In $\triangle ABC$, $AB=10$ and $AC=24$. Let $x$ be the length of the segment from $A$ to the midpoint of $\overline{BC}$, and define a function $f$ by letting $f(x)$ be the area of $\triangle ABC$. Then the domain of $f$ is an open interval $(p,q)$, and the maximum value $r$ of $f(x)$ occurs at $x=s$. What is $p+q+r+s$?

1. Let $M$ be the midpoint of $\overline{BC}$ and reflect $A$ through $M$ to $A'$. Then $ABA'C$ is a parallelogram, so $\triangle ABA'$ has sides $10$, $24$, $2x$ and the same area as $\triangle ABC$.
2. The triangle exists exactly when $24-10<2x<24+10$, so $7
3. $f(x)=\frac12\cdot10\cdot24\sin\angle BAC\le120$, with equality when $\angle BAC=90^\circ$. Then $BC=26$ and $x=AM=\frac12BC=13\in(7,17)$. So $r=120$ and $s=13$.
4. $p+q+r+s=24+120+13=\boxed{157}$.