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Olimpiadi della Matematica

Geometry Difficulty 4.3 AIME Find the answer Italy

Problem:

Let $ABC$ be a triangle and let $D$ be the foot of the bisector from vertex $A$ . Let $\omega$ be the circle tangent to $AC$ at $A$ and passing through $D$ . Let $P$ be the second intersection of $\omega$ with the line $BC$ . Given that $AC=54$ , $AD=63$ and $CP=108$ , find $AB$ .

Pick one

Solution

Solution:

The answer is $(\mathbf{C})$ . Since $CA$ is tangent to $\omega$ , we have $CP=\frac{AC^{2}}{CD}=\frac{54^{2}}{108}=27$ . Moreover $\angle DAC=\angle CPA$ , but $AD$ is the bisector of $\angle BAC$ , so $\angle BAD=\angle CPA$ . Consequently the circle circumscribed about triangle $ABP$ is tangent to the line $AD$ , so $DB=\frac{DA^{2}}{DP}=\frac{63^{2}}{108-27}=49$ . Now, by the angle bisector theorem, $\frac{AC}{CD}=\frac{AB}{BD}$ and therefore we can conclude that $AB=\frac{AC \cdot BD}{CD}=\frac{54 \cdot 49}{27}=98$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.