Let A, B be adjacent vertices of a regular n-gon (n ≥ 5) with center O. A triangle XYZ, which is congruent to and initially coincides with OAB, moves in the plane in such a way that Y and Z each trace out the whole boundary of the polygon, with X remaining inside the polygon. Find the locus of X.
Solution
Take AB = 2 and let M be the midpoint of AB. Take coordinates with origin at A, x-axis as AB and y-axis directed inside the n-gon. Let Z move along AB from B towards A. Let ∠YZA be t. Let the coordinates of X be (x, y). ∠YZX = π/2 - π/n, so XZ = 1/sin π/n and y = XZ sin(t + π/2 - π/n) = sin t + cot π/n cos t.
BY sin 2π/n = YZ sin t = 2 sin t. MX = cot π/n. So x = MY cos t - BY cos 2π/n + MX sin t = cos t + (cot π/n - 2 cot 2π/n) sin t = cos t + tan π/n sin t = y tan π/n. Thus the locus of X is a star formed of n lines segments emanating from O. X moves out from O to the tip of a line segement and then back to O, then out along the next segment and so on. x2 + y2 = (1/sin2π/n + 1/cos2π/n) cos2(t + π/n). Thus the length of each segment is (1 - cos π/n)/(sin π/n cos π/n).
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 1998