AN AUSTRALlAN MAT11EMAT1CS TRUST PUBLlCATlON
Published by A U S T R A L I AN MATHEMATICS T R U S T
Australian M athematics Trust University of Canberra ACT 2601 AUSTRALIA
Copyright
©
1998 Australian Mathematics Trust
Telephone: +61 2 6201 5137
AMTOS Pty Ltd ACN 058 370 559 National Library of Australia Card Number and lSSN Australian Mathematics Trust Enrichment Series lSSN 1326-0170 Polish and Austrian Mathematical Olympiads 1981-1995 ISBN 1 876420 02 2
THE A USTRALlAN MATHE MATICS T R U S T
ENRlCHMENT
SERlES
EDlTORlAL COMMlTTEE •
Chairman
GRAHAM H POLLARD, Canberra AUSTRALIA
•
Editor
PETER J TAYLOR, Canberra AUSTRALIA
WARREN J ATKINS, Canberra AUSTRALIA ED J BARBEAU, Toronto CANADA GEORGE BERZSENYl, Terra Haute USA RoN DUNKLEY, Waterloo CANADA WALTER E MIENTKA, Lincoln USA NIKOLAY KONSTANTINOV, Moscow RUSSIA ANDY L!U, Edmonton CANADA JORDAN B TABOV, Sofia BULGARIA JOHN WEBB, Cape Town SouTH AFRICA The books in this series are selected for the motivating, interesting and stimulating sets of quality problems, with a lucid expository style in their solutions. Ty pically, the problems have occurred in either national or international contests at the secondary school level. They are intended to be sufficiently detailed at an elementary level for the mathematically inclined or interested to understand but, at the same time, be interesting and sometimes challenging to the undergraduate and the more advanced mathematician. lt is believed that
these
mathematics
competition
problems
are
a
influence on the learning and enrichment of mathematics.
positive
THE AUSTRALIAN MATHE MATI CS TRUST
ENRlCHMENT BOOKS 1
1
1
2
SERlES
THE SERlES
lN
ALL THE BEST FROM THE AUSTRALIAN MATHEMATICS COMPETITION
JD Edwards, DJ King
Et
PJ O'Halloran
MATHEMATICAL TOOLCHEST
AW Plank
Et NH
Williams
3
TOURNAMENT OF TOWNS QUESTIONS AND SOLUTIONS 1984-1989
4
AUSTRALIAN MATHEMATICS COMPE11110N BOOK 2 1985-1991
PJ Taylor
PJ O'Halloran,
G
Pollard
Et
PJ Taylor
5
PROBLEM SoLVING VIA THE AMC
6
TOURNAMENT OF TOWNS QUESTIONS AND SOLUTIONS 1980-1984
7 8 9 1
10
I
11
1
12
W Atkins PJ Taylor
TOURNAMENT OF TOWNS QUESTIONS AND SOLUTIONS 1989-1993
PJ Taylor
THE AsiAN PACIFIC MATHEMATICS OLYMPIAD
H Lausch METHODS OF PROBLEM SOLVING BOOK 1
JB Tabov
Et
PJ Taylor
CHALLENGE! 1991-1995
JB Henry, J Dowsey, AR Edwards, U M ottershead, A N akos Et G Vardaro USSR MATHEMATICAL OLYMPlADS 1989-1992
AM Slinko
AUSTRALIAN MATHEMATICAL OLYMPIADS 1979-1995
H Lausch
Et
PJ Taylor
1
13
CHINESE MATHEMATICS COMPETITIONS AND 0LYMPlADS 1981-1993
1
14 PoLISH AND AusTRIAN MATHEMATICAL OLYMPIADS 1981-1995
A Liu
ME Kuczma
Et E
Windischbacher
FOREWORD
The traditions of National Mathematical Olympiads in many European countries dates back to about 1950 (in some cases further back). These Olympiads are used, inter alia, to select national teams to participate in the International Mathematical Olympiad. The traditions of the Polish and Austrian Mathematical Olympiads are particularly strong, and to a certain extent they are linked, since together they developed the Austrian-Polish Mathematical Olympiad, one of the world's strongest regional events. As the reader will determine, the problems in this book are quite exquis ite, having been hand-picked from the problems of many years. They are also noted for having multiple independent solutions, making the math ematics so much richer. There can be little more satisfying than finding a different, independent solution to a known one. Being mathematics, of course, the result is always the same after having taken a quite different route. The authors of this book have many decades of experience at this level. Of the two, I have only had the pleasure of personally knowing Dr Kuczma. Dr Kuczma has one of the world's highest reputations in prob lem creation. Indeed, he has had no less than four of his problems posed in International Mathematical Olympiads. Further, he has had many more reach the final preselection stage. He is also equally renowned as a problem solver. From mutual acquaintances and examination of the work in this book, Erich Windischbacher is held in no less regard. The Australian Mathematics Trust aims to set a high standard of mate rial and exposition in this Enrichment Series. The contents of this series involve pedagogical material in problem solving and instructive problems which have not appeared before in English. We are confident that this book achieves the high standards to which we have aimed. Peter Taylor Executive Director Australian Mathematics Trust Canberra 4 August 1998
PREFACE Mathematics Olympiads have a long tradition in Poland as well as in Austria, and they have many features in common in both these countries. Academic supervision comes from the Mathematical Societies and from university centres. Financial support is provided, in the greatest part, by the Ministries of Education (the exact official name of that institution, . in each country, has changed several times during the past decades). The effective running of the competitions relies on people (high school teachers and university teachers) whose enthusiasm and devotedness is practically the sole motive for their activities. The organizational format is much the same in the two countries. Con testants are high school students, most of them attending the last or the last but one grade. College students are not allowed to participate. The final round of the Austrian MO and of the Polish MO is a two day written exam, with three problems to be solved each day -just like the IMO. As regards earlier stages, there are some differences; but, anyhow, ea<::h elimination round consists of problem solving. All the problems posed at our olympiads are essay type; all steps of the reasoning have to be explained and justified by the solver - short answer questions or multiple choice questions are not used. In the late seventies, a bilateral agreement on cultural exchange was concluded between the Polish and the Austrian Ministers of Education. This resulted, in particular, in frequent visits of scientists and teachers, from one country to the other, and has led to exchange of experience for instance, in the org�nization of math olympiads (remember that, in those years, Austria and Poland pertained to distinct political zon� of Europe). It is also in that time that the Austrian-Polish Mathematics Competition was launched.1 The authors of the present book are just two of those "enthusiasts of the Olympic idea in mathematics", for many years involved in the running of the national mathematics olympiads in Poland and in Austria. It is quite a time ago that we first met. Soon the idea occurred to us to present, in book form, a selection of our countries' olympiad problems. As a guideline for the selection, we have decided to take the diversity of methods of solution. Accordingly, each problem in this book is presented 1A
compilation of all the problems posed at the first sixteen rounds of that compe
tition, with complete solutions, has appeared in book form:
ME Kuczma,
144 problems of the Austrian-Polish Mathematics Competition 1978-93,
Problems.
published
1994 by: The Academic Distribution Center, 1216 Walker Rd., Freeland, Maryland 21053, USA.
viii
Preface
with at least two solutions, and sometimes more than two; this feature of the book we consider important enough to be reflected in the sub-title. It is obvious that various ways of approach to any problem provide a better understanding of its nature, reveal several aspects of the relevant topics and teach various techniques. Now, it can always be questioned whether a different solution is a really different one. In some rare cases, it can be justly considered as such. In many cases, it cannot - and this is evident at first glance. And in most other cases- also not; the "second" method can use other sym bols, language, terminology, it may look quite unlike the "first" one, and still be, in fact, the same. For instance: is the Law of Cosines any thing else than operating with vectors and their inner products? Is the examination of divisibility of polynomials via manipulation in real do main anything essentially different from complex roots and factorization technique? Combinatorial arguments, when disguised in the language of polynomials (in fact, the generating functions of the quantities un der consideration) , do they really differ from the analogous arguments presented in pure form, without disguise? This list can be continued, of course. Viewed from a certain level of professionalism, all or almost all approaches to a particular olympiad style problem are just like dressing the same idea in a robe of one or another colour. What can be, however, immediately recognized by a mathematician, need by no means be evident to a young student who just makes the first steps in off-curricular areas of mathematics. Indeed, we think that - besides getting acquainted with various tools and tricks supplied by various methods - the reader's own discovery of the intrinsic uniformity hidden behind apparently distinct ways of ap proach is the true profit she or he can have from studying those solutions, and is the best we can offer her or him. Most of our solutions have been elaborated in detail. The intention was to make them accessible to a rather wide audience; some readers will find them unnecessarily lengthy, perhaps. We are sure that the readers' invention will often go further; no doubt, they will find yet other ways of resolving this or that problem, possibly more elegant or more general than the presented ones. So much the better! Satisfaction from a good job done is the solver's true reward. ( Another kind of satisfaction comes from detecting the authors' errors and mistakes; these are also very instructive! ) There is one more thing we must mention here. There should be no surprise if a problem turns out to be identical or very closely related to a question that had appeared at some other competition or in the problem section of some journal. It is no secret that problems "circulate" and
Preface
ix
are being "borrowed" from one competition to another. There is also nothing paradoxical in the fact that very similar ideas occur to people who independently devise olympiad stuff, in distant parts of the globe. Although we have tried to avoid the use of problems of which we knew to have been used elsewhere, we can by no means be sure. . . We are presenting 64 problems (a beautifully round number) from the two national olympiads, half from the Austrian, half from the Polish. 2 They are arranged more or less thematically; the rough rules are (ilasy to spot. Such rules can never be quite univalent; a problem may be difficult to classify; it can pertain to more than one thematic area, sometimes depending on the solution method. The arrangement has nothing to do with the level of difficulty; quite challenging problems often follow or are followed by trivially simple ones. The reader should not know "what to expect next". We will be happy to receive any feedback from the readers: comments, communication about mistakes, any suggestions. We wish all the readers joy, fun and pleasure in tackling the problems. Marcin E Kuczma
Erich Windischbacher
Institute of Mathematics University of Warsaw ul. Banacha 2 PL-02-097 Warsaw Poland
Bundesrealgymnasium KeplerstraBe 1 A-8020 Graz Austria
2Problems from the Austrian MO: 1, 4, 5, 7, 9-12; 16-19, 23, 28-30, 32-35, 37,
38, 43, 52, 53, 55-61.
Problems from the Polish MO: 2, 3, 6, 8, 13-15, 20-22, 24-27, 31, 36, 39-42, 44-51,
54, 62-64.
ACKNOWLEDGEMENTS
We are happy to see our book appearing as an AMT publication. Our sincere thanks go to Professor Peter Taylor, the executive director of the AMT, for his invitation to publish this book in the AMT Enrich ment Series and for his help in typesetting/ formatting; and to Dr Andrei Storozhev for producing the diagrams. MEK&EW May, 1998
CONTENTS •
FOREWORD
•
PREFACE
•
ACKNOWLED G EMENTS
•
PROBLEMS Arithmetic and Combinatorics Algebra G eometry
•
v
vii xi
3
6
11
SOLUTlONS Arithmetic and Combinatorics Algebra G eometry
15
57
117
a�-l + 1
th.
>0
for
i
==
1,
.
.
.
, n,
Problems: Arithmetic and Combinatorics 1. Show that 2<3n) + 1 is not divisible by
17, for any integer
n
0.
>
2.
Let a, b, c be positive integers with the properties: a3 is divisible by b, b3 is divisible by c, c3 is divisible by a. Show that (a+b+c)13 is divisible by abc.
3.
For every integer n divisible by 3. (The symbol
4.
>
2 show that the number
L ( - 1) k (3k ) is
ln/3J
n
k =O
Lx J denotes the Greatest Integer Function. )
=
Calculate the sum of all divisors of the form 2x · 3 Y (with x, y of the number N 19 88 - 1 .
>
0)
5 . Show that there do not exist four successive integers whose product
is
1993 less than a perfect square.
6.
Show that there are infinitely many positive integers n such that each one of the three numbers n - 1 , n , n + 1 can be represented as the sum of two perfect squares.
7.
Show that the following system of simultaneous equations has no solution in integers:
x 2 - 3xy + 3y 2 - z 2 -x 2 + 6yz + 2 z 2 x 2 + xy + 8z 2 8.
Solve the following equation in integers x,
9.
If x,
31 44
100. y:
y, z are integers, at least one of which is 1990, show that x 2 + y4 + z 6 > xy 2 + y 2 z 3 + xz3
.
10.
Consider the sequence
xo
=
0,
x1
=
1,
for n = 1, 2, 3 , . . . . Define Yn = x ; + 2n+2 . Show that Yn is the square of an odd integer, for every non-negative integer n . 11.
The sequence (an} is defined recursively by ao
=
1,
a1
=
2,
an
=
a ;_ _1 + an-2
1 for n
=
2, 3, 4,
.. . .
Show that each an is an integer. 12.
Find·all functions f mapping non-negative integers into non-negat ive integers and such that f (! (n )) + f (n) = 2n + 6 for every integer n
2:
0.
13.
Show: that ln.J3J is a power of 2 for infinitely many natural num bers n. (The symbol lx J denotes the Greatest Integer Function. )
14.
Four numbers are randomly chosen from the set { 1, 2, , 3n} (n is a fixed integer greater than 1). Compute the probability that the sum of those four numbers is divisible by 3.
15.
For what natural numbers n is it possible to tile the n x n-chess board with 2 x 2 and 3 x 3-squares?
16.
A triangular prism is a pentahedron whose two parallel faces ( "top base" and "bottom base" ) are congruent triangles and the remain ing three faces are parallelograms. We are given four non-coplanar points in space. How many distinct triangular prisms having the four given points as vertices are there?
1 7.
Consider the infinite chessboard with squares coloured white and black, in the usual manner. SupposeS is a set of 1976 squares such that every two squares in S can be connected by a path consisting of consecutively adjacent squares. (Two squares are adjacent if they have a common edge. ) Show that there are at least 494 white squares inS. Moreover, show that 494 is the exact bound.
.
.
.
Arithmetic a nd Combinatorics 18.
5
Consider an alphabet consisting of three symbols a, b, c. How many n-character words with the following properties (1) and (2) can be composed?
(1) the word should begin and end with an a; (2) neighbouring positions must be occupied by different symbols. 19.
Nine trucks follow one another, in a line, on a highway. At the end of a day's ride it turned out that each driver disliked the style of the driving of the one in front of him. They wish to rearrange themselves so that, next day, no truck would follow the same truck that it followed on the first day. How many such rearrangements are possible?
20.
We are considering paths (Po, Pt. . .. , Pn) of length n over lattice points in the plane (i.e., points (x, y) with integer coordinates); for each i, the points Pi-l and Pi are assumed to be adjacent on the lattice grid. Let F(n) be the number of those paths that begin in Po = (0, 0) and end in a point Pn lying on the line y = 0. Prove that F(n) = e:).
Problems: Algebra
21. Determine all real polynomials P (x) of degree not exceeding 5, such that P(x) + 1 is divisible by (x - 1) 3 and P(x)- 1 is divisible by (x + 1) 3. 22. Prove that the polynomial xn + 4 factors into the product of two polynomials of lower degrees with inte_ger coefficients if and only if n is divisible by 4. 23. Find all natural numbers n for which the polynomial Pn (x) = x2 n + (x + 1)2 n + 1 is divisible by the trinomial T(x) = x2 + x + 1. 24.
For every positive integer
k show that the polynomial
is divisible by the binomial x 5 + 1 . 25. Find all pairs of real numbers a, b such that the polynomials
have two distinct common real roots. 26. Let
a, x, y, z be real numbers such that cos x + cos y + cosz cos(x + y + z)
x + sin y + sinz = a. = sinsin(x + y +z )
Prove the equality: cos(y + z ) + cos(z + x) + cos(x + y)
= a.
27. If a, b, e are pairwise distinct real numbers, show that the value of the expression a-b b-e e-a -- + -- + -1 + ab 1 + be 1 + ea is never equal to zero.
7
Algebra
28. Solve the system of equations:
x + y + xy
=
19,
y + z + yz 11,
z + x + zx 14. =
=
29. Solve the system of equations
X l (X l - 1) X 2 (X 2 - 1)
=
X2 - 1 X3 - 1
in real numbers x 1 . . . . , X n· 30.
Solve the system of equations
2xy X 2 + Y 2 + -x + y = 1, in real numbers x, 31.
y.
Solve the system of equations
x 2 + Y 2 + z 2 2, in real numbers x, y, z. =
x + y + z = 2 + xyz
32. Let n � 3 be a fixed integer and let a, b, e be fixed real numbers with a + b + e = 0. Find all n-tuples (x1. . . . , x n ) of real numbers satisfying the system of simultaneous inequalities
ax i- l + bxi + ex i + l � 0 for where by definition xo = X n , X n+l = x 1 .
i
=
1, . . . , n,
33.
Let a,
b, e be the sides of a triangle. Show that b e a -- + -- + -- < 2. b+e e+a a+b
34.
Let a,
b, e, d be positive real numbers with abed = 1. Show that a 2 + b2 + e2 + d 2 + ab + ae + ad + be + bd + ed � 10.
Problems
8
35.
Let a, b be non-negative real numbers with a2 + b2 = 4. Show that
ab -
The real numbers ai, bi, ci, di are such that 0 � ci � ai � bi � di and ai + bi = Ci + di for i = 1,2, . . . , n. Prove the inequality n n n n + + bi i � ai C II II II i II di. i=l i=l i=l =l
37.
Prove the following inequality for all integers n > 1: 1 + (n + 1)n+l n-1 > 1 + nn n n +1 n +2
38.
39.
(
(
)
n n,. ) v' +l Let n ;::: 9 be an integer. Which one of the numbers (vr.:: and ( v'n + 1) ,;n is greater? Prove the inequality
C:) · 40.
)
ffn
< 4n for n 1,2,3,. .. . =
Prove that the inequality
holds for any real numbers a17 a2, ..., ar. Find conditions for equality. 41.
For a fixed integer n ;::: 1 find the least value of the sum x2 x3 xn Xl + _1. + __1_ + • · ·+ __!!: n ' 3 2 given that XI, .. . , Xn are positive numbers satisfying 1 1 1 + +···+-=n. Xn Xl X 2
g
Algebra
42.
On a given segment AD, find points B and C so as to maximize the product of the lengths of the six segments AB, AC, AD, BC, BD, CD.
43.
Find all functions j: JR.� JR. satisfying the equation x2 f(x) + f( 1-x) = 2x - x4 for xER
44.
Let A and B be real numbers different from zero. Prove that the function f ( x)= A sin x + B sin ( J2 · x) is not periodic.
45.
Find all monotonic functions j: JR.� JR. satisfying the equation f(4x) - f( 3 x) = 2x for xE JR..
46.
A sequence ao, a1. a 2 , .. . of real numbers different from zero is gen erated according to the rule: a n+l = (a� - 1)/ (2a n )· Show that it contains infinitely many positive terms and infinitely many neg ative terms.
47.
Four sequences of real numbers
satisfy the simultaneous recursions
for n =
48.
0, 1, 2, . . . . Suppose there exist integers k, r ;.::: 1 such that
The sequences xo, x1. x2 , . . . and yo, Y l . Y2 , . . . are defined by: Xo = YO= 1, Xn+l =
Xn + 2 Xn + 1'
y� + 2 Yn+l = --- for n = 2Yn
Show that Yn = X2 " - l for every integer n ;.:::
0.
0, 1, 2 , . . .
10
Problems
49.
Two sequences of integers all a2, a3, . . n and b1, b2, b3, . . are de fined uniquely by the equality (2 + v'3 ) = an + bn v'3. Compute ( ) nlim -+oo an/bn ·
50.
The sequence (xn) is defined by
.
1 X1= 2'
.
2n- 3 Xn= �· Xn-1 for n=2,3,4, ... .
Prove the inequality
x1 + x2 + 5 1.
·
·
·
+ Xn < 1 for n 1, 2, 3, ... . =
A sequence of real numbers ao, a1, a2, satisfies the recurrence ! ani = an-1 + an+1 for n = 1 , 2, 3, . . . . Show that an+9=an for all n. .
.
•
Problems: Geometry
52.
Construct a right triangle ABC with a given hypotenuse that two of its medians are perpendicular.
53.
Let ABC be a triangle, AC =/:. BC. Assume that the internal bi sector of angle AC B bisects also t_he angle formed by the altitude and the median emanating from vertex C. Show that ABC is a right triangle.
54.
If ABCDEF is a convex hexagon with AB = BC, CD = DE, EF = FA, prove that the altitudes ( produced ) of triangles BCD, DEF, FAB , emanating from vertices C, E, A , concur.
55.
Let ABCDEF be a regular hexagon with M and N points on diagonals CA and CE (respectively) such that AM = CN. If M , N and B are collinear, prove that AM = AB.
56.
Let ABC be an acute triangle with altitudes BD and CE. Points F and G are the feet of perpendiculars BF and CG to line DE. Prove that EF = DG.
57.
Consider the right triangle ABC with LC = 90 ° . Let A 1 and B 1 be two points on line AB (produced beyond A and B) such that AA 1 = AB = BB 1 and let N be the foot of the perpendicular from A 1 to line B 1 C. Show that the rectangle with sides B 1 C and CN has area twice as large as the square with side AB.
58.
Let ABC DE be a convex pentagon inscribed in a circle. The dis tances from A to lines BC, CD, DE, and BE are a, b, c, and d, respectively. Express d in terms of a , b, c .
59.
Let ABC be an isosceles triangle with base AB. Let U be its circumcentre and M be the centre of the excircle tangent to side AB and to sides C A and C B produced. Show that
2 ·CU < CM < 4·CU.
c
such
12
60.
Problems
The diagonals AC and BD of a convex quadrilateral ABCD inter sect in E. Let F1 , F2 and F be the areas of triangles ABE, CD E and quadrilateral ABCD , respectively. Show that
When does equality hold? 61.
Let P1 P2 be a fixed chord (not a diameter) of a circle k. The tangents to k at P1 and P2 intersect at Ao. Let P be a variable point on the minor arc P1 P2 . The tangent to k at P intersects lines AoP1 and AoP2 at A 1 and A 2 , respectively. Determine the position of P for which the area of triangle A 0 A 1 A 2 is a maximum.
62.
P be a point inside a parallelepiped whose edges have lengths a, b and c. Show that there is a vertex whose distance from P does not exceed � v'a 2 + b 2 + c 2 .
63.
Do there exist two cubes such that each face of one of them meets each face of the other one (possibly at an edge or a corner)?
64.
Let A 1 , A 2 , Aa, A 4 be points on the sphere circumscribed about the regular tetrahedron with edge 1 such that A i Aj < 1 fori =1- j. Prove that these four points lie on one side of a certain great circle of the sphere.
Let
�
Q(
- - cot 2 2
-�
tan
j3 2
-
7) 2
+ tan -
Solutions: Arithmetic and Combinatorics Problem 1
Show that 2W> +1 is not divisible by 17, for any integer n
>
0.
Problem 1, Solut ion 1
Assume, to the contrary, that 23" + 1 is divisible by 17 for a certain 1 ( we write just abe for aW>) . Let r be the remainder left by 23"-1 in division by 17. Thus
n�
by our assumption. This implies
0
=
r3 +1= (r + 1)(r2- r +1)
( mod 17).
Direct examination of all possible remainders shows that the second fac tor (r2- r+1) is never 0 ( mod 17); and since 17 is a prime, the .first factor must be 0 (mod 17), i.e., we haver= -1 ( mod 17). So we have shown that 23"
=
-1
17)
(mod
23"
forces
-1
-1
=
(mod 17).
Descending, we conclude inductively that 23"
=
-1
( mod 17)
for k = n , n - 1, n - 2, ... , 1, 0. This, however, is a contradiction because 230 = 2
=fi -1
( mod 17).
Problem 1, Solution 2
Fix an n � 1. The exponent 3 n is a number of the form 4 k + 1 or 4 k + 3. If 3 n = 4 k +1, then 23"
and if 3 n
=
= 24k+l
=
16k. 2
=
2(-1)k
( mod 17),
16k
=
8 (-1)k
( mod 17).
4 k + 3, then 23" = 24k+3
=
·
8
Note, however, that 2(-1)k
+ 1=
{
-1
for k odd, 3 for k even,
{-7 8(-1)k+1= 9
for k odd, for k even.
16
Solutions
n So 23 + 1 is never congruent to 0 (mod 17). 2 Let a, b, c be positive integers with the properties: a3 is divisible by b, b3 is divisible by c, c3 is divisible by a. Show that (a+ b + c)13 is divisible by abc. Problem
2, S o lution 1 The 13-fold product (a+ b + c)13, when multiplied out, splits into 313 summands of the form Problem
k, m, n
�
0 integers, k + m + n = 13 .
(1)
It will be enough to show that each of them is divisible by abc. This is evident when the exponents k, m, n are all positive. So it remains to consider the case where one of them is zero. Let e.g. n = 0. The product (1) then becomes k, m
�
0 integers, k + m= 13.
(2)
The conditions of the problem imply that a9 is divisible by c and b9 is divisible by a. Any number of the form (2) can now be represented as the product of four factors (separated by multiplication dots in the listing below): if k= 13 ' m=O if 10$k $ 12 , 1$ m$3 if 1$k$9 , 4$m$12 : if k=O , m= 13
akbm = a · a3· a9· 1; akbm = a ·b · a9·(ak-10bm-1); akbm = a ·b ·b3 · (ak-1 bm-4); akbm = b9·b ·b3·1;
in each case the first factor is divisible by a, the second by b, the third by c (and the fourth is an integer), and so the product abc is a factor of akbm. That does the job. 2, S o lution 2 Let p be any prime divisor of the product abc. Write
Problem
a = p01u,
(3)
where a, (3, 7 � 0 and u, v, w � 1 are integers non-divisible by p. Since a3 is divisible by b, the exponents a and (3 satisfy 3 a � (3. Likewise, 3 (3 � 7 and 3 7 � a; and hence 9a � 7, 9 (3 � a, 9 7 � (3. Let r = min( a, f3, 7 ). We obtain
Arithmetic and Combinatorics
17
The numbers a, b, c are divisible by pr. Thus ( a+ b + c ) 13 is divisible by p13r, hence by p01+/3+-r, in view of inequality (4). On the other hand, according to o: + (3+ 'Y is the exact power in which p enters the prime factorization of abc. Since p was an arbitrarily chosen prime factor of abc, we conclude that ( a+ b+ c ) 13 is divisible by abc. Problem 3 Ln/3J For every integer > 2 show that the number L ( is divisible by ( The symbol Lx denotes the Greatest Integer Function. )
(3),
3.
nk ) -1 )k ( k=O 3
n
J
Pr oblem 3, So lut ion 1
For any fixed non-negative integer
is valid for
(; )
n, the fundamental Binomial Identity (1)
eve ry integer j, if we agree that = 0 for j < 0 and for j
>
n.
(2)
Consider the following three sums:
An = Bn Cn =
�(- 1)kC:). �(- 1)k(3k':_ 1). �(- 1)k(3k':_2).
(3) k
Summation limits have not been indicated; we may assume that ranges from oo to oo That will cause no ambiguity because there are only finitely many non-zero terms in each of these sums. E.g., in An summa tion actually spreads from k = 0 to k = Thus An is exactly the number defined in the problem statement. We will show that -
.
Ln/3J. (4) An Bn Cn 0 ( mod 3) for n 3. ( It is only required to show that An= 0 ( mod 3); however, it proves practical to handle the assertion in this more general version. ) =
=
=
2:':
Using equation (1) we derive the following recursion formulas:
18
Solutions
=
�(-1)kC:)+ �(-1)k(3k�1) An+ Bn,
Bn+l
=
�)-1)k(n3k+-11) k �(-1)kck�1)+ �(-1)k(3k�2) Bn + Cn,
(5)
(6)
and
�)-1)k(n3k+-21) k �(-1)k(3:-2) + �(-1)kck�3) 2)-1)k(3kn-2)+I)-1)1+1(3ln) k �(-1)k(3k�2)- �(- 1)1(;) (7) Cn - An . And since A3 1- 1 = 0, B3 -3, G3 -3, obvious induction justi fies the claimed relations ( 4). Cn+l
=
=
l
=
=
=
Remark
It is not hard to derive from second order for the Ans:
(5), (6), (7) the recurrence equation of the
(8) An+2 3(An+l An) for n 1 (with the initial data A1 A2 1); we invite the reader to do that. Readers familiar with linear recurrences may like to work out an explicit formula (on the basis of the system (5), (6), (7) or of the single equation (8); compare Problem 10, Solution 3). We now show how to find that formula by a different method. =
-
�
=
=
Problem 3, Solution 2
(3) together with the convention (2) is preserved from Solution 1.Notation For any complex number we have by the Binomial Theorem z
Arithmetic and Com binatorics
where
fn(z)
=
Yn( z)
=
and hn( z)
=
19
�( (�) + (6j:3)z3)z6i, L (( . 6j-1)+ (6j+2)z3 ) z6j-l ' L ( ( 2)+ ( . 6j- 6j+1)z3) z6j-2 . J
n
n
n
n
J
J
-1
In particular, if w is any cubic root of (i.e., a complex number satis fying w3 then we have for any integer =
-1),
w6i
=
1,
j:
w -1 -w2, w6i-l j w6 -2 = w-2- -w· =
=
'
hence (compare
(3)),
and likewise
Equality (9) thus implies (1 +w)n =An- Bnw2- Cnw for w3 =-1 .
{1 0)
Setting w = -1 we hence obtain
Bn
=
An + Cn for n =
1, 2, 3, ....
(11)
Now consider the complex number
a
3
=
+
!+ !v'3i =cos(7r/3) i sin(7r/3),
1.
also satisfying a -1 , and moreover, a2 a For w a equalities (1 0) and yield (1 +a)n=An- Bn(a- 1 )- Cna =�An- ( !An+ Cn)v'3i.
(11)
=
=
1
-
=
{12)
On the complex plane, the numbers 0, and a represent the vertices of an equilateral triangle, which is completed to a parallelogram (rhombus)
Solutions
20
by the vertex 1 + a. Thus 1 + a =J3 (cos( 71"/6) + i sin( 71"/6)) , and by de Moivre's Theorem
(1 + a t=3n l2 (cos(
n1r
(6) + i sin(n7r/6)) .
(13)
Comparing the real parts of (12) and (13),
An= 2 3n/2-1 cos(n1r /6); ·
this is the explicit formula we have promised. If ln/2J =q, we can rewrite it as where
for n=2q, 2cos(n7r/6) 2VJcos(n71" . /6) for n= 2q+ 1.
For each n, Kn is an integer; in particular, K3 =0. Hence A3 = 0; and for n ;::: 4 we have q;::: 2, so An=3q-l Kn is divisible by 3. Problem 4
Calculate the sum of all divisors of the form 2z 3Y (with x, y number N =1988 - 1.
>
·
0) of the
Problem 4 , Solution 1
The only trouble is to determine the highest powers of 2 and 3 that divide
N . This can be done using the Binomial Theorem: 1 988
( 20- 1)88
c:) - c18) 20 + (8:) 202 - c:6 ) 203 +. . . + (::) 2088
1- 88 20 + (terms divisible by 2 ) ·
(we used the fact that ( 8:)
=
! 88 87= 22 3 · 1 1 29); and ·
·
·
·
(18 + 1 ) 88 88 88 88 3 . . . 88 88 2 88 18 18 + + 18 + 18 + = + 88 0 1 3 2 1 + 88 · 18 + (terms divisible by 3 4 )
() () ()
() .
()
Since 88 · 20= 25 5 1 1 and 88 18 = 32 2 4 11, these representations show that N =1988 - 1 is divisible by 25, but not by 2 6 , and is divisible by 32, but not by 33. ·
·
·
·
·
A rithmetic a nd Combinatorics
21
Consequently, the sum we are about to evaluate equals s
=
(x,y): x,y>O 2"' ·3Y dividing N
L
=
=
2"'· 3Y
xE{l,2,3,4,5} yE{l,2} 5 2 L::2"'·L::3y x=l y=l (2 + 4 + 8 + 16 + 32)(3 + 9) 744.
2 Let us inspect the powers of
Problem 4, Solut ion
1 92 = 361 -23, =
and
19 modulo 26 and modulo 33: 194 (-23)2 = 52 9 17 (mod 64 ), =
=
198 172 = 289 33
(1)
(mod 64);
=
=
while
(mod 27). (2) 1 92 = 361 10 The well-known theorem of Euler (sometimes referred to as ge n eralized Fermat's Theorem) asserts that if a, n are relatively prime natural num bers, then a(n) 1 (mod n ) , where =
=
18. Thus (mod 64) and 19 1 8 1
In particular, 4>(64) = 32 and 4>(27) =
(mod 27).
=
Raising the first of these relations to third power and the second one to fifth power, we get (mod 64) and
19 90 1
(mod 27);
=
or - which is exactly the same (mod 64) and
192 19 88 ·
=
1
(mod 27).
22
Solutions
Consequently, in view of
(1) and ( 2),
(3) 1988 "¢ 1 (mod 27) (if 19 88 were 1 (mod 64), the product 198 1988 would be 33 rather than 1 (mod 64); and the second relation of (3) is justified similarly). On the other hand, equation (1) shows that 19 8 1 (mod 3 2). Besides, 19 1 (mod 9) . If we raise the first relation to power 11 and the second (mod 64) and
·
=
=
to power 88, we obtain
(mod 3 2) and
1988
=
1
(mod
9).
(4)
Statements (3) and (4), combined, show that N is divisible by 3 2 and by 9, but not by 64 or 27. The concluding calculation is done as in Solution 1. P roblem 4, S o lution 3
The argument of Solution 2 can be carried out without resorting to Eu ler's Theorem and Euler's Function, in a fashion less sophisticated and more straightforward. Namely, upon arriving at formulas (1) and ( 2), we continue a:s follows. Since
332 = (3 2 + 1) 2
=
3 22 + 2 . 3 2 + 1
=
1
(mod 64),
(1) 19 88 = (198 ) 11 3311 = (332)5. 33 33
we obtain from
(mod 64).
=
=
And since by ( 2)
193 = 19 2 . 19
=
10 . 19 = 190 1 =
(mod 27),
we conclude that (mod 27). Claims (3) hence result. The remaining portion of the preceding solution has to be repeated without any changes, yielding the outcome: S = 744. P roblem 5
Show that there do not exist four successive integers whose product is 1993 less than a perfect square. P roblem 5, S o lut ion 1
Assume that the equation
x(x + 1) (x + 2)(x + 3) + 1993 = y2
(1)
Arithmetic a nd Combinatoric&
23
is fulfilled for some integers x and y. Examine equation (1) modulo 5. Either the product x(x + 1)(x + 2)(x + 3) is divisible by 5 or its four factors leave remainders 1, 2, 3, and 4, in which case the product equals 4 (mod 5). Anyhow, the expression on the left of (1) is either 3 or 2 (mod 5), and this is obviously a contradiction because a perfect square y 2 can only be 0, 1, or 4 (mod 5). P roblem 5, Solution 2 Assume equation (1) and transform the product under examination as follows:
x(x + 3) · (x + 1) (x + 2) = (x 2 + 3x) (x 2 + 3x + 2) = (z - 1) (z + 1) = z 2 - 1, where we have denoted by z the expression x 2 + 3x + 1; this quadratic trinomial has the minimum value (over the reals) equal to - 5 /4, and hence z;::: -1. Equation (1) now takes the form z 2 + 1992 = y 2 , i.e., (2) (y - z)(y + z) 1992. We see from (2) that z cannot be - 1 ; hence z ;::: 0. We may also assume (see (1)) that y ;::: 0. So the second factor in equation (2) is non-negative; =
consequently, both factors must be positive, the second one greater than the first. Both factors are integers of the same parity; their product is even, so they both are even. In view of the prime decomposition 1992 = 2 3 · 3 · 83, the prime factor 83 must enter y + z and we conclude that the pair (y - z, y + z) must be one of the following:
(2, 996 ), (4, 498 ), (6, 332), (12, 166 ). Accordingly, z equals 497, 247, 163, or 77, which means that the product (x + 1)(x + 2) equals 498, 248, 164, or· 78. However, it is easily verified
that no one of these four numbers is equal to the product of two successive integers. Contradiction ends the proof. Problem 6
Show that there are infinitely many positive integers n such that each one of the three numbers n- 1, n , n + 1 can be represented as the sum of two perfect squares. Problem 6, Solution 1
Define nk = (2k 2 + 1) 2 for k = 0, 1, 2, . . . . Then the sequence n1, n2 , n 3 , . .. is strictly increasing and each of its terms is equal to the sum of two squares: nk-
1 = (2k2 ) 2 + (2k) 2 , nk = (2k 2 + 1) 2 + 0 2 , nk + 1 = (2k 2 + 1) 2 + 1 2 .
24
Solutions
Problem 6, Solut ion
Now let n k
= 2m% + 1,
2 where m k
= k (k + 1). It is enough to notice that n k - 1 = m% + m %, n k = (k 2 + 2k ) 2 + (k 2 - 1) 2 , n k + 1 = (m k + 1) 2 + (m k - 1) 2 .
6, Solut ion 3 Define the sequences a 1 . a 2, a 3 , . .. and
Problem
b 1 . b2 , b3, . . . recursively by
ao = 4, bo = 3, and notice the equality a% + 2 = 2b% ( easy proof by induction ) . Hence, if we set n k = a% + 1, we are done because
7 Show that the following system of simultaneous equations has no solution in integers: Problem
x 2 - 3xy + 3y2 - z2 = 31 44 -x 2 + 6yz + 2z 2 2 2 100. x + xy + 8z 7, S olut ion 1 Since the terms x 2 and z2 appear in all the three equations, it is tempting to apply the method of elimination so as to get rid of them. If we multiply the first equation by a, the second by b, and the third by c, and add the resulting equations, we obtain an equation in which the coefficients of x 2 and z 2 are a - b + c and -a + 2b + 8c, respectively. Setting these expressions to be zero, we find that e.g. a = 10, b = 9 and c = -1 do the job, producing the equation Problem
10
i.e.,
·
( -3xy + 3y 2 ) + 9
·
6yz - xy
= 10 3 1 + 9 44 - 100,
y ( -31x + 30y + 54z)
·
=
·
606.
This yields the possible values of I YI: 1, 2, 3, 6, 101, 202, 303, 606. In a similar manner we can eliminate the terms x 2 and xy, multiplying the first, the second and the third equation of the system by suitable factors a, b, c; now we need that a - b + c and - 3a + c ( the coefficients
xy in the resulting equation ) should be zero. When we take 1, b = 4, c = 3, we obtain (3y 2 - z 2 ) + 4(6yz + 2z 2 ) + 3 8z 2 = 31 + 4 · 44 + 3 · 100,
A rithmetic a nd Combinatorics
25
of x2 and a=
·
i.e.,
31z 2 + 24yz + (3y 2 - 507) = 0.
Viewing this as a quadratic equation with the unknown its discriminant:
z, we compute
D = (24y) 2 - 4 · 31· (3y 2 - 507) = 4(51y 2 + 15717); then the roots z 1, z2 are: (- 12y ± ..fDJ4)/3l. Thus 51y 2 + 15717 ought to be a square number in order that z 2 be integers. Yet, for the previously found values of I YI this expression takes values 15768, 15921, 16176, 17553, 535968, 2096721, 4697976, 18744753, no one of which is a ZI,
perfect square. So the system has no integer solutions. P roblem 7, Solut ion 2 An astonishingly simple proof results from examination of the two outer equations modulo 5 ( the middle equation is not needed! ) . Multiplying the first equation by 8 and adding the third equation we get
9x 2 - 23xy + 24y 2 = 348,
which is
-x 2 + 2xy - y 2 3, i.e., (x - y) 2 2 (mod 5) . Yet the square of an integer can only be 0, 1 or 4 ( mod 5); the claim =
=
follows.
Problem 8
x, y: x 2 (y - 1) + y 2 (x - 1) = 1.
Solve the following equation in integers
P roblem 8, Solut ion 1
Set x =
u + 1, y = v + 1; the equation becomes (u + 1) 2 v + (v + 1) 2 u = 1;
equivalently:
u 2 v + 2uv + v + uv 2 + 2uv + u uv(u + v) + 4uv + (u + v) uv(u + v + 4) + (u + v + 4) (u + v + 4) (uv + 1)
1· 1 ,· 5 ·, 5.
,
26
Solutions
One of the factors must be equal to 5 or -5 and the other to 1 or -1 (respectively). This means that the sum u + v and the product uv have to satisfy one of the four equation systems:
u +v uv
1 0
u+v uv
=
u+v uv
-3
u+v uv
= -5 = -6
4
-9
-2
Accordingly, the numbers u and v have to be the roots of one of the four quadratic trinomials:
t 2 + 9t - 2 j
t2 + 5t
-6
The two trinomials in the middle (the second and the third) have no integer roots. The first one has roots 0, 1, and the last one has roots -6, 1. Thus ( u, v) must be one of these two pairs, up to permutation. Hence the final outcome: (x, y) = (u + 1, v + 1) must be one of the pairs: (1, 2), ( -5, 2), (2, 1), (2, -5).
2 The symmetric shape of the equation suggests introducing the funda mental symmetric forms s =x + y and q= xy. The equation, rewritten as xy (x + y) =x 2 + y 2 + 1, takes the form Problem 8, S olut ion
sq =s 2 - 2q + 1;
(1)
i.e., (s + 2)q =s 2 + 1. The factor (s + 2) cannot be zero, and division is admissible:
5 s2 + 1 =s - 2+--. (2) s +2 s +2 If this has to be an integer, the denominator s + 2 must be a divisor of 5, which means that s must be one of the numbers -7, -3, -1, 3. For each of these values of s , the corresponding value of q is computed from (2) and we arrive at the four possible systems of equations for s x + y, q = xy: q=
--
=
X
+y xy
x+y xy
=
-7
-10
x+y xy
-1 2
x +y xy
-3]
� =
10
(3)
(they correspond, in a certain order, to the four systems obtained in Solution 1). The numbers x and y must be the roots of the respective
Arithmetic and Combinatorics
27
quadratic trinomial
t 2 + 7t - 10 j
t 2 + 3t - 10 j
t 2 - 3t + 2 .
Of these, only the second and the fourth have integer roots; these are, respectively,. -5, 2 and 1, 2. So (x, y) is one of the pairs (-5, 2), (2, -5),
(1, 2), (2, 1).
Problem 8, Solut ion 3
Use the symmetric forms s = x + y, q = xy. The resulting relation (1) can be viewed as a quadratic equation with the unknown s and parameter
q:
s 2 - qs + (1 - 2q) = 0. Its discriminant equals D = q 2 + 4(2q - 1) = (q + 4) 2 - 20 and produces
the roots
(4) 8 2 = � (q - v'D). One of these roots has to be equal to x + y, an integer. Therefore D must be the square of an integer: D = d 2 ; d � 0. Then 20 = (q + 4) 2 - D = (q + 4 + d) (q + 4 - d), with both factors of same parity, the first factor greater than the second. There are only two factorizations of 20 that suit the need: 20 = 10 · 2 and 20 = (-2) · (- 10), giving rise to the equation systems
q + 4 + d = -2 q + 4 + d = 10 and q + 4 - d = -10, q+4-d=2 with solutions q = 2, d = 4 in the first .case and q = - 10, d = 4 in the second. Recall that d = v'D. Thus, in view of (4), the possible values of s are: 3, -1 ( if q 2) and -3, -7 (if q = - 10) . So we have obtained the systems of equations (3) from Solution 2. Repeating its final passage we determine the four integer pairs ( x, y) that make up the solution of the =
given equation.
Problem 8, So lut ion 4
The technique of inspecting the discriminant can be employed in a yet more straightforward manner, without introducing the forms s and q. Suppose a pair (x, y) is a solution. At least one of the integers x, y is greater than 1; otherwise the left-side expression would be nonpositive. In view of symmetry we may assume x > 1. Let us look at the given equation as a quadratic one with respect to variable y,
(5)
28
Solutions
with discriminant (6) D = x 4 + 4(x - 1)(x 2 + 1) = x 4 + 4x 3 - 4x 2 + 4x - 4, which must be a perfect square in order that equation (5) has an integer root y. Suppose x > 2. Then the following inequalities hold: D - (x2 + 2x - 4) 2 2 0(x - 1) > 0, 2 2 D - (x + 2x - 2) -4(x - 1 ) (x - 2) < 0, showing that D is strictly comprised between the squares of two skip consecutive integers x 2 + 2x - 4 and x 2 + 2x - 2. Therefore D has to be the square of x 2 + 2x - 3. This, however, cannot be the case, since this last number is of different parity than D (see (6)). The only possibility that remains is that x 2. Equation (5) then be comes y 2 + 4y - 5 0; equivalently; (y - 1)(y + 5) 0, and we get y = 1 or y = -5. So (2, 1) and (2, -5) are all pairs of integers ( x ,. y ) with x > 1, satisfying the equation. Symmetry yields two other pairs (1, 2) and ( -5, 2); and there are no more- as the argument shows. =
=
=
=
P roblem 8, Solut ion 5
Assume that the integers x, y satisfy the equation. Its left side is the sum of two addends, one of which must be 2:: 1 and the other one � 0. Let e.g. y 2 ( x - 1) 2:: 1, x 2 (y - 1) � 0. Then x 2::2, y # 0, y � 1. If y = 1, then of course x = 2 (just look at the equation). Assume y < 0 for the sequel (remember that y 0 has been excluded). Again let x + y = 8 and rewrite the equation in the form =
8x 2 - x 3 - x 2 + x 3 - 28x 2 + 8 2 x - x 2 + 28x - 8 2 1; x(8 + 2) (8 - x) = 8 2 + 1. The factor 8 - x y is negative; x is positive. Hence 8 + 2 must be negative, and so 8 � -3, whence 8 2 2:: 9. Rewrite the last equation as f(x) = 0, where by definition f(x) = [ - (8 + 2) ] x 2 + [ 8(8 + 2) ]x - [8 2 + 1 ] .
Expanding and regrouping,
=
=
Notice that the coefficients (in square brackets) are positive. Thus, in view of x 2:: 2, we get
f(x) 2::! (2)
=
-4(8 + 2) + 28 (8 + 2) - (8 2 + 1) 8 2 - 9 2::0. =
(7)
Arithmetic and Combin atorics
29
Equality f (x) = 0 implies that both inequalities in (7) must turn into equalities. Now, f(x) = ! (2) means that x = 2, while s2 = 9 means that s = -3. Hence y = s- x = -5. Recalling the case of y = 1 (mentioned at the beginning), we obtain the two solving pairs (x, y) with y� 1: (2, 1) and (2, -5). Interchanging the roles of x and y we get the other two pairs: (1, 2) and ( -5, 2) ; and these four pairs constitute the complete solution. P roblem 9
If x, y, z are integers, at least one of which is 1990, show that x2 +y4 +z6 > xy 2 +y2z3 +xz3.
Problem 9, So lut ion
1
This is in fact the Cauchy-Schwarz inequality for the triple of numbers x, y2, z3; it can be settled (in the weak form) as follows, using the arithmetic mean-geometric mean inequality for pairs of numbers: x2+y4 x2 +z6 y4 +z6 + + x2 +y4 +z6
2
> =
>
2
2
JxY+v'x2z6+� lxly 2 +lxllzl3+ Y21zl3 xy2 +y2z3 +xz3
(because lxl ;:::: x and lz l ;:::: z ). Equality would require that x2 = y4 = z6 and either y 0, xz ;:::: 0, or y =/= 0, x, z ;:::: 0. In the first case we get x = y = z 0, in contradiction to the "1990" condition. Regarding the second case, we now have z3 = y2 = x > 0. Since x, y, z have to be integers, z3 = y2 forces that z is itself a perfect square: z = u2, with u being a positive integer. Thus y = ±u3, x u6. By assumption, one of the numbers x = u6, y ±u3, z = u2 has to be 1990. And since 1990 is neither a square or cube or sixth power, equality cannot occur and the given inequality holds (in the strict form). =
=
=
=
P roblem 9, Solut ion 2
The proof can be also derived from the following transformations: (x2 +y4 +z6) (xy 2 +y2z3 +xz3) _
=
{
{
:
z6 (y2 +x)z3 + (y4 xy 2 + x2) x2 (z3 +y 2 )x+ (z6 y2z3+ y4) _
_
_
_
(z3- !(Y2 +x)) + i(y2- x)2 (x- !(z3+y2)) + i(z3- y2)2 .
(1)
30
Solutions
These expressions are non-negative. Now, x, y, z are integers, one of them being equal to 1990. If x = 1990, then y 2 =I x. If y = 1990 or z = 1990, then z 3 =I y 2. In each case one of the terms (y 2 - x) 2 and (z 3 - y 2) 2 is strictly positive, and so is the difference expressed by for mulas (1). P roblem 10 Consider the sequence xo = 0, x1 = 1,
X n+2 = 3x n+l - 2x n for n = 1, 2, 3, . . . . Define Yn = x� + 2 n+2 . Show that Y n is the square of an odd integer, for every non-negative integer n . P roblem 10, S o lut ion 1 The initial X n S are 0, 1, 3, 7, 15,
31, . . . , so it is natural to guess that (1) We prove this by induction. For n = 0 and n = 1, (1) holds. Assume that (1) holds for some two successive integers n and n + 1. Then 3x n+l - 2x n = 3(2 n+l - 1) - 2(2 n - 1) X n+2 = 3 . 2 n+l - 2 n+ l - 1 = 2 n+2 - 1, proving (1) for n + 2. Hence, formula ( 1) is true for all integers n � 0. From (1) we get (2 n - 1) 2 + 2 n+2 Yn = x ; + 2n+2 2 2n - 2 2 n + 1 + 4 2 n 2 2n + 2 2n + 1 = (2 n + 1) 2 , showing that Y n is the square of an odd integer, as asserted. =
·
·
·
10, Solut ion 2 The recursion formula for X n+2 can be rewritten as P roblem
X n+2 - X n+ l = 2x n+ l - 2x n for n 0, 1 , 2, . . . Thus, setting X n+l - X n = t n we have t n+l = 2t n for n = 0, 1, 2, . . . ; and since to = 1, we infer t k = 2 k , i.e., (2) for k = 0, 1, 2, . . . Fix an integer n � 1. Summing the equalities (2) over k = 0, 1, 2, . . , n 1 we obtain =
.
.
-
Arithmetic a nd Combinatorics
31
or, which is the same (in view of xo
=
0),
So we have formula (1) of Solution 1 (without guessing), and it remains just to repeat the last paragraph of that solution. Problem
10,
Solution 3
We are dealing with the homogeneous linear recursive equation of the second order
X n +2 - 3x n+l + 2x n = 0 for
n = 0,
1, 2, . . . .
The well-known method of handling such recursions is to solve the char acteristic equation, which in this case is
q2 - 3q + 2 0, (3) and to postulate X n = Aa n + B (3 n , where a and (3 are the roots of that equation (provided they are distinct). Now, equation (3) has roots a = 2 and (3 = 1, yielding X n = A 2 n + B. From the initial data xo 0, x 1 = 1 =
=
·
we get
A + B = 0, 2A + B = 1. Thus A = 1 and B = -1, i.e., X n 2 n - 1. As in the first solution, we hence obtain Yn (2 n + 1) 2 . =
=
Problem
10,
Solution 4
If one prefers (unwisely enough) to work out a recursive formula for the
Yn S, that is also possible. Squaring the equation that defines X n+2 we
obtain
x;+2 = 9x; + l - 12x n X n+ l + 4x ;, whence by setting x� Yn - 2 n+2 and denoting X n X n+l by Zn : =
This simplifies to
12zn = -Yn+2 + 9Yn+l + 4 yn - 18 · 2 n+2 . Consider
Zn+l : Zn+l = X n+1 X n+ 2 = X n+l (3x n+l - 2x n ) 3x�+ l - 2x n X n+l 3(Yn+l - 2 n +3) - 2zn . =
=
(4)
Solutions
32
Multiply this by 12 and insert expression sion for 12z n+l ) :
(4) ( and the analogous expres
-Y n+ 3 + 9y n +2 + 4Yn+l - 18 · 2 n+3 36(Yn+l - 2 n+3) - 2( -Y n +2 + 9Y n+l + 4yn - 18 · 2 n+2 ). The powers of 2 cancel out and we are left with
Apply the method described in Solution is
3.
The characteristic equation
q 3 - 7q 2 + 14q - 8 = 0 .
Its coefficients sum up to 0 , hence one of the roots is 1 and the equa tion factors into (q - 1) (q 2 - 6q + 8) = 0. The roots of the quadratic factor are found e.g. from the Viete's Formulas; they are 2 and 4. So we postulate (5) Y n = A 4n + B · 2 n + C. ·
The initial terms xo = 0 , x 1 = 1 , x 2 = 3 yield the initial terms of the sequence (Y n }: Yo = 4, Y l = 9, Y 2 = 25 . Setting these in (5) we obtain the system of linear equations for the constants A, B, C :
A +B+C
= 4,
4A + 2B + C 9, =
with the unique solution A
Problem
= 1, B = 2, C = 1.
= 25,
Therefore
11
The sequence
(a n } i s defined recursively by
ao = 1 ,
for
Show that each Problem
16A + 4B + C
11,
n = 2, 3, 4, . . . .
a n is an integer.
Solution
1
We proceed by induction. The first t hree terms ao = 1 , a1 = 2 and a 2 = 5 are integers. Fix n � 3 and assume that the aks are integers for all k :::; n ; we will show that a n+l i s a n integer also. According t o the defining formula, a n - 1 (a;_ 2 + 1)fa n -3 i thus =
Arithm etic a nd Combinatorics
33
The numbers an- 1 . an- 2 , an- 3 are integers, by the inductive assumption. The last formula shows that an-1 and an- 2 are coprime. Now,
a! + 1
=
= and hence
All the aks occurring in this equality are whole numbers. So the prod uct (a� + 1)a�_ 2 is divisible by an-1 · And since an- 2 and an-1 are coprime numbers, an-1 has to be a divisor of a� + 1. Consequently, an+l = (a� + 1)/an_1 is an integer. This completes the inductive step. P roblem
1 1 , S o lut ion 2 According to the definition,
Replacing
n
by
n
+ 1 we obtain
Subtracting the first equation from the second one,
This shows that the sequence ((an + l + an-1) /an) is constant . It begins with (a2 + ao)/a1 = (5 + 1)/2 = 3, and hence (an+l + an-d ian = 3 for all n ; equivalently,
an +1
=
3 an - an-1 for
n =
1, 2, 3, . . . .
Since ao = 1 and a1 = 2 , this forces that all the ans are integ�rs.
34
Solutions
Problem
11,
S o lut ion 3
few initial terms of the given sequence are: ao = 1, a1 = 2, a 2 = 5 , a 3 = 13 , a 4 = 34, a s = 89; the even-indexed Fibonacci numbers are immediately recognized. The Fibonacci sequence, defined by the recur sion
A
Fo =
1,
F1
=
1,
Fn = Fn- 1 + Fn-2 for
n
=
2 , 3 , 4, .
.. ,
(1)
begins with
( Fo , F1 , F2 , F3 , F4 , Fs , Fs , F7, Fa , . . . )
= (1, 1, 2 , 3, 5 , 8, 13, 2 1 , 34, . . . ) ,
so it is natural to guess that
= 0, 1, 2 , 3, . . . . 1, it will be enough
a n = F2n for
n
(2)
Since (2) holds for n = 0 and n = to prove that the sequence ( F2n ) fulfills the same recursion formula that defines the sequence (a n ) :
F2n
=
Fi( n - 1 ) + 1 F2 (n-2 )
n
for
=
2,
3, 4, . . .
equivalently,
F2nF2n- 4 - Fin - 2
=
1
for
n
=
2,
3, 4, . . . .
(3)
The Fibonacci numbers are expressed by the well-known equality where
a
1-1+v'5 v'5 , f3 = -= -2 2 .
( Readers not familiar with this expression may like to derive it from the recursion (1) , employing the techniques described in the solution to P roblem 10, this book. ) Notice that a + f3 = 1, a - f3 = v'5 , a /3 = -1. Thus
a4n -2
_
( a f3) 2n-3 ( a 4 + /3 4 )
+ f3 4n-2
5
a 4n -2
_
2 ( a f3) 2n - 1 5
+
f3 4n-2
Arithmetic a nd Combinatorics
35
4 ) 2 + ,8 4 = a - 2(a,B 5 =
(a
- ,8) 2 = 1 · ' 5
equality (3 ) results, proving our claim (2). It j ust remains to use the fact that the Fibonacci numbers are integers. Problem
12
Find all functions f mapping non-negative integers into non-negative integers and such that f (f ( n)) + f ( n) = 2n + 6 for every integer n � 0 . P roblem
12, Solut ion 1 Suppose f satisfies the given equation
J (f(n)) + f(n) = 2n + 6
n = 0, 1, 2, . . . . (1) n, m � 0 we get f(f(n)) = j(f(m)), for
Assuming f(n) = f(m) for some whence by (1) n = m. Thus f i s injective. Denote:
f(O) = a, f ( a ) = b, f ( b ) = c, f(c ) = d, f ( d ) = e. Setting in (1) n = 0 , a, b, c we obtain, respectively,
(2 )
b + a = 6, c + b = 2a + 6, d + c = 2b + 6, e + d = 2c + 6. (3) If a were zero, all the numbers in ( 2 ) would be zero, in contradiction to b + a = 6. So a =I= 0 , and by injectivity f(a) =I= f(O), i.e. , b =I= a. Since a + b = 6, we see that a =I= 3. Subtract the first equation of ( 3 ) from the second, the second from the third, and the third from the fourth:
c - a = 2a, d - b = 2b - 2a, e - c = 2c - 2b. By the first equation of
c = 3a,
(3), b = 6 - a.
d = 3b - 2a = 18 - 5a,
(4)
Relations (4) hence imply:
e = 3c - 2b = lla - 12.
(5)
All the values taken by f are non-negative integers; in particular, d � 0 and e � 0. This in view of (5) shows that �� :$ a :$ 1f . Since 3 has been excluded as a possible value of a, we infer a = 2.
Solutions
36
b = 4, and from equations (4) e = 10. The obvious guess is
Thus
( or
(5))
we compute:
c
= 6, d = 8,
(6) f(2k) = 2k + 2 for k = 0, 1, 2, . . . . This holds for small values of k. Assuming (6) holds for a certain k, we get from equation (1) f(2k + 2) = j(f(2k)) = 2(2k) + 6 - f(2k) = 2(k + 1) + 2, showing that (6) holds with k + 1 i n place of k. S o , equality (6) i s settled by induction. Now, let ! (1)
= q.
By equation
(1),
(7) f(q) + q = 8 . So q � 8. The numbers 2, 4, 6, 8 are values of f at 0, 2, 4, 6, respectively (see (6)). Injectivity forces that q !(1) must be one of the numbers: 0, 1, 3, 5, 7. We will show t hat 0, 1, 5, and 7 can be easily eliminated. If q = ! (1) 0 then, by (7) and ( 6 ) , f (O) = f(q) = 8 = ! (6), violating =
=
injectivity. If q = f(1) = 1, contradiction with equation (7) is evident. If q = !(1) = 5 then, by (7), ! (5) = 3. Setting in equation n = 5, and then n = 3, we obtain
(1),
first,
! (3) + f (5) = 16 and !(!(3)) + !(3) = 12; hence f (3) = 16 - ! (5) = 13 and f(f(3)) = 12 - f (3) = - 1, a contradic
7 then, by (7), ! (7) = 1. Equation (1) with n = 7 yields !(!(7)) + ! (7) = 20, i.e. , f(f(7)) = 19, in contradiction to !(!(7)) = ! (1) = 7. We are left with the only possible value f(1) = 3. Induction very similar to the proof of formula ( 6 ) shows that (8) f(2 k + 1) = 2k + 3 for k = 0, 1 , 2, . . . . Equalities ( 6 ) and (8) jointly result in f (n) = n + 2 for n = 0, 1 , 2, . . . . tion again. If q = f(1) =
and it is readily verified that this function indeed satisfies t he given equation (1).
Arithmetic and Combinatorics
37
P roblem 12, Solution 2
Choose and fix an integer non-negative integers:
n � 0.
Consider the following sequence of
ao = n, a 1 = f(n), a 2 = J (f(n)), . . .
'
superscript denoting iteration. In equation (1) set the result is a k +2 + a k +l = 2a k + 6. Subtracting
. . . , (9) f k (n) in place of n; (10)
2a k+1 from both sides we get
that is,
r k+l + 2r k - 6 = 0, where r k = a k +l - a k . Write r k = X k + 2; the equation becomes k = 0, 1, 2, . . . . The last equation obviously X k = ( -2 ) k xo. Consequently r k = 2 + (-2) k xo for k = 0, 1 , 2, . . . By telescoping, we obtain for every integer m � 1: m -1 am ao + L ( a k +l - a k ) k=O m- 1 = ao + L r k k=O m- 1 ao + 2m + L (-2) k xo k=O m = ao + 2m + 1 - (-2) (11) · xo. 3 Recall that all a m s are supposed t o b e non-negative. The exponential growth of I ( - 2) m xol can be in no way matched by the linear term 2m, un less xo 0. ( To be more precise: if xo > 0 then the expression obtained in (11) is negative for large even m ; and if xo < 0 then it is negative for large odd m.) Therefore xo must be 0, whence ro = 2; i.e. , a 1 - ao = 2. This i n view of definition (9) means that f(n) - n = 2.
All t hese relations hold for implies the explicit formula
=
=
=
Solutions
38
We have begun by choosing an integer n � 0 arbitrarily. The conclusion is that f (n) = n + 2 for n = 0, 1, 2, . . . . Problem 1 2 , S o lution 3
This is j ust a variation of Solution 2. Introduce the sequence of iterates (9) and write equation (10) . Note that (10) is an inhomogeneous linear recursion of the second order, with a constant free term. The method of solving such equations is algorithmic. One postulates a solution of the form aA, = Ck (ignoring the initial data) . In the case of equation (10) this yields C = 2; thus the sequence (2k) is a particular solution of (10). If ( ak ) is the sequence (9) we are looking for, then the difference
satisfies the homogeneous equation corresponding to
(10): (12)
This is solved by the standard method (see Problem 10, for example) : the characteristic equation .>. 2 + >. - 2 0 has roots 1 and - 2, and so C k = A ( -2) k + B is the general solution of (12). This implies =
(13) a k = A(-2) k + B + 2k, with unknown constants A and B; the explicit evaluation of those con stants has been carried out in the previous solution, formula (11), in terms of the data ao and xo ro - 2 a1 - ao - 2. But we do not need to know their values! Just note that if A # 0 then the term 2k is negligi ble alongside with A( -2) k , and so ak is negative for k sufficiently large, even or odd according as A < 0 or A > 0. And since it is required that ak f k (n) � 0 for all k, we conclude that A = 0. So a k = 2k + B for k 0, 1, 2, . . . . Hence by definition (9) f(n) - n = a 1 - a o = (2k + 2 + B) - (2k + B) = 2, =
=
=
=
and we arrive at the same result as in the two former solutions. Problem 1 2 , Solut ion 4
The ideas of the Solution 1 and Solutions 2/3 can be neatly combined to produce a fourth one. Consider the sequence of iterates (9) and their recursion equation (10):
Arithmetic a nd Co mbinatorics
39
Using this recursion we compute:
a3 a4 a5 as a7
=
3a l - 2ao, 6ao - 5a l + 18, lla 1 - 10ao - 12, 22ao - 21a l + 54, 43a l - 42ao - 72,
(15)
In the Solutions 2 and 3, the sequence (a k } was generated by an arbitrary initial term ao = n. Now let us take n = 0 and n = 1, and denote the resulting sequences (a k } by (P k } and (q k }:
P k = f k (O), qk = f k (1) for k = 0, 1, 2 , . ( thus P o = 0, qo = 1 ) . Equalities (15) yield, in particular, P 5 = llp l - 12, 54 - 21p l , P6 qs = 76 - 21% 43q l - 1 14. q7
.
.
These numbers have t o be non-negative. So w e get the two-sided estimates:
12 < p < 54 1 1 - l - 21 '
1 14 < q < 76 43 - l - 21 '
-
Each one of these intervals contains only one integer, and hence p 1 = 2, q 1 = 3. Formulas (15) applied to (a k } = (P k } and (a k } = (q k } now pro duce PO = 0, Pl = 2, P 2 = 4, P3 = 6 ( and so on ) and
1, q 1 = 3, q2 = 5, q3 = 7 ( and so on ) . The general rules P k = 2k and q k = 2k + 1 are easily guessed and equally easily proved by induction, based on the recursion formula ( 14 ) . Restate qo =
them more explicitly as:
f k (O) = 2k,
/ k (1) = 2k + 1 for k = 0, 1, 2, . . . This means that the function f acts as follows: In other words,
f is the function: f (n) = n + 2.
Solutions
40
P roblem 13
Show that Ln v'3J is a power of 2 for infinitely many natural numbers
n.
(The symbol Lx J denotes t he Greatest Integer Function.) Introductory Remark
There is nothing peculiar about t he number v'3. In fact, it can be re placed by any other number a with 1 < a < 2. We present three solu tions to the problem involving the sequence L na J , with an arbitrarily fixed a E (1, 2), plus a fourth solution in which a is additionally assumed to be irrational. So, let us fix an a with 1 < a < 2. Call an integer n nice if Lna J is a power of 2. We wish to show that there are infinitely many nice ns. P roblem 13, Solut ion 1
Assume this is not the case. Take an integer k with nice n. Let q be the (unique) integer such that qa < 2 k r = 2 k - qa; thus 0 < r ::5 a. There i s a (unique) integer j � 0 for which
Since (by assumption) obtain
a
<
2,
the number
a/2
exceeds
2 k > na for all :::; (q + 1)a. Set
a 1, -
and we
equivalently,
(2i q + 1 ) a 1 < 2H k :::; (2 i q + 1 ) a. Denoting 2iq + 1 by m we thus have Lmaj = 2i +k , and hence m i s nice. Note, however, that m satisfies the inequality -
which is impossible, according to the definition of k. Contradiction ends the proof. P roblem 1 3 , Solut ion 2
Clearly, 1 is nice (so the set of nice numbers is non-empty ) . Choose any nice number n. The product na represents as na = 2 k + r, k � 0 an integer, r E [0, 1). Consider three cases. Case 1. 0 :::; r < 1/2 . Then L2na J = 2 k+ 1 , hence 2n is nice. Case 2. a/2 :::; r < 1. (This case cannot occur for n = 1 ; indeed, if n = 1, then k = 0 and r = a - 1 < a/2.) Now we have
(2n - 1)a 2 k+1 + (2r - a), =
41
Arithmetic a nd Combinatoric&
[ 0, 1). Hence, L(2n - 1)aJ = 2 k+ l and so 2n - 1 is nice. - 1 > n.) Case 3. 1/2 ::; r < aj2. Define . �' xJ. �2 + a -2 1 ( 1 - � 2J ) or J 0 1 2 ) this is an increasing sequence, starting from xo 1/2 and converging to aj2. So there exists a ( unique) integer j � 1 such that x; - 1 � r · < x;. Since r = na - 2 k , we obtain the inequalities
with 2 r - a E ( Note that 2n
!I
=
!
!
! •
•
•
·
a - 1 ( 1 - -.1 - ) < na - 2 k < -1 + -a - 1 ( 1 - -:-1 ) 2 2 2J ' 2 2J -l =
1 2
- + -equivalent to
2 k+i +l + 2 - a � (2i + 1 n - 2j + 1)a < 2 k+i +l + 1. Denote the number i n parentheses by m . We see that l111 a J and hence m is nice. Evidently, m > n because l na J 2k .
=
2 k+i +l ,
=
In each of the three cases we have found a nice number m greater than n. It follows that there are infinitely many nice numbers. Problem
13,
S olut ion
3
Consider the binary representation of
�
a
=
(O.c1c 2 c3 . . . ) 2 with
1/a :
C k E {0, 1}
k = 1, 2, 3, . . . .
for
(1)
This representation is not unique if 1/a is a dyadic fraction ( e.g. , 3/ 4 can be written either as (0. 11) 2 or as (0. 1011 1 1 . . . ) 2 ) . In such a case, choose the infinite expansion. Thus, in the sequel we are only considering representations (1) with infinitely many cks equal to 1. Choose an index
Recalling that
k for which C k+ l = 1.
According t o
m a n integer,
T k E ( ! , 1] .
k 1 < a < 2, we get
with Hence
2 k a1 ·
-
�
(1),
T k .+ -a1 E (1, 2).
1 m k + 1 < (2 k + 1) · a , -
42
Solutions
showing that L(mk
+ 1)aJ
=
2k .
So mk
+ 1 is a nice number.
To distinct ks with ck+l = 1 there correspond distinct m k s (because mk = ( c 1 c 2 . . . ck ) 2 ). And since ck+l equals 1 for infinitely many ks, this proves that there are infinitely many nice numbers. P roblem
13,
S o lut ion 4
Here we assume that a E (1, 2) is an irrational number. Let b E (2, oo ) be the number determined from the equation
1 1 - + - = 1. a
(2)
b
The reasoning will be based on the well-known theorem which says that if a, b are any positive irrational numbers satisfying equation (2) then the sets A=
{ lna J : n E N }
and
B = { l nb J n E N } :
constitute a partition of the set N of positive integers ; this means that they are disjoint and their union exhausts all of N. (Reference: D. J. New man, A Problem Seminar, New York-Heidelberg-Berlin, 1 982; Problem 46, p. 8; Solution, p. 68. ) We will prove that
2k E A
for infinitely many ks;
this is equivalent to the assertion of the problem. And since N = A is a partition, it suffices to show that
2 k E B then 2 k+i E A for a certain j > Thus assume 2 k E B. So 2 k = LnbJ for some n E N: nb = 2 k + r, r E (0, 1 ) irrational. if
U
B
0.
There exists a (unique) exponent j ;:::: 1 such that 2ir E ( 1 , 2). We claim that 2 k+i E A . Suppose not ; then 2 k+i E B, i.e. , 2 k+i = Lm bJ for some m E N: s E (0, 1 ) irrational. m b = 2 k+i + s , Hence
(2i n - m ) b = (2i +k + 2i r) - (2 k+i + s ) = 2i r - s . The number on the right side belongs to the interval (0, 2) ; that on the left is a multiple of b > 2. This is obviously a contradiction. Thus, indeed, 2 k+i E A ; the proof is complete.
Arithmetic a nd Combinatorics
43
Remark
There exists numbers a > 2, arbitrarily close to 2, such that the set of integers "nice with respect to a" is finite. Take for instance a number whose reciprocal has the binary representa tion
1 -;; = (0.0 � � 1 � 1 � 1 . . . . . . ) 2 m where m � 1 and m i > m for each Consider the product u 2 k · (1/a), where k is an integer greater than �-
=
m. The first binary digit of u after the point is either a zero or a one
followed by a block of mi zeros (for some i). In either case, the "fractional part" of u satisfies the estimate
Note also that
1
1 . (0.0 � 01) 2 = 21 - 2 m1+ l + 2 m+ 3 m + (1/a) < 1, and consequently lu + (1/a)J =
-;; <
Hence u - LuJ luJ . As u is not an integer, this equality shows that there are no integers in the interval [u , u + (1/a)]. In other words, there is no integer n satisfying the inequalities 2 k ::; na ::; 2 k + 1. This shows that n o power 2 k , with any exponent k > m , i s equal t o the integer part of any product na. Clearly, a is close to 2 if m is large enough. The block lengths m 1 , m 2 , m 3 , . . . may form a periodic sequence or not ; accordingly, a can be made rational or irrational, as we please. Problem
14 Four numbers are randomly chosen from the set {1, 2, . . . , 3n} (n is a fixed integer greater than 1). Compute the probability that the sum of those four numbers is divisible by 3. Problem
14, Solut ion 1 The sum of four integers is divisible by 3 if and only. if their remainders modulo 3 constitute one of the following patterns:
(0, 0, 0, 0) ,
(0, 1, 1, 1) ,
(0, 2, 2, 2),
(0, 0, 1, · 2) ,
(1, 1, 2, 2)
- up to permutation (in each quadruple) . Enumerate these patterns 1 through 5, in the order as they are listed above. Suppose there are Ni
Solutions
44
four-element subsets of ( for i = 1, 2, 3, 4, 5).
{ 1, 2, . . . , 3n} corresponding to the i-t h pattern
Each residue class (0, 1 o r {1, 2, . . . , 3n}. Therefore
2
( mod
3))
i s represented by
n
numbers in
The probability p we are about to evaluate is equal to the fraction
NI D with numerator N = N 1 + N2 + N3 + N4 + Ns and denomina tor D = e:) , the number of all four-element subsets in the set under
consideration. It is a matter of a simple calculation to get the outcome p = N/D = 1/3. P rob lem 1 4 , Solut ion 2
Let :F be the family of all four-element subsets of {1, 2, . . . , 3n }. For each C E :F denote by r( C) the remainder the sum of elements of C leaves in division by 3. Clearly, :F = :Fo U :F1 U :F2 , where :Fi = {C E :F I r (C) = i} for i = 0, 1, 2. The probability sought equals
I :F_::.o l..; P = -:-l.ri=-o-=1 +--:F-:21 ' I :Fl i-+-:I-the symbol i :Fi l denoting the cardinality of family :Fi , i.e. , the number of sets in that family. Define the operation c �--+ c', acting in {1, 2, . . . , 3n}, by
c' = { c1 + 1
- the cyclic shift
Since
c < 3n, c = 3n ( mod 3n). To each set C E :F assign the set c ' = {c' I c E C}. if if
C consists of four numbers, i t follows that r (C) = 0 if and only if r(C ') = 1, r (C) = 1 if and only if r(C') = 2, r ( C) = 2 if and only if r(C') = 0. the assignment C C' maps :Fo onto :F1 , :F1 onto :F2 ,
Thus �--+ and :F2 onto :Fo ; hence the three families are equipotent ( consist of equally many members ) : I :Fo l = IF1 I = IF2 I , and so p = 1/3.
A rithmetic a nd Combinatorics
45
Remark
Suppose we choose a k-element set { 1 , 2, . . . , 3n }
C from
(k fixed, 1 :::; k :::; 3n) .
Are all values of r(C) equally probable, as in the case of k
= 4?
The method of the second solution yields an affirmative answer to this question, provided that k is not divisible by 3; indeed, r (C') = r (C) + k (mod 3) ; the assignment C 1-4 C' maps :Fo onto :F1 or onto :F2 (etc . ) , according as k = 1 o r 2 (mod 3) . The argument, however, breaks down when k is divisible by 3. For instance, the probability that the sum of 3 numbers randomly drawn from { 1 , 2, . . . , 9} is divisible by 3, equals 5/ 14 rather than 1/3. P roblem 1 5
For what natural numbers n is it possible to tile the with 2 x 2 and 3 x 3-squares?
n x
n-chessboard
P roblem 1 5 , Solution 1
If n is even, the tiling is trivially possible. Thus let n be odd and suppose the chessboard has been tiled as described. In each 3 x 3-tile, colour blue the three cells (unit squares) adj acent to its left edge, colour red the three cells adjacent to its right edge, and colour green the three cells in the middle; the 2 x 2-tiles remain uncoloured. Enumerate the columns (vertical lines) of the board 1 through n. S up pose there are b i blue cells, 9i green cells, ri red cells and u i uncoloured cells in the i-th column. Clearly, Ui is even, and the sum b i + 9i + ri + Ui is equal to n 2 , an odd number. Therefore
bi + 9i + ri = 1
(mod 2)
=
for
i
=
1, . . . , n.
(1)
The right neighbour of a blue cell is a green cell; the right neighbour of a green cell is a red one. Thus b i = 9 i + l ri + 2 and we restate relations ( 1 ) as
ri +2 + ri +l + ri = 1
(mod 2 )
for
i
=
or - which is the same -
ri +3 + ri+2 + ri+l = 1 Subtract (2) from (3) to obtain
(mod 2)
for
i
. - 2;
1, . . , n
= 0, . . . , n - 3.
(2)
(3)
46
Solutions
In the two leftmost columns of the board there are no red cells ; so = 0, and relations (4) imply
r1 = r 2
Ti 0 =
( mod 2 ) for i non-divisible by 3.
(5)
In the rightmost column there are no blue or green cells; so b n = 9 n = 0, whence by (1): T n = 1 ( mod 2 ) . This in view of conditions (5) shows that n must be divisible by 3. In conclusion, if the board can be tiled as required, then n is divisible by 2 or 3. The converse implication is obvious. Problem 1 5 , S o lution 2
A colouring argument can be used in a yet smarter manner. As in So lution 1 , assume n is odd. Colour all the columns of the board black and white alternately, in a "zebra" fashion. For n odd, the two outer columns are coloured alike - say, black; so there are n black cells more than white ones. Suppose the tiling is possible. Each 2 x 2-tile covers two white cells and two black cells. Each 3 x 3-tile covers three white cells and six black cells, or conversely. The difference between the number of black cells and the number of white cells covered by a single tile equals 3, -3 or 0. The total difference between the numbers of black and white cells ( in the whole board ) equals n. Thus n is the sum of some threes, some minus-threes, and some zeros - hence, it is a number divisible by 3. Conclusion as in Solution 1: a tiling in question is possible if and only if n is divisible by 2 or by 3. Remark
The easy "if" part results from "uniform" tilings, using tiles of only one of the two kinds. It is however worth noticing that if n > 6 is divisible by 3 or 2, then one can tile the n x n-board actually using at least one tile of either kind ( the reader may try to show that ) . Problem 1 6
A triangular prism is a pentahedron whose two parallel faces ( "top base" and "bottom base" ) are congruent triangles and the remaining three faces are parallelograms . We are given four non-coplanar points in space. How many distinct triangular prisms having the four given points as vertices are there? Problem 1 6 , Solut ion 1
The four points can be distributed between the two bases of a prism in two fashions: 3 + 1 or 2 + 2. First case ( 3 + 1): Choose three points out of four ( this can be done in 4 ways ) ; they span a triangle, which we take for the base of a prism . Link
A rithmetic a nd Combinatorics
47
the fourth point with one of the vertices of that base (3 possibilities) ; the connecting segment will be a side edge of the prism , which is thereby fully determined. Second case (2 + 2): S plit the given set of four points into two pairs (there are 3 ways to do that) ; label the points in one pair A, B and those in the other C, D . The points A, B are supposed to lie in one base of the prism under construction, and C, D in the other. Take one of the segments AC, AD, BC, CD to be a side edge of the prism (4 possibilities ) ; again, the prism is determined. Thus we can construct 4 3 12 prisms of the first type and 3 4 = 12 prisms of the second type, and this gives 24 as the final outcome. ·
Problem
16,
=
·
Solution 2
Imagine an arbitrary triangular prism. Any quadruple of its vertices nec essarily contains at least one pair of points belonging one to the bottom base, the other one to the top base, and connected by a side edge of the prism. Hence, if the four given points are to be vertices of a triangular prism, two of them (call them P, Q) must be the endpoints of a side edge. The other two points (R and S) cannot be joined by an edge, since they are not coplanar with P and Q . There are (�) = 6 ways t o choose the pair P, Q . This done, we can attach each of R, S to either P or Q . The resulting pairs of segments will be edges of the prism: PR , PS
or
Q R , QS
or
PR, QS
or
PS, Q R.
This defines four possible cases. We claim that in each case the prism is uniquely determined. In each one of the first two cases, we have already one base triangle (and the side edge P Q ) ; translate that triangle by the corresponding vector (PQ or QP) to get the other base. Consider the third case, with P Q , PR, QS being edges of the prism. Complete the p&allelograms P Q R'R and PQ SS' (note that R' =f. S and S' =f. R) ; the points R' and S' are the remaining two vertices of the prism. The fourth case is analogous. We see that, on the total, there exist 6 4 24 triangular prisms with four vertices in the given points. ·
Problem
=
17
Consider the infinite chessboard with squares coloured white and black, in the usual manner. S uppose S is a set of 1976 squares such that every two squares in S can be connected by a path consisting of consecutively adjacent squares. (Two squares are adj acent if they have a common
48
Solutions
edge. ) Show that there are at least 494 white squares in S. Moreover, show that 494 is the exact bound. Problem 17, S o lution 1
If every two squares in a certain set can be linked (within that set) by a path consisting of consecutively adjacent squares, we will say that the set is connected. We are going to prove a fact slightly more general than requested: For any positive integer n, the number of white squares in every con nected set of n squares is not smaller than (n - 1 )/4. This is trivially true for n = 1, 2. Fix an integer n > 2 and assume inductively that the claim holds for all positive integers smaller than n. Take any connected set S composed o f n squares. Define the distance between two squares as the minimum number of edges one has to cross while going from one square to the other, along an admissible path (within S) . Choose and fix a pair of squares A, B E S whose distance is a maximum; denote their distance by m . ( Since n > 2, m > 1.) Thus there exists a path CoC 1 . . . Cm - 1 C m , composed of squares Ci E S, with Co = A, Cm = B. Remove from S square Cm - 1 together with those squares adj acent to C m - 1 whose distance from A is exactly m. Denote by S' the set that remains.
Note that square Cm - 2 has not been removed (its distance from A is m - 2 and not m ). So we have removed at most four squares, and hence I S' I = n' � n - 4.
One of the squares C m - 1 and C m is white, and these two squares have been moved from S ; so there is at least one white square in the set S \ S'. We now show that S' is connected. Choose a square D E S'. There exists a path EoE 1 . . . E k- 1 E k in the set S, with Eo = A, E k = D ; of course, k :::; m (by the maximality of m ) .
Squares Eo, E 1 . . . . , E k- 2 obviously belong to S' (the distance from A to each of them is less than m - 1 , so they have not been removed from S in the formation of S') ; the question is whether E k- 1 also belongs to S'. S uppose not. The only removed square distant from A by less than m is C m - 1 · Hence Ek- 1 = C m - 1 and k = m. But then the square D = Ek , adjacent to Ek- 1 , i.e. , to Cm - 1 , has distance m from A ; and this means that this square should have been removed - contrary to its choice (D E S') . S o every square D E S' can be linked with A within S'. The connect edness of S' follows and the induction hypothesis applies: there are at least ( n' - 1 ) / 4 white squares in S'. And since S \ S' contains at least one white square, we conclude that the number of white squares in S is
A rithmetic a nd Combinatorics
not less than (n' - 1 ) /4 + 1 � precisely the induction claim.
49
(n - 4 - 1 ) /4 + 1
=
(n - 1) /4.
This is
The theorem formulated at the beginning is now proved. For n = 1976 it implies the required bound 494. To see that 494 is optimal, consider a horizontal 3 x 988 rectangle, with white squares removed from the upper row and from the lower row (but not from the middle one) . This is a connected set of 1976 squares, out of which exactly 494 are white. Problem 17, Solution 2
We will apply another inductive reasoning to prove the fact stated at the beginning of Solution 1 : if a connected set of n squares has w white squares, then w � (n - 1 ) /4. Assume this is true for all positive integers smaller than a fixed integer 1. Let S be a connected set of squares. Choose a white square W E S. From any other square in S a path (contained in S) leads to W , and this path necessarily passes through one of the four squares adjacent to W . Consequently, every square in S \ { W } is accessible from one of those four squares via a route omittipg W ; i.e., a route contained in S \ { W } . Thus the set S \ { W } is the union of at most four connected sets. Label these sets St , . . . , S1 (1 $ l$4) . Suppose consists of n i squares, of them being white. According to the inductive assumption, wi � 1 ) / 4 for i = 1, . . . , l . Therefore, denoting by w the number of white squares in S, we obtain
n>
n
si
Wi (ni I
i L i=l W
1+
w >
I
-1 L: n i 4
=
i= l
1
4 (n - 1 ) ,
completing the induction step. The claim i s proved. For an argument that for 1976 the bound w � 1) /41 = 494 is sharp, see the last paragraph of Solution 1 .
r(n -
n
=
Solution s
50
Problem
18
Consider a n alphabet consisting o f three symbols a, b, c. How many n-character words with the following properties ( 1) and (2) can be com posed?
(1)
the word should begin and end with an
a;
(2) neighbouring positions must be occupied by different symbols. Problem
18, Solution 1 Consider words of length n that begin with an a and satisfy condition (2) . Denote by a n , b n , e n the numbers of such words ending in a, b , c , respec tively. Attaching an a to a word of length n whose last character is b or c we obtain an admissible word of length n + 1. Hence a n+l = bn + en . Likewise, bn + l = a n + en and c n+l = a n + b n . Since the roles of symbols b and c are symmetric, we infer b n = en , and so
Consequently i.e. (3) The initial ans are: using (3) :
a 1 = 1, a 2 = 0 ; a few subsequent terms are computed
( �a n ) : ( . . . , � a 4, � a 5 , � a 6, � a 7, � a s, � a g, . . . ) = ( . . . , 3, 9, 15, 33, 63, 129, . . . ). The pattern becomes plain: � a n = 2 n -2 + (- 1) n - \ i.e. ,
A "roughly geometric" sequence? Consider
Once guessed, this equality is easily proved by induction. For n = 1 and n = 2 formula (4) gives the correct values, and the inductive step ( (n, n+1) -+ (n+2) ) follows immediately from equality (3) :
an +2 = H2 n-1 + (- 1) n ) + 2 . H2 n- 2 + (-1) n - 1 ) = H2 n + ( -1) n+l ) . Note that a n is the number we had to calculate. Its value is given by formula (4) .
A rithmetic a nd Com binatorics
51
Remark
The explicit formula (4) could be derived from (3) without guessing, by the usual method of solving linear recursions (compare Problem 10, Solution 3, for instance) . P roblem 1 8 , Solut ion 2
Assume n 2:: 3. Imagine a row of n empty cells. They have to be filled-in with symbols a, b, c, observing conditions ( 1 ) and (2). In the two outer cells, as must be placed; this is prescribed. Assume that character a oc curs k times inside the row. (Consider k to be fixed, for the while. ) There remain n - 2 - k cells to be filled with other symbols. The occurrences of a split those cells into k + 1 blocks of positive lengths {3 1 , . . . , f3 k+ l (positive, because the as never occur on neighbouring positions) .
There are ( n -� -k ) ways to represent the number n - 2 - k as a sum of k + 1 positive integers n 2 - k = {31 + · · · + f3k+ l (because the sums f3 t , f3 t +{3 2 , f3 t + fJ2 +f33 , . . . , f3 t + · · · +f3k can constitute an arbitrary subset of { 1 , . , n-3-k} ) . Every such representation determines the positions of a. Each of the k + 1 blocks must be filled with bs an d cs alternately, and that can be done in two ways. Summarizing, there are a n words satisfying conditions ( 1 ) and (2), where -
.
.
(5) applying the usual convention that ('; ) = 0 whenever (cf. the solution to Problem 3}, we may assume that formula (5) ranges over the set of all integers. To bring this sum to a closed form, note that
= =
= =
� (n �: � k ) 2 k+l
� (n - :
) 21+2
2 � n - : - l 2 1+1 - l
(
2a n-1
(for
)
n 2:: 4) ,
k < 0 or k > m k in the sum in
Solutions
52
and we arrive at the recurrence formula (3) of Solution 1. For n = 3, 4 the expression (5) yields aa = a4 = 2 . The explicit formula (4) is deduced in a standard way; see Solution 1 . Problem 1 9
Nine trucks follow one another, i n a line, o n a highway. A t the end of a day's ride it turned out that each driver disliked the style of the driving of the one in front of him. They wish to rearrange themselves so that , next day, no truck would follow the same truck that it followed on t he first day. How many such rearrangements are possible? P roblem 1 9 , Solution 1
Translated into mathematical terms, the problem is to calculate the num ber of permutations of the set { 1 , 2, . . . , 9} in which no one of the suc cessions 12, 23, 34, 45 , 56, 67, 78, 89 appears. Let F (n) be the num ber of permutations of { 1 , 2, . . . , n } with the analogous property (call them feasible) . We are going to derive a recurrence formula. Clearly, F ( 1 ) = F (2) = 1 . A permutation of { 1 , 2, . . . , n, n+ 1 } arises from a permutation
1r
of { 1 , 2 , . , . , n}
by inserting the element "n + 1" to one of n + 1 positions (n - 1 sockets be tween successive entries, plus two outer positions, one at the beginning and one at the end) . The arising permutation will be feasible if either 1r is a feasible permutation and the element "n+ 1" is placed on any po sition except the one immediately after the "n" or 1r is a non-feasible permutation, feasibility violated by a single forbidden pair k, k + 1 in di rect succession, and the element "n+ 1" is placed j ust so as to disconnect that pair. In the first case there are n possibilities of inserting the element "n+1" into one of F (n ) feasible permutations 1r of { 1 , 2 , . . . , n } . This yields the first summand in the recurrence formula ( 1 ) (below) . In the second case the address of "n+1" is determined (between "k" and "k+ 1" ) , while k can be any number out of 1 , 2 , . . . , n - 1 ; the permutation 1r with the only non-separated pair k, k + 1 can be identified with a feasible permutation of an (n - 1 )-element set, since we can regard the "brick" (k, k + 1) as a single entity. This yields the second summand in the formula we arrive at: (1) F (n + 1 ) = nF(n) + (n - 1 ) F (n - 1 ) . Using this formula and knowing the initial data F ( 1 ) = F (2) = 1 we easily compute F (9) 148329. =
Arithmetic a nd Com binatorics
Problem
19,
53
Solut ion 2
i = 1, 2, . . . , n , let Pi be the set of those permutations of the set {1, 2, . . . , n } in which the element "i" appears immediately after "i- 1" . Note that I Pi l = (n - 1)! ( the number of arrangements of the elements 1, 2, . . . , i-l, i + 1, . . . , n ; the position of "i" is determined ) . If K is a subset of {2, . . , n } with IKI = k, then For
.
I n Pi I = ( iEK
n - k)!
(2)
( such is the number of permutations of {1, 2, . . . , n } \ K; any such per mutation uniquely generates a permutation that belongs to n pi -
iEK
the elements of K have to be inserted, from smallest to greatest, to the appropriate places, fully determined ) .
n = n ! - IU Pi I · i=2
A permutation is feasible if and only if it does not belong to any Thus F (n )
By the Inclusion-Exclusion Principle, and in view of equality
Pi ·
(2),
n n = L I Pi l - L I Pi n P; l + . . . + (-1) n-1 1 n Pi I i= 2 i= 2 i
n-1 (n - 1)! L (- 1) k +l n k.� k . k=1 Consequently, F (n)
n-1 n-1 n-k n-k = (n - 1)! L( - 1) k -- . = n! - (n - 1)! L (-1) k + 1 1 k. k.1 k =1 k =O
For n = 9 this quantity evaluates to 148329. This is the number sought.
54
Solutions
Problem
20
We are considering paths ( Po, P1 , . . . , Pn ) of length n over lattice points in the plane (i.e. , points (x, y) with integer coordinates) ; for each i, the points Pi-1 and Pi are assumed to be adjacent on the lattice grid. Let F (n ) be the number of those paths that begin in Po = (0, 0) and end in a point Pn lying on the line y = 0. Prove that F (n) = e:) . Problem
20,
Solut ion 1
For any integer k , denote by f ( n , k ) the number of paths of length n (short, n-paths) beginning in (0 , 0) and ending in a point on line y = k . Thus F (n) = f (n , 0 ) . Every ( n - 1)-path ending either o n line y = k - 1 or on line y = k + 1 can be uniquely extended to an n-path ending on line y = k . An (n - 1 )-path ending on line y = k admits exactly two such extensions. Hence follows the recursion formula
1, k - 1) + 2f (n - 1 , k) + f (n - 1 , k + 1 ) .
f (n , k) = f (n -
{�
(1)
There exists only one feasible path of length 0 (both its endpoints coin ciding with the origin) . So f (O , k ) =
for k = 0, for k =f. 0.
(2)
Formulas (2) and (1) generate the table of values of f (n , k): k= n=O 1 2 3
.
.
-4 - 3 - 2 - 1
.
0 0 0 0
: : : : .
.
.
.
.
.
0 0 0 1 .
.
.
.
0 1 4 15
0 0 1
6
.
.
.
.
.
.
.
.
.
.
0
1
2
3
4
1 2
0 1 4 15
0 0 1
0 0 0
0 0 0 0
6
20 .
.
.
.
.
.
.
.
.
.
1
6
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
.
Even-numbered rows of Pascal's triangle are readily recognized. Hence we guess that f (n , k ) = g (n, k ) , where g (n, k ) =
( ) 2n n+k
for
n, lkl = 0, 1, 2, . . .
(3)
To prove this guess, it will be enough to show that the numbers g (n , k) obey the same recursion relation as the f (n , k ) s.
With the convention that (�) = 0 when r is smaller than 0 or greater than q, the fundamental Binomial Identity (;) = (�=D + (q � 1 ) holds, without any restriction, for every natural q and every integer r (co mp are
A rithmetic a nd Combinatorics
55
Problem 3, Solution 1 ) . Using this identity, we calculate
( ) ( ) ( ) ( � ;� ) ( � ;� ) C ;:) 2n n+k
2n - 1 n+k-1
2n - 1 n+k n + +2 n 2 n 1 n g (n - 1 , k - 1 ) + 2 g (n - 1 , k) + g (n - 1 , k + 1 ) ;
g (n, k)
=
and moreover, g (O, k ) =
{�
+
for k = 0 , for k #- 0 .
(4) (5)
The recursion formulas ( 1) and ( 4) for .f and g are the same, and so are the initial values (2) and (5) . So the guess (3) was correct. Hence, in particular, F (n) = .f (n, 0) = g (n, 0)
=
c:)
for
n = 0, 1 , 2, . . . .
20, Solution 2 An n-path (Po , P1 , . . . , Pn ) beginning in Po = (0, 0) can be encoded by a (2n)-string of zeros and ones ( q , c 2 , . . . , c 2n- 1 , C 2n ) , according to the
Problem
following rules:
if � = [1 , 0] , if � = [0, 1] , if Pj-lPj = [- 1 , OJ , if Pj - l Pj = [0, - 1] ,
then then then then
c 2j - l = 1 , c 2 j = 0 ; C 2j - 1 = 1 , c 2j = 1 ; C 2j-1 = 0, c 2j = 1 ; c 2j-l = 0, c 2 j 0 . =
To put this i n words: each couple o f successive symbols (c 2j-l , c 2j ) rep resents one step on the path; a step east is rendered by the coupling (10) ; a step north - by ( 1 1) ; a step west - by (01 ) ; a step south - by (00 ) . And conversely, every such (2n )-string represents a feasible path. To distinct paths there correspond distinct codes; the coding is bijective. Suppose a path consists of u steps north, v steps south, and (jointly) n - u - v steps east or west. Then the endpoint Pn lies on line y = k if and only if u - v = k ( "horizontal" steps are irrelevant) . The number of "ones" in the code of such a path equals 2u + (n - u - v ) , i.e. , n + k. Consequently, the paths ending on line y = 0 are encoded by binary (2n) strings with exactly n "ones" . As there are exactly e:) such steps, the result follows.
Solutions
56
P roblem
20,
Solut ion 3
As in Solution 2, we regard a path as consisting of steps north, south, east and west. A path ends on the line y = 0 if and only if there are equally as many steps north and south. Now, look at the polynomial (x 2 + 2 x + 1) n , written in the form
(x 2 + X + X + 1) (x 2 + X + X + 1) · · · (x 2 + X + X + 1)
(6)
n factors ) . Multiplying out, we obtain the sum of terms with exponents r = 0, 1, , 2n. Each of these terms results by an independent choice of one of the four entries from each factor ( the x 2 , the "first" x, the "second" x, the 1). Every such selection induces a feasible path. Namely, let u s agree that : ( product of
xr
.
.
.
if the term selected from the j-th factor is: first second
x2 x x 1
Paths ending on the line
then the j-th step is performed: north, west , east, south.
y =
0 correspond to those selections in which The product of the selected terms equals case it is different from x n . Hence, the number of paths ending on line y = 0 is equal to the coefficient of x n in the polynomial ( 6 ) . It remains j ust to notice that
x 2 is used as many times as 1. x n in that case; in any other
Accordingly, the coefficient of x n equals
e:) .
Thus
F(n) = e:) .
Solutions: Algebra Problem
21
Determine all real polynomials P ( x) o f degree not exceeding 5, such that P(x) + 1 is divisible by (x - 1) 3 and P(x) - 1 is divisible by (x + 1) 3 . Problem
Let
P(x)
2 1 , S olution 1 be as required; then
P(x) + 1 = (x - 1) 3 Q (x), P(x) - 1 = (x + 1) 3 R(x), with Q ( x) and R ( x) real polynomials of degree 2 ( at most ) . In each of them the leading coefficient is the same as in P(x) (zero not excluded ) , and thus Q(x) = ax 2 + bx + c, R(x) = ax 2 + px + q. The postulated identities result in P(x) + 1 = (x 3 - 3x 2 + 3x - 1) (ax 2 + bx + c), i.e. , P (x)+ 1 = ax 5 + (b-3a)x 4 +(c- 3b+3a)x 3 + ( -3c+3b-a)x 2 + (3c- b);z:- c and, analogously,
Hence, comparing the coefficients,
b - 3a c - 3b + 3a - 3c + 3b - a 3c - b -c
= = = = =
p + 3a, q + 3p + 3a, 3q + 3p + a, 3q + p, q + 2.
Luckily enough, this system of five equations with five unknowns is quite easy to solve and yields
a = _ 28 ' b - _ .!!8 ' c = - 1 , Thus
P (x)
P - .!! 8>
q = - 1.
is given by any one of the following two equalities:
P(x) = (x - 1) 3 (- i x 2 - �x - 1 ) - 1, P(x) = (x + 1) 3 (- ix 2 + �x - 1) + 1. Each of them produces the final formula P(x) = - ix 5 + defining the unique polynomial with properties a s needed.
�x 3 - 1;x,
58
Solutions
Problem
21,
Solut ion
2
The reasoning will be based on the well-known fact of algebra: a poly nomial F(x) is divisible by (x - x 0 ) k if and only if its derivative F'(x) is divisible by (x - xo) k- 1 , and moreover F(xo) = 0. Let P(x) be any polynomial of degree at most 5. Then each one of the statements 1-4 ( below ) is equivalent to the subsequent one ( we write G (x) IH(x) when G(x) divides H (x)) : Statement 1 . and (x + 1) 3 I P(x) This is j ust the condition of the problem. Statement 2.
(x - 1) 3 1 P(x) + 1
1.
(x - 1) 2 1 P'(x); (x + 1) 2 I P'(x); P(1) + 1 = 0 = P ( -1) - 1. Statements 1 and 2 are equivalent i n view o f the theorem formulated at the beginning, applied to F1 (x) = P(x) + 1 and F2 (x) = P (x) - 1. Statement
3.
(x 2 - 1) 2 I P '(x) ; Statements 2 and
P(1) = - 1 ;
P(- 1) = 1.
3 are equivalent because (x - 1) 2 and (x + 1) 2 are coprime polynomials, and hence P' ( x) is divisible by both of them if and only if it is divisible by their product. Statement 4 .
P'(x) = Ax 4 - 2Ax 2 + A; ( The degree of P(X) does Statement
P(1) = - 1 ;
P(- 1) = 1.
not exceed 5 . )
5.
P(x) = !Ax 5 - �Ax 3 + Ax + C;
P (1) = - 1 ;
P(-1) = 1.
The equivalence between statements 4 and 5 follows from the deriva tive/ antiderivative algorithm. Now, if P (x) satisfies the conditions of Statement 5, then by setting x = 1 and x = - 1 we obtain C + 18 A = -1, and C - 18 A = 1, whence C = 0 , 5 5 1 5 , and so A- 8 -
-
P(x) =
-
ix 5 + �x 3 - 1:x.
Conversely, if P(x) is given by this formula, then the conditions of state ment 5 are satisfied. And since statement 5 is equivalent to statement 1, we infer that the polynomial P(x) = - ix 5 + 1�x 3 - ';tx i s the unique solution to the problem. Problem
22 Prove that the polynomial x n + 4 factors into the product o f two polyno mials of lower degrees with integer coefficients if and only if n is divisible by 4.
A lgebra
59
Problem 2 2 , Solut ion 1
Assume
x n + 4 = F(x)G(x),
ai , b;
integers;
0 < k < n,
0<
m < n,
k
+ m = n.
Then aobo = 4, a k b m = 1, and hence a k = b m = ±1. It is convenient to set a i = 0 for i > k and b; = 0 for j > m. Let a be the least index for which a a is odd and let {3 be the least index for which bp is odd. Since a k bm = ±1, we have a 5 k and {3 5 m. In the product F (x)G(x), the coefficient of x a +/3 equals =
In this expression all summands except a a bp are even numbers ( because ais are even for i < a and b;s are even for j < {3). So the coefficient ( 1 ) o f x a +/3 i n F(x)G(x) = x n + 4 is odd, which means that a + {3 = n , and consequently a = k, {3 = m . Thus, according t o the definition o f a and {3, (2) ao, a � , . . . , a k - 1 , bo, b 1 , . . . , b m - 1 are even numbers. Since aobo = 4, we get ao = bo = ±2. We claim that k = m. Assume the contrary: let k < m, say. Write the product of F(x) and G(x) in the form (3) F(x)G(x) = P(x) + Q (x) + x n , where
P(x) = (ao + a 1 x + · · · + a k- 1 X k - 1 ) (bo + b 1 x + · · · + bm - 1X m - 1 ) , Q (x) (a k box k + a k b 1 x k+ 1 + + a k bm - 1X n -1 ) + (bmaox m + bma 1 x m+1 + · · · + bm a k- 1X n - 1 ) . In view of property (2) , all coefficients of P(x) are divisible by 4, while the coefficient of x k in Q (x) equals ±2 ( as a k = ±1, bo = ±2) . Thus the coefficient of x k in the polynomial (3) is an even integer, non-divisible by 4; and this is a plain contradiction with the equation F (x)G (x) = x n + 4. This settles the claimed equality k = m. Consequently n = 2k, and there fore (4) F(x)G(x) = x 2 k + 4 > 0. ·
·
·
It follows that F ( x) and G ( x) are real polynomials without real roots, hence of an even degree. So k is an even number, i.e. , n is divisible by 4.
60
Solutions
And conversely, if n
= 4l, l E N, then
is the desired factorization. P roblem
22,
Solution
2
The equation x n = -4 has n distinct complex roots z 1 , . . . , z n , each of an absolute value of 4 1 / n . If x n + 4 factors into a product F(x)G (x), then each o f zt, . . . , Z n must b e a root of either F(x) o r G (x). Assum � z 1 , . . , Z k are roots of F(x) and z 1 , . . . , Zn -k are roots of G (x ) ; then .
F(x) = A (x - z 1 ) · · · (x - z k ), G(x) = B (x - Z k+ t ) · · · (x - Zn ) , with non-zero constants A and B, satisfying A B = 1. The factors F ( x) and G ( x) are assumed to have positive degrees and integer coefficients. So 0 < k < "n and A = B = ±1. The free terms of F (x) and G (x), equal respectively to (-1) k Az 1 · · · z k and (-1) n -k Az k+ l · · · z n , should also be whole numbers. Their absolute values 4 k / n and 4 ( n -k ) j n are comprised strictly between 1 and 4 (as 0 < k < n) , and their product is equal to 4. Hence, each of them equals 2, which means that k n/2, and we arrive at the inequality ( 4) of Solution 1. Conclusion as before. =
Problem
23 Find all natural numbers n for which the polynomial
Pn (x) = x 2n + (x + 1) 2n + 1 is divisible by the trinomial T (x) = x 2 + x + 1 . Problem
23,
Solution
1
Note that T (x - 1) x 2 - x + 1 , and so x 6 - 1 = (x - 1)(x 2 + x + 1) (x + 1) (x 2 - x + 1) = (x 2 - 1) T (x) T( x - 1). =
Replacing
x by x + 1, (x + 1) 6 - 1 = (x 2 + 2x) T (x + 1) T (x).
Therefore the difference
x 2n+6 + (x + 1) 2n+6 x 2n (x + 1) 2 n x 2n (x 6 - 1) + (x + 1) 2n ( (x + 1) 6 - 1) _
_
T( x ) , for every integer n 2: 0. Since Po(x) 3, P1 (x ) = 2T (x) , P2 (x ) = 2T(x ) 2 ,
is divisible by
=
Algebra
61
obvious induction shows that is non-divisible by 3.
Pn (x) is divisible by T(x)
if and only if n
Problem 2 3 , Solut ion 2
A polynomial roots of T (� ) ,
P(x)
is divisible by
a = - ! + ! v'3 i
T(x)
if and only if the two complex
{3 = - ! - ! v'3i,
and
are also roots of P(x). These numbers satisfy the equalities:
Consequently, each of the two numbers
and
Pn (f3) = (f3 2 f + ((/3 + 1) 2 f + 1 an + {3 n + 1.
equals j ust V'lriting a natural number n in the form n 2, we obtain
= = =
Conclusion:
= 3 k + r , with r being 0 , 1 or
Pn (f3) a 3k +r + {3 3k+r + 1 a r + {3 r + 1
{3 0
if r if r
=
=
0,
1
or r
= 2.
T(x) divides Pn (x) if and only if r ¢. 0
(mod 3) .
P roblem 24
For every positive integer k show that the polynomial
is divisible by the binomial Problem 24, Solut ion
x 5 + 1.
1
Write for brevity
p(x) = x 3 - x 2 + x - 1,
q(x) = x 4 - p(x)
and notice that
x 4 - 1 = (x + 1)p(x),
x 5 + 1 = ( x + 1)q(x),
(1)
62
Solutions
and thus
The claim will be proved by induction. For k =
1,
{x 4 - 1) {x 3 - x 2 + x - 1 ) + (x + 1)x 3 x 7 - x 6 + x 5 + x 2 - x + 1 = (x 2 - x + 1) { x 5 + 1) . It is evident from (1) and (2) that the divisibility of Pk (x) by x 5 + 1 P1 (x)
=
is
equivalent to the divisibility of the polynomial
by
q(x ) .
And since, by the definition of q(x ) ,
p(x) k+2 + x 4k +3 p(x) k+2 + (p(x) + q(x))x 4 k - 1 p(x)Q k (x) + q(x)x 4 k - I , Q k +l (x) is divisible by q(x) whenever Q k (x) is. Q k + l (x)
=
we see that pletes induction. P roblem
Let
(1),
p(x)
24, Solut ion 2 and q(x) have the same meaning as in Solution
This com
1.
In view of
(x 4 - 1 ) q(x) = (x + 1)p(x)q(x) = (x 5 + 1)p(x). The k t h power o f p(x) = x 4 - q(x) equals ( by the Binomial Theorem ) x4 k plus the sum of terms divisible by q(x). Thus where R k (x) is a polynomial. Using these equalities together with (2) we obtain
Pk (x)
= =
showing that
{x 4 - 1) [q(x)R k (x) + x 4k ] + (x + 1 ) x 4 k - l (x 4 - 1)q(x)R k (x) + x 4k - l (x 5 - x) + (x + 1 ) x 4k-l (x 5 + 1 ) p(x )R k (x) + x 4k - l (x 5 + 1 ) ,
Pk (x) is divisible by x 5 + 1.
Problem
2 4 , Solut ion 3 It will be enough to show that each complex root of x 5 of Pk (x) . Obviously, xo = - 1 is a root of Pk (x) .
+ 1 is also a root
Algebra
63
Let now .A be any non-real root of x5 + 1 Then by the second equality of ( 1 ) , q(.A) = 0, and hence p(.A) = .A\ by the definition of q(x) . This inserted into equation (2) results in .
j ust as needed. Problem
25 Find all pairs o f real numbers
a, b such that the polynomials
P (x) = x 4 + 2ax 2 + 4bx + a 2 and Q (x) = x 3 + ax + b have two distinct common real roots. Problem
25, S o lut ion 1 Suppose x 1 and x 2 are real roots of P(x) and also the roots of
T(x)
=
P(x) - xQ (x)
=
Q (x) (x 1 =/= x 2 ) . They are
ax 2 + 3bx + a 2 •
Thus a =/= 0 and the discriminant D of the t rinomial tive: D 9b 2 - 4a3 > 0. By Viete's Formulas,
T (x) must be posi
=
X l + X 2 = -3bja, Since, by assumption, 0
x 1 =/= x 2 and Q (x l )
=
Q (x 2 ) = 0, we obtain
=
Xl - X2 x� - x� + ax1 - ax 2 = X l - X2 x 21 + x 1 x 2 + x 22 + a (x l + X 2 ) 2 - X l X 2 + a (-3bfa) 2 . =
So b = 0 and 0 < D And conversely, if b
= -4a3, i.e. , a < 0. 0 > a, then the polynomials
=
have the common roots � and those of the form (a, 0) with a < 0.
- �· Thus the pairs sought are
64
Solutions
Problem 2 5 , S o lut ion 2
The derivative of P (x) equals 4Q (x) . Thus if x 1 and x 2 are real roots of both P(x) and Q (x ) , then they are double roots of P(x ) , and conse quently Therefore 2x 1 + 2x 2 = 0, xix � = a 2 , and we conclude that one of the numbers xb x 2 must be equal to vial and the other to - vial (the condition x 1 =I= x 2 implies a =I= 0) . Now, Comparing this with the definition of P (x) we see that b = 0 and a = - la l < 0, and so the pair (a, b) must be of the form (a, 0) with a < 0. Conversely, every such pair satisfies the demands of the problem (see Solution 1 ) . Problem 26
Let a, x, y, z be real numbers such that + y+ cos(x + y + z)
COS X
COS
COS Z
sin x + sin y + sin z = a. sin(x + y + z)
Prove the equality: cos(y + z) + cos(z + x) + cos(x + y) = a. Problem 2 6 , Solut ion 1
In what follows, all sums are cyclic over the triple (x, y , z) (thus, e.g. , the symbol I: sin x denotes the sum sin x + sin y + sin z, etc.) . Write w = x + y + z. Then, by assumption,
2: cos x = a cos w ,
L sin x = a sin w .
Hence
2: cos(y + z)
L cos(w - x) L (cos w cos x + sin w sin x) (cos w) 2: cos x + (sin w ) 2:: sin x a cos 2 w + a sin 2 w = a.
A lgebra
65
26, S olut io n 2 Using the Euler Formulas
P roblem
cos x =
eix _ e-ix
sin x = ----
2
2i
we restate the assumptions in the form Adding and subtracting these two equalities, we obtain two new ones:
"'"' e -i:z: = ae -iw . L...,
Consequently,
2a
=
1 - e -ix -.e•1-w L e ix + -. e-•w L 2: e i(x-w> + 2: e i(w-x) L (ei(w-x) + e-i(w-x} ) 2 I: cos (w - x ) ,
and the claim results. Problem 27 If a, b, c are pairwise distinct real numbers, show that the value of the expression
b - e + -a - b + -c-a 1 + ab 1 + be 1 + ca
--
is never equal to zero. P roblem 27, Solution 1 Multiply the given expression by the product of the three denominators and denote the resulting expression by F(a, b, c):
F(a, b, c)
(a - b )(1 + bc) (1 + ca) + +(b - c) (1 + ca)(1 + ab) + (c - a) (1 + ab) (1 + be). (1) We have to show that if a =/= b =/= c =/= a, then F(a, b, c) =/= 0; and this fol =
lows directly from the transformation:
F(a, b, c) = (a - b) + (a 2 - b 2 )c + (ca - bc)abc +(b - c) + (b2 - c 2 )a + (ab - ca)abc +(c - a) + (c 2 - a2 )b + (be - ab)abc ca 2 - b 2 c + ab 2 - c 2 a + bc 2 - a 2 b = (a - b) (b - c) (c - a). =
Solutions
66
Problem 27, S olution 2
Consider a and b to be fixed and replace c in (1) by a variable x : F (a, b, x ) = (a - b) ( 1 + bx) (1 + ax) + (b - x ) ( 1 + ax) ( 1 + ab) +(x - a) (1 + ab) (1 + bx) . (2) For a, b fixed, (2) is a quadratic polynomial in x. The coefficient of x 2 in (2) equals (a - b)ba - a(1 + ab) + (1 + ab)b; this simplifies to b - a =f:. 0, showing that the polynomial ( 2 ) is not equal identically to zero. Setting in (2) x = a and x = b we get value 0 (easy verification). A non-zero polynomial of degree 2 cannot have a third root. Since a, b, c are three distinct numbers, we infer that the value of (2) for x = c is different from zero; i.e., F (a, b, c) =f:. 0, as needed. Problem 27, S o lut ion 3
Let a = tan a , b = tan ,B, c tan -y with a , ,B, -y E (-7r/2 , 7r/2) . Since a, b, c are pairwise distinct, so are a , ,8 , 'Y · From the equality =
tan(a - ,8 ) =
tan a - tan ,B a-b = 1 + tan a tan ,B 1 + ab
we see that the expression defined in the problem statement is equal to (3)
tan(a - ,B) + tan(,B - -y ) + tan('Y - a ) .
In the identity tan u + tan v (1 tan u tan v) tan(u + v) set for u and v the differences ,B - 'Y and 'Y - a; the sum (3) is seen to be equal to =
-
tan(a - ,B) + (1 - tan(,B - -y ) tan( 'Y - a) ) tan(,B - a) , simplifying to
(4) tan(,B - -y ) tan('Y - a) tan( a - ,B) . The (distinct) numbers a, ,B, 'Y lie in ( /2, /2) ; so the numbers ,B - 'Y , 'Y - a, a - ,B lie there, too, and are different from zero. It follows that the factors of the product (4) are different from zero, and this is j ust what we need to conclude the proof. -1r
1r
Problem 2 8
Solve the system of equations: x + y + xy
=
19,
y + z + yz
=
1 1,
z + x + zx
=
14.
67
Algebra
28, So lut ion 1 From the first and the second equation, Problem
x(y + 1) = 19 - y,
z(y + 1) = 11 - y.
Evidently, y cannot be -1, so we may divide by y + 1:
19 - y , x = -y + 1
11 - y . z = -y+1
Substitution into the third equation of the system yields
19 - y + -11 - y + (19 - y)(11 - y) - 14 , -2 y+1
y+1
(y + 1)
which is equivalent to
14(y + 1) 2 - (30 - 2y) (y + 1) - (19 - y)(11 - y) = 0, simplifying to y 2 + 2y - 15 = 0. The solutions of this quadratic are y 1 = 3 and Y = -5; the corresponding values of x and z are computed 2
from the previous formulas. So there are exactly two solution triples (x, y, z): (4, 3, 2) and ( 6 5 4 ) Problem 28, Solution 2 The system is equivalent to -
(x + 1)(y + 1) = 20,
,
-
,
-
.
(y + 1)(z + 1) = 12,
(z + 1)(x + 1) = 15. The sums x + 1, y + 1, z + 1 must be different from 0. Dividing the first equation by the second we obtain (x + 1)/(z + 1) = 5/3, i.e., x + 1 = i (z + 1). Inserting this into the third equation, (z + 1) 2 = 9. Thus z = 2 or z = -4; accordingly, x = i (z + 1) - 1 equals 4 or -6, and y is computed from any one of the first two equations of the system. Outcomes as in Solution 1. 28, Solution 3 Recast the equation system into the form as in Solution 2. Multiply these equations to obtain Problem
so the product P = (x + 1)(y + 1) (z + 1) is equal to either 60 or 6 0 Now, p -1 p x = (x + 1) - 1 = -1=-
(y + 1) ( z + 1)
12
.
68
Solutions
and analogously
p - 1, p = 15 z=20 - 1. Setting P = 60 and P = -60 yields the two triples (x, y, z ) ( -6, - 5 , -4) . y
=
( 4, 3, 2) and
29 Solve the system of equations
P roblem
Xt(Xt - 1) X2(X2 - 1)
= =
X2 - 1 X3 - 1
xt, . . . , Xn .
in real numbers P roblem 29, S o lut ion 1 Let be a solution. Note that � 1 =? � Assume = + ::; 0 for a certain io. Then � and all sub sequent ( cyclically ) are � in particular, � a contradiction. Thus all must be positive. Now the system implies that all the differences are simultaneously positive, negative or zero. In the first two cases we multiply all the equations and cancel the non-zero ( which appears on both sides ) , with the result that product n = This however contradicts the fact that all are greater than or they are all smaller than The only possibility that remains is that for all i. Clearly, this is a solution. Problem 29, Solution 2 Adding all the equations leads to
Xio
1
(x1, . . . , x n )
XiS
XiS Xi - 1 TI(xi - 1) Xi 1.
Xi Xi+t 1. Xio+l 1 Xi0(xi0 - 1) 1, 1; Xio 1, XiS
1.
Xi = 1
n n x � - xi) = L (xi - 1); �) i=l i=l equivalently, n (x� - 2xi + 1) 0, L i=l i.e., n (x� - 1) 2 = 0. L i=l Thus xt = · · · = X n 1. =
=
69
Algebra
30 Solve the system of equations
P roblem
2xy = 1, 2 + y 2 + -x+y in real numbers x, y. X
30, S o lut ion 1 If x, y are a solution, then clearly x + y get from the first equation Problem
1
2xy = X 2 + y 2 + -x+y
>
>
0. Assuming x + y > 1, we
x 2 + y 2 + -2xy = (x + y) 2 x+y x+y x+y
= X
+ y > 1,
a contradiction. A similar contradiction is yielded by assuming that
x + y < 1 ( the inequalities have to be reversed ) . Thus x + y must be 1, and the second equation of the system becomes 1 = x 2 - (1 - x ) , with roots x = 1 and x = -2 . So the system has two solutions ( x, y); these are: (1, 0) and ( - 2 , 3). 30, S o lut ion 2 Multiply the first equation by (x + y ): Problem
(x 2 + y 2 ) (x + y) + 2xy = x + y. Adding x 2 + y 2 to both sides of this equation, we are driven by standard manipulations to a nice factorization:
(x 2 + y 2 ) (x + y) + (x + y) 2 = (x 2 + y 2 ) + (x + y); (x 2 + y 2 ) (x + y - 1) + (x + y)(x + y - 1) = 0; (x 2 + y 2 + x + y) (x + y - 1) = 0. The first factor cannot be zero because the sum x + y is positive ( this is obviously implied by the system ) . So x + y - 1 must be zero. Inserting y = 1 - x into the second equation of the system we find the two solutions (x, y) = (1, 0) and (x, y) = ( 2 , 3) -
.
31 Solve the system of equations
Problem
x 2 + Y2 + z 2 = 2 , in real numbers x, y, z.
x + y + z = 2 + xyz
70
Solutions
3 1 , S o lution 1 Two equations and three unknowns? This is a clear indication that there must be some inequality hidden behind the problem statement. Suppose x, y, z satisfy the system. Write Problem
= yz,
zx,
s = x + y + z, q = xyz. From the first equation of the system we get x 2 + (y - z) 2 = 2 - 2 u . u
v=
w
= xy,
The left-side expression is a non-negative number. Hence u � 1; analo gously, v, w � 1 , and therefore
( 1 - u )( 1 - v ) ( 1 - w ) � 0.
(1)
By the definition of s, u , v , w ,
s 2 = (x + y + z) 2 = 2 + 2(u + v + w ) , and so u + v + w = !s 2 - 1. Moreover, vw + wu + uv = sq and uvw = q 2 • Thus we can transform the left side of ( 1 ) as follows: ( 1 - u ) ( 1 - v ) ( 1 - w ) 1 - ( u + v + w ) + ( vw + wu + uv ) - uvw = 1 - ( !s 2 - 1 ) + sq - q 2 = 2 - !(s - q) 2 - h 2 . =
According to ( 1 ) , this is a non-negative number. Hence follows the in equality
(2)
Therefore Is - ql � 2, while the second equation of the system says that - q = 2. This means that equality must hold in (2). Now, the right inequality in (2) arose from q 2 � 0; so it turns into equal ity only if one of the numbers x, y, z equals 0. Let e.g. z = 0; then u = v = 0 and x + y = = + q = 2. The left inequality in (2) comes from condition ( 1 ) ; equality in ( 1 ) , combined with z = 0, implies w = 1 , i.e., xy = 1. The unique solution o f the system x + y = 2, xy = 1 is 8
8
8
x = y = l.
This yields the Solution 1 triple (x, y, z) = ( 1, 1, 0 ) ( the only one with z = 0 ) . By symmetry, ( 1, 0, 1 ) and ( 0, 1 , 1 ) are two other solution triples; and there are no others. 31, Solution 2 Readers familiar with multivariable calculus can regard this as a maxi mization problem: inspect the extrema of Problem
f(x, y, z) = x + y + z - xyz,
Algebra
71
given that x 2 + y 2 + z 2 2. The last equation describes a sphere, hence a compact set. So a maximum must be attained at some point(s) of this sphere. By the Lagrange Mul�iplier Theorem, any extremum point is a critical point of the Lagrange function =
g(x, y, z) = f(x, y, z) - A(x 2 + y 2 + z 2 - 2) ; i.e. , a point such that agjax = agjay = agjaz 0. Thus, at an ex =
tremum point:
1 - yz = 2Ax,
1 - zx = 2Ay,
1 - xy = 2Az.
(3)
Multiplying the first of these equations by x, the second by y and the third by z, we get
xyz = x - 2Ax 2 = y - 2Ay 2 = z - 2Az 2 • (4) For any fixed value of A, the function 1/; (t) = t - 2At 2 can take the value xyz at two distinct points, at most; so the system (4) forces that two of x, y, z must be equal. Let e.g. x = y. Then the system (3), accompanied
by the equation of the sphere, becomes
1 - x 2 = 2Az, 2 - 2x 2 = z 2 . (5) The second and the third equation of (5) result in z 2 = 4Az. So we have either z 0 (yielding x = y ± 1) or A = !z, which inserted into the first equation gives 3xz 2. This together with the last equation of (5) is solved in a routine way. Outcomes: x = y = !z = ± i v'3 and x = y = z = ± i v'B. Now we have the complete list of points (x, y, z) at which f(x, y, z) might 1 - xz = 2Ax, =
=
=
be a maximum (up to permutation of :variables) :
(
0) , with = ±1. Comparing the values of f at these points we find out that 2 is the maximum value of f on the sphere in question, attained only at (1, 1, 0), (1, 0, 1) and (0, 1, 1 ) . Hence, these three triples are the only solutions (x, y, z) of the system under consideration. €, €, €
Prob lem 32
Let n 2: 3 be a fixed integer and let a, b, c be fixed real numbers with a + b + c = 0. Find all n-tuples (xt , . . . , x n ) of real - numbers satisfying
the system of simultaneous inequalities
axi - l + bxi + cxi+ l 2: 0 for i = 1, . . . , n, where by definition xo X n 7 X n+l = x1. =
72
Solutions
Problem 32, Solut ion 1
Suppose (x1, . . . , x n ) is a solution. Let s = x 1 + · · · + X n · The left side expressions of all the given inequalities are non-negative numbers and their sum equals as + bs + cs = 0. Thus all those numbers are equal to zero and the inequalities of the system are in fact equations:
axi - l + bxi + cxi+ l = 0 for i = 1, . . . , n . If a = b = c = 0 , then every n-tuple (xt , . . . , x n ) i s a solution.
(1)
Rejecting this trivial case from further considerations, assume that
a 2 + b 2 + c 2 > 0. Then at least two of the numbers a, b, c must be different from zero. Hence a 2 + c 2 > 0. Evidently, every constant n-tuple x 1 = · · = X n satisfies the system (1) . ·
Let us look for non-constant solutions. Since c = - (a + b) , equation (1) rewrites as
(2) which is further transformed into successively equivalent forms (each equation is valid for all i):
a(Xi+ l - Xi + Xi - Xi- l ) + b(x·i+ l - Xi) = 0; (a + b) (xi + l - Xi) + a(xi - Xi - l ) = O · a(x i - Xi - d = c(xi + l - xi) · (3) Set Yi = Xi - Xi-l · Clearly, Yl + · · + Yn = 0. Since not all the x i s are equal, there exists a Yi a =I 0. Equation (3) says that (4) This has to hold for all i. Since a2 + c 2 > 0, equation ( 4) can be solved either for Yi +l or for Yi ( Yi+ l = (a/c)yi or Yi = (cja) Yi+ l ) · If any one of the· numbers (y 1 , . . . , Yn ) were zero, we could infer (inducting forward or backward) that all the YiS are zero, in contradiction to Yia =I 0. Conse quently the product p = Yl Yn is different from zero. Multiplying the n equations (4) we obtain a np = c np, and hence a n = e n . If the numbers a and c were equal, equations ( 4) would force that all the being zero and product YiS are equal; and this is impossible, their sum non-zero. Therefore a =I c. The equality a n = e n then implies that n is an even number and c = -a ( =I 0) . Hence b = - (a + c) = 0. We go '
·
·
·
·
back to the system (1) , which becomes simply
Xi-l - Xi + l = 0 for i = 1, . . . , n .
(5)
Algebra
73
This means that the even-indexed x i s must be equal and the odd-indexed x i s must be equal; if they are, system (5) is satisfied. So we can formulate the answer: If a = b = c 0, the XiS can be arbitrary real numbers. If b 0, a = -c =f:. 0 and is even, the general solution of the system (1) is ( x t . . . . , x n ) = (u , v, u , v, . . . , u , v) with u , v arbitrary real numbers. In all the other cases, the general solution of the system (1) is
=
=
n
,
( x t . . . . , X n ) = ( t, . . . , t), with t an arbitrary real number. Problem 32, Solut ion 2
Begin as in Solution 1 and reduce the given system of inequalities to the system of equations (1). Assume ab =f:. 0. Rewrite the ith equation of (1) in the form (2): Squaring yields
a 2 x i2_1 + 2 a bXi -1Xi + b 2 x i2 = a 2 x i2+ 1 + 2 abx i2+ 1 + b 2 x i+2 1 . This holds for i = 1, Adding these equations we obtain n n n (a 2 + b 2 ) I.> � + 2ab L X i-1Xi (a 2 + b 2 + 2ab) L: x � , . . . , n.
n
i= 1
=
i= 1
i= 1
whence ( in view of the assumption ab =f:. 0)
n n 2 L X i-1Xi 2 L: x � . =
i= 1
Rewrite this as
n 2 L Xi - 1Xi i= 1
i.e.,
n
=
L: i= 1
(6)
i= 1
x� +
n
x i:. 1 , L: i =1
n
L ( Xi - Xi - 1 ) 2 i 1
0; = the equality x 1 = X n follows. (Another argument consists in notic ing that (6) is the instance of equality in the Cauchy-Schwarz Inequality 1 :�:::U i vi $ (L:::U D 11 2 (L:::V l ) 1 2 , applied to Ui X i 1 , Vi = Xii and this also implies x 1 · · · = x n . ) =
=
· · ·
=
=
-
74
Solutions
Now assume be =f 0; then (2) should be rewritten in the form
bx i + cx i+ l = (b + c)xi -1 · Repeating the reasoning of the previous case, we again conclude that
Xl =
.
.
.
= X n•
So, we are done with the cases where ab =f 0 or be =f 0. It remains to consider ab = be = 0. Then necessarily b = 0. (Indeed: assuming b =f 0, we would obtain a = c = 0, contrary to the condition that a + b + c = 0 ) The equality b = 0 implies c -a. Substitution into equations (1) then yields (7) a(x i -1 - Xi+I ) = 0 for i 1, . . . , n. If a = 0 then we have a b c 0 and all the equations of the system are satisfied trivially, for every choice of x 1 , . . , X n · And if a =f 0 then equations (7) reduce to Xi -1 = Xi + l for all i (compare equations (5) of Solution l ) , implying that n is even and the x i s assume some two values alternately (these values may be distinct or equal) . Summing up, we have the general form of n-tuples (xi . . . , x n ) satisfying the system; it is presented in detail in the final section of Solution 1. Problem 33 Let a, b, c be the sides of a triangle. Show that
= =
.
.
=
=
=
.
.
c <2 a + -b + --b+c c+a a+b . 33, S o lution 1 By the triangle inequality a < b + c we have
Problem
a b+c Likewise,
=
2a 2a (b + c) + (b + c) < ---a + b + c---
2b b ---c+a < a+b+c'
c 2c ---a + b < ---a + b + c---
Adding the three inequalities we obtain the required one. Problem 33, S olut ion 2 Let a + b + c = u, be + ca + ab v . The proposed inequality is equiva lent to =
i.e. , to L < R where L
R
=
b + -c < 2, a + --u-a u-b u-c
a(u - b) (u - c) + b(u - c) (u - a) + c(u - a)(u - b), 2(u - a) (u - b) (u - c) .
Algebra
75
Simple manipulations bring these expressions to the following forms: L
(a + b + c)u 2 - [a(b + c) + b(c + a) + c(a + b)]u + 3abe u 3 - 2uv + 3abe, 2[u 3 - u 2 (a + b + c) + u(bc + ca + ab) - abc] = 2uv - 2abe,
=
R
and the problem reduces to showing that
(1) -u 3 + 4uv - 5abe > 0. Since a, b, c are the sides of a triangle, the sum u = a + b + e exceeds each of 2a, 2b, 2e. This yields R-
L
=
0 < (u - 2a) (u - 2b)(u - 2e) =
u 3 - u 2 (2a + 2b + 2c) + u(4be + 4ca + 4ab) - Babe u 3 - u 2 2u + u · 4v - Babe, ·
i.e. ,
-u 3 + 4uv - Babe > 0. The claimed inequality (1) is immediately implied by (2).
Problem
(2)
34
Let a, b, e, d be positive real numbers with abed = 1. Show that
a 2 + b 2 + e 2 + d 2 + ab + ae + ad + be + bd + cd ;::: 10. 34, Solution 1 In view of the well-known inequality cause of abed 1, Problem
x
+ (1/x) ;::: 2 for x > 0 and be
=
1 + ae + -1 + ad + 1 > 6 . ab + ed + ae + bd + ad + be = ab + ad ae ab -
This combined with the AM-GM Inequality
a 2 + b 2 + e 2 + d 2 ;::: 4
·
�a 2 b 2 c 2 d2
=4·
�=
immediately results in the asserted inequality. Problem 34, Solut ion 2 Since and since
a + b ;::: 2Vab,
e + d ;::: 2�,
4
76
Solutions
we have
a 2 + b 2 + c2 + d 2 + ab + ac + ad + be + bd + cd = a2 + b2 + c2 + d 2 + (a + b) ( c + d) + ab + cd ;:::: 2ab + 2cd + 2-.rah · 2� + ab + cd 3ab + 3cd + 4 3 ( ab + :b ) + 4 ;:::: 10 . =
=
34, S olut ion 3 We have the chain of inequalities: Problem
a 2 + b2 + c 2 + d 2 -- a + b + c + d v abed ---J 4 - 4 >
>
4r;-;
=
1
(1)
(the root mean square, the arithmetic mean, and the geometric mean). Denote the sum a + b + e + d by 8. Then the left inequality of ( 1 ) implies a 2 + b 2 + e2 + d 2 ;:::: 8 2 f 4, and the right one says that 8 ;:::: 4. Therefore
a 2 + b2 + e 2 + d 2 + ab + ae + ad + be + bd + ed a 2 + b2 + c 2 + d 2 + (a + b + e + d) 2 2 2 2 82 8 ;:::: 8 + 2 58 2 5 16 = - > --
=
·
8 -
10 .
8
34, S o lut ion 4 The means inequality(-ies) can be used in various ways here, of which the following seems to be the fastest shortcut:
Problem
> =
=
a 2 + b2 + e 2 + d2 + ab + ae + ad + be + bd + cd 1 1 0 . {Ya 2 · b 2 · e 2 · d 2 · ab · ae · ad · be · bd · ed l 1 0 · {Ya5b5e5d5 l 1 0 . {YiS = 1 0 .
Problem 35 Let a, b be non-negative real numbers with a 2 + b 2
ab < y'2 _ 1 a+b+2 and determine when equality holds.
=
4.
Show that
A lgebra
77
P roblem 3 5 , Solut ion 1
Applying the AM-GM Inequality to the pairs of numbers a, b and a2, b 2 we have (1) a + b ;::: 2v;;h and 4 = a2 + b2 ;::: 2ab. The second inequality yields
ab :5 2, i.e.,
v;;b :5 vf'i.
(2)
Instead of examining the ratio from the problem statement, we consider its inverse, applying the first inequality of (1) and both inequalities of
(2):
(-1- _!_) (
)
1.
a + b + 2 2 -rab + 2 ;::: 2 2.. + � = vf2 + =2 + ;::: ab ab -lab ab vf2 2 (This requires ab =f. 0; clearly, if a = 0 or b = 0, the proposed inequality holds.) Inverting, we obtain 1 ab _ = vf'i - 1 . a + b + 2 <- _ v'2 + 1 The means inequality turns into an equality only when the averaged numbers are equal. Thus a = b = vf2 is the condition for equality. Problem 3 5 , Solution 2
If, for some reason, one prefers not to invert, one can start from inequal ities (1) and (2), and continue like this:
ab < ab ---a + b + 2 - 2 -rab + 2
y
2 = -2y + 2
'
where y stands for -lab; according to (2) , 0 :5 y :5 vf'i. It now suffices to show y2 (3) < - vl'i - 1 or, which is the same,
-2y + 2
y2 - 2 (v'2 - 1)y - 2 ( v'2 - 1) :5 0.
(4)
The roots of this quadratic trinomiai are y 1 = vf2 and Y 2 = vf2 - 2, and hence inequality (4) reduces to This holds because 0 :5 y :5 vf'i, so the second factor is positive, while the first one is negative or zero. Equality occurs for y = vf2 only; i.e., when (1) and (2) become equalities; and this is the case only for a = b = v'2.
78
Solutions
(Another option might be to examine the left-side expression of either (3) or (4) by calculus over y E [O , J2 ] .) Problem 35, S o lut ion 3 Starting with the second inequality of (1) and the resulting estimate (2) ( ...;;;b � v'2 ) , we transform the given expression as follows:
ab a+b+2
.;;;b .
( ..rai(
..rab
=
= (noting that for a = 0 or b GM Inequality implies
=
a +...; b +;;I; 2 2 -1 � + -b- + ...;;;b ...;;;I; ...;;;I; � + fF_ + 2 - 1
V "b
v�
...;;;I;
_
(5)
0 the claim holds trivially) . Now, the AM
/a + {F_ = a + b > 2
V "b
v�
) )
...;;;I; - ,
and so � � v'2.
(6)
Transformation (5) hence leads to the estimate
1 +1
v'2
=
V2 -
1,
just as required, equality holding (in (6) , hence in the claimed inequality) if and only if a = b = h. P roblem 35, S o lut ion 4 The case where a = 0 or b 0 is trivial. So we may assume ab > 0 and invert the proposed inequality: =
1 · a + b + 2 > -ab - J2 - 1 - '
---
equivalently,
2 -1 + -1 + a b ab -> h + l. The given condition a 2 + b 2 = 4 calls for setting a = 2 sin x, b 2 cos x, x E (0, 7r/2). =
The inequality ( 7) we are about to prove becomes
1 + -1 + 2 > v'2 + 1·' -2 sin x 2 cos x 4 sin x cos x -
(7)
Algebra
79
1 1 2 + -- + -- - 2v'2 + 2. sin x cos x sin 2x
equivalently,
>
(8)
cos x sin x 4 cos 2x + -- -cos sin2 x 2 x sin2 2x
Denote the left side by f (x) and examine the derivative: f'(x)
-
-=---
This is positive when sin x > cos x and negative when sin x < cos x; con sequently, f (x) decreases in (0, 11) 4] and increases in [1r I 4, 1r 12), attaining the minimum value
()
7r 1 1 2 + + f - = 4 sin(7rl4) cos(7rl4) sin(7rl2)
= vIn 2 + vIn2 +
2.
Inequality (8) is proved, and the condition for equality is x = 1r I 4, which corresponds to a b v'2. =
Problem
=
36
The real numbers a i , b i , q , di are such that 0 � c i � a i a i + bi ci + d i for i = 1, 2, . . . , n. Prove the inequality
�
bi
�
di and
=
n
n
n
n
i=l
i=l
i=l
i=l
II a i + II bi � II Ci + II di.
Problem
36,
S olution 1
We proceed by induction. For n = 1 we have a1 + b 1 = c1 + d b accord ing to assumption. Assume the claim holds for a certain n . Consider n + 1. Suppose a i , b i , c i , d i (for i = 1, . . . , n + 1) are numbers with 0 � Ci � a i � bi � di and ai + bi = ci + di. Set n
n
By the inductive hypothesis, A + B
�
C + D ; rewrite this as
A - C � D - B.
(1)
The following inequality is the content of the inductive claim: (2)
Solution s
80
In view of the conditions imposed on the numbers a i , b i , c i , di , we have
a n + l - Cn + l a n +l C
< <
d n + l - bn + l , b n+ l • D.
(3) (4)
(5)
Note that the differences that occur in inequalities ( 1) and (3) are non negative numbers. Multiplying (1) by (4) and (3) by (5) we obtain the inequalities
(A C ) a n+ l a ( n + l - Cn + l ) C -
< <
(D B ) bn+ b ( dn+ l - b n + l ) D. -
Adding them, we get
and this is exactly the inductive claim (2) . The assertion results by induction. Problem
36,
Solut ion 2
By the given conditions,
ri ai - Ci = di - bi ;:::: 0 for i = 1, . . . , n . =
Look at the product
n
II di
i=l
=
(bl
+
rt ) ( b 2 + r 2 ) · · · ( b n + rn ) ·
Multiplying out, we obtain the term b 1 b 2 · · · bn plus several summands of the form (0 :5 k :5
n-
1),
(6)
with distinct indices i 1 . . . . , i k from the set { 1 , . , n }, complemented by . . , in- k to the whole { 1 , . . , n } . Similarly, the product
h,
.
.
.
.
n
II Ci = ( a l - r 1 ) ( a 2 - r 2 ) · · · ( a n - rn ) i=l is equal to a 1 a 2 · · · a n plus the sum of terms a •1· · · a '"· ( -r31 ) · ( -rJn-/o ) (0 :5 k :5 n 1). ·
·
· ·
·
-
(7)
Algebra
81
Since 0 :::; a i :::; bi for i = 1, . . . , n, we see that each term (7) is dominated, in absolute value, by the corresponding term (6) . So the joint sum of all the numbers (6) and (7) is non-negative. Now, the sum of all numbers (6) equals II d i - II bi ; the sum of all numbers (7) equals II Ci II a i. Consequently, -
n
n
n
n
=1
=1
=1
=1
di - II bi + II Ci - II ai � 0, II i i i i as claimed. Problem 37 Prove the following inequality for all integers n > 1:
( 1 + (n + 1) n+1 ) n-1 > ( 1 + nn ) n · n+2
n+1
37, Solution 1 We show that the number n n(n-1 ) can be ( smartly enough ) put in be tween the two expressions we are about to compare: P roblem
( 1 + (n + 1) n+1 ) n- 1 > n n (n- 1 ) > ( 1 + nn ) n
(1) n+2 n+1 Taking roots of order n - 1 and n, respectively, we recast the left in equality and the right inequality of (1) into their equivalent forms: (n + 1) n+l + 1 > n n and n n - 1 > n n + 1 . (2) n+2 n+1 ___
---
The second inequality is immediate:
(n + 1)n n- 1 = n n + n n - 1 > n n + 1 for n > l. For the first one, apply the Binomial Theorem to obtain ( for n > 1)
(n n+ l + (n + 1)n n + · + 1) + 1 > n n+1 + (n + 1)n n > n n+l + 2n n n n (n + 2). Thus, both inequalities of (2), hence of (1), are proved. (n + 1) n+1 + 1
=
=
·
·
82
Solutions
2 Denote the left-side expression and the right-side expression by L and R , respectively. The sequence tends increasingly to There fore we have for � Problem 37, S o lut ion
((n + 1)/n) n
n 2
e.
( n + 1 ) n+ l � 4 n+2 (n n + 1)/nn 1 + n -n ::; 5/4 for n 2, and so n n + 1 - -45 n n for n 2.
Moreover,
9
>
and
e
>
�.
3
�
=
<
> _
We now estimate the ratio L j R from below:
((n + 1)n+1 + 1 ) n-1 ( nn + 1 ) -n R n+1 n+2 ( (n + 1 ) n+l ) n- 1 ( nn + 1 ) -n ; n+2 n+1 and we continue the estimate, using inequalities (4) and (3): en + 1)n+1 ) n- 1 ( � . � ) -n R n+2 4. n + 1 1 (n + 1)n2 +n - n n (� ) n (n + 2) n 1 ( n : 1 ) n2 (� ) n ( : :�) n� 1 = (� ) n (( n : 1 ) n ) n ( ::� n+ 1 nn ++ 21 2 -45 n -94 n 31 nn ++ 21 2 , 9 n 1 -5 . 3 ; and this number exceeds 1 for n 2. Thus R , as asserted. L
= >
L
>
2
=
=
>
>
(}
'
9
(3)
'
'
,
(}
�
L
>
Problem 3 7 , Solut ion 3
Taking logarithms on both sides, rewrite the claimed inequality as
(n - 1 ) ln 1 + (nn ++ 21)n+ l
>
n ln
1 + nn n+1
(4)
Algebra
83
or, which is the same,
(
� ln 1 + (n + 1)n+ 1 n n+2
)
>
So the problem reduces to showing that f (n + 1) f (x)
=
(
_1_ ln x"' + 1 x-1 x+1
)
=
)
(
_1_ ln 1 + n n n-1 n+1 · >
f (n) for n
�
2, where
ln(x"' + 1) - ln(x + 1) . x-1
This is done in a more or less routine way by calculus. Since (x"') ' = (e"' 1" "' ) 1 = e"' 1" "' (ln x + 1) = x"' (ln x + 1), we get f'(x)
1-) ( x"' (lnx x+ +1 1) - x - 1) - (ln(x"' + 1) - ln(x + 1) ) x+1 C "'
(x - 1) 2 _1_ x"' (ln x + 1) _ _1 _ _ ln(x"' + 1) - ln(x + 1) . (x - 1) 2 . X+1 X-1 Xx + 1
(
)
Since (x"' + 1) / (x + 1) < xx -1 for x played fraction fulfills the estimate
>
1, the numerator of the last dis
x"' + 1 ln(x"' + 1) - ln(x + 1) = In -- < ln x x -1 = (x - 1) ln x. x+1 Therefore -1 ln x x"' (ln x + 1) -1 - -!'(x) > -X+1 X-1 X "' + 1 X-1 (xx+ 1 - 1) - (x + 1) ln x (x - 1) (x "' + 1 ) (x + 1) ·
(
)
=
In view of the well-known inequality In x < x - 1 we obtain (xx+ l - 1) - (x + 1) ln x > (x"' + 1 - 1) - (x + 1)(x - 1) = x2 (x"' -1 - 1) > 0 for x > 1, and consequently !' ( x) > 0 for x > 1. This mel:j.ns that f ( x) is strictly increasing in (1, ) and hence f(n + 1) > f (n) for n � 2. Problem 38 Let n � 9 be an integer. Which one of the numbers ( v'n ) vn+ 1 and ( Vn+1 ) ,;n is greater? oo
,
84
Solutions
Problem 38, S olut ion 1
The answer is easy to guess, the first number is greater:
(1) To prove this guess, take logarithms on both sides: inequality (1) is equivalent to v'n + 1 · ln yn > yn ln v'n + 1, i.e., to l n ..;n l n v'n+l 9 � 10r n � . > r= n v n +1 y This holds because the function f (x) = (ln x)jx is strictly decreasing for x � v'9 = 3 (the derivative f'(x) = (1 - ln x)jx2 is negative for all x > e; and since. 3 > e, we are done) . Problem 3 8 , Solution 2 We will use the well-known relation ( 1 + �r < e, which holds for all positive integers n. Inequality (1) is equivalent (via squaring) to ·
£
Division by n ..;n transforms this into
( 1 + -n1 ) n+../n(n+l)
and raising both sides to the power � + ..;n brings this to the form
Qn
n
>
for
(2)
Qn ::; ( 1 .jii- ) n < 2.06n.
the last inequality is also equivalent to (1). The exponent = n + Jn (n + 1) is estimated as follows:
Hence
�
9
n (n + 1) ::; 19 n 2
( 1 + ;;1 ) qn < ( ( 1 + ;;1 ) n) 2 .0
n
proving (2) .
=}
°
=}
6
+
<
e2· 06 < 7.846 < 8 <
9
::;
n,
Algebra
85
Remark
J1i R
The use of a calculator can be avoided if we resort to another well-known inequality (1 + �f + 1 > e, holding for all n � 1 ; in particular, we have (18/ 1 7) 1 8 > e. Since < 1 + l8 , we get q n :$ ( 2 + l8 n for n � 9. Knowing just that e < 2.8, we obtain e 2 < 7.84, and so
)
=
( 1 n 2+1/1 8 1 + ( ) ( ;;: ) ) 1 1 + ;;:
q
n
<
<
e2 + 1/1 8
=
e 2 e l/ 1 8 < 7.84 .
18 < 9 :$ n . 17
39 Prove the inequality
Prob lem
c:) ·
ffn < 4n for
n = 1, 2, 3, . . . .
39, Solut ion 1 The following stronger inequality will be proved by induction:
Problem
(1) Obviously, the proposed inequality is immediately implied by (1). For n 1 , inequality (1) holds. Assume (1) holds for a certain integer n � 1 ; we must show that =
(2) This is done as follows:
(2nn ++ 12 ) . v'3n + 4
=
= <
(2nn ) . (2n(n + 1)1)(n(2n++1)2) , v'3n + 4 + 2(2n + 1) . (2n ) . v'3n + 1 . J 3n + 4 3n + 1 n n+1 2(2n + 1) . 4n . J 3n + 4 . 3n + 1
n+1
The claim (2) will be proved if we show that
2n + 1 n+1
. J 3n3n ++ 41 - 2_ <
(3)
86
Solutions
By squaring, relation (3) is equivalent to 1 )-'- 2 -'-( 3 n + 4--'-) < . -'--(2_n_+---4 ___ ( n + 1) 2,---( 3n + 1) - '
(4)
and this is recast into 12n3 + 28n2 + 19n + 4 ::; 12n3 + 28n2 + 20n + 4, holding trivially. Induction is complete. Remark
This is a very graceful example of an induction proof which requires a strengthening of the claim in order to carry out the induction step. ( In the inductive procedure, there is a moment in which "assertion becomes assumption" . ) An attempt to prove the inequality in its original form by induction, in a straightforward way, would fail! Problem 39, S o lut ion 2
We transform the asserted inequality into successively equivalent forms: (2n) ! 5n 1 · 2 3 · · (2n - 1) (2n) 5n 1 · 3 · 5 · · · (2n - 1) · 5n (1 · 3 · 5 · · · (2n - 1)) 2 (3n) ·
and finally
·
·
·
::; (2 n n!)2 , < (2 · 4 · 6 · · · (2n)) 2 , ::; 2 · 4 · 6 · · · (2n), < (2 · 4 · 6 · · · (2n)) 2 ,
( 23 ·· 45 ·· ·· ·· (2n(2n -- 2)1) ) 2 . 2n 1
Denote the left side of (5) by a n ; thus
2
::; 3
·
(5)
2n - 1 . --2n - 1 a n = 21 . 23 . -43 . -45 . -65 . -67 . . . --2n - 2 2n . To create a n+l out of a n , two further factors have to be attached at -
-
the end of : this product. Appending only one of them we obtain an expression, which we denote here by b n : 2n - 1 2n - 1 2n + 1 b n = -21 . -23 . -43 . -45 . 65 . 67 . . . --2n - 2 . --2n . --2n . -
Accordingly,
-
2n + 1 > b n = a n · --an . 2n
(6)
Algebra
87
On the other hand,
- - ---
-
4n 2 1 < bn - 1 · 4n 2 Thus b 1 , b 2 , ba , . . is a decreasing sequence. And since 1 3 3 5 5 7 7 9 9 11 2 bs 2 · 2 · 4 · 4 · 6 · 6 · 8 · 8 · 10 · 10 = 0 · 66618 · · · < 3 ' all the subsequent bn s are smaller than 2/3. Hence by inequality ( 6 ) , a n < 2 / 3 for n 5 , 6 , 7, . . . . Also the initial terms a1, a 2 , aa, a 4 are smaller than 2/3 , as can be verified directly. Thus estimate (5) is proved, and we are done. b
n
b
=
2n 1 2n + 1 n 1 · 2n 2n
.
---
·
= bn - 1
·
=
=
Remark
At calculus courses it is taught that the sequences ( a n ) and ( bn ) tend to a common limit, whose exact value is 2/7r ( the Wallis formula). P roblem 40 Prove that the inequality
holds for any real numbers at, a 2 , . . . , a r . Find conditions for equality. Problem 40, S olution 1 Denote the given expression by Fr (a l , . . . , a r ):
.
Fr (a1, . . . , a r ) =
t t ( :m:: ) · n=l m=l
(1)
Clearly, Fr ( O, . . , 0) 0. Now, for r 1 we have F1(a1) = a V 2 ;::: 0, with equality only for a 1 = 0. It is thus natural to conjecture, for each r, that the inequality (2) should hold for every r-tuple of real numbers ( a1, . . . , a r ) =/= (0, . . . , 0) . We will prove this guess by induction. The start ( r 1) has been done already. Fix r ;::: 2 and assume inductively =
=
=
Fr - 1(a1 , . . . , a r -d > 0 whenever (a1, . . . , a r-d =/= (0, . . . , 0) . (3) Consider r real numbers a1, . . . , a r -1, a r , not all zero. In the expression ( 1 ) , isolate the terms corresponding to n = r or m
= I: ( I: amann n=l m=l
m +
=
r:
+
aran
r+
n
)
88
Solutions
+
� mam+arr + a2r� . (m=1 )
(4)
If a 1 = = a r 1 = 0, then automatically a r =f 0, and therefore we have Fr (O, . . . , 0, a r ) = a�/ (2r) > 0, as needed. Thus assume that at least one of the numbers a1 , . . . , a r-1 is different · · ·
-
from zero. Consider the expression on the right side of (4) as a function of the variable ar, which we now denote by x: T (x) = Fr (at . . . . , U r- b x) = A x 2 + Bx + C ; according to (4) , the coefficients A , B , C are expressed by the formulas 1 A= -,
B-
2r
r- 1
r-1
r-1
ak a m 2 -U n + L -L L -r+k ' m +r r +n
n=1
m=1 k =1 UmUn · C= m n=1 m=1 + n Let us calculate the discriminant D = B 2 - 4A C of the quadratic trino mial T (x) :
�(�
)
hence
i.e., (5)
where
C mn
1
1
(m + r) (r + n ) 2r(m + n ) 2r(m + n) - (m + r)(r + n) (m + r)(r + n ) 2r(m + n ) (r - m) (r - n ) = 2r(m + r) (r + n ) (m + n ) ·
·
A lgebra
Writing
89
r k b k = -for k = 1 , . . . , r - 1 r+k -
we thus have
1 bmb n 2r m + n Inserting these expressions into formula (5) we obtain
C mn =
D = 4
-
-
·
---
_ ]_2r I: ( � bmmbn+amnan )
=
n=1 m=1 Fr-1 (a 1 b 1 , a 2 b 2 , . . . , a r-1 b r-1 )
=
Fr-1 (u l t u 2 , . . . , U r-1 )
2r
(6) 2r where U k = a k b k for k = 1 , . . . , r - 1 . Since the b k s are positive, there is a non-zero number among the numbers u 1 . . . . , U r-1 · Thus, by the inductive hypothesis (3) ( applied to the numbers u 1 , . . . , U r - 1 in place of a 1 . . . . , a r - 1 ), we have Fr - 1 (u l t u 2 , . . . , u r - 1 ) > 0. This, in view of the equality ( 6 ) , implies D < 0. A quadratic trinomial with a negative discriminant and a positive leading coefficient (A = 1 / (2r) > 0) assumes positive values only. We have thus shown that the inequality in (2) holds for every real num bers a 1 , . . . , a r - 1 , a r , not all equal to zero. This concludes the inductive step. Problem 4 0 , S olution 2
Consider the polynomial P(x) =
(P (x)) 2
=
=
=
r
L anxn .
n=l
(� an xn) (t, amX m) � an (t, amxm) x n � l�l am anxm+n) .
Now introduce the polynomial
Q(x) =
By squaring,
a m a n x m+n ) . t t ( m n=l m=l +n
(7)
Solutions
90
The assertion of the problem is that Q (1)
2:
0.
(8)
Claim (8) will be proved by examining the derivative of Q (x ), which equals
Comparing equation (7) , we see that
Q'(x) = .!.X (P( x)) 2 2: 0 for x > 0. (9) Notice that Q(O) = 0, by the definition of Q(x). The function Q (x) is continuoHs in [0, 1] and non-decreasing, in view of the inequality in (9) . Hence Q(1) 2: Q(O) = 0; claim (8) is settled. Equality holds in (8) if and only if Q(x) is constant, i.e. (see (9)), when P ( x) is the constant null function. And this is the case if and only if = ar = 0. a1 = ·
·
·
41 For a fixed integer n
P r oblem
2:
1 find the least value of the sum
given that x1 , . . . , X n are positive numbers satisfying 1 1 1 ++···+= n.
Xl
X2
Xn
41, S olution 1 Denote the sum under consideration by S ( = S(x1 . . . . , x n )). The follow ing inequality is the key to the solution: Problem
x k + 1 2: 1 + k1 for x > 0, -;;
k
k=
1 , 2, 3, . . . .
(1)
Four proofs of (1) are presented below! Now, taking inequality (1) for granted, we just match each term of S with the corresponding term of the constraint condition, to obtain
A lgebra
91
with equality for x1 = · · · = X n = 1. The "n" cancels and we get
1
1
1
1+-+-+···+ 2 3 n as the minimum value of S. It remains to prove inequality (1). First proof of (1). For x, k > 0 fixed (x real, k an integer) ,
(
k kx xk + -X1 - 1 - -k1 -
)
(2)
= x (x k - 1) + k(1 - x) =
k-1
( (�k xi ) - k)
x(x - 1) L x i + k(1 - x) i=O
= (x - 1)
k = (x - 1) L (xi - 1)
j=l
k
= L (x - 1)(x i - 1)
j=l
:2:
0
because the factors in each term of the last sum agree in sign. Second proof of {1). For fixed x, k consider the arithmetic mean and the geometric mean of the k + 1 numbers, one of them being xk and the others equal to 1/ x: 1/(k+ l ) = 1; _1_ xk + .!_ + . . . + :2: xk . .!_
k+1(
.!. ) (
X X �
k
. •
•
.!. )
X X --.,....._...,
k
hence xk + (kjx) ;:::: k + 1, which is just a restatement of {1). Third proof of {1). The Bernoulli Inequality {1 + a) k . ;:::: 1 + ka holds for all a :2: - 1 and every integer k ;:::: 1. Set a = x - 1, thus obtaining: xk ;:::: 1 + k (x - 1). Hence
(
)
x k 1 1 + k (x - 1) + -1 = x + -1 - 1 + -1 > 1 + -1 . - + -X > X X k
k
k
k
92
Solutions
Fourth proof of (1) . For a fixed integer k � 1 consider the left side of ( 1) as a function f ( x ) , defined on positive reals. Since
!' ( x ) x k - 1 =
_
x
-2
{
< >
0 for x E (0, 1) , 0 for x E (1 , oo ) ,
we see that f (x) takes at x = 1 its (global) minimum value 1 + (1/k). Proble m 4 1 , S o lution 2 The constraint imposed on the numbers x1 , . . . , X n is that their harmonic mean is equal to 1. So their geometric mean is at least 1; i.e. , we have (3) Take the sum (2) to the common denominator
1 1 m1 + m 2 + · · + m n m 1+-+..·+-= =, (4) , 2 n n. n! where m k . = (n!)/k for k = 1, 2, . . . , n, and consider the m positive num bers: . . . , x: , . . . ' x: . ·
------
Their arithmetic mean compares with the geometric mean: (5)
Since km k = n! for k = 1, . . . the product in the parentheses equals ( x 1x 2 · · x n ) n ! , and we get by inequalities (5) and (3) , n,
·
Division by n ! now yields (see (4)) Xl X� X� m 1 1 > - = 1 + - + . .· + - , -+-+..·+- (6) 1 2 n n! 2 n showing that the sum (2) is the least value that the expression in question can have.
Remark
The argument of the last solution shows that the inequality in ( 6 ) is a consequence of the condition (3) , actually weaker than that given in the problem statement (the geometric mean instead of the harmonic mean) .
Algebra
93
P r oblem 42
On a given segment AD, find points B and C so as to maximize the product of the lengths of the six segments AB, AC, AD, BC, BD, CD. Problem 4 2 , Solut ion
1
Without loss of generality, assume that B lies between A and C. Take the length of AD for a unit (AD = 1) and set
X = BC, y = AB - CD;
x E (0, 1 ] , y E (- 1, 1].
Then AB + CD = 1 - x, and hence
AB = ! (1 - x + y) , AC = ! (l + x + y) ,
CD = !(1 - x - y), BD = !(1 + x - y) .
The product under consideration equals
p
p(x, y) l6 x(1 - X + y) (1 - X - y)(1 + X + y) (1 + X - y) = l� x ((1 - x) 2 - y 2 ) ( (1 + x) 2 - y 2 ) .
Hence,
(1)
where
f(x) = x ( 1 - x 2 ) 2 = x 5 - 2x 3 + x; equality holds i n (1) i f and only i f y = 0 .
(2)
The derivative is positive for x E (0, ! .J5 ) and negative for x E ( ! .J5, 1) . Thus the maximum value of f(x) is attained at x = ! .J5. Consequently, the prod uct p( x, y) is maximized at x = ! .J5, y = 0. This corresponds to placing points B and C symmetrically with respect to the midpoint of AD, at the mutual distance BC = ! .J5. P roblem 42, Solution 2
above, the problem is reduced to the maximization of the polyno mial (2). This can be done without calculus. In the weighted AM-GM Inequality As
apbq :5 pa + qb for a, b � 0, p, q � 0, p + q = 1,
94
Solutions
set a = 4x 2 , b = 1 - x 2 , p = 1/5, q = 4/5, to obtain i.e. ,
4 1 / 5 f (x) 2/ 5 � t ' equality holding only when 4x 2 = 1 - x 2 ; that means, for x = i v'5 . Con
clusion as before.
Problem 42, S o lut ion 3
The means inequality can be used in a yet smarter manner. Assuming that B lies between A and C, set
AB = u, so
AC = u + x,
BC = x,
BD = X + v,
CD = v ; AD = u + X + v = 1 .
Denoting the product under investigation again by p , we have
(uxv (u + x) (x + v) ) 2 (u(x + v)) (u(u + x)) (v(u + x)) (v(x + v)) (x 2 ) .
(3)
The geometric mean of the five numbers
u(x + v) , u(u + x), v(u + x) , v(x + v), x 2
(4)
does not exceed their arithmetic mean. Thus the product on the right side of (3) does not exceed
( u(x + v) + u(u + x) + v5(u + x) + v(x + v) + x 2 ) 5
The numerator of the fraction in parentheses equals
ux + uv + u 2 + ux + vu + vx + vx + v 2 + x 2 = ( u + v + x ) 2 = 1 . Expression ( 3 ) for p now yields p � ( i ) 5 1 2 .
The means inequality becomes an equality only if the averaged quantities are equal. Since the arithmetic mean of the numbers listed in (4) is 1/5, the condition for equality takes the form
u(x + v) = u(u + x) = v(u + x) = v(x + v) = x 2 = � . The last system of equations ( together with u + v + x = 1) is fulfilled only for x = k v'5 and u = v = ! ( 1 - x). Conclusion as in Solution 1 .
95
Algebra
43 Find all functions f: JR. � JR. satisfying the equation
Problem
x 2 j (x) + f (1 - x) = 2x - x 4 for x E R 43, Solution In the given equation
Problem
1
x 2 f(x) + /(1 - x) = 2x - x 4 set 1 - x in place of x: (1 - x) 2 !(1 - x) + f (x) = 2(1 - x) - (1 - x) 4 . Multiply equation (1) by (1 - x) 2 : x 2 ( 1 - x) 2 f(x) + (1 - x) 2 /(1 - x) = (1 - x) 2 (2x - x 4 ).
(1) (2) (3)
Subtract equation (3) from (2): Rewrite this as
K (x)f(x) = (1 - x)M (x),
(4)
K ( x) and M ( x) denoting the polynomials in square brackets: 1 - x 2 (1 - x) 2 K (x) (1 - x + x 2 )(1 + x - x 2 ), M (x) 2 - (1 - x) ( (1 - x) 2 + 2x - x 4 ) = 2 - (1 - x) (1 + x 2 - x 4 ) 1 + x - x2 + x3 + x4 - x 5 ( 1 + x 3 )(1 + x - x 2 ) = = ( 1 + x)(1 - x + x 2 ) ( 1 + x - x 2 ) = (1 + x)K(x) . Equation (4) takes the form
K(x)f(x) = (1 - x 2 )K(x), implying f(x) = 1 - x 2 , unless K (x) = 0 . In this latter case . we get x 2 (1 - x) 2 = 1 (by the definition of K (x)), and this is equivalent to say ing that
96
Solutions
The first equation of roots:
(5)
has no real roots and the second one has two
a = ! (1 + ¥'5 ) and f3 ! (1 - ¥'5 ) ; (6) these roots of x 2 x + 1 satisfy a + /3 = 1, a /3 = - 1; (7) a 2 = a + 1, {3 2 = f3 + 1. In conclusion, f (x) = 1 - x 2 for all x =f a, {3. And for these two ex ceptional values of x equation (1) yields a 2 f (a) + / (/3 ) = 2 a - a 4 and {3 2 f(/3 ) + f(a) = 2{3 - {3 4 . {8) Using formulas (7) we calculate: a 4 (a 2 ) 2 (a + 1) 2 = a 2 + 2a + 1, and likewise {34 = {3 2 + 2{3 + 1. Equations (8) become a 2 f (a) + f ( /3 ) = -a 2 - 1 and {3 2 f(f3) + / (a) = -{3 2 - 1. (9) Multiplying the second equation of {9) by a 2 we obtain, by identities {7), f(/3 ) + a 2 f(a) -a 2 {3 2 - a 2 = - 1 - a 2 , =
=
=
=
=
which is j ust the first equation of (9) . Thus the two equations {9) do not yield a unique evaluation of f(a), f(f3) . In fact, f{a) can be any real number c, and then
f(f3) = - (a 2 + 1) - a 2 c = - (a + 2) - (a + 1)c. Thus the general solution of equation (1) is for x =f ! (1 ± ¥'5 ) , 1 - x2 f(x) c (arbitrary real number) for x = ! { 1 + ¥'5 ) , - ! {5 + ¥'5 ) - ! (3 + v'5 )c for x = ! (1 - ¥'5 ) .
{
=
(1 0 )
Problem 43 , Solution 2
A functional equation like {1), with polynomial coefficients and a polyno mial on its right side, is likely to have a polynomial solution. Evidently, if a polynomial f (x) satisfies equation (1) , it must be a polynomial of second degree. Postulating f(x) Ax 2 + Bx + C we get from (1) =
Ax 4 + Bx 3 + Cx 2 + A(1 - 2x + x 2 ) + B (1 x) + C = 2x - x 4 , whence A = - 1, B = 0 , C = 1 ; i.e. , f(x) = 1 - x 2 • Thus we have found the solution f(x) = 1 - x 2 by simple trial. It -
is unique in the class of polynomials; one is tempted to conjecture that it is unique in general. In an attempt to prove this guess, let us set
f(x) = 1 - x 2 + g(x) ;
{11)
A lgebra
97
{1)
that takes equation
equivalent to Replacing
to the form
x 2g(x) + g(1 - x) = 0.
{12)
(1 - x) 2 g(1 - x) + g(x) = 0.
{13)
x by 1 - x,
Viewing {12) and (13) as a homogenous system of two linear equations with unknowns g(x) and g(1 - x), compute its determinant:
I
1 x2 2 2 1 (1 - x) 2 = x (1 - x ) - 1.
I
{14)
If the determinant is different from zero, the system has only the triv ial solution g(x) = g(1 - x) = 0. If the determinant is zero, then the two equations {12) and {13) are linearly dependent ; g(x) may be gi¥en any value, and then g(1 - x) can be computed from any one of these equations. (We see that the uniqueness conjecture fails.) Now, the determinant {14) vanishes if and only if x satisfies one of the equations {5) from Solution 1; equivalently, if x is one of the numbers a, (3 given by {6). Equation {12) with x = a and with x = (3 becomes, respectively,
Since (a f3) 2 = 1, the two equations (just obtained) coincide. Denote g(a) by d; then g( f3) = -a 2 d. Hence by definition {11) (and formulas {7))
f(a) = 1 - a 2 + d = d - a, f(f3) = 1 - (3 2 - a 2 d = -(3 - (a + 1)d. Setting, further, d - a = c, we get f(a) = c and f(f3) = -f3 - {a + 1) {a + c) = -(3 - a 2 - a - (a + 1)c = -{a + 2) - (a + 1)c. So we have arrived at the same formula Solution 1.
(1 0 )
which we had found in
Problem 44
Let A and B be real numbers different from zero. Prove that the function f(x) = A sin x + B sin ( ../2 · x) is not periodic.
98
Solutions
Problem 44, S olut ion
1
Assume f is periodic with period T
>
0,
f (x + T) = f(x) = f (x - T) for all x E R Then of course f (x + T) - f (x - T)
=
0 for x E lR,
which in view of the definition of f (x) leads to the equality A sin(x + T) - A sin(x - T) + +B sin (v'2 · x + v'2 T) - B sin (v'2 · x - v'2 · T) 0, ·
2A cos x sin T + 2B cos ( v'2 · x) sin (v'2 · T) = O for x E R
(1)
Setting x = 0 we obtain (2)
2A sin T + 2B sin (v'2 · T) = 0 ; and setting in (1) x = 1rj2 we get 2B cos (1r ; v'2 ) sin ( v'2 · T) = 0.
(3)
Since B =I= 0, equality (3) shows that sin ( v'2 T) = 0. And since also A =I= 0, equality (2) now shows that sin T = 0. It follows that each one of the two numbers T and v'2 T has to be an integer multiple of 1r. This is however impossible, v'2 being irrational. Contradiction ends the proof. Problem 44, Solution 2 The function f (x) has derivatives of order one and two: ·
·
f'(x) = A cos x + B v'2cos (v'2 · x) ,
f"(x) = -A sin x - 2B sin (v'2 · x) .
We thus have the equations f (x) + f"(x) = -B sin (v'2 x) , ·
2f(x) + f"(x) = A sin x.
(4)
Assuming that f is periodic with period T > 0, we conclude that T is also the period of f' and f", hence also of f' + !" and 2!' + !". In view of the equations (4) and the condition A =I= 0, B =I= 0, we infer that T is a period of both sin x and sin ( v'2 · x) .
99
Algebra
Since each period of sin x is a multiple of 21r , and similarly that each period of sin ( V2 · x ) is a multiple of V2 · 21r, we are again led to a con tradiction with the irrationality of ¥2. Problem
45
Find all monotonic functions
f : lR
�
lR satisfying the equation
J(4x) - f(3x) = 2 x P roblem
45,
for
x E JR.
Solut ion 1
If a monotonic function
f satisfies the given equation, then
f(4x) - f(3x)
{
> 0
< 0
for for
x > 0, x < 0,
showing that f is strictly increasing in each one of the two intervals (-oo , O) and (O, oo). Hence, f is increasing on JR. Replacing 4x by x, rewrite the equation as
f(x) = !x + f (�x) For any real number x =f tion of formula (1) gives
f(x) =
=
0
for
x E JR.
(1)
and any natural number n , repeated applica
! x + f (�x) !x + ! · �x + f ( ( �) 2x) ! x + ! · �x + ! · ( i)2x + f ( ( �) 3 x) !x + ! · �x + ! · ( �) 2 x + · · · + ! · ( �) n - l x + f ( ( �) n x) 1 (3)n J(( n , !x · �-\ + �) x) 4
i.e. , (2) Now, keep x fixed and let n vary. The sequence ( ( i) n x ):= is decreasing l if x > 0 , and increasing if x < 0 ; the same is the behaviour of the sequence (f ( ( � t x ) ):=1; the value f (O) is its lower (upper) bound, not necessarily sharp. Thus there exists a finite limit (maybe, depending on x):
g(x)
=
n -+oo lim
f ( ( � t x) .
100
Solutions
Assume that an f with
g(u) < g(v) for some positive numbers u and v, and choose 0
<
e
<
! (g(v) - g(u)) .
(3)
By the definition of a limit, we have
f ( ( � r u) and
<
g ( u) + f
! ( (�rv) > g(v) -
e
for n large enough
(4)
for n large enough.
(5)
Pick an integer k so large that inequality ( 4) holds for n = k . Next, find an integer m so large that estimate (5) holds for n = m and, moreover, ( � ) mv < (�) k u. Since f is strictly increasing, and in view of condition (3), we hence obtain
0 < f ( ( � ) k u) - f ( (�) m v)
(g(u) + ) (g(v) - ) = g(u) - g ( v) + 2e < 0 - obviously a contradiction. This means that g ( u) = g ( v) for any posi tive numbers u and v. In other words, there exists a common limit lim f ((�rx) = for every X > 0. n->oo <
�::
e
-
C
Analogously, there exists a common limit lim
n->oo
! ( ( �rx) = a
for every
x < 0.
(2) , thus obtaining f (x) = 2x + c for x > O.
So we can pass with n to infinity in formula
f(x) = 2x + a for x < O and S ince f has to be increasing on lll, we arrive at the final result: 2x + a for x < 0, (6) f (x) = b for x = 0, 2x + c for x > 0, where a, b, c are arbitrary constants such that a � b � (Clearly, every
{
c.
such function satisfies the given equation. ) P ro blem
4 5 , S o lut ion 2 The reasoning becomes shorter if we resort to the well-known fact from calculus that every function, monotonic in some real-line interval, has one-sided finite limits (equal or not ) , at each point of that interval. Ac cordingly, if f is a function satisfying the conditions of the problem, then there exist the finite one-sided limits
a = xlim f(x), --+0-
c = X--+0+ lim f (x) .
Algebra
101
As in Solution 1, we observe that .f must be increasing on f ( O ) by b we have, as before, a ::::; b ::::; c. Define
R
Denoting
h (x) = f(x) - 2x. Then also
x-+lim0+ h(x) = c.
lim h(x) = a, :J:-+0-
The given functional equation, recast in terms of the function the form
h,
takes
(h(4x) + 2(4x)) - ( h(3x) + 2(3x)) 2x, i.e. , h(4x) h(3x) ; equivalently, h(x) = h ( ix) for all x E R So we have, for every x E lR and every n E N, =
=
When n tends to infinity, the sequence (h ( ( i t x) ):=l tends to a, c ·or b, according as x is negative, positive or zero. So we obtain, in limit,
h(x) =
{
a b c
for x < 0, for x = O, for x > 0,
which is nothing else than the formula ( 6 ) , worked out in the first solu tion. Problem 46
A sequence ao, a1, a 2 , . . . of real numbers different from zero is generated according to the rule: a n +l = (a; - 1)/ (2a n ) · Show that it contains infinitely many positive terms and infinitely many negative terms. Problem 46 , Solution
1 If we change the signs all terms of a certain sequence that obeys the given rule, we obtain another such sequence. Therefore it will be enough to show that any such sequence has infinitely many negative terms. Assume the contrary; that is, assume a n > 0 for n � no. Then
a n for n � no, 2 implying a no +k < 2 -k a n0 for k = 1, 2, 3, . . . . Thus, sooner or later, there must appear a term a m E (0, 1). The next term a m+l . equal to (a !, - 1)/(2a m ) , a n+ l
=
a; - 1 2a n
---
<
-
102
Solutions
is negative. The assumption that "almost all" terms are positive has driven us to a contradiction. P roblem 46, S olut ion
Let
d n = an+l - an .
2
By the recursion, 2anan+l = a � -
1, and so
for all n. Now suppose we have a block a N , aN+l • . . . , aN+r of consecutive positive ans. Thus, if N ::::; n < N + r, then
(1) which combined with the inequality d� > 1 implies: dn < - 1 . S o we have (2 ) aN > 1 + aN+l > 2 + aN+2 > · · · > r + aN+r > r,
yielding an upper bound on the length of the block ( r < aN ) . Conse quently, a block of consecutive positive terms cannot be infinitely long; a negative term must eventually occur. A similar argument shows that any block of negative terms necessarily has a finite length; j ust the in equalities in (1) and ( 2 ) have to be reversed. Problem 46 , S olution
3
The first two solutions do not differ in any essential way. The third one is different. If not so simple as the foregoing ones, this solution gives more insight into the nature of the problem. The sequence is determined by its initial term ao, so the information about the positions of positive and negative terms must be somehow encoded in that single number, and the present proof shows how to decode it. The clue observation is that the recursion formula imitates the well known trigonometric identity cot 2 (} - 1 cot 20 = ---2 cot (} Now, let t E (0, 1) be the unique number such that a0 = cot ?rt. Then, according to the above, a 1 = cot 27l't, a2 = cot 41l't, and by induction, n an = cot 2 1l't
for
n
=
0, 1, 2 , . . .
(3)
These values of the cotangent function are well defined. ( Indeed: assum ing n is the least index such that cot 2 n 1l't makes no sense, i.e. , 2 n t is n l an integer, we would get that 2 - t is an "integer and a half" , implying l n an - l = cot 2 - 7l't = 0, contrary to the condition that the sequence has
Algebra
103
non-zero terms. ) In other words, representation
t = (O.c1c 2 c3 . . . ) 2
with
t
is not a dyadic fraction; its binary
ci E {0, 1}
for
i = 1, 2, 3,
.
.
.
has infinitely many zeros and infinitely many ones. Notice that > 0 if L2xJ is even, co t 1rx < 0 if L2x J is odd;
{
therefore ( see (3)) a n is positive or negative according as L2 n + 1 tJ is even or odd. And since L2 n+l tJ = (c1 . . . C n + 1) 2 , we see that a n > 0 when C n +1 = 0 and a n < 0 when Cn +l = 1 . Thus the distribution of positive and negative terms in the sequence corresponds exactly to the distribution of zeros and ones in the binary expansion of the number t = (cot- 1 ao)/7r. P roblem 47 Four sequences of real numbers
satisfy the simultaneous recursions
for n =
0, 1, 2, . . . . Suppose there exist integers k, r
Prove that
;:::
1 such that
a1 = b1 = c 1 = d1 = 0.
Problem
47, S o lution 1 Define s n = a n + bn + C n + d n . Conditions imposed on the given se quences imply that S k+ r = S k and s n +1 = 2s n for n = 0, 1 , 2, . The last equality entails ( by induction ) s n = 2 n so for n = 0, 1, 2, . . . . Thus 2 k+r so = 2 k so, and hence so = 0. This yields s n = 0 for all n ;::: 0. Con sequently, we get for n ;::: 1: .
.
.
a n + Cn (a n - 1 + b n - 1) + (cn - 1 + d n - 1) = S n - 1 = 0. =
Wn+l
=
=
(an + bn ) 2 + (b n + Cn ) 2 + (en + dn ) 2 + (d n + a n ) 2 2(a; + b; + c; + d;) + 2(a n bn + bn Cn + Cn d n + dn a n ) 2w n + 2(a n + Cn ) (bn + d n ) ·
(1)
104
Solutions
For n ;::: 1 we have by (1): Wn + l = 2w n . Induction yields Wn = 2 n - l w 1 In particular (setting n = k + r and n = k ) , we obtain for n = 1, 2, 3, 2 k+r - 1 w 1 = 2 k - l w 1 , whence w1 = 0. And this is j ust enough to conclude a 1 = b1 = c1 = d1 = 0 . .
.
.
.
P roblem
4 7 , Solution 2 Introduce the polynomials
Pn (x) = a n x 3 + bn x 2 + CnX + dn
= 0, 1 , 2, . . . . The conditions of the problem imply that Pk+ r (x) = Pk (x) and Pn+l (x) (a n + b n )x 3 + (bn + cn )x 2 + (en + dn )x + (dn + a n ) = a n (x 3 + 1 ) + b n (x 3 + x 2 ) + cn (x 2 + x) + dn (x + 1) = (x + l) (a n (x 2 - x + l) + b n x 2 + cn x + dn ) ---' (x + 1) ( -a n (x 3 - x 2 + x - 1 ) + a n x 3 + b n x 2 + C n X + dn ) = (x + l) ( Pn (x) - an (x - l ) (x 2 + 1)) . for
n
x = 1, x = i, and x = - i (the imaginary units) , Pn+ l (l) = 2Pn (l), Pn+l (i) = (l+i)Pn (i), Pn+l (-i) = (1-i)Pn ( - i),
Hence, setting
and by induction:
= 0, 1, 2, . . The polynomials Pk and Pk+r coincide; so we get 2 k+r Po(l) = 2 k Po(l) and (l±i) k+r Po(±i) = (l±i) k Po(±i). Since r ;::: 1, we have 2 r =/= 1, (l+it =/= 1, (l - it =/= 1, and thus
for
n
.
.
Po(l) = 0, Po(i) = 0, Po(-i) = 0 . B y the definition o f Po(x), this means that ao + bo + co + do = 0 and (2) i(co - ao) = bo - do = i(ao - co) . The "real-imaginary" equations (2) force ao = co and bo = do, and con sequently the four numbers a 1 = ao + bo, b 1 = bo + co, c 1 = co + do, d 1 = do + ao are equal, their common value being
1 05
Algebra
P roblem
48
The sequences
Show that
xo, x�, x 2 , . . .
and
yo, Y l . Y 2 , .
.
. are defined by:
xo = YO = 1 , Xn + 2 Xn + 1 y; + 2 for n = 0 , 1 , 2 , . . . Yn+l 2yn Yn = X 2 " - 1 for every integer n ;:::: 0 .
P roblem
48, S olution 1 Define sequences ao, at , a 2 , . . . and
bo, b� , b 2 , . .
bn - Yn - ...J2 for Yn + ...J2 Denote the common value of ao and bo by .>. : a n = X n +- .../2 ' X n .../2
_
.>.
=
v'2 1 + v'2 1 -
---
.
by
n = 0, 1 , 2 , . . .
= ao = bo.
The recursion formulas that define the sequences analogous formulas for ( a n ) and ( bn ) :
(xn )
X n+l - ...J2 Xn+l + ...J2 x,.+2 ...j2 x,.+l = x,.+ x,.+l2 + ...j2 1 = X n .../2 . 1 - ...J2 Xn + .../2 + .../2 = >. a n , b n+l = Yn+l - ...J2 Yn+l + .../2 Y�y,.+2 ...j2 2 y�+2 + v'2 2y,. (Yn - .../2 ) 2 (Yn + .../2) 2
a n+l
_
-
_
and
( Yn ) yield the
Solutions
106
Hence by obvious induction
b n = .>. 2"
for
n = 0, 1 , 2, . . .
2 n - 1 in the first equality we obtain a 2 " -1 = .>, ( 2 " -1 ) +1 = .>. 2" = b n . By the definition of an and b n , this is equivalent to X 2 "- 1 ,.j2 = Yn - ,.j2 X 2 " -1 + ,.j2 Yn + ,.j2 '
Replacing n by
-
i.e. , to
2V2 = 1 - 2 V2 . X 2" -1 + ,.j2 Yn + ,.j2 X 2 " -1 = Yn follows !
1The claimed equality
48, S o lution 2 The inductive definition of the XnS can be rewritten as
Problem
where
f(x) =
X n +1 = f(x n ) for n = 0 , 1 , 2, . . . (x + 2)/ (x + 1 ) . Hence /(1), J X 2n = ...._ ___,_.... " 2 0
•
•· •
,
0
the circle denoting composition. For n = 1 and n = 2 we are dealing with the functions
g(x ) = f o f(x) x+2 +2 -= xX ++ 21 x+l +1 --
-3 x + 2 = 2 3 , x+2 f o f o f o f(x) = g o g(x) =
(I)
107
Algebra
-1217 x + 2 = x + -1217 Compare these expressions with the initial Yn S: � ) 2 + 2 17 ( = = 1, Yo Y2 2 . � -12 .
(2)
-
-
It is natural to guess that
+2 f o · · · o f(x) = YnX X + Yn .
(3)
______., 2"
Once guessed, this is without much trouble proved by induction; accord ing to equations (1) and equality (3) holds for n = 1 and n = Assume i t holds for a certain n. Then
(2),
�(x)
=
2.
� o � (x)
+2 +2 · YnX Yn x + = YnX + 2Yn x + y n + yn (y;, + 2)x + 4yn 2yn x + (y; + 2) y; + 2 x + 2 -= 2 yny;_ +' 2 x + 2 yn +2 = Yn+lX X + Yn+l ' showing that (3) holds with n replaced by n + 1 . By induction, claim (3) is true for every integer n ;:::: 1. Now, setting in (3) x 1 we get in view of representation (1) X 2 n Yn1 ++Yn2 The number on the left side equals f (x 2 n - l ); that on the right side equals f( Yn )· And since f(x) is strictly decreasing, we conclude that X2n-l = Yn· ---
=
--
=
= -- .
108
Solutions
Problem
48,
Set
Solution
3
Un = 1 + X n
n = 0, 1, 2, . . . .
1 1 Xn + 2 U n+ I = 1 + X n + I = 1 + --Un . X n + 1 = 2 + --Xn + 1 = 2 + Consider the sequence vo, VI , v 2 , . . . defined by vo = 1, Vn+ I = UnVn for n = 0, 1, 2, . . . So
uo = 2,
for
(4)
and
( 1 ) Vn+ I = 2vn+I + -Vn + I , Vn +2 = U n + I Vn+ I = 2 + Un Un
(5)
We obtain
i.e. ,
Vn +2 = 2vn + I + Vn
for
n = 0, 1, 2, . . . .
(6)
This is a homogeneous linear recursive equation of the second order. The sequel is routine (see Problem 10, Solution 3, for instance) : the general solution of equation (6) has the form
Vn = Aa n + B /3n for n = 0, 1, 2, . . . , where a = 1 + V2 and f3 = 1 - J2 are the roots of the characteristic polynomial t 2 - 2t - 1. The constants A, B have to be determined from the initial data vo = 1, VI = 2, thus creating the specific solution of equation (6) we are looking for. The result is:
Vn = � J2(a n +1 - 13n+1 )
for
n = 0, 1, 2, . . .
(simple calculations are omitted ) . Revisiting formulas find
(4) and (5) , we
Xn = U n 1 = VnVn+ I 1 = Vn+IVn- Vn 0 n + l (a _ 1) _ 13n + I ( /3 1) v'2 . 0 n + I + /3n + l = = 0 n + I 13n +I .' 0 n + I _ 13n + I _
_
_
_
this can be further rewritten as (7) Denote the number X 2 " - I j ust by Wn = Yn for all n. Since wo = xo
w n . We have to prove the equality = 1 = yo, it will be enough to show
Algebra
109
that the wns obey the same recursion that defines the we have to show that
Wn+ l According to formula
=
w n2 + 2 2 Wn
-
for
(7) , we have for n
n =
;::::
0,
12 ,
, .
.
Yn S·
That means,
(8)
.
1
where 'Y n stands for (a/ (3 f" ; the last equality is valid also for n verification) .
'Yn + l 7;. Consequently, 1 w n2 + 2 Wn + _ 2 Wn 2w n J2 'Yn + 1 1 · -'Yn - 1 = + · 2 'Yn - 1 J2 'Yn + 1 J2 . ('Yn + 1) 2 + bn - 1) 2 = 2 ('Yn - 1) ('Yn + 1) J2 . 'Y; + 1 'Y; - 1 In 'Y n +l + 1 v�· 'Yn + l - 1 - Wn+l for
Notice that
__
=
=
=
--
=
_
Equality (8) is settled, and the proof is complete. Problem 49
Two sequences of integers
and are defined uniquely by the equality
Compute
n =
0,
1, 2, . . .
0 (easy
Solutions
1 10
Problem 49 , S o lution 1
Expanding (2 + va t and ( 2 - va t binomially, we obtain in both ex pressions the same coefficients of terms involving the even powers of va, whereas those involving the odd powers of va differ in sign. Therefore the equality ( 2 + va t = U n + b n va implies
(2 - VJ ) n = a n - bn v'3 j this last sequence converges to zero because 2 - va is a number between
0 and 1 .
bn 2': 1 for all n 2': 1 . Therefore a n = bn v'3 + (2 - J3 t = y'3 + (2 - J3 ) n bn bn bn which is v'3 in limit. Note that
Problem 4 9 , S olut ion 2
The implicit definition of the ans and sive formulas. Since
(2 + J3 t +l (and since
bn s can be easily made into recur
(2 + v'3 t (2 + VJ ) = (a n + b n VJ ) (2 + VJ ) = (2a n + 3b n ) + (a n + 2bn ) v'3
va is irrational) , we infer (ao = 1 , bo 0) ;
in the matrix form,
( �::� ) = ( i � ) ( �: )
=
(I)
.
I t i s well-known that a pair o f sequences satisfying such a recurrence can be postulated to have the form
eigenvalues of the matrix i i.e., the roots of the characteristic polynomial (2 - >.) 2 - 3: p = 2 + v'3, q = 2 - v'3. The constants A , B, C, D are evaluated from the initial conditions ao = 1 , bo = 0 and a 1 = 2, b 1 = 1 , where p and
q
are the
( �),
A lgebra
111
which yield the system of four linear equations
A + B = 1,
C + D = 0,
Ap + Bq = 2,
Cp + Dq = 1,
with the solution
C = -D = i J3.
A = B = �. Thus
A + B (qjp) n C + D (q jp) n · Since 0 < qfp < 1, this ratio tends to AJC = J3 as n --+ Apn + Bqn Cp n + Dq n
an bn
oo .
P roblem 4 9 , Solut ion 3
We again use the recursive formulas Formulas (1) imply
(1). Denote the ratio
a n fbn
n + 3 (x ) x n+ l - 2aa n ++2b3b n - 2x n Xn + 2 - f n ' n _
_
where f (x)
by
Xn . (2)
_
+3 1 . = 2x x + 2 = 2 x +_2 ' _
_
note that f is an increasing function in the interval (0, oo ) . The initial terms of the sequence (x n ) are x 1 = 2, x 2 = 7/4. In view of (2) and the strict monotonicity of f , the inequality x 1 > x 2 implies by induction x n > Xn+ l for all n. So (x n ) is a decreasing sequence of non-negative numbers, hence convergent to a limit l ;:::: 0. Passing to the limit in the equality (2) we obtain the equation
l = 2ll ++23 , with the unique non-negative root l = J3. -
P roblem 5 0
The sequence
(x n )
is defined by
Xl = 21 ' Prove the inequality
2n - 3 · X 1 Xn = � n-
for
n=
2, 3, 4, . .
.
.
Solutions
112
Problem 50, S o lut ion 1
Consider the auxiliary sequence Yn = (2n - 1)x n · The recursion formula that defines the x n s yields the analogous formula for the Yn S:
2n - 3 (2n - 1) (2n - 3) . Yn - 1 , Yn - (2 n _ 1) · 2n . X n- 1 _2n 2(n - 1) - 1 i.e. ,
2n - 1 · Yn = � Yn -1 for n = 2, 3, 4, . . . . (1) This formula i s valid also for n = 1 i f we set yo = 1. Now, 2n Yn - 1 - Yn = 2n 1 · Yn - Yn = 2nYn_ 1 = X n for n = 1, 2, 3, . . . , (2) _
and therefore
X 1 + X2 +
·
·
·
+ X n = ( Yo - Y 1 ) + ( Y1 - Y2 ) + · · · + (Yn - 1 - Yn ) = YO - Yn = 1 - Yn < 1, (3)
as needed. P roblem 50, S olut ion 2
x n s are X2 = X 1 · l = � ! , The initial
X3 = X 2 � = � · ! · � � X4 = X 3 � = � · ! · � · � , and in general ( by induction) 2k - 3 for k = 1 , 2, 3, . . . . X k = 21 · 41 · 63 · 85 · 7 · · · 2k ·
•
•
10
Multiply the numerator and the denominator of this fraction by the product of even integers from 2 to 2k - 2: (1 · 3 . 5 -· 7 . . . (2k - 3)) (2 . 4 . 6 . . . (2k - 2))
Xk =
=
(2 . 4 . 6 · · · (2k - 2)) 2 (2k) (2k - 2)!
(
)
= 4 k-1 1 2kk -- 12 . 2k1 .
(2kk -- 12) . ( - 2k2k- 1 ) (2kk -- 12) - _4k-1_1 (2kk -- 12) (2k -4k1)2 (2k) (2kk -- 12) - 41k (2kk ) k = 1, 2, 3, . . . .
Continue the transformation as follows:
_ 1 X k = 4 k- 1 = _k1_ 4 -1 = 4 k-1 1
1
for
(4)
A lgebra
113
Fix an integer n 2: 1 and set in (4) k = 1 , 2, . . . , n; adding the equalities that result we obtain, by telescoping,
X l + X 2 + · · · + X n = .!_ 4°
(00)
-
(n )
_!_n 2n
4
=
1
-
()
_!_n 2n . 4 n
(5)
This number is smaller than 1 . Done. Remark
Solution 2 does not differ from Solution 1 in any essential way ; the Yn S of Solution 1 are expressed by the explicit formula Yn = 4 - n ( 2: ) ; equalities (2) and (3) closely correspond to (4) and (5). ( In fact, Solution 2 indicates how the idea of introducing the sequence ( Yn } in Solution 1 might have arisen. ) Problem 5 0 , Solution
It is obvious that all the recurrence as
3
XkS
are positive numbers. Rewrite the given
(6) 2kx k = (2k - 3)x k- 1 for k = 2, 3, 4, . . . . Fix n 2: 2 and set in (6) k = 2, 3, . . . , n, n+ 1; if we add all the resulting
equalities and cancel the summands that occur on both sides, we get
X 2 + X 3 + · · · + Xn + (2n + 2)x n + l = x 1 . Hence
X l + X 2 + X 3 + · · · + X n = 2x l - (2n + 2)x n+ l = 1 (2n + 2)x n + l · -
Since X n + l is a positive number, the value of this sum is smaller than 1 ; the proof i s complete. Problem 50, S olution
4 Assume the converse inequality to the asserted one:
X l + X 2 + X 3 + + Xn 2: 1 ( for a certain n 2: 2); equivalently: x2 + x 3 + · · · + Xn 2: X 2 + X-=3 + · · + Xn -> 1 - 1 = 1 Xl Xl ( as x 1 = !). We claim that then X k+ l + . + X n > 2k 1 Xk •
--
•
----
(7)
• •
-
1
-
X t.
i.e. , (8)
•
•
-
(9)
114
Solutions
for k = 1, 2, . . . , n- 1 . We are going to show this by induction. For k = 1 the inequality (9) coincides with (8) . Assume that the estimate (9) is true for some k (1 � k � n - 2) ; multiply (9) by X k /X k+ l and subtract 1 from both sides of the resulting inequality: X k+2 +
•
.
•
X k+ 1
+
Xn
1= k+ (2k - 1) � X k+ 1 - 2 1 j X k+l /x k = (2k - 1)/ (2k + 2)
2::
(10)
the last equality follows from ( the recur sion formula from the problem statement ) . The induction step ( from (9) to (10)) is done; thus inequality (9) holds for all k = 1, 2, . . . , n-1. Now set i n ( 9 ) k = n - 1:
(11) � >- 2n - 3. Xn 1 The fraction o n the left equals (2n - 3)/ (2n), according t o the definition of the sequence. So the estimate (11) yields 1/ (2n) ;:::: 1 - obviously a -
contradiction. Hypothesis the assertion is true.
(7) must have been wrong, which means that
Problem
51 A sequence of real numbers ao, a 1 , a 2 , . . . satisfies the recurrence a n - 1 + an +1 for n = 1, 2, 3, . . . . Show that a n+9 = a n for all n.
ia n I =
P roblem
5 1 , Solution 1 The sequence contains infinitely many non-negative terms. Choose one of them. By the given recurrence, it is equal to t he sum of the two neighbouring terms, one of which must be non-negative. So we have two non-negative terms in succession: a m ;:::: 0, a m +l ;;::: 0. Then
One of these two differences is non-negative. This gives us a third non negative term adj acent to the two already found. Now, we have three successive non-negative terms, the middle one equal to the sum of the other two. Denote them by a, a + b, b (a, b 2:: 0). Assume a � b. The recurrence determines the sequence fo rwa rd and backward, and we are able to control the signs of the four terms that follow the block a, a + b, b, as well as the signs of the four terms that precede that block. Hence, this is a piece of our sequence:
. . . , b, 2b - a, b - a, -b, a, a + b, b, -a, a - b, b, 2b - a, I n the case where a ;;::: b, the corresponding piece i s
1 15
Algebra
In either case, the two leftmost listed terms coincide with the two right most ones, the two pairs being separated by a block of length 7. This yields the desired periodicity. Problem 5 1 , S olut ion 2
Define a transformation of the coordinate plane IR2 into itself by the formula f (x, y) = (y , I Y I - x) ; its relevance to the problem is apparent in view of
f(a n - 1 , a n ) = (a n , a n + 1 ) · Fix a positive number r and consider the closed polygonal line
(1)
ABCDEFGHJA with vertices
A = (O, -r), B = (- r, O), C = (-r, r ), D = (O, r), E = (r, 2r) , F = (r, r), G = (2r, r), H = (r, O ) , J = ( r, -r) ; denote this line by Cr . Pick a point P from the segment AB ; t hus P = (x, y ) , with y = -r - x, -r ::; x ::; 0, and hence its image Q = J(P) = ( - r - x , 1 - r - x l - x) = ( - r - x , r) lies on the segment CD and partitions it in the same ratio as P partitions AB: CQ/QD = AP/PB . In an analogous fashion we show that CD is mapped by f onto the segment E F, and so on: every side of the polygon Cr is mapped onto the next-to-neighbouring side ( in the clockwise sense) :
AB ---+ CD ---+ EF ---+ GH ---+ JA ---+ BC ---+ DE ---+ FG ---+ H J ---+ AB. Restricted t o each particular side, the mapping f i s linear; that i s t o say, if a point divides a side in some proportion, then its image divides the corresponding (oriented ) side in the same proportion.
It follows that the nine-fold application of f maps each point P E Cr onto itself: j 9 (P) = P. And since the union of all polygons Cr ( as the parameter r varies ) covers the whole plane ( except the origin, which is obviously a fixed-point of f), we conclude that the ninth iterate ! 9 is the identity map. This in view of equation ( 1 ) proves that a n + 9 = a n . Problem 5 1 , Solut ion
3
The forward recurrence a n + l = l a n l - a n-1 yields the backward recur rence a n - 1 = l a n l - a n + l· Fix an index m � 0; consider the terms a m and a m + 9 , and write for brevity a m+4 = x , a m + 5 = y . Starting from the pair x , y and applying four times the recurrence in its forward form
Solutions
1 16
l l l i Y I - x i - y i - I Y I + x i - I IY I - x i + y, l l l i x l - Y l - x l - l x l + y l - l l x l - Y l + x.
and in its backward form, we express follows:
a m+9 am
=
=
a m and a m+9 through x and y as (2)
(3)
If we denote the expression on the right side of equation (2) by g(x, y ) , we get that the right side of equation (3) i s j ust g(y, x ) . S o the problem reduces to showing that g(x, y) g(y, x ). The verification of this identity is a more or less automatic task, rather tedious. (Without going too much into details, let us j ust observe that in view of the symmetry between the roles of x and y , one only needs consider three main cases: 0$x $y; x $ 0 $ y; x $ y $ 0 ; but then they split into subcases . . . ) =
Solutions: Geometry Problem
52 Construct a right triangle ABC with a given hypotenuse of its medians are perpendicular.
c such that two
Problem
52, S olution 1 Assume the right angle is at C and the two perpendicular medians are issued from the vertices A and C. Choose the coordinate system with origin at B and with A on the x-axis. Let R be the midpoint of AB and P the midpoint of BC. Suppose C has coordinates (u, v); thus
A = (c, O), B = (0, 0) , C = (u, v), The orthogonality condition A P ..lC R
P
= (u j2, vj2) , R = (c/2, 0).
i s restated in terms o f the inner
product of vectors:
Since
C lies on t he circle with centre R and radius c/2,
Substitute this into the former expression to get u = �c. So the point is the foot of the altitude from C. The method o f construction follows: draw t he semicircle with diameter AB of the given length c; partition this segment in the ratio
D = (�c, O)
AD : DB = 1 : 2. Draw the perpendicular to AB through D ; it will intersect the semicircle at C, the third vertex of the triangle sought. Problem 52, S olut io n 2 Assume the triangle ABC, right-angled at C, has its medians AP and CR perpendicular. They intersect at S, the centroid of ABC. Let Q be the midpoint of AC and let T be the projection of S on BC. Since S partitions A P in the ratio AS : SP = 2 : 1, the segment C P is partitioned by T in the same ratio. If C S P has to be a right angle, S must lie on the circle with diameter C P . This yields the following method o f construction: draw a n arbitrary segment BC, find its midpoint P and the point T on C P such that
1 18
Solutions
CT : T P 2 : 1. Draw the semicircle on C P as diameter; then draw the perpendicular to BC through T ; it cuts the semicircle at S, the centroid of the triangle. Find A as the point of intersection of P S and the line perpendicular to C B at C. Triangle ABC has the desired shape, but not the desired size. Transform everything using a suitable similarity to obtain AB = c . =
Problem 5 2 , Solut ion 3
Let P be the midpoint of side BC and S be the centroid of the triangle we wish to construct (right-angled at C, with perpendicular me dians from A and C). Let R and W be the midpoints of AB and BR, respectively.
ABC
Construction: Draw segment AB of length c and erect semicircles k, k 1 . k 2 and k 3 over AB, BR, AR and AW a s diameters (in one o f the two half-planes determined by AB). Since ACB and ASR are assumed to be right angles, points C and S have to lie on k and k 2 , respectively. Circle k1 is the image of k in the homothety with centre B and coefficient 1/2, and so P (the midpoint of BC) lies on k 1 . The centroid S divides A P s o that A P �AS. Therefore P lies on the circle k 3 , which i s the image of k2 in the .homothety with centre A and coefficient 3/2. =
A
R
w
B
Hence, P is obtained as the point of intersection of k1 and k3 . The vertex C is the point of intersection of line B P and circle k . Problem 5 3
Let ABC be a triangle, AC =I= BC. Assume that the internal bisector of angle AC B bisects also the angle formed by the altitude and the median emanating from vertex C. Show that ABC is a right triangle. Problem 5 3 , Solut ion 1
Denote by E the midpoint of AB and by H c the foot of the altitude dropped from C. The perpendicular bisector of AB intersects the cir cumcircle in two points, one of which lies on the other side of line AB than vertex C; denote this point by D .
Geometry
1 19
D Let 0 be the circumcentre of triangle ABC. S ince AD = BD, the arcs AD and BD are equal, and so they subtend equal angles ACD and BCD; thus ray CD is the bisector of angle C. According to assumption, it bisects angle H eC E; in other words, angles H eC D and DC E are equal. Lines CHe and fore
DE are parallel ( both are perpendicular to AB) . There
LEDC = LHeCD = LDCE, which means that CDE i s an isosceles triangle: CE = DE . Note that also triangle CDO is isosceles: CO = D O . Since 0 lies on line DE, EO = ± (DO - DE) = ±(CO - CE)
(1)
( plus sign if E lies on segment DO, minus sign otherwise) . Hence, points
E and
0 coincide; otherwise CEO would be a non-degenerate triangle and equality (1) could not hold. The circumcentre coincides with the midpoint of side AB only if AB is the diameter of the circumcircle. Thus LAC B = 90° .
Pro blem 5 3 , Solut ion
2
Let E and He have the same meaning as in Soh1tion 1 ; let a, b, c be the lengths of sides BC, CA, AB and let a, {3, 'Y be the sizes of angles A, B , C , respectively. By assumption, angles AC B and He C E have a common bisector, and this means that angles ACHe and BCE are equal:
LBCE = LACHe = 90° - L CAHe = 90° - a ;
120
Solutions
hence
LACE = LACE - LBCE = 7 - ( goo - a: ) = a + 7 - goo = goo - {3. Apply the Law of Sines to triangles
ACE and BC E:
sin LACE
AE CE
sin LCAE =
BE CE
sin ( goo - {3 ) sin a cos {3 sin a:
(2)
sin LBCE sin LCBE =
sin ( goo - a: ) sin {3 cos o: sin {3
(3)
Since E is the midpoint of AB, the left-side terms of ( 2 ) and (3) are equal. Equating the right-side terms we obtain sin a: cos a: = sin {3 cos {3, i.e. , (4) sin 2a: = sin 2{3. Sides AC and BC are not equal. Hence a # {3 , and so equation 2a: + 2{3 = 180 ° , which means that 7 = goo . Problem
( 4) yields
54
If ABCDEF is a convex hexagon with
AB = BC,
CD = DE, EF = FA ,
prove that the altitudes ( produced ) of triangles anating from vertices C , E , A , concur.
BCD, DEF, FAB, em
Problem
54, Solution 1 Consider three circles W I . w 2 , w3 , centred at D , F, B , respectively; W I passing through C and E ; w 2 passing through E and A; and w 3 passing through A and C . Let w 2 and w3 intersect at A and A'. Similarly, let w3 and W I intersect at C and C '. Finally, let W I and w2 intersect at E and
E'.
Points A and A' are symmetric across the line connecting the centres F and B of circles w2 and w3 ; thus AA' ..l F B . This means that the altitude of triangle F A B , dropped from A, is contained in line AA'. Analogously, the other two altitudes considered in the problem are contained in lines CC' and EE'.
121
Geometry
Now, line AA' i s the power axis o f circles W 2 and W 3 j lines c c ' and EE' are the power axes of the pairs w 3 , w 1 and w 1 1 w 2 . For a triple of pairwise intersecting circles whose centres are not collinear, it is a well-known fact that the three power lines determined by pairs of these three circles are concurrent. This proves the claim. Problem
S olut ion 2
54,
The altitudes ( produced ) of triangles F AB and BCD, issued from ver tices A and C, intersect at some point P ; so PAl. B F and PC l. B D . I t will b e enough t o show that also P E l. D F. I n terms of vectors ( and their inner products ) , we have to prove that the equalities
FA. . ifF = o
imply P"E F'D .
and
PC . i5B = o
= o.
Points B , D , F lie ( respectively ) o n the perpendicular bisectors o f seg ments AC, CE, EA, concurrent at 0 , the circumcentre of triangle ACE . Thus o:8 AC = 0, on . cE = 0, oF EA = 0, and hence .
o
.
oB · AC + on · GE + DF · EA = o:B (oc 01 ) + on . (DE - oc ) + oF (01 - DE ) = 01 . (oF - o:B ) + oc (DB - on ) + DE (on - "OF ) = 01 - ifF + OC · DB + DE · FD (FA. - PO ) . IfF + (PC - PO ) J5B + (P"E - PO ) · FD = PA · BF + PC · DB + PE · FD + PO · (lfD + DF + F"B ) . .
-
.
.
.
.
In the sum obtained, the first two summands vanish, according to as sumption; and the vector sum in the parentheses is the zero vector. Therefore PE · FD = 0, and this is j ust what we wished to prove. Problem
55
ABCDEF be a regular hexagon with M and N points on diagonals CA and CE ( respectively ) such that AM = CN . If M , N and B are collinear, prove that AM = AB. Let
Problem
55,
Solution
1
Triangles ABM and CDN are congruent, as AB = CD = a ( the length of the side of the hexagon ) , AM = CN ( by assumption ) , and
LM AB = LNCD = 30°. Therefore
LDNC = LBMA = LNMC,
122
Solutions
and consequently
LDNB
LDNC + LCNB = LNM C + LCNM = 180° - L M CN 180° - 60° 120° .
Also LDOB = 1 20° , where 0 denotes the centre of the hexagon. Since 0 lies on the circle with centre C and radius a, it follows that also N lies on this circle. Hence AM = CN = CO = a = AB. Prob lem
55,
Solution 2
Let AC = CE = 1 ; then AB = ! v'J. Denote the common length of AM and C N by x ; then C M = EN = 1 - x , and we have the vector equalities
eM = ( 1 - x) GA, ·
Since
B lies i n line with M and N , there exists a real number t such that cB = ( 1 - t) · CM + t · eN
O n the other hand, EB
= ( 1 - t) ( 1 - x) · cA + tx · GE.
= � ( EC + EA ) , whence
cB = GE + EB = C'E + � (- C'E + ( CA: - C'E ) ) = �
· GA - ! · C'E.
The representation o f a vector as a linear combination of cA and cE is unique. Thus, equating the coefficients in the formulas above we obtain ( 1 - t) ( 1 - x)
= �.
tx
= - 1.
These equations, combined, yield t + x = 0. Hence t -x and the sec ond equation becomes x 2 = l · Thus AM = x = l v'J = A B . =
Problem
55,
Solut ion 3
Let points A , B , C , D , E, M , N be represented by the complex numbers a, b, c , d, e, m , n, chosen as follows:
e= Writing
CN : CE = AM : AC m
= ( 1 - .X)a,
=
n
.X ,
we have
3-
iv'3 2
CM : CA = 1 - .X , so that
= .Xe (.X real positive) .
123
G eometry
The collinearity of B, M, N requires that the following quotient q be a real number: m-b (1 - >.)a - b (1 - >.)a - (a - 1) 1 - a>. = = q = >.e - b >.e - b e>. - b -b Since a = e and b = d ( bar denoting complex conjugation ) , the required equality q = 7j takes the form 1 - e>. 1 - a>. = -e>. - b a>. - d Substitution of the numerical values of a, b, d, e simplifies this to: 3 >. 2 = 1. Hence AM = >. · AC = >. · i a l = + i = 1 = AB . · n
--
/l Vl
56 Let ABC be an acute triangle with altitudes B D and CE. Points F and G are the feet of perpendiculars B F and C G to line DE. Prove that EF = D G . Problem 5 6 , S o lution 1 Since LBDC and LBEC are right angles, BCDE is a cyclic quadrilat eral, and hence Problem
LBCD = 180° - LBED = LBEF.
(1)
Therefore BEF and BCD are similar triangles, and we get EF CD = (2) B E BC Analogously, considering the similar triangles CD G and C BE, we have CD DG (3) = B E CB The asserted equality EF = DG follows immediately from relations (2) and (3) . Problem 56, S o lut ion 2 A slight variation of the previous solution: by equation ( 1 ) , EF = BE · cos ( LBEF ) = B E · cos C = BC · cos B · cos C,
(4)
where of course B and C are the angles of triangle ABC. The roles of the point systems B , E, F and C, D, G are symmetric, and therefore we may replace each character from the first system by the corresponding one from the second. Formula ( 4) thus yields DG = CB · cos C · cos B .
(5)
1 24
Solutions
Since the right sides of equations ( 4 ) and (5) are equal, so are the left sides. Problem 56, Solut ion 3 Let H and H1 be the midpoints of BC and FG. Quadrilateral B CGF is a trapezoid; thus H H 1 is parallel to BF and CG, hence perpendicular to DE. As in Solutions 1 and 2, notice that D and E lie on the circle with diameter B C. Line H H 1 passing through its centre H and perpen dicular to chord DE, must be the perpendicular bisector of that chord: Consequently H1 is the common midpoint of DE and FG, and therefore EF = DG. P roblem 56, S olut ion 4 The repeated use of the Pythagorean Theorem will also do the job: b.BFE : b.CGD : b.BEC : b.BDC : b.CGE : b.BFD :
EF 2 + BF 2 BE 2 ; CD 2 = CG 2 + DG 2 ; BE 2 + CE 2 = BC 2 ; BD 2 + CD 2 ; BC 2 2 2 CE 2 ; cc + EG 2 BF 2 + DF 2 . BD =
If we add these six equalities (and cancel the terms BF 2 , B E 2 , CD 2 , CG2 , CE 2 , BC 2 , BD 2 that appear on both sides) , we are left with Since EG = DE + DG, DF = DE + EF, this is equivalent to EF 2 + DE2 + DG 2 + 2 · DE · DG = DG 2 + DE 2 + EF 2 + 2 · DE · EF. Hence DE · DG = DE · EF, and consequently DG EF. Problem 57 Consider the right triangle ABC with LC = 90 ° . Let At and Bt be two points on line AB (produced beyond A and B ) such that =
AAt = AB = BBt and let N be the foot of the perpendicular from A 1 to line BtC. Show that the rectangle with sides B 1 C and C N has area twice as large as the square with side AB. Problem 57, Solution 1 Let CC1 be the altitude in triangle ABC and let N P be the altitude in triangle A 1 B1N .
1 25
Geometry
N
Write
We have to show that EtC · CN = 2 AB 2 ; that is, x y 2c2 • The right triangles BtCtC and BtN A t are similar ( LABtN being their common angle), and therefore BtCt : EtC = BtN : BtA t ; equivalently, (c + p) x (x + y) : (3c), or =
·
:
=
xy
=
3c2 + 3cp x 2 •
(2)
-
The altitude of the right triangle ABC satisfies the well-known equality h 2 p(c - p); hence, by the Pythagorean Theorem for triangle BtCt C , =
Inserting this into equation ( 2 ) we obtain xy =
(3c2 + 3cp) - (c2 + 3cp) 2c2 , =
as wished. 2 Segments AB and AtBt are diameters of two concentric circles whose common centre is M , the midpoint of AB . Since LACE and LAtNBt are right angles, points C and N lie on those circles. Line BtN cuts the smaller circle in two points (which can coincide, in the limit case) . Denote them by X and Y , with X lying closer to B t and Y closer to N ; point C coincides with either X o r Y . Problem 5 7, S olution
Solutions
1 26
Let S be the foot of the perpendicular from M to line B1N. Chords XY and B1N of the two circles are perpendicular to line MS. Since M is the common centre of those circles, M S is the common perpendicular bisector of X Y and B1N. Therefore SB1 = SN and SX = SY ; denote the common length of SX and SY by d. We get (3) B1X NY ( = B1S - d ) and B1Y = N X ( = B1S + d ) . Our task is to prove the equality =
(4) According as C = X or C Y , the product B1C · NC is either equal to B1X N X or to B1Y · NY . By virtue of (3) , we have =
·
B1C · NC = B1X · B1Y in each case. Considering segments intercepted by circle ( ABC ) on rays B1N and B1A 1 . we have by polarity B1X · B1Y = B1A · B1B.
And since B1A = 2 · AB and B1B = A B , claim (4) results. Problem 57, S o lut ion 3 This is a variation of Solution 1, from which we preserve notation (1). Moreover, let AC1 = q . The altitude CC1 = h of the right triangle ABC satisfies: h 2 = pq. The Pythagorean Theorem now implies: for triangle AtN Bt : A t B � = A1N 2 + NB ? , (5) A1C 2 for triangle A1NC : A1N 2 + N C 2 , (6) 2 2 2 A 1 C = A 1C 1 + C1C , for triangle A 1C1C : (7) B1C 2 = B1c r + C1c 2 . for triangle B1C1C : (8)
Geometry
127
Elimination of A 1N 2 and A1 C2 from equations (5), (6) , and (7) gives N B� - NC2 = A tB� - AtC2 = AtB� - (At C� + CtC2) , i.e. , (9) Equation (8) says that x 2 = (c + p) 2 + h 2 . Substituting this into equa tion (9) (whose left side reduces to 2xy + x2) we obtain 2xy + (c + p ) 2 + h2
=
9c2 - (c + q) 2 - h2 .
Hence, in view of h 2 = pq and p + q 2xy
=
c,
9c2 - ((c + p)2 + (c + q)2) - 2h2 9c2 - (2c2 + 2c(p + q) + (p2 + q2)) - 2pq = 9c 2 - 2c2 - 2c(p + q) - (p + q )2 = 4 c 2,
showing that xy = 2 c2 • Problem 58 Let ABC DE be a convex pentagon inscribed in a circle. The distances from A to lines BC, CD, DE, and B E are b, c , and d, respectively. Express d in terms of b, c . Problem 58, S o lut ion 1 Denote the feet of the perpendiculars from A to lines BC, CD, DE and B E by H, N , P and K, respectively; so AH , b = AN, c = AP, d = AK . Assume, for definiteness, that N lies on segment CD, K lies on segment BE, H lies on line C B produced beyond B , and P lies on line DE produced beyond E. Obviously, other configurations are also possible; the reasoning then requires but minor changes. The reader is invited to find out what cases can occur and to draw suitable diagrams. Now, in the case at hand: quadrilaterals AH B K , AH C N and AND P are cyclic, each of them having two right angles at opposite vertices ( at H , K , N , P). So a,
a,
a =
LH AK = 180° - LH BK = L CBE, LHAN = 180° - LHCN = 180° - LBCD, LN AP = 180° - LN DP.
(1) (2)
(3)
Since also BC DE is a cyclic quadrilateral ( inscribed in the given circle ) , LCBE = 180° - LCDE = 180° - LNDP, LBED 180° - LBCD.
(4) (5)
Solutions
1 28
Comparing equations (2) and (5) , we see that angle HAN equals BED, hence also BAD (inscribed angle sub tended by the same arc B D ) . There fore L H AB = LH AN - L BAN = LBAD - LBAN = LN AD.
= LN AP - LN AD = LDAP.
(6)
Comparing equations (3) and (4) , we see that LN AP = LCBE, and in view of (1) we get L H AK = LN AP; thus by ( 6 ) : L B A K = L H AK - LH AB
(7)
On account of relations ( 6 ) and (7) , we have the following pairs of similar right triangles: i::J. H AB "' b.N A D
,
b.BAK "' b.DAP.
Consequently AH B K and AN DP are similar quadrilaterals, which im plies that H AK and NAP are similar triangles. Thus AK AP = AH AN In other words, d/a = cfb, and we obtain the desired result: ac d= -. b•
58, S olution 2 Preserving notation of Solution 1, consider the angles:
Problem
¢ = LABE = LADE,
f =
LAEB = LACB
(¢ is the size of any angle subtended by arc EA, and f is the size of any angle subtended by arc AB) ; and let 0!
= LDEA = 180° - LACD,
the last equality following from the fact that quadrilateral AC DE is inscribed in the given circle. Assume for the while that the projection points H , N , P, K are situated as in Solution 1. Considering the right triangles AK B , AP D, AH C , A K E and A N C we see that AK = sin LABK AB = sin ¢ sin LADP
AP AD ' AH sin LACH AC = sin e sin LAEK AK AE ' AN sin LACN AC sin LACD sin(180° - a ) sin a .
(8)
=
=
(9)
=
=
( 10)
=
In the general case, each one of LABK and LADP might be equal either to ¢> or to 180° - ¢>. This however does not affect the validity of formulas (8) , as sin(180° - ¢>) sin ¢>; the same observation applies to the for mulas in line (9) , while in line (10) angle AC N can be either equaL or complementary to AC D, without affecting the formula. Thus, equalities (8) , (9) , (10) are true in any case. From (8) and (9) we have =
AB AD
=
AK AP
=
-d and AE c AC
=
AK AH
=
d a
Multiplying these equalities, d2 ac
AB · AE AC ·.AD
(11)
(a nice formula in itself) . Now, applying the Law of Sines to triangle AD E and using equations (8) and (10) we obtain AE AD
-
=
sin ¢> sin a
--
=
AK : AB AN : AC
AK · AC AN · AB
=
d AC -·b AB
Hence
AB · AE -d ( 12) AC · AD b Equations (11) and (12) result in d 2 f (ac) dfb, and s·o, finally, d = acfb. =
=
3 Denote the radius of the given circle by R. It is the circumradius of each triangle determined by any three points out of A, B , C , D , E. To Pro blem 5 8 , Solut ion
1 30
Solutions
express the area of any one of these triangles, we may apply either the formula: {product of sides) / {4R) or: {base times altitude)/2. And thus: area ABC =
AB · BC · A C B C · AH = 4R 2
=>
a=
area ACD =
AC · AD · CD 4R
CD · AN 2
=>
AD b = AN = AC · 2R '
area ADE
AD · AE · DE DE · AP = 2 4R
=>
c=
area ABE =
AB · A E · BE 4R
Therefore ac =
B E · AK 2
AC AH = AB · 2R '
AP = AD ·
AE , 2R
AE d = AK = AB · - . 2R
AB · AC · AD · AE = bd, 4R 2
implying d = ac fb . Remark
The last solution is shortest , easiest to comprehend (though not to invent, perhaps) , and it does not depend on any picture or assumption about the particular configuration (unlike Solution 1 and, to some extent, also the second one) . Moreover, it shows that A, B , C, D, E might be any five distinct points on a circle, not necessarily the consecutive vertices of a pentagon. Problem 5 9
Let ABC be an isosceles triangle with base A B . Let U be its circumcen tre and M be the centre of the excircle tangent to side AB and to sides CA and CB produced. Show that 2 · C U < CM < 4 · CU . Problem 5 9 , Solution
1
Let D be the intersection point of the circumcircle of triangle ABC and line CM . Denote the incentre of triangle ABC by I. Rays AI and AM are the internal and the external bisectors of angle A, hence they are perpendicular and I AM is a right triangle. Thus LIMA = LIAB = o/2 (where of course o = LCAB ) . The orthogonality relations AC .l.AD and AI .l.AM also yield the equality LM AD = LIAC = o/2. It follows that L M AD = LAM D, i.e. , DAM is an isosceles triangle and we have D M = D A < CD; the last inequal ity holds because CD is the diameter and AD is another chord of the circumcircle of ABC.
131
Geometry
CD < CM = CD + DM < 2 · CD. And since CD = 2 · C U , the claim results. P roblem 59, Solut ion 2
Let he be the altitude from C and let Pe be the exradius from M to the midpoint of AB . With the usual notation
BC = a, CA = b (= a), AB = c, a + b + c = 2s, R = UA = UB = UC, F = area ( ABC ) we restate the claim as
2R < he + Pe < 4R.
(1)
Using the well-known formulas
abc = a2 c R=4F 4F '
2F , he = c
F , Pe = -c 8 -
we recast inequalities (1) into the form
gp 2 4 F 2 2 2a 2 c < -c + -s - c < 4a c.
(2)
The area F is expressed by Heron's Formula
sc 2 (s - c) ; (3) F 2 = s(s - a)(s - b) (s - c) = s (s - a) 2 (s - c) = 4 we have used the fact that the triangle is isosceles (a = b), so that 2a +c c s - a = a + 2b + c - a = 2 -a= 2.
132
Solutions
In view of formula (3) , claim (2) becomes just
2a 2 < s (2s - c) < 4a 2 ; and since 2s - c = a + b = 2a, division by 2a reduces this inequality to
a < s < 2a. The left part holds trivially, and the right part follows, for instance, from: 2a - s = (a + b) - s = ( 2 s - c) - s s - c > 0. The claimed inequality (2) is thus proved. =
Problem 5 9 , S olut ion 3
Let H be the midpoint of side AC. Suppose the excircle in question touches side AB at T1 and the lines AC and BC at T2 and Ta, respec tively.
The segments AT1 and AT2 are. equal, as they are the tangents from A to the excircle. Thus AT1 = AT2 = c/2, and consequently CT2 = a + c/2 (with a and c standing for the lengths of BC and A B) . The right triangles CHU and CT2 M are similar, and hence
CT2 = a + c/2 = 2 + !:_ a CH a/2 The proposed inequality says that the ratio C M : CU should be com prised between 2 and 4, and so we are left with showing that c 0 < - < 2. a CM
CU
=
•
The lower estimate is evident, and the right one is so too, due to the triangle inequality c < a + b = 2a.
Geometry
1 33
60 The diagonals AC and BD of a convex quadrilateral ABCD intersect in E. Let F1 , F2 and F be the areas of trian�s ABE, CDE and quadri lateral ABC D, respectively. Show that y F1 + ..jF; :::; ../F. When does equality hold? Problem
60, Solut ion 1 Denoting the areas of triangles BC E and DAE by F3 and F4 , we have to show that ...jF; + M :::; VFl + F2 + F3 + F4 . D
Problem
A By squaring, this is equivalent to
B
(1) Let K and L be the feet of perpendiculars dropped to line AC from D and B , respectively. (They can lie on or outside segment AC . ) Write BL = b, DK d, AE = m, C E = n. Then =
The inequality (1) we are about to prove becomes
Vmb · nd :::; ! (nb + md) ; and this is just the inequality between the arithmetic mean and the geometric mean of the two products nb and md. To achieve equality, we need equality between the averaged quantities nb and md; and this is equivalent to
b : d = m : n.
(2)
1 34
Solutions
Lines BL and DK are parallel. So b : d = BL : DK = BE : DE, by the Intercept Theorem, and we can restate (2) as BE AE (3) DE CE By the (inverse) Intercept Theorem, equation (3) holds if and only if lines AB and CD are parallel, i.e., ABCD is a trapezoid with AB IICD. This is the condition for equality in (1). Problem 6 0 , S olut ion 2
Reduce the problem to inequality ( 1), as in Solution 1. Everything goes even faster if we write BE = p, DE = q (preserving the notation AE = m, CE = n) and express the areas Fi as Ft = �mp sin a,
}
r 2 = 2nq sm a , D
•
Fa = �np sin ,B,
F4 = �mq sin ,B,
where
,8 = LBEC = LDEA = 180° - a.
a = LAEB = LCED,
Hence sin a = sin ,B; denote this common value by 8 . Since a is a convex angle, 8 is a positive number. Inserting the trigonometric expressions for the Fi s into (1) we obtain the inequality Jm p 8 · nq8 :::; ! np8 + �mq8 (to prove) . Factor 8 cancels and we are left with Jmp nq :::; � (np + mq) , ·
the AM-GM Inequality for np and mq. Equality requires that np = mq, i.e. , pfq = m/n; and this is nothing else than equality (3) from Solution 1. Conclusion as before. Problem 6 1
Let P1P2 be a fixed chord (not a diameter) of a circle k . The tangents to k at Pt and P2 intersect at Ao. Let P be a variable point on the minor arc P1P2 . The tangent to k at P intersects lines AoPt and AoP2 at A t and A 2 , respectively. Determine the position of P for which the area of triangle AoA 1A 2 is a maximum. Problem 6 1 , Solut ion 1
Let M and r be the centre and the radius of k . Consider k as the excircle of triangle AoA1A 2 escribed at side AtA 2 . Denoting by 8 the semiperimeter of AoA 1A 2 , and by F its area, we have the formula
( 1)
Geometry
1 35
(Readers who have not encountered that formula are invited to provide a proof, which is not at all difficult - using, e.g., the more familiar F = ps , with p the inradius, plus a similarity argument.) The factor {s - A 1 A 2 ) in (1) is the distance from Ao to the point of contact of the incircle with side AoA 1 . To make it a maximum, the incentre should be chosen on ray AoM as far from Ao as possible; and this is the case (given the conditions of the problem) when P is the midpoint of arc P1 P2 . Problem 6 1 , Solut ion 2
Choose M, the centre of k, to be the origin of a coordinate system, with the radius of k as unit (r = 1). Thus the equation of k is x 2 + y 2 = 1. Choose P1 P2 parallel t o y-axis; i n coordinates, let
P1 = (u, v), P2 = (u, -v) , and let P (p, q). Assume u, v > 0, without loss of generality; then u < p :5 1. The lines t 1 , t 2 and t 3 , tangent to k at P1 . P2 and P, are =
described by the equations
v-q p-u ) A ( v+q u P ) A 1 -- ( pv-qu ' pv-qu ' 2 -- pv+qu ' pv+qu (Ao = t 1 n t 2 , A 1 = t1 n tg, A 2 = t 2 n tg) . The area F o f triangle AoA1A 2 is expressed by the determinant formula tg: px + qy = 1.
t1: ux + vy = 1;
( � . o),
They intersect pairwise at the points
Ao =
-
--
-
1 [1 ( p - u p-u ) 2 u pv - qu + pv + qu + q . u-p + v +q u-p ] + pvv -- qu pv + qu pv + qu pv - qu 1 ((p - u) f u) · 2pv + 2v(u - p) 2 (pv - qu)(pv + qu) v (p - u) 2 :;; · p 2 v 2 q 2 u 2 i
x i and Yi standing for the coordinates of A i · Substituting these coordi nates,
F
=
- -
· --"-
=
_
Solutions
136
-
here u, v are constants and p, q are variables satisfying p 2 + q 2 u2 + v2. Therefore p 2 v 2 - q 2 u 2 p 2 v 2 (1 - F 2 )u 2 p 2 - u 2 , and hence
.
=
.
=
(
)
=
1
=
v (p - u) 2 v p - u v 2u 1 � p2 - u2 � p + u = � - P + u . Recalling that 0 < u < p :::; 1 , we see that F is a maximum when 2u p+u is a minimum, i.e . , when p is a maximum, i.e. , when p 1. This corre sponds to P lying on the x-axis, hence coinciding with the midpoint of arc P1P2 . F
=
=
=
Problem 6 1 , S olut ion 3
Again, consider k to have centre M and radius
r =
1.
t2 Let the angles Ao, A 1 , A 2 of triangle AoA 1A 2 have sizes a , {3 , 'Y · Now, A1M is the bisector of LP1 A 1A 2 ; therefore L P1A1M 90° - (/3/2) , so that (in view of P1M 1) =
= r =
and similarly, Knowing AoP1 AoP2 cot (a/2) cot(a/2) , we can calculate the area F of triangle AoA 1A 2 from the trigonometric formula =
F
=
=
= r
.!. · AoA 1 · AoA 2 · sin a
2
=
Geometry
= =
) (cot �2 - tan 2 ) sin a (cot2 �2 - cot �2 (tan !!._2 + tan 2 ) + tan !!._2 tan I). 2 2 (
sin a cot � - tan !!._ 2 2 2
1 37
I I
Since
a f3 f3 1 1 a tan - tan - +tan - tan - +tan - tan - = 1 2 2 2 2 2 2 ' we can express the product of the numbers tan(/3 /2) and tan (I /2) by their sum: f3 a {3 + tan 21 . tan 2 tan 2I = 1 tan 2 tan 2
(
(
-
Hence
(
)
sin a a a a F = -- cot 2 2 + 1 - tan 2 + cot 2 2
) (tan 2{3 + tan 12 ) ) .
The angle a is constant. Consequently the area F is maximized when the sum tan(/3 /2) + tan(l /2) is minimized. Now, one can set 1 = 180° - a - {3 and examine this sum by calculus, as a function of the single variable {3 ; one can also use the convexity of tan x to deduce that this sum is a minimum when {3 = I· But we prefer to use a more elementary argument:
(
2
)
. {3 + I sm 2 2 = {3 2 cos - cos -I 2 2 2 sin
{3 + tan -I tan -
=
( � + �) cos ( � + �) + cos( � - �) a
2 cos 2 = /3 - 1 . a sm 2 + cos -2For a fixed a , this is a minimum when cos((/3 - 1)/2) = 1, i.e., when {3 1, and we arrive at the same conclusion as in the two previous solu tions. =
Problem 62
Let P be a point inside a parallelepiped whose edges have lengths a, and c. Show that there is a vertex whose distance from P does not exceed � Ja 2 + b 2 + c 2 . b
Solutions
138
62, S o lut ion 1 Consider the six planes containing the faces of the parallelepiped. Let be the plane whose distance from P is a minimum, let ABC D be the face contained in and let N be the foot of the perpendicular dropped from P to Then N lies within ABCD; otherwise the segment PN would intersect another face, less distant from P than contrary to the choice of Assume without loss of generality that the edges not parallel to have length c. Then P N $ c/2. Now, N is a point inside parallelogram ABC D , with sides of lengths a and b. Repeating the previous reasoning ( one dimension lower) , we find a side of ABC D whose distance from N is a minimum. Assume ( relabeling if necessary ) that this is side AB, with AB = a. Denote by K the foot of the perpendicular from N to line A B ; then K is a point of the segment AB . Note that N K $ b/2. We may also assume AK $ B K . Thus AK $· a/2. The three segments AK, K N , N P are the edges of a rectangular box and P A is its space diagonal. Hence, finally, Problem
1r
1r
1r .
1r ,
1r .
1r
PA <
VAK 2 + KN 2 + NP2 J(a/2) 2 + (b/2)2 + (c/2) 2 ! v'a2 + b2 + c 2 .
62, S o lut ion 2 Let 0 be the centre of the parallelepiped and let u, v, w be vectors of lengths a/2, b/2, cj2, parallel to the respective edges. The vertices can be labeled so that
P roblem
----+
-·
OAt = u + v + w, OA 2 = -u - v - w, OAa = -u + v + w, OA4 = u - v + w, OAs = u + v - w, OAs = u - v - w, OA7 = -u + v - w, OAs = -u - v + w. --+
--+
--,-.+
( In fact, vectors u, v, w provide a basis of a non-orthogonal coordinate system in the space. ) The vector OP determined by the given point P has representation --+
OP = xu + yv + zw with
- 1 $ x, y, z $ 1.
Consider the eight non-negative numbers -§ (1 + ix ) ( 1 + jy) (1 + kz) with i, j, k taking independently values +1 and - 1. Denote them by P l , . , ps according to the rule: .
---t
if OA m = iu + jv + kw then Pm =
(1 + ix) (1 + jy) (1 + kz ) 8
·
.
{1)
139
Geometry
Compute their sum: 8
L Pm = S1
m =l
L
(1 + i x + i y + k z + ij xy + ik xz +jk yz + ijk xyz ) .
i , j, k E { +l,- 1 }
When ( i, j, k) range over the set of the eight triples of plus-minus ones, then each one of the expressions i, j, k, ij , ik, jk, ijk takes values +1 and - 1 equally often. Therefore the sums L; i, L; j , L; k , L; ij , L; ik, L; jk, L; ijk are zero, and hence
t Pm � (8 + X ?: i + =
m=l
• ,J ,k
· · ·
)
+ xyz ?: ijk = 1. •,J,k
(2)
Choose an index m E { 1, 2, 3, 4, 5, 6, 7, 8}; it corresponds to a certain configuration of plus ones and minus ones, in agreement with (1) . For those values of i, j, k: ( 0A m - OP ) 2 [( i - x)u + (j - y)v + ( k - z)w] 2 = [i (1 - i x ) u + j (1 - j y)v + k (1 - kz)w] 2
PA�
(3) where
(4) Um = (1 - ix) 2 u 2 + 2jk (1 - jy) (1 - kz ) ( v • w) , Vm is obtained from Um by the cyclic shift i -+ j -+ k -+ i and the si multaneous shift x -+ y -+ z -+ x, and Wm arises from Vm in the same manner. By (1) and (4) ,
(1 - x 2 ) (1 - ix) (1 + jy) (1 + kz) u 2 + 8 (j k + ijkx ) ( 1 - y 2 ) (1 - z 2 ) V W . + ( • ) 4 Summing over m = 1, . . . , 8 ( that is, over all possible configurations of signs i, j, k ) we obtain 8
L Pm Um
m=l
=
[� ?: (1 - ix ) ( 1 + jy) (1 + kz)] (1 - x 2 )u2 + + [� ?: (jk + ij kx) ] (1 - y 2 ) ( 1 - z 2 ) (v 'I.,J, k
• ,,,k
·
w) .
Solutions
140
The first sum in square brackets equals 1, and the second one equals 0; see the argument preceding definition (2) . Thus 8
L PmUm = (1 - x 2 ) u2 .
m= l Analogously, by cyclicity,
8
8
L PmWm = (1 - z 2 ) w 2 .
L PmVm = (1 - y 2 ) v 2 ,
m=l m= l Equalities (3) , which hold for m = 1, , 8, now imply .
.
.
8
L P m · PA� = (1 - x 2 ) u 2 + (1 - y 2 ) v 2 + (1 - z 2 ) w 2
m =l
u2 + v 2 + w 2 a 2 + b2 + c2 = 4 In view of (2) , this sum is a weighted mean of the eight numbers <
PA�, . . . , PA�
(with weights P I , , P 8 ) · At least one of those numbers does not exceed the mean. Consequently, there exists an m such that a 2 + b 2 + c2 PA 2m < ' 4 and this is exactly what had to be proved. Problem 6 3 Do there exist two cubes such that each face of one of them meets each face of the other one (possibly at an edge or a corner)? Problem 6 3 , Solution 1 Suppose a cube C has vertices (±1, ±1, ±1) and let be a plane not passing through the origin 0 = (0, 0, 0) and having points in common with all the six faces of C. The equation of can be written in the general form ax + by + cz = k , with k ::J 0, a 2 + b2 + c 2 > 0. The distance from 0 to equals d l k l f v'a2 + b2 + c2 • In view of the standard symmetries of C, there is no loss of generality in assuming c � b � a � 0. By assumption, meets (in particular) the two faces of C, perpendicular to the z-axis. So there exist points P = (p, q, - 1) and U ( u, v , 1), both lying on with coordinates p, q, u , v E [- 1 , 1] . Consequently, .
.
.
·
1r
1r
=
1r
1r
=
1r ,
k = ap + bq - c :::; a + b - c :::; a and k = a u + bv + c � -a - b + c � -a;
141
Geometry
these two inequalities jointly imply a � J k J . So a d
=
>
0, and hence
a a Jk l < < = .ja 2 + b 2 + c 2 - .ja 2 + b 2 + c 2 - .ja 2 + a 2 + a 2
__!.._
v'3
<
1.
Thus if a plane meets all faces of a cube, its distance from 0 , the cube centre, is shorter than the distance of any face of that cube from 0 . Assuming that two opposite faces o f another cube C ' meet all faces o f C , we are led t o the conclusion that C ' has strictly smaller size than C. And since the roles of the two cubes in the problem statement are symmetric, the negative answer results.
Problem 63, Solution
2
Let C, C' be the two cubes. Assume C has edge length 1 and C' has edge length � 1. Choose two opposite faces of C'; visualize them horizontally and call them B and T (base and top) . Denote by 1{. the half-space consisting of all those points that lie below or on the plane of B. Suppose it contains at least two non-adjacent vertices A, B of C. Let M be the midpoint of AB. Clearly, M belongs to 1{. . I f AB i s a space diagonal o f C then M i s the centre o f C, and consequently every point of C lies within distance ! v'a from M . Since the distance between B and T is at least 1, the top face T is disjoint from C. If AB is a face diagonal of C then M is the centre of the corresponding face, whose all points lie therefore within distance � v'2 from M . Since also this number is smaller than 1 , the face in question (of C) cannot reach T. Now assume there are no two non-adj acent vertices of C in 71.. This means that 1{. contains either no vertex or exactly one vertex of C, or exactly two vertices of C, linked by an edge. Among the remaining (8 or 7 or 6) vertices of C one can find four points that span a face of C. As they are situated strictly above the plane of B, that face has no point in common with B. Thus, in any case, C has a face that does not meet either B or T. A pair of cubes with the proposed property does not exist.
Remark An analogous problem might be considered in the four-dimensional space: do there exist two 4-cubes in R.\ each 3-face of one cube meeting each 3-face of the other one? The answer, rather unexpectedly, is yes . Example: let C be the 4-cube whose vertices are the· 16 points (±1, ±1, ±1, ±1). Pick those points that have an even number (four, two o r none) o f co ordinates equal to 1 - there are eight of them - and adjoin to that
Solutions
142
set another eight points, each having one coordinate equal to 2 or -2, and the remaining three coordinates 0. These sixteen points also span a 4-cube C' with the property as needed: if you choose arbitrarily a 3-face of C and a 3-face of C', those two "faces" (3-cubes) will have at least one common vertex! (To verify this, without having to deal with too many cases, can be a nice challenging exercise in itself. ) The olympiad problem , discussed above, has been motivated by this four-dimensional example.
Problem
64
Let A 1 , A 2 , A a , A4 be points on the sphere circumscribed about the regular tetrahedron with edge 1 such that AiAj < 1 for i -:f. j. Prove that these four points lie on one side of a certain great circle of the sphere.
Problem
64,
Solution
1
To say that the points lie on a certain hemisphere is as much as to claim that the tetrahedron spanned by these points does not contain the centre of the sphere in its interior. Thus assume that 0 , the centre of the sphere, lies inside tetrahedron A 1 A2AaA4, which is therefore the union of four pyramids (numbered 1 through 4) , with a common vertex at 0 , the ith pyramid having for its base the face of A 1 A2AaA4 opposite to Ai. Let Vi be the volume of the ith pyramid and let di be its altitude issued from vertex 0 ; without loss of generality assume d 1 � d 2 ;:::: da ;:::: d 4 . Further, denote by hi (i = 1 , 2, 3, 4) the altitude of the "large" tetrahe dron A 1 A 2AaA4 dropped from vertex Ai to the opposite face and let V be the volume of A 1A2AaA4. Thus Vi/ V = d i/ hi (the ratio of volumes of those two pyramids equals the ratio of their altitudes, dropped to their common base) . The foot of altitude d4 of pyramid O A 1 A2Aa coincides with O', the circumcentre of t riangle A 1 A 2 A a . This triangle cannot be obtuse-angled; for if, say, L A a were obtuse, then the points O' and Aa would lie on distinct sides of line A 1 A 2 (within plane A 1A2Aa) , so 0 and O' would lie on distinct sides of plane A 1A2A4, and the distance from 0 to that plane would be smaller than 0 0 1 , in contradiction to da � d4 . Let Vm = min( V1 , V2 , Va , V4) . S ince V = V1 + V2 + Va + V4 , the volume Vm does not exceed � V . Note that h m � d m + R, where R denotes the radius of the sphere. Thus < h m < dm + R d m = Vmhm 4 4 V
whence (1)
Geometry
143
Now, visualize a regular tetrahedron inscribed into the given sphere so that one of its faces ( call it � ) is parallel to plane A 1 A A 3 . By hypoth 2 esis, all edges of � have length 1 . Point 0 , which is simultaneously the circumcentre, orthocentre and centroid of the regular tetrahedron, par titions each of its altitudes in ratio 3 : 1 . Hence, the distance from 0 to � is exactly ! R . Comparing this with inequalities ( 1 ) we see that the plane A 1 A A 3 is less ( or equally ) distant from 0 than ( as ) the plane of 2 �- Consequently the circumradius of triangle A 1 A A 3 , denote it by r, 2 is not smaller than the circumradius of � : (2) In triangle A 1A A3, let LAk be the greatest angle and let i, j be the 2 two remaining indices in { 1 , 2, 3 } . The triangle is not obtuse-angled, and so the size of LAk is comprised between 1r / 3 and 1r /2. As O ' is the circumcentre of A 1 A A3, the angle LAiO' Aj = 2 · L Ak is comprised 2 between 27r / 3 and 1r . Therefore cos ( LAiO'Aj ) ::::; - 1 / 2 and we obtain, by the Law of Cosines and by inequality (2) , (AiAj ) 2
=
(0' Ai)2 + (0' Aj ) 2 - 2 01 A i O' Aj · cos ( L AiO' Aj ) 2r2 (1 - cos ( L A i01Aj ) ) � 3r2 � 1 . ·
·
This is however impossible, according t o the condition o f the problem. Contradiction ends the proof.
Problem
64,
Solution
2
As in Solution 1 , denote by 0 and R the centre and the radius of the given sphere. Let T1T T3T4 be any regular tetrahedron inscribed in that 2 sphere. The conditions of the problem require that (3) Consider the following vectors:
Relations (3) can be translated into the language of inner products: =
....,....--.? ....,.-,-lo AiAj · AiAj (OAj - OAi) · (OAj - OAi)
=
(OAj ) 2 - 2 · 0Ai · OAj + (OAi ) 2 2(R2 - U i Uj ) •
and similarly
so that inequalities (3) become (4) Denote by w the sum
(5) The analogous sum of vectors Vk is the zero vector �
(6) this is j ust a restatement , in terms of vectors, of the fact that point 0 is the gravicentre of the system of equal point masses placed in the vertices of the regular tetrahedron. We now multiply, in the sense of inner product , both sides of equation (5) by vector Ut : u1 w •
=
u1 · ( ut
Similarly, multiplying equation
0
=
(6)
Vt · (vt
Subtracting equation (8) from U t · U t = Vt Vt = R 2 ) we get
+ u2 + ua + u4) .
(7)
by v1 we obtain
+ v2 + va + v4) .
(8)
(7) (and making use of the fact that
•
In view of estimates (4) , the three numbers in parentheses are positive. Therefore u1 · w > 0. The same argument may be repeated with any one of the vectors Uk in place of Ut . Thus Uk · w
>
0 for k = 1 , 2 , 3 , 4 .
(9)
The inner product of two vectors is positive if and only if they form an acute angle. Inequality (9) thus says that all the four vectors OA k are inclined to the vector w under acute angles. If we now draw through 0 the plane orthogonal to w , we get the four points A t , A 2 , Aa , A4 collected on one side of it - and this is exactly what we need.
Problem
64,
Outline Solution
3
Assume, contrary to assertion, that 0 is an interior point of tetrahedron A 1 A 2 A3A4 . The four solid angles OAiA;Ak dissect the sphere into four non-overlapping spherical triangles, of joint area 47r R 2 • Hence, at least one of them has area greater than or equal to 1r R 2 • Since the distance between any two points A i , A; is less than 1, their "angular distance"
Geometry
145
( size of angle AiOAj ) is smaller than the angular distance between any two vertices of a regular tetrahedron ( of edge 1 ) inscribed in the sphere.
The analogous construction involving solid angles, performed with use of a regular tetrahedron, produces a partition of the sphere into four congruent tr-iangular quarters ( spherical triangles ) , of area 1r R 2 each. We have shown that the largest of the spherical triangles AiAjAk has area at least 1r R 2 , whereas its "sides" ( circular arcs ) are strictly shorter than those of a triangular quarter. These two inequalities contradict each other. Intuitively, this is easy to believe; a rigorous proof is less easy!
M arcin
Kuczma
E
graduated with a PhD in
Pure Mathematics
from the University of
u
Warsaw. He is now a Senior
l nstru ctor
in
Pure Mathematics
in
the University's l nstit ute
of M athemati cs.
His research specialty is real analysis. He
has
much
experience
as
a
problem
composer and jury member of the Polish, and
Austri an/Polish
l n tern ational
M athematical Olympia d s and is an author of
a
book
on
the
Austrian -Polish
mathematical Olympiads. He has composed four
problems
Mathematical
in
the
Olympia d
ln tern ati onal
and
had
many
m ore shortlisted. Dr Kuczma is probl em contest editor in the journal Delta, is a frequent contributor to problem columns in several journals and in 1 99 2
was
Hilbert
awarded
Award
contribution
to
the WFNMC David
for
his
si gn i fi ca n t
the
enrich m ent
of
mathematics learning internationally.
Erich Windischbacher has worked as a high school
tea cher
in
Graz, Austria for more than 30 years. He has
(p
also taught at the Karl Franzens University in Graz
and
at
Peda gogical
the
lnstitute
for tea chers. Since 1 9 69 he has been very much engaged with mathematics Austri an
competitions and
M athem atical
Olympia d .
the Prof
Windischbacher co-authored several books e . g.
Os terreichisch e
0/ympiaden Mathematik A U STR A l l A N M ATH E M ATl C S T R U ST
E N R l C H M E N T
S E R l E S
1 9 70- 1 989
Mathema tik and
Wege zur
Anregungen
und
Vertiefungen.
l SBN 1 8 7 6 4 2 0 0 2 2
