The inscribed circles of the faces $SBC, SAC$, and $SAB$ of the triangular pyramid $SABC$ intersect pairwise and have radii $\sqrt{3}, \sqrt{5}$, and $\sqrt{7}$, respectively. Point $K$ is the point of tangency of the circles with side $SA$, and $SK=5$. Find the length of the segment $AK$, the perimeter, and the radius of the inscribed circle of triangle $ABC$.
Note that the inscribed circles of triangles $S B C, S A C$, and $S A B$ touch each other pairwise, and the points of tangency lie on the lateral edges of the pyramid. One of them is point $K$, and the points of tangency lying on the lateral edges $S B$ and $S C$ are denoted by $L$ and $M$ respectively. Let $D, E$, and $F$ be the points of tangency of the circles with the edges $A B, B C$, and $A C$ respectively. Since the segments of tangents drawn from a point to a circle are equal, $S L=S M=S K=5$. Denote

$$
A K=A D=A F=x, B L=B D=B E=y, C M=C E=C F=z
$$

By writing the area of each lateral face using Heron's formula and the formula $S=p \cdot r$, where $p$ is the semiperimeter and $r$ is the radius of the inscribed circle of the triangle, we obtain the system of equations

$$
\left\{\begin{array}{l}
\sqrt{5(y+z+5) z y}=(y+z+5) \sqrt{3} \\
\sqrt{5(x+y+5) x y}=(x+y+5) \sqrt{7} \\
\sqrt{5(x+z+5) x z}=(x+z+5) \sqrt{5}
\end{array}\right.
$$

Express $y$ and $z$ in terms of $x$ from the second and third equations respectively and substitute the found expressions into the first:

$$
\left\{\begin{array}{l}
5 z y = 3 ( y + z + 5 ) \\
y = \frac { 7 x + 3 5 } { 5 x - 7 } \\
z = \frac { x + 5 } { x - 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{5(x+5)}{x-1} \cdot \frac{7 x+35}{5 x-7}-3\left(\frac{7 x+35}{5 x-7}+\frac{x+5}{x-1}\right)=15 \\
y=\frac{7+55}{5 x-7} \\
z=\frac{x+5}{x-1}
\end{array}\right.\right.
$$

After obvious simplifications, we obtain the quadratic equation $19 x^2-98 x-245=0$, from which we find that $A K=x=7$. Then $B L=y=3, C M=z=2$. Therefore, the perimeter of triangle $A B C$ is $2(x+y+z)=24$, the semiperimeter $p=12, S_{\triangle A B C}=\sqrt{12 \cdot 7 \cdot 3 \cdot 2}=6 \sqrt{14}$, and the radius of the inscribed circle $r=\frac{S_{\triangle A B C}}{p}=\frac{\sqrt{14}}{2}$.

## Answer

$7 ; 24 ; \frac{\sqrt{14}}{2}$