I'd like to draw a triangle (given the side lengths) with the sides colored separately.

I tried Graphics[{FaceForm[Opacity[0]], EdgeForm[Blue], SSSTriangle[2, 2, 3]}]


my try

But I want to color the edges separately, eg Blue, Green and Red.

  • $\begingroup$ Pretty sure the only way is to make the graphic using individual lines. $\endgroup$
    – george2079
    Jun 28, 2016 at 21:09
  • $\begingroup$ What about using Graph? $\endgroup$
    – mgamer
    Jun 28, 2016 at 21:13

3 Answers 3


I was going to do the same as george2079 (+1). I'll post it on account of having used a rather different programming style.

colorTriangle[Triangle[{pt1_, pt2_, pt3_}], {c1_, c2_, c3_}] := {
  c1, Line[{pt1, pt2}],
  c2, Line[{pt2, pt3}],
  c3, Line[{pt3, pt1}]

  colorTriangle[SSSTriangle[2, 2, 3], {Red, Green, Blue}]

Mathematica graphics

Coloring it so that particular colors correspond to particular side lengths is more difficult, but here's one way:

SetAttributes[colorTriangle, HoldFirst]
colorTriangle[SSSTriangle[l1_, l2_, l3_], {c1_, c2_, c3_}] := Module[{allTriangles, wellOrderedQ, wellOrderedTriangles},
  allTriangles = SSSTriangle[l1, l2, l3] /. Triangle[{pt1_, pt2_, pt3_}] :> Permutations[{
       Line[{pt1, pt2}], Line[{pt2, pt3}], Line[{pt3, pt1}]

  wellOrderedQ = allTriangles /. {Line[{pt11_, pt12_}], Line[{pt21_, pt22_}], Line[{pt31_, pt32_}]} :> And[
      Norm[pt11 - pt12] == l1,
      Norm[pt21 - pt22] == l2,
      Norm[pt31 - pt32] == l3

  wellOrderedTriangles = Pick[allTriangles, wellOrderedQ, True];

  Graphics@Riffle[{Red, Green, Blue}, #] & /@ wellOrderedTriangles

Since there aren't always one unique way to color a triangle in this fashion the result is a list of up to six triangles:

Mathematica graphics

  • $\begingroup$ Looking good! But the edge lengths are in a different order to their colors! $\endgroup$
    – pdmclean
    Jun 28, 2016 at 21:29
  • 2
    $\begingroup$ @pdmclean That criterion complicates the problem significantly and should have been mentioned in the question... Anyway, I've posted an update about that now. $\endgroup$
    – C. E.
    Jun 28, 2016 at 23:29

Just to make this more difficult than it has to be,

 DiscretizeGraphics@SSSTriangle[2, 2, 3], {Style[{2, 1}, White]}~
  Join~({Style[{1, {#}}, Thick, {Red, Green, Blue}[[#]]] & /@ 

Mathematica graphics

  • $\begingroup$ +1 posted similar thing but was 1 min late :) I was struggling with 2D component because usually Style None works but here it breaks everything and White saves the day, $\endgroup$
    – Kuba
    Jun 28, 2016 at 21:43
  • $\begingroup$ @Kuba - I was trying for Style[.., Opacity[0]] but I couldn't get it to work.. $\endgroup$
    – Jason B.
    Jun 28, 2016 at 21:47
Graphics[Transpose[{{Red, Green, Blue}, SSSTriangle[2, 2, 3] /.
     Triangle[li_] :> Line /@ Partition[Append[li, li[[1]]], 2, 1] }]]

enter image description here

  • 3
    $\begingroup$ Vastly simpler: Graphics[Riffle[{Red, Green, Blue}, Map[Line, Partition[#, 2, 1, 1] & @@ SSSTriangle[2, 2, 3]]]] $\endgroup$ Jun 28, 2016 at 23:30

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