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Prove that S lies on the line AB

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anemone

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Feb 14, 2012
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Let $PQRS$ be a rhombus with $\angle Q=60^{\circ}$. $M$ is a point inside triangle $PSR$ such that $\angle PMR=120^{\circ}$. Let lines $QP$ and $RM$ intersect at $A$ and lines $QR$ and $PM$ intersect at $B$. Prove that $S$ lies on the line $AB$.
 

Pranav

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Nov 4, 2013
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Let $PQRS$ be a rhombus with $\angle Q=60^{\circ}$. $M$ is a point inside triangle $PSR$ such that $\angle PMR=120^{\circ}$. Let lines $QP$ and $RM$ intersect at $A$ and lines $QR$ and $PM$ intersect at $B$. Prove that $S$ lies on the line $AB$.

Since $PQRS$ is a rhombus, we have $\angle PQR=\angle PSR=60^{\circ}$.

It can be easily seen that $M$ is the circumcentre of $\Delta PSR$ and $\Delta PSR$ is an equilateral triangle.

Also, $QPA$ and $QRB$ are tangents to the circumcircle of $\Delta PSR$.

Hence, $\angle PSA=60^{\circ}$ and $\angle RSB=60^{\circ}$.

The angles $\angle PSA,\angle PSR$ and $\angle RSB$ sum to $180^{\circ}$, hence $S$ lies on$AB$.
 
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anemone

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Feb 14, 2012
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Since $PQRS$ is a rhombus, we have $\angle PQR=\angle PSR=60^{\circ}$.

It can be easily seen that $M$ is the circumcentre of $\Delta PSR$ and $\Delta PSR$ is an equilateral triangle.

Also, $QPA$ and $QRB$ are tangents to the circumcircle of $\Delta PSR$.

Hence, $\angle PSA=60^{\circ}$ and $\angle RSB=60^{\circ}$.

The angles $\angle PSA,\angle PSR$ and $\angle RSB$ sum to $180^{\circ}$, hence $S$ lies on$AB$.
Thanks for participating, Pranav!:)

But...I don't follow your reasoning because I don't understand why the equilateral triangle $PSR$ and that $QPA$ and $QRB$ are tangents to the circumcircle of $\Delta PSR$ imply $\angle PSA=60^{\circ}$ and $\angle RSB=60^{\circ}$.

I am definitely not saying your approach is wrong, it just that I don't get it, could you elaborate more on that, please? :eek:
 

magneto

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Jan 3, 2014
102
Given the line ${AB}$ is tangent to the circle at $S$, there is a theorem that states:

The angle between a tangent (${AB}$) and a chord (${PS}$) is equal to an angle subtended by the chord ($\angle PRS$).

Since $\angle PRS$ is 60 degrees since it is part of an equilateral triangle, so is angle $ASP$. The argument for angle $BSR$ being 60 degrees is analogous.
 
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anemone

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Feb 14, 2012
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Given the line ${AB}$ is tangent to the circle at $S$, there is a theorem that states:...
Thanks for replying, magneto but, as far as I can tell, if we know beforehand that $AB$ is a tangent to the circle at $S$, then S certainly lies on AB, and there is nothing to be proved...what do you think?(Smile)
 

Pranav

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Nov 4, 2013
428
I am definitely not saying your approach is wrong, it just that I don't get it, could you elaborate more on that, please? :eek:
Sure! :)

Do you agree that $M$ is the circumcentre of $PSR$? If so, it can be easily shown that the lines are tangent.

I hope that helps.
 
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anemone

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Feb 14, 2012
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Sure! :)

Do you agree that $M$ is the circumcentre of $PSR$? If so, it can be easily shown that the lines are tangent.

I hope that helps.
Yes, I can see QPA and QRB are tangent to the circle which has its center also a circumcenter of triangle PSR, I just don't understand why that implies the angles of ASP, RSB as 60 degree and hence ASB is a straight line. I admit that I don't see how this is obvious for me.(Tmi)
 

Pranav

Well-known member
Nov 4, 2013
428
Yes, I can see QPA and QRB are tangent to the circle which has its center also a circumcenter of triangle PSR, I just don't understand why that implies the angles of ASP, RSB as 60 degree and hence ASB is a straight line. I admit that I don't see how this is obvious for me.(Tmi)


(The diagram isn't too accurate, sorry about that. :eek: )

If we draw a tangent from S, it would form an acute angle $60^{\circ}$ with RS. Since the tangent at S and QRB forms the same angle with RS, they must intersect at the same point.

If you are not satisfied, I am not sure, you can assume that tangent at S intersect PM at B' and then prove BB'=0 or BM=B'M.

I hope it helps. :)
 
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anemone

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Feb 14, 2012
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Hey Pranav, I see it now.:eek: Thanks for all the clarification posts and thank you for participating!