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Mathematical Excalibur, Vol. 14, No. 4, Dec. 09-Mar. 10 Ping Ngai (La Salle College, Form 6), HUNG Ka Kin Kenneth (Diocesan Boys’ School) and Carlo PAGANO (Università di Roma “Tor Vergata”, Roma, Italy). Problem 324. (Due to FEI Zhenpeng, Northeast Yucai School, China) Let x,y ∊(0,1) and x be the number whose n-th digit after the decimal point is the nn-th digit after the decimal point of y for all n =1,2,3,…. Show that if y is rational, then x is rational. Solution. CHUNG Ping Ngai (La Salle College, Form 6), Since the decimal representation of y is eventually periodic, let L be the length of the period and let the decimal representation of y start to become periodic at the m-th digit. Let k be the least common multiple of 1,2,…,L. Let n be any integer at least L and nn ≥ m. By the pigeonhole principle, there exist i < j among 0,1,…,L such that ni ≡ nj (mod L). Then for all positive integer d, we have ni ≡ ni+d(j-i) (mod L). Since k is a multiple of j−i and n ≥ L > i, so we have nn ≡ nn+k (mod L). Since k is also a multiple of L, we have (n+k)n+k ≡nn+k ≡ nn (mod L). Then the n-th and (n+k)-th digit of x are the same. So x is rational. Other commended solvers: HUNG Ka Kin Kenneth (Diocesan Boys’ School) and Carlo PAGANO (Università di Roma “Tor Vergata”, Roma, Italy). Problem 335. (Due to Ozgur KIRCAK, Yahya Kemal College, Skopje, Macedonia) Find all a∊ℝ for which the functional equation f: ℝ→ ℝ f (x − f ( y)) = a( f (x) − x) − f ( y)
for all x, y ∊ℝ has a unique solution. Solution. LE Trong Cuong (Lam Son High School, Vietnam) Let g(x) = f(x)−x. Then, in terms of g, the equation becomes g(x−y−g(y))=ag(x)−x. Assume f(y)=y+g(y) is not constant. Let r, s be distinct elements in the range of f(y)=y+g(y). For every real x, g(x−r) = ag(x)−x = g(x−s). This implies g(x) is periodic with period T=|r−s|>0. Then ag(x) −x = g(x−y−g(y)) = g(x+T−y−g(y)) = ag(x+T) − (x+T) = ag(x)−x−T. This implies T=0, contradiction. Thus,
f is constant, i.e. there exists a real number c so that for all real y, f(y)=c. Then the original equation yields c=a(c−x)−c for all real x, which forces a=0 and c=0. Other commended solvers: LKL Problem Solving Group (Madam Lau Kam Lung Secondary School of MFBM).
Olympiad Corner
The inequality in the next example was very hard. It was proposed by Reid Barton and appeared among the 2003 IMO shortlisted problems. Example 6. Let n be a positive integer and let (x1, x2, …, xn), (y1, y2, …, yn) be two sequences of positive real numbers. Let (z1, z2, … , z2n) be a sequence of positive real numbers such that for all 1 ≤ i, j ≤ n, zi+j2 ≥ xiyj. Let M=max{z1, z2, …, z2n}. Prove that
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Problem 3. Let a,b,c be complex numbers such that for every complex number z with |z| ≤ 1, we have |az2+bz+c| ≤ 1. Find the maximum of |bc|. Problem 4. Let m,n be integers greater than 1. Let a1 < a2 < ⋯ < am be integers. Prove that there exists a subset T of the set of all integers such that the number of elements of T, denoted by |T|, satisfies
| T |≤ 1 +
am − a1 2n + 1
and for every i∊{1,2,⋯,m}, there exist t∊T and s∊[−n,n] such that ai=t+s. Problem 5. For n≥3, we place a number of cards at points A1, A2, ⋯, An and O. We can perform the following operations: (1) if the number of cards at some point Ai is not less than 3, then we can remove 3 cards from Ai and transfer 1 card to each of the points Ai−1, Ai+1 and O (here A0=An, An+1=A1); or (2) if the number of cards at O is not less than n, then we can remove n cards from O and transfer 1 card to each A1, A2, ⋯, An. Prove that if the sum of all the cards placed at these n+1 points is not less than n2+3n+1, then we can always perform finitely many operations so that the number of cards at each of the points is not less than n+1. Problem 6. Let a1, a2, a3, b1, b2, b3 be distinct positive integers satisfying (n +1)a1n + na2n + (n −1)a3n | (n +1)b1n + nb2n + (n −1)b3n
for all positive integer n. Prove that there exists a positive integer k such that bi=kai for i=1,2,3.
Max-Min Inequalities (continued from page 2)
⎛ M + z2 + L+ z2n ⎞ ⎛ x1 + L + xn ⎞⎛ y1 + L+ yn ⎞ ⎜ ⎟ ≥⎜ ⎟⎜ ⎟. 2n n n ⎝ ⎠ ⎝ ⎠⎝ ⎠
Solution. (Due to Reid Barton and Thomas Mildorf) Let X = max{x1, x2, …,xn} and Y = min{x1,x2,…, xn}. By replacing xi by xi’=xi/X, yi by yi’=yi/Y and zi by zi’= zi/(XY)1/2, we may assume X=Y=1. It suffices to prove M+z2+⋯+z2n ≥ x1+⋯+xn+y1+⋯+yn. (*) Then M + z2 +L+ z2n 1 ⎛ x1 +L+ xn y1 +L+ yn ⎞ ≥ ⎜ + ⎟, n n 2n 2⎝ ⎠
which implies the desired inequality by applying the AM-GM inequality to the right side. To prove (*), we will claim that for any r≥0, the number of terms greater than r on the left side is at least the number of such terms on the right side. Then the k-th largest term on the left side is greater than the k-th largest term on the right side for each k, proving (*). For r≥1, there are no terms greater than 1 on the right side. For r < 1, let A={i: xi>r}, B={j: yj>r}, A+B={i+j: i∊A, j∊B} and C={k: k>1, zk>r}. Let |A|, |B|, |A+B|, |C| denote the number of elements in A, B, A+B, C respectively. Since X=Y=1, so |A|, |B| are at least 1. Now xi>r, yj>r imply zi+j>r. So A+B is a subset of C. If A is consisted of i1<⋯<ia and B is consisted of j1<⋯<jb, then A+B contains i1+j1< i1+j2<⋯ < i1+jb < i2+jb <⋯ <ia+jb. Hence, |C| ≥ |A+B| ≥ |A|+|B|−1 ≥ 1. So zk>r for some k. Then M>r. So the left side of (*) has |C|+1≥ |A|+|B| terms greater than r, which finishes the proof of the claim.