BdMO National 2013: Higher Secondary 8
 FahimFerdous
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BdMO 2013 Higher Secondary Problem 8
$ABC$ is an acute angled triangle. Perpendiculars drawn from its vertices on the opposite sides are $AD$, $BE$, and $CF$. The line parallel to $DF$ through $E$ meets $BC$ at $Y$ and $BA$ at $X$. $DF$ and $CA$ meet at $Z$. Circumcircle of $XYZ$ meets $AC$ at $S$. Given, $\angle B=33^\circ$, find $\angle FSD$ with proof.
Your hot head might dominate your good heart!
 FahimFerdous
 Posts: 176
 Joined: Thu Dec 09, 2010 12:50 am
 Location: Mymensingh, Bangladesh
Re: BdMO 2013 Higher Secondary Problem 8
Hint:
How and why did I miss this? :'(
Your hot head might dominate your good heart!

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Re: BdMO 2013 Higher Secondary Problem 8
You spin my head right round right round,
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
Re: BdMO 2013 Higher Secondary Problem 8
Just another irony.sourav das wrote:viewtopic.php?f=25&t=916
This will help a lot. But my solution involves harmonic conjugate.
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 zadid xcalibured
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Re: BdMO 2013 Higher Secondary Problem 8
$A$,$X$,$C$,$Y$ cyclic.And $Z$,$X$,$S$,$Y$ cyclic.So
$EZ.ES$=$EX.EY$=$EA.EC$
Let $H$ be the orthocentre.$EA.EC=EH.EB$
But $ES.EZ=EH.EB$
so $H$ is the orthocentre of $\triangle ZBS$
So $ZH$ is perpendicular to $BS$.
Let $M$ be the midpoint of $AC$.
$ZH$ is the polar of $B$ with respect to circle $AFDC$. So $ZH$ is perpendicular to $BM$.
$EZ.ES$=$EX.EY$=$EA.EC$
Let $H$ be the orthocentre.$EA.EC=EH.EB$
But $ES.EZ=EH.EB$
so $H$ is the orthocentre of $\triangle ZBS$
So $ZH$ is perpendicular to $BS$.
Let $M$ be the midpoint of $AC$.
$ZH$ is the polar of $B$ with respect to circle $AFDC$. So $ZH$ is perpendicular to $BM$.
Re: BdMO 2013 Higher Secondary Problem 8
Let $M$ be the midpoint of $AC$
$A,X,C,Y$ cyclic.
$XE \cdot EY = AE \cdot EC = AM^2 ME^2$
Again, the polar of $Z$ on $\bigcirc AEDC$ is $BE$ (actually $BH$, $H=$ orthocenter of $\triangle ABC$).
And $ZM \cap BE = E$, so $E$ is the inverse of $Z$ WRT $\bigcirc AEDC$ as $M$ is the center of $\bigcirc AEDC$.
So, $ZM \cdot ME = AM^2$.
And thus $ZE \cdot ME = AM^2  ME^2 = XE \cdot EY$.
So, $X, Y, Z, M$ concyclic.
$A,X,C,Y$ cyclic.
$XE \cdot EY = AE \cdot EC = AM^2 ME^2$
Again, the polar of $Z$ on $\bigcirc AEDC$ is $BE$ (actually $BH$, $H=$ orthocenter of $\triangle ABC$).
And $ZM \cap BE = E$, so $E$ is the inverse of $Z$ WRT $\bigcirc AEDC$ as $M$ is the center of $\bigcirc AEDC$.
So, $ZM \cdot ME = AM^2$.
And thus $ZE \cdot ME = AM^2  ME^2 = XE \cdot EY$.
So, $X, Y, Z, M$ concyclic.
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Use $L^AT_EX$, It makes our work a lot easier!
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Re: BdMO 2013 Higher Secondary Problem 8
$X,A,Y,C$ is Cyclic $\Rightarrow EA.EC=EX.EY$ ; $X,S,Y,Z$ is Cyclic $\Rightarrow EX.EY=ES.EZ$.
So, $EA.EC=ES.EZ.....(1)$
Again, the Pencil of Rays $B(Z,C,E,A)$ is Harmonic.
So, $\displaystyle \frac{EA}{EC}=\frac{ZA}{ZC}.....(2)$
Using relation $(1)$ & $(2)$, we can deduce that, $SC=2AC$. So, $S$ is the midpoint of $AC$.
Therefore, $F,S,E,D$ is cyclic. So, $\angle{FSD}=\angle{FED}=180^{\circ}2\angle{B}=\boxed{114^\circ}$.
So, $EA.EC=ES.EZ.....(1)$
Again, the Pencil of Rays $B(Z,C,E,A)$ is Harmonic.
So, $\displaystyle \frac{EA}{EC}=\frac{ZA}{ZC}.....(2)$
Using relation $(1)$ & $(2)$, we can deduce that, $SC=2AC$. So, $S$ is the midpoint of $AC$.
Therefore, $F,S,E,D$ is cyclic. So, $\angle{FSD}=\angle{FED}=180^{\circ}2\angle{B}=\boxed{114^\circ}$.