Prove that one cannot choose more than $100n^{2/3}$ vertices with integer co-ordinates between 1 and n that form a convex polygon.
I tried this problem for a while ,looking at a maximal such polygon and arguing by looking at points outside and how it behaves it with three consecutive sides of the polygon(precisely for four consecutive vertices on the polygon,and one vertex outside we have at least one of the three angles obtuse due to maximality reasons) and tried simple double count which doesn't seem to work.
The proof is based on the edge vectors. We orient the polygon in trigonometric direction. In order for the polygon to be strictly convex, the edge vector angle must be a strictly increasing function, from $0$ (first edge, considered as origin of angles) to less than $2 \pi$ (last edge).
The edge vectors have integer coordinates. Let $p$ be the maximum number of edges that the polygon has in any of the $4$ quadrants. Let $k$ be the integer such that $k(k+1) \le p \lt (k+1)(k+2)$. To get a lower bound for the extent of the polygon in abscissa and ordinate, we can look at the extent in the first quadrant, because the edges in this quadrant cannot be mixed with the edges in any other quadrant; and any path with the edges in this quadrant can be realized symmetrically in the $3$ other quadrants.
Note that in $[1, k] \times [0, k]$ there are colinear vectors, so we do not fully use the fact that the edges cannot be colinear. Hence we could strengthen the bound further. However the reduction factor is at most $1$ minus the probability of two integers to be coprime, which is $6 / \pi^2$, i.e. ~$0.61$, so we would achieve a $0.39$ reduction, which is not much (considering that it is a constant factor) and the proof would be more difficult.
The vertices coordinates are integers all in $[1, n]$ as stated in the problem description. To get a relation between $k$ and $n$, we want to know how far in abscissa and ordinate the sum of the first quadrant edge vectors can get. The max between the extent in abscissa and the extent in ordinate, for a path with $k(k+1)$ edges in the first quadrant, is realized by edges all in $[1, k] \times [0, k]$: if one edge with angle in $[0,2 \pi[$ is not in $[1, k] \times [0, k]$, the extent is greater in abscissa and/or in ordinate. So the extent is upper-bounded by
$(k+1) \sum_{j=1}^k j$
$= \frac {k (k+1)^2} 2$
As the polygon extends twice that amount for abscissa (and same for ordinate), the polygon span is $\ge k (k+1)^2$. Our hypothesis is that the vertices coordinates are in $[1, n]$, so we have $k(k+1)^2 \le n$,
With this we get $k(k+1)$ vertices per quadrant, so $4 k(k+1)$ in total.
If $v$ is the total number of vertices, $v \le 4p \lt 4(k+1)(k+2)$.
$n \ge k(k+1)^2 \gt k^{3/2} (k+1)^{3/2}$ if $k \ge 1$.
$n^{2/3} \gt k(k+1) = \frac k {k+2} (k+2) (k+1)$
If $k \ge 1$, $\frac k {k+2} \ge \frac 1 3$, so
$n^{2/3} \gt \frac 1 3 (k+1)(k+2) \gt \frac 1 {12} v$
$v \lt 12 n^{2/3}$, with this bound tending to $4 n^{2/3}$ when $n$ grows.
Actually $4 n^{2/3}$ seems to be true everywhere. However I don't know how to prove it; and anyway we are already better that the $100 n^{2/3}$ which was asked.