Let's inscribe a cube in a given hemisphere such that four of its vertices lie on the surface of the hemisphere, and the other four lie on the boundary circle. (The problem should be solved both by construction and calculation.)
The four vertices of the cube on the circular base of the hemisphere form a face - because otherwise, choosing four vertices of the cube in a plane, the other four vertices would be distributed on both sides of the plane, not allowing four to lie on the surface of the hemisphere - let this face be the base of the cube. Thus, the four vertices on the surface of the sphere form the top face, parallel to it. The top square is inscribed in the circle $k$ cut by the plane from the sphere, and the center $K$ of this circle and the top face coincide. Drawing lines parallel to the edges through $K-\mathrm{n}$, we get an axis of the cube's face, and on the other hand, rotating around this, the circle $k$ shifts itself, so the hemisphere also shifts, which is why the center of the base of the cube coincides with the center $O$ of the hemisphere.

![](https://cdn.mathpix.com/cropped/2024_05_02_4335c7e43e99dfb4d0b4g-1.jpg?height=380&width=547&top_left_y=431&top_left_x=778)

Figure 1

Based on this, our computational task is to express $k$ and the radius $r$ of the circle circumscribed around the base square - half the diagonal of the base square - in terms of the given radius $R$ of the hemisphere.

Let one vertex of the base be $A$, and the endpoint of the edge starting from it on the surface of the hemisphere be $A^{\prime}$. Then the edge of the cube is $A A^{\prime}=a$, the mentioned radius is $r=O A=a \sqrt{2} / 2$, and from the right triangle $O A^{\prime} A$,

$$
a^{2}+\frac{a^{2}}{2}=R^{2}, \quad a=R \sqrt{\frac{2}{3}}, \quad r=R \frac{1}{\sqrt{3}}
$$

It is easy to show that the radius of the circle circumscribed around a regular triangle with side $R$ is exactly $r$. Based on this observation, one way to perform the construction after the auxiliary construction is: On the circular base of the hemisphere, draw a circle of radius $r$ around $O$, inscribe a square in it, erect perpendicular lines from its vertices to the base, and these will intersect the other four vertices of the cube from the surface of the hemisphere.

![](https://cdn.mathpix.com/cropped/2024_05_02_4335c7e43e99dfb4d0b4g-1.jpg?height=260&width=272&top_left_y=1439&top_left_x=688)

Remarks. 1. The construction can also be carried out using a similarity transformation based on the introductory description. Place a cube on the base of the hemisphere (either inward or outward) so that its center coincides with $O$. Then the four lines passing through the vertices of the opposite face and $O$ intersect the four vertices of the sought cube from the hemisphere, and their projections on the base are the other four vertices.

2. By reflecting the hemisphere and the cube in the plane of the base, we get a complete sphere and a 1:1:2 ratio rectangular prism (square column) inscribed in it. Starting from this, we can also solve the problem.