In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$, respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$. Suppose $XP=10$, $PQ=25$, and $QY=15$. The value of $AB\cdot AC$ can be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.
First we have $a\cos A=PQ=25$, and $(a\cos A)(c\cos C)=(a\cos C)(c\cos A)=AP\cdot PB=10(25+15)=400$ by PoP. Similarly, $(a\cos A)(b\cos B)=15(10+25)=525,$ and dividing these each by $a\cos A$ gives
$b\cos B=21,c\cos C=16$. 
It is known that the sides of the orthic triangle are $a\cos A,b\cos B,c\cos C$, and its angles are $\pi-2A$,$\pi-2B$, and $\pi-2C$. We thus have the three sides of the orthic triangle now.
Letting $D$ be the foot of the altitude from $A$, we have, in $\triangle DPQ$, 
\[\cos P,\cos Q=\frac{21^2+25^2-16^2}{2\cdot 21\cdot 25},\frac{16^2+25^2-21^2}{2\cdot 16\cdot 25}= \frac{27}{35}, \frac{11}{20}.\]
\[\Rightarrow \cos B=\cos\left(\tfrac 12 (\pi-P)\right)=\sin\tfrac 12 P =\sqrt{\frac{4}{35}},\]
similarly, we get  
\[\cos C=\cos\left(\tfrac 12 (\pi-Q)\right)=\sin\tfrac 12 Q=\sqrt{\frac{9}{40}}.\]
To finish, \[bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40})}=560\sqrt{14}.\]
The requested sum is $\boxed{574}$.
- crazyeyemoody907
Remark: The proof that $a \cos A = PQ$ can be found here: [http://www.irmo.ie/5.Orthic_triangle.pdf](https://artofproblemsolving.comhttp://www.irmo.ie/5.Orthic_triangle.pdf)