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Given an equilateral triangular pyramid (refer the below diagram) $\Delta ABC$ & $P$ is any point inside the triangle such that ${PA}^{2}={PB}^{2}+{PC}^{2}$, then $\angle BPC$ is -

enter image description here

I am unable to think of how to do this question

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  • $\begingroup$ This doesn't answer the question directly, but it might give you some good ideas: en.wikipedia.org/wiki/Pompeiu%27s_theorem $\endgroup$ – Joey Zou Jul 7 '16 at 16:18
  • $\begingroup$ Still No progress !! $\endgroup$ – Amritanshu Jul 7 '16 at 16:22
  • $\begingroup$ Is the theorem linked with the question $\endgroup$ – Amritanshu Jul 7 '16 at 16:28
  • $\begingroup$ There is an idea in the proof that you can apply here. $\endgroup$ – Joey Zou Jul 7 '16 at 16:29
  • $\begingroup$ Which idea?? Can you tell me please $\endgroup$ – Amritanshu Jul 7 '16 at 16:29
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Here's a complete solution. (I think)

Define an equilateral triangle on the coordinate system as follows:

$A=(0,{\sqrt3\over 2}),B=(0.5,0),C=(-0.5,0),P=(x,y)$

By the requirement of $P$$$y^2 +(x+0.5)^2+y^2+(x-0.5)^2=x^2+(y-{\sqrt3\over 2})^2$$$$\implies \left(y+{\sqrt3\over 2}\right)^2+x^2=1 $$

$\therefore$ The locus of $P$ is the circle with center $\left( 0,-{\sqrt3\over 2}\right)$ and radius $1$.

Let the center be $O$ So $\angle BPC=1/2(\angle BOC)=150^{\circ}$ ( Knowing the coordinates of $ B;O;C$) Problem SOLVED!!

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  • $\begingroup$ I like this approach, but your last sentence is off. Notice that $B$ and $C$ are also on this circle, so $\angle BPC$ is uniquely determined by the measure of the angle subtended from the arc going from $B$ to $C$. $\endgroup$ – Joey Zou Jul 7 '16 at 18:31
  • $\begingroup$ @JoeyZou Thanks. I didn't notice $\endgroup$ – Qwerty Jul 7 '16 at 18:52
  • $\begingroup$ @JoeyZou Problem Solved $\endgroup$ – Qwerty Jul 7 '16 at 18:58
  • $\begingroup$ @Amritanshu Check my solution $\endgroup$ – Qwerty Jul 7 '16 at 19:00
  • $\begingroup$ @Qwerty Can you generalize it ? $\endgroup$ – A---B Jul 7 '16 at 19:07
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Hint: rotate the triangle $60^{\circ}$ clockwise around $B$, so that $A$ is rotated onto $C$, and let $P'$ be the image of $P$ under this rotation. Can you show the following statements:

  1. $\angle PBP' = 60^{\circ}$
  2. $\triangle PBP'$ is equilateral
  3. $PP' = PB$
  4. $P'C = PA$
  5. $\triangle P'PC$ is a right triangle, with a right angle at $P$
  6. $\angle BPC = \angle BPP' + \angle P'PC$

If you can show these statements, then the answer should follow.

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  • $\begingroup$ can u upload an image too .... it then this question would be a piece of cake $\endgroup$ – Amritanshu Jul 7 '16 at 16:52
  • $\begingroup$ @Amritanshu I agree an image would make the above statements quite clear, but unfortunately I can't make one right now. You're going to have to make it yourself. Sorry! $\endgroup$ – Joey Zou Jul 7 '16 at 16:55
  • $\begingroup$ i am unable to understand the orientation of $P'$ $\endgroup$ – Amritanshu Jul 7 '16 at 17:02
  • $\begingroup$ Here is a picture of what I mean. Notice that the entire triangle has been rotated $60^{\circ}$ clockwise, with $C$ rotated onto $A$, $A$ rotated onto $A'$, and $P$ rotated onto $P'$. $\endgroup$ – Joey Zou Jul 7 '16 at 17:05

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