### Constructing a Regular Dodecagon by Cutting the Corners
off a Regular Hexagon

You can construct a regular dodecagon or twelve sided figure by
cutting off the corners of a regular hexagon. The question is how
much do you cut off of each corner. If we let the sides of the
regular hexagon each be one unit and we let the amount we cut off the
corners be x, then we get the following figure.

To find x, we note that x satisfies the following equation.

To solve for x we transpose the x terms to the same side of the
equation,

combine like terms,

and divide both sides by the coefficient of x.

Rationalize the denominator.

It is easy to find a 2 and a square root of 3 in the regular
hexagon.

Now that we have the distance we need to take off of the corners,
we can mark it off from each vertex.

To get a geometric proof that this construction works, we would
need to show that in a regular dodecagon, HIJKLMNPQRST, where G is
the point on the diameter, AD, whose distance from A is the same as
the length of the side of the regular dodecagon, that B and G are the
same distance from D. This is the same thing as saying that triangle
GBD is isosceles. Since __/__BDG is a 30^{o} angle, in
order for the triangle to be isosceles, it would suffice for
__/__DBG to be 75^{o}. Since DB __ | __ AB that means
that we need only show that __/__ABG is a 15^{o} angle.
Let us focus in on the top of the figure and draw in line segment HJ.

The isosceles 60^{o} triangles, AGH, AGI, and BJK are all
congruent equilateral triangles

where the length of each side is the length of the side of the
regular dodecahedron. It follows that HG is both congruent and
parallel to BJ. That makes quadrilateral HJBG a parallelogram since
we have a theorem
that a quadrilateral where one pair of opposite sides is both
congruent and parallel, the figure is a parallelogram.. Since HI and
IJ are both sides of the regular dodecagon, triangle HIJ is an
isosceles triangle whose vertex angle is 150^{o}. That makes
__/__IJH a 15^{o} angle. Since HJBG is a parallelogram, HJ
|| BG, and __/__ABG is a 15^{o} angle as required. QED.