AoPS - Problem

Source: 2018 Taiwan TST Round 1

Tags: geometry, geometric transformation, reflection, circumcircle

JustPostTaiwanTST

29.11.2019 18:27

Given a scalene triangle $ \triangle ABC $. $ B', C' $ are points lie on the rays $ \overrightarrow{AB}, \overrightarrow{AC} $ such that $ \overline{AB'} = \overline{AC}, \overline{AC'} = \overline{AB} $. Now, for an arbitrary point $ P $ in the plane. Let $ Q $ be the reflection point of $ P $ w.r.t $ \overline{BC} $. The intersections of $ \odot{\left(BB'P\right)} $ and $ \odot{\left(CC'P\right)} $ is $ P' $ and the intersections of $ \odot{\left(BB'Q\right)} $ and $ \odot{\left(CC'Q\right)} $ is $ Q' $. Suppose that $ O, O' $ are circumcenters of $ \triangle{ABC}, \triangle{AB'C'} $ Show that 1. $ O', P', Q' $ are colinear 2. $ \overline{O'P'} \cdot \overline{O'Q'} = \overline{OA}^{2} $

inversion exercise :c See that $AB \cdot AB' = AC \cdot AC' = AB \cdot AC$ means that $A$ lies on the radical axis of $(BB'P), (CC'P)$, i.e. $A, P, P'$ collinear, and similarly $A, Q, Q'$ collinear with $AP \cdot AP' = AQ \cdot AQ'= AB \cdot AC$. Thus we have a circle $\Gamma = (PP'QQ')$. Observe that $P, Q$ symmetric about $BC$ implies: $(APQ) \perp BC$ (1) and $\Gamma \perp BC$ (2). Now invert about $A$ with radius $\sqrt{AB \cdot AC}$, which fixes $\Gamma$ and maps $BC \leftrightarrow (AB'C')$. Then (1) becomes $P'Q' \perp (AB'C')$ and (2) becomes $\Gamma \perp (AB'C')$. These are enough to finish the problem: (1) implies $P', Q', O'$ collinear, and (2) implies $OA^2 = O'A^2 = \mathrm{Pow}_{O'}(\Gamma) = O'P' \cdot O'Q'$. $\Box$