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

AI Thread Summary
In triangle ABC, where angles ACB and ABC both measure 80 degrees, point P is located on segment AB such that BC equals AP. Participants in the discussion are attempting to find the measure of angle BPC, with some providing trigonometric and geometric proofs. There are disagreements regarding the correctness of certain proofs, particularly concerning the relationships between segments in the triangle. Several users express interest in elegant geometric constructions related to this problem, referencing external solutions for further insights. The conversation highlights the complexity and rich structure of the 80-80-20 triangle problem.
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In triangle ABC, $\angle ACB=\angle ABC=80^\circ$ and $P$ is on the line segment $AB$ such that $BC=AP$. Find $\angle BPC$.
 
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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 ) $
 
Amer said:
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...:(
 
anemone said:
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 :)
 
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.
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?
 
Trig proof:

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

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

Let $$\angle{BPC}=\theta$$. Then

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

Attachments

  • measure of Angle BPC=30.jpg
    measure of Angle BPC=30.jpg
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Albert said:
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. :)
 
johng said:
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.
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,
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
 
  • #10
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.
 
Last edited:
  • #11
Albert,
I believe you have made an error in your proof. Here's why I think this:

21c8qib.png
 
Last edited by a moderator:
  • #12
johng said:
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
 
  • #13
anemone said:
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.
 
  • #14
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.
 
  • #15
johng said:
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
 

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