De-Constructing the Regular Heptagon The analysis of the regular heptagon (7 sides) is similar to that for the trisection of an angle. Note that if the construction is possible, then we could also construct the cosine of 2/7. (Why?). So let's find a polynomial that has that number as a root. Step 1. Let = 2/7, and let x = cos(). We know that cos(7) = 1. So we want to express cos(7) as a polynomial in x. One way to do this is to introduce y = sin(), and use de Moivre's theorem: for any positive integer n, and any angle , cos(n) + i sin(n) = [cos() + i sin()]n In particular, with n = 7, and = , we know that cos(7) + i sin(7) = 1. So, substituting x and y for cos() and sin() leads to the equation 1 = (x + iy)7 And in particular, we can equate the real part of the complex expressions on each side of the equation. That is, 1 is equal to the real part of (x + iy)7 . So, in the space below, compute (x + iy)7 using the binomial theorem, and collect the real parts together. Remember that the coefficients come from the 7th row of Pascal's triangle. Step 2. Now you have an equation that involves both x and y. However, only even powers of y should appear. Reviewing the definition of x and y, you will see that y2 = 1 x2 so make that substitution in your equation in the space below, and simplify to a polynomial equation in x. You should now have the equation 64x7 - 112x5 + 56x3 - 7x - 1 = 0. (If you do not, recheck your algebra). Is this polynomial irreducible? If not, factor it as far as possible. [Hint: there is a simple integer root. Use that to get a linear factor.] Step 3. You should now have the equation (x-1)(64x6 + 64x5 - 48x4 - 48x3 + 8x2 + 8x + 1) = 0 Or perhaps you found that you could factor this expression further. In fact, the second factor is a perfect square. In the space below, verify that it can be factored as (8x3 + 4x2 - 4x - 1)2 Step 4. We have now shown that x = cos (2/7) is a root of the equation (x-1)(8x3 + 4x2 - 4x - 1)2 = 0 Clearly, x is not equal to 1. So that makes our x a root of the cubic factor above. Show that this cubic is irreducible over the rationals, and then argue that it cannot be constructed using compass and straight edge.

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# Constructing a Regular 7