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


(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 $d_T$?

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!

  • $\begingroup$ Here's a numerically computed picture for you: i.stack.imgur.com/0jj9x.png (blue is taxicab, red is Euclidean). There seem to be a lot of cases. $\endgroup$ – user856 Jul 13 '12 at 20:22
  • $\begingroup$ For $a=(0,0), b=(3,0)$ I get the solution to be a diamond with corners $(2,0), (3,3), (6,0), (3,-3)$, not two lines. $\endgroup$ – Ross Millikan Jul 13 '12 at 20:25
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    $\begingroup$ I think for points in general position, it would be useful to divide the plane into 9 regions like a tic-tac-toe board: those above both, between the two, and below both in $y$ and similarly in $x$. You should then be able to calculate the slope in each region. The plot Rahul Narain posted shows changes as you cross these boundaries. Your special cases may then correspond to those where the path crosses two lines at once. $\endgroup$ – Ross Millikan Jul 13 '12 at 20:33

EDIT: alright, got it. There is a special case, when $AB$ are both on a horizontal or a vertical line. Then the shape asked for is a convex kite shape, two orthogonal edges of equal length with slopes $\pm 1,$ the other two edges with slopes out of $3,-3, \frac{1}{3},\frac{-1}{3}, $ the whole figure symmetric across the $AB$ line, as one would expect. Otherwise, draw the rectangle with $AB$ as the endpoints of one diagonal. If this rectangle is a square, or the longer edge is no more than twice the length of the other, the shape is an isosceles trapezoid, as described below. If the longer edge is more than twice as long as the other, the shape is a nonconvex hexagon, as in Rahul's image. Furthermore, if we fix the longer edge and call it length $1,$ as the shorter edge goes to length $0$ the resulting hexagon shape comes to resemble the kite shape, which is its limit. Anyway, there are just the three possibilities. Note that, in the cases when $AB$ are not in a line parallel to the $x$ or $y$ axis, we can color the nine regions of the tic-tac-toe board as a chess or checkerboard; then regions the same color as the rectangle have segments (if any) of slope $\pm 1,$ while any segments in the regions that are the other color have slope among $3,-3, \frac{1}{3},\frac{-1}{3}. $ It is all pretty rigid.

ORIGINAL: A little to add to what Ross and Rahul pointed out. One segment is automatically there, the segment inside the rectangle passing through the 2/3 point along the $AB$ diagonal, but with slope $\pm 1,$ and closer to $B.$ This segment is part of a "circle" around $A$ as well as a "circle" around $B.$ There is another such with the same slope, passing through a point we might as well call $A',$ which is the reflection of $A$ along the line $AB,$ the same distance from $B$ as $A$ is but on the other side. This can be the longest edge involved, as it continues through one of Ross's tic-tac-toe regions.

There may also be common "circle" segments rotated $90^\circ$ from those, as in Rahul's diagram. So one ought to check for those first, in the nine regions, any segments with slope $\pm 1$ that are the overlap of a circle around $A$ and circle around $B.$ Rahul has shown that you can get three such segments, and there are automatically at least two, but I do not see how one could get four such. So I think the diagram you are asking about is likely to be either a quadrilateral (an isosceles trapezoid) or a non-convex hexagon, being an isosceles trapezoid with one vertex replaced by an extra triangle, one edge of which is orthogonal to the two parallel edges of the trapezoid part. For that matter, there are really only two genuinely distinct slopes allowed, $\pm 1$ and $3,-3, \frac{1}{3},\frac{-1}{3}. $

  • $\begingroup$ Now that I know how this works, it seems a good choice for computer animation, fix the point $A=(0,0)$ and let $B = (\cos t, \; \sin t).$ Then draw the rectangle with $AB$ as diagonal endpoints, edges parallel to the $x$ or $y$ axes in one color, and draw the varying solution shape in a different color. $\endgroup$ – Will Jagy Jul 13 '12 at 23:37
  • $\begingroup$ Thanks, folks! Your insight is much appreciated! $\endgroup$ – Glenn Jul 15 '12 at 3:19

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