circle has the property that the product of the distances from O to the two points where
the chord intersects the circle is constant, i.e. that
(OB)(OA) = (OC)(OD) = (OE)(OE)
This is similar to the better-known theorem about two chords that intersect inside a circle.
The proof is surprisingly short and easy.
This is explained very nicely on page 171 of Doerrie. You start with an equilateral triangle of side k, extend one side AC through A by distance k, and draw a line L1 from the end of the extended side through the third vertex B. Also extend side AB to form line L2. Now a line L3 from C cuts line L1 in point Q and line L2 in point P. With a marked ruler, arrange for the distance PQ to equal k. The segment CQ now has length k.
Doerrie refers to a "leg transversal theorem" (equivalent to topic 4) and to the theorem of Menelaus. An alternative to the "leg transversal theorem" is to apply the law of cosines to triangle PBC, where the angle at B has a cosine of -0.5.
Start from De Moivre's theorem and (again) show that
cos 3q = 4 cos3 q - 3 cos q
Now, following Bold, pp. 33-34, set x = 2 cos q
and show that the equation becomes x3 - 3x - 2 cos 3q = 0
Show that if 3q = p or 3q = p/2 (these are angles that we can trisect),
then this cubic equation can be factored, but that if 3q = p/3 the equation is irreducible (it has no rational roots). The argument on p. 34 of Bold is very similar to the proof that is irrational.
This is shown on p. 35 of Bold. The angle to be trisected is ASB. The trick is to align the
marked ruler (marked so that PQ equals the radius of the circle centered at P) so that
Q is on the line AS and P is on the circle.
Use the isosceles triangles PQS and SPB to show that the angle SQP (call it a) is one-third of
the angle ASB.
The construction is on the course Web site.
Lecture topic 8: A proof that the marked ruler construction leads to a cubic equation.
Here is a variant of the marked ruler construction where a line actually trisects the original angle at B.
Let the radius of the circle equal 1
The marked ruler is used to make DE also have unit length.
Then DEO is isosceles and its base OD = s = 2 cos a.
EBO is also isosceles and its base t = 2 cos 2a = 4 cos2 a - 2
Apply the Pythagorean theorem to the right triangle DFB.
(s + cos 3a)2 + (sin 3a)2 = ( t + 1)2 = (4 cos2 a - 1)2.
s2 + 2 s cos 3a + cos2 3a + sin2 3a = 16 cos4 a - 8 cos2 a + 1
Cancel out cos2 3a + sin2 3a against 1.
Replace s by its value 2 cos a
4 cos2 a + 4 cos a cos 3a = 16 cos4 a - 8 cos2 a
Factor out 4 cos a (if it's zero we can already trisect the angle 3a).
cos a + cos 3a = 4 cos3 a - 2 cos a
or cos 3 a = 4 cos3 a - 3 cos a
which is an irreducible cubic equation for cos a.