3. In a right triangle $ABC (\angle B=90^{\circ})$, a circle $\Gamma$ with center $I$ is inscribed, touching sides $AB$ and $BC$ at points $K$ and $L$ respectively. A line passing through point $I$ intersects sides $AB$ and $BC$ at points $M$ and $N$ respectively. Find the radius of the circle $\Gamma$ if $MK=144$, $NL=25$. Find $AC$ if it is additionally known that line $MN$ is parallel to $AC$.
Answer: $r=60, AC=390$.

Solution. Angles $KIM$ and $LNI$ are equal as corresponding angles when lines $BC$ and $KI$ are parallel, so right triangles $KIM$ and $LNI$ are similar. Therefore, $\frac{MK}{KI}=\frac{IL}{LN}$, or (if we denote the radius of the circle by $r$) $\frac{144}{r}=\frac{r}{25}$, from which $r=60$.

Then $BM=BK+KM=r+KM=204, BN=BL+LN=r+LN=85$, hence $MN^2=$ $204^2+85^2=17^2\left(12^2+5^2\right)=17^2 \cdot 13^2, MN=17 \cdot 13$. Let $h$ be the height of triangle $BMN$ drawn from the right angle vertex $B$. Then the area of triangle $BMN$ can be expressed in two ways: $2S_{\triangle BMN}=BN \cdot BM=MN \cdot h$, so $85 \cdot 204=17 \cdot 13 \cdot h, h=\frac{1020}{13}$.

If $MN \parallel AC$, then triangles $BAC$ and $BMN$ are similar, and the similarity coefficient $k$ is the ratio of the heights of these triangles drawn from vertex $B$. It remains to note that the height of triangle $ABC$ drawn from $B$ is equal to $r+h=60+\frac{1020}{13}=\frac{1800}{13}$. Therefore, $k=\frac{h+r}{h}=\frac{1800}{13} : \frac{1020}{13}=\frac{30}{17}$ and $AC=k \cdot MN=\frac{30}{17} \cdot 13 \cdot 17=390$.