Bicolor floor

@X X This is a great idea that works perfectly on any plane with length $l$ and width $w$ such that $\displaystyle \left \{ l, w \right \} \in \mathbb{R} \ge 10\sqrt{ 2 + \sqrt{3} }$ . But such a triangle couldn't exist on a plane where either $l$ or $w$ were less than $10\sqrt{ 2 + \sqrt{3} }$ (consider the square in which a triangle with side length 20 is inscribed) because it would extend beyond the boundaries of the plane.

For planes with length $l$ and width $w$ such that $10\sqrt{2} < l \le w < 10\sqrt{ 2 + \sqrt{3} }$ , you would need a different strategy. One way to think about it is to consider the points as squares on a chess board. Do you see what I mean?