Sorry about this but I have two objections to the post above: (1) Triangles always have at least two acute angles. However the binomial coefficient MC3 = 1/6M(M-1)(M-2) works in this case. More problematic is the quadrilateral described below. (2) it is not necessarily the case that an acute angled polygon skips n consecutive vertices in the M-gon. Consider a heptagon with vertices labelled 1, 2, 3, 4, 5, 6, 7. The quadrilateral drawn inside this with vertices 1, 3, 5, 6 of the heptagon is acute angled at vertex one AND vertex three, but no two vertices are separated by a gap as great as n = (7-1)/2 = 3. [/b]

From googleplex, lost in another thread (1) The total number of r-gons that can be constructed with vertices from an M-gon is the binomial coefficient MCr. (2) Triangles always have at least two acute angles, therefore MC3 is a constant term in the solution. (3) At least one pair of vertices of R separated by two edges of R must be separated by at least k + 1 edges in M if R has an acute angle. [This result comes from the fact that if two vertices of R differ by E edges from M then the edges from these two vertices to a third vertex not between them form an angle of 180/M at that vertex. So we require E.180/M < 90 for some vertices. My labelling is the from the opposite side of the polygon so you have >M/2 instead.] (4) As an immediate consequence no R-gon is acute-angled for R > k + 2. (5) Also squares in odd M-gons always have an acute angle since there is no way to assign their vertices to the 2k+1 vertices of M without creating a triplet whose end vertices differ by >= K + 1. (namely either pair of opposite vertices). So MC4 is another costant term in the solution. I hope that helps someone to make a bit more progress.

Well, can anybody post a complete solution?

pbornsztein wrote: (3) At least one pair of vertices of R separated by two edges of R must be separated by at least k + 1 edges in M if R has an acute angle. [This result comes from the fact that if two vertices of R differ by E edges from M then the edges from these two vertices to a third vertex not between them form an angle of 180/M at that vertex. So we require E.180/M < 90 for some vertices. My labelling is the from the opposite side of the polygon so you have >M/2 instead.] (4) As an immediate consequence no R-gon is acute-angled for R > k + 2. (5) Also squares in odd M-gons always have an acute angle since there is no way to assign their vertices to the 2k+1 vertices of M without creating a triplet whose end vertices differ by >= K + 1. (namely either pair of opposite vertices). So MC4 is another costant term in the solution. Actually, what are all your $k$'s?

I started working from JSteinhardt's idea. The gap that needs to be strictly greater than n isn't the gap between two adjacent vertices, but the gap between the vertices adjacent to the acute angle. So if we fix a location for $e$, then at most 1 of the $n$ vertices preceding $e$ can be used. This leads to a total of $(2n+1)(\binom{n}{m-1}+n\binom{n}{m-2})$ acute angles. Unfortunately, as googleplex pointed out, some polygons have more than one acute angle. However, a polygon can never have 2 nonadjacent acute angles (the two jumps of length $n+1$ have to overlap somewhere). This mean we need only subtract off the number of polygons which have two adjacent acute angles. Let us assume (by throwing in a factor of $2n+1$) that our first angle starts at vertex 1. Then we need to count the number of ways of placing the remaining $m-1$ vertices in order such that the following two conditions are satisfied: 1. Vertex #3 is placed in position $n+2$ or later. 2. If vertex #2 is placed in position $x$, vertex #4 is placed in position $x+n+1$ or later. Let $x$ be the location of vertex 2 (which must be somewhere between 2 and $n$). Let $y$ be the location of vertex 4 (which must be somewhere between $x+n+1$ and $2n+1$). We have $y-x-1$ choices for the placement of vertex 3, and $\binom{2n+1-y}{m-4}$ choices for the placement of the remaining vertices. This means the total overcounting we need to subtract off is $(2n+1)\sum_{x=2}^{n}\sum_{y=x+n+1}^{2n+1}(y-x-1) \binom{2n+1-y}{m-4}$. Does this look right, and if so, how do we evaluate this last sum?

kevinatcausa wrote: This leads to a total of $(2n+1)(\binom{n}{m-1}+n\binom{n}{m-2})$ acute angles. Well, I don't see how this sum comes. You're just counting the number of acute angles (any in the $2n+1$ gon)?

I got the sum $\sum_{k=1}^{n}(2n+1)(2n-k){k-1 \choose m-3}-\sum_{k=1}^{n}(2n+1)(n-k)^{2}{k-1 \choose m-4}$ but it seems tedious to evaluate. OK, after some calculations, without the factor of (2n+1) I get $n^{2}{n \choose m-3}-(2n+1){n+1 \choose m-2}(m-3)+{n+2 \choose m-1}(m-2)(m-3)+2n{n \choose m-2}-{n+1 \choose m-1}(m-2)$ ... hopefully it simplifies.

hmm I think the answer is something like $(2n+1) \left( \binom{n+1}{m-1}+(n-1) \binom{n}{m-2}-\binom{n+1}{m-1}-\binom{n}{m-1}\right)$ $=$ $(2n+1) \left((n-1) \binom{n}{m-2}-\binom{n}{m-1}\right)$ method: use PIE on the number of acute angles. there can be at most two acute angles and if there are two, they must be neighboring the first term is the number of one-acute angled things pick a vertex in 2n+1 ways and then find the number of ways to have that angle be acute; you get the first two terms above the second term is the number of two-acute angled things pick the length of the side connecting the two acute angles; it can go in 2n+1 possible places in the polygon, then get the second two terms above