HMMT-Nov 2014 guts, Problem 24

Let $A B C$ be a triangle with $A B=13, B C=14$, and $C A=15$. We construct isosceles right triangle $A C D$ with $\angle A D C=90^{\circ}$, where $D, B$ are on the same side of line $A C$, and let lines $A D$ and $C B$ meet at $F$. Similarly, we construct isosceles right triangle $B C E$ with $\angle B E C=90^{\circ}$, where $E, A$ are on the same side of line $B C$, and let lines $B E$ and $C A$ meet at $G$. Find $\cos \angle A G F$.

Solution: We see that $\angle G A F=\angle G B F=45^{\circ}$, hence quadrilateral $G F B A$ is cyclic. Consequently $\angle A G F+\angle F B A=180^{\circ}$. So $\cos \angle A G F=-\cos \angle F B A$. One can check directly that $\cos \angle C B A=\frac{5}{13}$ (say, by the Law of Cosines). $\fbox{-\frac{5}{13}}$.

Answer: -\frac{5}{13}