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Taxicab and Euclidean geometry differ a great deal, due to the modified metric function:

dT(A, B) = |xa - xb| + |ya - yb|

(Note that this means when measuring distance, it is not the length of the hypotenuse, but the sum of the legs of the same right triangle.)

My Main Problem

In Euclidean geometry, the answer to the question "Find the locus of points X such that: d(X, A) = 2 * d(X, B)" yields a regular, Euclidean circle. A little bit of algebra makes this very trivial.

But what is the answer to the same problem, but for dT?

What I Know So Far

This kind of geometry actually has a very interesting property, namely that as things rotate, their measures change. Consider the cases where points share either one of their coordinates. Many times, those situations yield the same answers as do their Euclidean counterparts.

Some things are noticeably different, though. For instance, a circle, as defined as the set of points a fixed distance from one point, actually comes out as a square, rotated 45 degrees. It is also trivial to illustrate that.

It did occur to me that the answer to this problem could be analogous to Euclidean geometry, and the solution may simply be a Taxicab circle (a square). But this didn't seem to work out. Plus, I worked out the solution for the points sharing an x or y coordinate, I end up with two mirror-image line segments. But the general case, where the two points are corners of any rectangle still eludes me. My second educated guess was that the solution could be a Euclidean circle, but this didn't work out either.

Lastly, some constructions seemed to differ depending on whether the points I chose formed the opposite diagonal corners of a general rectangle, or a square. E.g. (0,0) and (3, 3) seem to be a yet different type of exception.

Any thoughts on this problem would be greatly appreciated!

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closed as too localized by Will Jagy, Anton Petrunin, Andreas Blass, MTS, Gerry Myerson Jul 13 '12 at 22:12

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Your title asks for $d(P,A)=d(P,B)$, and the text of your question asks for $d(X,A) = 2 * d(X,B)$. –  Lee Mosher Jul 13 '12 at 21:00
asked and answered at math.stackexchange.com/questions/170440/… with corrected title –  Will Jagy Jul 13 '12 at 21:56

1 Answer 1

I believe you are seeking the bisector of points $A$ and $B$ in what is known as the $L_1$ metric: the locus of points equidistant from $A$ and from $B$ in that metric. This (and much more) is addressed in D.T. Lee's paper, "Two-Dimensional Voronoi Diagrams in the $L_p$-Metric" (ACM link). See Fig.1 below, and note especially that when $A$ and $B$ are corners of a square, the bisector (1b) consists of two regions and a line segment:

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Your insights are very much appreciated! However, I believe the bisector is the solution to a slightly simpler problem, namely the one where d(P, A) = d(P, B) for points A and B. In this case, I'm looking for d(P, A) = 2 * d(P, B). (Unless I'm missing something?) –  Glenn Habibi Jul 13 '12 at 20:17
Glenn: I think the title of your question might be confusing. –  Artie Prendergast-Smith Jul 13 '12 at 20:26
Joseph, see math.stackexchange.com/questions/170440/… where i corrected the title. It's either an isosceles trapezoid or a nonconvex hexagon, all segment slopes $\pm 1, \pm 3, \pm \frac{1}{3}.$ –  Will Jagy Jul 13 '12 at 21:55
Oh, I see: The original title was "$d(P,A)=d(P,B)$", to which I was responding, before Will corrected it to $d(P,A)=2d(P,B)$. Apologies for not checking MSE to see the corrections there. In any case, I do think the $L_1$ bisector is rather interesting. –  Joseph O'Rourke Jul 13 '12 at 23:25
Joseph, the MSE posting was long after your answer here, and my correction was after that. What's the bisector with metric $x^4 + x^2 y^2 + y^4?$ –  Will Jagy Jul 13 '12 at 23:42

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