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?
Equilateral Triangles In The Taxicab Space
geometrypacking-problemreal-analysis
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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}. $
Let us divide ⊿ABC into the following color coded diagram.
For simplicity, we further let [yellow] = 1h, [pink] = 1k, [⊿ABC] = 630 and [green] = x
Then, [orange] = 2h and [red] = 2k.
1 : 2 = [⊿ABG] : [⊿ACG] = (1h + 2h + 1k) : (x + 2k)
After simplification, x = 6h.
∴ [yellow] $= \frac {1}{1 + 6} \cdot \frac {1}{1 + 2} \cdot 630 = 30$
The areas of the other parts can then be found and the results are shown in fig. 2.
In fig.3, it is not difficult to see that [light green] = [green] – [yellow] = … = 150
Also, [⊿HIJ] = [blue] = [red] – [light green] = … = 90, showing that [⊿HIJ] : [⊿ABC] = 1: 7
Best Answer
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.