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, 2017

Geometry Difficulty 5.8 AIME, harder Prove it United States

Problem:
Let $\omega$ and $\Gamma$ be circles such that $\omega$ is internally tangent to $\Gamma$ at a point $P$ . Let $AB$ be a chord of $\Gamma$ tangent to $\omega$ at a point $Q$ . Let $R \neq P$ be the second intersection of line $PQ$ with $\Gamma$ . If the radius of $\Gamma$ is $17$ , the radius of $\omega$ is $7$ , and $\frac{AQ}{BQ} = 3$ , find the circumradius of triangle $AQR$ .

Solution

Solution:
Let $r$ denote the circumradius of triangle $AQR$ . By Archimedes Lemma, $R$ is the midpoint of arc $AB$ of $\Gamma$ . Therefore $\angle RAQ = \angle RPB = \angle RPA$ so $\triangle RAQ \sim \triangle RPA$ . By looking at the similarity ratio between the two triangles we have
$\frac{r}{17} = \frac{AQ}{AP}$
Now, let $AP$ intersect $\omega$ again at $X \neq P$ . By homothety we have $XQ \parallel AR$ so
$\frac{AX}{AP} = 1 - \frac{PQ}{PR} = 1 - \frac{7}{17} = \frac{10}{17}$
But we also know
$AX \cdot AP = AQ^2$
so
$\frac{10}{17} AP^2 = AQ^2$
Thus
$\frac{r}{17} = \frac{AQ}{AP} = \sqrt{\frac{10}{17}}$
so we compute $r = \sqrt{170}$ as desired.

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