15th Putnam 1955

O is the center of a regular n-gon P₁P₂ ... P_n and X is a point outside the n-gon on the line OP₁. Show that XP₁ XP₂ ... XP_n + OP₁ⁿ = OXⁿ.

Solution

Easy.

Evidently it is sufficient to prove the result for the case OP₁ = 1. So represent P_k by the complex number ω^k, where ω = e^{i 2π/n} an nth root of unity. Represent X by the real number r. Then we have to show that |r - 1||r - ω| ... |r - ω^n-1| = rⁿ - 1, but that follows immediately because rⁿ - 1 = (r - 1)(r - ω) ... (r - ω^n-1).

15th Putnam 1955

© John Scholes
jscholes@kalva.demon.co.uk
26 Nov 1999