# 2011 AMC 12A Problems/Problem 15

## Problem

The circular base of a hemisphere of radius $2$ rests on the base of a square pyramid of height $6$. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?

$\textbf{(A)}\ 3\sqrt{2} \qquad \textbf{(B)}\ \frac{13}{3} \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ \frac{13}{2}$

## Solution 1

Let $ABCDE$ be the pyramid with $ABCD$ as the square base. Let $O$ and $M$ be the center of square $ABCD$ and the midpoint of side $AB$ respectively. Lastly, let the hemisphere be tangent to the triangular face $ABE$ at $P$.

Notice that $\triangle EOM$ has a right angle at $O$. Since the hemisphere is tangent to the triangular face $ABE$ at $P$, $\angle EPO$ is also $90^{\circ}$. Hence $\triangle EOM$ is similar to $\triangle EPO$.

$\frac{OM}{2} = \frac{6}{EP}$

$OM = \frac{6}{EP} \times 2$

$OM = \frac{6}{\sqrt{6^2 - 2^2}} \times 2 = \frac{3\sqrt{2}}{2}$

The length of the square base is thus $2 \times \frac{3\sqrt{2}}{2} = 3\sqrt{2} \rightarrow \boxed{\textbf{A}}$

Consider a cross section of the pyramid such that it is a hemisphere inscribed inside of a triangle. Let $A$ be the vertex furthest away from the hemisphere. Let $P$ be a point where the hemisphere is tangent. Let $O$ be the centre of the hemisphere. Triangle $AOP$ is a right triangle, since $P$ is ninety degrees. Apply the pythagorean formula to find side $AP$ as $4\sqrt{2}.$ Since we know the answer must have $\sqrt{2}$ somewhere, we can remove choices B, D and E from selection. Choice C is simply the length of $AP$, which cannot be the desired length. Thus, the answer must be $\boxed{\textbf{A}}$.