$\begin{aligned} & {\left[\begin{array}{l}\text { Tangent Spheres }\end{array}\right]} \\ & \text { [ Sphere inscribed in a dihedral angle }]\end{aligned}$

At the base of the pyramid $S A B C$ lies an equilateral triangle $A B C$ with side $2 \sqrt{3}$, and $S A=S B=S C=\sqrt{7}$. A sphere $S 1$ is inscribed in the trihedral angle at vertex $C$. A sphere $S 2$, whose radius is three times that of sphere $S 1$, touches sphere $S 1$, the planes $S A C$ and $A B C$. The segment of the line $S B$ enclosed within sphere $S 2$ is $\frac{6}{\sqrt{7}}$. Find the radius of sphere $S 2$.
Let $SM$ be the height of the given regular pyramid $SABC$ (Fig.1), $K$ be the midpoint of edge $AC$. Then $M$ is the center of the equilateral triangle $ABC$, $BK$ is the height of this triangle,

$$
\begin{gathered}
BK=AC \cdot \frac{\sqrt{3}}{2}=2 \sqrt{3} \cdot \frac{\sqrt{3}}{2}=3 \\
BM=\frac{2}{3} BK=2, KM=\frac{1}{3} BK=1
\end{gathered}
$$

By the Pythagorean theorem from the right triangle $BMS$ we find that

$$
SM=\sqrt{BS^{2}-BM^{2}}=\sqrt{7-4}=\sqrt{3}
$$

Let $\beta$ be the angle between the lateral face of the pyramid and the plane of the base. From the right triangle $KMS$ we find that

$$
\tan \beta=\frac{SM}{KM}=\frac{\sqrt{3}}{1}=\sqrt{3}
$$

so $\beta=60^{\circ}$. The center of the sphere inscribed in a dihedral angle lies in the bisector plane of this angle. Let spheres $S1$ and $S2$ with centers $O1$ and $O2$ touch the plane of the base $ABC$ at points $P$ and $Q$ respectively, $P1$ and $Q1$ - orthogonal projections of points $P$ and $Q$ onto the line $AC$, and $r$ and $3r$ - the radii of the spheres (Fig.2). Then

$$
\begin{gathered}
PP1=r \cot \frac{\frac{9}{2}}{}=r \cot 30^{\circ}=r \sqrt{3} \\
QQ1=3r \cot \frac{\frac{9}{2}}{2}=3r \cot 30^{\circ}=3r \sqrt{3} \\
PQ=2 \sqrt{3r \cdot r}=2r \sqrt{3}
\end{gathered}
$$

Thus, $PP1+PQ=QQ1$. This means that points $P$ and $Q$ lie on the perpendicular $QQ1$ to the line $AC$, and point $P$ lies between $Q$ and $Q1$. Moreover, since the faces $ASC$ and $BSC$ form equal angles with the base plane, point $P$ lies on the bisector of angle $ACB$. Therefore,

$$
CQ1=PQ1 \cot 30^{\circ}=r \sqrt{3} \cdot \sqrt{3}=3r
$$

We draw a section of the pyramid by a plane passing through points $B$, $S$, and $K$ (Fig.3). We get a circle with center at some point $D$, lying on the bisector of angle $BSK$ of triangle $BSK$, and side $BS$ intersects this circle at chord $EF$, equal to $\frac{6}{\sqrt{7}}$. If $H$ is the orthogonal projection of point $D$ onto the base plane $ABC$, then $DH=O2Q=3r$. The distance $O2D$ from point $O2$ to the plane of the section is equal to the distance from point $Q$ to line $BK$, and since $QQ1 \parallel BK$,

$$
O2D=HQ=Q1K=|CK-CQ1|=|\sqrt{3}-3r|
$$

From the right triangle $O2DF$ we find that

$$
DE=\sqrt{O2E^2-O2D^2}=\sqrt{9r^2-(\sqrt{3}-3r)^2}=\sqrt{6r\sqrt{3}-3}
$$

Given that $r \geqslant \frac{1}{2\sqrt{3}}$. Let $G$ be the midpoint of chord $EF$. Then

$$
\begin{aligned}
DG & =\sqrt{DE^2-EG^2}=\sqrt{DE^2-\frac{1}{4}EF^2}= \\
& =\sqrt{6r\sqrt{3}-3-\frac{9}{7}}=\sqrt{6r\sqrt{3}-\frac{30}{7}}
\end{aligned}
$$

Moreover,

$$
HM=|HK-KM|=|QQ1-MK|=|3r\sqrt{3}-1|=3r\sqrt{3}-1,
$$

since $r \geqslant \frac{1}{2\sqrt{3}}>\frac{1}{3\sqrt{3}}$. This means that points $D$ and $K$ lie on opposite sides of line $SM$. If point $D$ lies inside triangle $BSM$ (Fig.4), then the area of triangle $BMS$ is equal to the sum of the areas of triangles $BDM$, $MSD$, and $BDS$. In this case,

$$
\frac{1}{2} BM \cdot DH + \frac{1}{2} MS \cdot HM + \frac{1}{2} BS \cdot DG = \frac{1}{2} BM \cdot SM
$$

or

$$
2 \cdot 3r + \sqrt{3}(3r\sqrt{3}-1) + \sqrt{7} \sqrt{6r\sqrt{3}-\frac{30}{7}} = 2\sqrt{3}
$$

After obvious transformations, we get:

$$
\Rightarrow r=\frac{\sqrt{3}}{3} \text{ or } r=\frac{19\sqrt{3}}{75}
$$

Both found roots do not satisfy the condition $3\sqrt{3}-15r \geqslant 0$. If point $D$ lies outside triangle $BSM$ (Fig.5) (this can only be the case when points $D$ and $M$ lie on opposite sides of line $BS$), then

$$
\frac{1}{2} BM \cdot DH + \frac{1}{2} MS \cdot HM - \frac{1}{2} BS \cdot DG = \frac{1}{2} BM \cdot SM
$$

or

$$
2 \cdot 3r + \sqrt{3}(3r\sqrt{3}-1) - \sqrt{7} \sqrt{6r\sqrt{3}-\frac{30}{7}} = 2\sqrt{3}
$$

After obvious transformations, we get:

$$
\begin{aligned}
\sqrt{4r\sqrt{3}-30} & = 15r - 3\sqrt{3} \Rightarrow 75r^2 - 44r\sqrt{3} + 19 = 0 \Rightarrow \\
& \Rightarrow r=\frac{\sqrt{3}}{3} \text{ or } r=\frac{19\sqrt{3}}{75} .
\end{aligned}
$$

Both roots satisfy the condition of the problem. Therefore, the radius of sphere $S2$ is $\sqrt{3}$ or $\frac{19\sqrt{3}}{25}$.

## Answer

$\sqrt{3} ; \frac{19\sqrt{3}}{25}$.