9. The base of the pyramid $T A B C D$ is a trapezoid $A B C D$, the length of the shorter base $B C$ of which is $4 \sqrt{11}$. The diagonals of the trapezoid $A C$ and $B D$ intersect at point $O$, the ratio of the areas of triangles $A O B$ and $B O C$ is $3: 2$, and the area of triangle $D O C$ is $132 \sqrt{2 / 25}$. All lateral edges of the pyramid $T A B C D$ are inclined to the base plane at an angle of $30^{\circ}$. Find the volume of the pyramid $T B M N C$, where points $M$ and $N$ are the midpoints of edges $T A$ and $T D$ respectively.

(16 points)
# Solution.

Let $T H$ be the height of the pyramid. Since all lateral edges are inclined to the base at the same angle, $H$ is the center of the circle circumscribed around the base. It follows that the trapezoid is isosceles. From the condition, $\frac{S_{A O B}}{S_{\text {ВОС }}}=\frac{3}{2}=\frac{A O}{O C}$. Triangles $A O D$ and $C O B$ are similar, and $\frac{A D}{B C}=\frac{3}{2}=\frac{A O}{O C}, A D=\frac{3 B C}{2}=6 \sqrt{11}$. Since

$\frac{O D}{B O}=\frac{3}{2}=\frac{S_{D O C}}{S_{B O C}}, S_{B O C}=\frac{2 S_{D O C}}{3}=\frac{88 \sqrt{2}}{5}$. Let $h_{1}$ and $h_{2}$ be the heights of triangles $B O C$ and $A O D$, drawn to the bases $B C$ and $A D$ respectively. Then

$h_{1}=\frac{2 S_{B O C}}{B C}=\frac{4 \sqrt{22}}{5}, \quad h_{2}=\frac{3 h_{1}}{2}=\frac{6 \sqrt{22}}{5}$. The height of the trapezoid $h=h_{1}+h_{2}=2 \sqrt{22}$. We find the lateral sides of the trapezoid

![](https://cdn.mathpix.com/cropped/2024_05_06_3205747dae2ad15e28c2g-05.jpg?height=742&width=583&top_left_y=543&top_left_x=1299)
$A B=C D=3 \sqrt{11}$. Let $Q$ and $R$ be the midpoints of sides $B C$ and $A D$ respectively. Denote $H R=x$, the radius of the circumscribed circle $r$. Then $r^{2}=B Q^{2}+Q H^{2}=44+(2 \sqrt{22}-x)^{2}$, and $r^{2}=A R^{2}+H R^{2}=99+x^{2}$. From this, we find $H R=\frac{3 \sqrt{22}}{8}$, $Q H=\frac{13 \sqrt{22}}{8}, \quad r=\frac{33 \sqrt{6}}{8}$. The height of the pyramid $T H$ is found from

![](https://cdn.mathpix.com/cropped/2024_05_06_3205747dae2ad15e28c2g-05.jpg?height=243&width=488&top_left_y=1335&top_left_x=1435)
the right triangle $T A H$ with angle $\angle T A H=30^{\circ}, \quad A H=r$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3205747dae2ad15e28c2g-05.jpg?height=394&width=440&top_left_y=1665&top_left_x=1485)

We have $T H=r \operatorname{tg} 30^{\circ}=\frac{33 \sqrt{2}}{8}$.

Let $T K$ be the height of the pyramid $T B M N C, T K$ be the perpendicular dropped from point $T$ to the line $Q S$, where $S$ is the midpoint of $M N$. We calculate the volume of the pyramid TBMNC:

$V_{\text {TBMNC}}=\frac{1}{3} \cdot \frac{M N+B C}{2} \cdot Q S \cdot T K$. The area of triangle $T Q S$ can be

![](https://cdn.mathpix.com/cropped/2024_05_06_3205747dae2ad15e28c2g-05.jpg?height=294&width=457&top_left_y=2206&top_left_x=1485)

calculated in two ways:

$S_{T Q S}=\frac{Q R \cdot T H}{4}, S_{T Q S}=\frac{Q S \cdot T K}{2}, \quad Q S \cdot T K=\frac{Q R \cdot T H}{2}$, and $V_{T B M N C}=\frac{1}{6} \cdot \frac{M N+B C}{2} \cdot Q R \cdot T H$.

From this, we get $V_{\text {TBMNC}}=105.875$.

Answer: 105.875.

## Solution for Variant #2