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Difference between revisions of "2017 USAMO Problems/Problem 5"

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We now prove that no labeling as desired exists for any <math>c\ge 2.</math> To do this, we will prove that labeling a <math>2^k</math>-by-<math>2^k</math> square grid of lattice points requires at least <math>k+3</math> labels for all natural numbers <math>k</math>; hence for a sufficiently large section of the lattice plane the number of labels required grows arbitrarily large, so the entire lattice plane cannot be labeled with finitely many labels. We will prove this using induction.
 
We now prove that no labeling as desired exists for any <math>c\ge 2.</math> To do this, we will prove that labeling a <math>2^k</math>-by-<math>2^k</math> square grid of lattice points requires at least <math>k+3</math> labels for all natural numbers <math>k</math>; hence for a sufficiently large section of the lattice plane the number of labels required grows arbitrarily large, so the entire lattice plane cannot be labeled with finitely many labels. We will prove this using induction.
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For the base case, <math>k=1,</math> we have four points in a square of side length <math>1.</math> The maximum distance between any two of these points is <math>\sqrt{2} < 2^1,</math> so all four points must have different labels. This completes the base case.
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Now, for the inductive step, suppose that labeling a <math>2^k</math>-by-<math>2^k</math> square grid of lattice points requires at least <math>k+3</math> labels for some natural number <math>k.</math> We will now prove that labeling a <math>2^{k+1}</math>-by-<math>2^{k+1}</math> square grid of lattice points requires at least <math>k+4</math> labels.

Revision as of 02:53, 3 May 2017

Problem

Let $\mathbf{Z}$ denote the set of all integers. Find all real numbers $c > 0$ such that there exists a labeling of the lattice points $( x, y ) \in \mathbf{Z}^2$ with positive integers for which: only finitely many distinct labels occur, and for each label $i$, the distance between any two points labeled $i$ is at least $c^i$.

Solution (INCOMPLETE)

For $c\le 1,$ we can label every lattice point $1.$ For $c\le 2^{1/4},$ we can make a "checkerboard" labeling, i.e. label $(x, y)$ with $1$ if $x+y$ is even and $2$ if $x+y$ is odd. One can easily verify that these labelings satisfy the required conditions. Therefore, a labeling as desired exists for all $0 < c\le 2^{1/4}.$

We now prove that no labeling as desired exists for any $c\ge 2.$ To do this, we will prove that labeling a $2^k$-by-$2^k$ square grid of lattice points requires at least $k+3$ labels for all natural numbers $k$; hence for a sufficiently large section of the lattice plane the number of labels required grows arbitrarily large, so the entire lattice plane cannot be labeled with finitely many labels. We will prove this using induction.

For the base case, $k=1,$ we have four points in a square of side length $1.$ The maximum distance between any two of these points is $\sqrt{2} < 2^1,$ so all four points must have different labels. This completes the base case.

Now, for the inductive step, suppose that labeling a $2^k$-by-$2^k$ square grid of lattice points requires at least $k+3$ labels for some natural number $k.$ We will now prove that labeling a $2^{k+1}$-by-$2^{k+1}$ square grid of lattice points requires at least $k+4$ labels.

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