MHB Prove three points P, A and C are collinear

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The discussion revolves around proving that points P, A, and C are collinear in a triangle PQR with point A inside it, given specific angle conditions and the relationship QA = 2AB. A typo in the original problem was acknowledged, which altered its meaning, but the corrected angles are now specified. The proof involves geometric relationships and the construction of angles at point A, demonstrating that the angles sum to 180 degrees, confirming collinearity. Visual aids, including diagrams, were shared to enhance understanding of the proof. The conversation highlights the collaborative nature of problem-solving in mathematics.
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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.
 
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anemone said:
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.
from the diagram it is clear :
[FONT=MathJax_Math-Web]P[FONT=MathJax_Main-Web],[FONT=MathJax_Math-Web]A[FONT=MathJax_Main-Web],[FONT=MathJax_Math-Web]C are collinear.

View attachment 4345
 

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

Hello Albert,

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
 
explanation of previous diagram:
$\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
 
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:

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

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

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Thank you Albert for your solution!

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 $\angle CAR=y$?:confused: I mean, of course $\angle CAR=y$, but, don't we have to prove that first before claiming such is the case?
 
  • #10
anemone said:
Thank you Albert for your solution!

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 $\angle CAR=y$?:confused: I mean, of course $\angle CAR=y$, but, don't we have to prove that first before claiming such is the case?
point A on circle C
AC=CR=QC, for
$\angle QAR=90^o$
 
  • #11
anemone said:
Thank you Albert for your solution!

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 $\angle CAR=y$?:confused: I mean, of course $\angle CAR=y$, but, don't we have to prove that first before claiming such is the case?
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 - - -
 
  • #12
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
 

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  • #13
Thanks Albert for your explanation and the second submission of the solution.

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

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.
 

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  • #14
anemone said:
Here is the solution of other that I want to share:

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]

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]
 

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