# How to plot Reuleaux triangle in rectangular coordinates?

I need to plot the Reuleaux triangle in rectangular coordinates.

I am not sure if there's a Reuleaux triangle equation. If yes, I can plot it by ContourPlot.

I saw link but I just need curve of Reuleaux triangle.

What's the best way to plot a Reuleaux triangle?

• The Reuleaux triangle has sharp corners, so ContourPlot would have problems to draw that. – Henrik Schumacher Mar 5 '18 at 9:12

## 2 Answers

A long time ago, I derived parametric equations for Reuleaux polygons.

Specializing the formulae in that post to the triangle case, we have

ParametricPlot[{(Sqrt[3] Cos[π/6 (2 SawtoothWave[u] - 1)] - 1) Cos[π/3 (2 Floor[u] + 1)] -
Sqrt[3] Sin[π/6 (2 SawtoothWave[u] - 1)] Sin[π/3 (2 Floor[u] + 1)],
Sqrt[3] Cos[π/3 (2 Floor[u] + 1)] Sin[π/6 (2 SawtoothWave[u] - 1)] +
(Sqrt[3] Cos[π/6 (2 SawtoothWave[u] - 1)] - 1) Sin[π/3 (2 Floor[u] + 1)]},
{u, 0, 3}]


After some prompting by Jens, I tried looking again for a polar representation of the Reuleaux triangle. I finally managed to find one, but it is not very pretty:

PolarPlot[Cos[Mod[θ, 2 π/3, 2 π/3]] + Sqrt[2 + Cos[Mod[θ, 2 π/3, 2 π/3]]^2],
{θ, 0, 2 π}]


• I'd be curious if there's a simple form that could be used with PolarPlot. Probably not... but maybe you've thought about it. – Jens Mar 5 '18 at 23:16
• @Jens, I haven't found one, but definitely not for lack of trying. – J. M.'s torpor Mar 5 '18 at 23:19

Not sure if the best but here is one way:

RegionIntersection @@ (Disk /@ CirclePoints[1/Sqrt[3], 3]) //
RegionBoundary //
Region


• I am impressed with the simplicity of your solution ! – yarchik Mar 5 '18 at 9:30
• @yarchik thanks, yes one can get lazy with Mathematica – Kuba Mar 5 '18 at 9:40