IMC 2026 · Day 2 · P10

We prove that the statement is true.

The standard one-dimensional Dirichlet approximation theorem says that, for every real $\alpha$ and positive integer $Q$, there are integers $p, q$ with $1 \leq q \leq Q$ and $|q \alpha - p| < 1/Q$. We need its simultaneous form, sometimes called multidimensional Dirichlet approximation. Its role here is to approximate all coordinates at once using the same multiplier $K$:

Lemma. For real numbers $\alpha_{1}, \ldots, \alpha_{m}$ and a positive integer $Q$, there exists an integer $K$ with $1 \leq K \leq Q^{m}$ such that $$\left| K \alpha_{j} - p_{j} \right| < \frac{1}{Q} \quad (j = 1, \ldots, m)$$ for suitable integers $p_{1}, \ldots, p_{m}$.

Proof. Consider the $Q^{m} + 1$ points $$\left( \left\{ k \alpha_{1} \right\}, \ldots, \left\{ k \alpha_{m} \right\} \right) \in [0,1)^{m}, \quad k = 0, 1, \ldots, Q^{m}.$$ Split $[0,1)^{m}$ into $Q^{m}$ half-open cubes of side length $1/Q$. By the pigeonhole principle, two of the $Q^{m} + 1$ points lie in the same cube; denote their indices by $k < \ell$. For $K = \ell - k$, subtracting the two coordinatewise gives integers $p_{1}, \ldots, p_{m}$ satisfying the required inequalities and proves the lemma.

Let the points be $\left( x_{1}, y_{1} \right), \ldots, \left( x_{N}, y_{N} \right)$ and apply the lemma with $m = 2N$ and $Q = 5$. We obtain an integer $1 \leq K \leq 5^{2N}$ and integers $p_{i}, q_{i}$ such that $$\left| K x_{i} - p_{i} \right| < \frac{1}{5}, \quad \left| K y_{i} - q_{i} \right| < \frac{1}{5} \quad (i = 1, \ldots, N).$$ Thus, after multiplication by the same integer $K$, every coordinate is “close” to an integer.

Take an axes-aligned checkerboard of size $$d = \frac{1}{2K} \leq \frac{1}{2} < 1,$$ translated by $d/2$ in each coordinate, i.e. with the origin being the center of one square and let us colour this square white. The grid lines then satisfy $x/d + 1/2 \in \mathbb{Z}$ and $y/d + 1/2 \in \mathbb{Z}$ and a point $(x, y)$ which does not belong to any grid line is white precisely when $$\left\lfloor \frac{x}{d} + \frac{1}{2} \right\rfloor + \left\lfloor \frac{y}{d} + \frac{1}{2} \right\rfloor \equiv 0 \quad (\bmod 2).$$

Write $K x_{i} = p_{i} + \delta_{i}$ with $\left| \delta_{i} \right| < 1/5$. The choice $d = 1/(2K)$ makes the main term $2 p_{i}$ even, while the translation by $d/2$ keeps the small error inside the same square. Indeed, $$\frac{x_{i}}{d} + \frac{1}{2} = 2 K x_{i} + \frac{1}{2} = 2 p_{i} + \underbrace{2 \delta_{i} + \frac{1}{2}}_{=: \varepsilon_{i}}.$$ Since $\left| 2 \delta_{i} \right| < 2/5$, we have $$0 < \varepsilon_{i} < 1,$$ so $$\left\lfloor \frac{x_{i}}{d} + \frac{1}{2} \right\rfloor = 2 p_{i},$$ which is even. The same argument gives $$\left\lfloor \frac{y_{i}}{d} + \frac{1}{2} \right\rfloor = 2 q_{i},$$ also even. Hence the sum of the two square indices is even, so every point lies in a white square. The strict inequalities $0 < \varepsilon_{i} < 1$ also show that no point lies on a grid line.