Hello to all i write with the hope that someone can help me to solve the problem currently faced. The problem is the following: taking an equilateral triangle, i must make another triangle that is 75% of the first and also rotate for example 6 or 15 degrees and put them in the same chart, this procedure must be done several times depending on the user's desire. I've tried to rotate and scale but without success, I hope someone has some idea in this regard would be grateful too.
Many thanks for helping me to solve this problem, the truth I saw that evil was facing this problem but with the help of the community vi how to solve it, I have no words to thank Dent de Lion which has invested too much of their time in helping me and share their solutions
Luis Ledesma
Given as Lines it is
Lines
Graphics[Table[ Line[closeL[{Re[#], Im[#]} & /@ (Exp[Pi I (n/6 + 1/2)] (9/16)^(n/2) Last[ Last[#]] & /@ Solve[x^3 == 1, x])]], {n, 0, 12}]]
Done with a Polygon it's
Polygon
Graphics[{EdgeForm[Thick], White, Table[Polygon[{Re[#], Im[#]} & /@ (Exp[Pi I (n/6 + 1/2)] (9/16)^(n/2) Last[ Last[#]] & /@ Solve[x^3 == 1, x])], {n, 0, 12}]}]
with other words
Please note that in contrast to your reference document
the third triangle does intersect the horizontal edge of the first one. Looking at the picture of the reference document (as shown above) it seems unlikely, that in a symmetrical situation the third triangle scratches 2 edges of first triangle, but the third one - not. Why should this hold? Was the reference document produced to let the learners grow on tackling inaccuracies?
I'm sorry, Luis, but I don't understand what you are trying to do. --David
Many thaks for you response David keith, but I need something like this: only put the example with three triangles but if going by turning each triangle "x" degree should have more but I can not do even with that amount of triangles,do you have your ideas on how to do that? I'm still trying without success, thanks in advance
Hi Luis, Maybe this?
t1 = Triangle[{{-1, 0}, {0, Sqrt[2]}, {1, 0}}]; Graphics[{Red, t1, Green, Scale[Rotate[t1, 6 Degree], .75]}]
Or even
Graphics[{Red, t1, Green, Translate[Scale[Rotate[t1, 6 Degree], .75], {2, 0}]}]