[ [sphere touching the edges of a tetrahedron]

A sphere touches the edges $A S, B S, B C$ and $A C$ of the triangular pyramid $S A B C$ at points $K, L, M$ and $N$ respectively.

Find the segment $K L$, if $M N=7, N K=5, L N=2 \sqrt{29}$ and $K L=L M$.

#
Let's prove that points $K, L, M$, and $N$ lie in the same plane (Fig.1). Denote

$$
A K=A N=a, B L=B M=b, C M=C N=c, S K=S L=d .
$$

If $c=d$, then lines $K N$ and $L M$ are parallel to line $S C$, so $K N \| L M$. Therefore, points $K, L, M$, and $N$ lie in the same plane. Let $c \neq d$. Therefore, points $K, L, M$, and $N$ lie in the same plane. Consider the section of the pyramid and the sphere by this plane (Fig.2). We obtain a quadrilateral $K L M N$ inscribed in a circle. Denote $K L=L M=x, \angle N K L=\alpha$. Then $\angle N M L=180^{\circ}-\alpha$. Further, we have:

$$
\left\{\begin{array} { l } 
{ N L ^ { 2 } = K N ^ { 2 } + K L ^ { 2 } - 2 K N \cdot K L \operatorname { c o s } \alpha } \\
{ N L ^ { 2 } = M N ^ { 2 } + M L ^ { 2 } - 2 M N \cdot M L \operatorname { c o s } ( 1 8 0 ^ { \beta } - \alpha ) }
\end{array} \left\{\begin{array}{l}
116=25+x^{2}-10 x \cdot \cos \alpha \\
116=49+x^{2}+14 x \cdot \cos \alpha
\end{array}\right.\right.
$$

Subtracting the first equation from the second term by term, we get $x \cos \alpha=-1$. Then from the first equation, we find that

$$
x^2=116-25-10=81
$$

Therefore, $K L=x=9$.

## Answer

9.00