\section*{Problem 3 - 191033}

Each cube has both a circumsphere (i.e., a sphere on which all vertices of the cube lie) and an insphere (i.e., a sphere that touches each face of the cube).

Similarly, each regular octahedron has both a circumsphere and an insphere.

Now, assume that the circumspheres of a cube and a regular octahedron have the same radius.

Determine the ratio \(r_{1}: r_{2}\), where \(r_{1}\) is the radius of the insphere of the cube and \(r_{2}\) is the radius of the insphere of the octahedron, under this assumption!
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The center \(M\) of a cube, i.e., the intersection of its body diagonals, is equidistant from all vertices and (since it is also the intersection of the connecting segments of the midpoints of opposite faces) equidistant from all faces. Therefore, it is the center of both the circumsphere and the insphere.

If \(a\) is the edge length of the cube, and thus \(a \sqrt{3}\) is the length of its body diagonal, the distance from \(M\) to the vertices, i.e., the circumsphere radius \(r=\frac{a}{2} \sqrt{3}\); the distance from \(M\) to the faces, i.e., the insphere radius, is \(r=\frac{a}{2}=\frac{r}{\sqrt{3}}\). Furthermore, let \(A B C D E F\) be a regular octahedron (see image).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1421.jpg?height=471&width=440&top_left_y=1255&top_left_x=311)

Its center \(N\), i.e., the intersection of \(A C, B D\), and \(E F\), is equidistant from all vertices and equidistant from all faces; therefore, it is the center of both the circumsphere and the insphere.

If \(b\) is the edge length of the octahedron, and thus \(B D=b \sqrt{2}\), the circumsphere radius \(r=N B=\frac{b}{2} \sqrt{2}=\frac{b}{\sqrt{2}}\). Furthermore, it follows that \(B C E N\) is a pyramid with an equilateral base \(B C E\); for its apex \(N\), \(N B=N C=N E\), so it is a regular pyramid.

The foot \(L\) of the perpendicular from \(N\) to the plane \(B C E\) is therefore the centroid of the base. Hence, \(L E=\frac{2}{3} \cdot \frac{b}{2} \sqrt{3}=\frac{b}{\sqrt{3}}\).

From this and from \(N E=\frac{b}{\sqrt{2}}\), it follows by the Pythagorean theorem that the insphere radius

\[
r_{2}=N L=\sqrt{N E^{2}-L E^{2}}=\sqrt{\frac{b^{2}}{2}-\frac{b^{2}}{3}}=\frac{b}{\sqrt{6}}=\frac{r}{\sqrt{3}}
\]

is. Therefore, \(r_{1}: r_{2}=1: 1\).