Problem 1230

Solution:

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 $OG = 8 + r$ and $O_{1}O_{2} = O_{2}O_{3} = O_{3}O_{1} = 2r$, implying that $O_{1}G = \frac{2r}{\sqrt{3}}$. Then $OO_{1} = \sqrt{OG^{2} + O_{1}G^{2}} = \sqrt{(8 + r)^{2} + \frac{4}{3} r^{2}}$.

Now draw the array $OO_{1}$, 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 $OP = \sqrt{15^{2} + 8^{2}} = 17 = OO_{1} + O_{1}P = \sqrt{(8 + r)^{2} + \frac{4}{3} r^{2}} + r$. We rearrange this to be
$$\begin{aligned} 17 - r &= \sqrt{(8 + r)^{2} + \frac{4}{3} r^{2}} \\ \Longleftrightarrow 289 - 34r + r^{2} &= \frac{7}{3} r^{2} + 16r + 64 \\ \Longleftrightarrow \frac{4}{3} r^{2} + 50r - 225 &= 0 \\ \Longrightarrow r &= \frac{-50 \pm \sqrt{50^{2} + 4 \cdot \frac{4}{3} \cdot 225}}{2 \cdot \frac{4}{3}} \\ &= \frac{15 \sqrt{37} - 75}{4} . \end{aligned}$$