is the inequality right? It seems i can be 1 and k can be n

Kingsbane2139 wrote: is the inequality right? It seems i can be 1 and k can be n My bad, fixed, thanks!

anybody can solve it?

Ok, here is a sketch. The construction part is a tricky one in my opinion because the problem statement is not super nice to work with, surprising problem. But after you get the construction for n=5, you can definitely think of the generalization. Answer: All $n$ odd work. proof that $n$ even doesn't work: Assume such points and a permutation exists. The permutation can be viewed as a sequence of permutation where two numbers are always swapped. Now during these switches you can easily check that in the case $n$ even The parity of the number of positive oriented triangles doesn't change. But the start (before permuting) and the end amount differ by 1. Contridiction Construction for $n$ odd: $n=3$ is obvious. For bigger $n$, first construct a regular $n-2$-gon $A_1A_2...A_{n-2}$ with midpoint $O$. Now, pick $P$ and $Q$ such that $PQ$ is parallel to $A_1A_2$, $O$ is the midpoint of $PQ$ and the length of $PQ$ is super small. We take $A_1, ..., A_{n-2}, P, Q$ as our points, and the permutation is $P$<->$Q$, $A_i$->$A_{i+\frac{n-1}{2}}$. It is easy to see that this works.