### 39th Putnam 1978

**Problem B1**

A convex octagon inscribed in a circle has 4 consecutive sides length 3 and the remaining sides length 2. Find its area.

**Solution**

Let the radius be R. The area is made up of four triangles sides 3, R, R and four triangles sides 2, R, R. The area is unchanged if we rearrange these triangles to form an octagon with sides alternately 2 and 3. But the new octagon is a square side 3 + 2√2 with four corners lopped off, each sides 2, √2, √2. Hence its area is (17 + 12√2) - 4 = 13 + 12√2.

39th Putnam 1978

© John Scholes

jscholes@kalva.demon.co.uk

30 Nov 1999