138. A regular tetrahedron of volume \( V \) is rotated about the line connecting the midpoints of its skew edges by an angle \( \alpha \). Find the volume of the common part of the given tetrahedron and the rotated one \((0<\alpha<\pi)\).
138. Let the edge of a regular tetrahedron \(ABCD\) be \(a\), and let \(R\) and \(L\) be the midpoints of the edges \(CD\) and \(AB\) (Fig. 23). We take a point \(M\) on the edge \(CB\),

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-081.jpg?height=427&width=446&top_left_y=1145&top_left_x=137)

Fig. 23. and draw a section through \(M\) perpendicular to \(KL\). Let \(|CM| = x\) and determine the value of \(x\) such that the rectangle formed in the section has an angle \(\alpha\) between its diagonals. Since the sides of the resulting rectangle are \(B^x\) and \(a - x\), the value of \(x\) can be found from the equation:

\[
\frac{x}{a-x} = \tan \frac{\alpha}{2}, \quad x = \frac{a \tan \frac{\alpha}{2}}{1 + \tan \frac{\alpha}{2}}
\].

If we take another point \(N\) on the edge \(BC\) such that \(|BN| = |CM| = x\) and draw a section through it perpendicular to \(KL\), we will get a second rectangle with an angle \(\alpha\) between its diagonals. Therefore, the plane \(BCD\) after rotating around \(KL\) by an angle \(\alpha\) counterclockwise will pass through the points \(K, P\), and \(N\). Thus, after the rotation, the plane \(BCD\) will cut off a pyramid \(K'PN C\) from the tetrahedron \(ABCD\), the volume of which is

\[
\frac{|KC|}{|CD|} \cdot \frac{|CP|}{|CA|} \cdot \frac{|CN|}{|CB|} V_{ABCD} = \frac{x(a-x)}{2a^2} V =
\]

\[
= \frac{\tan \frac{\alpha}{2}}{2(1 + \tan \frac{\alpha}{2})^2} V
\]

The same reasoning applies to any face of the tetrahedron. Therefore, the volume of the common part will be \(\frac{1 + \tan^2 \frac{\alpha}{2}}{(1 + \tan \frac{\alpha}{2})^2} V\).