Re: equidistant sets

In article <dmpmfs$rnm$1@xxxxxxxxxxxxxxxxxxx>, klewis@xxxxxxxxxxxxxxx says...
>
>
>quasi <quasi@xxxxxxxx> writes in article <9aqvo1l2v6v8dtkgdo4r7qnsf97ls9hvo3@xxxxxxx> dated Fri, 02 Dec 2005 01:40:11 -0500:
>>On Thu, 01 Dec 2005 22:49:41 -0500, quasi <quasi@xxxxxxxx> wrote:
>>I just realized -- that conjecture was already disproved by Robert
>>Sheskey in a prior reply. Robert's example was this:
>>
>>>Consider a rectangle with sides a, b such that a<=b.
>>>Let A be the line segment of length b-a centered in the rectangle parallel
>>>to the long sides. Let B be the union of the reflections of A in each of
>>>the sides. (And if you want B to be connected just connect the
>>>components with long paths that stay far away from the rectangle.)
>>
>>Somehow it failed to register that the above example had broken my
>>conjecture, but it's obvious now.
>>
>>Based on that, I withdraw my "smoothness" objection to your pentagon
>>construction, but I'll still need some time to review the construction
>>itself.
>
>Here's (hopefully) a simpler way to see what I base my concave construction
>on. Consider 2 parallel, aligned line segments of equal length. By that I
>mean they are opposite sides of a rectangle. Call them A1, B1.
>
>eqd(A1,B1) is a line.
>
>Let A2,B2 be a similar pair of line segments, not parallel to A1 and B1.
>
>eqd(A2,B2) is another line.
>
>Let P be the point where the 2 lines intersect. We already know
>dist(P,A1) = dist(P,B1)
>dist(P,A2) = dist(P,B2)
>
>If dist(P,A2) = dist(P,A1),
>then eqd({A1 u A2}, {B1 u B2}) is an angle (2 rays from P).
>
>My concave polygon construction is based on adjusting the lengths of the
>line segments so that this happens at every vertex.
>
>It could also be made to work with sets of points. But until somebody
>besides myself believes the line segment construction there is no point in
>explaining how.
>
> [snip]



As far as I can tell it seems to work perfectly. (But what do you mean
by saying 'it could be made to work with sets of points'?)

In fact in looks like with very little modification your method can be
extended to show the following:

Let G be a finite polygonal graph such that each vertex has even degree.
Then G is a eqd. set.

(Just follow your tactic of placing a line segment with the proper
endpoints on each side of each edge. We need vertices to have even degree
so that we can decide for each segment whether it goes in A or in B.)


RS



.

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