# Thread: spheres

1. ## spheres

Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of the larger sphere?

The answer is 3+ root 69/3.
I really can't think of an approach, yet...

2. The larger sphere would never be able to rest on two smaller spheres without falling... This question = FAIL!

3. Originally Posted by Veronica1999
Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of the larger sphere?

The answer is 3+ root 69/3.
I really can't think of an approach, yet...
1. The centers $\displaystyle (C_1, C_2, C_3)$of the smaller spheres form a isosceles triangle with the side-length 2.

2. The center M of the larger sphere is vertically above the centroid F of the isosceles triangle. The larger sphere is tangent to all three smaller spheres.

3. $\displaystyle MFC_1$ is a right triangle with
hypotenuse $\displaystyle |\overline{MC_1}| = 3$
horizontal leg $\displaystyle |\overline{FC_1}| = \frac 23 \cdot \sqrt{2^2-1^2}=\frac 23 \cdot \sqrt{3}$
vertical leg $\displaystyle |\overline{MF}| =\sqrt{3^2-\left(\frac 23 \cdot \sqrt{3} \right)^2} = \sqrt{\frac{23}{3}}$

4. You have to add one small radius and one large radius to the distance $\displaystyle |\overline{MF}|$ and you'll get the given result.

5. For additional information have a look here: Close-packing of spheres - Wikipedia, the free encyclopedia

4. Originally Posted by Prove It
The larger sphere would never be able to rest on two smaller spheres without falling... This question = FAIL!
some glue could be used.

5. Originally Posted by abhishekkgp
some glue could be used.
It would still be top heavy and still fall...

6. Thank you so much Earboth!

7. ## Re: spheres

also see Earboths solution of a four ball pile 12/22.10

8. ## Re: spheres

Thanks for the info. I really appreciate it.

### mutually tangent sphe

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