6. Find the volumes of the parts into which the plane, parallel to the diagonal $A C_{1}$ of the lateral face $A A_{1} C_{1} C$, passing through vertex $C$ and the center of symmetry of the lateral face $A A_{1} B_{1} B$, divides the regular triangular prism $A B C A_{1} B_{1} C_{1}$, if the area of the cross-section of the prism by this plane is 21, and the side of the base of the prism is $2 \sqrt{14} . \quad$ (20 points)
# Solution:

1) Construction of the section.

In the lateral face $A A_{1} C_{1} C$, draw a line $A_{2} C$ through point $C$ parallel to $A C_{1}$, where point $A_{2}$ is the intersection of $A A_{1}$ and $A_{2} C$, and $A A_{1}=A A_{2}$.

Let $D$ be the center of symmetry of the lateral face $A A_{1} B_{1} B$. Draw a line $A_{2} D$ through point $D$, which belongs to the plane of the section. Point $M$ is the intersection of $A B$ and $A_{2} D$, and point $N$ is the intersection of $A_{1} B_{1}$ and $A_{2} D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_52b2f7cff5dc0b15c76dg-06.jpg?height=400&width=260&top_left_y=191&top_left_x=1646)

Point $L$ is the intersection of line $B B_{1}$ and $A_{2} D$, $A M=N B_{1}=a / 3, B M=N A_{1}=2 a / 3, B B_{1}=B_{1} L=h$.

In the base plane $A_{1} B_{1} C_{1}$, draw a line $N K$ through point $N$ parallel to $M C$, where point $K$ is the intersection of $N K$ and $B_{1} C_{1}$.

The trapezoid $M N K C$ is the desired section.

2) Project the section onto the base plane $A B C$ of the prism. Let $N_{1}$ and $K_{1}$ be the projections of points $N$ and $K$ onto the plane $A B C$. Then $N_{1} K_{1} \| M C, B K_{1}=K_{1} C$. The projection of the section onto the base plane $A B C$ is the trapezoid $C M N_{1} K_{1}$, its area $S_{np}=S_{BMC}-S_{BN_{1}K_{1}}=S_{ABC}\left(\frac{2}{3}-\frac{1}{3} \cdot \frac{1}{2}\right)=\frac{1}{2} S_{ABC}=\frac{a^{2} \sqrt{3}}{8}=7 \sqrt{3}$.
3) Find the cosine of the angle $\alpha$ of the inclination of the section plane to the base plane of the prism. $S_{cey}=S_{np} / \cos \alpha \Leftrightarrow \cos \alpha=S_{np} / S_{cey}=1 / \sqrt{3}$.
4) Find the height of the prism $h$. Construct the plane $B H L$ passing through point $B$ and perpendicular to $M C$ (the line of intersection of the base and the section plane, $B H \perp M C, L H \perp M C$). The angle $\alpha$ of the inclination of the section plane to the base plane is equal to the angle $B H L, B L=2 h$.
5) $B H \cdot C M=B M \cdot C D$,

$B H=\frac{B M \cdot C D}{C M}=\frac{\frac{2 a}{3} \cdot \frac{\sqrt{3} a}{2}}{\sqrt{\left(\frac{\sqrt{3} a}{2}\right)^{2}+\left(\frac{a}{6}\right)^{2}}}=\frac{a \sqrt{3}}{\sqrt{7}}=2 \sqrt{6}$,

$\cos \alpha=1 / \sqrt{3}, \sin \alpha=2 / \sqrt{6}, \operatorname{tg} \alpha=\sqrt{2}$,

$2 h=B L=B H \operatorname{tg} \alpha=4 \sqrt{3}, h=2 \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_52b2f7cff5dc0b15c76dg-06.jpg?height=385&width=471&top_left_y=2049&top_left_x=1386)

6) $V_{L B M C}=\frac{1}{3} S_{BMC} \cdot B L=\frac{1}{3} \cdot \frac{2}{3} S_{ABC} \cdot B L=\frac{a^{2} \sqrt{3}}{18} \cdot B L=\frac{112}{3}$.
$V_{L B_{1} N K}=\frac{1}{8} V_{L B M C}, V_{B M C B_{1} N K}=\frac{7}{8} V_{L B M C}=\frac{98}{3}, V_{A B C A B_{1} C_{1}}=S_{ABC} \cdot B B_{1}=\frac{a^{2} \sqrt{3}}{4} \cdot h=84$,

![](https://cdn.mathpix.com/cropped/2024_05_06_52b2f7cff5dc0b15c76dg-07.jpg?height=100&width=463&top_left_y=321&top_left_x=274)

Answer: $\frac{112}{3}, \frac{154}{3}$.

## Solution for Variant 1 (11th Grade)