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?
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":