[ Radii of the inscribed, circumscribed, and exscribed circles (other) [ Area of a triangle (through the semiperimeter and the radius of the inscribed or exscribed circle).

Through the center $O$ of the inscribed circle $\omega$ of triangle $A B C$, a line parallel to side $B C$ is drawn, intersecting sides $A B$ and $A C$ at points $M$ and $N$, respectively. $S_{A B C}=\sqrt{15}, B C=2$, and segment $A O$ is four times the radius of $\omega$. Find the perimeter of triangle $A M N$.

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Let $r$ be the radius of $\omega, p$ be the semiperimeter of triangle $ABC, P$ be the point of tangency of $\omega$ with side $AB$. Then $AP^2 = AO^2 - OP^2 = 15r^2$, $AP = p - BC = AP = r\sqrt{15} + 2$,

$\sqrt{15} = S_{ABC} = pr = (r\sqrt{15} + 2)r$, from which $AP = r\sqrt{15} = 3, p = 5$.

The perimeter of triangle $AMN$ is $AB + AC$ (see the solution of problem $\underline{54896}$), that is, $2p - BC = 8$.

## Answer

8.