HMMT February

Geometry Difficulty 5.0 AIME, harder Find the answer

In triangle $A B C, \angle A B C$ is obtuse. Point $D$ lies on side $A C$ such that $\angle$ A B D $is right, and point$ E $lies on side$ A C $between$ A $and$ D $such that$ B D $bisects \angle E B C$ . Find $C E$ , given that $A C=35, B C=7$ , and $B E=5$ .

Your answer

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

Solution

Reflect $A$ and $E$ over $B D$ to $A^{\prime}$ and $E^{\prime}$ respectively. Note that the angle conditions show that $A^{\prime}$ and $E^{\prime}$ lie on $A B$ and $B C$ respectively. $B$ is the midpoint of segment $A A^{\prime}$ and $C E^{\prime}=$ $B C-B E^{\prime}=2$ . Menelaus' theorem now gives $\frac{C D}{D A} \cdot \frac{A A^{\prime}}{A^{\prime} B} \cdot \frac{B E^{\prime}}{E^{\prime} C}=1$ from which $D A=5 C D$ or $C D=A C / 6$ . By the angle bisector theorem, $D E=5 C D / 7$ , so that $C E=12 C D / 7=10$ .

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