[ [Sphere touching the edges or sides of the pyramid]

The base of the pyramid $S A B C$ is an equilateral triangle $A B C$ with side $2 \sqrt{2}$. Edges $S B$ and $S C$ are equal. A sphere touches the sides of the base, the plane of the face $S B C$, and the edge $S A$. What is the radius of the sphere if $S A=\frac{3 \sqrt{2}}{2}$?
Let a sphere of radius $R$ touch the side $BC$ of the base at point $D$ (Fig.1). Then $D$ is the only common point of the sphere with the plane of the face $SBC$, i.e., the sphere touches this plane at point $D$. If the sphere touches the edges $AB$ and $AC$ at points $M$ and $N$ respectively, then the center $O$ of the sphere lies in the plane passing through the midpoint of segment $MN$ perpendicular to $MN$. Since triangle $ABC$ is equilateral, this plane is also perpendicular to edge $BC$ and passes through the midpoint of $BC$, and since point $S$ is equidistant from the ends of segment $BC$, vertex $S$ lies in this plane. Since $SD \perp BC$, point $D$ is the midpoint of $BC$. The section of the sphere by the plane of the base of the pyramid is a circle inscribed in the equilateral triangle $ABC$ with side $a=2\sqrt{2}$ (Fig.2). If $r$ is the radius of this circle, then

$$
r=\frac{a \sqrt{3}}{6}=\frac{2 \sqrt{2} \sqrt{3}}{6}=\frac{\sqrt{6}}{3}
$$

If $E$ is the point of intersection of the incircle of triangle $ABC$ with segment $AD$, then

$$
DE=2r=\frac{2\sqrt{6}}{3}, \quad AE=AD-DE=\frac{2\sqrt{3}}{2}-2r=\sqrt{6}-\frac{2\sqrt{6}}{3}=\frac{\sqrt{6}}{3}
$$

Let the sphere touch the edge $SA$ at point $K$. Consider the section of the pyramid and the sphere by the plane $SAD$ (Fig.3).

We obtain a circle with center $O$ and radius $R$, passing through points $D, K, E$ and touching $SD$ and $SA$ at points $D$ and $K$. By the tangent-secant theorem,

$$
AK=\sqrt{AE \cdot AD}=\sqrt{\frac{\sqrt{6}}{3} \cdot \sqrt{6}}=\sqrt{2}
$$

Then

$$
SD=SK=SA-AK=\frac{3\sqrt{2}}{2}-\sqrt{2}=\frac{\sqrt{2}}{2}
$$

The center of the circle inscribed in an angle lies on the bisector of this angle, so $SO$ is the bisector of angle $ASD$. Let $\angle ASO=\angle DSO=\alpha$. By the cosine rule,

$$
\begin{aligned}
& \cos 2\alpha=\cos \angle ASD=\frac{SA^2+SD^2-AD^2}{2SA \cdot SD}= \\
& =\frac{\left(\frac{3\sqrt{2}}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2-(\sqrt{6})^2}{2 \cdot \frac{3\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}}=\frac{\frac{9}{2}+\frac{1}{2}-6}{3}=-\frac{1}{3}
\end{aligned}
$$

From the equation

$$
\frac{1}{-3}=\cos 2\alpha=\frac{1-\tan^2 \alpha}{1+\tan^2 \alpha}
$$

we find that $\tan \alpha=\sqrt{2}$. Therefore,

$$
R=OD=SD \tan \angle OSD=\frac{\sqrt{2}}{2} \cdot \tan \alpha=\frac{\sqrt{2}}{2} \cdot \sqrt{2}=1
$$

## Answer

1.00