Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
Posted: Sun May 09, 2021 4:28 pm
by Anindya Biswas
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 $\omega$.
Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
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 (313.79KiB)Viewed 7565 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 $A$) $\square$
Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
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 $\omega$
Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
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....
Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
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
Re: Problem - 01 - National Math Camp 2021 Geometry Test - "The intersection point lies on the circumcircle"
Posted: Thu Jun 17, 2021 8:31 pm
by Nayer_Sharar
SOL 1 ( Pascal):
Let $ AG \cap \omega =G_1 , AF \cap \omega = F_1 $