3. In an isosceles triangle $ABC$, the angle at the base is $\arcsin \frac{2}{3}$. A circle with its center at the midpoint of the base $BC$ intersects the line $AC$ at points $K$ and $L$, and the line $AB$ at points $M$ and $N$, such that segments $KM$ and $LN$ intersect. Find its radius if the radius of the circle passing through points $A, L$, and $N$ is 2.
# Problem 3.

Answer: $\frac{4 \sqrt{5}}{3}$.

Solution. Let point $O$ be the midpoint of the base $B C$. Without loss of generality, we assume that points $L$ and $M$ are closer to vertex $A$ than points $K$ and $N$ respectively. Due to symmetry with respect to the line $A O$, segment $M L$ is parallel to the base $B C$ and $\angle A M L=\angle A L M=\angle A B C=\arcsin \frac{2}{3}$. From the Law of Sines for $\triangle A L N$, we have: $L N=4 \sin \angle A=4 \sin \left(\pi-2 \arcsin \frac{2}{3}\right)=8 \cdot \frac{2}{3} \cdot \frac{\sqrt{5}}{3}=\frac{16 \sqrt{5}}{9}$. Then the desired radius of the circle with center $O$ is

$$
R=\frac{L N}{2 \sin \angle N M L}=\frac{16 \sqrt{5}}{9 \cdot 2 \cdot \frac{2}{3}}=\frac{4 \sqrt{5}}{3}
$$