The altitudes of an acute triangle ABC with AB<AC intersect at a point H, and O is the center of the circumcircle Ω. The segment OH intersects the circumcircle of BHC at a point X, different from O and H. The circumcircle of AOX intersects the smaller arc AB of Ω at point Y. Prove that the line XY bisects the segment BC. Proposed by A. Tereshin
Problem
Source: All-Russian MO 2024 9.6
Tags: geometry, circumcircle, geometry proposed
23.04.2024 06:37
Isn't O should be circumcenter?
23.04.2024 07:17
demmy wrote: Isn't O should be circumcenter? Yes, O is circumcenter and H is orthocenter Let D be A-humpty point. Then ∠DAO=∠DAC−90+∠B=∠DCB−(∠HCB)=∠DCH=∠DXOhence A,X,D,O cyclic.Let M be midpoint of BC. Do inversion about circle ω center M with radius MB,We know that MD.MA=MB2 hence (ADO) is orthogonal to ω from (AOD)↔(AOD) and (ABC)↔(HBC) we have X↔Y which give us Y,X,M are collinear.
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07.06.2024 02:05
Let M be the midpoint of BC and let Z be the reflection of Y about M. Since triangles ABC and BHC have equal circumradius Z lies on the circumcircle of BHC. We conclude with an angle chase: ∠HXZ+∠HXY=∠HCZ+∠YAO=90∘−∠ABC+∠YBC+∠YCB+90∘−∠ACB=180∘
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07.06.2024 03:50
A really interesting extension of the Humpty Point configuration which I'm surprised I didn't know before. We denote by HA and QA the A−Humpty and A-Queue points of △ABC respectively. We first deal with the following claim. Claim : X lies on (AHAO). Proof : Simply note that, ∡OXHA=∡HCHA=∡BCHA+∡HCB+∡BCHA+∡ACO=∡CAM+∡OAC=∡OAHAusing the well known facts that (AHAB) and (AHAC) are tangent to line BC and that H and O are isogonal conjugates. Now, we can also note the following. Claim : Quadrilateral QAHXY is cyclic. Proof : It is well known that QAH passes through the A−anitpodal point A′. Further A , O and A′ are collinear by definition. Thus, ∡YQAH=∡YQAA′=∡YAA′=∡YAO=∡YXO=∡YXHwhich finishes the proof of the claim. Now, we are simply left to note that by Radical Center on circles (QAHXY) , (AHAOY) and (AHAHQA) (it is well known that the last is a circle), it follows that QAH , AHA and YX concur. Since it is well known that QAH and AHA pass through the midpoint of BC, it follows that XY bisects the segment BC as desired.
13.06.2024 16:12
To begin with, let T be the reflection of H with respect to the midpoint M - it is well known that T lies on Ω (due to ∠BTC=∠BHC=180∘−∠BAC and is the antipode of A, in particular it lies on AO. Now, instead of chasing the collinearity of X, M, Y, let S be the reflection of X with respect to M and aim to show that Y, X, S are collinear. The advantage is that S lies on Ω, as ∠BSC=∠BXC=∠BHC=180∘−∠BAC. Moreover, XM=MS and HM=MT imply that HXTS is a parallelogram; in particular, OH∥TS. Therefore ∠AYX=180∘−∠AOX=180∘−∠ATS=180∘−∠ACS=∠AYS and we are done!
17.06.2024 14:01
Here is an alternative great solution from JBMO training. Let XY intersect BC at M. Having no other ideas and seen many similarity problems, we boldly claim that MB2=MX⋅MY, as then we would analogously have MC2=MX⋅MY and would conclude MB=MC. The equality MB2=MX⋅MY is equivalent to (by Power of a Point) ∠CBX=∠BYX. Without loss of generality treat O to be between H and X. We compute ∠CBX=180∘−∠XHC=180∘−∠OHC from the circumcircle of BXHC. On the other, ∠BYX=∠AYB−∠AYX=180∘−∠ACB−(180∘−∠AOX)=180∘−∠ACB−∠AOH. It remains to argue ∠OHC−∠AOH=∠ACB. For this, let CH∩AO=T. Then ∠TAC=∠OAC=90∘−∠ABC and ∠ACT=∠ACH=90∘−∠BAC, hence ∠OHC−∠AOH=∠OTH=∠ACB, as desired.