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Begin with the original diagram and add a few auxiliary lines. Angle \(ADG\) is a right angle since it is an inscribed angle subtending a diameter. Now \(GDA\) and \(DJA\) are similar right triangles. By our construction, \(|AG| = 6\) and \(|AD| = 1\), and by similar triangles (or trigonometry) \(|AG| / |AD| = |AD| / |AJ|\). Thus, \(|AJ| = 1 / |AG| = 1/6\). Since \(DJ\) is a perpendicular bisector of the chord \(AF\), we have \(|AF| = 2 |AJ|\) and so \(|AF| = 1/3\).

(Return to the 4 circles construction)

Beginning with points \(A\) and \(B\) and the line through them, we construct the circle with center at \(B\) and radius to \(A\). There are only two ways, up to symmetry, to construct a second circle.

The following outlines all the possibilities for three circles and a line. We use symmetry whenever possible; for instance, we only need to consider third circles with centers in the "first quadrant" (\(E\), \(D\) or \(C\)) for the case of the two congruent circles. We only draw lines from points in the top half to points in the bottom half since a line through an existing point on the \(x\)-axis does not create a new point on the \(x\)-axis.

We assume that \(B\) is at the origin and that \(A\) is at -1 and \(C\) is at 1 on the \(x\)-axis. Since we could equally well consider \(A\) or \(C\) to be the origin, we only consider the constructible values on the \(x\)-axis modulo 1, hence only need to list values between -0.5 and 0.5. Since we can flip the diagrams, we also can consider the constructible values on the x-axis modulo the +/- sign, hence we only need to list values between 0 and 0.5. For instance, in the third 3-circle-1-line construction below which adds the circle with radius \(DB\), the point \(M\) is actually at \(x =\) 2.6180339887 which modulo 1 is -0.3819660113 which modulo the +/- sign is 0.3819660113. This happens to be the value of \(K\) as well.

(Return to 3 Circles 2 Lines Construction)

We first consider the two ways we can construct two circles. The second case does not allow any new lines to be drawn so may be eliminated. For the case with two congruent circles, by symmetry we only need consider lines through the "first quadrant:": through the point \(C\), \(D\) or \(E\). The only possibilities are the blue lines. The only new points created are \(G\), \(H\), \(I\), \(J\). The only new lines that go through one of these new points and the original points off the \(x\)-axis are the two red lines which are parallel to the \(x\)-axis. In any case, the only constructible points on the \(x\)-axis are integral or half-integral.

Robert Styer (Villanova Univ.), "Trisecting a Line Segment (With World Record Efficiency!)," *Convergence* (February 2010), DOI:10.4169/loci003342