Svyatlovsky M.

On a plane, grasshopper Kolya and 2020 of his friends are sitting. Kolya is going to jump over each of the other grasshoppers (in any order) such that the starting and ending points of each jump are symmetric relative to the grasshopper being jumped over. We will call a point a finish point if Kolya can end up in it after the 2020th jump. For what largest number $N$ can there be an initial arrangement of grasshoppers such that there are exactly $N$ different possible finish points?
Evaluation. Method 1. Introduce a coordinate system on the plane so that Kolya is sitting at the point $(0,0)$. Let $\vec{a}_{1}, \vec{a}_{2}$, .., $\vec{a}_{2020}$ be the radius vectors of the grasshoppers in the order of their jumping over Kolya. It is not difficult to find the radius vector of the finish point $2\left(-\vec{a}_{1}+\vec{a}_{2}-\vec{a}_{3}+\ldots-\vec{a}_{2019}+\vec{a}_{2020}\right)$. The number of different sums is $C_{2020}^{1010}$ (the number of ways to choose 1010 terms in odd positions).

Method 2. Notice that if a grasshopper jumps over some point $A$ and then over some point $B$, it will ultimately move by the vector $2 A B=2(O B-O A)$.

Each finish point can be associated with a sequence in which Kolya jumped over his friends to end up in this point after 2020 jumps - that is, a permutation of the numbers from 1 to 2020 (consider all grasshoppers except Kolya to be numbered). Suppose Kolya at some point is at point $O$ and is about to jump over grasshoppers located at points $A, B, C$ (considering only three consecutive jumps). If Kolya makes these jumps in the order $A B C$, he will first move by the vector

$2OA$, and then over two jumps will move by the vector $2 B C=2(O C-O B)$, so in the end, after three jumps, he will move by the vector $2(\overrightarrow{O A}+\overrightarrow{O C}-\overrightarrow{O B})$. If Kolya makes these jumps in the order $C B A$, then, similarly, he will move by the vector $2(O C+O A-O B)$, that is, he will end up in the same point as in the previous case. Thus, in the sequence of Kolya's jumps, any two grasshoppers standing one apart can be swapped, and this will not change the finish point. Therefore, the finish point is characterized not by the permutation of the numbers from 1 to 2020 (of which there are 2020!), but only by which numbers in such a permutation stand in odd positions. Accordingly, there cannot be more than $C_{2020}^{1010}$ different finish points.

Example. Place the grasshoppers at points on the number line with coordinates 0 (at Kolya), 1, 3, 3², ..., $3^{2019}$. Prove that all sums of powers of three with coefficients $\pm 1$ are different. By adding a constant sum $1+3+3^{2}+\ldots+3^{2019}$ to such a sum, we get a sum with coefficients 0 and 2. It corresponds to a 2020-digit ternary number consisting of zeros and twos, and all such numbers are different.

## Answer

For $N=C_{2020}^{1010}$