4
$\begingroup$

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]}]

giving:

my try

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

Improve this question
$\endgroup$
2
  • $\begingroup$ Pretty sure the only way is to make the graphic using individual lines. $\endgroup$
    – george2079
    Commented Jun 28, 2016 at 21:09
  • $\begingroup$ What about using Graph? $\endgroup$
    – mgamer
    Commented Jun 28, 2016 at 21:13

3 Answers 3

Reset to default
6
$\begingroup$

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}]
  }

Graphics[{
  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

Improve this answer
$\endgroup$
2
  • $\begingroup$ Looking good! But the edge lengths are in a different order to their colors! $\endgroup$
    – pdmclean
    Commented 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. ♦
    Commented Jun 28, 2016 at 23:29
2
$\begingroup$

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

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

Mathematica graphics

Improve this answer
$\endgroup$
2
  • $\begingroup$ +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
    Commented 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.
    Commented Jun 28, 2016 at 21:47
1
$\begingroup$ 
Graphics[Transpose[{{Red, Green, Blue}, SSSTriangle[2, 2, 3] /.
     Triangle[li_] :> Line /@ Partition[Append[li, li[[1]]], 2, 1] }]]

enter image description here

Improve this answer
$\endgroup$
1
  • 3
    $\begingroup$ Vastly simpler: Graphics[Riffle[{Red, Green, Blue}, Map[Line, Partition[#, 2, 1, 1] & @@ SSSTriangle[2, 2, 3]]]] $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jun 28, 2016 at 23:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.