Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"

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Anindya Biswas
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Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"

Unread post by Anindya Biswas » Sun May 09, 2021 4:28 pm

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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Mahiir
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Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"

Unread post by Mahiir » Sun May 09, 2021 5:20 pm

Anindya Biswas wrote:
Sun May 09, 2021 4:28 pm
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 $\omega$.
$\textbf{Solution}$
Screenshot 2021-05-09 164608.png
Screenshot 2021-05-09 164608.png (313.79KiB)Viewed 5707 times
$Diagram$
By Angle chasing

$\angle{AFB}=\angle{AED}=\alpha+\angle{C}$ and
$\angle{AGC}=\angle{ADE}=\gamma+\angle{B}$

So $\Delta{AGF}\sim\Delta{ADE}$

Consider spiral similarity about $A$

$S_S(\Delta{AGF})=\Delta{ADE}$

So $S_S(\Delta{ADG})=\Delta{AEF}$

By the property of spiral similarity,
we can say that $DG$ and $EF$ meet a point which is $\odot{AGF}\cap\odot{ADE}(\omega)$ (other than $A$) $\square$

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Ohin01
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Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"

Unread post by Ohin01 » Tue May 11, 2021 10:15 pm

Anindya Biswas wrote:
Sun May 09, 2021 4:28 pm
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 $\omega$.
My solution is a bit lengthy. Still giving it anyway
$\textbf{Solution}$
Assume that $EF$ intersects $\omega$ at $P$.Connect $P,D$.$PD$ intersects $BC$ at $G'$. We will show $G=G'$
$\textbf{Claim 1}$: $FPDG'$ is cyclic
$\text{Proof}$: Given that, $\angle BAF$=$\angle CAD$
As $ACDB$ is cyclic,
$\angle CAD$=$\angle CBD$
Now we can show by angle chasing,
$\angle FAP$=$\angle BAF$+$\angle BAP$
=$\angle CBD$+$\angle BDP$
And $\angle FG'P$ is an external angle of $\triangle BDG'$
So, $\angle FG'P$=$\angle CBD$+$\angle BDP$
So, $\angle FG'P$=$\angle FAP$
So, $FPDG'$ is cyclic.
$\textbf{Claim 2}$: $\angle FAG'$=$\angle EAD$
$\text{Proof}$: As $FPDG'$ is cyclic,
$\angle FAG'$=$\angle FPG'$
Again in $APED$ cyclic quad,
$\angle FPG'$=$\angle EAD$
So, $\angle FAG'$=$\angle EAD$
$\textbf{Claim 3}$: $\angle BAG'$=$\angle CAD$
$\text{Proof}$: From Claim 2,
$\angle FAG'$+$\angle BAF$=$\angle EAD$+$\angle CAD$
So, $\angle BAG'$=$\angle CAD$
Which proves $G$ and $G'$ are the same point
So, DG and EF meet on $\omega$
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Mahiir
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Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"

Unread post by Mahiir » Wed May 12, 2021 12:52 am

Ohin01 wrote:
Tue May 11, 2021 10:15 pm
Anindya Biswas wrote:
Sun May 09, 2021 4:28 pm
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 $\omega$.
My solution is a bit lengthy. Still giving it anyway
$\textbf{Solution}$
Assume that $EF$ intersects $\omega$ at $P$.Connect $P,D$.$PD$ intersects $BC$ at $G'$. We will show $G=G'$
$\textbf{Claim 1}$: $FPDG'$ is cyclic
$\text{Proof}$: Given that, $\angle BAF$=$\angle CAD$
As $ACDB$ is cyclic,
$\angle CAD$=$\angle CBD$
Now we can show by angle chasing,
$\angle FAP$=$\angle BAF$+$\angle BAP$
=$\angle CBD$+$\angle BDP$
And $\angle FG'P$ is an external angle of $\triangle BDG'$
So, $\angle FG'P$=$\angle CBD$+$\angle BDP$
So, $\angle FG'P$=$\angle FAP$
So, $FPDG'$ is cyclic.
$\textbf{Claim 2}$: $\angle FAG'$=$\angle EAD$
$\text{Proof}$: As $FPDG'$ is cyclic,
$\angle FAG'$=$\angle FPG'$
Again in $APED$ cyclic quad,
$\angle FPG'$=$\angle EAD$
So, $\angle FAG'$=$\angle EAD$
$\textbf{Claim 3}$: $\angle BAG'$=$\angle CAD$
$\text{Proof}$: From Claim 2,
$\angle FAG'$+$\angle BAF$=$\angle EAD$+$\angle CAD$
So, $\angle BAG'$=$\angle CAD$
Which proves $G$ and $G'$ are the same point
So, DG and EF meet on $\omega$
Correction needed ! I think you wanna say $FPAG'$ is cyclic instead of $FPDG'$
Other things are alright..and by the way, nice work....

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Ohin01
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Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"

Unread post by Ohin01 » Tue May 18, 2021 9:09 am

Mahiir wrote:
Wed May 12, 2021 12:52 am
Ohin01 wrote:
Tue May 11, 2021 10:15 pm
Anindya Biswas wrote:
Sun May 09, 2021 4:28 pm
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 $\omega$.
My solution is a bit lengthy. Still giving it anyway
$\textbf{Solution}$
Assume that $EF$ intersects $\omega$ at $P$.Connect $P,D$.$PD$ intersects $BC$ at $G'$. We will show $G=G'$
$\textbf{Claim 1}$: $FPDG'$ is cyclic
$\text{Proof}$: Given that, $\angle BAF$=$\angle CAD$
As $ACDB$ is cyclic,
$\angle CAD$=$\angle CBD$
Now we can show by angle chasing,
$\angle FAP$=$\angle BAF$+$\angle BAP$
=$\angle CBD$+$\angle BDP$
And $\angle FG'P$ is an external angle of $\triangle BDG'$
So, $\angle FG'P$=$\angle CBD$+$\angle BDP$
So, $\angle FG'P$=$\angle FAP$
So, $FPDG'$ is cyclic.
$\textbf{Claim 2}$: $\angle FAG'$=$\angle EAD$
$\text{Proof}$: As $FPDG'$ is cyclic,
$\angle FAG'$=$\angle FPG'$
Again in $APED$ cyclic quad,
$\angle FPG'$=$\angle EAD$
So, $\angle FAG'$=$\angle EAD$
$\textbf{Claim 3}$: $\angle BAG'$=$\angle CAD$
$\text{Proof}$: From Claim 2,
$\angle FAG'$+$\angle BAF$=$\angle EAD$+$\angle CAD$
So, $\angle BAG'$=$\angle CAD$
Which proves $G$ and $G'$ are the same point
So, DG and EF meet on $\omega$
Correction needed ! I think you wanna say $FPAG'$ is cyclic instead of $FPDG'$
Other things are alright..and by the way, nice work....
Oops...I did prove $FPAG'$ is cyclic but accidentally wrote $FPDG'$. :/
But sadly you cant edit the post after someone replied :((
Anyways, thanks for correcting the mistake :)
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Nayer_Sharar
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Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"

Unread post by Nayer_Sharar » Thu Jun 17, 2021 8:31 pm

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.

$Q.E.D$

Nayer_Sharar
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Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"

Unread post by Nayer_Sharar » Thu Jun 17, 2021 8:46 pm

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$

$A.W.D$

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