12.241. A cone is inscribed in a sphere with radius $R$; a cylinder with a square axial section is inscribed in this cone. Find the total surface area of the cylinder, if the angle between the generatrix of the cone and the plane of its base is $\alpha$.
## Solution.

Consider the axial section of the given set of bodies. $BC$ is the generatrix of the cone, $FE$ is the generatrix of the cylinder, $D$ is the center of the base of the cone and the lower base of the cylinder, $K$ is the center of the upper base of the cylinder, $\angle BCD = \alpha$ (Fig. 12.108).

If the radius of the base of the cylinder is $r$, then $FE = 2r$, and the area of the complete surface of the cylinder is $S = 2\pi r^2 + 2\pi r \cdot FE = 2\pi r^2 + 4\pi r^2 = 6\pi r^2$.

The radius of the given sphere is the radius of the circle circumscribed around $\triangle ABC$.

From this, $AC = 2R \sin \angle ABC = 2R \sin (180^\circ - 2\alpha) = 2R \sin 2\alpha$,

$CD = R \sin 2\alpha$.

In $\triangle FEC (\angle FEC = 90^\circ)$:

$CE = FE \operatorname{ctg} \angle ECF = 2r \operatorname{ctg} \alpha, CD = CE + ED = 2r \operatorname{ctg} \alpha + r = r(2 \operatorname{ctg} \alpha + 1) \Rightarrow$ $\Rightarrow r(2 \operatorname{ctg} \alpha + 1) = R \sin 2\alpha \Leftrightarrow r = \frac{R \sin 2\alpha}{1 + 2 \operatorname{tg} \alpha}; S = \frac{6\pi R^2 \sin^2 2\alpha}{(1 + 2 \operatorname{tg} \alpha)^2}$.

Answer: $\frac{6\pi R^2 \sin^2 2\alpha}{(1 + 2 \operatorname{ctg} \alpha)^2}$.