5. In a given hemisphere of radius $R$, three spheres of equal radius are inscribed such that they touch each other and also touch the given hemisphere. Calculate the radius of the inscribed spheres.
## Solution.

Let $R$ be the radius of the hemisphere, $r$ the radius of the small spheres, and $d$ the distance from the center of the hemisphere to the point of tangency of a small sphere with the flat side of the hemisphere.

![](https://cdn.mathpix.com/cropped/2024_05_30_048b2e3055ac8a4932e8g-5.jpg?height=396&width=551&top_left_y=2114&top_left_x=717)

We need to determine the relationship between the radii $R$ and $r$. Notice that the center of the hemisphere, the center of one of the smaller spheres, and the corresponding point of tangency form a right triangle with legs $\tau, d$ and hypotenuse $R-\tau$.

(5 points)

![](https://cdn.mathpix.com/cropped/2024_05_30_048b2e3055ac8a4932e8g-6.jpg?height=490&width=526&top_left_y=475&top_left_x=755)

We see that the three points of tangency of the small spheres with the flat surface form a triangle congruent to the triangle formed by the centers of these circles, which is an equilateral triangle with side length $2r$. Therefore,

$$
d=\frac{2}{3} \cdot \frac{(2 r) \sqrt{3}}{2}=\frac{2 \sqrt{3}}{3} r
$$

Thus, from

$$
R-r=\sqrt{r^{2}+d^{2}}
$$

by substituting the obtained expression for $d$, we get

$$
\begin{gathered}
R-r=\sqrt{r^{2}+\frac{4}{3} r^{2}} \\
R=r\left(1+\sqrt{\frac{7}{3}}\right) \\
r=\frac{R \sqrt{3}}{\sqrt{3}+\sqrt{7}}=\frac{R(\sqrt{21}-3)}{4}
\end{gathered}
$$

(10 points)