Have a helper function

(* helpers *)
Clear[closeL]
closeL[l_List] := Join[l, {First[l]}]

an equilateral triangle is easiest generated from the solution of the cyclotomic equation of third degree

In[24]:= (* equilateral triangle *)
Solve[x^3 == 1, x]
Out[24]= {{x -> 1}, {x -> -(-1)^(1/3)}, {x -> (-1)^(2/3)}}

rotation by 30 Degree is multiplication by Exp[Pi I/6], so your first triangle is just:

  (* rotation by 30 Degree is multiplication by Exp[Pi I/6], so the first triangle is, keeping the zero in the center for practical reasons:*)
    Graphics[Line[closeL[{Re[#], Im[#]} & /@ (Exp[Pi I/2] Last[Last[#]] & /@ Solve[x^3 == 1, x])]]]

to get the solution of the cyclotomic equation upright, i.e. with one horizontal edge. How to proceed:

• shrink the roots by the right value (9/16)
• rotate the roots by multiplication with another Exp[Pi I/6]
• continue until you did that 13 times
• plot all solutions collected as shown above

Done with a Polygon it's

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

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}]}]