Question

A right circular cone has base radius $r$ and height $h.$ The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making $17$ complete rotations. The value of $h/r$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n.$

Answer

Solution:

The contact point of the base stays at distance $\ell = \sqrt{r^2 + h^2}$ (the slant height) from the fixed vertex, so it traces a circle of radius $\ell.$ Rolling without slipping, the cone makes one rotation for each base circumference of arc, so returning after exactly $17$ rotations means $2\pi\sqrt{r^2 + h^2} = 17 \cdot 2\pi r, \qquad \text{i.e.} \qquad \sqrt{r^2 + h^2} = 17r.$

Squaring gives $h^2 = 288r^2,$ so $h/r = \sqrt{288} = 12\sqrt{2},$ and $m + n = 12 + 2 = 14.$

2008 AIME I Problem 5

Solution:

Problem 5 in Other Years