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The following diagrams outline the geometric reasoning behind the Coble construction.

Begin with the original diagram and add several auxiliary circles and lines. Note the three congruent rhombuses, \(BHCA\), \(GBAD\), and \(JGDI\), with \(E\) at the intersection of the diagonals of the middle rhombus \(DGBA\), which is known to be at the midpoint of each diagonal. Note the similar triangles \(ACF\) and \(ICJ\).

Since \(|CA| = 1\), \(|CI| = 3\), and \(|IJ| =1\), ratios of similar triangles shows that \(|AF| = 1/3\).

A referee suggested a nice alternative outline of a proof. The triangles \(ADG\) and \(ABG\) are both equilateral triangles. Thus, the measure of angle \(DAE\) equals the measure of angle \(BAE\); by Side-Angle-Side, the triangles \(DAE\) and \(BAE\) are congruent. Then the length of the segments \(DE\) and \(BE\) are equal, and also the length of the segments \(CA\) and \(AD\) are equal, so the segments \(AB\) and \(CE\) are medians of the triangle \(BCD\). It is now a well known fact that the intersection \(F\) of the medians is at the centroid which is one-third of the distance along the median \(AB\).

(Return to the Coble construction)

Begin with the original diagram, add several auxiliary circles and lines, then note the similar triangles \(ACF\) and \(GCH\). Since \(|AC| = 1\), \(|GC| = 3\), and \(|GH| =1\), we see that \(|AF| = 1/3\).

Of course, the picture hides the hardest part of the proof: showing that the circle with center B and the two lines \(CH\) and \(DJ\) go through the point \(E\). Possibly the easiest way to see this is to use analytic geometry. We will find the point \(E_2\) on the intersection of the lines \(CH\) and \(DJ\), and then show this point is exactly distance one from \(B\), that is, the point \(E_2\) is the same as the point \(E\). For convenience, we will make \(B\) the origin, then one can easily verify the coordinates of these points:

- \(A\quad (-1,0)\)
- \(B\quad (0,0)\)
- \(C\quad (-1/2, \sqrt{3}/2)\)
- \(D\quad (-3/2, -\sqrt{3}/2)\)
- \(H\quad (-1, -\sqrt{3} )\)
- \(J\quad (1,0)\).

Then the slope of the line \(DJ\) is \(\sqrt{3}/5\) and the equation is \(y = (\sqrt{3}/5) x - (\sqrt{3}/5)\). The slope of the line \(CH\) is \(3 \sqrt{3}\) and the equation is \(y = 3 \sqrt{3} x + 2 \sqrt{3}\). Some algebra shows that the point \(E_2\) where these two lines intersect has \(x = -11/14\), so \(y = -5 \sqrt{3} / 14\). The point \(E_2\) \(( -11/14, -5 \sqrt{3}/ 14 )\) satisfies \((11/14)^2 + ( 5 \sqrt{3} / 14 )^2 = ( 121 + 75)/ 14^2 = 1\), thus showing it is the point \(E\) from our construction.

(Return to the 2 circle 3 line construction)

Begin with the original diagram, add a few auxiliary circles and lines, then note the similar triangles \(ADF\) and \(GDE\). Since \(|AD|=1\), \(|GD|=3\), and \(|GE| =1\), we see that \(|AF| = 1/3\).

Elegant! Beautiful! Satisfying!

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