# Largest possible right circular cone inside a sphere

The question I'am trying to solve is

If you fit the largest possible cone inside a sphere, what fraction of the volume of the sphere is occupied by the cone? (Here by “cone” we mean a right circular cone, i.e., a cone for which the base is perpendicular to the axis of symmetry, and for which the cross-section cut perpendicular to the axis of symmetry at any point is a circle.)

The problem here is there are two variables we can alter, the height of the cone 'h' and the radius of the cone 'r'. Let the radius of the sphere be 'R', I'am trying to establish the relationship between the three. But I'am unable to figure out how.

If you take a cross-section through the axis of the cone, you see an isoceles triangle inscribed in a circle. You can take one vertex at $(0,R)$, and the other two at $(\pm R\cos\theta,R\sin\theta)$. As $R$ is constant, this leaves a single parameter.
• Thanks for hints. If you set one of the corner of triangle at (h-R, r) you can apply pythagoras to get $R^2 = (h - R)^2 + r^2$ – Mayur Kulkarni Mar 13 '17 at 7:54