61st Putnam 2000 Problem A3 An octagon is incribed in a circle. One set of alternate vertices forms a square area 5. The other set forms a rectangle area 4. What is the maximum possible area for the octagon? Solution
: 3√5. Evidently the circle has radius √5/√2, and the rectangle has sides √2 and 2√2. The trick is to focus on the rectangle. The area of the octagon is the area of the rectangle plus the area of the four triangles on its sides. The area of each of these triangles is maximised by taking the vertex midway between the two vertices of the rectangle. But these four midpoints happen to form a square. The area of the two triangles with common base on a side of the rectangle, one with third vertex at the centre of the circle and the other at the midpoint of the arc is (radius x side)/2. So the total area of the octagon is (radius x sum of two adjacent sides of the rectangle) = 3√5. 61st Putnam 2000 © John Scholes jscholes@kalva.demon.co.uk 1 Jan 2001