3 [ Lateral Surface of a Parallelepiped ]

In a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$, the diagonals $A C$ and $B D$ of the base $A B C D$ intersect at point $M, \angle A M B=\alpha$. Find the area of the lateral surface of the parallelepiped if $B 1 M=b, \angle B M B 1=\beta$.
From the right triangle $M B B 1$ and the isosceles triangles $A M B$ and $B M C$, we find that

$$
B B 1=M B 1 \sin \angle B M B 1=b \sin \beta, M B=M B 1 \cos \angle B M B 1=b \cos \beta,
$$

$$
A B=2 M B \sin \frac{1}{2} \angle A M B=2 b \cos \beta \sin \frac{\frac{1}{2}}{2}, B C=2 M B \cos \frac{1}{2} \angle M C B=2 b \cos \beta \cos \frac{\frac{1}{2}}{2}
$$

Let $S$ be the lateral surface area of the parallelepiped $A B C D A 1 B 1 C 1 D 1$. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_d69082500b56170eb8ddg-09.jpg?height=63&width=1078&top_left_y=1784&top_left_x=496)

$$
\begin{aligned}
& =4 b \cos \beta \cdot b \sin \beta \cdot\left(\sin \frac{\frac{1}{2}}{\mathbf{2}}+\cos \frac{\mathbf{2}}{\mathbf{2}}\right)=4 b^2 \cos \beta \sin \beta \cdot \sqrt{2} \sin \left(\frac{\pi}{\mathbf{4}}+\frac{\mathbf{1}}{\mathbf{2}}\right)= \\
& =2 b^2 \sqrt{2} \sin 2 \beta \sin \frac{2 k+\pi}{4} .
\end{aligned}
$$

## Answer

$2 b^2 \sqrt{2} \sin 2 \beta \sin \frac{2 a+\pi}{4}$.