BMO 1966 Problem 4 A 1 , A 2 , A 3 , A 4 are consecutive vertices of a regular n-gon. 1/A 1 A 2 = 1/A 1 A 3 + 1/A 1 A 4 . What are the possible values of n? Solution
: n = 7. It is a nice question whether one has to show that n = 7 is a solution! [Does the question give as a fact that there exists a solution?] Clearly, there are no other possible solutions. Suppose the n-gon has side 1. Then A 1 A 3 and A 1 A 4 are obviously strictly increasing functions of n. So 1/A 1 A 3 + 1/A 1 A 4 is a strictly decreasing function of n, whereas 1/A 1 A 2 is always 1. So there can be at most one solution. Moreover, for n = 6, the lhs is 1/√3 + 1/2 > 1, whereas for n = 8, 1/√(2 + √2) + (2 - 1) < 1. So if there is a solution it is 7. If we are not allowed to assume that there is a solution, then we have to prove equality for n = 7. A kludgy (but easy) proof is as follows. A 1 A 3 = 2 cos π/7, A 1 A 4 = (1 + 2 cos 2π/7). Put c = cos π/7. The cos 2π/7 = 2c 2 - 1. So the result is true if 8c 3 - 4c 2 - 4c + 1 = 0. Put s = sin π/7. Then (c + is) 7 = - 1. Expanding and comparing real parts, we get: c 7 - 21c 5 (1 - c 2 ) + 35c 3 (1 - c 2 ) 2 - 7c(1 - c 2 ) 3 + 1 = 0, or 64c 7 - 112c 5 + 56c 3 - 7c + 1 = 0. But we know that c = -1 is a solution (not the one we want), so we can factor that out to get: 64c 6 - 64c 5 - 48c 4 + 48c 3 + 8c 2 - 8c + 1 = 0. But we know that this has three repeated roots (cos π/7 = cos (-π/7) etc), so it should factorize as the square of a cubic and indeed it does: (8c 3 - 4c 2 - 4c + 1) 2 = 0, so 8c 3 - 4c 2 - 4c + 1 = 0 as required. 2nd BMO 1966 © John Scholes jscholes@kalva.demon.co.uk 30 Aug 2000