3B. Given points $M$ and $N$ on sides $B D$ and $C D$, respectively, of square $A B C D$ such that the sum of the areas of triangles $A B M$ and $A N D$ is equal to the area of triangle $N M C$. If $\overline{D N}=3$ and $\overline{B M}=5$, what is the area of triangle AMN?
Solution. Let the side of the square be $a$. Then for the area of triangle $ABM$ we have $P_{1}=\frac{a \cdot \overline{BM}}{2}=\frac{5a}{2}$, for the area of triangle AND we have $P_{2}=\frac{a \cdot \overline{DN}}{2}=\frac{3a}{2}$, and for the area of triangle NMC we get

$$
P_{3}=\frac{\overline{NC} \cdot \overline{MC}}{2}=\frac{(a-3)(a-5)}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_86f0a3ff472e2cfe9073g-06.jpg?height=365&width=370&top_left_y=1714&top_left_x=1094)

According to the condition in the problem, $P_{1}+P_{2}=P_{3}$, so we have $8a=(a-3)(a-5)$, and from this we get the equation $a^{2}-16a+15=0$. Its solutions are 1 and 15, but the side of the square, according to the conditions, cannot be 1. Therefore, $a=15$ and for the area of triangle AMN we get:

$$
P=15^{2}-2 \cdot \frac{12 \cdot 10}{2}=225-120=105.
$$