23. Find the angle and the distance between two intersecting medians of two lateral faces of a regular tetrahedron with edge $a$.
23. The problem can be solved in the same way as problem 22. Let's propose another method for determining the distance between skew medians. Let \(ABCD\) be the given tetrahedron, \(K\) the midpoint of \(AB\), and \(M\) the midpoint of \(AC\). Project the tetrahedron onto a plane passing through \(AB\) and perpendicular to \(CK\). The tetrahedron will project into triangle \(ABD_1\), where \(D_1\) is the projection of \(D\). If \(M_1\) is the projection of \(M\) (with \(M_1\) being the midpoint of \(AK\)), then the distance between the lines \(CK\) and \(DM\) is equal to the distance from point \(K\) to the line \(D_1M_1\). This distance is easy to find, as \(D_1KM_1\) is a right triangle with legs \(D_1K\) and \(KM_1\) equal to \(a \sqrt{2/3}\) (the height of the tetrahedron) and \(a/4\), respectively.

The problem has two solutions. The second one we obtain by considering the medians \(CK\) and \(BN\), where \(N\) is the midpoint of \(DC\):

Answer: \(\arccos \frac{1}{6}, a \sqrt{\frac{2}{35}}\) and \(\arccos \frac{2}{3}, a \frac{\sqrt{10}}{10}\).