For each face of a regular octahedron with unit edge length, construct the pyramid defined by the plane of the given face and the planes of the adjacent faces. What is the surface area and volume of the resulting star-shaped solid?

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I. Solution. We will show that the pyramids erected on the faces of the octahedron are regular tetrahedra with unit edges. Due to the symmetry of the octahedron, the side faces of the pyramids are congruent isosceles triangles. These will form a regular tetrahedron if the angle they form with the base is equal to the dihedral angle of a regular tetrahedron. According to the construction, the angle that the side faces form with the base precisely complements the dihedral angle of the regular octahedron to $180^{\circ}$. Therefore, we need to prove that the sum of the dihedral angles of the regular tetrahedron and the regular octahedron is $180^{\circ}$.

First, let's calculate the dihedral angle of the regular octahedron. This angle can be determined using the isosceles triangle defined by the midpoint of one of the octahedron's edges and the two opposite vertices (Figures 1 and 2). Let the desired angle be $2 \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a9a162d1cec3ceea1cedg-1.jpg?height=659&width=590&top_left_y=516&top_left_x=754)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_a9a162d1cec3ceea1cedg-1.jpg?height=647&width=371&top_left_y=1319&top_left_x=866)

Figure 2

From the triangle $ABD$:

$$
\begin{gathered}
\sin \alpha=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}}=\sqrt{\frac{2}{3}}, \quad \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{\sqrt{3}}{3} \\
\sin 2 \alpha=2 \sin \alpha \cos \alpha=\frac{2 \sqrt{2}}{3}
\end{gathered}
$$

The dihedral angle of the regular tetrahedron can be determined based on Figures 3 and 4 (where $E$ and $F$ are two vertices, $G$ is the midpoint of $EF$, and $D$ is the midpoint of the edge opposite $EF$).

![](https://cdn.mathpix.com/cropped/2024_05_02_a9a162d1cec3ceea1cedg-2.jpg?height=442&width=460&top_left_y=126&top_left_x=819)

Figure 3

![](https://cdn.mathpix.com/cropped/2024_05_02_a9a162d1cec3ceea1cedg-2.jpg?height=461&width=445&top_left_y=710&top_left_x=821)

Figure 4

The values of the trigonometric functions of the desired angle $2 \beta$ can be calculated from the triangle $DEG$:

$$
\begin{gathered}
\sin \beta=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}, \quad \cos \beta=\sqrt{\frac{2}{3}} \\
\sin 2 \beta=2 \sin \beta \cos \beta=\frac{2 \sqrt{2}}{3}
\end{gathered}
$$

Since $\sin 2 \alpha=\sin 2 \beta$ and $2 \alpha \neq 2 \beta$, the dihedral angles indeed complement each other to $180^{\circ}$, so the pyramids are regular tetrahedra.

The solid is thus composed of a regular octahedron and 8 regular tetrahedra placed on its faces. The volume of the solid is the sum of the volumes of these solids, i.e.:

$$
V=V_{\text {oct. }}+8 \cdot V_{\text {tet. }}=\frac{\sqrt{2}}{3}+8 \cdot \frac{\sqrt{2}}{12}=\sqrt{2}
$$

The surface area is equal to the area of 24 unit-sided regular triangles:

$$
A=24 \cdot \frac{\sqrt{3}}{4}=6 \sqrt{3}
$$

Based on the work of Németh Sándor (Győr, Révai M. Gymnasium, 1st year)

II. Solution. The problem can also be solved without using trigonometry.

![](https://cdn.mathpix.com/cropped/2024_05_02_a9a162d1cec3ceea1cedg-3.jpg?height=626&width=618&top_left_y=126&top_left_x=735)

Figure 5

Consider a regular tetrahedron with a side length of 2 units (Figure 5). The midpoints of its edges form a regular octahedron with unit edges. In the problem, we "reconstruct" the pyramids that, together with the octahedron, form a regular tetrahedron with a side length of 2 units, which must therefore be regular tetrahedra. (Two "large" tetrahedra are built on the octahedron in total.)

The further calculations can be carried out in the same way as in the first solution.

Weisz Csaba (Kaposvár, Táncsics M. Gymnasium, 2nd year)