### 12.442 In a regular quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$

 $\left(A A_{1}\left\|B B_{1}\right\| C C_{1} \| D D_{1}\right)$ a plane is drawn through the midpoints of two adjacent sides of the base $D C$ and $A D$ and the vertex $B_{1}$ of the upper base. Find the angle between this plane and the plane of the base, if the perimeter of the section is three times the diagonal of the base.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-313.jpg?height=416&width=407&top_left_y=1189&top_left_x=220)

Let $B_{1} F N M E$ be the section of the given regular prism by a plane passing through $B_{1}$ and the midpoints $M$ and $N$ of the edges $A D$ and $D C$. Let point $K$ be the intersection of this plane with the extension of the edge $D_{1} D$. Then $B_{1} F K E$ is a parallelogram (Theorem 13). Moreover, it is a rhombus because $K F = K E$ (their projections $A D$ and $C D$ on the plane $A B C D$ are equal).

Let $L$ be the intersection of $B D$ and $M N$. $M N \perp B D \Rightarrow M N \perp B_{1} L$ (Theorem 12) $\Rightarrow \angle B_{1} L B$ is the desired angle.

Let $E K = a$. By the problem's condition, $3 \cdot B D = B_{1} F + B_{1} E + E M + F N + M N = 3 a + \frac{B D}{2} \left(M N = \frac{A C}{2} = \frac{B D}{2}, E M = F N = \frac{a}{2}\right) \Rightarrow a = \frac{5}{6} B D$. But $E F$ and $B_{1} K$ are the diagonals of the rhombus $\Rightarrow B_{1} K = 2 \sqrt{E K^{2} - \left(\frac{E F}{2}\right)^{2}} = 2 \sqrt{a^{2} - \frac{B D^{2}}{4}} = \frac{4}{3} \cdot B D \Rightarrow$

$$
\Rightarrow \cos \angle B_{1} L B = \frac{B L}{B_{1} L} = \frac{B L + L D}{B_{1} L + L K} = \frac{B D}{B_{1} K} = \frac{3}{4}
$$

Answer: $\arccos \frac{3}{4}$.