Prove that triangle PQR is isosceles

anemone

MHB POTW Director
Staff member
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

MHB Oldtimer
Staff member
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.

Messy solution using trigonometry (I hope someone can come with a more elegant approach):
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.

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anemone

MHB POTW Director
Staff member
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.

My solution:

 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.

solution :

Opalg

MHB Oldtimer
Staff member
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.

Albert

Well-known member
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.

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Opalg

MHB Oldtimer
Staff member
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.

Albert

Well-known member
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]

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Opalg

MHB Oldtimer
Staff member
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.

Opalg

MHB Oldtimer
Staff member
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.

Albert

Well-known member
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 )

Albert

Well-known member
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 ?

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Opalg

MHB Oldtimer
Staff member
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

Well-known member
MHB Math Scholar
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.

Deveno

Well-known member
MHB Math Scholar
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.

Albert

Well-known member
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