3 [ [sphere, circumscribed around a pyramid]

A regular quadrilateral pyramid is inscribed in a sphere. The radius of the sphere is 1. The plane angle at the vertex of the pyramid is $45^{\circ}$. Find the lateral surface area of the pyramid.

#
The center $O$ of the sphere is equidistant from the vertices of the base $A B C D$ of the regular pyramid $P A B C D$, so the point $O$ lies on the height $P Q$ of the pyramid (Fig.1). Let

$$
A B=B C=C D=A D=a, P A=P B=P C=P D=l
$$

By the cosine theorem,

$$
A B 2=P A 2+P B 2-2 P A \cdot P B \cos \angle A P B
$$

or

$$
a 2=2 l 2-2 l 2 \cos 45^{\circ}=12(2-\sqrt{2})
$$

from which $a=I \sqrt{2-\sqrt{2}}$. Then

$$
A C=a \sqrt{2}=\sqrt{4-2 \sqrt{2}}, A Q=\frac{1}{2} A C=\frac{1}{2} / \sqrt{4-2 \sqrt{2}}
$$

Let $\angle P A Q=\alpha$. From the right triangle $P A Q$, we find that

$$
\begin{gathered}
\cos \alpha=\frac{A Q}{A P}=\frac{1}{2} \cdot \frac{l \sqrt{1-2 \sqrt{2}}}{i}=\frac{1}{2} \sqrt{4-2 \sqrt{2}}=\sqrt{1-\frac{\sqrt{2}}{2}} \\
\sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-1+\frac{\sqrt{2}}{2}}=\sqrt{\frac{\sqrt{2}}{2}}
\end{gathered}
$$

Consider the section of the pyramid and the sphere by a plane passing through the points $A, P$ and $C$ (Fig.2). We obtain an isosceles triangle $A P C$ inscribed in a circle with center $O$ and radius $R=1$. By the sine theorem,

$$
A P=2 R \sin \angle A C P, \text { or } l=2 \sin \alpha=2 \sqrt{\frac{\sqrt{2}}{2}} \text {. }
$$

If $S$ is the lateral surface of the given pyramid, then

$$
\begin{aligned}
S & =4 S_{\triangle A P B}=4 \cdot \frac{1}{2} A P \cdot B P \sin \angle A P B= \\
& =212 \sin 45^{\circ}=2\left(2 \sqrt{\frac{\sqrt{2}}{2}}\right) 2 \cdot \frac{\sqrt{2}}{2}=4
\end{aligned}
$$

## Answer

4.00