A regular octahedron's one face is orthogonally projected onto the plane of the opposite face. What fraction of the opposite face's area is covered by the projection?

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Consider the cube whose face centers form a regular octahedron. Let one of the octahedron's faces be the triangle $ABC$, and the opposite face be the regular triangle $DEF$. Then, the faces of the cube containing points $A, B, C$ are $K$, and the faces containing points $D, E, F$ meet at the opposite vertex $L$ (Figure 1).

When the cube, and thus the octahedron, is rotated $120^{\circ}$ around the line $KL$, it maps onto itself. Therefore, the planes of $ABC$ and $DEF$ are perpendicular to the line $KL$. The triangles $ABC$ and $DEF$ are mirror images of each other with respect to the center $O$ of the cube. The line $KL$ also passes through $O$ and the centers of the triangles $ABC$ and $DEF$. If we project the triangle $ABC$ and $O$ perpendicularly—i.e., parallel to $KL$—onto the plane of $DEF$, then the image of $O$ will be $O'$, the center of the triangle $DEF$, and the images of $A, B$, and $C$ will be the mirror images of $D, E$, and $F$ with respect to $O'$ (Figure 2).

It is easy to see that the common part of the two triangles is a regular hexagon. The 12 small triangles visible in Figure 2 are clearly congruent, so the projection covers $\frac{6}{9}=\frac{2}{3}$ of the area of the face.

Based on the work of András Máthé (Budapest, ELTE Apáczai Cs. J. Gyak. Gimn., 12th grade)
![](https://cdn.mathpix.com/cropped/2024_05_02_8e61e6da0721637ba2a9g-1.jpg?height=1232&width=652&top_left_y=656&top_left_x=736)