Maths Olympiad Prep

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HMMT, 2014

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Point PP and line ℓ\ell are such that the distance from PP to ℓ\ell is 1212. Given that TT is a point on ℓ\ell such that PT=13PT = 13, find the radius of the circle passing through PP and tangent to ℓ\ell at TT.

Solution

Solution:

Answer: 16924\dfrac{169}{24}

Let OO be the center of the given circle, QQ be the foot of the altitude from PP to ℓ\ell, and MM be the midpoint of PTPT. Then since OM⊥PTOM \perp PT and ∠OTP=∠TPQ\angle OTP = \angle TPQ, △OMP∼△TQP\triangle OMP \sim \triangle TQP. Thus the OP=TP⋅PMPQ=13⋅13/212=16924OP = TP \cdot \dfrac{PM}{PQ} = 13 \cdot \dfrac{13/2}{12} = \dfrac{169}{24}

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