Track / Stage 5 / 11 of 400 #611 of 1964

Problem 611

AIME late

Geometry Difficulty 5.0 Find the answer

3. In $\triangle A B C$ , it is known that $A B=A C, D$ is the midpoint of side $B C$ , $B E \perp A C$ at point $E, B E$ intersects $A D$ at point $P$ . If $B P=3, P E=1$ , then $A E=(\quad)$ .

Pick one

Official solution

3. B.

Since $A D \perp B C, B E \perp A C$ , therefore, $P, D, C, E$ are concyclic.
Thus, $B D \cdot B C = B P \cdot B E = 12$ .
Also, $B C = 2 B D$ , so $B D = \sqrt{6}$ .
Hence, $D P = \sqrt{3}$ .
By $\triangle A E P \sim \triangle B D P \Rightarrow \frac{A E}{B D} = \frac{P E}{D P}$
$\Rightarrow A E = \frac{P E}{D P} \cdot B D = \frac{1}{\sqrt{3}} \times \sqrt{6} = \sqrt{2}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.