$2010$ $G1$, proving quadrilateral concyclic

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I was solving the following question, from $2010text IMO$ shortlist.

$G1.$ Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P$. The lines $BP$ and $DF$ meet at point $Q$. Prove that $AP=AQ$.
($L$ is $A$, $K$ is $B$, $P$ is $F$, and $Q$ is $P$)

While making the diagram on GeoGebra, I observed that $LQRP$, or $APQF$ in the question's labelling, is cyclic. I tried proving it, and I proved it in my mind. Then, as is characteristic of me, I forgot how I proved it. The only thing I remember is using $angle LQP=angle LMP=angle LNP$, then saying 'Wow, that was easy.'

I've tried to do that again now, but I've run into a problem. We have that $angle LQP=angle LNP$. If $LQRP$ is concyclic, then we must have $angle LRP=angle LQP$ as well. So we have $angle LNP=angle LRP$, which is absurd. However, it is obvious from the diagram that the quadrilateral indeed is concyclic.

Where did I make the mistake? And how can I prove that it is concyclic?

geometry contest-math

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asked Jul 25 at 18:14

MalayTheDynamo

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I was solving the following question, from $2010text IMO$ shortlist.

$G1.$ Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P$. The lines $BP$ and $DF$ meet at point $Q$. Prove that $AP=AQ$.
($L$ is $A$, $K$ is $B$, $P$ is $F$, and $Q$ is $P$)

While making the diagram on GeoGebra, I observed that $LQRP$, or $APQF$ in the question's labelling, is cyclic. I tried proving it, and I proved it in my mind. Then, as is characteristic of me, I forgot how I proved it. The only thing I remember is using $angle LQP=angle LMP=angle LNP$, then saying 'Wow, that was easy.'

I've tried to do that again now, but I've run into a problem. We have that $angle LQP=angle LNP$. If $LQRP$ is concyclic, then we must have $angle LRP=angle LQP$ as well. So we have $angle LNP=angle LRP$, which is absurd. However, it is obvious from the diagram that the quadrilateral indeed is concyclic.

Where did I make the mistake? And how can I prove that it is concyclic?

geometry contest-math

share|cite|improve this question

asked Jul 25 at 18:14

MalayTheDynamo

2,052833

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I was solving the following question, from $2010text IMO$ shortlist.

$G1.$ Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P$. The lines $BP$ and $DF$ meet at point $Q$. Prove that $AP=AQ$.
($L$ is $A$, $K$ is $B$, $P$ is $F$, and $Q$ is $P$)

While making the diagram on GeoGebra, I observed that $LQRP$, or $APQF$ in the question's labelling, is cyclic. I tried proving it, and I proved it in my mind. Then, as is characteristic of me, I forgot how I proved it. The only thing I remember is using $angle LQP=angle LMP=angle LNP$, then saying 'Wow, that was easy.'

I've tried to do that again now, but I've run into a problem. We have that $angle LQP=angle LNP$. If $LQRP$ is concyclic, then we must have $angle LRP=angle LQP$ as well. So we have $angle LNP=angle LRP$, which is absurd. However, it is obvious from the diagram that the quadrilateral indeed is concyclic.

Where did I make the mistake? And how can I prove that it is concyclic?

geometry contest-math

share|cite|improve this question

asked Jul 25 at 18:14

MalayTheDynamo

2,052833

I was solving the following question, from $2010text IMO$ shortlist.

$G1.$ Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P$. The lines $BP$ and $DF$ meet at point $Q$. Prove that $AP=AQ$.
($L$ is $A$, $K$ is $B$, $P$ is $F$, and $Q$ is $P$)

While making the diagram on GeoGebra, I observed that $LQRP$, or $APQF$ in the question's labelling, is cyclic. I tried proving it, and I proved it in my mind. Then, as is characteristic of me, I forgot how I proved it. The only thing I remember is using $angle LQP=angle LMP=angle LNP$, then saying 'Wow, that was easy.'

I've tried to do that again now, but I've run into a problem. We have that $angle LQP=angle LNP$. If $LQRP$ is concyclic, then we must have $angle LRP=angle LQP$ as well. So we have $angle LNP=angle LRP$, which is absurd. However, it is obvious from the diagram that the quadrilateral indeed is concyclic.

Where did I make the mistake? And how can I prove that it is concyclic?

geometry contest-math

share|cite|improve this question

asked Jul 25 at 18:14

MalayTheDynamo

2,052833

share|cite|improve this question

share|cite|improve this question

asked Jul 25 at 18:14

MalayTheDynamo

2,052833

asked Jul 25 at 18:14

MalayTheDynamo

2,052833

asked Jul 25 at 18:14

MalayTheDynamo

2,052833

2,052833

add a comment | 

add a comment | 

1 Answer
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$angle LQP$ is NOT equal to $angle LMP$. This would imply $LQMP$ is cyclic, but that would mean $P$ lies on the circumcircle of $LKM$ (the circle of $LQM$) but this is not true, as $P$ is the foot of an altitude.

Also note that $angle LQP$ is not equal to $LNP$, as that would imply $LQNP$ is cyclic, but you can see from your diagram that isn't true.

Finally, observe that proving $LQRP$ is cyclic would solve the original problem.

You can show this with inversion, or some synthetic length computing with that idea; it isn't too difficult..

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edited Jul 26 at 14:15

answered Jul 25 at 19:46

Faraz Masroor

7131720

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

$angle LQP$ is NOT equal to $angle LMP$. This would imply $LQMP$ is cyclic, but that would mean $P$ lies on the circumcircle of $LKM$ (the circle of $LQM$) but this is not true, as $P$ is the foot of an altitude.

Also note that $angle LQP$ is not equal to $LNP$, as that would imply $LQNP$ is cyclic, but you can see from your diagram that isn't true.

Finally, observe that proving $LQRP$ is cyclic would solve the original problem.

You can show this with inversion, or some synthetic length computing with that idea; it isn't too difficult..

share|cite|improve this answer

edited Jul 26 at 14:15

answered Jul 25 at 19:46

Faraz Masroor

7131720

add a comment | 

up vote
1
down vote



accepted

$angle LQP$ is NOT equal to $angle LMP$. This would imply $LQMP$ is cyclic, but that would mean $P$ lies on the circumcircle of $LKM$ (the circle of $LQM$) but this is not true, as $P$ is the foot of an altitude.

Also note that $angle LQP$ is not equal to $LNP$, as that would imply $LQNP$ is cyclic, but you can see from your diagram that isn't true.

Finally, observe that proving $LQRP$ is cyclic would solve the original problem.

You can show this with inversion, or some synthetic length computing with that idea; it isn't too difficult..

share|cite|improve this answer

edited Jul 26 at 14:15

answered Jul 25 at 19:46

Faraz Masroor

7131720

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

$angle LQP$ is NOT equal to $angle LMP$. This would imply $LQMP$ is cyclic, but that would mean $P$ lies on the circumcircle of $LKM$ (the circle of $LQM$) but this is not true, as $P$ is the foot of an altitude.

Also note that $angle LQP$ is not equal to $LNP$, as that would imply $LQNP$ is cyclic, but you can see from your diagram that isn't true.

Finally, observe that proving $LQRP$ is cyclic would solve the original problem.

You can show this with inversion, or some synthetic length computing with that idea; it isn't too difficult..

share|cite|improve this answer

edited Jul 26 at 14:15

answered Jul 25 at 19:46

Faraz Masroor

7131720

$angle LQP$ is NOT equal to $angle LMP$. This would imply $LQMP$ is cyclic, but that would mean $P$ lies on the circumcircle of $LKM$ (the circle of $LQM$) but this is not true, as $P$ is the foot of an altitude.

Also note that $angle LQP$ is not equal to $LNP$, as that would imply $LQNP$ is cyclic, but you can see from your diagram that isn't true.

Finally, observe that proving $LQRP$ is cyclic would solve the original problem.

You can show this with inversion, or some synthetic length computing with that idea; it isn't too difficult..

share|cite|improve this answer

edited Jul 26 at 14:15

answered Jul 25 at 19:46

Faraz Masroor

7131720

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 14:15

edited Jul 26 at 14:15

edited Jul 26 at 14:15

answered Jul 25 at 19:46

Faraz Masroor

7131720

answered Jul 25 at 19:46

Faraz Masroor

7131720

answered Jul 25 at 19:46

Faraz Masroor

7131720

7131720

add a comment | 

add a comment | 

 

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