Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3. Find the length of the segment $O D_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9620aefbc3ee8b62d5ddg-48.jpg?height=593&width=593&top_left_y=91&top_left_x=428)
Answer: 17.

Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$

![](https://cdn.mathpix.com/cropped/2024_05_06_9620aefbc3ee8b62d5ddg-48.jpg?height=595&width=591&top_left_y=841&top_left_x=431)

Fig. 10: to the solution of problem 11.6

drop a perpendicular $O X$ to this face, then point $X$ is the center of $\omega$ (point $O$ is equidistant from all points of $\omega$, so the projection of $O$ onto the plane of $\omega$ is also equidistant from them). Let $Y$ be an arbitrary point on $\omega$ (Fig. 10). Triangle $OXY$ is a right triangle; by the problem statement, $X Y=3$ and $O Y=10$. By the Pythagorean theorem, we get $O X^{2}=10^{2}-3^{2}=91$.

Similarly, we find the squares of the distances from point $O$ to the planes $A_{1} B_{1} C_{1} D_{1}$ and $A D D_{1} A_{1}$, which are both equal to $10^{2}-1^{2}=99$.

By the spatial Pythagorean theorem, the square of the length of segment $O D_{1}$ is equal to the sum of the squares of the distances from point $O$ to the three faces containing point $D_{1}$. Therefore, $O D_{1}^{2}=$ $91+99+99=289$, from which $O D_{1}=17$.