: 3√2/(3 + √2) = 0.9611. Easy. Let O be the apex of the cone. Let C be the center of the base. Let V be one of the vertices of the cube not on the base. V must lie in the sloping side of the cone. Continue OV to meet the base at X. Let OC meet the top side of the cube at K. Now consider the triangle OXC. XC = 1, OC = 3. OVK is similar and VK = x/√2, where x is the side length of the cube. KC = x, so OK = 3 - x. Since the triangles are similar, OK/OC = VK/XC leading to x = 3√2/(3 + √2) = 0.9611. Putnam 1998 © John Scholes jscholes@kalva.demon.co.uk 12 Dec 1998
Putnam 1998 Problem A1 A cone has circular base radius 1, and vertex a height 3 directly above the center of the circle. A cube has four vertices in the base and four on the sloping sides. What length is a side of the cube? Solution