CALIFORNIA STATE UNIVERSITY, NORTHRIDGE
EUCLIDEAK CONSTRUCTIONS:
ALTERNATE TOOLS TO THE TRADITIONAL COMPASS AND
STRAIGHTEDGE
A thesis submitted in partial satisfaction of the
requirements for the degree of Master of Science in
Mathematics
by
Sandra Kay Birrell
•
The Thesis of Sandra Kay Birrell is approved:
William Karush
Joel Zeitlin
Muriel 1/Jf_;i/g-ht
I
c;:Srmni ttee Chair
California State University,
Northridge
DEDICATION
I would like to express my sincere appreciation and
thanks to the many people who made this paper possible.
First, to my committee:
Dr. Joel Zeitlin,
patience, time,
Second,
Jr. High,
Dr. Muriel Wright,
and Dr. William Karush:
for their
and endurance of my persistance.
to the staff and students at Joe Walker
for their tolerance of a "grouch".
To my
Algebra class for excusing my mistakes and not complaining
too much about my delay in returning their tests.
Special
thanks to Nancy Clark and Allan Sacks for the times they
helped me out of a bind.
To Lynn,
for typing my "mess" into a beautiful
masterpiece.
To Grandma,
for baking me bread to help my spirits,
and for her company to see the wildflowers trying to
help me relax.
To Mom, Dad, Michelle, and Bob,
given to Kristyn and Kenneth.
for the love and care
Also to Mom and Dad,
for
the years of guidance and love that helped me to succeed.
~~lost
xeroxing,
of all to my husband,
typing,
and for the love,
S.teve,
for running errands,
giving up hours of much needed sleep,
understanding,
achieved my final goal.
and help to make sure I
I love you.
TABLE OF CONTENTS
DEDICATION
Chapter 1:
ABSTRACT
v
INTRODUCTION
1
ALTERNATE TOOLS FOR EUCLIDEAN
CONSTRUCTIONS:
Chapter 2:
iii
AN OVERVIEW
2
PROOFS OF ADEQUACY
Introduction
9
Rusty Compass and Straightedge
13
Compass Alone
16
Straightedge Alone, Given One
Fixed Circle and its Center
Double-edged Straightedge
50
Mira
69
Paper Folding
80
Chapter 3:
GEOMETROGRAPHY
Chapter 4:
"MORE POWERFUL TOOLS:
82
ANGLE
TRISECTION MADE POSSIBLE
Chapter 5:
30
96
CONCLUSION
108
BIBLIOGRAPHY
110
ABSTRACT
EUCLIDEfu~
CONSTRUCTIONS:
ALTERNATE TOOLS TO THE TRADITIONAL COMPASS AND
STRAIGHTEDGE
by
Sandra Kay Birrell
Master of Science in Mathematics
This paper examines the question whether there are
alternate tools to the traditional compass and straightedge capable of performing the same constructions.
tools considered are:
compass alone,
rusty compass and straightedge,
straightedge alone
and its center),
The
(given one fixed circle
double-edged straightedge with parallel
sides, Mira, and paper folding.
These tools are proved
to be adequate to perform all Euclidean constructions
(all constructions possible using a compass and straightedge).
The criteria for demonstrating this adequacy
are the "intersection criteria",
i.e., determining the
points of intersection of any two lines,
any line and
any circle, and any two circles.
The tools are then compared quantitatively to
determine the tool most easily used.
Geometrography
is used to make this comparison.
Finally the question of which tools are capable of
performing more constructions than are possible using
only a compass and straightedge is considered.
The
ability of the Mira and paper folding to trisect any
angle,
an impossible construction using compass and
straightedge, is demonstrated.
'I l.
INTRODUCTION:
Much of mathematics interests only the mathematician,
but geometric constructions seem to interest everyone.
Traditionally,
the compass and straightedge have been
used for geometrical constructions.
Euclid seemed to
make a conceptual decision rather than one based on
pedagogy when the choice of co:rr,pass and straightedge
was made.
Over the centuries many tools have been used
to perform constructions.
Could Euclid have made a
different choice and still accomplished the same
constructions?
The purpose of this paper is to prove that indeed
there are other tools which are capable of performing
all "Euclidean" constructions.
Further~ore,
some of
the tools are capable of performing even more powerful
constructions such as trisecting an angle,
with Euclid's choice of tools.
l
impossible
CHAPTER 1:
ALTERNATE TOOLS FOR EUCLIDEAN CONSTRUCTIONS:
AN OVERVIEW
Traditionally the compass and straightedge have
been used to perform all constructions met in Euclid's
Elements.
Yet it was Plato, not Euclid, who emphasized
the use of the compass and straightedge as the only
tools of pure geometry (Hallerberg, p. 231).
For the
Greeks the compass was a collapsible one, i.e.,
it
could not transfer distance as can a modern compass.
These two compasses are equivalent, a fact which will
be demonstrated later in this paper.
The modern compass
is much easier to use because fewer steps are required
to complete a construction than when a collapsible
compass is used.
There is some evidence that some
writers soon after Euclid did use compasses as devices
which could transfer distances
(Hallerberg, p. 232).
Over the centuries various other tools were used to
accomplish "Euclidean constructions",
constructions which are
and compass.
possib~e
i.e., all
using the straightedge
Many times the use of these tools began as
a practical solution to a problem.
Later,
they were used
to try to perform all Euclidean constructions.
now take a look at some of these tools.
2
vie will
3
One pair of tools is the straightedge and rusty
compass, a compass with a single fixed opening.
AbG'l-Wef@•s
(940-998) work on geometrical constructions
appears to be the first recorded attempt to use a rusty
compass
(Hallerberg, p. 232).
A possible reason for his
choice is that it was felt that fixed compass
constructions were more efficient, since additional
adjustments might cause inaccuracies,
possibly being more time consumidg.
in addition to
A most amusing
reason is that the compasses of the day were difficult
to adjust!
According to Hallerberg (p.
234),
however,
the most plausible explanation is that the fixed compass
constructions were developed as a practical need for
regular polygon constructions in art and architecture,
and also in the construction of scientific instruments.
AbG'l-Wefg had performed only some of his construetions using the rusty compass.
It was not until the
sixteenth century that mathematicians such as Ludovico
Ferrari or Nicolo Tartaglia gave £ixed compass
constructions for all the problems in Euclid's Elements
(Hallerberg, p.
237).
Then in 1674 Compendium Euclidis
Curiosi was published, anonymously.
In 1929, the author
was found to be Georg Mohr (Hallerberg, p.
presented only twenty-nine constructions
g;_l,~rr'-ents,
and he c;rave no proofs.
241).
fro~
Mohr
Euclid's
It was not Gntil the
4
Poncelet-Steiner Theorem (to be stated later), written
in the 1830's,
that it could be said that the rusty
compass and straightedge were proven to be capable of
performing all Euclidean constructions.
The compass alone is another tool which has received
attention.
As with the rusty compass and straightedge,
its use was an answer to some practical,
theoretical, needs,
instruments.
rather than
such as constructing astronomical
The constructions with the compass alone
are often called Mascheroni constructions, after Lorenzo
Mascheroni.
In 1797 Mascheroni wrote Geometria del
CQEf'tJ2asso, where he proved that all Euclidean constructions
were possible using only the compass
(Kostovskii, p. x)
Then in the 1920's Euclides Danicus, written by Georg
Mohr in 1672, was discovered.
Mohr also showed that
Euclidean constructions could be accomplished using the
compass alone.
Though Mascheroni wrote after Mohr,
it
is generally felt that Mascheroni was not aware of Mohr's
book.
While Mascheroni had used the modern compass in
his proofs,
in 1890 August Adler showed,
of inversions,
using the idea
that all Euclidean constructions could
also be done using a collapsible compass
(Cheney, p. 152)
The result of all this effort is the Mohr-Mascheroni
Theorem which states,
"·Any Euclidean construction,
as the given and required elements are points,
insofar
may be
5
accomplished with the Euclidean
alone."
(Eves,
(or modern)
compasses
History of Mathematics, p. 107).
With
this theorem the straightedge is no longer a necessity
in performing all Euclidean constructions.
as Carnahan (p.
388) states,
However,
"It cannot be urged that
the straightedge be rejected in making geometric
constructions for to do so would add materially to the
difficulty of many constructions,
but it is convenience
alone and not necessity that justifies its retention."
The next logical tool to consider is the straightedge alone.
As with the other tools,
the straightedge
constructions were of practical benefit,
for engineers and surveyors.
in this case
However with the straight-
edge alone i t is only possible to construct lengths
corresponding to rational numbers.
The compass and
straightedge can be used to construct all lengths corresponding to the rational numbers and square roots.
Consequently,
the straightedge alone is not capable of
performing all Euclidean constructions.
I
In 1822, Jean
I
1
Victor Poncelet wrote Traite des Proprietes Projectives
des Fiaures,
in which he showed that in the presence of
one fixed circle, given by its center and a point on the
circle,
all Euclidean constructions could be done using
only the straightedge (.i\rchi bald, p.
191) .
In 1833
~
6
Jacob Steiner gave a complete and systematic proof of
Poncelet's idea.
The result is the
Theorem which states,
£gnc~let-Steiner
"Any Euclidean construction,
insofar as the required and given elements are points,
may be accomplished with straightedge alone in the
presence of a given circle and its center,"
Survey of Geometry, p.
178).
(Eves,
In 1904 Severi proved
that only an arc of a circle, however small, with its
center, was needed with the straightedge to perform all
Euclidean constructions
p.
98).
crucial.
(Eves, History of Mathematics,
It is the center of the circle or arc,
that is
Without the center, we cannot use the straight-
edge alone to perform all Euclidean constructions
(Yancgihara, p.
47).
Another tool which has been used to perform Euclidean
constructions is the double-edged straightedge with
parallel sides.
Though there exist many references to
a double-edged straightedge whose sides are perpendicular,
like a carpenter's square,
little information was found
for the double-edged straightedge whose sides are
parallel.
Poncelet was the first to point out that
these two-edged straightedges could be used to perform
Euclidean constructions
(Archibald, p.
191).
Later,
in
1890, August Adler proved that all Euclidean constructions
7
could be accomplished, whether the edges of the
straightedge were intersecting or parallel
p.
(Cajori,
365).
Very little is written about the Mira and paper
folding,
in comparison to the other tools mentioned.
There are numerous books on paper folding,
such as
T. Sundara Row's Geometric Exercises in Paper-Folding,
but they are mainly concerned with how to perform the
construction, rather than why the construction works.
No sources were found which pursued the question of
whether all Euclidean constructions could be
by paper-folding.
perfor~ed
The Mira, a recent invention by
Scroggie and Gillespie, was developed for use in
transformational
geo~etry.
tool,
demo~strated
as will be
The Mira is a very powerful
in a later chapter.
There are other tools which have been used to
perform some or all of the Euclidean constructions,
such as a protractor.
Many were invented solely for
special constructions such as trisecting an angle,
an
impossible construction using the compass and straightedge.
When Plato made the restriction to compass and
straightedge for geometric constructions he did
mathematicians a favor.
Due to this restriction,
8
mathematicians and others tried alternate tools, both
for theoretical and practical reasons.
It was because
of the restriction that the three famous problems of
trisecting the angle, duplicating the cube, and squaring
the circle received so much notoriety, as they were
impossible to accomplish with a straightedge and compass.
It was not until the nineteenth century that the
impossibility of these constructions, using straightedge
and compass, was established and many important
mathematical discoveries were achieved trying to prove
why the constructions could or could not be done.
CHAPTER 2:
PROOFS OF
ADEQUAC~Y~=--~A=N~I=N~T~R~O~D~U~C~T~I~O=N~
The goal of this chapter is to prove that there are
alternate tools to the traditional straightedge and
compass, which are capable of performing all Euclidean
constructions,
i.e., all constructions that are possible
with a straightedge and compass.
These tools include
the rusty compass and straightedge, the compass alone,
the straightedge alone given one fixed circle with its
center,
the double-edged straightedge with parallel
sides, the Mira,
and paper folding.
In order to prove
these tools are adequate to perform all Euclidean
constructions we will use Poncelet's criteria
(Hallerberg, p.
243).
His criteria are based on the fact
that new points in a Euclidean construction are found
from old points by determining the points of intersection
of:
1)
any two lines
2)
any line and any circle
3)
any two circles
Note that compass and straightedge constructions are
made up of drawing a line, determined by two points,
along the straightedge,
given center,
given point) .
0,
and drawing a circle with a
and a given radius OP (where P is some
These operations yield new points which
10
are described in the above three ways.
constructions accomplished,
With these
the solution of all
construction problems merely becomes an application of
the steps listed above.
These criteria will be referred
to as the "intersection criteria"
(Hallerberg, p. 243).
"When dealing with a geometrical construction, one
must never forget that the problem is not that of
drawing figures in practice with a certain degree of
accuracy,
but of whether, by use of the straightedge
and compass alone
(or other tools)
be found theoretically,
have perfect precision,"
the solutions can
supposing our instruments to
(Courant and Robbins, p.
119)
Eence the question here is not how to do every Euclidean
construction with the different tools,
but is it
possible to accomplish all Euclidean constructions.
course,
Of
given the directions for a construction with a
straightedge and compass, we can follow these directions
with any of these other tools.
The directions are
merely a set of steps usinc; the "intersection criteria"
constructions,
we wish to
in an order particular to the construction
acco~plish.
be the best,
The question of which tool would
as far as efficiency is concerned for any
particular construction, will be discussed in the
chapter on c;eometrography.
11
Throughout the literature the term "equivalent" was
used when demonstrating that the tool in question could
be used to perform all Euclidean constructions.
the term
e~uivalent
However,
implies an equality, i.e., that the
tools are capable of constructing exactly the same
problems.
As will be shown later some of the tools are
"more powerful," i.e,.,
they are capable of performing
constructions that are impossible with a straightedge
and compass.
Consequently the term "adequacy" is used
rather than the term "equivalent"
(Wernick, p. 704).
There are alternate methods of proving that the
tools are adequate to perform all Euclidean constructions,
such as demonstrating that the tool can construct lengths
corresponding to the rational numbers and square roots.
The straightedge and compass are not capable of constructing all lengths corresponding to the real numbers, only
rational numbers and positive square roots,
shown in the chapter on trisecting an angle.
Klein
as will be
For example,
(p. 34-35) proves the adequacy of a straightedge
alone given a fixed circle with its center,
that the tool can solve every quadratic.
by proving
The postulates
for each system could also be proved equivalent.
of the literature found,
however,
Most
uses the "intersection
criteria" that will be used in this paper.
12
We now turn to prove the adequacy of these six
tools:
rusty compass and straightedge, compass alone,
straightedge alone
(given a fixed circle and its center),
double-edged straightedge with parallel sides, Mira,
and paper folding.
~
PROOF OF ADEQUACY:
RUSTY COMPASS AND STRAIGHTEDGE
Traditionally, Greek geometers used collapsible
compasses,
also referred to as Euclidean compasses,
their constructions.
The first natural change in
for
se~king
tools which are adequate to perform all Euclidean
constructions,
seems to be fixing the opening of the
compass to a constant width.
quite a strong restriction,
It would seem that this is
since apparently only circles
of this one radius can be constructed.
Eowever,
it is
possible to perform all Euclidean constructions with a
rusty,
or fixed,
compass and straightedge.
To prove
the adequacy, we assume Euclid's postulates with one
change in Postulate 3 that the compass opening is to
remain fixed at a constant width.
( Srni t h,
1)
p .
The postulates are
2 8 1) :
A straight line can be drawn from any point
to any point.
2)
A finite straight line can be produced
continuously in a straight line.
3)
A circle may be described with any center and
a fixed distance.
(This is the given opening
of the rusty compass.)
4)
All right angles are equal to one another.
5)
If a straight line falling upon two straight
14
lines makes the interior angles on the same
side together less than two right angles,
the
two straight lines, if produced infinitely,
meet on the same side which the angles are
together less than two right angles.
(Henceforth Euclid's Postulates are assumed for all
sections proving the adequacy of a tool.
Postulates
1 or 3 are changed appropriately, dependent upon
whether lines or circles cannot be drawn with the tool
being considered.)
Ferrari,
in Quinto Cartello, presented a complete
treatment of Euclid's Six Books, with the above change
in the third postulate
(Hallerberg, p. 237).
The
propositions were necessarily in an order different
from Euclid's,
but Ferrari was careful to only use
constructions and theorems previously established in
his development.
So Ferrari proved that the rusty
corr,pass and straightedge were indeed adequate to perforiT, ·
all Euclidean constructions.
A second approach to prove the adequacy of the
rusty compass and straightedge uses the Poncelet-Steiner
Theorem (Eves, A Survey of Geometry, p.
178) which states:
"If a single circle and its center are once drawn in a
plane, every construction possible with ruler and compass
15
can be carried out with ruler alone."
(The term ruler
used here should be changed to the term straightedge.
Numerous writers have inadvertently used "ruler" when
actually referring to a straightedge.
A ruler has
marks on it, marks which are not allowed by Euclid's
Postulates.)
Since an arbitrary fixed compass can be
used just once to draw the necessary circle,
it follows
that the rusty compass and straightedge must be adequate
to perform all Euclidean constructions.
the Poncelet-Steiner TheoreiD_,
The proof of
using the "intersection
criteria", will be given later in the chapter on using
the straightedge alone.
PROOF OF ADEQUACY:
COMPASS ALONE
The use of the compass alone to solve all Euclidean
constructions has probably received the most notoriety,
based on the volume of literature written on the subject.
After fixing the compass opening for the rusty compass
and straightedge constructions, using only the compass
seems to be a logical restriction for geometers to
consider.
We will show this tool is adequate to perform
all Euclidean constructions.
Recall that to prove this,
the following points of intersection must be determined:
1) any two lines, 2) any line and any circle, and 3) any
two circles.
Again all of Euclid's Postulates are
assumed, with the exception of a change in Postulate 1,
since without a straightedge lines cannot be drawn.
However, we will assume a line to be determined by any
two points.
The compass alone is adequate to perform all
Euclidean constructions, whether a Euclidean (collapsible)
compass or a modern (non-collapsible) compass is used.
A Euclidean compass allows circles to be drawn with any
given point as center and passing through any second
given point.
In practice we must deal with the fact that
some points are farther apart than the compass reaches,
but we do not concern ourselves with that here.
We are
not allowed to "carry" line segments, or lengths, from
lb
17
one position to another.
Hence a Euclidean compass
collapses the moment it is lifted from the paper.
Modern compasses, however, are allowed to "transfer"
lengths.
The opening can remain fixed, even after
being lifted from the paper.
It is well-known that the
Euclidean compass and the modern compass are equivalent.
Before proving the above statement, we define the
notation O(P) to mean the circle with center 0 and point
P on the circle, and therefore a radius of OP.
The
notation O(ST) will mean a circle with center 0 and a
radius equal in length to ST.
~
The notation AB will mean
~
the line as determined by points A and B; AB is the ray
from A through B; and AB is the segment with endpoints
A and B of length AB.
CONSTRUCTION 1:
Given a Euclidean compass, segment
and a point 0, construct a circle O(P), where OP
=
AB,
AB.
Construction:
Draw A(O) and O(A), intersecting in points Nand R.
Draw N(B).
O(P).
Draw R(B), intersecting N(B) also in P.
OP = AB.
Draw
(To aid in the proof, draw NR and OA,
intersecting in S.
Draw PB, intersecting NR in Q.)
18
Proof:
To prove OP
= AB we need to show that triangles ORP
and ARB are congruent.
RP
Since radii are equal, OR
= AR and
= RB, so we need only show that m2$_0RP = m4.-ARB.
First, since radii are congruent, triangles NOA and
ROA are congruent,
thus mLNOA
and RSO are now congruent, so
=
m~
mL\_ ROP...
Triangles NSO
NSO = mZ\._ RSO.
Since
angles NSO and RSO are supplementary and equal, they are
right angles.
Hence NR is perpendicular to OA.
Now tri-
angles SRO and SRA are congruent, thus m4- SRO = mL SRA.
Triangles NRP and NRB are congruent,
m;i_ NRB.
thus
m~NRP
=
Subtracting the measures of 4_ SRO and ,:i SRA from
the measures of ii. NRP and
4-
NRB yields m;'i__ORP
Hence triangles ORP and ARB are congruent,
= m+- ARB.
so OP = AB.
19
Hence the Euclidean compass is equivalent to the
modern compass.
Obviously,
using the Euclidean compass
is more cumbersome, and so for the sake of simplicity,
the modern compass will be used in the following proofs.
INTERSECTION CRITERION 3:
Determine the points of
intersection, X andY, of any two given circles, O(P)
and M( N) .
This problem is immediately solved, since we are
allowed to draw circles with the compass.
INTERSECTION CRITERION 2:
Determine the points of
intersection, X andY, of any given circle, O(P) and
~
~
any given line, AB, where AB is determined only by
points A and B.
There are two cases which must be considered:
1)
~>
AB and 0 are not collinear, and 2)
they are collinear.
(The following proofs are based upon Hlavaty's article,
"Mascheroni Constructions.")
20
Case 1:
A, B, 0 not collinear.
Construction:
Draw A(O)
and B(O), labeling their second point
of intersection,
E.
Draw E(OP),
given circle at points X andY.
required points of intersection.
intersecting the
X and Y are the
21
Proof:
By construction AO = AE and BO = BE, since radii of
a circle are congruent.
bisector of OE.
~
Therefore AB is the perpendicular
By construction OP = EX.
Since radii of
a circle are congruent, EX= EY and OP = OX= OY.
~
Substituting, EX = OX and EY
OY.
perpendicular bisector of OE.
Since AB and XY are both
Hence XY is the
~
~
~
perpendicular bisectors of OE, X and Y must lie on AB.
By construction, X andY lie on O(P).
Therefore X and Y
are the required points of intersection.
Q.E.D.
Anticipating Case 2, we must first show how to
locate the midpoint of an arc .
...----_
CONSTRUCTION 2:
r,
locate R,
Given arc,
NP, with center 0 and radius
.--...
the midpoint of NP.
Construction:
Using N and P as centers,
and radius r,
arcs passing through the center 0.
draw two
On one arc mark the
point S such the NP = OS; likewise mark point T on the
second arc such that NP
and S and T as centers,
that SP = SV = TV.
OT.
Using SP as the radius,
draw arcs intersecting at V,
Using OV as the radius and T as
so
22
,.......,_
center, draw an arc intersecting NP in R, so that
OV
= TR.
...--..-.
R is the required midpoint of NP.
S../
!
/
T
'
'
I
/
'
/
~~
I
/
\
\
\
/
I
(-
~
'
.
lo('\.
'
I
'
/"
I
/~
/
/
/
/
Proof:
By construction NP = OS; likewise NS = NO and ON =
OP,
and thus NS
= OP.
Since opposite sides are equal,
NPOS is a parallelogram, with OS parallel to
By construction,
are equal, PO
NP
= OT.
to NP.
Since radii of a circle
= PT and ON = OP, and thus ON
above, NPTO is a parallelogram,
Thus s,
0,
~P.
PT.
As
so TO is also parallel
and T are collinear and ST is parallel
23
to NP.
=
By construction OS
By construction SP
=
SV and SP
=
TV,
OT.
therefore SV
=
TV.
OV is perpendicular to chord NP.
OV also bisects NP and NP,
~
since OV passes through the
,......
center of O(P).
To prove that R is the midpoint of NP,
~
we must show that R lies on OV.
showing that
We
4
This will be proved by
ROT is a right angle.
see that
(1)
2
PT
= OT 2 +0P 2 -2(0T) (OP)
SP 2 = OS 2 +OP 2 -2 (OS) (OP) cos ( f f - ·><),
cosines and letting ~ =-iPOT.
Adding
and using the fact that cos (II- •::>"..:.)
2
2
sP +PT
= OT 2 +0S 2 +20P 2 .
OP.
Thus
(3)
=
coso<
and
using the law of
(1)
and (2),
-cosv~,
Since OS= OT,
2
2
2
SP +PT 2 = 20T +20P .
=
Since ST
~
,.....__
~
PT
=
OV is the perpendicular bisector of ST.
is parallel to chord NP,
(3)
OT, thus OS
~
Thus,
( 2)
=
NP and NP
we cJet
this becomes
By construction SP =TV and
2
2
2
2
becomes TV +OP = 20T +20P , so
E--'>
Since OV is perpendicular to ST,
2
2
2
2
Thus OV +0T = 20T +0P , or
(4)
ov 2
2
= OT 2 +0P .
By construction OV = TR.
~-
construction R lies on NP, and therefore OP
Hence
(4)
becomes TR
2
=
2
2
OT +0R .
Therefore
By
=
OR.
~ROT is
a right angle.
Angles VOT and ROT are right angles,
and
4
~
VOT are coincident, with R on OV.
..---...
and thus
4
ROT
Therefore R is
the midpoint of NP since it lies on the perpendicular
24
~
bisector OV, of chord NP, and by construction R lies on
,.......
NP.
Q.E.D.
Case 2:
Determine the points of intersection, X and Y,
~
~
of circle O(P) and line AO, where AO is determined only
by points A and 0.
Construction:
Draw A(WZ), where WZ is an arbitrary length such
that A(WZ) will intersect O(P)
in points Nand R.
Using
Construction 2, locate the midpoints, X and Y, of major
-......
arc NR and minor arc NR.
X and Y are the required points
,..~---
of intersection.
p
I
y
~R
25
Proof:
By construction AN
= AR and ON = OR.
the perpendicular bisector of chord NR.
~
~
Thus AO is
~
Hence AO bisects
~
NXR and NYR.
By construction 2, X andY are the midpoints of
~
~
NXR and NYR.
Then X and Y must lie on the perpendicular
~>
bisector of chord NR, which is AO.
they also lie on O(P)
By construction
Therefore X and Y are the
required points of intersection.
Q.E.D.
Anticipating Intersection Criterion 1, we first
show how to construct a symmetric (or reflection) point
for any point with respect to any line.
CONSTRUCTION 3:
Determine the sywmetric point P, of a
~
given point N, with respect to AB.
B
~
'
26
Draw A(N)
and B(N), intersecting in P.
P is the
required symmetric point.
Since BP
~
= BN and AP = AN, AB is the perpendicular
bisector of PN.
Hence P is the symmetric point of N.
INTERSECTION CRITERION 1:
Determine the point of inter-
section, X, of any line ~l (determined only by points
A and B)
C
(---0:>
and any line CD (determined only by points
and D) .
Construction:
First construct the symmetric point P, of B, with
~
.
respect to CD, us1ng Construction 3.
the symmetric point
~'
Likewise construct
~
of P, with respect to AB.
Third,
construct the symrr.etric pointE, of B, with respect to
~-
Now draw R(B)
and B(P)
intersecting in S and T.
Either point may be used to finish the construction;
S is chosen, without loss of generality.
P(B),
intersecting also in V.
intersecting S(B)
intersection.
in X.
Draw S(B) and
Last, draw V(R),
X is the required point of
~
27
i
!
I
I
I
/
/
Proof:
~
First we show that X lies on l-.B.
VX
=
VR.
By construction BP
Hence SB = PV and so S(B)
equal radii.
=
SB, SV
and P(V)
By construction,
=
SB, and BP
=
PV.
are circles with
Therefore chords VX and VR subtend equal
28
,...-....
;----..
arcs, VX and VR.
Since the measure of an inscribed
angle equals half the measure of the intercepted arc,
..-....
m~VBX
~m(VX)
m-tVBX
m~VBR.
and
=
m~VBR
;----..
~m(VR).
Substituting,
Hence the angles are coincident and
~
X lies on RB.
Since AP
AN and BP
bisector of PN.
P and N,
~
= BN, AB is the perpendicular
Likewise, R and B are equidistant from
~
and RB is the perpendicular bisector of PN.
~
~
<-')>
So RB and AB are the same line.
Since X lies on RB,
it
~
also lies on f'._B.
We now show that X lies on ~RB = RS and SX = SB.
By construction
Therefore triangles BRS and XSB
are isosceles with <j:, SBX in common.
Triangles BRS and
XSB are similar since their base angles are equal,
that BR/BS
= BS/BX.
becomes BR/BP
proportion,
PR = PB,
Since BS = BP,
= BP/BX.
so
this proportion
Since these sides are in
triangles RPB and PXB are similar.
triangle RPB is isosceles.
Since
Therefore triangle
PXB must be isosceles with BX = PX.
By construction DP
= DB.
Since X and D are equi-
~
distant from B and P, XD is the perpendicular bisector
of BP.
By construction CP
equidistant from B and P.
= CB.
Hence C and D are also
(------7
Therefore CD is also a
perpendicular bisector of BP.
~
~
So CD and XD are the same
29
line.
~
Therefore X lies on CD, and X is the required
point of intersection.
In conclusion,
Q.E.D.
since we have been able to determine
the points of intersection of 1) any two lines,
line and any circle, and 3)
2)
any
any two circles, the compass
alone is adequate to perform all Euclidean constructions.
A logical step then would be to throw away the straightedge in performing constructions.
However, as we shall
see in the chapter on Geometrography, the straightedge
does allow many constructions to be accomplished in
fewer steps.
PROOF OF ADEQUACY:
STRAIGHTEDGE ALONE, GIVEN ONE FIXED CIRCLE AND ITS CENTER
In this chapter we will show that it is possible to
perform all Euclidean constructions with a straightedge
alone, provided that one circle and its center is drawn.
These constructions are often called Poncelet-Steiner
constructions, named after the men who proved the
adequacy of this tool.
To prove the adequacy of this
tool we will again use the "intersection criteria".
INTERSECTION CRITERION 1:
Determine the point of inter-
section of two lines.
This is clearly solvable, since we can draw
straight lines with the straightedge.
However, we cannot draw circles with a straightedge,
so a circle will be given by its center and a point on
the circle (thus a radius is also given).
\'ie
given, however, one circle and its center,
for all
constructions.
(Alone,
are
the straightedge can only
construct lengths corresponding to the rational numbers
and not to square roots).
:50
31
Steiner was the first to prove adequacy for all
Euclidean constructions in the present case (pp. 189-264).
However, we shall follow Yanagihara's approach which is
easier to understand and is based on the three lemmas
below.
Anticipating Lemma 1, we must prove Ceva's
Theorem.
CEVA'S THEOREM:
If three concurrent lines are drawn from
the vertices A, B,
and C of triangle ABC to meet the
opposite sides in G,
(1)
D, and E, respectively, then
AE/EB · BG/GC · CD/DA
= 1
(Taylor, p. 55).
32
Proof:
To prove Ceva's Theorem we use the fact that the
areas of triangles with equal altitudes,
to the bases of the triangles.
of triangle ABC.
are proportional
Let (ABC) mean the area
Then AE/EB = (ACE)/(ECB) = (AEF)/(BEF)
It follows that AE/EB
=
ITACE)-(AEF~ I [EcB)-(BEFTI =
(AFC)/(CFB).
Likewise BG/GC = (AFB)/(AFC)
(CFB)/(AFB).
Multiplying, AE/EB · BG/GC · CD/DA =
EAFC)/(CFB)] · [AFB)/(AFCD · ~CFB)/(AFBD.
AE/EB · BG/GC · CD/DA = 1.
LEMtvlA 1:
and CD/DA =
Hence
Q.E.D.
Given a segment, AB and its midpoint E, we can
draw a line parallel to AB, and passing through a given
point D by straightedge alone.
Construction:
~
Draw AD.
between A and C.
~
Draw AF and BC,
parallel line.
~
Choose a point C on AD such that D is
Draw BD and CE,
intersecting in G.
intersecting in F.
~
Draw ~G, the required
~
33
B
E
A
Proof:
By construction AG,
Ceva's Theorem,
=
AE/EB
AB.
AE/EB · BG/GC · CD/DA
= DA/CD.
=
1.
By
Since
~
Therefore DG is parallel to
Q.E.D.
LEMMA 2:
AB,
BG/GC
1,
BD, and CE are concurrent.
Given a segment, AB,
we can find the midpoint,
alone.
and a line m parallel to
C, of AB by straightedge
34
Construction:
Take any point D, not on AB or line m, or in between
Draw ~ and ~'
AB and m.
E and F, respectively.
G.
~
intersecting line m in points
Draw AF and BE, intersecting in
~
Draw DG,
intersecting EF in H and AB in C.
c is the
midpoint of AB.
I
I
I
m
E
H
F
Proof:
Since AB is parallel to m,
triangles EGH and BGC are
similar and triangles FGH and AGC are similar.
=
EH/BC
GH/GC and FH/AC
FH/AC, or (1)
Again,
BC/AC
=
GH/GC,
from which EH/BC
FH/EH.
=
= EH/FH.
triangles DCA and DHE are similar and
triangles DCB and DHF are similar.
and DC/DH
Hence
Hence AC/EH
= DC/DH
= BC/FH, from which AC/EH = BC/FH or (2) BC/AC
35
Multiplying
(1)
2
2
and (2) we get BC /AC = 1, or
BC/AC=l.
Hence c is the midpoint of AB.
LEMVill 3:
Given two parallel lines, m and n, we can draw
a line t parallel to m and n,
given point D,
Q.E.D.
and passing through a
using straightedge alone.
D
t
n
m
A
c
B
Take any two points, A and B, on line m.
we can locate C,
the midpoint of AB.
required parallel line,
By
Le~IDa
By Lemma 1 the
t,can be drawn through D.
We turn to six constructions used by Yanagihara
(pp. 46-47)
to solve Intersection Criteria 2 and 3.
Briefly stated they are:
2
36
1)
Construct a line through a given point, parallel
to a given line.
2)
Bisect a given line segment.
3)
Through a given point,
construct a line
perpendicular to a given line.
4)
Copy a line segment.
5)
Construct the fourth proportional to three
given lengths.
6)
Construct the mean proportional to two given
lengths.
For each construction we assume given the fixed circle
with center F,
in addition to the straightedge.
CONSTRUCTION 1:
Through a given point A,
draw a line
parallel to a given line m.
Construction:
Through point A draw any line n,
intersecting line
m in B and the given circle in S and T.
SS'
and TT'
m in C.
B'
of the given circle.
Draw
13f'
and
d
and C', respectively.
coincides with A,
Draw diameters
(
)
Draw S'T',
intersecting
intersecting ~''I,> and
!T in
b•c;> is parallel to m.
the problem is finished.
points
If C'
If C' and A
37
are distinct points, a line t,
through A, is constructed
using Lemma 3.
t
r..
Proof:
Since S'T' = ST and S'T = ST', STS'T'
parallelogram,
and ST is parallel to S'T'.
is a
It follows
triangles BSF and B'S'F are congruent as are triangles
CT'F and C'TF.
Thus BS = B'S' and CT' = C'T,
so BC' = B'C.
38
Now points B and C
<
lie on S T
1
)
1
•
Hence BC
1
~
lie on ST and points 8
1
is parallel to B 1 C.
BC B C is a parallelogram, and B C
1
1
1
m.
By Lemma 3 line t,
m.
Q.E.D.
CONSTRUCTION 2:
through A,
1
1
and C
Thus,
is parallel to line
is parallel to line
Bisect a given segment AB.
D
A
m
c
B
By Construction 1 draw a line m parallel to AB.
through any given point D.
By Lemma 2, AB can be
bisected, with C as the midpoint.
CONSTRUCTION 3:
Through a given point A,
t perpendicular to a given line m.
draw a line
39
A
F
s
n
m
t
By Construction 1, draw a line n parallel to m,
such that line n intersects the given circle in R and S.
Draw diameter ST; then
m
perpendicular to line n.
parallel to RT,
point A,
90°.
Hence RT is
Now draw line t,
using Construction 1.
perpendicular to line m,
CONSTRUCTION 4:
4-- TRS
Then line t
is
the desired result.
Given line m passing through a given
and segment BC,
such that line m and BC are not
collinear, we can find a point E on line m,
AE = BC.
through A,
so that
~
40
Construction:
Using Construction 1, draw two diameters, SS' and
TT', of the given circle, such that SS'
is parallel to
BC and TT' is paralllel to line m.
Now form parallelogram ABCD, using Construction 1,
by drawing a line n through A, parallel to BC,
t through C, parallel to AB.
of line n and line t
parallel to ST.
is D.
and a line
The point of intersection
Through D draw line r
Let r intersect m in E; then AE is the
desired segment.
We can locate a second point, E',
so that AE'
= BC.
Through D draw a line w, parallel to ST', and let E'
its intersection with m.
be
Then AE' = BC.
(See diagram on following page.)
Proof:
Through E' draw line z parallel to ED.
Through E,
draw line v parallel to DE' and let its intersection with
z be D'
m
<):TST'
We prove that EDE'D'
90°,
parallel to ST,
ST is perpendicular to ST'.
and DE! is parallel to ST'.
perpendicular to DE'.
EDE'D'
is a rectangle.
is a rectangle.
Since
Now DE is
Hence :UE is
Since opposite sides are parallel
Hence AD= AE = AE'.
41
Now ABCD is a parallelogram with BC
AE
AE'
= BC, the desired results.
= AD.
Thus
Q.E.D.
Note that using this construction we can add or
subtract the lengths of any two segments.
s
S'
;---'=8"-------/ c
n
r
z
v
t
w
m
42
CONSTRUCTION 5:
Given segments of lengths a,
construct a segment of length d such that ad
b, and c,
= be.
Construction:
Draw any line m and mark any point A on line m.
Through A draw any other line n.
Using Construction 4
copy the longer segment of a and b,
such that AC
C,
= b.
say b, onto line n
Likewise copy length a onto line n at
on the side which A is, such that BC
Construction 1 draw a line,
line m.
r,
a.
Using
through B, parallel to
Using Construction 4 copy length c onto line r
such that BD = c.
-E---7
Draw CD,
intersecting line m in E.
AE is the desired length d.
a
b
c
r
m
n
43
Proof:
Since r is parallel to m,
the desired result.
BC/AC
BD/AE, or ad
be,
Q.E.D.
Note that using this construction we can multiply
and divide the lengths of any two segments, if we have
a segment of length 1.
CONSTRUCTION 6:
Given segments of lengths a and b,
construct a segment with length c such that c 2
= ab.
Construction:
Draw RS,
any diameter of the fixed circle.
any line m through any point D, parallel to RS,
Construction 1.
= b, using Construction 4.
and FS, intersecting in G.
H.
~
Draw GE,
~
Draw bR
<E----?
intersecting RS in
By Construction 3, draw a line through H perpendicular
to RS,
intersecting the fixed circle in T.
~
a point B on HR,
HB
by
On line m locate points E and F such
that DE = a and EF
~
Draw
Now locate
on the same side of H as R is,
= a, using Construction 4.
so that
Using Construction 1 draw
~
a line through B, parallel to RT,
E-'>
intersecting HT in C.
CH is the required mean proportional length c.
44
•\
I
1
.
/;l_ c)
1 ~i!
\\
;·I
j\_
p
!'~~II
,
II
D
"'
I
I
EI
('.
-
b
\I
E'\
-
m
45
Proof:
Since line m is parallel to RS triangles GRH and GDE
are similar; likewise triangles GHS and GEF are similar.
Thus, letting a'
= RH and b' = HS, we get (1) a/a' = b/b'.
Triangle RTS is a right triangle, so a'/c'
or (2)
a'b'
to RT,
( 3)
Thus ab
= (c') 2 , where c' = HT.
a/ a '
= (c ' ) 2
=
c/ c ' .
( c/ c '
Now ab
= c'/b',
Since BC is parallel
= a'b' (a/a'
· c/c'), or ab
= c2.
b/b I) •
Q.E.D.
(Please note that there is an error in Yanagihara's
article for Construction 6
the required."
(p.
47).
For his notation,
It states,
II
CL is
it should read CN
instead of CL.)
Note that if a + b equaled the diameter of the fixed
circle our construction is simplified.
a and b onto the diameter,
First copy
and then construct a
perpendicular at the point at which the two lengths
meet.
We would have a right triangle inscribed in a
semi-circle.
Hence a/c
= c/b, or c 2 = ab.
46
We now turn to Intersection Criteria 2 and 3, using
the leruF.as and constructions previously presented.
INTERSECTION CRITERION 2:
Determine the points of
intersection, X and Y, of a given line m and a given
circle O(P), where O(P)
is given only by its center 0
and point P on the circle (and thus a radius OP).
m
Construction:
Using Construction 3, draw a perpendicular to line
m, at 0,
intersecting line m in A.
If X is one of the required points, and letting
a
= OP,
We can determine
4.
=
a
2
0?,
(a + OA)
2
(a
.
and
(a - OA)
+ OA) (a -
OA)
using Construction
By Construction 6 we can determine the r;.eac
47
proportional,
c, between (a + OA)
and (a - OA).
such that AX
=
Then c
=
is copied onto line m,
at A,
c and AY
using Construction 4.
X and Y are the required points.
c,
The proof is evident from the construction.
INTERSECTION CRITERION 3:
intersection,
and M(N).
X andY,
O(P)
P on the circle.
Determine the points of
of any two given circles O(P)
is given only by its center 0 and point
M(N)
is given only by its center M
and point N on the circle.
(See diagram on following page.)
Construction:
We first assume the radii are not equal,
loss of generality,
and without
take MN to be· greater than OP.
Construction 2 determine the midpoint L, of MO.
X is one of the required points,
then line m,
is constructed perpendicular to MO.
Theorem XQ 2
=
OX 2 -
This yields
(1)
ox 2 = ~Q 2
OQ)
(~:Q
- OQ 2 .
+ OQ) .
Assuming
through X,
By the Pythagorean
OQ 2 and likewise XQ 2
MX 2 -
Using
=
MX 2 - MQ 2 .
Now
c':ow ( 3)
:-!Q
+ OQ
48
and MQ - OQ
=
Hence (4) MQ -
=
OQ
2QL.
(MX - OX)/QL
ox 2 =
2MO
QL + LO - OQ.
Substituting ( 3)
= 2MO . QL.
into ( 2) we get MQ2 - OQ2
2
becomes MX -
=
QL + LM - OQ
and (4)
Hence
( 1)
. QL, which yields
= 2MO/(MX + OX).
I
:lx
.,
/
I\
~
/ I ''
\'
.,p
L
I'~1
0
Q
\
y
rn
An analysis of the problem,
lengths
(MX + OX) ,
(MX - OX)
and 2MO can be
the length a, where a = QL.
Then using Construction 4
we copy length a onto MO at L,
= a.
perpendicular to MO,
stated above,
deter~ined
By Construction 5 we can determine
using Construction 4.
L as 0, so that QL
shows that segments with
and on the same side of
At Q construct a line m
using Construction 3.
contains points X andY.
Line m,
X and Y are
as
49
given by the intersection of line m and either circle,
using Intersection Criterion 2.
Now consider the case when the radii are equal.
Then Intersection Criterion 3 quickly reduces to
Intersection Criterion 2.
OM,
Locate the midpoint L of
construct a perpendicular to OM at L, and find
the intersection with either circle.
In conclusion,
since we have been able to determine
the points of intersection of 1) any two lines, 2)
line and any circle, and 3)
straightedge alone
any
any two circles, the
(given one fixed circle and its center)
is adequate to perform all Euclidean constructions.
PROOF OF ADEQUACY:
THE DOUBLE-EDGED STRAIGHTEDGE
In the previous section the straightedge alone was
proved to be adequate, provided that one circle with its
center was drawn.
However,
if both edges of a double-
edged straightedge are used, whether the edges are
intersecting of parallel, we can perform all Euclidean
constructions without a circle being given.
In this
section we will restrict ourselves to discussing only
a double-edged straightedge with parallel sides.
Again,
to prove the adequacy of this tool we must determine the
points of intersection of 1)
and any circle, and 3)
any two lines,
2)
any line
any two circles.
INTERSECTION CRITERION 1:
Determine the point of
intersection of two lines.
Since a double-edged straightedge can be used to
draw lines, this is clearly solved.
Since a double-edged straightedge cannot be used to
draw circles,
a circle will be glven by only its center
and a point on the circle.
Before turning to Intersection
Criteria 2 and 3, we present the following preliminary
50
51
constructions based on the article by Wernick,
Constructions:
1)
"Geometric
The Double-Edged Straightedge":
Construct any number of equally spaced parallel
lines.
2)
Double a given segment, AB.
3)
Bisect a given segment, AB.
4)
Construct a line through a given point A and
parallel to a given line m.
5)
Construct a line through a given point A and
perpendicular to a given line m.
6)
Bisect a given angle, { ABC.
7)
Copy a given segment AB.
CONSTRUCTION 1:
Construct any number of equally spaced
parallel lines.
Construction:
Along the given edges of the straightedge, e
e , draw two lines,
2
m and n.
1
and
Continue drawing as many
lines as necessary by placing e
1
along a previously
drawn line, and drawing another line along e
proof of this construction is self-evident.
2
.
The
52
CONSTRUCTION 2:
Double a given segment AB.
\
/
A
J
G
/----~----
;
/
/
c /
I
E.
-·---7-------+-D
m
Construction:
Extend AB.
to AB.
By Construction 1 draw line m parallel
Choose any point C on line m.
By Construction 1
mark any two equal lengths on m so that CD = DE.
0A
c
and ~
DB,
in G.
AG
intersecting in F.
= 2AB.
~
Draw FE,
Draw
~
intersecting AB
The proof of the construction follows
from similar triangles.
(Note that this construction can be repeated to
give any integral multiple of AB.)
53
Bisect a given segment, AB.
CONSTRUCTION 3:
'\
\
'\
m
Construction:
Using Construction 1 draw line m parallel to AB,
and determine points C, D, and E on line m such that
CD = DE.
·~
~
Draw CA and EB, intersecting in F.
intersecting AB in G.
~
Draw FD,
G is the midpoint of AB.
The
proof follows from similar triangles.
(Note that we can generalize this construction to
divide a segment into n equal parts.)
CONSTRUCTION 4:
line m,
Construct a line parallel to a given
through a given point A not on m.
~
54
Construction:
Using Construction 1, on line m mark points B, C,
and D such that BC
= CD.
Draw
tt
and mark any point
~
E on BA such that A is between B and E.
intersecting in F.
~
Draw AG.
~
Draw BF and ED,
Draw EC and AD,
intersecting in G.
~
AG is parallel to m.
(Note that this construction is similar to the
construction of Lemma 1 in the section on the straightedge alone.)
rn
D
Proof:
By Ceva's Theorem, BC/CD · DG/GE · EA/AB
Since BC/CD =1, DG/GE = AB/EA.
are in proportion,
Hence,
= 1.
since the sides
~
AG is parallel to line
ffi.
Q.E.D.
55
CONSTRUCTION 5:
A,
a)
Construct a line through a given point
and perpendicular to a given line m.
Point A is on line m.
m
A
~
n
p
1
i
t
r
I
Construction:
Using Construction 1, mark two points B and C, on
= AC.
line m at point A,
so that BA
BECD by placing e
and e 2 of the double-edged straight-
1
Construct rhombus
edge so that they lie on points B and C in a "positive
slope" direction.
Draw lines n and p along the edges.
Likewise draw lines rand t,
Let E
= n n t and D = p
to m at A.
rhombus.
1
r.
but with "negative slope".
~
Draw ED.
~
ED is perpendicular
The proof follows from BECD being a
56
b)
A is not on line m.
A
B
t
m
n
Construction:
Mark any point B on line m.
Using Construction Sa,
draw line n perpendicular to line m through B.
Construction 4 draw a parallel line t
A.
Then t
to line n,
is obviously perpendicular to m.
Using
through
57
CONSTRUCTION 6:
Bisect a given angle,
~ABC.
Construction:
~
Using Construction 1, draw m parallel to AB and n
~
parallel to BC, so that a rhombus is formed.
m n n.
~
Then BD is the bisector of
~
ABC.
follows from a rhombus being constructed.
Let D
The proof
=
58
CONSTRUCTION 7:
-------3>
AC of
4-
Copy a given segment AB from A along
BAC.
m
Construction:
Using Construction 6,
Using Construction Sb, draw a
(-----'>
AD,
AF
through B.
~
AB.
Let E
~
draw AD to bisect
p~rpendicular
~
~
BAC.
line m, to
~
= AD n m and F = AC n m.
Then
The proof follows from congruent triangles.
We have now presented the construction for
establishing Intersection Criteria 2 and 3.
For each of
these problems we are given circles as determined only
by their centers and a point on the circle.
one point is given on the circle,
Though only
it is possible to
construct any number of other points on the circle as
follows.
First double the given radius,
OP,
to diameter
59
PR.
Then through P draw any number of lines.
construct perpendiculars to these lines.
From R,
The intersect-
ions of the perpendiculars will be points on the circle.
This basic construction will work for all sections in
which we are not allowed to draw a circle:
straightedge
alone, double-edged straightedge, Mira, and paper folding.
We now turn to Intersection Criteria 2 and 3.
\
'\
\
INTERSECTION CRITERION 2:
Determine the points of
intersection, X and Y, of a given line m and a given
circle 0 (P) .
Construction:
~
Using Construction 4 draw OD parallel to line m,
' .
60
through 0.
Using Construction 7,
that OP
= OR.
side of
bD
~
copy OP onto OD such
Draw line n parallel to OD, on the same
as line m, using Construction 1.
a perpendicular line, s,
to m,
through R,
using
Let F = s n m and B = s (' n.
Construction Sb.
intersecting line n in E.
Construct
~
Draw bF,
Using Construction 4, draw
~
a line through E, parallel to RF,
~
intersecting OD in G.
(Note that OG/OR
=
greater than RF,
the perpendicular distance from OD to
m,
RB/RF.
Since OR,
the radius,
is
<':---4
OG is greater than RB,
the width of the double-edged
straightedge.)
Construct a rhombus by placing the edges of the
double-edged straightedge on points 0 and G in both a
Draw
"positive slope" and "negative slope" direction.
lines p,
H
r and v, w,
p(':v, K = nnr,
p
nn,
and N
=
w
along the edges.
X= mnr, L
nn.
points of intersection.
Let J
= r
0 w,
= vfln, Y = vf1m,
X and Y are the required
Q
61
---
s
w
/
/
\
/
i
J
v
\
D
\
m
x;
(
I
Proof:
By construction m is parallel to n,
so n divides
the sides of triangles OXY and OYF proportionally.
Therefore OX/OK
= OY/OL and OY/OL = OF/OE.
~
~
since GE is parallel to RF,
(1)
OF/OE
Likewise
OR/OG.
Hence
OY/OL = OR/OG.
OX/OK
OGMK and OGNL are obviously parallelograms and,
in fact,
w,
are rhombi since all altitudes are given by
the width of our straightedge,
Hence OK
=
OR = OP,
the points X and Y lie on 0 (P)
OL
=
and thus are equal.
OG,
so (1)
becomes OX
construction X and Y lie on m.
Q.E.D.
=
OY
=
By
OR.
Since
62
INTERSECTION CRITERION 3:
Determine the points of
intersection, X andY, of circle O(P)
and circle M(N).
Construction:
Using Construction 4 draw line m,
parallel to MN.
m so that ON'
OM as N.
Using Construction 7,
through 0,
construct N' on
= OP, where N' lies on the same side of
~
E---4
Draw NN' and OM,
intersecting in A.
He now
E--------7
locate R and R' , the other points in which N'N intersects
O(P)
and M(N),
using Intersection Criterion 2.
Using Intersection Criterion 2 determine the
points of intersection, T', S', T,
O(P)
andM(N).
in D.
~
Draw chords N'S'
and S, of OM and
and RT,
~
intersecting
Draw T'R' and SN, intersecting in C.
and D determine the radical axis,
Points C
the line through
the desired points of intersection.
X and Y are then
found using Intersection Criterion 2.
Note that if the two circles have the same radius,
~
NN'
"'---';>
is parallel to OM, but the radical axis is then the
perpendicular bisector of OM.
'
I~
'
\
/
64
Proof:
We show that T'R'NS and N'S'TR are each concyclic,
i.e.,
they are quadralaterals whose vertices lie on a
circle.
~
By construction, ON' is parallel to MN. Hence by
= m4 2 (refer to diagram) .
corresponding angles, mi.. 1
Since triangles S'ON' and SMN are isosceles with equal
vertex angles, m4- 3
=
m;:i 4.
is inscribed in a circle,
Since quadralateral T'R'N'S'
its opposite angles are
supplementary, thus m4- 3 + m;;_ 5
.
m u 4 + mlL 5
= 18 0 0 .
= 180°.
Therefore
Since opposite angles are
supplementary, quadralateral T'R'NS is concyclic.
Quadralateral TRNS is inscribed in a circle and
therefore m4- 4
= m:::'+. 6
supplementary, m4- 4
one side (N'T)
angles
( 4 3 and
(RandS'),
Since angles 6 and 7 are
mti.. 7.
Hence miL 3
= m4 7.
Since if
of a quadralateral subtends congruent
4
7)
at the two nonadjacent vertices
the quadralateral is concyclic.
(Briefly,
this can be proved by an indirect argument that shows R
cannot lie in the interior or the exterior of the circle,
through T, S' and N' ,and thus must lie on the circle.)
Therefore N'S'TR is also concyclic.
~
We need to show that CD is the radical axis of O(P)
and M(N), where the radical axis is the line through
the points of intersection of the two circles.
To
65
accomplish this we first define the "power of a point"
with respect to a circle as the product of its distances
from any two points on the circle that are collinear
with the given point (Shively, p. 69).
Refering to the
following diagram, the power of P with respect to O(R)
is PQ · PR or PS · PT.
The "power" of a point with
respect to a circle is constant.
To prove this consider
the following diagram.
p
s
T
4
Since
QRS and
~
QTS intercept the same arc, triangles
PRS and PTQ are similar.
PS · PT = PQ · PR.
Hence PS/PQ
=
PR/PT, so
(Note the power of a point is constant
whether P is inside (see second diagram) or outside the
circle.
If a point lies on the circle its power is zero).
The power of a point may also be defined in terms of
the distance, d, from the point, P, to the center of the
66
If the line goes through the center, we can
circle.
clearly see that the power of a point on the line is
+ r) (d - r), or d
(d
2
- r
2
.
Since the power of a point
with respect to a circle is constant,
this expression
holds for all lines drawn through P.
If the powers of a point with respect to two circles
are equal,
then the point lies on the radical axis.
To prove this first consider the following diagrams,
I
p,
#
/\
'
" /I,\ \
a /II ' " ' ,
p
// . \
d
I .\
ft;\~~
(
0;',/
\~ '-:·01 \
\
\ /_____ _
/
where,
the center of circle 0, with radius r,
coordinates
r
1
,
(a,b); the center of circle
is given by
is given by
o 1 , with radius
(a ,b ); and P is given by
1 1
(x,y).
If the
power of P with respect to circle 0 is equal to the power
o1 ,
of P with respect to circle
Now d 2
Hence
=
(x-a)
2
+ (y-b)
( x-a ) 2 + ( y- b) 2 -
2
and
r2
=
then d
2
- r
2
= a1 2 - r 1 2
2
2
(x-a )
+ (y-b ) .
1
1
( x-a ) 2 + ( y-o
' )2 - r 2.
1
1
1
a12 =
67
Solving,
this becomes:
or Ax + By + C
= 0.
Hence the locus of points with equal
powers with respect to the two circles is a line.
Now consider the case when two circles intersect.
One point of intersection, X, which lies on the radical
axis,
is given by the coordinates
circle 0, with radius r,
(x,y).
is given by
The center of
(a,b); and the center
of circle o , with radius r , is given by (a ,b ).
The
1
1
1 1
2
2
equation of circle 0 is (x -a)
= r 2 , and for
+ (y- b)
circle
o1
the equation is
The simultaneous solution of the 2 equations yields:
,-
2
2
(2a -2a)x + (2b -2b)y +[(a -a ) +
1
1
1
or Ax+ By+ C = 0,
(r
2
1
-r
2
-j
)_:
0
the equation of the radical axis.
Hence the equation of the radical axis is identical to the
equation of the line whose points have equal powers with
respect to both circles,
i.e.,
if the powers of a point
with respect to two circles are equal,
the radical axis.
the point lies on
68
-E----7
We now show that CD is the radical axis.
of C with respect to O(P)
to M(N)
is CR'
the power of C is CN ·
is concyclic, CR'
The power
CT' and with respect
cs.
· CT' = CN · CS.
However since T'R'NS
Hence the power of
C with respect to both circles is equal, and so C lies
on the radical axis.
Likewise since N'S'TR is concyclic
the power of Dis DT · DR= DN'
Therefore D lies
"'-0
on the radical axis.
Hence CD is the radical axis, and
by Intersection Criterion 2,
X and Y,
· DS'.
the points of intersection,
of the radical axis and either circle may now
be determined.
Q.E.D.
In conclusion,
since we have been able to determine
the points of intersection of any two lines, any line and
any circle, and any two circles,
the double-edged
straightedge is adequate to perform all Euclidean
constructions.
PROOF OF ADEQUACY:
MIRA
The Mira, a transparent plastic reflector,
recent tool,
Gillespie.
is a
invented by George Scroggie and N. J.
We will show that the Mira is adequate to
perform all Euclidean constructions by determining the
points of intersections of 1) any two lines,
line and any circle,
and 3)
INTERSECTION CRITERION 1:
2)
any
any two circles.
Determine the point of
intersection of any two lines.
The
edg~
of a Mira acts like a straightedge, so
Intersection Criterion 1 is iiTmediately solved.
Since a Mira cannot be used to draw a circle,
a
circle is given by its center and a point on the circle.
It is very easy, however, to construct any number of
points on the circle using the Mira.
placed on the center of the circle.
The ;H ra is
Then the reflected
image, P', of the given point P on the circle,
the circle.
lies on
As will be shown later, the line drawn
along the Mira's edge is the perpendicular bisector of
70
pp
I
•
Since the line drawn passes through the center of
the circle, PP' is a chord of the given circle.
P'
Hence
is on the circle.
The Mira can be used to locate the reflection
image A' of A with respect to a line mas follows:
place the drawing edge of the Mira on line m and mark
the image A' of A,
as observed through the face of the
The notation r
Mira.
m
the reflection point.
only if r
m
(A')
(A)
= A' will be used to denote
Note that r
m
(A)
= A' if and
=A.
Anticipating Intersection Criteria 2 and 3, we
shall perform the following constructions:
1)
Given a segment,
copy
it onto another given
line.
2)
Given a line m and any point C,
construct a
line perpendicular to m through C.
3)
Given a segment,
construct its perpendicular
bisector.
4)
Given two segments, add their lengths.
5)
Given two segments,
6)
Given three lengths,
proportional.
subtract their lengths.
construct the fourth
71
CONSTRUCTION 1:
D on
tr
Given segment AB and line
so that AB
tJt,
locate
= CD.
B
~
E
m
B'
n
Construction:
Draw line m that reflects A onto C and mark 8'
such that r
m
(B)
= 8'.
Now place the Mira's edge on
C and locate the reflecting line, n,
is an element of CE.
AB
Then AB
= CD.
such that r
n
(B')
This follows from
= CB' and CB' = CD.
(Note that by observing the reflected image from
opposite sides of the Mira, we can choose D to be on
either side of c.)
D
72
CONSTRUCTION 2:
Given any line
XB and
any point C,
~
construct a perpendicular line, m,
to AB, through C.
c
,.
A
i
)D
•B
.
tc
!
I
A
B
I
I
I
I
m
m
Construction:
Place the Mira on C so that r
line is reflected
o~to
itself.
~
Mira's edge and let D = AB
n m.
~
line m is perpendicular to AB,
~
m
(AB)
Draw
~
= AB, i.e., the
li~e
m along the
Then it is obvious that
through C.
(Note the
~
construction applies whether C lies on or off AB.)
CONSTRUCTION
3:
Given any segment AB,
perpendicular bisector,
line m,
to AB.
construct the
73
B
A
m
Construction:
Place the Mira so that r
along the Mira's edge.
bisector of AB,
CONSTRUCTION 4:
m
(A)
= B.
Draw line m
Line m is the perpendicular
as is obvious.
Given segments AB and CD, determine
EF on given line m,
such that EF
= AB + CD.
(See diagram on following page)
Construction:
Using Construction 1, copy AB onto line m such
that AB
= EG.
Again,
using Construction 1,
copy CD
74
;.;A:.__ _ __.;B
c
D
m
onto m such that GF
of G from E.
= CD, and F is on the opposite side
Then EF
CONSTRUCTION 5:
F
G
E
= AB + CD.
Given segments AB and CD, with AB
determine EF on given line m, such that EF
A
C
>
= AB - CD.
B
D
E
F
G
m
CD,
75
Construction:
Using Construction 1, copy AB onto line m such
that EG
= AB.
Again, using Construction 1, copy CD
= CD, and F is between E and
onto line m such that FG
G.
Then EF
= AB - CD.
CONSTRUCTION 6:
Given segments of lengths a,
construct a segment of length d such that a/b
b,
= c/d.
b
a
c
n
I
I
I
i
c
c
Bl
D
d
b
t
A
I
m
I
I
r
c,
76
Construction:
Draw any line m and choose any point A on m.
Using Construction 1 and 4, copy the segments with
lengths a and b onto line m such that AB + BC
Draw any other line, n,
through C.
= b + a.
Copy the segment
Draw BD.
with length c onto n such that CD = c.
Through any point on BD construct a perpendicular
line, r,
to BD, using Construction 2.
Throuc;:rh A, draw
,L___>
line t perpendicular to line r.
DE is the desired length d.
Let E =
CD n
t,
then
The proof follows from
the parallelism of BD and t.
(Note that in the presence of a unit length, if
a= 1,
then d =be,
of two segments.
i.e., we can multiply the lengths
If c
b
= 1 then d =a'
i.e., we can
divide the lengths of two segments.)
We now turn to Intersection Criteria 2 and 3,
using Construct·ions 1 - 6.
INTERSECTION CRITERION 2:
Determine the points of
intersection, X and Y, of given line m and given
circle 0 (P) .
77
Construction:
Draw a line n through 01
the given line m; let X
0
1
that reflects P onto
= r n (P).
that reflects X onto m; let Y
-\
Draw line t
through
= rt(X).
i
/
-
' '
... p
q
I
I
\\0
-
n
\
m
/
t
Proof:
Line n is the perpendicular bisector of PX and
passes through the center 0.
Hence PX is a chord of
0 (P)
I
so X lies on 0 (P)
Likewise XY is a chord of
O(P)
I
soY lies on O(P)
Hence X and Y are the required
points of intersection of line m and circle O(P).
INTERSECTION CRITERION 3:
intersection,
X andY.
Determine the points of
of given circles O(P)
and M(N)
78
p
'
•
/
I
"
'
'
/'
;
\
/
' N
,,
IS
'
'
0
'
Q
M
1,'
I
'
'
ly
'
Ij
II
As in the case of the straightedge alone and the
double-edged
straighted~e.
this problem is reduced to
Intersection Criterion 2 by determining the radical
axis of O(P)
and M(N),
the line through the desired
points of intersection.
To construct the radical axis we locate the point
Q,
the point of intersection of OM and the radical axis.
At Q we can then construct a perpendicular line to
obtain the radical axis.
Q is determined as follows:
Theorem,
( 1)
OX
2
-
OQ
2
By the Pythagorean
= MX 2 - MQ 2 .
If Q is between
79
0 and M, MQ
= MO - OQ.
(If Q is not between 0 and M
a similar expression to the following may be obtained.)
Substituting into ( 1) and solving for 20Q we obtain:
2
2
ox 2 + M0MX
20Q
MO
MO
MO
Constructions 4 - 6 show that we may construct this
function of the given lengths, as desired.
Since we have been able to determine the points
of intersection of two lines, a line and a circle,
and two circles, the Mira is adequate to perform all
Euclidean constructions.
We note that if we are given
a unit length, the Mira is adequate to perform
multiplication, division and formation of square roots.
PROOF OF ADEQUACY:
PAPER FOLDING·
Paper folding is another construction tool which is
adequate to perform all Euclidean constructions.
only materials needed are pencil and paper.
(p. 1)
The
Johnson
found that heavy wax paper is the most appropriate
paper to use,
as a crease becomes a distinct white line,
and its transparency simplifies superposition.
However,
white tissue is the most appropriate to use as it is
also possible to easily write on the paper.
An advantage
to doing some Euclidean constructions by paper folding is
that congruence is easier to "see" and verify, as it is
for the Mira.
In paper folding the following assumptions and
definitions are made (Olson, p. 2):
1)
Paper can be folded so that the crease formed
is a straight line.
2)
Paper can be folded so that the crease passes
through one or two given points.
3)
Paper can be folded so that a point can be
made coincident with another point on the
same sheet.
4)
Paper can be folded so that a point on the paper
can be made coincident with a given line on the
same sheet,
and the resulting crease made to
20
81
pass through a second given point which is not
in the interior of a parabola that has the first
point as focus and the given line as directrix.
(This restriction applies because the parabola
is the "envelope" of all the folds that bring
the first point into coincidence with the given
line.)
d
5)
Paper can be folded so that straight lines on
the same sheet can be made coincident.
6)
Lines and angles which can be made coincident
by folding paper are congruent.
The fact that paper folding is adequate to perform
all Euclidean constructions now follows from the fact that
all Mira operations can be duplicated by paper fold1ng.
We need only replace the instruction of "placing the Mira"
by "fold and crease".
CHAPTER 3:
GEOMETROGRAPHY
Alternate tools of constructions to the traditional
compass and straightedge have been presented for the
performance of all Euclidean constructions.
natural question arises:
to use?
The
Which tool is the best one
Among other things,
the answer depends on the
number of "actions" needed to complete the construction,
i.e.,
the number of steps in the construction.
In
this chapter we shall take a particular construction,
namely,
bisecting a line segment,
and compare the
different tools for this construction.
In 1907 Emile Leomoine devised geometrography
to compare different geometric constructions using
the straightedge and compass.
following operations
s1:
Leomoine considered the
(Hobson, p.
113):
make the straightedge pass through one'
given point,
s2:
rule a straight line,
c1 :
make one compass leg coincide with a given
point,
c2 :
make one compass leg coincide with any given
point of a given line
(Note the distinction
between cl and c2 is that for c2 the point
22
83
of coincidence lies on a given line),
c3 :
describe a circle.
If the above "actions" are performed m , m , n , n , n
1
2
2
3
1
respectively, then m s + m s + n c + n c +
1 1
2 2
1 1
2 2
times,
n c is the "symbol" of the construction; m + m +
3 3
1
2
n
1
+ n
2
+ n
\
is the "coefficient of simplicity".
3
The
"coefficient of exactitude", -i.e., the total number
of alignments rather than drawings,
(Hobson, p.
113-114).
is m
1
+ n
"As Leomoine remarks,
1
+ n
2
these
coefficients really measure, not the simplicity and
the exactitude, but the complication and chance of
error."
(Hobson, p. 114).
For this paper the term
"complexity" will be used rather than "simplicity'',
as it better conveys the meaning of the problem.
The symbol indicates only the final result of
all constructions, not the order in which they occur.
Therefore it is possible for two different constructions
to have the same symbol.
As an example of these terms
consider drawing a straight line through two points.
The symbol is 2s
1
exactitude is 2.
+ s
2
, the complexity is 3, and the
Another example would be to draw a
circle given the center and a point on the circle.
The symbol is
2c 1
+
exactitude
2.
Note that if one compass leg is made
lS
c3 ,
the complexity is 3, and the
84
coincident with a given point on a given line the symbol
is
c 1 + c2 + c 3 .
If both legs are made coincident with
given points on given lines,
the symbol is
2c 2 + c 3 .
Leomoine made a comparison of geometric constructions
based only on the number of "actions" needed to complete
the construction.
however.
These should not be the only criteria,
Other questions to consider are the accuracy
of the tool or its ease of use in relationship to the
goal of the construction.
For example, one of the
purposes in finding constructions that used only the
straightedge was for surveying, where a compass is not
easily used.
Likewise the compass alone was often chosen
for the purpose of astronomical instrument
~aking.
Although Leomoine's geometrography was invented
for the compass and straightedge, we will use it to
compare the different tools for the bisection of a
line segment.
(Note the following constructions
appear to be the most efficient for each tool being
considered, so that we may compare the efficiency of
the tools for the same construction.)
85
Bisect a given segment AB,
CONSTRUCTION 1:
using the
straightedge and compass.
Construction:
Draw A(B)
and B(A)
each circle is 2c
T.
Draw ST (2S
midpoint.
1
2
(the symbol for constructing
+ c ) intersecting in points S and
+ s
3
2
),
intersecting AB in C,
the
The proof follows from congruent triangles.
(Note that the symbol for each circle drawn is
2c
2
+ c , not 2c + c , since the compass legs were made
3
3
1
coincident with points of AB.)
..
-------
1
I
I
I
-~
The symbol for this construction is 2s
4c
2
1
+ s
2
+
+ 2c , with a complexity of 9 and an exactitude of 6.
3
86
CONSTRUCTION 2:
Bisect a given segment AB, using the
rusty compass and straightedge.
Construction:
Using an arbitrary fixed compass opening, XY, mark
off equal lengths on AB, starting at both A and B, until
the distance between F and G is less than XY.
(Each arc
drawn is represented by the symbol:
Draw F(XY)
and G(XY), intersecting in S and T (2 (c
(2s
1
2
+ c )).
3
Draw ST
+ s ), intersecting FG in C, the desired midpoint.
2
The proof follows from congruent triangles and segment
addition.
s
T
The symbol depends upon the length of XY; the smaller
the fixed compass opening, the greater the complexity.
For the lengths AB and XY shown, the symbol is 2s
6c
2
1
+ s
2
+
+ 6C , the complexity is 15, and the exactitude is 8.
3
If the fixed compass opening is greater than AB,
The
co~plexity
the
is 7,
less than that of the straightedge and compass in
Construction 1.
~
.
87
Bisect a given segment AB
CONSTRUCTION 3:
compass alone.
1
using the
AB is given only by its points A and B.
Construction:
Draw B(A) and A(B)
(2(2c
point of intersection as P.
B(A)
starting at
I
PI
compass opening (2(2c
(2(c
+ c
1
2
intersection is C
3
I
labeling either
Mark off two more arcs on
+ c )).
3
1
Label the points of
Draw S(A)
in T and V.
+ c ))
3
+ c ))
and using AB as the width of the
intersection RandS.
intersecting A(B)
2
(c
1
+ c
2
Draw T(A)
+ c )
3
1
and V(A)
I
where their second point of
1
the midpoint of AB.
Proof:
By construction,
equilateral.
triangles ABP, PBR,
Therefore angles ABP,
60°, ~ABS is a straight angle,
S are collinear.
triangles,
coincident.
PBR,
and RBS are
and RBS are
and points A,
B, and
SVA and VAC are similar isosceles
since a pair of their base angles are
Hence AS/AV = AV/AC.
AS= 2AB this becomes AC = AB/2.
midpoint of AB.
Q.E.D.
Since AV = AB and
Hence C is the
88
(
T
\
The symbol for this construction is 7c
with complexity of 21 and exactitude 14.
1
+ 7c
2
+ 7c ,
3
Eliminating
the straightedge from the compass and straightedge,
1
increases the complexity by a factor of 23.
89
CONSTRUCTION 4:
Bisect a given segment AB,
using only
the straightedge and a given fixed circle with center F.
Construction:
we will consider only the first of two cases:
a)
A,
B,
F not collinear, and b) A,
Through F,
~
~
draw AF and BF,
circle in R and S, and P and Q,
B, F collinear.
intersecting the fixed
as shown.
~
Draw QR
-('--'>
and SP,
~
EF,
intersecting AB in D and E respectively.
~
intersecting QR in H.
in G.
~
Draw DF,
Draw
~
intersecting SP
~
Draw GH.
~
Referring now to the second diagram, draw AH and
___,.
BG,
intersecting in J.
in K.
~
Draw JK,
-E:-->
~
Draw AG and BH, intersecting
~
intersecting GH in L and AB in C.
C is the midpoint of AB.
90
)~
,(:'
/I;
/
/
I \
;
/ I
I
II
H/I
I
;·
L
I..
\
I
I
\
. :G
,.f
/ _\--kN.
~\. // \\·.
// /il
i \
"·.
!• \
iF
I\\
////
I
/ \\ / /
/
,
1//v\
/
A'lt.
'!
i
I
\
\
\
B
91
Proof:
The proof is as follows.
Since PQ and RS bisect
each other, QRPS ia a parallelogram and PS is parallel
=
Thus m 4 FQH
to RQ.
are congruent.
m 4- FPE,
Thus FH
and triangles FQH and FPR
= FE and FG = FD, so DG and EH
Therefore DEGH is a parallelogram,
bisect each other.
and HG is parallel to DE,
so HG is also parallel to
Thus triangles JHL and JAC are similar and
to AB.
triangles JLG and JCB are similar.
and JL/JC
Hence HL/AC = JL/JC
= GL/BC, so (1) AC/BC = HL/GL.
Trianlges
HLK and BCK are similar and triangles GLK ar.d ACK are
similar.
so (2)
Therefore BC/HL
AC/BC
= GL/HL.
= CK/LK and AC/GL = CK/LK,
Multiplying
AC 2 /BC 2
= 1 or AC/BC = 1.
of AB.
Q.E.D.
(1) and (2) we get
Hence C is the midpoint
There were 12 lines drawn through two points each
with a symbol of 2s
construction is 24S
1
+ s2.
1
+ 12s 2 , the complexity is 36, and
the exactitude is 24.
for the given circle)
four times.
Hence the symbol of this
Eliminating the compass
(except
increases the complexity by
92
CONSTRUCTION 5:
Bisect a given segment AB, using a
double-edged straightedge with parallel sides.
Construction:
Make one edge of the double-edged straightedge
coincident with AB and draw line m along the other edge
Draw a line n through A,
intersecting line
~
For efficiency use AD as alignment of
e , and using the ''width" 'Of the double-edged straight1
edge, determine points E and F on m such that DE
~
= EF
~
+ s ), intersecting AD in G.
2
-E----4
Draw GE (2S + s ), intersecting AB in C, the midpoint
1
2
(2Sl + 2s ).
2
of AB.
Draw FB
(2S
1
The proof follows from similar triangles.
n
m
The symbol of this cbnstruction is 9s
complexity of 15,
and exactitude of 9.
the straightedge alone,
1
+ 6S , a
2
In comparison to
the double-edged straightedge
reduces the number of "actions" by slightly more than
half.
93
CONSTRUCTION 6:
Bisect a given segment AB, using the
Mira.
Place the Mira across AB such that B
= r m (A) .
Draw
line m along the Mira's edge, intersecting AB in C.
By reflection C is the midpoint of AB.
B
rn
Leomoine's geometrography must be modified slightly
for the Mira.
Drawing along the Mira's edge is
analogous to using a straightedge.
B coincident, through reflection,
Making points A and
is analogous to making
a straightedge or compass coincident with a point.
arbitrarily label this "action" as
for this construction is
and the exactitude is 1.
s1.
We
Hence the symbol
s 1 + s 2 , the complexity is 2,
p '
94
CONSTRUCTION 7:
Bisect a given segment AB, by paper
folding.
Fold the paper so points A and B are coincident.
Crease the paper along the fold,
intersection of m and AB is
c.
forming line m.
The
By superposition, C
is the midpoint of AB.
rn
The same modifications apply to paper folding as
they did for the Mira.
Hence the symbol is
s1 + s2.
For problems other than the one just considered,
this hierarchy of complexity could easily change.
Also
the choice of tool should not depend solely on the
complexity,
desired,
but also on factors such as the accuracy
the purpose of the construction, and the tools
95
that are available.
"Further,
the construction which is
simplest for a certain problem standing alone may not be
the simplest when the same problem occurs as part of
another; for a construction less simple in itself may
fit in better with the context, by making use of lines
which are already drawn or which can be used later on in
the course of the construction of the more complicated
problem."
(Hobson, p.
114).
The importance of this chapter is not just to
compare constructions quantitatively,
but to show there
are various methods to perform any given construction.
These alternate approaches need to be stressed with any
geometry class,
for too often students feel that the
only method of construction is with a compass and
straightedge.
P..s Bussey states
(p.
2 80) ,
"If college
teachers do not give their students more of the material
of which I have been talking
than they have in the past,
(using different tools)
the next generation of high
school teachers will be as fully bound by the classical
tradition as their predecessors; and they will have
missed a lot of interesting subject matter that is very
definitely related to their job as teachers of
elerr,en tary geometry. "
CHAPTER 4:
"MORE POWERFUL TOOLS":
ANGLE TRISECTION MADE POSSIBLE
Some of the tools presented in this paper are "more
powerful",
in the sense. they can achieve constructions
that are impossible using the straightedge and compass.
We will show that the Mira and paper folding are capable
of trisecting any angle,
a construction impossible with
straightedge and compass.
For two thousand years mathematicians and others
tried,
in vain,
and compass.
to trisect an angle with straightedge
The key to finally proving the impossibi-
lity of this construction with Euclid's tools,
lies in
translating the geometrical problem into the language
of algebra.
-~
Using the straightedge,
and given lengths a,
b,
and a unit of length 1, we can construct lengths
corresponding to a + b, a -
b,
ab,
a/b, with the
construction of the first two expressions being obvious.
The multiplication and division of lengths is based on
similar triangles,
following page.
as shown in the figures on the
97
a+b
a
'----------,
a
b
1
b
, ___...,a -------..,.
a-b
b
b
Therefore, starting with the unit length, we can construct
all rational numbers using these constructions.
that the sum, difference, product,
rational numbers is rational,
form a field,
We know
and quotient of two
and so the set of rationals
F .
0
We now show that the use of the straightedge alone
can never yield lengths for numbers outside F .
0
If we
are given a set of lines which have equations with
rational coefficients and points with rational
coordinates, we construct new points and lines by 1)
two points already given to form a line, or 2)
using
considering
the intersection of two lines to be a new point for us
to use.
We now see that each of these new objects will
continue to have rational coefficients or coordinates.
98
The line between two points
by x - a
f(b-
d)
I
(a,
b)
and (c, d)
is given
(y - b) , which
(a- c)]
simplifies to
x(a- c)
If a, b,
+ y(d- b)
+b(b- d) + a(c- a)
c, d are rational,
0.
the coefficients of the
equation for our new line will be rational.
Now the
simultaneous solution of two linear equations with
ax + by + c = 0 and
rational coefficients,
dx + ey + f = 0 yields,
X
(bf - ec)/(ae - bd)
andy
(cd - af)/(ae - bd).
The point of intersection of these two lines has
coordinates that are clearly rational functions of the
coefficients in the equations of the lines.
Therefore
the use of the straightedge leads to no operations
other than rational operations on given lengths.
If we add the use of the compass to the use of the
straightedge we may form an extension field of F
diameter.
AB
Draw a semi-circle with (1 + c)
as the
and erect a perpendicular at the point where
and the unit length meet.
triart<=cJles.
First
·vC, if we are given a segment
we show how to construct
AB of length c.
0
1/z = z/c.
so
x
=
Based
;,/c .
99
Now all numbers of the type a + b
butv~ ¢
compass.
F
0
I
Jc,
where a,
b,
c E F0
I
can be constructed using straightedge and
Performing the rational operations on quantities
of the form a + b 1/c also yield quantities of this form
and therefore numbers of the form a + b
extension field,
v7 form
an
F .
1
Now repeated use of the compass and straightedge
allows us to build complicated constructions like
-J a
+ \(-;;;-;--c where these quanti ties are constructible
if a,
b,
c,
are given lengths.
Complicated as these
expressions may seem they are only a series of square
roots of constructible lengths.
A + B
'./c to
We will use the symbol
represent these quanti ties, where A,
B,
c
are constructible quantities and ~Jc is a quadratic
irrationality of higher order than A and B.
F
1
is the
set of numbers arising from any finite combination of the
rational and square root operations.
We will call them
quadratic numbers for ease of description.
Now we must show analytically that the use of the
straightedge and compass will not yield lengths other
than those composed of r2tional numbers and expressions
of the form A + B '/C.
intersection of 1)
To accomplish this we look at the
a line and a circle,
and 2)
two circles.
100
First, the simultaneous solution of a linear equation
and a quadratic equation of a general circle, given by
ax+ by+ c
=
0 and
(x- h)
2
+ (y- k)
2
=
r 2 yields:
(a 2 +b 2 )x 2 + 2(ac-hb 2 +abk)x + [2
c + 2bck + (h 2 +k 2 -r 2 )b
or Ax
A,
2
+ Bx + C,
JB 2 -
B-vc_
b,
(If
not intersect.)
4AC)/2A, where
c and h, k,
JB
2
-
41\C
<
r,
0,
and x is of the form
then the circle and line do
The same process will yield y of
similar form.
We now consider the points of intersection of two
2
circles given by (x
h)
(x- f)2 + (y- g)2
= s2.
+ (y - k)
2
= r 2 and
The simultaneous solution
yields:
-2 2
2 2 2 2]
(2f-2h)x + (2g-2k)y + h -f + k-g +s -r
l
or Ex + Fy + G
Hence,
0.
=
0
tr:e intersection. of ·two
circles yields a linear equation, where E,
F, G are
clearly rational functions of the given quantities
h,
0,
B, C are clearly rational functions of the given
quantities a,
A+
= (-B +
so that x
2] =
k,
r, and f,
g,
s.
The problem now reduces to
determining the points of intersection of this line with
either circle, which has previously been demonstrated.
~
'
101
Analyzing the points of intersection of a line and a
circle or 2 circles, whose coefficients are rational,
yields only points with coordinates in F .
1
In summary we have demonstrated that the straightedge
and compass together are capable only of constructing
lengths that are obtainable by a finite number of the
operations of addition, subtraction, multiplication,
division,
and the extraction of positive square roots
of given lengths.
We will show that trisecting an angle would produce
points with coefficients not of the forD A +
By
C,
and
therefore it is impossible to trisect an arbitrary angle
using the straightedge and compass.
Consider an angle of 60°.
cos 38
9
SO
=
From the identity
= 4cos 3 G - 3cos8, we obtain, letting 39 = 60°.
0
20 ,and X= 2cos8:
x
0
3
-
3x -
1
Suppose the equation had a rational root in the field
F
0
.
= a/b, where a and b are relatively prime
Let x
integers and b
a3 -
3 a b2
=
-1.
Then
0.
= b3.
3
divides b .
a
*
(a/b)
3
-
3(a/b)
-
1 = 0.
Therefore a
Since a and b are relatively prime,
Similarly,
or
a
3
'o 3 + 3ab 2 , or a 3 =
o2 ( b
a = 1 or
+ 3a) .
~
102
Thus b divides a
prime,
=
b
3
Again,
.
1 or b
=
since a and b are relatively
Hence the only possible rational
-1.
x, is 1 or -1, and we can check that these are not
root,
roots.
= A +
Consider further a root of the form x
with A,
C,
B,
C f
B
-Jc,
F , but with A and B of lower order than
1
i.e., involving fewer applications of the square root.
Since the coefficients in the equation x
are rational,
the conjugate, A -
B
lc-,
3
-
3x -
1
=
0
should also be a
Now in a cubic equation if
root of the equation.
Hence the sum of the roots is the coefficient of the x
t.erm.
Thus,
for x
3
3x -
-
1
=
0,
which
0,
yields x
Kow
= -2A.
3
A' + B V C
1
and
must either be rational or of the form
~
1
,
where A'
and B' are of lower order than
is of lower
is of the form A
I
+ B rc~
I
craer than
If A
B \}C.
then a repeat of the
preceeding argument will yield x3
=
either rational or of the form A" +
and D" are of lower
+
A
order than
-
2
tJ
It
L>
>
!..---
I
.
~< o~,.r
Jc"
\.' r~" ar.C'~
!'.
I
is
'.111e re i\11
\/'(":_-:.
1s
of
2
103
lower order than A 1 + B 1
\/
c
1
•
Continuing this chain of
reasoning leads us to the conclusion that there is a
rational root.
However,
this has been shown to be
impossible and hence A + B
,;c
is not a root of x 3 -3x-l=O.
The following picture shows that if we could
construct a 20° angle,
could construct x,
then 2·cos 20°
x, so that we
a number shown to be not in F .
1
straightedge and compass
co~structions
Since
only produce
points with coordinates in F , we deduce that we cannot
1
trisect a 60° angle.
We now show that the Mira and paper folding are
more powerful than the straightedge and compass, by
demonstrating that they can trisect any angle.
Martin
(p. 205)
In fact,
indicates that the Mira can be used to
perform all constructions whose analytic representation
produces a cubic or quartic equation.
The construction and proof for trisecting any angle
using the Mira is based on an article by Lott and Dayoub
(pp.
398 -- 399).
They give credit to Pappus for the idea
for the construction.
Pappus'
in today'::::; notdtior: is
(Lott c::.nc:i
Theore~
(A.D.
:Jayo~_:;.:;,
p.
300),
398):
stated
104
----?
Given4.BAC, on one side, AB, mark off an
arbitrary line segment AD.
Through D draw two
---?-
lines, n parallel to AC and m perpendicular to
____,.
AC.
If a line segment HF of length 2AD is
fitted between these two lines m and n so that
its extension beyond H passes through A,
m 4 BAC
then
= 3m . ~FAC.
It is the "fitting" that cannot be accomplished using
only the straightedge and compass.
This ''fitting", or
"verging", as it is also called, is more precisely
described and accomplished using the Mira's ability to
We now present the construction and proof.
reflect.
CONSTRuCTION 1:
Trisect a given angle,
4
BAC,
using
a Mira.
Construction:
~
Choose any point D on AB.
----4
reflect A onto AB; E = r
----4
perpendicular to AC.
m
(A).
Place the Mira on D and
Through D draw line m
Also through D draw line n
----4
perpendicular to m.
Thus AC is parallel to n.
Locate a reflecting line s that reflects E onto
line m and reflects h onto line n (this is "verging")
Let F = r
~
s
(A)
and G
=
rr(E)
0
Draw AF,
intersecting m
105
in H.
~
AF trisects 4- BAC.
(Segments EG and FG have
been drawn only to aid in the proof.)
B
n
H
\
\
\
\
c
s
106
Proof:
Let A
F
4 =
4
4
1
FAD, F
3
=
AFD, and
~
DFG.
Since reflection preserves measure, m(A )
2
Since EG is parallel to AF, m(A )
2
=
m
~
= m(F 3 +F 4 ).
AEG, thus tri-
angles ADH and EDG are congruent, and DH
= DG.
It
follows that right triangles GDF and HDF are congruent
Thus
-------"'
Therefore AF
trisects
4 BAC.
Q.E.D.
We now turn to trisecting an angle by paper folding.
Since both a Mira and paper folding use the idea of
reflections in their constructions, and the Mira can
trisect any angle,
co~ld
it is reasonable that paper folding
do the same.
Cot·~STRCCTION
2:
Trisect a given angle,
~
BAC,
by paper
folding.
Construction:
~
Choose any point D on AB.
Fold the paper at D, so
·~
that AB is superimposed upon itself; E
~
rfold(A) ·
AC onto itself such that the resulting crease,
passes through D.
Fold
line m,
Thus ~AC is perpendicular to m.
Fold
107
m onto itself such that the resulting crease, line n,
passes through D.
Thus n is perpendicular to m.
Hence
~
n is parallel to AC.
Now fold the paper so that E is coincident with m and
n passes through A.
F
=
r
s
(A) and G
=
r
Crease the resulting line, s;
s
(E) .
----
Draw ~
AF; AF trisects
4
BAC.
This is the same construction as with the Mira, so
the previous diagram and proof hold.
It is also possible to trisect an angle using a
double-edged straightedge with perpendicular sides, such
as a carpenter's square.
Yates (Mathematical Sketch Book,
p. 158) performs this construction, along with doubling a
cube, another famous problem which is impossible with a
straightedge and compass.
George Martin demonstrates
how to duplicate the cube using the Mira, so it would seem
that paper folding can duplicate a cube.
It seems that
the ability of the Mira, paper folding, and the doubleedged straightedge, to trisect any angle or duplicate a
cube,
lies in their ability to construct the conic
sections.
The ancients knew that by the use of the conics
these two famous problems could be solved (Lott, p. 394).
In fact, if a rational conic is used in place of the
given fixed circle, the straightedge alone is capable of
solving problems nf the third and fourth degree, e.g. it
could trisect any angle (Archibald, p. 192).
CHAPTER 5:
CONCLUSION
The adequacy of 1) the rusty compass and straightedge,
2)
the compass alone,
3)
the straightedge alone
given one fixed circle with its center,
4)
edged straightedge with parallel sides,
5) the Mira, and
6) paper folding,
the double-
to perform all Euclidean constructions
has clearly been demonstrated.
As Jacob Steiner wrote
(Archibald, p. 252-253):
I take this opportunity to add also the
following observation:
It appears that in
general up to now too little care has been
bestowed upon geometrical constructions.
The
customary method, handed down. to us by the
ancients, according to which indeed problems
are r~garded as solved as soon as it is
indicated by what means they may be reduced
to others previously treated, is a great
hindrance to the accurate, critical examination
of what their complete solution requires.
Thus
it happe~s that often in this way constructions
are given, which, if it were necessary to
carry out actually and exactly all that they
include, would soon be given up, since thereby
one would speedily be convinced that it is a
very different matter actually to carry out
the constructions, i.e., with the instruments
in the hand, than it is to carry them through,
if I may use the expression, simply by means
of the ton~ue.
It is indeed easy to say; I
shall do that, and then that and then this;
but the difficulty, and we may say in certain
cases the impossibility, of really completing
constructions which are hi~hly complicated
requires us to weigh carefully in a proposed
problem which of the various processes is the
simplest for the complete construction, or
which is the most suitable under special
circumstances.
10?
109
The point made in this quote is valid,
for most of
the constructions and proofs in this paper could be
understood by a high school geometry student.
However,
it is not the actual performance of a construction
that is important.
The wonder and importance lie in
the fact that with these other tools,
some of which are
incapable of drawing lines or circles, you can perform
all constructions that are possible using both a
straightedge and compass.
paper folding,
In the case of the Mira,
and a double-edged straightedge with
perpendicular sides, one can even trisect an angle.
What other tools are adequate to perform all
Euclidean constructions?
could be considered.
That is a question which
BIBLIOGRAPHY
Archibald, R.C., ed.
"Steiner's Geometrical
Constructions," tr. by M.E. Stark.
Scripta
Matematica, 14 (1948). pp. 189 - 264.
Austin, Joe Dan.
"Constructions with an Unmarked
Protractor." Mathematics Teacher, 75 (April 1982),
pp. 291 - 295.
Bussey, H.H.
"Geometric Constructions Without the
Classical Restriction to Ruler and Compasses."
American Mathematical Monthly, 43 (May 1936),
pp. 265 - 279.
Cajori, Florian.
"l\ Forerunner of Mascheroni."
Jl,merican Mathematical f\1onthly, 36 (Aug. - Sept.
pp. 364 -365.
1929),
Carnaham, Walter H.
"Compass Geometry."
School Science
and Mathematics, 32 (April 1932), pp. 384 - 390.
Cheney, Hm. Fitch, Jr.
"Can We Outdo fv1ascheroni?"
Mat~ematics Teacher,
46 (March 1953), pp. 152 -
156.
Courant, Richard ana Robbins, Herbert.
v~hat~
Mathematics? Jl.n Elementary Approach __to J;deas and
Methods.
New York:
Oxford University Press, 1941.
Coxeter, H.S.M. and Greitzer, S.L.
Geometry Revisited.
New York:
Random House, 1967.
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