6.15. The edge of a regular tetrahedron $ABCD$ is equal to $a$. The vertices of the spatial quadrilateral $A_{1} B_{1} C_{1} D_{1}$ lie on the corresponding faces of the tetrahedron ($A_{1}$ - on the face opposite to $A$, and so on), and its sides are perpendicular to the faces of the tetrahedron: $A_{1} B_{1} \perp BCD, B_{1} C_{1} \perp CDA, C_{1} D_{1} \perp DAB$ and $D_{1} A_{1} \perp ABC$. Calculate the lengths of the sides of the quadrilateral $A_{1} B_{1} C_{1} D_{1}$.
6.15. Drop perpendiculars $A_{1} K$ and $B_{1} K$ to $C D$, $B_{1} L$ and $C_{1} L$ to $A D$, $C_{1} M$ and $D_{1} M$ to $A B$, $D_{1} N$ and $A_{1} N$ to $B C$. The lengths of these perpendiculars are equal to the cosine of the dihedral angle at the edge of a regular tetrahedron, i.e., they are equal to $1 / 3$ (see problem 2.14). Since the sides of the quadrilateral $A_{1} B_{1} C_{1} D_{1}$ are perpendicular to the faces of the regular tetrahedron, their lengths are equal (see problem 8.25). Therefore, $A_{1} K = B_{1} L = C_{1} M = D_{1} N = x$ and $B_{1} K = C_{1} L = D_{1} M = A_{1} N = 3x$. Consider the net of the tetrahedron (Fig. 46). The edges of the tetrahedron are divided by points $K, L, M$, and $N$ into segments of length $n$ and $m$. Since $x^{2} + n^{2} = D_{1} B^{2} = 9x^{2} + m^{2}$, we have $8x^{2} = n^{2} - m^{2} = (n + m)(n - m) \Rightarrow a(n - m)$. Let the ray $B D_{1}$ intersect the side $A C$ at point $P$; $Q$ and $R$ are the projections of point $P$ onto the sides $A B$ and $B C$. Since $P R : P Q = 1 : 3$, then $C P : P A = 1 : 3$. Therefore, $\frac{B R}{C B} = \frac{1}{2} + \frac{3}{8} = \frac{7}{8}$ and $\frac{B Q}{A B} = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}$. Consequently, $n : m = B R : B Q = 7 : 5$, and thus, $x = a / 4 \sqrt{3}$. The lengths of the sides of the quadrilateral $A_{1} B_{1} C_{1} D_{1}$ are $2 \sqrt{2} x = a / \sqrt{6}$.