Dr. Wilson

After looking at the methods for constructing regular octagons, dodecagons, and decagons by cutting off the corners of regular squares, hexagons, and pentagons, respectively, the question becomes, can these methods be generalized to a generic method which will work with any regular polygon? While our constructions of the octagon and dodecagon were fairly simple and interesting and could be derived either algebraically or geometrically, the geometric verification for the dodecagon was more complicated that the one for the ocatgon. The construction of the decagon was much more complicated than the octagon or dodecagon, and, as far as I know, there is no known method for verifying it geometrically. These considerations are discouraging in the search for a generic method. However, there is a generic method which will work for all regular polygons. We can circumvent these discouraging considerations by using a method which is different than the ones mentioned above.

All of the methods mentioned above start off by considering a regular polygon whose edges all have a length of one unit.

Let x denote the length of the part of the side which is cut off. r is the ratio of the length of the base of the isosceles triangle that is cut off to the congruent sides. If the resulting figure is regular we get the following equation.

1 - 2x = rx

Transpose

1 = 2x + rx

1 = (2 + r)x

Now divide both sides by the coefficient of x.

At this point, since r involved a square root, the methods mentioned above all involve rationalizing the denominator. While that yielded a quick method that could be verified geometrically in the case of the square and hexagon, it proved more complicated in the case of the pentagon. The difficulty in generalizing the method is due to the fact that different things happen after you rationalize the donominator in the different figures. If we don't bother to rationalize the denominator at this point, we can construct the desired distance. We can find an r in the figure.

By similar triangles the distance from A to C is r. We can then get a 2 + r and find its reciprocal

by using similar trianlges.

Note the unit circles centered at A and C. That makes the distances from A to D and from C to E one unit. As a result, the distance from D to E is 2 + r. Let EF be parallel to AG, and have length 1. Then triangles DEF and DAG are similar triangles and by the ratio

since AD and EF both have length 1, this gives us

If we cut off a distance as long as AG from each of the corners, we get a regular polygon with twice the number of sides

no matter how many sides our original regular polygon has.

The figure for the equilateral triangle, square, regular pentagon and regular hexagon look like

Turning an equilateral triangle into a regular hexagon is the most trivial of the constructions. It simply requires dividing the sides of the equilateral triangle into thirds, which this construction does. The constructions where we rationalized the denominator were simpler and more intersting for the square and hexagon. Note that in the construction for the hexagram, point H, where DF crosses the bottom of the hexagram, is the vertex of the dodecagon, because you can show that, if H is the vertex of the dodecagon, then triangles DCH and DAG are similar. As a result, DF does go through the point where the circles centered at A and C intersect, because O is the center of the hexagon. The construction we have here is actually simpler for the pentagon than the method we got after rationalizing the denominator, leading one to suspect that this method is probably simpler for the rest of the regular polygons as well.