Firstly, I would like to say that I don't have any idea if this problem is known or not, if it was asked before or not, it came in my teacher's mind during the class. The problem is the following:

All the points of the plane are colored in two colors, let's say blue and red, but it is not mono-colored. Let $ABCD$ be an arbitrary quadrilateral (we don't care about its position or the colors of its vertices).

  1. Is it true that there exists a similar quadrilateral to $ABCD$ such that all of its vertices have the same color, i.e. blue or red?

  2. What about a similar quadrilateral with $ABCD$ such that 3 vertices have the same color, and the other one the other color?

Two polygons are similar if corresponding sides taken in the same sequence (even if clockwise for one polygon and counterclockwise for the other) are proportional and corresponding angles taken in the same sequence are equal in measure.

What I have tried: Before this, we solved another problem (instead of quadrilateral, we considered a random triangle, and the question was " Is it true that there exists a similar triangle to the given triangle such that all of its vertices have the same color, i.e. blue or red?" and that led to question this about a quadrilateral. One solution can be found here (it is the same thing for a random triangle). Using this, I obtained that there exists a triangle $A'B'C'$ similar to $ABC$. Next, I considered $D'$ such that $A'B'C'D'$ is similar to $ABCD$. From here it is obvious that we have either situation 1 or 2 in some cases, but this is not a proof for neither of them.


Certainly 2. may not hold, since the plane could be mono-colored. To show that 1. always holds when $ABCD$ is a parallelogram, construct a $14 \times 14$ lattice of repetitions of $ABCD$, and use the known fact that a two-colored $15\times 15$ grid must contain a mono-colored square (Bacher and Eliahou, Extremal binary matrices without constant 2-squares). The reference allows $2\times 2$ squares only, so in fact this shows that there is a mono-colored translation of a parallelogram $ABCD$. There may well be a smaller construction that proves the conclusion for similar parallelograms.

Finally, when $ABCD$ is an arbitrary quadrilateral, consider the mapping $f:\mathbb{Z}^3\rightarrow \mathbb{R}^2$ given by $$f((i,j,k))= i \cdot AB + j \cdot AC + k \cdot AD.$$ The Gallai-Witt theorem says that any 2-coloring of $\mathbb{Z}^3$ includes a monochromatic copy of $\{(0,0,0),(1,0,0),(0,1,0),(0,0,1)\}$, possibly scaled by an integer and translated (i.e., a homothetic copy). In particular this is true for the coloring of $\mathbb{Z}^3$ consistent under $f$ with our coloring of ${\mathbb{R}}^2$. But any such homothetic copy maps to a quadrilateral similar to (indeed, a homothetic copy of) $ABCD$.

(Note that this argument immediately extends to any finite set of points and any dimension... given any 2-coloring of $\mathbb{R}^d$ and any finite set $A\subset \mathbb{R}^d$, there is a homothetic and monochromatic copy of $A$.)

  • $\begingroup$ The plane is not mono-colored, I forgot to mention that. I edited the post, so 2. could still hold $\endgroup$ – razvanelda Dec 15 '18 at 15:05

This is an answer to 2 for the case of squares: If the plane is not monocolored, there exists a square with exactly three same-colored vertices.

Assume there is no square with exactly three same-colored vertices.

Take two points A and B that are of different color (call A blue, and B red). Take their midpoint C, and without loss of generality, assume C also blue (the case of red C is symmetric).

Let D and E be such that ADBE is a square. A and B are differently colored, therefore D and E must also be differently colored. We can assume D is blue and E is red (the other case is symmetric).

Consider square AHCI. A and C are the same color, so H and I must also be the same. Similarly, from square ICFD, I and F must be the same color.

Similarly, from squares CGBF and HEGC we can conclude that F and G are differently colored, and so are H and G.

Thus, G has a different color from F, H and I, and so square FGHI has exactly three same-colored vertices.

enter image description here

Unfortunately, unlike the proof for triangles, this doesn't directly translate to general quadrilaterals (or even parallelograms).


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