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From Euclid to Spigotry
Assignment 8 (outline 8)
Last modified: September 15, 2004
You can do everything on this piece of paper and hand it in.
1. An alternative construction of the regular heptagon starts with the same diagram as in lecture topic 2. But instead of constructing CQ, construct the perpendicular bisector of BP to get a segment of length y/2 = cos 2p/7. Construct a circle of radius 1 centered at P.The perpendicular bisector of BP intersects this circle in two vertices of the heptagon, and the intersection of the circle with BP is another vertex. Now you have three consecutive vertices and can easily construct the other four.  The diagram below will get you to the marked-ruler step.

2. Johnson's proof of the heptagon construction uses the fact that the angle q = p/14
satisfies sin 4q = cos 3q. It also satisfies cos 4q = sin 3q. Show that this second approach also
leads to the equation 8 sin3 q - 4 sin2 q - 4 sin q + 1 = 0. In factoring the fourth-degree equation, you will want to notice that q = -p/2 (or 3p/2) is a solution to the equation that has nothing to do with the heptagon.