Putnam 1998, Problem A-1

A right circular cone has base of radius 1 and height 3. A cube is inscribed in the cone so that one face of the cube is contained in the base of the cone. What is the side-length of the cube?

Solution:
Consider the plane containing both the axis of the cone and two opposite vertices of the cube's bottom face. The cross section of the cone and the cube in this plane consists of a rectangle of sides \( s \) and \( s\sqrt{2} \) inscribed in an isosceles triangle of base \( 2 \) and height \( 3 \), where \( s \) is the side-length of the cube. (The \( s\sqrt{2} \) side of the rectangle lies on the base of the triangle.) Similar triangles yield \( \frac{s}{3} = \frac{1 - s\sqrt{2}/2}{1} \), or \( s = \frac{9\sqrt{2} - 6}{7} \).