Mock AIME 2 2006-2007 Problems/Problem 7

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 $m=51$ , $n=3$ , $m+n=\boxed{054}$ .

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Page

Toolbox

Search

Mock AIME 2 2006-2007 Problems/Problem 7

Problem

Solution

See Also