IMOC 2017 G5 (<A=120 => E, F, Y,Z are concyclic, incenter related)

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IMOC 2017 G5 (<A=120 => E, F, Y,Z are concyclic, incenter related)

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Source: https://artofproblemsolving.com/community/c6h1740077p11309077

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#1 h Mar 20, 2020, 9:28 AM • 2 Y

Y by Mango247, AlexCenteno2007

We have $\vartriangle ABC$ with $I$ as its incenter. Let $D$ be the intersection of $AI$ and $BC$ and define $E, F$ in a similar way. Furthermore, let $Y = CI \cap DE, Z = BI \cap DF$ . Prove that if $\angle BAC = 120^o$ , then $E, F, Y,Z$ are concyclic.

https://1.bp.blogspot.com/-5IFojUbPE3o/XnSKTlTISqI/AAAAAAAALd0/0OwKMl02KJgqPs-SDOlujdcWXM0cWJiegCK4BGAYYCw/s1600/imoc2017%2Bg5.png

This post has been edited 3 times. Last edited by parmenides51, Mar 21, 2020, 8:31 PM

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Mathematicsislovely

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Mathematicsislovely

#2 h Sep 1, 2020, 4:05 AM

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$\textbf{\textcolor{red}{CLAIM:}}$ $AFZI$ and $AEYI$ are cyclic quadrilateral.
$\textbf{proof:}$
At first observe that
$\frac{AF}{FB}=\frac{AC}{CB}=\frac{sin\angle B}{sin 120^{\circ}}=\frac{sin\angle B}{sin 60^{\circ}}=\frac{AD}{BD}$ .

So, $DF$ is angle bisector of $\angle ADB$ .So $Z$ is incentre of $\triangle BAD$ .Hence, $\angle FAZ=30^{\circ}$ .
On the other hand, $ZIF=180^{\circ}-\angle BIC=30^{\circ}$ .So, $AFZI$ is concyclic.
Similarly, $AEYI$ is concyclic $\blacksquare$

Now we have, $DZ\times DF=DI\times DA=DY\times DE$ .Hence, $FZYE$ is a cyclic quadrilateral $\blacksquare$

This post has been edited 3 times. Last edited by Mathematicsislovely, Sep 1, 2020, 10:36 AM

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#3 h Yesterday at 3:46 PM

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I wonder if there is a trig-free solution.

My solution is almost the same as @above but I finished with
$\angle FZE=\angle FZI=60^{\circ}=\angle IYE=\angle FYE$

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#4 h Yesterday at 5:34 PM • 1 Y

Y by MrdiuryPeter

hehe synthetic solution

Let $AI$ intersect $(ABC)$ at $G$ and let $L$ be a point on line $AB$ such that $BC=BL$
$\implies BC=BL=BG$
Let $M=GL\cap (ABC)$ and $K=CM\cap AG$
$\angle GBL=60^{\circ}+\angle C$
$\implies BGL=60^{\circ}-\angle C/2=\angle BCM$
But $\angle CDK=60^{\circ}+\angle C$ which $\implies DC=DK$
Apply pascals theorem to $CMGGAB$ but as tangent $G$ will be parallel to $BC$
$\implies KL \parallel BC$
$\implies \frac{AB}{BC}= \frac{AB}{BL}=\frac{AD}{DK}=\frac{AD}{DC}$
$\therefore ED$ bisects $\angle CDA$
and rest follows as @above

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#5 h Yesterday at 9:28 PM

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Actually it is quite simple with the excenter argument ...

$E$ is simply excenter of $\triangle ABD$ implying $\angle ZED=30^{\circ}$ . Analogously $F$ is excenter of $ACD$ implying $\angle YFD=30^{\circ}$ . Thus, $EFYZ$ is cyclic as desired.

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