Find m-n

Let $Y = MN \cap AQ$ . $\frac {BQ}{QC} = \frac {NY}{MY}$ since $\triangle AMN \sim \triangle ACB$ . Since quadrilateral $AMPN$ is cyclic, $\triangle MYA \sim \triangle PYN$ and $\triangle MYP \sim \triangle AYN$ , yielding $\frac {YM}{YA} = \frac {MP}{AN}$ and $\frac {YA}{YN} = \frac {AM}{PN}$ . Multiplying these together yields $\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)$ .

$\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}$ . Also, $P$ is the center of spiral similarity of segments $MD$ and $NE$ , so $\triangle PMD \sim \triangle PNE$ . Therefore, $\frac {PN}{PM} = \frac {NE}{MD}$ , which can easily be computed by the angle bisector theorem to be $\frac {145}{117}$ . It follows that $\frac {BQ}{CQ} = \frac {15}{13} \cdot \frac {145}{117} = \frac {725}{507}$ , giving us an answer of $725 - 507 = \boxed{218}$ .

After I did this problem using the Synthetic Solution, I decided to try a bash lol. Since the synthetic way is already posted, I'll post the bash.

We will use barycentric coordinates. We know that $a = 14 , b = 13 , c = 15$ . $A = (1, 0, 0)$ $N = (1 : 1: 0)$ $M = (1: 0 : 1)$ $E = (a:b : 0)$ $D = (a:0:c)$ Using this information, you can find the equations of the circles.

You know that circle $AMN$ is $- a^2yz - b^2xz - c^2xy + (x+y+z)(\frac{c^2}{2}y + \frac{b^2}{2}z) = 0$ You know that circle $ADE$ is $- a^2yz - b^2xz - c^2xy + (x+y+z)(\frac{ac^2}{a+b}y + \frac{ab^2}{a+c}z) = 0$

We can solve set both of these equations equal to each other to find the intersection. After simplification, you get $\frac{15^2}{27}y = \frac{13^2}{29} z$ which implies that $P = (x_0 : \frac{27}{15^2} : \frac{29}{13^2})$ .

Line $AP$ has an equation in the form $y = kz$ . Plugging in the values for $P$ , we find that the equation of line $AP$ is $y = \frac{27 \cdot 13^2}{29 \cdot 15^2} z$

Since $Q$ lies on line $BC$ , we know that $Q = (0 : 27 \cdot 13^2 : 29 \cdot 15^2 ) = (0 , \frac{27 \cdot 13^2}{27 \cdot 13^2 + 15^2 \cdot 29} , \frac{29 \cdot 15^2}{27 \cdot 13^2 + 15^2 \cdot 29})$

It is known that given the displacement vector $PQ$ , $|PQ|^2 = -a^2yz - b^2xz - c^2xy$

We know that $BQ = (0 , \frac{29 \cdot 15^2}{27 \cdot 13^2 + 15^2 \cdot 29} , - \frac{29 \cdot 15^2}{27 \cdot 13^2 + 15^2 \cdot 29} )$ , the $x = 0$ immediately kills a bunch of terms, which implies $|BQ|^2 = 14^2 \cdot (\frac{15^2 \cdot 29}{27 \cdot 13^2 + 15^2 \cdot 29})^2$

Similarly, $|CQ|^2 = 14^2(\frac{27 \cdot 13^2}{27 \cdot 13^2 + 15^2 \cdot 29})^2$ $(\frac{BQ}{CQ})^2 = (\frac{15^2 \cdot 29}{27 \cdot 13^2 + 15^2 \cdot 29} \cdot \frac{27 \cdot 13^2 + 15^2 \cdot 29}{27 \cdot 13^2})^2$ $= (\frac{725}{507})^2 \implies \frac{BQ}{CQ} = \frac{725}{507} \implies m - n = \boxed{218}$