## Section8.1Geometric Constructions

In ancient Greece, three classic problems were posed. These problems are geometric in nature and involve straightedge-and-compass constructions from what is now high school geometry; that is, we are allowed to use only a straightedge and compass to solve them. The problems can be stated as follows.

1. Given an arbitrary angle, can one trisect the angle into three equal subangles using only a straightedge and compass?

2. Given an arbitrary circle, can one construct a square with the same area using only a straightedge and compass?

3. Given a cube, can one construct the edge of another cube having twice the volume of the original? Again, we are only allowed to use a straightedge and compass to do the construction.

After puzzling mathematicians for over two thousand years, each of these constructions was finally shown to be impossible. We will use the theory of fields to provide a proof that the solutions do not exist. It is quite remarkable that the long-sought solution to each of these three geometric problems came from abstract algebra.

First, let us determine more specifically what we mean by a straightedge and compass, and also examine the nature of these problems in a bit more depth. To begin with, a straightedge is not a ruler. We cannot measure arbitrary lengths with a straightedge. It is merely a tool for drawing a line through two points. The statement that the trisection of an arbitrary angle is impossible means that there is at least one angle that is impossible to trisect with a straightedge-and-compass construction. Certainly it is possible to trisect an angle in special cases. We can construct a $30^\circ$ angle; hence, it is possible to trisect a $90^\circ$ angle. However, we will show that it is impossible to construct a $20^\circ$ angle. Therefore, we cannot trisect a $60^\circ$ angle.

### Worksheet8.1.1Activity: Geometric Constructions

Our first application of algebra (to other mathematics) will be to questions of classical geometry. We will look at geometry as it was done in ancient Greece, except that we will use GeoGebra for our constructions.

Our main question is, what can you construct using reasonable, fundamental tools. The tools are: an unmarked straightedge and a compass.

In GeoGebra, there are many more tools than these. Make sure you only use the “new point” tool (to place points at the intersections of lines and circles), the “line through two points” tool, and the “compass” tool (under the circle menu). You can also use the arrow to drag things around if you need to.

To get a feel for the sorts of things you can construct, and maybe things you cannot, here are a few challenges.

###### 1.

Can you construct a $60^\circ$ angle? A $30^\circ$ angle? If you have constructed any angle at all, can you construct an angle half its measure? That is, can you bisect an given angle?

Solution

If you can construct an equilateral triangle (draw two circles with the same radius with centers at the two points on a line, and look for the intersections), then the angles at the vertices will be $60^\circ\text{.}$

To bisect an angle, create a circle centered at the vertex of the angle. Call the two points of intersection of this circle and the rays of the angle $A$ and $B\text{.}$ Draw a circle centered at $A$ with radius $\overline{AB}\text{,}$ and another circle centered at $B$ with radius $\overline{AB}\text{.}$ Create a line from the vertex of the angle to the intersection of these two circles (either or both).

###### 2.

Can you construct a square? Can you double the square? That is, if you can construct a square, can you construct a square of twice the area? Careful: this is not a square whose side length is twice the side length of the original.

Solution

Constructing the square can be accomplished by creating perpendicular lines. You can make the length of all the sides equal using the compass.

To “double the square” of side length 1 say, you need to find a line segment of length $\sqrt{2}$ off which to build a square. Take the diagonal of the original square.

###### 3.

Can you double the circle? That is, can you construct a circle and then construct a second circle of twice the area?

Solution

Call the radius of the first circle 1, so you want to find a circle with radius $\sqrt{2}\text{.}$ If you can double the square, you should be able to double the circle!

###### 4.

Here are three much harder, but related challenges. For each, play around enough to convince yourself these are really hard, if not impossible:

1. Can you trisect and angle? That is, given a constructed angle, can you construct an angle $1/3$ its measure?

2. Can you double the cube? That is, if you can constructed a cube (or at least a line segment which is the length of the edge of a cube), can you construct cube with twice the volume of the original?

3. Can you square the circle? That is, if you have constructed a circle, can you construct a square that has the same area as the circle?

Solution

None of these are possible, although it will take us a while to prove it (less than the 2000 years it took humanity).

### Worksheet8.1.2Activity: Constructible Numbers

We have considered what geometric shapes we could or could not construct. We are left with three big questions: is it possible to trisect an angle, to double a cube, or to square a circle. To answer these questions, we must “algebratize” geometric constructions.

Start with two constructible points $0$ and $1$ one unit apart. We define constructible recursively from this base case:

1. A constructible line is a line passing through two constructible points.

2. A constructible circle is a circle whose radius is a constructible number and whose center is a constructible point.

3. A constructible point is the intersection of two constructible lines, two constructible circles, or a constructible line and a constructible circle.

We say a number $a$ is constructible provided $a = 0$ or there are two constructible points distance $|a|$ apart. So far we have constructible numbers 0, 1, and $-1\text{.}$ What else is constructible?

For this activity, use GeoGebra. Use only the “new point” tool (to place points at the intersections of lines, circles, or both), the “line through two points” tool, and the “compass” tool (under the circle menu). You can also use the arrow to drag things around if you need to.

###### 1.

Show that the numbers 2 and 4 are constructible. Then show that the number $3 = 4 - 1$ is constructible.

###### 2.

If $a$ and $b$ are constructible numbers, are the numbers $a+b$ and $a-b$ also constructible?

###### 3.

Suppose $a$ and $b$ are constructible. Construct a triangle containing a base of unit length adjacent to a side of length $a\text{.}$ Construct a similar triangle with where the side corresponding the unit length side now has length $b\text{.}$ What is the length of the side corresponding to the $a$-length side? ###### 4.

Explain how you can modify the above construction to prove that if $a$ and $b$ are constructible, then $a/b$ is constructible.

###### 5.

Given constructible number $c\text{,}$ explain how you can construct the figure below. The vertical line should be perpendicular to the horizontal line, which is the diameter of the circle. What is the length of the vertical line?

###### 6.

Let $\mathfrak C$ be the set of all constructible numbers. What sort of set is this? Is it a group? A ring? A field? Is it one of these we know about already?

### Subsection8.1.3Constructible Numbers

A real number $\alpha$ is constructible if we can construct a line segment of length $| \alpha |$ in a finite number of steps from a segment of unit length by using a straightedge and compass.

###### Proof.

Let $\alpha$ and $\beta$ be constructible numbers. We must show that $\alpha + \beta\text{,}$ $\alpha - \beta\text{,}$ $\alpha \beta\text{,}$ and $\alpha / \beta$ ($\beta \neq 0$) are also constructible numbers. We can assume that both $\alpha$ and $\beta$ are positive with $\alpha \gt \beta\text{.}$ It is quite obvious how to construct $\alpha + \beta$ and $\alpha - \beta\text{.}$ To find a line segment with length $\alpha \beta\text{,}$ we assume that $\beta \gt 1$ and construct the triangle in Figure 8.2 such that triangles $\triangle ABC$ and $\triangle ADE$ are similar. Since $\alpha / 1 = x / \beta\text{,}$ the line segment $x$ has length $\alpha \beta\text{.}$ A similar construction can be made if $\beta \lt 1\text{.}$ We will leave it as an exercise to show that the same triangle can be used to construct $\alpha / \beta$ for $\beta \neq 0\text{.}$

###### Proof.

In Figure 8.4 the triangles $\triangle ABD\text{,}$ $\triangle BCD\text{,}$ and $\triangle ABC$ are similar; hence, $1 /x = x / \alpha\text{,}$ or $x^2 = \alpha\text{.}$

By Theorem 8.1, we can locate in the plane any point $P =( p, q)$ that has rational coordinates $p$ and $q\text{.}$ We need to know what other points can be constructed with a compass and straightedge from points with rational coordinates.

###### Proof.

Let $(x_1, y_1)$ and $(x_2, y_2)$ be points on a line whose coordinates are in $F\text{.}$ If $x_1 = x_2\text{,}$ then the equation of the line through the two points is $x - x_1 = 0\text{,}$ which has the form $a x + by + c = 0\text{.}$ If $x_1 \neq x_2\text{,}$ then the equation of the line through the two points is given by

\begin{equation*} y - y_1 = \left( \frac{y_2 - y_1}{x_2 - x_1} \right) (x - x_1)\text{,} \end{equation*}

which can also be put into the proper form.

To prove the second part of the lemma, suppose that $(x_1, y_1)$ is the center of a circle of radius $r\text{.}$ Then the circle has the equation

\begin{equation*} (x - x_1)^2 + (y - y_1)^2 - r^2 = 0\text{.} \end{equation*}

This equation can easily be put into the appropriate form.

Starting with a field of constructible numbers $F\text{,}$ we have three possible ways of constructing additional points in ${\mathbb R}$ with a compass and straightedge.

1. To find possible new points in ${\mathbb R}\text{,}$ we can take the intersection of two lines, each of which passes through two known points with coordinates in $F\text{.}$

2. The intersection of a line that passes through two points that have coordinates in $F$ and a circle whose center has coordinates in $F$ with radius of a length in $F$ will give new points in ${\mathbb R}\text{.}$

3. We can obtain new points in ${\mathbb R}$ by intersecting two circles whose centers have coordinates in $F$ and whose radii are of lengths in $F\text{.}$

The first case gives no new points in ${\mathbb R}\text{,}$ since the solution of two equations of the form $a x + by + c = 0$ having coefficients in $F$ will always be in $F\text{.}$ The third case can be reduced to the second case. Let

\begin{gather*} x^2 + y^2 + d_1 x + e_1 y + f_1 = 0\\ x^2 + y^2 + d_2 x + e_2 y + f_2 = 0 \end{gather*}

be the equations of two circles, where $d_i\text{,}$ $e_i\text{,}$ and $f_i$ are in $F$ for $i = 1, 2\text{.}$ These circles have the same intersection as the circle

\begin{equation*} x^2 + y^2 + d_1 x +e_1 x + f_1 = 0 \end{equation*}

and the line

\begin{equation*} (d_1 - d_2) x + b(e_2 - e_1)y + (f_2 - f_1) = 0\text{.} \end{equation*}

The last equation is that of the chord passing through the intersection points of the two circles. Hence, the intersection of two circles can be reduced to the case of an intersection of a line with a circle.

Considering the case of the intersection of a line and a circle, we must determine the nature of the solutions of the equations

\begin{align*} a x + by + c & = 0\\ x^2 + y^2 + d x + e y + f & = 0\text{.} \end{align*}

If we eliminate $y$ from these equations, we obtain an equation of the form $Ax^2 + B x + C = 0\text{,}$ where $A\text{,}$ $B\text{,}$ and $C$ are in $F\text{.}$ The $x$ coordinate of the intersection points is given by

\begin{equation*} x = \frac{- B \pm \sqrt{B^2 - 4 A C} }{2 A} \end{equation*}

and is in $F( \sqrt{\alpha}\, )\text{,}$ where $\alpha = B^2 - 4 A C \gt 0\text{.}$ We have proven the following lemma.

So what is the field of all constructible numbers? This will definitely contain $\Q\text{,}$ but then we can repeatedly add square roots of elements. And not just elements from $\Q\text{,}$ but also square roots of elements found in extension fields we got by taking square roots previously.

This brings us to the idea of an itterated extension: you get an extension field by adding a new number, then get another extension field by adding a new number to the first extension field. And we can keep going as far as we like.

So what we have now is that every constructable number is an element of some extension field of some extension field of some extension field of … some extension field of $\Q\text{,}$ each extension of the form $F(\sqrt{\alpha_i})$ where $\alpha_i \in F$ (but the $F$s are changing as we go up the line). We can state this more cleanly as follows.

Now if we want the field of all constructible numbers, we need to extend this process forever. As we will soon see, this shows that the field of constructible numbers is not a “finite” extension of $\Q\text{,}$ although it is still an algebraic extension. (We will define these terms in the upcoming sections.)

What does all of this have to do with solving the problems of doubling the cube, squaring the circle, and trisecting an angle? To answer this question we need to develop a notion of size of a field extension, which we will call its degree. It will turn out that the size of an extension is intimately tied to the degree of the polynomials which have roots in the extension. But we are getting ahead of ourselves. In the next section we will re-introduce some basics of fields and their relation to factoring polynomials.

### Exercises8.1.4Collected Homework

###### 1.

Write up a careful argument for question 4 from the activity Constructible Numbers: Prove that if $a$ and $b$ are constructible numbers, with $b \ne 0\text{,}$ then $a/b$ is also constructible.

Hint

As suggested in the worksheet, you will modify the argument that $a\cdot b$ is constructible when $a$ and $b$ are. That is, you should use similar triangles. Make sure you explain why you can construct the similar triangles you need, and also why the side you find really is $a/b\text{.}$