Where is my incenter [self-made]

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Phlembac Adib Hasan
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Where is my incenter [self-made]

Unread post by Phlembac Adib Hasan » Wed Apr 03, 2013 9:25 pm

Suppose $\omega$ is a circle with center $O$ and $DE$ is a chord of it. The tangents of $\omega$ at $D$ and $E$ intersect each other at $P$. $M$ is the midpoint of $DE$. A line through $M$ intersects $\omega$ at $X$ and $Y$. Prove that the incenter of $\triangle PXY$ lies on $\omega$.
Hint:
Draw the circumcircle of $\triangle PXY$

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FahimFerdous
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Re: Where is my incenter [self-made]

Unread post by FahimFerdous » Wed Apr 03, 2013 10:28 pm

Me and Zadid did this one in Zadid's home. I used total symmetry and Zadid used the killer lemma (you know which one). :-D
Zadid's solution was more beautiful than mine though. :-/
Your hot head might dominate your good heart!

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FahimFerdous
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Re: Where is my incenter [self-made]

Unread post by FahimFerdous » Wed Apr 03, 2013 10:31 pm

On a second thought, as this one's your self made one, me and Zadid may have solved very similar one to this one. But the ideas are same. Sorry for the inconvenience. :-/
Your hot head might dominate your good heart!

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Tahmid Hasan
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Re: Where is my incenter [self-made]

Unread post by Tahmid Hasan » Thu Apr 04, 2013 12:15 am

বড় ভালবাসি তোমায়,মা

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Phlembac Adib Hasan
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Re: Where is my incenter [self-made]

Unread post by Phlembac Adib Hasan » Thu Apr 04, 2013 8:39 am

সব ভুইলা যাই। :evil:
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Re: Where is my incenter [self-made]

Unread post by photon » Tue Apr 09, 2013 1:09 pm

In $ODPE$ , $\angle ODP=\angle OEP=90^o$ , $ODPE$ is cyclic quadrilateral.
$OM.MP=DM.ME=XM.MY$ , therefore $O,Y,P,X$ are concyclic.
$OX=OY \Rightarrow \angle OXY=\angle OYX\Rightarrow \angle OPY=\angle OXY$ ; let $OP$ intersects $\omega $ at $J$. $JP$ bisects $\angle XPY$.
on the other hand , $\angle XYP=\angle XOP=2\angle XYJ$. $JY$ bisects $\angle XYP$.
Then it is proved $J$ is the incenter of $\Delta PXY$.
Try not to become a man of success but rather to become a man of value.-Albert Einstein

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