4 Spheres all touching each other??

Question

If there are 4 spheres all touching each other and 3 of them have diameters 4, 6 and 12 what is the diameter of the fourth one? I imagine it like 3 balls on a flat table touching each other and then we are supposed to put another one on top of them but in my imagination the top sphere could be any size basically right?

Answer 1

Not any size. If the fourth sphere is sufficiently small, it can fit in the hole in the middle, also resting on the table, without touching any of the other three spheres.

I’m guessing this problem is asking for the minimum diameter of the fourth sphere which guarantees contact with all of the other three.

Answer 2

Just complementing the answer just posted. If the new sphere is sufficiently small, it will fit in the hole in the middle, so we have a lower bound for the size of the new sphere. Similarly, if the new sphere is large enough, it might also make one of the previous spheres' size to be "sufficiently small", thus there might be an upper bound.

Answer 3

Imagine all three balls touch each other. The fourth one must exactly fit in the middle of these three balls. Only then each ball touches every other ball. (of course the assumption here is, that it's all in 2D).

If we say, that speres are allowed inside other spheres or these are more than 2D spheres, then there are many other possibilities, too...

Answer 4

Hint

The word cirsphe might refer to a circle, a sphere of a hypersphere.

Descartes' Theorem for $n\geq 2$ dimensions tells us that we need $n+1$ cirsphes in order to be able to determinate the radius of the $n+2$th cirsphe...

Now, how to do that?

Define the curvature $k_d$ of the $d$th cirsphe with radius $r_d$ as $$k_d=\pm\frac{1}{r_d}$$ (wheter plus or minus dependes on wheter the cirsphe is externally or internally tangent)

Descartes' Theorem for higher dimensions tells us now that

$$\bigg(\sum _{d=0}^{n+2}k_d\bigg)^2=2·\sum_{d=0}^{n+2}k_d^2$$

And knowing $k_1, k_2,...,k_{n+1}$ you can determine the curvature of the $n+2$th cirsphe and hence its radius.

There's even a poem regarding this formula!

Answer 5

If the table is mandatory, then a large enough fourth sphere could be prevented (by the table) from touching the smallest other sphere. And if gravity is not ruled out, a large one might fall off its perch. :-)