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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. $\endgroup$ – Joseph O'Rourke Nov 2 '13 at 21:08
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    $\begingroup$ As noted by @JosephO'Rourke the answer for the case of triangular cells can be obtained from the paper by Yoshio Agoka and Yukako Ueno: the answer is no to both questions. I don't know the answer for the case of non-triangular cells. $\endgroup$ – Fiktor Nov 2 '13 at 21:54
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    $\begingroup$ @JosephO'Rourke : Thank you. The open-access 78-page article is available @ projecteuclid.org/… $\endgroup$ – Wlodek Kuperberg Nov 2 '13 at 22:12
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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." $\endgroup$ – Douglas Zare Nov 3 '13 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 '14 at 20:28
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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:
             tritile2

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    $\begingroup$ Indeed, beautiful patterns and fantastic graphics. This one could be used for jewelry design or a Christmas tree ornament. $\endgroup$ – Wlodek Kuperberg Nov 3 '13 at 2:34

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