Let $ABC$ be an acute triangle with circumcentre $O$. Let $\mathit{\Gamma}_B$ be the circle through $A$ and $B$ that is tangent to $AC$, and let $\mathit{\Gamma}_C$ be the circle through $A$ and $C$ that is tangent to $AB$. An arbitrary line through $A$ intersects $\mathit{\Gamma}_B$ again in $X$ and $\mathit{\Gamma}_C$ again in $Y$. Prove that $|OX|=|OY|$.
Problem
Source: Benelux Mathematical Olympiad 2015, Problem 2
Tags: geometry, circumcircle
TelvCohl
10.05.2015 12:37
My solution : Let $ XY $ cut $ \odot (O) $ again at $ Z $ . Since $ \angle XBA=\angle ZAC=\angle ZBC, \angle BAX=\angle BCZ $ , so $ \triangle BAX \sim \triangle BCZ \Longrightarrow \triangle BAC \sim \triangle BXZ \Longrightarrow XZ:AC=BZ:BC $ . $ (\star ) $ Similarly, we can prove $ \triangle CAY \sim \triangle CBZ \Longrightarrow AY:BZ=AC:BC $ , so combine $ (\star) $ we get $ AY=XZ \Longrightarrow $ the perpendicular bisector of $ XY $ pass through $ O $ . i.e. $ OX=OY $ Q.E.D
buratinogigle
10.05.2015 12:59
An easy extension Let $ABC$ be a triangle with circumcenter $O$. $E,F$ lie on $CA,AB$. $EF$ cuts circles $(AEB),(AFC)$ again at $M,N$. Prove that $OM=ON$.
TelvCohl
10.05.2015 13:31
buratinogigle wrote: An easy extension Let $ABC$ be a triangle with circumcenter $O$. $E,F$ lie on $CA,AB$. $EF$ cuts circles $(AEB),(AFC)$ again at $M,N$. Prove that $OM=ON$. My solution : Let $ T=BM \cap CN $ . Since $ \angle TMN=\angle BAC=\angle TNM $ , so we get $ TM=TN $ and $ \angle MTN=180^{\circ}-2\angle BAC=180^{\circ}-\angle BOC \Longrightarrow T \in \odot (BOC) $ , hence from $ \angle CTO=\angle CBO=90^{\circ}-\angle BAC \Longrightarrow TO \perp MN \Longrightarrow O $ lie on the perpendicular bisector of $ MN $ . i.e. $ OM=ON $ Q.E.D
Soph11
21.09.2017 15:43
Would it be useful to invert around A?
Gomes17
21.09.2017 17:51
Soph11 wrote: Would it be useful to invert around A? Yes. Notice the problem is equivalent to $$OX^{2}-R^{2}=XZ.XA=YZ.YA=OY^{2}-R^{2}$$(Power of a point). Invert around $A$ and use Thales' Theorem and distance formula.