[Math] How to find the number of faces of a rhombicosadodecahedron

combinatoricsgraph theory

I need to use the Euler's formula. I know there are $62$ faces…first, how do I find the number of vertices it has. From there, I can get the amount of edges, which will then in turn get me the number of faces. The question states as follows: A rhombicosadodecahedron is a polyhedron whose every vertex is incident to one triangular, one pentagonal, and two (opposite) quadrilateral faces. Find the number of faces.

Best Answer

Let $v,e,f,p,q,t$ represent the total number of vertices, edges, faces, pentagons, quadrilaterals and triangles in the polyhedron respectively.

We know that $v-e+f=2$ due to the Euler characteristic of the sphere.

We also know $f=p+q+t$

Also $e = \frac{5p+4q+3t}{2}$ as each pentagon contributes five edges, each quadrilateral contributes four edges, etc... The division by two comes from each edge being adjacent to two faces simultaneously

Next, $v = \frac{5p+4q+3t}{4}$ as again each pentagon contributes five vertices, etc... but each vertex is adjacent to a total of four faces.

So, putting this information together, we have $2 = v-e+f = \frac{5p+4q+3t}{4}-(\frac{5p +4q+3t}{2}) + p+q+t$

Combining fractions, we have:

$8 = 5p+4q+3t-10p-8q-6t+4p+4q+4t = t-p$

So, we know that there must be eight more triangles than there are pentagons.

To reach a final conclusion on the specific number of each type of face, we look at a picture of how each shape interacts:

enter image description here

We see that from the perspective of the pentagons, there are five neighboring triangles. So, $5p$ would overcount the number of triangles. Each triangle however is counted by three pentagons. So, we have the relation $\frac{5}{3}p = t$.

Similarly, we have $\frac{5}{2}p=q$

Combining this with the information we had before, that $t=p+8$, we have $\frac{5}{3}p=p+8$ and thus $\frac{2}{3}p=8$ and $p=12$.

Then, $t=20$ and $q=30$

There must then be $20$ triangles, $30$ quadrilaterals and $12$ pentagons, making a total of $p+q+t=62$ faces.

We could also go back and find the total number of vertices and edges if we desired now that we have found the number of each type of face.

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