3

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

Amritanshu
  • 961

2 Answers2

4

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!!

Qwerty
  • 6,165
  • 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 07 '16 at 18:31
  • @JoeyZou Thanks. I didn't notice – Qwerty Jul 07 '16 at 18:52
  • @JoeyZou Problem Solved – Qwerty Jul 07 '16 at 18:58
  • @Amritanshu Check my solution – Qwerty Jul 07 '16 at 19:00
  • @Qwerty Can you generalize it ? –  Jul 07 '16 at 19:07
  • @ritwiksinha Generalise for what? – Qwerty Jul 07 '16 at 19:27
  • @Qwerty Without knowing the coordinates of the vertices. –  Jul 07 '16 at 19:34
  • @ritwiksinha i don't think. This property is that of an equilateral triangle, and I have just placed it adequately to make calculations easy If you make the vertices more general , Calculations will be difficult. Try it on your own if you want to. I didn't want to make things unnecessarily cumbersome. – Qwerty Jul 07 '16 at 19:38
  • @Qwerty But evidence is not a proof. But i will surely try to do with general equilateral triangle. –  Jul 07 '16 at 19:56
3

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.

Joey Zou
  • 8,466
  • can u upload an image too .... it then this question would be a piece of cake – Amritanshu Jul 07 '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 07 '16 at 16:55
  • i am unable to understand the orientation of $P'$ – Amritanshu Jul 07 '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 07 '16 at 17:05