# Thread: A pyramid off three balls

1. ## A pyramid off three balls

Three spheres with the radius R are lying on a plane, so that they all tangent each other. A forth Sphere, also with the radius of R, are lying upon the three others, (so that a pyramid arises). Compute the length between the middle point of the fourth sphere to the underlying plane, (whereas we assume the friction to be big enough to prevent the speheres from gliding).

Anyone?
This is for our repetetition course of the high school-math given at the university, so I thought this was the right place even though I was somewhat unseure.

2. Hello, a4swe!

I'll give you my game plan.
I'd give you a complete solution, but it's getting late.
I can work on it tomorrow . . . *yawn*

Three spheres with the radius R are lying on a plane, so that they all tangent each other.
A fourth sphere, also with the radius of R, are lying upon the three others,
so that a pyramid os fomed.
Compute the length between the center of the fourth sphere to the underlying plane.

The centers of the four spheres form a regular tetrahedron of edge $\displaystyle 2R.$

The top vertex of the tetrahedron is the center of the fourth sphere.

The bottom vertices of the tetrahedron are the centers of the lower spheres.
These centers are $\displaystyle R$ units from the plane.

We need only the altitude of the regular tetrahedron.

Can you finish it now?

3. Side view of the situation (lousy sketch I know)

4. Your problem is equivalent to this 2D problem with circles and trying to find the height from the bottom. If you are unsure why just say and I will try to explain further.

Each circle has radius R so the lines connecting the centres of the circles all have length 2R. They are straight lines as either side of the tangent the line is at right angles to the tangent.

This means that the triangle formed has equal sides and is therefore equilateral

We can find the height of the triangle by splitting the triangle into 2 right angles triangles and using pythagoras

Hypotenuse 2R
Base R

Height = $\displaystyle \sqrt{(2R)^2 - R^2} = \sqrt{3R^2} = \sqrt{3}R$

Add the radius to this height and you get the height of the top circle so get answer

$\displaystyle \sqrt{3}R + R = (\sqrt{3} + 1)R$

5. Thank you Glaysher.
I got it just by taking a glance at your drawing and after that solving it was just a few seconds away.
I realy have to train my skill of making pictures of my thoughts and vice versa.

6. I just did it the glaysher way and got the same answer as him after calculationg it, yay I thought and went to answer page in my textbook just to find I was 0.09R off!
The right answer was $\displaystyle (1+\frac{2}{3}\sqrt{6})R$.
I will now do it the Soroban way (if it is not the same) to see if I can get a better answer.
If not, anyone here know why I am off?

Edit: And now that I have looked throught that solution I can say that I find 'em essentialy the same and therefore one of the following are true:
1.My textbook are wrong.
2.The answers in this thread is wrong.
Witch is it?

7. Option 3. Soroban is right, I'm not.

My diagram misleads as the circles in the 2D diagram where they touch would not be the same size. Soroban's method will lead to the correct answer. Mine would be right if the 3 spheres were balanced on each other and there was no 4th sphere

Tertrahedron each side of 2R with same reasoning as the 2D version.

8. Well since there where not essentialy the same my deduction of the options where off one where true was of course not valid.

Well what did I do wrong when a I said your methods where the same?
How do I compute the height (for that is what I need right?) of a tertrahedron?
I have checked my tables and formulas book and found nothing of intrest, also checked the websites at hand.

9. Well...I found it, calculated it and got i right.
But now I also know that I have to know some stuff about platonic solidsthis saturday, and platonic solids aren't intresting at all.
Thanks all!