Problem 1357

1. Let $O$ be the center of the circle $\mathcal{C}$ with radius $R$. Given that $\angle AOB = \frac{\pi}{2}$, points $A$ and $B$ are on the circle $\mathcal{C}$ such that the arc $AB$ subtends a right angle at the center $O$.

2. Let $O_1$ and $O_2$ be the centers of the circles $\mathcal{C}_1$ and $\mathcal{C}_2$ respectively, with radii $R_1$ and $R_2$. These circles are internally tangent to $\mathcal{C}$ at points $A$ and $B$ respectively, and externally tangent to each other.

3. Since $\mathcal{C}_1$ and $\mathcal{C}_2$ are tangent to $\mathcal{C}$ at $A$ and $B$, the distances $O_1A$ and $O_2B$ are equal to $R - R_1$ and $R - R_2$ respectively.

4. The circles $\mathcal{C}_1$ and $\mathcal{C}_2$ are externally tangent to each other, so the distance $O_1O_2$ is equal to $R_1 + R_2$.

5. Consider the rectangle $O_2OO_1D$ where $D$ is the fourth vertex. Since $\angle AOB = \frac{\pi}{2}$, the points $O_1, O, O_2$ form a right triangle with $\angle O_1OO_2 = \frac{\pi}{2}$.

6. In the right triangle $\triangle O_1OO_2$, we have:
$$O_1O_2 = R_1 + R_2$$
$$OO_1 = R - R_1$$
$$OO_2 = R - R_2$$

7. By the Pythagorean theorem in $\triangle O_1OO_2$:
$$O_1O_2^2 = OO_1^2 + OO_2^2$$
$$(R_1 + R_2)^2 = (R - R_1)^2 + (R - R_2)^2$$
$$R_1^2 + 2R_1R_2 + R_2^2 = R^2 - 2RR_1 + R_1^2 + R^2 - 2RR_2 + R_2^2$$
$$2R_1R_2 = 2R^2 - 2R(R_1 + R_2)$$
$$R_1R_2 = R^2 - R(R_1 + R_2)$$

8. Since $D$ is the fourth vertex of the rectangle $O_2OO_1D$, it lies on the circle $\mathcal{C}$ and is equidistant from $O_1$ and $O_2$.

9. The circle $\mathcal{C}_3$ is tangent to $\mathcal{C}_1$ and $\mathcal{C}_2$ at points $P$ and $R$ respectively, and internally tangent to $\mathcal{C}$ at point $S$. The center of $\mathcal{C}_3$ is $D$.

10. Since $D$ is the center of $\mathcal{C}_3$, the distances $DP$, $DR$, and $DS$ are equal to the radius of $\mathcal{C}_3$, which is $R - (R_1 + R_2)$.

11. The angle $\angle PSR$ is the angle subtended by the arc $PR$ at the center $D$ of $\mathcal{C}_3$. Since $\mathcal{C}_3$ is tangent to $\mathcal{C}_1$ and $\mathcal{C}_2$ at $P$ and $R$ respectively, and these points are symmetric with respect to $D$, the angle $\angle PSR$ is $45^\circ$.

$$\boxed{45^\circ}$$