Svatoovsky M.

On a plane, there is a grasshopper Kolya and 2020 of his friends. 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 finishing point if Kolya can reach it after the 2020th jump. For what largest number $\$ N \$$ can there be an initial arrangement of grasshoppers for which there are exactly $\$ N \$$ different possible finishing points?
Evaluation. Method 1. Introduce a coordinate system on the plane so that Kolya is sitting at the point \$(0,0) \$. Let \(\mid \operatorname{vec}\{\mathrm{a}\}_1 \), \(\mid \operatorname{vec}\{\mathrm{a}\}_2 \), \(\ldots\), \(\mid \operatorname{vec}\{\mathrm{a}\}_{2020} \) 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 \(2\left(-\operatorname{vec}\{\mathrm{a}\}_1 + \mid \operatorname{vec}\{\mathrm{a}\}_2 - \mid \operatorname{vec}\{\mathrm{a}\}_3 + \ldots + \mid \operatorname{vec}\{\mathrm{a}\}_{2019} + \mid \operatorname{vec}\{\mathrm{a}\}_{2020}\right)\). The number of different sums is \(C_{2020} \text{ choose } 1010\) (selecting 1010 terms from the 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 \(2\overrightarrow{AB} = 2(\overrightarrow{OB} - \overrightarrow{OA})\). Each finish point can be associated with a sequence in which Kolya jumps over his friends to end up in this point after 2020 jumps—i.e., a permutation of the numbers from 1 to 2020 (consider all grasshoppers, except Kolya, numbered). Suppose Kolya at some point is at point \(0\) 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 \(ABC\), he will first move by the vector \(2\overrightarrow{OA}\) and then by the vector \(2\overrightarrow{BC} = 2(\overrightarrow{OC} - \overrightarrow{OB})\), so in total, after three jumps, he will move by the vector \(2(\overrightarrow{OA} + \overrightarrow{OC} - \overrightarrow{OB})\). If Kolya makes these jumps in the order \(CBA\), he will similarly move by the vector \(2(\overrightarrow{OC} + \overrightarrow{OA} - \overrightarrow{OB})\), ending 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 without changing 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 stand in the odd positions in such a permutation. Accordingly, the number of different finish points cannot exceed \(C_{2020}^{1010}\).

Example. Place the grasshoppers at points on the number line with coordinates 0 (Kolya's position), 1, 3, \(3^2\), ..., \(3^{2019}\). Prove that all sums of powers of three with coefficients \(\pm 1\) are different. By adding the 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.

## [ Examples and counterexamples. Constructions ]  Problem 67098 Topics: [Polyhedra and polygons (other) $]$  [ Rearrangement of areas $\quad]$  Difficulty: $4-$  Classes: $8,9,10,11$  Author: Saghafian $M$.  On the plane, there are eight points in general position. In a row, write down the areas of all 56 triangles with vertices at these points. Prove that between the written numbers, you can place the signs “\$+\$” and “\$-\$” so that the resulting expression equals zero.

## Solution

Notice that the statement of the problem is true for any four points. Indeed, if the points form a convex quadrilateral \(ABCD\), then \(S_{ABC} + S_{ACD} - S_{BCD} - S_{ABD} = 0\). If point \(D\) lies inside triangle \(ABC\), then \(S_{ABC} - S_{ABD} - S_{ACD} - S_{BCD} = 0\).

Now, associate the given points with the vertices of a cube \(ABCDA'B'C'D'\) and consider the following 14 sets of four vertices: the six faces of the cube, the six sections passing through opposite edges, and the two inscribed tetrahedra \(A'CD\) and \(A'BC'D\). Any two of these sets have no more than two points in common, so each of the 56 triangles belongs to exactly one set. Therefore, by placing the signs for each set, we obtain the desired arrangement.