Let $ABC$ be an arbitrary triangle and $O$ is the circumcenter of $\triangle {ABC}$.Points $X,Y$ lie on $AB,AC$,respectively such that the reflection of $BC$ WRT $XY$ is tangent to circumcircle of $\triangle {AXY}$.Prove that the circumcircle of triangle $AXY$ is tangent to circumcircle of triangle $BOC$.
Problem
Source: TST3,day1,P2
Tags: geometry, circumcircle, geometric transformation, reflection
baopbc
16.07.2016 07:03
$\textbf{Proof.}$ Let $\Omega$ be the symmetric of $\odot (AXY)$ through $XY$. Hence, $\Omega$ tangents to $BC$ at $R$. Let $T$ be the intersection of $\odot (BRY)$ and $\odot (CRX)$. We have : $\angle BTC=\angle BYR+\angle RXC$ But from $\Omega$ is the symmetric of $(AXY)$ through $XY$ so $\angle XRY=\angle XAY.$ Hence $\angle BYR+\angle RXC=2\angle BAC=\angle BOC$ which gives $B,T,O,C$ are concyclic. Now, by simple angle chasing and we are done. $\blacksquare$
Attachments:
andria
16.07.2016 22:11
The iranians can see my solution in the following pdf :
Attachments:
proof.pdf (27kb)
Ismael10
26.03.2017 02:21
Can you clarify andria in your proof why the angle condition imply the tangent result . Thanks in advance.
andria
26.03.2017 09:25
Ismael10 wrote: Can you clarify andria in your proof why the angle condition imply the tangent result . Thanks in advance. Because of the following lemma:
In the figure the circumcircles of $\triangle AXB$ and $\triangle CXD$ are tangent at $X$ if and if only:
$$\angle XAB+\angle XCD=\angle BXD$$[asy][asy]
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[/asy][/asy]
example for the lemma
Ismael10
26.03.2017 18:10
To prove the lemma we can consider the tangent to one of the circle at X . For the implication it is clear . For the reciproqual, we find that the condition and the tangency for one of the circle imply the tangency to the other circle.
nixon0630
22.10.2021 16:23
A really hard problem, I wouldn't have solved it without geogebra. Here's my solution: Let the reflection of $\odot{(AXY)}$ over $\overline{XY}$ touches $\overline{BC}$ at $T$ and let $Q$ be the Miquel point of $\{X, T, Y\}$ w.r.t. $\triangle{ABC}$. Claim. $Q$, $B$, $C$, $O$ are concyclic.
Note that $$\measuredangle{YAX} = \measuredangle{XTY}$$by simmetry, therefore: $$\measuredangle{CQB} = \measuredangle{CQT} + \measuredangle{TQB} = \measuredangle{CYT} + \measuredangle{TXB} = 2\measuredangle{YAX} = 2\measuredangle{CAB} = \measuredangle{COB}$$as claimed. $\square$
Finally: $$\measuredangle{BCQ} + \measuredangle{QYX} = \measuredangle{TCQ} + \measuredangle{QYX} = \measuredangle{TYQ} + \measuredangle{QYX} = \measuredangle{TYQ} = \measuredangle{TYX} = \measuredangle{BTX} = \measuredangle{BQX}$$hence, $\odot{AXY}$ and $\odot{BOC}$ are tangent to each other at $Q$. $\blacksquare$
Attachments:
JAnatolGT_00
27.10.2021 18:54
Oh no, seems like I spent too much time on this, thinking that it would be harder... Let reflection of $(AXY)$ wrt $XY$ tangents to $BC$ at $S.$ By the Miquel $(AXY),$ $(BSX),$ $(CSY)$ concur at some point $Z.$ $$\measuredangle BZC=\measuredangle BZS+\measuredangle SZC=\measuredangle XAY+\measuredangle YSX=2\measuredangle BAC\implies Z \in (BOC),$$$$\measuredangle XZB=\measuredangle XSB=\measuredangle XYS=\measuredangle XYZ+\measuredangle ZYS=\measuredangle XZY+\measuredangle ZYS.$$Hence $(AXY),$ $(BOC)$ tangent at $Z.$
guptaamitu1
27.05.2022 14:15
One motivation of considering the Miquel configuration is the following: Suppose $\overline{BC}$ is tangent to reflection of $\odot(AXY)$ in $\overline{XY}$ at $T$ ; and $Q$ is the required tangency point of $\odot(BOC)$ and $\odot(AXY)$ (notations are same as in the diagram above, post #8).If points $B,C,Y,X$ are concyclic, we are just directly done by inversion at $A$ swapping $\{B,X\}$ and $\{C,Y\}$, as it also swaps $$\odot(BOC) \longleftrightarrow \odot(XYT) \qquad , \qquad \overline{BC} \longleftrightarrow \odot(AXY)$$In that case, $Q$ is just the image of $T$ under inversion, and the only possible way to define $Q$ without inversion or point $A$, is that $$ Q = \odot(BXT) \cap \odot(CYT) \ne T $$So this motivates us to define $Q$ in general like that too. Luckily, we are just done after that.