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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?

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  • $\begingroup$ I think that's right if the three balls on the table have the same diameter. I wonder if it's true in any other case. $\endgroup$
    – saulspatz
    Mar 3, 2019 at 15:11
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    $\begingroup$ There is more than one possibility, if the fourth sphere is not necessarily lying on the table with three others. en.m.wikipedia.org/wiki/Descartes%27_theorem $\endgroup$
    – user376343
    Mar 3, 2019 at 15:19
  • $\begingroup$ If the spheres are, indeed, "lying on a table" beware they are not concentric. $\endgroup$ Mar 3, 2019 at 20:56
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    $\begingroup$ There is also the (smartass) scenario where a fourth (larger) sphere touches the other three internally, i.e. circumscribes them. $\endgroup$
    – smci
    Mar 4, 2019 at 4:09

5 Answers 5

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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.

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    $\begingroup$ The limits can be computed from the four kissing circles problem. If the spheres are at the limit, all the tangent points will be in a plane. $\endgroup$ Mar 3, 2019 at 15:30
  • $\begingroup$ For some combination of sizes for the initial three spheres it's also possible for the fourth sphere to be too big. I haven't calculated whether the fourth sphere can be too big given the specific combination of the initial three. $\endgroup$
    – kasperd
    Mar 3, 2019 at 22:37
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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.

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  • $\begingroup$ There certainly is an upper bound for that reason for some initial three radii, but I don't think it'll happen with the OP's data. I guess the condition is roughly whether there's no plane tangent to the three first balls. $\endgroup$ Mar 3, 2019 at 16:01
  • $\begingroup$ No, there is NO upper bound. The table itself, where in the OP the 3 other balls are lying on can be considered as being the 4th ball with infinite radius! $\endgroup$ Mar 8, 2019 at 21:30
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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!

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    $\begingroup$ just call it a n-sphere $\endgroup$
    – qwr
    Mar 3, 2019 at 21:46
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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. :-)

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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...

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    $\begingroup$ If it is in 2D, you would usually say "circle", not "sphere". $\endgroup$ Mar 4, 2019 at 0:46
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    $\begingroup$ @PaŭloEbermann Could also be a disk ;-) $\endgroup$
    – Déjà vu
    Mar 4, 2019 at 5:07
  • $\begingroup$ Sphere to circle, ball to disk. One is boundary only, the other is filled space. $\endgroup$
    – Nij
    Mar 18, 2019 at 9:12

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