The unit volume regular tetrahedron, $T$, is magnified by a factor of two from a point $P$ on one of its symmetry axes. The volume of the common part of the original and the magnified tetrahedron is $1 / 8$. How far is the center of $T$ from $P$? (H)
A regular tetrahedron has two types of symmetry axes. One type connects the midpoints of two opposite edges - reflecting the tetrahedron over this line as an axis, its image will be itself. The other type connects a vertex of the tetrahedron with the center of the opposite face - rotating around this line by $120^{\circ}$, the image of the tetrahedron will be itself. The problem setter (and most solvers) thought of the second type of symmetry axis, but this is not clear from the problem statement. We accepted as a 5-point solution any work that solved the problem for one of the axes. Due to space constraints, we only present the solution for the second type of axis. (For the first type, the calculation is more tedious, and the final result is $\sqrt[3]{3}$.)

We can assume that $P$ lies on the symmetry axis $OA$ of the regular tetrahedron $ABCD$ passing through $A$. $P$ cannot be an interior point of the tetrahedron, because then the enlarged tetrahedron would completely contain $T$. By the properties of central similarity, the faces of the enlarged tetrahedron are parallel to the corresponding faces of $T$, and the two tetrahedra share one symmetry axis. Therefore, the intersection of the two tetrahedra is also a regular tetrahedron. The ratio of the volumes of similar solids is equal to the cube of the similarity ratio, so $\sqrt[3]{\frac{1}{8}}=\frac{1}{2}$ implies that the linear dimensions of the intersection are half of the corresponding dimensions of $T$. After this, we need to distinguish two cases depending on whether $P$ and $A$ are on the same side of the plane $BCD$ or not.

If they are (1st figure), then the image $A'$ of $A$ under the enlargement will be the vertex of the intersection that does not lie on the face $BCD$. By the previous, $A'$ will be half as far from the face $BCD$ as $A$. It is known that $O$ is the point one quarter of the way along the height of the tetrahedron from $BCD$, so $OA'=\frac{1}{2}AA'$. By the enlargement, $PA=AA'$, so

$$
PO=PA+AA'+A'O=\frac{5}{2}AA'=\frac{5}{4}m
$$

where $m$ denotes the height of $T$. The volume of $T$ is 1, so $m=\frac{2}{3}\sqrt[6]{243}$, and thus $PO=\frac{5}{6}\sqrt[6]{243}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d60e17ed50fea02f1423g-1.jpg?height=845&width=616&top_left_y=1152&top_left_x=302)

![](https://cdn.mathpix.com/cropped/2024_05_02_d60e17ed50fea02f1423g-2.jpg?height=848&width=715&top_left_y=129&top_left_x=1125)

If the plane $BCD$ separates $P$ and $A$ (2nd figure), then $A$ will be one vertex of the intersection, and the opposite face of the intersection will be cut out of $T$ by the image $S'$ of the plane $BCD$. Therefore, the plane $S'$ bisects the height of $T$ corresponding to $A$, and by the enlargement, $S'$ is twice as far from $P$ as the plane $BCD$. Thus, the planes $BCD$ and $S'$ trisect the segment $PA$. Therefore, $PA=\frac{3}{2}m$. But $OA=\frac{3}{4}m$, so

$$
PO=PA-OA=\frac{3}{4}m=\frac{\sqrt[6]{243}}{2}
$$

Thus, the center of the tetrahedron is either $\frac{5}{6}\sqrt[6]{243} \approx 2.0817$ or $\frac{1}{2}\sqrt[6]{243} \approx 1.0409$ units away from $P$.

Based on the work of Barnabás Hegyi (Zalaegerszeg, Zrínyi M. Gymnasium, 3rd grade)