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\chapter{Mechanical and Other Solutions to the Construction Problems}

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The search for solutions to the three famous problems fueled the mathematical enterprise of the ancient Greeks.  Yet, early on in their attempts, the Greeks realized the restrictions of using only the Euclidean tools were hindering their abilities to develop solutions.\footnote{Burton states that ``early investigators must have suspected that the allowable means were inadequate,'' (p.116).}  Many mathematicians, therefore, developed their own techniques to solve the construction problems, using methods and tools beyond the use of straightedge and compass.  In this regard, these geometers were successful, as attempts to trisect the angle, duplicate the cube, and square the circle persisted through the Dark Ages and into the early modern period.  

The first mechanical method used to solve the construction problems was developed by Hippias of Elis in the fifth century BCE. The only surviving sources on the life of Hippias come from the dialogues of Plato, in which he is cast in a mostly negative light.\footnote{Burton notes that Plato describes Hippias as ``an arrogant, boastful buffoon,'' (p.125).}  Born about 460 BCE, Hippias was part of a group of itinerant teachers, called ``sophists,'' who would travel throughout Greek cities, sharing their mathematical and philosophical knowledge in exchange for money.  While many sophists were well-informed and reputable scholars serving as effective tutors, others were mere imposters seeking money, and thus ``accusations of shallowness directed against the sophists'' were, in some ways, warranted.\footnote{Boyer, p.68.}  Perhaps not surprisingly, Plato and other skeptics castigated the sophists, and the term acquired a negative, cynical connotation.

Leaving aside accusations of his morals, it is true that Hippias' quadratrix would become a critical mathematical contribution to the solutions of the three geometric problems.  The construction of the quadratrix curve can be described as follows:\footnote{The definition of the quadratrix curve can be found on Burton, p.126 and Boyer, p.68-9.} 

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\includegraphics[width=4in,height=4in]{Quadratix.png}
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Given a square $ABCD$, let segment $AB$ move down with a constant velocity toward segment $DC$.  At the same moment that $AB$ leaves its initial position, let side $DA$ rotate clockwise with a constant velocity, so that both segments coincide at exactly the same time with $DC$.  At any given moment, define the positions of the two moving lines to be $A'B'$ and $DA''$, intersecting at point $P$.  The locus of these intersection points forms the quadratrix curve $APQ$.  This definition, however, does not locate any point on $DC$.  If the moving segment $A'B'$ and the rotating radius $DA''$ coincide with $DA$ simultaneously, then they intersect everywhere along $DC$.  The point on the quadratrix, labelled as $Q$, can only be defined by the modern notion of a limit of points $P$ on the sector of the curve. 

The quadratrix is therefore the first curve beyond the straight line and circle curve ``that could not be drawn by the traditionally required straightedge and compass'' but rather must be plotted point by point.\footnote{Burton, p.125.  Boyer also states that that we owe Hippas ``the introduction into mathematics of the first curve beyond the circle and straight line,'' (p.68).}  Hippias had also developed an apparatus which could effectively draw a quadratrix curve, leading many geometers to further reject its use on the grounds of its mechanical origins.  Plutarch, for example, in his work \textit{Quaestiones Convivales}, cites Plato as chastising mechanical constructions, for they make ``sensible things more powerful over us than intelligible, and by forcing the understanding to determine rather according to passion than reason.''\footnote{citation?}  To Plato, geometry was a divine gift, and it is through mechanical operations that one ``loses that instrument and light of the soul, which is worth a thousand bodies, and by which alone the Deity can be discovered.''\footnote{Ibid. For Plato, ``God always plays the geometer.''} 

The quadratrix curve of Hippias can easily be used to trisect a given angle.  But, it is important to first consider the case for trisecting a given line segment, which is used in the proof.  This construction, however, does not follow immediately from any proposition Euclid provides; rather, it uses the result of a construction he gives in Proposition 10 of Book IV and facts about similar triangles.  Given a segment $AB$ to be trisected, draw another line, $CA$, intersecting $AB$ at $A$, forming an $\angle{BAC}$.  Construct a circle with center $A$ that intersects both $AB$ and $AC$, and denote by $X$ the intersection with $AC$.  Now, draw another circle with center $X$ and radius $AX$, and let $Y$ denote the point of intersection with $AC$. Draw a third circle with center $Y$ and radius $YX$ and call $Z$ the point of intersection with $AC$.  Now, join $Z$ and $B$.  Construct a line from $Y$ to $AB$ parallel to $ZB$, with $S$ as the point of intersection.  Finally, draw a line from $X$ to $AB$ parallel to both $YS$ and $ZB$, with $R$ being the point of intersection with $AB$.  The claim is that $AR=RS=SB=\frac{1}{3}AB$, and this result follows from the facts that lines $ZB$, $YS$, and $XR$ are all parallel, $AX=XY=YR$, and properties of similar triangles.        

With his construction in hand, the trisection of the angle can be described as follows: suppose, for example, that $\angle{PDC}$ is the angle to be trisected.  Given line $A'B'$ that contains $P$, trisect lines $A'D$ and $B'C$ at points $R$, $S$, $T$, and $U$.  Lines $TR$ and $US$ intersect the quadratrix at $V$ and $W$, and lines $VD$ and $WD$ divide $\angle{PDC}$ in three equal parts.\footnote{Boyer, p.69.}

The use of the quadratrix to square the circle---which ultimately gave the curve its name---is given by Dinostratus in the mid-fourth century BC.  This solution, however, requires the use of the point $Q$.  If this point is assumed to be found, Pappus provides a proposition involving ``three straight line segments and the circular arc $AC$,'' stated as 
$$\overarc{$AC$}/AB=AB/DQ.\footnote{Boyer, p.97}$$  
This is proved by a double \textit{reductio ad absurdum} argument, where the alternatives to the proposition are proven to be false.\footnote{Burton notes that this is one of the ``earliest examples in Greek mathematics of the indirect method of reasoning Euclid used so extensively,'' (p.127).}  Assume that $$\overarc{$AC$}/AB=AB/DR,$$ where $DR>DQ$.  Let the circle with center $D$ and radius $DR$ intersect the quadratrix at $S$ and side $AD$ at $T$.  From $S$ draw $SU$ perpendicular to $DC$.  As Burton states, ``since corresponding arcs of a circle are proportional to their radii,'' $$\overarc{$AC$}/AB=\overarc{$TR$}/DR,$$ and from the hypothesis, it follows that $$\overarc{$TR$}=AB.\footnote{Burton, p.127}$$  Yet it is known that $$\overarc{$TR$}/\overarc{$SR$}=AB/SU$$ from the definition of the quadratrix.  Since $$\overarc{$TR$}=AB,$$ it follows that $$\overarc{$SR$}=SU,$$ which cannot be true, since ``the perpendicular is shorter than any other line or curve from point $S$ to $DC$.''\footnote{Boyer, p.97}.  A similar argument will show that $DR$ cannot be less than $DQ$, thus $DR=DQ$ and $$\overarc{$AC$}/AB=AB/DQ.$$  

Dinostratus' brother Menaechmus was also a famed mathematician of the fourth century BCE, eventually becoming a tutor to Alexander the Great.\footnote{Burton, p.123.}  Prior to Menaechmus, Hippocrates of Chios made considerable progress in solving the duplication of the cube, reducing it finding two mean proportionals of $a$ and $2a$; that is, lengths $x$ and $y$ satisying

\begin{eqnarray} 
\label{newformula}
\frac{a}{x}=\frac{x}{y}=\frac{y}{2a}
\end{eqnarray} 

While Hippocrates himself did not find the mean proportionals using the Euclidean tools, it was nevertheless a ``significant achievement'' to reduce ``a problem in solid geometry to one in plane geometry.''\footnote{Burton goes on to say: ``From this time on, the duplication of the cube was always attacked in the form in which Hippocrates stated it,'' (p.123).}   Menaechmus, however, anticipating work with conic sections, discovered that ``the cutting of a right circular cone by a plane perpendicular to an element of the cone'' produces a ``family of appropriate curves'' that satisfy the property expressed by Hippocrates.\footnote{Boyer, p.93.}.  These curves, in modern terms, would become known as the ellipse, parabola, and hyperbola, which are names given by Apollonius.  
	
With modern algebraic notation, the discovery articulated by Menaechmus can provide a theoretical solution to the duplication of the cube.\footnote{A thorough explanation of this solution can be found on Burton, p.123 and Boyer, p.95.}  Given a cube of edge $a$, construct two parabolas, one with latus rectum $a$ and the other with latus rectum $2a$, sharing a common vertex and perpendicular axes.  Thus, the two equations corresponding to the parabolas are 

\begin{eqnarray}
x^2=ay,\\
y^2=2ax
\end{eqnarray}

Isolating $x$ in (3) gives 

\begin{eqnarray*}
x=\frac{y^2}{2a}
\end{eqnarray*}

which substituted into (2) yields 

\begin{eqnarray*}
\frac{y^4}{4a^2}=ay
\end{eqnarray*}

or 

\begin{eqnarray*}
y=\sqrt[3]{4}a
\end{eqnarray*}

Substituting this result into (3) yields 

\begin{eqnarray*}
(\sqrt[3]{4}a)^2=2ax
\end{eqnarray*}

or 

\begin{eqnarray*}
x=\sqrt[3]{2}a
\end{eqnarray*}

Therefore, the $x$-coordinate of the intersection of the two parabolas is the desired edge of the cube sought.  This solution, however, is not mechanical in the same sense as the quadratrix curve, which uses moving segments to construct an entirely new curve than the Euclidean tools provide.  Menaechmus' solution, rather, is purely theoretical, and therefore the desired curves and segments cannot be constructed with a striaghtedge and compass.

About a century after Menaechmus, the third-century BCE geometer Nicomedes attempted to find solutions to the problems of the angle trisection and cube duplication using a \textit{neusis} construction.  The word \textit{neusis} is Greek, and it roughly means sloping or verging towards.  Bos defines the \textit{neusis} construction as follows: ``Given two lines, a point $O$ and a segment $a$, to draw a straight line through $O$ intersecting the two lines in points $A$ and $B$ such that $AB=a$.''\footnote{Bos, p.168.}  Therefore, the mechanical nature of this type of construction lies in marking a straightedge to transfer a segment of length $a$, which was in violation of the use of Euclidean tool.    

The use of the \textit{neusis} to trisect an angle is provided by Burton.\footnote{Burton, p.123.}  Let $\angle{AOB}$ be a given angle.  Through $B$, draw a line perpendicular to segment $OA$ at $C$ and another line parallel to $OA$.  Using a straightedge, mark the length $a=2OB$ and slide the straightedge so that it passes through $O$, with endpoints of segment a as $P$ on $BC$ and $Q$ on $BD$ (so $PQ=a$).  If $M$ is the midpoint of $PQ$, then $OB=\frac{a}{2}$ by construction, $PM=MQ=\frac{a}{2}$ by definition of a midpoint, and $BM=\frac{a}{2}$ since ``the midpoint of the hypotenuse of a right triangle is equidistant from the endpoints of its sides.''\footnote{Burton, p.123.}  Therefore, $\angle{MOB}=\angle{BMO}$, since the base angles of an isosceles triangle are equal.  In $\triangle{BMQ}$, $\angle{BMP}=\angle{MBQ}+\angle{MQB}$ from the exterior angle theorem, and thus $\angle{BMP}=2\angle{MQB}$ since $\triangle{MQB}$ is isosceles.  Since $BD$ and $OA$ are parallel, the interior angles $\angle{AOQ}$ and $\angle{BQO}$ are equal.  Thus, since $\angle{AOB}=\angle{AOQ}+\angle{QOB}$, $\angle{AOB}=3\angle{AOQ}$, and therefore $\angle{AOQ}=\frac{1}{3}\angle{AOB}$, and the angle is trisected.

Nicomedes also approached the duplication of the cube by means of finding the two mean proportionals, much like Menaechmus.  His solution is described as follows:\footnote{See Burton, p.123.} construct a rectangle $ABCD$ with side $AB=a$ and $AD=2a$, with $M$ being the midpoint of $AD$ and $N$ the midpoint of $AB$.  Extend the segments $CM$ and $BA$ to meet in $G$.  Choose the point $F$ on the perpendicular $FN$ to be such that $FB=a$.  Draw a segment $BH$ parallel to $GF$ and $a$ segment $FP$ to cut segment $AB$ at $p$, with $P$ chosen such that $HP=a$.  Extend the segments $PC$ and $AD$ until they meet at $Q$.  Let $DQ=x$, $BP=y$, and $FH=z$. 

With this construction established, one can see that $\triangle{PAQ}$, $\triangle{PBC}$, and $\triangle{CDQ}$ are similar.  Each triangle contains a right angle, $\angle{PCB}=\angle{PQA}$ since they are alternate interior angles, and $\angle{APC}=\angle{DCQ}$ since they are also alternate interior angles.  This similarity establishes the relationship 

\begin{eqnarray}
\frac{a}{x}=\frac{y}{2a}=\frac{a+y}{2a+x}.
\end{eqnarray} 
Likewise, $\triangle{PBH}$ and $\triangle{PGF}$ are similar, since both contain $\angle{P}$, $\angle{PBH}=\angle{PGF}$, and $\angle{PHB}=\angle{PFG}$ by definition of alternate interior angles.  This similarity establishes the relationship
\begin{eqnarray*}
\frac{a}{y}=\frac{z+a}{y+2a}.
\end{eqnarray*}
Cross-multiplication yields $ay+2a^2=yz+ay$.  Subtracting $ay$ from both sides leaves $2a^2=yz$, or $\frac{a}{z}=\frac{y}{2a}$.  Substituting this into (4) shows that $\frac{a}{x}=\frac{a}{y}$, and thus $x=z$.  Since $\triangle{FNB}$ and $\triangle{FNP}$ have $FN$ as a common side, the Pythagorean Theorem produces 
\begin{eqnarray*}
a^2-\frac{a}{2}^2=(x+a)^2-\left(y+\frac{a}{2}\right)^2,
\end{eqnarray*}
or 
\begin{eqnarray*}
\frac{x}{y}=\frac{y+a}{x+2a}.
\end{eqnarray*}  
From (4), this yields 
\begin{eqnarray*}
\frac{a}{x}=\frac{x}{y}=\frac{y}{2a}
\end{eqnarray*}
which is precisely the equation (1) that Hippocrates of Chios discovered in the fourth century BCE.  Thus, the segments $DQ=x$ and $BP=y$ are the two mean proportionals between $a$ and $2a$, which implies that given a cube of edge $a$, $x$ is the edge of the duplicated cube. 


The third century BCE also saw the flourishing of Archimedes, considered by many scholars as the greatest mathematician of all antiquity.  Few details of his life are known, though it is speculated that he was born about 287 BCE in Syracuse, a Greek city on the coast of Sicily.\footnote{Burton, p.186.  Burton also notes that, according to Plutarch, Archimedes' family was of the same royalty as King Hieron II, ruler of Syracuse.}  While it is likely that he studied at the Museum in Alexandria under former students of Euclid, he spent most of his ``productive years'' in Syracuse, where he devoted his time to studying and experimenting.\footnote{Ibid, p.187.}  Tradition has it that Archimedes was killed in 212 BCE during a siege of Syracuse by Roman troops in the Second Punic War. 
	
In addition to mathematics, Archimedes' interests also included ``astronomy, hydraulics, mechanics and general engineering.''\footnote{Hollingdale, p.65.}  It was indeed his mechanical inventions that earned Archimedes great fame throughout his life.  His Archimedean screw, for example, a device still in use today, serves to raise water, primarily in irrigation systems.  He also invented a number military defenses used to fight off the siege of the Romans, and his work with pulleys and levers led him to boast: ``Give me a place to stand and I will move the earth.''\footnote{Burton, p.187}.  
	
Despite his mechanical achievements, Archimedes ``remained firmly within the Greek philosophical tradition'' in championing theoretical studies and abstract thought.\footnote{Hollingdale, p.66.}  He was, therefore, attracted to the three famous construction problems of geometry, and invented a new type of curve that would provide solutions to both the angle trisection and quadrature of the circle.  The curve, known as the Archimedean spiral, is described in his treatise \textit{On Spirals}:
\begin{quote}
If a straight line [half-ray] one extremity of which remains fixed be made to revolve at a uniform rate in the plane until it returns to the position from which it started, and if, at the same time as the straight line is revolving, a point moves at a uniform rate along the straight line, starting from the fixed extremity, the point will describe a spiral in the plane.\footnote{Burton, p.196.}
\end{quote}

As Boyer explains, the spiral is ``the plane locus of a point which, starting from the end point of a ray or half line, moves uniformly along this ray while the ray in turn rotates uniformly about its end point.''\footnote{Boyer, p.126.}  Yet, while Archimedes' spiral was, like many of his developments, inspired by practical considerations, it was nevertheless considered a mechanical solution.  Like the quadratrix, the spiral is a curve constructed from means other than a straightedge and compass, and therefore could not provide a legitimate solution to the construction problems.        

However, with the use of the Archimedean spiral, the trisection of an angle can be carried out with ease.\footnote{A description of this solution can be found on Boyer, p.126 and Burton, p.200}  Take $\angle{POA}$ to be trisected.  Position the angle so that the vertex and initial side of the angle coincide with initial point $O$ of the spiral and the initial position $OA$ of the rotating ray.  The point $P$ is given to be the intersection of the terminal side of the angle and the spiral.  Trisect $OP$ at $Q$ and $R$, and construct two circles with centers at $O$ and $OQ$ and $OR$ as radii.  If these points intersect the spiral at points $U$ and $V$, lines $OU$ and $OV$ trisect $\angle{POA}$.   

The Archimedean spiral also provides a clever solution to the squaring of the circle.  Given a circle with center $O$ and radius $OA$, construct a spiral from $O$ such that the end of its first revolution coincides with $A$.  Construct a tangent to the spiral at $A$, and extend a perpendicular to $OA$ from $O$, where the tangent and the perpendicular intersect at $B$.  Using Cartesian coordinates, label the point $O$ as the origin, $(0,0)$, and call the radius of the circle $x$, where $A$ is labelled as $(x,0)$.  The slope of the tangent line can be derived using a parameterization: $$\alpha
(\theta)=\left(\frac{\theta x}{2\pi}\cos(\theta),\frac{\theta x}{2\pi}\sin(\theta)\right).$$  The tangent line, therefore, has a parameterization $$\alpha'(\theta)=\left(\frac{x}{2\pi}\cos(\theta)-\frac{\theta x}{2\pi}sin(\theta),\frac{x}{2\pi}\sin(\theta)+\frac{\theta x}{2\pi}\cos(\theta)\right),$$ and thus $$\alpha'(2\pi)=\left(\frac{x}{2\pi},x\right).$$ The slope of the tangent at $2\pi$ is given as $$\frac{x}{\frac{x}{2\pi}},$$ which is $2\pi$.  Hence, the coordinates of $B$ must be (0,-2$\pi x$), and the length of the segment $OB$ is given as 2$\pi x$, which equals the circumference of the circle.  The area of $\triangle{OAB}$, thererefore, is given as $\frac{1}{2}(x)(2\pi x)$ or $\pi x^2$.  The area of $\triangle{OAB}$ is thus equal to the area of the circle.\footnote{Heath, in his work \textit{The Thirteen Books of The Elements}, cites a quote by Proclus: ``Archimedes actually proved that any circle is equal to the right-angles triangle which has one of its sides about the right angle [the perpendicular] equal to the radius of the circle and its base equal to the perimeter of the circle.  But of this elsewhere,'' (p.347).}

To transform the triangle into a square requires some algebraic manipulation.  First, construct a line segment, $AB$ equal in length to the area of the triangle and circle, say $y$.  From point $B,$ extend the segment a unit length to produce a new segment, $AC,$ with length $y+1$.  At $B,$ erect a perpendicular $BD$ to $AC$ with length $x.$  Bisect segment $AC$ at $O$ and construct a circle with center $O$ and radius $OA$, with $D$ on the circle.  From propositions given in Book I of the \textit{Elements}, $\triangle{ADC}$ makes a right triangle, which is similar to $\triangle{ABD}$ and $\triangle{CBD}.$  This similarity establishes the relationship $$\frac{x}{1}=\frac{y}{x},$$ or $x^2=y.$  A square with side length $x$ thus has the same area as the triangle and circle, and the squaring of the circle is accomplished.  

It should be noted, however, that this algebraic technique would not be how the Greeks---Euclid and Archimedes in particular---would approach the problem.  Heath, in his translation of the \textit{Elements}, cites two propositions from Euclid that allow one to transform the triangle into a square with equal area: Proposition 45 from Book I, which enables one to construct a parallelogram equal to a given rectilinear figure, and Proposition 14 from Book II, which states that one can construct a square equal to a given rectilinear figure.\footnote{Heath, p.347. Heath adds a comment by Proclus: ``I conceive that it was in consequence of this problem that the ancient geometers were led to investigate the squaring of the circle as well.''}    

The search for solutions for the construction problems did not end with the fall of the Greek world.  Rather, it took two thousand years for valid proofs of impossibility to be published, cementing the claim that the three geometric problems of antiquity could not be solved by straightedge and compass alone.  The mathematical developments in the early modern period that led to the breakthroughs of algebra and calculus would, ultimately, provide the missing link that was needed for the proofs.  

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