Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $\omega_1$ , $\omega_2$ , and $\omega_3$ be three circles passing through the origin $O$ of the coordinate plane but not tangent to each other or to either axis. Denote by $(x_i, 0)$ and $(0, y_i)$ , $1 \leq i \leq 3$ , the respective intersections (besides $O$ ) of circle $\omega_i$ with the $x$ and $y$ axes. Prove that $\omega_1$ , $\omega_2$ , and $\omega_3$ have a common point $P \neq O$ if and only if the points $(x_1, y_1)$ , $(x_2, y_2)$ , and $(x_3, y_3)$ are collinear.

Solution

Solution:

First of all, note that if a circle passes through $(0,0)$ , $(x_i, 0)$ , and $(0, y_i)$ , its center must be $\left(\frac{x_i}{2}, \frac{y_i}{2}\right)$ , the midpoint of the side opposite the right angle at $O$ . Also note that the three points $\left(\frac{x_i}{2}, \frac{y_i}{2}\right)$ are related to $(x_i, y_i)$ by a dilation about $O$ ; thus the former three will be collinear if and only if the latter three are. It suffices to prove that the circles have a second common point if and only if their centers are collinear.

If the centers are collinear, all three circles are symmetric about the line of centers. Thus the reflection of $O$ about this line is a second common point of the three circles. Conversely, assume that the circles have two common points, $O$ and $P$ . Then all three centers lie on the perpendicular bisector of $OP$ .

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