IMO 1990 SL 16

Origin: NET

Problem

Is there a 1990-gon with the following properties (i) and (ii)? (i) All angles are equal; (ii) The lengths of the 1990 sides are a permutation of the numbers 12, 22, . . . , 19892, 19902.

Solution

Let A0A1 . . . A1989 be the desired 1990-gon. We also define A1990 = A0. Let O be an arbitrary point. For 1 \leqi \leq1990 let Bi be a point such that −−\to OBi = −−−−\to Ai−1Ai. We define B0 = B1990. The points Bi must satisfy the fol- lowing properties: \angleBiOBi+1 = 2\pi 1990, 0 \leqi \leq1989, lengths of OBi are a permutation of 12, 22, . . . , 19892, 19902, and 1989 i=0 −−\to OBi = −\to0 . Conversely, any such set of points Bi corresponds to a desired 1990-gon. Hence, our goal is to construct vectors −−\to OBi satisfying all the stated properties. Let us group vectors of lengths (2n −1)2 and (2n)2 into pairs and put them diametrically opposite each other. The length of the resulting vec- tors is 4n −1. The problem thus reduces to arranging vectors of lengths 3, 7, 11, . . ., 3979 at mutual angles of 2\pi 995 such that their sum is −\to0 . We partition the 995 directions into 199 sets of five directions at mutual an- gles 2\pi 5 . The directions when intersected with a unit circle form a regular pentagon. We group the set of lengths of vectors 3, 7, . . ., 3979 into 199 sets of five consecutive elements of the set. We place each group of lengths on directions belonging to the same group of directions, thus constructing five vectors. We use that −−\to OC1 + \cdot \cdot \cdot + −−\to OCn = 0 where O is the center of a regular n-gon C1 . . . Cn. In other words, vectors of equal lengths along directions that form a regular n-gon cancel each other out. Such are the groups of five directions. Hence, we can assume for each group of five lengths for its lengths to be {0, 4, 8, 12, 16}. We place these five lengths

in a random fashion on a single group of directions. We then rotate the configuration clockwise by 2\pi 199 to cover other groups of directions and re- peat until all groups of directions are exhausted. It follows that all vectors of each of the lengths {0, 4, 8, 12, 16} will form a regular 199-gon and will thus cancel each other out. We have thus constructed a way of obtaining points Bi and have hence shown the existence of the 1990-gon satisfying (i) and (ii).