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The problem is this:

Let $G$ be a connected, planar graph whose number of vertices is a multiple of $8$. $5/8$ of the vertices have degree $3$, $1/4$ have degree $4$, and $1/8$ have degree $5$. All faces of $G$ are triangles or quadrilaterals. Find the number of triangular faces, the number of quadrilateral sides, the number of vertices and the number of edges of $G$. Draw at least one such graph.

It is a problem from an old exam. (I'm trying to prepare for mine.)

I've seen people solve problems like this using a formula that had some kind of a weighted sum in it (and I think the weights were degrees, probably of vertices), and it looked like it was linked to Euler's formula. I remember people saying that they were no-brainers, with an algorithmic procedure for solving them. Unfortunately, I didn't understand the formula then, and I can't find it now, either in my notes or on the internet.

I've been trying to derive something useful from what I know about Euler's theorem, but I'm failing badly.

Could you please help me with this?

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up vote 7 down vote accepted

From counting vertex-edge incidences, we have $$\tag1 2e = 3\cdot\frac58v+4\cdot\frac14v+5\cdot \frac18v=\frac72v.$$ From counting face-edge incidences, we have $$\tag2 2e = 3f_3+4f_4.$$ From Euler's $v+f=e+2$, we have $$\tag3 v+f_3+f_4=e+2.$$ Thus $7(3)$ gives us using $(1)$: $ 7e+14=7v+7f_3+7f_4=4e+7f_3+7f_4=6e+4f_3+3f_4$ or $$ e=4f_3+3f_4-14.$$ Using $(2)$ again, we find $3f_3+4f_4=2e=8f_3+6f_4-28$ or $$ 5f_3+2f_4=28.$$ As $f_3,f_4\ge0$, this allows only $(f_3,f_4,e,v)\in\{(0,14,28,16),(2,9,21,12),(4,4,14,8)\}$. An example with the tlast tuple is obtained by taking a cube graph and adding two face diagonals with a common end point (that assume degree 5 this way). Rereading the original problem statement (as opposed to equation $(1)$ we obtained from it), we see that $v$ must be a multiple of $8$, hence $(2,9,21,12)$ does not work.

Alright, "pics, or it didn't happen": enter image description here

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Thank you very much! It didn't occur to me to use the handshaking lemma for faces and edges. To have a complete solution though, shouldn't I prove or disprove that each of the three tuples can be realized by a planar connected graph? To do that, do I just have to find examples, or is it possible to prove that such graphs exist in some other, simpler way? – Bartek Jun 10 '13 at 22:12
@Bartek Yes, for completeness the existence of realizations needs to be shown (or disproved). I did that for one and a half case only. – Hagen von Eitzen Jun 11 '13 at 6:27

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