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My question is related to kind of problems, called "rational distances problem"(at least by wolfram mathworld). I couldn't find a specific solution, so it would be a real help if you have an idea or link about the solution. Here is the problem :

Given $n>2$ points on the plane, not all on a line, is there another point, call $X$, which its distance from all the given points is rational number?

Cna Mrz
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  • A Reddit post talks about how some solutions are found for specific cases – WaveX Mar 15 '19 at 21:35
  • If you do not confine yourself to being $among$ the $n$ points and there being an infinite number of distance combinations from $outside$ the points, there is certainly a finite (albeit among large numbers) number of rational distance sets. e.g. similar question – poetasis Mar 15 '19 at 21:36
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    Answer definitely seems to be no, could you elaborate on your question? with just n = 2 $(0,0)$ and $(\pi, e)$ it should be impossible. (although it is on a line) – Matthew Liu Mar 15 '19 at 21:37
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    @MatthewLiu : The OP is asking "does there exist" such a point. Non-examples cannot be considered a proof in this case. Your example does not satisfy the assumption too! It has to be proven for all possible choices of $n$ points in a plane. – Yadati Kiran Mar 15 '19 at 21:52
  • @MatthewLiu Yes its better to think about n>2(or even n>4 beacause of "unit square" problem). Being non-colinear doesn't make sense when number of points is less than three,but anyway, I'll edit the post. – Cna Mrz Mar 15 '19 at 22:00
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    @Yaditi Kiran He has not specified any restriction on these $n$ points, which kind of makes it a pointless question. – Matthew Liu Mar 15 '19 at 22:07
  • @MatthewLiu It seems to me the conditions of the question allow $X$ to have irrational coordinates as long as its distance to each given point is rational. So construct circles of radius $10$ around each of your points and let $X$ be either point if intersection of the circles. My hunch is the answer for $n>2$ is "not in general," but no idea how to show it. – David K Mar 15 '19 at 22:22
  • @DavidK, I have a hunch you might find my answer of interest. – Barry Cipra Mar 16 '19 at 01:06
  • @BarryCipra Yes, I think that was the idea I was looking for. – David K Mar 16 '19 at 01:19

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The Wolfram Mathworld article (see here) clearly states that the problem is unsolved in the very specific case where $n=4$ and the four points are the vertices of a unit square. This makes it immediately clear that asking the question in such generality as you have is completely pointless, and that it is extremely likely that there exist configurations for which no such $X$ exists.

Indeed it is very easy to find configurations for which no such point exists. For the following configuration of $5$ points the algebra is particularly simple; let $$P_1:=(1,0),\qquad P_2:=(-1,0),\qquad P_3=(0,1),\qquad P_4=(0,-1),\qquad P_5=(\pi,0).$$ Suppose $X=(x_0,y_0)\in\Bbb{R}^2$ is such that $r_i:=d(X,P_i)$ is rational for all $i$. Then $$d(X,P_1)^2=x_0^2-2x_0+1+y_0^2=r_1^2,$$ $$d(X,P_2)^2=x_0^2+2x_0+1+y_0^2=r_2^2,$$ from which it follows that $x_0$ is rational because $$4x_0=r_2^2-r_1^2=(r_2-r_1)(r_2+r_1).$$ By symmetry also $y_0$ is rational. But then $$d(X,P_5)^2=(x_0-\pi)^2+y_0^2,$$ is not rational, a contradiction. Hence no such point $X\in\Bbb{R}^2$ exists.

Servaes
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The OP asks, "Given $n\gt2$ points on the plane, not all on a line, is there another point, call $X$, which its distance from all the given points is rational number?"

The answer is, sometimes yes and sometimes no.

To see that the answer is sometimes yes, consider any $n$ points on a circle of radius $1$. They are obviously all at the rational distance $1$ from the center of the circle.

To see that the answer is sometimes no, start with any two points, and draw all circles of rational radius around each point, then take the set of intersections of pairs of these circles. This set is countable, and it constitutes the set of all points where the future point $X$ must lie. Now around each of these countably many points, draw all circles of rational radius. The union of all these circles is a set of measure $0$ (because each circle is a set of measure $0$, and there are only countably many of them). So if your third point (when $n\gt2$) is not in this set of measure $0$, it is at an irrational distance from all the points which are at a rational distance from both of the first two points. Roughly speaking, if you pick your $n$ points "at random," then, with probability $1$, any three of them will fail to be simultaneously at a rational distance from any other point.

Barry Cipra
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