exupero's blog

Angle bisection in taxicab geometry

In the previous post we saw how angles in taxicab geometry vary in size depending on their direction. Given that behavior, how can one bisect angle?

In Euclidean geometry, an angle can be bisected with a compass and straightedge. If we do the same construction with taxicab geometry's circles, we can get something like this:

The smaller circles centered on the rays have an infinite number of points of intersection, rather than just two, and it's not clear where to draw the angle bisector. When the angle does allow ray-circles to have only two points of intersection, those two points max not align with the vertex of the angle, as in this example:

A line from the angle's vertex to the circle intersection closest to it clearly wouldn't produce an angle bisector, but would a line from the vertex to the far intersection? That's not so obvious.

The most reliable way I've found to get an angle bisector in taxicab geometry is to find the intersection of lines parallel to an angle's rays. Here are two parallel lines:

And here's a line from the vertex to their intersection, which is very plausibly a bisector:

As a check of whether this does produce a bisector, let's use parallel lines to find a triangle's incenter. Here's an arbitrary triangle and three interior parallel lines:

And here are the angle bisectors suggested by the intersections of the parallel lines:

The first indication that we're in luck is that all three bisectors intersect at the same point. A circle centered on that point and given the correct radius does appear to touch all three sides of the triangle, as an incircle should:

This is hardly proof, however, especially given taxicab geometry's proclivity for edge cases. The triangle above is acute; let's try an obtuse triangle:

So far, so good. For my purposes, I don't need a proof, but if you have a counter-example or edge case to consider, let me know.