Recently I've encounter a problem:
Every space point is colored either red or blue. Show that among the squares with side 1 in this space there is at least one with three red vertices or at least one with four blue vertices.
The solution is somehow hard-to-think-of (at least for me):
Denote by B the property that there is a unit square with four blue vertices. Case 1: All points of space are blue ⇒ B. Case 2: There exists a red point P1. Make of P1 the vertex of a pyramid with equal edges and the square P2P3P4P5 as base. Case 2.1: The four points Pi, i = 2, 3, 4, 5 are blue ⇒ B. Case 2.2: One of the points Pi, i = 2, 3, 4, 5 is red, say P2. Make of P1P2 a lateral edge of an equilateral prism, with the remaining vertices P6, P7, P8, P9. Case 2.2.1: The four points Pj, j = 6, 7, 8, 9 are blue ⇒ B. Case 2.2.2: One of the points Pj, j = 6, 7, 8, 9 is red, say P6. Then P1, P2, and P6 are three red vertices of a unit square.
But here comes the question: shouldn't red and blue have some sort of symmetry? Can we assert from the above-mentioned proof and from symmetry that there is at least one with FOUR red vertices or at least one with four blue vertices? I fail to figure out whether this is true or false, but I simply find it wierd to differentiate between red and blue.