12.457 The height of a regular truncated square pyramid is $\boldsymbol{H}$. The lateral edge forms an angle $\alpha$ with the base, and the diagonal of the pyramid forms an angle $\beta$ with the base. Find the area of the section of the pyramid by a plane passing through the diagonal of the pyramid parallel to the non-intersecting diagonal of the base.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-323.jpg?height=427&width=625&top_left_y=636&top_left_x=428)

Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be the given regular truncated pyramid, and $O$ be the projection of $B_{1}$ onto the lower base $A B C D$. Then $\angle B_{1} D O=\beta, \angle B_{1} B O=\alpha$ (by the problem statement), and $O$ lies on the diagonal $B D$, since $B_{1} D_{1} D B \perp A B C D$.

Now let $P$ be the plane passing through $B_{1}, D$ and parallel to $A C$, and $B_{1} L D K$ be the section of the pyramid by the plane $P$. Then the lines $M N$ and $M_{1} N_{1}$ of intersection of $P$ with the bases of the pyramid will be parallel to $A C$ and $A_{1} C_{1}$ (Theorem 14). (The points $M, N$, $M_{1}, N_{1}$ lie on the extensions of the corresponding sides of the bases.) But then $K L \| M N$ (Theorem 14).

Since $A C \perp B_{1} D_{1} D B$ (the pyramid is regular), then $A C \perp B_{1} D \Rightarrow K L \perp B_{1} D \Rightarrow S_{B_{1} L D K}=\frac{1}{2} K L \cdot B_{1} D$. From the similarity of triangles, we have $(Q$ is the intersection point of $K L$ and $B_{1} D): \frac{K L}{M N}=\frac{B_{1} Q}{B_{1} D}, \frac{K L}{M_{1} N_{1}}=\frac{D Q}{D B_{1}} \Rightarrow K L\left(\frac{1}{M N}+\frac{1}{M_{1} N_{1}}\right)=1$.

But \quad M_{1} N_{1}=2 \cdot A_{1} C_{1}, M N=2 \cdot A C \Rightarrow K L=\frac{2 \cdot A C \cdot A_{1} C_{1}}{A C+A_{1} C_{1}}.

Further, $A C=B D=B O+O D=H \cdot \operatorname{ctg} \alpha+H \cdot \operatorname{ctg} \beta$.

Similarly, $A_{1} C_{1}=B_{1} D_{1}=H \cdot \operatorname{ctg} \beta-H \cdot \operatorname{ctg} \alpha$. Moreover, $B_{1} D=\frac{H}{\sin \beta}$.

In the end, we have: $S_{B_{1} L D K}=\frac{H^{2}\left(\operatorname{ctg}^{2} \beta-\operatorname{ctg}^{2} \alpha\right)}{2 \sin \beta \cdot \operatorname{ctg} \beta}$.

Answer:

$$
S_{B_{1} L D K}=\frac{H^{2} \sin (\alpha+\beta) \cdot \sin (\alpha-\beta)}{\sin ^{2} \alpha \cdot \sin \beta \cdot \sin 2 \beta}
$$