2
$\begingroup$

Is it possible to color each point of the plane red or blue so that no square with unit side length and monochromatic vertices is formed?

EDIT: I think this is possible,for each point in the plane color the points that are unit length apart with an opposite color.This way it is guaranteed that no monochromatic vertices of unit length apart is formed and so no square with unit side length and monochromatic vertices is formed.

EDIT: The point I was trying to make was that there must be at least one point in the unit length square with a different color, if not it violates the coloring defined above. Hence, it is possible to color points such that no square with unit side length and monochromatic vertices is formed.

$\endgroup$
  • $\begingroup$ What are your thoughts on the problem? What have you tried and where did you get stuck? Where did you find this problem? $\endgroup$ – Servaes Mar 14 at 23:12
  • $\begingroup$ @Servaes updated my question with some thoughts, as to where I found this: A Walk Through Combinatorics $\endgroup$ – thetraveller Mar 14 at 23:41
  • 1
    $\begingroup$ It's not possible to do that. Color a point red. You're saying to color all points in the unit circle around that point blue. But there are chords on the unit circle of length exactly $1$. That means it's not possible to color the plane in such a way that all points that are a unit length apart have different colors. Hell, just consider the vertices of an equilateral triangle with sides of length $1$. $\endgroup$ – Robert Shore Mar 15 at 3:50
  • 1
    $\begingroup$ @RobertShore you appear to be trying to avoid a pair of same-color points at unit distance from each other, but that's a much stronger restriction than the restriction being proposed in the question, which only wants to avoid the corners of a unit square all having the same color. $\endgroup$ – Gregory J. Puleo Mar 15 at 4:01
  • 1
    $\begingroup$ I was responding to the addition in the EDIT, which attempts to satisfy the much stronger condition (a task that can't be accomplished). $\endgroup$ – Robert Shore Mar 15 at 6:13
2
$\begingroup$

Horizontal bands with height exactly $1$, closed on the bottom, open on the top, provide such a coloring. Thus, $(x, y)$ is red if $2|\lfloor y \rfloor$ and blue otherwise.

Number the square's vertices in order of height and let $\theta$ be the angle between a horizontal line passing through vertex $1$ and the side of the square $\overline{12}$. Without loss of generality, $0 \leq \theta \leq \frac{\pi}{2}$. (Otherwise perform a reflection.) Let $y_1$ be the $y$-coordinate of vertex $1$. Then the $y$-coordinates of the four vertices are:

$$y_1\\ y_2 = y_1+\sin \theta\\ y_3=y_1+ \cos \theta \\ y_4=y_1+\sin \theta + \cos \theta = y_1+\sqrt{2}\sin (\theta + \frac{\pi}{4}).$$

The maximum height difference between any two vertices is $\sqrt{2} \lt 2$ so a monochromatic square is possible only if the height difference between some two consecutive vertices is strictly greater than $1$.

But that can't possibly happen. It obviously can't happen between $1$ and $2$, $3$ and $4$, $2$ and $4$, or $3$ and $1$. And $y_3-y_2= \cos \theta - \sin \theta = \sqrt{2} \cos (\theta + \frac{\pi}{4})$ so $0 \leq \theta \leq \frac{\pi}{2} \Rightarrow |y_3-y_2| \leq 1$, completing the proof.

Edited to respond to comments requesting additional detail.

$\endgroup$
  • 1
    $\begingroup$ -1 "I don't think there's any way" is far from a proof, as your previous answer shows. Without any argument this seems more like a comment than an answer. $\endgroup$ – Servaes Mar 15 at 11:13
  • $\begingroup$ People use answers all the time to outline solutions without filling in all the details. Let's leave a little something (in this case basic trigonometry) for the original poster to do here. $\endgroup$ – Robert Shore Mar 15 at 15:32
  • 1
    $\begingroup$ @Servaes I've edited to fill in the gaps. $\endgroup$ – Robert Shore Mar 15 at 19:34
  • $\begingroup$ Great! By the way; since the vertices are ordered by height, without loss of generality $\theta\leq\tfrac{\pi}{4}$ even. $\endgroup$ – Servaes Mar 15 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.