# What is the number of intersections of diagonals in a convex equilateral polygon?

Question: [See here for definitions]. Consider an arbitrary convex regular polygon with $$n$$-vertexes ($$n\geq 4$$) and the $$n$$-sequence $$\langle \alpha_i~|~i of its angles which $$\alpha_i$$ is the angle corresponding to vertex labeled by $$i$$ and vertexes labeled by successive numbers are neighbors. What is the number of intersection points of diagonals of such a shape with respect to $$n,\alpha_1,\cdots,\alpha_n$$?

Note that it is always less than or equal to $$\frac{n(n-1)(n-2)(n-3)}{24}$$ because in an ideal case when each intersection point is corresponding to a unique pair of diagonals (not two or more pairs) then by choosing $$4$$ vertexes of $$n$$ we will have two diagonals and this pair of diagonals determines an intersection point and this point is not an intersection point for any other pair of diagonals.

But the difficulty is that even in the simplest case which our convex equilateral polygon is a convex regular polygon there are cases which more than one pairs of diagonals share a same point as an intersection and this decreases the total number of intersection points of diagonals. For example diagonals of a regular convex polygon with $$6$$ vertexes have only $$13$$ intersection points but $$\frac{6\times 5\times 4\times 3}{24}=15$$ because three pairs of diagonals shared a single point in the center as their intersection. Note that in more complicated regular convex polygons with $$n\geq 6$$ these shared points will occur in many other places as well as center of the shape. Thus:

Any special formula for calculating the number of intersections of diagonals of a regular convex $$n$$-polygon is also welcome. For regular polygons this is OEIS A006561, where for odd $n$ it is $(n^4-6n^3+11n^2-6n)/24$. For even $n$ there are corrections of subtracting ${k \choose 2}-1$ for each $k$ crossing, which implies there are no 3-crossings in odd polygons except the center point. The best paper referenced seems to be here in arXiv

• Very useful! Thanks. Are you aware of any known result about the general form of the question? – user180918 Nov 8 '14 at 3:44
• No, I know nothing about more general polygons. Amusingly, the number in OEIS, $6561=3^8$ – Ross Millikan Nov 8 '14 at 3:49

If $$n$$ is odd, therefore the diagonals don't intersect each other at the center. Note that if the two diagonals intersect, it will create one intersection. So the problem is similar to find how many quadrilaterals can be formed by connecting the vertices of the polygon, which is $$^nC_4$$. If $$n$$ is even, the diagonals will intersect at the centre. So we need to find the numbers of diagonals which pass through the centre. Note further that we just need to find one point, and find another point which is symmetry to the first point. Connect these two points will give us a diagonal. Thus the number of diagonals pass through the centre is $$^{\frac{n}{2}}C_1$$. But one of them is to be included, so, the total intersection is $$^nC_4 - ^{\frac{n}{2}}C_1 + 1$$.

A formula for a regular polygon's (n$$\ge$$3) pts of intersection is

The image of formula

Here $$\delta_m(n)=1$$ if n is a multiple of m, $$0$$ otherwise.

Note that for odd m, it is simply $$^nC_4$$.