62nd Ukrainian National Mathematical Olympiad

Consider any point $A_i$ and count the number of pairs of points $(A_j,A_k)$ such that $\angle A_i{A}_j{A}_k=90^{\circ }$. For each point $A_j$ there exists at most one point $A_k$ (because on the line through $A_j$ perpendicular to $A_i{A}_j$ there can be at most one point other than $A_j$). Also note that if $X$ is the point at the largest distance from $A_i$ then there can be no point $A_k$ with $\angle A_{i}X{A}_k=90^{\circ }$, because then we would have $A_i{A}_k>A_{i}X$.

Thus, each point can be a vertex of the hypotenuse in at most 2020 right triangles at these points. Since the hypotenuse of each triangle has two vertices, the total number of these triangles does not exceed $\frac{2022\cdot 2020}{2}=2022\cdot 1010$.

This number can be achieved because, for example, we could take 2022 points on a circle so that they are divided into 1011 pairs so that in each pair the points form a circle diameter. Note that for each diameter there will be exactly 2020 points that form a right triangle with it. Then we have at least $1011\cdot 2020$ different right angles.