Prove three points P, A and C are collinear

[anemone]

Let $PQR$ be a triangle and let $A$ be an interior point such that $\angle QAR=90^{\circ}$, $\angle QBA=\angle QRA$.

Let $B,\,C$ be the midpoints of $PR,\,QR$ respectively. Suppose $QA=2AB$, prove that $P,\,A,\,C$ are collinear.

 

[Albert1]

Let $PQR$ be a triangle and let $A$ be an interior point such that $\angle QAR=90^{\circ}$, $\angle QBA=\angle QRA$.

Let $B,\,C$ be the midpoints of $PR,\,QR$ respectively. Suppose $QA=2AB$, prove that $P,\,A,\,C$ are collinear.

Spoiler

from the diagram it is clear :
[FONT=MathJax_Math-Web]P[/FONT][FONT=MathJax_Main-Web],[/FONT][FONT=MathJax_Math-Web]A[/FONT][FONT=MathJax_Main-Web],[/FONT][FONT=MathJax_Math-Web]C[/FONT] are collinear.

View attachment 4345

 

[MarkFL]

Spoiler

from the diagram it is clear :
[FONT=MathJax_Math-Web]P[/FONT][FONT=MathJax_Main-Web],[/FONT][FONT=MathJax_Math-Web]A[/FONT][FONT=MathJax_Main-Web],[/FONT][FONT=MathJax_Math-Web]C[/FONT] are collinear...

Hello Albert,

Spoiler

I don't think it is enough to post a diagram, and state that based on the diagram, the result is "clear." A diagram can be a useful aid, but it should at most be a supplement to a proof, not the proof itself.

I think the diagram should be accompanied by stated reasoning that inexorably leads to the desired conclusion. :D

 

[Albert1]

explanation of previous diagram:

Spoiler

$\triangle GAH , \triangle QAC $ both are equilateral triangles
$\therefore \angle GAH=\angle AQC=60^o$
in $\triangle PQC : \angle Q=90^o, \angle PCQ=2x=60^o$
$\therefore \angle QPC=30^o$
also in $\triangle PQA$ :$\angle PAQ=180^o-30^o-30^o=120^o$
and we have :$\angle QAC +\angle PAQ=60^o+120^o=180^o$
so points $P,A,C $ are collinear

 

[anemone]

Words and tears aren't enough to tell how sorry I am as I just realized I have been posted a problem with a typo and that consequently changed the structure and meaning of the original problem entirely. This problem actually is a good India Olympiad Math Contest problem, but I bungled it with a typo. I am truly sorry.(Sadface)

I have certainly wasted a good time of those who attempted at this problem, for that I apologize. (Tmi)

The problem should read:

Let $PQR$ be a triangle and let $A$ be an interior point such that $\angle QAR=90^{\circ}$, \(\displaystyle \color{yellow}\bbox[5px,purple]{\angle QPA=\angle QRA}\).

Let $B,\,C$ be the midpoints of $PR,\,QR$ respectively. Suppose $QA=2AB$, prove that $P,\,A,\,C$ are collinear.

 

[Albert1]

please upload your original diagram

 

[anemone]

I owe this thread a diagram, so here goes:View attachment 4346

 

[Albert1]

solution :

Spoiler

View attachment 4348

 

[anemone]

Thank you Albert for your solution!

Spoiler

But, I don't think the given info suffices to deem the point $A$ as the barycenter of the triangle $PQR$; also, I don't quite follow how $\angle CAR=y$? I mean, of course $\angle CAR=y$, but, don't we have to prove that first before claiming such is the case?

 

[Albert1]

Thank you Albert for your solution!

Spoiler

But, I don't think the given info suffices to deem the point $A$ as the barycenter of the triangle $PQR$; also, I don't quite follow how $\angle CAR=y$? I mean, of course $\angle CAR=y$, but, don't we have to prove that first before claiming such is the case?

Spoiler

point A on circle C
AC=CR=QC, for
$\angle QAR=90^o$

 

[Albert1]

Thank you Albert for your solution!

Spoiler

But, I don't think the given info suffices to deem the point $A$ as the barycenter of the triangle $PQR$; also, I don't quite follow how $\angle CAR=y$? I mean, of course $\angle CAR=y$, but, don't we have to prove that first before claiming such is the case?

Spoiler

yes,you are right ,point A does not have to be the barycenter of triangle $PQR$ , but always lie on segment PC
(the prove and diagram is simiar to post #8)
in special case it will switch to its barycenter

- - - Updated - - -

 

[Albert1]

Spoiler

View attachment 4350

$\triangle PQC $ and $\triangle RQI$
$\theta=\beta+y$
$\angle CAR=\angle ARC= \angle QPA=\angle PAI=y$
for $\triangle PAI $ is similar to $\triangle RAC$
in this diagram point $A'$ is the barycenter of $\triangle PQR$
and point $A$ lies on segment PC ,is what we want
$\therefore$ point $P,A,C $ collinear

 

[anemone]

Thanks Albert for your explanation and the second submission of the solution.

Here is the solution of other that I want to share:

Spoiler

View attachment 4352

Extend $RA$ such that $RA=AS$. Let $\angle QRA=\angle QPA=\theta$.

Observe that $QA$ is the perpendicular bisector of $RS$. Hence $QR=QS$ and $QRS$ is an isoceles triangle. Thus, $\angle QSA=\theta$. But then $\angle QSA=\angle QPA=\theta$. This implies that $Q,\,A,\,P,\,S$ all lie on a circle. In turn, we conclude that $\angle QAS=\angle QPS=90^{\circ}$.

View attachment 4353

Since $A$ is the midpoint of $RS$ (by construction) and $B$ is the midpoint of $PR$ (given), it follows that $AB$ is parallel to $SP$ and $SP=2AB=QA$. Thus, $QAPS$ is an isosceles trapezium and $SQ$ is parallel to $AP$.

We hence get $\angle SAP=\angle QPA=\angle QRA=\angle CAR$.

The last equality follows from the fact that $\angle QAR=90^{\circ}$, and $C$ is the midpoint of $QR$ so that $CA=CR=CQ$ for the right-angled triangle $QAR$.

It follows that $P,\,A,\,C$ are collinear.

 

[Opalg]

Here is the solution of other that I want to share:

Spoiler

Extend $RA$ such that $RA=AS$. Let $\angle QRA=\angle QPA=\theta$.

Observe that $QA$ is the perpendicular bisector of $RS$. Hence $QR=QS$ and $QRS$ is an isoceles triangle. Thus, $\angle QSA=\theta$. But then $\angle QSA=\angle QPA=\theta$. This implies that $Q,\,A,\,P,\,S$ all lie on a circle. In turn, we conclude that $\angle QAS=\angle QPS=90^{\circ}$.
Since $A$ is the midpoint of $RS$ (by construction) and $B$ is the midpoint of $PR$ (given), it follows that $AB$ is parallel to $SP$ and $SP=2AB=QA$. Thus, $QAPS$ is an isosceles trapezium and $SQ$ is parallel to $AP$.

We hence get $\angle SAP=\angle QPA=\angle QRA=\angle CAR$.

The last equality follows from the fact that $\angle QAR=90^{\circ}$, and $C$ is the midpoint of $QR$ so that $CA=CR=CQ$ for the right-angled triangle $QAR$.

It follows that $P,\,A,\,C$ are collinear.

That is a beautiful proof! I drew this diagram to help me understand it:
[sp]

View attachment 4354​

The aim is to show that the blue and red line segments are collinear. The green elements are those constructed for the proof. At the point $A$, the angle between $AP$ and $AS$ is $\theta$, the angle between $AS$ and $AQ$ is $90^\circ$, and the angle between $AQ$ and $AC$ is $90^\circ - \theta$. Their sum is $180^\circ$, as required. [/sp]