HMMT-Nov 2012 team, Problem 8

$A B C$ is a triangle with $A B=15, B C=14$, and $C A=13$. The altitude from $A$ to $B C$ is extended to meet the circumcircle of $A B C$ at $D$. Find $A D$.

Solution: Let the altitude from $A$ to $B C$ meet $B C$ at $E$. The altitude $A E$ has length 12; one way to see this is that it splits the triangle $A B C$ into a $9-12-15$ right triangle and a $5-12-13$ right triangle; from this, we also know that $B E=9$ and $C E=5$. Now, by Power of a Point, $A E \cdot D E=B E \cdot C E$, so $D E=(B E \cdot C E) / A E=(9 \cdot 5) /(12)=15 / 4$. It then follows that $A D=A E+D E=63 / 4$. $\fbox{\frac{63}{4}}$.

Answer: \frac{63}{4}