On the unit $2$-sphere ${\mathbb S}^2$ furnished with the geodesic distance, a subset homeomorphic to a planar disk is called a cell. A finite family of cells is a tiling if their interiors are mutually disjoint and their union is the whole sphere. My main question is:

(1) Can the sphere be tiled by congruent cells of an arbitrarily small diameter? If not, how small can the diameters of the cells be?

An obvious example of a tiling with arbitrarily many congruent cells is obtained by cutting the sphere into $n$ sectors by $n$ uniformly spaced great semicircles, each connecting the North and the South poles. Since the cells' diameter is $\pi$ - the same as the diameter of the whole sphere, they cannot be called small by any means.

A somewhat less obvious example is constructed as follows. Consider the $4k$-faceted polyhedron inscribed in the sphere, consisting of a $2k$-faceted antiprism ($k\ge3$) capped off by two pyramids, as shown below for $k=18$. With the properly chosen altitude of the antiprism, all $4k$ (isosceles-triangular) facets become congruent by design.

${\qquad\qquad\qquad}$The inscribed $4k$-faceted polyhedron.

The central projection of the facets to the sphere produces a tiling of the sphere with $4k$ congruent, isosceles-triangular cells of diameter considerably smaller than $\pi$, but greater than $\pi/3$ and converging to $\pi/3$ as $k\to\infty$.

In the special case of $k=5$, the inscribed polyhedron is the regular icosahedron. In this case, if each of its $20$ equilateral triangular facets is barycentrically partitioned into 6 triangles, the central projection to the sphere yields a tiling with $120$ congruent, triangular cells of diameter well below $\pi/3$. No better examples are known to me, which raises the following, specific two questions:

(2) Is there a tiling of the sphere with an arbitrarily large number of congruent tiles, each of diameter $d\le\pi/3$?

(3) Is there a tiling of the sphere with congruent cells of diameter smaller than that in the subdivided-dodecahedral $120$-cell tiling described above?

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    $\begingroup$ For triangle tiles, there is this paper which I cannot access at the moment: "Classification of tilings of the 2-dimensional sphere by congruent triangles." Yoshio Agaoka and Yukako Ueno. Hiroshima Math. J.. Volume 32, Number 3 (2002), 463-540. $\endgroup$ Nov 2, 2013 at 21:08
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    $\begingroup$ @JosephO'Rourke : Thank you. The open-access 78-page article is available @ projecteuclid.org/… $\endgroup$ Nov 2, 2013 at 22:12
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    $\begingroup$ @JosephO'Rourke : True even for non-convex tiles, only define a vertex as a point at which at least three tiles meet; then use the Euler characteristic for the sphere. $\endgroup$ Nov 3, 2013 at 1:59
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    $\begingroup$ By the way, the hyperbolic plane can be tiled by congruent copies of polygons with arbitrarily small diameters using "horobricks." $\endgroup$ Nov 3, 2013 at 18:32
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    $\begingroup$ The first questin is Problem 60 in the Scottish Cafe book of problems: Can one, for every $\varepsilon>0$, represent the surface of a sphere as a sum of a finite number of regions which are smaller in diameter than $\varepsilon$, closed, connected, congruent, and have no interior point in common? We assume that the boundaries of these sets are: (a) polygons, (b) curves of finite lengt, (c) sets of measure zero. (RUZIEWICZ) $\endgroup$
    – juan
    Feb 25, 2014 at 20:28

3 Answers 3


Not an answer. But permit me to draw attention to Robert J. MacG. Dawson's website on congruent sphere tilings, including this beautiful tiling by triangles:

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    $\begingroup$ Indeed, beautiful patterns and fantastic graphics. This one could be used for jewelry design or a Christmas tree ornament. $\endgroup$ Nov 3, 2013 at 2:34
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    $\begingroup$ Joseph, add adjective "non-congruent". :) $\endgroup$
    – Wlod AA
    Oct 14, 2022 at 20:56
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    $\begingroup$ When I click the link security software says the connection is not private, so I cannot get there. How to solve? $\endgroup$ Sep 20 at 10:02
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    $\begingroup$ @OscarLanzi The Wayback Machine has saved a snapshot of the page, which might be helpful. $\endgroup$ Sep 20 at 10:58
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    $\begingroup$ @OscarLanzi, the certificate is for a different domain. Looks like they're migrating, because the page is also present on the domain covered by the certificate: cs.smu.ca/~dawson/images4.html#Swirl $\endgroup$ Sep 20 at 13:33

Here is a partial result that says that if the sphere can be partitioned into small congruent polygons, then the polygons must be "thin" in some sense. Specifically:

Theorem Suppose a sphere can be partitioned into congruent spherical $n$-gons of diameter $\varepsilon>0$ in an edge-to-edge tiling. Then the area $A$ of the $n$-gon is $O_n( \varepsilon^{2 + c_n} )$ for some constant $c_n>0$ depending on $n$.

In particular, the area is much less than the disk of radius $\varepsilon$, when $\varepsilon$ is small; geometrically, this asserts that the $n$-gon is "thin" compared to its circumscribing disk. This is broadly consistent with the examples provided by the OP of tiling by thin triangles.

Probably the restriction on edge-to-edge tiling can be removed by a variant of the argument. In the convex case, Euler characteristic arguments give $n \leq 5$ as noted already in comments, while the triangular tiling case has already been classified, so it "only" remains to rule out the case of small thin quadrilaterals and small thin pentagons, which seems doable in principle (the thinness does place significant constraints on the angles and lengths of these $n$-gons, which would be difficult to reconcile with the tiling hypothesis, especially if one imposes an edge-to-edge tiling condition. For instance, a thin pentagon cannot have all five sidelengths equal to each other.) but there appears to be quite a lot of combinatorial cases to check.

Proof We can take $\varepsilon$ to be small, and write the area $A$ as $\delta \varepsilon^2$. We allow all implied constants to depend on $n$. Let $r$ be a real number with $1 \ll r \ll 1/\varepsilon$, and consider which copies of the $n$-gon intersect a disk of radius $r \varepsilon$. Let $N_r$ denote the number of copies that intersect the disk, and $M_r$ the number that intersect but are not contained in the disk. Standard volume packing arguments (similar to those used to establish the trivial bound of $\pi r^2 + O(r)$ in the Gauss circle problem of estimating the number of lattice points in a disk of radius $r$) show that $N_r \asymp r^2/\delta$ and $M_r = O(r/\delta)$. (Because we are in the regime $r \ll 1/\varepsilon$, the fact that we have a spherical geometry instead of a plane geometry does not cause significant distortion to these estimates.) On the other hand, from Gauss-Bonnet, the angles $\alpha_1,\dots,\alpha_n$ of each $n$-gon add up to $(n-2)\pi + \delta \varepsilon^2$. Summing up over all the $N_r$ copies of the $n$-gon and double-counting, we conclude that $$ N_r ((n-2)\pi + \delta \varepsilon^2) = 2\pi V_r + \sum_{i=1}^n c_{r,i} \alpha_i$$ where $V_r$ are the number of vertices of the packing that lie inside the disk, and $c_{r,i}$ are the number of angles $\alpha_i$ corresponding to one of the $M_r$ boundary $n$-gons that lie outside the disk. Clearly the $c_{r,i}$ are natural numbers with $0 \leq c_{r,i} \leq M_r$. In particular $$ 2V_r - (n-2)N_r = O( N_r \delta \varepsilon^2 + M_r )$$ $$ = O( r^2 \varepsilon^2 + r/\delta ) = O(r/\delta)$$ since $r \ll 1/\varepsilon \leq \frac{1}{\delta \varepsilon^2}$. We have thus created linear relations $$ N_r = \sum_{i=0}^n c_{r,i} \beta_i \tag{1}$$ for any $1 \ll r \ll 1/\varepsilon$, where $\beta_i = \alpha_i / \delta \varepsilon^2$ for $i=1,\dots,n$, $\beta_0 = \pi / \delta \varepsilon^2$, and $c_{r,0} = 2V_r - (n-2)N_r$. Thus the $c_{r,i}$ are integers of size $O(r/\delta)$ and $N_r$ is an integer of magnitude $\asymp r^2/\delta$. The key point here is that the integer $N_r$ is significantly larger than any of the coefficient integers $c_{r,i}$. This turns out to be exploitable information even without knowing much about the magnitudes of the variables $\beta_i$.

Now let $1 < r_1 < r_2 < \dots < r_{n+2}$ be a sequence of real numbers defined recursively by $r_1 := C$ and $$ r_{j+1} = C r_j^2 r_{j-1} \dots r_1 / \delta^j$$ for $j=1,\dots,n+1$, where $C$ is a large constant to be chosen later. Then $$ r_{n+2} \asymp_C \delta^{-O(1)}.$$ If $r_{n+2} \geq 1/\varepsilon$ then $\delta = O( \varepsilon^{c_n} )$ for some $c_n>0$ and we are done, so we may suppose that $r_{n+2} < 1/\varepsilon$. In particular we have the linear relations $$ N_{r_j} = \sum_{i=0}^n c_{r_j,i} \beta_i \tag{2}$$ for $j=1,\dots,n+2$.

It turns out that this system of $n+2$ linear equations on the $n+2$ variables $1, \beta_0,\dots,\beta_{n+1}$ is non-singular and so (2) cannot be satisfied (it would imply $1=0$, which is absurd). To show this, we first claim inductively that for each $1 \leq j \leq n+2$, there is a $j \times j$ minor $$ \begin{pmatrix} N_{r_1} & c_{r_1,i_1} & \dots & c_{r_1,i_{j-1}} \\ \vdots & \vdots & \ddots & \vdots \\ N_{r_j} & c_{r_j,i_1} & \dots & c_{r_j,i_{j-1}} \end{pmatrix} $$ which is nonsingular for some $0 \leq i_1 < \dots < i_{j-1} \leq n$. This is clear for $j=1$ since $N_{r_1}$ is non-zero. Now suppose the claim is already established for some $1 \leq j \leq n$. Using (2) and multilinearity of the determinant in the columns, we conclude that there is a $j \times j$ minor $$ \begin{pmatrix} c_{r_1,i'_1} & \dots & c_{r_1,i'_j} \\ \vdots & \ddots & \vdots \\ c_{r_j,i'_1} & \dots & c_{r_j,i'_j} \end{pmatrix} $$ which is nonsingular for some $0 \leq i'_1 < \dots < i'_{j+1} \leq n$. The determinant is obviously an integer. By cofactor expansion, the $j+1 \times j+1$ minor $$ \begin{pmatrix} N_{r_1} & c_{r_1,i'_1} & \dots & c_{r_1,i'_j} \\ \vdots & \vdots & \ddots & \vdots \\ N_{r_{j+1}} & c_{r_{j+1},i'_1} & \dots & c_{r_{j+1},i'_j} \end{pmatrix} $$ then has determinant of magnitude $$ \gg N_{r_{j+1}} - O( \sum_{i=1}^j N_i \prod_{1 \leq k \leq j+1: k \neq i} (r_k/\delta) )$$ $$ \gg r_{j+1}^2/\delta - O( r_{j+1} r^2_j r_{j-1} \dots r_1 / \delta^{j+1} )$$ $$ > 0$$ by the construction of $r_{j+1}$ (if $C$ is large enough), closing the induction.

Applying the inductive claim at $j=n+2$, we see that the $n+2 \times n+2$ matrix $$ \begin{pmatrix} N_{r_1} & c_{r_1,0} & \dots & c_{r_1,n+1} \\ \vdots & \vdots & \ddots & \vdots \\ N_{r_{n+2}} & c_{r_{n+2},0} & \dots & c_{r_{n+2},n+1} \end{pmatrix} $$ is nonsingular, but this contradicts (2) as discussed previously. $\Box$

UPDATE: The following recent papers, when put together, form a complete classification of monohedral edge-to-edge tilings of the sphere by convex polygons:

Wang, Erxiao; Yan, Min, Tilings of the sphere by congruent pentagons. I: Edge combinations $a^2b^2c$ and $a^3 bc$, Adv. Math. 394, Article ID 107866, 36 p. (2022). ZBL1484.52016.

Wang, Erxiao; Yan, Min, Tilings of the sphere by congruent pentagons. II: Edge combination $a^3b^2$, Adv. Math. 394, Article ID 107867, 68 p. (2022). ZBL1484.52017.

Yohji Akama, Erxiao Wang, Min Yan, Tilings of the Sphere by Congruent Pentagons III: Edge Combination $a^5$

Hoi Ping Luk, Min Yan, Tilings of the Sphere by Congruent Pentagons IV: Edge Combination $a^4b$

Ho Man Cheung, Hoi Ping Luk, Min Yan, Tilings of the Sphere by Congruent Quadrilaterals or Triangles

Most of the tilings in the classification have a bounded number of faces (and hence diameter bounded below), but there are some families of tilings with unboundedly many faces, mostly based around the "earth map" tiling construction, which when unwrapped looks something like one of these two figures (taken from Figure 3 of https://arxiv.org/abs/2307.11453):

Figure 3 from https://arxiv.org/pdf/2307.11453.pdf

In each of these tilings, the upper end converges to the north pole, the bottom end converges to the south pole, and the left and right ends are glued together. The "height" of the tiling is bounded, but the "width" is unbounded. In these constructions, the individual n-gons have small area, but still large diameter (it only takes a bounded number of them to reach from one pole to another); they are quite thin as the number of faces goes to infinity, consistent with the above result. So there are no counterexamples to Q1 of the OP arising from edge-to-edge tilings of convex polygons. I suspect that some of the examples in the linked papers may also provide a positive answer to Q2 or Q3, though one would have to check each of the tilings in these papers separately to do so. (Perhaps the simplest thing would be to contact the authors of these papers for followup questions.)

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    $\begingroup$ I think this argument mixes up the sum of the external angles of a polygon, which is close to $2\pi$, and the sum of internal angles, which is close to $\pi (n-2)$ and which has the property that its sum over the polygons in a region of a tiling is $2\pi$ times the number of vertices plus an edge contribution. But I don't think this causes any real problem. $\endgroup$
    – Will Sawin
    Sep 21 at 18:52
  • $\begingroup$ Oops, this should now be fixed now. $\endgroup$
    – Terry Tao
    Sep 21 at 19:02

One possibility for a 120-way division with (slightly) smaller- diameter tiles uses a snub dodecahedron.

Start by inscribing the snub dodecahedron in the sphere and radially projecting the edges of the polyhedron onto the spherical surface. On the sphere, select any of the 60 vertices and connect the centers of the surrounding faces in rotational order; these segments enclose a pentagon surrounding the selected vertex. Below is a picture of the snub dodecahedron with one of the above pentagons constructed.

enter image description here

Source of the snub dodecahedron

The snub dodecahedron does not have any mirror planes globally, but the pentagon formed by the construction does have one locally and it may be divided into two quadrilaterals along this mirror plane.

The resulting tile is a quadrilateral, 120 congruent copies of which tile the sphere. This quadrilateral is fully described by noting its angles in rotational order ($A=36°,B=120°,C=120°,D=90°$) and identifying a $2:1$ ratio between arcs $BC$ and $BD$. Compared with the equal-area right triangle formed by dividing the faces of a regular icosahedron or a regular dodecahedron (both of these give the same end result), the quadrilateral will have a slightly shorter maximal vertex-to-vertex connection and the diameter will follow suit. Precise calculations reveal that the diameter of the smallest circle containing the quadrilateral is approximately $\color{blue}{0.620394}$ radian or $\color{blue}{35°32'45''}$, which is actually shorter than the long side of the triangle obtained after dividing regular icosahedral or dodecahedral faces ($0.652358$ radian, or $37°22'39''$).

Like the division pictured by Joseph O'Rourke, this one is chiral. Also the quadrilateral is indeed relatively "thin" because of the $36°$ vertex angle, which cannot be avoided in an icosahedrally symmetric 120-cell tiling.

  • $\begingroup$ The pentagons you constructed overlap... so then also the quadrilateral. Maybe they can be all be trimmed the same way, so as to not overlap. Probably so. But I think a picture here would be worth 1000 words. $\endgroup$ Sep 20 at 13:59
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    $\begingroup$ Ah, sorry, I misunderstood you pentagons as something larger: 4 adjacent triangles and the the slice of pentagon joining the vertices of the 1st and 4th triangle. So nothing wrong (except my comment). Nice construction by the way. $\endgroup$ Sep 20 at 19:31
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    $\begingroup$ So the pentagonal hexecontahedron? $\endgroup$ Sep 21 at 10:15
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    $\begingroup$ A distorted version thereof. The constructed spherical pentagon is not planar. We would not exoect such planarity when constructing a loop around an Archimedean-solid vertex except if the loop were triangular. The pentagonal hexecontahedron with planar faces does not have all its vertices on one sphere. $\endgroup$ Sep 21 at 10:40

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