THE GREEN BOOK OF MATHEMATICAL PROBLEMS
Kenneth Hardy and Kenneth S. Williams Carleton University, Ottawa
DOVER PUBLI...
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THE GREEN BOOK OF MATHEMATICAL PROBLEMS
Kenneth Hardy and Kenneth S. Williams Carleton University, Ottawa
DOVER PUBLICATIONS, INC. Mineola, New York
Copyright Copy right © 1985 by Kenneth Hardy and Kenneth S. Williams. All rights reserved under Pan American and International Copy right Conventions. Published in Canada by General Publishing Company, Ltd., 30 Les mill Road, Don Mills, Toronto, Ontario. Published in the United Kingdom by Constable and Company, Ltd., 3 The Lanchesters, 162-164 Fulham Palace Road, London W6 9ER.
Bibliographical Note This Dover edition, first published in 1997, is an unabridged and slightly corrected republication of the work first published by Integer
Press, Ottawa, Ontario, Canada in 1985, under the title The Green Book: 100
Practice Problems for UndergradU(lte Mathematics Competitions. Library of Congress Cataloging-in-Publication Data
Hardy, Kenneth. [Green book] The green book of mathematical problems / Kenneth Hardy and Kenneth S. Williams. cm. p. Originally pUblished: The green book. Ottawa, Oot., Canada : Integer Press, 1985. Includes bibliographical references. ISBN 0-486-69573-5 (pbk.) 1 . Mathematics-Problems, exercises, etc. Kenneth S.
I. Williams,
II. Title.
QA43.H268 51O/.76-dc21
1997 96-47817 CIP
Manufactured in the United States of America Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y. 11501
PREFACE
There is a famous set of fairy tale books, each volume of which is designated by the colour of its cover: The Red Book, The Blue Book, The Yellow Book, etc. We are not presenting you with The Green Book of fairy stories, but rather a book of mathematical problems. However, the conceptual idea of all fairy stories, that of mystery, search, and discovery is also found in our Green Book. It got its title simply because in its infancy it was contained and grew between two ordinary green file covers. The book contains 100 problems for undergraduate students training for mathematics competitions, particularly the William Lowell Putnam Mathematical Competition. Along with the problems come useful hints, and in the end Gust like in the fairy tales) the solutions to the problems. Although the book is written especially for students training for competitions, it will also be useful to anyone interested in the posing and solving of challenging mathematical problems at the undergraduate level. Many of the problems were suggested by ideas originating in articles and problems
in mathematical journals such as Crux Mathematicorum, Mathematics Magazine, and the American Mathematical "'1onthly, as well as problems from the Putnam competition itself. Where possible, acknowledgement to known sources is given at the end of the book. We would, of course, be interested in your reaction to The Green Book, and invite comments, alternate solutions, and even corrections. We make no claims that our solutions are the "best possible" solutions, but we trust you will find them elegant enough, and that The Green Book will be a practical tool in the training of young competitors. We wish to thank our publisher, Integer Press; our literary adviser; and our typist, David Conibear, for their invaluable assistance in this project.
Kenneth Hardy and Kenneth S. Williams Ottawa, Canada May, 1985
(iii)
Dedicated to the contestants of the William LoweD Putnam Mathematical Competition
(v)
To Carole with love KSW
(vi)
CONTI!:NTS Page Notation ...... . The Problems The Hints The Solutions. Abbreviations References
.. .. .. .. .. .. .. .. ;0> .. .... ..
............
.. .. .. .. ..
..
.... ..
.. .......... .... .. .. ........
.. ..
..
.. .. ......
..
..
..
..
.. ..
.... .. .. .. ..
..
..
.. .. ..
..
.. .. .. .. .. .. .. .. ...... ..
..
.. .. .. .. ............ ..
.. ..
.. ..
.. ..
..
.. .. .. .... .. .. ..
.1 . 25 .41 169 171
.. .. ......................
............ .. .. .. .. ............ .. .. .. ...... .. ........ .. .. ................ .. .. .. ................ .. .... .. .. .. .. .. .. .. .. .. .... .. ..
.. ..
..
.. .. .. ..
IX •
..
..
....
..
..
..
"
.. .. .. .. .. .. ..
.. .. .. ..
.. .. .. .. .. ..
..
..
..
.. ..
.. .. .. .. ..
.. .. .. ..
.. ..
..
.. ..
..
..
....
.. ..
....
.. .... ..
.. .. ..
..
..
.... ..
.. .. .. .. ...... .. .. .. .. ..
.. .. .. ..
.. ........ ..
(vii)
.. ....
..
..
.. .. ..
.. .. .. ..
..
..
..
..
..
.... .. ..
.. .. .. ..
..
..
..
................
........ .. ..
.. .. .... .. .. .... ........ ..
..
....
.. .. ........
.. ..
.. .. .. ..
.. .. ..
.. .. .. ..
..
..
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..
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....
..
NOTATION
�
[xl
denotes the greatest integer x is a real number.
{x}
denotes the fractional part of the real number x, that is, {x} - x - [xl.
In
x
x, where
denotes the natural logarithm of x.
exp x
denotes the exponential function of x.
�(nl
denotes Euler's totient function defined for any natural number n.
GCO(a,bl
denotes the greatest common divisor of the integers b. a and
n k
denotes the binomial coefficient nl/k!(n-kl!, are non-negative integers where n and k (the symbol having value zero when n < k l. denotes a matrix with entry.
det A
as the (i,j)th a ij
denotes the determinant of a square matrix A.
(Ix)
THE PROBT,EMS Problems, problems, problems all day long. Will my problems work out right or wrong? The Everly Brothers
{b : n = 0,1,2, } n numbers, prove that the series
1.
If
.
.
.
is a sequence of non negative real
b
(1.0)
L
n
n=O converges for every positive real number a.
2.
Let a,b,c,d be positive real numbers, and let a(a+b)(a+2b)...(a+(n-1)b) . 11 (a,b,c,d) = c(c+d)(c+2d) (c+(n-1)d)
o
.
Evaluate the limit
3. (3.0)
•
.
L = lim Q (a,b,c,d). n n+a>
Prove the following inequality:
in
x
,
3 x -1
1
x
>
0,
x
..
1.
2
PROBLEMS (4-12)
Do there exist non-constant polynomials
4.
variable
such that
z
Ip (z)1
p (z) is monic and of degree
5.
Let
f (x)
a
For
E >
°
R> °
R ,where
be a continuous function on [O,a], where
f (x) :-:- '( '-: � -;:f'x( -') + f a
o
and
[O,a] .
a
>
0,
Evaluate the
x . d ' )
evaluate the limit x+1 2 sin (t )dt
lim
7.
=
in the complex
n?
integral
•
Izl
on
f (x) + f (a-x) does not vanish on
such that
6
Rn
2k .
recursively by
(8. 0)
X
n
=
x
n-1
Prove that lim n .... ex> exists and determine its value.
x
n
,til
+
k
x
n-1
,
n
=
1,2,3, .. .
•
PROBLEMS
9. b
(4-12)
3
denote a fixed non-negative number. and let
Let
be positive numbers satisfying vb
Define
x n
a
2vb .
>
ac
•
Evaluate 2 2 max (ax + 2bxy + cy ) . x.y e; R x2 +y2 =1
11.
Evaluate the sum [n/2]
(1 1.0)
for
n
12. ( 12.0)
..
s
I
n(n-l)
r=O
.
(n-(2r-l» 2 (r! ) •
•
n-2r 2
a posltlve lnteger. "
.
Prove that for
m = 0. 1.2 •. .
S (n)
is a polynomial in
m
n(n+l).
•
2m+l + ... + n
•
a
and
4
13.
PROBLEMS
Let
be positive integers such that
a,b,c
GCD(a,b) Show that
=
GCD(b,c) .. GCD(c,a)
l .. 2abc - (bc+ca+ab)
is insolvable in non negative integers
Determine a function
=
1 .
is the largest integer such that
bc x + ca y + ab z
14.
(13-21)
..
l
x,y,z.
such that the n
f(n)
th
term of the
sequence
(14.0)
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, . . .
is given by
15. zero .
[f(n)] .
be given real numbers, which are not all a' ' a , , a l 2 n Determine the least value of Let
•
.
•
•
where
x ' l
. . . •
x n
•
•
2 + x n
,
are real nnmbers satisfying + ... + a x .. 1 n n
16.
Evaluate the infinite series S ..
17.
•
+ ...
•
F(x) is a differentiable function such that F'(a-x)" F'(x) a for all x satisfying 0 � x � a. Evaluate f F(x)dx and give an o example of such a function F(x).
PROBLEMS
18.
5
(13-21)
(a) Let
r,s,t,u
be the roots of the quartic equation
4 2 x + Ax3 + Bx + Cx + D .
Prove that if (b) Let
rs
=
tu
then
a,b,c,d
2 A D
2
a
=
.
2 c .
be the roots of the quartic equation 2 py + qy
Use (a) to determine the cubic equation (in terms of p , q,r) whose roots are ac - bd a + c - b - d '
ab - cd a + b c - d ' -
19.
where
Let
n
p(x)
ad - bc a + d - b - c '
be a monic polynomial of degree
D
is a non negative integer and
=:
d dx
m � 1 , and set
denotes differentia-
tion with respect to x. f (x) is a pOlynomial in x of degree n UiIt-n . l.n f (x) Determine the ratio of the coefficient of x n constant term in f (x) n Prove that
( lUU
to the
•
20.
Determine the real function of
x
whose power series is
15 3 9 x x x + + ;:';:-:'" + .. . 9! 15 ! 3!
21. (21.0)
•
Determine the value of the integral I
1T
n
=
a
Sl.n nx i Sl.n x •
for all positive integral values of
n .
dx
,
- n).
6
22.
PROBLEMS (22-31)
During the year 1985, a convenience store, which was open
7 days a week, sold at least one book each day, and a total of 600 books over the entire year.
Must there have been a period of consec
utive days when exactly 129 books were sold?
23. that as
Find a polynomial m
with rational coefficients such
run through all positive integral values,
n
and
f(x,y)
f(m,n)
takes on all positive integral values once and once only.
24.
Let
m
be a positive squarefree integer. Give a condition involving
positive integers.
tees that there do not exist rational numbers
Let
R,S,m
R,S
be
which guaran
x,y,z and w
such that
(24.0)
25.
Let
k
and
h
1 � k
be integers with
3k - 5k , for k 2,3,... n 35.
•
n
-
n(k)
•
be a sequence of real ntmbers {p : n " 1,2,3, } n p � 1 for n " 1,2,3, Does the series n [pn ]-1 •
•
•
•
.
•
.
00
L
(35.0)
n=l
converge? 36. degrees
Let
f(x) , g(x)
n+1 , n
be polynomials with real coefficients of
respectively, where
n
�
0
,
and with positive
9
(32-40)
PROBLEMS
A , B
leading coefficients
x L
in terms of
37. where
A, B and
=
Iim x .... '"
n
g(x)
e
f(t) -f(x)
•
The lengths of two altitudes of a triangle are h" k .
and
k ,
DeterUline upper and lower bounds for the length h
and
k .
n,r
= P
n,r
(x)
(
=
n+l 1_x
n+2 n+r ) ) (1_x )(1_x r 2 (1 - x)(1-x ) (1-x )
of degree
is a polynomial in
x
negative integers.
(When P
and we have
1
Let
r = D
n,D
A, B, C, D, E
Prove that the ntmber
N
= 1
•
.
.
39.
h
Prove that P
to be
dt ,
D
of the third altitude in terUlS of
38.
Evaluate
respectively.
.
nr , where
.
•
n
and
r
are non
the empty product is understood for all
n
�
D .)
be integers such that
of pairs of integers
B" D
(x,y)
and
such that
2 Ax + Bxy + Cx + Dy + E = D ,
(39. D) satisfies
N "
where, for integers divisors of
n
1 , d(n)
Evaluate
L
k=l
,
denotes the number of positive
n . n
40.
�
2d( I F I)
k •
10
41 m
PROBLEMS
denote the sum of all possible products of
p =P (n) m m
Let
•
different integers chosen from the set
formulae for
42.
For
P (n) 2
a
>
and
.
Find
b > 0 , evaluate the integral ax bx - e e � --'::=-dx . - a x-- b x(e +1) +1) (e
o
For integers
•
{1,2,...,n}
P (n) . 3
(42.0)
43
(41-50)
n � 1 , determine the sum of
n
terUlS of the
series 2n
2n(2n-2) (2n-4) �� L�:l!::::'�� + + ... + - --'=.t.� _ "_ 2n-l 3 )(2n -S ) (2n- 1)(2n -3) 2n l )
(43.0)
44. be
2n(2n-2)
Let
k (�l)
m
be a fixed positive integer and let
45.
s=m,m+l,m+2,
•
) - (a A n ij
•
•
,m+k-1.
be the
a ij
where
46. tions
o
... +
Let
•
•
.
,z
k
complex numbers such that
(44.0) for all
z ,z , l 2
•
x > 2 .
Evaluate
Must
nXn
,
z =0 1
•
•
•
,k?
matrix where
x ,
if
i=j ,
1 ,
if
I i-j I =1
0 ,
otherwise,
D =det A n n
for i=1,2,
,
•
Deterilline a necessary and sufficient condition for the equa
PROBLEMS
(41-50)
11
x
+
y
+
z
=
A ,
2 + x
2 2 + z Y = 3 + 3 3 x y + z =
(46.0)
B , C
to have a solution with at least one of
47
•
Let
S
be a set of
n 0 -1 , where 1,2,3, ... ,1
(4 7.0)
n
0
prove that
,
(b .
l
o
1
x
21
Evaluate � +
m
78.
a+b 2a+b + + m m
Let a 1.···.an be 2 S = a + ... + 1
n (>1) 2 an
•
.
•
•
(m-1)a+b + m
distinct real numbers. = min (a lSi<jsn i
M
Prove that .§. M
> -
n (n-1) (n+1) 12
Let x1,···,x be n real n n L I�I = 1 , k=l Prove that n� S (79.0) L k k=l 79
•
•
•
numbers such that n
t
� k 1 2
-
1 2n
o
-
•
•
•
Set
PROBLEMS (75-84)
80
19
Prove that the sum of two consecutive odd primes is the
,
product of at least three (possibly repeated) prime factors.
81.
Let·
be an integrable function on the closed interval
f(x)
[TI/2,TIJ and suppose that TI f(x) sin kx dx
0
-
1 ,
TI/2
i f(x) i
Prove that
82,
n
For
(82.0) where
=
s
a
n
> -
0, 1
n
,
2,
n
n+1
.
-
n
.
•
•
on a set of posl.tl.ve measure. •
, let
+
1
+
n- 2
+ .. ,
+
exists and determine its
lim s n '" n-+
Show that
•
k
•
bl 2
11
=
6n+1 =
1
1 S k S n-1 ,
,
value.
83,
Let
be a non-negative strictly increasing function on
f(x)
a
the interval [a,bJ, where the curve so that Let
y
A(a)
for all
84 .
.
be a function such that (x'
a s x
Let
Let
a
denote the area below
A(x)
0 , let
N
00
s; 2e:
•
Prove that
L
L =
be a positive
k +1 (_l) � satisfies
k=1
the inequality
IL
(85.0)
86.
(85-94)
< e:
-
•
DeteImine all positive continuous functions
defined
f(x)
on the interval CO,n] for which
n f(x) cos
(86.0)
nx
n = (-1) (2n+1) ,
dx
n = 0,1,2,3,4
•
o
87.
P
Let
pI
and
circular ellipse
E
be points on opposite sides of a non
such that the tangents to
E
through
P
and
pI
respectively are parallel and such that the tangents and nOImals
to
E
.
at
P
and
pI
DeteImine the equation of
E
of
maximum area.
with respect to a rectangular coordin
ate system, with origin at the centre of
E
and whose y-axis is
R .
parallel to the longer side of
88.
R
deteImine a rectangle
If four distinct points lie in the plane such that any three
of them can be covered by, a disk of unit radius, prove that all four points may be covered by a disk of unit radius.
89
•
Evaluate the sum
S =
00
00
L
L
m=1 n=1 m nP'n
1 2
- n
2
•
(85-94)
PROBLEMS
90
If
•
21
is a positive integer which can be expressed in the
n
a,b,c
n =
form
k,
that, for each positive integer 2
form
+
A
91.
B2
Let
relations
92.
2 , where C
+
3 aba = b
satisfying
can b e expressed in the
are positive integers.
A,B,C
b5 = 1 .
and
{a : n = 1,2,3, n 0
n
2k
be the group generated by
G
Let
are positive integers, prove
a n
0 ,
suggested by the inequality (3.0), is increasing. Apply Rouche's theorem to the polynomials
4.
Rouch€'s theorem anaLytic within and on a simpLe Ig(z)1 < If(z) I on C , where and fez) + g(z) have the same
g(z)
5.
=
p(z).
and
x
=
a - t
to
a
-
o
6.
=
states that if fez) and g(z) are cLosed contour C and satisfy fez) does not vanish, then fez) number of zeros inside c .
Apply the change of variable I
f(z)
n -z
f(x) dx f(x) + f(a- x)
•
Integrate x+1 I
..
x
by parts and obtain an upper bound for
II I .
7.
Consider (7.0) modulo 16.
8.
By squaring (8.0), obtain the lower bound
;'zk+ n
Using this bound in (8.0) obtain an upper bound for
x n
for •
x. n
HINTS
9.
(3-14)
27
Assume that the required limit exists, and use (9.0) to det
ermine its value 10.
Either set
function of 2
L.
6 ,
x
2
A(x + y ) - (Bx +Cy) 11.
y " sin 6
cos 6 ,
=
and maximize the resulting
2
2
ax +2bxy +cy
or express
2
Ixn - LI .
Again use (9.0)to estimate
in the form A,B,C .
for appropriate constants
Consider the coefficient of
n x
in both sides of the identity
-
12.
Express
•
k k (l.(l.+l» - (Cl. - Ol.) ,
(k
1, 2 , polynomial'in l. , then sum over l. k (n(n +1 » as a linear combination of =
.
•
•
=
1,2,3 , .. )
, n
.
as a
to obtain
Complete the argument using induction. 13.
Prove that the equat�on •
bc x+ca y +ab z
=
2abc - (bc +ca+ ab)+k
is solvable in non-negative integers
x,y,z
Then show that the equation with k " 0 �ntegers •
for every integer
k � 1.
is insolvable in non-negative
x,y,z,
th term of the sequence (14.0) and show 1 4 . Let un be the n (kl) k k-l , 0, 1, 2 , +1 +l., l. that u - k for n 2 n 1 and deduce that k S; 2(1 + /8n-7) < k + 1 =
=
•
•
.
"
HINtS (15-27)
28
15.
Use Cauchy's inequality to prove that n
L and choose the
16.
,
1
so that equality holds.
x.
1
Express
2 a.
-1
(n +
1) 3
in the form
An(n - l)(n - 2) + Bn(n - 1) + Cn + D for suitable constants
A,B,C,D .
17.
Integrate
1 8.
For part (b), find the quartic equation whose roots are
a -
z,
b
-
z,
c
F' (a-x) - F' (x)
-
z,
d -
tWlce. •
and use part (a) to ensure that the
z,
product of two of these roots is equal to the product of the other two.
19.
Differentiate
f (x) n
to obtain the difference-differential
equatlon •
•
20.
Consider
21.
Show that
f'(x) - p'(x) f (x) n n
sinh x + sinh wx
I - I n-l n
-
'If
o
•
2 + sinh w x , where
sin(2n i Sln x
1) x ··dx
,
1 w - -(-1 + 1-3) 2
n i: 2
•
,
and then use a similar idea to evaluate the integral on the right side.
HINTS (15-27)
29
a. (i 1, 2, , 365) denote the number of books sold � th day inclusive. during the period from the first day to the i
22. Let
=
"
'
Apply Dirichlet's box principle to
23.
Show that a polynomial of the required type is (x + y -
f(x,y) by showing that
1)(x + y - 2) + x , 2 '
f(x,y) · k , where
k
unique solution in positive integers in terms of the integers
r
1} r (r 2.:.. --= �=----=..::": ..:. _
of the fotUl
is a prime of the form
Then, assuming a � n , 2 2 obtain a contradiction b y considering the factor a +q of b =
0
and
41
a ;;: a 1 > 0 . nn
SOLUTIONS (1-2)
42
n � 1 , we may deduce that
Now, for
b
n
( a +c. O+b 1+ ' . . +bn
=
3 /2 )
'"
a - a n n-1
/2 3 a
1
n
a n-1
-
1 /2 a
/2 3 a
n
=
n
a 1 n-
1
1 /2 a
1
-
n
=
a 1 n-
1
1/2 a
-
1/2 a n-1
m �
1
n
1 /2 a
n
Hence , for
1
1 /2 a
1 2 / 1 a
-
m
1/2 a
-
1/2 a n-1
1 1 /2 a n
1
-
1 /2 a n
•
n
we have
m s
1
1
n-1
n
n- 1
,
. /2 1 a
1/2 a n- 1
1 +
1/2 a 2
+
1 /2 a n- 1
n
1
m
=
+ n=O
2 L n=l
b =
O
3 72 a
+
O.
are bounded , the infinite series
43
SOLUTIONS (1 -2)
2.
Let a , b , c , d be positive real numbers , and let (a+(n-l)b) a(a+b ) (a+2b ) Qn (a " b c , d) - c(c+d)(c+2d) . . . (c+(n- l)d) . .
_
.
•
Eval uate the limit L = lim � (a ,b ,c ,d) . n + OO Solution : Considering the five cases specified in THE HINTS , we show that +00 , in cases (a) , (b ) , L = o , in cases (c ) , (d) , 1 , in case (e) . c We set k = d + 1 , so that c < kd •
(a) When b
>
d , as c + j d
(j + k)d , we have for n n-1 a(n-l ) ! b Q (a,b ,c , d ) � k(k+I) . . . (k+(n-l) drl n n-1 ! b a(k-l) �!::. !. .:. .. ..:....:. �-= n (k+n-l) (k+n-2) n d
c we have
n- l a + Q,b Qn ( a ,b , c , d). = II c + Q,b Q,=O n-l a c = II 1 + c + )',b Q, = O n-l >
II )',=0
>
n-l a - c L b x'=0
Q.
1 +k
,
co.
SOLUTIONS (3-4)
44
which tends to infinity as n tends to infinity since the harmonic series diverges, showing that L = + co
•
(c) (d) By considering the reciprocal of Qn (a ,b ,c ,d) , we obtain from the above results L 0 , if b < d , or if b d and a < c =
-
•
(e) Clearly � (a,b,c ,d) '" 1 in this case , so that
L =
1.
3. Prove the following inequality :
.in x
(3 . 0)
3 x -1
Solution : For x
>
0, x
;I!
1.
0 we define
3 (x -1) (x+l) F(x) '" 3.in x , 3 (x +x) so that - 3 .in x
F(x) '"
•
Differentiating F (x) with respect to x , we obtain 2 3 2 3 3 4 ( -x-l) (3x +3x -l) (x +x) (x +x 4x +1. ) 3 F' (x) '" ' 2 3 x (x +x) -
x that is
6 5 4 3 2 x +3x -2x +3x -3x+1 -3x (3 . 1 ) 3 2 (x +x) The polynomial p (x) in the numerator on the right in (3 . 1 ) has the 1 6 -3 property that p (x) '" x p (-x ) , and so x p (x) can be written as a cubic polynomial in x + l/x . •
SOLUTIONS (3-4)
We have
As
45
F' (x )
�
3 3 x !.) (x+ 2 x +x 3 (x )
_
3 2 X -3 X+4 " (x..-bl) ( X-2 ) we obtain
2 2 t1 ) 1 +x+ (x x ) 2 ( F' (x) 2 2 x + (x l ) 2 2 x ( t �) + i) (x-I ) 2 2 x 2(x +1 ) ..
•
so that F'(x ) > 0 for x > 0 , while F'(l) .. O . Thus F(x) 18 a strictly increasing funct ion of x for all x > 0 . Hence 1n part1cular we have F(x) > F ( l ) for x > 1 , and so .lnx > for x 1 , (3 . 2 ) 3 x -1 1 Rep lacing x by x in (3. 2 ), we obtain .in x for o < x < 1 , (3 . 3 ) 3 x -1 •
•
•
,
•
•
Inequalities (3 . 2 ) and ( 3 . 3 ) give the required inequality . 4. Do there exist non-constant polynomials p ( z ) in the complex n variable z such that I p (z) 1 < R on I zl - R , where R > 0 and p(z) is monic and of degree n? Solution : We show that no such polynomial p (z) exists , for suppose there exists a non-constant polynomial + an- l such that
I p (z ) 1
0 , o b + YO + bu - bv > 0 ,
Zo + cv - ct
> 0
•
Set x = a-I + x + at - au , o y '" b-l + Y O + bu - bv
,
Z '"
so that
x,y,z
+ Z
-1
o
+ cv - ct ,
are non-negatlve lntegers . •
•
Moreover bc x + ca y + ab z =
2abc - (ab + bc + ca) + k
as required . Proof of ( b ) :
Suppose the equation is solvab l e , then 2abc = bc (x+l ) + ca(y+l) + ab ( z+l) ,
where
x+l , y+l , z+l
are poslt lve lntegers . •
•
•
such
SOLUTIONS (13-14)
Clear ly , as divide s
59
GCD ( a , b )
x+l , b
GCD (b , c )
=
divides
are posit ive integers x+l
=
=
GCD ( c , a)
y+l , and
r,s,t
c
, we have that
divides
z+l .
a
Thus there
such that
y+1 = bs ,
ar ,
1
=
z+l· ct .
Hence we have 2abc = abc ( r + s + t ) , that is 2
=
r + s + t >. 3 ,
which is impos s ib le. This comp l e tes the solution .
14.
Determine a function
f (n)
such that the n
th
term of the
sequence (14.0)
1 , 2 , 2 , 3 , 3 , 3 , 4 , 4, 4 , 4. 5, . . .
is given by Solut ion :
[fen) 1 .
Let
u
n lnteger •
be the n
th
term of the sequence ( 1 4 . 0) .
f ir s t occurs in the sequence when
k
n = 1 + 2 + 3 + . . . + (k-1) + 1 Hence ( 14 . 1 )
u = k n
=
(k-l ) k + 1 2
for
n =
;
(k- ) k
+ 1 + t.
t = 0 , 1 .2
• • . . •
From (14 . 1 ) we ob tain (kl ) k o :0 n 2
_
1 :0 k 1 -
and s o ( 1 4 . 2)
2 k - k + 2 :0 n 2
•
k-l .
•
The
60
SOLUTIONS (15-16)
Mul t i p l y ing (14 . 2) by S and comp let ing the square , we have (2k-1) (2k-1)
2
2
� Sn - 7
�
�
�
(2k+1 )
( 2k+1)
2
- 1 ,
0 , we have
=
I
•
I u = - , so that v
I , we set
I(u) ..
I uI
0
and
B>
0
0
, and with posit ive
Evaluate
f ( t) - f (x)
A , B and n . As
,
, we have
dx
,
SOLUTIONS (35-36)
87
li m x + oo and li m x + oo
More ove r
x f ( t) e d t .. + 00 f (x) e
lim
•
) x f( d e dx g(x)
x + co
..
- + 00
2 (x) li m f ' (x) g�x) - g ' (x) x + oo
2 2n B x + = li m n x + oo AB( n+l)i +
. • .
..
so,
by
B ( n+OA
• • •
'
L 'H8pi ta 1's ru le , we have L ..
=
..
lim
li m x + oo
x d e f ( t)d t dx Uo d e f x !g (x) dx
B (n+l)A
•
SOLUTIONS (37-38)
88
The lengths of two al titudes of a trian gle a re h and k , whe re h � k . De te rmine uppe r and lowe r b ounds f or the length of the third altitude in ter ms of h and k . jJ,
Soluti on: We show tha t hk < l < hk h+k J
(37 . 1 )
•
Le t the poin ts P, Q, R be chosen on the sides BC, CA, AB ( possibly extended ) re spec tively of the trian gle ABC s o tha t AP,BQ, CR a re the alti tudes of the trian gle . Se t a IBCI, b = I CAI , c = I ABI, h = IAPI , k IBQ I l ICRI . Clea rly ' ah bk = c l , s o tha t b l a =l c h' c k · =
=
=
=
-
-
:;:
-
Wi thout loss of gene ral i ty we may s uppose tha t h < k . F rom the ineq uality a b > 0 , evaluate the integral co
(42 . 0)
ax bx e ::" -=e:"' --': - ax-- b'-x- dx o x ( e +1 ) (e + 1 )
•
•
SOLUTIONS (42-43)
94
Solut ion :
C , the func tion
For any constant
f (x )
=
x e --'--- + C , eX + 1
X ::: 0 ,
i s such that f ( ax) - f (bx) t > 0
For
we have t
J ( t)
=
ax bx e - e -------7--- , ax bx ( e +0 ( e +l )
=
o t -
o
t
ax
bx e - e ---=---.::,-- dx bx ax x ( e +0 (e +1)
t
f ( ax) dx x
f ( ax) - f (bx) dx x
= o
f (bx) dx , x
o
provided b o th the l a t ter integral s eXl. s t , f ey) ' eXl. s t s , which holds if and only This l.S guaranteed if I l.m y -+ o y ,
,
•
if
C
=
-
1
-
2
With the cho ice
,
J(t)
at
=
o
f ey) dy y
-
t l im f (x) x -+ OO
real number
If
=
Xo
1:. 2
=
so that g iven any x ( e: ) O
such that
x > x o
.
t > x /b , then O
>
=
1
we have 2 '
-
bt
f e y) dy .::.-'.-"c y
at
Now
C
-
o
f ey) dy y
�.:...
,
e: > 0
there exi s t s a po s i t ive
1 1 - e: < f (x) < ' + e: , 2 2
-
at > bt > X
o
' and so we have
95
SOLUTIONS (42-43)
1 ( 2
-
- e:)
a t (!. 2 ...c::..
a bt - b
_
e:)
y
_ _
t
dy
< J(t) at (.!. + e:) 2 --=--::dy y t
2 . Evaluate D n
=
det An
•
Solution : Expanding Dn by the first row, we obtain Dn x Dn-l - Dn- 2 ' so that =
(45 . 1 )
Dn - x Dn- 1
+
Dn- 2 " 0
•
The auxiliary equation for this difference equation is 2 t - xt
+
1 = 0 ,
which has the distinct real solutions ,
as x > 2 . Thus the solution of the difference equation (45 . 1 ) is given by for some constants A and B . We now set for convenience a so that , as
=
t(x + Ix2 -4)
t(x + Ix2 -4) t(x •
1 a
-
=
Ix2 -4)
=
, 1
1-(x - IX2 _4 ) , 2
,
SOLUTIONS (45-46)
which give s Now
99
n -n Dn .. Aa + Ba , n D1
and
so that (45 . 2 )
•
x
•
•
1,2,3, . . . .
1 B a + -a .. Aa + -a
2 2 1 1 - 1 = a + -a - 1 = a + 2 + 1 a
a
,
4 4 2 a A + B .. a + a + 1
•
Solv in g (45 . 2 ) for A and B y ield s 2 a 1 :... . . -= A .. ....,2=, B " ...". 2 ' a - 1 a - 1' _ _
so that
tha t is
46 .
t�on s
2 n a Dn .. 2 a a -1 2n+2 a 1 Dn .. n 2 n " 1, 2 , 3 , a (a -1 ) '
• • •
•
Determ ine a nece ssary and suff icient cond it ion for the equa-
•
(46 . 0 )
x + y + 2 x +l+ 3 3 x + y +
z A , 2 z .. B , 3 C , z •
..
to have a solut ion with at least one of x, y,z equal to zero .
SOLUTIONS (47-49)
100 •
So lut ion : Let x , y , z be a solution of (46 . 0) . Then, from the ident ity 2 2 + 2 (xy+yz+zx) , (x+y+z) = x and (46 . 0) , we deduce that xy + yz + zx
(46 . 1 )
=
.!. (A2 - B) 2
•
Next , from the identity
we obtain us ing (46 . 1 ) 3- AB 2
C - 3xyz so that (46 . 2)
Hence a solut ion (x, y , z) of (46 . 0) has at least one of x , y , z zero if and only if xyz = 0 , that is, by (46 . 2) , if and only if the 3 condition A - 3AB + 2C = 0 holds . 47 .
Let S be a set of k distinct integers chosen from n 1 , 2 , 3 , , 10 -1 , where n �s a pos�t�ve �nteger. Prove that if •
.
(47 . 0)
•
•
•
•
.
n < .en
(k+O 2
.en
10 ,
it is possible to find 2 disjoint subsets of S whose members have the same sum. n Solution : The integers in S are a l l :> 10 -l . Hence the sum of the integers in any subset of S �s •
SOLUTIONS
(47-49)
101
The number of non-empty subsets of
S
is
k
2 _l .
From
(47 . 0)
we have
n 1 k 2 - 1 > k · l 0 - Z k ( k+1) , and so , by Dir ichlet ' s box princip le , there mus t ex ist at least two which have the same sum. ' 2 l are d i s j o int the problem is so lved .
diff erent sub s e t s of If
S
l
and
S
S , say
S
and
S
2 If not , remova l of the common elements from S' I
two new sub s e t s 48 .
Let
n
and
S
and
S
2 l with the required property .
S2
I s it po s s ib l e for
be a positive integer .
yie lds
6n
distinct straight l ines in the Eucl idean plane to b e situated so as 2 to have at l east 6n -3n point s where exactly three of these l ines intersect and at least
6n+l
points where exactly two of thes e
l ines inter sect? So lut ion :
Any two distinct l ines in the po �nt . •
l ine s .
There are altogether
mo st one '" 3n ( 6n- 1 )
A triple intersection account s for
pairs of
of the s e pairs of
l ine s , and a s imp le intersection accounts for one pair .
As
2
( 6n -3n) 3 + (6n+I ) 1 2 .. 1 8n - 3n + 1 > 3n (6n - 1 ) the conf iguration is impo s s ib l e . 49 .
Let
S
be a set with
n ( �1 )
exp l icit formu la for the number cardina lity is a mul t i p l e of
3 .
A (n)
e l ement s .
Determine an
of subsets of
S
whose
SOLUTIONS (49-50)
102
Solution :
� 3
, l = 0 , 1 , 2 , . . . , [ n/3 ]
Let
[ n /3 ]
L
A (n) =
(49 . 1 ) .!. w = 2 (-1
+
i/"3)
contalnlng
S
The number of subsets of
•
3�
l=O
Thu s , we have n
n k
L
=
•
k=O k==O (mod 3 )
so that 3
'TN
and , for
3l e l ements i s
•
•
= 1 ,
r = 0 , 1 , 2 , define
Then , by the binomial theorem, we have .
(49 . 2 )
=
( 1 + w)
(49 . 3 )
n
=
n
L
k=O n
L
k=O n
L
(4 9 . 4 )
k=O
n k
= S
+ o
n k w = S o k n 2k W = So k
Adding (49 . 2 ) , (49 . 3 ) , (49 . 4) , we obtain , as
1 +
1.1
2
+ 1.1
= 0 ,
so that n 2 n 1 n A (n) = - (2 + ( -1.1 ) + (-1.1) ) 3
•
Hence we have A(n) =
i (2
n
n
+ 2(-l) ) ,
.!. ( 2n 3
(_ l )
n
) ,
if
n == 0
(mod 3) ,
if
n 1. 0
(mod 3) .
SOLUTIONS (49-50)
103
For each integer n � 1 , prove that there is a polynomial pn (x) with integral coefficients such that 2 ( 1+x ) pn (x) Define the rational number an by 50 .
•
n-l 1 _ ( l) an = n- 1 Pn (x) dx , n = l , 2 , . . . 0 4
(50 . 0)
.
Prove that an satisfies the inequal ity , n = 1,2, . . .
•
Solut ion (due to L . Smith) : Let Z denote the domain of rat ional integers and Z [ i] = {a+b i : a , b £ Z} the domain of gaussian integers . For n = 1 , 2 , 3 , set .
.
•
(50 . 1 ) so qn (x) £ Z [x] . As qn (±i) = 0 , qn (x) is divisible by x+i and 2 x-i in Z [ i ] [x] , and so pn (x) = qn (x) /Cx +1) £ Z[ i] [x] . However pn (x) £ R[x] , and as R[x] n Z [ i ] [x] Z[x] , we have pn (x) £ Z [ x ] . This proves the first part of the question. For the second part , we note that m
, so that 1 dx
Now, as
1 dx 1T 4 = o 1+X 2
'
=
n n pn (x) dx + (-1) 4
we have us ing (50 . 0)
1
SOLUTIONS (51-52)
104
1
I'IT - an I = n-l 4
dx . o
Now 1
' 4n 4 as x ( l-x)
< -
2:. , and thus we have
4
I'IT
-
an I
0 ( 53 . 1)
b s nx � dx - l im x. y + 0 b s �n x dx x
s �n x dx x •
•
�
b �xy s � x !l dx e x •
s x �n max x O:oxlOb �xy b e M(b) �y 0 �by ) � e ( 1 - M(b) y :0 M(b) y :0
•
,
b � e XYdx
•
•
Let ting y + a > n .
be n ( � l ) integers such that a ,a " " ,a n l Z < an , o < a l < aZ < a . ( 1 � j < i :ii n) are di stinct , a l l the differences a . 1. J a . - a (mod b ) ( 1 � i � n) , where a and b are po s l. t 1.ve 1. 1.ntegers such that 1 � a � b-l
Let
•
•
•
-
•
-
•
-
•
•
Prove that n
L ar
r=l Soluti'on :
Let
r
be an integer such that Z � r � n For r I I � j < i � r there are = - r (r- l ) d i s t inct diffZ Z erences a . - a . , and thes e are a l l divisible by b . Thus the l ar1. J gest difference among these , namely a - a ' must be at l east r l r (r- l ) , that is •
!
and so a
r
�
Z � r � n .
SOLUTIONS (56-57)
111
As a 1 a (mod b) there 1S an integer t such that a 1 " a + bt 1 then a 1 " a + bt :i a - b < -1 , which loS impossible as If t ;;; Hence we have t Ii: 0 and so a l Ii: a glov1ng al > 0 -
•
•
•
-
-
•
,
ar
r
:0
n .
The inequality (56 . 1 ) clearly holds for r ·
1
.
(56 . I )
n
I ar r=l
�
2
an
:0
•
•
Thus we have
n b2" dr-I) .. an I r"l
+
,
so that n
z: a r r=1 57 .
Let An
-
-
-'!r
O
Let
.
•
Around every point in the xy-plane with integral E
co-ordina t e s draw a circle of radius
.
Prove that every straight
l ine through the origin mu s t intersect an inf inity of these circles .
Solution :
Let b
s a t i s fy ing
If be a l ine through the or1.g1.n with slope b k are 1.ntegers , where k and 1.S rat iona l , say b L
•
•
•
•
•
.t
>
-
1
and
- 1 (k , .t) -
= I ,
then
passes through the
L
centres of the inf init y of circles a l l o f radius l a t t ice p o in t s
(.tt , kt)
, wh ere
t
E
1.S an 1.nteger . •
•
,
centred a t
SOLUTIONS (62)
120
If b is irrational , then by Hurwitz ' s theorem there are infini tely many pairs of integers (m, n) with n � 0 and GCD (m,n) = 1 for which m < 1 -n - b 2 n •
Choosing only those pairs
(m,n) for which ,
we see that there are infinitely many pairs (m ,n) for which m -n - b
,
n
that is , for which
> _
�
1 ,
1 then ( 7 1 . 1 ) clearly hold s ,
1W+1 1 + I wl
1
>
•
Let z l , z 2 be the roots of the given quadratic chosen so that As a and b are non-zero , z 2 ;1O 0 and z l+z 2 ;1O 0 , I Zl l � I Z 2 1 and setting w z l /z 2 (;10-1) in ( 7 1 . 1 ) we obtain I zl z2 1 + I z +z 1 1 2 •
=
•
The inequality ( 7 1 . 0) follows as
-
b and a
-
=
c a
-
•
135
SOLUTIONS (71-73)
Determine a monic polynomial f (x) with integral coefficients such that f (x) = 0 (mod p) is solvab le for every prime p but f(x) = 0 is not solvable with x an �nteger . 72 .
•
2 Solution : If p 2 or p = 1 (mod 4) the congruence x + 1 = 0 (mod p) � s solvable . If p 3 (mod 8) the congruence 2 x + 2 0 (mod p) � s solvable . If p 7 (mod 8) the congruence 2 x - 2 0 (mod p) � s solvable . Set �
•
-
-
•
-
-
-
-
-
-
•
-
Clearly f (x) is a monic polynomial with integral coefficient s such that f (x) = 0 is not solvable with x an �nteger . •
73 .
Let n be a f ixed positive integer . DeterlDine
M=
max L I x �. -xJ. I . 1 < ' < J' 1 . Evaluate (m1 ) a + b � + a+b + 2a+b + + m m m m 77 .
.
•
•
.
•
Solution : Our starting point is the identity k-1
L
(77 . 1 )
x=O
�+
e
�
[ ek] ,
where k is any positive integer and e is any real number . As {y} = Y [y] , for any real y , (77 . 1 ) becomes -
(77 . 2 )
k-1
1 L .!k + e = -2 (k-1) + {ek} . x=O
For fixed k and e , .!k + e is periodic in x with period k . If c is chosen to be an integer such that GCD ( c , k) = 1 , the map ping x .... cx is a bij ection on a complete res idue system modulo k . Applying this biject ion to (77 . 2) , we obtain (77 . 3)
k- 1
�
x=O
cx + e = "21 (k-1) + { ek } k
•
We now choose c = a/GCD (a,m) and k = m/GCD (a,m) , so that GCD ( c , k) = 1 , and e = b /m . Then (77 . 3) becomes (keeping k �n place of m/GCD (a,m) where convenient) •
143
SOLUTIONS (77-78)
k-:l ax+b b 1 m 1 = + ::-TL a , m """" ) ( -= G-= CD GCD (a,m) 2 m x=O and so
,
t
GCD (at)-l k m-l ax+b + b a (y+kz ) L m m z�O y-O x=O GCD (af m) -l k-l ay+b = L m z=O y=O 1 b - 2' (m - GCD(a ,m) + GCD(a ,m) GCD(a ,m) Finally, we have .
•
ax+b = �(m- 1 ) + b - 1 (m-GCD(a,m) - GCD(a ,m) b � ";:' � , 2' :-:: :'::: '::' m GCD (a , m'"')' 2 that is m-l L x=O
m
b = j(am - a - m + GCD (a ,m) + GCD (a,m) GCD(a ,m)
•
Remarks : The identity (77 . 1) is given as a problem (with hint s) on page 40 in Number Theory by J . Hunter , Oliver and Boyd , 1964 . 78.
Let a l ,
• • .
, an be n (>1) distinct real numbers . Set +
Prove that
.
•
.
2 2 + an , M min (a . - a . ) J l:iOi <j :iOn '"
•
1.
�
M
�
n(n- 1 ) (n+l ) 12
•
Solution : Relabel ing the a ' s , so that preserves the values of S and M .
n •
-
2 a. J
where J loS a fixed subscript . Then , we have > J , a . > 0 , for 1. a 1.. < 0 , for 1. < J •
,
•
•
1.
•
•
•
•
min ( a . +1 - a 1.. ) = I min ( a . +1 - a . ) 2 = I mi� (a. - a . ) 2 1. 1. 1. 1. J lSi�n-1 lSiSn- 1 l�i<j �n
Next , =
r. "1'1
•
Def ine b . = a . + 1M ( i - j ) , i = 1 , 2 , 1. J b i = b + IM ( i - 1) . 1 Then , for i > j , we have a . = ( a1.. - a . - 1 ) + (a . - 1 - a . - 2 ) + 1. 1. 1. 1. � 1M ( i - j ) + a . J = b.
.
•
.
•
.
, n , so that
+ (aj+1 - a . ) + a . J J
•
1.
a . > - a 1.. , J that is a 1.. � b . ;;: - al.. , S imilarly we have �
-
1.
1. •
> J •
•
1. •
Thus , we obtain n 2 S = � a . ;;: .1. = 1 1.
(i = 1 , 2 ,
.
.
•
< J
•
•
, n) , and so
n = . �(b 1 1.= 1 ) 2 1 (n-1)n (n+1) (n ; ) + = n (b 1 + 1M M 12 (n1 ) n(n+1) >_ M . 12
SOLUTIONS (78-79)
145
79. Let x l " " ,xn be n real numbers such that ,
Prove that (79 .0)
;$
12
-
1 2n
•
2n Solution : For 1 :;; k :;; n+l we have 1 1 - n :;; 1 - n ' 2n and for n+l :;; k :;; n we have o :;;
-
1n '
so that 2- - 1 - -1 :;; 1 - -1 , n n k
( 79 . 1) Thus , as
n L � = 0 , we have
an+1 - an+2 > O . For any E > 0 , let N be a positive
85 .
•
.
.
'"
'"
integer such that � the inequality
S
2E
•
k+l _ sat isfies Prove that L '" L ( l) � k=1
N k+l I L - L (_l) ak I < E k=1
(85 . 0)
•
n k+l Solution : For n a pos itive integer, we define Sn = L (_I) . � k=1 We have L = Sn + and 00
n-l \" L = Sn-l + (-1) L (an+r+l - a n+r ) . r=1 As a n+r-l - an+r
>
an+r - an+r+l , we have
(85 . 1 ) Since have
and
L
lies between Sn-l and Sn , we
(85 . 2) Taking n = N , where aN < 2E , we obtain I SN - L I < E as required .
SOLUTIONS (86-87)
152
86. Determine all positive continuous functions f (x) defined on the interval [ O , lT ] for which IT
(81j . 0)
o
f (x) cos
nx
dx
=
n (-l) (2n+1) , n
=
0,1,2,3,4
.
So lut ion : We begin with the identities cos 2x
=
cos 3x
=
cos 4x
=
2 2 cos x - 1 , 3 4 cos x - 3 cos x , 2 4 8 cos x - 8 cos x + 1
•
Hence cos 4x + 4 cos 3x + 16 cos 2x + 28 cos x + 23 2 3 4 8 cos x + 16 cos x + 24 cos x + 1 6 cos x + 8 2 2 8(cos x + cos x + 1) , =
=
and so IT
2 2 8 f (x) (cos x + cos x + 1) dx o 9 + 4 (-7) + 16(5) + 28 (-3) + 23 ( 1 ) =
=
0 ,
which is impossible as f (x) is positive on [ O , lT] . Hence there are no pos itive functions f (x) satisfying (86 .0) . 87 . Let P and p I be points on opposite sides of a non circular ellipse E such that the tangents to E through p and respective lY are parallel and such that the tangents and normals p' to E at P and p I determine a rectangle R of maximum area . Determine the equation of E with respect to a rectangular coordin ate system, with origin at the centre of E and whose y-axis is parallel to the longer s ide of R .
SOLUTIONS (86-87)
153
Solution : We choose initially a coordinate system such that the X 2 y2 The po ints equation of E 1.S a2 + b2 - 1 a > b > 0 Q (a cos t b sin t) (0 $ t $ 2iT) and Q ' - (-a cos t , -b sin t) l ie on E and the tangents to E through Q and Q ' are parallel . We treat the case 0 $ t $ iT/2 as the other cases iT/2 $ t $ iT , iT $ t < 3 iT/2 , 3iT/2 $ t $ 2 iT can be handled by appropriate ref lect1.ons . Let the normal s through Q and Q' meet the tangents through Q ' and Q at T and T ' respectively . Our first aim is to choose t so that the area of the rectangle QTQ ' T ' 1.S maX 1.mllm. The s lope -b cos t . of the tangent to E at Q ' 1.S a S1.n . t , and so the equations of the l ines Q' T and QT are respectively b cos t x + a sin t y + ab = 0 2 2 and a sin t x - b cos t y - (a _b ) sin t cos t = O . Thus the lengths I QT I and I Q ' T I are given by •
-
=
,
•
,
-
•
•
•
2ab � ,g, !ib: , I QT I = �';;= == =; := ":' + 2 =: 2 := '=' 2 2 = = = a s in t b c 0 s t The area of the rectangle QTQ ' T '
•
is clearly
b is attained when tan t = a
whose maximllm value case
-
•
In this
2 2 QT = a +b > 1 , Q 'T a 2_b 2 so that R 1.S not a square . Thus and the s lope of the tangent at P through iT/4 clockwise by means of (x,y) (X , y ) , where X 1.. (x-y) ' v'2 tion of the required ellipse is •
..
=
2 2 a b P is the po int -/a2 +b2 ' la2 +b2 1.S -1 . Rotating the axes the orthogonal transformation y =� (x+y) , we find the equa•
SOLUTIONS (88-89)
154
1
.
88. If four distinct po ints l ie in the plane such that any three of them can be covered by a disk of unit radius , prove that all four points may be covered by a disk of unit radius . Solut ion : We first prove the following special case of He l ly ' s 4 2 theorem: If D . (i 1 , , 3 , ) are four disks in the plane such that any three have non-empty intersect ion then all four have non-empty intersection . Choose points W,X, Y , Z in D l'l D{ D3 ' respectively . We consider two cases D l'1 D2r! D4 ' D l' � D3�D4 ' D f D 3' � D4 according as one of the po ints W,X,Y,Z is in or on the (possibly degenerate) triangle formed by the other three points , or not . In the first case suppose that Z is in or on triangle WXY . Then the l ine segments WX ,WY ,XY belong to respectively, so that tr iangle WXY belongs to D l ' and thus belongs to D l . Hence Z is a point of D l D2 n D3 " D4 . In the second case WXYZ is a quadrilateral whose diagonal s intersect at a point C ins ide WXYZ . Without loss of general ity we may suppose that C is the intersect ion of WY and XZ . Now the l ine segments WY and XZ belong to D l D3 and D2 " D4 respectively. Thus C is both in D l n D3 and in D2 n D4 and so in D I ll D2 I) D3 " D4 . To solve the problem let A , B , C ,D be the four given distinct points . Let G be the centre of the unit disk to which A , B , C belong . Clearly the distances AG , BG , CG are all less than or equal to 1 , and so G belongs in the three unit disks UA ,UB ,UC centred at A , B , C respectively . Thus any three of the four disks UA ,UB ,UC , UD have a non-empty intersection , and so by the first result there is a 1.
=
f1
"
'
SOLUTIONS (88-89)
point A, B ,
155
The unit disk centred at
P C
89 .
and
P
contalns
D .
Evaluate the sum 1 2 2 · m - n
Solut ion : For positive integers m and N with N > m , we have N
N 1 1 1 1 A(m, N) = I -;2�2 = - 2m I n-m - n+m n=1 n=1 m -n n;z!m n;z!m =
3 - S (N-m» ) - 2 ' 4m
� ( S ( N + m ) 2
where for r = 1 , 2 , 3 , . . . we have set 1 -k = ln r + c + E(r) , S (r) = k=1 c denoting Euler ' s constant and the error term E (r)
sat isfying
for some absolute cons tant A . Then lim ( S (N+m) - S (N-m» ) N + oo = lim N + oo 0
N+m .en N-m + E(N+m) - E (N-m)
=
and so lim A(m,N) = N + oo
,
•
SOLUTIONS (�92)
156
and thus S = lim
M + oo
..
M
L lim A(m,N) m=l N + oo
3 lim 4
-
-
•
If n is a positive integer which can be expressed in the 2 c , where a , b , c are pos �t �ve �ntegers , prove form n .. 2k can be expressed in the that , for each positive integer k , n 2 2 2 form A + B + C , where A , B , C are positive integers . 90.
•
•
•
Solut ion: We begin by showing that if m = x , y , z are positive integer s , then where X , Y , Z are pos�t�ve �ntegers . Without loss of generality we may choose x � y � z . Then the required X,Y , Z are given by •
•
•
2 - z , Y .. 2xz , Z
..
2yz
•
r Letting 2k .. 2 (2s+1 ) , where r � 1 , s � 0 , we have and applying the above arg1!ment successively we obtain
where X, Y , Z are pos �t �ve �nteger s . •
91 .
•
•
be the group generated by a and b subj ect to the S 3 relations aba " b and b " 1 . Prove that G is abel ian . Let
G
SOLlITIONS (90-92)
Solut ion :
157
It suffices to show that aba .. b
3
b
g�ve s •
-1
-2
b
a
b
and
The relation
COIlDDute .
2 -1 , and so ab .. b a
ab
2
•
•
•
-1
b b
-1
(b
-1
ab ) b
2 -1 (b a ) b
2
-1 -1 b (b a b ) 2
.. b (b
-1
ab )
-1
2 -1 -1 = b (b a ) 2
2
.. b ab
-2 ,
81v�ng •
•
ab Hence , as
92 ,
b
S
4
4
.. b a
•
4
S
ab " b ab .. b (b a) b .. ba
= 1 , we obtain
•
be a sequence of real numbers "" diverge s � a sati s fying 0 < a < 1 for a l l n and such that n n n=l 2 � an converge s . Let f (x) be a funct ion def ined on [ 0 , 1 1 while n=l � f (an) If such that f " (x) ex� s t s and is bounded on [ 0 , 1 1 ne1 also converges . converge s , p rove that {a : n = 1 , 2 , 3 , n
Let
.
.
.
}
""
""
•
•
""
Solution :
Applying the extended mean value theorem to interval [ O , a 1 , there exi s t s n
and
w n
such that
f
on the 0
< wn < an
""
If
�
n=l
f (a ) n
converge s , then we must have
l im n ....
""
f ( a ) .. 0 , and so n
SOLUTIONS (93-94)
158
Next , as I f" (x) 1 � M , 0 x 1 , we have b y cont1.nu1.ty f eO) = 0 2 2 2 co co co a f " (w ) ) a f" (w ) w C a " f . 2 n n n n n n !:! and a that both , so L L I I 2 2 n=l n 2 2 n=l n=l n=1 co a 2 f " (w ) co co n n converges , and converge . Hence f ' (0) L an L f (an ) - L 2 n=l n=l n=l
a , and so
on [ 1 ,�) , contrary to as sumption .
Let
where
By the above argument we deduce
on [ a , c ] , and thus
This shows that
.. 0
rv
be
n (�2)
equa l ly spaced
Evaluate the sum
•
deno tes the distance between the points Without loss of genera l ity we may take to be the po int
I Pj Pk I
2
..
0
�
l exp ( 2rrj i/n)
j
j
and