A convex pentagon has rational sides and equal angles. Show that it is regular.
Solution
Use complex numbers. Let the vertices be the complex numbers v1, v2, ... , v5. Let z1 = v2 - v1, z2 = v3 - v2, z3 = v4 - v3, z4 = v5 - v4, z5 = v1 - v5. Then we have z1 + z2 + z3 + z4 + z5 = 0.
Let wi = zi/z1. We now have w1 = 1, and w1 + w2 + w3 + w4 + w5 = 0. Since the pentagon had equal angles, we have (relabeling if necessary), w2 = a2ω, w3 = a3ω2, w4 = a4ω3, w5 = a5ω4, where ω = exp(i2π/5) and the ai are all rational. So ω satisfies the rational polynomial 1 + a2ω + a3ω2 + a4ω3 + a5ω4 = 0.
But the minimal polynomial for ω over the rationals is 1 + ω + ω2 + ω3 + ω4 = 0. Hence a2 = a3 = a4 = a5 = 1, hence all |wi| = 1, hence all |zi| = |z1|, so the pentagon is regular.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002