10. All edges of the regular quadrilateral pyramid $S A B C D$ are equal to $b$. The height of the regular quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ is equal to the height of the specified pyramid. The base $A B C D$ is common to both. (1) A plane $\alpha$ is drawn through points $A$ and $C_{1}$ parallel to the line $B D$. Find the area of the polygon formed by the intersection of plane $\alpha$ and the pyramid $S A B C D$.

(2) A plane $\beta$ is drawn through points $B$ and $D_{1}$. Find the minimum possible area of the polygon formed by the intersection of plane $\beta$ and the pyramid $S A B C D$.

- $b^{2} \sqrt{5} / 6$. This value is also equal to the smallest area of the polygon specified in the problem conditions.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_f2f4694f581fbf9d7159g-6.jpg?height=652&width=828&top_left_y=622&top_left_x=131)

(1) $B D=b \sqrt{2}, D Q=Q S, B P=P S, P Q \| B D$. Triangles $S B D$ and $S P Q$ are similar with a similarity ratio of $2: 1: P Q=B D / 2=b / \sqrt{2}$.

(2) $A C_{1}=b \sqrt{5 / 2}, H N=N S$. $A N=N C_{1}, A N=A C_{1} / 2$, triangles $S M N$ and $C M C_{1}$ are similar with a similarity ratio of $1: 2, N M=A C_{1} / 6, A M=(2 / 3) A C_{1}$;

(3) $S_{A Q C_{1} P}=\frac{1}{2} A M \cdot Q P=\frac{1}{2} \cdot \frac{2}{3} \cdot b \sqrt{\frac{5}{2}} \cdot b \sqrt{\frac{1}{2}}=b^{2} \cdot \frac{1}{3} \cdot \sqrt{\frac{5}{4}}=b^{2} \frac{\sqrt{5}}{6}$.

(4) Theorem 1. The length of the angle bisector of a triangle is less than the semi-sum of the two adjacent sides.

Proof. The length of the angle bisector is $l_{c}=\sqrt{a b-c_{a} c_{b}}=\sqrt{(\sqrt{a b})^{2}-c_{a} c_{b}} \leqslant \sqrt{\left(\frac{a+b}{2}\right)^{2}-c_{a} c_{b}}$ $=\frac{a+b}{2} \sqrt{1-\frac{4 c_{a} c_{b}}{(a+b)^{2}}}<\frac{a+b}{2}$

(5) Let line $l_{A}$ be parallel to $B D_{1}$ and pass through point $A$, line $l_{S}$ be parallel to $B D_{1}$ and pass through point $S$, line $l_{C}$ be parallel to $B D_{1}$ and pass through point $C$. Let plane $\beta$ pass through points $B, D_{1}$ and intersect $A S$ at point $E$, intersect $C S$ at point $F$; intersect $S D$ at point $G$. Let line $l_{F}$, parallel to $B D_{1}$ and pass through point $E$, line $l_{F}$ parallel to $B D_{1}$ and pass through point $F$. Then the area of polygon $B E G F$ is half the product of the length of segment $G B$ and the distance between lines $l_{F}$ and $l_{F}$. From Theorem 1, it now follows that the smallest possible distance between lines $l_{F}$ and $l_{F}$ occurs if and only if $E F \| A C$.