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
Center the pyramid's base square at the origin with half-side $a/2$, apex at $(0,0,6)$. One lateral face's plane, through the base edge $x=\frac a2$ and the apex, has equation $12x+az=6a$ (after simplification).
Distance from the origin (hemisphere's center) to this plane equals the hemisphere's radius $2$: $\frac{6a}{\sqrt{144+a^2}}=2\Rightarrow 9a^2=144+a^2\Rightarrow a=3\sqrt2$. The answer is (\textbf{A}).
The circular base of a hemisphere of radius $3$ rests on the base of a square pyramid of height $5$. 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}$ 