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Is there a set of seven points in the plane such that, among any three of these points, there are two, $P, R$, which are distance $1$ apart?

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  • $\begingroup$ Any set of at most two points. Or the vertices of an equilateral triangle with side length $1$. Or two such triangles sharing an edge. Or a chain o fthree such triangles. $\endgroup$ – Hagen von Eitzen Jul 9 '15 at 8:41
  • $\begingroup$ You can also take the vertices of a regular pentagon with sides of length 1. Among any set of 3 points among the pentagon's vertices, there is at least 2 which are adjacent vertices (thus with distance 1 between them). $\endgroup$ – Nihl Jul 9 '15 at 8:44
  • $\begingroup$ Sorry I have edit the question $\endgroup$ – Bless Jul 9 '15 at 8:54
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Yes. Normally I like to give some more motivation for an answer, but in this case it's probably hard to say anything suggestive that the illustration below does not, besides perhaps that after experimenting one might guess that the four-point "diamond" configurations are helpful (and probably necessary) in constructing such an arrangement.

enter image description here

(I remember seeing this problem, by the way, during a mail-in high school mathematics competition circa 1999; probably the widespread availability of the Internet makes it impractical to hold this sort of competition today.)

Edit I've posed a follow-up question, asking whether this is the unique configuration up to Euclidean motions. Servaes wrote an excellent, detailed answer showing that it is.

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  • $\begingroup$ I competed in a mail-in math competition in high school: I sat in a room with the questions and a proctor for two or three hours, and they mailed the results in. This was prior to the WWW but would work as well today if the proctor is vigilant about cell phones. Either you trust people under an honor system or you don't; cheating was always possible if there was no proctor. $\endgroup$ – David K Jul 9 '15 at 13:43
  • $\begingroup$ In the competition in which this problem appeared, the questions were sent periodically to one's home. There were no large prizes (if any at all); cheating would certainly have been possible, but not quite as easy as it would be now, if the frequency with which active contest problems seem to be posted here is any indication. $\endgroup$ – Travis Jul 9 '15 at 13:46
  • $\begingroup$ I think the reason you wouldn't do a mail-in competition today is that it's far more convenient to disseminate the questions and collect the responses on-line. (This also makes cheating easier to detect, if it occurs, because it can be spotted by people not affiliated with the contest--as often happens here.) $\endgroup$ – David K Jul 9 '15 at 13:59
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    $\begingroup$ for the record, this is the Moser spindle, also here, and is related to the chromatic number of the plane and Hadwiger–Nelson problem, and unit distance graphs $\endgroup$ – Mirko Jul 25 '15 at 18:46
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EDIT: This answer (which starts with a bold "No.") is wrong as can be seen from Travis's answer (which starts with a justified "Yes."). In principle, I should delete it, but - for the moment - I'll leave it here because somebody may find it helpful in other ways, such as seeing how to arrive at some neccessary conditions (cf. some comments)


No.

Assume $d(A,B)$, $d(A,C)$, $d(A,D)$, $d(A,E)$ are all $\ne 1$. Then all distances amoing $B,C,D,E$ must be $=1$. However, because the tetrahedron cannod be "flattened",

Among any four points there must be two of distance $\ne 1$.

We conclude that $A$ can have at most three points at distance $\ne 1$, hence

Every point hast at least three points at distance $=1$.

Hence if we pick any two points at distance $\ne1$, then one of the remaining five points must be one of the intersection points of the $1$-circles around them, i.e.,

For any two points $A,B$ with $d(A,B)\ne 1$ there exists $C$ with $d(A,C)=d(B,C)=1$.

Can you take it from here?

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    $\begingroup$ All of the highlighted statements are true in Travis's answer. $\endgroup$ – David K Jul 9 '15 at 13:37
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    $\begingroup$ @David D'oh. Next time I get stuck in the middle of an answer for half an hour, I should refrain from posing the remainder as an exercise for the reader :) $\endgroup$ – Hagen von Eitzen Jul 9 '15 at 13:43
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    $\begingroup$ I wonder if this line of reasoning will lead to a proof of uniqueness as asked for in the OP's followup question math.stackexchange.com/questions/1355105/… $\endgroup$ – Lee Mosher Jul 9 '15 at 13:58
  • $\begingroup$ (@LeeMosher The followup question is actually mine, not OP's.) $\endgroup$ – Travis Jul 9 '15 at 14:06
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    $\begingroup$ @Hagen von Eitzen, your answer is very misleading (you claim the answer is 'No'), though the efforts you make are certainly helpful in answering the question in the affirmative. Could you please edit it to avoid confusion for future readers? $\endgroup$ – Servaes Jul 9 '15 at 18:40

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