Solutions and Comments

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25.

Let a, b, c be non-negative 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 Cauchy-Schwarz 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

E - 1 4 £ 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.

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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 a₂ º s_i (mod m + 1). If i were to exceed 1, then we would have a contradiction, since then s_i - a₂ would be a multiple of m + 1. Therefore, a₂ º s₁ = a₁, so that a₂ º a₁ (mod m + 1), whence a₂ = a₁. By cyclic rotation of the a_k, we can argue that all of the a_k are equal.

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27.

 Find the least number of the form |36^m - 5ⁿ | where m and n are positive integers.

Solution. Since the last digit of 36^m is 6 and the last digit of 5ⁿ is 5, then the last digit of 36^m - 5ⁿ is 1 when 36^m > 5ⁿ and the last digit of 5ⁿ - 36^m is 9 when 5ⁿ > 36^m. If 36^m - 5ⁿ = 1, then

5n = 36m - 1 = (6m + 1)(6m - 1)

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

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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, a₁ = f(a), and, for n ³ 2, a_n = f(a_n-1). Consider the sequence { x_n } with

xn = |an+1 - an |

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

xk = |ak+1 - ak | > |f(ak+1) - f(ak) | = |ak+2 - ak+1 | = xk+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 a₁ and a₂ in A for which M = f(a₁) and m = f(a₂). Hence

M - m = |f(a1) - f(a2) | < |a1 - a2 | £ |M - m | = M - m

which is a contradiction. Therefore, f(A) Ì A.

Note that A Ê f(A) Ê f² (A) Ê ¼ Ê fⁿ(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 f2 (A) É ¼ É fn (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) = x².

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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

ëa1/2 û Î A ,   ëëa1/2 û1/2 û Î A ,   ¼   , ë¼ëa1/2 û1/2 û1/2 ¼û Î A ,    ¼ .

Also, the following inequalities are true

1 £ ëa1/2 û £ a1/2 ,   1 £ ëëa1/2 û1/2 û £ a1/22  ,    ¼ ,  1 £ ë¼ëa1/2 û1/2 û1/2 ¼û £ a1/2n ,

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 £ ë¼ëa1/2 û1/2 û1/2 ¼û £ a1/2k £ 1.5 ,

and thus

ë¼ëëa1/2 û1/2 û1/2 ¼û = 1 ,

(ii) We next prove that 2ⁿ Î A for n = 1, 2, ¼. Indeed, since 1 Î A, we obtain that, for each positive integer n, 2²ⁿ Î A so that 2ⁿ = ëÖ{2²ⁿ} 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^p_k £ m^2k < 2^p_k+1 (we can take p_k = ëlog₂ m^2k û). For k sufficiently large, we have the inequality

æ ç è 1 + 1 m ö ÷ ø 2k   ³ 1 + 1 m ·2k > 4 .

This combined with the foregoing inequality produces

2pk £ m2k < 2pk + 1 < 2pk + 2 < (m + 1)2k .      (*)

Since 2^{2(p_k + 1)+1} Î A, we have that

ë Ö   22(pk + 1) + 1   û = ë2pk + 1 Ö2 û Î A .

Hence, with k+1 brackets,

ë¼ëë2(pk + 1)Ö2 û1/2û1/2 ¼û Î A .

On the other hand, using (*), we get

m2k < 2pk + 1 £ ë2(pk + 1) Ö2 û < (m + 1)2k

and, then, with k+1 brackets,

m £ ë¼ëë2(pk + 1)Ö2 û1/2û1/2 ¼û < m+1 .

Thus,

m = ë¼ëë2(pk + 1)Ö2 û1/2û1/2 ¼û Î A .

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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 n-gon A₁ A₂ ¼A_n. Choose a system of coordinates centred at O (the circumcentre) such that

Ak ~ æ ç è r cos 2kp n , r sin 2kp n ö ÷ ø  ,     r = ||OAk ||

for k = 1, 2, ¼, n. Then

n å k = 1  OAk = 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[(2p)/n] + isin[(2p)/n] and using DeMoivreß Theorem, we have that zⁿ = 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  zk = z æ ç è 1 - zn 1 - z ö ÷ ø = 0 ,

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

n å k = 1  ||MAk || = 1 r n å k = 1  ||OAk -OM|| ||OAk || ³ 1 r n å k = 1  (OAk -OM) ·OAk = 1 r æ ç è n å k = 1  ||OAk||2 - ||OM || n å k = 1  OAk ö ÷ ø = n å k = 1  r = n å k = 1  ||OAk || .

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