Temp Directory?: /var/www/html/mathimages/imgUpload/tmp Compass & Straightedge Construction and the Impossible Constructions - Math Images

# Compass & Straightedge Construction and the Impossible Constructions

{{Image Description Ready |ImageName=Creating a regular hexagon with a ruler and compass |Image=HexagonConstructionAni.gif |ImageIntro=This image shows the step by step construction of a hexagon inscribed in the circle using a compass and a unmarked straightedge. |ImageDescElem=Let's assume we only have a compass and a unmarked straightedge. What can we construct and how can we construct them? That were the problems that Euclid pondered not only because those were probably the only instruments that he had at his time but also he wanted to build his theorems with as few assumptions, or axioms, as possible.[1] With these two simple tools, he managed to build myriad of theorems in both plane and solid geometry .[2] These theorem are as good as they were some 2000 years ago and it is this enduring quality of Euclid's work that inspired this page. In the main image of this page, we want to divide the circle into six equal arcs and then connect consecutive points to form the hexagon. It seems to be a fairly simple construction but you should be prompted to ask two questions: Are other polygons constructible and is every polygon constructible, that is able to be constructed using only compass and straightedge? To extend the question, what are constructible and what are not? That is the problem that is resolved in this page. |ImageDesc===What is Compass & Straightedge Constructions==

## Contents

### Introduction

We start by familiarizing ourselves with Euclid's three Postulates in his books Elements.

 Let it be granted 1. that a straight line may be drawn from any one point to any other point; 2. that a line segment may be extended into a straight line; 3. that given any straight line segment, a circle may be described having the segment as radius and one endpoint as center.[3]
 It should be carefully noted that Euclid started with two given points and produced a line segment, from where he could extend into a straight line if he pleased. Then ONLY from the original two points, he could use one point as center and ONLY spread the legs of compass the distance of the line segment to produce a circle. Even though many translations say that a circle can be described at any center with any radius, we must take it with a pinch of salt. He could not specify any points and lengths other than what was already given, that is to say he could not claim "I wanted to spread the legs of the compass $\pi$ centimeters apart (or any specified denominations) with the center of the circle half way between the two given points". In all, Compass and Straightedge Constructions only allow us to start with points (and hence lengths) we have been given(or constructed from given points), and create the ones we don't. Thus, we define Compass & Straightedge Construction as the construction of points, lengths, angles, and circles using only ideal straightedge and compass. A straightedge is infinite in length, has no markings on it and only one edge. A compass has two legs, one end of which is fixed on the plane of construction and the other end is of given distance away and maintains the distance throughout the construction. It collapses when lifted from the page, so may not be directly used to transfer distances. However, it turns out that this restriction makes no difference due to the Compass Equivalence Theorem which was stated as Proposition II of Book I of Euclid's Elements. It stated that from a given point, it was possible to construct a line segment equal to a given line segment using collapsible compass in any desirable direction. Euclid's proof for the Compass Equivalence Theorem will be presented after the section of Basic Construction. Since Euclid has proven this using only the three postulates, then he did not have to use a collapsible compass any more.[4]

### Some Basic Constructions

The constructions below are some basic ones from where many more constructions are possible and they are by no means exhaustive. In the figures below, what we are given are in blue; intermediate steps are in dotted black; the resulting products are in red. The proofs for these constructions are relatively simple and only require the knowledge of congruent triangles. Euclid derived the theories on congruency and congruent triangles directly from his Postulates. Try proving the theorems yourself!

#### Line Segment Bisection

Given points ${\color{Gray}A}$ and ${\color{Gray}B}$ and the straight line passing through it. Construc [...]

Given points $A$ and $B$ and the straight line passing through it. Construct a line that bisects line segment $AB$.

1. Draw a circle centered at point $A$ with radius equals $AB$.
2. Next, draw a circle centered at point $B$ with the same radius.
3. Where the two circles intersect, call those points $C$ and $D$.
4. Draw a line through points $C$ and $D$.

Line $CD$ intersects line segment $AB$ at the midpoint. It should be noted that $CD \perp AB$ as well. Line $CD$ is the perpendicular bisector of line segment $AB$.

#### Angle Bisection

Given angle ${\color{Gray}\angle AOB}$, construct a line that bisects the angle.

Given angle $\angle AOB$, construct a line that bisects the angle.

1. Construct a circle centered at point $O$ with radius $OA$. This circle intersects $OA$ and $OB$ at point $A$ and $B$.
2. Keeping the same radius, draw a circle at point $A$ and $B$ respectively. Where they intersect each other, call it point $C$.
3. Draw a line through point $C$ and $O$. This line bisects $\angle AOB$.

#### Perpendicular Through a Point

Given a point ${\color{Gray}M}$ on a line ${\color{Gray}AB}$, construct a line that is perpendicular to [...]

Given a point $M$ on a line $AB$, construct a line that is perpendicular to the given line through $M$.

1. Draw a circle centered at $M$ with radius $MB$ (or $MA$. It is your choice).
2. Where the circle intersects the original line, construct a perpendicular bisector (see the line segment bisector construction above).

#### Parallel Line

Given two points, ${\color{Gray}A}$ and ${\color{Gray}B}$ and the straight line passing through them, c [...]

Given two points, $A$ and $B$ and the straight line passing through them, construct a line that is parallel to the given line through another given point $C$.

1. Draw a circle at $A$, crossing $C$. Where the circle $A$ intersects $AB$, call the point $D$.
2. Centered at points $C$ and $D$, draw circles crossing point $A$. Where these two circles intersect each other, call it $E$.
3. Draw a line through point $C$ and $E$.

$CE$ is parallel to $AB$

#### Tangent Line to a Circle

Given a circle centered at ${\color{Gray}O}$ and another given point, ${\color{Gray}A}$, construct a li [...]

Given a circle centered at $O$ and another given point, $A$, construct a line that is tangent to the circle.

1. Draw a line through point $A$ and the center of the circle $O$
2. Let $M$ be the midpoint of $OA$(construction omitted since we know how to construction mid point). Draw the circle centered at $M$ going through $A$ and $O$.
3. Let the point where the two circles meet be $C$. Draw line segment $AC$.

To see that $AC$ is tangent to the circle, connect $OC$. $\angle ACO$ is an inscribed angle in the circle about $M$, so the inscribed $\angle ACO$ is $90^\circ$. So $OC$ is perpendicular to $AC$. Since the tangent is perpendicular to the radius to the point of tangency, by the uniqueness of the line through $C$ perpendicular to $OC$, $AC$ is tangent to the circle.

#### Euclid's Proof of Compass Equivalence Theorem

This part refers back to the previous section about the issue of compass being collapsible. Euclid's original proof is presented. Additional comments are contained in the parenthesis.

From a given point to draw a line segment equal to a given line segment.

From a given point to draw a line segment equal to a given line segment. Let $A$ be the given point, and $BC$ the given straight line : it is required to draw from the point $A$ a straight line equal to $BC$.

1. From the point $A$ to $B$ draw the straight line $AB$ and on it describe the equilateral triangle $DAB$(this is done the same way as bisecting a line segment), and produce the straight lines $DA$, $DB$ to $E$ and $F$.
2. From the center $B$, at the distance $BC$(with the radius equals to $BC$), describe the circle $CGH$, meeting $DF$ at $G$.
3. From the center $D$, at the distance $DG$(with the radius equals to $DG$), describe the circle $GKL$, meeting $DE$ at $L$.
4. $AL$ shall be equal to $BC$.

Because the point $B$ is the center of the circle $CGH$, $BC$ is equal to $BG$. And because the point $D$ is the center of the circle $GKL$, $DL$ is equal to $DG$ and $DA$, $DB$ parts of them are equal therefore the remainder $AL$ is equal to the remainder $BG$.

But it has been shewn that $BC$ is equal to $BG$ ; therefore $AL$ and $BC$ are each of them equal to $BG$. But things which are equal to the same thing are equal to one another. Therefore $AL$ is equal to $BC$.

Wherefore from the given point $A$ a straight line $AL$ has been drawn equal to the given straight line $BC$.∎[5]

It should be noted that from $AL$, we could "duplicate" $AL$ in all directions by construct a circle centered at $A$ with radius $AL$.

## Algebraicization of Compass & Straightedge Constructions

### What is Algebraicization?

From the few basic constructions, you would have probably realized that the different possibilities seems infinite. Hence, mathematician are curious to find out what are constructible and what aren't and for this purpose, the language of pure geometry seems to have "limited vocabulary"[6]. Back in ancient times, mathematicians had limited algebraic knowledge and were more familiar with geometry. But in modern times, the reverse is true. Hence, today's mathematicians go back to their familiar realm of Algebra and try to find the link between geometry and algebra.

Algebraicization is the translation of any problem statements into algebraic problems. In the case of Compass & Straightedge construction, we algebraicize each step of a straightedge and compass construction, and consequently obtaining general results about the nature of constructibility. Hilda P. Hudson put it aptly in his lecture Ruler & Compasses,

"each step of a ruler and compass construction is equivalent to a certain analytical process; it is found that the power to use a ruler corresponds exactly to the power to solve linear equations, and the power to use compasses to the power to solve quadratics...... Since each step of a ruler and compass construction is equivalent to the solution of an equation of the first or second degree, we consider that these algebraic processes can lead to , when combined in every possible way, and that enables us to answer the question before us......"[7]

Hudson lectured on this in the early 20th century and certain phrases of his could potentially cause confusion. The take-away from this paragraph is that in order to algebraicize straightedge and compass construction, we begin by designating a given point as the origin and the coordinates of another given point (we are given two points at least) as $(1,0)$ or $(0,1)$. Thus we have established the Cartesian Coordinates. Then, every time we construct a straight line or a circle, we think of it instead as adding a new equation into a system of equations. These equations represent the coordinates of all the points on the line or circle, but that is easy since we all know the expression for a line and a circle as $y = ax + b$ and $(x-m)^2 + (y-n)^2 = r^2$. However, the only times we can pinpoint a point (and find its coordinates as a result) is when a line intersects with a line, or a circle, or a circle intersects with another circle in which case we can pinpoint 2 points. We then conclude that only those coordinates of the points of intersections are constructible. In this way, a geometric process is translated into an algebraic process.

### A Simple Derivation

Firstly, we define "1" on a straight line as stated previously. Then, once you have chosen that point to be $(1,0)$, you have to stick to this specification throughout your construction. Next, it is very obvious that we could construct all the integers, that is $\cdots -3,-2,-1,0,1,2,3,\cdots$ (or $x$ = Failed to parse (lexing error): \{x |other=A little Geometry and Some Abstract Algebra |AuthorName=Wikipedia |AuthorDesc=Wikipedia, Powerpoint and Flash |SiteName=Compass and Straightedge Constructions |SiteURL=http://en.wikipedia.org/wiki/Compass_and_straightedge_constructions |Field=Algebra |FieldLinks=*):http://planetmath.org/ *):http://hptgn.tripod.com/ *):http://en.wikipedia.org/wiki/Compass_and_straightedge_constructions *):http://www.mathopenref.com/tocs/constructionstoc.html *):http://mathforum.org/library/drmath/view/66052.html *):http://mathforum.org/library/drmath/view/52601.html *):http://www.ams.org/notices/200004/fea-hartshorne.pdf =Notes= <references/> |References=#Peterson, D. (2003, November 21). Collapsible Compass. Retrieved from The Math Forum: http://mathforum.org/library/drmath/view/66052.html #Hartshorne, Robin . (2000). Teaching geometry according to euclid. NOTICES OF THE AMS, 47(4), 460-465. #Taylor, H. M. (1895). Euclid's elements of geometry. Cambridge: Cabridge University Press. #Hudson, H. P. (1916). Ruler & compass. London: Longmans Green & Company, Inc.. #Bryant, John, & Sangwin, Christopher. (2008). How Round is your circle?. Princeton & Oxford: Princeton Univ Pr. #Herstein, I. N. (1975). Topics in algebra. John Wiley & Sons Inc. |InProgress=No }}