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Let there be a regular n-sided polygon. A "minimalist" triangle is a triangle which has all vertices on vertices of n. let p be a point on this polygon. What is the minimal number of correct triangles which contain this point p? By contain I mean the point can be on the triangle or in the area it contains.

If it is on one side (not vertex) then it is n-2 but I don't know how to prove this is the least number.

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Are you allowing the point to be inside the polygon? or must it be "on" the polygon, that is, a vertex or on an edge? – Gerry Myerson Feb 24 '13 at 2:52
it can be on an edge, in fact I think that is the minimum but I can't prove it. For an edge it is n-2 – Carry on Smiling Feb 24 '13 at 2:55
My question wasn't whether it can be on an edge, but whether it can be in the interior of the polygon. – Gerry Myerson Feb 24 '13 at 4:18
Oh, duh, sorry yes it can be on or in – Carry on Smiling Feb 24 '13 at 4:20
up vote 2 down vote accepted

Label the vertices $1,2,...,n$ (clockwise) and call a pair $(i,j)$ of vertices clockwise adjacent if $j=i+1 \mod n$ so the clockwise adjacent pairs are $(1,2),(2,3),...,(n-1,n),(n,1).$ [We'll refer to these by $i,i+1$, understanding subscripts mod $n$] So the clockwise adjacent pairs are the oriented sides of the regular $n$-gon. As noted by Khromonkey, if $P$ lies on the interior of one of the edges of the $n$-gon then it lies in exactly $n-2$ correct triangles (correct meaning vertices of the triangle are also vertices of the $n$-gon, see OP). If the side $P$ lies on is $(i,i+1)$ then these triangles are those including any of the $n-2$ other vertices other than $i,i+1$ along with these two.

Now if $P$ lies at a vertex $k$ of the regular $n$-gon then $P$ lies on at least the $n-2$ correct triangles containing both $k,k+1$, as well as on a triangle containing $k-1,k$, so the count of correct triangles at a vertex $P=k$ of the $n$-gon is greater than $n-2$.

Now suppose $P$ lies in the interior of the $n$-gon, and happens not to lie on any side of any correct triangle. We can produce a special collection of correct triangles in the following way: For each vertex $k$ of the triangle, there will be exactly one clockwise adjacent pair $(i,i+1)$ so that $P$ lies in the triangle $k,i,i+1$. We notate this triangle as $[k,(i,i+1)]$ to emphasize it was made starting with vertex $k$ and subsequently the clockwise adjacent pair $(i,i+1)$ was chosen. This is formally a list of $n$ "special" triangles, and we claim that at least $n-2$ of them are distinct.

Coincidences occur only when $P$ happens to lie in a triangle $(k,k+1,k+2)$, i.e. formed by two adjoining clockwise adjacent pairs. Such a triangle may be regarded as coming from vertex $k$ as $[k,(k+1,k+2)]$ or from vertex $k+2$ as $[k+2,(k,k+1)]$ So any such triangle in the list appears twice in the list. Now there are cases where $P$ appears in two of these at once, so that $P$ lies in the intersection of the two cyclic triangles $$(k-1,k,k+1),\ (k,k+1,k+2).$$ This happens when $P$ is obtained by moving a small amount away from a point interior to side $(k,k+1)$ of the $n$-gon. In these cases, there are two cases of concidence in the formal list of $n$ triangles above, leaving the count of distinct special triangles at $n-2$.

Note now that, for any $P$ interior to the triangle and not on any segment joining two vertices of the $n$-gon, we have produced a subcollection of the set of correct triangles containing $P$ which has at least $n-2$ triangles in it.

Finally if $P$ is interior to the triangle but lies on one or more segments joining two vertices of the $n$-gon, then taking a point $P'$ near $P$ which does not lie on such a segment, we get at least the $n-2$ distinct correct triangles containing $P'$, as well as possibly more. The point $P'$ here is to be taken in one of the open regions inside the $n$-gon upon removing all the segments joining two vertices, in such a way that the given point $P$ lies on the boundary of that open region. A bit long-winded, but a few pictures make things clear.

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Excellent it took me a while to understand exactly what you meant but It convinces me and I think it is correct. Thank you very much! – Carry on Smiling Feb 24 '13 at 14:13
@Khromonkey : If you say what parts of the argument seem unclear, I can try for better explanations. – coffeemath Feb 24 '13 at 18:14
i get it now, thanks – Carry on Smiling Feb 24 '13 at 18:27
Nice.${}{}{}{}$ – Gerry Myerson Feb 24 '13 at 22:35

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