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Topic:
Comments on some IMO problems
Replies:
6
Last Post:
Jul 24, 1996 3:14 PM




Comments on some IMO problems
Posted:
Jul 21, 1996 5:04 PM


Thank you for posting here the IMO problems! I solved only the first day and doubt if I try the second, coz I'm not very impressed. I refer to Wonder Worm's comments and solutions. 1. Let ABCD be a rectangular board, with AB = 20 and BC = 12. The board is divided into 20x12 squares. Let r be a given positive integer. A coin can be moved from one square to another iff the distance between the centres of the two squares is sqrt(r). The task is to find a sequence of moves taking the coin from the square with A as a vertex to the square with B as a vertex. a) Show that the task cannot be done if r is divisible by 2 or 3. b) Prove that the task can be done if r = 73. c) Can the task be done when r = 97? >c) The answer is no. 97 = 9^2 + 4^2 is the only way of expressing 97 as > the sum of two (nonnegative) squares, since it is prime. So we can > only change coordinates by (+4, +9) or (+9, +4). I constructed a > table for each point showing the points reachable from it that were > legal (i.e. inside the given rectangle). Actually, not for each point, > but for each reachable point. There are only 60 such points, and (20,1) > is not one of them. I also made a diagram for this part, but mine contains only 58 possible reachable fields! Here it is: D  C  X X X X  X X X X X X   X X X X  X X X X X   X X X X   X X X X X X 12  X X X X   X X X X X  X X X X  X X X X X X   X X X X  X X X X X X  A  B 20 where X denotes the reachable field for the coin in part (c). I don't claim it's correct. Has anyone made a similar diagram to compare? > I was quite disappointed with this as a question; part a yields to standard > technique, while parts b and c can be done fairly mindlessly and reasonably > quickly by the aforementioned table idea (although it is easier to construct > the solution in case b). Questions which come in parts are always (IMHO) > a bad idea for olympiads, and 4 cases is too many. I entirely agree. I learned nothing from this problem and had no pleasure "working" on it... I wonder how many IMO contestants made the above diagram during the competition and how many coordinates enjoyed correcting such papers! > 3. Let S = {0, 1, 2, ...}. Find all functions f defined on S taking their > values in S such that f(m + f(n)) = f(f(m)) + f(n) for all m and n in S. > Let f(1) = a. Then f(a) = f(f(1)) = f(1) = a. [...] > In general, m is not equal to n, so we must have a = 0 or 1. These correspond > to the functions f(x) = 0 for all x, and f(x) = x for all x, both of which are > easily seen to be solutions. I'm afraid that's not all... How about f(0)=0, f(2k1)=f(2k)=2k for k=1,2,3,... ? Generally, fix a positive integer a and let r denote the remainder of division n by a. Then f(n) = n + ag(r)  r, where g: {0,1,...,a1} > S is any function with g(0)=0, is also a solution! Contrary to all geometry ignorants, I'll say that I really enjoyed solving problem 2, which I find beautiful! I found a nice solution, and my friend found another, also very nice. Since I'm kinda tired of sitting at the computer today, I'm not posting these solutions now. If there is a need of seeing them, I'll do it in a couple of days  just let me know. Waldemar Pompe



