The lateral edge of a regular quadrilateral pyramid is $b$, and the plane angle at the vertex is $\alpha$. Find the radius of the sphere circumscribed about the pyramid.

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Let $P A B C D$ be a regular quadrilateral pyramid with vertex $P, M$ - the center of the base $A B C D, K$ - the midpoint of edge $A B$. Denote $A B=a, \angle M A P=\phi$. Since $P K$ is the height of the isosceles triangle $A P B$, then

$$
a=A B=2 A K=2 A P \sin \angle A P K=2 b \sin \frac{\sqrt{2}}{2} \text {. }
$$

From the right triangle $A M P$, we find that

$$
\cos \phi=\cos \angle M A P=\frac{A M}{A P}=\frac{a \sqrt{2}}{2 b}=\sqrt{2} \sin \frac{2}{2}
$$

Then

$$
\sin \phi=\sqrt{1-\cos ^{2} \varphi}=\sqrt{1-2 \sin ^{2} \frac{2}{2}}=\sqrt{\cos \alpha}
$$

(angle $MAP$ is acute). Let $R$ be the desired radius. Consider the section of the pyramid and the sphere circumscribed around it by a plane passing through points $A, P$, and $C$. We obtain an isosceles triangle $A P C$ inscribed in a circle of radius $R$. By the sine theorem,

$$
R=\frac{A P^{\prime}}{2 \sin \angle A C C^{P} P}=\frac{b}{2 \sin \varphi \varphi}=\frac{b}{2 \sqrt{\operatorname{coser}}}
$$

## Answer

$\frac{b}{\sqrt[2]{\text { Cokir2 }}}$