Annual Harvard-MIT November Tournament

Solution:
Answer: $245 \sqrt{3}$

If $A$ is the area of the triangle, the sides are $\frac{2A}{15}$, $\frac{2A}{21}$, and $\frac{2A}{35}$. So the triangle is similar to a $\frac{1}{15}$, $\frac{1}{21}$, $\frac{1}{35}$ triangle, which is similar to a $3, 5, 7$ triangle. Let the sides be $3k$, $5k$, and $7k$. Then the angle between the sides of length $3k$ and $7k$ is $120^{\circ}$, so the area is $\frac{15 \sqrt{3}}{4} k^2$. But the area can also be calculated as $\frac{(3k)(35)}{2} = \frac{105k}{2}$. Setting these values equal, $k = \frac{14 \sqrt{3}}{3}$ and the area is $245 \sqrt{3}$.