12.463 The base of the pyramid FABC is an isosceles triangle $ABC$, where the angle between the equal sides $AB$ and $AC$ is $\alpha (\alpha < \pi / 2)$. A triangular prism $AED A_{1} E_{1} D_{1}$ is inscribed in the pyramid; points $A_{1}, E_{1}$, and $D_{1}$ lie on the lateral edges $AF, CF$, and $BF$ of the pyramid, respectively, and the side $ED$ of the base $AED$ passes through the center of the circle circumscribed around triangle $ABC$. Find the ratio of the volume of the prism to the volume of the pyramid.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-327.jpg?height=321&width=381&top_left_y=716&top_left_x=192)

By Theorem 14, $B C\|D E\| D_{1} E_{1}$. Let $h_{1}$ and $h_{2}$ be the heights of the prism $T_{1}=A E D A_{1} E_{1} D_{1}$ and the pyramid $T_{2}=F A B C$. From the similarity of the pyramids $F A_{1} D_{1} E_{1}$ and $T_{2}$, it follows that $\frac{h_{1}}{h_{2}}=\frac{B D_{1}}{B F}$.

$$
\text { But } \frac{B D_{1}}{B F}=\frac{B D}{B A} \Rightarrow \frac{h_{1}}{h_{2}}=\frac{B D}{B A} \text {. }
$$

Moreover, $\frac{S_{A D E}}{S_{A B C}}=\frac{D A^{2}}{B A^{2}}$ (Theorem 7). Therefore, $\frac{V_{T_{1}}}{3 V_{T_{2}}}=\frac{h_{1} \cdot S_{A D E}}{h_{2} \cdot S_{A B C}}=\frac{B D \cdot A D^{2}}{B A^{3} .}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-327.jpg?height=209&width=364&top_left_y=1147&top_left_x=206)

A

Now consider the base $A B C$ of the pyramid $T_{2}$. Let $A K$ be the height of $\triangle A B C$, and $O$ be the center of the circle circumscribed around $\triangle A B C$ ( $O$ lies inside $\triangle A B C$, since $\angle A=\alpha<\frac{\pi}{2}$ ). Then $\angle B O K=\alpha$

$$
(\angle B O K=\angle O B A+\angle O A B=2 \cdot \angle O A B) \Rightarrow
$$

$\Rightarrow O K=O B \cdot \cos \alpha, \frac{A O}{A K}=\frac{A O}{A O+A O \cdot \cos \alpha}=\frac{1}{1+\cos \alpha}(A O=B O)$.

But from the similarity of triangles, it follows that $\frac{A O}{A K}=\frac{A D}{A B}=\frac{1}{1+\cos \alpha}$.

In the end, $\frac{V_{T_{1}}}{3 V_{T_{2}}}=\frac{A D^{2}(A B-A D)}{A B^{3}}=\left(\frac{A D}{A B}\right)^{2} \cdot\left(1-\frac{A D}{A B}\right)=\frac{\cos \alpha}{8 \cos ^{6} \frac{\alpha}{2}}$.

Answer: $\frac{\cos \alpha}{8 \cos ^{6} \frac{\alpha}{2}}$.