Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of the larger sphere? $\mathrm{(A) \ } 3+\dfrac{\sqrt{30}}{2} \qquad$ $ \mathrm{(B) \ } 3+\dfrac{\sqrt{69}}{3} \qquad$ $ \mathrm{(C) \ } 3+\dfrac{\sqrt{123}}{4} \qquad$ $ \mathrm{(D) \ } \dfrac{52}{9} \qquad$ $ \mathrm{(E) \ } 3+2\sqrt{2}$
Solution
The centers of the three radius-$1$ spheres form an equilateral triangle of side $2$ at height $1$; its circumradius is $\frac2{\sqrt3}$.
The center of the big (radius $2$) sphere lies above the centroid at height $H$ with $\left(\frac2{\sqrt3}\right)^2+(H-1)^2=3^2$, giving $H=1+\frac{\sqrt{69}}3$.
The top of the big sphere is at height $H+2=3+\frac{\sqrt{69}}3$. The answer is (\textbf{B}).
Three mutually tangent spheres of radius $3$ rest on a horizontal plane. A sphere of radius $1$ rests on top of them, tangent to all three. What is the distance from the plane to the top of the smaller sphere?
