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\centerline{\bf \Huge Classical Geometric Construction}
\vskip .1in
\centerline{\Large Tom Davis}
\centerline{tomrdavis@earthlink.net}
\centerline{http://www.geometer.org/mathcircles}
\centerline{October 17, 2000}
\vskip .3 in

\section{Example}
\label{sec:example}

The best introduction to classical geometric construction is to show
an example.

In Section~\ref{sec:rules}, the ``official'' rules for geometric construction
will be presented, but we'll begin with just an intuitive idea of what can be
done with a straightedge and compass.

What we would like to do is to begin with a line segment, and using only
a straightedge and compass, divide that line into two equal segments.  In
other words, if we are given a line segment $AB$, we would like to construct
the midpoint $M$ of that segment.  Then we have bisected the segment, since
$AM = MB$.

\begin{figure}[ht]
\includegraphics [scale = .5]{Bisect1.eps}
\caption{Bisecting a segment---steps 1 and 2}
\label{fig:bis1}
\end{figure}

Figure~\ref{fig:bis1} shows the first two steps.  A compass is used with
center at $A$ to draw a circle passing through $B$, and in the same way,
another circle is drawn centered at $B$ and passing through point A.

\begin{figure}[ht]
\includegraphics [scale = .5]{Bisect2.eps}
\caption{Bisecting a segment---steps 3 and 4}
\label{fig:bis2}
\end{figure}

In Figure~\ref{fig:bis2}, two new points are determined---the intersections
of the two circles that were drawn in step 1.  Label these two points $C$ and
$D$.

\begin{figure}[ht]
\includegraphics [scale = .5]{Bisect3.eps}
\caption{Bisecting a segment---steps 5 and 6}
\label{fig:bis3}
\end{figure}

The final two steps are shown in Figure~\ref{fig:bis3}.  The straightedge
is used to connect points $C$ and $D$ with a new line, and finally, the
intersection $M$ of that line with the original segment $AB$ is determined.
$M$ is the required midpoint.  (The other segments in the figure, such as
$CA$, $CB$, $DA$, and $DB$ are not part of the construction, but we will
use them below.)

It may seem obvious from the figure that $M$ is the midpoint, but part of
any construction is a proof that the required point has been found.  In
this case it is very easy to do---notice that all the lengths $CA$, $CB$,
$DA$, and $DB$ are equal, since they are radii of two circles with radius
$AB$.  Therefore the quadrilateral $ACBD$ is a parallelogram, and we know
that the diagonals of a parallelogram bisect each other.  There are dozens
of other easy proofs.

\section{Who Cares?}

Why should anyone care about what figures can be constructed using only
these two specialized tools?  Why not choose different tools, or more
tools?

One reason for the choice is that the fundamental objects in classical
plane Euclidean geometry are points, lines, and circles, and a
straightedge and compass are the minimal set of tools that will allow
any circle or any line to be drawn.  So it is obviously interesting to
look at what can be drawn (constructed) using only the two tools that
are obviously necessary to draw all the sorts of figures that are
interesting.

The other reason for the choice of a straightedge and compass is that
this minimal set of tools enables one to make an amazingly large
collection of constructions.  Also, for thousands of years, many
people thought that any ``reasonable'' construction could be done with
just those two tools.  There were, however, 3 famous constructions
(see Section~\ref{sec:impossible}) that nobody seemed to be able to
do (using just a straightedge and compass, that is).

Finally, doing classical straightedge and compass constructions provides
a wonderful selection of problems that helps to reinforce important
geometric concepts from all parts of geometry.

\section{Rules}
\label{sec:rules}

OK, so what are the ``official rules'' for straightedge and compass
constructions?  For example, can marks be made on the straightedge?  Can
you hold the compass against the straightedge to use it as a measuring
device for one particular length?  If you have a circle drawn already,
can you put a straightedge against it so that it just touches (is tangent
to) the circle?  Et cetera.

\subsection{The Official Rules}
\label{sec:official}

The rules below are actually a little more flexible than the ``official''
rules, but they are easy to understand, and the difference between these
rules and the official rules is explained in the footnote.  Here they are:

A construction problem begins with a set of given points, lines, and
circles, and with some desired point, line or circle to construct, based
on the given objects.  In the example in Section~\ref{sec:example}, we
were given two points and the segment between them, and the object of the
construction was to find the point that lies on the segment midway between
the two original points.

Notice that we could reduce the statement above to require that only
certain points be constructed.  Obviously, if you want a line, you can
simply require that two different points be found on it, or if you want
a circle you can require that the center and a point on the circle be
found.

At any stage in the construction, you may do any of the following
things to obtain additional points, lines or circles:
\index{straight-edge and compass!rules for use}
\begin{enumerate}
\item You may draw a straight line of any length through two existing
points.  (This means, of course, that the straightedge is as long
as you need it to be, so it is better than a real ruler in that sense.)

\item You may find a new point at the intersection of two lines, two
circles, or of a line and a circle.  When you are given a segment, of
course, you are given the two points at its ends, so you can certainly
use those.

\item You may construct a circle centered at any existing point having
a radius equal to the distance between any two existing points.  In
other words, you can set the size of the compass from any two points
$A$ and $B$, and then you can move the point of the compass to another
point $C$ without changing the setting and draw a circle of radius
$AB$ about the point $C$\footnote{This is a ``modern compass''.
Originally, most construction problems were stated in terms of a
``Euclidean compass'' which collapses after each circle is drawn.  In
other words, a Euclidean compass cannot be used to copy a length.  But
it is possible to do any construction with a Euclidean compass that
can be done with a modern compass.  In fact, all constructions that
can be done with a straight-edge and compass can be done with a
compass alone---this is called the Mohr-Mascheroni
Theorem.}.  (Of course this includes drawing a circle given its
center and a point on the edge---you use the center and the edge to
set the compass size, and then you re-use the center point as the
center of the circle.)  As with the straightedge, there is no limit to
the size of a circle that can be drawn, so the mathematical compass is
better than any real one could be.

\item You may choose an arbitrary point on a line, circle,
or on the plane.  (And of course you can also choose a point
\emph{not} on a line or circle as in ``pick any point not on segment
$AB$.'')
\end{enumerate}

\subsection{Other Possible Rules}

In fact, constructions using other sets of tools have been examined,
but we will not go into that here.  For example, you might be interested
in what can be constructed using only a straightedge, or just a compass,
or perhaps a straightedge and a ``rusty compass''---a compass that is
frozen into one particular setting, or even constructions using more
interesting tools.

Or what if you are allowed to put marks on the straightedge, effectively
turning it into a ruler?  What if the straightedge is only a fixed
finite length?  What if one circle has already been drawn, but you
no longer have a compass?  There are any number of other possibilities.

\subsection{Some Final Warnings}

Obviously, with a physical straightedge, compass, and pencil, no
construction can be carried out to infinite accuracy; what is required
here is a construction that would be to infinite accuracy, assuming
that all the tools work to infinite accuracy.

In other words, it is not good enough to get an answer that is accurate
to one part in a million or to one part in a trillion---an infinitely
accurate result is required.  As we will see later, if you ``only'' need
accuracy to one part in a trillion, you can construct anything with a
straightedge and compass.

Finally, the construction has to be completed in a finite number of steps;
you cannot require an infinite set of operations to construct a point
that is the limit of these operations.  If an infinite number of steps were
allowed, again, anything can be constructed.  Simply construct points to
an accuracy of one in a million, then one in a billion, then one in a
trillion, and continue forever.  The limit of these points will be the
exact solution, but only after an infinite number of steps.

\section{Basic Constructions}

Here are a list of elementary constructions that can be carried out
with a straightedge and compass.  They are arranged roughly in order
of difficulty, and if you can do most of these, you can do most
standard geometric constructions.

\begin{enumerate}
\item Copy a segment.  In other words, mark off a segment that
exactly matches the length of a given segment on a different
straight line.

\item Copy an angle.  Given an angle, make another angle of exactly
the same size somewhere else.

\item Bisect a segment.  This problem was already solved in
Section~\ref{sec:example}

\item Bisect an angle.  Given an angle, find a line through the
vertex that divides it in half.

\item Construct a line perpendicular to a given line through a point
on the given line.

\item Construct a line perpendicular to a given line and passing
through a point not on the given line.

\item Given a line $L$ and a point $P$ not on $L$, construct a new
line that passes through $P$ and is parallel to $L$.

\item Construct an angle whose size is the sum or difference of
two given angles.

\item Given three segments, construct a triangle whose sides have the
same lengths as the segments.

\item Construct the perpendicular bisector of a line segment.

\item Given three points, construct the circle that passes through
all of them.\label{prob:circumcircle}

\item Given a circle, find its center.

\item Given a triangle $T$, construct the inscribed and circumscribed
circles.  The inscribed circle is a circle that fits inside the
triangle and touches all three edges; the circumscribed circle is
outside the triangle except that it touches all three of the vertices
of the triangle.

\item Construct angles of $90^{\circ}$, $45^{\circ}$, $30^{\circ}$,
$60^{\circ}$, and if you want a challenge, $72^{\circ}$.

\item Construct a regular pentagon.  (A regular pentagon is a five
sided figure all of whose sides and angles are equal.)

\item Given a point $P$ on a circle $C$, construct a line through $P$
and tangent to $C$.

\item Given a circle $C$ and a point $P$ not on $C$, construct a line
through $P$ and tangent to $C$.

\item Given two circles $C_1$ and $C_2$, find lines internally and
externally tangent to both.

\end{enumerate}

\section{Algebraic Equivalent}
\label{sec:algebra}

The original construction problems began with the Greeks, and for
thousands of years, the methods were the same.  Problems would be stated,
a construction would be found, and then a standard geometric proof
was supplied to show that the construction in fact behaved as advertised.

This worked great for thousands of years, except that it did not
provide any method to show that certain constructions were impossible
(see Section~\ref{sec:impossible}).  But nobody thought that impossible
constructions existed, so there was no real reason to do so.

Now we know that (in a sense) ``almost all'' constructions are impossible.
In this section we will not prove that fact, but we will provide some
overwhelming evidence.  But to do so, we'll have to look at geometric
construction from an algebraic point of view.

We will  simplify the problem of construction to be that of finding only
points.  As we stated in Section~\ref{sec:official}, this is good enough,
since any lines or circles that you might want can be identified in terms
of a couple of points.

\subsection{Arithmetic with Straightedge and Compass}

One way to look at arithmetic in terms of geometry is to let the lengths
of line segements represent numbers.  If we have two different line segments
(or in our simplification, two different sets of points), we have two
lengths, and it is easy to see how to add or subtract the lengths, given
only a straightedge and compass (see the arithmetic exercises in
Section~\ref{sec:arithmeticexercises}).

Addition and subtraction is easy, but a problem occurs when we talk
about multiplication---a problem having to do with units.  If you have
two segments, for example that are one meter long, if you multiply them,
you'll get something that is a square meter.  How can you compare the
area of a square with the length of a line?  In fact, you cannot---if
you think the answer is 1, since $1\times 1 = 1$, why isn't the answer
10000, since the lines are 100 centimeters long, and $100\times 100 = 10000$?
In fact, an equally good case can be made for any such result.  Geometrically,
you can only compare lengths with lengths, areas with areas, volumes
with volumes, et cetera.

A similar problem arises with division, square roots, cube roots, et
cetera.  There are a couple of ways to get around the problems.  The
easiest is when something like multiplication or square roots occurs
is to give an additional segment whose length is defined to be 1, and
to work in terms of that.  Another is so make sure that the units are
right, so it is possible to construct a segment of length $AB/C$ or
$\sqrt{AB}$, if $A$, $B$, and $C$ are lengths of given segments.

\subsection{Arithmetic Exercises}
\label{sec:arithmeticexercises}

In the exercises below, assume that you are given pairs of points whose
distance apart represents a length.

\begin{enumerate}
\item \textbf{Addition and subtraction.} Given segments of lengths $A$
and $B$, construct a segment of length $A+B$ or $A-B$.
\item \textbf{Multiplication.}  Given segments of lengths $A$, $B$, and
$1$, construct a segment of length $AB$.
\item \textbf{Division.}  Given segments of lengths $A$ and
$1$, construct a segment of length $1/A$.  Using the result of the problem
above, show that if you also have a segment of length $B$, you can
construct a segment of length $B/A$.
\item \textbf{Square roots.}  Given segments of lengths $A$ and $B$,
construct a segment whose length is $\sqrt{AB}$.  Clearly, if one of
the segments has length $1$, you can construct a segment of length
$\sqrt{A}$ using the same technique.
\item \textbf{Square a rectangle.}  Given a rectangle, construct a
square with exactly the same area.
\item \textbf{Construct a pentagon.}  Construct a regular pentagon.
For a hint, see the footnote\footnote{The cosine of $72^\circ$ is
$(\sqrt{5}-1)/4$}.
\end{enumerate}

\subsection{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:

\begin{itemize}
\item 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 \ldots.
\end{itemize}

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:
\begin{itemize}

\item \textbf{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
$(x_2, y_2)$:
\[ (y - y_1) = \frac{(y_2 - y_1)}{(x_2 - x_1)}(x - x_1). \]

\item \textbf{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:
\[ (x - x_1)^2 + (y - y_1)^2 = r^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2. \]

\item \textbf{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 $C_2$.

\item \textbf{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.

\item \textbf{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.
\end{itemize}

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:

\begin{enumerate}
\item You can make all the integers: $1, 2, 3, \ldots$.
\item You can make all the fractions (usually called ``rational
numbers'': $i/j$, where $i$ and $j$ are integers.
\item 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.\label{item:s1}
\item 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~\ref{item:s1}.
\label{item:s2}
\item 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~\ref{item:s2}.
\label{item:s3}
\item 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~\ref{item:s3}.
\item Et cetera\ldots
\end{enumerate}

Thus, numbers like the following are ``easily'' obtainable:
\begin{equation}
\sqrt{\frac{314159265}{2718281828}+
\frac{\frac{17}{35{\sqrt{\sqrt{\sqrt{1111}}}}}
- 5\sqrt{46}}{1 - \sqrt{\sqrt{1/13}} + \frac{81}{91}\sqrt{
\frac{3}{1198}\sqrt{\frac{11}{17} - \sqrt{2}}}}}.
\label{exp:e1}
\end{equation}

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~\ref{exp:e1}, 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 &=& 17 + 4\sqrt{\frac{3}{5}} \\
(x - 15)^2 - 17 &=& 4\sqrt{\frac{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\ldots).  So if the number you
would like to construct involves a cube root (or fifth root, or sixth
root, \ldots), you are out of luck---it cannot be constructed using
the methods described in Section~\ref{sec:rules}.

\section{Three Impossible Constructions}
\label{sec:impossible}

The three classical construction problems of antiquity are known as
``squaring the circle'', ``trisecting an angle'', and ``doubling a
cube''.  Here is a short description of each of these three problems:
\begin{itemize}
\item \textbf{Squaring the Circle.} Given a circle, construct a square that
has exactly the same area as the circle.
\item \textbf{Trisecting an Angle.} Given an angle, construct an angle whose
measure is exactly 1/3 the measure of the original angle.
\item \textbf{Doubling a Cube.} Given the length of the side of a
(three-dimensional) cube, constuct a length so that a cube with an
edge of this length will have exactly double the area of the original
cube.
\end{itemize}

It is beyond the scope of this paper to prove that these three problems are
impossible to solve, but given the ideas that were presented in 
Section~\ref{sec:algebra}, we can at least indicate how the formal
proofs work.

The easiest to show (to be impossible) is the problem of doubling the
cube.  Clearly if the volume of the required cube is double the volume
of the original cube, if the length of the side of the original cube
is $L$, the desired length is $\sqrt[3]{2}L$.

In other words, from an algebraic point of view, we can construct the
cube root of 2, or equivalently, we can solve the following equation:
\[ L^3 = 2, \]
which is irreducible over the rationals, and is not going to be in any
quadratic extension of the rationals.

The problem of trisecting an angle is similar, but what is usually
done is to show that there is a particular angle that cannot be
trisected, and that angle is typically chosen to be $60^\circ$.  If
we can't trisect this particular one, then we know that not all angles
can be trisected.  (Remember that a solution to the problem requires a general
method that will work for any angle.  Certain angles, such as $90^\circ$
angles, can obviously be trisected, since we can construct a $30^\circ$
angle from scratch.  Also, we can construct a $60^\circ$ angle from scratch,
so if this one cannot be trisected, the general problem of trisection is
clearly unsolvable.)

If we can trisect a $60^circ$ angle, that is equivalent to constructing
a $20^\circ$ angle from scratch, which is equivalent to constructing the
cosine of $20^\circ$.

Now the cosine of $60^\circ$ is $1/2$, so
\[ \cos 60^\circ = \cos (20^\circ + 20^\circ + 20^\circ) = 1/2. \]
Using the standard formulas for the sine and cosine of sums of angles,
this equation above can be converted to:
\[4\cos^3 20^\circ - 3\cos 20^\circ = 1/2, \]
or if $x = \cos 20^\circ$, to
\[ 8x^3 - 6x - 1 = 0,\]
which is also an irreducible cubic equation that cannot possibly have
a solution in any quadratic extension of the rationals.

Much more difficult is to show that a circle cannot be squared.  If
the original circle has radius 1, its area will be $\pi$, so squaring
a circle is equivalent to the construction of a length equal to
$\sqrt{\pi}$.  It turns out (but is not easy to prove) that $\pi$ is
transcendental---it is not the solution to \emph{any} polynomial equation,
so it is surely not the solution to any combination of quadratic
polynomials.



\section{Problems}

Here are a few more construction problems to try your hand at.

\begin{enumerate}
\item Given a semicircle centered at a point $C$ with diameter
$AB$, find points $I$ and $J$ on $AB$, and points $H$ and $G$ on
the semicircle such that the quadrilateral $GHIJ$ is a square.

\item Given a quadrant of a circle (two radii that make an
angle of $90^\circ$ and the included arc), construct a new circle
that is inscribed in the quadrant (in other words, the new circle
is tangent to both rays and to the quarter arc of the quadrant).

\item Given a point $A$, a line $L$ that does not
pass through $A$, and a point $B$ on $L$, construct a circle passing
through $A$ that is tangent to $L$ at the point $B$.

\item Given two points $A$ and $B$ that both lie on the same
side of a line $L$, find a point $C$ on $L$ such that $AC$ and
$BC$ make the same angle with $L$.\label{prob:goat}

\item Given two points $A$ and $B$ that both lie on the same
side of line $L$, find a point $C$ on $L$ such that $AC + BC$ is
as small as possible. (Hint: This problem is related to the
construction problem \ref{prob:goat}.  Also remember that the
shortest distance between two points is a line.)

\item Given two non-parallel lines $L_1$ and $L_2$
and a radius $r$, construct a circle of radius $r$ that is tangent to
both $L_1$ and $L_2$.
\end{enumerate}


\end{document}
