Circles $\Gamma_1$ and $\Gamma_2$ have centers $O_1$ and $O_2$ and intersect at $P$ and $Q$. A line through $P$ intersects $\Gamma_1$ and $\Gamma_2$ at $A$ and $B$, respectively, such that $AB$ is not perpendicular to $PQ$. Let $X$ be the point on $PQ$ such that $XA=XB$ and let $Y$ be the point within $AO_1 O_2 B$ such that $AYO_1$ and $BYO_2$ are similar. Prove that $2\angle{O_1 AY}=\angle{AXB}$. Author: Matthew Brennan
Problem
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Tags: geometry, geometric transformation, reflection, rotation, dilation, perpendicular bisector, geometry proposed
Vo Duc Dien
22.07.2011 18:06
Is there something wrong with the formulation of the problem? My computer graphic does not agree with the problem.
armpist
22.07.2011 19:07
Vo Duc Dien wrote: Is there something wrong with the formulation of the problem? My computer graphic does not agree with the problem. Dear Vo Problem statement is correct. You probably found the point Y incorrectly. M.T.
Vo Duc Dien
22.07.2011 20:02
OK. I guess the problem does not work for any arbitrary AB. I verified it by drawing angle O1AY = 1/2 angle AXB and Y is totally outside AO1O2B. I am busy to try another configuration where the problem may work.
Vo Duc Dien
22.07.2011 20:15
OK. The problem works for a different configuration where I have the first circle on the left with a radius equal approximately 2/3 of the one on the right and angle O1AB approximately 60 degree. I will try to prove it.
SnowEverywhere
23.07.2011 00:49
Just to clarify: the fact that triangles $AYO_1$ and $BYO_2$ are similar is intended to mean that $A$, $Y$, $O_1$ correspond to $B$, $Y$ and $O_2$, respectively. These triangles are inversely similar.
Vo Duc Dien
02.08.2011 20:11
Actually, I haven't been able to get my graphic to work for the problem.
Vo Duc Dien
31.12.2011 19:16
This problem is fairly easy once you nail down the graphic just like I have done. The two parallel segments AB, CD cause the triangles to be similar and thus the result.
SnowEverywhere
02.01.2012 06:48
Vo Duc Dien wrote: This problem is fairly easy once you nail down the graphic just like I have done. The two parallel segments AB, CD cause the triangles to be similar and thus the result. Wait, what are $C$ and $D$? Can you share your solution? I was under the impression that this problem was slightly harder than that...
Vo Duc Dien
15.02.2012 21:50
Let me have your e-mail. I can mail it to you. My solution involves too much graphic and syntaxes I cannot translate it into the format to fit in here.
polya78
24.05.2013 18:47
Let $M$ be the midpoint of $AB$ and $Z$ the reflection of $Y$ in $BO_2$. Since $\triangle AO_1Q \sim \triangle BO_2Q$, there is a homothetic transformation $f$ (a rotation and dilation about $Q$) such that $f(O_2)=O_1, f(B)=A$. But $\triangle AO_1Y \sim \triangle BO_2Y \sim \triangle BO_2Z$, it must be that $f(Z)=Y$. Thus $\triangle YQZ \sim \triangle AQB$. Let $g$ be the homothetic transformation about $Q$ such that $g(Y)=A,g(Z)=B$. Then $g(B)$ lies on the perpendicular bisector of $AB$, namely $XM$. We can easily see that $\angle QBO_2 =\angle QAO_1 = \angle PXM$, so in fact, $g(B)=X$, which means that $\triangle YBZ \sim \triangle AXB$ and the desired result follows immediately.
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