let $ w_1 $ and $ w_2 $ two circles such that $ w_1 \cap w_2 = \{ A , B \} $ let $ X $ a point on $ w_2 $ and $ Y $ on $ w_1 $ such that $ BY \bot BX $ suppose that $ O $ is the center of $ w_1 $ and $ X' = w_2 \cap OX $ now if $ K = w_2 \cap X'Y $ prove $ X $ is the midpoint of arc $ AK $
Problem
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Tags: IGO, geometry
andria
03.09.2015 19:56
Trivial problem: Let $BX\cap \omega_1=T$. then $Y,O,T$ are collinear and $AX'BX$ is cyclic $\Longrightarrow \angle AX'O=\angle ABT=\angle AYT\Longrightarrow AYX'O$ is cyclic thus $\angle OAY=\angle KX'X=\angle AYO=\angle ABT\Longrightarrow XK=XA$. Q.E.D
acupofmath
03.09.2015 20:04
let $ AX'X = \alpha $ and $ XX'K = \alpha ' $
we must prove $ \alpha = \alpha ' $
see also : $ \angle ABX = \alpha $ and $ \angle YX'O = \alpha ' $
we have : $ \angle ABY = 90^{0} - \alpha $
this can help us to move $ \alpha $ in the other circle $ ( $ $ w_1$ $ ) $
now $ \angle AOY = 180 - 2\alpha $
but $ OA = OY \Rightarrow \angle OYA = \angle OAY = \alpha $
remember we have $ \angle YX'O = \alpha ' $ and we must prove $ \alpha = \alpha ' $
It means we have to prove $ OX'AY $ is cyclic but it is obvius because $ \angle AYO = \angle AX'X = \alpha $
So we have done!
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acupofmath
03.09.2015 20:06
andria wrote: Trivial problem: Let $BX\cap \omega_1=T$. then $Y,O,T$ are collinear and $AX'BX$ is cyclic $\Longrightarrow \angle AX'O=\angle ABT=\angle AYT\Longrightarrow AYX'O$ is cyclic thus $\angle OAY=\angle KX'X=\angle AYO=\angle ABT\Longrightarrow XK=XA$. Q.E.D Yes that's true! it's the easiest problem.
MRF2017
20.09.2015 17:40
proposed by Davoud Vakili, Iran.
anantmudgal09
02.09.2016 19:02
Let $BX$ meet $\omega_1$ again at point $X_1$. It is clear that points $Y,O_1,X_1$ are collinear. Moreover, $\angle AYO_1=\angle AYX_1=\angle ABX_1=\angle ABX=\angle AX'X=180^{\circ}-\angle AX'O_1$, therefore, the points $A,Y,O_1,X'$ are concyclic, say, on a circle $\omega$. Since $A$ is the second intersection point of $\omega_1$ and $\omega$, other than $X'$, we see that $A$ is the centre of a spiral similarity sending $YO_1$ to $XK$. Since $\triangle AO_1Y \sim \triangle AXK$, we see that $X$ is the midpoint of arc $AK$.