HMMT November

Let $O$ be the center of the large sphere, and let $O_{1}, O_{2}, O_{3}$ be the centers of the small spheres. Consider $G$, the center of equilateral $\triangle O_{1} O_{2} O_{3}$. Then if the radii of the small spheres are $r$, we have that $O G=8+r$ and $O_{1} O_{2}=O_{2} O_{3}=O_{3} O_{1}=2 r$, implying that $O_{1} G=\frac{2 r}{\sqrt{3}}$. Then $O O_{1}=\sqrt{O G^{2}+O O_{1}^{2}}=\sqrt{(8+r)^{2}+\frac{4}{3} r^{2}}$. Now draw the array $O O_{1}$, and suppose it intersects the large sphere again at $P$. Then $P$ is the point of tangency between the large sphere and the small sphere with center $O_{1}$, so $O P=\sqrt{15^{2}+8^{2}}=17=O O_{1}+O_{1} P=\sqrt{(8+r)^{2}+\frac{4}{3} r^{2}}+r$. We rearrange this to be $$\begin{aligned} 17-r & =\sqrt{(8+r)^{2}+\frac{4}{3} r^{2}} \\ \Longleftrightarrow 289-34 r+r^{2} & =\frac{7}{3} r^{2}+16 r+64 \\ \Longleftrightarrow \frac{4}{3} r^{2}+50 r-225 & =0 \\ \Longrightarrow r & =\frac{-50 \pm \sqrt{50^{2}+4 \cdot \frac{4}{3} \cdot 225}}{2 \cdot \frac{4}{3}} \\ & =\frac{15 \sqrt{37}-75}{4} \end{aligned}$$