Dividing circles into arcs. There are $2022$ points on a circle. These points are distributed uniformly (distance between $2$ neighbor points is the same) and this circle is divided into arcs where the endpoints of these arcs are points on the circle such that,
(i) No two arcs overlap.
(ii)  The lengths of arcs is pairwise different.
(iii)  The smallest arc doesn't intersect with the biggest arc i.e they don’t share the same endpoint.
Let $k$ be the maximal number of arcs we can put in the circle. Find $k$ for $2022$ points.
So the first thing I did is try small cases, here is what I found
\begin{array} {|r|r|}\hline n & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\ \hline k & 1 & 1 & 1 & 2 & 2 & 3 & 3 & 3 & 4 & 4 & 4 & 4 \\ \hline  \end{array}
Where $n$ is the number of points on the circle.
The numbers are weird ; they kind of repeat and then jump by $1$ and actually I noticed that $k$ is exactly the biggest arc in the circle but I have no way to prove this. I have an idea that I'm not sure of, we can transform this problem into something like that $$n=(2+1+3+4+...+k)+r$$
Where $0\le r\le k$ and notice that I put $2$ first intead of $1$ so that we can avoid the biggest arc intersecting with the smallest arc.
 A: Notice that if we let the lengths of the arcs to be $a_1<\cdots <a_k$, then we have that $a_1+\cdots+a_k \leq 2022$. Moreover, since $a_i<a_{j}$  for appropriate $i<j$, if we let $b_i=a_i-i$, then $b_i\leq b_j$ for $i<j$.
The question then just boils down to the existence of a non-negative integer solution to $x_1+\cdots+x_k\leq 2022-\frac{k(k+1)}{2}$ such that $x_i\leq x_j$ for $i<j$. Whenever, there is a solution to $x_1+\cdots+x_k\leq 2022-\frac{k(k+1)}{2}$, we can just put it in non-decreasing order and we can get such a required solution.
A solution to $x_1+\cdots+x_k\leq 2022-\frac{k(k+1)}{2}$ exists whenever the RHS is non-negative, therefore, we want the largest $k$ such $2022-\frac{k(k+1)}{2}\geq 0$.
Since $2016=\frac{63\times 64}{2}$, $k$ must be $63$.
As for the actual lengths, take a partition of $6$, say $3+1+1+1$. Now, add them to $63,62,61,60$ respectively to get $1+2+\cdots+59+61+62+63+66=2022$
A: Because the sums of $k$ distinct positive integers is at least $1+2+\cdots+k=\frac{k(k+1)}2,$ if $\frac{k(k+1)}2>2022,$ you certainly can't solve it.
So you need $$\frac{k(k+1)}2\leq n$$
Multiplying by $8$ and adding one, we see this is equivalent to $(2k+1)^2\leq 8n+1$ or: $$k\leq\frac{\sqrt{8n+1}-1}2.\tag1$$
But if $k>3,$ we can use arcs of minimal lengths in circular order: $1,2,k,3,\dots,k-1,$ to satisfy the third condition.
(When $k\leq 3,$ the third condition makes it harder, which is why the largest such $k$ doesn't work form $n=3,4,6.$)
So you need the largest $k$ such is: $$\left\lfloor\frac{\sqrt{8\cdot2022+1}-1}2\right\rfloor=63$$
We can actually use the whole circle, by replacing the $63$ arc with the remaining length, $69.$ So you get:
$$1+2+69+3+4+\cdots+62=2022.$$
