Number of Diagonal Crossings in Regular Polygon I have the following problem that I'm unsure how to approach.  It reads, let $P$ be an $n$-sided regular polygon such that every diagonal of $P$ lies inside $P$.  count the number of pairs of diagonals of $P$ that cross.
It seems that there should be some pattern here similar to patterns for the number of diagonals, $\tfrac12 n(n-3)$, but after looking over a few basic examples, none has come to mind.  Could someone please help me get started?
 A: A start: (and nearly a finish) Choose any $4$ vertices. Then exactly one pair of the diagonals determined by these vertices meets inside the polygon. 
Note that this counts the number of intersecting pairs, and not the number of intersection points. That problem is quite complicated, because of symmetries. 
A: Mr. Nicolas' proof is better than mine for its conciseness and elegance of reasoning - but since sometimes it helps to see more than one way to prove something, here is they way I got the answer:
Number the vertices from 1 to N.
Note there is no diagonal from a vertex K to itself or K-1 or K+1.  Likewise for the special cases of K=1 or N there is no diagonal from vertex 1 to N.
Note also that for arbitrary vertices A and B, a diagonal CD will cross AB iff C is between A and B going around the polygon in one direction and D is between A and B going in the other direction.
If B>A then there are B-A-1 vertices between A and B going in one direction and N+A-B-1 going in the other.  This makes (B-A-1)(N+A-B-1) diagonals that cross AB.
Now we can count all the intersecting pairs by iterating over A and B in a way that will "travel around" the polygon.  This will count the pairs 4 times, once for each end of each intersecting diagonal.  This gives us:
(Sum
    (for A=1 to N)
    of (Sum
        (for B=A+2 to N+A-2)
        of (B-A-1)*(N+A-B-1)
    ))/4
Note the terms of the product only depend on (B-A) or (A-B), not on A or B individually.  They start at 1 and iterate through N-3 and the iteration is symmetrical with a pattern like i*(Q-i).  So we can reindex the inner sum to make:
(Sum (for i=1 to N-3) of i*(N-2-i))
At this point the inner sum is independent of the outer one, so the outer sum is the same as just multiplying once by N.  This makes sense because symmetry makes all the vertices of a regular polygon equivalent.  We now have:
N * (Sum (for i=1 to N-3) of i*(N-2-i))/4
Which refactoring = N * (  N*(Sum of i) - 2*(Sum of i) - (Sum of i^2)  ) / 4
At this point we apply the identities:
(Sum (from 1 to M) of i) = M*(M+1)/2
and 
(Sum (from 1 to M) of i^2) = M*(M+1)*(2M+1)
to get N * [ (N-3)(N-2)/2 - (N-3)(N-2) - (N-3)(N-2)(2n-5)/6 ] / 4
Factoring out the (N-3)(N-2) and making 6 the common denominator inside, we get
N*(N-1)(N-2)(N-3)/24
Which = (Ways to choose 4 from N) as according to Mr. Nicolas' proof.
