Let three circles $ \Gamma_1, \Gamma_2, \Gamma_3$, which are non-overlapping and mutually external, be given in the plane. For each point $ P$ in the plane, outside the three circles, construct six points $ A_1, B_1, A_2, B_2, A_3, B_3$ as follows: For each $ i = 1, 2, 3$, $ A_i, B_i$ are distinct points on the circle $ \Gamma_i$ such that the lines $ PA_i$ and $ PB_i$ are both tangents to $ \Gamma_i$. Call the point $ P$ exceptional if, from the construction, three lines $ A_1B_1, A_2 B_2, A_3 B_3$ are concurrent. Show that every exceptional point of the plane, if exists, lies on the same circle.
Problem
Source: APMO 2009 Q.3
Tags: radical axis, geometry unsolved, geometry
Russian_Man
15.03.2009 18:25
circle's center is radical center O this 3 circles and R=OK, K - is point in circle $ G_i$ OK is both tangent.
jgnr
14.02.2010 05:10
Let $ O_i$ and $ r_i$ be the center and the radius of $ \Gamma_i$. Consider an exceptional point $ P$, such that $ A_1B_1, A_2B_2, A_3B_3$ concur at $ Q$. Consider circle $ \Gamma$ with diameter $ PQ$, center $ O$ and radius $ r$. Let $ A_1B_1$ intersect $ PO_1$ at $ X_1$. We have $ r_1^2=O_1X_1\cdot O_1P=O_1O^2-r^2$. So we get $ r^2=O_1O^2-r_1^2=O_2O^2-r_2^2=O_3O^2-r^2$. So $ O$ must be the radical center of $ \Gamma_1,\Gamma_2,\Gamma_3$ and $ r$ does not depend on $ P$. Thus all exceptional points lie on $ \Gamma$.
kapilpavase
22.01.2016 12:15
Let the point of concurancy be $Q$.Dilate with centre $P$ and ratio $1/2$.This maps $A_iB_i$ to the radical axis of point circle $P$ & $\Gamma_i$.Since these radical axes are concurrent,we get them midpt of $PQ$ say $M$ to be radical centre of 3 circles and $P$ point circle.Now since the power of $M$ wrt point circle $P$ is fixed we get $MP^2$ is fixed .Done
CT17
06.03.2024 20:53
Wow. If the three polars concur, then the radial axes of the point circle at $P$ with the three circles concur, so $P$ lies on the circle centered at the radical center of $\Gamma_1, \Gamma_2, \Gamma_3$ with radius equal to the square root of the common power.
khina
22.08.2024 18:25
Whoa this is cool. Given an exceptional $P$, let $Q$ be the point of concurrency. Then the midpoint of $PQ$ is the radical center of the three circles, and $(PQ)$ is orthogonal to all three circles. This uniquely determines $(PQ)$.