# Thread: Platonic Solids question out of fractions sum

1. ## Platonic Solids question out of fractions sum

I have been asked to help with this question and I don't get what it is asking. I have a good knowledge of solid shapes and plane shapes and Euler's Rule but don't get the link to the fractions question. Any ideas anyone?

For each of the following sums of reciprocals, state whether or not it corresponds to a regular or semiregular
tiling or a Platonic solid or Archimedean solid.
If it does, give the formal name of the tiling or solid and briefly explain (in just a few words) what the
name means. If it does not, state briefly why not.

a) 1/3 +1/12 + 1/12 =

j) 1/4 + 1/6 +1/0 =

2. Originally Posted by slatteryp
I have been asked to help with this question and I don't get what it is asking. I have a good knowledge of solid shapes and plane shapes and Euler's Rule but don't get the link to the fractions question. Any ideas anyone?

For each of the following sums of reciprocals, state whether or not it corresponds to a regular or semiregular
tiling or a Platonic solid or Archimedean solid.
If it does, give the formal name of the tiling or solid and briefly explain (in just a few words) what the
name means. If it does not, state briefly why not.

a) 1/3 +1/12 + 1/12 =

j) 1/4 + 1/6 +1/0 =
It looks as though those fractions correspond to the external angles of regular polygons. The exterior angle of a triangle is 1/3 of a complete revolution, for a dodecagon it is 1/12, and so on.

For problem (a), 1/3 + 1/12 + 1/12 = 1/2, and 1/2 of a complete revolution is 180º. If a triangle and two dodecahedra meet at a point then their external angles add up to 180º, so the sum of the internal angles is 360º. That tells you that the three polygons lie in a plane. In this case they form part of a tessellation of the plane known as a truncated hexagonal tiling.

For problem (j) there's presumably a misprint in the question. I assume that 1/0 should be 1/10. Then 1/4 + 1/6 + 1/10 > 1/2, so the sum of the internal angles is less than 360º in this case. A square, a hexagon and a decagon can be arranged to meet at a vertex of an Archimedean solid known as a truncated icosidodecahedron.