(C) Edward Ordman 2020 (feel free
to copy for any educational purpose)
Why does a soccer ball have twelve
pentagons?
Strangely, we can prove that there are twelve pentagons
starting with only these facts: (1)The surface is divided into hexagons and
pentagons
(2)Exactly
three
faces meet at each corner.
It does, however, require some fun algebra and an interesting
lemma.
Don’t worry if you don’t know what a lemma is, we’ll do it
anyway.
First – lets
call
the regions on the soccer ball (the pentagons and hexagons)
“faces”,
the lines
separating them “edges”, and the corners, well, “corners”.
Let’s call the number of faces F, the number of edges E,
and the number of corners C.
I want to
convince you of this
unlikely-sounding fact:C
– E + F = 2.
In fact, this formula holds for lots of shapes, almost
anything you can draw on the
surface of a hollow rubberball
or balloonthat
divides it into “faces” by drawing
edges
between “corners”, as long as all the faces could, if cut out of
the ball, be
stretched out flat on the ground with no holes in them. Or almost anything you can
draw on a piece of
paper with “edges” that meet at “corners” and don’t cross each
other, as long as it is one big drawing and not several little
ones.
Suppose the ball is made of soft rubber. Cut out the center
of one face, pull it and
stretch the whole thing flat on the ground. We haven’t
changed C or E or even F, as
long as we count the big area outside as one face. In order to follow the next bit
of
reasoning, make almost any kind of sketch you want – as simple
as a box with a
diagonal drawn in it will do, that picture has four corners, five
edges,
and three faces
(remember to count the outside!)
Now start erasing edges, one at a time. There are two
circumstances
to watch for.
(a)An edge separates two faces. When you
erase the
edge, you reduce the
number of edges by one and the number of faces by
one.So C – E + F
does not
change.Do this
as many times as you can.
(b)There are no faces left except the big
outside
one, but there are edges and
corners. Find an edge that has one corner that is
an “end”, that is, no other edges are
at that corner. Erase that edge and corner. (Never take out a “middle” edge,
keep the picture connected.) You reduce E
by one
and C by one,
so C – E + F does not change. Do this until just one edge is
left.
Now E = 1, C = 2, F
=1, so C – E + F = 2.
Since we never changed it while erasing things,
C – E + F
=2 for the original drawing – and for the
corners, edges, and faces on the
soccer ball.
Now, back to the original soccer ball.
Instead of just “faces”, we see that there are Pentagons and
Hexagons.F = P + H,
the number of
pentagons and the number of hexagons. HenceC – E + P + H = 2
How many edges are there?Each Pentagon has five edges, each Hexagon 6,
but if we add up 5 times P
plus six times H we have counted each edge twice,
since each edge is on two
faces.
That is,5P +
6H =
2E
We can do the same for corners – but as each corner is on
three faces,
adding up by faces makes each corner be counted three times.
So5P + 6H = 3C.
OK, here is what we now know:
(a)C – E + P + H = 2
(b)5P + 6H = 2E
(c)5P + 6H = 3C
We see from (b) and (c) that 2E = 3C, so also 4E = 6C
Multiply formula (a) by 6:6C – 6E + 6P + 6H = 12
Subtract formula (b) from that on both sides:6C – 6E + P = 12 – 2E
Simplify:6C -
4E +
P = 12
But 6C = 4E; they cancel.Hence P = 12. The soccer ball has 12 pentagons!
Incidentally, now we might ask, how many hexagons? For this, we need another fact about the soccer
ball: Each corner touches one pentagon,
and two hexagons.
In other words,5P =
C,but 6H = 2C
Since P = 12, C = 60 and H = 20. (Oh, if you still care, E = 3C/2 = 60 and F = 12 + 20 = 32)
Isn’t it amazing what we can do without counting?
By the way, the fact that C - E + F = 2 was the "lemma", the
intermediate step.
It is called Euler's Polyhedral Formula and was probably
first noticed around 1537
but popularized by Leonhard Euler about 1758.
