## A Construction of a Regular Octagon

There are several ways to construct a regular octagon. One of the
most efficient would be to take a square

and cut off the corners

This is a little tricky because the diagonal sides, which are
hypoteneuses to the right triangles on the corners, have to be as
long as what is left of the original sides of the squares after the
triangles are removed. If we let the side of the square be 1 and use
the Pythagorean theorem to find the diagonals of the triangles on the
corners, that gives us the following equation

which we can solve for x. Transpose the x terms to the left side

Factor out the x.

so

When we see an expression like this, we are in the habit of
rationalizing the denominator.

When we divide a polynomial by a monomial, we divide the monomial
on the bottom into all the terms on top.

1 is the length of the side of the square, and half the square
root of 2 is the length of half of a diagonal of a square. We can
subtract this off from the side of the square by scribing an arc
centered at a corner of the square and going through the point in the
middle where the diagonals meet and seeing where the arc intersects
the sides of the square.

We can get all of the points of the octagon this way.

While we were able to verify algebraically that this construction
works, the question arises as to whether there is a simple way of
seeing that it works. Let us take a look as some angles in the
figure.

Assume that EFGHIJKL is a regular octagon. Then __/__LOK is
45^{o}, because it is 1/8 of a full circle. If we draw in the
line segment from O perpendicular to LK, Since our figure is a
regular octagon, triangle LOK will be isosceles and the line from the
vertex perpendicular to the base will bisect the vertex angle.

__/__DLO will then be 67.5^{o}. Since __/__MOD is
45^{o},

/LOD will be 67.5o just like /OLD, and triangle LOD will be
isosceles which says that the distance from D to O will be the same
as the distance from D to L.

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