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If the three points are on a line, the answer is obviously "yes".
If the three points determine an equilateral triangle, their coordinates can be (0,0), (1,0) and (½,½√3), so taking the fourth point at (52/49, (12√3)/49) solves the problem.
Finally, if the points form any other triangle, we can set A(0,0), B(c,0) y C(p,q), with b=AC and a=BC, having a≠b. We find p=(c²+b²-a²)/2c. Taking p'=c-p we find D(p',q) such that DA=a, BD=b, and DC=|p-p'|. |
