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On the sides $B C$ and $A D$ of a regular quadrilateral pyramid $\operatorname{SABCD}$ ( $S$ - vertex), points $P$ and $Q$ are taken. The sections of the pyramid $S A B C D$ by two mutually perpendicular planes $\alpha$ and $\beta$, passing through the line $P Q,$ are trapezoids with equal bases. The face $S A B$ forms an angle $\frac{\pi}{4}$ with the intersecting plane, and the angle between the faces $S A B$ and $A B C D$ is $\operatorname{arctg} 2$. Find the areas of the sections of the pyramid by the planes $\alpha$ and $\beta$, if $P Q=13$.
Let $\alpha$ be the plane that intersects the face $S A B$, and $K L-$ the segment where they intersect (Fig.1). Points $K$ and $L$ do not lie on the edge $A B$ (otherwise the planes $\alpha$ and $A B C$ would coincide). Let points $K$ and $L$ lie on the edges $S A$ and $S B$ respectively. If the lines $L P$ and $K Q$ were parallel, the planes $A S D$ and $B S C$ would be parallel (since two intersecting lines $K Q$ and $A D$ of one would be respectively parallel to two intersecting lines $P L$ and $B C$ of the other), which means $K L \| P Q$. Planes $\alpha$ and $A S B$ pass through parallel lines $P Q$ and $K L$ and intersect along the line $A B$, so $P Q \| A B$. Therefore, $A B = P Q = 13$. Similarly, we obtain that the plane $\beta$ intersects the face $S C D$ along the segment $R T$, and $R T \| C D$. By the problem's condition, $K L = R T$. This means that points $K, L, R$, and $T$ divide the lateral edges of the pyramid in the same ratio. Therefore, they lie in a plane parallel to the base of the pyramid. Let $E$ and

$F$ be the midpoints of the edges $A B$ and $C D$ respectively, $O$ be the center of the square $A B C D$, and $M, N$, and $G$ be the midpoints of the segments $P Q, K L$, and $R T$ respectively. The plane $E S F$ is perpendicular to the faces $A S B, C S D$, and $A B C D$, since $A B \perp E S F, C D \perp E S F$, and $S O \perp A B C D$. Therefore,

$$
\angle S E F = \angle S F E = \operatorname{arctg} 2, \angle M N E = \frac{\pi}{4}, \angle N M G = \frac{\pi}{2}
$$

From the triangle $E N M$ (Fig.2), we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-34.jpg?height=149&width=1330&top_left_y=1815&top_left_x=370)

and since $N G \| E F$, then $\operatorname{tg} \angle M N G = 3$. Let $U$ be the point of intersection of the height $S O$ of the pyramid with the segment $N G$, and $M M 1$ be the height of the triangle $N M G$. Denote $N M 1 = x$. Then

$$
\begin{gathered}
M M 1 = N M 1 \operatorname{tg} \angle M N M 1 = 3 x, M 1 G = \frac{M M_{1}^{2}}{N M_{1}} = \frac{9 x^{2}}{x} = 9 x, N G = x + 9 x = 10 x, \\
S U = N U \operatorname{tg} \angle S N U = 2 N U = 10 x, S O = S U + U O = S U + M M 1 = 10 x + 3 x = 13 x, \\
\text{ and since } S O = 2 E O = E F = 13, \text{ then } x = 1. \text{ Then } \\
M N = \sqrt{M M_{1}^{2} + M_{1} N^{2}} = \sqrt{9 x^{2} + x^{2}} = x \sqrt{10} = \sqrt{10}, M G = 3 M N = 3 \sqrt{10}
\end{gathered}
$$

- the heights of the trapezoids $K L P Q$ and $P R T Q$ with bases $P Q = 13$ and $K L = R T = L R = N G = 10$. Therefore,

$$
\begin{aligned}
& S_{K L P Q} = \frac{1}{2}(P Q + K L) \cdot M N = \frac{1}{2}(13 + 10) \cdot \sqrt{10} = \frac{23 \sqrt{10}}{2}, \\
& S_{P R T Q} = \frac{1}{2}(P Q + R T) \cdot M G = \frac{1}{2}(13 + 10) \cdot 3 \sqrt{10} = \frac{69 \sqrt{10}}{2}.
\end{aligned}
$$

## Answer

Through the midpoint of the edge $B C$ of a regular triangular pyramid $S A B C$ ( $S$ is the vertex), planes $\alpha$ and $\beta$ are drawn, each forming an angle $\operatorname{arctg} \frac{1}{2}$ with the plane $A B C$. Find the areas of the sections of the pyramid $S A B C$ by the planes $\alpha$ and $\beta$, if these sections have a common side of length 3, lying in the face $A B C$, and the plane $\alpha$ is perpendicular to the edge $S C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-35.jpg?height=172&width=240&top_left_y=477&top_left_x=923)

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