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, 2014

Geometry Difficulty 5.7 AIME, harder Prove it United States

Problem:

Two circles are said to be orthogonal if they intersect in two points, and their tangents at either point of intersection are perpendicular. Two circles $\omega_{1}$ and $\omega_{2}$ with radii $10$ and $13$ , respectively, are externally tangent at point $P$ . Another circle $\omega_{3}$ with radius $2 \sqrt{2}$ passes through $P$ and is orthogonal to both $\omega_{1}$ and $\omega_{2}$ . A fourth circle $\omega_{4}$ , orthogonal to $\omega_{3}$ , is externally tangent to $\omega_{1}$ and $\omega_{2}$ . Compute the radius of $\omega_{4}$ .

Solution

Solution:

Let $\omega_{i}$ have center $O_{i}$ and radius $r_{i}$ . Since $\omega_{3}$ is orthogonal to $\omega_{1}$ , $\omega_{2}$ , $\omega_{4}$ , it has equal power $r_{3}^{2}$ to each of them. Thus $O_{3}$ is the radical center of $\omega_{1}$ , $\omega_{2}$ , $\omega_{4}$ , which is equidistant to the three sides of $\triangle O_{1} O_{2} O_{4}$ and therefore its incenter.

For distinct $i, j \in\{1,2,4\}$ , $\omega_{i} \cap \omega_{j}$ lies on the circles with diameters $O_{3} O_{i}$ and $O_{3} O_{j}$ , and hence $\omega_{3}$ itself. It follows that $\omega_{3}$ is the incircle of $\triangle O_{1} O_{2} O_{4}$ , so $8 = r_{3}^{2} = \frac{r_{1} r_{2} r_{4}}{r_{1} + r_{2} + r_{4}} = \frac{130 r_{4}}{23 + r_{4}} \Longrightarrow r_{4} = \frac{92}{61}$ .

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