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?
Given three coplanar points, either the three points form a triangle, or they are colinear.
If they are colinear, we can simply pick another point on the same line that is a rational distance from one of our given points, and it will be a rational distance from all of them.
If the points are not cloinear, they form a triangle. Pick one of the three points arbitrarily, and call that the origin (O). Pick one of the two remaining points, call it A and assign it the cartesian coordinates (a, 0), where a is the distance from O to A. This determines a coordinate system for the plane, and the remaining point (B) has the coordinates (bx, by). We know that the points are all a rational distance apart, which gives us the following (by the Pythagorean theorem):
- i) a is rational
- ii) bx^2 + by^2 is rational
- iii) (a-bx)^2 + by^2 is rational
Expanding iii), we get:
- iv) a^2 - 2a*bx + bx^2 +by^2 is rational
Which implies (using i) and ii)) that 2a*bx is rational, so:
- v) bx is rational, and thus
- vi) by^2 is rational.
Define a new point C at (2bx, 0). Now we have the following distances:
- |OC| = |2bx| (rational)
- |AC| = |a - 2bx| (rational)
- |BC| = sqrt(bx^2 + by^2) = |OB| (rational)
The only wrinkle is if C is not distinct from O or A [note: it must be distinct from B because by can't be zero since O, A, and B are non-colinear].
If C = O, then angle BOA is a right angle, and the solution is simple (eg. let C1 = (-a,0)).
If C = A, it's trickier. If O, A, and B don't form an equilateral triangle, we can start over choosing a different O and the problem will be averted.
If O, A, and B form an equilateral triangle, I haven't been able to find a point thats a rational distance from the original three, or a proof that one doesn't exist. As it stands, I've proven that such a point exists in all cases except the equilateral triangle case.
not quite a solution
