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在柏拉圖立體#拓扑证明中用到 欧拉公式 $V - E + F = 2$ 和$$mV=nF=2E\tag1\label1$$
其中 $m$ 为每个顶点出发的棱的数目, $n$为每个面上的棱的数目.
2017 年，数学普及者 Matt Parker 发起了一项请愿，将英国交通标志条例更新为几何正确的足球。
见英国政府和议会 - 第202305号请愿 (22,536 个签名)
您可能没有注意到此类标志的不准确性，也许只是很高兴知道您正在驶往足球比赛。 但很明显（图 5）标志上的足球与 Matt 所持的真实足球不同。 足球的表面由五边形和六边形组成，日常足球的形状称为截断二十面体（truncated icosahedron 可以通过在二十面体的汇于每个顶点的五条棱的1/3处刨削而得到）这个标志令 Matt 讨厌的原因不是标志的足球画得很糟糕，而是这个几何体实际上不存在。
不可能通过将标志上所示的六边形缝合在一起来制成球体。使用式\eqref{1}，这将是一个 $m=3,n=6$ 的正多面体的例子，但我们可以用这种方式覆盖平面——这可能是为什么这个标志乍一看似乎是合理的。
由欧拉公式可知足球上有多少个五边形和六边形。回顾如何截断二十面体，我们看到五边形与原来的顶点一样多（12个），六边形与原来的面一样多（20个），下面用欧拉公式证明这是构成足球的唯一方法。假设足球有 $P$ 个五边形面和 $H$ 个六边形面。然后，在将它们粘合在一起之前，我们有 $5P + 6H$ 条未粘合的边和相同数量的未粘合顶点。从 Matt 手中的球可见
(i) 需要两个未粘合的边才能在足球上形成边，
(ii) 三个未粘合的顶点构成一个顶点，
(iii) 两个六边形和一个五边形在一个顶点相遇；
从 (iii) 我们看到六边形上的“未粘合”顶点的数量是五边形上的两倍。所以
\[F=P+H,\quad E=(5P+6H)/2,\quad V=(5P+6H)/3,\quad2×5P=6H.\]
将这些代入欧拉公式 $V-E+F=2$，我们发现
\[(5P+6H)/3-(5P+6H)/2+P+H=2\]
化简为 $P =12$. 从 $10P = 6H$ 得出 $H =20$. 如果我们遵循规则 (i)、(ii)、(iii)，这就是制作足球的唯一方法。
原文:
Earl Richard, Topology: A Very Short Introduction (Oxford, 12 Dec. 2019)
In 2017 the mathematics popularizer Matt Parker began a petition seeking to get road signs to football stadia corrected.
You may not have noticed the inaccuracy of such signs in the past, perhaps being happy just to know you’re travelling the right way for the game. But it’s clear (Figure 5) that the sign’s football does not resemble the actual football that Matt is carrying. A football’s surface is made from pentagons and hexagons and the everyday football is more formally known as a truncated icosahedron. (It can be created from an icosahedron by planing down, around each vertex, the five edges meeting there. If we plane down one-third of each of those five edges, we create a new pentagonal face and continued planing eventually shrinks all the triangular faces to hexagons.) I think though the irksome principle for Matt was not that the sign’s football was badly drawn, it was in fact impossibly drawn.
There is no way that a sphere can be made by stitching together hexagons as shown on the sign. That would be an example where $m=3$ and $n=6$, using the previous notation, and whilst we can cover the plane in this way—which may be why the sign looks plausible at first glance—making a football this way is mathematically impossible.
In fact, Euler’s formula shows us how many pentagons and hexagons there are on a football. Recalling how to truncate an icosahedron we see there are as many pentagons as original vertices (12) and hexagons as original faces (20), but Euler’s formula can show this is the only way to construct such a football. Say a football has $P$ pentagonal faces and $H$ hexagonal faces. Then, before gluing these together, we have $5P + 6H$ unglued edges and the same number of unglued vertices. Looking at Matt’s ball we can see that (i) two unglued edges are needed to make an edge on the football and (ii) three unglued vertices make a vertex with (iii) two hexagons and one pentagon meeting at a vertex; from (iii) we see there are twice as many ‘unglued’ vertices collectively on the hexagons as on the pentagons. So
\[F=P+H,\quad E=(5P+6H)/2,\quad V=(5P+6H)/3,\quad2×5P=6H.\]
If we put these values into Euler’s formula $V-E+F=2$ we find that
\[(5P+6H)/3-(5P+6H)/2+P+H=2\]
which simplifies and rearranges to $P =12$ and the equation $10P = 6H$ yields $H =20$. This, then, is the only way to make a football if we follow the rules (i), (ii), (iii). |
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