\section*{Problem 3 - 011233}

A cube with edge length \(a\) has a tetrahedron and an octahedron inscribed in it.

a) How do the volumes of the three bodies relate to each other?

b) A sphere is inscribed in the tetrahedron. Explain why this sphere also touches the octahedron, and express the volume of this sphere as a function of \(a\).
}

Since the cube has six square faces and the tetrahedron has six edges, we take six suitable face diagonals as the edges of the tetrahedron (Figure a), which are of course of equal length and thus form a regular tetrahedron.

The octahedron has six vertices, so the centers of the cube's faces can be chosen as the vertices of a regular octahedron (Figure b).
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1737.jpg?height=306&width=668&top_left_y=772&top_left_x=700)

a) The tetrahedron cuts off four triangular pyramids from the cube with volume \(V_{W}=a^{3} 4\), where the volume of each of these pyramids can be calculated using \(V=\frac{1}{3} A h\).

For this, we choose the isosceles right triangle with leg length \(a\) as the base, so \(A=\frac{1}{2} a^{2}\). The height is also \(h=a\), so \(V=\frac{1}{6} a^{3}\) follows. From this, the volume of the tetrahedron is

\[
V_{T}=a^{3}-4 \frac{1}{6} a^{3}=\frac{1}{3} a^{3}
\]

For the octahedron, the edge length is \(\frac{\sqrt{2}}{2} a\); it can be composed of two square pyramids. The base sides of these pyramids (a square) have an area of \(A=\frac{1}{2} a^{2}\), so the octahedron has a volume of

\[
V_{O}=2 \frac{1}{6} a^{2} \cdot \frac{1}{2} a=\frac{1}{6} a^{3}
\]

Thus, the volumes are in the ratio \(V_{W}: V_{T}: V_{O}=6: 2: 1\).

b) To determine the radius of the sphere inscribed in the tetrahedron, we first consider the circumsphere of the tetrahedron. This is clearly identical to the circumsphere of the cube, so the half-space diagonal of the cube is equal to the circumsphere radius: \(R=\frac{\sqrt{3}}{2} a\).

The insphere with radius \(r\) touches the tetrahedron faces from the inside, where each tetrahedron face spans a tangent plane to the insphere at the point of contact. Thus, the cube vertex, the point of contact, and the center lie on a straight line, and it holds that \(R=h+r\) or \(r=R-h\) with \(h\) as the height of the triangular pyramids considered under a) with the (now equilateral) triangle as the base. This height is calculated from the volume formula above with

\[
V=\frac{1}{6} a^{3} \quad \text { to } \quad h=\frac{3 V}{A}=\frac{\frac{a^{3}}{2}}{\sqrt{3} \frac{a^{2}}{2}}=\frac{a}{\sqrt{3}}
\]

From this, it follows that \(r=\frac{1}{2 \sqrt{3}} a=\frac{1}{3} R\).