Begin with the original diagram and add a few auxiliary lines. The angle \(CDG\) is a right angle since this angle subtends a diameter of the circle. Then \(CDG\) and \(CHD\) are similar right triangles. By our construction, \(|CD| = 2\) and \(|CG| = 6\), and by similar triangles (or trigonometry) \(|CH| / |CD| = |CD| / |CG|\) so \(|CH| = 2/3\). Since \(DH\) is a perpendicular bisector of the chord \(CF\), we have \(|CF| = 4/3\). Since \(|CA| = 1\), we have \(|AF| = |CF| - |CA| = 1/3\).
We have seen a trisecting construction using seven circles.
We prove six circles is not enough by using a Maple program that creates all points that can be constructed with a given number of circles (Maple file, pdf). Essentially, we construct all four-circle constructions (there are 14 up to symmetry) and note that none of these go through the point \((1/3, 0)\) or \((-1/3, 0)\). If a six circle construction exists, the next two circles added must go through the desired trisecting point. It is then easy to verify that no fifth circle goes through the desired \((1/3,0)\) or \((-1/3,0)\) point, hence one requires at least seven circles.
We summarize by giving the number of points generated by N circles by our Mohr-Mascheroni construction.
For convenience, we assume we begin with the two points, \((1,0)\) and \((-1,0)\). By symmetry, we need only list those points in the first quadrant. Here are the number of new points in the first quadrant generated by \(N\) circles:
In order to see the essence of the proof, we draw a more general picture: the key assumptions are that \(|OA| = |AB|\) and that \(|OG| = |GH|\).
Beginning with Martin's four points, we have seen a trisecting construction using eight lines.
The proof that seven lines is not enough is contained in this Maple file (pdf). This worksheet generates all points that can be constructed by line-only constructions up to ten lines. A list of all points that can be constructed by two through ten lines is given in this Maple file (pdf). Note that seven lines can generate the point \((2/3, 0)\) so seven lines does trisect a line segment, though not the segment \(AB\).
We summarize by giving the number of points generated by \(N\) lines with Martin's starting configuration.
We begin with Martin's four points \((1,0)\), \((2,0)\), \((0,1)\) and \((0,2)\). If we use \(N\) lines, we have \(M\) possible new points where