Estonian Mathematical Olympiad

a. Each side and diagonal connects vertices that are $1$ to $1012$ side lengths apart as you move along the polygon. Exactly $2$ sides or diagonals of a given length meet at each vertex. We color the sides and diagonals that have an odd number of sides between their endpoints red and all other diagonals blue. Since there are an equal number of odd and even lengths, there are also an equal number of $1012$ red and $1012$ blue segments meeting at each vertex.

b. Suppose we can color the diagonals as required. Then each vertex has $1011$ red (and $1011$ blue) diagonals. Adding up the numbers of red diagonals that meet at each vertex, we get $1011\cdot 2025$, which is an odd number. On the other hand, each diagonal has been counted exactly $2$ times (once at each of its endpoints), so the sum should be even. The contradiction shows that the situation is not possible.

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Alternative solution.

Consider a graph whose vertices are the vertices of a given polygon and whose edges are its colored sides and diagonals.

a. We note that each vertex has $2024$ adjacent edges, which is an even number. Hence there is an Euler cycle in this graph. In total, there are $\frac{2025\cdot 2024}{2}=2025\cdot 1012$ edges in this graph, which is also an even number. We can color the edges in the resulting Euler cycle alternately red and blue. Then there will be one red and one blue edge at each vertex when we pass through that vertex. Since we eventually pass through all the edges and return to the starting vertex, the numbers of blue and red edges at each vertex will be equal.

b. Now there are $\frac{2025\cdot 2022}{2}=2025\cdot 1011$ edges in total, which is odd. If the same number of red and blue line segments meet at each vertex then the total numbers of red and blue segments must be the same. So the total number of edges should be even, a contradiction.