# Mock AIME 2 2006-2007 Problems/Problem 7

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

A right circular cone of base radius $17$cm and slant height $51$cm is given. $P$ is a point on the circumference of the base and the shortest path from $P$ around the cone and back is drawn (see diagram). If the length of this path is $m\sqrt{n},$ where $n$ is squarefree, find $m+n.$

## Solution

"Unfolding" this cone results in a circular sector with radius $51$ and arc length $17\cdot 2\pi=34\pi$. Let the vertex of this sector be $O$. The problem is then reduced to finding the shortest distance between the two points $A$ and $B$ on the arc that are the farthest away from each other. Since $34\pi$ is $1/3$ of the circumference of a circle with radius $51$, we must have that $\angle AOB=\frac{360^{\circ}}{3}=120^{\circ}$. We know that $AO=OB=51$, so we can use the Law of Cosines to find the length of $AB$: $$AB=\sqrt{AO^2+OB^2-2AO\cdot OB\cdot\cos{120^{\circ}}}=\sqrt{51^2+51^2+51^2}=51\sqrt{3}.$$ Hence $m=51$, $n=3$, $m+n=\boxed{054}$.