Solutions and Comments
 25.

Let a, b, c be nonnegative numbers such that
a + b + c = 1. Prove that

ab c + 1

+ 
bc a + 1

+ 
ca b + 1

£ 
1 4

. 

When does equality hold?
Solution. It is straightforward to show that the inequality
holds when one of the numbers is equal to zero. Equality holds
if and only if the other two numbers are each equal to 1/2.
Henceforth, assume that all values are positive.
Since a + b + c = 1, at least one of the numbers is less than
4/9. Assume that c < 4/9. Let E denote the left side of the
inequality. Then
E = abc 
æ ç
è


1 a

+ 
1 b

+ 
1 c

 
1 a + 1

 
1 b + 1

 
1 c + 1


ö ÷
ø

. 

Since

1 a + 1

+ 
1 b + 1

+ 
1 c + 1

= 
1 4

[(a+1) + (b+1) + (c+1)] 
é ê
ë


1 a + 1

+ 
1 b + 1

+ 
1 c + 1


ù ú
û

³ 
9 4

, 

(by the CauchySchwarz Inequality for example), we have that
E £ abc 
æ ç
è


1 a

+ 
1 b

+ 
1 c

 
9 4


ö ÷
ø

= ab + bc + ca  
9 4

abc . 

On the other hand, (1  c)
^{2} = (a + b)
^{2} ³ 4ab, whence
ab
£ ^{1}/
_{4}(1  c)
^{2} . Therefore


£ ab + c(a + b)  
9 4

abc  
1 4


 
= ab 
æ ç
è

1  
9 4

c 
ö ÷
ø

+ c(1  c)  
1 4


 
£ 
1 4

(1  c)^{2} 
æ ç
è

1  
9 4

c 
ö ÷
ø

+ c(1  c)  
1 4


 
=  
1 16

c (3c  1)^{2} £ 0 . 

 

Equality occurs everywhere if and only if a = b = c = 1/3.
 26.

Each of m cards is labelled by one of the numbers
1, 2, ¼, m. Prove that, if the sum of labels of any
subset of cards is not a multiple of m + 1, then each card is
labelled by the same number.
Solution. Let a
_{k} be the label of the kth card, and
let s
_{n} =
å_{k = 1}^{n} a
_{k} for n = 1, 2,
¼, m. Since the
sum of the labels of any subset of cards is not a multiple of
m + 1, we get different remainders when we divide the s
_{n} by
m + 1. These remainders must be 1, 2,
¼, m in some order.
Hence there is an index i
Î { 1, 2,
¼, m } for which
a
_{2} º s
_{i} (mod m + 1). If i were to exceed 1, then
we would have a contradiction, since then s
_{i}  a
_{2} would be
a multiple of m + 1. Therefore, a
_{2} º s
_{1} = a
_{1}, so that
a
_{2} º a
_{1} (mod m + 1), whence a
_{2} = a
_{1}. By cyclic rotation of the
a
_{k}, we can argue that all of the a
_{k} are equal.
 27.

Find the least number of the form 36^{m}  5^{n} 
where m and n are positive integers.
Solution. Since the last digit of 36
^{m} is 6 and the last digit
of 5
^{n} is 5, then the last digit of 36
^{m}  5
^{n} is 1
when 36
^{m} > 5
^{n} and the last
digit of 5
^{n}  36
^{m} is 9 when 5
^{n} > 36
^{m}. If 36
^{m}  5
^{n} = 1, then
5^{n} = 36^{m}  1 = (6^{m} + 1)(6^{m}  1) 

whence 6
^{m} + 1 must be a power of 5, an impossibility.
36
^{m}  5
^{n} can be neither 1 nor 9. If 5
^{n}  36
^{m} = 9,
then 5
^{n} = 9(4·36
^{m1} + 1), which is impossible.
For m = 1 and n = 2, we have that
36
^{m}  5
^{n} = 36  25 = 11, and this is the least number of the given
form.
 28.

Let A be a finite set of real numbers which contains at
least two elements and let f : A ® A be a function such
that f(x)  f(y)  < x  y  for every
x, y Î A, x ¹ y. Prove that there is a Î A for which
f(a) = a. Does the result remain valid if A is not a finite set?
Solution 1. Let a
Î A, a
_{1} = f(a), and, for n
³ 2,
a
_{n} = f(a
_{n1}). Consider the sequence { x
_{n} } with
x_{n} = a_{n+1}  a_{n}  

where n = 1, 2,
¼. Since A is a finite set and each a
_{n}
belongs to A, there are only a finite number of distinct
x
_{n}. Let x
_{k} = min
_{n ³ 1} { x
_{n} }; we prove
by contradiction that x
_{k} = 0.
Suppose if possible that x_{k} > 0. Then
x_{k} = a_{k+1}  a_{k}  > f(a_{k+1})  f(a_{k})  = a_{k+2}  a_{k+1}  = x_{k+1} . 

But this does not agree with the selection of x
_{k}. Hence,
x
_{k} = 0, and this is equivalent to a
_{k+1} = a
_{k} or
f(a
_{k}) = a
_{k}. The desired result follows.
Solution 2. We first prove that f(A) ¹ A.
Suppose, if possible, that f(A) = A. Let M be the largest
and m be the smallest number in A. Since f(A) = A, there
are elements a_{1} and a_{2} in A for which M = f(a_{1}) and
m = f(a_{2}). Hence
M  m = f(a_{1})  f(a_{2})  < a_{1}  a_{2}  £ M  m  = M  m 

which is a contradiction. Therefore, f(A)
Ì A.
Note that A Ê f(A) Ê f^{2} (A) Ê ¼ Ê f^{n}(A) Ê ¼. In fact, we can
extend the foregoing argument to show strict inclusion
as long as the sets in question have more than one element:
A É f(A) É f^{2} (A) É ¼ É f^{n} (A) É ¼ . 

(The superscripts indicate multiple composites of f.)
Since A is a finite set, there must be a positive integer
m for which f
^{m} (A) = { a }, so that f
^{m+1} (A) = f
^{m} (A). Thus, f(a) = a.
Example. If A is not finite, the result may fail.
Indeed, we can take A = (0, ^{1}/_{2}) (the open interval of
real numbers strictly between 0 and ^{1}/_{2}) or
A = { 2^{2n} : n = 1, 2, ¼} and f(x) = x^{2}.
 29.

Let A be a nonempty set of positive integers such that
if a Î A, then 4a and ëÖa û
both belong to A. Prove that
A is the set of all positive integers.
Solution. (i) Let us first prove that 1
Î A. Let a
Î A.
Then we have
ëa^{1/2} û Î A , ëëa^{1/2} û^{1/2} û Î A , ¼ , ë¼ëa^{1/2} û^{1/2} û^{1/2} ¼û Î A , ¼ . 

Also, the following inequalities are true
1 £ ëa^{1/2} û £ a^{1/2} , 1 £ ëëa^{1/2} û^{1/2} û £ a^{1/22} , ¼ , 1 £ ë¼ëa^{1/2} û^{1/2} û^{1/2} ¼û £ a^{1/2n} , 

where there are n brackets in the general inequality.
There is a sufficiently large positive integer k for which
a
^{1/2k} £ 1.5, and for this k, we have, with k
brackets,
1 £ ë¼ëa^{1/2} û^{1/2} û^{1/2} ¼û £ a^{1/2k} £ 1.5 , 

and thus
ë¼ëëa^{1/2} û^{1/2} û^{1/2} ¼û = 1 , 

(ii) We next prove that 2^{n} Î A for n = 1, 2, ¼.
Indeed, since 1 Î A, we obtain that, for each positive
integer n, 2^{2n} Î A
so that 2^{n} = ëÖ{2^{2n}} û Î A.
(iii) We finally prove that an arbitrary positive integer m is
in A. It suffices to show that there is a positive integer
k for which m^{2k} Î A. For each positive integer k, there
is a positive integer p_{k} such that 2^{pk} £ m^{2k} < 2^{pk+1}
(we can take p_{k} = ëlog_{2} m^{2k} û). For k sufficiently
large, we have the inequality

æ ç
è

1 + 
1 m


ö ÷
ø

2^{k}

³ 1 + 
1 m

·2^{k} > 4 . 

This combined with the foregoing inequality produces
2^{pk} £ m^{2k} < 2^{pk + 1} < 2^{pk + 2} < (m + 1)^{2k} . (*) 

Since 2
^{2(pk + 1)+1} Î A, we have that
ë  Ö

2^{2(pk + 1) + 1}

û = ë2^{pk + 1} Ö2 û Î A . 

Hence, with k+1 brackets,
ë¼ëë2^{(pk + 1)}Ö2 û^{1/2}û^{1/2} ¼û Î A . 

On the other hand, using (*), we get
m^{2k} < 2^{pk + 1} £ ë2^{(pk + 1)} Ö2 û < (m + 1)^{2k} 

and, then, with k+1 brackets,
m £ ë¼ëë2^{(pk + 1)}Ö2 û^{1/2}û^{1/2} ¼û < m+1 . 

Thus,
m = ë¼ëë2^{(pk + 1)}Ö2 û^{1/2}û^{1/2} ¼û Î A . 

 30.

Find a point M within a regular pentagon for which the sum of
its distances to the vertices is minimum.
Solution. We solve this problem for the regular ngon
A
_{1} A
_{2} ¼A
_{n}. Choose a system of coordinates centred at
O (the circumcentre) such that
A_{k} ~ 
æ ç
è

r cos 
2kp n

, r sin 
2kp n


ö ÷
ø

, r = OA_{k}  

for k = 1, 2,
¼, n. Then

n å
k = 1

OA_{k} = 
n å
k = 1


æ ç
è

r cos 
2kp n

, r sin 
2kp n


ö ÷
ø

= 
æ ç
è

r 
n å
k = 1

cos 
2kp n

, r 
n å
k = 1

sin 
2kp n


ö ÷
ø

= (0, 0) . 

Indeed, letting
z = cos[(2
p)/n] + isin[(2
p)/n]
and using DeMoivreß Theorem, we have that
z^{n} = 1 and


n å
k = 1

cos 
2kp n

+ i 
n å
k = 1

sin 
2kp n



= 
n å
k = 1


æ ç
è

cos 
2p n

+ i sin 
2p n


ö ÷
ø

k


 
= 
n å
k = 1

z^{k} = z 
æ ç
è


1  z^{n} 1  z


ö ÷
ø

= 0 , 

 

whence
å_{k = 1}^{n} cos(2
pk/n) =
å_{k = 1}^{n} sin(2
pk/n) = 0.
On the other hand,


= 
1 r


n å
k = 1

OA_{k} OM OA_{k}  
 
³ 
1 r


n å
k = 1

(OA_{k} OM) ·OA_{k} 
 
= 
1 r


æ ç
è


n å
k = 1

OA_{k}^{2}  OM  
n å
k = 1

OA_{k} 
ö ÷
ø


 
= 
n å
k = 1

r = 
n å
k = 1

OA_{k}  . 

 

Equality occurs if and only if M = O, so that O is the desired point.