Competition problem — Geometry — Maths Olympiad Prep

Library / /192 of 348

Geometry Difficulty 4.9 AIME Find the answer

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 $\cos \angle A G F$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

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).

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.