6th Eötvös 1899

ABCDE is a regular pentagon (with vertices in that order) inscribed in a circle of radius 1. Show that AB·AC = √5.

Solution

AB subtends 72^o at the center and AC subtends 144^o at the center. So AB·AC = 4 sin 36^o sin 72^o.

Using e^iθ = cos θ + i sin θ or otherwise, we get sin 5k = sin k (16 sin⁴k - 20 sin²k + 5) = 0. The roots of sin 5k = 0 are k = 0, ±36^o, ±72^o, so the roots of 16s² - 20s + 5 = 0 are sin²36^o and sin²72^o. Hence sin²36^o sin²72^o = 5/16, so 4 sin 36^o sin 72^o = √5.

6th Eötvös 1899

© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03