An n-gon has all angles equal and the lengths of consecutive sides satisfy a1 ≥ a2 ≥ ... ≥ an. Prove that all the sides are equal.
Solution
For n odd consider the perpendicular distance of the shortest side from the opposite vertex. This is a sum of terms ai x cosine of some angle. We can go either way round. The angles are the same in both cases, so the inequalities give that a1 = an-1, and hence a1 = ai for all i < n. We get a1 = an by repeating the argument for the next shortest side. The case n even is easier, because we take a line through the vertex with sides a1 and an making equal angles with them and look at the perpendicular distance to the opposite vertex. This gives immediately that a1 = an.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
21 Sep 1998
Last corrected/updated 24 Sep 2003