Given 2021 different points in a circle. Let S be the number of acute triangles that are formed using 3 points from 2021 above. Find max S Given 2021 different points in a circle. Let S be the number of acute triangles that are formed using 3 points from 2021 above. Find max S
This is a problem in a summer camp in my country. This problem I find it difficult because the number of triangles is so big and calculate several particular situations such as 2021 points make up a polygon can’t help because I don’t know whether these specific situation is relevant or not (by the way if 2021 points make up a polygon, S = 2021C3 -2021*1010C2)
 A: When you don’t know how to solve a problem, try writing down what you do know and see whether that gets you to your destination.  That’s the approach I took to reach the following solution.
Since all of the points are on a circle, no set of $3$ points is collinear.  Therefore, we always have $\binom{2021}{3}$ triangles, so it’s enough to minimize the number of non-acute triangles.
We start by observing that a triangle is acute if and only if every diameter of the circle passes through its interior. Next, we observe that we can assume that no two points are on the same diameter.  If they were, every triangle using those two points would be a right triangle, which is not acute.  We could therefore increase $S$ (or at least be sure we’re not decreasing it) by moving one of the points a non-zero angular distance smaller than the minimum angular distance between any two of our $2021$ points and their antipodes.
To make the notation easier (and more general), define $n=2021$.  Choose an arbitrary diameter and assume $x$ points are on one side of that diameter, so that $n-x$ points are on the other side.  Then we have at least $\binom x3 + \binom {n-x}{3}$ obtuse triangles (possibly more, if a different diameter divides the points in a different way), and we want to minimize that sum in terms of $x$.
When you do the algebra, you discover that the sum is $\dfrac 12 (n-2)(x^2-nx+C)=\dfrac 12 (n-2)(x- \dfrac n2)^2+C_1$, where $C, C_1$ are constants that depend only on $n$.  Therefore, the sum has an absolute minimum at $x=\dfrac n2$, and the farther you get from that minimum (in either direction), the larger the sum becomes.
That means that a minimum occurs if every diameter that doesn’t hit a point has $1011$ points on one side and $1010$ points on the other, if that situation can be arranged.  But it can, by simply spacing the points evenly around the circle.  Thus,
$$\max S = \binom {2021}{3}-\binom{1011}{3}-\binom {1010}{3}.$$
