Geometry · Packing combinatorics
How many equal circles can you press right up against one central circle, so that no two of them overlap? In the plane the answer is elementary and provable in five lines. The same-looking question in space is the subject of a dispute between Newton and Gregory that took 259 years to settle.
In 1694, Isaac Newton and the Scottish astronomer David Gregory argued over how many equal balls could be arranged so that each one touches one central ball of the same radius, without any two of the outer balls overlapping. Newton insisted no more than twelve would fit. Gregory believed there might still be room for a thirteenth. Neither could prove his claim — not from carelessness, but because the question turned out to be genuinely hard.
This number is called the kissing number: touching spheres are said to "kiss," and the name stuck in mathematics.
The kissing number $K(n)$ in dimension $n$ is the largest number of radius-$1$ balls in $\R^n$ that can be arranged so that each one touches a fixed central ball of radius $1$, while no two of the outer balls share an interior point.
Two radius-$1$ balls touch when the distance between their centers equals $2$; they overlap (share interior points) when that distance is less than $2$.
On a line ($n=1$) this is trivial: the central "ball" (a segment) has exactly two directions, left and right — $K(1)=2$. It gets more interesting from here.
Take the planar case: circles of equal radius can be dragged freely around the plane — but not through each other, and not through the central one. As soon as a circle bumps into a neighbor, it pushes that neighbor aside and stops there itself, exactly like solid objects would. While a circle touches the central one, a green arrow lights up inside it, pointing straight at the point of tangency; pull the circle away and the arrow disappears. When a circle gets close to another one, it "sticks" to it a little, as long as that would not overlap anything else — makes it easier to line them up exactly. Three circles are already placed below; add more and feel out, by hand, how many fit around the central one at once without overlapping.
The widget above hints at the mechanism: for a circle to touch the central circle of radius $r$, its center must sit at distance exactly $2r$ from the center — on one and the same ring (the "touching ring," the same dashed line you see in the figure). For all the circles that touch the center at once, the only parameter left is the angle $\theta$ on that ring, and the whole problem reduces to how close in angle two such circles can stand without overlapping.
Let two circles sit on the touching ring (radius $2r$ from the center $O$) at angle $\theta$ from each other. The triangle with vertices at the center $O$ and at the two circles' centers is isosceles, with two sides of length $2r$ and the angle $\theta$ between them. The distance between the two circles' centers (the triangle's base) equals $2\cdot(2r)\sin(\theta/2)$.
The circles fail to overlap exactly when this distance is at least $2r$ (the sum of their radii):
So: two circles on the touching ring avoid overlapping if and only if the angle between them is at least $60^\circ$. This is exactly the condition the widget above checks on every drag.
Suppose $m$ circles stand on the ring. Going around, they split the full $360^\circ$ into $m$ gaps between neighbors, and the gaps sum to exactly $360^\circ$. If $m\ge7$, the average gap is $360^\circ/m\le360^\circ/7\approx51.4^\circ$ — less than $60^\circ$. And since the average gap is below $60^\circ$, at least one individual gap must also be below $60^\circ$ (otherwise the sum of all gaps would be at least $7\cdot60^\circ=420^\circ>360^\circ$ — a contradiction). That neighboring pair therefore violates the angular condition — they overlap. Conclusion: $m\le6$.
It remains to exhibit an actual arrangement of exactly six circles without overlap — which is exactly what the "Spread evenly" button does with six circles on the ring: angles $0^\circ,60^\circ,120^\circ,180^\circ,240^\circ,300^\circ$. Every neighboring pair is exactly $60^\circ$ apart — the angular condition holds with equality (the circles touch each other but do not overlap). Six circles genuinely fit.
$K(2)=6$.
Notice that the optimal planar arrangement is not merely one option among many — it is essentially unique (up to rotating the whole ring) and completely rigid: in the regular hexagon of six circles there is no slack at all, and each one touches both of its neighbors. That rigidity is exactly what makes the proof elementary. In space, as it turns out next, there is no such rigidity — and that is exactly why things got so hard.
In $\R^3$ the argument starts out exactly the same way. The direction from a ball's center to the center of each ball touching it is a unit vector; two such directions correspond to non-overlapping balls exactly when the angle between the vectors is at least $60^\circ$ — the very same $4\sin(\theta/2)\ge2$ condition, the same conclusion, just now stated for points on the sphere of directions rather than on a circle.
A $60^\circ$ angle between directions means a spherical "cap" of angular radius $30^\circ$ (half of $60^\circ$) can be drawn around each direction, and different balls' caps do not overlap. The area of such a cap on the unit sphere is $2\pi(1-\cos30^\circ)\approx0.842$, and the area of the whole sphere is $4\pi\approx12.566$. A simple division of areas gives:
So the area argument immediately gives an upper bound of $K(3)\le14$ — and gets stuck there. Unlike the planar case, it cannot tell $14$ apart from $13$ or from the real answer, $12$ — because the area argument never accounts for whether the caps can actually be packed geometrically consistently on the sphere, only for how much total area they cover.
What's more, twelve balls really do fit around one — and in more than one way. You can place them at the vertices of a regular icosahedron: then there is a visible gap between every pair of neighbors, and none of the twelve touch each other. Or — as in the cubic or hexagonal densest sphere packing (the very one from the Kepler conjecture) — you can arrange them so each one touches four more of its own neighbors among the twelve. Both give exactly $12$, but this is not a single rigid configuration like the regular hexagon in the plane — it's a whole family of different arrangements with real slack. That slack is exactly what fed Gregory's confidence: since there is visibly free room, it feels like a thirteenth ball ought to fit — even though it never does.
Because of this, the same summed-angles argument that settled the plane case in five lines is not enough in space. A rigorous proof that a thirteenth ball never fits, no matter the arrangement, was only given in 1953, by Kurt Schütte and Bartel van der Waerden — 259 years after the Newton-Gregory dispute. A simpler proof was found in 1956 by John Leech.
$K(3)=12$ — Newton was right.
The kissing number $K(n)$ makes sense in every dimension $n$, but it is known exactly surprisingly rarely — the area method and the rigid constructions that worked for $n=2$ and (with great effort) for $n=3$ stop being sufficient on their own from here on.
Dimensions $8$ and $24$ are known exactly for a real reason: it is precisely there that the $E_8$ and Leech lattices achieve not only a record kissing number but also — as Maryna Viazovska and her coauthors proved in 2016–2017, by applying Delsarte's linear-programming method with specially constructed "magic functions" from the theory of modular forms — the densest possible sphere packing; the same method settles both problems at once. Viazovska received the Fields Medal in 2022 for solving the sphere-packing problem in dimensions 8 and 24.
But $K(5)$, $K(6)$, and $K(7)$ are still not known exactly today — only upper and lower bounds, with a gap between them that remains open. A problem you can fully and honestly settle for balls on a table stays, past the first few dimensions, an active area of research right now.