The Secret Life of Triangles

Christopher ThomsonID: 4265645

Supervisor: Trevor Hawkes

A report presented in the Department of Mathematics and Physics, Coventry University, for thedegree of Bachelor of Science in Mathematics, April 2015.

Abstract

This report explores various theorems and properties related to the Euclidean geometry of triangles,as well as providing proofs and figures to further enhance the explanation. This report containsfour chapters, each with a different key focus.

The first chapter, entitled Three Lines Through a Point, explains and proves theorems relating tothree lines that pass through a point (concurrent), with a particular emphasis on so called ’trianglecentres’, which are listed in Kimberling’s Encylopedia of Triangle Centres. These include, amongstothers, the incentre, the centroid, the orthocentre and the circumcentre.

The second chapter, called Three Points Through a Line, explains and proves four theoremsthat involve a line passing through three points (collinear). The first, Menelaus’ Theorem, is usedto prove Desargues’ and Pappus’s Theorems.

The third chapter, entitled Unexpected Equilateral Triangles, gives detail on two theorems thatinvolve three points forming equilateral triangles. This includes Morley’s Theorem with an extensiveproof, and Napoleon’s Theorem.

The final chapter talks about the nine-point circle and the Euler line, as well as properties andtheorems relating to them.

Each chapter gives detailed figures to provide a visual representation of the theorems. Themajority of these figures have been drawn in GeoGebra. Each section is also cited, with referenceslisted in the appendices at the end of the report.

Page: 1

Contents

Introduction 4

Literature Review 5

1 Three Lines Through A Point 7

1.1 Ceva’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 The Centre of a Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 Trilinear Co-ordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4 The Incentre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.5 The Centroid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.6 The Orthocentre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.7 The Circumcentre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.8 The Napoleon Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.9 The Exeter Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2 Three Points On A Line 22

2.1 Menelaus’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.2 Desargues’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.3 Pappus’s Hexagon Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.4 The Simson Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3 Unexpected Equilateral Triangles 29

3.1 Morley’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2 Napoleon’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2

4 The Nine-Point Circle and The Euler Line 37

4.1 The Nine-Point Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.2 The Euler Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Conclusion 43

References 44

Appendices 46

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Introduction

What is a triangle? The simple definition of a triangle is a polygon with three sides and threevertices. However, there is a lot more to a triangle than that. For example, the internal angles addup to 180 degrees, and the external angles add up to 360 degrees; a triangle with three equal sidesand three equal angles is called an equilateral triangle; and in a right-angled triangle, the squareof the hypotenuse equals the sum of squares of the other sides (Pythagoras’s Theorem). Thereare many well-known properties and theorems relating to triangles, however there are many moreless-known and complex properties and theorems that we can explore.

For example, where is the centre of a triangle? How do you find it, and is there a single centre?This report aims to answer this question, and many more, in order to discover the secrets of triangles.In four themed chapters, this report will provide detailed explanations of various properties andtheorems relating to the Euclidean geometry of triangles that have been discovered throughouthistory, as well as providing mathematical proofs and detailed diagrams for the majority, to furtherenhance the explanation. I will also research into the history of the person responsible for thediscovery and/or proof of the mentioned theorem.

To complete this report, I will undertake research from various sources, gathering informationabout a wide range of theorems related to triangles. In the first chapter we will answer the questionabout the centre of a triangle, detailing a few possible candidates. In the second chapter we willfocus on theorems that involve three collinear points. In the third chapter we will discuss theoremsthat involve three points making an equilateral triangle. And in the final chapter we will discussfurther findings that do not fit into the themes of the previous chapters.

As detailed further in the literature review, I will use a variety of sources for my research, rangingfrom books to on-line articles, each with their own strengths and weaknesses. Each source providesa different use for my research. For example, some sources provide mathematical proofs for statedtheorems, whilst others do not. I hope you will now read on with interest and find this report veryinformative!

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Literature Review

This report involved researching various topics relating to the Euclidean geometry of triangles. Asthis is a purely desk-based project, the content of this report is based on 100% research. I researchedvarious properties, theorems and proofs relating to triangle geometry. I have found my literaturethrough the library and internet searching and chosen them based on the specific topics they coveras well as the quality of explanations for each. This literature review examines the content of themain literature sources I used, how I used them, and provides comparisons between each based onhow useful they were to my research.

The first book I used for my research was Geometry Revisited by Coxeter, H.S.M. and Greitzer,S.L. (1967). This book explains, in detail, a wide range of topics relating to Euclidean geometry thatwas beneficial to my research. The first chapter is purely based on geometry relating to triangles,and was my main area of reading. The third chapter is about collinearity and concurrency, whichalso proved useful for my report. The book introduces each topic with a brief history of the personwho discovered and/or proved the theorem that is then mentioned, followed by a full proof of thetheorem. There are also plenty of figures/images that give a visual representation of the theorems.

Another book I used for my report was Exploring Advanced Euclidean Geometry with GeoGebraby Venema, G.A. (2013). Like Coxeter and Greitzer (1967), this book covers a wide range of topicsrelating to Euclidean geometry. There was a larger emphasis on triangle geometry, with seven ofthe fourteen chapters related to triangles. Compared to Coxeter and Greitzer (1967), each sectionstarts with a generally more extensive summary of the history of the person responsible for eachtheorem. However, the book often does not provide a proof for each theorem, and the majority ofthe time it leaves proving a theorem as a task for the reader in the form of exercises. This means Ihad to use multiple sources for a topic to research the theorem and the proof. Each section includesa figure/image alongside the theorem, often in much better detail than Coxeter and Greitzer (1967).

An on-line resource that contains a high amount of useful information is Wolfram MathWorld,an extensive mathematical resource created by Eric Weisstein (1995), as well as many other contrib-utors. This website contains several pages on almost every area of mathematics, including geometry,which was needed for this report. Each page about a specific topic contains a brief description, aswell as many properties and information relating to the topic. There are also helpful figures/imagesto give a visual representation to support my findings. Unlike Coxeter and Greitzer (1967), thereare no proofs of any theorems that are included, so this resource was only used for information andnot proofs. There is also little to no information on the history of the theorem or person responsiblefor discovering it, so other sources had to be used for this information.

As well as Wolfram Mathworld, I used many more on-line resources for my research, mostly for

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single topics in this report. For example, I used the website MacTutor and Encylopaedia Brittanicato research the history on some mathematicians. I also researched various on-line webpages, articles,PDFs and extracts to find information on specific topics.

All resources I used give an image/figure to provide a visual representation of the topic mentionedin the section or article. However, I decided, in most cases, to recreate the figures in GeoGebramyself, to demonstrate that I have understood the topic. GeoGebra is an interactive geometry,graphing and algebra application, where the user can freely draw geometrical figures. I learnt howto use this software through practice, as well as the GeoGebra Wiki and various on-line help forums.

I believe I have used a wide range of resources for my research, and have utilised the bestelements from each.

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Chapter 1

Three Lines Through A Point

In geometry, if three or more lines intersect each other at a single point, they are said to beconcurrent. A single triangle has many ways of drawing three lines that are concurrent, and in mostcases, this point of intersection is called a centre. Before looking further into these centres, we willfirst look at a very important theorem.

1.1 Ceva’s Theorem

Giovanni Ceva (c.1647 - c.1734) was an Italian mathematician and professor at the University ofPisa until becoming the Professor of Mathematics at the University of Mantua in 1686. His mainarea of study was geometry, and published many works throughout his life, including Opusculamathematica and Geometria Motus (O’Connor and Robertson 2012). However, in his workDe lineis rectis, he published an important theorem:

Theorem 1.1.1 (Ceva’s Theorem). Let ABC be a triangle and D, E and F be points on lines BC,AC and AB respectively. The lines AD, BE and CF are concurrent iff (if and only if)

BD

DC· CEEA· AFFB

= 1 (1.1)

(Coxeter and Greitzer, 1967: p4-7)

Any line that passes through the vertex of a triangle and the opposite side is called a cevian.This theorem is important for proving that three cevians are concurrent, and will be used extensivelyin this chapter to prove special sets of cevians are concurrent.

Proof. First we show that equation (1.1) is necessary for the three cevians to be concurrent. Assumethey meet at a point G. Let α = BD/DC. We will show that α = ABG/CAG. Draw altitudesh1 perpendicular from BC to A and h2 perpendicular from BC to G (See Figure 1.1). The area ofABD is 0.5h1BD and the area of ACD is 0.5h1DC. If we divide ABD and ACD we get

ABD

ACD=

0.5h1BD

0.5h1DC

7

Figure 1.1: Ceva’s Theorem Proof

Cancelling out 0.5h1 from the fraction on the right hand sides leaves

ABD

ACD=BD

DC

Now do the same using altitude h2

BGD

CGD=

0.5h2BD

0.5h2DC=BD

DC

ThereforeBD

DC=ABD

ACD=BGD

CGD=ABD −ACDBGD − CGD

Leaving

α =BD

DC=ABG

CAG(1.2)

Using the same process, we can show that

β =CE

EA=BCG

ABG(1.3)

and

γ =AF

FB=CAG

BCG(1.4)

Multiplying equations (1.2), (1.3) and (1.4) together, we have

αβγ =BD

DC· CEEA· AFFB

=ABG

CAG· BCGABG

· CAGBCG

By cancelling terms on the right hand side we obtain

BD

DC· CEEA· AFFB

= 1 (1.5)

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as desired.

We can also prove the inverse of the theorem to show that (1.1) is sufficient. Assume the ceviansAD and BE intersect at point G, and the third cevian that passes through this point is CF ′. Then

BD

DC· CEEA· AF

′

F ′B= 1

But from theorem 1.1.1 we know that

BD

DC· CEEA· AFFB

= 1

From this we can show thatAF ′

F ′B=AF

FB

So therefore F ′ and F are the same point, therefore AD, BE and CF are concurrent. QED

(Wilson, n.d.)(Coxeter and Greitzer, 1967: p4-7)

1.2 The Centre of a Triangle

Where is the centre of a triangle? There is no single way to find the centre, thus there is no singlepoint that can be considered the centre of a triangle. Some of the candidates for the centre of atriangle are derived from various geometric ideas. The most common process is by constructingthree lines in the same way from each vertex and/or edge of the triangle, and by using Ceva’sTheorem we can prove the three lines are concurrent. This single point becomes one of the manycentres of the triangle.

Clark Kimberling (1942-), a Mathematics professor at the University of Evansville, USA, main-tains an extensive list of triangle centres that have been discovered and proved by many othermathematicians over the past few millennia. Currently, the list contains over 7,500 triangle centres,with more being discovered over time. Listed below are a few select centres, most of which will bediscussed further in this report:

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Ref Sym Centre Definition

X1 I Incentre Intersection of the three angle bisectors, and the centre ofthe incircle (Chapter 1.4)

X2 G Centroid Intersection of the three medians (Chapter 1.5)

X3 O Circumcentre Intersection of the three perpendicular bisectors, and thecentre of the circumcircle (Chapter 1.7)

X4 H Orthocentre Intersection of the three altitudes (Chapter 1.6)

X5 N Nine-Point Centre Centre of the nine-point circle (Chapter 4.1)

X6 K Symmedian Point Intersection of the three symmedians (reflection of mediansover corresponding angle bisectors)

X7 Ge Gergonne Point Symmedian point of contact triangle (made from pointswhere incircle meets triangle)

X8 Na Nagel Point Intersection of lines from each vertex to the correspondingsemiperimetre point

X9 M Mittenpunkt Symmedian Point of the triangle formed by the centres ofthe three excircles

X10 Sp Spieker Centre Centre of the Spieker circle (incircle of the medial triangle)

X11 F Feuerbach Point Point where incircle is tangent to the nine-point circle(Chapter 4.1)

X13 F Fermat Point Point where total distance from the three vertices is mini-mum

X17 N1 First Napoleon Point Intersection of the lines from vertices to centroids of exter-nally drawn equilateral triangles (Chapter 1.8)

X18 N2 Second Napoleon Point Intersection of the lines from vertices to centroids of inter-nally drawn equilateral triangles (Chapter 1.8)

X20 L de Longchamps Point Reflection of orthocentre around the circumcentre

X21 S Schiffler Point Concurrence of the Euler Lines of ABC, IBC, IAC andIAB, where I is the incentre (Chapter 4.2)

X22 Ex Exeter Point (See Chapter 1.9)

(Kimberling, c.1994)

Figure 1.2 gives a representation of the many triangle centres that have been discovered. How-ever, it does not show all of them, as only so many can be shown on one image.

1.3 Trilinear Co-ordinates

If we have a triangle ABC, we may need to know where a particular point, P say, in the sameplane is. This position is determined by the ratios of the distances from P to the sides of triangleABC (See Figure 1.3). These ratios are the trilinear co-ordinates of P and are written in the formα : β : γ, where α, β and γ are the directed ratio of the distances from X to BC, AC and ABrespectively. For example, the vertices of triangle ABC have co-ordinates A = 1 : 0 : 0, B = 0 : 1 : 0and C = 0 : 0 : 1. (Kimberling, 1994)

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Figure 1.2: Triangle Centres (Weisstein, ”Kimberling Center”)

Figure 1.3: Trilinear Co-ordinates

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1.4 The Incentre

Definition 1.4.1 (Internal Angle Bisector). A line that passes through a vertex, so as to cut thevertex in half internally.

Theorem 1.4.2 (Angle Bisector Theorem). The three internal angle bisectors of a triangle areconcurrent.

Definition 1.4.3 (Incentre). The point of intersection of the internal angle bisectors is called theincentre.

The incentre is often labelled as I, or X1. The incentre lies equal distance from each of thethree sides of the triangle. The incentre is also the centre of the incircle, a circle that fits insidethe triangle so that the three sides form tangents to the circle. The distance from each side of thetriangle to the incentre is known as the inradius. (See Figure 1.4) (Coxeter and Greitzer, 1967:p11-14)

The Cartesian co-ordinates of the incentre can be determined by

(xI , yI) =

(axA + bxB + cxC

a+ b+ c,ayA + byB + cyC

a+ b+ c

)where a, b and c are the lengths of sides BC, AC and AB respectively.

The trilinear co-ordinates of the incentre are 1 : 1 : 1, due to it being equidistant from each sideof the triangle. (Weisstein, ”Incenter”)

Figure 1.4: Incentre and Incircle

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Proof. Let ABC be a triangle, D, E and F be points on BC, AC and AB respectively, and AD,BE and CF be the angle bisectors of ABC.

Lemma 1.4.4. Let ABC be a triangle and the angle bisector of ∠CAB intersect BC at D. Then

AB

AC=BD

CD

Using the above lemma for each fraction in Ceva’s Theorem

BD

DC=BA

AC

CE

EA=CB

BA

AF

FB=AC

CB

This leads toBD

DC· CEEA· AFFB

=BA

AC· CBBA· ACCB

= 1

Therefore the three angle bisectors of triangle ABC are concurrent. QED

(Faucette, 2007)

Excircles and Excentres

Excircles, also known as escribed circles, are circles that lie fully outside the triangle. An excircleis drawn tangent to one side of the triangle, externally, and the two other sides when they areextended outwards. Each triangle has three distinct excircles, one for each side. The centre ofeach excircle is called the excentre. These centres can be found by the intersection of the internalangle bisectors of the opposite angle, and the external angle bisectors (perpendicular to the internalangle bisectors) of the other two angles (See Figure 1.5). Excentres are often labelled Ji, where iis the vertex of ABC opposite the excentre. The trilinear co-ordinates of each of the excentres are−1 : 1 : 1, 1 : −1 : 1 and 1 : 1 : −1. (Weisstein, ”Excircles”)

1.5 The Centroid

Definition 1.5.1 (Median). A line that passes through a vertex of a triangle and the midpoint ofthe opposite side is called a median.

Theorem 1.5.2 (Median Concurrence Theorem). The three medians of a triangle are concurrent.

Definition 1.5.3 (Centroid). The point of intersection of the medians is called the centroid.

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Figure 1.5: Excircles and excentres

A median effectively cuts the area of a triangle in half, therefore the three medians of a trianglewill cut it into six smaller triangles of equal area. The centroid is often labelled G, or X2. Thecentroid of a triangle is also known as the centre of mass. That is, if the triangle was a physical,2D object, it would balance on this point (See Figure 1.6). (Coxeter and Greitzer, 1967: p7-8)

The Cartesian co-ordinates of the centroid can be determined by

(xG, yG) =

(xA + xB + xC

3,yA + yB + yC

3

)There are three ways to calculate the trilinear co-ordinates of the centroid:

bc : ca : ab = 1/a : 1/b : 1/c = csc(A) : csc(B) : csc(C)

(Weisstein, ”Centroid”)

Proof. Let ABC be a triangle, D, E and F be the midpoints of BC, AC and AB respectively, andAD, BE and CF be the medians of ABC. The median from vertex A intersects the midpoint, D,of BC. Therefore BD is equal to DC, and BD/DC = 1. The same follows for the other medians.So CE/EA = 1 and AF/FB = 1. Substituting these into Ceva’s theorem

BD

DC· CEEA· AFFB

= 1 · 1 · 1 = 1

Therefore the three medians of triangle ABC are concurrent. QED

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Figure 1.6: Centroid

(Faucette, 2007)

1.6 The Orthocentre

Definition 1.6.1 (Altitude). A line that passes through a vertex of a triangle and the oppositeside, so that the line is perpendicular to the side, is called an altitude.

Theorem 1.6.2 (Altitude Concurrence Theorem). The three altitudes of a triangle are concurrent.

Definition 1.6.3 (Orthocentre). The point of intersection of the altitudes is called the orthocen-tre.

The orthocentre is often labelled H, or X4. If the triangle is obtuse (contains an angle largerthan 90 degrees), then the orthocentre will lie outside the triangle. When drawing the altitudes,the two edges either side of the obtuse angle must be extended outside the triangle. If the triangleis right-angled (contains an angle of 90 degrees), then the orthocentre coincides with the right angle(See Figure 1.7). (Coxeter and Greitzer, 1967: p9)

The Cartesian co-ordinates of the orthocentre can be determined by

(xH , yH) =

(xA tan(A) + xB tan(B) + xC tan(C)

tan(A) + tan(B) + tan(C),yA tan(A) + yB tan(B) + yC tan(C)

tan(A) + tan(B) + tan(C)

)where A, B and C are the sizes of the respective angles.

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The trilinear co-ordinates of the orthocentre are sec(A) : sec(B) : sec(C). (Weisstein, ”Ortho-center”)

Figure 1.7: Orthocentre

Proof. Let ABC be a triangle, D, E and F be points on BC, AC and AB respectively, and AD,BE and CF be the altitudes of ABC. Take the right-angled triangles BEC and ADC. Because∠BCE is acute, ∠BCE = ∠ACD. This angle together with ∠CBE equals 90 degrees, similarlythis angle with ∠CAD equals 90 degrees. Therefore BEC and ADC are similar triangles whereCE/DC = BC/AC. A similar process with triangles FAC and EAB leads to AF/EA = AC/AB,and with triangles BFC and BDA leads to BD/FB = AB/BC.

BD

DC· CEEA· AFFB

=CE

DC· AFEA· BDFB

=BC

AC· ACAB· ABBC

= 1

Therefore the three altitudes of triangle ABC are concurrent. QED

(Faucette, 2007)

1.7 The Circumcentre

Definition 1.7.1 (Perpendicular Bisector). A line that passes through the midpoint of a side of atriangle, and is perpendicular to the side, is a perpendicular bisector.

Theorem 1.7.2 (Perpendicular Bisector Concurrence Theorem). The three perpendicular bisectorsof a triangle are concurrent.

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Definition 1.7.3 (Circumcentre). The point of intersection of the perpendicular bisectors is calledthe circumcentre.

The circumcentre is often labelled O, or X3. The circumcentre lies equidistant from the threevertices of the triangle. It is also the centre of the circumcircle of the triangle, which is a circle thatpasses through all three vertices (See Figure 1.8). (Coxeter and Greitzer, 1967: p7)

The Cartesian co-ordinates of the circumcentre can be determined by the following process:

D = 2(xA(yB − yC) + xB(yC − yA) + xC(yA − yB))

Ux = ((x2A + y2A)(yB − yC) + (x2B + y2B)(yC − yA) + (x2C + y2C)(yA − yB))/D

Uy = ((x2A + y2A)(xC − xB) + (x2B + y2B)(xA − xC) + (x2C + y2C)(xB − xA))/D

(xO, yO) = (Ux, Uy)

The trilinear co-ordinates of the circumcentre are cos(A) : cos(B) : cos(C). (Weisstein, ”Cir-cumcenter”)

Figure 1.8: Circumcentre

Proof. Draw the perpendicular bisectors of AC and BC. The point of intersection is at point O.As any point on the perpendicular bisector of AC is equidistant from A and C

OA = OC

Likewise for the bisector of BCOB = OC

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ThereforeOA = OB

Because any point equidistant from the end points of a segment lies on its perpendicular bisector,O must lie on the perpendicular bisector of AB, proving the existence of the circumcentre.Since OA = OB = OC, point O is equidistant from A, B and C. Also, this means that there existsa circle with centre at O that passes through all three vertices of the triangle, therefore proving theexistence of the circumcircle.

QED

1.8 The Napoleon Points

Napoleon points are another type of triangle centres, found from the intersection of three lines.However these points are found slightly differently. The method is to construct an equilateraltriangle on each of the three sides of triangle ABC, and connect each vertex of ABC to the centreof the equilateral triangle connected to the opposite side with a line. These lines are concurrent at apoint known as a Napoleon Point. As the equilateral triangles can be drawn outwardly or inwardly,there are two Napoleon points, called First and Second Napoleon points.

Figure 1.9: First Napoleon Point

To find the First Napoleon point for triangle ABC, outwardly construct equilateral triangles on

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sides BC, CA and AB, as in Figure 1.9. Label these triangles BCD, ACE and ABF . Let X, Yand Z be the centres of these triangles respectively. Construct a line that passes through AX, BYand CZ. These lines are concurrent, and meet at the point N1, known as the first Napoleon point.

Finding the Second Napoleon point, or N2, follows the exact same process, but with constructingthe triangles BCD, ACE and ABF inwardly. (Weisstein, ”Napoleon Points”)

Generalisation

When defining Napoleon points, we always draw equilateral triangles on each sides of the triangleand consider their centres. We can consider the centres of these equilateral triangles as the vertexangle of an isosceles triangle, connected to the sides of triangle ABC, and the base angles equal to30 degrees. A generalisation of the Napoleon points determines which other triangles constructed ofthe sides of triangle ABC, have concurrent lines when joining the external vertices and the verticesof ABC.

If the three triangles XBC, Y CA and ZAB, constructed on each side of triangle ABC, haveequal base angles, θ, then the three lines AX, BY , CZ are concurrent at N (See Figure 1.10). Thetrilinear co-ordinates of X, Y , Z and N are:

X = − sin(θ) : sin(C + θ) : sin(B + θ)

Y = sin(C + θ) : − sin(θ) : sin(A+ θ)

Z = sin(B + θ) : sin(A+ θ) : − sin(θ)

N = csc(A+ θ) : csc(B + θ) : csc(C + θ)

(Weisstein, ”Napoleon Points”)

Figure 1.10: Generalisation of the Napoleon Point

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Proof. The intersections of AX and BC is at A′, BY and AC is at B′ and CZ and AB is at C ′

(See Figure 1.10). First we find the areas of BCZ and ACZ.

BCZ = BC ·BZ · sin(B + θ)

ACZ = AC ·AZ · sin(A+ θ)

As BZ = AZACZ

BCZ=AC · sin(A+ θ)

BC · sin(B + θ)

Due to ACZ and BCZ having the same base, CZ, the altitudes from B and C to CZ have thesame ratios as the areas, therefore

AC ′

BC ′=AC · sin(A+ θ)

BC · sin(B + θ)

We can use the same process to find the ratios for BA′/CA′ and AB′/CB′. Multiply them to get

AC · sin(A+ θ)

BC · sin(B + θ)· AB · sin(B + θ)

AC · sin(C + θ)· BC · sin(C + θ)

AB · sin(A+ θ)= 1

From Ceva’s Theorem, it follows that AA′, BB′ and CC ′ are concurrent. It then follows that AX,BY and CZ must also be concurrent. QED

(Floor, 1998)

If the three triangles are similar (same sized angles), then the three lines AX, BY , CZ areconcurrent at N .

Furthermore, if the two base angles connected to each vertex of ABC are equal, ie ∠CBX =∠ABZ, ∠ACY = ∠BCX and ∠BAZ = ∠CAY , then the lines AX, BY , CZ are concurrent at N .

1.9 The Exeter Point

The Exeter point is another special point, and is classed as a triangle centre. To find the Exeterpoint of triangle ABC, draw the circumcircle, the circle that meets all three vertices of ABC.Extend the medians from A, B and C, so they intersect the circumcircle at points A′, B′ and C ′

respectively. Let DEF be a triangle formed from the tangents of the circumcircle at points A, Band C, with D being opposite A, E being opposite B and F being opposite C. The lines thatpass through DA′, EB′ and FC ′ are concurrent, and meet at the Exeter Point (See Figure 1.11).(Weisstein, ”Exeter Point”)

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Figure 1.11: The Exeter Point

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Chapter 2

Three Points On A Line

2.1 Menelaus’ Theorem

Menelaus’ theorem dates back to approximately AD 100. Menelaus of Alexandria (c.70-140), aGreek Mathematician and astronomer, published the treatise Sphaerica, focusing on the geometryof the sphere. In Book III of this treatise, he introduces a special theorem in spherical geometry thatlater became Menelaus’ theorem. He uses this theorem, that we will state and prove, as a lemmato prove a spherical version. It is unknown who actually discovered the theorem. (EncylopaediaBrittanica, 2015)

Theorem 2.1.1 (Menelaus’ Theorem). Let ABC be a triangle, and let X, Y and Z be points onthe sides BC, CA and AB respectively. X, Y and Z are collinear iff

BX

XC· CYY A· AZZB

= −1 (2.1)

Equation (2.1) uses signed lengths for each segment. For example, AB will be directed positivelyfrom A to B, and BA = −AB. Orient the sides of the triangle, so that AB, BC, and CA are positive.Then, for instance, the ratio BX/XC will be positive iff X lies strictly between B and C. As aresult, there will always be either one or three of the points X, Y and Z outside the triangle, theproduct on the left of equation (2.1) will always be negative. (Coxeter and Greitzer, 1967: p66-67;Venema, 2013: p78-80)

Proof. We first prove that the equation (2.1) is a necessary condition for collinearity. Assuming X,Y and Z are collinear, construct three segments, h1, h2 and h3, perpendicular to the line XZ andending at A, B and C respectively, as in Figure 2.1. If these segments are situated on opposite sidesof XZ, treat the lengths on one side as positive and the other as negative. Using similar triangles,we get the equations

h2h3

=BX

XC(2.2)

h3h1

=CY

Y A(2.3)

22

Figure 2.1: Menelaus’ Theorem

h1h2

=AZ

ZB(2.4)

Multiply (2.2), (2.3) and (2.4) to get

h2h3· h3h1· h1h2

=BX

XC· CYY A· AZZB

= 1

which shows that X, Y and Z are collinear. Since either one or three of these points lie outside thetriangle, the left-hand side must be negative. Therefore, the equation becomes

BX

XC· CYY A· AZZB

= −1

We can also prove that equation (2.1) is a sufficient condition by proving the converse. We firstassume that equation (2.1) is true. Let X, Y and Z be points such that

BX

CX· CYAY· AZBZ

= 1

Let AB and XY meet at Z ′BX

CX· CYAY· AZ

′

BZ ′= 1

ThereforeAZ

BZ=AZ ′

BZ ′

Z ′ coincides with Z, therefore X, Y and Z are collinear. QED

(Coxeter and Greitzer, 1967: p66-67)

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2.2 Desargues’ Theorem

Girard Desargues (1591 - 1661) was a French architect, engineer and mathematician and one of thefounders of projective geometry. He discovered a theorem that has many applications in perspectivedrawing. (Field, 1995)

Definition 2.2.1 (Centre of Perspectivity). Two triangles ABC and A′B′C ′ are perspective cen-trally if the three lines AA′, BB′ and CC ′ are concurrent at point O. This point is called thecentre of perspectivity.

Definition 2.2.2 (Axis of Perspectivity). Two triangles ABC and A′B′C ′ are perspective axiallyif the intersections of the lines AB and A′B′ (D), lines AC and A′C ′ (E) and lines BC and B′C ′

(F ) are collinear. The line through D, E and F is called the axis of perspectivity.

Theorem 2.2.3 (Desargues’ Theorem). Iff two triangles are perspective from a single point (cen-trally), then they are perspective axially, providing the pairs of corresponding sides intersect.

Figure 2.2: Desargues’ Theorem

If any two corresponding sides of the triangles are parallel, then there is no point of intersection.Therefore Theorem 2.2.3 includes the condition that the pairs of corresponding sides must intersect.(Coxeter and Greitzer, 1967: p70-72; Venema, 2013: p86-87)

Proof. To prove this theorem, we use Menelaus’ Theorem on the triads of points DB′A′, EC ′A′ and

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FC ′B′ on the sides of the triangles OBA, OAC and OBC respectively. This leads to the equations

BD

DA· AA

′

A′O· OB

′

B′B= −1 (2.5)

AE

EC· CC

′

C ′O· OA

′

A′A= −1 (2.6)

CF

FB· BB

′

B′O· OC

′

C ′C= −1 (2.7)

Multiply equations (2.5), (2.6) and (2.7) together, and cancel out like terms to get

BD

DA· AEEC· CFFB

= −1

indicating that D, E and F are collinear. QED

(Coxeter and Greitzer, 1967: p70-72)

2.3 Pappus’s Hexagon Theorem

Pappus of Alexandria (c. 290 - 350 AD) is widely considered the last and most important ’geometersof antiquity’. Pappus’s Hexagon theorem was first proved around AD 300 by Pappus, but its im-portance in projective geometry wasn’t realised until the sixteenth century. (Coxeter and Greitzer,1967: p67-70; Venema, 2013: p91-93)

Theorem 2.3.1 (Pappus’s Hexagon Theorem). Let A, B, C, D, E and F be any six points. Let Ybe the intersections of CD and AF , Z be the intersections of BF and CE and X be the intersectionsof AE and BD. If A, B and C lie on one line and D, E and F lie on another line, then X, Y andZ are collinear.

(Venema, 2013: p91-93)

Proof. Let the lines CD, BF and AE form a triangle UVW , as in Figure 2.3. Now use Menelaus’Theorem on the triads of points AY F , CZE, BXD, ABC and DEF . This leads to the equations

V Y

YW· WF

FU· UAAV

= −1 (2.8)

V C

CW· WZ

ZU· UEEV

= −1 (2.9)

V D

DW· WB

BU· UXXV

= −1 (2.10)

V C

CW· WB

BU· UAAV

= −1 (2.11)

V D

DW· WF

FU· UEEV

= −1 (2.12)

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Figure 2.3: Pappus’s Hexagon Theorem

Take the product of equations (2.8), (2.9) and (2.10), and divide the result by the product of (2.11)and (2.12) and cancelling gives

V Y

YW· WZ

ZU· UXXV

= −1

Showing that X, Y and Z are collinear, as desired. QED

(Coxeter and Greitzer, 1967: p67-70)

2.4 The Simson Line

Let ABC be any triangle and P a point distinct from A, B and C. Let X be the unique point onthe line AB such that PX and AB are perpendicular. Similarly drop perpendiculars from P ontoBC and AC intersecting Y and Z respectively. These are the closest points on each side of thetriangle to P . The triangle XY Z is known as the pedal triangle and point P is known as the pedalpoint.

Theorem 2.4.1 (The Simson Line). If P lies on the circumcircle of triangle ABC, then the threeclosest points to P on lines AB, BC and AC are collinear.

The line that is formed from these points is called the pedal line, or the Simson line (See Figure2.4), named after Robert Simson (1687-1768), a Scottish mathematician. This line can also be

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described as a degenerate pedal triangle (a triangle with collinear vertices and an area of zero).(Venema, 2013: p93-96)

Figure 2.4: The Simson Line

Proof. This proof assumes that the pedal point P lies on the circumcircle between B and C. Thesame can be proved if P lies between two other points by changing the labels of the points either inthe diagram or in the proof.

Because X, Y and Z are perpendicular, P is a point on the circumcircles of XAZ, XY B andCY Z. Therefore

∠CPB = 180◦ −A = ∠ZPX

By subtracting ∠CPX, we find that

∠XPB = ∠ZPC

Because points X, B, P and Y lie on a circle

∠XPB = ∠XY B

and because points C, Y , P and Z lie on a circle

∠ZPC = ∠ZY C

Therefore∠XY B = ∠ZY C

so the points X, Y and Z are collinear. QED

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(Coxeter and Greitzer, 1967: p40-41)

There are many interesting properties that relate to the Simson Line. First, as you move thepedal point P around the circumcircle, the Simson Line traces out, or envelopes, a deltoid. Thisdeltoid is known as the Steiner deltoid.

Also, if you include another pedal point, Q, diametrically opposite P on the circumcircle, anddraw its Simson Line, the two Simson Lines will be perpendicular. Furthermore, if you move thetwo pedal points around the circumcircle together, so that they remain diametrically opposite, theintersection of the Simson Lines will trace out the nine-point-circle (see Chapter 4.1). (Frank andWeisstein, ”Simson Line”)

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Chapter 3

Unexpected Equilateral Triangles

3.1 Morley’s Theorem

Frank Morley (1860-1937) was an English mathematician who lived in the USA since 1887. Whilsthe was a professor at the Johns Hopkins University, USA, he edited the American Journal ofMathematics, as well as several other books. In 1899, Morley discovered a theorem, published in1929, which is now known as Morley’s Theorem. (Jost and Maor, 2014: p160-163)

Theorem 3.1.1 (Morley’s Theorem). Let ABC be a triangle with internal angles 3α, 3β and 3γ atvertices A, B and C respectively. Let X be the intersection of the trisectors from A and B adjacentto side AB. Define Y and Z similarly. Points X, Y and Z form an equilateral triangle.

The resulting equilateral is known as the first Morley triangle, or simply the Morley triangle(See Figure 3.1). The length of each side of the Morley triangle is equal to

XY = Y Z = XZ = 8r sin(α) sin(β) sin(γ)

where r is the circumradius of triangle ABC. The area of the Morley triangle can be found by

Area = 16√

3r2 sin2(α) sin2(β) sin2(γ)

The trilinear co-ordinates of each vertex of the Morley triangle are

X = 2 cos(β) : 2 cos(α) : 1

Y = 1 : 2 cos(γ) : 2 cos(β)

Z = 2 cos(γ) : 1 : 2 cos(α)

(Weisstein, ”Morley’s Theorem)

For the proof of theorem 3.1.1, we will need two lemmas, which we now state and prove beforeproving the theorem.

29

Figure 3.1: Morley’s Theorem

Lemma 3.1.2. Let 0 < α < π. The function f(0, π − α)→ R defined by

f(t) = sin(t)/ sin(t+ α)

in the range 0 < t < π − α is injective.

Proof. We can prove Lemma 3.1.2 by using the contrapositive of the injection proof. Assumef(x) = f(y), where f(x) = sin(x)/ sin(x+ α) and f(y) = sin(y)/ sin(y + α)

sin(x)

sin(x+ α)=

sin(y)

sin(y + α)

Using the identity sin(A+B) = sin(A) cos(B) + cos(A) sin(B) on the denominator of each fractiongives

sin(x)

sin(α) cos(x) + cos(α) sin(x)=

sin(y)

sin(α) cos(y) + cos(α) sin(y)

Cross multiply the equation

sin(x) sin(α) cos(y) + sin(x) cos(α) sin(y) = sin(y) sin(α) cos(x) + sin(y) cos(α) sin(x)

We now need to start eliminating or simplifying some terms each side so that we are left with x = y.Start by dividing the equation by sin(x) and then sin(y) (we’re assuming that sin(x), sin(y) 6= 0)

1

sin(y)sin(α) cos(y) + cos(α) =

1

sin(x)sin(α) cos(x) + cos(α)

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We can cancel out cos(α) from both sides, and can also use the definition sin(A)/ cos(A) = cot(A)to give

cot(y) sin(α) = cot(x) sin(α)

Now divide the equation by sin(α) to leave

cot(y) = cot(x)

As 0 < x < π − α and 0 < y < π − α, we can then conclude that

y = x

Thus proving the lemma to be true. QED

Lemma 3.1.3. For all real values of u, the following identity holds

sin(3u) = 4 sin(u) sin(π/3 + u) sin(π/3− u) (3.1)

Proof. To prove Lemma 3.1.3, we need to show that the left hand side is equal to the right handside. For the left hand, we use De Moivre’s theorem, which is given by

(cos(x) + i sin(x))n = cos(nx) + i sin(nx)

Substitute x = u and n = 3.

(cos(u) + i sin(u))3 = cos(3u) + i sin(3u)

Use binomial expansion on the left hand side

cos3(u) + 3 cos2(u)i sin(u) + 3 cos(u)i2 sin2(u) + i3 sin3(u) = cos(3u) + i sin(3u)

Separate real and imaginary parts (taking i2 = −1 and i3 = −i)

cos3(u)− 3 cos(u) sin2(u) + i(3 cos2(u) sin(u)− sin3(u)) = cos(3u) + i sin(3u)

Equate the imaginary parts

3 cos2(u) sin(u)− sin3(u) = sin(3u)

Use the identity cos2(u) + sin2(u) = 1

3(1− sin2(u)) sin(u)− sin3(u) = sin(3u)

Substitute s = sin(u)3(1− s2)s− s3 = sin(3u)

Simplify3s− 3s3 − s3 = sin(3u)

3s− 4s3 = sin(3u) (3.2)

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For the right-hand side, we use the trigonometric identities

sin(A+B) = sin(A) cos(B) + cos(A) sin(B)

sin(A−B) = sin(A) cos(B)− cos(A) sin(B)

Substituting A = u and B = π/3 into each identity

sin(π/3 + u) = sin(π/3) cos(u) + cos(π/3) sin(u)

sin(π/3− u) = sin(π/3) cos(u)− cos(π/3) sin(u)

Take the product of these to give

(sin(π/3) cos(u) + cos(π/3) sin(u))(sin(π/3) cos(u)− cos(π/3) sin(u))

Expand the brackets

sin2(π/3) cos2(u)−sin(π/3) cos(π/3) sin(u) cos(u)+cos(π/3) sin(π/3) cos(u) sin(u)−cos2(π/3) sin2(u)

The second and third terms cancel out leaving

sin2(π/3) cos2(u)− cos2(π/3) sin2(u)

Substituting cos2(u) = 1− sin2(u), sin(u) = s, and calculating the known functions leaves

sin(π/3 + u) sin(π/3− u) =3

4(1− s2)− 1

4s2 (3.3)

Now we substitute equations (3.2) and (3.3) into (3.1)

3s− 4s3 = 4s(34(1− s2)− 14s

2)

Simplify and rearrange the equation to make both sides equal

3s− 4s3 = 4s(34 −34s

2 − 14s

2)

3s− 4s3 = 3− 3s3 − s3

3s− 4s3 = 3s− 4s3

The left hand side equals the right hand side, thus proving the lemma is always true. QED

We will now prove Theorem 3.1.1, using the above lemmas. For convenience, the theorem hasbeen repeated here:

Theorem 3.1.1. Let ABC be a triangle with internal angles 3α, 3β and 3γ at vertices A, B and Crespectively. Let X be the intersection of the trisectors from A and B adjacent to side AB. DefineY and Z similarly. Points X, Y and Z form an equilateral triangle.

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Figure 3.2: Morley’s Theorem (with θ and µ)

Proof. Let a = BC, b = AC and c = AB and let θ = ∠AXZ and µ = ∠BXY , as in Figure 3.2. Itwill be enough to show that

θ =π

3+ β (3.4)

andµ =

π

3+ γ (3.5)

because if this holds, then∠Y XZ = 2π − (θ + µ+ ∠AXB)

= 2π − (β + π/3 + α+ π/3 + π − α− β)

= 2π − (2π/3 + π)

= π/3

We can do the same to show that ∠XZY and ∠XY Z are also equal to π/3, thus proving that XY Zis an equilateral triangle. We can now prove equation (3.5). If we prove that equation (3.5) is true,it will follow that equation (3.4) is also true, which will conclude the proof.

It will be sufficient to prove that

f(θ) = f(β + π/3) (3.6)

because f is injective from Lemma 3.1.2. From the sine rule we have

AZ

sin(θ)=

AX

sin(θ + α)

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because sin(∠XZY ) = sin(π − θ − α) = sin(θ + α). We can rearrange this to give

AZ

AX=

sin(θ)

sin(θ + α)= f(θ)

To complete the proof we must show that the ratio AZ/AX is also equal to f(π/3 + β). Because3α+ 3β + 3γ = π, we know that π − (α+ γ) = 2π/3− β. By using the sine rule again we have

AZ

sin(γ)=

b

sin(π − α− γ)=

b

sin(2π/3 + β)=

b

sin(π/3− β)(3.7)

Using the same process with AX we get

AX

sin(β)=

c

sin(π/3− γ)(3.8)

Dividing equation (3.7) by equation (3.8) we get

f(θ) =AZ

AX=b sin(γ) sin(π/3− β)

c sin(β) sin(π/3− γ)(3.9)

Setting t = π/3 + β, we have sin(t+ α) = sin(π/3 + γ), therefore

f(π/3 + β) =sin(π/3 + β)

sin(π/3 + γ)(3.10)

and by using the sine rule we getb sin(3γ) = c sin(3β) (3.11)

Using the identity in Lemma 3.1.3 on equation (3.11), with u = β on the left-hand side andu = γ on the right-hand side, we calculate

b sin(π/3 + γ) sin(π/3− γ) sin(γ) = c sin(π/3 + β) sin(π/3− β) sin(β)

We can rearrange this to give

sin(π/3 + β)

sin(π/3 + γ)=b sin(γ) sin(π/3− γ)

c sin(β) sin(π/3− β)

The left-hand side is equal to equation (3.10) and the right hand side is equal to equation (3.9),concluding that

f(π/3 + β) = f(θ)

As required from equation (3.6). QED

(Barbara, 1997: p447-450)

More Morley Triangles

Theorem 3.1.1 trisects the internal angles of triangle ABC, however, the theorem holds true whentrisecting the external angles of ABC too (pink triangle in Figure 3.3). The sides of this new Morleytriangle are parallel to the sides of the first Morley triangle. Furthermore if we take the intersectionsof the pairs of remaining trisectors with the extended sides of this new Morley triangle, we get threemore equilateral triangles (red triangles in Figure 3.3). (Jost and Maor, 2014: p160-163)

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Figure 3.3: More Morley Triangles

3.2 Napoleon’s Theorem

Let ABC be a triangle. Construct equilateral triangles (BCD, ACE, ABF ) on each side of ABC,all facing outwards. The centroids of these triangles, G, H and I have a special property (See Figure3.4).

Theorem 3.2.1. If an equilateral triangle is constructed on each side of any triangle ABC, alloutwardly or all inwardly, then the centroids of these constructed triangles form themselves anequilateral triangle.

This theorem is named after the French emperor Napoleon Bonaparte (1769-1821). He was anamateur mathematician with an interest in geometry. However, there is much doubt as to whetherit was actually Napoleon who discovered this theorem. The earliest known reference to Napoleon’stheorem is an entry from the 1825 edition of the Ladies’ Dairy written by William Rutherford(1798-1871). He did not prove the theorem, but others provided a proof in the following year’sedition. Neither edition mentioned Napoleon, he wasn’t attributed to the theorem until 1911 inElementi di Geometria. (Coxeter and Greitzer, 1967: p60-65)

Proof. Because angle IAC = GAB = 30◦, we can use the law of cosines to show that

GI2 = AI2 +AG2 − 2 ·AI ·AG · cos(A+ 60◦) (3.12)

The centroid lies on the median 2/3 of the distance from vertex to the side of the triangle. Thereforewe get

AG = (2/3)√

3/2 ·AB = AB/√

3

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AI = (2/3)√

3/2 ·AC = AC/√

3

and equation (3.11) becomes

3 ·GI2 = AC2 +AB2 − 2 ·AC ·AB · cos(A+ 60◦) (3.13)

By using the identity cos(A + B) = cos(A) cos(B) sin(A) sin(B), and that cos(60◦) = 1/2 andsin(60◦) =

√3/2, we get

3 ·GI2 = AC2 +AB2 −AC ·AB · cos(A) +√

3 ·AC ·AB · sin(A)

Using the law of cosines on ABC

BC2 = AC2 +AB2 − 2 ·AC ·AB · cos(A)

and the law of sinesArea(ABC) = (1/2)AC ·AB · sin(A)

we can write equation (3.12) as

3 ·GI2 = (1/2)(BC2 +AC2 +AB2) + 2√

3 ·Area(ABC)

Since this is symmetrical in BC, AC and AB, it follows that the triangle connecting the centroidsis equilateral. QED

(Bogomolny, n.d.)

Figure 3.4: Napoleon’s Theorem

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Chapter 4

The Nine-Point Circle and The EulerLine

4.1 The Nine-Point Circle

In 1821, Jean-Victor Poncelet (1788-1867) and Charles-Julien Brianchon (1783-1864), discoveredand proved a special property relating to nine particular points of any triangle.

Theorem 4.1.1 (Nine-Point Circle Theorem). If ABC is a triangle, then the midpoints of eachsides, the feet of the three altitudes, and the midpoints between each vertex and the orthocentre ofABC all lie on a single circle.

(Coxeter, 1961: p18-20; Coxeter and Greitzer, 1967: p20-22; Venema, 2013: p57-59)

Definition 4.1.2 (Nine-Point Circle). The circle in Theorem 4.1.1 is known as the nine-pointcircle.

(See Figure 4.1)

We will now prove that the nine points from Theorem 4.1.1 lie on a single circle.

Proof. Let ABC be a triangle. Let D, E and F be the midpoints of BC, AC and AB respectively,J , K and L be the feet of the altitudes on BC, AC and AB respectively, H the orthocentre ofABC, and X, Y and Z be the midpoints of AH, BH and CH respectively. We will now prove thatD, E, F , J , K, L, X, Y and Z lie on a single circle.

In ABC, since E and F are the midpoints of AC and AB, EF must be parallel to BC. InBCH, Y and Z are the midpoints of BH and HC, therefore Y Z and BC must be parallel. Itfollows that EF and Y Z are also parallel.

In BAH, since Y and F are midpoints of BH and BA, Y F is parallel to HA. In CAH, E andZ are midpoints of AC and HC, so EZ is parallel to HA. Therefore it follows that Y F and EZare parallel. Since Y F is parallel to HA, and HA lies on AJ , Y F is also parallel to AJ .

37

As AJ is perpendicular to BC (since AJ is the altitude from A to BC), and BC and EF areparallel, then AJ is also perpendicular to EF . Since Y F is parallel to AJ , Y F is perpendicular toEF . Similarly, EZ is also perpendicular to EF . We now have a rectangle EFY Z, which can beinscribed in a circle.

We can use a similar process as above to get the rectangle DFXZ, which can also be inscribedin a circle. Because the two rectangles share opposite points F and Z, the two circles share thesame diameter FZ, and therefore coincide. From this D, E, F , X, Y and Z all lie on a commoncircle.

Since AJ is an altitude of ABC from A to BC and XJ lies on AJ , ∠XJD is a right angle. AsXD is a diagonal of DFXZ, it is a diameter of the circle, so it follows that J must also lie on thecircle. Similarly for K and L.

Therefore all nine points D, E, F , J , K, L, X, Y and Z lie on the circle. This circle is thenine-point circle.

QED

(Umberger, n.d.)

Definition 4.1.3 (Nine-Point Centre). The centre of the nine-point circle is called the nine-pointcentre of ABC.

The nine-point centre is the midpoint of the line segment connecting the orthocentre and cir-cumcentre of ABC. It is also considered a triangle centre, and appears in Kimberling’s Encylopediaof Triangle Centres as X5, and is often labelled as N . The radius of the nine-point circle is equalto half the radius of the circumcircle. (Coxeter, 1961: p18-20; Coxeter and Greitzer, 1967: p20-22)

The trilinear co-ordinates of the nine-point centre are

cos(B − C) : cos(C −A) : cos(A−B)

or, by using the sides of ABC,

bc(a2(b2 + c2)− (b2 − c2)2) : ca(b2(c2 + a2)− (c2 − a2)2) : ab(a2(a2 + b2)− (a2 − b2)2)

Feuerbach’s Theorem

In 1822, Karl Wilhelm Feuerbach (1800-1834) published a theorem relating to the nine-point circleand the four equicircles. These equicircles are the incircle and the three excircles.

Theorem 4.1.4 (Feuerbach’s Theorem). The Nine-Point Circle of triangle ABC is tangent to theincircle and the three excircles of ABC.

The point where the nine-point circle and the incircle are tangent to each other is known as theFeuerbach Point. This point is a triangle centre and is listed in the encylopedia as X11, and oftenlabelled F . The three points where the nine-point circle and the excircles are tangent to each otherforms the Feuerbach Triangle (See Figure 4.2). (Venema, 2013: p60-61; Weisstein, ”Feuerbach’sTheorem”)

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Figure 4.1: Nine-Point Circle and Nine-Point Centre

4.2 The Euler Line

Leonhard Euler (1707-1783) was a Swiss mathematician. Born in Switzerland, he spent the majorityof his life in Berlin, Prussia (now Germany) and St Petersburg, Russia. Despite many tragedieshappening throughout his later life, including the death of his wife, the loss of his home and librarythrough fire and becoming completely blind, his creative and academic abilities continued until hisdeath. By this time, he completed many works and discovered many formulas and theorems coveringnearly every area of mathematics, as well as science. In 1765, Euler discovered a relationship betweenthree particular triangle centres. (Jost and Maor, 2014: p123-125)

Theorem 4.2.1 (Euler Line Theorem). For any non-equilateral triangle, the centroid (G), thecircumcentre (O) and the orthocentre (H) are collinear. G lies between H and O and HG = 2GO.

Definition 4.2.2 (Euler Line). The line passing through G, H and O of any non-equilateral triangleis called the Euler Line.

(See Figure 4.3)

However, since Euler discovered this line, it has been shown that other centres lie on the EulerLine as well, most notably the Nine-Point Centre and the Exeter Point. Other points on the EulerLine include the Schiffler Point, the de Longchamps Point and many more. (Coxeter, 1961: p17-18;Coxeter and Greitzer, 1967: p18-20; Venema, 2013: p27-29)

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Figure 4.2: Feuerbach’s Theorem

Figure 4.3: The Euler Line

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Any point with trilinear co-ordinates α : β : γ lies on the Euler line if∣∣∣∣∣∣α β γ

cos(A) cos(B) cos(C)cos(B) cos(C) cos(C) cos(A) cos(A) cos(B)

∣∣∣∣∣∣ = 0

which can also be written as

α cos(A)(cos2(B)− cos2(C)) + β cos(B)(cos2(C)− cos2(A)) + γ cos(C)(cos2(A)− cos2(B)) = 0

similarlyα sin(2A) sin(B − C) + β sin(2B) sin(C −A) + γ sin(2C) sin(A−B) = 0

(Weisstein, ”Euler Line”)

Right-Angled Triangle

If the triangle is right-angled, the Euler Line passes through the right-angled vertex and the midpointof the hypotenuse. This is because the orthocentre of the triangle coincides with the right-angledvertex and the circumcentre coincides with the midpoint. Therefore, the Euler line also coincideswith the median through the right-angled vertex.

Isoceles Triangle

If the triangle is isoceles, the Euler line coincides with the single axis of symmetry of the triangle.Also, the Euler passes through the Incentre (I) of the triangle, which it doesn’t otherwise.

Equilateral Triangle

If the triangle is equilateral, all the triangle centres coincide. The Euler Line is therefore notdefined in this case, which is why Theorem 4.2.1 and Definition 4.2.2 states that the triangle mustbe non-equilateral.

Schiffler’s Theorem

Theorem 4.2.3 (Schiffler’s Theorem). For any triangle ABC with incentre I, the Euler Lines ofthe four triangles BCI, ACI, ABI and the original triangle ABC are concurrent at S.

Definition 4.2.4 (Schiffler Point). The point of concurrency in Theorem 4.2.3 is called the SchifflerPoint.

The Schiffler Point, S, is an entry in Kimberling’s Encylopedia of Triangle Centres as X21.

The trilinear co-ordinates of the Schiffler Point are

=1

cos(B) + cos(C):

1

cos(C) + cos(A):

1

cos(A) + cos(B)

Page: 41

=b+ c− ab+ c

:c+ a− bc+ a

:a+ b− ca+ b

(Emelyanov and Emelyanova, 2003)

Figure 4.4: Schiffler’s Theorem

Page: 42

Conclusion

In this report, I have discussed many theorems relating to triangle geometry. I have talked aboutthe centre of a triangle, and the fact that there are multiple centres for a single triangle, includingthe incentre, centroid, orthocentre and circumcentre. I have explained the processes to find eachcentre along with a justification of its existence through a proof, and have also provided an imagefor each. I also stated and proved Ceva’s Theorem and used it to prove the concurrency of certainsets of three lines that determine some of the triangle centres.

I then moved on to focus on proving three sets of special points are collinear in four differenttheorems, including Menelaus’ Theorem, Desargues’ Theorem and Pappus’s Hexagon Theorem. Ithen went on to investigate certain equilateral triangles associated with an arbitrary triangle inMorley’s Theorem and Napoleon’s Theorem. Then finally, I discussed the nine-point circle and theEuler line to round up my report.

Before I started working on this project, the last time I learnt any Euclidean geometry was inschool at GCSE level. So it has been a long time since I have learnt anything new in this area.Whilst working on this project, I learnt a lot of new information about triangle geometry that Inever would have thought of before. For example, I never considered before where the centre of thetriangle might be or whether there was more than one candidate for it. I found undertaking thisproject both challenging, instructive and interesting.

To complete this project I used LaTeX to complete it, and GeoGebra to create the figures. Ihave had experience of word-processing in LaTeX for the past few years, using it to write up nearlyall of my coursework at university. GeoGebra was new to me and turned out to be a relatively easyprogram to learn. It proved very useful for drawing detailed and clear diagrams for this report.

There were, however, many more theorems relating to triangle geometry that I did not includein this report. They include plenty more triangle centres, for instance the Fermat point, Gergonnepoint and Torricelli point. There are also the medial and orthic triangles, as well as Miquel’s theoremand the Miquel Points. This report is in no way a full account of everything related to trianglegeometry, and only scratches the surface of a well-developed and intriguing area of mathematics.

43

References

.

Barbara, R. (1997). The Mathematical Gazette, Volume 81. London: Bell and Sons

Bogomolny, A. (n.d.). Napoleon’s Theorem, Two Simple Proofs. [on-line] Available at http:

//www.cut-the-knot.org/proofs/napoleon.shtml [Accessed 26 January 2015]

Coxeter H.S.M. (1961). Introduction to Geometry. New York: John Wiley & Sons, Inc

Coxeter, H.S.M. and Greitzer, S.L. (1967). Geometry Revisited. Washington DC: MathematicalAssociation of America

Emelyanov, L. and Emelyanova, T. (2003). A Note on the Schiffler Point. [PDF] Available athttp://forumgeom.fau.edu/FG2003volume3/FG200312.pdf [Accessed 1 April 2015]

Encyclopaedia Brittanica on-line. (2015). Menelaus of Alexandria. [on-line] Available athttp://www.britannica.com/EBchecked/topic/374905/Menelaus-of-Alexandria [Accessed 25January 2015]

Faucette, W.M. (2007). Ceva’s Theorem and Its Applications. [PDF] Available at: http:

//www.westga.edu/~faucette/research/CevaApps.pdf [Accessed 23 November 2014]

Field, J.V. (1995). Girard Desargues. [on-line] Available at http://www-history.mcs.st-andrews.ac.uk/Biographies/Desargues.html [Accessed 25 January 2015]

Floor (1998). The Napoleon Point and More. [on-line] Available at http://mathforum.org/

library/drmath/view/55042.html [Accessed 8 March 2015]

Jackson, F. and Weisstein, E.W. (n.d). ”Simson Line.” From MathWorld. Available at http:

//mathworld.wolfram.com/Simson-line.html [Accessed 20 January 2015]

Jost, E. and Maor, E. (2014). Beautiful Geometry. Woodstock: Princeton University Press

Kimberling, C. (c 1994). Clark Kimberling’s Encyclopedia of Triangle Centers. [on-line] (Up-dated 2015) Available at: http://faculty.evansville.edu/ck6/encyclopedia/ETC.html [Ac-cessed 1 March 2015]

Kimberling, C. (1994) Central Points and Central Lines in the Plane of a Triangle. Mathemat-ics Magazine, Vol. 67, No. 3. [on-line] Available at: http://www.jstor.org/stable/2690608

[Accessed 21 March 2015]

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O’Connor, J.J. and Robertson, E.F. (2012). Giovanni Benedetto Ceva. [on-line] Available athttp://www-history.mcs.st-andrews.ac.uk/Biographies/Ceva_Giovanni.html [Accessed 12 April2015]

Umberger, S. (n.d). Proof of the Nine-Point Circle. [on-line] Available at http://jwilson.coe.uga.edu/emt668/EMAT6680.2000/Umberger/EMAT6680smu/Assign4smu/nineptproof.html [Accessed8 April 2015]

Venema, G.A. (2013). Exploring Advanced Euclidean Geometry With Geogebra. WashingtonDC: Mathematical Association of America

Weisstein, E.W. (n.d). ”Centroid.” From MathWorld. Available at http://mathworld.wolfram.com/Centroid.html [Accessed 29 November 2014]

Weisstein, E.W. (n.d). ”Circumcenter.” From MathWorld. Available at http://mathworld.

wolfram.com/Circumcenter.html [Accessed 29 November 2014]

Weisstein, E.W. (n.d). ”Euler Line.” From MathWorld. Available at http://mathworld.

wolfram.com/EulerLine.html [Accessed 11 March 2015]

Weisstein, E.W. (n.d). ”Excircles.” From MathWorld. Available at http://mathworld.wolfram.com/Excircles.html [Accessed 8 April 2015]

Weisstein, E.W. (n.d). ”Exeter Point.” From MathWorld. Available at http://mathworld.

wolfram.com/ExeterPoint.html [Accessed 9 March 2015]

Weisstein, E.W. (n.d). ”Feuerbach’s Theorem.” From MathWorld. Available at http://mathworld.wolfram.com/FeuerbachsTheorem.html [Accessed 10 March 2015]

Weisstein, E.W. (n.d). ”Incenter.” From MathWorld. Available at http://mathworld.wolfram.com/Incenter.html [Accessed 28 November 29 November 2015]

Weisstein, E.W. (n.d). ”Kimberling Center.” From MathWorld. Available at http://mathworld.wolfram.com/KimberlingCenter.html [Accessed 1 March 2015]

Weisstein, E.W. (n.d). ”Morley’s Theorem.” From MathWorld. Available at http://mathworld.wolfram.com/MorleysTheorem.html [Accessed 7 March 2015]

Weisstein, E.W. (n.d). ”Napoleon Points.” From MathWorld. Available at http://mathworld.wolfram.com/NapoleonPoints.html [Accessed 30 November 2014]

Weisstein, E.W. (n.d). ”Napoleon’s Theorem.” From MathWorld. Available at http://mathworld.wolfram.com/NapoleonsTheorem.html [Accessed 27 January 2015]

Weisstein, E.W. (n.d). ”Orthocenter.” From MathWorld. Available at http://mathworld.

wolfram.com/Orthocenter.html [Accessed 28 November 2014]

Wilson, J (n.d). Proof of Ceva’s Theorem and its Converse. [on-line] Available at http://

jwilson.coe.uga.edu/emt668/EMAT6680.2001/Mealor/EMAT6690/Ceva’s%20theorem/cevaproof.

html [Accessed 25 October 2014]

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Appendices

Approximately 9,460 words

46

Page: 47

Diary

.

Meeting 2nd October 2014: This was my first meeting with supervisor, Trevor Hawkes. Weintroduced ourselves to each other, including what modules I am studying on my course. Trevorstarted explaining some basic information about the project, giving some examples of interestingfacts about triangles.

2nd - 9th October: Was tasked to research at least three interesting facts about triangles,including theorems, proofs, etc, with the aim of finding something that Trevor did not know about.Facts found included medians, Apollonius Theorem, Napoleon’s Theorem and Centres (includingEuler Line).

Meeting 9th October: I showed my fact sheet to Trevor. The project was then discussed inmore detail, with some background information on Ceva’s Theorem.

9th - 16th October: Began research on Ceva’s Theorem. This mainly involved researchingand writing up a proof of the theorem in my own words, as if writing for the report. This was forpractice in researching and in writing with good structure and clarity for report.

Meeting 16th October: Trevor read through my write-up of Ceva’s Theorem proof, andsuggested improvements that can be made.

16th - 21st October: Improve the write-up using Trevor’s suggestions.

Meeting 21st October: Trevor looked over the improvements and discussed further improve-ments that can be implemented to finalise the section on Ceva’a Theorem. He also gave an intro-duction to Morley’s Theorem.

21st October - 11th November (No meetings due to illness): Finished off the proof of Ceva’sTheorem. Started researching Morley’s Theorem, using notes provided by Trevor as informationfor proofs. Started writing up a proof of the theorem in my own words. Trevor suggested a possiblestructure for the report, providing ideas of various information to research and include in the report.

Meeting 11th November: Trevor read through my write-up of the Morley’s Theorem proofsand suggested improvements.

11th - 20th November: Made improvements to Morley’s Theorem proofs, and also proved ofa Lemma involved in the proof by calculations, using hints provided by Trevor. Also used Ceva’sTheorem to prove certain line triples are concurrent, using a mixture of research and my ownknowledge.

Meeting 20th November: Trevor looked through work from over the week. He then suggestedthat I start writing the report proper. Provided certain topics to research and include as part ofchapter 1 of the report.

20th - 27th November: Started writing chapter 1, and also including work done over theterm so far.

Meeting 27th November: Looked through current work of chapter 1 and suggested improve-

Page: 48

ments

27th November - 4th December: Made improvements to work so far and continued withchapter 1. Created a structure in the report, including chapter headers, a contents page and a titlepage. Made a start with the Literature Review.

Meeting 4th December: Quick meeting explaining current progress

4th - 10th December: Prepared for End-Of-Term Report.

Meeting 10th December: Handed in End-Of-Term Report.

10th December 2014 - 9th January 2015 (Christmas break): Started chapter 2. Wroteproofs for three theorems.

Meeting 9th January: Discussed work done over Christmas break. Suggested further im-provements that can be made to Morley’s Theorem section.

9th - 16th January: Made improvements, and wrote a proof of a lemma. Changed layout forsection

Meeting 16th January: Suggested further improvements.

16th - 23rd January: Made improvements. Completed section for The Simson Line

Meeting 23rd January: Read through Chapter 2 and suggested improvements. Talk aboutlayout and numbering of theorems and definitions. Suggested mentioning bits of maths history ineach section.

23rd - 30th January: Added more detail to Chapter 2, including information about thefounders of each theorem. I also changed Menelaus’ Theorem to take into account the direction ofthe lines. Started Napoleon’s Theorem and drew an image in GeoGebra.

Meeting 30th January: Suggested some improvements to be made regarding the layout ofthe report, including making page numbers appear on each page. Also recommended a webpageabout the Fermat Point to read through

30th - 6th February: Improved the layout of the report as per supervisor’s recommendations.Read through webpage on Fermat Point and highlighted important information.

Meeting 6th February: Suggested further improvements to the layout of the report.

6th February - 5th March (No meetings due to supervisor’s annual leave and illness): Addedautomatic numbering to the theorems, lemmas and definitions. Also reformatted the proofs, to makeit more in-line with standard report styles. Started on Chapter 4 with the Nine-Point Circle. Alsoadded a new section to Chapter 1 regarding the Encyclopaedia of Triangle Centres.

Meeting 6th March: Catch up on work completed over past month

6th - 20th March (No meeting due to Poster Presentation): Researched the Euler Line, aswell as little things for rest of report

Meeting 20th March: Found a proof for Napoleon Points

20th - 27th March: Wrote a proof for generalisation of Napoleon Points and included trilinear

Page: 49

points throughout report.

Meeting 27th March: Looked through Chapter 2. Supervisor suggested improvements to bemade on explanation and proofs for Menelaus’ and Desargues’ Theorems.

27th - 2nd April: Improved Menelaus’ and Desargues’ Theorem sections to improve explana-tion. Started working through all diagrams to ensure all are done and consistent in design. Starteda list of references.

2nd - 10th April: Worked through the list of references and included what topics come fromwhich source. Fleshed out Chapter 3 and 4.

Meeting 10th April: Looked through Chapter 4. Found more information for the SchifflerPoint.

10th - 17th April: Finished main body and added images. Added references to bibliographyand citations in report. Started writing literature review.

Meeting 17th April: Read through literature review and suggested improvements.

17th - 24th April: Completed literature review, abstract, introduction and conclusion, withsupport from supervisor. Sent report to family for proof reading. Finished report ready for sub-mission on Friday 24th April.

Page: 50

Certificate of Ethical Approval

Student:

Christopher Thomson

Project Title:

The Secret Life of Triangles

This is to certify that the above named student has completed the Coventry

University Ethical Approval process and their project has been confirmed and

approved as No Risk

Date of approval:

16 October 2014

Project Reference Number:

P27558