2. Points $M$ and $N$ divide side $A C$ of triangle $A B C$ into three parts, each of which is equal to 5, such that $A B \perp B M, B C \perp B N$. Find the area of triangle $A B C$.
# Problem 2.

Answer: $3 \sqrt{3}$ (Option IV-1), $27 \sqrt{3} / 4$ (Option IV-2), $12 \sqrt{3}$ (Option IV-3), $75 \sqrt{3} / 4$ (Option IV-4).

Solution. Points $M$ and $N$ are located on side $A C$ in the following order: $A, N$, $M, C$ (otherwise, angle $B$ would be greater than $\left.180^{\circ}\right)$.

Let $A N=N M=M C=a$. Drop a perpendicular from vertex $B$ to $B H$. Let $N H=$ $x \Rightarrow D M=a-x$ (point $H$ lies between $N$ and $M$). Then, from the two right triangles $A B M$ and $N B C$, express the height: $B H^{2}=(a+x)(a-x)=x(2 a-x)$. From this, $x=a / 2$ (i.e., triangle $A B C$ is isosceles), the height $B H$ is equal to $a \sqrt{3} / 2$, and therefore, the area of $\triangle A B C$ is equal to $3 a^{2} \sqrt{3} / 4$.