Math CentralQuandaries & Queries


Question from Peter, a student:

hi, i know that if i use a compass to draw a circle, i can use the same setting to mark 6 intervals along the circle circumference, how would i go about marking out 21 equally spaced points around the circumference?

if there an online calculator i can us, or some other interesting trick?




Constructing things with compasses and straightedge is basically equivalent to solving quadratic equations (or just taking square roots.) If you imagine a simple calculator that can add, subtract, multiply, divide, and do square roots (but to infinite precision!) the points whose coordinates you can compute with it are exactly the ones you can construct with the classical tools.

Why? You have five operations:

  • drawing the line through given points,

  • drawing a circle with given center through a given point,
and finding (if they exist) the intersections of
  • two lines

  • a line and a circle

  • two circles.

It can be shown that, in Cartesian coordinates, you can compute the formulae of the line or circle form the coordinates of the points, or vice versa, using only the operations listed above.

If you can construct a 21-gon, you can construct a regular heptagon by skipping two of every three vertices - and if you can construct a heptagon, you can construct a 21-gon by constructing an equilateral triangle with one vertex in common and the same center. The two closest other vertices of this diagram will be 1/21 of a circle apart. So your problem reduces to constructing a regular heptagon.
Here's where it gets technical. Using complex numbers, the points of a heptagon are the seven seventh roots of 1. If one of the non-real ones is z, we have $1 + z+z^2+z^3+z^4+z^5+z^6 = 0. $

Divide out $z^3: z^{-3} + z^{-2} + z^{-1} + 1 + z + z^2 + z^3 = 0.$

Let $w = z + z^{-1}.$ A bit of algebra shows that $w^3 + w^2 - 2w - 1 = 0$. So constructing a heptagon is equivalent to solving this cubic equation.

Now, if a cubic equation has integer coefficients, we can show that either it has a rational solution or none of its solutions can be expressed using only +,-,x,/, and square roots. I'll skip the details on this one.

We can show fairly easily that the equation above does not have any rational solutions. Suppose there is a rational solution. It must have a lowest-terms form, a/b. Plug that in and multiply by $b^3:$

$a^3 + a^2b -2ab^2 -b^3 = 0.$

Now, a cannot be odd. For then the first term is odd, the second and fourth terms have the same parity, and the third term is obviously even, so their sum can't be 0.

So a is even. Then the first three terms are even and $b^3,$ hence b, is even as well. But that contradicts our lowest-terms assumption.

So - we've shown that the heptagon cannot be constructed using the "classical tools." Similar methods are used to show the the 9-gon cannot be constructed (which proves angle-trisection is impossible, as a 120-degree angle can be constructed) and that the "duplication of the cube" (construction of a segment whose length is the cube root of 2 times a give one) is impossible.

So, what can you do? You can of course approximate the solution using a decimal approximation to $\sin(π/7)$ which is about 0.43388373911755812047576833284836, far more precise than needed for any practical purpose. But this is still not an exact mathematical solution!

There are various other tools (such as given curves in the plane) that allow all cubics to be solved exactly by geometric construction using those tools. Origami construction is also powerful enough to solve all cubics.

Good Hunting!

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