Inside a right circular cone with base radius $5$ and height $12$ are three congruent spheres with radius $r$. Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is $r$? $\textbf{(A)}\ \frac{3}{2} \qquad$ $\textbf{(B)}\ \frac{90-40\sqrt{3}}{11} \qquad$ $\textbf{(C)}\ 2 \qquad$ $\textbf{(D)}\ \frac{144-25\sqrt{3}}{44} \qquad$ $\textbf{(E)}\ \frac{5}{2}$
Solution
The three sphere centers, at height $r$ above the base, lie on a circle of radius $d=\frac{2r}{\sqrt3}$ about the axis (since consecutive centers are a distance $2r$ apart).
In the axial cross-section, the lateral surface is the line through $(5,0)$ and $(0,12)$: $12x+5z=60$. Tangency of a sphere center $(d,r)$ to this line at distance $r$ gives $\frac{60-12d-5r}{13}=r$, so $d=5-\frac32r$.
Combining $\frac{2r}{\sqrt3}=5-\frac32r$ gives $r=\dfrac{90-40\sqrt3}{11}$. The answer is (\textbf{B}).
Inside a right circular cone with base radius $2\sqrt3$ and height $6$ are three congruent spheres of radius $r$. Each sphere is tangent to the other two spheres and also tangent to the base and to the lateral surface of the cone. What is $r$?
$\textbf{(A)}\ \frac{3}{2} \qquad$ $\textbf{(B)}\ \frac{90-40\sqrt{3}}{11} \qquad$ $\textbf{(C)}\ 2 \qquad$ $\textbf{(D)}\ \frac{144-25\sqrt{3}}{44} \qquad$ $\textbf{(E)}\ \frac{5}{2}$ 