combinatoricsgeometry
In a $21$ sides regular polygon, how many points inside it are
intersection of its diagonal?
I found that a polygon with $n$ sides has $\dfrac{n(n – 3)}{2}$ diagonals, but I feel this is not so useful to the problem solution. I've been trying for $3$ hours without success.
What's the correct solution?
This is part of a contest that is already finished (the solutions have not been released yet).
We interpret diagonal as meaning a line segment that joins non-adjacent vertices.
There are $\dbinom{18}{4}$ ways to choose $4$ vertices. If the polygon is strictly convex, then exactly one pair of the lines joining the vertices meets in the interior of the polygon. So the required number of intersection points of diagonals is $\dbinom{18}{4}$.
As stated the problem can have many solutions.
For example:
Let the quadrilateral $ABCD$ such that $AD=DB=BC=x$ and $AC=\sqrt{3}x$. See figure 1.
Figure 1
The angle $b$ depends on $a$ as the expression:
$$b = \frac{a}{2}- \frac {\pi}{2}+ \arccos(- \frac{1}{2 \sin {\frac{a}{2}}}+ \sin {\frac{a}{2}})$$
So we have many solutions, for example: $a=\frac{\pi}{3}$ and $b=\frac{\pi}{3}$ or $a=\frac{\pi}{4}$ and $b=\frac{\pi}{2}$ as shown in figure 2.
Figure 2
Best Answer
You may find the following idea useful. Take a convex $n$-gon. Suppose that there is no point inside the $n$-gon at which three diagonals meet. Then there are $\binom{n}{4}$ intersection points of diagonals inside the $n$-gon.
There are various ways to get at this result, but only one simplest one. Choose $4$ vertices. Exactly one of the pairs of lines determined by these $4$ points meets in the interior of the $n$-gon, and therefore the total number of intersection points in the interior of the $n$-gon is $\binom{n}{4}$.