25) Let $Q$ be a cube and $S$ be a sphere that has its center at one of the vertices of $Q$ and a radius equal to the side of $Q$. The volume of the intersection between $Q$ and $S$ is:

(A) one eighth of the volume of the sphere, (B) one quarter of the volume of the sphere,

(C) one sixth of the volume of the cube, (D) one quarter of the volume of the cube,

(E) half of the volume of the cube.
25. The answer is $(\mathbf{A})$.

The cube $Q$ is defined by the edges $O A, O B, O C$ of length equal to the radius of the sphere. The center of the sphere thus coincides with the vertex $O$ of the cube. By symmetry, the plane containing the points $B, O, C$ divides the sphere into two equal volumes. The cap thus obtained is in turn divided into two equal parts by the plane passing through $A, O, B$. Finally, the plane $C, O, A$ divides the spherical sector $A B C D$ into two equal parts. Consequently, the volume of the part of the sphere contained in $Q$ is one-eighth of the volume of the sphere we started with.

![](https://cdn.mathpix.com/cropped/2024_04_17_4701f996ee50400fb9c5g-6.jpg?height=480&width=408&top_left_y=734&top_left_x=1418)