From Euclid to Spigotry #7
Beyond Ruler and Compass
Last modified: September 15, 2004
By relaxing the strict Greek rules for ruler-and-compass constructions, it becomes possible
to achieve constructions that trisect an arbitrary angle and "duplicate the cube."  These methods "cheat" by using a marked ruler (Euclid allows only an unmarked straightedge).
References:
Bold, Chapters 4 and 5
Doerrie, 100 Great Problems, pages 170 - 177 

Preliminaries for duplicating the cube:
Lecture topic 1: the Law of Sines

By writing the length h of the "altitude" 
of the triangle in two different ways,

show that A sin b = b sin A

Conclude that

for an arbitrary triangle.

Lecture topic 2: the Law of Cosines

By applying the Pythagorean Theorem
to each of the two right triangles into
which the large triangle is divided, and
then eliminating h, show that

c2 = a2 + b2 - 2 a b cos C.

You can think of this as a generalization
of the Pythagorean theorem.  If C = p/2
it simplifies to c2 = a2 + b2

Lecture topic 3: The theorem of Menelaus

The theorem of Menelaus concerns a triangle ABC that is cut by a "transversal" DEF which intersects its three sides, at least one of which (AB in the diagram) must be extended. It states that 
AD BE CF = BD CE AF.

Here is one of many proofs.
Apply the Law of Sines to triangle CEF to get
CE/ sin F = CF /sin E
Apply the Law of Sines to triangle ADF to get
AF/ sin D = AD / sin F
Apply the Law of Sines to triangle BDE to get
BD/ sin E = BE / sin D

Multiply the three equations, and cancel the common factor of sin D sin E sin F from the denominator to get
CE AF BD = CF AD BE

Lecture topic 4: A substitute for Doerrie's elusive "leg transversal theorem (p. 172)

Euclid, Book III, proposition 36 says that any chord of the circle through point O outside the
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. 
Construct segment QP, of length x.
From right triangle OQP,

From right triangle DQP,

So  
or

Notice also that  where e is the length of the tangent from O to the circle.

Lecture topic 5: Duplicating the cube
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.  

Applying the theorem of Menelaus to the triangle APC, cut by the transversal DBQ, you get
AD CQ BP = PQ AB CD or
k x y = k k 2k or

Applying the Law of Cosines to the triangle BPC, you get

or

Alternatively, construct a circle centered at C with A and B on the circumference and apply
Euclid's theorem about lines from P that intersect the circle. The lengths in the diagram for topic 4 are now a = x + k, c = k, b = y + k, d = k,
so the conclusion that (a-c)(a+c) = (b-d)d becomes
x(2k + x) = y (y+k).
Multiply both sides by x2 to obtain x2(2k + x) = xy (xy+kx).
But xy = 2k2, so x2(2k + x) = 2k2 (2k2+kx) = 2k3 (2k+x).

cancel out a factor of 2k + x, and you find that
 so that x =  k
The construction is on the course Web site.

Lecture topic 6: The algebra of angle trisection
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.
Lecture topic 7: Archimedes' marked ruler construction

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.