# What could be a homeomorphism from the circle to a triangle?

I'm looking for a bijection from a circle to a triangle that is continuous with a continuous inverse. What could be one?

Here is an animation I made exhibiting such a homeomorphism:

Mathematica code:

CirclePoint[t_] := {Cos[2 Pi t], Sin[2 Pi t]}

TrianglePoint[t_] :=
If[0 <= t < 1/3,
{((1-(3t))+(-1/2) (3t)), -(1/Sqrt[3])(((1-(3t))+(-1/2)(3t))+1/2)+Sqrt[3]/2},
If[1/3 <= t < 2/3,
{-1/2, (1-(3t-1))Sqrt[3]/2+(3t-1)(-Sqrt[3]/2)},
{((-1/2)(1-(3t-2))+(3t-2)),
(1/Sqrt[3])(((-1/2)(1-(3t-2))+(3t-2))+1/2)-Sqrt[3]/2}]]

CircleTriangle[s_] :=
ParametricPlot[(1-s) CirclePoint[t] + s TrianglePoint[t], {t, 0, 1},
PlotRange -> {{-1.1, 1.1}, {-1.1, 1.1}}, Ticks -> None, Axes -> False]

Export["animation2.gif",
Join[Table[CircleTriangle[Max[s, 0]], {s, -0.1, 1, 0.02}],
Table[CircleTriangle[Min[s, 1]], {s, 1.1, 0, -0.02}]],
"DisplayDurations" -> 0.1]

• This is a sweet animation, but downvoted because it outlines a homotopy, not a homeomorphism per se. Jun 18, 2013 at 6:22
• @dfeuer: Did you downvote eccstartup's answer for giving a triangle inscribed in a circle, which is technically something other than a homeomorphism? Or is it possible that giving a visual intuition for what the homeomorphism is at least as useful as describing it explicitly but opaquely? Incidentally, your answer is exactly the same homeomorphism as all other answers provided so far (you can even look at my code to see exactly how it was made). Jun 18, 2013 at 6:58
• @ZevChonoles, no, it is not. It matches up vertices of the triangle to corresponding points on the circle in the same way if the triangle is equilateral, but it behaves differently in between. eccstartup's answer I saw as an illustration of André's. Jun 18, 2013 at 7:10
• Don't get me wrong... I appreciate your answer. It just doesn't match this particular question. Jun 18, 2013 at 7:11
• It seems that the point of @dfeuer was that your animation shows a homotopy and not a homeomorphism. This means, your animation shows a function $F(x,t)$ where $F(\cdot,0):[0,1]\to \mathbb{R}^2$ parameterizes a circle and $F(x,1)$ a triangle. You could add after the animation, however, that one can obtain the homeomorphism $f:C\to T$ by mapping the points $C:=\{ F( x ,0) | 0\le x\le 1 \}$ of the circle to the points of the triangle $T= \{ F(x,1) | 0\le x \le 1 \}$, so defining $f(x):=F(x,1)$ is the desired homeomorphism. You can then argue that the homotopy is continuous and so also $f$. Jun 17, 2020 at 16:24

I will assume that by circle you mean the curve, not the disk, and also that by triangle you mean the "curve" consisting of three line segments. In a remark at the end, we mention how to modify things if we are dealing with a disk and a triangle-with-interior.

We are given a circle $C$ and a triangle $T$. Draw the incircle $C^\ast$ of $T$. There is a natural bijection $\varphi$ from $C$ to $C^\ast$ (scale and translate).

We will produce a bijection from $C^\ast$ to $T$. Let $O$ be the centre of $C^\ast$, and let $P$ be any point on $C^\ast$. Draw the half-infinite ray $OP$. This ray meets the triangle at a uniquely defined point. Let $\psi(P)$ be that point. Then $\psi$ is a continuous bijection with continuous inverse from $C^\ast$ to $T$, so $\psi\circ\varphi$ is a continuous bijection from $C$ to $T$, with continuous inverse.

Remark: Using a similar idea, we can find a continuous bijection with continuous inverse from any disk $C$ to the figure made up of a triangle and its interior. All we need to do is to modify the definition of $\psi$. Let $O$ be the centre of the disk $C^\ast$, and let $P$ be a point on the boundary of that disk. Suppose that the ray $OP$ meets the boundary of the triangle at $Q$, and let $X$ be a point in the disk and on the line segment $OP$. Map $X$ to the point $Y$ on the line segment $OP$ such that $\frac{OY}{OX}=\frac{OQ}{OP}$.

• umm... what about obtuse triangles? You'll need to tweak the bijection a little. Jun 18, 2013 at 10:00
• @JanDvorak: Yes, thanks, will need to be away for a while, but fix is easy, will be made in few hours. Jun 18, 2013 at 11:04

I think it may be something like this.

I'm just answering to add a different homeomorphism: match arc length along the circle to cumulative perimeter along the triangle.