Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

Circle Ω\Omega has radius 1313. Circle ω\omega has radius 1414 and its center PP lies on the boundary of circle Ω\Omega. Points AA and BB lie on Ω\Omega such that chord ABAB has length 2424 and is tangent to ω\omega at point TT. Find AT⋅BTAT \cdot BT.

Solution

Solution:

Let MM be the midpoint of chord ABAB; then AM=BM=12AM = BM = 12 and by the Pythagorean theorem on triangle AMOAMO we have MO=5MO = 5.

Note that ∠AOM=∠AOB/2=∠APB=∠APT+∠TPB\angle AOM = \angle AOB / 2 = \angle APB = \angle APT + \angle TPB, or tan⁡(∠AOM)=tan⁡(∠APT+∠TPB)\tan(\angle AOM) = \tan(\angle APT + \angle TPB). Applying the tangent addition formula,
AMMO=ATTP+BTTP1−ATTP⋅BTTP=AB⋅TPTP2−AT⋅BT \begin{aligned} \frac{AM}{MO} &= \frac{\frac{AT}{TP} + \frac{BT}{TP}}{1 - \frac{AT}{TP} \cdot \frac{BT}{TP}} \\ &= \frac{AB \cdot TP}{TP^2 - AT \cdot BT} \end{aligned}
from which
AT⋅BT=TP2−AB⋅TP⋅MOAM=142−24⋅14⋅512=56. AT \cdot BT = TP^2 - \frac{AB \cdot TP \cdot MO}{AM} = 14^2 - \frac{24 \cdot 14 \cdot 5}{12} = 56.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.