\section*{Problem 4 - 021124}

From a regular tetrahedron, the 4 corners are to be cut off such that the remaining side faces are regular hexagons.

The volume and surface area of the resulting body are to be calculated.
}

Regular hexagons on the side faces of the tetrahedron with edge length \(a\) only arise if its edges are trisected, and thus four small regular tetrahedra with edge length \(\frac{1}{3} a\) are cut off.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1690.jpg?height=428&width=368&top_left_y=1882&top_left_x=841)

Each of these small tetrahedra has a volume of \(\left(\frac{1}{3}\right)^{3}=\frac{1}{27}\) of the original tetrahedron, so the remaining body has a volume of \(V^{\prime}=\left(1-\frac{4}{27}\right) V=\frac{23}{27}\) of the original volume \(V\). Using the volume formula of a tetrahedron \(V=\frac{\sqrt{2}}{12} a^{3}\), we get \(V^{\prime}=\frac{23 \sqrt{2}}{324} a^{3}\).

On each side face, three small equilateral triangles of edge length \(\frac{1}{3} a\) are removed by the cutting, but a new one of these triangles is formed at each corner. The surface area of the remaining body is therefore

\[
A^{\prime}=A-8 \cdot\left(\frac{1}{3}\right)^{2} \cdot \frac{1}{4} A=\frac{7}{9} A \quad \text { or } \quad A^{\prime}=\frac{7}{9} \sqrt{3} a^{2}
\]