Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

There are circles ω1\omega_{1} and ω2\omega_{2}. They intersect in two points, one of which is the point AA. BB lies on ω1\omega_{1} such that ABA B is tangent to ω2\omega_{2}. The tangent to ω1\omega_{1} at BB intersects ω2\omega_{2} at CC and DD, where DD is the closer to BB. ADA D intersects ω1\omega_{1} again at EE. If BD=3B D=3 and CD=13C D=13, find EB/EDE B / E D.

Solution

Solution:

> [diagram]

By power of a point, BA=BDBC=43B A=\sqrt{B D \cdot B C}=4 \sqrt{3}. Also, DEBDBAD E B \sim D B A, so EB/ED=BA/BD=43/3E B / E D=B A / B D=4 \sqrt{3} / 3.

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