Every point in the plane was colored in red or blue. Prove that one the two following statements is true:
$\bullet$ there exist two red points at distance $1$ from each other;
$\bullet$ there exist four blue points $B_1$, $B_2$, $B_3$, $B_4$ such that the points $B_i$ and $B_j$ are at distance $|i - j|$ from each other, for all integers $i $ and $j$ such as $1 \le i \le 4$ and $1 \le j \le 4$.
Being unable, as a new user, to post pictures, I'll just sketch a proof.
* If they exist at all, the blue points B_{1,2,3,4} are collinear. (Prove by drawing circles of radii 1,2,3 around each B_i.)
* If there's a finite number of red points, everything to the right of one of them is blue and finding B_{1,2,3,4} in that region is trivial.
* Otherwise, suppose there is no unit-spaced collinear B_{1,2,3,4} and no red points a distance 1 apart. We derive a contradiction as follows.
** Tile the plane with equilateral triangles, putting a lattice point at each vertex with WLOG one of them at (0,0). Colour the (0,0) point red, thus the six around it are each blue. Now apply two rules for adjacent nodes in the lattice: (i) ?BB? => RBB? or ?BBR; (ii) if two segments BBx and BBx make a 60 degree angle at their intersection x then x=B. Applying rules (i) & (ii) repeatedly, you get a lattice of BBR triangular tiles.
** The same construction applies at any rotation of the lattice, generating a red circle of radius sqrt(3) around (0,0). Clearly, this circle contains pairs of red points a distance 1 apart: contradiction.