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

  • $\begingroup$ 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

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!!

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.