10,11

The volume of a regular quadrilateral pyramid $SABCD$ is $V$. The height $SP$ of the pyramid is an edge of a regular tetrahedron $SPQR$, the plane of the face $PQR$ of which is perpendicular to the edge $SC$. Find the volume of the common part of these pyramids.
Let $A B=a, \angle S C P=\alpha$. Suppose the plane passing through the line $B D$ perpendicular to the edge $S C$ of the pyramid $S A B C D$ intersects this edge at point $H$ (Fig.1). The height $S P$ of the pyramid $S A B C D$ is an edge of a regular tetrahedron $S P Q R$, and the height of the tetrahedron lies on the line $S C$. Therefore, the plane of the face $P Q R$ of the tetrahedron passes through point $P$ perpendicular to $S C$, which means it coincides with the plane $B D H$. Thus, $S H$ is the height of the regular tetrahedron $S P Q R$. The segment $P H$ is the height of the right triangle $S P C$, drawn from the vertex of the right angle, so $\angle S P H=\angle S C P=\alpha$, and since $S P H$ is the angle between the edge $S P$ of the regular tetrahedron and the face $P Q R$, we have

$$
\cos \alpha=\cos \angle S P H=\frac{1}{\sqrt{3}}, \sin \alpha=\sqrt{1-\frac{1}{3}}=\frac{\sqrt{2}}{\sqrt{3}}, \operatorname{tg} \alpha=\sqrt{2}
$$

From the right triangle $S P C$, we find that

$$
S P=C P \operatorname{tg} \alpha=\frac{a \sqrt{2}}{2} \cdot \sqrt{2}=a
$$

i.e., the edge of the regular tetrahedron $S P Q R$ is equal to $a$, and since $H$ is the center of the equilateral triangle $P Q R$, then $P H=$ $\sqrt{\sqrt{3}}$

$\frac{\frac{a \sqrt{3}}{3}}{}$. Consider the section of the pyramid and the tetrahedron by the plane $B D H$. We obtain an isosceles triangle $B D H$

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-31.jpg?height=97&width=2033&top_left_y=1847&top_left_x=1)

$$
\operatorname{tg} \gamma=\frac{B P^{\prime}}{H P^{\prime}}=\frac{\frac{a \sqrt{2}}{a \sqrt{3}}}{3}=\frac{\sqrt{6}}{2}, \cos \gamma=\frac{1}{\sqrt{1+g^{2} \gamma}}=\frac{1}{\sqrt{1+\frac{1}{2}}}=\sqrt{\sqrt{2}}, \sin \gamma=\frac{\sqrt{3}}{\sqrt{5}} .
$$

Let $E$ and $F$ be the points of intersection of the lines $P Q$ and $P R$ with the segments $B H$ and $D H$ respectively. Since $P H$ is the bisector of the angle $Q P R$ (Fig.2), triangles $H E P$ and $H F P$ are equal by side and two adjacent angles, so $H F=H E$. By the Law of Sines,

$$
\frac{H E^{\prime}}{\sin \angle H P^{P} E^{\prime}}=\frac{H P^{\prime}}{\sin \angle P^{\prime} E H H}, \frac{H E^{\prime}}{\sin 30^{9}}=\frac{H P^{\prime}}{\sin \left(180^{\circ}-\left(30^{\beta}+y\right)\right)},
$$

from which

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-31.jpg?height=154&width=938&top_left_y=2381&top_left_x=560)

Moreover,

$$
\sin \angle P E H=\sin \gamma=\frac{\sqrt{3}}{\sqrt{5}}>\frac{1}{2}=\sin 30^{\circ}=\sin \angle H P E
$$

therefore

$$
P E<H E=\frac{a \sqrt{15}}{3(3+\sqrt{2})}<a=P Q
$$

Thus, point $E$ lies on the segment $P Q$. Therefore, the common part of the pyramid $S A B C D$ and the tetrahedron $S P Q R$ is a quadrilateral pyramid $S P E H F$ with vertex $S$. Let $V 1$ be the volume of this pyramid. Then

$$
=\frac{1}{3} S_{S P E H F} \cdot S H=\frac{1}{3} \cdot 2 S_{\triangle H E P} \cdot S H=\frac{1}{3} \cdot 2 \cdot \frac{1}{2} H P \cdot H E \sin \gamma \cdot S H=
$$

$$
=\frac{1}{3} \cdot \frac{a \sqrt{3}}{3} \cdot \frac{a \sqrt{15}}{3(3+\sqrt{2})} \cdot \frac{\sqrt{3}}{\sqrt{5}} \cdot a \sqrt{\frac{2}{3}}=\frac{a^{3} \sqrt{2}}{9(3+\sqrt{2})}
$$

Since the volume of the pyramid $S A B C D$ is $V$, then

$$
V=\frac{1}{3} S_{S A B C D} \cdot S P=\frac{1}{3} a 2 \cdot a=\frac{1}{3} a 3
$$

from which $a 3=3 V$. Therefore,

$$
V 1=a 3 \cdot \frac{\sqrt{2}}{g(3+\sqrt{2})}=3 V \cdot \frac{\sqrt{2}}{g(3+\sqrt{2})}=\frac{V \sqrt{2}}{3(3+\sqrt{2})}=\frac{3 \sqrt{2}-2}{21} V
$$

## Answer

$\frac{3 \sqrt{2}-2}{21} V$