Simple Algebraic Concepts

Now imagine that the given points and the required points are drawn in some nice Cartesian coordinate system. Then finding new points is equivalent to finding their coordiantes. In other words, a construction problem can be restated in an algebraic form something like this:

• Given a set of pairs of coordinates: $(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)$, construct other pairs of coordinates $(X_1, Y_1), (X_2, Y_2), \ldots, (X_m, Y_m)$ using the following algebraic transformations ....

As we'll see below, those ``algebraic transformations'' are pretty simple to describe, and such a description will allow us to identify the sorts of sets of coordinates that can be constructed using a straightedge and compass.

Here's the idea--we just need to know how to how each of the allowed geometric constructions translates into algebra. Let's take a look:

• You can draw a line through two points. Well, it is easy to write down the equation of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$:
  
  $$(y - y_1) = \frac{(y_2 - y_1)}{(x_2 - x_1)}(x - x_1).$$
  
  

• You can draw a circle with any point as center and any other point on the boundary. If $(x_1, y_1)$ is the center, and $(x_2, y_2)$ is a point on the boundary, we know the radius $r$ is given by:
  
  $$r = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}.$$
  
  
  Then the equation of a circle centered at $(x_1, y_1)$ and having radius $r$ is given by:
  
  $$(x - x_1)^2 + (y - y_1)^2 = r^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2.$$
  
  

• You can find the intersection of any two lines. Again, this is easy--if the given lines have the equations $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$, a simultaneous solution of these two equations is simply a matter of algebra, and the result is just an equation in the variables $A_1, B_1, C_1, A_2, B_2,$ and $C_2$.

• You can find points that are the intersection of a line and a circle. In this case, suppose the line has equation $Ax + By + C = 0$, and that the circle has equation $(x - D)^2 + (y-E)^2 = R^2$. Then we can solve the equation of the line for $x$, giving $x = -(By + C)/A$, and substituting this into the equation for the circle. Solving will involve a quadratic equation, but the solution will at worst require the quadratic formula, and thus a square root of some numbers we already have. Note that there are two, one, or zero roots, depending on whether the line cuts the circle, is tangent to it, or misses it entirely.

• Finally, we need to be able to find the intersections of two circles. In this case, suppose that the equations of the two circles are given by: $(x- A_1)^2 + (y - B_1)^2 = R_1^2$ and $(x- A_2)^2 + (y - B_2)^2 = R_2^2$. But if you multiply these out, you get two eqauations that look like this:
  $$\begin{eqnarray*} x^2 + y^2 + ax + by + c &=& 0 \\ x^2 + y^2 + dx + ey + f &=& 0. \end{eqnarray*}$$
  
  
  
  
  
  If you subtract them, you can see that the system above is equivalent, algebraically, to the system:
  $$\begin{eqnarray*} x^2 + y^2 + ax + by + c &=& 0 \\ (a-d)x + (b-e)y + (c-f) &=& 0. \end{eqnarray*}$$
  
  
  
  
  
  This can be solved exactly as we did in the line-circle case, giving up to two solutions, and requiring nothing more than a square root in the quadratic formula.

So if you look at the algebraic result of any of the above constructions, you can see that nothing is much worse than a combination of addition, subtraction, multiplication, division, and square roots of the numbers (coordinates) you started with before the construction.

It is also not hard to show that any such arithmetic operation can be emulated with a proper geometric construction, so all of those sorts of numbers (coordinates) can be constructed. So here's the idea of where you can get after a certain number of steps, beginning only with the number 1:

1. You can make all the integers: $1, 2, 3, \ldots$.
2. You can make all the fractions (usually called ``rational numbers'': $i/j$, where $i$ and $j$ are integers.
3. You can make anything that looks like $q_1 + q_2\sqrt{q_3}$, where $q_1$, $q_2$ and $q_3$ are any rational number.
4. You can make anything that looks like $x_1 + x_2\sqrt{x_3}$, where $x_1$, $x_2$ and $x_3$ are any numbers you generate in step 3.
5. You can make anything that looks like $y_1 + y_2\sqrt{y_3}$, where $y_1$, $y_2$ and $y_3$ are any numbers you generate in step 4.
6. You can make anything that looks like $z_1 + z_2\sqrt{z_3}$, where $z_1$, $z_2$ and $z_3$ are any numbers you generate in step 5.
7. Et cetera...

Thus, numbers like the following are ``easily'' obtainable:

$$\sqrt{\frac{314159265}{2718281828}+ \frac{\frac{17}{35{\sqrt... ...1}{91}\sqrt{ \frac{3}{1198}\sqrt{\frac{11}{17} - \sqrt{2}}}}}.$$ (1)

Another way of looking at it is that any number that can be expressed as a ``tower'' of square roots and arithmetic operations can be constructed with a straightedge and compass.

But let's look at an example that's a little simpler than expression 1, above. Consider:

$$x = 15 + \sqrt{17 + 4\sqrt{\frac{3}{5}}},$$

where we've called the constructed number $x$.

The equation above can be converted as follows:

$$\begin{eqnarray*} x - 15 &=& \sqrt{17 + 4\sqrt{\frac{3}{5}}} \\ (x - 15)^2 &=& ... ...3}{5}} \\ ((x - 15)^2 - 17)^2 &=& 16\frac{3}{5} = \frac{48}{5}. \end{eqnarray*}$$

The final equation above is just (if you multiply it out) an equation containing powers of $x$ that are powers of 2, and exactly the same thing can be done with any such constructable number.

It may be obvious (but it certainly requires proof, and the proof is beyond the scope of this paper) that no matter how many square roots you apply, you can never get to a cube root (or a fifth root, or sixth root, or seventh root, or ninth root...). So if the number you would like to construct involves a cube root (or fifth root, or sixth root, ...), you are out of luck--it cannot be constructed using the methods described in Section 3.