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.
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