Calculating Pyramid Height from 3 Circles

[Mentallic]

Given 3 circles each of radius r, stacked up into a triangular pyramid shape, find the height of the entire structure. This might be expressed more clearly in a graphical form:

http://img206.imageshack.us/img206/3646/stackcircleseb0.png
http://g.imageshack.us/img206/stackcircleseb0.png/1/

I haven't been able to answer this question for years! Any suggestions as to how to find the height h would be appreciated.

 

[daniel_i_l]

Draw a triangle connecting the centers of the circles. Calculate the height of the triangle using trivial trigonometry. Can you proceed from here?

 

[Mentallic]

Ahh it would form an equilateral triangle, giving me the angles I have searched long and hard for to find!

[tex]h=\frac{6+\sqrt{3}}{2}r[/tex]

That one line:

Draw a triangle connecting the centers of the circles.

all it took to build that bridge I've been longing for. Thankyou very much, I greatly appreciate that one line of help

 

[KnowPhysics]

I too worked out some and i got this:
h=2r+sqrt((2r*2r)-(r*r));

 

[D H]

Ahh it would form an equilateral triangle, giving me the angles I have searched long and hard for to find!

[tex]h=\frac{6+\sqrt{3}}{2}r[/tex]

You are correct that the triangle is an equilateral triangle. Your derived height is however incorrect. If you arrived at this result we can help find the error in your reasoning.

 

[Mentallic]

Seems I was too excited and skipped a whole lot of rational thinking.
I knew it would've been a stupid mistake, and while I don't know what I did wrong yesterday (no point in trying to find out), I have the correct answer now.

[tex](2+\sqrt{3})r[/tex]

which is equivalent to KnowPhysics' after a bit of manipulation:

[tex]2r+\sqrt{(2r)(2r)-(r)(r)}[/tex]

[tex]2r+\sqrt{4r^2-r^2[/tex]

[tex]2r+\sqrt{3r^2}[/tex]

[tex]2r+\sqrt{3}r[/tex]

Thanks