It is known that among all quadrilaterals that can be inscribed in a given circle, the square has the largest area. Is it true that among all eight-vertex solids that can be inscribed in a given sphere, the inscribed cube has the largest volume?
Not true. The space diagonal of a cube inscribed in a sphere of radius $r$ is $2r$, so the length of its edge is $\frac{2r}{\sqrt{3}}$, and thus its volume is $\frac{8}{3 \sqrt{3}} \cdot r^{3}$. Among the polyhedra inscribed in a sphere, there are many whose volume is larger than this. Below, we provide an example.

Take one of the great circles of the sphere and inscribe a regular hexagon in it. The vertices of the hexagon give six vertices of the polyhedron, and the other two vertices are the points on the sphere that are farthest from the plane of the hexagon. These two points are the intersections of the sphere with the line passing through the center of the sphere and perpendicular to the plane of the hexagon. The resulting polyhedron is composed of two straight hexagonal pyramids. The area of the base hexagon of the pyramids is $6 \cdot r^{2} \cdot \frac{\sqrt{3}}{4}$, and their height is $r$, so their volume is $\frac{1}{3} \cdot 6 r^{2} \cdot \frac{\sqrt{3}}{4} r = \frac{\sqrt{3}}{2} r^{3}$. Therefore, the volume of the two pyramids is $\sqrt{3} r^{3}$, which is indeed larger than the volume of the cube, since $\sqrt{3} > \frac{8}{3 \sqrt{3}}$.

Based on the work of Gergely Kiss (Fazekas M. Főv. Gyak. Gimn., 12th grade)
![](https://cdn.mathpix.com/cropped/2024_05_02_010ac076ef9c3c9f15d1g-1.jpg?height=828&width=444&top_left_y=743&top_left_x=843)