Subfield: Combinatorics


Question: There are given $2^{500}$ points on a circle labeled $1,2, \ldots, 2^{500}$ in some order. Prove that one can choose 100 pairwise disjoint chords joining some of these points so that the 100 sums of the pairs of numbers at the endpoints of the chosen chords are equal.

Solution:
The proof is based on the following general fact.

Lemma. In a graph $G$ each vertex $v$ has degree $d_{v}$. Then $G$ contains an independent set $S$ of vertices such that $|S| \geq f(G)$ where

$$
f(G)=\sum_{v \in G} \frac{1}{d_{v}+1}
$$

Proof. Induction on $n=|G|$. The base $n=1$ is clear. For the inductive step choose a vertex $v_{0}$ in $G$ of minimum degree $d$. Delete $v_{0}$ and all of its neighbors $v_{1}, \ldots, v_{d}$ and also all edges with endpoints $v_{0}, v_{1}, \ldots, v_{d}$. This gives a new graph $G^{\prime}$. By the inductive assumption $G^{\prime}$ contains an independent set $S^{\prime}$ of vertices such that $\left|S^{\prime}\right| \geq f\left(G^{\prime}\right)$. Since no vertex in $S^{\prime}$ is a neighbor of $v_{0}$ in $G$, the set $S=S^{\prime} \cup\left\{v_{0}\right\}$ is independent in $G$.

Let $d_{v}^{\prime}$ be the degree of a vertex $v$ in $G^{\prime}$. Clearly $d_{v}^{\prime} \leq d_{v}$ for every such vertex $v$, and also $d_{v_{i}} \geq d$ for all $i=0,1, \ldots, d$ by the minimal choice of $v_{0}$. Therefore

$$
f\left(G^{\prime}\right)=\sum_{v \in G^{\prime}} \frac{1}{d_{v}^{\prime}+1} \geq \sum_{v \in G^{\prime}} \frac{1}{d_{v}+1}=f(G)-\sum_{i=0}^{d} \frac{1}{d_{v_{i}}+1} \geq f(G)-\frac{d+1}{d+1}=f(G)-1 .
$$

Hence $|S|=\left|S^{\prime}\right|+1 \geq f\left(G^{\prime}\right)+1 \geq f(G)$, and the induction is complete.

We pass on to our problem. For clarity denote $n=2^{499}$ and draw all chords determined by the given $2 n$ points. Color each chord with one of the colors $3,4, \ldots, 4 n-1$ according to the sum of the numbers at its endpoints. Chords with a common endpoint have different colors. For each color $c$ consider the following graph $G_{c}$. Its vertices are the chords of color $c$, and two chords are neighbors in $G_{c}$ if they intersect. Let $f\left(G_{c}\right)$ have the same meaning as in the lemma for all graphs $G_{c}$.

Every chord $\ell$ divides the circle into two arcs, and one of them contains $m(\ell) \leq n-1$ given points. (In particular $m(\ell)=0$ if $\ell$ joins two consecutive points.) For each $i=0,1, \ldots, n-2$ there are $2 n$ chords $\ell$ with $m(\ell)=i$. Such a chord has degree at most $i$ in the respective graph. Indeed let $A_{1}, \ldots, A_{i}$ be all points on either arc determined by a chord $\ell$ with $m(\ell)=i$ and color $c$. Every $A_{j}$ is an endpoint of at most 1 chord colored $c, j=1, \ldots, i$. Hence at most $i$ chords of color $c$ intersect $\ell$.

It follows that for each $i=0,1, \ldots, n-2$ the $2 n$ chords $\ell$ with $m(\ell)=i$ contribute at least $\frac{2 n}{i+1}$ to the sum $\sum_{c} f\left(G_{c}\right)$. Summation over $i=0,1, \ldots, n-2$ gives

$$
\sum_{c} f\left(G_{c}\right) \geq 2 n \sum_{i=1}^{n-1} \frac{1}{i}
$$

Because there are $4 n-3$ colors in all, averaging yields a color $c$ such that

$$
f\left(G_{c}\right) \geq \frac{2 n}{4 n-3} \sum_{i=1}^{n-1} \frac{1}{i}>\frac{1}{2} \sum_{i=1}^{n-1} \frac{1}{i}
$$

By the lemma there are at least $\frac{1}{2} \sum_{i=1}^{n-1} \frac{1}{i}$ pairwise disjoint chords of color $c$, i. e. with the same sum $c$ of the pairs of numbers at their endpoints. It remains to show that $\frac{1}{2} \sum_{i=1}^{n-1} \frac{1}{i} \geq 100$ for $n=2^{499}$. Indeed we have

$$
\sum_{i=1}^{n-1} \frac{1}{i}>\sum_{i=1}^{2^{400}} \frac{1}{i}=1+\sum_{k=1}^{400} \sum_{i=2^{k-1+1}}^{2^{k}} \frac{1}{i}>1+\sum_{k=1}^{400} \frac{2^{k-1}}{2^{k}}=201>200
$$

This completes the solution.