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It is well known that three points on the plane determine a circle uniquely. Is there a similar statement for regular hexagons?

It is obvious that if we have two points that are vertices, there are just two posibilities; thus, a third point (a non-colinear one) would determine the regular hexagon uniquely. Would that be true if the points were not vertices (or we did not know it)? How could one construct that hexagon?

EDIT: Without any other assumption, three points are not enough. It seems four points might suffice, if we know which edge they belong to. Does any set of four non-colinear points define a unique hexagon?

AugSB
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    "It is obvious that if we have two points that are vertices, there are just two possibilities." Not quite. The points could be adjacent vertices, or opposite, etc; altogether, two vertices provide for five possible hexagons. However, it remains true that a third non-collinear point distinguishes a unique hexagon from this collection. – Blue Feb 20 '15 at 17:07
  • You are right. I was just thinking about contiguous vertices. – AugSB Feb 24 '15 at 08:10

2 Answers2

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My attempt:

The hexagon has four degrees of freedom so that four points should suffice, but you can expect a finite number of solutions.

Take two of the points and make an hypothesis on the edges they belong to. This gives you the angle aperture between these edges. Knowing the two points and the angle aperture, you know that the apex of the angle belongs to a circular arc that joins the point and you can determine its center.

The center of the hexagon lies on the line from the apex to the center of the arc, so it belongs to a circle of unknown radius with the same center as the arc. Repeating the construction for three point pairs, the hexagon center is the point equidistant to the three centers.

In the case of two points on the same edge, the locus of the hexagon center is a straight line parallel to that formed by the two points, and the solution will be found as the point equidistant to the line and two points, or two lines and a point.

Other combinations are degenerate, such as two pairs of points belonging to two parallel sides.

UPDATE: the statement "The center of the hexagon lies on the line from the apex to the center of the arc" is wrong, so that the locus of the center is another curve.

  • A more thorough discussion is necessary but this shows a way. –  Feb 20 '15 at 10:43
  • The steps to determine the first arc are clear for me. But I am bit confused after that. Why does the center of the hexagon lie on the line from the apex to the center of the arc? My impression is that it should lie on the bisector of the angle. – AugSB Feb 20 '15 at 12:30
  • Yes that line is the bisector. –  Feb 20 '15 at 13:51
  • Your method uses the line from the apex to the center of the arc defined by two of the points and the angle aperture, right? Then, unless the apex is the midpoint of the arc, that line is not the bisector of the angle. – AugSB Feb 20 '15 at 14:34
  • You are right, my bad. Then the locus of the center of the hexagon is not a circle. –  Feb 20 '15 at 15:58
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If we get to specify which sides the given points are on, and if no two points lie on the same side (unless they are adjacent vertices), then it indeed appears that four points suffice. Here's an as-yet-unfinished proof.

In what follows, I'll consider a point on the edge of a hexagon if it lies on the line containing that edge of the hexagon.

Let $A$, $B$, $C$, $D$ be the points. There are three arrangements to check, which we can denote $$[A, B, C, D, \;\cdot\;, \;\cdot\;] \qquad [A, B, C, \;\cdot\;, D, \;\cdot\;] \qquad [A, B, \;\cdot\;, C, D, \;\cdot\;]$$

Case $[A, B, C, D, \;\cdot\;, \;\cdot\;]$.

enter image description here

Let $\stackrel{\frown}{AB}$, $\stackrel{\frown}{BC}$, $\stackrel{\frown}{CD}$ be circular arcs of points that make $120^\circ$ angles with the endpoints of corresponding segments $\overline{AB}$, $\overline{BC}$, $\overline{CD}$. (Note: There are actually two such arcs per segment. I don't yet rule-out the possibility that different choices among these arcs lead to distinct solutions in some cases.)

Choose $P$ on $\stackrel{\frown}{AB}$. Let $\overrightarrow{PB}$ meet $\stackrel{\frown}{BC}$ at $Q$, and then let $\overrightarrow{QC}$ meet $\stackrel{\frown}{CD}$ at $R$.

Claim. There is at most one choice of $P$ for which $\overline{PQ}\cong\overline{QR}$. (That $P$, and the corresponding $Q$ and $R$, are then vertices of the target hexagon.)

Case $[A, B, C, \;\cdot\;, D, \;\cdot\;]$.

enter image description here

We take $\stackrel{\frown}{AB}$ and $\stackrel{\frown}{BC}$ as before; this time, $\stackrel{\frown}{CD}$ is the (well, an) arc of points making a $60^\circ$ angle with $C$ and $D$.

Point $P$ on $\stackrel{\frown}{AB}$ leads to $Q$ and $R$ on $\stackrel{\frown}{BC}$ and $\stackrel{\frown}{CD}$. However, $R$ is not a candidate vertex; rather $M$, the midpoint of $\overline{QR}$ is; and so is $N$, the point on $\overrightarrow{RD}$ such that $\overline{MQ}\cong\overline{MN}$.

Claim. There is at most one choice of $P$ for which $\overline{PQ}\cong\overline{MQ}$. (That $P$, and the corresponding $Q$, $M$, $N$, are then vertices of the target hexagon.)

Case $[A, B, \;\cdot\;, C, D, \;\cdot\;]$.

enter image description here

Here, $\stackrel{\frown}{BC}$ makes $60^\circ$ angles, and $\stackrel{\frown}{CD}$ makes $120^\circ$ angles. The candidate vertices are $P$, $M$ (the midpoint of $\overline{PQ}$), and $N$ (the point on $\overline{QC}$ such that $\overline{MP}\cong\overline{MN}$).

Claim. There is at most one choice of $P$ for which $\overline{PQ}\cong\overline{QM}$. (That $P$, and the corresponding $M$, $N$, $R$, are then vertices of the target hexagon.)

The three Claims about $P$ seem to follow from a straightforward continuity argument: For $P$ close to $A$, some segments in question are shorter than we want; for $P$ close to $B$, those segments are longer; by continuity, there must be a $P$ that's "just right" (unless no $P$ at all works for the given $A$, $B$, $C$, $D$). However, I haven't yet explicitly demonstrated that the length changes are in fact monotonic; I plan to do so in a later edit.

Blue
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