[ Cross-sectional area $]$ $[\underline{\text { Regular Pyramid }}]$

In a regular quadrilateral pyramid $S A B C D$, the side length of the base $A B C D$ is $a$, and the height is $2 a \sqrt{2}$. A plane is drawn through vertex $A$ parallel to the diagonal $B D$ of the base such that the angle between line $A B$ and this plane is $30^{\circ}$. Find the area of the cross-section.
Let $S H=2 a \sqrt{2}$ be the height of the pyramid, $O$ be the point of intersection of the cutting plane with the height $S H$. The plane $S D B$ is drawn through the line $B D$, parallel to the cutting plane, and intersects this plane along some line $l$ passing through the point $O$, which means that the line $l$ is parallel to the diagonal $B D$ of the base. Let the line $l$ intersect the lateral edges $S D$ and $S B$ at points $M$ and $K$ respectively, and the line $A O$ intersects the lateral edge $S C$ at point $L$. Then the quadrilateral $A M L K$ is the section mentioned in the problem. Since the line $B D$ is parallel to the cutting plane, all points on this line are at the same distance from the cutting plane, so the length of the perpendicular dropped from point $B$ to the cutting plane is equal to the length of the perpendicular $H P$ dropped from point $H$ to the line $A L$. Let $O H=x$. From the right triangle $A H O$, we find that

$$
A O=\sqrt{A H^{2}+O H^{2}}=\sqrt{\left(\frac{a \sqrt{2}}{2}\right)^{2}+x^{2}}=\sqrt{\frac{a^{2}}{2}+x^{2}}
$$

Then

$$
H P=\frac{A H \cdot O H}{A O}=\frac{\frac{a \sqrt{2} \cdot x}{2}}{\sqrt{\frac{a^{2}}{2}+x^{2}}}=\frac{a x \sqrt{2}}{2 \sqrt{\frac{a^{2}}{2}+x^{2}}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_786127613a7b2f99b094g-03.jpg?height=132&width=1372&top_left_y=1286&top_left_x=0)

$$
S O=S H-O H=S H-X=2 a \sqrt{2}-\frac{a \sqrt{2}}{2}=\frac{3 a \sqrt{2}}{2}
$$

$$
\frac{S O}{O H}=3, M K=\frac{S O}{S H} \cdot B D=\frac{3}{4} \cdot a \sqrt{2}=\frac{3 a \sqrt{2}}{4} .
$$

Let $N$ be the orthogonal projection of point $L$ onto the plane of the base. Then $L$ lies on the diagonal $A C$ of the base. Consider the section of the pyramid by the plane $A S C$. Draw a line through the vertex $S$ parallel to $A C$, and extend $A L$ to intersect this line at point $T$. From the similarity of triangles $S O T$ and $H O A$, it follows that $S T=$ $A H \cdot \frac{S O}{O H}=\frac{3 a \sqrt{2}}{2}$, and from the similarity of triangles $S L T$ and $C L A$,

$$
\frac{S L}{L C}=\frac{S T}{A C}=\frac{\frac{3 a \sqrt{2}}{2}}{a \sqrt{2}}=\frac{3}{2}
$$

Thus,

$$
\begin{aligned}
& N H=\frac{3}{5} H C=\frac{3 \sqrt{2}}{10}, A N=A H+H N=\frac{a \sqrt{2}}{2}+\frac{3 \sqrt{2}}{10}=\frac{4 a \sqrt{2}}{5} \\
& L N=\frac{2}{5} S H=\frac{2}{5} \cdot 2 a \sqrt{2}=\frac{4 a \sqrt{2}}{5}=A N, A L=A N \sqrt{2}=\frac{5 a}{5}
\end{aligned}
$$

By the theorem of three perpendiculars, $A L \perp B D$, so $A L \perp M K$. Therefore,

$$
S_{A M L K}=\frac{1}{2} A L \cdot M K=\frac{1}{2} \cdot \frac{a}{5} \cdot \frac{3 a \sqrt{2}}{4}=\frac{3 a^{2} \sqrt{2}}{5}
$$

## Answer

$\frac{3 a^{2} \sqrt{2}}{5}$.

In the base of a triangular prism lies an equilateral triangle with side $a$. A line connecting one of the vertices of the upper base with the center of the lower base is perpendicular to the planes of the bases. It is known that a sphere can be placed inside this prism, touching all its faces. Find the lateral edge of the prism.

## Solution

Let $H$ be the center of the base $A B C$ of the prism $A B C A 1 B 1 C 1$ (Fig.1), $A 1 H$ be the height of the prism, $O$ be the center of the given sphere, $r$ be the radius of the sphere, $M$ and $M 1$ be the midpoints of the edges $B C$ and $B 1 C 1$ respectively, $\angle A 1 S H=\alpha$ be the angle of the lateral edge of the prism with the plane of its base, $P$ be the point of tangency of the sphere with the plane of the base $A B C$. The sphere touches the parallel planes $A B C$ and $A 1 B 1 C 1$, so

$$
r=\frac{1}{2} A 1 H=\frac{1}{2} A H \operatorname{tg} \alpha=\frac{1}{2} \cdot \frac{a \sqrt{3}}{3} \operatorname{tg} \alpha=\frac{a \sqrt{3}}{6} \operatorname{tg} \alpha
$$

The sphere is inscribed in the dihedral angle with edge $A A 1$, so its center lies in the bisector plane of this angle, i.e., in the plane $A M M 1 A 1$. Consider the section of the prism and the sphere by this plane (Fig.2). We get a parallelogram $A M M 1 A 1$ and a circle with center $O$ and radius $r$, touching the sides $A M, M M 1$ and $A 1 M 1$, and the side $A M$ at point $P$. From the right triangle $O M P$, we find that

$$
M P=O P \operatorname{tg} \angle O M P=r \operatorname{ctg} \frac{1}{2}(180^{\circ}-\alpha)=r \operatorname{tg} \frac{\alpha}{2}
$$

Let $\angle H M A 1=\beta$ be the angle of the lateral face of the regular triangular pyramid $A 1 A B C$ with the plane of its base $A B C$. Then

$$
\operatorname{tg} \beta=\frac{A 1 H}{H M}=\frac{A H \operatorname{tg} \alpha}{H M}=\frac{A H}{H M} \cdot \operatorname{tg} \alpha=2 \operatorname{tg} \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_786127613a7b2f99b094g-04.jpg?height=129&width=1164&top_left_y=1233&top_left_x=1)

$$
\operatorname{tg} \frac{\alpha}{2}=\frac{\sqrt{1+4 \operatorname{tg}^{2} \alpha-1}}{2 \operatorname{tg} \alpha}=\frac{\sqrt{1+4 \operatorname{tg}^{2} \alpha}}{2 \operatorname{tg} \alpha}, \operatorname{ctg} \frac{\alpha}{2}=\frac{2 \operatorname{tg} \alpha}{\sqrt{1+4 \operatorname{tg}^{2} \alpha}-1}=\frac{\sqrt{1+4 \operatorname{tg}^{2} \alpha}+1}{2 \operatorname{tg} \alpha