3. In a circle of radius 3, triangles $A B C$ and $A M N$ are inscribed. The line $A M$ passes through the midpoint $E$ of segment $B C$, and the line $B N$ passes through the midpoint $F$ of segment $A C$. Find the perimeter of triangle $A B C$, if $A M: A E=2: 1$ and $B N: B F=17: 13$.
# Problem 3.

Answer: $\frac{72}{5}$.

Solution. According to the intersecting chords theorem: $A E \cdot E M = B E \cdot E C$. But since by the condition $A E = E M$ and $B E = E C$, then $A E = B E = C E$, i.e., $E$ is the center of the circumscribed circle around $\triangle A B C$. Therefore, angle $A$ is a right angle.

Similarly, $A F \cdot F C = B F \cdot F N$ and by the condition: $B F = 13 x, F N = 4 x$, then $A F = F C = 2 \sqrt{13} x$. Consequently, from the right triangle $A B F: A B^{2} = B F^{2} - A F^{2} = 9 \cdot 13 x^{2}$, and from triangle $A B C: 9 \cdot 13 x^{2} + 16 \cdot 13 x^{2} = 36 \Rightarrow x = \frac{6}{5 \sqrt{13}}$.

Therefore, the perimeter is $7 \sqrt{13} x + 6 = \frac{72}{5}$.