Consider a planar graph with 2n vertices which has two faces that are n-gons, and all remaining faces are triangles.

Consider a planar graph with 2n vertices which has two faces that are n-gons, and all remaining faces are triangles. How many edges does this graph have?

• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos May 15 '18 at 16:14
• What have you tried? You might draw some graphs for small $n$ and count them. You might use the Euler characteristic $V+F-E=2$ – Ross Millikan May 15 '18 at 16:16
• @JoséCarlosSantos Thank you! I was trying to express faces as f = 2n + 3x , where 2- number of n- gons , n - number of vertices n-gon have , 3 - number of vertices triangle have, and x - number of triangles, but I feel like it's a wrong path to go. Also, we all know about double counting and planar graph, means 2*#e = deg*#v, but I'm not sure if it's applicable here. As well,I tried to express #e using faces: #e = p/2 *#f, where p - number of corners of each face. That is all I have for now( – J.Doe May 15 '18 at 16:26
• I think you mean $E=2n+3x$, and that's almost right, but each edge will appear in two faces, so that expression counts each edge twice. Furthermore, how can you express $x$ in terms of $F$? – Kevin Long May 15 '18 at 17:41
• No, I meant F = 2n + 3x, meaning that number of faces consist of 2 n-gons and unknown number of triangles, which is x. And using the formula #e = p/2 *#f, where p - number of corners of each face, we can get E = E1 + E2 = n^2 + 9x/2. But I don't know how to get rid of x? – J.Doe May 15 '18 at 18:04

Draw two concentric $n-$gons with the inner one clocked slightly so that its vertices are between the vertices of the outer one. You have accounted for all $2n$ vertices, so you can't have any more. Now draw the band of obvious triangles between the $n-$gons. Each triangle has a base on one of the $n-$gons and two sides that it shares with other triangles, so there are $2n$ triangles. There are $n$ edges in each $n-$gon and one between each pair of neighboring triangles, for $4n$ in total. A version of the figure with $n=8$ is below. • Yes, we have $V=2n, F=2n+2,$ so $E=4n$ from the Euler Characteristic that I mentioned in my first comment. – Ross Millikan May 15 '18 at 22:56
• You start with the two $n-$gons, so have $2n$ edges and the region between the $n-$gons needs to be cut into triangles. Each new side needs to go between two existing vertices. You can compute the number of triangles by the total angle. The interior $n-$gon contributes $2\pi n-\pi(n-2)$ in exterior angles and the exterior $n-$gon contributes $pi(n-2)$ in interior angles. That totals $2\pi n$ so it takes $2n$ triangles – Ross Millikan May 15 '18 at 23:27