### 28th Putnam 1967

**Problem B1**

A hexagon is inscribed in a circle radius 1. Alternate sides have length 1. Show that the midpoints of the other three sides form an equilateral triangle.

**Solution**

Each side length 1 forms an equilateral triangle with the centre. Let the other three sides subtend angles 2A, 2B, 2C at the centre. The midpoints of the sides subtending 2A and 2B are distances cos A and cos B from the centre and the line joining them subtends an angle A+B+60 at the centre. So, using the cosine formula, the square of the distance between them is cos^{2}A + cos^{2}B - 2 cos A cos B cos(A+B+60) (*). But A + B + C = 90, so cos(A+B+60) = cos(150-C) = -√3/2 cos C + 1/2 sin C. Hence we may write (*) as (cos^{2}A + cos^{2}B + cos^{2}C) + √3 cos A cos B cos C - cos A cos B sin C - cos^{2}C. But we can write cos C as sin(A + B) = sin A cos B + cos A sin B and hence cos^{2}C as sin A cos B cos C + cos A sin B cos C. Thus (*) has the symmetrical form (cos^{2}A + cos^{2}B + cos^{2}C) + √3 cos A cos B cos C - (cos A cos B sin C + cos A sin B cos C + sin A cos B cos C), which establishes that the triangle is equilateral.

28th Putnam 1967

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002