Orthocenter and Parallelogram

Let $D$ be the intersection of lines $AH$ and $BC$ and let $X$ be the foot of the perpendicular from $M$ to $BC$ . Then $MX\parallel ND$ . Since $BHCM$ is a parallelogram, we have $\angle{HBD}=\angle{MCX}$ , which follows that $\triangle{HBD}\cong\triangle{MCX}$ by hypotenuse-leg theorem, so we have $BD=CX$ and $HD=MX$ .

Using Heron's Formula, we get $[ABC]=84$ , where $[ABC]$ is the area of $\triangle{ABC}$ and $AD=12$ . Also, by Pythagorean Theorem, we get $BD=CX=5$ and $CD=9$ , so $DX=CD-CX=4$ . Furthermore, we have $AH=2R\cos A$ (see note below for the derivation of this formula), so by Cosine Law, we have $\cos A=\dfrac{13^2+15^2-14^2}{2\cdot 13\cdot 15}=\dfrac{33}{65}, R=\dfrac{13\cdot 14\cdot 15}{4\cdot 84}=\dfrac{65}{8}$ and $AH=2\cdot\dfrac{65}{8}\cdot\dfrac{33}{65}=\dfrac{33}{4}$ . Since $N$ is the midpoint of $AH$ , we deduce that $AN=\dfrac{33}{8}$ and $ND=AD-AN=12-\dfrac{33}{8}=\dfrac{63}{8}$ . We also have $MX=HD=AD-AH=12-\dfrac{33}{4}=\dfrac{15}{4}$ . Note that $MX\parallel ND$ gives $\triangle{MPX}\sim\triangle{NPD}$ (by AA similarity theorem), so we get $\dfrac{XP}{PD}=\dfrac{MX}{ND}=\dfrac{\frac{15}{4}}{\frac{63}{8}}=\dfrac{10}{21},$ and with $DX=4$ , we get $XP=4\cdot\dfrac{10}{31}=\dfrac{40}{31}$ and $PD=4\cdot\dfrac{21}{31}=\dfrac{84}{31}$ . Thus, we get $\dfrac{BP}{PC}=\dfrac{BD+DP}{CX+XP}=\dfrac{5+\frac{84}{31}}{5+\frac{40}{31}}=\dfrac{155+84}{155+40}=\dfrac{239}{195},$ and $m+n=239+195=\boxed{434}$ .

Note: The derivation of $AH=2R\cos A$ is as follows: let $M$ and $N$ be the foot of the altitudes of $A$ and $B$ , respectively. Then, $ANMB$ is cyclic, so $\triangle{ANH}\sim\triangle{BNC}$ , which implies that $AH=\dfrac{AN\cdot BC}{BN}$ . But considering triangle $ANB$ we have $BN=AN\tan A$ and by Sine Law, $BC=2R\cos A$ . Thus, by substitution, we have $AH=\dfrac{AN\cdot 2R\sin A}{AN\tan A}=2R\cos A.$