Subfield: Combinatorics


Question: On a circle, Alina draws 2019 chords, the endpoints of which are all different. A point is considered marked if it is either

(i) one of the 4038 endpoints of a chord; or

(ii) an intersection point of at least two chords.

Alina labels each marked point. Of the 4038 points meeting criterion (i), Alina labels 2019 points with a 0 and the other 2019 points with a 1 . She labels each point meeting criterion (ii) with an arbitrary integer (not necessarily positive).

Along each chord, Alina considers the segments connecting two consecutive marked points. (A chord with $k$ marked points has $k-1$ such segments.) She labels each such segment in yellow with the sum of the labels of its two endpoints and in blue with the absolute value of their difference.

Alina finds that the $N+1$ yellow labels take each value $0,1, \ldots, N$ exactly once. Show that at least one blue label is a multiple of 3 .

(A chord is a line segment joining two different points on a circle.)

Solution:
First we prove the following:

Lemma: if we color all of the points white or black, then the number of white-black edges, which we denote $E_{W B}$, is equal modulo 2 to the number of white (or black) points on the circumference, which we denote $C_{W}$, resp. $C_{B}$.

Observe that changing the colour of any interior point does not change the parity of $E_{W B}$, as each interior point has even degree, so it suffices to show the statement holds when all interior points are black. But then $E_{W B}=C_{W}$ so certainly the parities are equal.

Now returning to the original problem, assume that no two adjacent vertex labels differ by a multiple of three, and three-colour the vertices according to the residue class of the labels modulo 3. Let $E_{01}$ denote the number of edges between 0 -vertices and 1-vertices, and $C_{0}$ denote the number of 0 -vertices on the boundary, and so on.

Then, consider the two-coloring obtained by combining the 1-vertices and 2-vertices. By applying the lemma, we see that $E_{01}+E_{02} \equiv C_{0} \bmod 2$.

$$
\text { Similarly } E_{01}+E_{12} \equiv C_{1}, \quad \text { and } E_{02}+E_{12} \equiv C_{2}, \quad \bmod 2 \text {. }
$$

Using the fact that $C_{0}=C_{1}=2019$ and $C_{2}=0$, we deduce that either $E_{02}$ and $E_{12}$ are even and $E_{01}$ is odd; or $E_{02}$ and $E_{12}$ are odd and $E_{01}$ is even.

But if the edge labels are the first $N$ non-negative integers, then $E_{01}=E_{12}$ unless $N \equiv 0$ modulo 3 , in which case $E_{01}=E_{02}$. So however Alina chooses the vertex labels, it is not possible that the multiset of edge labels is $\{0, \ldots, N\}$.

Hence in fact two vertex labels must differ by a multiple of 3 .
As before, colour vertices based on their label modulo 3.

Suppose this gives a valid 3-colouring of the graph with 2019 0s and 2019 1s on the circumference. Identify pairs of 0-labelled vertices and pairs of 1-labelled vertices on the circumference, with one 0 and one 1 left over. The resulting graph has even degrees except these two leaves. So the connected component $\mathcal{C}$ containing these leaves has an Eulerian path, and any other component has an Eulerian cycle.

Let $E_{01}^{*}$ denote the number of edges between 0 -vertices and 1-vertices in $\mathcal{C}$, and let $E_{01}^{\prime}$ denote the number of such edges in the other components, and so on. By studying whether a given vertex has label congruent to 0 modulo 3 or not as we go along the Eulerian path in $\mathcal{C}$, we find $E_{01}^{*}+E_{02}^{*}$ is odd, and similarly $E_{01}^{*}+E_{12}^{*}$ is odd. Since neither start nor end vertex is a 2 -vertex, $E_{02}^{*}+E_{12}^{*}$ must be even.

Applying the same argument for the Eulerian cycle in each other component and adding up, we find that $E_{01}^{\prime}+E_{02}^{\prime}, E_{01}^{\prime}+E_{12}^{\prime}, E_{02}^{\prime}+E_{12}^{\prime}$ are all even. So, again we find $E_{01}+E_{02}$, $E_{01}+E_{12}$ are odd, and $E_{02}+E_{12}$ is even, and we finish as in the original solution.