[Correct tetrahedron]  [ Properties of sections ]

The edge of a regular tetrahedron is $a$. Find the sides and the area of the section parallel to two of its skew edges and at a distance $b$ from the center of the tetrahedron, where $0 < b < a \sqrt{2} / 4$.
The secant plane $P E F K$ is parallel to the line $S C$, and therefore intersects the faces in which the edge $S C$ lies, along lines parallel to the edge $S C: E F|| S C, P K|| S C$ (fig.). For the same reason, the secant plane intersects the faces containing the edge $A B$, along lines parallel to this edge: $P E \| A B, K F|| A B$. Thus, a parallelogram is obtained in the section. Since non-intersecting edges in a regular tetrahedron are mutually perpendicular ($A B \perp S C$), a rectangle $P E F K$ is obtained in the section. Draw $S M \perp$ plane $A B C$. The plane $S C M$ is perpendicular to the base plane; $A B \perp$ plane $S C D$. Therefore, $K F \perp$ plane $S C D$, since $K F|| A B$. Further, plane $S C D \perp$ plane $P E F K$, in which lies a line perpendicular to the first plane. These planes intersect along the line $N T|| S C$; draw $D J \perp S C, D J \perp$ plane $P E F K$, since $D J \perp K F$ and $D J \perp N T$, $K F|| A B, N T|| S C$. $O$ is the center of the tetrahedron, as it is the point of intersection of the height of the tetrahedron $S M$ with the perpendicular bisector of the edge $S C$ (or the bisector of the angle $S D C$). $O L=b$, where $L$ is the point of intersection of $D J$ with the section plane. $D C=S D=a \sqrt{3} / 2, M C=a \sqrt{3} / 3, S M=a \sqrt{6} / 3$ (from triangle $M C S$). From the isosceles triangle $S D C$, we find $D J=a \sqrt{2} / 2$. From the similarity of triangles $M S C$ and $J S O$, we find $O J: S J / O J=M S / M C$, from which $O J=a \sqrt{2} / 4$. Hence, $L J=O J-O L=a \sqrt{2} / 4-b, D L=D J-L J=a \sqrt{2} / 2-(a \sqrt{2} / 4-b)=(a \sqrt{2}+4 b) / 4$. Now we can determine the height of the rectangle $N T$ from the similarity of triangles $S C D$ and $N T D: N T / S C=D L / D J, N T=(a+2 \sqrt{2} b) / 2$. The base of the rectangle $P E F K$ will be determined from the similarity of triangles $A B C$ and $K F C$, for which we first find $T C=D C-D T$. $D T$ is determined from the right triangle $D L T$:

$$
\begin{gathered}
D T=\sqrt{D L^{2}+L T^{2}} \\
, \\
T C=\left(\sqrt{3}_{(}(a-2 \sqrt{2} b)\right) / 4, K F=(a-2 \sqrt{2} b) / 2
\end{gathered}
$$

The area of the rectangle is $N T \cdot K F=\left(a^{2}-8 b^{2}\right) / 4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6a190adc6812227a77adg-54.jpg?height=620&width=680&top_left_y=-1&top_left_x=689)

## Answer

The section is a rectangle with sides $(a+2 \sqrt{2} b) / 2, (a-2 \sqrt{2} b) / 2$ and area $\left(a^{2}-8 b^{2}\right) / 4$.

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