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What is the least number of internal diagonals a simple $n$–gon may have? (For a fixed $n$)

I know that any simple polygon has at least one internal diagonal.

The main problem is with the concave polygons, how do I generalize for them?

Any help would be truly appreciated.

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  • $\begingroup$ See math.stackexchange.com/questions/83580/… for some information. $\endgroup$ Commented Jan 6, 2016 at 16:31
  • $\begingroup$ I suspect the answer is $n-3$. $\endgroup$ Commented Jan 6, 2016 at 17:41
  • $\begingroup$ @6005: It's certainly possible to create an $n$-gon with only $n-3$ diagonals. (For $n>3$, create increasingly-"C"-shaped figures.) I don't know if we can do better. $\endgroup$
    – Blue
    Commented Jan 6, 2016 at 18:54

1 Answer 1

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The answer is $n-3$.

If there are $n-3$ consecutive concave vertices and only $3$ convex ones then all internal diagonals must start from the middle convex vertex.

For the lower bound: it is well-known that it is possible to divide the polygon into $n-2$ triangles, using only internal diagonals. To do this we need exaclty $n-3$ diagonals...

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