Prove that triangle PQR is isosceles

[anemone]

A triangle PQR has the following property:

There is an interior point $A$ such that $\angle APQ=10^{\circ}$, $\angle AQP=20^{\circ}$, $\angle ARP=30^{\circ}$ and $\angle APR=40^{\circ}$.

Prove that the triangle PQR is isosceles.

 

[Opalg]

A triangle PQR has the following property:

There is an interior point $A$ such that $\angle APQ=10^{\circ}$, $\angle AQP=20^{\circ}$, $\angle ARP=30^{\circ}$ and $\angle APR=40^{\circ}$.

Prove that the triangle PQR is isosceles.

View attachment 1320​

Messy solution using trigonometry (I hope someone can come with a more elegant approach):
[sp]Call the angles of the triangle $\alpha,\,\beta,\,\gamma$, at $P,\,Q,\,R$ respectively. By the sine rule in triangle $APQ$, $\dfrac{PQ}{AP} = \dfrac{\sin30^\circ}{\sin20^\circ}.$ By the sine rule in triangle $APR$, $\dfrac{PR}{AP} = \dfrac{\sin70^\circ}{\sin30^\circ}.$ Therefore $\dfrac{PQ}{PR} = \dfrac{\sin30^\circ}{\sin20^\circ}\,\dfrac{\sin30^\circ}{\sin70^\circ} = \dfrac1{4\sin20^\circ\sin70^\circ}.$

Now by a multiple angle formula, $\sin20^\circ\sin70^\circ = \frac12(\cos(70^\circ - 20^\circ) - \cos(70^\circ + 20^\circ)) = \frac12\cos50^\circ.$ So $\dfrac{PQ}{PR} = \dfrac1{2\cos50^\circ}.$

By the sine rule in triangle $PQR$, $\dfrac{PQ}{PR} = \dfrac{\sin\gamma}{\sin\beta}.$ Hence $\dfrac1{2\cos50^\circ} = \dfrac{\sin\gamma}{\sin\beta},$ or $\sin\beta = 2\cos50^\circ \sin\gamma.$

But $\beta = 180^\circ - \alpha - \gamma$, and $\alpha = 10^\circ + 40^\circ = 50^\circ$. So $\beta = 130^\circ - \gamma$, and $$\sin\beta = \sin130^\circ\cos\gamma - \cos 130^\circ\sin\gamma = \sin50^\circ\cos\gamma + \cos 50^\circ\sin\gamma.$$ Therefore $\sin50^\circ\cos\gamma + \cos 50^\circ\sin\gamma = 2\cos50^\circ \sin\gamma,$ so $\sin50^\circ\cos\gamma = \cos 50^\circ\sin\gamma.$ Thus $\tan50^\circ = \tan\gamma$, so $\gamma = 50^\circ = \alpha$, which shows that $PQR$ is isosceles.[/sp]

 

[anemone]

Thanks for participating, Opalg!

I too used a trigonometric approach, but my solution is slightly different than yours. So I will show my work here.

Spoiler

My solution:
View attachment 1324

By applying the Sine Rule to the triangle APR, we get:	By applying the Sine Rule to the triangle PQA, we get:
\(\displaystyle \frac{PR}{\sin 110 ^{\circ}}=\frac{PA}{\sin 30 ^{\circ}}\) \(\displaystyle PA=\frac{PR}{2\sin 110 ^{\circ}}\)	\(\displaystyle \frac{PA}{\sin 20 ^{\circ}}=\frac{PQ}{\sin 150 ^{\circ}}\) \(\displaystyle PA=2PQ\sin 20 ^{\circ}\)	This implies \(\displaystyle \frac{PR}{PQ}=4\sin 20 ^{\circ}\sin 110 ^{\circ}=4\sin 20 ^{\circ}\cos 20 ^{\circ}=2\sin40^{\circ}\)---(*)

Now, apply the Sine Rule to the triangle PQR, we obtain:

\(\displaystyle \frac{PR}{\sin (20+x)^{\circ}}=\frac{PQ}{\sin (110-x)^{\circ}}\)

\(\displaystyle \frac{PR}{PQ}=\frac{\sin (20+x)^{\circ}}{\sin (110-x)^{\circ}}=\frac{\sin (20+x)^{\circ}}{\cos (20-x)^{\circ}}\)---(**)

By equating the equation (*) and (**) and solve them for $x$, we get

\(\displaystyle \frac{\sin (20+x)^{\circ}}{\cos (20-x)^{\circ}}=2\sin40^{\circ}\)

\(\displaystyle \sin (20+x)^{\circ}=2\sin40^{\circ}\cos (20-x)^{\circ}\)

\(\displaystyle \sin (20+x)^{\circ}=\sin (60-x)^{\circ}+\sin (20+x)^{\circ}\)

\(\displaystyle \sin (60-x)^{\circ}=0\)

It's obvious that $x$ must take the value $60^{\circ}$ and it gives $\angle QPR=50^{\circ}$ and $\angle QRP=50^{\circ}$ and we prove hereby that triangle PQR is isosceles.

 

[Albert1]

solution :

View attachment 1328

 

[Opalg]

solution :

View attachment 1328

That looks like a neat construction, but I do not understand it. It starts by asserting that the points $P,C,B,R$ lie on a circle with centre $O$ on the line $PR$. In that case, $O$ must be the midpoint of $PR$, and you are saying that the circle with diameter $PR$ meets the line $QR$ at a point $B$ which is collinear with $P$ and $A$. I cannot see why that should be true. If it is true, then the problem is immediately solved without any further work, because then $\angle PBR$ is a right angle (angle in a semicircle) and therefore $\angle BRP = 90^\circ - \angle BPR = 50^\circ$, which shows that triangle $PQR$ is isosceles.

 

[Albert1]

That looks like a neat construction, but I do not understand it. It starts by asserting that the points $P,C,B,R$ lie on a circle with centre $O$ on the line $PR$. In that case, $O$ must be the midpoint of $PR$, and you are saying that the circle with diameter $PR$ meets the line $QR$ at a point $B$ which is collinear with $P$ and $A$. I cannot see why that should be true. If it is true, then the problem is immediately solved without any further work, because then $\angle PBR$ is a right angle (angle in a semicircle) and therefore $\angle BRP = 90^\circ - \angle BPR = 50^\circ$, which shows that triangle $PQR$ is isosceles.

the key point here is that $ \angle PAR=110^o$ as given ,and we may construct $\angle PAD=50^o$,and we get $ \overset{\frown} {PD}=80^o$
extend PA and meet circle O at point B,indeed point P,C,B,R lie on a circle with centre [FONT=MathJax_Math]O[/FONT] on the line [FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]R[/FONT].in that case [FONT=MathJax_Main]∠[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Math]R[/FONT] is a right angle (angle in a semicircle) and therefore [FONT=MathJax_Main]∠[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]90[/FONT][FONT=MathJax_Main]∘[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]∠[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]50[/FONT][FONT=MathJax_Main]∘[/FONT], which shows that triangle [FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]Q[/FONT][FONT=MathJax_Math]R[/FONT] is isosceles.

 

[Opalg]

the key point here is that $ \angle PAR=110^o$ as given ,and we may construct $\angle PAD=50^o$,and we get $ \overset{\frown} {PD}=80^o$
extend PA and meet circle O at point B,indeed point P,C,B,R lie on a circle with centre [FONT=MathJax_Math]O[/FONT] on the line [FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]R[/FONT].in that case [FONT=MathJax_Main]∠[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Math]R[/FONT] is a right angle (angle in a semicircle) and therefore [FONT=MathJax_Main]∠[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]90[/FONT][FONT=MathJax_Main]∘[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]∠[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]50[/FONT][FONT=MathJax_Main]∘[/FONT], which shows that triangle [FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]Q[/FONT][FONT=MathJax_Math]R[/FONT] is isosceles.

I am still not convinced by this argument. You have not explained why P,C,B,R lie on a circle with centre on the line [FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]R[/FONT]. Also, you mention the arc $ \overset{\frown} {PD}$ before explaining how the circle is constructed.

 

[Albert1]

1. extend line PA and meet QR at point B
2. line AC and BO meet at point D (here O is the midpoint of PR ,point C on PQ,and between Pand Q)
3 $\angle BPC=\angle BRC=10^o$
4. triangle PCE and triangle RBE are similar (here point E is the intersection of BP and RC
4 if $\angle PAD=50^o$ ,it is easy to construct it, then $\angle BDC=10^o$ (for arc PD and arc BR=$80^o)$
now can we say P,C,B,R lie on a circle with centre on the line [FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]R ?
[/FONT]

 

[Opalg]

Thank you for your patience, but I am no closer to understanding this.

1. extend line PA and meet QR at point B

No problem about that. It defines the position of $B$ on the line $QR$.

2. line AC and BO meet at point D (here O is the midpoint of PR ,point C on PQ,and between Pand Q)

That defines $O$, and it defines the line $BO$. But $C$ has not yet been defined. As far as I know at this stage, it could be any point on $PQ$. Consequently, the line $AC$, and therefore the position of $D$, does not seem to have been fixed yet.

3 $\angle BPC=\angle BRC=10^o$

Where does that come from? The only way I can interpret this is that the condition $\angle BRC=10^\circ$ defines the position of $C$ on the line $PQ$, because this is the first statement that pins down the position of $C$. If you define the position of $C$ in this way, then it is already clear that $P,\,C,\,B,\,R$ lie on a circle (because the angles $\angle CPB$ and $\angle CRB$ are in the same segment). But it does not appear to imply that $O$ is the centre of this circle.

4. triangle PCE and triangle RBE are similar (here point E is the intersection of BP and RC

True, given what has gone before, but I don't see the relevance of it.

4 if $\angle PAD=50^o$ ,it is easy to construct it, then $\angle BDC=10^o$ (for arc PD and arc BR=$80^o)$

If $C$ is fixed by the fact that $\angle BRC=10^\circ$, then the direction of the line $CA$ is already fixed, and so $\angle PAD$ is determined. But I do not see why it should be $50^\circ$. Alternatively, if you define the position of $C$ by the condition $\angle PAD=50^\circ$, then I do not see where the condition $\angle BRC=10^\circ$ came from.

now can we say P,C,B,R lie on a circle with centre on the line [FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]R ?[/FONT]

I still don't see why the centre should be at $O$.

Sorry to keep on about this. I would very much like to have a geometric proof of this result, but I am far from seeing how this approach works.

 

[Albert1]

View attachment 1337

 

[Opalg]

View attachment 1337

This just seems to get worse. You are now defining $A$ as the point where the lines $CD$ and $PB$ meet. But there is nothing to explain why this should be the same point as the $A$ defined in the statement of the problem. Unless you can prove that these two points are the same, I do not think that your proof is valid.

 

[Albert1]

Giving point P is an inner point of triangle ABC

If $\angle APB=X ,\angle APC=Y ,\angle BPC=Z$

Of course X+Y+Z=$360^o$

How many points will satisfy this restriction ?(this I real want to know )

 

[Albert1]

This just seems to get worse. You are now defining $A$ as the point where the lines $CD$ and $PB$ meet. But there is nothing to explain why this should be the same point as the $A$ defined in the statement of the problem. Unless you can prove that these two points are the same, I do not think that your proof is valid.

point A is an inner point of triangle PQR giving $\angle PAQ=150^o, \angle QAR=100^o ,
\angle PAR=110^o$
I think there is only one point which will meet this restriction
$A$ as the point where the lines $CD$ and $PB$ meet (this must hold )
in my proof ,at first I found the location of this point A,and then assure the point satisfy all the condition given ,also I prove $\triangle PQR$ is isosceles
in fact the original post did not define the location of point A ,it only said A is an inner point of triangle PQR , and my definition of point A also meets
if I adjust anemone's post as following :

A triangle PQR has the following property:

There is an interior point $A$ such that $\angle APQ=10^{\circ}$, $\angle AQP=20^{\circ}$, $\angle ARP=30^{\circ}$ and $\angle APR=40^{\circ}$.

Prove that the triangle PQR is isosceles.and try to find the location of the point A

Do you think my solution will do or not ?

 

[Opalg]

I think we shall have to agree to differ on this question. I agree that there is a unique point $A$ inside the triangle $PQR$ such that the lines from $A$ to the vertices make angles of $100^\circ,\,110^\circ,\,150^\circ$. I also accept that your construction does in fact pinpoint this point $A$. But I do not believe that you have proved this. In your construction, you define the points $B$ and $C$ by drawing a circle with diameter $PR$, and you define a point that you call $A$ as the intersection of the lines $PB$ and $CD$. But you do not say anything to prove that your point $A$ makes those angles $100^\circ,\,110^\circ,\,150^\circ$ with the vertices of the triangle. So I do not see that you can claim that your point $A$ coincides with the point $A$ as described in the statement of the problem.

 

[Deveno]

Albert, Opalg is not the only one confused by your post. I spent several hours trying to figure out why PB and PD must be perpendicular (that is, why arc PD = 80°).

Alternatively, if we define angle OPD to be 50, then I do not see how we can deduce that AD is perpendicular to PR.

Once either one of these is established, it is easy to show that triangle PQR is isosceles.

 

[Albert1]

View attachment 1339

 

[Deveno]

I still have a couple of questions.

I think its easy to establish angle PAR = 110, after all we are given angles ARP and APR.

Similarly, since we are given angles APQ and AQP, it is clear angle PAQ = 150.

So I am willing to accept angle QAR = 100.

Here is where I have a problem:

To establish that angle ARQ = 20, I think we need to know that PA is perpendicular to RQ, which is equivalent to establishing A is collinear with PB. Can you convince me of this?

In fact (and this troubles me greatly), if your proof WAS correct, we must have:

angle AQR = 60, implying that:

angle PQR = 80, which certainly is NOT the same as angle QRP = 50.

 

[Albert1]

I still have a couple of questions.

I think its easy to establish angle PAR = 110, after all we are given angles ARP and APR.

Similarly, since we are given angles APQ and AQP, it is clear angle PAQ = 150.

So I am willing to accept angle QAR = 100.

Here is where I have a problem:

To establish that angle ARQ = 20, I think we need to know that PA is perpendicular to RQ, which is equivalent to establishing A is collinear with PB. Can you convince me of this?

In fact (and this troubles me greatly), if your proof WAS correct, we must have:

angle AQR = 60, implying that:

angle PQR = 80, which certainly is NOT the same as angle QRP = 50.

please look at post #10 and post #16 carefully the definition of points ,and all your questions will be answered