In particular, we can draw the parabola y = x2.
Consider the circle that is centered at (a,b) and goes
through the origin. This circle has the equation
(x - a)2 + (y - b)2 = a2 + b2.
If a point (x, y), other than the origin, also lies on the parabola , y = x2 then
(x - a)2 + (y - b)2 = a2 + b2,
x2 - 2ax + x4 - 2bx2 = 0,
|x3 + (1 - 2b)x - 2a = 0.||
The regular 7-gon has a solid construction, a fact that was known to
Archimedes. To construct it, we must construct x = e2pi/7, which satisfies the equation
x7 - 1 = (x - 1)(x6 + x5 + x4 + x3 + x2 + x + 1) = 0.
Let w = x + x-1 = 2cos(2p/7). Then
w3 = x3 + 3x + 3x-1 + x-3,
w2 = x2 + 2 +x-2,
w3+ w2 - 2w - 1 = 0.
To exploit (1), we complete the cube. That is, we make the substitution z = w + 1/3, so that
(z - 1/3)3 + (z - 1/3)2 - 2(z - 1/3) - 1 = 0,
which simplifies to
z3- (7/3)z - 7/27 = 0.
We therefore choose (a,b) = (7/54,5/3) and proceed as follows (following along in the figure): We construct the circle centered at (7/54,5/3) that goes through the origin. We drop a perpendicular from a point of intersection of this circle with the parabola y = x2, to find the point (w + 1/3,0). We find the point (w/2,0) and the perpendicular to the x-axis through this point. This perpendicular intersects the unit circle at two of the seven points of a regular 7-gon. We use those points to find the rest of the vertices.