12.234. The side of the base of a regular quadrilateral pyramid is equal to $a$; the lateral face forms an angle $\alpha$ with the base plane. Find the radius of the circumscribed sphere.
## Solution.

Let $S O$ be the height of the regular pyramid $S A B C D$ (Fig. 12.101), $A B=a$, and $M$ be the midpoint of $C D$. Then $O M \perp C D, S M \perp C D, \angle S M O=\alpha$.

The radius $R$ of the sphere will be found as the radius of the circle circumscribed around $\triangle A S C$.

If $\angle A S C=x$, then

$$
R=\frac{A C}{2 \sin \angle A S C}=\frac{a \sqrt{2}}{2 \sin x}
$$

In $\triangle S O M\left(\angle S O M=90^{\circ}\right): S O=O M \operatorname{tg} \angle S M O=\frac{a}{2} \operatorname{tg} \alpha$.

In $\triangle S O A\left(\angle S O A=90^{\circ}\right): \operatorname{tg} \angle A S O=\frac{A O}{S O} \Rightarrow$

$\Rightarrow \operatorname{tg} \frac{x}{2}=\frac{a \sqrt{2}}{2}: \frac{a}{2} \operatorname{tg} \alpha=\sqrt{2} \operatorname{ctg} \alpha \Rightarrow \sin x=\frac{2 \operatorname{tg} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}}=\frac{2 \sqrt{2} \operatorname{ctg} \alpha}{1+2 \operatorname{ctg}^{2} \alpha} \Rightarrow$

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

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

Answer: $\frac{a(3+\cos 2 \alpha)}{4 \sin 2 \alpha}$.