Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

Points AA and BB lie on circle ω\omega. Point PP lies on the extension of segment ABA B past BB. Line \ell passes through PP and is tangent to ω\omega. The tangents to ω\omega at points AA and BB intersect \ell at points DD and CC respectively. Given that AB=7A B=7, BC=2B C=2, and AD=3A D=3, compute BPB P.

Solution

Solution:

Say that \ell is tangent to ω\omega at point TT. Observing equal tangents, write

CD=CT+DT=BC+AD=5. C D = C T + D T = B C + A D = 5.

Let the tangents to ω\omega at AA and BB intersect each other at QQ. Working from Menelaus applied to triangle CDQC D Q and line ABA B gives
1=DAAQQBBCCPPD=DABCCPPC+CD=32CPPC+5, \begin{aligned} -1 & = \frac{D A}{A Q} \cdot \frac{Q B}{B C} \cdot \frac{C P}{P D} \\ & = \frac{D A}{B C} \cdot \frac{C P}{P C + C D} \\ & = \frac{3}{2} \cdot \frac{C P}{P C + 5}, \end{aligned}
from which PC=10P C = 10. By power of a point, PT2=APBPP T^{2} = A P \cdot B P, or 122=BP(BP+7)12^{2} = B P \cdot (B P + 7), from which BP=9B P = 9.

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