Subject: Re: Circle cutting
Date: Sat, 23 Oct 2000 20:45:59 -0500
From: Brent H. Hoerman

I first encountered this problem in my undergrad abstract algebra course. In one edition of John B. Fraleigh's "A first course in abstract algebra" he started the chapter on math induction by setting up the circle cutting problem. After showing how our intuition can prove wrong ( 2^(n-1) is not correct!) my instructor placed a wager. An "A" in the course plus exemption from all future tests. Since it was only the beginning of the semester this was a great reward.

I stumbled upon the correct equation,

(1/24)n4 - (1/4) n3 + (23/24)n2 - (3/4)n + 1,

by using induction, since that was the chapter we had been studying. My instructor was skeptical about the correctness of the equation and said he would concede defeat only if I could do one of two things:

  1. make the construction for n=10 and count all the areas, or
  2. make the proof directly. Counting 256 areas and making sure no lines intersected didn't sound like fun so I eventually ended up making the direct proof and received the reward.

After meeting that challenge my instructor asked if I could prove that a minimum number of areas existed for the construction if we relaxed the constraint on only two lines meeting at a point in the circle's interior. Further, what is that minimum number of areas and what does the construction look like. I worked pretty hard on this problem, but decided to give up as my current research takes more than enough time.

I have been able to show how many areas are eliminated by adding other lines to the intersection points. I have a hypothesis that the minimal number of areas is achieved only in the highest symmetry arrangement: for n even, the points are evenly separated around the circumference; for n odd, the n points are placed at positions where (n+1) points would be evenly separated. Can anyone prove this and end my dilemma?

Good luck, Brent Hoerman


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