A Solid Construction of the Regular 7-gon

In solid constructions, we allow for the use of a (possibly only hypothetical) conic drawing tool. Given a constructible point A, a constructible line l, and a constructible length e, we are permitted to draw the conic section with focus A, directrix l, and eccentricity e.  The constructible points in this context are then the points where a pair of (distinct) constructible lines, circles, or conics intersect.

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.  
(1)                                                                                      
Thus, given real values a and b, we can construct the real roots of the cubic in (1).

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,
so
        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.

-Arthur Baragar, November 16th, 2003.