Where do soccer balls come from?

Geometry Level 2

The dodecahedron has 12 pentagonal faces. Therefore it has 12 × 5 2 = 30 \frac{12 \times 5}{2} = 30 edges.

The icosahedron has 20 equilateral triangular faces. Therefore it has 20 × 3 2 = 30 \frac{20 \times 3}{2} = 30 edges.

How many edges does a truncated icosahedron have?

180 60 90 120

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Clara Blackstone
Oct 16, 2015

The truncated icosahedron is like the offspring of a icosahedron and a dodecahedron. (It's also the standard pattern for a soccer ball.) One way to make a truncated icosahedron is to slice each of the 12 corners of an icosahedron off to reveal 12 pentagons each surrounded by 5 hexagons. Or you can slice each of the 20 corners off a dodecahedron to reveal 20 hexagons, each touching 3 pentagons.

Either way, you get a truncated icosahedron with 20 hexagonal faces and 12 pentagonal faces. 20 × 6 + 12 × 5 2 = 90 \frac{20 \times 6 + 12 \times 5}{2}\ = \fbox{90} is therefore the number of edges.

Why? Because the 20 hexagons have 6 × 20 = 120 6 \times 20 = 120 edges among them. And the 12 pentaogns have 5 × 12 = 60 5 \times 12 = 60 edges among them. Which naively gives 180 edges total. But each of those edges is shared by 2 polygons. Therefore, each was double-counted so in reality there are only 90 edges total.

The maths in the question is wrong. 12x5/2 is 30, not 60. Similarly 20x3/2 is 30, not 60.

Jeremy Child - 5 years, 3 months ago

Log in to reply

Correct that 12x5/2 is 30. The second part is not 20x3/2 but 20x6/2 because the slicing means the flat surfaces are now hexagons (with 6 edges) not triangles (with 3 edges). So 20x6/2 works out to 60. Add that to the 30 edges from the pentagons and you get 90 edges in total.

Angela Sullivan - 5 years ago

Another way to do this would be to use Euler's formula. We know that the vertex configuration for the shape is (6,6,5). At each vertex, 3 edges meet so E = 3V/2 . At each vertex, one pentagon meets so F = V/5. And two hexagons so F = 2V/6 = V/3. Putting it into Euler's formula, V+F = E+2, we get V equal to 60, E = 90 and F = 32. Whats amazing is that this is true for any polyhedra !

Atharva Vankundre - 5 months, 3 weeks ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...