Kepler's Spheres and Rubik's Cube
Kepler's Spheres and Rubik's Cube
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开普勒球体和魔方
DOI：
10.1080/0025570x.1988.11977378
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发表时间：
1988
影响因子：
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通讯作者：
J. Propp
中科院分区：
文献类型：
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作者：
J. Propp
“半径r的球形r可以同时与同一半径的固定球体相切？”这个问题至少可以追溯到1694年，当时艾萨克·牛顿（Isaac Newton）和他的当代戴维·格雷戈里（David Gregory）考虑了这个问题。约翰·开普勒（Johann Kepler）在1611年[8]中已经表明，可以在中央球体周围安排十二个外球，现在格雷戈里声称可以增加第十三个。牛顿不同意，但没有人证明他的主张，直到1874年才找到证据，证明了牛顿的猜想。 （R. Hoppe的原始证据在[2]中描述了；最近，Giinter [7]和Schiitte和van der Waerden [10]发现了证据。也许最优雅的证据是John Leech [ 9]有关问题的历史，请参见[6]。没有人知道如何在中央大肠杆菌周围安排超过二十四个超球，但也没有证明无法添加二十五。在较高的维度中，情况甚至不太了解，除了尺寸8和24的显着外，最大的接触数是精确知道的[1]。原始的格雷戈里·纽顿问题的难度（以及解决了180年的原因）是几乎有第十三领域的空间。可以以各种方式推动和拉动的十二个开普勒球体，并且可以很可信的是，某种摆弄可以创建一个足够大的空间以容纳额外的球体。本说明将以Erno Rubik的精神处理一个相关的问题：如果我们将十二个球标记并将其滚动在内部球体的表面上，那么可以实现哪些排列？答案令人惊讶，证明只需要分析几何和群体理论的基础。该企业的一个边缘好处是，它导致了常规二十面体的顶点的自然协调化。通过仅考虑球体的中心来解决问题的等效表述：我们想象十二个顶点，可以在固定半径r = 2r的球体上自由移动，但要受到固定的起源，但要受到约束，没有两个顶点可能永远不会是比R更近。我们会发现在两个配方之间来回切换很方便。为了确定性，请put r = 1，1 = 1。要开始，我们必须规定外部球体的初始配置。想到的弹簧的一种可能性是安排十二个顶点以形成常规的二十面体。正如我们将看到的那样，刻有单位球体的Icosahedron的任何两个不同的顶点至少是
"How many spheres of radius r may simultaneously be tangent to a fixed sphere of the same radius?" This question goes back at least to the year 1694, when it was considered by Isaac Newton and his contemporary David Gregory. Johann Kepler had already shown in 1611 [8] that twelve outer spheres could be arranged around a central sphere, and now Gregory claimed that a thirteenth could be added. Newton disagreed, but neither man proved his claim, and it was not until 1874 that a proof was found, substantiating Newton's conjecture. (R. Hoppe's original proof is described in [2]; more recently, proofs have been found by Giinter [7] and by Schiitte and van der Waerden [10]. Perhaps the most elegant proof known is the one given by John Leech [9]. For a history of the problem, see [6].) It's worth noting that the 4-dimensional version of this problem is still unsolved; no one knows how to arrange more than twenty-four hyperspheres around a central hypersphere, but neither has it been shown that a twenty-fifth cannot be added. In higher dimensions the situation is even less well understood, with the remarkable exception of dimensions 8 and 24, for which the maximum contact numbers are precisely known [1]. The source of difficulty in the original Gregory-Newton problem (and the reason it took 180 years to be solved) is that there is almost room for a thirteenth sphere; the twelve spheres of Kepler can be pushed and pulled in all sorts of ways, and it's credible that some sort of fiddling could create a space big enough to accommodate an extra sphere. This note will deal with a related question, in the spirit of Erno Rubik: if we label the twelve spheres and roll them over the surface of the inner sphere at will, what permutations are achievable? The answer is surprising, and the proof requires only the rudiments of analytic geometry and group theory. One fringe benefit of this enterprise is that it leads to a natural coordinatization of the vertices of the regular icosahedron. An equivalent formulation of the problem is gotten by considering only the centers of the spheres: we imagine twelve vertices, free to move on a sphere of fixed radius R = 2r about a fixed origin but subject to the constraint that no two vertices may ever be closer together than R. We will find it convenient to switch back and forth between the two formulations. For definiteness, put r = 1, 1 = 1. To begin, we must prescribe an initial configuration for the outer spheres. One possibility that springs to mind is to arrange the twelve vertices to form a regular icosahedron. As we'll see, any two distinct vertices of the icosahedron inscribed in the unit sphere are at least