3.87. The height of a regular truncated triangular pyramid is $H$ and is the mean proportional between the sides of the bases. The lateral edge makes an angle $\alpha$ with the base. Find the volume of the pyramid.
3.87. Given $A B C A_{1} B_{1} C_{1}$ is a regular truncated triangular pyramid, $O O_{1}=H, \quad O O_{1}=\sqrt{B C \cdot B_{1} C_{1}}$, $\angle A_{1} A O=\alpha$ (Fig. 3.87). Let $B C=a$, $B_{1} C_{1}=b$; then $O A=\frac{a \sqrt{3}}{3}, O_{1} A_{1}=\frac{b \sqrt{3}}{3}$.

Draw $A_{1} K \| O O_{1};$ from $\triangle A_{1} K A$ we find $A K=H \operatorname{ctg} \alpha$ and since $A K=A O-K O=$

$$
=A O-A_{1} O_{1}=\frac{a \sqrt{3}}{3}-\frac{b \sqrt{3}}{3}=\frac{\sqrt{3}}{3}(a-b), \quad \text { then } \frac{\sqrt{3}}{3}(a-b)=H \operatorname{ctg} \alpha .
$$

Moreover, from the condition it follows that $a b=H^{2}$. As a result, we get

$$
\begin{gathered}
V_{\text {truncated pyramid}}=\frac{1}{3} H\left(\frac{a^{2} \sqrt{3}}{4}+\frac{a b \sqrt{3}}{4}+\frac{b^{2} \sqrt{3}}{4}\right)=\frac{H \sqrt{3}}{12}\left(a^{2}+a b+b^{2}\right)= \\
=\frac{H \sqrt{3}}{12}\left((a-b)^{2}+3 a b\right)=\frac{H \sqrt{3}}{12}\left(3 H^{2} \operatorname{ctg}^{2} \alpha+3 H^{2}\right)= \\
=\frac{H^{3} \sqrt{3}}{4}\left(\operatorname{ctg}^{2} \alpha+1\right)=\frac{H^{3} \sqrt{3}}{4 \sin ^{2} \alpha}
\end{gathered}
$$