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Reflected point lies on radical axis
Mahdi_Mashayekhi   3
N Apr 19, 2025 by Mahdi_Mashayekhi
Source: Iran 2025 second round P4
Given is an acute and scalene triangle $ABC$ with circumcenter $O$. $BO$ and $CO$ intersect the altitude from $A$ to $BC$ at points $P$ and $Q$ respectively. $X$ is the circumcenter of triangle $OPQ$ and $O'$ is the reflection of $O$ over $BC$. $Y$ is the second intersection of circumcircles of triangles $BXP$ and $CXQ$. Show that $X,Y,O'$ are collinear.
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Mahdi_Mashayekhi
Apr 19, 2025
Mahdi_Mashayekhi
Apr 19, 2025
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Reflected point lies on radical axis
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geometry
circumcircle
reflection
power of a point
radical axis
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Source: Iran 2025 second round P4
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Mahdi_Mashayekhi
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Mahdi_Mashayekhi
#1 h  Apr 19, 2025, 11:01 AM
Y by
Given is an acute and scalene triangle $ABC$ with circumcenter $O$. $BO$ and $CO$ intersect the altitude from $A$ to $BC$ at points $P$ and $Q$ respectively. $X$ is the circumcenter of triangle $OPQ$ and $O'$ is the reflection of $O$ over $BC$. $Y$ is the second intersection of circumcircles of triangles $BXP$ and $CXQ$. Show that $X,Y,O'$ are collinear.
This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Apr 19, 2025, 11:25 AM
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ItzsleepyXD
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ItzsleepyXD
#2 h  Apr 19, 2025, 11:26 AM
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Angle chasing :
$\angle BPX = \angle BPA + \angle APX = 180^{\circ} - \angle BAP - \angle ABO + 90^{\circ} + \angle ACB = \angle BAC + 90^{\circ} = 180^{\circ} - \angle O'BC $
and from $OX//BC$
Let $O'B \cap OX=E , O'B \cap OX = F$
so $\angle BEX + \angle BPX = 180^{\circ}$
thus $E,B,P,X$ concyclic .
in the same way $F,C,P,X$ concyclic.
so $O'$ is on the radical axis of $(BXP),(CXQ)$
implies that $X,Y,O'$ collinear . done $\square$
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gghx
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gghx
#3 h  Apr 19, 2025, 11:35 AM • 1 Y
Y by blackidea
Define $Y'$ as the second intersection of $O'X$ and $(BO'C)$. We claim that $Y=Y'$. Note that $$\angle BY'X=\angle BCO'=\angle BCO=90^\circ-\angle PQO=\angle XPO,$$hence $BPYX'$ is concyclic. Similarly, $CXQY$ is concyclic, and we are done.
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Mahdi_Mashayekhi
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Mahdi_Mashayekhi
#4 h  Apr 19, 2025, 4:34 PM • 2 Y
Y by Parsia--, sami1618
Note that $\angle XYC = \angle XQC =\angle XQO = 90 - \frac{\angle QXO}{2} = 90 - \angle QPO = 90 - (90 - \angle C + 90 - \angle B) = 90-\angle A$ and $\angle XYB = \angle XPO = 90 - \frac{\angle PXO}{2} = 90 - \angle PQO = 90 - (90 - \angle C + 90 - \angle B) = 90 - \angle A$ so $X$ lies on angle bisector of $BYC$.
Now note that $\angle BYC = 2(90 - \angle A) = 180 - 2\angle A = 180 - \angle BOC = 180 - \angle BO'C$ so $BYCO'$ is cyclic and since $BO'=CO'$ then $O'$ also lies on angle bisector of $BYC$ so $Y,X,O'$ are collinear as wanted.
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