12.230. Two cones have concentric bases and the same angle, equal to $\alpha$, between the height and the slant height. The radius of the base of the outer cone is $R$. The lateral surface area of the inner cone is half the total surface area of the outer cone. Find the volume of the inner cone.
Solution.

Let $\triangle A B C$ be the axial section of the outer cone (Fig. 12.97), $\triangle M L K$ be the axial section of the inner cone, $O$ be the common center of their bases, $O A=R, \angle A B O=\angle M L O=\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0979.jpg?height=450&width=454&top_left_y=405&top_left_x=96)

Fig. 12.98

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0979.jpg?height=759&width=603&top_left_y=63&top_left_x=623)

Fig. 12.99

The slant height of the outer cone $l_{1}=A B=\frac{R}{\sin \alpha}$, its total surface area $S_{1}=\pi R\left(R+l_{1}\right)=\pi R\left(R+\frac{R}{\sin \alpha}\right)=\frac{\pi R^{2}(1+\sin \alpha)}{\sin \alpha}=\frac{\pi R^{2}\left(1+\cos \left(\frac{\pi}{2}-\alpha\right)\right)}{\sin \alpha}=$ $=\frac{2 \pi R^{2} \cos ^{2}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}{\sin \alpha}$.

If the radius of the inner cone $O M=r$, then its slant height $l_{2}=M L=\frac{r}{\sin \alpha}$, and its lateral surface area $S_{2}=\pi r l_{2}=\frac{\pi r^{2}}{\sin \alpha}$. According to the condition $S_{2}=\frac{1}{2} S_{1} \Rightarrow \frac{\pi r^{2}}{\sin \alpha}=\frac{\pi R^{2} \cos ^{2}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}{\sin \alpha} \Rightarrow r=R \cos \left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$. The height of the inner cone $h=L O=r \operatorname{ctg} \alpha$, and its volume

$$
V=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi r^{3} \operatorname{ctg} \alpha=\frac{1}{3} \pi R^{3} \cos ^{3}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) \operatorname{ctg} \alpha
$$

Answer: $\frac{1}{3} \pi R^{3} \cos ^{3}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) \operatorname{ctg} \alpha$.