smc

Place the rectangle on a coordinate grid, with diagonal vertices $(0,0)$ and $(x,y)$. Each horizontal segment of the rectangle will have length $\frac{x}{m}$, while each vertical segment of the rectangle will have length $\frac{y}{n}$. The center of this rectangle will be $(\frac{x}{2},\frac{y}{2})$. Triangle $A$ has a base length of $\frac{y}{n}$, one of the vertical segments. It has an altitude of $\frac{x}{2}$, which is the perpendicular distance from the center of the square to the left side of the square. Thus, the area of triangle $A$ is $\frac{1}{2}\cdot \frac{y}{n}\cdot \frac{x}{2}=\frac{xy}{4n}$. Triangle $B$ has a base length of $\frac{x}{m}$, one of the horizontal segments. It has an altitude of $\frac{y}{2}$, which is the perpendicular distance from the center of the square to the bottom side of the square. Thus, the area of triangle $B$ is $\frac{1}{2}\cdot \frac{x}{m}\cdot \frac{y}{2}=\frac{xy}{4m}$. The ratio of areas is $\frac{\frac{xy}{4n}}{\frac{xy}{4m}}=\frac{m}{n}$, which is answer $\boxed{m/n}$.