## Angle Bisectors, Concurrency Problem $\underline{115562}$ topics: [Area of a Triangle (using semiperimeter and radius of inscribed or exscribed circle [ Inscribed, circumscribed, and exscribed circles; their radii

A circle with center $O$, inscribed in triangle $A B C$, touches its sides $A B$ and $A C$ at points $M$ and $N$. A circle with center $Q$ is inscribed in triangle $A M N$. Find $O Q$, if $A B=13, B C=15$ and $A C=14$.

#
Let $r$ be the radius of the circle inscribed in triangle $ABC$, and $p$ be the semiperimeter of triangle $ABC$. Then,

$$
p=\frac{AB+BC+AC}{2}=\frac{13+14+15}{2}=21
$$

By Heron's formula,

$$
S_{\triangle ABC}=\sqrt{p(p-AB)(p-BC)(p-AC)}=\sqrt{21 \cdot 8 \cdot 6 \cdot \bar{i}}=84
$$

Thus,

$$
r=\frac{S_{\triangle A}+BC}{p}=\frac{84}{21}=4
$$

We will prove that the center $Q$ of the circle inscribed in triangle $AMN$ lies on the inscribed circle of triangle $ABC$. Indeed, let $Q'$ be the midpoint of the smaller arc $MN$ of the inscribed circle of triangle $ABC$. From the theorem about the angle between a tangent and a chord, it follows that

$$
\angle AMQ'=\angle MNQ'=\angle NMQ'
$$

Therefore, $MQ'$ is the angle bisector of $\angle AMN$. Similarly, $NQ'$ is the angle bisector of $\angle ANM$, so $Q'$ is the point of intersection of the angle bisectors of triangle $AMN$, i.e., the center of the inscribed circle of this triangle. Thus, point $Q'$ coincides with point $Q$.

Therefore, $OQ=r=4$.

Answer

## 4.