Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
- Anindya Biswas
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Let $\triangle ABC$ be a triangle inscribed in a circle $\omega$. $D,E$ are two points on the arc $BC$ of $\omega$ not containing $A$. Points $F,G$ lie on $BC$ such that \[\angle BAF = \angle CAD, \angle BAG = \angle CAE\] Prove that the two lines $DG$ and $EF$ meet on $\omega$.
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Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
$\textbf{Solution}$Anindya Biswas wrote: ↑Sun May 09, 2021 4:28 pmLet $\triangle ABC$ be a triangle inscribed in a circle $\omega$. $D,E$ are two points on the arc $BC$ of $\omega$ not containing $A$. Points $F,G$ lie on $BC$ such that \[\angle BAF = \angle CAD, \angle BAG = \angle CAE\] Prove that the two lines $DG$ and $EF$ meet on $\omega$.
$Diagram$
Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
My solution is a bit lengthy. Still giving it anywayAnindya Biswas wrote: ↑Sun May 09, 2021 4:28 pmLet $\triangle ABC$ be a triangle inscribed in a circle $\omega$. $D,E$ are two points on the arc $BC$ of $\omega$ not containing $A$. Points $F,G$ lie on $BC$ such that \[\angle BAF = \angle CAD, \angle BAG = \angle CAE\] Prove that the two lines $DG$ and $EF$ meet on $\omega$.
$\textbf{Solution}$
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Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
Correction needed ! I think you wanna say $FPAG'$ is cyclic instead of $FPDG'$Ohin01 wrote: ↑Tue May 11, 2021 10:15 pmMy solution is a bit lengthy. Still giving it anywayAnindya Biswas wrote: ↑Sun May 09, 2021 4:28 pmLet $\triangle ABC$ be a triangle inscribed in a circle $\omega$. $D,E$ are two points on the arc $BC$ of $\omega$ not containing $A$. Points $F,G$ lie on $BC$ such that \[\angle BAF = \angle CAD, \angle BAG = \angle CAE\] Prove that the two lines $DG$ and $EF$ meet on $\omega$.
$\textbf{Solution}$
Other things are alright..and by the way, nice work....
Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
Oops...I did prove $FPAG'$ is cyclic but accidentally wrote $FPDG'$. :/Mahiir wrote: ↑Wed May 12, 2021 12:52 amCorrection needed ! I think you wanna say $FPAG'$ is cyclic instead of $FPDG'$Ohin01 wrote: ↑Tue May 11, 2021 10:15 pmMy solution is a bit lengthy. Still giving it anywayAnindya Biswas wrote: ↑Sun May 09, 2021 4:28 pmLet $\triangle ABC$ be a triangle inscribed in a circle $\omega$. $D,E$ are two points on the arc $BC$ of $\omega$ not containing $A$. Points $F,G$ lie on $BC$ such that \[\angle BAF = \angle CAD, \angle BAG = \angle CAE\] Prove that the two lines $DG$ and $EF$ meet on $\omega$.
$\textbf{Solution}$
Other things are alright..and by the way, nice work....
But sadly you cant edit the post after someone replied (
Anyways, thanks for correcting the mistake
A dream doesnt become reality through magic;it takes sweat,determination and hard work
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Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
SOL 1 ( Pascal):
Let $ AG \cap \omega =G_1 , AF \cap \omega = F_1 $
$ \angle CBD =\angle CAD = \angle BAF = \angle BAF_1 =\angle BDF_1 \Longrightarrow F_1D \parallel BC$
Similarly $ EG_1 \parallel BC \parallel DF_1 \Longrightarrow EG_1 \parallel DF_1$
Let $ P_{\infty}$ be the point at infinity along $BC$ then $ EG_1 \cap DF_1 = P_{\infty}$
Let $ DG \cap \omega = Q$
Then By Pascals theorem on hexagon, $QDF_1AG_1E,$
$ QD\cap AG_1=G , DF_1 \cap G_1E =P_{\infty}, QE \cap AF_1$ are collinear .
Since , $G,P_{\infty} \in BC$ therefore $QE \cap AF_1$ must also lie on $BC$
Now $ AF_1 \cap BC = F$ which means $ Q,E,F$ are collinear.
$Q.E.D$
Let $ AG \cap \omega =G_1 , AF \cap \omega = F_1 $
$ \angle CBD =\angle CAD = \angle BAF = \angle BAF_1 =\angle BDF_1 \Longrightarrow F_1D \parallel BC$
Similarly $ EG_1 \parallel BC \parallel DF_1 \Longrightarrow EG_1 \parallel DF_1$
Let $ P_{\infty}$ be the point at infinity along $BC$ then $ EG_1 \cap DF_1 = P_{\infty}$
Let $ DG \cap \omega = Q$
Then By Pascals theorem on hexagon, $QDF_1AG_1E,$
$ QD\cap AG_1=G , DF_1 \cap G_1E =P_{\infty}, QE \cap AF_1$ are collinear .
Since , $G,P_{\infty} \in BC$ therefore $QE \cap AF_1$ must also lie on $BC$
Now $ AF_1 \cap BC = F$ which means $ Q,E,F$ are collinear.
$Q.E.D$
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Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
SOL 2 (Moving points) :
Fix $\triangle ABC, E,G$
Let $D$ be a moving point on $ \omega$ $ . AF$ is just the reflection of $AD$ over the angle bisector of $ \angle BAC$
$\therefore D\rightarrow AD \rightarrow AF \rightarrow F$ is projective .
Let $ EF \cap \omega = P_2$ then $D \rightarrow F \rightarrow P_2$ is projective.
Now let $ DG \cap \omega = P_1 $ and $ D \rightarrow P_1$ is clearly projective.
$\therefore$ It suffices to check 3 positions of $D$.
$1) D=C \Longrightarrow F=B \Longrightarrow EF=EB , DG=BE \Longrightarrow P_1=P_2=B$
$ 2) D=B $ is similar.
$ 3) D=E \Longrightarrow F=G \Longrightarrow P_1=P_2$
$A.W.D$
Fix $\triangle ABC, E,G$
Let $D$ be a moving point on $ \omega$ $ . AF$ is just the reflection of $AD$ over the angle bisector of $ \angle BAC$
$\therefore D\rightarrow AD \rightarrow AF \rightarrow F$ is projective .
Let $ EF \cap \omega = P_2$ then $D \rightarrow F \rightarrow P_2$ is projective.
Now let $ DG \cap \omega = P_1 $ and $ D \rightarrow P_1$ is clearly projective.
$\therefore$ It suffices to check 3 positions of $D$.
$1) D=C \Longrightarrow F=B \Longrightarrow EF=EB , DG=BE \Longrightarrow P_1=P_2=B$
$ 2) D=B $ is similar.
$ 3) D=E \Longrightarrow F=G \Longrightarrow P_1=P_2$
$A.W.D$