Is it possible to put pairwise disjoint open 3d-balls with radii $\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots$ inside a unit ball?

not an original question, I found it somewhere in the internet once, but without any answer.


marked as duplicate by Alex M., Jeremy Rickard, Watson, user91500, gebruiker Jun 25 '16 at 8:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The answer is certainly yes. You can already do this in two dimensions, as discussed (with lovely images) here: math.stackexchange.com/questions/1491754/… $\endgroup$ – mjqxxxx May 27 '16 at 14:47
  • $\begingroup$ I think the analogue for area in this case is volume so. Volume of unit ball in $\Bbb R^3$ is ${4\over3}\pi$, and the volume of the spheres to be fitted is ${4\over3}\pi(\zeta(3)-1)$. With something like 3.34 units of volume left over. $\endgroup$ – snulty May 27 '16 at 14:57

As pointed out by @mjqxxxx, this is possible.

In fact, it is possible to pack two families of open balls of radii $\frac12, \frac13, \ldots$ into unit sphere.
The picture below illustrate one possible configuration (with $n$ up to $300$):

$\hspace1in$ Two families of balls in a ball

In above picture, the two families are related to each other by a reflection with respect to origin.

Let us concentrate on the families of brown balls. The brown ball with radius $\frac12$ is centered at $(0,0,-\frac12)$. For $n \ge 3$, the brown ball with radius $\frac1n$ is centered at

$$(x_n,y_n,z_n) = (r_n\sin\theta_n\cos\phi_n,r_n\sin\theta_n\sin\phi_n, r_n\cos\theta_n)$$

where $\displaystyle\;r_n = 1 - \frac1n,\;\theta_n = \frac{\pi}{4}\left(\frac{n+1}{n-1}\right)\;$ and $\phi_n$ is defined recursively by

$$ \phi_n = \begin{cases} 0, & n = 3\\ \phi_{n-1} + \rho_n + \rho_{n-1}, &n > 3 \end{cases} \quad\text{ where }\quad \rho_n = \sin^{-1}\left(\frac{1}{(n-1)\sin\theta_n}\right) $$ If one orthogonal project the ball with radius $\frac1n$ onto $xy$-plane, $\rho_n$ will be the half-angle subtended by the ball's image with result to origin. The condition $\phi_{n} - \phi_{n-1} = \rho_{n} + \rho_{n-1}$ guarantee the balls with radii $\frac{1}{n}$ and $\frac{1}{n-1}$ are disjoint from each other.

  • For $n = 2$, the brown ball with radius $\frac12$ is touching the blue ball with radius $\frac12$ and the brown ball with radius $\frac13$. However, their interiors are disjoint.

  • For small $n$, one can build a 3d model of above configuration and verify by eye the balls are disjoint from each other. Above picture contains balls with $n$ upto $300$, balls with $2 < n \le 300$ have been inspected and they are disjoint from each other.

  • For larger $n$, the balls form a spiral converging to some sort of limit cycle at $\theta = \frac{\pi}{4}$. For a typical ball there, its projection onto $xy$-plane will subtend an angle $2\rho_n \sim \frac{2\sqrt{2}}{n}$ with respect to the origin. Furthermore, the balls nearby will be arranged info "arms". Let $\frac{1}{Kn}$ be the radius of nearest ball on the next arm of spiral (i.e. the arm closer to the limit cycle immediately above the current arm). The "angles" between the ball and its nearest neighbor on next arm will be around $2\pi$. This means $$\int_{n}^{Kn} \frac{2\sqrt{2}}{n} dn = 2\sqrt{2}\log K \approx 2\pi \quad\implies\quad K \approx e^{\pi/\sqrt{2}}$$

    The distance between this pair of balls is around $\theta_n - \theta_{Kn} \approx \frac{\pi}{2Kn} - \frac{\pi}{2n} = \frac{\pi}{2n}\left(\frac1K-1\right)$.
    We can compare it with the required separation $\frac{1}{n} + \frac{1}{Kn} = \frac{1}{n}\left(1 + \frac{1}{K}\right)$ between them.
    Since the ratio $\frac{\pi}{2}\left(\frac{K-1}{K+1}\right) = \frac{\pi}{2}\left(\frac{e^{\pi/\sqrt{2}}-1}{e^{\pi/\sqrt{2}}+1}\right) \approx 1.263418 > 1$, when $n$ is large, the ball with radius $\frac1n$ is disjoint from the one on next arm of spiral.

To bridge the analysis for small and large $n$, we need to check by the time $n$ reaches $300$, whether we can switch to use the result for large $n$ or not.

I don't have a rigorous proof. However, when we increase $n$ to around $100$, the relative error between $2\rho_n$ and its approximation $\frac{2\sqrt{2}}{n}$ already falls below $0.55\%$. The relative error in $K$ should be around $\frac{\pi}{\sqrt{2}} \cdot 0.55\% \sim 1.3\%$. As far as I can see, other approximations should generator relative errors of same order. Since the ratio $1.263418$ above is tens of percents larger than $1$. It seems $n = 100$ is already large enough.

Combining all these analysis, a safe bet is all the balls in above two families are disjoint from each other.


See if I did understand it right:

Given any two 3D-balls (and they are open), none can interset with each other. That is $$ B_n \cap B_k = \oslash$$ for any $k$ and $n$.

It is wanted to know if you can fit them all in a ball of radius $1$, while each $B_n$ has radius $\frac 1 n$ with $n >1$.

If the answer is yes, then you can fit them all without intersections. That is, the sum of their volume must be lesser than $\frac {4 \pi} 3$.

$$ \sum V_{olume} < \frac {4 \pi} 3 $$

We have to check if the above inequality is true.

Does the serie belows converge? To what number?

$$ \sum V_{olume} = \sum_{n=2}^\infty \frac {4 \pi} 3 \frac 1 {n^3}$$

And yes, it does converge to (approximately) $$\sum_{n=2}^\infty \frac {4 \pi} 3 \frac 1 {n^3} \approx 0.202157 \ \frac {4 \pi} 3$$

Which confirms that the inequality is true.

Hence, yes. You can fit all those 3D-balls inside a ball of radius $1$.

Based on Post new answer or significantly edit old wrong answer?

I deleted the old wrong answer and am posting this new one.

  • $\begingroup$ Just because a solid $X$ has smaller volume than a solid $Y$, it does not follow that $X$ fits inside $Y$. For example, three disjoint balls of radius $\frac{1}{2}$ do not fit inside a unit ball. That is, this argument seems incomplete. $\endgroup$ – Andrew D. Hwang May 28 '16 at 3:40
  • $\begingroup$ If two solids $X$ and $Y$ have the same shape; it follows that if the volume of $X$ is greater than the volume of $Y$, then the solid $Y$ fits inside $X$. In my argument all the solids have the same shape (a sphere). If I have $n$ of them and want to put they all inside a solid $S$ (without intersecting with one another), I must see if their volume summed is lesser than that of $S$. --Remember they are open (every point of it is an interior point). The border of a open ball is not part of it; then they can 'touch' one another without intersecting. $\endgroup$ – Pedro H. N. Vieira May 28 '16 at 21:37

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