Question 149, In a sphere of radius $R$ are placed four smaller spheres, each of radius $r$. The maximum possible value of $r$ is $\qquad$ _.
Question 149, Solution: Let the centers of the four small spheres be $\mathrm{O}_{\mathrm{i}}(1 \leq \mathrm{i} \leq 4)$, then $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$ form a regular tetrahedron. Let the center of the large sphere be 0, and let $P$ be the centroid of $\triangle O_{1} \mathrm{O}_{2} \mathrm{O}_{3}$, then 0 is at $\frac{3}{4}$ of $\mathrm{O}_{4} \mathrm{P}$. Since $\mathrm{O}_{1} \mathrm{O}_{4}=2 \mathrm{r}$, $\mathrm{O}_{1} \mathrm{P}=\frac{2 \sqrt{3} \mathrm{r}}{3}$, thus $\mathrm{OO}_{4}=\frac{3 \mathrm{PO}_{4}}{4}=\frac{\sqrt{(2 \mathrm{r})^{2}-\left(\frac{2 \sqrt{3} r}{3}\right)^{2}}}{4}=\frac{\sqrt{6} \mathrm{r}}{2}$, hence $R=0 \mathrm{O}_{4}+\mathrm{r}=\frac{(2+\sqrt{6}) \mathrm{r}}{2}$, solving for $\mathrm{r}=(\sqrt{6}-2) R$.