Snow Cones and Spheres: A Volume Solution and a New Geometry Puzzle

Sawyer introduces an extension of the previous Real Numbers challenge and first solves that earlier problem. The goal was to design a cone-shaped snow-cone holder whose rim fits the equator of a sphere of flavored ice and whose volume equals the lower hemisphere.

A cone is determined by its base radius and height. The sphere has radius 5 cm, so the matching rim gives the cone a base radius of 5 cm. The remaining unknown is the height that makes the two volumes equal.

The hemisphere’s volume is half of 4πr³/3, or 250π/3 cm³ for r = 5 cm. The cone’s volume is one third of its base area times its height: 25πh/3 cm³. Setting them equal gives h = 10 cm, the diameter of the original sphere.

The new challenge is to reshape the lower hemisphere into the inverted cone of radius 5 cm and height 10 cm while moving as little slush as possible. Ice is shaved from the portion of the sphere wider than the cone and packed into the cone’s tip below the sphere.

What is the radius of the lower circle where the sphere and cone intersect? As a bonus for readers familiar with integral calculus, Sawyer asks for the total volume of shaved ice that must be displaced. The episode leaves both questions for viewers to solve.