12.269. Two vertices of an equilateral triangle with side $a$ lie on the circumference of the upper base of a cylinder, while the third vertex lies on the circumference of the lower base. The plane of the triangle forms an angle $\alpha$ with the generatrix of the cylinder. Find the lateral surface area of the cylinder.
## Solution.

Let the side $BC$ of the equilateral triangle $ABC$ (Fig. 12.133) lie on the upper base of the cylinder, and the vertex $A$ on the circumference of the lower base, $BC=a, AM$ is the generator of the cylinder, $K$ is the midpoint of $BC$. Then $BM=CM, MK \perp BC, AK \perp BC$. Therefore, the line $BC$ is perpendicular to the plane $AMK$, which means the planes $AMK$ and $ABC$ are perpendicular.

In the plane $AMK$, a perpendicular $ML$ is drawn to $AK$. Then $AK$ is the projection of the line $AM$ on the plane $ABC$ and $\angle MAK=\alpha$.

$$
AK=\frac{a \sqrt{3}}{2}
$$

In $\triangle AMK\left(\angle AMK=90^{\circ}\right):$

$AM=AK \cos \angle MAK=\frac{a \sqrt{3}}{2} \cos \alpha ; MK=AK \sin \angle MAK=\frac{a \sqrt{3}}{2} \sin \alpha$.

In $\triangle AMB\left(\angle AMB=90^{\circ}\right):$

$$
\begin{aligned}
& BM^{2}=AB^{2}-AM^{2}=a^{2}-\frac{3 a^{2}}{4} \cos ^{2} \alpha=\frac{a^{2}}{4}\left(4-3 \cos ^{2} \alpha\right)= \\
& =\frac{a^{2}}{4}\left(1+3 \sin ^{2} \alpha\right)
\end{aligned}
$$

The radius of the circle circumscribed around $\triangle BMC$:

$R=\frac{BM \cdot CM \cdot BC}{4 S_{\triangle BMC}}=\frac{BM^{2} \cdot BC}{4 \cdot \frac{1}{2} BC \cdot MK}=\frac{BM^{2}}{2 MK}=\frac{\frac{a^{2}}{4}\left(1+3 \sin ^{2} \alpha\right)}{a \sqrt{3} \sin \alpha}=$ $=\frac{a\left(3 \sin ^{2} \alpha+1\right)}{4 \sqrt{3} \sin \alpha}$.

The lateral surface area of the cylinder

$$
S=2 \pi R \cdot AM=2 \pi \cdot \frac{a\left(3 \sin ^{2} \alpha+1\right)}{4 \sqrt{3} \sin \alpha} \cdot \frac{a \sqrt{3}}{2} \cos \alpha=\frac{1}{4} \pi a^{2} \operatorname{ctg} \alpha\left(3 \sin ^{2} \alpha+1\right)
$$

Answer: $\frac{1}{4} \pi a^{2} \operatorname{ctg} \alpha\left(3 \sin ^{2} \alpha+1\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1024.jpg?height=777&width=749&top_left_y=25&top_left_x=316)

Fig. 12.134