Math 150


A Construction of a Regular Octagon

Dr. Wilson

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.


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 45o, 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 67.5o. Since /MOD is 45o,

/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.