10,11

In the pyramid $ABCD$, the plane angles $DAB, ABC, BCD$ are right angles. The vertices $M, N, P, Q$ of a regular tetrahedron are located on the edges $AC, BC, AB, BD$ of the pyramid $ABCD$ respectively. The edge $MN$ is parallel to the edge $AB$. Find the ratio of the volumes of the regular tetrahedron $MNPQ$ and the pyramid $ABCD$.
Since $M N P Q$ is a regular tetrahedron (Fig.1), $P Q \perp M N$, and since $M N \| A B$, then $P Q \| A D$, which means $P Q \perp A B$. Let $Q O$ be the height of the regular tetrahedron $M N P Q$, and $D H$ be the height of the pyramid $A B C D$. Then the points $B, O$, and $H$ lie on the same line - the projection of the inclined line $B D$ onto the plane $A B C$. From the theorem of three perpendiculars, it follows that $A H \perp A B$ and $C H \perp B C$, so $A B C H$ is a rectangle. Its diagonals $A C$ and $B H$ are equal and are bisected by their point of intersection $E$. Let the edge of the regular tetrahedron $M N P Q$ be $a$. The vertices $M$ and $N$ of the equilateral triangle $M N P$ lie on the hypotenuse $A C$ and the leg $B C$ of the right triangle $A B C$, respectively, and the vertex $P$ lies on the leg $A B$, with $M N \| A B$ (Fig. 2), so

$$
\angle B P N=\angle M N P=60^{\circ}, B P=\frac{1}{2} N P=\frac{a}{2}, B N=\frac{a \sqrt{3}}{2}
$$

Extend the segment $P O$ to intersect $M N$ and $A C$ at points $G$ and $F$, respectively. Denote

$$
\angle E O F=\angle B O P=\angle O B C=\angle E B C=\angle E C B=\angle E F O=\alpha
$$

From the right triangle $B P O$, we find that

$$
\operatorname{tg} \alpha=\frac{B P}{P O}=\frac{\frac{a}{2}}{\frac{a \sqrt{3}}{3}}=\frac{\sqrt{3}}{2}
$$

$$
\cos \alpha=\frac{1}{\sqrt{1+\operatorname{tg}^{2} \alpha}}=\frac{1}{\sqrt{1+\frac{3}{4}}}=\frac{2}{\sqrt{7}}
$$

$$
\begin{gathered}
B O=\frac{O F}{\cos \alpha}=\frac{\frac{a \sqrt{3}}{\frac{2}{2}}}{\sqrt{7}}=\frac{a \sqrt{21}}{6}, F G=\frac{M G}{\operatorname{tg} \angle M F G}=\frac{M G}{\operatorname{tg\alpha }}=\frac{\frac{\frac{a}{2}}{\sqrt{3}}}{2}=\frac{a \sqrt{3}}{3} \\
O F=O G+F G=\frac{a \sqrt{3}}{6}+\frac{a \sqrt{3}}{3}=\frac{a \sqrt{3}}{2}
\end{gathered}
$$

From the isosceles triangle $O E F$, we find that

$$
O E=\frac{O F}{2 \cos \alpha}=\frac{\frac{a \sqrt{3}}{4}}{\frac{2}{\sqrt{7}}}=\frac{a \sqrt{21}}{8}
$$

Therefore,

$$
\begin{aligned}
& A C=2 B E=2(B O+O E)=2\left(\frac{a \sqrt{21}}{6}+\frac{a \sqrt{21}}{8}\right)=\frac{7 a \sqrt{21}}{12}, \\
& B C=A C \cos \alpha=\frac{7 a \sqrt{21}}{12} \cdot \frac{2}{\sqrt{7}}=\frac{7 a \sqrt{3}}{6}, A B=B C \operatorname{tg} \alpha=\frac{7 a \sqrt{3}}{6} \cdot \frac{\sqrt{3}}{2}=\frac{7 a}{4}, \\
& S_{\triangle A B C}=\frac{1}{2} B C \cdot A B=\frac{1}{2} \cdot \frac{7 a \sqrt{3}}{6} \cdot \frac{7 a}{4}=\frac{49 a^2 \sqrt{3}}{48}
\end{aligned}
$$

From the similarity of the right triangles $Q O B$ and $D H B$ (Fig.1), we find that

$$
\frac{O Q}{D H}=\frac{B O}{B H}=\frac{B O}{A C}=\frac{\frac{a \sqrt{21}}{6}}{\frac{7 a \sqrt{21}}{12}}=\frac{2}{7}
$$

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_786127613a7b2f99b094g-31.jpg?height=197&width=1246&top_left_y=1754&top_left_x=406)

## Answer

$\frac{24}{343}$