Let ABC be an acute triangle with circumcentre O. Let ΓB be the circle through A and B that is tangent to AC, and let ΓC be the circle through A and C that is tangent to AB. An arbitrary line through A intersects ΓB again in X and ΓC again in Y. Prove that |OX|=|OY|.
Problem
Source: Benelux Mathematical Olympiad 2015, Problem 2
Tags: geometry, circumcircle
10.05.2015 12:37
My solution : Let XY cut ⊙(O) again at Z . Since ∠XBA=∠ZAC=∠ZBC,∠BAX=∠BCZ , so △BAX∼△BCZ⟹△BAC∼△BXZ⟹XZ:AC=BZ:BC . (⋆) Similarly, we can prove △CAY∼△CBZ⟹AY:BZ=AC:BC , so combine (⋆) we get AY=XZ⟹ the perpendicular bisector of XY pass through O . i.e. OX=OY Q.E.D
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.
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∩CN . Since ∠TMN=∠BAC=∠TNM , so we get TM=TN and ∠MTN=180∘−2∠BAC=180∘−∠BOC⟹T∈⊙(BOC) , hence from ∠CTO=∠CBO=90∘−∠BAC⟹TO⊥MN⟹O lie on the perpendicular bisector of MN . i.e. OM=ON Q.E.D
21.09.2017 15:43
Would it be useful to invert around A?
21.09.2017 17:51
Soph11 wrote: Would it be useful to invert around A? Yes. Notice the problem is equivalent to OX2−R2=XZ.XA=YZ.YA=OY2−R2(Power of a point). Invert around A and use Thales' Theorem and distance formula.