Re: equidistant sets

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.

Now I'll throw out a conjecture of my own.

If B is the reflection of A over line L, eqd(A,B) contains L.

--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
.

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