# Equilateral Triangles In The Taxicab Space

It's fairly common to represent a unit circle in the Taxicab space ($$1$$-normed metric space) as a diamond in $$\mathbb{R}^2$$ with extreme points $$(1,0), (0,1), (-1,0), (0,-1)$$. What will an equilateral triangle of edge length $$1$$ 'look like' under this norm? As a followup, how many equilateral triangles (of edge length $$1$$) can be packed in the unit circle?

• @Clayton Why not define a triangle to be $3$ points? Then an equilateral triangle is $3$ points such that the distance between and pair of them is equal. – Alex Kruckman Jan 25 at 1:16

In this answer, I'm assuming the following definition of an equilateral triangle in a metric space $$(X,d)$$: a set of $$3$$ points $$\{O,P,R\}$$ such that $$d(O,P) = d(O,R) = d(P,R)$$. This common distance is the side-length of the triangle. When I talk at the end about packing triangles inside the unit circle, I'm identifying the triangle with the standard convex hull of the three points in $$\mathbb{R}^2$$, so the triangle defined by three points really looks like a standard triangle. I'm assuming this is what you had in mind when you asked the question - if not, please clarify.

You can find an equilateral triangle of side-length $$1$$ in $$\mathbb{R}^2$$ with the taxicab metric as follows: Pick a point $$O$$ (for simplicity, let's assume it's the origin) and draw the unit circle (which looks like a diamond) around $$O$$. Now pick a point $$P$$ on that unit circle, and draw the unit circle (diamond) around $$P$$. If these two unit circles intersect in a point $$Q$$, then $$OPQ$$ forms an equilateral triangle.

If you try this, with $$O = (0,0)$$, you'll quickly find that there are two cases.

Case 1: $$P = (\pm 1/2, \pm 1/2)$$. Let's call a point like this "special". Then there are infinitely many possible points $$Q$$. For example, if $$P = (1/2, 1/2)$$, then $$Q$$ can be any point on the line segment between $$(-1/2,-1/2)$$ and $$(0,1)$$, or any point on the line segment between $$(1/2,-1/2)$$ and $$(1,0)$$.

Case 2: $$P$$ is any other point. Then there are two possible points $$Q$$. For example, if $$P$$ is $$(3/4, 1/4)$$, then $$Q$$ can be $$(1/4, 3/4)$$ or $$(1/2,-1/2)$$. Note that one of the choices for $$Q$$ is always a "special" point.

Speaking informally, you can think of the possible equilateral triangles containing the point $$(0,0)$$ as "sliding" around the unit circle (diamond) as follows: Start with $$P = (1,0)$$ and $$Q = (1/2,1/2)$$. Then $$P$$ and $$Q$$ slide at equal speed along the line segment from $$(1,0)$$ to $$(0,1)$$ until $$P$$ reaches $$(1/2,1/2)$$ and $$Q$$ reaches $$(0,1)$$. Then $$P$$ stays constant and $$Q$$ slides from $$(0,1)$$ to $$(-1/2,1/2)$$. Then $$Q$$ stays constant and $$P$$ slides from $$(1/2,1/2)$$ to $$(0,1)$$. Then $$P$$ and $$Q$$ slide at equal speed along the lines segment from $$(0,1)$$ to $$(-1,0)$$, and the process repeats. I hope that attempt at visualization helps.

Once you see what's going on here, it's not too hard to show that every equilateral triangle of edge length $$1$$ has area $$1/4$$ (where area is the standard area in $$\mathbb{R}^2$$), since every such triangle can be viewed as a triangle with base and height $$\sqrt{2}/2$$. The unit circle (diamond) has (standard) area $$2$$, so you can't pack more than $$8$$ equilateral triangles into the unit circle. And the bound is tight, because $$8$$ equilateral triangles with corners at $$(0,0)$$, $$(0,\pm 1)$$, $$(\pm 1, 0)$$, and $$(\pm 1/2, \pm 1/2)$$ pack into the unit circle.

• What shape are the edges (shortest paths between vertices)? I'm having a hard time visualising it (and haven't tried drawing it.) – timtfj Jan 25 at 1:51
• @timtfj I've edited to clarify the definition of equilateral triangle that I'm using: a triangle is just three points, there are no explicit edges. In the taxicab metric, the shortest path between two points is not unique. – Alex Kruckman Jan 25 at 2:06
• Thanks! I did wonder about the uniqueness. Though actually it's obvious: any path between two points such that neither its vertical nor horizontal component ever reverses direction will have the same length. – timtfj Jan 25 at 2:25