Download 36. Trisection of an Angle ϕ θ

36. Trisection of an Angle
To divide an arbitrary angle into three equal angles.
This famous problem cannot be solved with compass and straightedge alone. See the
supplement.
The simplest solution is by means of the following paper strip construction of
Archimedes. Let 2 be the given angle and S be its vertex. Let S be the center of a circle of
radius r that intersects the legs of the angle at A and B. Mark off a segment of length r on
the edge of a paper strip, and then place the edge on the figure in such a way that it passes
through B and the marked-off segment coincides with PQ with P on the circle and Q (outside
the circle) on the extension of AS. Refer to the figure below:
B
P
r
Q
Then 0PQS I 1
3
r
r
θ
ϕ
S
A
2.
Proof. Since PS PQ r, dPQS is isosceles, and 0PSQ I, while the external angle
0SPB 2I. Since dSPB is isosceles, 0SBP 2I too. Finally the external angle (of
dBQS) 2 0BQS 0QBS I 2I 3I, or I 13 2. R
Note 1. The figure above illustrates that the method works for acute angles; it also works
for obtuse angles, but then B might lie between P and Q.
Note 2. The quadratrix of Hippias (ca. 420 B.C.) was invented to solve the trisection
problem. See The History of Mathematics by Burton.
Note 3. Dörrie describes a second method from Pappus (a. 300 A.D.) that uses a fixed
hyperbola.
Note 4. It is also possible to trisect an angle using the conchoid of Nicomedes.
Note 5. The limaçon, discovered by Etienne Pascal (1588-1640), father of the
better-known Blaise Pascal, can also be used to trisect angles. See The History of
Mathematics by Burton.
Note 6. Most geometry programs, e.g. Geometer’s Sketchpad, allow the user to build
tools to trisect arbitrary angles. (These tools however, use things like measurements
and rotations through marked angles not permitted in compass and straightedge
1
constructions.)
Supplement to Nos. 35, 36 and 37.
[The language in this sections seems somewhat old fashioned and hard to follow. It’s
basically about extension fields, but that terminology is not used here. A course in Abstract
Algebra nowdays usually reaches the same results in a more easily understood fashion.
The reader might want to refer to A First Course in Abstract Algebra by John B. Fraleigh.]
Theorem. A complex number z is constructible from the rationals Q if the degree of the
QŸz over Q is a power of 2.
Idea. If z is constructible from Q, z is arrived at by a succession of intersections of two
lines, a line and a circle or two circles. The second and third of these intersections
might require the introduction of square roots, i.e., a quadratic extension. Repeated
R
applications mean that the degree of QŸz over Q must be a power of 2.
I. It is impossible to double a cube with compass and straightedge alone.
Proof. This requires construction 3 2 , but x 3 " 2 is an irreducible polynomial over Q, and
R
so the degree of Q 3 2 over Q is 3, not a power of 2.
II. It is impossible to trisect an arbitrary angle with compass and straightedge alone.
Proof. Let 2 60 ( . Then I 1
3
2 20 ( . Since
cos 2 cosŸ2I I cos 2I cos I " sin 2I sin I
Ÿ2 cos 2 I " 1 cos I " 2 sin I cos I sin I
Ÿ2 cos 2 I " 1 cos I " 2 cos IŸ1 " cos 2 I 4 cos 3 I " 3 cos I
I cos 20 ( is a root of the equation 12 4x 3 " 3x or 8x 3 " 6x " 1 0. It can be shown
that 8x 3 " 6x " 1 is an irreducible polynomial of degree 3 over Q, and hence none of
its roots, especially, cos 20 ( is constructible. It follows that a 20 ( angle is not
constructible, for if it were we could construct the following triangle:
1
20°
cos20°
2