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25.
Let a, b, c be non-negative numbers such that a + b + c = 1. Prove that
 abc + 1 + bca + 1 + cab + 1 £ 14 .
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 æç è 1a + 1b + 1c - 1a + 1 - 1b + 1 - 1c + 1 ö÷ ø .
Since
 1a + 1 + 1b + 1 + 1c + 1 = 14 [(a+1) + (b+1) + (c+1)] éê ë 1a + 1 + 1b + 1 + 1c + 1 ùú û ³ 94 ,
(by the Cauchy-Schwarz Inequality for example), we have that
 E £ abc æç è 1a + 1b + 1c - 94 ö÷ ø = ab + bc + ca - 94 abc .
On the other hand, (1 - c)2 = (a + b)2 ³ 4ab, whence ab £ 1/4(1 - c)2 . Therefore
 E - 14
 £ ab + c(a + b) - 94 abc - 14
 = ab æç è 1 - 94 c ö÷ ø + c(1 - c) - 14
 £ 14 (1 - c)2 æç è 1 - 94 c ö÷ ø + c(1 - c) - 14
 = - 116 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 ak be the label of the kth card, and let sn = åk = 1n ak 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 sn 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 si (mod m + 1). If i were to exceed 1, then we would have a contradiction, since then si - 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 ak, we can argue that all of the ak are equal.

27.
Find the least number of the form |36m - 5n | where m and n are positive integers.
Solution. Since the last digit of 36m is 6 and the last digit of 5n is 5, then the last digit of 36m - 5n is 1 when 36m > 5n and the last digit of 5n - 36m is 9 when 5n > 36m. If 36m - 5n = 1, then
 5n = 36m - 1 = (6m + 1)(6m - 1)
whence 6m + 1 must be a power of 5, an impossibility. 36m - 5n can be neither -1 nor 9. If 5n - 36m = 9, then 5n = 9(4·36m-1 + 1), which is impossible. For m = 1 and n = 2, we have that 36m - 5n = 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, a1 = f(a), and, for n ³ 2, an = f(an-1). Consider the sequence { xn } with
 xn = |an+1 - an |
where n = 1, 2, ¼. Since A is a finite set and each an belongs to A, there are only a finite number of distinct xn. Let xk = minn ³ 1 { xn }; we prove by contradiction that xk = 0.

Suppose if possible that xk > 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 xk. Hence, xk = 0, and this is equivalent to ak+1 = ak or f(ak) = ak. 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 fm (A) = { a }, so that fm+1 (A) = fm (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.

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 a1/2k £ 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 m2k Î A. For each positive integer k, there is a positive integer pk such that 2pk £ m2k < 2pk+1 (we can take pk = ëlog2 m2k û). For k sufficiently large, we have the inequality

 æç è 1 + 1m ö÷ ø 2k ³ 1 + 1m ·2k > 4 .
This combined with the foregoing inequality produces
 2pk £ m2k < 2pk + 1 < 2pk + 2 < (m + 1)2k .      (*)
Since 22(pk + 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 .

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 ¼An. Choose a system of coordinates centred at O (the circumcentre) such that
 Ak ~ æç è r cos 2kpn , r sin 2kpn ö÷ ø ,     r = ||OAk ||
for k = 1, 2, ¼, n. Then
 n å k = 1 OAk = n å k = 1 æç è r cos 2kpn , r sin 2kpn ö÷ ø = æç è r n å k = 1 cos 2kpn , r n å k = 1 sin 2kpn 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 2kpn + i n å k = 1 sin 2kpn
 = n å k = 1 æç è cos 2pn + i sin 2pn ö÷ ø k
 = n å k = 1 zk = z æç è 1 - zn1 - z ö÷ ø = 0 ,
whence åk = 1n cos(2pk/n) = åk = 1n sin(2pk/n) = 0. On the other hand,
 n å k = 1 ||MAk ||
 = 1r n å k = 1 ||OAk -OM|| ||OAk ||
 ³ 1r n å k = 1 (OAk -OM) ·OAk
 = 1r æç 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.
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