Here is the problem I've faced with.
In a convex n-gon (n≥4), no three diagonals intersect at the same point. Prove that the number of all intersection points of the diagonals is equal to the number of quadrilaterals whose vertices are all vertices of a given n-gon.
I've tried to search the solution in a browser and actually I've been trying to solve this problem for 2 days, but I can't. So, how to solve that, and what is the algorithm to the solution ?
Be aware that this is only 8-grade problem. And this one is from the "Finite sets. Mutually unambiguous correspondence" topic.
Pick any 4 points in the n-gon. Call them A, B, C, and D going around the n-gon clockwise. Note that there is exactly one quadrilateral defined by these 4 points, ABCD, and there is exactly one intersection caused by the segments AB, BC, CD, DA, AC, and BD, namely the intersection between AC and BD. Clearly, there is a bijection between any 4 points and the quadrilateral they define, and because of the no-3-lines-intersecting rule there is also a bijection between any intersection point and the 4 endpoints of the 2 lines that intersect there, so there is a one-to-one correspondence and we are done.