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Source. The last two paragraphs of the lecture, spanning pp. 53--54, of P. Erdős, Combinatorial problems in geometry, Math. Chronicle 12 (1983), 35--54, the transcript of an invited address at the 17th New Zealand Mathematics Colloquium (Dunedin, 17--19 May 1982), as named on the source card. The lecture numbers none of its statements; the pages are the journal's own.

Statement

Question (p. 53). Can one find $n$ points in the plane, no three on a line and no four on a circle, which determine $n-1$ distinct distances, so that the $i$-th distance occurs $i$ times? The ordering of the distances is free ("in any order you wish").

Four points (pp. 53--54). An isosceles triangle $ABC$ with $AC=BC$ and its centre $O$, equidistant from the three vertices, gives four points and three distances: $OA=OB=OC$ three times, $AC=BC$ twice and $AB$ once. The print says "isosceles triangle and you take its centre"; that $O$ is the circumcentre is read from the counts.

Five points (p. 54, Pomerance; the print spells the name "Pommerance" [sic]). Take a unit equilateral triangle $OAB$, its circumcentre $C$, and the point $D$ where the perpendicular bisector of $CB$ meets the unit circle about $O$. The print says only "you bisect one of these lines ($CB$) and here is the fifth point ($D$)"; the placement of $D$ on that bisector at distance $1$ from $O$ is read from the figure and the counts. The lecture states that no three of the points are on a line and no four on a circle, and that $OA=AB=OB=OD$ occurs four times, $OC=CA=CB$ three times, $CD=BD$ twice, and $AD$ once.

Reported further (p. 54). Erdős says he had mistakenly asserted that he did not believe the configuration possible for $n>4$. A Hungarian high school student showed it can be done for six points; Erdős is not sure about seven.

Read depth. Claims checked: the passage, its two figures and the distance counts were read clause by clause on the page images of the print. A second reader checked the statement, hypotheses, label and page against the print.

Proof pointer

The lecture calls the conditions for the five-point construction easy to see and gives no verification; none is supplied here. The four-point count follows from the stated equalities, provided the three values $OA$, $AC$ and $AB$ are distinct.

Dependencies

None.

Bears on

• Problem 217: the question is the problem's question. The lecture answers it yes for $n=4$ by Erdős's own example and for $n=5$ by Pomerance's construction, reports that a Hungarian high school student did six points, and leaves $n\ge7$ open; it says nothing about general $n$.