Find $\angle BPC$ in Triangle ABC with $\angle ACB=\angle ABC=80^\circ$

[anemone]

In triangle ABC, $\angle ACB=\angle ABC=80^\circ$ and $P$ is on the line segment $AB$ such that $BC=AP$. Find $\angle BPC$.

 

[Amer]

Spoiler

Let $AB = 2x = AC$ then $AP = PB = x$, use the cosine law in the triangle $APC$ we found that
$PC^2 = x^2 + 4x^2 - 2(x)(2x) \cos 20 = x^2 ( 5 - 4\cos 20)\Rightarrow PC = x \sqrt{5 - 4\cos 20} $

again

$AC^2 = PC^2 + AP^2 - 2(PC)(AP) \cos \angle APC$

$3x^2 = x^2 ( 5 -4 \cos 20) + - 2x^2 \sqrt{5 - 4 \cos 20} \cos \angle APC $

$3 = 5 - 4 \cos 20 - 2 \sqrt{5 - 4 \cos 20} \cos \angle APC $

$\cos APC = \frac{1 - 2\cos 20}{\sqrt{5 - 4 \cos 20}}= - 0.789$

$\angle BPC = 180 - \arccos( - 0.789 ) $

 

[anemone]

Spoiler

Let $AB = 2x = AC$ then $AP = PB = x$, use the cosine law in the triangle $APC$ we found that
$PC^2 = x^2 + 4x^2 - 2(x)(2x) \cos 20 = x^2 ( 5 - 4\cos 20)\Rightarrow PC = x \sqrt{5 - 4\cos 20} $

again

$AC^2 = PC^2 + AP^2 - 2(PC)(AP) \cos \angle APC$

$3x^2 = x^2 ( 5 -4 \cos 20) + - 2x^2 \sqrt{5 - 4 \cos 20} \cos \angle APC $

$3 = 5 - 4 \cos 20 - 2 \sqrt{5 - 4 \cos 20} \cos \angle APC $

$\cos APC = \frac{1 - 2\cos 20}{\sqrt{5 - 4 \cos 20}}= - 0.789$

$\angle BPC = 180 - \arccos( - 0.789 ) $

Hi Amer! Thanks for participating but I'm sorry...your answer isn't correct...:(

 

[Amer]

Hi Amer! Thanks for participating but I'm sorry...your answer isn't correct...:(

I understand the question in a wrong way I will edit my post :)

 

[johng1]

I'm quite certain I know the answer, but I can't prove it. If the following is the way you went, I'd appreciate a hint.

Spoiler

When I carefully draw the figure, I see that the circum center O of triangle PBC has the property that triangle OBC is equilateral. From this the desired angle is obvious, but why is OBC equilateral?

 

[Greg]

Trig proof:

Spoiler

WLOG, let \(\displaystyle \overline{BC}=1\), so \(\displaystyle \overline{AP}=1\). Construct segment \(\displaystyle \overline{PD}\) such that \(\displaystyle D\) lies on \(\displaystyle \overline{AC}\) and \(\displaystyle \overline{PD}\) is parallel to \(\displaystyle \overline{BC}\).

\(\displaystyle \overline{PD}=2\cos80\), so \(\displaystyle \dfrac{1}{\overline{PB}+1}=2\cos80\implies\overline{PB}=\dfrac{1-2\cos80}{2\cos80}\)

Let \(\displaystyle \angle{BPC}=\theta\). Then

\(\displaystyle \begin{align*}\tan(\theta-10)&=\dfrac{\dfrac{1+2\cos80}{2}}{\dfrac{\cos10-2\cos10\cos80}{2\cos80}} \\
&=\dfrac{(1+2\cos80)\cos80}{\cos10-2\cos10\cos80} \\
&=\dfrac{\cos80+2\cos^280}{\cos10-\cos70} \\
&=\dfrac{\cos80+2\cos^280}{\sin40} \\
&=\dfrac{\cos80+1-\cos20}{\sin40} \\
&=\dfrac{1-\sin50}{\sin40} \\
&=\csc40-\cot40 \\
&=\tan20 \\
\theta-10&=20 \\
\theta=30^\circ\end{align*}\)

 

[Albert1]

Geometric Solution:

Spoiler

View attachment 4896

 

[MarkFL]

Geometric Solution:

Albert, please post attachments inline so that you can hide them using the spoiler tags. Also, I find your solution to be very hard to follow without the supporting computations. :)

 

[anemone]

I'm quite certain I know the answer, but I can't prove it. If the following is the way you went, I'd appreciate a hint.

Spoiler

When I carefully draw the figure, I see that the circum center O of triangle PBC has the property that triangle OBC is equilateral. From this the desired angle is obvious, but why is OBC equilateral?

Hi johng,

Spoiler

I appreciate you taking the time to try this challenge out and I'm sorry as the solutions that I have don't use the method as you proposed so I think you have to rely on your own to figure out why OBC is equilateral...:)

And thanks to greg1313 for your trigonometric proof and Albert, I actually have hope to see your geometric proof to be back up by facts. :D

 

[Albert1]

Spoiler

Two circles with centers C and E
We have BC=CE=EP=AP
then $\angle A=\angle 1=20^o=2\angle 2$
$\therefore \angle 2=10^o,\,\, and \,\,\angle APE=140^o$
$\angle APC=150^o$ , so we get
$\angle BPC=30^o$
I am sure my geometric proof is back up by facts.

 

[johng1]

Albert,
I believe you have made an error in your proof. Here's why I think this:

Spoiler

 

[Albert1]

Albert,
I believe you have made an error in your proof. Here's why I think this:

from my diagram point E is on segment AC
no error I am sure,please check again

 

[Opalg]

In triangle ABC, $\angle ACB=\angle ABC=80^\circ$ and $P$ is on the line segment $AB$ such that $BC=AP$. Find $\angle BPC$.

The 80-80-20 triangle has a very rich structure, and is the source of many good problems. I remember coming across some of them when I was a teenager, 60 years ago. I found this particular problem on the internet, here, with links to seven different solutions. They are all interesting, especially Solutions #1–#3, which all use different (ingenious and elegant) geometric constructions.

 

[johng1]

Albert,
My apologies. In your diagram, I didn't understand why |PE|=|BC| (I still can't see this). I then got all excited and drew the wrong diagram.
OpalG, I read the first three solutions in your link and I agree completely that they are elegant and ingenious. Thanks much for the link.

 

[Albert1]

Albert,
My apologies. In your diagram, I didn't understand why |PE|=|BC| (I still can't see this). I then got all excited and drew the wrong diagram.
OpalG, I read the first three solutions in your link and I agree completely that they are elegant and ingenious. Thanks much for the link.

please see my post #10