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 $MK=225$, $NL=64$. Find $AC$ if it is additionally known that line $MN$ is parallel to $AC$.
Answer: $R=120, AC=680$.

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 $r$) $\frac{225}{r}=\frac{r}{64}$, from which $r=120$.

Then $BM=BK+KM=r+KM=345, BN=BL+LN=r+LN=184$, hence $MN^2=$ $345^2+184^2=23^2\left(15^2+8^2\right)=23^2 \cdot 17^2, MN=23 \cdot 17$. Let $h$ be the height of triangle $BMN$, drawn from the vertex of the right angle $B$. Then the area of triangle $BMN$ can be expressed in two ways: $2S_{\triangle BMN}=BN \cdot BM=MN \cdot h$, so $184 \cdot 345=17 \cdot 23 \cdot h, h=\frac{2760}{17}$.

If $MN \| AC$, then triangles $BAC$ and $BMN$ are similar, and the coefficient of their similarity $k$ 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 $r+h=120+\frac{2760}{17}=\frac{4800}{17}$. Therefore, $k=\frac{h+r}{h}=\frac{4800}{17}: \frac{2760}{17}=\frac{40}{23}$ and $AC=k \cdot MN=\frac{40}{23} \cdot 23 \cdot 17=680$.