From dsavitt@unixg.ubc.ca Wed Jul 20 12:44:44 MET DST 1994 Article: 65433 of sci.math Path: news.win.tue.nl!tuegate.tue.nl!news.nic.surfnet.nl!sun4nl!EU.net!howland.reston.ans.net!europa.eng.gtefsd.com!newsxfer.itd.umich.edu!nntp.cs.ubc.ca!unixg.ubc.ca!unixg.ubc.ca!dsavitt From: dsavitt@unixg.ubc.ca (David Savitt) Newsgroups: rec.puzzles,sci.math Subject: Re: International Math Olympiad papers Date: 20 Jul 94 00:19:38 GMT Organization: The University of British Columbia Lines: 194 Message-ID: References: NNTP-Posting-Host: unixg.ubc.ca Keywords: IMO Xref: news.win.tue.nl rec.puzzles:27364 sci.math:65433 Here are the problems and results from this year's IMO. This post is a copy of the original post in sci.math, I don't recall by whom. ---------------------------------------------------------------------------- Problems for the 35th International Mathematical Olympiad: (held in Hong Kong, July 12-19, 1994) --------------------------------------------------------- First Day (9am-1:30pm, July 13, 1994; Each problem is worth 7 points) Problem 1 (proposed by France) Let m and n be positive integers. Let a_1, a_2, ..., a_m be distinct elements of {1, 2, ..., n} such that whenever a_i + a_j <= n for some i, j, 1 <= i <= j <= m, there exists k, 1 <= k <= m, with a_i + a_j = a_k. Prove that a_1 + a_2 + ... + a_m n + 1 ------------------------- >= --------- m 2 Problem 2 (proposed by Armenia/Australia) ABC is an isosceles triangle with AB=AC. Suppose that (i) M is the midpoint of BC and O is the point on the line AM such that OB is perpendicular to AB; (ii) Q is an arbitrary point on the segment BC different from B and C; (iii) E lies on the line AB and F lies on the line AC such that E, Q and F are distinct and collinear. Prove that OQ is perpendicular to EF if and only if QE=QF. Problem 3 (proposed by Romania) For any positive integer k, let f(k) be the number of elements in the set {k+1, k+2, ..., 2k} whose base 2 representation has precisely three 1s. (a) Prove that, for each positive integer m, there exists at least one positive integer k such that f(k)=m. (b) Determine all positive integer m for which there exists exactly one k with f(k)=m. -------------------------------------------------------------------------- Second Day (9am-1:30pm, July 14, 1994; Each problem is worth 7 points) Problem 4 (proposed by Australia) Determine all ordered pairs (m,n) of positive integers such that n^3 + 1 ------------- mn - 1 is an integer. Problem 5 (proposed by United Kingdom) Let S be the set of real numbers strictly greater than -1. Find all functions f: S -> S satisfying the two conditions: (i) f(x+f(y)+xf(y)) = y+f(x)+yf(x) for all x and y in S; (ii) f(x)/x is strictly increasing on each of the intervals -1 < x < 0 and 0 < x. Problem 6 (proposed by Finland) Show that there exists a set A of positive integers with the following property: For any infinite set S of primes there exist two positive integers m in A and n not in A each of which is a product of k distinct elements of S for some k >= 2. **************************************************************************** Result of the 35th International Mathematical Olympiad (held in Hong Kong, July 12-19, 1994) Rank Teams Score Gold Silver Bronze 1 United States 252 6 0 0 2 China 229 3 3 0 3 Russia 224 3 2 1 4 Bulgaria 223 3 2 1 5 Hungary 221 1 5 0 6 Vietnam 207 1 5 0 7 United Kingdom 206 2 2 2 8 Iran 203 2 2 2 9 Romania 198 0 5 1 10 Japan 180 1 2 3 11 Australia 173 0 2 3 12 Chinese Taipei 170 0 4 1 12 Republic of Korea 170 0 2 4 14 India 168 0 3 3 15 Ukraine 163 1 1 2 16 Hong Kong 162 0 2 4 17 France 161 1 1 3 18 Poland 160 2 0 3 19 Argentina 159 0 3 1 20 Czech Republic 154 0 2 2 21 Slovakia 150 1 1 2 22 Germany 145 1 2 3 23 Belarus 144 0 1 4 24 Canada 143 1 0 3 24 Israel 143 0 1 4 26 Colombia 136 0 2 2 27 South Africa 120 0 0 3 28 Turkey 118 0 0 4 29 New Zealand 116 0 0 4 29 Singapore 116 0 2 0 31 Austria 114 1 0 0 32 Armenia/5 110 0 0 4 33 Thailand 106 0 0 3 34 Belgium 105 0 0 2 34 Morocco 105 0 0 2 36 Italy 102 0 0 2 37 Netherlands 99 0 0 2 38 Latvia 98 0 0 3 39 Brazil/5 95 0 2 0 39 Republic of Georgia 95 0 0 2 41 Sweden 92 0 0 1 42 Greece 91 0 0 1 43 Croatia 90 0 0 2 44 Estonia/5 82 0 0 1 45 Norway 80 0 1 1 46 Macau 75 0 1 0 47 Lithuania 73 0 0 1 48 Finland 70 0 0 0 49 Ireland 68 0 0 0 50 Macedonia/4 67 0 0 1 51 Mongolia 65 0 1 0 52 Trinidad & Tobago 63 0 0 0 53 Philippines 53 0 0 0 54 Chile/2 52 0 1 0 55 Moldova 52 0 0 1 55 Portugal 52 0 0 0 57 Denmark/4 51 0 0 2 58 Cyprus 48 0 0 0 59 Slovenia/5 47 0 0 0 60 Indonesia 46 0 0 0 61 Bosnia/5 44 0 0 1 62 Spain 41 0 0 0 63 Switzerland/3 35 0 0 1 64 Luxembourg/1 32 0 1 0 65 Iceland/4 29 0 0 0 66 Mexico 29 0 0 0 67 Kryghyzstan 24 0 0 0 68 Cuba/1 12 0 0 0 69 Kuwait/5 12 0 0 0 Remarks:1. A full team consists of 6 students. Armenia/5 means that Armenia sent only 5 students. 2. The maximum possible score is 252 points: 6 students x 6 problems/student x 7 points/problem = 252 points. 3. For Problem #2, the "if" part was proposed by Australia and the "only if" part was proposed by Armenia. It is interesting that two countries that are in two different hemispheres, with different languages, without any communication, would propose the same problem such that their proposed problems complement each other. Some statistics (from the IMO committee): Number of participating teams: 69 Number of participating contestants: 385 Number of Gold Medals (scored 40-42): 30 Number of Silver Medals(scored 30-39): 64 Number of Bronze Medals(scored 19-29): 98 Future host countries for the IMO: 1995: Canada 1996: India 1997: Argentina 1998: Taiwan 1999: Romania 2000: South Korea --------------------------------------------------------------------------- -Dave -- ------------------------------------------------------------------------------ Dave Savitt | 3rd year mathematics | AKA Little Dave | Go VooDoo! dsavitt@unixg.ubc.ca | University of B.C. | AKA Goliath | Go Jays! --- "I am not a crook" - Richard Nixon ----- "I am not your cook" - my mom ---