In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
Because $M$ is the midpoint of $BC$, following from the Stewart's theorem, $AM = 2 \sqrt{37}$.
Because $A$, $B$, $C$, and $P$ are concyclic, $\angle BPA = \angle C$, $\angle CPA = \angle B$.
Denote $\theta = \angle PBQ$.
In $\triangle BPQ$, following from the law of sines,
\[ \frac{BQ}{\sin \angle BPA} = \frac{PQ}{\angle PBQ} \]
Thus,
\[ \frac{BQ}{\sin C} = \frac{PQ}{\sin \theta} . \hspace{1cm} (1) \]
In $\triangle CPQ$, following from the law of sines,
\[ \frac{CQ}{\sin \angle CPA} = \frac{PQ}{\angle PCQ} \]
Thus,
\[ \frac{CQ}{\sin B} = \frac{PQ}{\sin \theta} . \hspace{1cm} (2) \]
Taking $\frac{(1)}{(2)}$, we get
\[ \frac{BQ}{\sin C} = \frac{CQ}{\sin B} \]
In $\triangle ABC$, following from the law of sines,
\[ \frac{AB}{\sin C} = \frac{AC}{\sin B} . \hspace{1cm} (3) \]
Thus, Equations (2) and (3) imply
\begin{align*} \frac{BQ}{CQ} & = \frac{AB}{AC} \\ & = \frac{13}{15} . \hspace{1cm} (4) \end{align*}

Next, we compute $BQ$ and $CQ$.
We have
\begin{align*} BQ^2 & = AB^2 + AQ^2 - 2 AB\cdot AQ \cos \angle BAQ \\ & = AB^2 + AQ^2 - 2 AB\cdot AQ \cos \angle BAM \\ & = AB^2 + AQ^2 - 2 AB\cdot AQ \cdot \frac{AB^2 + AM^2 - BM^2}{2 AB \cdot AM} \\ & = AB^2 + AQ^2 -  AQ \cdot \frac{AB^2 + AM^2 - BM^2}{AM} \\ & = 169 + AQ^2 - \frac{268}{2 \sqrt{37}} AQ .  \hspace{1cm} (5) \end{align*}
We have
\begin{align*} CQ^2 & = AC^2 + AQ^2 - 2 AC\cdot AQ \cos \angle CAQ \\ & = AC^2 + AQ^2 - 2 AC\cdot AQ \cos \angle CAM \\ & = AC^2 + AQ^2 - 2 AC\cdot AQ \cdot \frac{AC^2 + AM^2 - CM^2}{2 AC \cdot AM} \\ & = AC^2 + AQ^2 -  AQ \cdot \frac{AC^2 + AM^2 - CM^2}{AM} \\ & = 225 + AQ^2 - \frac{324}{2 \sqrt{37}} AQ .  \hspace{1cm} (6) \end{align*}
Taking (5) and (6) into (4), we get $AQ = \frac{99}{\sqrt{148}}$
Therefore, the answer is $99 + 148 = \boxed{\textbf{(247) }}$
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)