12.123. The angle at the vertex of the axial section of a cone is $2 \alpha$, and the sum of the lengths of its height and slant height is $a$. Find the volume of the cone.
## Solution.

By the condition $S A+S O=a \Rightarrow S O=a-S A$ (Fig. 12.123). The volume of the cone $V=\frac{1}{3} \pi R^{2} H$. From $\triangle S O A \quad S O=S A \cos \alpha$. Then $S A \cos \alpha=a-S A$. $S A=\frac{a}{1+\cos \alpha}$ and $H=S O=\frac{a \cos \alpha}{1+\cos \alpha}$, and $A O=R=S A \sin \alpha=\frac{a \sin \alpha}{1+\cos \alpha}$. Then we finally get:

$$
V=\frac{1}{3} \pi \frac{a^{2} \sin ^{2} \alpha}{(1+\cos \alpha)^{2}} \cdot \frac{a \cos \alpha}{(1+\cos \alpha)}=\frac{a^{3}}{3} \pi \frac{\cos \alpha \cdot 4 \sin ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\alpha}{2}}{\left(2 \cos ^{2} \frac{\alpha}{2}\right)^{3}}=
$$

$$
=\frac{a^{3} \pi \cos \alpha \cdot 4 \sin ^{2} \frac{\alpha}{2}}{3 \cdot 8 \cos ^{4} \frac{\alpha}{2}}=\frac{a^{3} \pi \cos \alpha \cdot \sin ^{2} \frac{\alpha}{2}}{6 \cos ^{4} \frac{\alpha}{2}}
$$

Answer: $\frac{a^{3} \pi \cos \alpha \cdot \sin ^{2} \frac{\alpha}{2}}{6 \cos ^{4} \frac{\alpha}{2}}$.