# Finding angle in an equilateral triangular pyramid

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 -

I am unable to think of how to do this question

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

$A=(0,{\sqrt3\over 2}),B=(0.5,0),C=(-0.5,0),P=(x,y)$
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$) Problem SOLVED!! • I like this approach, but your last sentence is off. Notice that$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$. – Joey Zou Jul 7 '16 at 18:31 • @JoeyZou Thanks. I didn't notice – Qwerty Jul 7 '16 at 18:52 • @JoeyZou Problem Solved – Qwerty Jul 7 '16 at 18:58 • @Amritanshu Check my solution – Qwerty Jul 7 '16 at 19:00 • @Qwerty Can you generalize it ? – 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$If you can show these statements, then the answer should follow. • can u upload an image too .... it then this question would be a piece of cake – Amritanshu Jul 7 '16 at 16:52 • @Amritanshu I agree an image would make the above statements quite clear, but unfortunately I can't make one right now. You're going to have to make it yourself. Sorry! – Joey Zou Jul 7 '16 at 16:55 • i am unable to understand the orientation of$P'$– Amritanshu Jul 7 '16 at 17:02 • 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'\$. – Joey Zou Jul 7 '16 at 17:05