Let $\triangle ABC$ be an equilateral triangle with side length $1$ and $M$ be the midpoint of side $AC$. Let $N$ be the foot of the perpendicular from $M$ to side $AB$. The angle bisector of angle $\angle BAC$ intersects $MB$ and $MN$ at $X$ and $Y$, respectively. If the area of triangle $\triangle MXY$ is $\mathcal{A}$, and $\mathcal{A}^2$ can be expressed as a common fraction in the form $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, find $m + n$.

Let's think step by step and output the final answer within \boxed{}.
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