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From what I read, buckminsterfullerene is composed of 60 carbon atoms, $\ce{C_{60}}$ and it is made up of 20 hexagons and 12 pentagons. So, I decided to do a back of the hand calculation with the knowledge that each of these were tri-atomic junctions. Since one hexagonal tri-atomic site would comprise a $\ce{C-C}$ bond angle of $\frac{2\pi}{3}$, thus just $\frac{1}{3}$rd of an atom and hence two atoms in a hexagon. Since there are 20 hexagons, this gives me 40 atoms. Now the $\ce{C-C}$ bond angle at the pentagon site is $\frac{3\pi}{5}$ which makes it $\frac{3}{10}$ atoms per site and hence $\frac{3}{2}$ atoms per pentagon. With 12 pentagons, it results in 18 atoms. So the total number of atoms according to this calculation amounts to 40+18=58. Where did I go wrong?

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    $\begingroup$ Now that's a nice trick indeed. Think of it this way: forget the angles, just count the atoms, as if there were six in each hexagon and five in each pentagon. That would make 180. But wait, we've just counted each atom more than once. Like you said, each atom is on the junction of three polygons, hence it must have been counted three times, so ${180\over3}=60$. $\endgroup$
    – Ivan Neretin
    Commented Aug 10, 2016 at 8:56
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    $\begingroup$ Because if you count them as $3\over10$, then each atom (which, BTW, is a junction of two hexagons and a pentagon) is counted as ${1\over3}+{1\over3}+{3\over10}={29\over30}$, rather than a full solid $\mathbf1$. $\endgroup$
    – Ivan Neretin
    Commented Aug 10, 2016 at 9:23
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    $\begingroup$ Surely the angles need not add up to $2\pi$. The fractions, however, do need to add up to 1. This is a certain place in the molecule; it is occupied by one atom. You can't split atoms. $\endgroup$
    – Ivan Neretin
    Commented Aug 10, 2016 at 9:38
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    $\begingroup$ Let's try counting the vertices of a cube this way. Each face has 4 corners with an angle of $\frac\pi2$, which is $\frac14$ of a full angle, so 1 vertex per face. There are 6 faces so a total of 6 vertices, but obviously a cube has 8 vertices... clearly this method doesn't work correctly. $\endgroup$
    – f''
    Commented Aug 10, 2016 at 14:01
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    $\begingroup$ Fun fact, Descartes' theorem of angular defect states that if you do this on any polyhedron without holes, you'll always be off by 2. $\endgroup$
    – f''
    Commented Aug 11, 2016 at 0:10

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