BdMO National 2013: Higher Secondary 8

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FahimFerdous
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BdMO 2013 Higher Secondary Problem 8

Unread post by FahimFerdous » Mon Feb 11, 2013 3:43 pm

$ABC$ 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.
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FahimFerdous
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Re: BdMO 2013 Higher Secondary Problem 8

Unread post by FahimFerdous » Mon Feb 11, 2013 4:06 pm

Hint:
Just use ratios and a little Trig to show that if $G$ is the midpoint of $AC$, then $G=S$. Then some angle chasing and the answer comes along.
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Re: BdMO 2013 Higher Secondary Problem 8

Unread post by sourav das » Mon Feb 11, 2013 5:29 pm

viewtopic.php?f=25&t=916

This will help a lot. But my solution involves harmonic conjugate.
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Re: BdMO 2013 Higher Secondary Problem 8

Unread post by *Mahi* » Mon Feb 11, 2013 9:54 pm

sourav das wrote:viewtopic.php?f=25&t=916

This will help a lot. But my solution involves harmonic conjugate.
Just another irony.
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Re: BdMO 2013 Higher Secondary Problem 8

Unread post by zadid xcalibured » Mon Feb 11, 2013 10:11 pm

$A$,$X$,$C$,$Y$ 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 $BM$.

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Re: BdMO 2013 Higher Secondary Problem 8

Unread post by *Mahi* » Mon Feb 11, 2013 11:17 pm

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.
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Re: BdMO 2013 Higher Secondary Problem 8

Unread post by sowmitra » Sun Sep 08, 2013 7:40 pm

$X,A,Y,C$ 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 mid-point of $AC$.
Therefore, $F,S,E,D$ is cyclic. So, $\angle{FSD}=\angle{FED}=180^{\circ}-2\angle{B}=\boxed{114^\circ}$.
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