Berkeley Math Circle 2001-2002

Kiran Kedlaya

**The straightedge and compass operations**.

- (a)
- Given two points, we may construct the line through them.
- (b)
- Given two points, we may construct the circle centered at one point passing through the other point.
- (c)
- Given two lines, two circles, or a line and a circle, we may construct their intersection points.

- the perpendicular bisector of the segment joining two points;
- the angle bisectors of two given lines;
- the circle through three given points;
- a segment of length , given segments of length 1 and ;
- the regular hexagon, pentagon and 17-gon (!);

- the trisectors of an arbitrary angle;
- a regular heptagon (7-gon) or nonagon (9-gon);
- a segment of length , given segments of length and .

- the number 1 is constructible;
- if and are constructible, then so are , , and ;
- if is constructible, so is ;
- if is constructible, so are ;
- a number is constructible if and only if it can be written in terms of rational numbers using only addition, subtraction, multiplication, division and square roots.

**Warning**. The next part of the discussion belongs to the subject of
abstract algebra, so may involve some ideas you may not have seen before;
but I hope the intuition will be clear.
(A subset of that is what is called ``linear algebra''; that's pretty much
all I'm really using.)
Given a finite set of algebraic numbers, the set of complex numbers
obtained from and the rational numbers
by addition, subtraction, multiplication and division forms what is called a
*number field*. If is a number field, we can always find
in such that

**Example: constructing the 17-gon**.
In case you've never seen the construction of the 17-gon (due to Gauss), it's
worth a look, in part because it's a very simple example of a notion
from abstract algebra called a ``Galois group''.
Here's the idea: let
be a primitive 17th root of unity.
Note that is a primitive root modulo 17, so that
cover all the nonzero residue classes modulo 17. Now write

Gauss realized that each can be written as the root of a quadratic polynomial in terms of the previous ones. I'll leave it to you to check my arithmetic:

How do you guess this? I don't know how Gauss did it, but they are predicted by

**The marked straightedge operations**.
In using the marked straightedge to trisect angles, Archimedes assumed an
additional axiom: (here is the length marked on the straightedge)

- (d)
- Given a line , a circle , and a point , we can construct all pairs of points and , with on and on , such that the line passes through and the distance is equal to .

- (e)
- Given circles and and a point , we can construct all pairs of points and , with on and on , such that the line passes through and the distance is equal to .

**The ``solid'' operations**.
The ancient Greeks considered a geometric object to be ``solid'' constructible
if if could be obtained using the ordinary straightedge and compass operations
plus the ability to draw conic sections. One can phrase this as follows:

- (d)
- Given five points, we may construct the conic section (parabola, hyperbola, ellipse, or degenerate conic) through them.
- (e)
- Given a conic section and a line, circle or another conic, we may construct their intersection points.

- is algebraic, i.e., for some polynomial with integer coefficients;
- for some such , the number field generated by
*all*of the roots of has degree a power of 2 times a power of 3. (It's enough to check this for the polynomial of smallest degree, called the*minimal polynomial*of .)

**The origami operations**.
The Japanese art of paper folding gives an entirely different approach to
constructing geometric figures. Here is one set of axioms for ``origami
geometry''; there may be others in the literature. (This one is from an
``origami mathematics'' web site.)

- (a)
- Given two points, we may construct the line through them.
- (b)
- Given two points, we may construct the perpendicular bisector of the segment joining the two points (i.e., we may fold one point onto another).
- (c)
- Given two lines, we may form the angle bisector of either angle between them (i.e, we may fold one line onto another).
- (d)
- Given a line and a point, we may construct the perpendicular to the line through the point.
- (e)
- Given points and a line , we may construct the line through such that the reflection of across lies on .
- (f)
- Given points and and lines and , we may construct the line(s) such that the reflection of across lies on , and the reflection of across lies on .

**Challenge problems**.
Some of these are discussed in the article by Baragar mentioned above,
which I again heartily recommend; he also includes many more challenge
problems. Warning: the characterizations of
constructible numbers involves the notion of a ``Galois group'', which
we haven't talked about.

- Find a construction of a regular 7-gon using marked straightedge.
- You might wonder why I didn't state a version of the marked straightedge axiom (d) with two lines instead of a line and a circle. Prove that actually the resulting construction can already done with straightedge and compass.
- Prove that origami constructions (a)-(e) can also be accomplished with straightedge and compass. (The only subtle one is (e).)
- Find constructions for trisecting a given angle and doubling a cube using origami. (This will imply that origami construction (f) cannot be accomplished with straightedge and compass.)
- Find a regular polygon that cannot be constructed using origami. Even better, find a characterization of origami constructible numbers. (I believe the latter is unsolved!)