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?

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    $\begingroup$ @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. $\endgroup$ – Alex Kruckman Jan 25 '19 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.

  • $\begingroup$ What shape are the edges (shortest paths between vertices)? I'm having a hard time visualising it (and haven't tried drawing it.) $\endgroup$ – timtfj Jan 25 '19 at 1:51
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    $\begingroup$ @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. $\endgroup$ – Alex Kruckman Jan 25 '19 at 2:06
  • $\begingroup$ 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. $\endgroup$ – timtfj Jan 25 '19 at 2:25

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