So I have this question:
"An icosidodecahedron is a convex polyhedron built out of P regular pentagons and T equilateral triangles, for suitable $P, T \in \mathbb{N}$, with the rule that every pentagon can only share an edge with a triangle, and every triangle can only share an edge with a pentagon"
Then I need to prove that $E=5P=3T$ and that every vertex must neighbour exactly two triangles, and two pentagons.
This is what I thought so far, each pentagon shares one edge with a triangle and one triangle will share one edge with a pentagon. Then each pentane neighbours five triangles and each triangle neighbours three pentagons.
But this is all the info I can see and can't see how I can get to either of the proofs? Does anyone know how I could do this? And any techniques for these type of problems as for me it seems like guess work the whole time? Thank you!
Best Answer
Consider a vertex of a triangle. It is adjacent to two edges of the triangle. Those edges must be adjacent to two pentagons. So, the vertex is now adjacent to at least two pentagons and at least one triangle. Now, the two pentagons each have edges that are adjacent to the vertex, but not adjacent to the original triangle. They must be adjacent to a triangle. Now, we have at least two triangles and at least two pentagons adjacent to that vertex.
Each interior angle of a pentagon is $108^\circ$. Each interior angle of an equilateral triangle is $60^\circ$. So, with a minimum of two pentagons and two triangles adjacent to the vertex, we have a minimum of $336^\circ$ degrees of angles adjacent to the vertex. Even if the vertex were on a plane, we could not add another triangle or pentagon as that would increase the sum of adjacent angles to more than $360^\circ$.
It is the same argument for any vertex of a pentagon. This proves that every vertex must neighbour exactly two triangles and two pentagons.
And you have already proven $5P=3T$. Next, you just need to show $E=5P=3T$, which should be easy since no two pentagons are adjacent to each other.