Given three points on a plane, at rational distances from each other, can you always find at least another point in the same plane such that the distances to the other three are also rational?
In my previous comment, I still needed to prove it for the case when the three points you're given form an equilateral triangle.
Suppose we have three points:
- O: (0,0)
- A: (2,0)
- B: (1,sqrt(3))
This forms an equilateral triangle with sides of length 2. Since the rational numbers are closed under scalar multiplication, it suffices to show the result for this triangle.
Put a point C at (5/4,0). Clearly, C is a rational distance from both O and A. It remains to show |CB| is rational.
- |CB| = sqrt((1/4)^2 + sqrt(3)^2)
- = sqrt(1/16 + 3)
- = sqrt(1/16 + 48/16)
- = sqrt(49/16)
- = 7/4 rational!
Therefore, this together with my previous post prove that indeed for any three coplanar points at rational distances from each other, there exists at least one other point in the plane that is a rational distance from all three.
the rest of my solution
