Surprising Geometric Constructions

Surprising Geometric Constructions

Moti Ben-Ari

http://www.weizmann.ac.il/sci-tea/benari/

© 2019–20 by Moti Ben-Ari.

This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/3.0/ or send a letter to

Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA.

Contents

1 Introduction 7

1.1 Constructions with straightedge and compass . . . . . . . . . . . . . . . . . . 7

1.2 Constructions with origami . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

I Straightedge and Compass 9

2 Help, My Compass Collapsed! 10

2.1 Fixed compasses and collapsing compasses . . . . . . . . . . . . . . . . . . . . 10

2.2 Euclid’s construction for copying a line segment . . . . . . . . . . . . . . . . . 11

2.3 An incorrect construction for copying a line segment . . . . . . . . . . . . . . 12

2.4 A “simpler” construction for copying a line segment . . . . . . . . . . . . . . . 13

2.5 Don’t trust a diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 How to Trisect an Angle (With a Little Help from a Friend) 16

3.1 Trisection using the neusis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.2 Trisection using the quadratrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.3 Squaring the circle using the quadratrix . . . . . . . . . . . . . . . . . . . . . . 18

4 How to (Almost) Square a Circle 20

4.1 Approximations to π . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4.2 Kochansky’s construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.3 Ramanujan’s first construction . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.4 Ramanujan’s second approximation . . . . . . . . . . . . . . . . . . . . . . . . 26

5 A Compass is Sufficient 29

5.1 Reflection of a point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

5.2 Construct a circle with a given radius . . . . . . . . . . . . . . . . . . . . . . . 30

5.3 Addition and subtraction of line segments . . . . . . . . . . . . . . . . . . . . . 31

5.4 Construct a line segment relative to three other line segments . . . . . . . . . 36

3

5.5 Find the intersection of two lines . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.6 Finding the intersection of a line and a circle . . . . . . . . . . . . . . . . . . . 40

6 A Straightedge (with Something Extra) is Sufficient 41

6.1 Construct a line parallel to a given line . . . . . . . . . . . . . . . . . . . . . . . 42

6.2 Construct a perpendicular to a given line . . . . . . . . . . . . . . . . . . . . . 46

6.3 Copy a line segment in a given direction . . . . . . . . . . . . . . . . . . . . . . 46

6.4 Construct a line segment relative to three other line segments . . . . . . . . . 47

6.5 Construct a square root . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

6.6 Construct the points of intersection of a line and a circle . . . . . . . . . . . . . 49

6.7 Construct the points of intersection of two circles . . . . . . . . . . . . . . . . . 49

II Origami 51

7 Axioms 52

7.1 Axiom 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

7.2 Axiom 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

7.3 Axiom 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

7.4 Axiom 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

7.5 Axiom 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

7.6 Axiom 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

7.7 Axiom 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

8 Trisecting an Angle 70

8.1 Abe’s trisection of an angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

8.2 Martin’s trisection of an angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

9 Doubling a Cube 74

9.1 Messer’s doubling of a cube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

9.2 Beloch’s doubling of a cube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

10 Lill’s Method for Finding Roots 79

10.1 Magic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

10.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

10.3 Multiple roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

10.4 Paths that do not lead to roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

10.5 Specification of Lill’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4

10.6 Negative coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

10.7 Zero coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

10.8 Non-integer roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

10.9 The cube root of two . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

10.10Proof of Lill’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

11 Beloch’s Fold and Beloch’s Square 87

11.1 The Beloch fold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

11.2 The Beloch square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5

6

Chapter 1

Introduction

Geometric constructions are fundamental to Euclidean geometry and appear in secondary-school textbooks [6]. Most students of mathematics will also know that three constructionsare impossible: trisecting angle, squaring a circle and doubling a cube. There are manyinteresting and surprising geometrical constructions that are probably unknown to studentsand teachers. This document presents these constructions in great detail using only secondary-school mathematics.

The LATEX source can be found at https://github.com/motib/constructions.

Part I presents constructions with the familiar straightedge and compass, as well as construc-tions known to the Greeks that use extensions of the straightedge and compass. In recentyears, the art of origami—paper folding—has been given a mathematical formalization asdescribed in Part II. It may come as a surprise that constructions with origami are morepowerful than constructions with straightedge and compass.

1.1 Constructions with straightedge and compass

Chapeter 2, the collapsing compass The modern compass is a fixed compass that maintainsthe distance between its legs when lifted off the paper. It can be used to construct a linesegment of the same length as a given segment. The compass used in the ancient worldwas a collapsing compass that does not maintain the distance between its legs when liftedfrom the paper. Euclid showed that any construction that can be done with a fixedcompass can be done with a collapsing compass. Numerous incorrect proofs have beengiven based on incorrect diagrams [17]. In order to emphasize that a proof must notdepend on a diagram, I “prove” that every triangle is isoceles.

Chapter 3, trisecting an angle Trisecting an arbitrary angle is impossible, but the Greeksknew that any angle can be trisected using extensions of the straightedge and compass.This chapter presents constructions that trisect an angle using a neusis and a quadratrix.The quadratrix can also be used to square a circle.

Chapter 4, squaring a circle To square a circle requires the construction of a line segmentof length π. This chapter presents three constructions of approximations to π, one byAdam Kochansky and two by Ramanujan.

Chapter 5, construction with only a compass Are both a straightedge and a compass neces-sary? Lorenzo Mascheroni and Georg Mohr showed that a compass only is sufficient.

Chapter 6 ,construction with only a straightedge Is a straightedge sufficient? The answeris no because a straightedge can “compute” only linear functions, whereas a compasscan “compute” quadratic functions. Jacob Steiner proved that a straightedge is sufficientprovided that somewhere in the plane a single circle exists.

7

1.2 Constructions with origami

Chapter 7, the axioms of origami The seven axioms of origami can be formalized in math-ematics. This chapters derives formulas for the axioms, together with numericalexamples.

Chapter 8, trisecting an angle Two methods for trisecting an angle with origami are given.

Chapter 9, doubling a cube Two methods for doubling a cube with origami are given.

Chapter 10, finding roots This chapter explains Eduard Lill’s geometric method for findingreal roots of any polynomial (actually, verifying that a value is a root). We demonstratethe method for cubic polynomials.

Chapter 11, constructing a cube root Margharita P. Beloch published an implementation ofLill’s method that can find a root of a cubic polynomial with one fold.

8

Part I

Straightedge and Compass

9

Chapter 2

Help, My Compass Collapsed!

2.1 Fixed compasses and collapsing compasses

A modern compass is a fixed compass: the distance between the two legs can be fixed so thatit is possible to copy a line segment or a circle from one position to another. I have seengeometry textbooks that present the construction a perpendicular bisector to a line segmentas follows: construct two circles centered at the ends of the line segment such that the radiiare equal and greater than half the length of the segment (left diagram):

A B

C

D

A B

C

D

Euclid used a collapsing compass whose legs fold up when the compass is lifted off the paper.Teachers often use a collapsing compass consisting of a piece of chalk tied to a string. It isimpossible to maintain a fixed radius when the chalk and the end of the string are removedfrom the blackboard. The right diagram above shows how to construct a perpendicularbisector with a collapsing compass: the length of the segment AB is, of course, equal to thelength of the segment BA, so the radii of the two circles are equal.

The proof that the line constructed is the perpendicular bisector is not at all elementarybecause relatively advanced concepts like congruent triangles have to be used. However,the proof that the same construction results in an equilateral triangle is very simple (rightdiagram below). The length of AC equals the length of AB since they are radii of the samecircle, and for the same reason the length of BC is equal to the length of BA. We have:

AC = AB = BA = BC .

A B

C

D

A B

C

D

10

The left diagram above shows that for the construction with the fixed compass the trianglewill be isosceles, but not necessarily equilateral.

This construction of an equilateral triangle is the first proposition in Euclid’s Elements. Thesecond proposition shows how to copy a given line segment AB to a segment of the samelength, one of whose end points is a given point C. Therefore, a fixed compass adds noadditional capability. Toussaint [17] showed that many incorrect constructions for thisproposition have been given. In fact, it was Euclid who gave a correct construction! Thefollowing section presents Euclid’s construction and the proof of its correctness. Then I showan incorrect construction that can be found even in modern textbooks.

2.2 Euclid’s construction for copying a line segment

Theorem: Given a line segment AB and a point C, a line segment can be constructed (using acollapsing compass) at C whose length is equal to the length of AB:

A

B

C A

B

C

D

Construction:

Construct the line segment from A to C.

Construct an equilateral triangle whose base is AC (right diagram above). Label the thirdvertex D. By Euclid’s first proposition, the triangle can be constructed using a collapsingcompass.

Construct a ray that is a continuation of DA and a ray that is a continuation of DC (leftdiagram below).

Construct a circle centered at A with radius AB. Label the intersection of the circle and theray DA by E (right diagram below).

A

B

C

D

A

B

C

D

E

Construct a circle centered at D with radius DE. Label the intersection of the circle and theray DC by F:

11

A

x B

C

D

yy

EF

xx

Claim: The length of the line segment CF is equal to the length of AB.

Proof: DC = DA because 4ACD is equilateral. AE = AB because they are radii of thesame circle centered at A. DF = DE because they are radii of the same circle centered at D.Therefore, the length of the line segment CF is:

CF = DF− DC = DE− DC = DE− DA = AE = AB .

2.3 An incorrect construction for copying a line segment

Construction([13]):

Construct a circle centered at A with radius AB:

A

B

C A

B

C

Construct a circle centered at A with radius AC and a circle centered at C with radiusAC = CA. Label the intersections of the two circles E, F. Label the intersection of the circlecentered at C and the circle centered at A with radius AB by D:

A

B

C

D

E

F

Construct a circle centered at E with radius ED. Label the intersection of this circle with thecircle centered at A with radius AC by G:

12

A

B

C

D

E

F

G

Claim: The length of the line segment GC is equal to the length of AB.

Proof: CD = CE are radii of the circle centered at C. AE = AG are radii of the (larger)circle centered at A. CD = CE = AE = AG since the radii of the two circles are AC = CA.EG = ED are radii of the circle centered at E. Therefore,4EAG ∼= 4DCE by side-side-sideso 6 GEA = 6 DEC.

6 GEC = 6 GEA− 6 CEA = 6 DEC− 6 CEA = 6 DEA. Therefore, 4ADE ∼= 4CGE by side-angle-side. AB = AD are radii of the smaller circle centered at A, so CG = AD = AB.

Is there an error in the proof? No! But there is a problem because the equality AB = GC holdsonly when the length of AB is less that the length of AC. In contrast, Euclid’s constructionand proof are true, independent of the relative lengths of AB and AC, and independent ofthe position of the point C relative to the line segment AB [17].

2.4 A “simpler” construction for copying a line segment

Given a line segment AB and a point C, if we can build a parallelogram with these threepoints as its vertices, we obtain a line segment with C at one end whose length is equal to thelength of AB (left diagram):

xA B

C

y

x

y

D

A B

CD

E

This construction can be found in [18, pp. 207–208].

Construction (right diagram):

Construct the line segment from B to C. Construct an altitude from C to the line containingthe line segment AB. Label the intersection by E. Construct an altitude to the line segmentCE at C. This line is parallel to AB. Use a similar method to construct a line parallel to BCthrough A. Label the intersection of the two lines by D.

AD‖BC, AB‖DC and by definition ABCD is a parallelogram, so AB = CD as required.

13

Construction with a collapsing compass: It is possible to construct an altitude to the line lthrough a given point C with a collapsing compass. Construct a circle centered at C witha radius that is greater than the distance of C from l. Label the intersections with l by D, E.Construct circles centered at D, E with radii DC = EC. The line connecting the intersectionsC, F of the circles centered at D, E is an altitude through C.

l

C

D E

F

The proof the correctness of this construction is much more difficult than Euclid’s proof ofhis construction.

2.5 Don’t trust a diagram

We can prove that all triangles are isosceles!

P

B C

A

α α

EF

Da a

Given an arbitrary triangle4ABC, let P be the intersection of the angle bisector of 6 BAC andthe perpendicular bisector of BC. Label by D, E, F the intersections of the altitudes from P tothe sides BC, AB, AC. 4APF ∼= 4APE because they are right triangles with equal angles α

and a common side AP.

4DPC ∼= 4DPB by side-angle-side because PD is a common side, 6 PDB = 6 PDC are rightangles and BD = DC = a because PD is the perpendicular bisector of BC. 4EPB ∼= 4FPCby side-side-angle in a right triangle, because EP = PF by the first congruence and PB = PCby the second congruence. By combining the equations we get that4ABC is isoceles:

AB = AE + EB = AF + FC = AC .

14

The problem with the proof is that the diagram is incorrect because point P is outside thetriangle, as can be seen from the following diagram:

A

α α

B C

15

Chapter 3

How to Trisect an Angle (With a Little Help from a Friend)

The reason is that it is impossible to trisect an arbitrary angle is that it requires the construc-tion of cube roots, but the compass and straightedge can only construct lengths that areexpressions built from the four arithmetic operators and square roots.

Greek mathematicians discovered that if other instruments are allowed, angles can be tri-sected [20]. Section 3.1 presents a construction of Archimedes using a simple instrumentcalled a neusis [19]. Section 3.2 shows a more complex construction of Hippias using thequadratrix [22]. As a bonus, Section 3.3 shows that the quadratrix can square a circle.

3.1 Trisection using the neusis

The term straightedge is used instead of “ruler” because a straightedge has no marks on it.The only operation it can perform is to construct a straight line between two points, while aruler can measure distances. To trisect an angle all we need is a straightedge with two marksthat are a fixed distance apart, called a neusis. We define the distance between the marks as 1:

1

Let α be an arbitrary angle 6 ABE within a circle with center B and radius 1. The circlecan be constructed by setting the compass to the distance between the marks on the neusis.Extend the radius EB beyond the circle. Place an edge of the neusis on A and move it until itintersects the extension of EB at D and the circle at C, using the marks so that the length ofthe line segment CD is 1. Draw the line AD:

Bα

A

D

C

E

1

16

Draw line BC and label the angles and line segments as shown:

Bα δ β

1

A

γ

β D1

Cγε

E1

Both 4ABC and 4BCD are isoceles: AB = BC since both are radii and BC = CD byconstruction using the neusis. A computation (using the facts that the angles of a triangleand supplemenary angles add up to π radians) shows that β trisects α:

ε = π − 2β

γ = π − ε = 2β

δ = π − 2γ = π − 4β

α = π − δ− β

= 4β− β

= 3β .

3.2 Trisection using the quadratrix

The following diagram shows a quadratrix compass: two (unmarked) straightedges connectedby a joint that constrains them to move together. One straightedge moves parallel to thex-axis from DC to AB, while the second straightedge is allowed to rotate around the originat A until it lies horizontally along AB. The curve traced by the joint of the two straightedgesis called the quadratrix curve or simply the quadratrix.

A B

CD

17

As the horizontal straightedge is moved down at a constant velocity, the other straightedgeis constrained to move at a constant angular velocity. In fact, that is the definition of thequadratrix curve. As the y-coordinate of the horizontal straightedge decreases from 1 to 0, theangle of the other straightedge relative to the x-axis decreases from 90◦ to 0◦. The followingdiagram shows how this can be used to trisect an arbitrary angle α:

Dθ

A B

C

P2

P1

F

E

t/3

t

1 − t

α

P1 is the intersection of the line defining the angle α relative to DC and the quadratrix. Thispoint has y-coordinate 1− t, where t is the distance that the horizontal straightedge hasmoved from its initial position DC. Now trisect the line segment DE to obtain point F. (It iseasy to trisect a line segment using Thales theorem.) Let P2 be the intersection of a line fromF parallel to DC and the quadratrix. By the equality of the velocities, we have:

θ

α=

t/3t

θ = α/3 .

3.3 Squaring the circle using the quadratrix

Aθ

B

CD

y

x F

E

t

1 − t

P

G

18

Suppose that the horizontal straightedge has moved t down the y-axis to point E and therotating straightedge forms an angle of θ with the x-axis. P is the intersection of the quadratrixand the horizontal straightedge, and F is the projection of P on the x-axis. What are thecoordinates of P? Clearly, y = PF = EA = 1− t. On the quadratrix, θ decreases at the samerate that t increases:

1− t1

=θ

π/2

θ =π

2(1− t) .

Check if this makes sense: when t = 0, θ = π/2 and when t = 1, θ = 0.

The x-coordinate of P follows from trigonometry:

tan θ =yx

.

which gives:

x =y

tan θ= y cot θ = y cot

π

2(1− t) = y cot

π

2y .

We usually express a function as y = f (x) but it can also be expressed as x = f (y).

As the horizontal straightedge moves down, it will merge with AB and P with lie on a pointon the x-axis, which we denote G. Let us compute the x-coordinate of the point G. We can’tsimply plug in y = 0 because cot 0 is not defined, but we might get lucky by computing thelimit of x as y goes to 0. First, multiply and divide by π/2:

x = y cotπ

2y =

2π· π

2y cot

π

2y .

For convenience, perform a change of variable z =π

2y and compute the limit:

limz→0

z cot z = limz→0

z cos zsin z

= limz→0

cos zsin z

z

=cos 0

1= 1 ,

using the well-known fact that limz→0

sin zz

= 1. Therefore, as y→ 0:

x → 2π· lim

y→0

π

2y cot

π

2y =

2π· 1 =

2π

.

Using the quadratrix we have constructed a line segment AG whose length is x =2π

.With an ordinary straightedge and compass it is easy to construct a line segment of length√

2x=

√2

2/π=√

π and then construct a square whose area is π.

19

Chapter 4

How to (Almost) Square a Circle

4.1 Approximations to π

To square a circle the length√

π must be constructed, however, π is transcendental, meaningthat it is not the solution of any algebraic equation.

This chapter brings three constructions of approximations to π. The following table shows theformulas of the lengths that are constructed, their approximate values, the difference betweenthese values and the value of π, and the error in meters that results if the approximation isused to compute the circumference of the earth given that its radius is 6378 km.

Construction Formula Value Difference Error (m)

π 3.14159265359 − −

Kochansky

√403− 2√

3 3.14153338705 5.932× 10−5 756

Ramanujan 1355113

3.14159292035 2.667× 10−7 3.4

Ramanujan 2(

92 +192

22

)1/4

3.14159265258 1.007× 10−9 0.013

Kochansky’s construction from 1685 can be found in [2].

Ramanujan’s constructions from 1913 can be found in [14, 15].

In the constructions in this chapter we need to divide a line segment into three parts; here weshow how the division of two lengths can be constructed. Given a line segment of length 1and line segments of lengths a, b, by similar triangles:

1b=

ODa

,

so OD =ab

.

OA

C

B

D

b

1

aa/b

20

4.2 Kochansky’s construction

4.2.1 The construction

Construct three circles:

1. Construct a unit circle centered at O, let AB be a diameter and construct a tangent tothe circle at A.

2. Construct a unit circle centered at A. Its intersection with the first circle is C.1

3. Construct a unit circle centered at C. Its intersection with the second circle is D.

Construct OD and denote its intersection with the tangent by E.

From E construct F, G, H, each at distance 1 from the previous point; then AH = 3− EA.

Construct BH.

Claim: BH =

√403− 2√

3 ≈ π.

O

A

B

1

1

C

D

E F G H

1 1 1

1For the second and third circles, the diagram only shows the arc that intersects the previous circle.

21

4.2.2 The proof

Extract the following diagram from the first one. Dashed line segments have been added.Since all the circles are unit circles, it is easy to see that the length of each dashed line segmentis 1. It follows that AOCD is a rhombus so its diagonals are perpendicular to and bisect eachother at K and AK = 1

2 .

60◦30◦1/2

1/√

3

O

A

B

C

D

E F G H

K

The diagonal AC forms two equilateral triangles 4OAC,4DAC so 6 OAC = 60◦. Sincetangent forms a right angle with the radius OA, 6 KAE = 30◦. Now:

1/2EA

= cos 30◦ =

√3

2

EA =1√3

AH = 3− EA =

(3− 1√

3

)=

3√

3− 1√3

Returning to the first diagram, we see that4ABH is a right triangle:

BH2= OB2

+ AH2

= 4 +9 · 3− 6

√3 + 1

3=

403− 2√

3

BH =

√403− 2√

3 ≈ 3.141533387 ≈ π .

22

4.3 Ramanujan’s first construction


Construct a unit circle centered at O and let PR be a diameter.

H bisects PO and T trisects RO.

Construct a perpendicular at T that intersects the circle at Q.

Construct a chord RS = QT.

Construct PS.

Construct a line parallel to RS from T that intersects PS at N.

Construct a line parallel to RS from O that intersects PS at M.

Construct the chord PK = PM.

Construct the tangent at P of length PL = MN.

Connect the points K, L, R.

Find point C such that RC is equal to RH.

Construct CD parallel to KL that intersects LR at D.

Claim: RD2=

355113≈ π.

OP R

H T1/2 1/2 2/3 1/3

a

Q

a

S

N

M

K

L

C

D

23

4.3.2 The proof

By Pythagoras’ theorem on4QOT:

QT =

√12 −

(23

)2

=

√5

3.

4PSR is a right triangle because it subtends a diameter. By Pythagoras theorem:

PS =

√√√√22 −(√

53

)2

=

√4− 5

9=

√313

.

4MPO ∼ 4SPR so:

PMPO

=PSPR

PM1

=

√31/32

PM =

√316

PK = PM =

√316

.

4NPT ∼ 4SPR so:

PNPT

=PSPR

PN5/3

=

√31/32

PN =5√

3118

MN = PN − PM =√

31(

518− 1

6

)=

√319

PL = MN =

√319

.

4PKR is a right triangle because it subtends a diameter. By Pythagoras’s theorem:

RK =

√√√√22 −(√

316

)2

=

√1136

.

4PLR is a right triangle because it PL is a tangent. By Pythagoras’s theorem:

RL =

√√√√22 +

(√319

)2

=

√3559

.

24

RC = RH =13+

23+

12=

32

. Since CD is parallel to LK, by similar triangles:

RDRC

=RLRK

RD3/2

=

√355/9√113/6

RD =

√355113

.

Here is the construction with line segments labeled with their lengths:

OP R

H T1/2 1/2 2/3 1/3

√5/3

S

√31/3

N

M

√31/6

√31/9

√31/6

K

√319

L

RC = 3/2

RK =√

113/6

C

D

RD =√

355/113

RL =√

355/9

The value355113

could be constructed by constructing two line segments of length 355 and 113and then using the division construction shown in Section 4.1, but that is rather tedious!

25

4.4 Ramanujan’s second approximation


Construct a unit circle centered at O with diameter AB, and let C be the intersection of theperpendicular at O with the circle.

Trisect AO so that AT = 1/3 and TO = 2/3.

Construct BC and find points M, N such that CM = MN = AT = 1/3.

Construct AM and AN and let P be the point on AN such that AP = AM.

From P construct a line parallel to MN that intersects AM at Q.

Construct OQ and then from T construct a line parallel to OQ that intersects AM at R.

Construct a line segment AS tangent to A of length AR.

Construct SO.

Claim: 3√

SO =

(92 +

192

22

) 14

≈ π.

1/3 2/3 1

1/3

1/3

√2−2/3

OA B

45◦

C

T

M

NP

Q

R

S

26

4.4.2 The proof

4COB is a right triangle, OB = OC = 1, so by Pythagoras’ theorem CB =√

2 and NB =√2− 2/3. The triangle is isoceles so 6 NBA = 6 MBA = 45◦.

We use the law of cosines on4NBA to compute AN:

AN2= BA2

+ BN2 − 2 · BA · BN · cos 6 NBA

= 22 +

(√2− 2

3

)2

− 2 · 2 ·(√

2− 23

)·√

22

=

(4 + 2 +

49− 4)+√

2 ·(−4

3+

43

)=

229

AN =

√229

.

Similarly, we use the law of cosines on4MBA to compute AM:

AM2= BA2

+ BM2 − 2 · BA · BM · cos 6 MBA

= 22 +

(√2− 1

3

)2

− 2 · 2 ·(√

2− 13

)·√

22

=

(4 + 2 +

19− 4)+√

2 ·(−2

3+

23

)=

199

AM =

√199

.

By construction QP‖MN so4MAN ∼ 4QAP, and by construction AP = AM giving:

AQAM

=APAN

=AMAN

AQ =AM2

AN=

19/9√22/9

=19

3√

22.

By construction TR‖OQ so4RAT ∼ 4QAO giving:

ARAQ

=ATAO

AR = AQ · ATAO

=19

3√

22· 1/3

1=

199√

22.

27

By construction AS = AR and4OAS is a right triangle. By Pythagoras’ theorem:

SO =

√12 +

(19

9√

22

)2

3√

SO = 3(

1 +192

92 · 22

) 14

=

(34 +

34 · 192

92 · 22

) 14

=

(92 +

192

22

) 14

≈ 3.14159265262 ≈ π .

Here is the construction with line segments labeled with their lengths:

1/3 2/3 1

1/3

1/3

√2−2/3

AM =√

19/9

AN =√

22/9

AQ = 19/3√

22

AR = 19/9√

22

AS = 19/9√

22SO =

√1 +

(192

92 · 22

)

OA B

45◦

C

T

M

NP

Q

R

S

28

Chapter 5

A Compass is Sufficient

In 1797 the Italian mathematician Lorenzo Mascheroni proved that any construction carriedout with a compass and straightedge can be carried out with the compass alone! Later itcame to light that the construction had already been discovered by the Danish mathematicianGeorg Mohr 1672. The theorem is now called the Mohr-Mascheroni Theorem.

In this chapter I present a proof of the theorem based on the proof in problem 33 of [4] andreworked by Michael Woltermann [5].1 A different proof can be found in [8].

What does it mean to perform a construction with only a compass? The right diagram belowshows the construction of an equilateral triangle using a straightedge and compass. How canwe construct a triangle without the line segments AB, AC, BC? In fact, there is no need to seethe lines. A line is defined by two points, so it is sufficient to construct the points in order toobtain a construction equivalent to the one with a straightedge (left diagram):

A B

C

A B

C

In the diagrams we will draw lines, but they are used only to understand the construction andthe proof of its correctness. It is important that you convince yourself that the constructionitself uses only a compass.

A construction by straightedge and compass is a sequence taken from these three operations:

• Find the point of intersection of two straight lines.

• Find the point of intersection of a straight line and a circle.

• Find the point(s) of intersection of two circles.

It is clear that the third operation can be done with only a compass. We need to show that thefirst two operations can be done with a compass alone.

Notation:

• c(O, A): the circle with center O through point A.

• c(O, r): the circle with center O and radius r.

• c(O, AB): the circle with center O and radius the length of line segment AB.

First we will solve four preliminary problems (Sections 5.1–5.4). Next, we show the construc-tion for finding the intersection of two lines (Section 5.5), and finally the construction of theintersection of a line and a circle (Section 5.6).

1I would like to thank Woltermann for permission to use his work.

29

5.1 Reflection of a point

Given a line AB and a point C not on AB, build a point C′ which is a reflection of C aboutAB.

C′ is a reflection about a line segment AB if AB (or the line containing AB) is the perpendicularbisector of the line CC′.

Construct a circle centered on A passing through C and circle centered on B passing throughC. The intersection of the two circles is the point C′ which is the reflection of C.

A B

C

C′

Proof: 4ABC ∼= 4ABC′ by side-side-side since AC, AC′ are radii of the same circle as areBC, BC′, and AB is a common side. Therefore, 6 CAB = 6 C′AB so AB is the angle bisector of6 CAC′. But4CAC′ is an isosceles triangle and the angle bisector AB is also the perpendicularbisector of the base of4CAC′. By definition C′ is the reflection of C around AB.

5.2 Construct a circle with a given radius

Given points A, B, C, construct c(A, BC).

Construct c(A, B) and c(B, A) and let X and Y be their points of intersection.

A

B

C

XY

30

Construct C′, the reflection of C about line XY as described in Section 5.1.

A

B

C

C′

XY

c(A, C′) is the desired circle.

A

B

C

C′

XY DE

Proof: A is the reflection of B around XY (since4YAX ∼= 4YBX) and C′ is the reflection of Caround the XY. By definition, XY is the perpendicular bisector of CC′ and AB, so C′E = EC,AD = DB and 6 DEC = 6 DEC′ are right angles. 4DEC ∼= 4DEC′ by side-angle-side, soDC = DC′ and 6 ADC′ = 6 BDC (they are complementary to 6 EDC′, 6 EDC). Therefore,4ADC′ ∼= 4BDC by side-angle-side, so AC′ = BC.

5.3 Addition and subtraction of line segments

Given line segment PQ of length a and line segment RS of length b, construct line seg-ments QT, QU such that PUQT is a line segment, where the length of PU is a− b and thelength of PT is a + b.

P QU Tb

Rb

Sa

a − ba + b

31

Constructing an isosceles trapezoid

Let H be any point on c(Q, b). Construct H′, its reflection about PQ. h is the length of HH′:

a QP

b

H

H′

h

Construct the circles c(H, b), c(Q, h). Let K be the intersection of the circles and construct K′,the reflection of K about line PQ:

QP

b

H

H′

h

Q′h

b

K

K′

PQ is the perpendicular bisector of both HH′ and KK′, so these line segments are parallel.KH = b since it is the radius of the circle centered on H. K′, H′ are reflections of K, H andit is not hard to show that K′H′ = KH (4QQ′H ∼= 4QQ′H′ and then4KQH ∼= 4K′QH′).Therefore, KHH′K′ is an isosceles trapezoid whose bases are HH′ = h, KK′ = 2h. Let d bethe length of the diagonals K′H = KH′:

QP

H

H′

h

K

h

b

K′

h

b

d

d

32

Circumscribing the trapezoid by a circle

We want to proof that it is possible to circumscribe KHH′K′ by a circle. We will provethat: if the opposite angles of a quadrilateral are supplementary, then the trapezoid canbe circumscribed by a circle, and that in an isosceles trapezoid the opposite angles aresupplementary. Geometry textbooks give the simple proof that the opposite angles of aquadrilateral circumscribed by a circle are supplementary, but it is hard to find a proof of theconverse, so I present both proofs here.

If a quadrilateral can be circumscribed by a circle then the opposite angles are supple-mentary: An inscribed angle equals half the subtended arc, so 6 DAB is half of the arc DCBand 6 DCB is half of the arc DAB. The two arcs subtend the entire circumference of thecircle, so their sum is 360◦. Therefore, 6 DAB + 6 DCB =

12· 360◦ = 180◦, and similarly

6 ADC + 6 ABC = 180◦

A

B

C

D

A quadrilateral whose opposite angles are supplementary can be circumscribed by a cir-cle: Any triangle can be circumscribed by a circle. Circumscribe 4DAB by a circle andsuppose that C′ is a point such that 6 DAB + 6 DC′B = 180◦, but C′ is not on the circumfer-ence of the circle. Without loss of generality, let C′ be within the circle:

A

BC

D

C′

Construct a ray that extends DC′ and let C be its intersection with the circle. ABCD iscircumscribed by a circle so:

6 DAB + 6 DCB = 180◦

6 DAB + 6 DCB = 6 DAB + 6 DC′B

6 DCB = 6 DC′B ,

which is impossible if C is on the circle and C′ is inside the circle.

33

Finally, we show that the opposite angles of an isosceles trapezoid are supplementary.

A

B

D

C

x

y

x

y

B′

x

Construct the line AB′ parallel to CD. AB′CD is a parallelogram and4ABB′ is an isoscelestriangle, so 6 C = 6 ABB′ = 6 AB′B = 6 B. Similarly, 6 A = 6 D. Since the sum of the internalangles of any quadrilateral is equal to 360◦:

6 A + 6 B + 6 C + 6 D = 360◦

2 6 A + 2 6 C = 360◦

6 A + 6 C = 180◦ .

and similarly 6 B + 6 D = 180◦.

Ptolemy’s theorem

Ptolemy’s theorem relates the lengths of the diagonals and the lengths of the sides of aquadrilateral that is circumscribed by a circle:

e f = ac + bd .

A

B

C

D

a

b

c

d

ef

There is a geometric proof of the theorem (see Wikipedia), but I will present a simpletrigonometric proof. The law of cosines for the four triangles 4ABC, 4ADC, 4DAB,4DCB gives the following equations:

e2 = a2 + b2 − 2ab cos 6 B

e2 = c2 + d2 − 2cd cos 6 D

f 2 = a2 + d2 − 2ad cos 6 A

f 2 = b2 + c2 − 2bc cos 6 C .

34

6 C = 180◦ − 6 A and 6 D = 180◦ − 6 B because they are opposite angles of a quadrilateralcircumscribed by a circle, so:

cos 6 D = − cos 6 B

cos 6 C = − cos 6 A .

We can eliminate the cosine term from the first two equations and from the last two equations.After some messy arithmetic, we get:

e2 =(ac + bd)(ad + bc)

(ab + cd)

f 2 =(ab + cd)(ac + bd)

(ad + bc).

Multiply the two equations and simplify to get Ptolemy’s theorem:

e2 · f 2 = (ac + bd)2

e f = (ac + bd) .

Using Ptolemy’s theorem

For the construction on page 32, the diagonals are of length d, the legs are of length b, and thebases are of lengths h and 2h, so Ptolemy’s theorem gives d · d = b · b + h · 2h or d2 = b2 + 2h2.

Let X be the point on line PQ that extends PQ by b. (We will eventually construct X; nowwe’re just imagining it.) Define x = K′X. Since4QK′X is a right triangle, x2 = b2 + h2:

QP

H

H′

K

K′

h

X

x

b

From the computation of Ptolemy’s theorem theorem above:

d2 = b2 + 2h2

= (x2 − h2) + 2h2

= x2 + h2 .

Don’t look for a right triangle in the diagram. We are claiming that it is possible to construct atriangle with sides x, h, d.

Let us construct the point S as the intersection of the circles c(K, d), c(K′, d):

35

QP

K

K′

h

S

d

X

We obtain a right triangle4QSK′. By Pythagoras’ theorem QS2+ h2 = d2, so:

QS2= d2 − h2 = x2 ,

and QS = x.

It is possible to construct the point X as the intersection of the circles c(K, x), c(K′, x):

QP

K

K′

SXX′

x

x

b b

PQ = a

h

Recall that we want to extend PQ of length a by a length b, or decrease its length by b. Sincethe length of QX is

√x2 − h2 = b, the length of PX is a + b and the length of PX′ is a− b.

5.4 Construct a line segment relative to three other line segments

Given line segments of length n, m, s, construct a line segment of length:

x =nm

s .

Construct two concentric circles c1 = c(Z, m) and c2 = c(Z, n), and chord AB = s on c1. (Achord can be constructed using only a compass as shown in Section 5.2.)

We assume that m > n. If not, exchange the notation.

36

Z

AB

c1

c2

s

mn

m

n

s

We also assume that s does not intersect c2. If not, use the construction in Section 5.3 tomultiply m, n by a number k so that the chord does not intersect the circle. Note that this

does not change the value that we are trying to construct since x =knkm

s =nm

s.

Choose any point H on circle c2. Label the length of AH by w. Construct point K on c2 suchthat the length of BK is w.

Z

AB

c1

c2

s

wH

w

K

m

n

s

w

4AHZ ∼= 4BZK by side-side-side: ZA = ZB = m, the radius of c1, ZH = ZK = n, theradius of c2, and AH = BK = w by construction.

Z

AB

c1

c2

s

wH

w

Km

n

From 4AHZ ∼= 4BZK, we get 6 AZB = 6 HZK. It is difficult to see this equality fromthe diagram, but the following diagram clarifies the relation among the angles. Defineα = 6 AZH = 6 BZK and β = 6 BZH. It is easy to see that 6 AZB = 6 HZK = α− β.

37

A

Z

B

H

Kα

α

β α − β

α − β

4ZAB ∼ 4ZHK by side-angle-side, since both are isosceles triangles and we have shownthat they have the same vertex angle.

Z

AB

c1

c2

s

H

Km

n

x

Label HK by x. Then:

ms

=nx

x =nm

s .

5.5 Find the intersection of two lines

Given two lines containing the line segments AB, CD, it is possible to construct theirintersection using only a compass.

Let C′, D′ be the reflections of C, D around AB. S, the point of intersection of CD and C′D′,lies on AB, because 4CZS ∼= 4C′ZS by side-angle-side: CZ = C′Z, 6 CZS = 6 C′ZS areright triangles and ZS is a common side. Therefore, C′S = CS and similarly D′S = DS.

A B

C

D

C′

D′

Sc dx

e− x

CD = C′D′ = e

Z

38

Label x = CS, c = CC′, d = DD′, e = CD. 4CSC′ ∼ 4DSD′ are similar sox

e− x=

cd

.

Solving the equation for x gives x =c

c + de.

If C, D are on the same side of AB:

A B

C

D

C′

D′

Sc d

e

e

x− e

x− e

CS = C′S = x

4CSC′ ∼ 4DSD′ givesx

x− e=

cd

. Solving for x gives x =c

c− de.

Construct the circles c(C′, d), c(D, e) and label their intersection by H. The sum of the linesegments CC′, C′H is c + d. We have to show that H is on the extension of CC′ so that CH isa line segment of length c + d. (CH = c− d in case D is on the same side of AB as C.)

A B

C

DC′

D′

S

e

c d

CD = C′D′ = DH = e

H

d

e

From the definition of H as the intersection of c(C′, d), c(D, e), we get DH = e, C′H = d, butC′D′ = e, D′D = d, so the quadrilateral C′D′DH is a parallelogram, since the lengths of bothpairs of opposite sides are equal. By construction, the line segment DD′ is parallel to CC′, soC′H is parallel to DD′ is also parallel to CC′. Since one of its end points is C′, it must be onthe line containing CC′.

The lengths c, d, e are given and we proved in Section 5.3 that a line segment of length

c + d can be construction and in Section 5.4 that a line segment of length x =c

c + de can be

constructed. S, the intersection of c(C′, x) and c(C, x), is the intersection of AB, CD has beenconstructed.

39

A B

C

DC′

D′

S

x

x

c d

CD = C′D′ = DH = e

5.6 Finding the intersection of a line and a circle

Given a circle k = C(M, r) and a line AB, construct their intersections using only a com-pass.

Construct M′, be the reflection of M about AB and the circle k′ = c(M′, r). The points ofintersection of k, k′ are the points of intersection of the line AB and the circle k. This can beshown by congruent triangles, as indicated by the dotted lines in the diagram.

A B

M

M′

r

r

This construction cannot be done if M is on the line AB. In that case, extend and shortenAM by length r as described in Section 5.3. The end points of the extended and shortenedsegments are the intersections of k with AB.

A BMr

AM + rAM − r

40

Chapter 6

A Straightedge (with Something Extra) is Sufficient

Can every construction with straightedge and compass be done with only a straightedge?The answer is no because lines represent linear equations and cannot represent quadraticequations like circles. In 1822 the French mathematician Jean-Victor Poncelet conjectured thata straightedge only is sufficient, provided that one circle exists in the plane. This was provedin 1833 by the Swiss mathematician Jakob Steiner. In this chapter I present a proof of thetheorem based on the proof in problem 34 in [4] as reworked by Michael Woltermann [5].1

Every step of a construction with straightedge and compass is one of these three operations:

• Find the point of intersection of two straight lines.

• Find the point(s) of intersection of a straight line and a circle.

• Find the point(s) of intersection of two circles.

It is clear that the first operation can be performed with a straightedge only.

What does it mean to perform a construction with straight-edge alone? A circle is defined bya point O, its center, and a line segment whose length is the radius r, one of whose endpointsis the center. If we can construct the points labeled X and Y in the diagram below, we canclaim to have successfully constructed the intersection of a given circle with a given lineand of two given circles. The circles drawn with dashed lines in the diagram do not actuallyappear in the construction. In this chapter, the single existing circle is drawn with a regularline, and the dashed circles are only used to help understand a construction and its proof.

O

r

X

Y O1 O2

r1r2

X

Y

First we present five auxiliary constructions (Sections 6.1–6.5), and then show how to find theintersection of a line with a circle (Section 6.6) and the intersection of two circles (Section 6.7).

1I would like to thank Woltermann for permission to use his work.

41

6.1 Construct a line parallel to a given line

Given a line l defined by two points A, B and a point P not on the line, construct a linethrough P that is parallel to AB.

There are two cases:

• A “directed line”: the midpoint M of AB is given.

• Any other line.

Case 1, directed line: Construct a ray that extends AP and choose any point S on the raybeyond P. Construct the lines BP, SM, SB. Label by O the intersection of BP with SM.Construct a ray that extends AO and label by Q the intersection of the ray AO with SB.

A B

S

P Q

O

M

Claim: PQ is parallel to AB.

Proof: We will use Ceva’s theorem that we proof later.

Theorem (Ceva): Given three line segments from the vertices of a triangle to the oppositeedges that intersect in a point (O in the diagram, but M is not necessarily the midpoint of theside), the lengths of the segments satisfy:

AMMB· BQ

QS· SP

PA= 1 .

In the construction above, M is the midpoint of AB, soAMMB

= 1, so the equation becomes:

BQQS

=PASP

=APPS

. (6.1)

We will prove that4ABS ∼ 4PQS, so that PQ is parallel to AB because 6 ABS = 6 PQS.

BS = BQ + QS

BSQS

=BQQS

+QSQS

=BQQS

+ 1

AS = AP + PS

ASPS

=APPS

+PSPS

=APPS

+ 1 .

42

Using Equation 6.1:BSQS

=BQQS

+ 1 =APPS

+ 1 =ASPS

,

so that4ABS ∼ 4PQS.

Proof of Ceva’s theorem: Examine the following diagrams:

A B

S

QO

A B

S

QO

If the altitudes of two triangles are equal, their areas are proportional to the bases. In bothdiagrams, the altitudes of the gray triangles are equal, so:2

4BQO4SQO

=BQQS

,4BQA4SQA

=BQQS

.

By subtracting the areas of the indicated triangles, we get the proportion between the graytriangles:

A B

S

Q

O

4BOA4SOA

=4BQA−4BQO4SQA−4SQO

=BQQS

.

This might look strange at first. We explain it using a simpler notation:

cd

=ab

ef

=ab

c− e =adb− a f

b=

ab(d− f )

c− ed− f

=ab

.

2We use the name of a triangle as a shortcut for its area

43

Similarly, we can prove:

AMMB

=4AOS4BOS

,SPPA

=4SOB4AOB

,

so:AMMB

BQQS

SPPA

=4AOS4BOS

4BOA4SOA

4SOB4AOB

= 1 ,

because the order of the vertices in a triangle makes no difference:

4AOS = 4SOA, 4BOA = 4AOB, 4SOB = 4BOS .

End of the proof of Ceva’s theorem

Case 2, any other line: Label the line by l and the existing circle, which we will call the fixedcircle, by c, where the center of c is O and its radius is r. P is the point not on the line throughwhich it is required to construct a line parallel to l. Convince yourself that the constructiondoes not depend on the location of the center of the circle or its radius.

Choose M, any point on l, and construct a ray extending MO that intersects the circle at U, V.

O

lM

U

V

r

r

cP

This line is a directed line because O, the center of the circle, bisects the diameter UV. Choosea point A on l and use the construction for a directed line to construct a line parallel to UV,which intersects the circle at X, Y.

O

lM

U

V

A

X

Yc

P

Construct the diameters XX′ and YY′. Construct the ray from X′Y′ and label by B itsintersection with l.

44

O

lM

U

V

A

X

Yc

PX′

Y′

B

l′ZZ′

l is a directed line because M is the bisector of AB, so a line can be constructed through Pparallel to l.

Proof: OX, OX′, OY, OY′ are all radii of the circle and 6 XOY = 6 X′OY′ since they are verticalangles. 4XOY ∼= 4X′OY′ by side-angle-side. Define (not construct!) l′ to be a line through Oparallel to l that intersects XY at Z and X′Y′ at Z′. 6 XOZ = 6 X′OZ′ because they are verticalangles, so4XOZ ∼= 4X′OZ′ by angle-side-angle and ZO = OZ′. AMOZ and BMOZ′ areparallelograms (quadrilaterals with opposite sides parallel), so AM = ZO = OZ′ = MB.

Corollary: Given a line segment AB and a point P not on the line. It is possible to construct aline through P that is parallel to AB and whose length is equal to the length of AB. In otherwords, it is possible to copy AB parallel to itself with P as one of its endpoints.

Proof: We have proved that it is possible to construct a line m through P parallel to AB and aline n through B to parallel to AP. The quadrilateral ABQP is a parallelogram so oppositesides are equal AB = PQ.

mP Q

lA B

n

45

6.2 Construct a perpendicular to a given line

Construct a perpendicular line segment through a point P to a given line l (P is not on l).

Construct (Section 6.1) a line l′ parallel to l that intersects the fixed circle at U, V. Constructthe diameter UOU′ and chord VU′.

c

l

l′

O P

U V

U′

6 UVU′ is an angle that subtends a semicircle so it is a right angle. Therefore, VU′ is perpen-dicular to UV and l. Construct (Section 6.1) the parallel to VU′ through P.

6.3 Copy a line segment in a given direction

The corollary at the end of Section 6.1 shows that it is possible to copy a line segment parallelto itself. Here we show that it is possible to copy a line segment in the direction of anotherline. The meaning of “direction” is that the line defined by two points A′, H′ defines an angleθ relative to some axis. The task is to copy PQ to AS so that AS will have the same angle θ

relative to the same axis. In the diagram PQ is on the x-axis but that is of no importance.

P Q

A′

H′

S

A

θ

θ

By Section 6.1 it is possible to construct a line segment AH so that AH‖A′H′ and to constructa line segment AK so that AK‖PQ.

P Q

A′

H′

H

KA

θ

θ

46

6 HAK = θ so it remains to find a point S on AH so that AS = PQ.

P Q

A′

H′

A

H

c

O

K

U

VS

θ

θθ

Construct two radii OU, OV of the fixed circle which are parallel to AH, AK, respectively,and construct a ray through K parallel to UV. Label its intersection with AH by S.

Claim: AS = PQ

Proof: By construction, AH‖OU and AK‖OV, so 6 SAK = θ = 6 UOV. SK‖UV and4SAK ∼ 4UOV by angle-angle-angle,4UOV is isosceles, because OU, OV are radii of thesame circle. Therefore,4SAK is isosceles and AS = AK = PQ.

6.4 Construct a line segment relative to three other line segments

Given line segments of lengths n, m, s, construct a segment of length x =nm

s.

The three line segments are located at arbitrary positions and directions in the plane.

sm

n

Choose a point A and construct two rays AB, AC. By the construction in Section 6.3 it ispossible to construct points M, N, S such that AM = m, AN = n, AS = s. Construct a linethrough N parallel to MS which intersects AC at X and label its length by x. 4MAS ∼4NAX by angle-angle-angle so:

mn

=sx

x =nm

s .

A C

B

m

M

N

S Xs

AN = nAX = x

x

n

47

6.5 Construct a square root

Given line segments of lengths a, b, construct a segment of length√

ab.

We want to express x =√

ab in the formnm

s in order to use the result of Section 6.4.

• For n we use d, the diameter of the fixed circle.

• For m we use t = a + b which can be constructed from a, b as shown in Section 6.3.

• We define s =√

hk where h, k are defined as expressions on the lengths a, b, t, d, and wewill show how it is possible to construct a line segment of length

√ab.

Define h =dt

a, k =dt

b and and compute:

x =√

ab =

√thd

tkd

=

√(td

)2

hk =td

hk =td

s .

We also compute:

h + k =dt

a +dt

b =d(a + b)

t=

dtt= d .

By Section 6.3 construct HA = h on a diameter HK of the fixed circle. From h + k = d wehave AK = k:

c

OH K

A

S

s

h k

d2

d2− k

By Section 6.2 construct a perpendicular to HK at A and label the intersection of this line

with the circle by S. OS = OK =d2

because they are radii of the circle and OA =d2− k. By

Pythagoras’s theorem:

s2 = SA2=

(d2

)2

−(

d2− k)2

=

(d2

)2

−(

d2

)2

+ 2dk2− k2

= k(d− k) = kh

s =√

hk .

Now x =td

s can be constructed by Section 6.4.

48

6.6 Construct the points of intersection of a line and a circle

Given a line l and a circle c(O, r), construct their points of intersection.

c

Or

l

By Section 6.2 it is possible to construct a perpendicular from the center of the circle O tothe line l. Label the intersection of l with the perpendicular by M. M bisects the chord XY,where X, Y are the intersections of the line with the circle. 2s is the length of the chord XY.Note that s, X, Y in the diagram are just definitions: we haven’t constructed them yet.

c

Or

R

l

X

Y

r

M

t

ss

t

t

4OMX is a right triangle, so s2 = r2 − t2 = (r + t)(r− t). r is the given radius of the circleand t is defined s the length of OM, the line segment between O and M. By Section 6.3 it ispossible to construct a line segment of length t from O in the two directions OR and RO. Theresult is two line segments of length r + t, r− t.

Section 6.5 shows how to construct a line segment of length s =√(r + t)(r− t). By Sec-

tion 6.3 it is possible to construct line segments of length s from M along the given line l inboth directions. Their other endpoints are the points of intersection of l and c.

6.7 Construct the points of intersection of two circles

Given two circles c(O1, r1), c(O2, r2), construct their points of intersection.

With a straightedge it is possible to construct a line segment O1O2 that connects the twocenters. Label its length t.

49

O1 O2

r1 r2

c1 c2

t

t = O1O2

Label by A be the point of intersection of O1O2 and XY, and label the lengths q = O1A,x = XA. Note that we have not constructed A, but if we succeed in constructing the lengthsq, x, by Section 6.3 we can construct A at length q from O1 in the direction O1O2.

O1 O2

X

Y

r1 r2

c1 c2

Aq

x

t

t = O1O2

q = O1A

x = XA

By Section 6.2 a perpendicular to O1O2 at A can be constructed and again by Section 6.3it is possible to construct line segments of length x from A is both directions along theperpendicular. Their other endpoints are the points of intersection of the circles.

Constructing the length q: Define d =√

r21 + t2, the hypotenuse of a right triangle. It can be

constructed from r1, t, the known lengths of the other two sides. On any line construct a linesegment RS of length r1, then a perpendicular to RS through R, and finally, a line segmentRT of length t through R on the perpendicular. The length of the hypotenuse ST is d. Thisright triangle can be constructed anywhere in the plane, not necessarily near the circles.

By the law of cosines for4O1O2X:

r22 = r2

1 + t2 − 2r1t cos 6 XO1O2

= r21 + t2 − 2tq

q =(d + r2)(d− r2)

2t.

By Section 6.3 these lengths can be constructed and by Section 6.4 q can be constructed fromd + r2, d− r2, 2t.

Constructing the length x: 4AO1X is a right triangle, so x =√

r21 − q2 =

√(r1 + q)(r1 − q).

By Section 6.3 h = r1 + q, k = r1 − q can be constructed, as can x =√

hk by Section 6.5.

50

Part II

Origami

51

Chapter 7

Axioms

Each axiom states that a fold exists that will place given points and lines onto points and lines,such that certain properties hold. The term fold comes from the origami operation of foldinga piece of paper, but here it is used to refer to the geometric line that would be created byfolding the paper.

The axioms are called the Huzita-Hatori axioms [21], although their final form resulted from thework of several mathematicians. Lee [10, Chapter 4] is a good overview of the mathematicsof origami, while Martin [11, Chapter 10] is a formal development. The reader should beaware that, by definition, folds result in reflections. Given a point p, its reflection around a foldl results in a point p′, such that l is the perpendicular bisector of the line segment pp′:

l

p

p′

In the diagrams, given lines are solid, folds are dashed, auxiliary lines are dotted, and dottedarrows indicate the direction of folding the paper.

52

7.1 Axiom 1

Axiom Given two distinct points p1 = (x1, y1), p2 = (x2, y2), there is a unique fold l thatpasses through both of them.

0 1 2 3 4 5 6 7 8

1

2

3

4

5

6

l

p1

p2

Derivation of the equation of the fold

The equation of fold l is derived from the coordinates of p1 and p2: the slope is the quotientof the differences of the coordinates and the y-intercept is derived from p1:

y− y1 =y2 − y1

x2 − x1(x− x1) . (7.1)

Example

Let p1 = (2, 2), p2 = (6, 4). The equation of l is:

y− 2 =4− 26− 2

(x− 2)

y =12

x + 1 .

53

7.2 Axiom 2

Axiom Given two distinct points p1 = (x1, y1), p2 = (x2, y2), there is a unique fold l thatplaces p1 onto p2.

0 1 2 3 4 5 6 7 8

1

2

3

4

5

6

l

p1

p2


The fold l is the perpendicular bisector of p1 p2. Its slope is the negative reciprocal of the slopeof the line connecting p1 and p2. l passes through the midpoint between the points:

y− y1 + y2

2= − x2 − x1

y2 − y1

(x− x1 + x2

2

). (7.2)

Example

Let p1 = (2, 2), p2 = (6, 4). The equation of l is:

y−(

2 + 42

)= −6− 2

4− 2

(x−

(2 + 6

2

))

y = −2x + 11 .

54

7.3 Axiom 3

Axiom Given two lines l1 and l2, there is a fold l that places l1 onto l2.

0 1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

l1

l2

pi

α

α

β β

l f1

l f2

If the lines are parallel, let l1 be y = mx + b1 and let l2 be y = mx + b2. The fold is the line

parallel to l1, l2 and halfway between them y = mx +b1 + b2

2.

If the lines intersect, let l1 be y = m1x + b1 and let l2 be y = m2x + b2.

Derivation of the point of intersection

pi = (xi, yi), the point of intersection of the two lines, is:

m1xi + b2 = m2xi + b2

xi =b2 − b1

m1 −m2

yi = m1xi + b1 .

Example

Let l1 be y = 2x− 2 and let l2 be y = −x + 8. The point of intersection is:

xi =8− (−2)2− (−1)

=103

yi = 2 · 103− 2 =

143

.

55

Derivation of the equation of the slope of the angle bisector

The two lines form an angle at their point of intersection, actually, two pairs of vertical angles.The folds are the bisectors of these angles.

If the angle of line l1 relative to the x-axis is θ1 and the angle of line l2 relative to the x-axis is

θ2, then the fold is the line which makes an angle of θb =θ1 + θ2

2with the x-axis.

tan θ1 = m1 and tan θ2 = m2 are given so mb, the slope of the angle bisector, is:

mb = tan θb = tanθ1 + θ2

2.

The derivation requires the use of two trigonometric identities that we derive here:

tan(θ1 + θ2) =sin(θ1 + θ2)

cos(θ1 + θ2)

=sin θ1 cos θ2 + cos θ1 sin θ2

cos θ1 cos θ2 − sin θ1 sin θ2

=sin θ1 + cos θ1 tan θ2

cos θ1 − sin θ1 tan θ2

=tan θ1 + tan θ2

1− tan θ1 tan θ2.

We use this formula to obtain a quadratic equation in tan(θ/2):

tan θ =tan(θ/2) + tan(θ/2)

1− tan2(θ/2)

tan θ (tan(θ/2))2 + 2 (tan(θ/2)) − tan θ = 0 ,

whose solutions are:

tan(θ/2) =−1±

√1 + tan2 θ

tan θ.

First derive ms, the slope of θ1 + θ2:

ms = tan(θ1 + θ2) =m1 + m2

1−m1m2.

Then derive mb, the slope of the angle bisector:

mb = tanθ1 + θ2

2

=−1±

√1 + tan2(θ1 + θ2)

tan(θ1 + θ2)

=−1±

√1 + m2

sms

.

56

Example

For the lines y = 2x− 2 and y = −x + 8, the slope of the angle bisector is:

ms =2 + (−1)

1− (2 · −1)=

13

mb =−1±

√1 + (1/3)2

1/3= −3±

√10 .


Let us derive equation of the fold l f1 with the positive slope; we know the coordinates of the

intersection of the two lines mi =

(103

,143

):

143

= (−3 +√

10) · 103

+ b

b =44− 10

√10

3

y = (−3 +√

10)x +44− 10

√10

3≈ 0.162x + 4.13 .

57

7.4 Axiom 4

Axiom Given a point p1 and a line l1, there is a unique fold l perpendicular to l1 that passesthrough point p1.

0 1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

l1

p1

l


Let l1 be y = m1x + b1 and let p1 = (x1, y1). l is perpendicular to l1 so its slope is − 1m1

. Since

it passes through p1, we can compute the intercept b and write down its equation:

y1 = − 1m

x1 + b

b =(my1 + x1)

m

y = − 1m

x +(my1 + x1)

m.

Example

Let p1 = (2, 6) and let l1 be y = 2x− 4. The equation of the fold l is:

y = −12

x +2 · 6 + 2

2= −1

2x + 7 .

58

7.5 Axiom 5

Axiom Given two points p1, p2 and a line l1, there is a fold l that places p1 onto l1 and passesthrough p2.

0 1 2 3 4 5 6 7 8 9

1

2

3

4

5

6

7

8

9

l1

p1

p2

p′1

p′′1

l f2

l f1

For a given pair of points and a line, there may be zero, one or two folds.

Derivation of the equations of the reflections

Let l be a fold through p2 and p′1 be the reflection of p1 around l. The length of p1 p2 equalsthe length of p′1 p2. The locus of points at distance p1 p2 from p2 is the circle centered at p2

whose radius is the length of p1 p2. The intersections of this circle with the line l1 give thepossible points p′1.

Let l1 be y = m1x + b1 and let p1 = (x1, y1), p2 = (x2, y2). The equation of the circle centeredat p2 with radius the length of p1 p2 is:

(x− x2)2 + (y− y2)2 = r2 , where

r2 = (x2 − x1)2 + (y2 − y1)

2 .

Substituting the equation of the line into the equation for the circle:

(x− x2)2 + ((m1x + b1)− y2)

2 = (x− x2)2 + (m1x + (b1 − y2))

2 = r2 ,

59

we obtain a quadratic equation for the x-coordinates of the possible intersections:

x2(1 + m21) + 2(−x2 + m1b−m1y2)x + (x2

2 + (b21 − 2b1y2 + y2

2)− r2) = 0 . (7.3)

The quadratic equation has at most two solutions x′1, x′′1 and we can compute y′1, y′′1 ) fromy = m1x + b1. The reflected points are p′1 = (x′1, y′1), p′′1 = (x′′1 , y′′1 ).

Example

Let p1 = (2, 8), p2 = (4, 4) and let l1 be y = −12

x + 3. The equation of the circle is:

(x− 4)2 + (y− 4)2 = r2 = (4− 2)2 + (4− 8)2 = 20 .

Substitute the equation of the line into the equation of the circle and simplify to obtain aquadratic equation for the x-coordinates of the intersections (or use Equation 7.3):

(x− 4)2 +

((−1

2x + 3

)− 4)2

= 20

5x2 − 28x− 12 = 0

(5x + 2)(x− 6) = 0 .

The two points of intersection are:

p′1 =

(−2

5,

165

)= (−0.4, 3.2) , p′′1 = (6, 0) .

Derivation of the equations of the folds

The folds will be the perpendicular bisectors of p1 p′1 and p1 p′′1 . The equation of a perpendicu-lar bisector is given by Equation 7.2, repeated here with for p′1:

y− y1 + y′12

= − x′1 − x1

y′1 − y1

(x− x1 + x′1

2

). (7.4)

Example

For p1 = (2, 8) and p′1 =

(−2

5,

165

), the equation of the fold l f1 is:

y− 8 + (16/5)2

= − (−2/5)− 2(16/5)− 8

(x− 2 + (−2/5)

2

)

y = −12

x + 6 .

For p1 = (2, 8) and p′′1 = (6, 0), the equation of the fold l f2 is:

y− 8 + 02

= −6− 20− 8

(x− 2 + 6

2

)

y =12

x + 2 .

60

7.6 Axiom 6

Axiom Given two points p1 and p2 and two lines l1 and l2, there is a fold l that places p1 ontol1 and p2 onto l2.

−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7

1

2

3

4

5

−6

−5

−4

−3

−2

−1

p1

p2

p′1

p′2

p′′1

p′′2

l1

l2

For a given pair of points and pair of lines, there may be zero, one, two or three folds.

If a fold places pi onto li, the distance from pi to the fold is equal to the distance from li to thefold. The locus of points that are equidistant from a point pi and a line li is a parabola withfocus pi and directrix li. A fold is any line tangent to that parabola. A detailed justification ofthis claim is given below.

For a fold to simultaneously place p1 onto l1 and p2 onto l2, it must be a tangent common tothe two parabolas.

The formula for an arbitrary parabola is quite complex, so we limit the presentation toparabolas with the the y-axis as the axis of symmetry. This is not a significant limitationbecause for any parabola there is a rigid motion that moves the parabola so that its axis ofsymmetry is the y-axis. An example will also be given where one of the parabolas has thex-axis as its axis of symmetry.

61

Derivation of the equation a fold

Let (0, f ) be the focus of a parabola with directrix y = d. Define p = f − d, the signed lengthof the line segment between the focus and the directrix.1 If the vertex of the parabola is on

the x-axis, the equation of the parabola is y =x2

2p. To move the parabola up or down the

y-axis so that its vertex is at (0, h), add h to the equation of the parabola: y =x2

2p+ h.

x-axis

y-axisdirectrix y = d is y = −2

(0, f ) is (0, 4)focus

vertex(0, 1)

p = 6h = 1

Define a = 2ph so that the equation of the parabola is:

y =x2

2p+

a2p

x2 − 2py + a = 0 .

The equation of the parabola in the diagram above is:

x2 − 2 · 6 y + 2 · 6 · 1 = 0

x2 − 12y + 12 = 0 .

Substitute the equation of an arbitrary line y = mx + b into the equation for the parabola toobtain an equation for the points of intersection of the line and the parabola:

x2 − 2p(mx + b) + a = 0

x2 + (−2mp)x + (−2pb + a) = 0 .

1We have been using the notation pi for points; the use of p here might be confusing but it is the standardnotation. The formal name for p is one-half the latus rectum.

62

The line will be a tangent to the parabola if and only if this quadratic equation has exactly onesolution if and only if its discriminant is zero:

(−2mp)2 − 4 · 1 · (−2pb + a) = 0 ,

which simplifies to:m2 p2 + 2pb− a = 0 . (7.5)

This is the quadratic equation with variable m for the slopes of tangents to the parabola.There are an infinite number of tangents because for each m, there is some b that makes theline a tangent by moving it up or down.2

To obtain the common tangents to both parabolas, the equations for the two parabolas havetwo unknowns and can be solved for m and b.

Example

Parabola 1: focus (0, 4), directrix y = 2, vertex (0, 3), p = 2, a = 2 · 2 · 3 = 12. The equation ofthe parabola is:

x2 − 2 · 2y + 12 = 0 .

Substituting into Equation 7.5 and simplifying:

m2 + b− 3 = 0 .

Parabola 2: focus (0,−4), directrix y = −2, vertex (0,−3), p = −2, a = 2 · −2 · −3 = 12. Theequation of the parabola is:

x2 − 2 · (−2)y + 12 = 0 .


m2 − b− 3 = 0 . (7.6)

The solutions of the two equations:

m2 + b− 3 = 0

m2 − b− 3 = 0 ,

are m = ±√

3 ≈ ±1.73 and b = 0. There are two common tangents that are the folds:

y =√

3x , y = −√

3x .

Example

Parabola 1 is unchanged.

Parabola 2: focus (0,−6), directrix y = −2, vertex (0,−4), p = −4, a = 2 · −4 · −4 = 32. Theequation of the parabola is:

x2 − 2 · (−4)y + 32 = 0 .

2Except of course for a line parallel to the axis of symmetry.

63


2m2 − b− 4 = 0 .

The solutions of the two equations (using Equation 7.6 for the first parabola):

m2 + b− 3 = 0

2m2 − b− 4 = 0 ,

are m = ±√

73≈ ±1.53 and b =

23

. There are two common tangents that are folds:

y =

√73

x +23

, y = −√

73

x +23

.

Example

Let us now define a parabola whose axis of symmetry is the x-axis.

Parabola 1 is unchanged.

Parabola 2: focus (4, 0), directrix x = 2, vertex (3, 0), p = 2, a = 2 · 2 · 3 = 12. The equation ofthe parabola is:

y2 − 4x + 12 = 0 .

This is an equation with y2 and x instead of x2 and y, so we can’t use Equation 7.5 and wemust perform the derivation again.

Substitute the equation for a line:

(mx + b)2 − 4x + 12 = 0

m2x2 + (2mb− 4)x + (b2 + 12) = 0 ,

set the discriminant equal to zero and simplify:

(2mb− 4)2 − 4m2(b2 + 12) = 0

−3m2 −mb + 1 = 0 .

If we try to solve the two equations (using Equation 7.6 for the first parabola):

m2 + b− 3 = 0

−3m2 −mb + 1 = 0 ,

we obtain a cubic equation with variable m:

m3 − 3m2 − 3m + 1 = 0 . (7.7)

Since a cubic equation has at least one and at most three (real) solutions, there can be one, twoor three common tangents. There can also be no common tangents if the two equations haveno solution, for example, if one parabola is “contained” with another: y = x2, y = x2 + 1.

64

The formula for solving cubic equations is quite complicated, so I used a calculator on theinternet and obtained three solutions:

m = 3.73 , m = −1 , m = 0.27 .

Choosing m = 0.27, b = 3−m2 = 2.93, and the equation of the fold is:

y = 0.27x + 2.93 .

From the form of Equation 7.7, we might guess that 1 or −1 is a solution:

13 − 3 · 12 − 3 · 1 + 1 = −4

(−1)3 − 3 · (−1)2 − 3 · (−1) + 1 = 0 .

Divide Equation 7.7 by m− (−1) = m + 1 to obtain the quadratic equation m2 − 4m + 1whose roots are 2±

√3 ≈ 3.73, 0.27.

Derivation of the equations of the reflections

We derive the position of the reflection p′1 = (x′1, y′1) of p1 = (x1, y1) around some tangentline lt whose equation is y = mtx + bt. The derivation is identical for any tangent and for p2.

To reflect p1 around lt, we find the line lp with equation y = mpx + bp that is perpendicularto lt and passes through p1:

y = − 1mt

x + bp

y1 = − 1mt

x1 + bp

y =−xmt

+

(y1 +

x1

mt

).

Next we find the intersection pt = (xt, yt) of lt and lp:

mtxt + bt =−xt

mt+

(y1 +

x1

mt

)

xt =

(y1 +

x1

mt− bt

)

(mt +

1mt

)

yt = mtxt + bt .

The reflection p′1 is easy to derive because the intersection pt is the midpoint between p1 andits reflection p′1:

xt =x1 + x′1

2, yt =

y1 + y′12

x′1 = 2xt − x1 , y′1 = 2yt − y1 .

65

Example

p1 = (0, 4), l1 is y =√

3x:

xt =

(4 +

0√3− 0)

(√3 +

1√3

) =√

3

yt =√

3√

3 + 0 = 3

x′1 = 2xt − x1 = 2√

3− 0 = 2√

3 ≈ 3.46

y′1 = 2yt − y1 = 2 · 3− 4 = 2 .

A fold is any tangent to a parabola

Students are usually introduced to parabolas as the graphs of second degree equationsy = ax2 + bx + c. However, parabolas can be defined geometrically: given a point, the focus,and a line, the directrix, the locus of points equidistant from the focus and the directrix definesa parabola.

The following diagram shows the focus—the large dot at p = (0, f ), and the directrix—thethick line d whose equation is y = − f . The resulting parabola is shown as a dotted curve. Itsvertex p2 is at the origin of the axes.

x-axis

y-axisdirectrix d : y = − f

(0, f )focus

p

p2

p′

p3

p4

p5

p1

a2

a2

a3

a3

a4

a4

a5

a5

a1

a1

l1

We have selected five points pi, i = 1, . . . , 5 on the parabola. Each point pi is at a distance ofai both from the focus and from the directrix. Drop a perpendicular from pi to the directrixand let p′i be the intersection of the perpendicular and the directrix. Using Axiom 2, constructthe line li by folding p onto p′i. Since pi is on the parabola, p′pi = pi p = ai. The diagramshows the fold l1 through p1.

66

Theorem The folds are tangents to the parabola.

Proof (Oriah Ben Lulu) In the following diagram, the focus is p, the directrix is l, p′ is a pointon the directrix and m is the fold that places p on p′. By definition, m is the perpendicularbisector of pp′. Let s be the intersection of pp′ and m; then ps = p′s = a and m ⊥ pp′.

l

p

p′

r

q

p′′

b

b

c

d

ms

ca

a

Let r be the intersection of a perpendicular to l through p′ and the fold m. Then4psr ∼= 4p′srby side-angle-side, since ps = p′s, 6 psr = 6 p′sr = 90◦ and rs is a common edge. It followsthat pr = p′r = b and therefore r must be on the parabola.

Choose a point p′′ on the directrix that is distinct from p′ and suppose that m is also the foldthat places p on p′′. Let q be the intersection of the perpendicular to l through p′′ and thefold m. As before, we can prove that pq = p′q = c. Let qp′′ = d. If q is on the parabola thend = qp′′ = qp = c. But c is the hypotenuse of the right triangle4qp′′p′ and cannot be equalto one of its sides d.

We have proved that m intersects the parabola in only one point so it is a tangent to theparabola.

67

7.7 Axiom 7

Axiom Given one point p1 and two lines l1 and l2, there is a fold l that places p1 onto l1 andis perpendicular to l2.

0 1 2 3 4 5 6 7 8 9

1

2

3

4

5

6

7

8

p1

p′1

l1

l2

lp

l

pm


Let p1 = (x1, y1), let l1 be y = m1x + b1 and let l2 be y = m2x + b2.

Since the fold l is perpendicular to l2, and the line lp containing p1 p′1 is perpendicular to l, itfollows that lp parallel to l2:

y = m2x + bp .

lp passes through p1 so y1 = m2x1 + bp and its equation is:

y = m2x + (y1 −m2x1) .

p′1 = (x′1, y′1), the reflection of p1 around the fold l, is the intersection of l1 and lp:

m1x′1 + b1 = m2x′1 + (y1 −m2x1)

x′1 =y1 −m2x1 − b1

m1 −m2

y′1 = m1x′i + b1 .

68

The midpoint pm = (xm, ym) of lp between p1 and p′1 is on the fold l:

(xm, ym) =

(x1 + x′1

2,

y1 + y′12

).

The equation of the fold l is the perpendicular bisector of p1 p′1. First compute the intercept ofl which passes through pm:

ym = − 1m2

xm + bm

bm = ym +xm

m2.

The equation of the fold l is:

y = − 1m2

x +

(ym +

xm

m2

).

Example

Let p1 = (5, 3), let l1 be y = 3x− 3 and let l2 be y = −x + 11.

x′1 =3− (−1) · 5− (−3)

3− (−1)=

114

y′1 = 3 · 114

+ (−3) =214

pm =

5 +114

2,

3 +214

2

=

(318

,338

).

The equation of the fold l is:

y = − 1−1· x +

33

8+

318−1

= x +

14

.

69

Chapter 8

Trisecting an Angle

8.1 Abe’s trisection of an angle

This construction is based upon the presentation in [12]. The second proof is based upon [1].


P

Q R

p

q

r

A

B

a

a

D

A′

Q′

B′

l

Given an acute angle 6 PQR, let p be the perpendicular to QR at Q. Let q be a perpendicularto p that intersects PQ at point A, and let r be the perpendicular to p at B that is halfwaybetween Q and A.

Using Axiom 6, construct a fold l that places A at A′ on PQ and Q at Q′ on r. Let B′ be thereflection of B around l.

Draw the lines QB′ and QQ′. We claim that 6 PQB′, 6 B′QQ′ and 6 Q′QR are a trisection of6 PQR.

70

8.1.2 First proof

P

Q R

p

q

r

A

B

a

a

D

A′

Q′

B′

l

αα

α

α

α

Since A′, B′, Q′ are all reflections around the same line l of the points A, B, Q on one line DQ,they are all on one line DQ′. By construction, AB = BQ, BQ′ is perpendicular to AQ; BQ′ isa common side, so4ABQ′ ∼= 4QBQ′ by side-angle-side. Therefore, 6 AQ′B = 6 QQ′B = α,since Q′B is the perpendicular bisector of the isoceles triangle4AQ′Q.

By alternating interior angles, 6 Q′QR = 6 QQ′B = α.

By reflection,4AQ′Q ∼= 4A′QQ′.1

The fold l is the perpendicular bisector of both AA′ and QQ′; drop perpendicularsfrom A and A′ to QQ′; then AQ = A′Q′ follows by congruent right triangles.AA′Q′Q is an isoceles trapezoid so its diagonals are equal AQ′ = A′Q.

Therefore, QB′, the reflection of Q′B, is the perpendicular bisector of an isoceles triangle and6 A′QB′ = 6 B′QQ′ = 6 QQ′B = 6 Q′QR = α.

1The two triangles have been emphasized using different patterns of dashes and dots, as well as using color.

71

8.1.3 Second proof

P

Q R

p

q

r

A

B

a

a

D

A′

Q′

B′

l

U

V

b

b

α

α

α

α

Since l is a fold, it is the perpendicular bisector of QQ′. Denote the intersection of l withQQ′ by U, and its intersection with QB′ by V. 4VUQ ∼= 4VUQ′ by side-angle-side sinceVU is a common side, the angles at U are right angles and QU = Q′U = b. Therefore,6 VQU = 6 VQ′U = α and then 6 Q′QR = 6 VQ′U = α by alternating interior angles.

As in Proof 1, A′, B′, Q′ are all reflections around l, so they are all on one line DQ′, andA′B′ = AB = BQ = B′Q′ = a. Then4A′B′Q ∼= 4Q′B′Q and 6 A′QB′ = 6 Q′QB′ = α.

72

8.2 Martin’s trisection of an angle


P

QR

M

p

q

lP′

Q′

V

U

W

a

a

b

b

c

c

γ

γ

β

β

α

α

α

Given the acute angle 6 PQR, let M be the midpoint of PQ. Construct p the perpendicular toQR through M and construct q perpendicular to p through M. q is parallel to QR.

Using Axiom 6, construct a fold l that places P at P′ on p and Q at Q′ on q. More than onefold may be possible; choose the one that intersects PM.

Draw the lines PP′ and QQ′. Denote the intersection of QQ′ with p by U and its intersectionwith l by V. Denote the intersection of PQ and P′Q′ with l by W.2

8.2.2 Proof

4QMU ∼= 4PMP′ by angle-side-angle: 6 P′PM = 6 UQM = β by alternate interior angles;QM = MP = a since M is the midpoint of PQ; 6 QMU = 6 PMP′ are vertical angles.Therefore, P′M = MU = b.

4P′MQ′ ∼= 4UMQ′ by side-angle-side: we have shown that P′M = MU = b; the angles atM are right angles; MQ′ is a common side. Since the altitude of the isoceles triangle4P′Q′Uis the bisector of 6 P′Q′U, so 6 P′Q′M = 6 UQ′M = α.

4QWV ∼= 4Q′WV by side-angle-side: QV = VQ′ = c; the angles at V are right angles sincethe fold is the perpendicular bisector of QQ′; VW is a common side. Therefore, 6 WQV =

β = 6 WQ′V = 2α. By alternate interior angles 6 Q′QR = 6 MQ′Q = α. We have 6 PQR =

β + α = 2α + α = 3α so 6 Q′QR is one-third of 6 PQR.

2It is not immediate that both PQ and P′Q′ intersect l at the same point. 4PP′W ∼ 4QQ′W so the altitudesdivide the vertical angles 6 PWP′, 6 QWQ′ similarly and thus must be on the same line.

73

Chapter 9

Doubling a Cube

9.1 Messer’s doubling of a cube

To double a cube we need to construct 3√

2. This construction is based on [12, 10].

9.1.1 Dividing a length into thirds

Lang [9] shows efficient constructs for rational fractions of the length of the side of a square(piece of paper). Here, we need to divide the side of the square into thirds.

First, fold the square in half to locate the point J = (1, 1/2). Next, draw the lines AC and BJ.

A = (0, 1)

B = (0, 0) C = (1, 0)

D = (1, 1)

I = (0, 1/2) J = (1, 1/2)

y = 1 − x

y =12

x

K =(2/3, 1/3)

E = (0, 1/3) F = (1, 1/3)

G = (0, 2/3) H = (1, 2/3)

The coordinates of their point of intersection K are obtained by solving the two equations:

y = 1− x

y =12

x .

The result is x = 2/3, y = 1/3.

Construct the line EF perpendicular to AB through K, and construct the reflection GH of BCaround EF. The side of the square has been divided into thirds.

74

9.1.2 Building 3√

2

A

B C

D

E F

G H

C′α

α′

F′

α′

Jα

l

(a + 1)− b

(a + 1)− bb

a + 13

a + 13

a

1

a− a + 13

a + 13

Label the side of the square by a + 1. The construction will show that a = 3√

2.

Using Axiom 6 place C at C′ on AB and F at F′ on GH. Denote by J the point intersection ofthe fold with BC and denote by b the length of BJ. The length of JC is (a + 1)− b.

When the fold is performed, the line segment JC is reflected onto the line segment JC′ ofthe same length, and CF is folded onto the line segment C′F′ of the same length. A simplecomputation shows that the length of GC′ is:

a− a + 13

=2a− 1

3. (9.1)

Finally, since 6 FCJ is a right angle, so is 6 F′C′ J.

4C′BJ is a right triangle so by Pythagoras’ theorem:

12 + b2 = ((a + 1)− b)2

a2 + 2a− 2(a + 1)b = 0

b =a2 + 2a

2(a + 1).

75

6 GC′F′ + 6 F′C′ J + 6 JC′B = 180◦ since they form the straight line GB. Denote 6 GC′F′ by α.

6 JC′B = 180◦ − 6 F′C′ J − 6 GC′F′ = 180◦ − 90◦ − α = 90◦ − α ,

which we denote by α′. The triangles4C′BJ,4F′GC′ are right triangles, so 6 C′ JB = α and6 C′F′G = α′. Therefore, the triangles are similar and using Equation 9.1 we have:

b(a + 1)− b

=

2a− 13

a + 13

.

Substituting for b:a2 + 2a

2(a + 1)

(a + 1)− a2 + 2a2(a + 1)

=2a− 1a + 1

a2 + 2aa2 + 2a + 2

=2a− 1a + 1

.

Simplifying results in a3 = 2 and a = 3√

2.

76

9.2 Beloch’s doubling of a cube

Margharita P. Beloch formalized Axiom 6 (Section 7.6) and showed that it could be used tosolve cubic equations. Here we give her construction for doubling the cube. The solution ofcubic equations is discussed in Chapters 10, 11.


Place point A at (−1, 0) and point B at (0,−2). Let p be the line with equation x = 1 and letq be the line with equation y = 2. Using Axiom 6 construct a fold l that places A at A′ on pand B at B′ on q. Denote the intersection of the fold and the y-axis by Y and the intersectionof the fold and x-axis by X.

O(0, 0)

A(−1, 0)

B(0,−2)

y-axis

x-axis

p : x = 1

q : y = 2

A′

B′

Y

X

l

77

9.2.2 Proof

Let us extract a simplified diagram:

OA α

B

α′

A′

B′

Y

αα′

Xα

α′

The fold is the perpendicular bisector of AA′ and BB′. Therefore, 6 AYX and 6 YXB are rightangles and AA′ is parallel to BB′. By alternate interior angles 6 YAO = 6 BXO = α. If anacute angle in a right triangle is α, the other acute angle must be 90◦ − α, which we denote α′.The labeling of the angles in all the triangles in the diagram follows immediately.

We have three similar triangles4AOY ∼ 4YOX ∼ 4XOB. OA = 1, OB = 2 are given, so:

OYOA

=OXOY

=OBOX

OY1

=OXOY

OY2= OX

OY1

=2

OX

OY2= OX =

2OY

,

resulting in OY3= 2 and OY = 3

√2.

78

Chapter 10

Lill’s Method for Finding Roots

10.1 Magic

Construct a path consisting of four line segments {a3, a2, a1, a0} of lengths:

{a3 = 1, a2 = 6, a1 = 11, a0 = 6} ,

starting from the origin, first along the positive direciton of the x-axis and turning 90◦

counterclockwise between segments. Construct a second path as follows: draw a line fromthe origin at an angle of 63.4◦ and mark its intersection with a2 by P. Turn left 90◦, draw aline and and mark its intersection with a1 by Q. Turn left 90◦ once again, draw a line andnote that it intersects the end of the first path at (−10, 0).

−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2

1

2

3

4

5

6

7

a3 = 1

a2 = 6

a1 = 11

a0 = 6

P

63.4◦

Q

Let p(x) = a3x3 + a2x2 + a1x + a0 = x3 + 6x2 + 11x + 6. Compute tan 63.4◦ = 2, the tangentof the angle at the start of the second path. Then:

p(− tan 63.4◦) = (−2)3 + 6(−2)2 + 11(−2) + 6 = 0 .

Congratulations! You have found a root of the cubic polynomial x3 + 6x2 + 11x + 6.

79

10.2 Introduction

This example demonstrates a method discovered by Eduard Lill in 1867 for graphicallyfinding (or more accurately, verifying) the real roots of any polynomial [3, 7, 16]. We limit thepresentation to cubic polynomials. Lill’s method has seen renewed interest because it can beimplemented using origami, as we shall see in Chapter 11.

In Sections 10.3–10.4 we continue the initial example to find additional roots and to showthat if an angle α is such that (− tan α) is not a root, then the construction doesn’t work.Section 10.5 presents the full specification of Lill’s method. Special cases of the method aredemonstrated by the examples in Sections 10.6–10.8. Since Lill’s method can find a real rootof any cubic polynomial, it can be used to trisect an angle. By computing 3

√2 as a root of

x3 − 2, it can double a cube (Section 10.9). Section 10.10 gives a proof that Lill’s method canfind the real roots of any cubic polynomial. The proof for arbitrary polynomials is similar.

10.3 Multiple roots

Let us continue the example above. The polynomial p(x) = x3 + 6x2 + 11x + 6 has threeroots −1,−2,−3. Computing the arc tangent of the negation of the roots gives:

α = − tan−1(−1) = 45◦, β = − tan−1(−2) = 63.4◦, γ = − tan−1(−3) = 71.6◦ .

In the diagram below we see that for each of the three angles, the second path intersects theend of the first path.

−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2

1

2

3

4

5

6

7

1

6

11

6

P1

α

Q1

P2

β

Q2

P3

γ

Q3

80

10.4 Paths that do not lead to roots

Perhaps the second path intersects the end of the first path for any initial angle, for example,56.3◦. In the following diagram, the second path intersects the extension of the line seg-ment for the coefficient a0, but not at (−10, 0), the end of the first path. We conclude that− tan 56.3◦ = −1.5 is not a root of the equation.

−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2

1

2

3

4

5

6

7

1

6

11

6

P

56.3◦

Q

81

10.5 Specification of Lill’s method

Examining the examples below will help understand the details.

• Start with an arbitrary cubic polynomial: p(x) = a3x3 + a2x2 + a1x + a0.

• Construct the first path as follows:

– For each coefficient a3, a2, a1, a0 (in that order) draw a line segment starting at theorigin O = (0, 0) in the positive direction of the x-axis. Turn 90◦ counterclockwisebetween each segment.

• Construct the second path as follows:

– We use the symbol for a coefficient ai to also denote the corresponding side of thefirst path.

– Construct a line from O at an angle of θ with the positive x-axis that intersects a2

at point P.

– Turn ±90◦ and construct a line from P that intersects a1 at Q.

– Turn ±90◦ and construct a line from Q that intersects a0 at R.

– If R is the end point of the first path, then − tan θ is a root of p(x).

• Special cases:

– When drawing the line segments of the first path, if a coefficient is negative, drawthe line segment backwards.

– When drawing the line segments of the first path, if a coefficient is zero, do notdraw a line segment but continue with the next ±90◦ turn.

• Notes:

– “Intersects ai” means “intersects the line that contains the line segment ai”.

– When building the second path, choose to turn left or right by 90◦ so that there isan intersection with the next segment of the first path.

82

10.6 Negative coefficients

Section 7.6 gave an example of the use of Axiom 6 that resulted in the polynomial p(x) =x3 − 3x2 − 3x + 1 with negative coefficients.

We start by drawing a segment of length 1 to the right. Next we turn 90◦ to face up, but thecoefficient is negative, so we draw a segment of length 3 down. After turning 90◦ to the left,the coefficient is again negative, so we draw a segment of length 3 to the right. Finally, weturn down and draw a segment of length 1.

We start the second path with a line angled 45◦ with the x-axis. It intersects the extension ofthe line segment for a2 at (1, 1). Turning −90◦ (to the right), the line intersects the extension ofthe line segment for a1 at (5,−3). Turning −90◦ again, the line intersects the end of the firstpath at (4,−4).

Since − tan 45◦ = −1, a real root of the polynomial is −1:

p(−1) = (−1)3 − 3(−1)2 − 3(−1) + 6 = 0 .

The loosely dashed lines in the diagram will be discussed in Section 10.8.

0 1 2 3 4 5 6

−5

−4

−3

−2

−1

1

2

1

a2 = −3

a1 = −3

1

P

Q

83

10.7 Zero coefficients

a2, the coefficient of the x2 term in the polynomial x3 − 7x − 6 = 0, is zero. For a zerocoefficient, we “draw” a line segment of length 0, that is, we do not draw a line, but we stillmake the ±90◦ turn before “drawing” it, as indicated by the arrow pointed up at point (1, 0)in the diagram. Next make an additional turn and draw a line of length −7, that is, of length7 backwards, to point (8, 0). Finally, turn again and draw a line of length −6 to point (8, 6).

There are three second paths that intersect the end of the first path. They start with angles of:

α = 45◦, β = 63.4◦, γ = −71.6◦ .

We conclude that there are three real roots:

− tan 45◦ = −1, − tan 63.4◦ = −2, − tan(−71.6◦) = 3 .

Check:(x + 1)(x + 2)(x− 3) = x3 − 7x− 6 .

0 1 2 3 4 5 6 7 8 9 10 11

−3

−2

−1

1

2

3

4

5

6

7

O

A

1 −7

−6

P1

Q1

P2

Q2

P3

Q3

84

10.8 Non-integer roots

Consider the polynomial p(x) = x3 − 2x + 1. The first segment is from (0, 0) to (1, 0) andturns up. The coefficient of x2 is zero so no segment is drawn and turns left. The nextcoefficient is negative so the segment it goes backwards from (1, 0) to (3, 0) and turns right.Finally, a segment is drawn from (3, 0) to (3,−1). Clearly, 1 is a root of p(x) and since− tan−1(−45◦) = 1, there is a path OP1Q1A.

If we divide p(x) by x− 1, we obtain the quadratic polynomial x2 + x− 1 whose roots are−1±

√5

2 ≈ 0.62, −1.62. There are two additional second paths: one starting at −31.8◦ since− tan−1 0.62 = −31.8◦, and one starting at 58.3◦ since − tan−1 1.62 = 58.3◦.

Similarly, the polynomial in Section 10.6 has roots 2±√

3 ≈ 3.73, 0.27. The correspondingangles are −75◦ and −15◦, because − tan(−75◦) ≈ 3.73 and − tan(−15◦) ≈ 0.27.

0 1 2 3 4

−1

1

2

O

A

1 −2

1

P1

Q1

P2

Q2

P3

Q3

10.9 The cube root of two

3√

2 is a root of the cubic polynomial x3 − 2. In the first path, we turn left twice withoutdrawing any line segments, because a2 and a1 are both zero. Then we turn left again (to facedown) and draw backwards because a0 = −2 is negative. The first segment of the secondpath is drawn at an angle of −51.6◦ and − tan(−51.6◦) ≈ 1.26 ≈ 3

√2.

0 1 2 3

−1

1

2

1

−2

P1

Q1

85

10.10 Proof of Lill’s method

We limit ourselves to monic cubic polynomials p(x) = x3 + a2x2 + a1x + a0.1 In the diagrambelow, segments of the first path are labeled with coefficients and with b2, b1, a2 − b2, a1 − b1.

Since the sum of the angles of a triangle is 180◦, in a right triangle if one acute angle is θ, theother is 90◦ − θ. Therefore, the angle above P and the angle to the left of Q are equal to θ. Wenow derive a sequence of formulas for tan θ:

tan θ =b2

1= b2

tan θ =b1

a2 − b2=

b1

a2 − tan θ

b1 = tan θ(a2 − tan θ)

tan θ =a0

a1 − b1=

a0

a1 − tan θ(a2 − tan θ).

Simplifying the last equation gives:

(tan θ)3 − a2(tan θ)2 + a1(tan θ)− a0 = 0

−(tan θ)3 + a2(tan θ)2 − a1(tan θ) + a0 = 0

(− tan θ)3 + a2(− tan θ)2 + a1(− tan θ) + a0 = 0 .

We conclude that − tan θ is a real root of p(x) = x3 + a2x2 + a1x + a0.

−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2

1

2

3

4

5

6

7

1

a2

a1

a0

P

b2

a2 − b2

Qa1 − b1 b1

θ

θ

θ

1If the polynomial is not monic, divide it by a3 and the resulting monic polynomial has the same roots.

86

Chapter 11

Beloch’s Fold and Beloch’s Square

11.1 The Beloch fold

Margharita P. Beloch discovered a remarkable connection between origami and Lill’s methodfor finding roots of polynomials [7]. She found that one application of the operation oforigami Axiom 6 (Section 7.6) applied to the first path of Lill’s method can obtain a real rootof any cubic polynomial. The operation is often called the Beloch fold.

Consider the polynomial p(x) = x3 + 6x2 + 11x + 6 (Section 10.1). In the following diagramwe have emphasized the second path and renamed some vertices. To solve the equation weperform a Beloch fold to simultaneously place the points P, Q at P′, Q′ on the line segmentsof lengths a2, a1, respectively. Unfortunately, if you perform the fold, the path does not solvethe equation: Q′ is way off to the right, so the angles at P′ and Q′ are not right angles.

−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2

1

2

3

4

5

6

7

PQ

a2

a1

P′?

Q′?

87

Recall that a fold is the perpendicular bisector of the line segment between any point andits reflection around the fold. We want P′?Q′? to be a fold so that it will be perpendicularto both QQ′ and PP′. The fold is the perpendicular bisector of QQ′ and PP′, so P′, Q′, thereflections of P, Q, must be the same distance away from the fold as P and Q, respectively.With some change of notation we have the following diagram.

−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3

1

2

3

4

5

6

7

8

9

10

11

12

13

PQ

a2

a1 a′2

a′1

R

S

P′

Q′

A line a′2 is drawn so that it is parallel to a2 and the same distance from a2 as a2 is from P.Similarly, line a′1 is drawn so that it is parallel to a1 and the same distance from a1 as a1 isfrom Q. Apply Axiom 6 to simultaneously place P at P′ on a′2 and to place Q at Q′ on a′1. Thefold RS is the perpendicular bisector of the lines PP′ and QQ′; therefore, the angles at R andS are right angles as required.

88

Let us try the Beloch fold on the polynomial x3 − 3x2 − 3x + 1 from Section 10.6. a2 is thevertical line segment of length 3 whose equation is x = 1, and its parallel line is a′2 whoseequation is x = 2, because P is at a distance of 1 from a2. a1 is the horizontal line segmentof length 3 whose equation is y = −3, and its parallel line is a′1 whose equation is y = −2because Q is at a distance of 1 from a1. The fold RS is the perpendicular bisector of both PP′

and QQ′. The line PRSQ is the same as the second path in Section 10.6.

0 1 2 3 4 5 6

−4

−3

−2

−1

1

2

P

Q

a2

a1

P′

Q′

R

S

a′2

a′1

89

11.2 The Beloch square

This construction in the previous section can be expressed in terms of a Beloch square: Giventwo points P, Q and two lines a2, a1, construct a square ARSB, such that:

• One side is RS where R lies on a2 and S lies on a1;

• P lies on RA and Q lies on SB.

The following diagram extends the construction for x3 + 6x2 + 11x + 6 to show the Belochsquare. The length of RS is

√80 = 4

√5 ≈ 8.94. We can construct the square by adding three

sides of this length.

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3

1

2

3

4

5

6

7

−6

−5

−4

−3

−2

−1

a2

a1

B

A

R

S

P

Q

90

Bibliography

[1] Oriah Ben-Lulu. Angle trisections in various axiom systems. Weizmann Institute ofScience, 2020. (in Hebrew).

[2] Benjamin Bold. Famous Problems of Mathematics: A History of Constructions with StraightEdge and Compass. Van Nostrand, 1969.

[3] Phillips Verner Bradford. Visualizing solutions to n-th degree algebraic equa-tions using right-angle geometric paths. Archived May 2, 2010, at the WaybackMachine, https://web.archive.org/web/20100502013959/http://www.concentric.net/~pvb/ALG/rightpaths.html, 2010.

[4] Heinrich Dorrie. 100 Problems of Elementary Mathematics: Their History and Solution.Dover, 1965.

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