3.29. The sides of the spatial quadrilateral $K L M N$ are perpendicular to the faces of the tetrahedron $A B C D$, and their lengths are equal to the areas of the corresponding faces. Find the volume of the tetrahedron $K L M N$, if the volume of the tetrahedron $A B C D$ is $V$.
3.29. The existence of such a spatial quadrilateral $K L M N$ for any tetrahedron $A B C D$ follows from the assertion of problem 7.19; there are several such quadrilaterals, but the volumes of all the tetrahedra they define are equal (problem 8.26).

Using the formula from problem 3.2, it is easy to prove that $V^{3}=(a b c / 6)^{3} p^{2} q$, where $a, b$ and $c$ are the lengths of the edges emanating from vertex $A$; $p$ is the product of the cosines of the dihedral angles at vertex $A$, and $q$ is the product of the cosines of the dihedral angles of the trihedral angle at vertex $A$. Dropping perpendiculars from an arbitrary point $O$ inside the tetrahedron $A B C D$ to the faces emanating from vertex $A$, and laying off segments $O P, O Q$, and $O R$ equal to the areas of these faces. From the solution of problem 8.26, it follows that the volume $W$ of the tetrahedron $O P Q R$ is equal to the volume of the tetrahedron $K L M N$. The plane (respectively dihedral) angles of the trihedral angle $O P Q R$ complement the internal (respectively

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$a$

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6

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$\sigma$

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2

Fig. 30

plane) angles of the trihedral angle $A B C D$ (see problem 5.1). Therefore, $W^{3}=\left(S_{1} S_{2} S_{3} / 6\right)^{3} q^{2} p$, where $S_{1}, S_{2}, S_{3}$ are the areas of the faces emanating from vertex $A$. Since $S_{1} S_{2} S_{3}=(a b c)^{2} p / 8$, then $W^{3}=\left(\frac{1}{6}\right)^{3}\left(\frac{1}{8}\right)^{3}(a b c)^{6} p^{4} q^{2}=\left(\frac{3}{4} V^{2}\right)^{3}$, i.e., $W=\frac{3}{4} V^{2}$.