IMO 2002

Volume 7, Number July 2002 2002 – October 2002 2002 Olympiad Corner The 43rd International Mathematical Olympiad 2002 Problem Let n be a positive integer Let T be the set of points (x, y) in the plane where x and y are non-negative integers and x + y < n Each point of T is colored red or blue If a point (x, y) is red, then so are all points (x′, y′ ) of T with both x′ ≤ x and y′ ≤ y Define an X-set to be a set of n blue points having distinct x-coordinates, and a Y-set to be a set of n blue points having distinct y-coordinates Prove that the number of X-sets is equal to the number of Y-sets Problem Let BC be a diameter of the circle Γ with center O Let A be a point on Γ such that 0° < ∠AOB < 120° Let D be the midpoint of the arc AB not containing C The line through O parallel to DA meets the line AC at J The perpendicular bisector of OA meets Γ at E and at F Prove that J is the incentre of the triangle CEF Problem Find all pairs of integers such that there exist infinitely many positive integers a for which (continued on page 4) Editors:  ý (CHEUNG Pak-Hong), Munsang College, HK ɧ  (KO Tsz-Mei) ௚ (LEUNG Tat-Wing)  ᓸ (LI Kin-Yin), Dept of Math., HKUST ന (NG Keng-Po Roger), ITC, HKPU Artist: Mathematical Games (I) Kin Y Li An invariant is a quantity that does not change A monovariant is a quantity that keeps on increasing or keeps on decreasing In some mathematical games, winning often comes from understanding the invariants or the monovariants that are controlling the games black pieces in the row or in the column will become – k, a change of (8 – k) – k = – k black pieces on the board Since – k is even, the parity of the number of black pieces stay the same before and after the move Since at the start, there are 32 black pieces, there cannot be black piece left at any time Example (1974 Kiev Math Olympiad) Numbers 1, 2, 3, l , 1974 are written on a board You are allowed to replace any two of these numbers by one number, which is either the sum or the difference of these numbers Show that after 1973 times performing this operation, the only number left on the board cannot be Example Four x’s and five o’s are written around the circle in an arbitrary order If two consecutive symbols are the same, then insert a new x between them Otherwise insert a new o between them Remove the old x’s and o’s Keep on repeating this operation Is it possible to get nine o’s? Solution There are 987 odd numbers on the board in the beginning Every time the operation is performed, the number of odd numbers left either stay the same (when the numbers taken out are not both odd) or decreases by two (when the numbers taken out are both odd) So the number of odd numbers left on the board after each operation is always odd Therefore, when one number is left, it must be odd and so it cannot be ࿎ ፱ (YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Math Dept., HKUST for general assistance On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students With your submission, please include your name, address, school, email, telephone and fax numbers (if available) Electronic submissions, especially in MS Word, are encouraged The deadline for receiving material for the next issue is November 2, 2002 For individual subscription for the next five issues for the 01-02 academic year, send us five stamped self-addressed envelopes Send all correspondence to: Dr Kin-Yin LI Department of Mathematics The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong Fax: (852) 2358 1643 Email: makyli@ust.hk Example In an × board, there are 32 white pieces and 32 black pieces, one piece in each square If a player can change all the white pieces to black and all the black pieces to white in any row or column in a single move, then is it possible that after finitely many moves, there will be exactly one black piece left on the board? Solution No If there are exactly k black pieces in a row or column before a move is made to that row or column, then after the moves, the number of Solution If we let x = and o = – 1, then note that consecutive symbols are replaced by their product If we consider the product P of the nine values before and after each operation, we will see that the new P is the square of the old P Hence, P will always equal after an operation So nine o’s yielding P = – can never happen Example There are three piles of stones numbering 19, and 9, respectively You are allowed to choose two piles and transfer one stone from each of these two piles to the third piles After several of these operations, is it possible that each of the three piles has 12 stones? Solution No Let the number of stones in the three piles be a, b and c, respectively Consider (mod 3) of these numbers In the beginning, they are 1, 2, After one operation, they become 0, 1, no matter which two piles have stones transfer to the third pile So the remainders are always 0, 1, in some order Therefore, all piles having 12 Page Mathematical Excalibur, Excalibur Vol 7, No 3, Jul 02- Oct 02 stones are impossible Example Two boys play the following game with two piles of candies In the first pile, there are 12 candies and in the second pile, there are 13 candies Each boy takes turn to make a move consisting of eating two candies from one of the piles or transferring a candy from the first pile to the second The boy who cannot make a move loses Show that the boy who played second cannot lose Can he win? Solution Consider S to be the number of candies in the second pile minus the first Initially, S = 13 – 12 = After each move, S increases or decreases by So S (mod 4) has the pattern 1, 3, 1, 3, l Every time after the boy who played first made a move, S (mod 4) would always be Now a boy loses if and only if there are no candies left in the second pile, then S = – = So the boy who played second can always make a move, hence he cannot lose Since either the total number of candies decreases or the number of candies in the first pile decreases, so eventually the game must stop, so the boy who played second must win Example Each member of a club has at most three enemies in the club (Here enemies are mutual.) Show that the members can be divided into two groups so that each member in each group has at most one enemy in the group Solution In the beginning, randomly divide the members into two groups Consider the number S of the pairs of enemies in the same group If a member has at least two enemies in the same group, then the member has at most one enemy in the other group Transferring the member to the other group, we will decrease S by at least one Since S is a nonnegative integer, it cannot be decreased forever So after finitely many transfers, each member can have at most one enemy in the same group (Continued on page 4) IMO 2002 Kin Y Li The International Mathematical Olympiad 2002 was held in Glasgow, United Kingdom from July 19 to 30 There were a total of 479 students from 84 countries and regions participated in the Olympiad The Hong Kong team members were Chao Khek Lun (St Paul’s College) Chau Suk Ling (Queen Elizabeth School) Cheng Kei Tsi (La Salle College) Ip Chi Ho (St Joseph’s College) Leung Wai Ying (Queen Elizabeth School) Yu Hok Pun (SKH Bishop Baker Secondary School) The team leader was K Y Li and the deputy leaders were Chiang Kin Nam and Luk Mee Lin The scores this year ranged from to 42 The cutoffs for medals were 29 points for gold, 24 points for silver and 14 points for bronze The Hong Kong team received gold medal (Yu Hok Pun), silver medals (Leung Wai Ying and Cheng Kei Tsi) and bronze medals (Chao Khek Lun and Ip Chi Ho) There were perfect scores, two from China and one from Russia After the perfect scores, the scores dropped to 36 with students! This was due to the tough marking schemes, which intended to polarize the students’ performance to specially distinguish those who had close to complete solutions from those who should only deserve partial points The top five teams are China (212), Russia (204), USA (171), Bulgaria (167) and Vietnam (166) Hong Kong came in 24th (120), ahead of Australia, United Kingdom, Singapore, New Zealand, but behind Canada, France and Thailand this year One piece of interesting coincidence deserved to be pointed out Both Hong Kong and New Zealand joined the IMO in 1988 Both won a gold medal for the first time this year and both gold medallists scored 29 points The IMO will be hosted by Japan next year at Keio University in Tokyo and the participants will stay in the Olympic village Then Greece, Mexico, Slovenia will host in the following years Addendum After the IMO, the German leader Professor Gronau sent an email to inform all leaders about his updated webpage http://www.Mathematik-Olympiaden.de/ which contains IMO news and facts Clicking Internationale Olympiaden on the left, then on that page, scrolling down and clicking Top-Mathematikern, Die erfolgreichsten IMO-Teilnehmer in blue on the right, we could find the following past IMO participants who have also won the Fields medals, the Nevanlinna prizes and the Wolf prizes: Richard Borcherds (1977 IMO silver, 1978 IMO gold, 1998 Fields medal) Vladmir Drinfeld (1969 IMO gold, 1990 Fields medal) Tim Gowers (1981 IMO gold, 1998 Fields medal) Laurent Lafforgue (1984 IMO silver, 1985 IMO silver, 2002 Fields medal) Gregori Margulis (1959 IMO member, 1962 IMO silver, 1978 Fields medal) Jean-Christoph Yoccoz (1974 IMO gold, 1994 Fields medal) Alexander Razborov (1979 IMO gold, 1990 Nevanlinna prize) Peter Shor (1977 IMO silver, 1998 Nevanlinna prize) László Lovász (1963 IMO silver, 1964 IMO gold, 1965 IMO gold, 1966 IMO gold, 1999 Wolf prize) Page Mathematical Excalibur, Excalibur Vol 7, No 3, Jul 02- Oct 02 Problem Corner We welcome readers to submit their solutions to the problems posed below for publication consideration Solutions should be preceded by the solver’s name, home (or email) address and school affiliation Please send submissions to Dr Kin Y Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon The deadline for submitting solutions is November 2, 2002 Problem 156 If a, b, c > and a + b + c = 3, then prove that 1 + + ≥ + ab + bc + ca Problem 157 In base 10, the sum of the digits of a positive integer n is 100 and of 44n is 800 What is the sum of the digits of 3n? Problem 158 Let ABC be an isosceles triangle with AB = AC Let D be a point on BC such that BD = 2DC and let P be a point on AD such that ∠BAC = ∠BPD Prove that ∠BAC = ∠DPC Problem 159 Find all triples (x, k, n) of positive integers such that 3k − = x n Problem 160 We are given 40 balloons, the air pressure inside each of which is unknown and may differ from balloon to balloon It is permitted to choose up to k of the balloons and equalize the pressure in them (to the arithmetic mean of their respective pressures.) What is the smallest k for which it is always possible to equalize the pressures in all of the balloons? ***************** Solution CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form 5), Antonio LEI (Colchester Royal Grammar School, UK, Year 12), LEUNG Chi Man (Cheung Sha Wan Catholic Secondary School, Form 6), Poon Ming Fung (STFA Leung Kau Kui College, Form 5), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7), Tsui Ka Ho (CUHK, Year 1), Tak Wai Alan WONG (University of Toronto) and WONG Wing Hong (La Salle College, Form 5) Let a m = m (m+1) /2 This is the sum of 1, 2,  , m and hence the sequence a m is strictly increasing to infinity So for every integer n greater than 2, there is a positive integer m such that a m ≤ n < a m + Then n is the sum of the m positive integers 1, 2,  , m – 1, n – m ( m – 1) / Assume A(n) > m Then a m + = + + ⋯ + ( m + 1) ≤ n , a contradiction Therefore, A(n) = m Solving the quadratic inequality a m = m ( m + 1) / ≤ n , we find m is the greatest integer less than or equal to − + n + / ( ) Other commended solvers: CHU Tsz Ying (St Joseph’s Anglo-Chinese School, Form 7) Problem 152 Let ABCD be a cyclic quadrilateral with E as the intersection of lines AD and BC Let M be the intersection of line BD with the line through E parallel to AC From M, draw a tangent line to the circumcircle of ABCD touching the circle at T Prove that MT = ME (Source: 1957 Nanjing Math Competition) Solution CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form 5), CHU Tsz Ying (St Joseph’s Anglo-Chinese School, Form 7), Antonio LEI (Colchester Royal Grammar School, UK, Year 12), Poon Ming Fung (STFA Leung Kau Kui College, Form 5), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7), TANG Sze Ming (STFA Leung Kau Kui College), Tsui Ka Ho (CUHK, Year 1) and WONG Wing Hong (La Salle College, Form 5) So ME = MD ⋅ MB By the intersecting chord theorem, also MT = MD ⋅ MB Therefore, MT = ME Problem 153 Let R denote the real numbers Find all functions f : R → R such that the equality f ( f (x) + y) = f (x² – y) + f (x) y holds for all pairs of real numbers x, y (Source: 1997 Czech-Slovak Match) Solution CHU Tsz Ying (St Joseph’s Anglo-Chinese School, Form 7) and Antonio LEI (Colchester Royal Grammar School, UK, Year 12), Setting y = x², we have f ( f (x) + x² ) = f ( ) + x² f ( x ) Setting y = – f ( x ), we have f ( ) = f ( f ( x ) + x² ) + f ( x ) ² Comparing these, we see that for each x, we must have f (x) = or f (x) = x² Suppose f (a) = for some nonzero a Putting x = a into the given equation, we get f ( y ) = f ( a² – y) For y ≠ a² / 2, we have y² ≠ ( a² – y ) ² , which will imply f (y) = Finally, setting x = 2a and y = a² / 2, we have f ( a² / ) = f ( a² / 2) = So either f (x) = for all x or f (x) = x² for all x We can easily check both are solutions Comments: Many solvers submitted incomplete solutions Most of them got ∀ x (f (x) = or x²), which is not the same as the desired conclusion that ( ∀ x f (x) = 0) or ( ∀ x f (x) = x²) Problem 154 For nonnegative numbers a, d and positive numbers b, c satisfying b + c ≥ a + d, what is the Solutions ***************** Problem 151 Every integer greater than can be written as a sum of distinct positive integers Let A(n) be the maximum number of terms in such a sum for n Find A(n) (Source: 1993 German Math Olympiad) Since ME and AC are parallel, we have ∠MEB = ∠ACB=∠ADB=∠MDE Also, ∠BME=∠EMD So triangles BME and EMD are similar Then MB / ME = ME / MD minimum value of b c ? + c+d a+b (Source: 1988 Olympiad) All Soviet Math Solution Without loss of generality, we may assume that a ≥ d and b ≥ c From b + c ≥ a + d, we get Page Mathematical Excalibur, Excalibur Vol 7, No 3, Jul 02- Oct 02 b + c ≥ ( a + b + c + d ) / Now b c + c+d a+b = b + c − c  −  c+d a+b c+d ≥ a+b+c+d (c + d )   − ( c + d ) −  c + d a + b = a + b c + d + − 2(c + d ) a + b ≥ = a + b c + d ⋅ − (c + d ) a + b 2− , a i Hence b is divisible by M Since M is a multiple of b, so b = M, a contradiction to having 1997 distinct integers To get an example of pairwise relatively prime integers among them, let p n be the n-th prime number, a i = p i (for i = 1, 2, l , 8), a = p p 10 m p 1988 and bi = p1 p m p1988 / p i for i = 1, 2, l , 1988 It is easy to see that the a i ’s are pairwise relatively prime and any 10 of these 1997 numbers have the same least common multiple Other commended solvers: SIU Tsz Hang (STFA Leung Kau Kui College, Form 7) Olympiad Corner (continued from page 1) Problem (cont.) where the AM-GM inequality was used to get the last inequality Tracing the equality conditions, we need b + c = a + d, c = c + d and a + b = c So the minimum − / is attained, for example, when a = + 1, b = − , c = 2, d = Other commended solvers: CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form 5) and SIU Tsz Hang (STFA Leung Kau Kui College, Form 7) Problem 155 We are given 1997 distinct positive integers, any 10 of which have the same least common multiple Find the maximum possible number of pairwise relatively prime numbers among them (Source: 1997 Hungarian Math Olympiad) Solution Antonio LEI (Colchester Royal Grammar School, UK, Year 12) and WONG Wing Hong (La Salle College, Form 5) The answer is Suppose there were 10 pairwise relatively prime numbers a , a , l , a 10 among them Being pairwise relatively prime, their least common multiple is their product M Then the least common multiple of b , a , l , a10 for any other b in the set is also M Since a is relatively prime to each of a , l , a 10 , so b is divisible by a Similarly, b is divisible by the other am + a −1 an + a2 −1 is an integer Problem Let n be an integer greater than The positive divisors of n are d , d , l , d k where = d < d < m < d k = n Define D = d d + d d + m + d k −1 d k (a) Prove that D < n (b) Determine all n for which D is a divisor of n Problem Find all functions f from the set ℝ of real numbers to itself such that ( f ( x ) + f ( z ) )( f ( y ) + f ( t ) ) = f ( xy − zt ) + f ( xt + yz ) for all x, y, z, t in ℝ Problem Let Γ1 , Γ , l , Γn be circles of radius in the plane, where n ≥ Denote their centers by O1 , O , l , O n respectively Suppose that no line meets more than two of the circles Prove that ( n − 1)π ≤ 1≤ i < j ≤ n O i O j ∑ Mathematical Games (I) (Continued from page 2) Remarks This method of proving is known as the method of infinite descent It showed that you cannot always decrease a quantity when it can only have finitely many possible values Example (1961 All-Russian Math Olympiad) Real numbers are written in an m × n table It is permissible to reverse the signs of all the numbers in any row or column Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative Solution Let S be the sum of all the mn numbers in the table Note that after an operation, each number stay the same or turns to its negative Hence there are at most mn tables So S can only have finitely many possible values To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum If no such line exists, then we are done Otherwise, reverse the sign of all the numbers in the line Then S increases Since S has finitely many possible values, S can increase finitely many times So eventually the sum of the numbers in every line must be nonnegative Example Given 2n points in a plane with no three of them collinear Show that they can be divided into n pairs such that the n segments joining each pair not intersect Solution In the beginning randomly pair the points and join the segments Let S be the sum of the lengths of the segments (Note that since there are finitely many ways of connecting 2n points by n segments, there are finitely many possible values of S.) If two segments AB and CD intersect at O, then replace pairs AB and CD by AC and BD Since AB + CD = AO + OB + CO + OD > AC + BD by the triangle inequality, whenever there is an intersection, doing this replacement will always decrease S Since there are only finitely many possible values of S, so eventually there will not be any intersection Volume 8, Number August 2003 – October 2003 利用 GW-BASIC 繪畫曼德勃羅集的方法 Olympiad Corner 梁子傑老師 The 2003 International Mathematical Olympiad took place on July 2003 in Japan Here are the problems Problem Let A be a subset of the set S = {1, 2, …, 1000000} containing exactly 101 elements Prove that there exist numbers t1, t2, … , t100 such that the sets aj = {x + tj | x∊A } for j = 1, 2, … , 100 香港道教聯合會青松中學 已知一個複數 c0，並由此定義一 個複數數列 { cn }，使 cn+1 = cn2 + c0， 其中 n = 0, 1, 2, ……。如果這個數列 有界，即可以找到一個正實數 M，使 對於一切的 n，| cn | < M，那麼 c0 便屬 於曼德勃羅集（Mandelbrot Set）之內。 are pairwise disjoint 170 180 190 200 210 220 230 240 250 Problem Determine all pairs of positive integers (a, b) such that a2 ab − b + 260 is a positive integer 270 Problem A convex hexagon is given in which any two opposite sides have the following property: the distance between their midpoints is / times the sum of their lengths Prove that all the angles of the hexagon are equal (A convex hexagon ABCDEF has three pairs of opposite sides: AB and DE, BC and EF, CD and FA.) 以下是這程序的解釋： 可以將以上定義寫成一個 GW-BASIC 程序（對不起！我本人始 終都是喜歡最簡單的電腦語言，而且 我認為將 GW-BASIC 程序翻譯成其他 電腦語言亦不難），方法如下： W 紀錄在電腦畫面上將要畫出 圖形的大小。現將 W 設定為 360（見 第 10 行），表示打算在電腦畫面上一 個 360 × 360 的方格內畫出曼德勃羅集 （見第 50 及 60 行）。 10 LEFT 是繪圖時左邊的起點，TOP 是圖的最低的起點（見第 210、250 及 260 行） 。注意：在 GW-BASIC 中，畫 面坐標是由上至下排列的，並非像一 般的理解，將坐標由下至上排，因此 要以 “TOP − Y” 的方法將常用的坐標 轉換成電腦的坐標。 (continued on page 4) Editors: 張 百 康 (CHEUNG Pak-Hong), Munsang College, HK 高 子 眉 (KO Tsz-Mei) 梁 達 榮 (LEUNG Tat-Wing) 李 健 賢 (LI Kin-Yin), Dept of Math., HKUST 吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Math Dept., HKUST for general assistance On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students With your submission, please include your name, address, school, email, telephone and fax numbers (if available) Electronic submissions, especially in MS Word, are encouraged The deadline for receiving material for the next issue is November 30, 2003 For individual subscription for the next five issues for the 03-04 academic year, send us five stamped self-addressed envelopes Send all correspondence to: Dr Kin-Yin LI Department of Mathematics The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong Fax: (852) 2358 1643 Email: makyli@ust.hk IF J >= AND J < THEN COLOR : REM BLUE IF J >= AND J < 12 THEN COLOR : REM GREEN IF J >= 12 AND J < 15 THEN COLOR 15 : REM WHITE IF J >= 15 THEN COLOR 12 : REM RED PSET (X + LEFT, (TOP – Y)* M) NEXT X NEXT Y COLOR 15 : REM WHITE LINE (LEFT, (TOP – W / 2) * M) – (W + LEFT, (TOP – W / 2) * M) LINE (W / + LEFT, (TOP – W) * M) – (W / + LEFT, TOP * M) END 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 LEFT = 150 : TOP = 380 : W = 360 : M = 833 R = 2.64 : S = * R / W RECEN = : IMCEN = SCREEN : CLS FOR Y = TO W FOR X = TO W REC = S * (X – W / 2) + RECEN : IMC = S * (Y – W / 2) + IMCEN RE = REC : IM = IMC RE2 = RE * RE : IM2 = IM * IM : J=0 WHILE RE2 + IM2 = AND J < THEN COLOR 14 : REM YELLOW 由於電腦畫面上的一點並非正 方形，橫向和縱向的大小並不一樣， 故引入 M（= ）來調節長闊比（見 第 10、210、250 及 260 行）。 留意 W 祇是「畫面上」的大小， 並非曼德勃羅集內每一個複數點的實 際坐標，故需要作出轉換。R 是實際 的數值（見第 20 行） ，即繪畫的範圍 實軸由 −R 畫至 +R，同時虛軸亦由 −R 畫至 +R。S 計算 W 與 R 之間的比例， 並應用於後面的計算之中（見第 20 及 70 行）。 Page Mathematical Excalibur, Vol 8, No 4, Aug 03- Oct 03 RECEN 和 IMCEN 用來定出中心 點的位置，現在以 (0, 0) 為中心（見 第 30 行）。我們可以通過更改 R、 RECEN 和 IMCEN 的值來移動或放 大曼德勃羅集。 第 40 行選擇繪圖的模式及清除 舊有的畫面。 程序的第 50 及 60 行定出畫面上 的坐標 X 和 Y，然後在第 70 行計 算出對應複數 c0 的實值和虛值。 注意：若 c0 = a0 + b0 i，cn = an + bn i， 則 cn+1 = cn2 + c0 = (an + bn i)2 + (a0 + b0 i) = an2 − bn2 + 2anbn i + a0 + b0 i = (an2 − bn2 + a0) + (2anbn + b0)i。 所以 cn+1 的實部等於 an2 − bn2 + a0， 而虛部則等於 2anbn + b0。 將以上的計算化成程序，得第 110 及 120 行。REC 和 IMC 分別是 c0 的實值和虛值。RE 和 IM 分別是 cn 的實值和虛值。RE2 和 IM2 分別 是 cn 的實值和虛值的平方。 IMO 2003 T W Leung The 44th International Mathematical Olympiad (IMO) was held in Tokyo, Japan during the period - 19 July 2003 Because Hong Kong was declared cleared from SARS on June 23, our team was able to leave for Japan as scheduled The Hong Kong Team was composed as follows Chung Tat Chi (Queen Elizabeth School) Kwok Tsz Chiu (Yuen Long Merchants Assn Sec School) Lau Wai Shun (T W Public Ho Chuen Yiu Memorial College) Siu Tsz Hang (STFA Leung Kau Kui College) Yeung Kai Sing (La Salle College) Yu Hok Pun (SKH Bishop Baker Secondary School) Leung Tat Wing (Leader) Leung Chit Wan (Deputy Leader) Two former Hong Kong Team members, Poon Wai Hoi and Law Ka Ho, paid us a visit in Japan during this period 執行本程序所須的時間，要視乎 電腦的速度，以現時一般的電腦而 言，整個程序應該可以 分鐘左右 完成。 The contestants took two 4.5 Hours contests on the mornings of July 13 and 14 Each contest consisted of three questions, hence contest composed of Problem to 3, contest Problem to In each contest usually the easier problems come first and harder ones come later After normal coordination procedures and Jury meetings cutoff scores for gold, silver and bronze medals were decided This year the cutoff scores for gold, silver and bronze medals were 29, 19 and 13 respectively Our team managed to win two silvers, two bronzes and one honorable mention (Silver: Kwok Tsz Chiu and Yu Hok Pun, Bronze: Siu Tsz Hang and Yeung Kai Sing, Honorable Mention: Chung Tat Chi, he got a full score of on one question, which accounted for his honorable mention, and his total score is point short of bronze) Among all contestants three managed to obtain a perfect score of 42 on all six questions One contestant was from China and the other two from Vietnam 參考書目 Heinz-Otto Peitgen, Hartmut Jürgens and Dietmar Saupe (1992) Fractals for the Classroom Part Two: Introduction to Fractals and Chaos NCTM, Springer-Verlag The Organizing Committee did not give official total scores for individual countries, but it is a tradition that scores between countries were compared This year the top five teams were Bulgaria, China, U.S.A., Vietnam and Russia J 用來紀錄第 100 至 140 行的循 環的次數。第 100 行亦同時計算 cn 模的平方。若模的平方大於 256 或 者循環次數多於 15，循環將會終 止。這時候，J 的數值越大，表示 該數列較「收斂」，即經過多次計 算後，cn 的模仍不會變得很大。第 150 至 200 行以顏色將收斂情況分 類，紅色表示最「收歛」的複數， 其次是白色，跟著是綠色、藍色和 黃色，而最快擴散的部分以黑色表 示。第 210 行以先前選定的顏色畫 出該點。 曼德勃羅集繪畫完成後，以白色 畫出橫軸及縱軸（見第 240 至 260 行） ，以供參考。程序亦在此結束。 respectively The Bulgarian contestants did extremely well on the two hard questions, Problem and Many people found it surprising On the other hand, despite going through war in 1960s Vietnam has been strong all along Perhaps they have participated in IMOs for a long time and have a very good Russian tradition Among 82 teams, we ranked unofficially 26 We were ahead of Greece, Spain, New Zealand and Singapore, for instance Both New Zealand and we got our first gold last year But this year the performance of the New Zealand Team was a bit disappointing On the other hand, we were behind Canada, Australia, Thailand and U.K Australia has been doing well in the last few years, but this year the team was just point ahead of us Thailand has been able to quite well in these few years IMO 2004 will be held in Greece, IMO 2005 in Mexico, IMO 2006 in Slovenia IMO 2007 will be held in Vietnam, the site was decided during this IMO in Japan For the reader who will try out the IMO problems this year, here are some comments on Problem 3, the hardest problem in the first day of the competitions Problem A convex hexagon is given in which any two opposite sides have the following property: the distance between their midpoints is / times the sum of their lengths Prove that all the angles of the hexagon are equal (A convex hexagon ABCDEF has three pairs of opposite sides: AB and DE, BC and EF, CD and FA.) The problem is hard mainly because one does not know how to connect the given condition with that of the interior angles Perhaps hexagons are not as rigid as triangles It also reminded me of No 5, IMO 1996, another hard problem of polygons The main idea is as follows Given a hexagon ABCDEF, connect AD, BE and CF to form the diagonals From the given condition of the hexagon, it can be proved that the triangles formed by the diagonals and the sides are actually equilateral triangles Hence the interior angles of the hexagons are 120 o Good luck Page Mathematical Excalibur, Vol 8, No 4, Aug 03- Oct 03 Problem Corner ***************** We welcome readers to submit their solutions to the problems posed below for publication consideration The solutions should be preceded by the solver’s name, home (or email) address and school affiliation Please send submissions to Dr Kin Y Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon The deadline for submitting solutions is November 30, 2003 Problem 186 (Due to Fei Zhenpeng, Yongfeng High School, Yancheng City, Jiangsu Province, China) Let α, β, γ be complex numbers such that α + β + γ = 1, nonnegative integer k Taking square root, we get the desired inequality Solutions **************** Problem 181 (Proposed by Achilleas PavlosPorfyriadis, AmericanCollege of Thessaloniki “Anatolia”, Thessaloniki, Greece) Prove that in a convex polygon, there cannot be two sides with no common vertex, each of which is longer than the longest diagonal Proposer’s Solution Suppose a convex polygon has two sides, say AB and CD, which are longer than the longest diagonal, where A, B, C, D are distinct vertices and A, C are on opposite side of line BD Since AC, BD are diagonals of the polygon, we have AB > AC and CD > BD Hence, α + β2 + γ2 = 3, AB + CD > AC + BD α3 + β3 + γ3 = By convexity, the intersection O of diagonals AC and BD is on these diagonals By triangle inequality, we have Determine the value of α21 + β21 + γ21 AO + BO > AB and CO + DO > CD Problem 187 Define f (n) = n! Let a = f (1) f (2) f (3) … In other words, to obtain the decimal representation of a write the numbers f(1), f (2), f (3), … in base 10 in a row Is a rational? Give a proof Problem 188 The line S is tangent to the circumcircle of acute triangle ABC at B Let K be the projection of the orthocenter of triangle ABC onto line S (i.e K is the foot of perpendicular from the orthocenter of triangle ABC to S) Let L be the midpoint of side AC Show that triangle BKL is isosceles Problem 189 2n + segments are marked on a line Each of the segments intersects at least n other segments Prove that one of these segments intersect all other segments Problem 190 (Due to Abderrahim Ouardini) For nonnegative integer n, let ⌊x⌋ be the greatest integer less than or equal to x and f ( n) = [ n + n +1 + − [ 9n + ] n+2 ] Find the range of f and for each p in the range, find all nonnegative integers n such that f (n) = p So AC + BD > AB + CD, a contradiction Other commended solvers: CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form5), John PANAGEAS (Kaisari High School, Athens, Greece), POON Ming Fung (STFA Leung Kau Kui College, Form 6), SIU Tsz Hang (CUHK, Math Major, Year 1) and YAU Chi Keung (CNC Memorial College, Form 6) Problem 182 Let a0, a1, a2, … be a sequence of real numbers such that an+1 ≥ an2+ 1/5 for all n ≥ Prove that an + ≥ an2− for all n ≥ (Source: 2001 USA Team Selection Test) Solution CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form5) and TAM Choi Nang Julian (Teacher, SKH Lam Kau Mow Secondary School) Adding an+1 – an2 ≥ / for nonnegative integers n = k, k + 1, k + 2, k + 3, k + 4, we get k +4 a k + − ∑ (a n2 n = k +1 − a n ) − a k2 ≥1 Observe that x − x + / = ( x − / 2) ≥ implies / ≥ −( x − x) Applying this to the inequality above and simplifying, we easily get a k + ≥ a k2 for nonnegative integer k Then a k +10 ≥ a k2 + ≥ a k4 for Other commended solvers: POON Ming Fung (STFA Leung Kau Kui College, Form 6) and SIU Tsz Hang (CUHK, Math Major, Year 1) Problem 183 Do there exist 10 distinct integers, the sum of any of which is a perfect square? (Source: 1999 Russian Math Olympiad) Solution Achilleas Pavlos PORFYRIADIS (American College of Thessaloniki “Anatolia”, Thessaloniki, Greece) and SIU Tsz Hang (CUHK, Math Major, Year 1) Let a1, a2, …, a10 be distinct integers and S be their sum For i = 1, 2, …, 10, we would like to have S – = ki2 for some integer ki Let T be the sum of k12, …, k102 Adding the 10 equations, we get 9S = T Then = S – (S – ai) = (T/9) – ki2 So all we need to is to choose integers k1, k2, …, k10 so that T is divisible by For example, taking ki = 3i for i = 1, …, 10, we get 376, 349, 304, 241, 160, 61, –56, –191, –344, –515 for a1, …, a10 Other commended solvers: CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form 5) Problem 184 Let ABCD be a rhombus with ∠B = 60o M is a point inside ∆ADC such that ∠AMC = 120o Let lines BA and CM intersect at P and lines BC and AM intersect at Q Prove that D lies on the line PQ (Source: 2002 Belarussian Math Olympiad) Solution John PANAGEAS (Kaisari High School, Athens, Greece), and POON Ming Fung (STFA Leung Kau Kui College, Form 6) Since ABCD is a rhombus and ∠ABC = 60o , we see ∠ADC, ∠DAC, ∠DCA, ∠PAD and ∠DCQ are all 60o Now ∠CAM+∠MCA = 180o –∠AMC = 60o and ∠DCM + ∠MCA =∠DCA = 60o imply ∠CAM = ∠DCM Since AB || CD, we get ∠APC = ∠DCM = ∠CAQ Page Mathematical Excalibur, Vol 8, No 4, Aug 03- Oct 03 Also, ∠PAC = 120o = ∠ACQ Hence ∆APC and ∆CAQ are similar So PA/AC = AC/CQ Since AC = AD = DC, so PA/AD = DC/CQ As ∠PAD = 60 o = ∠DCQ, so ∆PAD and ∆DCQ are similar Then ∠PDA + ∠ADC + ∠CDQ =∠PDA + ∠PAD + ∠APD = 180 o Therefore, P, D, Q are collinear Other commended solvers: CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form 5), Achilleas Pavlos PORFYRIADIS (American College of Thessaloniki “Anatolia”, Thessaloniki, Greece), SIU Tsz Hang (CUHK, Math Major, Year 1), TAM Choi Nang Julian (Teacher, SKH Lam Kau Mow Secondary School) Problem 185 Given a circle of n lights, exactly one of which is initially on, it is permitted to change the state of a bulb provided one also changes the state of every d-th bulb after it (where d is a divisor of n and is less than n), provided that all n/d bulbs were originally in the same state as one another For what values of n is it possible to turn all the bulbs on by making a sequence of moves of this kind? Olympiad Corner Solution Let ω = cos (2π/n) + i sin (2π/n) and the lights be at 1, ω , ω , …, ω n −1 with the one at on initially If d is a divisor of n that is less than n and the lights at ω a , ω a + d , ω a + d ,L, ω a +( n−d ) have the same state, then we can change the state of these n/d lights Note their sum is a geometric series equal to a n d ω (1 − ω ) /(1 − ω ) = So if we add up the numbers corresponding to the lights that are on before and after a move, it will remain the same Since in the beginning this number is 1, it will never be + ω + ω + L + ω n −1 = Therefore, all the lights can never be on at the same time Comments: This problem was due to Professor James Propp, University of Wisconsin, Madison (see his website http://www.math.wisc.edu/~propp/ ) and was selected from page 141 of the highly recommended book by Paul Zeitz titled The Art and Craft of Problem Solving, published by Wiley (continued from page 1) Problem Let ABCD be a cyclic quadrilateral Let P, Q and R be the feet of the perpendiculars from D to the lines BC, CA and AB respectively Show that PQ = QR if and only if the bisector of ∠ABC and ∠ADC meet on AC Problem Let n be a positive integer and x1, x2, … , xn be real numbers with x1 ≤ x2 ≤ L ≤ xn (a) Prove that ⎞ ⎛ n n ⎜ ∑ ∑ xi − x j ⎟ ⎟ ⎜ i =1 j =1 ⎠ ⎝ ≤ 2(n − 1) n n ∑ ∑ ( xi − x j ) i =1 j =1 (b) Show that equality holds if and only if x1, x2, …, xn is an arithmetic sequence Problem Let p be a prime number Prove that there exists a prime number q such that for every integer n, the number n p − p is not divisible by q The 2003 Hong Kong IMO team from left to right: Wei Fei Fei (Guide), Leung Chit Wan (Deputy Leader), Chung Tat Chi, Siu Tsz Hang, Kwok Tsz Chiu, Yu Hok Pun, Yeung Kai Sing, Lau Wai Shun, Leung Tat Wing (Leader) Volume 9, Number August 2004 – September 2004 IMO 2004 Olympiad Corner th The 45 International Mathematical Olympiad took place on July 2004 Here are the problems Day Time allowed: hours 30 minutes Problem Let ABC be an acute-angled triangle with AB ≠ AC The circle with diameter BC intersects the sides AB and AC at M and N, respectively Denote by O the midpoint of the side BC The bisectors of the angles BAC and MON intersect at R Prove that the circumcircles of the triangles BMR and CNR have a common point lying on the side BC Problem Find all polynomials P(x) with real coefficients which satisfy the equality P(a-b)+P(b-c)+P(c-a) = 2P(a+b+c) for all real numbers a, b, c such that ab + bc + ca = Problem Define a hook to be a figure made up of six unit squares as shown in the diagram (continued on page 4) Editors: ஻ Ի ஶ (CHEUNG Pak-Hong), Munsang College, HK ଽ υ ࣻ (KO Tsz-Mei) గ ႀ ᄸ (LEUNG Tat-Wing) ‫ ؃‬୊ ፱ (LI Kin-Yin), Dept of Math., HKUST ֔ ᜢ ‫( ݰ‬NG Keng-Po Roger), ITC, HKPU Artist: ྆ ‫( ़ ؾ‬YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Math Dept., HKUST for general assistance On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students With your submission, please include your name, address, school, email, telephone and fax numbers (if available) Electronic submissions, especially in MS Word, are encouraged The deadline for receiving material for the next issue is October 20, 2004 For individual subscription for the next five issues for the 03-04 academic year, send us five stamped self-addressed envelopes Send all correspondence to: Dr Kin-Yin LI Department of Mathematics The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong Fax: (852) 2358 1643 Email: makyli@ust.hk T W Leung The 45th International Mathematical Olympiad (IMO) was held in Greece from July to July 18 Since 1988, we have been participating in the Olympiads This year our team was composed as follows Members Cheung Yun Kuen (Hong Kong Chinese Women’s Club College) Chung Tat Chi (Queen Elizabeth School) Kwok Tsz Chiu (Yuen Long Merchant Association Secondary School) Poon Ming Fung (STFA Leung Kau Kui College) Tang Chiu Fai (HKTA Tang Hin Memorial Secondary School) Wong Hon Yin (Queen’s College) Cesar Jose Alaban (Deputy Leader) Leung Tat Wing (Leader) I arrived at Athens on July After waiting for a couple of hours, leaders were then delivered to Delphi, a hilly town 170 km from the airport, corresponding to more hours of journey In these days the Greeks were still ecstatic about what they had achieved in the Euro 2004, and were busy preparing for the coming Olympic Games in August Of course Greece is a small country full of legend and mythology Throughout the trip, I also heard many times that they were the originators of democracy, their contribution in the development of human body and mind and their emphasis on fair play After receiving the short-listed problems leaders were busy studying them on the night of July However obviously some leaders had strong opinions on the beauty and degree of difficulty of the problems, so selections of all six problems were done in one day Several problems were not even discussed in details of their own merits The following days were spent on refining the wordings of the questions and translating the problems into different languages The opening ceremony was held on July 11 In the early afternoon we were delivered to Athens After three hours of ceremony we were sent back to Delphi By the time we were in Delphi it was already midnight Leaders were not allowed to talk to students in the ceremony Contests were held in the next two days The days following the contests were spent on coordination, i.e leaders and coordinators discussed how many points should be awarded to the answers of the students This year the coordinators were in general very careful I heard several teams spent more than three hours to go over six questions Luckily coordination was completed on the afternoon of July 15 The final Jury meeting was held that night In the meeting the cut-off scores were decided, namely 32 points for gold, 24 for silver and 16 for bronze Our team was therefore able to obtain two silver medals (Kwok and Chung) and two bronze medals (Tang and Cheung) Other members (Poon and Wong) both solved at least one problem completely, thus received honorable mention Unofficially our team ranked 30 out of 85 The top five teams in order were respectively China, USA, Russia, Vietnam and Bulgaria In retrospect I felt that our team was good and balanced, none of the members was particularly weak In one problem we were as good as any strong team Every team members solved problem completely Should we did better in the geometry problems our rank would be much higher Curiously geometry is in our formal school curriculum while number theory and combinatorics are not In this Olympiad we had two geometry problems, but fittingly so, after all, it was Greece Page Mathematical Excalibur, Vol 9, No 3, Aug 04- Sept 04 Extending an IMO Problem Hà Duy Hung Dept of Math and Informatics Hanoi Univ of Education In this brief note we give a generalization of a problem in the 41st International Mathematical Olympiad held in Taejon, South Korea in 2000 IMO 2000/5 Determine whether or not there exists a positive integer n such that n is divisible by exactly 2000 different prime divisors, and 2n + is divisible by n The answer to the question is positive This intriguing problem made me recall a well-known theorem due to O Reutter in [1] as follows Theorem If a is a positive integer such that a + is not a power of 2, then a n + is divisible by n for infinitely many positive integers n We frequently encounter the theorem in the case a = The theorem and the IMO problem prompted me to think of more general problem Can we replace the number in the IMO problem by other positive integers? The difficulty partly lies in the fact that the two original problems are solved independently After a long time, I finally managed to prove a generalization as follows Theorem Let s, a, b be given positive integers, such that a, b are relatively prime and a + b is not a power of Then there exist infinitely many positive integers n such that • n has exactly s different prime divisors; and • a n + b n is divisible by n notations r | s means s is divisible by r and u ≡ v (mod m) means u – v is divisible by m Lemma Let a, b be distinct and relatively prime positive integers, and p an odd prime number which divides a + b Then for any non-negative integer k, p k +1 | a m + b m , where m = p k k k Let x = a p and y = b p Since p −1 x + y = ( x + y )∑ (−1) x p i a +b a+b n is an odd integer ≥ , equality if and only if n = or a = b = The proof of Lemma is simple and is left for the reader Also, we remind readers the usual p −1− i yk = b p −1 + ∑ [(a p ) 2i (b p ) p −1− 2i p k ( p −1) k k i =1 k − (a p ) 2i −1 (b p ) p − 2i ] k > bp + ap = xk k ≥ p k +1 It follows that yk ≥ pk > p i y , it suffices to show that the whole summation is divisible by p Since x ≡ − y (mod p k +1 ), we have p −1 ∑ (−1) x i p −1− i yi By Lemma 1, p −1 ≡ ∑ (−1) 2i x p −1 ≡ px p −1 (mod p k +1 ) prime numbers {qk }k =1 satisfying the +∞ completing the proof In the rest of this note we shall complete the proof of Theorem Proof of Theorem Without loss of generality, let a > b Since a + b is not a power of 2, it has an odd prime factor p For natural number k, set xk = a x + b , yk = k +1 xk p −1 (i) gcd ( xk , qk ) = (ii) gcd ( p, qk ) = (iii) qk | xk +1 (iv) xk | xk +1 We shall now show that the sequence {qk }k =1 +∞ yk = ∑ (−1)i (a p ) p −1−i (b p )i k numbers and is thus infinite Indeed, if k0 < k1 ≡ ∑ (−1) 2i (a p ) p −1 k p −1 are positive integers and qk0 = qk1 , then i =0 (mod p k +1 yk is a positive p ) qk1 = qk0 | xk0 +1 | ⋅⋅⋅ | xk1 by properties (iii) and (iv) But this contradicts property (i) Next, set n0 = p s q1 qs −1 and nk+1 = integer Also, we have yk p k p −1 ≡b ( ) p consists of distinct prime k i =0 which implies that following properties pk Then yk is a positive integer and ≡ px yk p We now have a sequence of odd i =0 pk yk is an odd positive p integer, so we can choose an odd prime divisor qk of i =0 p −1 n for k = 1, 2, … Moreover, we have xk ≥ p k This leads us to i =0 We give a proof of Theorem below We shall make use of two familiar lemmas Lemma Let n be an odd positive integer, and a, b be relative prime positive integers Then ⎛y ⎞ gcd ⎜ k , p k ⎟ = ⎝ p ⎠ k Proof We prove the lemma by induction It is clear that the lemma holds for k = Suppose the lemma holds for some non-negative integer k, and we proceed to the case k + p for k = 1, 2, … By Lemma 2, we also have ⎛ xk ⎜ mod p ⎝ ⎞ ⎟, ⎠ pnk for k = 0, 1, 2, … It is evident that {nk }k =0 is a strictly increasing sequence +∞ so that ⎛x y ⎞ gcd ⎜ k , k ⎟ = ⎝ p p⎠ (continued on page 4) Mathematical Excalibur, Vol 14, No 2, May-Sep 09 Likewise, AQ·QB = (R+OQ)(R−OQ) = R2−OQ2 These force OP2 = OQ2, or OP = OQ, done! Problem Suppose that s1, s2, s3, … is a strictly increasing sequence of positive integers such that the subsequences ss1 , ss2 , ss3 , and ss1+1, ss2 +1, ss3 +1, are both arithmetic progressions Prove that the sequence s1, s2, s3, … is itself an arithmetic progression This was one of the two hard problems (3 and 6) Fortunately, it turned out that it was still within reach One trouble is of course the notation Of course, ss1 stands for the s1th term of the si sequence and so on Starting from an arithmetic progression (AP) with common difference d, then it is easy to check that both ss1 , ss2 , ss3 , and ss1+1, ss2+1, ss3+1, are APs with common difference d2 The question is essentially proving the “converse” So the first step is to prove that the common differences of the two APs ssi and ssi +1 are in fact the same, say s It is not too hard to prove and is intuitively clear, for two lines of different slopes will eventually meet and cross each other, violating the condition of strictly increasing sequence The next step is the show the difference between two consecutive terms of si is indeed s , (thus s is a square) One can achieve this end by the method of descent, or max/min principle, etc Problem Let ABC be a triangle with AB = AC The angle bisectors of ∠ CAB and ∠ ABC meet the sides BC and CA at D and E, respectively Let K be the incenter of triangle ADC Suppose that ∠ BEK = 45° Find all possible values of ∠ CAB This problem was also relatively easy It is interesting to observe that an isosceles triangle can be the starting point of an IMO problem With geometric software such as Sketchpad, one can easily see that ∠ CAB should be 60° or 90° To prove the statement of the problem, one may either use synthetic method or coordinate method One advantage of using the coordinate method is after showing the possible values of ∠ CAB, one can go back to show these values work by suitable substitutions Some contestants lost marks either because they missed some values of ∠ CAB or forgot to check the two possible cases work Problem Determine all functions f from the set of positive integers to the set of positive integers such that, for all positive integers a and b, there exists a non-degenerate triangle with sides of lengths a, f(b) and f(b+f(a)−1) (A triangle is non-degenerate if its vertices are not collinear.) The Jury worried if the word triangle may be allowed to be degenerate in some places But I supposed all our secondary school students would consider only non-degenerate triangles This was a nice problem in functional inequality (triangle inequality) One proves the problem by establishing several basic properties of f Indeed the first step is to prove f(1)=1, which is not entirely easy Then one proceeds to show that f is injective and/or f(f(x)) = x, etc, and finally shows that the only possible function is the identity function f(x) = x for all x Problem Let a1, a2, …, an be distinct positive integers and let M be a set of n−1 positive integers not containing s=a1+a2+⋯+an A grasshopper is to jump along the real axis, starting from the point O and making n jumps to the right with lengths a1, a2, …, an in some order Prove that the order can be chosen in such a way that the grasshopper never lands on any point in M It turned out that this problem was one of the most difficult problems in IMO history Only three of the 564 contestants received full scores (Perhaps it was second to problem posed in IMO 2007, for which only contestants received full scores.) When I first read the solution provided by the Problem Committee, I felt I was reading a paper of analysis Without reading the solution, of course I would say we could try to prove the problem by induction, as the cases of small n were easy The trouble was how to establish the Page induction step Later the Russians provided a solution by induction, by separating the problem into sub-cases M < an or M ≥ an, and then applying the principle-hole principle, etc judiciously to solve the problem Terry Tao said (jokingly) that the six problems were easy But in his blog, he admitted that he had spent sometime reading the problem and he even wrote an article about it (I have not seen the article.) The two hard problems (3 and 6) were more combinatorial and/or algebraic in nature I had a feeling that this year the Jury has been trying to avoid hard number theory problems, which were essentially corollaries of deep theorems (for example, IMO 2003 problem by the Chebotarev density theorem or IMO 2008 problem by a theorem of H Iwaniec) or hard geometry problem using sophisticated geometric techniques (like IMO 2008 problem 6) The Germans ran the program vigorously (obstinately) They had an organization (Bildung und Begabung) that looked after the entire event They had also prepared a very detailed shortlist problem set and afterwards prepared very detailed marking schemes for each problem The coordinators were very professional and they studied the problems well Thus, there were not too many arguments about how many points should be awarded for each problem Three of the problems (namely 1, and 4) were relatively easy, problems and were not too hard, so although problem was hard, contestants still scored relatively high points This explained why the cut-off scores were not low, 14 for bronze, 24 for silver and 32 for gold It might seem that we still didn’t the hard problems too well But after I discussed with my team members, I found that they indeed had the potential and aptitude to the hard problems What may still be lacking are perhaps more sophisticated skills and/or stronger will to tackle such problems (continued on page 4) Page Mathematical Excalibur, Vol 14, No 2, May-Sep 09 Problem Corner We welcome readers to submit their solutions to the problems posed below for publication consideration The solutions should be preceded by the solver’s name, home (or email) address and school affiliation Please send submissions to Dr Kin Y Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong The deadline for sending solutions is October 3, 2009 Problem 326 Prove that 34 + 45 is the product of two integers, each at least 102009 Problem 327 Eight pieces are placed on a chessboard so that each row and each column contains exactly one piece Prove that there is an even number of pieces on the black squares of the board (Source: 1989 USSR Math Olympiad) Problem 328 (Due to Tuan Le, Fairmont High School, Anaheim, Ca., USA) Let a,b,c > Prove that a3 + b3 b3 + c3 c3 + a3 + + a2 + b2 b2 + c2 c2 + a2 ≥ 6(ab + bc + ca) (a + b + c) (a + b)(b + c)(c + a) Problem 329 Let C(n,k) denote the binomial coefficient with value n!/(k!(n−k)!) Determine all positive integers n such that for all k = 1, 2, ⋯, n−1, we have C(2n,2k) is divisible by C(n,k) Problem 330 In ΔABC, AB = AC = and ∠ BAC = 90° Let D be the midpoint of side BC Let E be a point inside segment CD and F be a point inside segment BD Let M be the point of intersection of the circumcircles of ΔADE and ΔABF, other than A Let N be the point of intersection of the circumcircle of ΔACE and line AF, other than A Let P be the point of intersection of the circumcircle of ΔAMN and line AD, other than A Determine the length of segment AP with proof (Source: 2003 Chinese IMO team test) ***************** Solutions **************** Problem 321 Let AA’, BB’ and CC’ be three non-coplanar chords of a sphere and let them all pass through a common point P inside the sphere There is a (unique) sphere S1 passing through A, B, C, P and a (unique) sphere S2 passing through A’, B’, C’, P If S1 and S2 are externally tangent at P, then prove that AA’=BB’=CC’ Solution NGUYEN Van Thien (Luong The Vinh High School, Dong Nai, Vietnam) and Jim Robert STUDMAN (Hanford, Washington, USA) Consider the intersection of the spheres with the plane through A, A’, B, B’ and P M B' A N A' Let MN be the common external tangent through P to the circle through A, B, P and the circle through A’, B’ P as shown above We have∠ABP = ∠APM = ∠A’PN = ∠A’B’P = ∠A’B’B = ∠BAA’ = ∠BAP Hence, AP=BP Similarly, A’P = B’P So AA’ = AP+A’P = BP+B’P = BB’ Similarly, BB’ = CC’ Other commended solvers: CHUNG Ping Ngai (La Salle College, Form 6) and LAM Cho Ho (CUHK Math Year 1) Problem 322 (Due to Cao Minh Quang, Nguyen Binh Khiem High School, Vinh Long, Vietnam) Let a, b, c be positive real numbers satisfying the condition a+b+c = Prove that a2 (b + 1) b2 (c + 1) c2 (a + 1) + + ≥ a + b + ab b + c + bc c + a + ca Solution CHUNG Ping Ngai (La Salle College, Form 6), NGUYEN Van Thien (Luong The Vinh High School, Dong Nai, Vietnam) and the proposer independently Observe that a (b + 1) ab =a− a + b + ab a + b + ab (*) Applying the AM-GM inequality twice, we have ab ab ab a + b + ≤ = ≤ 2 a + b + ab a b By (*), we have Adding two other similar inequalities and using a+b+c = on the right, we get the desired inequality Other commended solvers: LAM Cho Ho (CUHK Math Year 1), Manh Dung NGUYEN (Special High School for Gifted Students, HUS, Vietnam), Paolo PERFETTI (Math Dept, Università degli studi di Tor Vergata Roma, via della ricerca scientifica, Roma, Italy), Stefan STOJCHEVSKI (Yahya Kemal College, Skopje, Macedonia), Jim Robert STUDMAN (Hanford, Washington, USA) and Dimitar TRENEVSKI (Yahya Kemal College, Skopje, Macedonia) Problem 323 Prove that there are infinitely many positive integers n such that 2n+2 is divisible by n P B a (b + 1) a + b + 8a − b − ≥a− = 9 a + b + ab Solution CHUNG Ping Ngai (La Salle College, Form 6), LAM Cho Ho (CUHK Math Year 1) and WONG Ka Fai (Wah Yan College Kowloon, Form 4) We will prove the stronger statement that there are infinitely many positive even integers n such that 2n+2 is divisible by n and also that 2n+1 is divisible by n−1 Call such n a good number Note n = is good Next, it suffices to prove that if n is good, then the larger integer m = 2n+2 is also good Suppose n is good Since n is even and m = 2n+2 is twice an odd integer, so m = nj for some odd integer j Also, the odd integer m−1 = 2n+1 = (n−1)k for some odd integer k Using the factorization ai+1 = (a+1)(ai−1−ai−2+⋯+1) for positive odd integer i, we see that 2m+2 = 2(2(n−1)k+1) = 2(2n−1+1) (2(n−1)(k−1)−⋯ +1) is divisible by 2(2n−1+1) = m and 2m+1 = 2nj+1 = (2n+1)(2n(j−1)−⋯+1) is divisible by 2n+1= m−1 Therefore, m is also good Problem 324 ADPE is a convex quadrilateral such that ∠ ADP = ∠ AEP Extend side AD beyond D to a point B and extend side AE beyond E to a point C so that ∠ DPB = ∠ EPC Let O1 be the circumcenter of ΔADE and let O2 be the circumcenter of ΔABC If the circumcircles of ΔADE and ΔABC are not tangent to each other, Page Mathematical Excalibur, Vol 14, No 2, May-Sep 09 then prove that line O1O2 bisects line segment AP Solution Jim Robert STUDMAN (Hanford, Washington, USA) ∠ MQN = ∠ MQD + ∠ DQN = ∠ NQE + ∠ DQN = ∠ DQE = ∠ DAE = ∠ MAN Let the circumcircle of ΔADE and the circumcircle of ΔABC intersect at A and Q Comments: Some solvers used a bit of homothety to simplify the proof Observe that line O1O2 bisects chord AQ and O1O2⊥AQ Hence, line O1O2 bisects line segment AP will follow if we can show that O1O2 || PQ, or equivalently that PQ⊥AQ Other commended solvers: CHUNG Ping Ngai (La Salle College, Form 6), LAM Cho Ho (CUHK Math Year 1), NG Ngai Fung (STFA Leung Kau Kui College, Form 7) Problem 325 On a plane, n distinct lines are drawn A point on the plane is called a k-point if and only if there are exactly k of the n lines passing through the point Let k2, k3, …, kn be the numbers of 2-points, 3-points, …, n-points on the plane, respectively A Q O1 D M P E O2 N B C Let points M and N be the feet of perpendiculars from P to lines AB and AC respectively Since ∠ANP = 90° = ∠AMP, points A, N, P, M lie on a circle Г with AP as diameter We claim that ∠MQN =∠MAN This would imply Q is also on circle Г, and we would have PQ⊥AQ as desired Since we are given ∠ ADP = ∠ AEP, we get ∠ BDP = ∠ CEP This combines with the given fact ∠ DPB = ∠ EPC imply ∆DPB and ∆EPC are similar, which yields DB/EC = DP/EP=DM/EN Since A,E,D,Q are concyclic, we have ∠ BDQ =180°− ∠ ADQ =180°− ∠ AEQ = ∠ CEQ This and ∠ DBQ=∠ABQ =∠ACQ = ∠ ECQ imply ∆DQB and ∆EQC are similar So we have QD/QE=DB/EC Combining with the equation at the end of the last paragraph, we get QD/QE=DM/EN Using ∆DQB and ∆EQC are similar, we get ∠ MDQ = ∠ BDQ = ∠ CEQ = ∠ NEQ These imply ∆MDQ and ∆NEQ are similar Then ∠ MQD = ∠ NQE Finally, for the claim, we now have Determine the number of regions the n lines divided the plane into in terms of n, k2, k3, …, kn (Source: 1998 Jiangsu Province Math Competition) Solution LAM Cho Ho (CUHK Math Year 1) Take a circle of radius r so that all intersection points of the n lines are inside the circle and none of the n lines is tangent to the circle Now each line intersects the circle at two points These 2n points on the circle are the vertices of a convex 2n-gon (call it M) as we go around the circle, say clockwise Let the n lines partition the interior of M into P3 triangles, P4 quadrilaterals, ⋯, Pj j-gons, ⋯ These polygonal regions are all convex since the angles of these regions, which were formed by intersecting at least two lines, are all less than 180° By convexity, no two sides of any polygonal region are parts of the same line So we have Pj = for j>3n Consider the sum of all the angles of these regions partitioning M On one hand, it is 180°(P3+2P4+3P5+⋯) by counting region by region On the other hand, it also equals 360°(k2+k3+⋯+kn)+(2n−2)180° by counting all the angles around each vertices of the regions Cancelling 180°, we get P3+2P4+3P5+⋯=2(k2+k3+⋯+kn)+(2n−2) Next, consider the total number of all the edges of these regions partitioned M (with each of the edges inside M counted twice) On one hand, it is 3P3+4P4+5P5+⋯ by counting region by region On the other hand, it is also (4k2+6k3+⋯2nkn)+4n by counting the number of edges around the k-points and around the vertices of M The 4n term is due to the 2n edges of M and each vertex of M (being not a k-point) issues exactly one edge into the interior of M So we have 3P3+4P4+5P5+⋯=4k2+6k3+⋯2nkn+4n Subtracting the last two displayed equations, we can obtain P3 + P4 + P5 +L= k2 + 2k3 + (n −1)kn + n +1 Finally, the number of regions these n lines divided the plane into is the limit case r tends to infinity Hence, it is exactly k2+2k3+⋯+(n−1)kn+n+1 Other commended solvers: CHUNG Ping Ngai (La Salle College, Form 6) and YUNG Fai Remarks on IMO 2009 (continued from page 2) As I found out from the stronger teams (Chinese, Japanese, Korean, or Thai, etc.), they were obviously more heavily or vigorously trained For instance, a Thai boy/girl had to go through more like 10 tests to be selected as a team member Another thing I learned from the meeting was several countries were interested to host the event (South-East Asia countries and Asia-Minor countries) In fact, one country is going to host three international competitions of various subjects in a row for three years Apparently they think hosting these events is good for gifted education The first IMO was held in Romania in 1959 Throughout these 51 years, only one year IMO was not held (1980) To commemorate the fiftieth anniversary of IMO in 2009, six notable mathematicians related to IMO (B Bollabas, T Gowers, L Lovasz, S Smirnov, T Tao and J C Yoccoz) were invited to talk to the contestants Of course, Yoccoz, Gowers and Tao were Fields medalists The afternoon of celebration then became a series of (rather) heavy lectures (not bad) They described the effects of IMOs on them and other things The effect of IMO on the contestants is to be seen later, of course! Volume 16, Number November 2011 -January 2012 Remarks on IMO 2011 Olympiad Corner Below are the problems of the 2011 International Math Olympiad Problem Given any set A={a1, a2, a3, a4} of four distinct positive integers, we denote the sum a1+a2+a3+a4 by sA Let nA denote the number of pairs (i,j) with 1≤i n > 1, we eventually get (m,n) = (3,2) or (4,2) Finally we get (a1, a2, a3, a4) = (k, 5k, 7k, 11k) or (k, 11k, 19k, 29k), where k is a positive integer As the derivation of the answers is rather straight-forward, it does not pose any serious difficulty For problem 4, it is really quite easy if one notes the proper recurrence relation Indeed the weights 20, 21, 22, …, 2n−1 form a “super-increasing sequence”, any weight is heavier than the sum of all lighter weights Denote by f(n) the number of ways of placing the weights We consider first how to place the lightest weight (weight 1) Indeed if it is placed in the first move, then it has to be in the left pan However if it is placed in the second to the last move, then it really doesn’t matter where it goes, using the “super-increasing property” Hence altogether there are 2n−1 possibilities of placing the weight of weight Now placing the weights 21, 22, …, 2n−1 clearly is the same as placing the weights 20, 21, …, 2n−2 There are f(n−1) ways of doing this Thus we establish the recurrence relation f(n) = (2n−1)f(n−1) Using f(1) = 1, by induction, we get f(n) = (2n −1)(2n −3)(2n −5)⋯1 In my opinion problem is the easier of the pair Indeed we may without loss of generality assume a1 < a2 < a3 < a4 So if the sum of one pair of the ai’s divides sA, then it will also divide the sum of the other pair But clearly a bigger pair cannot divide a smaller pair, so it is impossible that a3 + a4 dividing a1 + a2, nor is it possible that a2 + a4 dividing a1 + a3 Therefore the maximum possible value of nA can only be To achieve this, it suffices to consider divisibility conditions among the other pairs The problem becomes a mere exercise of recurrence relation if one notices how to place the lightest weight (minimum principle) It is slightly harder if we consider how to place the heaviest weight Indeed if the heaviest weight is to be placed in the ith move, then it has to be placed in the ⎛n−1⎞ ⎟ ways of ⎝ i −1⎠ left pan There are ⎜ Page Mathematical Excalibur, Vol 16, No 3, Nov.11 -Jan 12 choosing the previous i−1 weights and there are f(i−1) ways of placing them After the heaviest weight is placed, it doesn’t matter how to place the other weights, and there are (n−i)!×2n−i ways of placing the remaining weights Thus n ⎛ n − 1⎞ n −i f (n) = ∑ ⎜ ⎟ f (i − 1)(n − i)!2 i =1 ⎝ i − ⎠ Replacing n by n −1 and by comparing the two expressions we again get f(n) = (2n−1)f(n−1) We have no serious difficulty with this problem Problems and In my opinion both problems and were of similar flavor Both were “functional equation” type of problems Problem was slightly more involved and problem more number theoretic One can of course put in many values and obtain some equalities or inequalities But the important thing is to substitute some suitable values so that one can derive important relevant properties that can solve the problem In problem 5, indeed the condition f(m−n) | (f(m) − f(n)) (*) poses very serious restrictions on the image of f(x) Putting n=0, one gets f(m) | (f(m) − f(0)), thus f(m) | f(0) Since f(0) can only have finitely many factors, the image of f(x) must be finite Putting m=0, one gets f(−n) | f(n), and by interchanging n and −n, one gets f(n) = f(−n) Now f(n) | (f(2n) − f(n)), hence f(n) | f(2n), and by induction f(n) | f(mn) Put n = into the relation One gets f(1) | f(m) The image of f(x) is therefore a finite sequence f(1) = a1< a2< ⋯ < ak = f(0) One needs to show | ai+1 To complete the proof, one needs to analyze the sequence more carefully, say one may proceed by induction on k But personally I like the following argument Let f(x) = and f(y) = ai+1 We have f(x−y) | (f(y) − f(x)) < f(y) and f(y) − f(x) is positive, hence f (x − y) is in the image of f(x) and therefore f(x−y) ≤ ai= f(x) Now if f(x−y) < f(x), then f(x) − f(x −y) > Thus f(y) = f(x−(x−y)) | (f(x) − f(x −y)) In this case the right-hand side is positive We have f(y) ≤ f(x) − f(x −y)) < f(x) < f(y), a contradiction So we have f(x−y) = f(x) Thus f(x) | f(y) as needed It seems that Problem is more involved However, by making useful and clever substitutions, it is possible to solve the problem in a relatively easy way The following solution comes from one of our team members Put y = z−x into the original equation f(x+y) ≤ yf(x) + f(f(x)), one gets f(z) ≤ z f(x) − xf (x) + f( f(x)) By letting z = f(k) in the derived inequality one gets f(f(k)) ≤ f(k) f(x) − xf (x) + f( f(x)) Interchanging k and x one then gets f(f(x)) ≤ f(k) f(x) − kf (k) + f( f(k)) Hence f(x+y) ≤ y f(x) + f( f(x)) ≤ f(x)f(k) −kf(k) + f(f(k)) Letting y =f(k) − x in the inequality, we get f(f(k)) ≤ f(k) f(x) − xf (x) + f(k) f(x) − kf (k) + f( f(k)) or ≤ f(k) f(x) − xf(x) −k f(k) Finally letting k = f(x) and simplifying, we arrive at the important and essential (hidden) inequality ≤ −xf(x) This means for x > 0, f(x) ≤ 0, and for x < 0, f(x) ≥ But if there is an x0 < such that f(x0) > 0, then putting x = x0 and y = into the original equation, we gets 0 0, then f(f(x0)) ≤ 0, hence a contradiction This means for all x < 0, f(x) = Finally one has to prove f(0) = We suppose first f(0) > Put x = and y < sufficiently small into the original equation, one gets f(y) < 0, a contradiction Suppose f(0) < Take x, y < We get = f(x+y) ≤ y f(x) + f( f(x)) = y f(x) + f(0) = f(0) < 0, again contradiction! This implies f(0) = Problem To me, problem was one of a kind The problem was considered as “intermediate” and should not be too hard However at the end only 21 out of 564 contestants scored full marks It was essentially a problem of computational geometry We know that if there is a line that goes through two or more of the points and such that all other points are on the line or only on one side of the points, then by repeatedly turning angles as indicated in the problem, the convex hull of the point set will be constructed (so-called Jarvis’ march) Therefore some points may be missed So in order to solve the problem, we cannot start from the “boundary” Thus it is natural that we start from the “center”, or a line going through a point that separates the other points into equal halves (or differ by one) Indeed this idea is correct The hard part is how to substantiate the argument Many contestants found it hard Induction argument does not work because adding or deleting one point may change the entire route The proposer gives the following “continuity argument” We consider only the case that there are an odd number of points on the plane Let l be a line that goes through one of the points and that separates the other points into two equal halves Note that such line clearly exists Color one half-plane determined by the line orange (for Netherlands) and the other half-plane blue The color of the plane changes accordingly while the line is turning Note also that when the line moves to another pivot, the number of points on the two sides remain the same, except when two points are on the line during the change of pivots So consider what happen when the line turns 180°, (turning while changing pivots) The line will go through the same original starting point Only the colors of the two sides of the line interchange! This means all the points have been visited at least once! A slightly modified argument works for the case there are an even number of points on the plane Problem This was the most difficult problem of the contest (the anchor problem), only out of more than 564 contestants solved the problem Curiously these solvers were not necessarily from the strongest teams The problem is hard and beautiful, and I feel that it may be a known problem because it is so nice However, I am not able to find any further detail It is not convenient to reproduce the full solution here But I still want to discuss the main idea used in the first official solution briefly Lc Lb A H B' B" A' C' C" La γ C L B T A" From ΔABC and the tangent line L at T, we produce the reflecting lines La, Lb, and Lc The reflecting lines meet at A”, B” and C” respectively Now from A, we draw a circle of radius AT, meeting the circumcircle γ of ABC at A’ Likewise we have BT=BB’ and CT=CC’ (see the figure) (continued on page 4) Page Mathematical Excalibur, Vol 16, No 3, Nov.11 -Jan 12 Problem Corner We welcome readers to submit their solutions to the problems posed below for publication consideration The solutions should be preceded by the solver’s name, home (or email) address and school affiliation Please send submissions to Dr Kin Y Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong The deadline for sending solutions is February 28, 2012 Problem 381 Let k be a positive integer There are 2k balls divided into a number of piles For every two piles A and B with p and q balls respectively, if p ≥ q, then we may transfer q balls from pile A to pile B Prove that it is always possible to make finitely many such transfers so as to have all the balls end up in one pile Problem 382 Let v0 = 0, v1 = and vn+1 = 8vn−vn−1 for n = 1,2,3,… Prove that is divisible by if and only if is divisible by Problem 383 Let O and I be the circumcenter and incenter of Δ ABC respectively If AB≠AC, points D, E are midpoints of AB, AC respectively and BC=(AB+AC)/2, then prove that the line OI and the bisector of ∠CAB are perpendicular Problem 384 For all positive real numbers a,b,c satisfying a + b + c = 3, prove that a2 + 3b2 b2 + 3c2 c2 + 3a2 ≥ + + 2 ab (4 − ab) bc (4 − bc) ca (4 − ca) Problem 385 To prepare for the IMO, in everyday of the next 11 weeks, Jack will solve at least one problem If every week he can solve at most 12 problems, then prove that for some positive integer n, there are n consecutive days in which he can solve a total of 21 problems ***************** Solutions **************** Problem 376 A polynomial is monic if the coefficient of its greatest degree term is Prove that there exists a monic polynomial f(x) with integer coefficients such that for every prime p, f(x) ≡ (mod p) has solutions in integers, but f(x) = has no solution in integers Solution Alumni 2011 (Carmel Alison Lam Foundation Secondary School), Maxim BOGDAN (“Mihai Eminescu” National College, Botosani, Romania), Koopa KOO and Andy LOO (St Paul’s Co-educational College) Let f(x)=(x2−2)(x2−3)(x2−6) Then f(x) = has no solution in integers For p = or 3, f(6) ≡ (mod p) For a prime p > 3, if there exists x such that x2 ≡ or (mod p), then f(x) ≡ (mod p) has solutions in integers Otherwise, from Euler’s criterion, it follows that there will be x such that x2 ≡ (mod p) and again f(x) ≡ (mod p) has solutions in integers Comments: For readers not familiar with Euler’s criterion, we will give a bit more details For c relatively prime to a prime p, by Fermat’s little theorem, we have (c(p−1)/2−1)(c(p−1)/2+1) = cp−1−1≡ (mod p), which implies c(p−1)/2 ≡ or −1 (mod p) If there exists x such that x2 ≡ c (mod p), then c(p−1)/2 ≡ xp−1 ≡ (mod p) Conversely, if c(p−1)/2 ≡ (mod p), then there is x such that x2 ≡ c (mod p) [This is because there is a primitive root g (mod p) (see vol 15, no 1, p of Math Excalibur), so we get c ≡ gi (mod p) for some positive integer i, then gi(p−1)/2 ≡ (mod p) Since g is a primitive root (mod p), so i(p−1)/2 is a multiple of p−1, then i must be even, hence c ≡ (gi/2)2 (mod p).] In above, if and are not squares (mod p), then 6(p−1)/2=2(p−1)/23(p−1)/2 ≡ (−1)2 =1 (mod p), hence is a square (mod p) Problem 377 Let n be a positive integer For i=1,2,…,n, let zi and wi be complex numbers such that for all 2n choices of ε1, ε2, …, εn equal to ±1, we have n ∑ε z i =1 Prove that i i ≤ n ∑ε w i =1 n n i =1 i =1 i i ∑ | zi |2 ≤ ∑ | wi |2 Solution William PENG and Jeff PENG (Dallas,Texas, USA) The case n = is clear Next, recall the parallelogram law |a+b|2+|a−b|2=2|a|2+2|b|2, which follows from adding the + and − cases of the identity (a ± b)(a ± b ) = aa ± ab ± ba + bb For n = 2, we have |z1+z2|≤|w1+w2| and |z1−z2|≤|w1−w2| Squaring both sides of these inequalities, adding them and applying the parallelogram law, we get the desired inequality Next assume the case n=k holds Then for the n=k+1 case, we use the 2k choices with ε1 = ε2 to get from the n=k case that | z1 + z |2 + | z3 |2 +L+ | z k +1 |2 ≤ | w1 + w2 |2 + | w3 |2 + L+ | wk +1 |2 Similarly, using the other 2k choices with ε1 = −ε2, we get | z1 − z | + | z | + L+ | z k +1 | ≤ | w1 − w2 | + | w3 | + L+ | wk +1 | Adding the last two inequalities and applying the parallelogram law, we get the n=k+1 case Other commended solvers: Alumni 2011 (Carmel Alison Lam Foundation Secondary School),Maxim BOGDAN (“Mihai Eminescu” National College, Botosani, Romania), O Kin Chit, Alex (G.T.(Ellen Yeung) College) and Mohammad Reza SATOURI (Bushehr, Iran) Problem 378 Prove that for all positive integers m and n, there exists a positive integer k such that 2k −m has at least n distinct positive prime divisors Solution William PENG and Jeff PENG(Dallas,Texas, USA) For the case m is odd, we will prove the result by inducting on n If n=1, then just choose k large so that the odd number 2k −m is greater than Next assume there exists a positive integer k such that j = 2k −m has at least n distinct positive prime divisors Let s= k+φ(j2), where φ(j2) is the number of positive integers at most j2 that are relatively prime to j2 Since j is odd, by Euler’s theorem, s − m ≡ k × − m = j (mod j ) Then 2s − m is of the form j+tj2 for some positive integer t Hence it is divisible by j and (2s − m)/j is relatively prime to j Therefore, 2s − m has at least n+1 distinct prime divisors For the case m is even, write m=2ir, where i is a nonnegative integer and r is odd Then as proved above there is k such that 2k − r has at least n distinct prime divisors and so is 2i+k − m Page Mathematical Excalibur, Vol 16, No 3, Nov.11 -Jan 12 Other commended solvers: Maxim BOGDAN (“Mihai Eminescu” National College, Botosani, Romania) Problem 379 Let ℓ be a line on the plane of ∆ABC such that ℓ does not intersect the triangle and none of the lines AB, BC, CA is perpendicular to ℓ Let A’, B’, C’ be the feet of the perpendiculars from A, B, C to ℓ respectively Let A’’, B”, C” be the feet of the perpendiculars from A’, B’, C’ to lines BC, CA, AB respectively Prove that lines A’A”, B’B”, C’C” are concurrent Solution William PENG and Jeff PENG (Dallas, Texas, USA) and ZOLBAYAR Shagdar (9th Grade, Orchlon Cambridge International School, Mongolia) C" C B" A' Note that the set T={(a,b): a∊A and b∊B} has |A|×|B| ≥ 3999 elements Also, the set W={a+b: a∊A and b∊B} is a subset of {2,3,…4000} If W = {2,3,…,4000}, then and 4000 in W imply sets A and B both contain and 2000 This leads to A−A and B−B both contain 1999 If W≠{2,3,…4000}, then W has less than 3999 elements By the pigeonhole principle, there would exist (a,b) ≠ (a’,b’) in T such that a+b=a’+b’ This leads to a−a’=b’−b in both A−A and B−B B' C' A" Solution Maxim BOGDAN (“Mihai Eminescu” National College, Botosani, Romania) and William PENG and Jeff PENG (Dallas,Texas, USA) B A l Problem 380 Let S = {1,2,…,2000} If A and B are subsets of S, then let |A| and |B| denote the number of elements in A and in B respectively Suppose the product of |A| and |B| is at least 3999 Then prove that sets A−A and B−B contain at least one common element, where X−X denotes {s−t : s, t ∈ X and s ≠ t} (Source: 2000 Hungarian-Israeli Math Competition) D Olympiad Corner O Let lines B’B” and C’C” intersect at D To show line A’A” also contains D, since ∠ CA”A’ = 90°, it suffices to show ∠CA”D = 90° Let lines BC and B’B” intersect at O We claim that ΔDOA” is similar to Δ COB” (Since ∠ OB”C = 90°, the claim will imply ∠OA”D = 90°, which is the same as ∠CA”D = 90°.) For the claim, first note ∠AC”D = 90° = ∠AB”D, which implies A,C”,B”,D are concyclic So ∠C”AB”=∠B”DC” Next, ∠BC”D = 90° =∠DA”B implies B,C”,A”,D are concyclic So ∠C”BA” =∠A”DC” Then ∠ODA”=180°−(∠A”DC”+∠B”DC”) =180°− (∠C”BA”+∠C”AB”) =∠ACB =∠OCB” This along with ∠DOA”=∠COB” yield the claim and we are done Other commended solvers: Alumni 2011 (Carmel Alison Lam Foundation Secondary School) and Maxim BOGDAN (“Mihai Eminescu” National College, Botosani, Romania) (continued from page 1) Problem Let f : ℝ → ℝ be a real-valued function defined on the set of real numbers that satisfies f(x+y) ≤ y f(x) + f( f(x)) for all real numbers x and y Prove that f(x) = for all x ≤ Problem Let n > be an integer We are given a balance and n weights of weigh 20, 21, …, 2n−1 We are to place each of the n weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all the weights have been placed Determine the number of ways in which this can be done Problem Let f be a function from the set of integers to the set of positive integers Suppose that, for any two integers m and n, the difference f(m)−f(n) is divisible by f(m−n) Prove that, for all integers m and n with f(m)≤f(n), the number f(n) is divisible by f(m) Problem Let ABC be an acute triangle with circumcircle γ Let L be a tangent line to γ, and let La, Lb and Lc be the line obtained by reflecting L in the lines BC, CA and AB, respectively Show that the circumcircle of the triangle determined by the lines La, Lb and Lc is tangent to the circle γ Remarks on IMO 2011 (continued from page 2) The essential point is to observe that A”B”C” is in fact homothetic to A’B’C’, with the homothetic center at H, a point on γ , i.e A”B”C” is an expansion of A’B’C’ at H by a constant centre This implies the circumcircle of A”B”C” is tangent to γ at H A lot of discussions were conducted concerning changing the format of the Jury system during the IMO At present the leaders assemble to choose six problems from the short-listed problems There are issues concerning security and also financial matter (to house the leaders in an obscure place far away from the contestants can be costly) Many contestants need good results to obtain scholarships and enter good universities and the leaders have incentive for their own good to obtain good results for their teams For me I am inclined to let the Jury system remains as such The main reason is simply the law of large numbers, a better paper may be produced if more people are involved Indeed both the Problem Selection Group and the leaders may make mistakes But we get a better chance to produce a better paper after detailed discussion In my opinion we generally produce a more balanced paper The discussion is still going on Perhaps some changes are unavoidable, for better or for worse Here are some remarks concerning the performance of the teams We keep our standard or perhaps slightly better than the last few years I am glad that some of our team members are able to solve the harder problems Although the Chinese team is still ranked first (unofficially), they are not far better than the other strong teams (USA, Russia, etc) In particular, the third rank performance of the Singaporean team this time is really amazing Volume 17, Number May-August 2012 IMO 2012 (Leader Perspective) Olympiad Corner Tat-Wing Leung Below are the problems of the 2012 International Math Olympiad Problem Given triangle ABC the point J is the centre of the excircle opposite the vertex A This circle is tangent to the side BC at M, and to the lines AB and AC at K and L, respectively The lines LM and BJ meet at F, and the lines KM and CJ meet at G Let S be the point of intersection of the lines AF and BC, and let T be the point of intersection of the lines AG and BC Prove that M is the midpoint of ST (The excircle of ABC opposite the vertex A is the circle that is tangent to the line segment BC, to the ray AB beyond B, and to the ray AC beyond C.) Problem Let n ≥ be an integer, and let a2,a3,…,an be positive real numbers such that a2a3⋯an =1 Prove that (1+a2)2(1+a3)3⋯(1+an)n > nn Problem The liar’s guessing game is a game played between two players A and B The rules of the game depend on two positive integers k and n which are known to both players As leader, I arrived Mar del Plata, Argentina (the IMO 2012 site) four days earlier than the team Despite cold weather, jet lag and delay of luggage, I managed to get myself involved in choosing the problems for the contest Once the “easy” pair was selected, the jury did not have much choice but to choose problems of possibly other topics for the “medium” and the “difficult” pairs The two papers of the contest were then set We had to decide the various official versions and the marking scheme of the contest After that, I just had to wait for the contestants to finish the contest and get myself involved in the coordination to decide the points obtained by our team Here I would like to discuss the problems (Please see Olympiad Corner for the statements of the problems.) A G F B S M T C K L (continued on page 4) w Editors: 張 百 康 (CHEUNG Pak-Hong), Munsang College, HK 高 子 眉 (KO Tsz-Mei) 梁 達 榮 (LEUNG Tat-Wing) 李 健 賢 (LI Kin-Yin), Dept of Math., HKUST 吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU Artist: J 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU The editors welcome contributions from all teachers and students With your submission, please include your name, address, school, email, telephone and fax numbers (if available) Electronic submissions, especially in MS Word, are encouraged The deadline for receiving material for the next issue is Septembert 20, 2012 For individual subscription for the next five issues for the 09-10 academic year, send us five stamped self-addressed envelopes Send all correspondence to: Dr Kin-Yin LI, Math Dept., Hong Kong Univ of Science and Technology, Clear Water Bay, Kowloon, Hong Kong Fax: (852) 2358 1643 Email: makyli@ust.hk © Department of Mathematics, The Hong Kong University of Science and Technology the external bisector of ∠ABC, so is BK and BM Now SM = SB+BM = AB+BK =AK Similarly, TM=AL So SM=TM It is relatively easy to tackle the problem using coordinate geometry For instance, we can let the excircle be the unit circle with J=(0,0), M=(0,1), BC is aligned so that B=(b,1) and C=(c,1) Coordinates of other points are then calculated to verify the required property But one must be really careful if he tries to use coordinate method It was somehow decided that if a contestant cannot get a full solution using coordinate method, then he will be “seriously penalized”! Problem As it turned out, this problem caused quite a bit of trouble and many students didn’t know how to tackle the problem at all More sophisticated inequalities such as Muirhead not work, since the expression is not “homogeneous” The Japanese leader called the problem a disaster There were trivial questions such as “why is there no a1?” A more subtle issue is how to isolate a2,a3,…,an Clearly(1+a2)2 ≥ 22a2 by the AM-GM inequality But how about (1+a3)3? Indeed the trick is to apply AM-GM inequality to get for k=2 to n−1, Acknowledgment: Thanks to Elina Chiu, Math Dept., HKUST for general assistance On-line: http://www.math.ust.hk/mathematical_excalibur/ Thus ∠LFJ =∠LAJ Hence, F lies on ω By the same token, so is G Now AB and SB are symmetric with respect to Problem Really problem is quite easy, merely a lot of angle chasings and many angles of 90° (tangents) and similar triangles, etc, and no extra lines or segments needed to be constructed First note that ∠AKJ =∠ALJ = 90°, hence A,K,L,J lie on the circle ω with diameter AJ The idea is to show that F and G also lie on the same circle Looking at angles around B, we see that 4∠MBJ +2∠ABC = 360° Thus∠MBJ = 90°− ½∠ABC Also,∠BMF=∠CML = ½∠ACB (as CM=CL) Then ∠LFJ = ∠MBJ −∠BMF = ½∠BAC ⎞ ⎛1 (1 + ak +1 ) k +1 = ⎜ + L + + ak +1 ⎟ k k ⎠ ⎝ k +1 k +1 ⎛ a ⎞ (k + 1) k +1 a k +1 ≥ ⎜⎜ (k + 1) k +1 k k+1 ⎟⎟ = k ⎠ kk ⎝ By multiplying the inequalities, the constants cancelled out and we get the final inequality That the inequality is strict is trivial using the conditions of AM equals GM The above inequality can also be used as the inductive step of proving the equivalent inequality (1+a2)2(1+a3)3⋯(1+an)n > nna2a3⋯an Page Mathematical Excalibur, Vol 17, No 1, May-Aug 12 Problem Comparing with problem 6, I really found this problem harder to approach! Nevertheless there were still contestants who completely solved the problem Among them three were from the US team That was an amazing achievement! caused more trouble because of the disputes about the marking scheme First, by putting a=b=c=0, one gets f(0)=0 By putting b=−a and c=0, one gets f(a) = f(−a) More importantly, by putting c=−(a+b) and solving f(a+b) = f(−(a+b)) as a quadratic equation of f(a) and f(b), one gets We can deal with this combinatorial probabilistic problem as follows Ask repeatedly if x is 2k If A answers no k+1 times in a row, then the answer is honest and x≠2k Otherwise B stops asking about 2k at the first time answer yes He then asks, for each i=1,2,…,k, if the binary representation of x has a in the i-th digit Whatever the answer is, they are all inconsistent with a certain number y in the set {0,1,2,⋯, 2k−1} The answer yes to 2k is also inconsistent with y Hence x≠y Otherwise the last k+1 answers are not honest and that is impossible So we find y and it can be eliminated Or we can eliminate corresponding numbers with nonzero digits at higher end Notice we may need to some re-indexing and asking more questions about the indices of the numbers subsequently With these questions, we can reduce the size of the set that x lies until it lies in a set of size 2k f (a + b) = f (a ) + f (b) ± f (a) f (b) Part makes use of a function so that using the function, A can devise a strategy (to lie or not to lie, but lying not more than k times consecutively) so that no extra information will be provided to B and hence B cannot eliminate anything for sure Due to limit of space, I cannot provide all details here It was decided that part answered correctly alone was worth points and part alone worth points But altogether a problem is worth at most points So + = 7! At the end it really did not matter After all, not too many students did the problem right The problem is noted to be related to the Lovasz Local Lemma See N Alon et al, The Probabilistic Methods, Wiley, 1992 In the book it seems that there is an example that deals with similar things One may check how the lemma and the problem are related! Problem Despite being regarded as an easy problem, this problem is not at all easy It is much more involved than expected Also this problem eventually Putting a=b and c=−2a into the original equation, one gets f(2a)=0 or f(2a)=4f(a) Now the problem becomes getting all possible solutions from these two relations Using the two conditions, one checks that there are four types of solution: (ii) f2(x) = kx2 , (i) f1(x)≡0 , (iii) f ( x) = ⎧⎨0, x even ⎩ k , x odd and n + a2 + + an = 1, a1 3 The “k” in the solutions is essentially f(1) Indeed if f(1)=0, then f(2)=0, one then show by induction f(x)=0 for all x (Or by showing f(x) is periodic of period 1.) Now if f(1)=k, using the condition f(2a)=0, one can show again by induction f(x) is k for x odd and is for x even Now if f(1)=k and f(2)=4k, then f(4)=0 or 16k In the first case we get a function with period and arrive at the solution f4(x) In the second case we get f2(x) (One needs to verify the details.) By checking the values of a, b and c mod or 4, or other possible forms, one can check the solutions are indeed valid Eventually if a contestant claimed that all the solutions are easy to check, but without checking, one point would be deducted If a contestant says nothing about the solutions satisfy the functional equation and check nothing, then two points would be deducted! H X K A M D one gets x1+2x2+⋯+nxn=3a, where x1,x2,…,xn are non-negative integer powers of Taking mod 2, one gets n(n+1)/2 ≡ 1(mod 2) This is the case only when n ≡ 1, (mod 4) The hard part is to prove the converse also holds The cases n=1 or are easy By trials, for n=5, (a1,…,a5)=(2,2,2,3,3) works The official solution gave a systematic analysis of how to obtain solutions by using identities 1/2a=1/2a+1+1/2a+1 and w/3a=u/3a+1+v/3a+1, where u+v=3w For n=4k+1≥5, one can arrive at the solution a1=2=a3, a2=k+1, a4k = k+2 = a4k+1 and am = [m/4]+3 for 4≤m 0, ending the induction Problem 393 Let p be a prime number and p ≡ (mod 4) Prove that there exist integers x and y such that x2 − py2 = − Solution AN-anduud Problem Solving Group (Ulaanbaatar, Mongolia), Kevin LAU (St Paul’s Co-educational College, S.3), Simon LEE (Carmel Alison Lam Foundation Secondary School), Andy LOO (St Paul’s Co-educational College), Corneliu MĂNESCU-AVRAM (Dept of Math, Transportation High School, Ploiesti, Romania), Alice WONG (Diocesan Girls’ School) and ZOLBAYAR Shagdar (9th grader, Orchlon International School, Ulaanbaatar, Mongolia) Let (m,n) be the fundamental solution (i.e the least positive integer solution) of the Pell’s equation x2 − py2 = (see Math Excal., vol 6, no 3, p.1) Then m2 − n2 ≡ m2 − pn2 = 1(mod 4) Then m is odd and n is even Since m −1 m +1 ⎛n⎞ ⋅ = p⎜ ⎟ 2 ⎝2⎠ and (m−1)/2, (m+1)/2 are consecutive integers (hence relatively prime), either m −1 m +1 = pu , = v , n = 2uv 2 or m − = u , m + = pv , n = 2uv 2 for some positive integers u and v In the former case, v2−pu2=1 with < v ≤ v2 = (m+1)/2 < m and < u = n/(2v) < n This contradicts the minimality of (m,n) So the latter case must hold, i.e u2−pv2 = −1 Problem 394 Let O and H be the circumcenter and orthocenter of acute ΔABC The bisector of ∠BAC meets the circumcircle Γ of ΔABC at D Let E be the mirror image of D with respect to line BC Let F be on Γ such that DF is a diameter Let lines AE and FH meet at G Let M be the midpoint of side BC Prove that GM⊥AF Solution AN-anduud Problem Solving Group (Ulaanbaatar, Mongolia), Kevin LAU (St Paul’s Co-educational College, S.3), MANOLOUDIS Apostolos (4° Lyk Korydallos, Piraeus, Greece), Mihai STOENESCU (Bischwiller, France), ZOLBAYAR Shagdar (9th grader, Orchlon International School, Ulaanbaatar, Mongolia), Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania) Page Mathematical Excalibur, Vol 17, No 1, May-Aug 12 Other commended solvers: Simon LEE (Carmel Alison Lam Foundation Secondary School), and Alice WONG (Diocesan Girls’ School) A F G O E B H M D C X As AD bisects ∠BAC, D is the midpoint of arc BC Hence, FD is the perpendicular bisector of BC Thus, (1) FE || AH Let line AH meet Γ again at X Since ∠BCX=∠BAX=90°−∠ABC=∠BCH, H is the mirror image of X with respect to BC Therefore, ∠HED=∠XDE= ∠AFE Thus, (2) AF || HE By (1) and (2), AFEH is a parallelogram Hence, G is the midpoint of AE As M is also the midpoint of DE, we get GM || AD Since DF is the diameter of Γ, AD⊥AF, hence GM⊥AF Solution Andy LOO (St Paul’s Co-educational College) Place the figure on the complex plane and let the circumcircle of ΔABC be the unit circle centered at the origin Denote the complex number representing each point by the respective lower-case letter Without loss of generality we may assume a = and that the points A, B and C lie on the circle in anticlockwise order Let b =u2 and c=v2, where |u|=|v|=1 Then d=uv and hence f =−uv Next, E is the mirror image of D with respect to BC means e−b ⎛ d −b ⎞ , =⎜ ⎟ c−b ⎝ c−b ⎠ giving e = u2 − uv + v2 By the Euler line theorem, h=a+b+c=1+u2+v2.Now G on lines AE and FH means g − a g − a and g − f g − f = = e−a e −a h− f h − f Solving these simultaneously for G, we get g = (u2−uv+v2+1)/2 Also, m = (b+c)/2 = (u2+v2)/2 To show GM⊥AF, it suffices to prove that (m−g)/(f−a) is an imaginary number Indeed, m − g = ⋅ − uv and f −a + uv 1 ⎛ m − g ⎞ 1 − u ⋅ v uv − m−g = ⋅ =− ⎜ ⎟= ⋅ f −a ⎝ f − a ⎠ + ⋅ uv + u v as desired Problem 395 One frog is placed on every vertex of a 2n-sided regular polygon, where n is an integer at least At a particular moment, each frog will jump to one of the two neighboring vertices (with more than one frog at a vertex allowed) Find all n such that there exists a jumping of these frogs so that after the moment, all lines connecting two frogs at different vertices not pass through the center of the polygon Solution Kevin LAU (St Paul’s Co-educational College, S.3), Simon LEE (Carmel Alison Lam Foundation Secondary School), LI Jianhui (CNEC Christian College, F.5) and Andy LOO (St Paul’s Co-educational College) If n ≡ (mod 4), say n=4k+2, then label the 2n=8k+4 vertices from to 8k+4 in clockwise direction For j ≡ or (mod 4), let the frog at vertex j jump in the clockwise direction For j ≡ or (mod 4), let the frog at vertex j jump in the counter-clockwise direction After the jump, the frogs are at vertices 2, 6, …, 8k+2 and 3,7, …, 8k+3 No two of these vertex numbers have a difference of the form (mod 4) So no line through two different vertices with frogs will go through the center If n ≢2 (mod 4), then assume there is such a jump We may exclude the cases all frogs jump clockwise or all frogs jump counter-clockwise, which clearly not work Hence, in this jump, there is a frog, say at vertex i, jumps in the counter-clockwise direction, then the frog at vertex i+m(n−2) (mod 2n) must jump in the same direction as the frog at vertex i for m=1,2,… If n is odd, then gcd(n−2,2n) = So there are integers a and b such that a(n−2) + b(2n) = For every integer q in [1,2n], letting m = (q−i)a, we have i+m(n−2) ≡ q (mod 2n) This means all frogs jump in the counter-clockwise direction, which does not work If n is divisible by 4, then gcd(n−2,2n) = So there are integers c and d such that c(n−2)+d(2n)=2 Letting m=nc/2, we have i+m(n−2)≡i+n (mod 2n) Then frogs at vertices i and i+n jump in the counter-clockwise direction and the line after the jump passes through the center, contradiction Therefore, the answer is n ≡ (mod 4) Other commended solvers: Alice WONG (Diocesan Girls’ School) Olympiad Corner (continued from page 1) Problem (Cont.) At the start of the game A chooses integers x and N with 1≤x≤N Player A keeps x secret, and truthfully tells N to B Player B now tries to obtain information about x by asking player A questions as follows: each question consists of B specifying an arbitrary set S of positive integers (possibly one specified in some previous question), and asking A whether x belongs to S Player B may ask as many such questions as he wishes After each question, player A must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any k + consecutive answers, at least one answer must be truthful After B has asked as many questions as he wants, he must specify a set X of at most n positive integers If x belongs to X, then B wins; otherwise, he loses Prove that: If n ≥2k, then B can guarantee a win For all sufficiently large k, there exists an integer n ≥1.99k such that B cannot guarantee a win Problem Find all functions f: Z → Z such that, for all integers a, b, c that satisfy a+b+c=0, the following equality holds: f (a)2 + f (b)2 + f (c)2 = 2f (a) f (b) + 2f (b) f (c) + 2f (c) f (a) (Here Z denotes the set of integers.) Problem Let ABC be a triangle with ∠BCA = 90°, and let D be the foot of the altitude from C Let X be a point in the interior of the segment CD Let K be the point on the segment AX such that BK=BC Similarly, let L be the point on the segment BX such that AL=AC Let M be the point of intersection of AL and BK Show that MK=ML Problem Find all positive integers n for which there exist non-negative integers a1,a2,…,an such that 1 1 n + a2 + + an = a1 + a2 + + an = a1 2 3 Volume 17, Number Olympiad Corner Below are the problems of the 2012 IMO Team Selection Test from Saudi Arabia Problem In triangle ABC, points D and E lie on sides BC and AC respectively such that AD⊥BC and DE⊥AC The circumcircle of triangle ABD meets segment BE at point F (other than B) Ray AF meets segment DE at point P Prove that DP/PE = CD/DB Problem In an n×n board, the numbers through n2−1 are written so that the number in row i and column j is equal to (i−1)+n(j−1) where 1≤i,j ≤n Suppose we select n different cells of the board, where no two cells are in the same row or column Find the maximum possible product of the numbers in the n cells Problem Let ℚ be the set of rational numbers Find all functions f :ℚ→ℚ such that for all rational numbers x,y, f (f (x)+ x f (y)) = x + f (x) y Problem Find all pairs of prime numbers p, q such that p2−p−1 = q3 Editors: 張 百 康 (CHEUNG Pak-Hong), Munsang College, HK 高 子 眉 (KO Tsz-Mei) 梁 達 榮 (LEUNG Tat-Wing) 李 健 賢 (LI Kin-Yin), Dept of Math., HKUST 吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Math Dept., HKUST for general assistance On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students With your submission, please include your name, address, school, email, telephone and fax numbers (if available) Electronic submissions, especially in MS Word, are encouraged The deadline for receiving material for the next issue is November 20, 2012 For individual subscription for the next five issues for the 09-10 academic year, send us five stamped self-addressed envelopes Send all correspondence to: Dr Kin-Yin LI, Math Dept., Hong Kong Univ of Science and Technology, Clear Water Bay, Kowloon, Hong Kong Fax: (852) 2358 1643 Email: makyli@ust.hk © Department of Mathematics, The Hong Kong University of Science and Technology October 2012 IMO 2012 (Member Perspective) Andy Loo This year’s International Mathematical Olympiad (IMO) has been of considerable significance to Hong Kong At the 1997 IMO held in Mar del Plata, Argentina, shortly after our official transfer of sovereignty, the Hong Kong delegation accomplished the special mission of elucidating Article 149 of its Basic Law in light of Annex I of the Sino-British Joint Declaration, thereby consolidating the legitimacy of its participation in the IMO This July, following the 15th anniversary of the establishment of the Special Administrative Region, this annual event returns to Argentina, in exactly the same city as last time’s In addition to battling in the examination hall, the Hong Kong team was endowed with the invigorating task of bringing the IMO to Hong Kong again in 2016 Joined by 542 young brains from 99 countries, the Hong Kong team comprised the following personnel: Dr Leung Tat Wing (leader), Mr Leung Chit Wan (deputy leader) and the team members were Kevin Lau Chun Ting (St Paul’s Co-educational College), Andy Loo (St Paul’s Co-educational College), Albert Li Yau Wing (Ying Wah College), Jimmy Chow Chi Hong (Bishop Hall Jubilee School), Kung Man Kit (SKH Lam Woo Memorial Secondary School) and Alice Wong Sze Nga (Diocesan Girls’ School) This contest bestows certain personal touch upon me, for it not only marks my unprecedented landing on the continent of South America, but is also my first and, in all probability, my last IMO, an ultimate platform for me to display my years of Mathematical Olympiad endeavor in my high school career Having represented Hong Kong at both the International Physics Olympiad (IPhO) and the IMO is a great responsibility which I feel extremely grateful to have had the unique chance to shoulder July and Our flights from Hong Kong to Frankfurt and from Frankfurt to Buenos Aires, each over 12 hours long, were predominantly occupied by sleep and math exercises, considering the disappointing fact that our planes turned out to be two of the very few Boeing-74748 models of Lufthansa that lack in-flight entertainment systems Our amazement at a German flight attendant, who spoke more than fluent Mandarin Chinese, as well as a cozy conversation with a Slovakian neighbor, highlighted the otherwise uneventful journey We arrived at the Argentinean capital city early in the morning of July (in winter!), and, after being transported to the domestic airport, employed a timeconsuming conglomeration of Google Translate effort and sign language to manage to purchase a couple of SIM cards at a tiny store, where the shopkeeper knew literally no English A Maradona-like bus driver kindly offering us a free ride, we embarked on a tour around the city and enjoyed a beef-dominated meal before returning to the airport in the late afternoon to catch our flight to Mar del Plata, on which I, being absolutely exhausted, slept from the first to the last minute July The major event of this day was the Opening Ceremony It was held in the Radio City I met British team member Josh Lam and congratulated him on his mother’s recent promotion to Chief Secretary of Hong Kong If I were to describe the entire ceremony in one word it would definitely be “Spanish” Almost all the speeches were delivered in Spanish, albeit accompanied by English interpretation To most, the more exciting parts of the ceremony included the IMO anthem, the parade of nations and the distant waves from the leaders, who were forbidden to communicate with us before the contest as they took part in problem selection Page Mathematical Excalibur, Vol 17, No 2, Oct 12 July 10 On this first day of the contest, we had problems to solve in 4.5 hours Because questions could only be raised in the first 30 minutes, I had to understand all the problems quickly Problem Given triangle ABC the point J is the center of the excircle opposite the vertex A This excircle is tangent to the side BC at M, and to the lines AB and AC at K and L, respectively The lines LM and BJ meet at F, and the lines KM and CJ meet at G Let S be the point of intersection of the lines AF and BC, and let T be the point of intersection of the lines AG and BC Prove that M is the midpoint of ST I decided to use my favorite method – complex numbers Indeed, denote the complex number representing each point by the corresponding small letter Setting j=0 and m=1, I found s=2k/(k+l) and t=2l/(k+l) after a straightforward computation, and the result followed Problem Let n ≥ be an integer, and let a2,a3,…,an be positive real numbers such that a2a3⋯an =1 Prove that (1+a2)2(1+a3)3⋯(1+an)n > nn Inequalities were once among the hottest topics on the IMO but totally disappeared in the last three years due to the rising popularity of brute force techniques, e.g Muirhead’s inequality and Schur’s inequality But my firm belief in the revival of inequalities has never been shaken, and instead was only strengthened by Problem of APMO 2012 Consequently I had done appreciable preparation in this area before the Olympiad In IMO history, this problem was quite unique For one, it is an n-variable inequality For the other, it has no equality case Both features are unparalleled according to my memory I spent about an hour attempting to solve the problem using induction or analysis, with no avail In despair, I took logarithm and applied Jensen’s inequality by appealing to concavity of the log function Miraculously, it gave precisely the inequality in the problem! After checking that equality case cannot satisfy the condition a2a3⋯an =1, I was basically done Then on a second thought, I realized that I could actually convert my proof into a logarithm-free one that involves the AM-GM inequality only So I rewrote my solution in this new form and marked the original as an alternative solution It turned out that Alice was also able to solve this problem with the AM-GM inequality Problem The liar’s guessing game is a game played between two players A and B The rules of the game depend on two positive integers k and n which are known to both players At the start of the game A chooses integers x and N with 1≤x≤N Player A keeps x secret, and truthfully tells N to B Player B now tries to obtain information about x by asking player A questions as follows: each question consists of B specifying an arbitrary set S of positive integers (possibly one specified in some previous question), and asking A whether x belongs to S Player B may ask as many such questions as he wishes After each question, player A must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any k+1 consecutive answers, at least one answer must be truthful After B has asked as many questions as he wants, he must specify a set X of at most n positive integers If x belongs to X, then B wins; otherwise, he loses Prove that: If n ≥ 2k, then B can guarantee a win For all sufficiently large k, there exists an integer n ≥1.99k such that B cannot guarantee a win This problem was not only long, but also terribly difficult In the end, only contestants managed to solve it Despite my effort, the only thing I was able to was proving the k = case in Part 1, with the hope of getting slim partial credits Finally Day of the contest was over Our team aced Problem As for Problem 2, Alice and I should be able to get 7’s while Albert’s partial analytic solution would be subject to vigorous debate Kit also finished the k = case in Part of Problem Overall I was satisfied with my performance on Day July 11 The six IMO problems are usually partitioned into the four categories (algebra, combinatorics, geometry and number theory) in the fashion of {1,5}, {2,4}, {3} and {6} (up to permutation) Judging from this pattern I would face an easy algebraic problem, an intermediate geometric problem and a hard number theoretic problem on Day I figured that I would plausibly get a Gold medal for solving two of them, a Silver medal for one and a Bronze medal for none My strategy was to guarantee Problem and then aim to get Problem by hook or by crook To my astonishment, Problem was much more involved than I had expected On the other hand I felt I could Problem with analytic tools: Problem Let ABC be a triangle with ∠BCA = 90°, and let D be the foot of the altitude from C Let X be a point in the interior of the segment CD Let K be the point on the segment AX such that BK=BC Let L be the point on the segment BX such that AL=AC Let M be the point of intersection of AL and BK Show that MK=ML I proceeded to coordinate geometry, only to find out I was doomed after almost one hour The reason was as follows The expressions were quadratic in nature (as lengths took part in the formulation of the problem), leading to the prevalence of square roots (As a side note, this also deterred me from using complex numbers, where one may have difficulty in selecting the correct roots of the quadratic equations.) As the old Chinese saying goes, one should “drop his cleaver and become a Buddha (放下屠刀，立地成佛)” I decided to abandon Problem for a moment and to reconsider Problem 4: Problem Find all functions f: Z → Z such that, for all integers a, b, c that satisfy a + b + c = 0, the following equality holds: f (a)2 + f (b)2 + f (c)2 = 2f (a) f (b) + 2f (b) f (c) + 2f (c) f (a) (Here Z denotes the set of integers.) This was a problem with unusual answers It took me quite a while to write up a tidy solution and to ensure that no point could sneak away from my hands Thus it was 2.5 hours into Day I still had Problems and left Problem Find all positive integers n for which there exist non-negative integers a1,a2,…,an such that 1 1 n + + + an = a1 + a2 + + an = 2a1 2a2 3 I quickly determined that Problem was hopeless Turning to Problem again, I spent all the remaining time expanding everything I was finally able to convince myself that my proof was complete (continued on page 4) Page Mathematical Excalibur, Vol 17, No 2, Oct 12 Problem Corner We welcome readers to submit their solutions to the problems posed below for publication consideration The solutions should be preceded by the solver’s name, home (or email) address and school affiliation Please send submissions to Dr Kin Y Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong The deadline for sending solutions is November 20, 2012 Problem 401 Suppose all faces of a convex polyhedron are parallelograms Can it have exactly 2012 faces? Please provide an explanation to your answer Problem 402 Let S be a 30 element subset of {1,2,…,2012} such that every pair of elements in S are relatively prime Prove that at least half of the elements of S are prime numbers Problem 403 On the coordinate plane, 1000 points are randomly chosen Prove that there exists a way of coloring each of the points either red or blue (but not both) so that on every line parallel to the x-axis or y-axis, the number of red points minus the number of blue points is equal to −1, or Problem 404 Let I be the incenter of acute ∆ABC Let Γ be a circle with center I that lies inside ∆ABC D, E, F are the intersection points of circle Γ with the perpendicular rays from I to sides BC, CA, AB respectively Prove that lines AD, BE, CF are concurrent Problem 405 Determine all functions f,g: (0,+∞) → (0,+∞) such that for all positive number x, we have f ( g( x)) = x x and g( f ( x)) = xf ( x) − xg( x) − ***************** Solutions **************** Problem 396 Determine (with proof) all functions f : ℝ→ℝ such that for all real numbers x and y, we have f (x2 + xy + f (y)) = (f(x))2 + xf(y) + y Solution AN-anduud Problem Solving Group (Ulaanbaatar, Mongolia), CHEUNG Ka Wai (Munsang College (Hong Kong Island)), CHEUNG Wai Lam (Queen Elizabeth School), Dusan DROBNJAK (Mathematical Grammar School, Belgrade, Serbia), Kevin LAU (St Paul’s Co-educational College, S.4), Simon LEE (Carmel Alison Lam Foundation Secondary School), Mohammad Reza SATOURI (Persian Gulf University, Bushehr, Iran) and Maksim STOKIĆ (Mathematical Grammar School, Belgrade, Serbia) Call the required equation (*) For x=0, we get f(f(y))=y+f(0)2 for all y Call this (**) The right side may be any real number, hence f is surjective By (**), y = f(f(y))− f(0)2 If f(y) = f(y’), then the last equation implies y=y’, i.e f is injective Putting x=−y in (*), we get f(f(y)) = (f(−y))2−yf(y)+y for all y Call this (***) Now f surjective implies there exists z such that f(z)=0 Let x=y=z, then (*) yields f(2z2)=z Putting (x,y)=(0,2z2) in (*), we get 0=2z2+f(0)2 Then z=0 and f(0)=0 So (**) reduces to f(f(y))=y for all y Putting y = in (*), since f(0) = 0, we get f(x2) = (f(x))2 The last two sentences reduce (***) to y = (f(y))2−yf(y)+y This simplifies to f(y) = or f(y) = y for every y Since f is injective and f(0) = 0, we get f(y) = y for all y Conversely, a quick check shows f(y) = y for all y satisfies (*) Other commended solvers: Tobi MOEKTIJONO (National University of Singapore) Problem 397 Suppose in some set of 133 distinct positive integers, there are at least 799 pairs of relatively prime integers Prove that there exist a,b,c,d in the set such that gcd(a,b) = gcd(b,c) = gcd(c,d) = gcd(d,a) = Solution CHEUNG Ka Wai (Munsang College (Hong Kong Island)), Dusan DROBNJAK (Mathematical Grammar School, Belgrade, Serbia), Kevin LAU (St Paul’s Co-educational College, S.4), Simon LEE (Carmel Alison Lam Foundation Secondary School), Andy LOO (Princeton University), Tobi MOEKTIJONO (National University of Singapore) and Maksim STOKIĆ (Mathematical Grammar School, Belgrade, Serbia) Let S={n1, n2,…,n133} be the set of these 133 positive integers From i=1 to 133, let Xi be the set of all nk in S such that k≠i and gcd(ni,nk)=1 Denote by |X| the number of elements in set X For k≠i, gcd(ni,nk)=1 implies ni∈Xk and nk∈Xi Then N = |X1| + |X2| + ⋯ + |X133| ≥ 2×799 = 1598 Define f(x) = x(x−1)/2 In a set X with j elements, there are exactly j(j−1)/2 = f(|X|) pairs of distinct elements Since f(x) is concave on ℝ, by Jensen’s inequality, 133 ∑ f (| X i =1 i ⎛ N ⎞ ⎛ 1598 ⎞ |) ≥ 133 f ⎜ ⎟ ≥ 133 f ⎜ ⎟ ⎝ 133 ⎠ ⎝ 133 ⎠ > 133 f (12) = f (133) = f(|S|) Since every pair of distinct element in Xi is also a pair of distinct element in S, the inequality above implies in counting pairs of distinct elements in the Xi’s, there are repetitions, i.e there are Xi, Xk with i≠k sharing a common pair of distinct elements a,c Let b=ni and d=nk Then a,b,c,d satisfy gcd(a,b) = gcd(b,c) = gcd(c,d) = gcd(d,a) = Problem 398 Let k be positive integer and m an odd integer Show that there exists a positive integer n for which the number nn−m is divisible by 2k Solution AN-anduud Problem Solving Group (Ulaanbaatar, Mongolia), Dusan DROBNJAK (Mathematical Grammar School, Belgrade, Serbia), KWAN Chung Hang (Sir Ellis Kadoorie Secondary School (West Kowloon)), Kevin LAU (St Paul’s Co-educational College, S.3), Simon LEE (Carmel Alison Lam Foundation Secondary School), Andy LOO (Princeton University), Tobi MOEKTIJONO (National University of Singapore) and Maksim STOKIĆ (Mathematical Grammar School, Belgrade, Serbia) For k=1, let n=1 Suppose it is true for case k (i.e there exists n such that 2k | nn−m) Now m odd implies n odd For case k+1, if 2k+1 | nn−m, then the same n works for k+1 Otherwise, nn−m=2kl for some odd integer l Let v=2k By binomial theorem, (n+v)n+v = nn+v+(n+v)nn+v−1v+v2x = nn+v+vnn+v+v2y for some integers x,y By Euler’s theorem, since n is odd and φ(2k+1)=2k, k n v = n ≡ (mod k +1 ) Since l+nn is even, we have (n+v)n+v = nn+v+vnn+v+v2y ≡ nn+2knn = m+2k(l+nn) ≡ m (mod 2k+1) So n+v works for k+1 Problem 399 Let ABC be a triangle for which ∠BAC=60° Let P be the point of intersection of the bisector of ∠ABC and the side AC Let Q be the point of intersection of the bisector of ∠ACB and the side AB Let r1 and r2 be the radii of the incircles of triangles ABC and APQ respectively Find the radius of the circumcircle of triangle APQ in terms of r1 and r2 with proof Page Mathematical Excalibur, Vol 17, No 2, Oct 12 Solution Dusan DROBNJAK (Mathematical Grammar School, Belgrade, Serbia), Kevin LAU (St Paul’s Co-educational College, S.4), Andy LOO (Princeton University), MANOLOUDIS Apostolos (4° Lyk Korydallos, Piraeus, Greece), Tobi MOEKTIJONO (National University of Singapore) and Maksim STOKIĆ (Mathematical Grammar School, Belgrade, Serbia) A Q F E S P I B C Let I and S be the incenters of ΔABC and ΔAPQ respectively (Note A,S,I are on the bisector of ∠BAC.) Now ∠PIQ =∠CIB = 180°−(∠CBI +∠BCI) = 180°− ½(∠CBA+∠BCA) = 120° using ∠BAC=60° So APIQ is cyclic Applying sine law to ΔAPI, we get IP/(sin∠IAP) = 2R So R = IP By a well-known property of incenter, we have IP=IS (see vol.11, no.2, p.1 of Math Excal.) Let the incircles of ΔABC and ΔAPQ touch AC at E and F respectively Then R=IP=IS = AI−AS = IE/sin30°−SF/sin30° = 2r1−2r2 Other commended solvers: AN-anduud Problem Solving Group (Ulaanbaatar, Mongolia), Ioan Viorel CODREANU (Secondary School Satulung, Maramure, Romania), Simon LEE (Carmel Alison Lam Foundation Sec School) and Mihai STOENESCU (Bischwiller, France) Problem 400 Determine (with proof) all the polynomials P(x) with real coefficients such that for every rational number r, the equation P(x) = r has a rational solution Solution Tobi MOEKTIJONO (National University of Singapore), Maksim STOKIĆ (Mathematical Grammar School, Belgrade, Serbia) and TAM Ka Yu (MIT) We will show P(x) satisfies the desired condition if and only if P(x)=ax+b, where a,b ∈ℚ and a ≠ For the if-part, P(x) = r ∈ℚ implies x = (r−b)/a ∈ℚ Conversely, let P(x) satisfy the desired condition and let n=deg P For each r = 0,1,…, n, let P(xr)=r for some xr∈ℚ By the Lagrange interpolation formula, n ⎛ x − xi ⎞ ⎟ P( x) = ∑ ⎜⎜ r ∏ ⎟ r = ⎝ ≤ r ≤ n , r ≠ i xr − xi ⎠ Expanding the right side, we see P(x) has rational coefficients Letting M be the product of the denominators, we see Q(x)=MP(x) has integer coefficients Let k be the leading coefficient of Q(x) and c be the constant term of P(x) Let p1, p2, p3, … be the sequence of prime numbers Let P(x)=c+pi/M has solution ti∈ℚ Then Q(x)−(cM+pi) has k as the leading coefficient and −pi as constant term Now Q(ti)=0, which implies ti=1/di or pi∕di for some (not necessarily positive) divisor of k Since P(ti)’s are distinct, so the ti’s are distinct Hence, ti=1/di for at most as many time as the number of divisors of k So there must exist a divisor d of k such that there are infinitely many times ti=pi∕d This imply that P(x)−(c+dx/M)=0 has infinitely many solutions So the left side is the zero polynomial Then P(x)=ax+b with a=d/M≠0 and b=c rational Other commended solvers: Simon LEE (Carmel Alison Lam Foundation Secondary School) IMO 2012 (Member Perspective) (continued from page 2) The arrival of Dr Leung stirred up much happiness after the contest We reported on how we did Albert and Jimmy shone on Day 2, solving Problems and Kit was also comfortable with Problem while Kevin had some technical troubles in one particular case Nobody achieved anything substantial on Problem We celebrated that evening at a Chinese restaurant It was especially memorable that our deputy leader raised a couplet (對 聯), which he regarded as an open puzzle for millenniums (千古絕對): 望江樓，望江流 望江樓上望江流 江樓千古，江流千古 It took me nearly an hour to come up with a so-so solution: 觀雨亭，觀雨停 觀雨亭下觀雨停 雨亭四方，雨停四方 July 12 It was the contestants’ turn to have fun and the leaders’ turn to work hard At night, Dr Leung briefed us on the progress of the first day of coordination In addition to our previous expectations, Albert pocketed one point for proving the necessary condition on Problem Regretfully, Kit lost one point on Problem for not having verified the feasibility of the functions obtained Dr Leung had refused to sign Alice’s and Kevin’s scores on Problem in order to bargain later July 13 The marking scheme stipulated that any solution of Problem with coordinate geometry would score a if not a Despite our leaders’ relentless effort, the coordinators were able to detect a fatal error of mine So my Problem was destined to be a On another note, Dr Leung succeeded in getting point for Alice on Problem 4, which in his words was “an achievement” Kevin’s Problem was finalized with a score of July 14 We got up early in the morning to enjoy the sunrise scene at the seaside Kevin had a pitiful blunder His shoes and trousers were wetted by a sudden strike of waves That morning the last coordination on Problem was done Albert was awarded marks for his analytic struggle The uncertainties of our results then shifted from our actual scores to the medal cutting scores We went shopping for souvenirs in the afternoon and as soon as we got back to the hotel, I learned from the Chinese leaders that the cutting scores for Gold, Silver and Bronze Medals were 28, 21 and 14 respectively, all being multiples of I breathed a sigh of relief as my Silver Medal was ultimately secure July 15 In the afternoon we had the Closing Ceremony followed by a chain of photo-taking We won three Silver Medals (Albert, Jimmy and me), one Bronze Medal (Alice) and two Honorable Mentions (Kit and Kevin) July 16, 17 and 18 The six-hour bus journey from Mar del Plata to Buenos Aires passed rapidly in our dreams Then after a long flight, we were finally home in one piece and me with several bonus pimples In conclusion I shall stress one point – succinctly but with all the strength that I command – one can never pay sufficient tribute to our IMO trainers, who have so selflessly devoted countless hours of their own time to Mathematical Olympiad over the years I can find no words to thank them the way they truly deserve “Ask not what your country can for you; ask what you can for your country.” With this John F Kennedy exclamation I urge you all to support the 2016 Hong Kong IMO by whatever means you can, so that together we can make it an all-time success ... (1977 IMO silver, 1978 IMO gold, 1998 Fields medal) Vladmir Drinfeld (1969 IMO gold, 1990 Fields medal) Tim Gowers (1981 IMO gold, 1998 Fields medal) Laurent Lafforgue (1984 IMO silver, 1985 IMO. .. IMO 2004 will be held in Greece, IMO 2005 in Mexico, IMO 2006 in Slovenia IMO 2007 will be held in Vietnam, the site was decided during this IMO in Japan For the reader who will try out the IMO. .. gold, 1990 Nevanlinna prize) Peter Shor (1977 IMO silver, 1998 Nevanlinna prize) László Lovász (1963 IMO silver, 1964 IMO gold, 1965 IMO gold, 1966 IMO gold, 1999 Wolf prize) Page Mathematical