A sphere is inscribed in a regular triangular pyramid $S A B C$ ($S$ - vertex), and also inscribed in a right triangular prism $K L M K_1 L_1 M_1$, where $K L = K M = \sqrt{6}$, and the lateral edge $K K_1$ lies on the line $A B$. Find the radius of the sphere, given that the line $S C$ is parallel to the plane $L L_1 M_1 M$.
Through a straight line that has no common points with a sphere, exactly two planes can be drawn that are tangent to the sphere, so the planes of the faces $ABC$ and $KMM_1K_1$ coincide, as well as the planes of the faces $ASB$ and $KLL_1K_1$. The center $O$ of the sphere inscribed in a regular pyramid lies on its height (Fig.1). Consider the section of the pyramid and prism by a plane passing through the height $SH$ of the pyramid and its lateral edge $SC$. In this plane lies the center $O$ of the sphere and the midpoint $K_2$ of the edge $AB$. The cutting plane is drawn through the line $SC$, parallel to the plane $LL_1M_1M$, so the cutting plane intersects the plane $LL_1M_1M$ along a line parallel to $SC$. Therefore, if $L_2$ and $M_2$ are the points where the cutting plane intersects the lateral edges $LL_1$ and $MM_1$ of the prism, then $L_2M_2 \| SC$. In this case, the section of the sphere by the drawn plane is a circle inscribed in the triangle $K_2L_2M_2$ (Fig.2), in which $K_2L_2=K_2M_2=\sqrt{6}$, and since $L_2M_2 \| SC$, then $SK_2=CK_2$. The height $SK_2$ of the isosceles triangle $ASB$ is equal to the height $CK_2$ of the equilateral triangle $ABC$, so $ABS$ is also an equilateral triangle, and the regular pyramid $SABC$ is a regular tetrahedron. Let $\angle SK_2C=\beta, AB=SC=a$. From the right triangle $SHK_2$ we find that

$$
\operatorname{tg} \beta=\frac{SH}{HK_{2}}=\frac{\frac{a \sqrt{\frac{2}{3}}}{\sigma \sqrt{3}}}{6}=2 \sqrt{2} \text{. }
$$

Then

$$
\cos \beta=\frac{1}{\sqrt{1+g^{2} \beta}}=\frac{1}{\sqrt{1+8}}=\frac{1}{3}, \sin \beta=\frac{2 \sqrt{2}}{3}, \sin \frac{g}{2}=\sqrt{\frac{1-\cos \beta}{2}}=\sqrt{\frac{1-\frac{1}{3}}{2}}=\frac{1}{\sqrt{3}}
$$

$$
L_2M_2=2K_2L_2 \sin \frac{9}{2}=2\sqrt{6} \cdot \frac{1}{\sqrt{3}}=2\sqrt{2}
$$

Let $r$ be the radius of the sphere inscribed in the regular tetrahedron $SABC$. Then $r$ is the radius of the circle inscribed in the triangle $K_2L_2M_2$. Therefore,

$$
r=\frac{2 S_{\Delta K_{2} L_{2} M_{2}}}{K_{2} L_{2}+K_{2} M_{2}+\dot{L}_{2} M_{2}}=\frac{2 \cdot \frac{1}{2} K_{2} L_{2} \cdot K_{2} M_{2} \sin \beta}{K_{2} L_{2}+K_{2} M_{2}+L_{2} M_{2} M_{2}}=
$$

$$
=\frac{2 \cdot \frac{1}{2} \sqrt{6} \cdot \sqrt{6} \cdot \frac{2 \sqrt{2}}{2 \sqrt{2}}}{2 \sqrt{6}+2 \sqrt{2}}=\frac{2}{\sqrt{3}+1}=\sqrt{3}-1
$$

## Answer

$\sqrt{3}-1$.