10,11
$[\quad$ Area of the cross-section $\quad]$

[ Angles between lines and planes ]

The side of the base $ABC$ of a regular triangular pyramid $ABCD$ is 6, and the dihedral angle between the lateral faces is $\arccos { }^{7} / 32$. Points $A_{1}$ and $B_{1}$ are the midpoints of edges $AD$ and $BD$ respectively, and $BC_{1}$ is the height in triangle $DBC$. Find:

1) the angle between the lines $AB$ and $B_{1}C_{1}$;
2) the area of triangle $A_{1}B_{1}C_{1}$;
3) the distance from point $B$ to the plane $A_{1}B_{1}C_{1}$;
4) the radius of the sphere inscribed in the pyramid $A_{1}B_{1}C_{1}D$.
Let $D H$ be the height of the pyramid, $M$ be the midpoint of $A B$. The line $C D$ is perpendicular to two intersecting lines $A B$ and $B C 1$ in the plane $A C 1 B$, so $A C 1 B$ is the linear angle of the dihedral angle between the faces $A D C$ and $B D C$.

Denote $\angle A C 1 B=\gamma$. According to the problem, $\cos \gamma=\frac{7}{32}$. Additionally, $B C 1 \perp C D$ and $A C 1 \perp C D$, so the right triangles $D C 1 B$ and $D C 1 A$ are equal by the leg and hypotenuse, making $A C 1 B$ isosceles.

![](https://cdn.mathpix.com/cropped/2024_05_06_78a338f936ec2a37fae8g-51.jpg?height=115&width=1726&top_left_y=1023&top_left_x=0)
We find that $\operatorname{tg} \frac{\gamma}{2}=\frac{5}{\sqrt{39}}$. From the isosceles triangle $A C 1 B$, we get

$$
M C 1=\frac{\frac{B M}{1 g_{2}^{7}}}{\ln }=\frac{3 \sqrt{39}}{5}
$$

Let $\angle D C H=\alpha$. Then

$$
\begin{aligned}
& \sin \alpha=\frac{M C_{1}}{M C}=\frac{\frac{M C_{1}}{A B \sqrt{3}}}{2}=\frac{\frac{3 \sqrt{39}}{\frac{9}{3}}}{2}=\frac{\sqrt{13}}{5}, \\
& \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\sqrt{1-\frac{13}{25}}=\frac{2 \sqrt{3}}{5}, \operatorname{tg} \alpha=\frac{\sin 12}{\cos \alpha 1_{2}}=\frac{\sqrt{13}}{2 \sqrt{3}} .
\end{aligned}
$$

From the right triangles $C D H$ and $B D C 1$, we find that

$$
\begin{aligned}
& D H=C H \operatorname{tg} \alpha=2 \sqrt{3} \cdot \frac{\sqrt{13}}{2 \sqrt{3}}=\sqrt{13}, D C=\frac{H C}{\operatorname{OABCI2}}=\frac{\frac{A B \sqrt{3}}{3}}{\frac{3}{\operatorname{Oon} \sqrt{2}}}=\frac{\frac{\sqrt{3}}{2 \sqrt{3}}}{3}=5 \text {, } \\
& B C 1=\sqrt{M C_{1}^{2}+B M^{2}}=\sqrt{\left(\frac{3 \sqrt{39}}{5}\right)^{2}+9}=\frac{24}{5}
\end{aligned}
$$

Then

$$
D C 1=\sqrt{B D^{2}-B C_{1}^{2}}=\sqrt{25-\left(\frac{24}{5}\right)^{2}}=\frac{7}{5}
$$

Since $B 1 C 1$ is the median of the right triangle $B C 1 D$,

$$
B 1 C 1=\frac{1}{2} D B=\frac{1}{2} D C=\frac{5}{2}
$$

Triangle $A 1 D C 1$ is equal to the isosceles triangle $B 1 D C 1$ by two sides and the angle between them, so $A 1 C 1=B 1 C 1=\frac{5}{2}$. Let $\angle A 1 B 1 C 1=\phi$. Since $A 1 B 1 \| A B$, the angle between the lines $A B$ and $B 1 C 1$ is equal to the angle between the lines $A 1 B 1$ and $B 1 C 1$. From the isosceles triangle $A 1 B 1 C 1$, we find that

$$
\cos \phi=\frac{\frac{A_{1} B_{1}}{\frac{2}{2}}}{B_{1} C_{1}}=\frac{3}{\frac{3}{2}}=\frac{3}{5}
$$

Then $\sin \phi=\frac{\mathbf{4}}{5}$ and

$$
S_{\triangle A} 1 B 1 C 1=\frac{1}{2} A 1 B 1 \cdot B 1 C 1 \sin \phi=\frac{1}{2} \cdot 3 \cdot \frac{5}{2} \cdot \frac{4}{5}=3
$$

Let $h$ be the distance from point $B$ to the plane $A 1 B 1 C 1$. Since $B 1$ is the midpoint of segment $B D$, the distance from point $D$ to the plane $A 1 B 1 C 1$ is also $h$. Let $V$ and $V 1$ be the volumes of the pyramids $A B C D$ and $A 1 B 1 C 1 D$ respectively. Then

$$
\begin{gathered}
V=\frac{1}{3} S_{\triangle A B C} \cdot D H=\frac{1}{3} \cdot \frac{A B^{2} \sqrt{3}}{4} \cdot \sqrt{13}=\frac{1}{3} \cdot \frac{36 \sqrt{3}}{4} \cdot \sqrt{13}=3 \sqrt{39} \\
V 1=\frac{D A_{1}}{D A} \cdot \frac{D B_{1}}{D B} \cdot \frac{D C C_{1}}{D C} \cdot V=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{7}{5} \cdot 3 \sqrt{39}=\frac{21 \sqrt{39}}{100},
\end{gathered}
$$

and since the height of the pyramid $A 1 B 1 C 1 D$ is $h$,

$$
h=\frac{3 V_{1}}{S_{\Delta A_{1} B_{1} C_{1}}}=\frac{\frac{3 \cdot 21 \sqrt{39}}{109}}{3}=\frac{21 \sqrt{39}}{100}
$$

Since $B C 1$ is the height of triangle $B D C$,

$$
S_{\triangle B D C}=\frac{1}{2} D C \cdot B C 1=\frac{1}{2} \cdot 5 \cdot \frac{24}{5}=12
$$

Then

$$
S_{\triangle B} 1 D C 1=\frac{D B_{1}}{D B} \cdot \frac{D C_{1}}{D C} \cdot S_{\triangle B D C}=\frac{1}{2} \cdot \frac{7}{25} \cdot 12=\frac{42}{25}
$$

Let $S$ be the total surface area of the pyramid $A 1 B 1 C 1 D$, and $r$ be the radius of the sphere inscribed in this pyramid.

$$
S=2 S_{\triangle A} 1 B 1 C 1+2 S_{\triangle B} 1 D C 1=2\left(3+\frac{42}{25}\right)
$$

| Problem 110530 Topics: | Angles between lines and planes | Difficulty: 4  Grades: 10,11 |
| :---: | :---: | :---: |
|  | Volume helps solve the problem |  |
|  | Regular pyramid |  |
|  | Area of the section |  |

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_78a338f936ec2a37fae8g-52.jpg?height=146&width=834&top_left_y=1219&top_left_x=612)

## Answer

 $\arccos \frac{3}{5} ; 3 ; \frac{21 \sqrt{39}}{100} ; \frac{7 \sqrt{39}}{104}$