Question 897902
{{{drawing(400,250,-4,4,-3,2,
line(0,0,0,-1),line(0,0,-0.5,0.866),
line(-1.366,1.366,-0.5,0.866),line(-1.366,1.366,-2.366,1.366),
line(-2.366,1.366,-3.232,0.866),line(-3.732,0,-3.232,0.866),
line(-3.732,0,-3.732,-1),line(-3.732,-1,-3.232,-1.866),
line(-2.366,-2.366,-3.232,-1.866),line(-2.366,-2.366,-1.366,-2.366),
line(-1.366,-2.366,-0.5,-1.866),line(0,-1,-0.5,-1.866),

line(0,0,0.5,0.866),
line(1.366,1.366,0.5,0.866),line(1.366,1.366,2.366,1.366),
line(2.366,1.366,3.232,0.866),line(3.732,0,3.232,0.866),
line(3.732,0,3.732,-1),line(3.732,-1,3.232,-1.866),
line(2.366,-2.366,3.232,-1.866),line(2.366,-2.366,1.366,-2.366),
line(1.366,-2.366,0.5,-1.866),line(0,-1,0.5,-1.866),
line(0.5,0.866,-0.5,0.866),line(0.5,-1.866,-0.5,-1.866),
green(arrow(0,0,0,2)),
green(arrow(0,0,-1,1.732)),green(arrow(0,0,1,1.732)),
green(arc(0,0,3.5,3.5,-90,-60)),green(arc(0,0,3.4,3.4,-120,-90)),
locate(-0.6,1.7,green(30^o)),locate(0.2,1.7,green(30^o))
)}}}
The angles between the two 12-sided polygons are made up of 2 exterior angles for the 12-sided polygons.
The measure of each of those exterior angles is {{{360^o/12=30^o}}} ,
so the regular polygon that fits in that space is one with angles measuring {{{2*30^o=60^o}}} ,
and that is an equilateral triangle.
 
Probably the expected solution involves more formulas and less understanding, but the shorter, simpler solution, involving understanding of the concepts is better.