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Do there exist $6$ points in plane such that no three of them are collinear, no four of them are cyclic and for each $1\le i \le 5$, there exists a number such that it is counted exactly $i$ times as the distance between two points of those $6$ points?

I saw a question similar to this one, such that in that question, instead of $6$ points, the problem asks for $5$ points in plane such that no three of them were collinear, no four of them were cyclic and for each $1\le i \le 4$, there exists a number such that it was counted exactly $i$ times as the distance between two points of those $5$ points...

In that question, the answer was yes, and there exists such an example, but could anyone tell me what is the answer of the question for $6$ points?

EDIT: As requested, for the easier version, consider $3$ points $A,B,C$ as vertices of an equilateral triangle of side lenght $a$, point $D$ such that $DA=a$ and $120<\angle DAB<180$, and consider $E$ as the third vertex of the equilateral triangle that has one side $DB$.

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  • $\begingroup$ Is there any chance you could post the solution to the easier case, to give us all a head start on answering this one? $\endgroup$
    – user22805
    Apr 13, 2012 at 7:00
  • $\begingroup$ I've written how to construct that example. $\endgroup$
    – Goodarz Mehr
    Apr 13, 2012 at 15:18
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    $\begingroup$ When placing $n$ points, we have $2n-4$ degrees of freedom (4 are used up for position, orientation, and scale -- in other words, we can assume the first two point locations are fixed). The distance equality constraints then take away $(n-2)(n-1)/2$ degrees of freedom, leaving $(n-2)(2-(n-1)/2)$ degrees of freedom in a solution. So for $n$=5 points, we expect solutions to be isolated (rather than coming in families, like "all isosceles triangles" for $n$=3). For $n$=6 points, any solution would require a fortuitous coincidence (which the "general position" constraint is trying to prevent). $\endgroup$
    – Matt
    Apr 22, 2012 at 18:18
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    $\begingroup$ Would you please explain more? I couldn't Undrestand what you said. Thanks $\endgroup$
    – Goodarz Mehr
    Apr 23, 2012 at 12:25

3 Answers 3

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The answer is yes.

With some help of Mathematica, I have found the following solution:

$$\begin{array}\\T_1=(1,0)\\T_2=(-1,0)\\T_3=\left(\frac18(-5+\sqrt{73}),-\frac14\sqrt{\frac12(7+13\sqrt{73})}\right)\\T_4=\left(-\frac18(-5+\sqrt{73}),\frac14\sqrt{\frac12(7+13\sqrt{73})}\right)\\T_5=\left(\frac14(3+\sqrt{73}),\frac12\sqrt{\frac12(-5+\sqrt{73})}\right)\\T_6=\left(-\frac14(3+\sqrt{73}),-\frac12\sqrt{\frac12(-5+\sqrt{73})}\right)\end{array}$$

The distances are: $$\begin{array}\\d_1=\sqrt{2(9+\sqrt{73})}\\d_2=\sqrt{\frac12(3+\sqrt{73})}\\d_3=\sqrt{7+\sqrt{73}}\\d_4=\frac12\sqrt{25+3\sqrt{73}}\\d_5=2\end{array}$$ where $d_i$ is the distance that appears $i$ times.

With some calculation, one can check that no three points are collinear and no four points cyclic.

Added: a picture:

enter image description here

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  • $\begingroup$ Thanks a lot. And would you please place a picture of the figure you say in your post? $\endgroup$
    – Goodarz Mehr
    Apr 27, 2012 at 14:06
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I think this works, though I haven't computed everything.

Let $A,B,C$ be vertices of an equilateral triangle of side $s$. Drop a perpendicular from $B$ to $AC$ and extend it beyond $AC$ to a point $D$ with $BD$ of length $s$. Similarly, drop a perpendicular from $C$ to $AB$ and extend it beyond $AB$ to a point $E$ with $CE$ of length $s$. There are two places where you could put a point $F$ to make $DEF$ equilateral; choose the one closer to $A$.

You get $AB=AC=BC=BD=CE$; $AD=CD=AE=BE$; $DE=DF=EF$; $BF=CF$. If there aren't any other equalities, then this solves the problem. It's clear that $AB\ne AD$, $AB\ne BF$, $AB\ne BF$, $AD\ne DE$, $AD\ne BF$, $DE\ne BF$, $DE\ne AF$, $BF\ne AF$, but there is still a bit more to check.

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  • $\begingroup$ It only remains to check that $AF< AE$. This follows from the observation that $\angle(EAD)=90^\circ$ and not $=120^\circ$. $\endgroup$
    – Christian Blatter
    Apr 17, 2012 at 8:07
  • $\begingroup$ But then we get that $BEDC$ is a cyclic quadrilateral! $\endgroup$
    – Goodarz Mehr
    Apr 17, 2012 at 16:46
  • $\begingroup$ Oops.${}{}{}{}$ $\endgroup$
    – Gerry Myerson
    Apr 18, 2012 at 1:52
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Let the points be $(-3,0), (-1,0), (1,0), (3,0), (0,\sqrt{\frac{1}{3}}), (0,- \sqrt{\frac{1}{3}})$. For distances, $6, 4, 2, \frac{\sqrt{28}}{3}, \frac{2}{\sqrt{3}}$ appear 1, 2, 3, 4, 5 times respectively.

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    $\begingroup$ The question asks for no three points collinear. $\endgroup$
    – Gerry Myerson
    Apr 17, 2012 at 12:56
  • $\begingroup$ Ok, didn't notice that earlier. $\endgroup$
    – Wonder
    Apr 17, 2012 at 13:01

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