we can write Q S P R We conclude tat quadrilateral quadrilateral P Q S R is a parallelogram, parallelogram, and ence P Q / I R S. Tis sows sow s tat all arrows arrows representing te same vector are are equal and parallel, as we claimed c laimed onversely, onversely, it is not ard ard to see tat two arrows arrows tat are equal, parallel and poit in te same rater tan in opposite direcions correspond correspond to equal vectors . Finally, we mention tat te geometric signicance si gnicance of multiplic multiplication ation of a vecor vecor v by a positive scalar sc alar z is tat an arrow representin representingg z points in te same s ame direction as as an arrow representing V but it is sorter s orter tan, equal to, to , or longer tan te original arrow according accordi ng to weter z is less l ess tan equal equal to, or great greater er tan tan More precis precisely, ely, te lengt of zv is exactly exactly z times te lengt of v If te te scalar sc alar z is negative, te direction of te vector is reversed, but oterwise , we w e get te same srinng or stretcing stretcing eect as wit a positive scalar. sc alar. For example, an arrow representing representin g 3 v as tree times te leng of an arrow arrow represe representing nting v , but it points in te opposite direction. direction . To see some so me furte furterr examples examples,, suppose su ppose tat P, Q, R and S are four four points point s lying lyi ng in tat order order along a line and assume assum e tat tey tey are equally spaced so tat P Q Q R = R S. and = Ten S ome of te oter equation s tat we can write 3P in tis situation s ituation are and =
5B
Vecto ect o and Geoety Geoe ty
For convenienc in applying vector vector tecniques to geometry, geometry, we introduce a notational ortcut We uppoe uppo e tat ome point , wic we cll te gn, as been selected in te pln and we old ti ti point xed we will se, it i not usually nece ss ssar aryy to ow te actual poition of ti point, but sometimes it i po p osible sible to simplify a proof proof by
B VECTRS AND GEMETRY
19
coosing coos ing te origin in some s ome particularly particularly clever way Te notational sortcut to wic we referred referred is tat a vector v ector of o f te form form wit tail at point , will simply s imply be written as In oter words, words , wenever we never a vector is named by a single point rater tan by a pair of points , we assu as sume me tat te tail of te corresponding arrow arrow is at te origin and tat te te ead is at te named point point + Given two points A and B in te plane, we know know tat and using B From tis, te notational sortcut just described, we can rewrit rewritee tis as + tis, B  and ence any vector named we get named by two points can c an be described de scribed as a dierence dierence of two two vectors, vectors , eac of wic is named by a single si ngle point poi nt Notice tat te correct way way to do tis is always a lways ead nus tail tai l Te vector from from P to Q , for example, example, is Q  P We mention tat one way to prove tat two two points point s P and Q are actually actual ly identical identi cal tiss is te zero zero vector precisely precis ely wen is to sow so w tat P ut P Q  P, and ti Q P. In oter words, to sow tat P and Q are te same point, it sufces to sow tat te vectors P and Q corresponding corresponding to tese points are equal
O
If point M is te midpoint of line segment AB sow ow to expres expresss te vector in terms of and B.
(.1) PROBLEM.
To get to M from A we need nee d to travel exactly alf of te way from A to B Tis can be expressed in vector language by writing and using usi ng our notational sortcut, we can rewrit rewritee tis as Tus (B and a bit of algebraic algebraic simplication simpli cation yields + (B + B) •
utn.
Te result of roblem 5 is useful us eful and easy to remember It says say s tat te vector corresponding to te midpoint M of line segmet A B is exactly te average of te vectors and B, correspondin correspo ndingg to te endpoints of te segment segment nd note tat we can make make tis statement witout knowing wic point was selected sel ected to be te te origin origi n s our o ur rst example of o f a geometry geometry proof using usi ng vectors, we give anoter argument argument tat sows tat te tree tree medians media ns of a triangle triangle are concurrent
The medians of of A B C are onurrent at a point G that lies two thirds of the way along eah median moving from a vertex to the midpoint of the opposite pos ite side. Furthermor Furthermo re +B+
(.) THEORM.
C
We compute co mpute te vector corresponding correspon ding to te point G tat lies two tirds of te way along median AM were M is te midpoint of BC. y roblem roblem 5 we know tat (B + and we ave
Prf.
C,
2 2 2 G  A A G A M M  A B + C  A . 3 3 3 2
C
Some easy ea sy algebra now yields tat oter words, words , te vector + B + In oter corresponding to te point po int two tirds of te te way along median A M is te average average of te te tree vectors corresponding to te vertices vertices of te te tringle Similar Si milar reasoning sows tat te vector corresponding to te point two tirds of te te wy along eac of te oter two medins must lso l so be te avrag avrag of te tre tre vectors corresponding correspo nding
160
CAPTER
VECTOR METODS O PROO
to te vertice verticess Te vectors corresponding corresp onding to te points poi nts two tirds of te way along te tree medians are terefore equal, and it follows tat tese tree points re identical Tis completes co mpletes te proof • One disadvantage dis advantage of tis metod of proof proof tat te medians of a triangle are concur rent is tat in order to use it, we ad to know in advance tat te point of concurrence lies two tirds of te way along eac median If we adn t already already known or guessed guesse d tis fact, fact, we could co uld not ave found found tis proof On te oter and, once o nce we know wat we are trying trying to prove, te metod works works purely mecanically; we do not need to be clever We simply compute te vectors corresponding to te tree points tat are two tirds tirds of te te way along al ong te tree medians medians,, and tese tree vectors turn out to be equal equal If te teorem is true, ten in some som e sense, sen se, tis metod of proof proof must work ere is anoter easy example et ABCD be any quadrilate quadrilateral ral and let le t W, X, Y, and Z be te midpoints of A B , B C, CD, and D A, as sown in Figure 5 Give a vector proof tat W X Y Z is a parallelogram
(5.3) PROBLEM.
A
D
B
Figure 52
ecall tat tis is a fact fact we ave seen s een before before Our previous previous proof required one ittle trick: raw AC Ten WX is a line joining te midpoints of two sides side s of LABC, and tus W X A C and WX = iAC. Similarly, Z Y I AC and Z Y = iAC, and ence W X and Z Y are are equal and and parallel, parallel, and te result follows
We want to sow s ow tat since tat will imply tat = W X is bot bot parallel and equal to Z Y , and te te result will follow Te given data are are te arbitrary four points A, B C and D and so we will work to express and in terms of and D We expect tat we will get ge t te same express expre ssion ion for bot and and ifif we do, tat tat will complete complet e te proof and X = We ave = + Tus
utn t Prbm 5.3.
C, i
i C
1
1 ( ) 1 ( ) WX=XW= B+C  A+B = CA . �
Similarly,
ZY = Y Z = 
(
.


)
.


2 C+D D+A = 2 CA , 1
(


)
1
(
.
)
1
(


)
B VECTORS AND GEOMETRY
and since sinc e tis is i s te same expres expression sion tat we obtained for for plete
161
te proof is com
•
We give give one more example of o f a problem tat can be solved solv ed by tis metod meto d Given ABC we construct con struct RST by tang points R S and T on te sides of te original triangle, triangle, as follows follows oint oint R lies one tird of te way from A to B along AB point S lies one tird of te way from from B to C along BC and point T lies one tird of te way from C to A along CA Now repeat tis process starting wit wit R ST and obtain XY Z, as sown in Figure 53 Sow tat XY Z r C AB and sow tat te corresponding corresponding sides of tese two t wo triangles are are parallel
(5.4) PROBLEM.
A
S
Figure 53
To solve sol ve roblem 4 and oter related problems, it is convenient to ave a general metod for determining te vector corresponding to te point obtained by moving a specied fraction of te way along alon g a given line segment segme nt A B
y
y
y
et be a real number numb er with 0 < < 1 and suppose that X is the point poin t lying lying ofthe way from A to B along seg s egme ment nt A B Then he n X ( 1 + B
(5.5) LEMMA.
y
y y
y
For example, suppose suppos e tat 1 /2 Ten X is te point tat lies alf of te te way from A to B and so X is te midpoint of segment segment A B. In tis tis case, te lemma asserts as serts tat tat X i + i B and not surprisingly, surprisingl y, tis agrees wit our earlier earlier formula formula for dpoints dpoint s lso ls o note tat as approaces 0, point X approaces point A and so X sould approac approac and tis is consistent con sistent wit te formula formula given in te lemma since 1 approaces 1 as approaces O Similarly, as approaces approaces 1 , we see tat X approaces B and tis is also als o consistent consi stent wit te lemma
y
y
Prf f Lmma 5.5.
compute tat as required required
y
We see tat
y
y and tus X y (B We
•
162
CAPTER
VECTOR METODS O PROO
Te given data are te points A, B, and C, and so our strategy strategy is to express expre ss te vectors along al ong te sides sid es of X Y Z in terms of B, and First, since s ince R is i s one tird of te way way from from A to B , we see by emma emma 55 tat Si nce X lies l ies one tird of te way from from R R + B Silarly, B Since to , emma 55 yields
utn t Pbm 5.4.
C A
S
A
C
A bC
little elementary algebra now yields X + B + Now tat we ave a formula for X we can see wat te foula for Y must be witout doing do ing any real real work, wic is i s te preferred preferred metod To get te formula formula for Y, simply move around te tria triangle ngle and replace replace A by B, B , B by C, and C by A Tis gives Y B + + and we ave 4 4 1 4 4 1 XY Y  X B + c + A  A + B + C 9 9 9 9 9 9 1 1 1 C  3 A 3 AC 3
C b A,
X
Since Sinc e te vector I is one tird of te vector we know tat tat te corresponding arrows are parallel paralle l and tat te former former as one tird te lengt of o f te latter Tus XY CA an and XY C A Similarly, Similarly, eac side of X Y Z is parallel parallel to te te corre corre sponding side of CA C A B , and eac eac side side of X Y Z as lengt lengt equal to one ti tird rd of te lengt of te correspondin correspondingg side of CA B Tus X Y Z CA B by SSS, and te proof is complete •
We mention tat if we ave ave any two triangles suc tat te tree sides side s of one are parallel to te tee tee sides side s of te oter, ten ten te triangles are automatically automati cally simi s imilar lar Tis follows fairly fairly easily by
xercses 5B 5B.l
5B.
5B3
Give a vector proo prooff of te fact fact tat tat if X and Y are are points on sides s ides A B and AC of A BC and and AX/ AB A Y / AC, ten X Y BC In te situation si tuation of roblem roblem 5.4 we saw tat te te sides s ides of X X Y Z were were eac eac one tird of te sides of te te similar si milar C A B , and it follows follows tat te area area of X Y Z is one nint of te area area of te te original original A A B C ecall ecall tat we obtained obtained X Y Z by applying a cetain procedure procedure twice to te starting triangle triangle Te rst application yielded R R , and ten ten wen te te proced procedure ure was was applied to to R , te result result was X Y Z Tis suggests sugge sts,, but does not prove, tat tat te procedure always yields a triangle wit one tird of te te area of te te triangle to wic it is applied rove rove tat tis is true Fix a number wit 0 < < 1 Start Starting ing wit wit A B C, construct construct R by taking R to be of te way from A to B on AB A B , and similarly, similarly, take take and and to
C DOT PRODUCTS
y
y
16
be of te way along sides sid es B C and CA. If 1 /3 , terefo terefore, re, tis is exactly exactly backward version version"" of tis process to te process of roblem 54) Next, apply a backward R to obtai obtainn XY Specically, Speci cally, take take X to be of te way way from from to R on R , and similarly, take take Y and Z to be of te way from T to and and of te way from R to T , respectively Note tat even even if 1 /3 , tis does does not not yield te te same w e ad in roblem 4) Sow tat X Y Z is similar s imilar to te original original XY Z as we triangle wit an appropriate ordering of te vertices vertices and tat correspon corres ponding ding sides s ides are parallel parallel
y y
5C
y
y
Dot Do t Product
Most readers of tis tis book are probably familiar familiar wit te dot product · of two vectors vecto rs and If we write for te ordered pair (a b) and for te ordered pair ( d) ten te dot product · is dened to be te te scalar ot products products,, wic are sometimes called c alled scalar sc alar products, can c an also be dened in tree tree or more more dimensions dimensi ons,, and te rule is te same in all cases: Multiply te corresponding coordinates and ten add te results. results . Tus in tree tree dimensions , if (a a2 , a3) and (bl b2 b3) we ave · albl a2 b 2 a3b3.) It is easy to ceck tat te commutative and distributive dis tributive laws old for dot products In oter words, words, if , and are any tree tree vectors, we ave ave te following following : · • and also · ) · · Note tat tat in te last equation, te plus plu s sign si gn on te left left represents represents vector addition, but te plus sign on te rigt represents ordinary scalar sc alar addition etuing to vectors in te plane, we investigate te geometric geometric signi s ignicance cance of te te dot produc productt First Firs t we consider cons ider te dot product of a vector vector wit itself. its elf. If (a (a b), b) , we see tat · a 2 b 2 , and we sould recognize tis as te square sq uare of te lengt of an arro arrow w representing representing Of course, we are appealing to te ytagorean ytagorean teorem ere) ere ) It is custom cu stomary ary to to use us e te absolute absolu te value notation to represent te lengt of a vector, vector, and tus we can write V · v 1 2 . Note tat if P and Q are points, points , and if as usual us ual,, we write te line segment segm ent tey determine, determine, we w e can write P P Q P Q to denote te lengt of te Now consider ABC and, as usual, let a , b , and denote te te lengts of sides BC AC and AB respectively ecall tat te law o f cosines co sines tells us u s tat a 2 b2 2 cos (A). Write and so tat we ave • 1 2 (Ac) 2 b 2 , and simila similarlrly, y, 2 . Since we see tat W and tus  )  ) 2 B C ) 2 a 2 If we compute te left side of te previous equation using te commutative and distributive dis tributive laws for for dot products, products , we get • · · w w w w w
X
X
164
CAPTER
VECTOR METODS O PROO
We now ave cos A) a 2 b + e • ) b2 + e 2 be cos and we see tat · be cosA) cosA ) . Since Since b and e , we ave ave a geoetr geoetric ic interpretation of te dot product of two two vectors : It is te product of teir lengts ties t ies te cosine of o f te te angle between te. In particular particular,, if and are are perpendicular perpendicular vectors vectors ten t en since cos co s (90) 0 we see tat · O onversely onversely,, if and are are nonzero, nonzero, ten ten 0 , and and tus tus if · 0 te only only possibili poss ibility ty is tat tat cosA) cos A) O Nonzero vectors are perpendicular, terefore, if and only if teir dot product is zero. (5.6) PROBLEM.
concurrent.
se vector etods to sow tat tat te te altitudes of A B C ust be
et H be te intersection of te altitudes fro A and fro B so tat our task is to sow tat H also als o lies on te altitude altitude fro fro C. C . If H is te point C tere isis noting to prove, and so we can c an assue assu e H is dierent fro fro C, and we w e need to sow tat tat line C H is perpend perpendicu icular lar to A B . Since te vector vectorss e and are nonzero, it sufces sufces to sow tat e · 0 and so we want c · B i O Since Sinc e we d o not ave a forula forula tat tat expresses in ters ters of i, B, and C, we cannot coplete te proof siply by plugging and coputing, and so we need to be a bit ore clever. y te distributive law, te equation we need to prove is equivalent to · B B i C · B i . ut we know tat H lies on te altitude fro A, and fro tis infoation, we deduce tat i) · (B C) 0 and so · B B c i · B c Since Sin ce H also als o lies on te altitude altitude fro B, B , siilar siil ar reasoning yields · C i B · C i. If we we add tese tes e two equations equa tions and use te distributive law a few few ore ties and te coutative law twice, we get B i = i B B c + B c c i =iBic+BcBl B. C i. C = CBi as desired.
utin.
8
8
8
8
8 8
8.
se vectors to sow tat tat te circucenter circucenter of A A B C is collinear wit te ortocenter and te centroid centroid..
(5.7) PROBLEM.
C DOT PRODUCTS
16
=
ecall tat te circumcenter te centroid G and te te ortocenter H actually actu ally lie on te Euler line, line , and te point lies on te opposite side of G from H , and H G 2G . us e tis advance knowledge of te relati relative ve positions position s of H, H , G and to nd a We will use proof, proof, but we stress s tress tat te te proof can stand alone alone Since we ave te freedom to allow our origin to be any point poi nt in te plane, plane , we can c an take take it to be be te point tat we we know will tu out to be te circumcenter. ut of course, course , we do not assume tat tis point, wic w ic we will call is te circumcenter; we ave to prove tat. tat. Following tis strategy, strategy, we coose co ose te origi originn as follows. follows. If H and and G are te te same point, we let be tis point too Oterwise, Oterwise, we coose on line H G on te opposite side of G from H, and alf collinear wit wit H and G it sufces to sow s ow tat tat as far from G as H is Since is collinear distanc es actually is te circumcenter We need to sow terefore, tat te tree distances A, B , and and C are are all all equal. equal. ecause eca use of te te way we constructe con structedd te origin we ave 3 . sing si ng our notational sortcut, we can rewrite rewrite tis tis equation as 3 + + were te second equality follows from Teorem .2. Wit tis tis particular particular coice of origin origin,, terefore, we ave perpendicul cular ar to B C + Since AH is perpendi tis yields · o • • + c · c • · and tus and we ave ut recall tat C, and so we ave and ence is te distance B . Similarly, Similarly, proved tat B C, and tus is equidistan equidistantt from from B and C. S imilarly imilarly,, is equidistant equi distant from from A and C , and tus really is te circumcenter of ABC as • desired.
utin t Prbm 5.7.
= H= =A
H  A = E C = H A C  E = E E = E E  C C = = = = I E 2 E E C C I C 2 , l E C I E = = E C =
xercses 5 se.1
E C,
In Figure 5.4 two squares sare vertex and line segments AC and BD are drawn connecting vertices of te two squares. Te midpoint P of AC is constructed, constructed, and line P is drawn drawn and extended extended to meet BD at Q rove rove tat P is perpendicular perpendic ular to BD H Take te origi ori ginn to be at and sow tat · ompute P · ( ctually, tere tere is more going on ere We sall see in Exercise Exercise F. F.11 tat, in fact, BD 2P.
E
A
=
B
Q Figure 54
D
= E C
166
CAPTER
5D
VECTOR METODS O PROO
Checkerboard
So ar, ar, we ave done do ne almost almos t noting wit vector tecniques of o f proof tat could not easily be done witout vectors. vectors . (ltoug one must be clever cle ver to to nd a vectorfree vectorfree proof o Exercise Exercise 5. 5 . 1 , once one nds it, it is not long long or difcult. difcult.)) In tis section, and even even more in Section F, we will demonstrate ow powerful tese vector metods o proof really are. A
A
A
D
D
D B
B
c
c
B
c
Figure 55
ABC D be a convex quadrilateral. In oter words, all of its angles are less et ABCD tan tan 1 80 Now divide divide eac side of AB C D into n equal parts parts,, were n is some xed positive pos itive integer, integer, and join jo in te corresponding corresponding points, as in Figure 5 . , to orm a cris crisscross scross patte tat we sall call an n n chckrbard (Tis is denitely not standard nomenclature.) In Figure ., for example, we ave drawn 2 2 3 3, and and 4 4 ceckerboards, ceckerboards, all based on te same quadrilateral. quadrilateral. onsider a 2 2 ceckerboard. We know tat te midpoints midpoi nts of te te four sides of te quadrilateral ABCD are te vertices vertices of a parallelogram parallelogram by roblem 55 3 , for example. example. ABC D are Te two crossing line segments of te 2 2 ceckerboard ceckerboard are te diagonals diagonal s of tis parallel parallelogram, ogram, and ence tey bisect bis ect eac oter, oter, and tus eac of te six line l ine segments tat make up a 2 2 ceckerboard ceckerboard is cut into two equal pieces . We are refe referr rring, ing, of course, course, to te four four sides of te original quadrilater quadrilateral al plus te two crisscros s lines lin es.. Similar Simil arly, ly, but muc less les s obviously, obviou sly,__ eac of te te eigt line segments tat make make up a 3 3 ceckerboard is divided divided into tree equal pieces . (We will prove prove tis in a moment.) moment .) More generally, generally, an n n ceckerboard is made up of 4 + 2(n  1 ) line line segments, segments, and it tus tus out tat tat eac of tese tese segments s egments is divided into into n equal pieces. Of course, by te denition of a ceckerboard, ceckerboard, we know tat eac side s ide of te original quadrilateral quadrilateral is divided di vided into n equal pieces te surprise ere is tat te 2n 2 crisscross segments are also equally divided. Eah o o the 2n + 2 line segments that omprise an n n heker board is ut into n equal piees pie es
(5.8) THEOREM.
ecall ecal l tat eac of te four sides of quadril quadrilatera aterall ABC D is divided into into n equal parts. oint P in Figure .6 . 6 is one of te te division division points on side AB, and Q is te corresponding division divisi on point on side DC Ten P Q is one of te te crisscross line segments, segments , and w ave A P AB kn D Q DC, were k is some s ome integer suc tat 0 < k < n Silarly Silarly,, R and and S are corresponding corresponding division points on sides s ides A D
Prf.
D CHECKRBOARDS
67
Figure 5.6
=
and BC, and tus R S is a crisscros crissc rosss line segment and we we ave AR AD = ln BS BN, were is an integer wit 0 < < n . We need to sow tat P Q cuts R S at a point tat tat lies exactly kn o te way rom R to S as we move along R S and tat tis intersection point lies exactly n o te way rom P to Q along P Q O course, we ave to prove prove tis or all coices o k and l For notational simplicity, s implicity, let u s write a kn and f ln. y emma emma 5 5, tereor tereore,e, we w e obtain te ollowing vector descriptions o te te points P, Q, R, and S = (1 a) + aB Q (1 a) + aC R = (1 f) + f D = (1 f)B + fC Now let X be te point on R S tat we expect is te point poi nt were P Q crosses tat lies lie s o te way wa y rom rom R to S along RS RS In oter words, X is te point tat Similarly, Similarly, let Y b e te te point o n P Q tat tat we expect lies on R S so tat Y lies f o te way rom P to Q along P Q Our goal is to sow tat X and Y are are te te same same point, and we proceed by nding ormulas tat express X and Y in terms o te given points A, B, C and D . s is usual us ual wit tese tese vector proos, we simply s imply need to do some calculations c alculations and ten ten compare compare te results results . We expect, expect, o course, course , to get identical identic al ormulas ormulas or X and or Y We compute X = (1 a) R + as = (1 a) (1 f) + f + (1 f ) B + fC f C = ( 1 a ) ( 1 f ) + a (1 f) B + afC + (1 a) f Similarly, Y = (1  f ) + f Q = (1 f) (1  a) + aB aB + f (1 ( 1 a) + aC (1 a) (1 f) + a (1 f) B + afC + (1 a) f us X as w xpecte, and nce X and Y are te same point point Since S ince tis te proo is complete • point poin t must be te point o intersection o P Q and R S, te
=
S
Y
=
=
168
CHAPTER
VECTOR METHODS O PROO B v
A
c
Figure 5.7
p
B
D
Figure 5.8
Now consider con sider wat appens if we remove remove one row and one column colu mn of boxes from C D , and we a ceckerboard. ceckerboard. In Figure 5. 7, we ave drawn drawn a 5 5 ceckerboard A B CD focus on te smaller quadrilateral U V C , as indicated. y Teorem 5 8 , we know know tat tat all of te te pieces on eac crisscross line of te origin original al ceckerboard ceckerboard are equal, and so s o we see tat U V and U are are eac divided into four four equal pieces. It follows follows tat tat U V C is a 4 4 ceckerboard. Te same ting clearly cl early works in general: general : We We can create an ( 1 ) ( 1 ) ceckerboard ceckerboa rd from from an ceckerboard ce ckerboard by delting delting te rst row and rst column of boxes. boxes . Now No w we come co me to te amazing part of te teory teory of ceckerboards. ceckerboards. To introduce introduce tis topic, let us rst consider a 2 2 ceckerboard. ABC D be a 2 2 ceckerboard, et ABCD ceckerboard, as sown in Figure Figure 5 8 , were two of te four four boxes boxe s ave been saded. saded . Sow S ow tat te saded area is exactly e xactly alf of te te total area of te te ceckerboard. ce ckerboard.
(5.9) PROBLEM.
ABC D, as et P, Q, R, and S be te midpoints of te sides of quadrilateral quadril ateral ABCD, sown, and let be te point poi nt were P R meets Q S. raw te line segments joining X ABC D to A, B, C, and D and note tat tis partitions te total area of quadrilateral quadril ateral ABCD into four triangular pieces AXB, BXC, CXD, and DXA It sufces, teref terefore, ore, to t o ssow ow tat exactly alf of te area of eac of tese tese four triangles triangles is is saded ut A P P B, and tus A P X and B P X ave equal bases base s A P and P B, and tey ave ave equal altitudes altitudes.. It follows tat A P X and B P X ave equal areas, AX B is saded. and tis poves p oves tat exactly alf of te te area of AXB s aded. similar simil ar argum argument ent ABC D , and works for eac of te te oter tree triangle triangless tat comprise quadrilateral quadril ateral ABCD • it follows tat exactly alf of te te total area of te te quadrilateral quadrilateral is saded. saded .
otion.
Now for te surprise: Tere is a nice generalization of roblem 5.8 tat olds for all ceckerboards and and not just ju st in te case were 2 . If we sade te boxes along te diagonal diag onal of any ceckerboard, ceckerboard, we will prove tat te total area a rea of te te saded boxes is exactly 1 / of te area of te entire ceckerboard. Of course, we ave saded exactly one t of te te boxes, boxes , but since in general, general, te boxes do not all all ave equal areas, tis certainly certainly does not sow tat we ave ave saded one t t of te area; area; someting more subtle is going on ere. ere. Of course, te t e case 2 of tis fact is exactly roblem roblem 5.9, 5. 9, and te te case 1 is a triviality triviality wit no content
D CHECKRBOARDS
16
ctually, someting even more amazing is true: We need not restrict ourselves to diagonal boxes. boxes . If we sade any n of te n boxes, sub subject ject only to te condition tat no two of te te saded s aded boxes lie in te same row or column, ten exactly one t of te entire entire area will be saded. sade d. We omit te proof of tis, tis , owever, because becau se to give a proof would carry carry us u s too far far from from our goal, go al, wic is i s to demonstrate demonstrate te utility of vectors in geometry proofs. THEOREM. Suppose that we are given an arbitra n n hekerboard A B C D with with area area KA of the n boxes along alo ng KABB C D . Writing d to denote the total area of the diagonal diagon al of of this heke hekerboa rboarrd we have h ave d � KA B C D . Te teorem certainly certainly olds wen n 1 , and ence ence we can c an assume ass ume tat n 2 . lso l so,, we can suppose su ppose tat tat te te teorem as a s already already been establised est ablised for all all smaller values of n . In particular, particul ar, terefore, terefore, we can c an assume ass ume tat te area of te n 1 diagonal boxes of any any (n  1 ) (n 1 ) ceckerboard is exactly 1 / (n (n  1 ) of te total area of tat ceckerboard. We are proceeding by matematical induction, and te fact tat te result olds wen n is replaced by n  1 is referred to as te inductive ypotesis. et P Q and R S be te leftmost and uppermost of te te crisscros cris scrosss lines o f te n n ceckerboard ABCD ABC D and let X be te point were were tese lines line s meet, as sown in Figure Figure 5.9 5 .9 . Tus quadri qu adrilatera laterall A P X R is te uppermost of te te n diagonal boxes wose total area d we need to compute. We ave saded tis box in te gure, and we are to imagine tat tere are n  1 more saded boxes, all of wic lie inside quadrilateral quadrilateral X SC Q. In fact, fact, quadrilateral quad rilateral X SC Q is an ( n 1 ) (n 1 ) ceckerboa ceckerboard. rd. (ecall tat tat tis i s a consequen cons equence ce of Teore Teorem m 5 . 8 , wic was proved proved using usi ng vector metods.) metods .) It follows follows from our inductive inductive ypotesis ypotesi s tat te area of te n  1 diagonal boxes boxe s inside ins ide quadrilateral quadrilateral X SC Q is K , were we are using our standard K notation to indic ate te area of of a gure. gure . We We conc lude tat te total ABC D is given by te formula saded diagonal area d of te entire ceckerboard ABCD formula K d KAPXR + n1 We want to sow tat d � KA B C D and so we need to prove tat nd accompli s tis, we join joi n X to eac of o f te te points A, B, C, and D Since KA B C D . To accomplis
Proof. Proof.
c
Figure 5.9
HAPTER
170
AP
=
VETOR METHODS O PROO
l A B and A R
=
l A D , we we see th see that at K X
=
l K x and K X
=
l K x .
dding dd ing tese equations and multiplying multiplying by n , we obtain K · n K = K Similarly, since Q C = l Q D and SC = BC, we ave K = K tese equations and multiply by n , we get and K = K If we add tese n K (n ) K Now by combining our equations, we get n K nd = n K + n (n ) K = K K + n = K K + K = K , as required.
xercses xercse s 5 SD.1
Figu Figure re 5 5 0 sows a 2 3 part of some so me ceckerboard. Sow tat te sum of te te areas of te two boxes labeled l abeled is equal to te sum of te areas areas of te two boxes labeled 2 . c orresponding esponding result about te four corner corner boxes of any par of a NOTE: Te corr ceckerboard ceckerboard bounded by two two orizontal and two vertical vertical lines is also als o valid Tis is te key to a proof of te generalization of Teorem 50 concerning notnecessarilydiagonal sets of n boxes of an n n ceckerboard.
Figre 50
5E
A Bit of igonometry
In Section Sec tion F, we sall need to refer refer to te socalled soc alled addition formula formulass for sine and cosine co sine,, and so we digres s briey to review review tese formulas formulas ere. The fo fo llowing llowi ng fo fo rmulas hold for for all angles ang les and . a. cos( + ) = cos() cos cos ()  sin() sin() sin(). sin(). b. sin( + ) = sin() cos() cos() + cos() sin(). sin().
(5.11) THEOREM.
E A BT O TRGONOMETRY
171
Te ollowing easy e asy proo uses dot products o vectors vectors and coordinate geometry. geometry. In te coordinate plane, let be te origin, let P be te point (0 ) , and and let let A and B be te points o te unit circle suc su c tat L P A and L P B . Since te coordinates o A are are (cos () sin sin ()) () ) and and te te co ordina ordinates tes o o B are are (cos () , sin()) sin( )) , we can writ writee = (cos() sin()) ad ad (cos(), (cos(), sin()). sin()). ecall tat we sowed tat te dot produc o two vectors is equal to te product o teir lengts times te angle between tem. Since A and and B and te angle angle between between tese vectors is  , we see tat tat dot product, owever, owever, we ave ave cos (  ) . y te denition o te dot · · cos() cos() + sin() sin( ) sin() sin( ) , and tus tus we coclude tat tat cos( cos(  ) cos() cos() + sn() sn( ) sin( sin()) . ddition ormula (a) ollows rom tis equation by substituting or (ecall (ecall tat tat cos (  ) cos(), but but sin  )  sin(). sin( ). ) Final Finally, ly, to to prov provee (b), we compute tat sin( + ) cos (90   ) sin( ) cos()) + sin (90  ) sin() cos (90  ) cos( cos(90 cos () + cos() sin() , sin() cos() were were te second equality ollows ollows by substituting sub stituting 90  or in te equation o te previous previous paragrap. paragrap.
Proof of Thorm 5.11.
Tere is anoter more purely geometric proo o te addition ormulas ta we cannot resist presenting For brevity, owever, we will only consider te case were o < + < 90 , so tat tat Figur Figuree 5. appli applies. es. In te gure, we w e started started wit line , and te we drew B and O A so tat LBO and L A O B . Next, we droppe droppe perpendiculars A and A V rom A to B and , respectively, ad ten we dropped perpendiculars X and U rom to A V and , respectivey We see tat L W A P and L V P are are complementary compl ementary to te equal vertical angles an gles LA P W and LOPV, and tus L W A P L V O P , as indicated. indicated.
Atr Atrati ativ v Proof of Thorm Thorm 5. 11 .
A
B
V
U
Figure 511
172
CHAPTER
VECTOR METHODS O PROO
We can assume tat te lengt a A is unit, and tus a v cos ( + ) and and we see tat O V a v We ave O W cos() cos () and a OW co(), and tus a cos() cos() cos() cos() . lso lso A W sin() and W X j A W sin() , and tis yields W X sin sin () sin() sin () . ut W X V U since X W U V is a rectangl rectanglee and ence we ave cos( + )
aV
a  VU
sin() sin() sin() , a W X cos() cos()  sin()
as required Now W U O W sin(), and tus we ave W U sin() cos( ) lso, A X/ A W cos () , an andd enc encee A X cos() cos () sin( ) . We We now see see tat sin( + ) A V
X V + A X W U + A X sin() sin() cos( ) + cos() cos () sin() sin() ,
and te te proo is complete. co mplete.
SF
Linear Linea r Operators Ope rators
ecors become muc more powerul as a tecnique o proo i we add one more ingredient: linear operators operators n oprator is a unctio unctionn T tat yields a vector wenever we plug in a vector In oter words, i is any vector, ten T is some s ome vector determined by v according to some specic spe cic rule. n easy, but not especially especi ally interesting interesting,, example o an an operator is given by te ormula T . In tis tis case, ca se, o course, cours e, we can tink o T as te operation o reversing te direction o all arrows representing vectors, or equivalently: T rotates rotates all arrows arrows by 80 • More generally, given any number , we can consider consi der te te operator T tat roates arrows repr repres esenting enting vectors counterclockwise troug degrees degrees We sould really call tis operator someting like Te so as to empasize tat we ave a particular amount o rotaion in mind, but we preer preer not not to clutter our notation wit inessential ines sential subs subscri cripts pts lso, we mention tat tere is noting especially important about our coice o coun terclockwise terclockwise or te direction o rotation rotation it is important, owever, to to be specic speci c in tis regard It is not dicult to give an explicit explici t ormula ormula describing tis rotation operator, altoug alt oug we sall not really need to use te ormula we are about to derive. I (a b) we want to express te coordinates and d o te vector T ( , d) in terms o te coordinates a and b and te angle o rotation rotation . For tis purpos purpose,e, we can suppose suppos e tat sown in in Figu Figure re 5 2 P and T so tat L Q P R , as sown I is te angle between P and te orizontal vector (0, ), we see tat a co () and and b sin() sin () , were I PQ Te angle between and te orizontal is + , and te lengt P R P Q (Tis is because our rotation operator T does not cange te lengts lengts o vectors vectors ) It I t ollows ollows tat (, (, d) T
( c o s( s( + ), sin( n( + )) ,
LNEAR OPERATORS
17
R
Figure 5.12
and tus by Teorem 5 1 1 , we get formulas for te coordinates of T (v): (cos() cos() cos() sin() sin() sin()) sin()) a cos() b sin() cos( + 0 (cos() sin() + sin() d sin( + 0 (cos() sin() sin() cos()) a sin() + b cos () For te te bnt bn t of radrs radrs familiar wit matrix matrix notation, we can write write tese tes e trans formation ormation equations equati ons as cos() sin() (, ( , d) (a b) sin() cos() and tus T(v) v A wr A is t matrix matrix of sines and cosine co siness in te preceding form formula ula In otr words, t oprator T simply multiplies a vector (on te rigt) by te matrix A W sall sa ll sow s ow tat t oprator T tat tat rotats rotats all vctors countrclockwise countrcloc kwise troug a giv angle 0 is, is , in fact, a linear oprator, oprator, wic means tat te operator respects respe cts bot addition and scalar scal ar multiplication multiplic ation of vctors vctors More precisly, precis ly, we say s ay tat an operator T is nar if T(v ) T(v) + T(w) and T(zv) zT(v) for all vectors vecto rs v and and and and or all scalars sc alars z o cck tat our rotation rotation oprator is linar linar w coos coos arbitrar arbitraryy vctors v and , and w comput tat T (v w) (v w) A vA wA T (v) + T () as requird r, A is t matrix of sins and cosins cos ins,, as arlir, arlir, and we are using te distributiv distributiv law of matr matrix ix multiplication lso, lso , i z and v ar an arbitrary scalar and an arbitrary vctor, w av T(zv) (zv (z v ) A z (vA) zT(v) as w wantd s two simpl calculations c alculations sow tat t t rotation rotation oprator oprator T realy is a linar oprator, as w assrtd ass rtd It is also possibl to s gomtrcally wy t rotation operator T is linear To sow tat T rspcts addition, it is convnint to to tink o te gometric signicance signi cance of vctor addition in a way tat is sligtly s ligtly dierent rom our prvious approac appro ac Suppo S uppose se tat w rprsnt t vctors v W and tir sum v w as arrows all aving te same tail P If v and w and w write v + w w ask ow te point S is dtrmind from a know kno wldg ldg of o f te tree points poi nts P , Q , and R
(
)
174
VECTOR METHODS O PROO
CHAPTER
s
s
p
p
Figure 5.13
onsider ons ider te te let diagram in in Figure Figure 5 3 We We drew drew a line troug troug parallel to P R and a line troug R parallel paral lel to P , and we let be te te point were were tese tes e two lines meet Ten P R is a parallelogram, and ence we ave We see now tat P + P + Tis s ows ow to add two vectors represented by arrows wit a common tail: It suces to complete te parallelogram Te arrow (wit te same tail) along te diagonal o te te parallelogram parallelogram represents te sum o te two original vectors vectors In te te rigt rigt diagram o Figure Figure 5 5 3 , we drew drew anoter anoter copy o parallelogram P Q R , and ten ten w e rotated rotated tis entire entire gure counterclockwise troug degrees about point P Te result result o tis tis rotation rotation is parallelogra parallelogram m P R, R , and it sould so uld be clear cl ear tat tat ( P and n) We can now s ee tat ) + ( P + ( ( P + ( n and tu tu s te operator respects respe cts vector vector addition addition (ecall (ec all tat tis is one o te two two tings we must sow to establis establis tat an operator operator is linear li near ) To To see tat te rotation operator operator also als o res respec pectsts scalar sc alar multiplicati multip lication, on, and ence ence is linear, it is enoug to observe obs erve tat tat i we stretc a vector and ten rotate it, te result is te same as would ave been obtained ad we rotated te vector rst and ten stretced it We can now demonstrate ow powerul linear operators are as a tecnique o geometric proo. In te ollowing , te point poi nt we reer reer to to as a s te center o a squar sq uaree is i s te unique point tat is equidistant equ idistant rom rom te our our vertices vertices Tis point, o course, cours e, is also als o te intersection o te te diagonals diag onals o te te squar sq uaree
Outwardacing squares are drawn on te sides o an arbitrary ABC D , as sown in Figu quadrilateral ABCD Figure re 5 I P, , R , and are are te te centers centers o o tese our squares, squ ares, as sown, sown , prove tat tat line line segments P R and are equal and perpendicular
(5.12) PROBLEM.
et be te linear operator operator corresponding corresponding to a 90 9 0 counterclockwise counterclockwis e rota rota tion. I we can sow s ow tat ) it will wil l ollow tat P R and and also tat tat P R is i s perpendicular to , and tat will complete te proo To decide exactly wat vector equation we need to prove, we must examie te diagram careully so as to avoid conusing counterclockwise wit clockwise In tis problem we
outo.
LNEAR OPERATORS
75
denitely must sow s ow tat T equals S te proo proo would not work i we careles sly replaced wit Te independent variables in tis problem are te our vertices A, B, C, and D o te given quadrilateral, because once tose points are known, it is clear tat te oter relevant points P, Q, R, and S are unambiguously deter deter mined. We attempt, tereore, to express , R, and in terms o , B, C, and We begin by nding a ormula or in Figure 5.14 terms o and B To do tis, we let U be te midpoint o AB, and we note tat P U A U and P U is perpendicular to AU (Tese observations are are immediate consequence co nsequencess o te act act tat P is te center o te square wit side AB) It ollows tat T and tus tus,, using u sing te linearity linearity o te operator T , we ave
T  T T T
,
and tis yields yiel ds + T() T() Since U is te midpoint o AB, we know tat �( + B), and we can substitute sub stitute tis into in to our ormula ormula or sing te linearity o T again, we obtain
( 1 + I T l + T I
p ( 1 + I + T l + T I T l l
Now we must nd similar expressi expressions ons or R, and s usual, we can do ts wit almost no work; we simply marc around te quadrilateral, replacing A by B, B by C, C by D , and D by A We repeat our ormula or and modiy mo diy it to get te tree oter ormulas we need:
p ( 1 + I T l l + T I 2
R C + D TC + T D
Q I C T I + TC
l 2 D + 1 T D + T l
176
CHAPTER
VECTOR METHODS O PROO
Next, we compute n= R p
=
2 c +  T C + T  i + T i  T
We need to compute compu te T(P and so we ave to apply te linear operator T to te rigt side o ti tiss equation . For tat tat purpos purpos e, we need to know ow to compute itrary vector. ut tis tis is easy. Since Si nce T is a 90 rotation, we T (T v, were v is an arbitrary see tat applying it twice yields a 80 80 rotation, and it ollows tat T (T (n v sing tis act, ogeer ogee r wit te linearity o T w e obtain obtain T n
2 T C + T + C   T i T  i + B
Finally, w obsrv obs rv tat tat 
2 C  T T C   T � T i
and tis is idnti i dntical cal,, xcpt xc pt or a rarrangmnt rarrangmnt o t trms, wit our ormula ormula or and tus tus lin sgmnts P R and S ar T(n) W conclud conclu d tat T(n) qual and prpndicular, prpndicular, as dsird • ollowing is anotr xampl o wat wat would woul d b a dic dicult ult problm probl m i on ad to rely on clvrns clvrnsss . It can b don by a mr computation, computation, owv ow vrr using us ing vctors and linar oprators. PROM In Figur Figur , uilatr uilatral al triangl triangls s � P A �PC D , and � P F sar a vrtx. rmaining six vrtics o ts tr triangls triangl s ar j oind in pairs by lin sgmnts s gmnts FA C and D, as sown, and points Y and ar t midpoints o ts tr sgmnts rov tat tat � x Y is uilatral caus caus bot t givn givn data data and t dsird dsird conclusion conclu sion involv quilatral quilatral triangls , it sould so uld b clar tat t linar oprator oprator tat rotats rotats vctors countrcl c ountrclock ock wis wis troug 0 is rlvant and w call tis oprator oprator T lso ls o it is convnint c onvnint but
outon
B
Figure 515
LNER OPERTORS
"
177
F
certainly not necessar necess ary, y, to coose coos e our origin origin at P so tat, for example, we can wrte simply as Tus B = T() = T(), and = T(E) S ince we can view te points A, C, and E as te independent variables variables in tis problem, our strategy strategy will be to express express X, Y and in terms of and E To complete compl ete te argument, it will sufce s ufce to prove tat T ( Z) = since tat will sow tat 6X Y Z is isosceles wit base X Y and vertex vertex angle equal to 60 . It follows follows tat eac base angle angl e is tus equal to 60 , and so s o 6X Y Z is equiangular, and ence is equilateral, as desired desired s is i s usual us ual wit tis type of proof, tis tis metod must work If, wen we compute tis sows tat we ave made an error in T ( ) it tus out to be unequal to computation or in setting up te equations equations.. Since X is te midpoint of AF, we can write X = ( + ut since sub stitute to obtain obtain = T (E) we can substitute
Z I
F)
F
s usual, us ual, we can c an marc marc around te gure to obtain obtain te correspon corresponding ding formulas formulas :
It follow followss tat
r = c + T Z = E + T ( () ()) ZX = X  Z = A + T E  E T C
and we ave to apply T to tis and ceck tat te te result is equal e qual to ZY = Y  Z = C + T A  E  T C 2 To accompli acco mplis s tis, ti s, it is useful u seful to ave a formula for T (T (v » were v is an arbitrary arbitrary vector To To obtain te correct formula, formula, consi c onsider der Figure 5 . 6 . In tis gure, gu re, we assume ass ume tat and = v = T (v ) and = T(T( v » Ten PR = P and L R P = 60 , and it follows follows tat tat R P is equilatera equilaterall Tus R = R P = P and L R P = 60 = L R P We conc lude tat R is i s parallel and equal e qual to P , =  v We tus ave and ence R = T(T(v T(T( v » = TT = = + T (v )  v
( () ())
p
Figure 5.16
=
178
CHPTER
VECTOR METHODS O PROO
We can now compute compu te tat
= T + T E  E  T = 2 T + T T  T E  TT = 2 T + T E  E  T E  T + = 2 T E  T + , and indeed indee d tis does d oes agree wit our formula formula for for Te proof is now complete T ( Z)
Te foll followin owingg teorem, wic is sometimes so metimes attributed attributed to Emperor Emperor Napoleon ona on a parte, parte, also a lso involves invol ves equilateral eq uilateral trangles, trangles , and ere ere too, our vector proof depends on te 60 rotation linear line ar operator. ou tward dpointing pointing equilateral equi lateral triangles on the THEOREM we construct outwar three sides sid es of of an arbitra arbi tra given giv en triangle tri angle then the triangle formed or med by the centr cen troo ids of the three equ e quilate ilateral ral triangles is equilateral equ ilateral
In Figure 5.7, we are given ABC and we ave constructed equilateral BC P , C A Q, and and A B R , wit centroids X, Y, and Z , respectively respec tively Our strategy strategy will be to express X, Y, and in terms of te given data, wic w ic are , and We will ten compute c ompute tat T were T is te operator tat rotates vectors 60 counterclockwise co unterclockwise.. It will follow follow tat tat X Y Z is equilateral, as required gain, we know tat tis metod me tod of proof proof must work; all tat is required is to car c arry ry out te computations witout error. error. Since Since BC P is equilateral, we see tat T(c) and tus  T  ) wit a bit of algebra yields yiel ds te equation ) Te linearity of T togeter wit
Proof
t ( n
8
8
R
Q
p
Figure 5.17
LNER OPERTORS
179
P = C  T C) + T B), and since we know by Teorem 5 . 2 tat X = B + C + P) , we get
s usual, us ual, we can c an avoid work by marcing marcing around around te diagram, and we obtain
y= � t = 3
( C + 2 1 T (1 + T(C
and
( 1 + 2 T ( B + T ( 1
Tese equations yield I=YX =
=
� (C
+
2 1  T ( 1) + T ( C)   2 C + T ( C)  T ( ) )
� ( 2 1   C  T ( 1)  T ( ) + 2 T ( ) ) 3
,
and = zx =
=
� (1 2 T ( ) T (1)  2C + T (C) T ( B) ) � (1 +  2 C + T ( 1)  2T ( ) + T (C) ) +
+
.
To compute T I), we will wil l use us e te fact fact tat tat we derive derivedd previously : T T v) = T v) v for an arbitrary vector v We ave I ) T I)
=� �
T(C T ( 1 (2T(1 (2T( 1 T( T( T(C
+ T ( B + B + 2T(C 2 C
( 1 +  2 C + T (1 2T( 2T( + T(C We see tat, as expected, we obtained exactly te same formulas for T I) and , and tis completes comple tes te proof Since equilateral triangles appear bot in te statement and in te conclusion of Napoleons teorem (Teorem 5), and since a special property of te 60 rotation operator was was used in te proof, proof, it seems surprising surprising tat, in fact, fact, tere tere is a generalization of Napoleons Napoleon s teorem tat as noting to do wit 60 angles or wit equilateral triangles triangles
180
CHPTER
VECTOR METHODS O PROO
Sup Suppose that three similar outwardpo outwardpointing inting triangles t riangles ar a re con con structed on the sides of an arbitra 6ABC as shown in Figure where 6PCB 6PC B 6C 6CQA QA 6BAR X Y and Z are respectively the centroids of these three similar sim ilar triangles then 6 X Y Z is simila s imilarr to each of of them
(1) THEOREM.
Of course, if te tree tree similar trangles trangles on te sides side s of 6ABC appen to be equi lateral, lateral, ten ten Teorem Teorem 5 5 5 tells us u s tat tat 6 X Y Z is also equilateral, e quilateral, and tus tus we w e see tat Napoleon Napoleonss teorem is included in Teorem Teorem 5 5 We close tis capter capter wit a proof proof of tis result tat uses vector tecniques and linear operat operators, ors, but wic does doe s not seem to be quite so mecanical or lacking in cleveess as are our oter oter vector vector proofs proofs
P
l
T
Our rst task is i s to t o dene an a n appropriate appropriate linear operator operator We would like to ave and ltoug = = = CB it may seem tat tis tis is asking for for too muc, we sall see tat because 6 P CB 6C QA 6B A R, it is poss ible to dene so tat its eect on C and is as desired Note tat L C P B L Q C A = LAB R , and so we want our linear operator to rotate rotate vectors counterclockwise by tis angle, wic we call ut ut we do not want want to be a pure rotation; it sould also stretc or srnk vectors appropriately We dene terefore, to be te resu resultlt of a counterclockwis counterclockw isee rotation troug troug followed by multiplication by te scalar z = P B PC rotatio n operator is PC We know tat a rotation linear, linear, and it follows easily tat our operator operator dened by a rotation troug some xed angle followed by multiplication by some s ome xed scalar scal ar is also als o linear y te = 6PC B 6CQA, and denition of we see se e tat as wanted wanted ut ut 6PCB tus P C C Q PBCA, and we see tat = PBPC = CACQ It follows tat if we rotate te vector C counterclockwise troug te angle and ten multiply by te scalar z te result is te vector Tus C and R BA , we see tat similarly, since we al so ave z B R as desired = Our goal will be to sow tat Wen we establis tis, it will = XY = z = follow tat L Y X Z = = L C P B We sall also know tat X Z XY P B P C, and ence PC XY P B XZ , and tus tus 6XYZ 6 PC B by te SS similarty similarty crteron crteron
Proof Proof of Tor Torm m 1
T
T
T T
T
T
P
T
T
U.
R
p
Figure 5.18
T
TT l
T
LNEA OPETORS
181
We ave te following equations : T ( C)  T ( P) T ( ) P B  P T  T T =  T (A)  T (B) T ( ) J R  B , Since 3 = + + , 3 = + + and 3 2 = + + R, we see tat by adding te preceding tree tree equations and multiplying by 1 /3, we obtain obtain T  T = 2  and tus tus T() as desired =
=
=
=
=
=
x
xercses SF SF1
SF2
In te situation of o f Exercis Exercisee 5, use us e an appropriate appropriate linear operator to sow tat B D is perpendicular to 0 P and tat B D = 0 P ABC D i s a parallelo In te situation s ituation of roblem 5 1 , ass assume ume tat quadrilateral ABCD parallelo gram Sow S ow tat P Q R S is a square
CHP ER S X
Geometric Constructions
6A
Rules of the Game
Wat information information are we allowed to use wen we are trying to determine weter or not some statement about triangles or circles is true? true? re we permitted to use everyting everyting tat we know to be correct? Obviously, Obviou sly, tat depends on our goal. go al. If we simply want to decide weter or not te te asserti ass ertion on is true, true, it would certainly certainly be permissible permiss ible to consult cons ult an advanced geometry text or to ask an expert wom we trust trust Even if we are unable to nd the specic speci c ass assertion ertion in te book or if te expert appens to be unfamiliar wit wit it, we gt neverteles nevertelesss lear le arnn some s ome oter relevant relevant facts facts from from tese sources, sources , and we migt ten be able to use us e tese to prove or disprove our original statement statem ent irecty or indirectly, indirectly, terefore, terefore, books and expert knowledge knowledge can often be used us ed to determine determine the trut trut or falsity falsity of some given giv en assert ass ertion ion ut suppose suppos e tat te assertion as sertion wose trut trut we are trying trying to establis is i s one of te te exercis exercises es in tis book bo ok Or suppose suppos e tat tat we are are teacing teacing a geometry class, clas s, and we are are trying to demonstrate te deductive metod In suc circumstances, it clearly would violate te rules rules of te game to prove someting by an appeal to autority because, becau se, as we ow, tose rules permit us to use only previously previously establised e stablised teorems or explicitly stated axioms We stress tat te proibition proibition of appeals appeals to autority is not based on a fear fear tat suc appeals are likely to yield wrong answers; answers ; appeals to autority autority are excluded because they simply s imply ave no place in te game of deductive matemat matematic icss Te situation is i s somewat s omewat analogou analogouss wen we consider consi der wat tools we are permit permitted ted to use to draw geometric ge ometric gures gures If te purpo purpose se of te te drawing drawing is simply to obtain an accurate diagram, tere are a number of eective tools tool s tat could be used use d Te diagrams in tis book were drawn wit te aid of a computer, for example, but older drawing tools include inclu de rulers rulers,, compas ses ses,, protractors, protractors, and various various oter devices devices designed desi gned to dra draw w particular particular angles or line segments wit particular particular lengts lengts Te classi cla ssical cal Greek geometers, owever, owever, establised es tablised certain rules for for te game game of geometric geometric constructions constructio ns s we will explain, tese tes e Greek rules limit us to just ju st two tools: tools : a straigtedge straigtedge for drawing lines and a compass for drawing drawing circles Te rules also require tat we use tese tools too ls in certain specied ways w ays Oter tools are not necessarily less accurate; tey are excluded simply because tey violate te rules of te game 182
RULES O THE GME
18
Te only legal le gal use u se fo a straigtedge is to draw te line determined by two points tat ave ave been previously previous ly maked on ou paper. paper. (In practice, practice, of course, we can c an draw draw only a segment of that line.) Te ofcial regulation straigtedge is not a uler it as no distance dist ance makings on it, and an d furtemore, furtemore, we ae not allowed to make any marks marks on it it We We shall sh all see se e tat if we are allowed to mak mak our straigtedge, ten it is poss po ssible ible to do cetain constuctions constuction s tat would wou ld oterwise be impossible imposs ible.. lso, ls o, we mention tat while most ulers ave two parallel straigt edges, te ofcial ofcial staigtedge as a single usable us able edge. Wit Wit a straightedge alone, terefor terefore,e, we cannot draw draw two parallel lines, line s, altoug as we sall see, se e, iti t is easy to draw paallel lines using bot a staigtedge staigtedge and a compass. compas s. We mention a furter furter proibited use us e of a straigtedge: straigtedge: drawing drawing tangents to t o circles circles Suppose, fo example, tat a circle as been drawn on our paper and tat a point P is maked on te paper outside of te circle. If we wish to draw one of te two ines trough P that ae tangent tangent to te circle, it is i s tempting to place te straigtedge so tat it uns throug te point P and ten to otate it slowly about P until it just ju st touces te circe. Tis will work, w ork, of course, course , but it is not allowed allowed by te ules of te game. game . (s ( s we sall see, it is not no t ard to describe des cribe a legal constuction con stuction of te te tangents to a cicle from from an outside point if we use bot a straigtedge and and a compass compass ) Similarly S imilarly,, if i f we we wis wi s to daw daw a tangent tangent to a given circle troug troug a given point Q on te cicle, it is not not permissible permiss ible simply to place te straigtedge so tat it touces te circle at te point Q only. s an altenative, altenative, we migt try to coose coos e a second sec ond point R on te circle, circle, near Q . We We cetainly can draw te secant sec ant line Q R , and of course, we know tat tat the desired tangent line is te limit of tis secant line as R approace approacess Q ut ut it is not legal legal to draw draw te tangent tangent line troug troug Q by taking limits. Te ules ules require tat a constuction must ave only nitely nitely many steps, steps, and so we canno c annott let R get arbit arbitrai raily ly close to Q . espite te rules tat limit te ways a straigtedge can be used, tere are a few tings we can do wit this ideal tool tat would be impractical wit ordinary mundane drawing drawing instuments. ins tuments. Wit an ofcial ofcial straigtedge straigtedge for example, we w e can draw te te line detemned by any two dierent points no matte ow far apart tey are or ow close togete togete Wit an actual uler, of couse, we could not draw the line joining jo ining two points p oints tat ae farte farte apart tan te lengt of the ruler, ruler, and it would be b e extemely difcult difcult to daw daw accuately te line lin e joining jo ining two points po ints tat ae ae very close clo se toget tog eter. er. Te ofci ofcial al Geek Geek compass compas s does te following following and nothing else : Given two two distinct di stinct points P and Q on our paper, it draws te unique circle centeed at P and passing troug troug Q, or it daws daws an appopriate appopriate ac of tat circle circle Te T e only requirement is tat te points P and Q should be b e dieent dieent it does not matter matter ow close togeter or fa apa apatt tey ae ae n actual pysical compass c ompass,, of course, is a device tat tat as a inged pair of ams ams wose tips ae eld at a xed but adjus adjustable table distance r apat. One of tese tes e tips is sap and is usually usu ally made of metal, metal, and te oter is a pen or pencil point In use, use , te metal tip punctures te pape, and so it is eld stationay sta tionay at at some speci sp ecied ed point P , wile te pencil draws a cicle or an arc arc centered centered at P and of radius r To use us e an actual compass to draw te circle centered ce ntered at P and passing troug Q, te rst step is to place te metal metal tip at P Next, Next, te separatio s eparationn between te metal tip tip and te pencil point po int is caefully caefully adjusted so tat te pencil point just touces Q, and nally, after tis adjustment is made, te circle can be drawn.
184
CHPTER
GEOMETRC CONSTRUCTONS
Suppose Supp ose tat we are are given three three points P, Q, and R on our paper, and we wis wis to draw the circle centered at R tat as radius equal to P Q With a pysical compass, tis is easy First, place te metal tip at P and adjust te opening of the compass so tat tat te pencil point is at Q Next, lift te te compass compas s from from te paper and ten, witout canging te distance between the metal tip tip and te pencil point, place pl ace te metal tip at R and draw draw te circle Tis Ti s procedure is not legal, legal , owever, according to te ofcial ofcial Greek Greek construction rules ecall tat, tat, ofcially, all we can do wit a compass comp ass is draw a circle wit a specied spec ied center and going throug throug a specied point poi nt One way to think think of tis is to imagine im agine that an actual pysical pys ical compass compa ss as a memory, wereas te ofci ofcial al Greek Greek compass compas s will wil l forget forget te distance between te metal metal tip and te pencil pen cil point as soon as te metal tip lose lo sess contact wit te paper paper To use us e a Greek Greek compass compa ss to draw te te circe wit radius radius P Q centered at R, we rst need to construct some point S suc tat centere d at R and running troug S To R S P Q, and then we could draw te circle centered con co n struct suc a point S using usin g only a regulation compass and a regulation straigtedge, it is convenient to be able to construct parallel lines Given a line m and a point P not on te line, construct te te line troug P tat is para p arallel llel to m
(61) PROBLEM.
oose a point A on line m and draw an arc troug P , centered at A and meeting m at point poi nt B , as sown in Figure Figure . Next, draw draw arcs troug troug A centered at P and at B and let Q be te point oter tan A were were te corr co rrespo esponding nding two circles circles meet Ten line P Q is the desired parallel to m throug P We need to prove tat P Q m, and for for tis purpose, we observe ob serve tat B Q = B A P A P Q by construction, constructi on, and tus quadrilateral quadril ateral A P Q B is a parallelogram parallel ogram because it has two pairs of equal opposite sides side s (In fact, fact, A P Q B is a rombus, but tis is irrelevant) irrelevant) It follows follows that PQAB, and since line A B is our original line m , we are done
outon.
We can now sow ow to do wit a straigtedge and Greek compass compa ss wat we can do wit an ordinary compass compa ss tat remember rememberss its setting wen its metal tip is lifted from from te paper paper Given distinct points P, Q, and R , construct con struct te circle centered at R and having radius P Q
(62) PROBLEM.
Figure 6.1
RULES O THE GE
18
First, assume tat P, Q, and R are not collinear raw lines P Q and P R and then, using roblem 6., construct te line troug Q parallel to P R and te line troug R parallel to P Q and let S be te point were were tese lines line s meet Ten Q S R P is a parallelogram, and so P Q RS We can now draw the circle troug S centered at R , and we see tat its radius is equal to P Q, as required required Since te t e actual point Q is irreleva irrelevant nt ere, and only te distance di stance P Q is signif icant, we can replace Q by any other convenient point Q' on te circle troug Q centered at P In particular, in te case were te given given points poin ts P, Q, and R appen to be collinear, colli near, we can replace Q by anoter point Q' on tis tis circle, were Q ' i s not collinear colline ar with P and R We can ten carry out te construction construc tion of te previous paragrap
outon.
We see now tat it really does not matter that an ofcial Greek compass does not remember remember its distance dis tance setting wen it is lifted lifted from from te paper paper sing s ing te cons c onstructio tructionn of roblem roblem 6. 2, we can pretend wenever wenever it is convenient to do do so s o tat our compass compa ss does do es remember remember its setting, setting , and so hencefort hencefort we will cange te rules of our game and allow a compass wit a memory To demonstrate demonstrate some of what we can do with such an enanced, powerful powerful compass, compa ss, we present a few few easy but useful constructions (63) PROBLEM.
onstruct the perpendicular bisector of a given line segment
oose oo se an arbitrary arbitrary point P on te given segment AB, closer to B tan to A , and draw an arc arc troug P centered at A , as i n Figu Figure re 6.2. 6. 2. Now, Now, using roblem roblem 6.2 t o pretend pretend tat our compass has a memory, memory, draw an arc arc of o f radius A P centered centere d at B , and let X and Y b e te te two points were tese tese two arcs cross cros s We claim tat line X Y i s te desired perpendicular bisector bis ector To see wy, observe tat A X A P BX and A Y A P BY, and tus eac of X and Y is equidi eq uidistant stant from from both A and B It follows follows by Teor Teorem em 0 tat tat eac of X and Y lies on te perpendicular bisector of segment A B, and tus te line troug X and Y is, in fact, fact, te perpendicular bisector
outon.
Once we have have constructed co nstructed te perpendicular perpendicular bisector bise ctor of a line segment, we can get te midpoint of o f te segment segmen t for for free free:: It is just ju st te intersection of te perpendicular perpendicu lar bisector bis ector wit te original segment. S ince we can construct midpoints midpoints of line segments, we clearly can construct c onstruct te medians of o f a given given triangle, triangle, and a nd tus we can construct co nstruct te
A
p
Figure 6.2
B
186
CAPTER
GEOMETRC CONSTRUCTONS
centroid centroid of the triangle, triangle, which is the intersection intersection of te te medians. medians . Of course, cours e, we need to construct just ju st two of te te tree medians medians to nd teir intersection intersection.. lso, l so, we can c an construct te circumcenter of a given triangle triangle since s ince is te intersection of te perpendicular perpendicular bisectors bi sectors of te sides. side s. nd of course, course , it is enough to con struct struct j ust two of tese tese tree perpendicular bisectors. Once we ave te circumcenter, it is easy to costruct te circumcircle: Just draw te circle centered at troug troug any one of te te vertices of te triangle. It is also als o easy eas y to drop a perpendicular perpendic ular from from a point to a line, and so we can c an construct cons truct te altitudes of o f a given trangle, and tus we can construct con struct te ortoce ortocenter. nter. (64) PROBLEM.
Given a point P and a line m construct te te line troug P tat tat is
perpendicular to m
Note tat tat in roblem roblem , , point P may or may may not lie on line m ltoug te construction is es e s sentially the same in tes tesee two cases, cas es, te language that is customarily cu stomarily used is slightly s lightly dierent. If point P does not lie on line m we drop te perpendicular perpendicu lar from P to m but if P does do es lie on m we erect te perpendicular perpendic ular to m at P. First, we construc cons tructt a pair of points A and B on m suc tat A and B are equidistant from P. Tis can be done by drawing a circle centered at P, wit a radius large enoug so tat te circle meets m in two points. y roblem 3, we can construct te perpendicular bisector b of line segmen segmentt AB AB. Ten b i s perpendicular perpendicular to to line m A B , and and b goes trough trough P , as required, required, since P i equidistant equidi stant from A and and B. B.
outon to Probm 64
(6) PROBLEM.
onstruct ons truct te te bisector of a given angle.
=
Given Given A BC , as sown in Figu Figure re 3, construct construct points points P and Q on AB AB and A C suc tat B P B Q . o ts by drawing drawing any circle circle centered at at B and letting P and Q be te points point s were tis circle meets te sides side s of te angle. angle . Next, construct cons truct a point R , equidistant equidi stant from P and Q and dier dierent ent from B . n easy way to do tis is i s to draw te te circle troug troug P centered at Q and te te circle troug Q centered centered at P and to let R be one of te points points of intersection of tese tese two circles. Ten P R P Q Q R, as desired ine B R is i s te desired angle bisector. To To see wy tis is true, note tat B P R B QR by SSS and and tus tus P B R Q B R , as as requ requred red..
outon.
=
c
Figure 6.3
B RECONSTRUCTNG TRANGLES
87
S nce te nc enter of a trangle s te ntersecton nterse cton of te tree tree angle bs b sectors ectors,, we now see tat t s poss po ssble ble to construct cons truct te ncenter ncenter I of a gven ABC Of course, course, to nd I t sufces to construct construct just jus t two of te angle bsectors. bsectors. Once we ave te center of te nscrbed crcle, ow can we construct cons truct te crcle tself? It clearly clearly sufces to construct any one pont tat tat we know know must le on te te crcle. onsder te pont P of tangency of te te nscrbed crcle of A B C wt sde A B We know tat radus I P s perpendcular to tangent AB, and t follows tat we can construct P by droppng te perpendcular from I to A B ont P s, s , of course, course, te foot of ts perpendcular: te pont were te perpendcular perpendcular lne meets A B Once we ave constructed cons tructed P , we can draw te crcle troug P centered center ed at I and tat wll be te nscrbed crcle of BC
xecses 6 6A.l
6A.2 6A.3 6A.4 6A.5
6A.6
6B
Gven Gven a lne segment, construct c onstruct an equlateral equlateral trangle trangle wose sdes s des ave lengts equal to te lengt of te gven segment. Gven a lne segment, construct a square wos wosee sdes sde s ave legts equal to te lengt of te gven segment. segmen t. Gven a crcle, construct c onstruct ts center. Gven two lne segments, construct a rombus wose dagonals ave lengts equal to te lengts len gts of te two gven gven segments. segments . Gven ABC, were C s obtuse, construct P BC avng te same area, were P 90 on struct a crcle crc le wt dameter AB HINT onstruct raw a crcle centered centere d at P and coose a pont A on te crcle. ext, draw te crcle wt te same radus centered at A and let B be a pont were ts crcle meets te orgnal org nal crcle. crcle . ow draw te crcle wt te same radus and centered at B and let C be te pont po nt oter tan tan A wee wee ts crcle meets te orgnal crcle. raw te crcle crc le wt te same s ame radus centered centered at C and let D be te t e pont oter tan B were ts ts crcle meets te te orgnal crcle. ontnung lke ts, consruct con sruct ponts E, F, and G . rove tat G s te same pont as A and sow tat exagon ABCDEF s regular regular
Reconstructing Recons tructing iangles
Gven Gven a trangl trangle,e, we ave seen ow o w to construct ts medans, medan s, alttudes alttudes,, and angle bsecors bs ecors.. more more dfcu dfcultlt and nterestng nterestn g type of problem s to reverse ts ts procedure pro cedure:: Gven some data about a trangle, trangle, but wtout beng gven te trangle tself, we want to reconstruct te orgnal trangle trangle For example, example, we mgt want to reconstruct A B C f we are gven te lengts b A C and A B of two sdes of te trangle, trangle, and we are also als o gven te lengt m of medan AM, were M s te mdpont of sde BC Of course, cours e, te best
88
CAPTER CAPTER
GEOMETRC CONSTRUCTONS
we can ope for is to constct a triangle ongruent to te original triangle tere is obvously no way wa y to reconstct te te actual actual A B C from from te given lengts lengts efore we proceed to solve tis and similar problems we sould clarify wat it means to be given te tree lengts G b, and m Wat Wat we are actually given is i s tree line segments drawn on a piece of paper and we are told tat tat te lengts of tese segments are G b, and m We are also told wic segment as wic lengt. lengt . PROBLEM. econstct eco nstct a triangle given te lengts len gts of two of its sides s ides and te te median median to te tird tird side. si de. x
A
\\ B
\
I / \ I // / '"P ,
I
\\ \\ 1Z \
C
/
Q Figure 64
=
To explain our constction we pretend tat tat te te original (idden triangle left diagram diagram of Figure 6, 6 , and we assume ass ume tat tat lengts b AC is A B C in te left are given were were i s te te midpoint of B C. omplete omplete AB and m A are te parallelog paralle logram ram B A C P by drawing B P parallel p arallel to A C and C P parallel to A B and draw draw diagonal di agonal A P . Since w e know tat tat te diagonals o f a parallelogram parallelogram bisect bise ct eac oter oter it follows follows tat A P goes troug troug te dpoint of BC B C and tat tat A P 2 A 2m lso lso we see tat P B A C b In particu particular lar te lengts of te te sides sides of A A B P are b, and 2m, and terefore b + 2m ow to begin te actual actual constction con stction draw a line and using a compass co mpass wit wit memory memory mark o segments X and Q on tis line eac of tem tem aving lengt le ngt m (See (Se e te rigt diagram o f Figure 6 6 )) gain using a compass wit memory memory draw a draw a circle centered at Q and circle centered at X and aving radius A B and draw Since b + 2m, tese two circles must mus t intersect and we aving radius b AC . Since select one of o f te points of intersection and and call it Y as in te gure. We now ave ave A B P . X Y Q by SSS ext complete parallelogram Y X Z Q b y constcting cons tcting X Z parallel t o Y Q and and Q Z parallel parallel to Y X. (ecall tat tat we know know ow ow to do tis tis by roblem roblem 6 ) We We ave ave now now constcted constcted X Y Z and and we claim tat tat X Y Z . A B C as desir desired. ed. We ave XZ Y Q b AC and XY AB. lso LBAC LBAP + L P A C LBAP + L AP B and and simila similarlrly y L YX Z L Y X Q + LXQY. utLBAP LYXQandLAPB LXQYsinceweknowtatABP . XYQ. It follows ollows tat tat L B A C L YXZ and and tus tus AB C . XY Z by SS
utn.
=
=
= =
= =
=
== = = = = =
ext we set ourselve ours elvess a arder task task
=
= ==
=
B RECONSTRUCTNG TRANGLES
(6.7) PROBLEM.
89
econstct a triangle given given te lengts of its tree medians medians..
To do tis constction con stction we need to be able to constct con stct te te point two tw o tirds of te te way from from X to Y along a given given line line segment XY X Y . In fact fact a more more general construction is available available and we digress to present it before before we retu retu to to te solution of o f roblem 6.7. 6. 7. (6.8) PROBLEM
positive integer.
ivide a given line segment into equal parts were is any A
B
C
D
E
Figure 65
et X Y be b e te given segment segme nt and draw some s ome line li ne X A different different from from X Y as sown sown in i n Figur Figuree 6 .5 . Select some point point P o n X A and using a compass compas s mark mark off off additional segments P Q, QR , RS, etc. along XA all of tem tem equal equal to XP . (We (We ave illustrated illu strated te case 5 in Figure 6. 5 . ow j oin te far far end of te te t equal segment to Y as sown in te diagram diagram and draw draw P B QC R D etc para parallllel el to tis line. line . n easy argument wit similar triangles triangles now sows s ows tat te pieces XB BC CD etc. into wic te original line segment XY is divided all ave equal lengts.
utn
In te original original A A B C suppose suppo se tat te te tree tree medians are A B and CP and te point were tey meet te centroid of ABC is . ow consider B C . Sides S ides B and C C of tis tis triangle triangle ave ave lengts B and and C P respectively and since we are gi given ven line line segments wit wit lengts B and CP , we can use roblem roblem 6. 6. to constct line segments segments wit wit lengts lengts B and C C ow segment segment is a medi median an of B C and since since 1 A and we are given a segment of lengt A we can construc cons tructt a segment wose lengt is equal to tat of median of B C . We ave now constcted te lengts of sides B and C and of median of o f B C and tus tus by roblem 66 6 6 we can reconstruct recon struct a triangle triangle congruent to B C and in particular particular we can constct cons tct te lengt of B C wic is i s a side s ide of te te original triangle. triangle. We can similarly constct a line segment wose lengt is i s B A and since s ince we are given given a segment wose lengt is i s tat of median follows from roblem 6.6 6. 6 tat we can reconstct te te original triangle. B Q it follows
utn t Pbm 6.7.
We will do d o one more fairly fairly difcult reconstction reconstc tion problem. problem . (69) PROBLEM
econstruc econs tructt a triangle given te lengts of its tree altitudes .
9
CHAPTER CHAPTER
GEOMETRC CONSTRUCTONS
et h , h , and h be te given lengts of te altitudes from A, B, and C , respectively in ABC, and as usual u sual let a, b, and be te lengts of sides BC , AC, and A B of tis triangle. ote tat by computing te area of ABC in tee tee ways we get te equations a h = bh h, and tus h h a and h h ab Our strategy strategy will be b e rst to t o construct construct a triang triangle le similar to A B C and ten to rescale it so as to obtain a triangle triangle tat is actually congruent to ABC First draw line segments P X and P Z of lengts h and h , respectively forming L P as sown in te te left left diagram diagram of Figure 66 6 6 oose oo se point Y on P X (extended (extended in some convenient convenie nt but arbitr arbitrary ary positio po sitionn as sown s own and draw draw Y X Z . ext we do a similar construction construction as a s sown in i n te middle diagram diagram of Fig Fig ure 66 Tis time we let QR h and Q h, and we coose on QR (extended (extended so tat tat R R = X Y We ten ten draw draw R . Te next step is i s to draw draw line segment J H s o tat tat J H X Y = R and ten to coose point K so tat J K Z and and H K . Tis of course course is done done by taking K to be one of o f te points of intersection of te circle centered c entered at J and aving radius radius Z and te te circle centered centered at H and aving radius radius . (To be sure tat tese tese circles really really do intersect we need to establis tat Z > J H We will explain later wy tis tis inequality is guaranteed to old. old . Wat ave ave we accomplis accomp lised ed?? We We see tat JH XY PX h a JK Z P Z h b and J H R QR h a H K Q h It follow followss tat JH JK HK a b and tus K H J ABC by SSS SS S Tis also expl explain ainss wy wy Z > H Tes esee lengts are are proportional to te lengts of te sides of te original triangle. triangle. We ave now constructed cons tructed a triangle similar simi lar to our original triangle . s sown in te rigt diagram of Figure 66, 6 6, construct J D perpendicular to J H, wit J D h , and ten ten construct cons truct D E parallel to J H et be te point of intersection of line J K wit line D E and construct M parallel to K H , were M lies on J H It follows Furtermore we w e ave arranged matters so s o tat tat M J K H J ABC Furtermore te altitude from of M J as lengt equal to J D h , wic wic is te lengt lengt of
oution.
U
D
L
E
W
x
y
Q
R
Figure 6.6
S
J
H
M
C TAGENTS
191
te alttude from A n �ABC It s easy ea sy to see se e from from ts tat �M J and �ABC are are not j ust us t smlar sm lar tey are n fact congent cong ent as requred. requ red.
xercses 6B
Gven two sdes and te alttude to te trd sde of an acut angled trangle reconstruct te trangle trangle.. Gven one sde s de te t e alttude alttude to tat tat sde and te medan to tat tat sde of a trangle trangle reconstct te trangle. 6B.3 Gven te ypotenuse and te alttude to te ypotenuse of a rgt trangle reconstruct te trangle. 6B.4 Gven te crcumradus one sde and te alttude to tat sde of a trangle reconstct te trangle. trangle. 6B.l
6C
Tanents
We mentoned earler tat t was not legal to use a stragtedge to constct cons tct a tangent to a crcle smply by placng pl acng te stragtedge so tat t just ju st touces te crcle crcl e ow ten can we constct cons tct tangents to a crcle crcle?? (6.10) PROBLEM.
tat pont.
Gven a pont on a crcle construct c onstruct te tangent to te crcle at
To solve so lve roblem 6 0, we need to construct te center of a gven crcle crcle an so we dgress brey brey to dscu ss ts easy problem problem wc appeared appeared as Exercse Exercse 6. 3 . (6.11) PROBLEM.
Gven a crcle construc cons tructt ts center. center.
oose tree ponts A, B, and C on te crcle raw cords A B ad A C and constct te perpendcular bsectors o f tese two cords. cords . Te pont p ont were te two perpendcular bsectors meet s equdstant from from A, B, and C, and ence ence t s te center of te crcle. c rcle.
utn.
et P be te gven pont on te crcle onstruct te center of te crcle and draw radus P ow erect te perpendcular to P at P and note tat tat tat perpendcular s te desred tangent.
utin t Pbm 6.10.
S omewat more more nterestng nterestng s te followng problem. Gven a crcle and a pont outsde of te te crcle construct cons truct te two two tangents to te crcle c rcle from from te pont.
(6.12) PROBLEM.
192
CAPTER CAPTER
GEOMETRC CONSTRUCTONS
et P be te gven pont and constct te center of te te gven crcle. raw lne seg ment P and construct ts mdpont M (ote tat M may be outsde outs de of te te gven crcle as t s n Fgure 67 6 7,, or t may may be on or nside te crcle. raw te crcle troug P and cen tered at M, and note tat tat ts crcle also goes troug pont and tat segment P is a Figure 67 dameter. et A and B be te ponts were ts crcle meets te orgnal crcle. To avod clutter n Fgure 6 7, we ave drawn only small arcs of te crcle centered at M, but of course tese arcs are sufcent to dene te ponts A and B We clam cla m now tat lne P A, and smlarly smlarly also P B, s tangent tangent to te gven crcle. To see wy observe tat t sufces to sow tat L P A 90• Ts s true owever snce L P A s nscrbed n te crcle centere c enteredd at M, and lne segment P is a dameter of ts crcle.
utn.
Te constcton of a lne tat s smultaneously smultaneou sly tangent to two gven crcles depends on a subtler trick. (ote tat te te number of common tangent lnes to t o two gven crcles can be zero one two tree t ree or four dependng on ow te two crcles crcle s are are arranged arranged.. We wll not attempt attempt to be completely general and n partcular partcular we w e wll ass assume ume tat te te two gven crcles ave unequal rad. lso ls o we wll construct cons truct only a common tangent wt wt te property property tat t does not separat sep aratee te centers of te two crcles. crcle s.
PROBLEM.
onstruct a common tangent tangent lne to two gven crcles. crcles .
onstruct o nstruct te centers and V of te two crcles as sown n Fg T ure 68, and coose some conve nent pont Q on te crcle centered at V. onstruct te correspondng pont P on te crcle centered ce ntered at by drawng radus P parallel to V Q , were P and Q le on te same sde of te lne V jonng j onng te centers of te two crcles et X be te pont were were Figure 6.8 meetss V . P Q meet We wll sow tat te pont X s ndependent of te coce of Q To see wy ts ts s so observ ob servee tat �X �X P �XV Q , and and tus us X/ X V P / V Q Snce P and V Q are rad of te two gven crcle s te rato P / V Q s clearly c learly ndependent of te coce of te te pont Q , and t follows follows tat te rato rato X / X V s also ndependent ndependent of te te coce of o f Q ut t s easy to see tat te rato X / X V s a monotoncally decreasng functon functon of te pont X X as X moves along lne V from from nntely far away on te left n Fgure Fgure 6 6 8 toward . It follows tat t at tere tere cannot be two derent ponts X and Y on lne V bot to te left of and suc s uc tat
utn.
C TANGENTS
19
X U / X V Y U/ V V Ts sows tat te pont X really really s ndependent of Q Q as clamed. We observe ob serve next tat te common tangent lne ST, wc s te lne we are tryng tryng to constct crosses cros ses U V at X Ts is because be cause rad rad U S and V T are bot perpendcular to ST, and ence tese two rad are parallel. In oter words words f we we took te te point Q to be T , ten te pont P would be S ut by te reasonng reason ng n te prevous paragrap we ow tat for every every coce of Q te lne lne P Q goes go es troug X , and tu tuss n partcular partcul ar S T goes troug X We ave constcted cons tcted te pont X were were te common tangent lne l ne S T crosses centers U V It follows follows tat ff we use roblem roblem 6. 6 . 1 2 to construc c onstructt te te lne of centers tangent X S from te pont po nt X to te crcle centered at U , ten tat tat lne also must mu st go toug T , and tus t s te desred common tangent lne.
more dfc dfcult ult and very very general general type of problem conceng conce ng tangents tangen ts was studed s tuded by pollonus of erga about 2200 years ago. Tese pollonan problems ave te followng form: Gven tree objects wc can be crcles lnes or ponts constct a crcle tangent to all tree objects objects.. Wat we mean by a crcle tangent to a pont n ts context s tat te crcle sould go troug te gven pont. If te gven objects are are tree tree ponts ponts we see tat te correspondng correspondng pollonan problem s ess e ssentally entally tat of constctng te crcumcrcle of a gven trangle. trangle. If te te gven obje objects cts are tree lnes on te oter and te correspondng pollonan problem s te constcton of te nscrbed crcle of a gven gven trangle. Of course we ave already already solved tese two easy problems. problems . In te remander of ts secton s ecton we dscuss dsc uss several more dfcult dfcult pollonan pollo nan problems. ut we sall not present a soluton to te most dfcult suc problem te constcton cons tcton of a crcle crcle tangent to tree tree gven gven crcles. crcles . Gven two lnes l nes and a pont constct a crcle troug te given pont and tangent to te two gven lnes
(6.14) PROBLEM.
We gve up a lttle generalty and assume tat te two gven lnes are not parallel and we let R be ter pont of ntersecton. In Fgure 6. 9, te two gven lnes are R A and R B, and te gven pont s P ote tat tere are actually two crcles crc les tat go troug P and are tangent to lnes lne s R A and R B, but we ave drawn drawn only one on e
utn
A
Figure 69
94
CHAPTER
GEOMETRC CONSTRUCT CON STRUCTONS ONS
of tem n Fgure 6.9 6 .9.. Our mmedate mmedate goal wll be to constct te center U of te desred des red crcle. crcle . Te pont U wll be equdstant from from lnes R A and RB, and t foll follows ows tat U must le on te bsector of L A R B We constct ts angle angle bsector bs ector and coose any any convenent pont V on t. t . We ten ten drop te perpendcular perpendc ular V T from V to R A, and we draw draw te crcle c rcle centered at V and passng pas sng troug te foot T of ts perpendcu per pendcular lar.. Snce R V s te bsector bse ctor of LA R B, we see tat ts crcle s tangent to bot R A and R B ut of course ts crcle probably does does not pass troug te pont P ; we stll stl l ave ave more work to do ext we coo c oose se one of te te two ponts po nts were were lne R P meets te crcle centered at V, and we call ts pont Q (s coce determnes determnes wc of te two crcles troug P tat ta t are are tangent to RA and R B we eventually obtan. We construct te lne troug P parallel to Q V, and we let U be te pont pon t were ts lne meets te angle angle bsector bs ector R V, as sown so wn n te dagram dagram Snce by constcton constcton U les on te bsector of LA R B, we ow tat te crcle centered at U and tangent to R A wll also als o be tangent to to R B We construct con struct ts crcle by droppng te perpendcular U S from U to R A and drawng te crcle troug S and centered at U To prove tat ts crcle crc le also als o goes toug toug pont po nt P , as requred we must sow tat U P US ow V T I I U S snce bot V T and U S are perpendcular to AR, and also con stcton. It I t follows follows tat � TV R r � SUR and also tat � Q V R V Q U P by constcton. � P U R We tus ave V Q VR VT UP UR US U S V Q VT ut V Q V T snce Q and T le on te same and tus U P US crcle centered cente red at V , and ence U P US, as requred. ow tat we ave solved roblem 6, a related pollonan problem becomes farly farly easy. eas y. Gven two lnes and a crcle constct a circle tangent to te tree gven objects.
(6.15) PROBLEM.
Te two two gven gven lnes app appear ear n Fgure Fgure 6. 0 as R A and RB, wc we are assu as suming ming are not parallel parallel Te gven circle s te one centered at P n te dagam and we wll say tat te radus of ts crcle s s r . onstct lne SX parallel to RA, as sown s own were te dstance between R A and SX s equa equall to r To carry ou te constcton of S X, several several easy steps are needed. Frst erect a perpendcular to to R A at some pont (say (s ay at R), and ten usng a compas co mpasss wit memory mark o on ts perpendcular lne a dstance d stance from from R equal to r . Fnally draw S X parallel to R A troug te te marked pont. To do ts one must of course rst rst nd te dstance r by constructng cons tructng te pont P , wc s te t e center of te gven crcle. Smlarly S mlarly consct SY as sown were SY s paral p arallel lel to R B and te dstance between tese parallel lnes is also als o equal to r
utn.
C TANGENTS
195
ow usng us ng te constructon co nstructon of robl roblem em 6 6 , nd te center U of a crcle tangent to S X and S Y and nnng troug pont P Ten te dstances from U to SX, SY, and P are all equal and te dstance from U to te gven crcle s U P r We see s ee tat U P r s also te dstance ds tance from from U to eac of te te lnes R A and R B, and t follows tat U s te center of te crcle tat we seek. pont lyng on ts t s crcle s te ntersecton of lne lne segment U P wt te gven crcle crcle and so we can complete co mplete our constcton cons tcton by drawng te crcle crcle troug troug tat pont and centered at U
/ //
/
/ //
/
// / 
y
Figure 60
A
B
C
Figure 6
efore efore we present our nal pollonan poll onan problem we dgress brey to develop a tool tat we wll need Gvn tee quanttes x y , and , we say tat y s te man 2 pptna between x and f x / y y / , or equvalently y X (6.16) PROBLEM.
lengts
onstct a lengt equal to te mean proportonal of two gven
We can measure o te two gven gven lengts usng us ng a compass compas s wt w t memory memory as adjacent adjacent segments of some lne and so n Fgure 6. , we assume assum e tat tat A B and C B are te two gven lengts. raw te crcle wt dameter A C and erect a perpendcular B P to A C at B , were P les on te crcle. s sown so wn n te dagram dagram we really need only te semcrcle. We clam tat B P s te mean proportonal between A B and C B To see wy ts s te obser ob serve ve tat LAPC 90 L C B P, and tus L P AB 90 LPCA LPC A L C P B Snce also L A B P = 90 LPBC, we see tat �ABP � P B C by and ence A B / P B P B / C B Tus P B s te mean proportonal between A B and CB, as desred.
utn.
We can now present and solve our nal pollonan poll onan problem. (6.17) PROBLEM
lne.
onstct a crcle troug two gven ponts and tangent to a gven
We assume tat te lne determned by te gven ponts P and Q s not parallel to te gven lne AB, and we constct con stct te ntersecton pont po nt U of lne P Q
utn
196
CHAPTER CHAPTER
GEOMETRC CONSTRUCTONS
U
V
Figure 612
wt lne AB, as sown n Fgu Fgure re 6 2 Our mmedat mmedatee goal s to constct te te point of tangency V of te te desired desir ed crcle wt lne AB Once V s found we can constct cons tct te crcle crcle as te crcumcrcle crcumcrcle of P Q V To nd V , of course course t sufces sufces to constct te dstance U V ecall from Teorem 32 tat we ave U U Q (UV) (See (S ee also Te Te orem 35.) Tus U P / UV U V / U Q, and so te requred dstance U V s te mean proportonal between U P and U Q We can constct cons tct ts dstance ds tance by rob lem 6 7
xercses 6 6C.l
6C.2
6C.3 6C.4
6C.5
6D
Gven a crcle and a lne ln e construct co nstruct a lne tangent to te given crcle crcl e and parallel to te gven lne. Gven Gven two crcles wose nterors nterors ave ave no ponts in common c ommon construct cons truct a lne tangent to bot crcles and suc tat te centers of te crcles le on opposte sdes of te te lne. Gven a square and a postve po stve nteger construct a square wose area ss exactly exactly squ are. tmes te area of te gven square. Gven two lnes and a pont on one of tem but but not on te oter oter cons c onstruct truct a crcle troug troug te pont tat s tangent to bot lnes. lnes . onsd onsder er bot bot te case cas e were te gven lnes are parallel parallel and te case were tey ntersect ntersect onstct a crcle c rcle tangent to two gven crcles and avng a radus equal to te lengt of a gven lne segment. segmen t. ssume s sume tat te gven gven lne segment s egment s long lo ng enoug to make make ts possble. possbl e.
h ad Pom
Te ancent Greek Greek geometers proposed propose d tree tree constcton problems tat became noto rious for ter df dfculty: culty : squarng a crcle doublng doublng a cube cube and trsectng an angle. In fact tese problems remaned unresolved untl comparatvely comparatvely mode tmes tmes but t s now own own tat tat eac of tese tree tree constctons s mpos sble. sble . In tis secton se cton we w e wll dscuss dsc uss tese tree tree classcal clas scal ard constructon constructon problems problems and n te te followng followng secon we wll try try to ndcate nd cate wtout gvng detaled forma formall proofs proofs wy tey tey are mpossble mpos sble..
D TREE ARD PROBLEMS
197
Wat Wat does it mean to square a circle circle?? In general, to qua a geometric gure means to construct a square sq uare wose area is equal to te area of of te te given gure gure 6.18) PROBLEM.
Squar Squ aree a given triangle triangle
esignate es ignate one side s ide of te given triangle triangle as te base and construct its mid point If is i s te lengt of te te base, teref terefore, ore, we ave constructed con structed a line segment wose lengt is i s /2 /2 ow construct con struct te altitude of of te given triangle triangle perpendicular perpendicul ar to te designated desig nated base and let h be its eigt ei gt We know tat te area of te triangle i s h, and tus our task is to construct c onstruct a square square aving tis tis same area If s is te lengt leng t of te side si de of te desired square, terefore, terefore, we w e ave s h, and tus s is te mean mean proportional between /2 / 2 and h We ave ave line segments wit lengts /2 and and h, and by roblem 6. 6, we know ow to construct construct a line segment wose wose lengt is te t e mean proportional of tese two lengts leng ts We can terefore terefore construct a line segment s egment of lengt s and once we ave tat, tat, it is easy to construct cons truct a square aving tat tat line segment as a side side
outio.
2=
Wat gures oter oter tan triangles triangles can c an be squared? If we wis wis to squar squ aree a quadr qu adrilateral, ilateral, for example, we could draw a diagonal diagonal to divide it into two triangles triangles,, and ten using roblem roblem 6 8 , we could constr co nstruct uct two two squares, eac aving te te same area as one of te te two component triangle triangless of te given quadrilater quadrilateral al It sufces, terefor terefore,e, to be able to cnstruc a square squ are wose area is equal to te sum of te te areas of two given squares sq uares Given Given two squar s quares, es, construct construct a square wose area is te sum of te areas areas of te two given squares squares
6.1) PROBLEM.
Tis is i s very easy Simply construct co nstruct a rigt triangle wose arms ave lengts equal to te sides of te two given squares and ten ten construct a square wose side is te ypotenuse ypotenu se of tis rigt triangle triangle y te ytagorean ytagorean teorem, te area of te square on te ypotenuse is te sum of te te areas of te two given squares squares
outio.
Similar Si milar reasoning reasoni ng allows us to square any any polygon Suppose, Suppo se, for example, we ave ave a polygon polygo n wit wit sides sid es,, were 3 is an integer integer If we work by matematical matematical induction, we can suppose suppo se tat we already know ow to square polygons wit w it fewer fewer tan sdes To square te te given gon, gon, draw draw a diagonal tat divides divides te gure into two polygon s, ec aving aving fewer fewer tan tan sides, and square eac of tese. Ten use roblem roblem 6. 9 to construct construct a square square wose w ose area area is te sum s um of te areas areas of te two squares just construct c onstructed ed ctually, we are ceating a little ere ow do we know know tat it is always alway s possible pos sible to nd a diagonal of a polygon tat divides divides it into two smaller polygons polygon s ? If te original polygon polygo n is convex, c onvex, ten it is easy to see tat any of its ( 3) /2 diagonals diagonal s will work (ecall tat tat a polygon polyg on is covx if none none of its angles exceeds exceeds 80 , and in tis tis situation, situation, all of te diagonals lie inside te gure ut if te polygon is not convex, convex, ten at least some of o f its diagonals di agonals do not lie inside in side te gure everteless everteless,, even a nonconvex polygon always as at least one interior interior diagonal, and so tat diagonal can be used to subdivide te polygon polygo n into two smaller s maller polygons polyg ons Te fact fact tat tat suc an interior interior diagonal
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EOMETRC CONSTRUCT CON STRUCTONS ONS
must always exist exi st is not obvious o bvious,, owever, owever, especially e specially if te polygon polyg on as a very large large number number of sides side s Since circles are are surely surely te simplest nonpolygonal nonpolygonal gures, and since sinc e we now know know ow to square any polygon, polyg on, iti t is reasonable to try try to nd a tecnique tat will square s quare a circle ut in fact, fact, iti t is not possible pos sible to do tis tis It is a teorem tat tat no no construction co nstruction using only a straigtedge and compass, compas s, and obeying te ofcial ofcial rules, rules , can succeed suc ceed in squari squ aring ng a circle circle We str s tres esss tat we w e are saying s aying muc more tan tat no one knows ow ow to square a circle circle It is denitely de nitely known known tat squaring a circle circle is impossi impo ssible ble nfort nfortunately unately tis distnction is sometimes sometime s misunderstood, misunderstoo d, and tere tere are people wo continue to attempt attempt to solve tis and te oter oter two impossible impos sible Greek construction problems problems ere are, are, in fact some individuals individu als wo w o insist insis t tat tey ave succeeded in solving so lving one or more more of tese problems and wo refuse refuse to accept te fact fact tat tat teir solutions solution s cannot be b e correct niversity niversity matematics departments occasionally receive letters from from selfproclaimed circle squarers and cube doublers, doublers , and most often, often, from from angle trise trisectors ctors ese people usually usu ally seek recognition, recogn ition, and sometimes so metimes even money, money, for teir teir accomplisment" accompli sment" su ally, te proposed, and necessarily incorrect, construction fails eiter because it does not ollow ollow te ofcial ofcial rules or because even ev en if it could be carried carried out perf p erfectly, ectly, using u sing an ideal straigtedge and an ideal compass, compas s, it would res result ult in only an approximate approximate solution to te problem (Of course, any actual actual construction, carried out wit real real pysical pysi cal tools, can at best yield only an approximate solution, but te correct constructions tat we ave been discuss disc ussing ing,, if tey tey could be carried carried out ou t wit perf perfect, ideal tools too ls,, would always yield exact exact solutions solutions In roblem roblem 6. 6 . 9, we saw s aw ow ow to use us e te ytagor ytagorean ean teore teorem m to construct a square wose wo se area is equal eq ual to te sum of te te areas of two given squares sq uares In particular particul ar,, we can construct a square squ are wose area is double tat of a given square robably, robably, te easiest eas iest way to do tis is to construct con struct te te square wose side is a diagonal of te original square We cannot resist presenting a very slick proof tat tis tis works is proof does not n ot rely on an appeal to te ytagorean teorem In Figure 6.3, te given square is ABCD and we ave drawn square AEFC, wose side s ide is diagonal A C of te original original square It I t is obvious in i n te gure gure tat tat te te small squar squ aree is composed compo sed of two two congruent copies of ABC wile te large square s quare is composed c omposed of o f four four copies copie s of tis triangle It follows tat te te large large square as exactly double te area of te small square, s quare, as claimed Of course, it is necess nec essar aryy to prove tat tat te ve small triangl triangles es in Figure Figure 6. 6 . 3 actually are congruent, but tat tat is easy, e asy, and we omit omit te details We now know tat it is easy eas y to constct a square wose wo se area is exactly double te area of a given square nalogously, te Greek geometers posed te problem of constructin constructingg a cube wose volume is double te te volume of a given cube Of course, course, since a cube is i s a treedimensiona treedimensionall object, we cannot cannot actually actually be given a cube c ube on our piece of paper paper Wat we are really really given is a line segment wose lengt equals te edge lengt of te given cube, cube , and te task is to construc con struc a line segment segm ent equal to te edge of a cube wit double te volume volume If te lengt lengt of our given line segment s egment is s , ten te volume of te original cube is s3, and double tat is 2s3 e edge lengt of te doubled cube is , terefore, terefore, doubl ing a cube �s In oter words, te problem of doubling
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D TREE AR PROBLEMS PROBL EMS
199
E
A
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F
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Figure 6.13
Figure 6.14
requires tat tat we construct con struct wit straigtedge straigtedge and compass a line segment s egment wose wos e lengt squ aring a circle circle,, tis is is times te lengt of a given segment. s was te case wit squaring provably provably impossibl impos sible.e. It is a teorem tat tat no no legal construct cons truction ion can accomplis accompli s tis feat feat Te tird tird classic clas sical al ard ard construction problem is tat of of trisecting a given angle. angle . s s we ave ave seen, it is easy to bisect a given angle and by roblem roblem 6 8, we know tat tat we can trisect trisect a given line segment, s egment, and so it certainly certainly seems reasonable tat one sould soul d be able to trisect trisect a given angle. angle . S ome angles, angles , in fact, fact, can be trisected. trisected. To trise trisect ct a 90 angle, for example, it sufce sufcess to construct a 30 30 angle, angl e, and tis is very very easy: easy : Simply Si mply construc cons tructt an equilateral equilateral triangle and bisect bis ect one of its angles angles ut it is imposs impo ssible ible to tris trisect ect an arbitra given angle angle using usin g only a straigtedge and compas compass.s. In fact, fact, an even stronger impossibility impos sibility teorem teorem is true. Not only does tere not es t a straigtedge straigtedge and compas c ompasss procedure tat will trisect trisect an arbitra arbitrary ry angle, but it is actually impossible impos sible to trisect a certain particula particularr angle, namely, 60 If it were were possibl pos siblee to trisect trisect a 60 angle, ten te n we could easily construc cons tructt a 0 angle : Simply construct co nstruct an an equilateral equilateral triangle and trisect trise ct one of its angles, angle s, tereby dividing it into a 20 angle and a 0 angle. angl e. s we sall explain, oweve owever,r, it is not possible pos sible to construct a 0 0 angle, and tus tus it is not possible pos sible to trise trisect ct a 60 angle. It is, terefor terefore,e, certainly impos impossible sible to trisect an arbitr arbitrary ary angle. We know ow to construct a 90 angle and a 60 angle, and we sall see ow to construct a 3 6 angle, angle , but it is impossible impos sible to construct a 0 0 angle. Wat is going on ere? good way to tink about tis is to consider te following general problem: ivide ivid e a circle into equal arcs, were were is some positive integer. integer. Of O f course, course , we require tat tis tis division of a circle circle sould s ould be carried out u sing only a straigtedge and compass and by following following te ofcial construction cons truction rules rules.. onstruct Te circle division divisio n problem is trivia triviall if = and it is very easy eas y if = 2 onstruct te center of te circle and draw a diameter. diameter. Te two points were te diameter meets te circle circle divide te circle into two 80 arcs arcs.. Te case = is also easy: eas y: raw two perpendicular perpendicular diameters to obtain four four 90 arcs. Once te center of te te circle as been constructed, it is also easy to divide te te circle into six equal arcs. In fact, fact, tis can be done wit a compass compas s witout witou t memory and witout u sing a straigtedge. s sown in Figure 6 , we mark mark an arbit arbitra rary ry point A on te circle and ten draw te circle troug centered at A and meeting te original circle at points B and F Next, we draw te circle troug and centered at B , and we note tat since A B = A = B, tis circle goes troug A and also meets te original circle at
00
CAPTER
EOMETRC CONS TRUCTONS
Simlarly, we draw the circles through centered at F and B to obtain two new intersection points E and C Finally, Finally, we construct con struct the point by drawing the circle through and centered at C
D
milarly, each each and sim that A L A 0 B = 60, and si equilateral, we see that S ince �A 0 B is equilateral, consists of 360 of 360 whole circle circle consists 60• Since the Since the whole E and of arcs thus the points points A, B, C D, E, and F al so be 60 60 , and and thus o arc, arc, it follows follows that f m ust also
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divide the circle circl e into into six eual arc arc s . Now that we have solved the circle division problem for 6, we see se e that that we get for for free free a solutio sol utionn for 3 Just take points A, C and E in Figure Figure 6. 5 . More genera generaly, ly, we see that if we can c an solve sol ve the circle division divisio n problem for any any even integer 2m, then we can also solve sol ve it for for m by taking taking alternate alternate division points poi nts.. onversely, f we can solve s olve the circle division divi sion problem for for some integer m , then we can also als o solve s olve it for for this , it sufces sufces to construct the the midpoint of a given arc, arc, and and this thi s is easy, m . To do this, since we can c an bisect bisec t the corresponding central central angle. In particular, particular, we now see that the the circe division divisi on problem can c an be solved s olved for all all integers of the orm orm and all integers integers of the form 3 , where e o I a circle circle is i s divided into eual arcs, arcs, then of of course, each of these arcs will measure 360 / , and ssoo each of the corresponding corresponding central central angles wil also als o measure 360 3 60 / . It ollows that we can divide a circle into eual arcs i and and only if we can c an construct cons truct an angle angle eual eual to 3 60 / Since 360/9 3 60/9 0, we see that the impossibility imposs ibility of constructing a 0 angle is euivalent to the assertion ass ertion that that the the circle division problem cannot be solved so lved when when 9 . We mention that if we divide a circle into eual arcs, with 3 , and and we join adj adj acent divis division ion points with w ith line lin e segments, we obtain a regular regular gon g on.. (ecall that a polygon polygo n is ua if all o its sides sid es are eual and all o its angles angle s are eual. eual . ) We have seen that the circle division problem can be solved when wh en is one o the numbers , , 3, , 6, 8, , 6, 2, etc. , and we have have said that that it cannot be solved when 9. The obvious obvi ous uestion, ue stion, of course, is What exactly exactly is the full full set se t o positive integers for which the problem can be solved? s olved? The complete answer was found about about 200 years ago by ar arll Gauss, Gaus s, who was one of the greatest greatest mathematicians mathematicians who ever lived. To describe the set of integers integers for which the circle division problem can be soved, we need nee d to digress briey brie y into a discussion discus sion of prime numbers This T his seems s eems appropriate appropriate in this book on Euclidean geometr geo metryy since Euclid, Eu clid, himself, hims elf, made one of the earliest signicant signi cant contributions contribution s to the theory of prime numbers e proved that there there are innitely innit ely many o them. ecall ecal l that that an integer p is said to be pim i its only divisors divis ors are and itsel. nd the the number number is, is , by denition, not prime. prime. The rst sever several al prime prime numbers numbers are 2, 3 , 5 , 7, , 3 , and 7, and it should be clear clear that that wit withh the except exception ion o the the prim primee number number , al prime numbers are odd. The mathematician . Fermat Fermat consi con sidered dered the uestio ue stionn o which odd prime numbers can be written writ ten in the form form p + , where e is a positive posi tive integer, integer, and in his honor, such s uch primes are called call ed mat pim I we search s earch for Fermat primes by computing comp uting the numbers of the orm + for the rst dozen or so positive p ositive integers e, we nd the following Fermat primes 3 + , + , 7 + , and + The number + is prime when the exponent e is one of the numb numbers ers ,
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D TREE ARD PROBLEMS
01
2, , or 8, and the apparent apparent patte that we see here suggest sugg estss that perhaps we might also als o get a prime when e = 1 In fact, fact, we do; do ; the number 1 + 2 = 5537 is indeed indee d prime. prime. It is not hard to prove that the only way tht the number 1 + 2 can c an possibl pos siblyy be prime is when e is a power po wer of 2, and thus all al l Fermat primes must mus t have the form form 1 + 2 for integers a O The numbers F 1 + 2 are called mat numb and although it is true that every Fermat prime prime is i s a Fermat Fermat number, number, it is certainly not true that every every Fermat number number is prime. prime . We have F = 3 , F = 5 , F = 17 , F 257, and F 5537, and of course, course , Fermat knew knew that these ve v e numbers numbers are prime. Fermat Fermat was wa s unable to factor F = 29967 29 967297 297,, but it was later found that that this number is not prime; it is a multiple of 41 It is now no w known that none of the next several Fermat numbers is prime, prime , and in fact, fact, no one on e has found any Fermat primes other than the ve that were known to 257, and 5537, but it is Fermat. The only known kn own Fermat primes, primes , therefore, are 3, 5,5, 1 7, 257, not no t known whether whether or not no t these are all al l of the the Fermat Fermat primes. We can now state Gauss ' theorem, but unfo unfortun rtunately ately we will not n ot be able to give the proof proof in this book. bo ok. The cir ci rcle division di vision problem is solvable for an integer intege r n and only n 2 m where e 0 is an integer integer and m is either equal to 1 or else m is a produc prod uctt of derent erent Fermat primes
6.0) THEORM.
In other words, to decide whether or not the circle division problem is solvable for some integer n , rst factor factor out o ut from from n as many factors factors of 2 as possible pos sible and write number. Next, factor factor into n = 2 m, where e is a nonnegative integer and m is an odd number. prime numbers and check that all of the prime factors of n are distinct and that all of them are Fermat primes primes.. Since Sin ce there are only ve ve known Fermat Fermat primes, primes, w e see that there are exactly 3 2 known numbers that can play the role of m in Gauss ' theorem. theorem. The rst several several of these these 5 , 1 7, 7 , 5 1 3 17, 85 85 = 5 17, 2 5 5 3517, and 257 In particular, are 1, 3,3, 5, 15 3 5, particul ar, because Gauss Gaus s ' theorem does not not allow the the possibility poss ibility that that m 9, it follows tha the circle division problem cannot be solved for n = 9, and thus a 0 angle cannot be constructed and hence a 60 angle cannot be trisected. trisected. Thus Gauss Gaus s ' theorem implies that it is impos sible to trisect trisec t an arbit arbitrar raryy given angle with straightedge and compass. compas s. ecause ec ause Gauss ' theorem allows the possibility pos sibility that that m = 5 , it follows that the circle division problem can c an be solved s olved for for n 5, and this fact fact was known to the anciens. anciens . It was not own, however, however, that the problem problem could be solved when n 17, and hence that it was possible pos sible to construct con struct a regular 1 7 gon with straight straightedge edge and compass. compass . Gauss Gaus s was the rst person to accomplish accomplis h that feat. feat. We close this section by showing how to solve the circle division problem with n = 10 y taking alternate points, this also solves the circle division problem for n = 5 , and if we join these ve division points, we will have constructed a regular pentagon. 6.) PROBLEM.
tagon.
ivide a circle into ten eual arcs and a nd construct a regular pen
0
CAPTER
EOMETRC CONSTRUCTONS CON STRUCTONS
We want to construct an arc o 3° in our circle, which we can assume has radius 1 Our rst task will be to determine the length x o the chord that will wil l cut o an arc measuring 3°, and or or this purpose, purpos e, we consider co nsider two radii radii A B and A C that orm a 3° central angle LBAC Then �ABC is isosceles, where radii A B 1 A C and LBAC = 3°, and the base B C is the chord whose length x we are trying trying to compute. compute . Note that each o the base angles o �ABC is (1 80° 80°  3°)/2 3°)/2 72° et B D be the bisector o LABC, as shown in Figure 15, and observe that LABD 3° Thus �ABD is isosceles, is osceles, and and we have have B D = AD lso, since LDBC 3° and L C 72°, we see that LBDC = 72°, and hence �BDC is isosceles and B D B C = x It ollo ollows ws that A D x and thus DC 1 x as indicated indicated in the gure gure ut �ABC � �BDC by , and it ollows that x B D DC x 1 AB BC x and thus x 1  x We have x + x 0, and when when we solve s olve this euation using usi ng the uadratic uadratic ormula, ormula, we obtain x (1 ± ) /2 O course, the length x cannot be negative, and hence x = ( 1 + ) ) /2 To car c arry ry out the desired construction con struction now, now, it su su ces to construct co nstruct a line segment o thi thiss length. length . We start with a circle and assume that the the radius is 1 unit onstruct the center A , choose a point B on the circle, draw radius A B, construct a perpendicular radius A P, construct cons truct the the midpoint M o A P, and draw M B ll o this this is shown in Figure 1 Now A B = 1 and AM 1/2, and thus B M /2 by the ythagoean theorem Now swing s wing an arc through A centered at M and let Q be the point po int where this this arc crosses crosse s segment M B, as shown in the gure. Then B Q B M Q M B M A M = /2 1/2, 1/2, and so the length o B Q is the number x that we calculated calcu lated previousl previously.y. Next, swing s wing an arc through through Q and centered centered at B and let C be a point where this arc cuts the circle Then chord B C has length B Q = x ° 3° Next, and so s o this chord c hord sub tends a central angle o 3°, and and w e have we construct co nstruct the point D, where the circle through B and centered at C meets the
oution.
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Figure 6.15
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I Figure 6.16
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E CONSTRUCTBLE NUMBERS
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° 3°, and we continue like this, moving around the original circle. Then circle in jumps jump s of 3° and thereby thereby dividing the circle into ten eual arcs. arcs . If we join jo in altern alternate ate divisi division on points with line segments, segments, as shown sho wn in the gure, we get a regular regular pentagon
xecses 6 6D.l
6D. 6D.3
6D.4
6E
ivide a circle into into 5 eual arcs arcs onstruct an euilater euil ateral al triangle whose area is eual to that of a given sua su aee Suppose Supp ose that is a positive po sitive integer and that p = + is a prime number rove that must be a power power of HINT: I f e is not a power power of , it is i s possible po ssible to factor actor e ab, where a is odd and < a et m = + and note that p = (m la + is a mltiple of m This yields yiel ds a contradiction. contradiction. ompute the length of a diagonal of a regula regularr pentagon whose s ide has length
otut Num
In Section , we asserte ass ertedd without proof that that it is impossible impos sible to nd straightedge and compass constructions that will suare a circle, double a cube, or trisect an angle lthough we argued argued that the the impossibili impos sibility ty of the trisectio trisectionn problem is a conse c onseuence uence of the the fact fact that one cannot construct cons truct a 40° angle, we did not really explain why the construction of such an angle is impossible (That a 40° angle cannot be constructed co nstructed is a conseuence con seuence of the much more general general theorem theorem of Gauss Gauss on circle circle divisio divi sionn but we certainly did not explain why Gauss ' theorem is true.) true. ) We We will now try to indicate some of the ideas that that underlie the impossibility imposs ibility proofs proofs for for the three three classical class ical impos im possible sible problems ecall that if we could double d ouble a cube, then we could solve solv e the following following probem probem Given an arbitrary line segment, construct a segment whose length is � tims the length of the given segment. segment . In fact, this is precisely what what the cube doubling doublin g problem reuires us to do. With this in mind, we dene what we shall refer to as the ratio cotructio cotructio probm for a given positive po sitive number a : Given any any line segment, construc cons tructt a segment whose length length is a times the length of the given segment For example example the ratio construction problem is completely trivial when when a = , and it is very very easy when integer lso, l so, since it is possible po ssible to double a suare, it follows follows that the a is a positive integer ratio ratio construction con struction problem problem is solvable when a = and by roblem .8, we know that the ratio ratio construction problem is also solvable when a / where is any postive integer Suppose now that there exists a construction techniue that will suare a circle Given any any line segment, se gment, we can draw a circle having having that segment as a radius , and hen we can suare that circle to obtain a suare with side s (ecall that we are are assuming as suming it is possible pos sible to suare a circle.) circle. ) If the the length of the given line segment is r, then the
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area of the the circle is n r and since si nce the suare and the the circle have eual areas, areas , we have who se length is s = r . Thus s r and we have constructed a line segment whose times the length of the given segment. In other words, if we could suare s uare a circle, we coud cou d solve sol ve the the ratio construction cons truction problem with with a = Finally, suppose suppo se that it isis possible pos sible to construct a 0 0 angle. an gle. Given Given any line segment, we can construct cons truct a right �AB, where the hypotenuse B is the given segment and where L B = 0. Then A B / B cos(0), cos( 0), and thus the length of segment AB is cos (0) (0 ) times the length of the given segment. This shows that if it were possible pos sible to trisect angles, and thereby to construct a 0 angle, then we could solve the ratio construction problem for a = cos(0). We see s ee now that to prove that the three three constructions we have been discus di scus sing are are impossible, impos sible, it sufc sufces es to show that that the ratio ratio construction problem problem cannot be solved s olved when a is any of o f the numbers �, or cos (0) . We We want, therefore, therefore, to try to understand understand exactly for which positive p ositive numbers the ratio ratio construction con struction problem problem is solvable. solvabl e. We say s ay that such numbers are contuctib and our goal is to show that none of the three numbers �, and cos (0) is construct constructible. ible. For technical reasons, reasons , it is i s conveni c onvenient ent to expand our denition of o f constructible real real numbers to include 0 and to include include those thos e negative negative numbers whose abs olute values are constructib co nstructible le.. With With this understanding, understan ding, it is easy to show that the set se t consisting consi sting of all constructible numbers is closed clos ed under addition addition and and subtraction subtraction and multiplication. It is only slightly s lightly harder to show that the the constructi cons tructible ble number set is also closed close d under under division, division , but of course, course , we must mus t not divide divide by zero. In other words words,, is what is called c alled a ubd of the eld of real numbers. numbers . This Thi s explains why the branch of abstract algebra called eld theory is the basic tool for establishing that a particular particular number is or is not constructible. lthough we shall omit o mit the proofs proofs of most of the facts facts that we as sert about , we remind the reader reader that these are indeed theorems, and to emphasize emphas ize this point, p oint, we will prove a few few of them. lso, although we will not no t go deeply into eld theory, we will wi ll try try to give the reader at least the avor of what is involved . We begin begi n by proving one of the facts facts that we mentioned previousl previou sly.y.
THEOREM.
The set s et of of constructible construct ible numbe nu mbers rs is closed under division. divisio n.
Suppose that a and lie in and assume, as we may, that a and are positive. We are given a line segment segme nt AB, and our our goal is to constr cons truct uct a line segment whose length length is A B
Poof.
B
Figure 6.17
E CONSTRUCTBLE NUMBERS
205
s in i n Figure 17, we draw a line through through A making some s ome convenient angle with the given line A B Since S ince is constructible, con structible, we can construct a line segment s egment of of length length A B , and so using a compass with memory memory,, we can construct the point , as in the gure, gure, with A A B lso, since is constructible, constructible, we w e can construct construct the point on A , as shown, shown, with A AB Next, draw B and then construct C B, where C lies on line AB It follows that that A A / A C A / A B , and thus A·AB (AB) AB ' AC A AB and hence line segment segme nt AC A C has the desired length.
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Not only is C clo c losed sed under the ordinary arithmetic arithmetic operations of addition additi on,, subrac tion multiplication, and division, division , but it is also als o closed under the extraction extraction of real suare roots.
THEOREM.
is in C and
0 then also lies in C
We are given a line lin e segment segme nt of length x and we know that we can construct co nstruct a segment of length . . Since x is the mean proportional between bet ween x and , , we can use roblem 1 to construct con struct a line lin e segment of length x and this shows that is construct cons tructible ible,, as reuired.
Poof.
We now kow that complic comp licated ated looking numbers such su ch as
3 � 2 lie in C More precis ely, any number that that can be formed formed by starting with integers and 2
repeatedly (but nitely often) often) adding, subtracting subtracting,, multiplying, multiplying , dividing, dividing , and extracting sure roots lies in C Of O f course, we are only allowed to extract extract suare roots roots of o f positive pos itive numbers, numbers , and we are not allowed to divide divide by zer It is a deeper fact fact that every every member of C can be formed in thi thi s wy wy by repeatedly using ordinary arithmetic operations together with the extraction of real suare roots Field Field theoretically, what we are saying here is that C is the uniue smallest smalle st subeld sub eld of the real numbers that is closed under the extraction of real suare roots. It is also true that C is i s exactly exactly the real subeld of the the suareroot closure of the rational rational numbers in the complex numbers, but we shal not n ot need this slightly sl ightly deeper fact. fact. though we will not provide any details of the proof here, iti t is easy e asy to understand the underlying reason for the assertion ass ertion of the previous previous paragraph. paragraph. t every step in a lega geometric construction, we are either taking the intersection of two lines, the intersection of two circles, or the intersection of a line and a circle. If we were to work out the the coordinates of a point obtained in this way, we would have to solve so lve some s ome euations, but these euations can never be worse than uadratic Essentially, this is because the euation of a circle contains no power higher than the second ut because
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EOMETRC CONS TRUCTONS
of the uadratic formula, we know that linear and uadratic euations can be solved by the processes of ordinary arithmetic together with the extraction of suare roots. It can be proved from from this that every every number in C is i s formed formed by repeatedly applying these operations. To show that suaring a circle circle,, doubling doubli ng a cube, and constructing a 0 angle angle are impossible, we need to establish that �, and cos(0) do not lie in C, and so we need to show that none of these numbers can be formed by arithmetic operaions together ith ith suare root extraction. extraction. To understand the th e underlying principle of the proof, proof, we recall that if f (x) is a polynomal, then a oot of the polynomal f is a number x such that f (x) 0. For example, the numbers 2/3, �, and + are are roots roots of the polynomials 3  2, x 2, an and x 2x + , respectively. real or complex number is said to be abaic if it is a root of some nonzero polynomial having integer coefcients, and thus, for example, each of the the numbers 2/3 , �, and  + is algebraic algebraic ut u t not not every number is algebraic, algebraic, and those that are are not algebraic are are said to be tancndnta If one knows the rick, it is not very hard to prove that transcendental numbers actually exist. (The trick, trick, discovered discovered by G antor ( 85 85 9 8), 8) , is to show that that the the set of real number numberss is uncountably large, while the se t of algebraic algebraic numbers is merely countably innite. ) What is much more difcult is to establish that some particular number is not algebra algebraic. ic. It is a famous famous theorem of . indema indemann nn ( ( 852 85 2 939), 93 9), for for example, example, that 1 is transc transcendental, endental, and this result is considere cons ideredd a major major accomplishment accompli shment of 9cent 9 century ury mathematics. (It is also known that is transcendental, but to demonstrate how hard such problems can be, we mention that it is still unknown whether or not the number 1 + is algebr algebraic. aic.)) One can ca n show that any number built from from algebraic numbers numbe rs by repeated addition, addition , subtraction, multiplication, division, and extraction of roots is itself algebraic, and so in particular, particular, the members of C are all algebraic algebraic.. If I f a circle could be suared, then would lie in C, C , and thus 1 would also lie in C since C is close c losedd under multiplication. ut by indemann' indemann'ss theorem, 1 is not no t algebrai algebraic,c, and hence it cannot lie in C This contradiction shows that circles circles cannot be suared. ecall that the d of a nonzero polynomial f (x) is the highest highes t power of x that appears appears with a nonzero coefcient in the polynomial. Thus, Thus , for example, the polynomial poly nomial number, then by denition, there is some 2x 4x + 5 has degree 3. If is an algebraic number, nonzero polynomial f(x) with integer coefcients coefcients such that f() 0, and we dene the d of to be the smallest sm allest possible pos sible degree of such a polynomial. polynomial . The key fact fact here, proved by elementary elem entary eld theory, is that since the members of C can all be built using us ing arithmetic and the extraction of suare roots roots,, it follows follow s that the degree of every every member of C is a power pow er of 2. To prove prove that some algebraic algebrai c number does not lie in C, theref therefore, ore, iti t sufces t o show that the degree degree of is not a power of 2, and so iti t would be usef us efll to be able to t o compute this degree. (We (We should shoul d mention, however ho wever,, that not every every real algebraic algebraic number whose degree degree is a power 2 actually lies in C In other words words the condition condi tion that the degree degree is a power of 2 is necess nece ssar aryy but it is not sufcient sufci ent.).) Suppose that f() 0, where f is a nonzero nonzero polynomial whose whos e degree euals the degree of the algebraic number In other words, words, we are assuming assumin g that the polynomial po ssible degree such that f() 0. We claim that in this case, f(x) f has the smallest possible cannot be factored factored as g(x)h(x) where neither g (x) nor hx) i s constant. cons tant. To To see se e why, why,
=
=
CONSTRUCTB NUMBRS
27
observe that if f (x ()() and neither () nor ) is constant, then each of () and ) ) must have have degree degree smaller tha thann the the degree degree of () ( ) ut by the minimalit minimalityy of the the degree degree of , , we cannot have have () () 0 or or () 0, and this is a contradiction contradiction since 0 () ()() This converse c onverse of this this fact is also true, true, but it is a bit harder harder to prove prove If () ( ) 0 and the nonzero nonzero polynomial polynomi al has no factorization factorization into two two nonconstant polynomials polyn omials with inteer coefcient coefcients,s, then the degree of is equal to the degree of This provides a tool that we can use us e to compute c ompute the degrees of certain certain algebraic numbers onsider, onsi der, for example, the number = � learly, is a root of the degree 3 polynomial (x) = x 2, and so our task task is to deterine whether or not this polynomial poly nomial can be factored factored as a product of two nonconstant polynomials with integer coefcients ssuming ss uming that there there is such a factorizat factorization, ion, we can write write  2 (ax +bx +c)(dx +e), where a, b, c, d, and e are integers It foll follows ows that a d = 1 and c e 2 2 and thus d = ± 1 and e has one of the four four values values ± 1 or ±2 Since dx + e is a factor factor of () ( ) however, however, it follows that ( e d) = O ut e d is one of the numbers ± 1 or ±2, ±2 , and it is a triviality to check that none of these four numbers is a root of This contradiction shows that our assumed as sumed factorization factorization cannot exist, and thus the degree of the algebraic number � is 3 Since 3 is not a power of 2, we conclude that that � is not constructible, and hence it is impos sible to double a cube using only a straightedge straightedge and compass compas s Finally, we deal with the number = cos(40) cos (40) We want to know whether or not is alebraic, alebrai c, and an d if it is, is , we want to deterine its its degree degr ee The key to this is the following tripleangle formula formula for cosines cosine s With a little manipulation of trigonometric trigonometric identities, we have cos(3) = cos( + 2) = cos() cos(2)  sin) sin(2) sin ())  sin()(2 sin()(2 sin) sin) cos() cos() ) cos()(cos ()  sin co s ()  3 cos() sin () = cos = 4 cos ()  3 cos() where where we obtained the the last equality by substituting 1  cos () for sin () Since Since cos(120) =  1 /2, /2 , we can apply this this formula formula with with € = 40° 40° to obta obtain4 in4  3 = 1/2, and and thus thus 8  6 + 1 O It follows that is a root of the polynomial ( x ) = 8 x  6 + 1 , and in particular particular,, is algebraic algebraic bx + c)(dx c) (dx + e), e) , where a , Suppose Suppo se that that there exists a factorizati factorization on f(x) = (ax + bx b, c, d, and e are integers integers Then a d = 8 and e c = 1 , and we see s ee that there there are are only eight possibilities pos sibilities for the number ed, and and thes thesee are are ±1 , ± 1 /2, ± 1 /4, and and ± 1 /8 ut since since dx + e is a factor of f(x), we kow that ed must be a root routine check shows that none none of these eight eig ht possibilities poss ibilities for for e d actually is a root of , and it foll follows ows that the assumed factorization of cannot exist y our earlier remarks, we conclude that the degre degreee of cos (40) (40 ) is 3, 3 , and thus thus cos (40) (40 ) is not a constructible number since 3 is not a power of 2 We conclude that it is indeed impossible impos sible to construct cons truct a 40° 40° angle, and hence it is impos i mpossible sible to trisect an arbitr arbitrary ary given given angle using only a straightedge and compass
08
CAPTR CAPTR
GOMTRC CONSTRUCTONS
xercses 6 6E.l
6E.
6F
If and lie in the set C of constructible cons tructible numbers prove that that is constructible con structible Is it possible pos sible to construct con struct with straightedge straightedge and and compass a line lin e segment whose length is equal to the th e circumference circumference of a given circle? circle ? Explain how you know
hgg th Ru
In this section, s ection, we discuss discus s some of the constructi constructions ons that can be done using nonstandar nonstandardd tools or by using the usual tools and following nonstandard rules We begin wth a pretty pretty angle angle tris trisection ection construction attribu attributed ted to to rchimedes rchimedes (287 (2 87 ?21 ?2 1 2 B C ) lthough rchimedes trisection construc cons truction tion uses only a straightedge and compass, compas s, it does not follow the the us usual ual ofcial rules because b ecause iti t requires requires marks to be placed on the straightedge straightedge Suppose that the angle we want to trisect trisect is LABC, as shown in Figu Figure re 6 1 8 hoose a point P on side B C of the the angle, angl e, draw draw the circle through P and centered at B , and extend side AB, as sshown hown in the the gure The next task is to draw the line labeled P R in the gure where R lies on the line A B and where P R meets the circle at a point Q such suc h that Q R is equal eq ual to the radius radius of the circle lthough ltho ugh it is imposs imp ossible ible to construct this this line if we limit ourselves to the usual usu al tools and rules rules it is easy with the following following procedure procedure First, we make two mar marks ks on our straightedge straighted ge positioned pos itioned so that the distance between them is equal to the radius of the the circle We then place our marked straightedge straightedge on the paper in such a way tat it runs through the point P and so that t hat the mark mark farthest farthest from from P is at the point where the the straightedge crosses cross es line A B. The dashed dashed lne lne in Figure Figure 6 1 8 shows s hows a possible poss ible postion postion for the straightedge, where the two marks are indicated by the arrowheads We then slowly turn and slide the straightedge so that it continues to run through the poit P , while whil e the mark mark that started on line A B remains on that line and moves toward the point B . n easy way to picture picture this tuandslide process process i s to imagine that that a pin i s stuck into the paper at P We move the mark along line li ne A B while keeping the straightedge press pressed ed against the pin We continue this tuing and sliding of the straightedge until the second mark mark just ju st touches touche s the circle (Note that if the straightedge starts in the position po sition indicated by the dashed line in i n the gure we want to turn turn it counterclockwi counterclockwise se and slide it upward c
R
Figue 6.18
CHANGNG TH RUS
29
so as to attain our goal.) Now, with the straightedge passing through P , and with one mark on line A B and the other mark on the circle, we draw line P R. We now claim that L A R P = LAB C , so that we will have our desired angle trisector trisector by constructing cons tructing the line through B paralle to R P
�
In Figure 6.18, point B is the center of the circle, R lies on line A B, and secant R P is drawn, meeting the circle at Q , where Q R is equal to the radius of the circle circle Show S how that LAR P = LABC.
(6.4) PROBLEM.
�
Since L P Q B i s an exterior exterior angle of isos is osceles celes � R Q B , w e see that L P Q B = LPRB + L QBR = 2LPRB ut � B Q P is also isosceles, and so L R P B = L P QB, and thus L R PB P B = 2LP RB Finally, L P B A is an exterior angl anglee of �B P R, and hence LP BA = LP RB + LR P B = 3 L R P B , as desired. desired.
oution.
B
c
D
Figue 6.19
et another angle angletrtrisec isection tion scheme s cheme is based ba sed on a tool that we will call a hatchet. The essential ess ential components of a hatchet hatchet are indicated indicated in the left left diagram diagram of Figure 6 1 9, while an actual actual hatchet hatch et that can be cut from cardboard and and used use d to trise trisect ct angles angle s is shown on the right. ine segment AC is divided into three three equal parts, and one of the trisection points , whch is not shown in the gure, is the center of the semicircle. The other trisection point is B, where the circle meets AC, as shown. ine B D is drawn perpendicular to to AC, and hence it is tangent to the semcircle. semcircle . We will use only point po int A , the semicircle, and line B D to trisect angles , but these three objects objects need to be held together somehow and that is the purpose purpos e of o f the the gadget shown on the right. The reader is urged to manufactu manufacture re such a device device and use it, as we w e are are about to explain explain to trisect some angle s. Suppose we w e want to trisect trisect LX Y Z lace the hatchet so that point A lies on side Y X of the angle, line B D passes through the vertex Y , and the semcircle is tangent to side Y Z (This is shown in Figure 6. 20 ) We claim that in this this situation line B D is a trisector trisector of LX Y Z
21
CAPTR CAPTR
GOMTRC CONSTRUCTONS x z
D
Figue
.20
In Figure Figure 6.20, line Y Z is tangent tangent to the circle at T and point A lies on Y X lso, lso, is the center of the circle, the circle meets A at its midpoint B , and B Y i s perpendicul perpendicular ar to A Show that LX Y B 1 L X Y Z
(6.5) PROBLEM.
raw Y and note that since B Y is the perpendicular bisector of A , we have Y A YO, and hence �YAB r � Y O B From From this, this , we see that LAYB = LBYO Furthermore, � Y O B r � Y O T by since O B OT, LOBY = 90° L O T Y , and O Y OY It follo follows ws that L B Y O L T Y O We see now that LX Y B = LXYZ, as required
oution.
1
What happens if we modify the standard construction rules rul es so as to make them even even more restrictive restrictive than usual? Suppose Supp ose for for example, example, that we we allow allow the use of a compass, compass , but not a straightedge straightedge and suppose suppos e furth further er that that we insist in sist that our compass compas s be b e o the ofcial ofcial class cl assical ical Greek type, type, without a memory memory What constructions are are pos sible under these rules? mazingly, virtually everything everything that that is pos sible with with a straightedge and compass compas s can be done with a compass alone Of course, we cannot draw draw a straight line without a straightedge, straight edge, but any construction con struction that can be done with a straightedge straighte dge and compass compa ss and that does not explicitly require us to draw a line can in fact, be done with a compass alone For example, given two points, we might be asked to construct the mdpoint of the ine segment they determine, or given three three noncolinear points A, B, and C, we might be asked to construct con struct the point D so that ABC D is a parallelogram, parallelogram, or we might be asked to construct the circumcircle of �A B C Each of these constructions co nstructions can c an be done with a compass alone and more generall generally, y, every every construc cons truction tion that is pposs ossible ible ith a straightedge and compass and that requires us to construct c onstruct a point or a circle, bu not a line, can be done with a compass compas s alone lthough we will not present a proof of the general theorem that a straightedge is never necessary, we will give several compassonly constructions, and these will demonstrate demonstrate some s ome of o f the techniques that that are available available We begin with a comparatively comparatively easy problem
CANGNG T RUS
211
Given two points A and B, construct with a compass alone the point C such suc h that that B is the midpoint of of A C
(6.6) PROBLEM.
raw the circle through B centered at A and the circle through throu gh A centered cente red at B and let P be one of the points where where these circles intersect Next, draw the the circle through B centered at P and, as shown in Figure 621, le Q be the point other than than A where this this circle meets the circle centered centered at B Finally, draw the the circle through B centered at Q and let C be the point other than P where this circle meets the circle centered centered aatt B Since points A and C each lie on the circle centered at B, we clearly have A B = B C, and so what remains is to show that A B C = 1 80 It is easy to see rom rom our construction, however, that AB A P B P B C = B Q = P Q = C Q, P , P B Q, and B QC Q C i s equilateral, equilateral, and each o f the angles and thus each o f A B P, of these triangles equals eq uals 60 • It follows follows that A B C A B P + P B Q + Q B C = 1 80 , as as requir required ed
oution.
x
Figue 6.21
B
y
Figue 6.22
Given points A and B , and using only a compass compas s construc the circle centered at A and having radius equal to A B
(6.7) PROBLEM.
First, use the previous problem to construct points X and Y, as shon in Figure Figure 6.22, 6. 22, where where A is the midpoint of X B and B is the midpoint of A Y , and thus XA = A B B Y Next Next construct construct point P so that A P A Y and and Y P = AB This This is easy: easy : Take Take P to be a point of intersection intersection of the the circle through Y centered at at A and the circle circl e through B centered at Y Y To avoid clutter in Figure Figure 622 6 22,, howeve, we have not drawn these circles Now draw an arc through A centered at X and an arc through P centered at B, and let Q be the point of intersection of these arcs, as shown That these circles really do intersect is clear from from the diagram diagram or by actually carrying out the construct cons truction ion lso ls o we will comment later on another another way to see this We will prove prove that A Q = A B , and thus the circle through Q centered at A solves so lves the probem probem To avoid avoid clutter in Figure 6.2 6 .222 we have not not drawn this this circle circl e either First, we argue that XQB r ABP We have XQ X A = AB and X B 2A B A Y = A P lso l so B Q = B P, and thus the triangles triangles are congruent by
oution.
212
CAPTR
GOMTRC CONSTRUCTONS
SSS SS S It follows that that Q X B = B A P, and we can rewr rewrititee this this as Q XA = PAY Now AY = 2AB = 2XA and and AP = A Y = 2XA = 2X Q, and thus thus A Y/X A = 2 = A P / X Q y the SS similarit si milarityy criterion, criterion, this proves proves that � Q X A r �PAY and thus the ratio Y P / A Q is also als o equal to t o 2, and we have A Q ! Y P ut Y P = AB an andd thu thuss A Q = ! A B as wanted wanted It is not hard to calcul cal culate ate the distance dist ance B P in Figure 6 6 22, 22 , and this provides prov ides an alteative argument argument to show why the two circles whose intersection denes the point Q realy do intersect For simplicity, let us suppose that A B = 1 Then Then also Y P = B Y = 1 and AP = 2 y Stewart's theorem (Theorem 220) applied in �APY, we have 4 1 = 2 + 2, where = B P It follows that B P = = and thus B A = 1 < B P < 2 = B X Thus Thu s A lies li es inside ins ide the circle through P centered centere d atat B, and out side of this circle c ircle It is now clear that the circle through A centered at X u u st X lies outside meet the circle circl e through P centered centered at B , as we saw s aw in our construction We can now construct the midpoint of the line segment deterined by two given points Given Given points A and B, B , construct with with a compass compas s alone the id id point of line segment A B
(6.8) PROBLEM.
In some sense this is i s now very very easy Just use roblem roblem 6 27 twice to constr cons truct uct circles circles centered about each of A and B and having radii equal to A B These Thes e two circles will be tangent, of course, course , and the unique point where they meet will be the desired d esired midpoint This seems to be cheating a little, however, and so instead of worrying about whether or not we have described legal construct con struction, ion, we present an alteative alteative whose leglity leg lity should not be in doubt et P and Q be the two points of intersection of the circle through A centered at at B and the the circle through B centered at A Then A P B Q is is a parallelogram and so he point we seek, the midpoint of A B , is also al so the midpoint of P Q, Q , and hence it lies li es on the circle centered at A with radius A B and also on the circle centered at P with radius ! P Q Each of these these circles c ircles can be constructe cons tructedd by roblem roblem 6 27, and we see that one of the the two points points of intersect intersection ion of these cicles is the desired des ired midpoint If there there is any doubt about which of thes thesee two intersection points is the correct one, draw the circle centered at B with radius ! A B That That circle will pas p asss through just jus t one of the the two intersection points, and that is the point we want
oution to Probm 6.8.
!
It is now easy to do d o the completetheparallelogram completetheparallelogram problem, mentioned mentione d earlier earlier Given three noncollinear points A, B and C construct wth a compass alone the point D such tha thatt AB C D is a parallelogram
(6.) PROBLEM.
First, use roblem 628 to construct the midpoint M of AC and then then use roblem roblem 62 6 266 to construct construct the point D such suc h that M is the th e midpoint midpoint of B D. Then
oution.
CANGNG T RUS RUS
21
diagonals diagonal s AC and B D of quadrilate quadrilateral ral AB A B C D have the common comm on mdpoint M, and so these diagonals bisect each other. It follows by Theorem 1.9 that ABCD ABC D is a parallelogram as desired. It is a consequence of robem 629 that given three points A, B, and C and working with a compas com passs alone, alon e, we w e can draw a circle of radius A B cenered at C. To do this, construct D as in roblem roblem 629 6 29 and draw draw the circle circle through through D and centered at at C C Even for for compass compa ssalone alone constructions, co nstructions, therefo therefore, re, we can assume that our compass has a memory. We close with a problem that is an exampe of another variation on the theme of straightedge and compass constructions Since the ideal Greek construction tools are allowed to use the entire Euclidean plane, we w e have have not, up to now, considered con sidered the size or shape of the paper on which we are are working. ut ut consider cons ider this this problem. proble m. PROBLEM. Given Given nonparall nonparalle e line segments A B and C D on a rectangular rectangular piece of paper paper and a point P lying between them, let X be the point where lines A B and C D meet, but do not assume ass ume that X lies lie s on the paper. onstruct ons truct a segment of the line P X, using usin g only a straightedge and compass, compass , but keep keep all of the construct cons truction ion work on the paper p aper
D
R Figue 6.23
The enclosing enclos ing rectangle in Figure Figure 6.23 6. 23 represents the paper paper on which we are are working. hoose hoo se arbitrar arbitraryy points Q and V on AB and R on CD , as shown shown raw lines Q P and Q R and construct lines through V parallel to these two lines In particular, particular, this thi s determines the point W on CD C D such that that V W Q R. Next draw cons truct the line line through W parallel to R P s shown in Figure Figure 6.23, 6.2 3, this R P and construct determines determines a point U such that W U "R P and V U " Q P Finally, draw P U, which is the th e desired line segment. s egment. The Th e fact fact that line P U pass passes es through through X is a consequence of Theorem Theorem 6 3 1 , which we state and prove prove separate separately. ly.
oution.
Suppose that corres corresponding side sidess of of P Q R and U V W are THEOREM. Sup parallel parallel Then lines Q V P U and R W are are eiher concurrent or parallel
214
CHAPTR CHAPTR
GOMTRC CONS TRUCTONS TRUCTONS
If the thee lines line s are parallel parallel there is nothing to prove, and so we w e assume ass ume that two of them (say, (s ay, Q V and R W) meet at some point, and we will use vectors to show that point also als o lies on P U . = a P for some scalar a , and First, since P Q " U V , we can write = and = Q for s ome sc alars and Since = simlarly, and = P + Q we have +
Proof.
V y y V + Q = = = + V = a + yQ y
and thus thus ( ( a) P =  ) ut P Q is not parall parallel el to Q R, and so the only way that a scalar scal ar multiple of the vector P can equal a s calar multiple of the vector vector Q is if both scalars are O In particular, we conclude that a = 0 =  , = a and = a Q In particular, if and and so = = a and we have have and thus Q V WR parallelogram am This is not W R i s a parallelogr = that a = 1 we see that the case, however h owever,, s ince we assumed ass umed that lines Q V and R W have a common point and we conclud e that a 1 We cho ose the point P to be our origin so that by the notational convent c onvention ion of hapter 5 , we can write Q = and R = . lso,
y
V,
V
y
and similarly since = a we have
w = + = + a = + aR .
Since a 1 , we can can writ writee = 1/(1 a) , and we consider the point X on = line Q V such that We have
X=Q+ = Q + = Q + A(V Q) = Q + ( + a Q Q) =
where where the last equality follows follows since 1 + a = O Similarl Simil arly y if Y is the point on = , we compute that Y = It follows that X = Y, line R W such that and thus thus X and Y are the same point, and we conclude that that X is the intersection intersection point of lines Q V and R W = X = = , and this shows that X lies We know that tha t lie s on line PU Since X also al so lies on both Q V and R W, it follows that lines P Q V, and R W are concurrent as required We mention mention that Theorem 63 6 3 1 can be thought of of as a special s pecial case cas e of the the converse of esargu es argues es'' theorem theorem in pro projj ective geometry geometry (See ( See Exercis Exercisee 47) 47 )
CHANGNG THE RUES RUES
xercses 6F
215
Suppose that our only construction tool is a twoedged straightedge where the two edges are are parallel. Show how to construct the the perpendicular bise bisector ctor of a given line segment IN: o the case cas e rst where the length of the given segment exceeds the dis dis tance between the two edges edg es of our straightedge. straightedge. egin by placing the straightedge straightedge on the paper so s o that one of its parallel edges runs though though and the other edge runs through 6F. sing only only a twedged straightedge, erect a perpendicular to a given line through a given point on the line. 6F.3 sing only a twoedged straightedge, construct a line parallel to a given line though though a given point. p oint. IN: egin eg in by drawing drawing three three parallel parallel equally spaced line s where where one of these these lines lin es goes goe s through the given point. onsider the points where these three lines meet the given line. 6F.4 sing s ing only a twoedged twoed ged straightedge, straightedg e, drop a perpendicular perpendi cular to a given line from a given point not on the line. line . 6F.5 sing only only a twoedged straightedge, constr cons truct uct a regular octagon. 6F.6 Suppose Suppo se that we work on lined paper, paper, where the printed printed lines line s are parallel and equally spaced. sing only an ordinary straightedge, construct a line through a given point parallel to a given given line. line . IN: This Thi s problem is much easier eas ier if the given point lies on o n one of the prited lines. o that that case rst 6F.7 Given points and and using only a compass, construct the circle centered at with radius radius 6F.l
1
Some Fur Further Reading Readi ng
Needless to say for a subj subj ect as old as geometry, geometry, there are are countless books on the suject dating dating from from Euclid s Elements up to the present. The purpo purpose se of this brief bibiogrphy is to introduce a few works that we believe mght be of interest to readers of this this book. b ook. We begin, of course, course , with Euclid, himself. Euclid, The Thrteen Books of the Elements (3 vols.) vols .),, transl translated ated with an introduction introduction and commentary commentary by Sir S ir Thomas . eath. over, New ork, 1 956 95 6 This is probably not the best source if you really want to lea le a more geometry, geometry, but it is wonderul wonderul to see the book that started started it all. all. In these volumes , one can c an also nd an extensive and and scholarly s cholarly historical commentary. commentary. eath eath s translation translation was rst published in 1 908 and a revised version appear appeared ed in 1925 1 925 There There is vastly more to geometry than than we have discus sed or even mentioned in this book boo k samplin s amplingg of some of this other materi material al can be found found in the following work by one of the most distingui di stinguished shed geometers of mode mode times. times . Geomet Wiley . S . M. oxeter, oxeter, Introducon to Geomet Wiley,, New New ork, ork, 1 96 1 In the rst 25 or so s o pages of oxeter's beautiful beautiful book, readers will nd a number of familiar familiar topics and theorems: the Euler line, line, the ninepoint circle Fermat' Fermat'ss minimzation problem, and Morleys theorem, to name some of them. ut ut oxeter goes on to discus s completely diff different geometric geometric ideas . These Thes e include symmet sym metry ry groups, proj proj ective geom etry nonEuclidean nonEuc lidean geometry, dierentia dierentiall geometry geometry topology, and higher dimensional geometry geometry oxeter oxeterss book bo ok should not be thought of as a compendium of theorems theorems it is a ouquet of ideas somewhat loosely loo sely tied together by the theme of geometry. geometry. s oxeter's Introducto demonstrates there is much more to geome Introductonn to Geomet Geom et demonstrates try than yet another theorem" Nevertheless Nevertheless , it is also true that that there are many beatiful beatiful theorems theorems that we have not been able to include inc lude in this book. Some So me of these can found in the olowing olowing work which seems well on o n its way to becoming a classi clas sic.c. • . S. M. oxeter and S. . Greitzer, Greitzer, Geomet Revsted New Mathematical Mathematic al ibrary, Mthemaic Mthemaica a ssociati ss ociation on of merica, merica, Washingto Washington n .,., 1 967 967 Geomet Revsted incudes inc udes many of the topics we have discus discu s sed and a numbe that we have have not. It also includes some interesting historical commentary commentary and a large number of exercises with hints and answers. The large overlap in topics between Geomet 216
SOME URTHER URTHER RADNG
217
Revisited and this tet is not a coincidence coi ncidence since, sin ce, in fact, I used u sed oeter and Greitzer Greitzer as a source of ideas for for the geometry geometry course upon which this book was based. base d. Still more really neat theorems can be found in the following following work. • onsberger, Episdes in Nineteenth nd Twentieth Centu Eucliden Gemet. New Mathematical Mathematical ibrar ibrary, y, Mathematical Mathematical ssociation ssoc iation of merica, merica, Washington, ashington , . . , 1995 s the title title suggests sugg ests,, the focus of this this book is on some so me of the newer discoveries dis coveries in geometry. geometry. mong mo ng these, these , we cannot resist mentioning an especially especi ally pretty pretty result about isogonal conjug conjugat ates es:: If P and Q are isogonal is ogonal conj conj ugates in the interior of 6ABC and perpendiculars perpendic ulars are dropped from from each of these points poin ts to the sides side s of the the trian triangle gle,, then the si feet of these thes e perpendiculars perpendi culars lie on o n a circle. circl e. The proofs in this book are clearly written and easy to follow, althoug a lthoughh not all of the amazing theorems that the author menions are proved. ere is the answer for for readers who feel feel that we did not provide enough, or hard enough, eercise eercisess in this book. • . S. osamentier Gem et over, New osamentier and . T. Sal S alkind, kind, Chllenging Prblems in Gemet ork, 1996 In this reprint of a book rst published in 1970, there are solutions as well as problems. Finally, Finally, we come to what can only be described as an encyclopedic collection of theorems in Euclidean Eucli dean geomet geo metry. ry. New ork ork 1 952. 95 2. • N. ltshillerourt, Cllege Gemet. es and Noble, New This book, which is a reprint of a work rst published in 1925, contains a vast number of interesting theorems and an equally vast ollection of eercise eercisess . lthough this book was written as a tet, many of today's today' s college students might nd nd ltshiller ourt's ourt' s proofs proofs somewhat difcult difcult to follow. Neverth Nevertheles eless,s, this is i s the place to go to look up some obscure geometric g eometric fact. fact. nfortun nfortunatel ately, y, this book b ook has long been out of o f print, print, but it should be b e available in a good library.
Index
AA similarity crterion acute angle addition formula formula addition foulas or sine and cosine addition rincie for roortions 2 adj acent oint of set algebraic algebraic number number 2 alteate eterior angles 2 alteate alteate interor interor angles 2 altitude and circumcircle altitude altitude of triang triangle le altitude altitude oot oot of 2 altitudes o equal length altitudes altitudes concur concuren ence ce of angle between Simson lines angle between vectors ange bisecto bisectorr 2 and circumcircle ormula rom trangle sides of edal triangle angle bisector bisector construction of angle bisectors of equal length angle bisectors concurence of2 of 2 2 angle trsection 2 angle trsection Archimedes method 2 angle trisection with hatchet 2 angle trisecto in Morley's theorem 2 angles determined by trangle sides angles o ara arallelogram llelogram angles angles of olygon olygon angular Cevian roduct Aolonian construction roblems Aolonius o Perga aology 2 aothegm o olygon arc 2 2 arc subtended by angle 2 Archimedes 2 area area and checkerboard
area area maimiation maimiation of inscribed olygon area area of trangle trangle 22 fom side lengths 22 rom rom sid es and angle 2 areas of similar gures arithmetic arithmetic of vectors vectors arm of rght trangle arows and vectors ASA congruence crteron aio aiom m aiomatic method ais Lemoine ais radical 2 22 balance eeriment base base angles in isosceles triangle base in area foula between bisector o angle 2 bounded bounded set buttery roerty roerty Buttery theore theorem m Cantor G 2 center of circle construction of o mass 2 o nineoint circle of square central angle 2 centroid and circum center center coincid co incid e and incenter coincide using vectors Ceva's theorem 2 general fo Ceva G 2 Cevian roduct 2 and angles Cevian Cevians s conc concur ure enc ncee of 2 218
NDEX Cevian interior and eterior eterior checkerboard and area chord 2 chords angle between 2 circle 2 circle squaring 2 circle Buttery theorem for division into into equa equall arcs arcs 2 circles radical ais of 2 22 circes radical center of 2 circumcenter circumenter and centroid coincide and incenter distance between circumcirce 2 2 and altitude and ange biector of eda eda triange triange 2 and erendicular biector reections of circumradiu 2 determined by side ength and inradius comarison circumcribed 2 cosed set cocircuar cocircuar oints cross ratio ratio for for 2 coine coinear ar oint oint 2 2 cros ratio for and Menelau' theorem common chord 2 common common ecant ecant ine comm common on tan tange gent nt line line 2 common tangent ine construction of 2 comac comactt et et comass and straightedge straightedge construction construction 2 Greek 2 memo memory ry of 2 2 comasony construction 2 2 comlementary angles 2 concurence of altitudes of angle bisectors 2 2 of Cevi Cevian anss 2 of common tangents of diagonal of inscribed inscribed heagon of Euler ine of isogona Cevian of media dians of nineoint circle of er erendicular endicular bisector bisector 2 of radical ae 2
of Simon line of symmedian 2 concurent circe conc concur urre rent nt ine ine 2 congruence criterion congruent gure contructibe number 2 construction imosibe of radica ai 2 rue for 2 on ae aerr 2 2 without without comas comas 2 without straightedge 2 contradiction a method of roof conve olygon coordinate geometry 2 coresonding angle for for arael line oint in congruence sides for simiar trange coine addition addition formua formua for cosines aw of crisscros cros ratio for cocircuar cocircuar oint 2 for collinear oints and and ine ine cube doubling 2 cycic quadrilateral degree of oynomial and agebraic number 2 Desarg Desargues' ues' theo theorem rem 2 Deargue G Descartes diagona diagonall of olygon olygon diagonals of incribed heagon of arale araleogram ogram of quadrilater quadrilateral al 2 of traeoid diagram information from 2 diameter diameter 2 2 diameter diameter of nineoint circle distance between incenter and circumcenter to line division of circe into eqal arc 2 of line line egment into equa equa arts dot roduct doubling a cube 2 droing and erecting erendicuars
19
220
INDEX
elipse, Bu ttery ttery theorem theorem or or 1 1 6 elliptic geometry 5 equa in degrees, for arcs 25 equilateral trangle 34, 38, 60, 79, 89, 130 in Morey's theorem 82 in Napoeon's Napoeon's theoem theoem 93, 17 8 equilateral trangles sharing verex 176 Eucid 1, 4, 200 Euler line 60, 66, 165 Euler point 62 Euler point, distance to vertex 66 Euer, L. 60, 77 excrcle, tangency points or 80, 130 experiment o nd centroid 57 experiment to nd Feat point 92 exradii and inadius, equation relating 80 exscibed cicle cic le (see also excicle) 80 extend extended ed law o sines 5 2 exterior exterior angle o f triangle triangle 12 exteior exteior Cevian 1 3 1 Fermat point 89, 140 Fermat prme 200 foot o altitude 62, 88, 94 foundation o geometry 3 four points on cicle cicle 67, 104, 1 12 Gauss, C 5, 200 general position 100 Gergonne point 25, 130 Gree Greek k compas compasss 8 3, 2 0 H A congruence congruence criterion criterion 9 hatchet 209 height, in area ormula ormula 18 Heron o Alexandria 22 Heron's ormula 22, 68 hexagon 187 concurrent diagonal o 144 inscribed inscribed 19, 44 hexagonal Cevian product 145 Hilbet, D 3 hyperbolic geometry 5 hypotenue 9 ideal point 32 if and only i 30 impossible constr constructi uction on 19 8 incenter 59, 142 and centroid coincide 79 and circumcenter, ditance between 77 and orthocenter coincide 79 incident point and line lin e 50
incircle 73 incirce tangency points 75, 125 innity, point at 132 inradius 73, 77 and circumradius, comparison 79 determined by side lengths 74 and exradii, equation relating 80 inscribed angle 25 circle 73 hexagon, diagonals o 144 polygon 24, 34 quadrilaeral 26, 55, 72 insect 105 inter intero o Cevian Cevian 1 3 1 isogonal Cevian 141 conjugate points 142, 143, 146, 217 isosceles trapezoid 17 triangle 7, 8, 57, 7 law o cosines 67, 76, 163 o ines 20, 42 o sines, exended 52 of tangents 76, 81 Lemoine axis 49 Lemoine point 142, 146, 149 length o vector 158 length f vector, rom dot poduct 163 Lindemann, C 206 line egment egment 2 line segment, division ino equal pats 189 linear operato 172 lined paper, constuction on 25 Lobachevsk, N 5 locu locuss 16, 18, 49 , 6 6 , 7 3, 3 , 20 20, 1 21 21 mas points 57, 60, 29 maxmin problems 34, 36, 85 mean mean poportional, poportional, constuction o 95 medial triangle 59, 13 cicumc cicumcircl irclee o 63 median median 8, 2 , 23, 43, 1 3 0 median of equal length 0, 57 concurence of 48, 56, 5 6, 159 memory o co c omp mpaas 83, 210, 23 Menelaus' theorem and collinear points 147 midpoint midpoint using vecrs 159 midpoint, compa compassonly ssonly constuction constuction 2 12 Morley conguration 83 Morley, F 82
NDEX mutua mutually lly tan tangen gentt circ circle le Naoleon nineoint center circle circle and Simson lines radius nonEuclidean geometry nonmetric geometry notation hortcut for vector obtue angle oerator linear 2 otiation origin or vector notaion shortcut orthic quadrule orthic triangle (see alo edal trangle) 2 orth orthoc ocen ente terr 2 and incenter coincide and and Simon lines reection of aer contruction on 2 ae aer r lined 2 Paus' theorem ara arall lle e Cevia Cevians ns arallel line to side o triangle 2 lines lines ostulate aralelogram angle of area rea of diagona of Paca Paca B edal triange 2 angle bisector o midoint midoint of ides erimeter erimeter o entagon 2 2 erimeter minimiation of eda triangle of olygon olygon erendicuar biector and circumcircle circumcircle construction o erendicular erendicular bisector concurence concurence of 2 erendicular Simon lines erendicuar erendicuar vectors vectors erendiculars to sids o edal triangle ersective hotocoy
hysic 2 2 i 2 2 2 2 oint adjacent to set oint at innity innity 2 oints of tangency of ecircle of tangency of incircle ole ole for for Simson ine olygon 2 squarng of olynomial 2 on asinorum converse ostulat ostulat ower o oint 2 2 rime number 2 rojection ro jection oint oint rojective geometry lane 2 roof by contradiction roo ro o use o "sim ilarly in roof writing o roortional ength Ptolemy' ormula 2 Pyth Pythag agor orea eann theo theorem rem 2 2 QED quadrilateral 2 2 center of ma of cyclic diagonals diagonals of 2 incribed 2 midoints o side square on ides of radical ai 2 22 construction of 2 radical center 2 2 radiu 2 radiu o nineoint circle rairoad tracks ratio contruction roblem 2 rational number reconstruction reconstruction of triangle rectangle area o ree reect ctio ionn ree angle reguar reguar heagon heagon contruction of reguar entagon 2 regular entagon contruction of 2 regular oygon 2 remote interior ange 2
1
222
NDEX
rhombus rght angle 2 2 rght trangle (see also Pythagoean theorem) rght trangle congruence crteron o rgor 2 rotaton of vectors 2 SAA conguence crteron SAS congruence crteron SAS smlarty crteron scaar roduct scale factor seca secant nt lne 2 segment segment o ne 2 semcrcle seme semem mete eterr 22 shortcut notaton o vectors sdes o arallelogram trangle n ormula or angle bsector trngle n omula omula for angles angles trangle n ormula fo area 22 trngle n ormula or crcuadus trangle n ormula or nradus smlar gures n general smlar smlar trangle trangless smlarty and area smlaty crteron Smson lne arallel to sde Smson lnes angle between ntersecton of Smson's theorem converse Smson R sne addton formula or n area ormula 2 n Cevan roduct n cross cross rat ratoo oula oula snes etended law o 2 law o 2 2 squae squae center of squares wth common vete on sd es o quadrlate quadrlateal al squang a crcle 2 oygon
angle SSS congruence cteron smlarty crteon Stewart's theorem straght angle 2 straghtedge and comass constructon 2 wth two edges 2 straghtedgeonly constructon 2 sub eld 2 subtended arc 2 subtracton rncle for roorons 2 sulementay angles 2 symmedans concurence o 2 symetrc onts 2 angency onts of ecrcle ncrcle ncrcle 2 tangent crcle tangent lne const constuct ucton on o tangent lnes to crcumccle tangents law law o The Element Elementss 2 transcendenta transcendentall numbe 2 transormatonal geometry transv transves esal al traeod trangle area o 22 reconstuct reconstucton on of squarng of an angl gles es on sdes sdes o tra trang ngle le tgonometc unctons sectng an angle 2 Achmedes method 2 wth hatchet 2 trsector o angle 2 trtangent crcle twocolumn roo ormat twoedged staghtedge 2 vector dot roduc notaton shotcut leng length th of vectors angle between erendcular vercal angles 2
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