Problem 1337

1. Given that $\triangle ABC$ is isosceles with $AB = AC$ and $\angle BAC$ is obtuse, we need to find the length of $BC$.
2. Points $E$ and $F$ lie on sides $AB$ and $AC$ respectively, such that $AE = 10$ and $AF = 15$. The area of $\triangle AEF$ is $60$.
3. Using the area formula for a triangle, we have:
$$\frac{1}{2} \cdot AE \cdot AF \cdot \sin(\angle EAF) = 60$$
Substituting the given values:
$$\frac{1}{2} \cdot 10 \cdot 15 \cdot \sin(\angle BAC) = 60$$
Simplifying, we get:
$$75 \cdot \sin(\angle BAC) = 60 \implies \sin(\angle BAC) = \frac{4}{5}$$
4. Since $\sin(\angle BAC) = \frac{4}{5}$, and $\angle BAC$ is obtuse, we have $\cos(\angle BAC) = -\sqrt{1 - \sin^2(\angle BAC)} = -\sqrt{1 - \left(\frac{4}{5}\right)^2} = -\frac{3}{5}$.
5. The area of quadrilateral $BEFC$ is $102$, so the total area of $\triangle ABC$ is:
$$\text{Area}(\triangle ABC) = \text{Area}(\triangle AEF) + \text{Area}(BEFC) = 60 + 102 = 162$$
6. Using the area formula for $\triangle ABC$:
$$\frac{1}{2} \cdot AB \cdot AC \cdot \sin(\angle BAC) = 162$$
Let $AB = AC = x$. Then:
$$\frac{1}{2} \cdot x \cdot x \cdot \frac{4}{5} = 162 \implies \frac{2x^2}{5} = 162 \implies x^2 = \frac{162 \cdot 5}{2} = 405 \implies x = \sqrt{405} = 9\sqrt{5}$$
Therefore, $AB = AC = 9\sqrt{5}$.
7. Using the Law of Cosines to find $BC$:
$$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)$$
Substituting the known values:
$$BC^2 = (9\sqrt{5})^2 + (9\sqrt{5})^2 - 2 \cdot (9\sqrt{5}) \cdot (9\sqrt{5}) \cdot \left(-\frac{3}{5}\right)$$
Simplifying:
$$BC^2 = 405 + 405 + 2 \cdot 405 \cdot \frac{3}{5} = 405 + 405 + 486 = 1296$$
Therefore:
$$BC = \sqrt{1296} = 36$$

The final answer is $\boxed{36}$.