HMMT February

Geometry Difficulty 5.0 AIME, harder Find the answer

Consider triangle $A B C$ with $\angle A=2 \angle B$ . The angle bisectors from $A$ and $C$ intersect at $D$ , and the angle bisector from $C$ intersects $\overline{A B}$ at $E$ . If $\frac{D E}{D C}=\frac{1}{3}$ , compute $\frac{A B}{A C}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let $A E=x$ and $B E=y$ . Using angle-bisector theorem on $\triangle A C E$ we have $x: D E=A C: D C$ , so $A C=3 x$ . Using some angle chasing, it is simple to see that $\angle A D E=\angle A E D$ , so $A D=A E=x$ . Then, note that $\triangle C D A \sim \triangle C E B$ , so $y:(D C+D E)=x: D C$ , so $y: x=1+\frac{1}{3}=\frac{4}{3}$ , so $A B=x+\frac{4}{3} x=\frac{7}{3} x$ . Thus the desired answer is $A B: A C=\frac{7}{3} x: 3 x=\frac{7}{9}$ .

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