HMMT-Nov 2016 team, Problem 6

Let $A B C$ be a triangle with $A B=5, B C=6$, and $A C=7$. Let its orthocenter be $H$ and the feet of the altitudes from $A, B, C$ to the opposite sides be $D, E, F$ respectively. Let the line $D F$ intersect the circumcircle of $A H F$ again at $X$. Find the length of $E X$.

Solution: Since $\angle A F H=\angle A E H=90^{\circ}, E$ is on the circumcircle of $A H F$. So $\angle X E H=\angle H F D=\angle H B D$, which implies that $X E \| B D$. Hence $\frac{E X}{B D}=\frac{E Y}{Y B}$. Let $D F$ and $B E$ intersect at $Y$. Note that $\angle E D Y=180^{\circ}-\angle B D F-\angle C D E=180^{\circ}-2 \angle A$, and $\angle B D Y=\angle A$. Applying the sine rule to $E Y D$ and $B Y D$, we get \[ \frac{E Y}{Y B}=\frac{E D}{B D} \cdot \frac{\sin \angle E D Y}{\sin \angle B D Y}=\frac{E D}{B D} \cdot \frac{\sin 2 \angle A}{\sin \angle A}=\frac{E D}{B D} \cdot 2 \cos \angle A \] Next, letting $x=C D$ and $y=A E$, by Pythagoras we have \[ \begin{aligned} & A B^{2}-(6-x)^{2}=A D^{2}=A C^{2}-x^{2} \\ & B C^{2}-(7-y)^{2}=B E^{2}=B A^{2}-y^{2} \end{aligned} \] Solving, we get $x=5, y=\frac{19}{7}$. Drop the perpendicular from $E$ to $D C$ at $Z$. Then $E D \cos \angle A=$ $E D \cos \angle E D Z=D Z$. But $A D \| E Z$, so $D Z=\frac{A E}{A C} \cdot D C=\frac{95}{49}$. Therefore \[ E X=\frac{E Y}{Y B} \cdot B D=2 E D \cos \angle A=2 D Z=\frac{190}{49} \] $\fbox{\frac{190}{49}}$.

Answer: \frac{190}{49}