A New Way to Think About Triangles

Adam Carr, Julia Fisher, Andrew Roberts, David Xu, and advisor Stephen Kennedy

June 6, 2007

1 Background and Motivation

Around 500-600 B.C., either Thales of Miletus or Pythagoras of Samos introduced the Western worldto the forerunner of Western geometry. After a good deal of work had been devoted to the field, Euclidcompiled and wrote The Elements circa 300 B.C. In this work, he gathered a fairly complete backboneof what we know today as Euclidean geometry. For most of the 2,500 years since, mathematicianshave used the most basic tools to do geometry—a compass and straightedge. Point by point, line byline, drawing by drawing, mathematicians have hunted for visual and intuitive evidence in hopes ofdiscovering new theorems; they did it all by hand. Judging from the complexity and depth from thegeometric results we see today, it’s safe to say that geometers were certainly not suffering from a lackof technology.

In today’s technologically advanced world, a strenuous effort is not required to transfer the ca-pabilities of a standard compass and straightedge to user-friendly software. One powerful exampleof this is Geometer’s Sketchpad. This program has allowed mathematicians to take an experimentalapproach to doing geometry. With the ability to create constructions quickly and cleanly, one is ableto see a result first and work towards developing a proof for it afterwards. Although one cannot claimproof by empirical evidence, the task of finding interesting things to prove became a lot easier withthe help of this insightful and flexible visual tool.

One of the most incredible features of Geometer’s Sketchpad is its ability to produce dynamicconstructions. In other words, one can create a construction with interdependent elements, alter one ofthe elements, and observe the resulting movement in the rest of the construction. These motions couldnot be easily represented otherwise (without the use of a flip book and a huge number of painstakinglyaccurate drawings). Because of this, Geometer’s Sketchpad has enabled modern mathematicians toask questions that their predecessors were not able to formulate. On paper, geometry is static. WithGeometer’s Sketchpad, it becomes dynamic.

With this tool in hand, let us begin our project with a few questions: Is there a way to naturallyrelate triangles? Is there a logical method to define “families” using intrinsic properties of triangles?Can we easily construct these families? What would the triangle space that results from this con-struction look like topologically? How could we characterize motions through this space? In short,our task is to find a natural and meaningful way to name and relate triangles.

Recognizing there are many ways to approach this task, we need a starting point. Since we wantto find a way to characterize families of triangles that utilizes intrinsic triangle properties, it seemslogical to begin with triangle centers. Thus, let us start with a brief overview.

2 Basic Triangle Concepts

First, we should define a few very well-known triangle centers. Although we are not including anyproofs of their existence in our paper, these are simple to reproduce and can be found in any basicgeometry text.

Definition 1. A median is straight line joining a triangle vertex to the midpoint of the opposite side.

Proposition 1. The three medians intersect at a point. This point is known as the centroid and willbe denoted by G .

Definition 2. A perpendicular bisector is a straight line that bisects a line segment at a right angle.

Definition 3. The circumcircle is the circle determined by a triangle’s three vertices.

Definition 4. The circumradius is the radius of the circumcircle.

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Definition 5. The circumcenter is the center of the circumcircle, denoted by O henceforth.

Proposition 2. The three perpendicular bisectors of a triangle meet at the circumcenter.

Definition 6. An altitude is a straight line going through a triangle vertex, perpendicular to theopposite side or extension of the opposite side.

Proposition 3. The three altitudes of a triangle intersect at a point. This point is known as theorthocenter and will be denoted by H .

In addition to the few presented here, there are a huge number of known triangle centers, many ofwhich are constructed through quite elaborate processes. Interestingly, many of these triangle centersare collinear, lying on a line known as the Euler line. A few of these include less common centerssuch as the nine-point center, the de Longchamps point, the Schiffler point, the Exeter point, and thefar-out point. Most importantly for our purposes, the centroid, circumcenter, and orthocenter all lieon the Euler line. Additionally, it is a well-known fact that G always lies 1

3 of the way from O to H,a relationship of which we will frequently make use. Because of its connection to triangle centers (andthe fact that April 15th 2007 is Euler’s 300th birthday), it seems natural to use the Euler line as thebasis of our construction. Such a construction would take full advantage of much work already done,since it is exploits the work already done on triangle centers.

3 Classifying Triangles Based on the Euler line

Now, we can begin our search for a method of constructing a triangle based on a given Euler line.In the end, we want to develop a general method to construct all possible triangles. So, we beginby classifying all triangles that share the same Euler line. Consider the following question: Giventhe circumcenter, the centroid, and one vertex of a triangle, can you construct a triangle? Is thetriangle with these three points unique? Are there other triangles with different vertices but the samecircumcenter and centroid? In order to address the first question, we need the known result below.Our own proof of it is provided.

A'

G

Ma

Mc

Mb

A

B

C

Figure 1: An illustration of Proposition 4.

Proposition 4. Given 4ABC and its centroid G , let Ma be the midpoint of segment BC , Mb themidpoint of segment AC , and Mc the midpoint of segment AB . Then, AG = 2GMa , BG = 2GMb ,and CG = 2GMc .

2

We will prove this statement for segment BG . The proofs for segments AG and CG are identical.

Proof. Extend segment AMa past Ma and mark point A′ such that A′Ma = AMa . Since byconstruction BMa

∼= MaC and by the Vertical Angle Theorem ∠BMaA′ ∼= ∠CMaA , it follows

from SAS Congruence that 4A′MaB ∼= ∆AMaC . Thus, ∠BA′Ma∼= ∠CAMa . By Proposition

27 of Euclid’s Elements, this implies that AC ‖ BA′ . So, ∠AMbG ∼= ∠A′BG . Again by theVertical Angle Theorem, ∠AGMb

∼= ∠A′GB . Thus, ∆BA′G ∼ ∆MbAG by AAA Similarity. Since

BA′ = AC = 2AMb , this and the previous triangle similarity imply thatBG

GMb=

A′BAMb

=AC

AMb= 2 .

Therefore, BG = 2GMb .

With this information in hand, we will show that with a given circumcenter, centroid, and onevertex of a triangle, we can construct a triangle.

3.1 Triangle Construction

Let line l and a point O on l be given. Choose a point A not on l and create circle c centeredat O with radius OA . O will be the circumcenter of our triangle and A one of its vertices. Thus,the other two vertices of our triangle will also lie on circle c . Let M and N be the intersections ofl with c . Now, choose a point G 6= O on l such that G lies between M and N . G will be thecentroid of our triangle. Construct point H on l such that OG = 1

3OH . By construction, H mustbe the orthocenter of the triangle. Now, construct the line that passes through A and G and theline that passes through A and H . Since it was just proven that the distance from one vertex of atriangle to its centroid is twice the distance from the centroid to the midpoint of the opposite side,construct point Ma on AG such that GMa = 1

2GA and such that Ma is not between A and G .Then, construct line m through Ma such that m ⊥ AH . Let B and C be the intersections of mwith circle c . Since H is the orthocenter of the triangle, AH must be the altitude from vertex A .Moreover, we constructed line m to be perpendicular to AH . Thus, line m must be the side of thetriangle opposite vertex A , and B and C will be the triangle’s remaining two vertices. Therefore,we have constructed a 4ABC from a given Euler line l , two points on l , and a point A not on l .

l

c

m B

C

Ma

GM NHO

A

Figure 2: The elements of the triangle construction.

Removing the condition that the specific point A must be a vertex of the triangle producesan infinite number of triangles that have circumcenter O and orthocenter H . So, we will choose toclassify triangles according to their circumcenter, centroid, and a given circumradius. As a convention,we will orient ourselves with the Euler line as the traditional x-axis unless otherwise stated.

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Definition 7. Let 4ABC with circumcenter O and centroid G be given, and let T be the spaceof all triangles. Then, define a mapping F : T→ R3 , where F (4ABC) = (θ, g, r) with θ , g , and rdetermined as follows:

θ = m∠GOP, where vertex P lies on the “one-vertex side”,g = OG,

r = OA.

r

gθ

C

B

HO

A

G

Figure 3: OG = g , OA = r , and ∠GOP = θ where P is the vertex on the “one-vertex” side; in thediagram above, P = A

Note that in the event that one vertex falls on the Euler line we will have two “one-vertex sides,”and in that case θ will be the smaller of the angles.

3.2 Special Cases of the Construction

The natural question to ask now is, “is this construction well-defined?” Before we address this, wemust first discuss some special cases of our construction. These will help us prove that the constructionis in fact well-defined.

Mc

Ma

Mb

C

B

O

A

G

Figure 4: When g = 0 , O and G are coincident.

Proposition 5. 4ABC is equilateral if and only if g = 0 .

4

Proof. (⇒) Let 4ABC be equilateral. Let Ma and Mb be the midpoints of sides BC and AC ,respectively. Then, construct lines AMa and BMb . Since ∠ACB ∼= ∠ABC , BMa

∼= CMa , andAB ∼= AC , it follows from SAS congruence that 4ABMa

∼= 4ACMa . Therefore, m∠AMaB =m∠AMaC = π . Thus, AMa is the perpendicular bisector of BC . Consequently, the circumcenterO of 4ABC lies on AMa . Moreover, by construction, the centroid G of 4ABC lies on AMa .

By the same argument, O and G also lie on BMb . Since AMa and BMb only intersect once,the only way that this can occur is if O = G . Thus, OG = g = 0 .

(⇐) Let 4ABC be given with OG = 0 . Let Ma and Mb be defined as above. Now, constructAG . Since G is the centroid, Ma must lie on AG . Since O = G , ∠AMaB ∼= ∠AMaC . Thus, sinceAMa is common, by SAS congruence, 4AMaB ∼= 4AMaC . This implies that AB ∼= AC . By thesame argument, BC ∼= AB . Therefore, AB ∼= AC ∼= BC , and 4ABC is equilateral.

Note that an equilateral triangle does not in fact have an Euler line, since the triangle centersdetermining the Euler line collapse a single point. Also, since G and O are the same point the angleformed by O , G , and a vertex does not technically exist. Thus, we will define θ = π

2 for reasonsthat will become clear later.

Proposition 6. Let 4ABC with orthocenter H be given. If H lies on the circumcircle of 4ABC ,then H is coincident with one vertex of 4ABC .

Proof. (⇒) Let circle c centered at O be given, and let H and A be points on c such that H 6= A .We will construct 4ABC with circumcenter O , circumcircle c , and orthocenter H and show thatH must coincide with either B or C .

To begin, connect A to the centroid G , which lies on the Euler Line one third of the way fromO to H . Extend AG beyond G one-half the length of AG , and call the endpoint of this extendedsegment Ma . By Proposition 4, Ma must be the midpoint of BC . Since H is the orthocenter of4ABC , AH must be perpendicular to BC . Thus, if we construct the line MaP , where P is thepoint on AH such that MaP ⊥ AH , this line will contain vertices B and C . We will show thatP = H .

c

Hc

C

B

Ma

GH

O

A

P

Figure 5: P lies past H on ray ~AH .

To do this, suppose to the contrary that P 6= H . Then, P lies between A and H , past H onthe ray

−−→AH , or past A on the ray

−−→HA .

Case 1: (Refer to Figure 5 above.) Suppose P lies past H on−−→AH . Then, vertex B , the

intersection of the circumcircle and the line MaP lies on the opposite side of the Euler line from

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A . By construction, m∠APC = π2 . Now, construct CH . CH is the altitude from vertex C

and thus meets AB perpendicularly at Hc . Consider 4ACHc . By angle subtraction, m∠ACH +m∠BAC = π

2 . Consider ∆ACP . Again by angle subtraction, m∠ACP + m∠CAP = π2 . However,

π2 = m∠ACP + m∠CAP > m∠ACH + m∠BAC = π

2 . This is a contradiction. Thus, P cannot liepast H on

−−→AH .

c

Hc

C

B

Ma

G HO

A

P

Figure 6: P lies between A and H .

Case 2: (Refer to Figure 6 above.) Suppose P lies between A and H . Then, vertex B , theintersection of the circumcircle and the line MaP lies on the same side of the Euler line as A . As inCase 1 , since AH is the altitude from A , m∠APC = π

2 . Thus, m∠CAP +m∠ACP = π2 . Construct

CH . Since CH is the altitude from C to AB , it meets AB perpendicularly at Hc . It follows thatm∠CAHc + m∠ACHc = π

2 . However, π2 = m∠CAP + m∠ACP < m∠CAHc + m∠ACHc = π

2 . Thisis a contradiction. Thus, P cannot lie between A and H .

c

Hc

C

B

Ma

G HO

A

P

Figure 7: P lies past A on ray ~HA .

Case 3: (Refer to Figure 7 above.) Suppose P lies past A on the ray−−→HA . As in the previous two

cases, m∠APC = π2 . Thus, m∠BAP +m∠ABP = π

2 . This implies that m∠ABC = m∠BAP + π2 >

π2 . Construct CH . Since CH is the altitude from C to AB , it meets AB perpendicularly at Hc .Thus, m∠HcBC = m∠ABC < π

2 . This is a contradiction. Thus, P cannot lie past A on ray−−→HA .

Therefore, P is coincident with H . By the initial discussion, MaH contains the vertices B andC of 4ABC . Since these vertices must lie on the circumcircle c , and H is on the circumcircle, it

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follows that H must coincide with either B or C .

Corollary 1. Let 4ABC with orthocenter H be given. Then, H is coincident with vertex B ifand only if m∠ABC = π

2 .

Proof. (⇒) Assume that H is coincident with vertex B . It follows immediately from the work inProposition 6 that m∠ABC = π

2 .(⇐) Assume that m∠ABC = π

2 . Then, BC is the altitude from vertex C to AB , and AB isthe altitude from vertex A to BC . Since the altitudes of a triangle coincide at H , it follows that Band H are coincident.

Proposition 7. Triangle 4ABC is a right triangle if and only if g = r3 .

Proof. (⇐ ) If g = r3 , then OH = r . That is, H lies on the circumcircle. Therefore, by Proposition 6

and Corollary 1, 4ABC is a right triangle.(⇒ ) If 4ABC is a right triangle, then by Corollary 1, H is coincident with one vertex. Without

loss of generality, let that vertex be vertex B . Thus, H is on the circumcirle, so OH = r . Since Gis one-third of the way from O to H , it follows that g = r

3 .

C

B

HO

A

G

Figure 8: Right triangles occur when g = r3 and H lies on the circumcircle.

To follow Proposition 7, we have the next two propositions which deal with the cases in whichg > r

3 and g < r3 .

Proposition 8. Triangle 4ABC is an obtuse triangle if and only if g > r3 .

Proof. We will begin with the reverse direction.(⇐ ) Let g > r

3 . Then, since OH = 3OG = 3g > 3( r3) = r , it follows that the orthocenter H lies

outside the circumcircle. Construct the altitude from vertex A . Without loss of generality, assumethat H lies past vertex B on ray

−−→CB . Since m∠AHB = π

2 , it follows that m∠ABH < π2 . Thus,

m∠ABC > π2 . Therefore, 4ABC is obtuse.

(⇒ ) Let 4ABC be obtuse. Without loss of generality, assume that m∠ABC > π2 . Construct

the altitude from vertex A . It will intersect side BC either past B on ray−−→CB or past C on

ray−−→BC by the following: Suppose that the altitude from vertex A intersects segment BC . Let

Ha be its intersection with BC . Then, m∠AHaB = π2 . By assumption, m∠ABC > π

2 . Thus,m∠AHaB + m∠ABC + m∠BAHa > π . This is a contradiction. Thus, the altitude from vertex A

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intersects an extention of segment BC . By the same argument, the altitude from vertex C intersectsan extention of segment AB . Now, the extentions of BC and AB must lie outside the circumcircleof 4ABC . Therefore, the orthocenter H , the intersection of the altitudes from vertices A and C ,must lie outside the circumcircle. Since OH = 3OG = 3g , and since we now know that OH > r , itfollows that 3g > r or g > r

3 .

Proposition 9. Triangle 4ABC is an acute triangle if and only if g < r3 .

Proof. This follows from a process of elimination using Propositions 7 and 8.

The three propositions above are nice because they indicate that the construction cleanly separatesacute, right, and obtuse triangles. We will next examine another type of triangles — isosceles triangles.

Proposition 10. If 4ABC is isosceles, one of its vertices lies on the Euler line.

Proof. Assume that 4ABC is isosceles with AB ∼= AC . Let Ma be the midpoint of BC . Then,AMa contains the centroid G of 4ABC . Moreover, since AB ∼= AC , BMa

∼= MaC , and AMa isshared, it follows that 4AMaB ∼= 4AMaC . Thus, ∠AMaB ∼= ∠AMaC . Since they are supplemen-tary, m∠AMaB = m∠AMaC = π

2 . This implies that AMa is the altitude of 4ABC from point A .Thus, the orthocenter H of 4ABC must lie on AMa . Therefore, AMa coincides with the Eulerline.

C''

B''

C'

B'

C

B

AO G G' G''

Figure 9: A few different isosceles triangles.

Proposition 11. If 4ABC is not a right triangle and has a vertex lying on the Euler line, then itis isosceles.

Proof. Assume that the Euler line l of 4ABC intersects vertex A . Since the centroid of 4ABC lieson l , it follows that l must intersect BC at the midpoint M of BC . Thus, BMa

∼= CMa . Also, theorthocenter of 4ABC lies on l and is not coincident with A . Thus, m∠AMaB = m∠AMaC = π

2 .Finally, by SAS congruence, 4AMaB ∼= 4AMaC (since side AMa is shared). This implies thatAB ∼= AC .

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3.3 Showing that the Construction is Well-Defined

We are finally ready to show that the construction is well-defined.

Proposition 12. The mapping F : T→ R3 is well-defined.

O'

G'

Mf

Md

Me

O

G

Mc

Ma

Mb

A

C

B E

F

D

Figure 10: Showing the construction is well-defined.

Proof. To show our construction is well-defined we must show that given two triangles 4ABC and4DEF , 4ABC ∼= 4DEF if and only if the coordinates (θ, g, r) of 4ABC equal the coordinates(θ′, g′, r′) of 4DEF . The reverse direction is covered since given two sets of equal parameters, weknow that the corresponding triangles will be congruent by construction. Thus,we only have to showthat two congruent triangles will produce the same parameters.

Assume 4ABC ∼= 4DEF with A corresponding to D , B corresponding to E , and C corre-sponding to F . Construct the centroids and circumcenters of the given triangles. Since the trianglesare congruent it must be that corresponding vertices lie on the same side of the Euler line with respectto the other vertices.

Case i) Without loss of generality let A and B be on the same side of the Euler line. Then, Dand E must also lie on the same side of the Euler line. Also, it must be that r = r′ , since otherwisethe triangles would have different circumcircles.

Therefore, OC ∼= O′F in the diagram. Also we know the corresponding medians must be congruentby simple SAS arguments. Now, GC = 2CMc

3 and G′F = 2FMf

3 . But CMc∼= FMf because they are

corresponding medians. Therefore, GC ∼= G′F . By a similar argument, GMa∼= G′Md . So, by SSS

4CGMa∼= 4FG′Md , which gives us ∠GCMa

∼= ∠G′FMd . Also, by a hypotenuse-leg right-trianglecongruence argument, 4OCMa

∼= 4O′FMd . Therefore, ∠OCMa∼= ∠O′FMd . By subtraction,

∠GCO ∼= ∠G′FO′ . So, by ASA congruence, 4GCO ∼= 4G′CO′ and GO ∼= G′O′ . Thus, it must bethat g = g′ . It also follows that ∠GOC ∼= ∠G′O′F , so θ ∼= θ′ .

We now have to consider the cases in which we have a vertex on the Euler line. First, we willconsider the right triangle case.

Case ii) If we let m∠ABC = π2 = m∠DEF , then r = AC

2 = DF2 = r′ . Since we have right

triangles, g = r3 = g′ . Also 4OAB ∼= 4O′DE by SSS. So ∠BOA ∼= ∠EO′D . If ∠BOA ≤ π

2 , thenθ = m∠BOA = θ′ . Otherwise, ∠BOA > π

2 , which means θ = π −m∠BOA = θ′ .Case iii) If we have congruent isosceles triangles, the arguments for cases i) and ii) hold, except

that it does not matter which vertex you choose to form θ as long as it is not the vertex on the Eulerline.

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Case iv) When we have congruent equilateral triangles, r must equal r′ , θ (likewise θ′ ) does notactually exist or have any affect in determining the triangle and g = 0 = g′ .

4 The Topology of Triangle Space

As mentioned in the introduction, one of our goals is to figure out what triangle space looks like. Onecould argue that this can be done by considering triangles in the traditional sense: defining a triangleby the lengths of its sides. In either case, you are given three numbers, and those three numberscorrespond to a point in R3 . This point corresponds to the triangle defined by the three numbers.The problem with considering triangles in the traditional sense is that every triangle is corresponds tosix different points (one for each permutation of the three sides). It is not a trivial problem to definethe space in such a way that every triangle is represented only once. By defining a triangle the way wehave, we have ensured that every triangle is represented by exactly one ordered triplet. However, notevery point in R3 represents a triangle. It will be our next task to determine exactly which values ofθ , g , and r give us legitimate triangles.

To help us reach our goal, let us first examine similar triangles.

Theorem 1. Two triangles 4ABC defined by (θ, g, r) and 4A′B′C ′ defined by (θ′, g′, r′) aresimilar if and only if θ = θ′ , g = kg′ , and r = kr′ for k > 0 .

Proof. (⇐) Let 4ABC be similar to 4A′B′C ′ .

Let G and G′ be the centroids of and O and O′ the circumcenters of 4ABC and 4A′B′C ′ ,respectively. Also, let Ma and Ma

′ be the midpoints of segments BC and B′C ′ , respectively. LetMb and Mb

′ be the midpoints of AC and A′C ′ , respectively. Since the two triangles are similar,it follows that AB

A′B′ = BMa

B′Ma′ = k . Therefore, 4ABMa ∼ 4A′B′Ma

′ . So, AMa

A′Ma′ = k . Now, G

and G′ must lie on AMa and A′Ma′ ,respectively. Moreover, GMa = 1

3AMa and G′Ma′ = 1

3A′Ma′ .

So, GMa

G′Ma′ = AMa

A′Ma′ . Also, BC

B′C′ = MbCMb

′C′ = k . So, 4BCMb ∼ 4B′C ′Mb′ . So, BMb

B′Mb′ = k . But,

GMb = 13BMb and G′Mb

′ = 13B′Mb

′ . Thus, GMb

G′Mb′ = k . This implies that 4BMbMa ∼ 4G′Mb

′Ma′

by SSS similarity. Thus, ∠GMbMa∼= ∠G′Mb

′Ma′ . Now,

m∠O′Mb′G′ + m∠G′Mb

′Ma′ + m∠Ma

′Mb′C ′ = π

m∠GMbMa = m∠G′Mb′Ma

′

m∠MaMbC = m∠BMbC −m∠GMbMa = m∠B′Mb′C ′ −m∠G′Mb

′Ma′ = m∠Ma

′Mb′C ′

m∠OMbG = m∠O′Mb′G′

Thus, by SAS similarity, 4OMbG ∼= 4O′Mb′G′ . This implies that OG

O′G′ = GMb

G′Mb′ = k . So,

OG = kO′G′ . But, O′G′ = g′ and OG = g . Thus, g = kg′ .

Now, we know that MbCMb

′C′ = MaCMa

′C′ = k . So, 4MbMaC ∼ 4Mb′C ′Ma

′ by SAS similarity.Therefore, ∠Mb

′Ma′C ′ ∼= ∠MbMaC , ∠MaMbC ∼= ∠Ma

′Mb′C ′ , and MMa

M ′Ma′ . Thus, by subtraction,

∠OMaM ∼= ∠O′Ma′Mb

′ and ∠OMbMa∼= ∠O′Mb

′Ma′ . This implies that 4OMbMa ∼ 4O′Mb

′Ma′ .

So, OMb

O′Mb′ = MbMa

Mb′Ma

′ = k . Thus, 4OMbC ∼ 4O′Mb′C ′ . Therefore, OC

O′C′ = k . But, OC = r andO′C ′ = r′ . So we know r = kr′ .

Now, construct lines OG , O′G′ , and segments OC and O′C ′ . We have shown that the segmentsconnecting the centroids to the vertices are proportional by k , and OG = kO′G′ . Thus, 4OGC ∼4O′G′C ′ by SSS similarity. This implies that ∠GOC ∼= ∠G′O′C ′ . So, since ∠GOC = θ and∠G′O′C ′ = θ′ , θ = θ′ .

10

b

a

g

r

b/k

a/k

g/k

r/k

θθ

C'

B'

A'

G'

C

BO G

A

O

Figure 11: Angle θ is the same in both triangles.

(⇒) Let 4ABC defined by (θ, g, r) and 4A′B′C ′ defined by (θ′, g′, r′) be two triangles withthe property that r = kr′ , g = kg′ , and θ = θ′ .

First, let Ma be the midpoint of CB . Now, we can see that 4AOG ∼ 4A′O′G′ by SASsimilarity. Thus, AG

A′G′ = k . So,

AMaA′M ′

a=

32AG

32A′G′ = AG

A′G′ = k

Then, GMaG′M ′

a= k and GMa = kG′M ′

a . Also, ∠AGO ∼= ∠A′G′O′ and thus ∠OGMa∼= ∠O′G′M ′

a .Therefore, 4OGMa ∼ 4O′G′M ′

a by SAS similarity. So, OMaO′M ′

a= k and OMa = kO′M ′

a . Also,∠OMaG ∼= ∠O′M ′

aG′ . Because O and O′ lie on the perpendicular bisectors to sides BC and B′C ′

respectively, we know,

CM2a = (kr)2 −OM2

a

(C ′M ′a)

2 = r2 − (O′M ′a)

2

CM2a = (kr)2 − (kO′M ′

a)2

CM2a = k2(r2 − (O′M ′

a)2)

CM2a = k2(C ′M ′

a)2

CMa = kC ′M ′a

Also, BC = 2CMa = 2kC ′M ′a = kB′C ′ . Thus, 4ACMa ∼ 4A′C ′Ma

′ by SAS similarity. Thus,∠ACMa

∼= ∠A′C ′Ma′ and 4ABC ∼ 4A′B′C ′ by SAS similarity.

Let us now begin looking at the bounds on θ , g , and r .

Proposition 13. If (θ, g, r) represents a triangle, then r > 0 .

Proof. The only acceptable values of r are r > 0 because r = 0 would mean that the circumcircle isonly a point, which does not allow for any triangles. Moreover, r < 0 is absurd. Also, if r is bounded

11

above, then we restrict ourselves to triangles with side lengths less than 2r , so r can be any positivenumber.

That being said, Theorem 1 implies that r is just a scaling variable. Therefore we can get a clearpicture of what triangle space looks like by examining cross-sections of R3 parallel to the θg -plane.

Proposition 14. If (θ, g, r) represents a triangle, then 0 ≤ g < rcos[12(θ + sin−1(1

2 sin θ))]cos[12(θ − sin−1(1

2 sin θ))]

θθθθ

C'

B'B

C

OGmin G'max

A

Figure 12: Note that vertices B′ and C ′ are coincident.

Proof. As motivation for this proof, imagine being given a triangle inscribed in a circle. Take thecentroid G of the triangle, which lies on the Euler line, and start moving it away from the circumcentertoward the closest edge of the circle. Since G is the intersection of the lines connecting the verticesof the triangle with the midpoints of the opposite sides, it follows that as G approaches the edge ofthe circle, two vertices of the triangle (say B and C ) begin to come closer together. This is becauseG is approaching one side of the triangle (say AB ), and as it gets closer and closer to that side, theline connecting vertex A and the midpoint of BC , Ma moves closer and closer to the edge of thecircumcircle. Thus, when G reaches side AB , BC and consequently 4ABC disappears. It seemsthat this point is the limit for OG or g . Therefore, we first identify a way to calculate this limit.

Let Ma again be the midpoint of the side opposite vertex A. We want to find the value of gso that Ma lies on the the circumcircle. Now, consider 4OAMa . Let A be the vertex of 4ABC

inscribed in the circle and Ma be the point on ray−→AG such that AG = 2GMa . Let y = GMa and

g = OG .

Then, since OMa and OA are both radii of the circle, it follows that 4OAM is isosceles. Thus,∠OAMa

∼= ∠OMaA . Let α = ∠MaOG and ∠GOA = θ . Now, by the Law of Sines, sin θ2y = sin ∠OAMa

g

in 4OAG . Also, sin αy = sin OMaA

g = sin ∠OAMag in 4MaOG . Therefore, sin θ

2y = sin αy , which implies

that sinα = sin θ2 . This in turn implies that α = sin−1 1

2 sin θ . Now, π−(θ+α)2 = β . If we define

γ = ∠OGA , then γ = π − θ − β = π − θ − π2 + θ

2 + α2 = π

2 − θ2 + α

2 . Finally, again by the Law of

Sines,sin π

2 − θ2 − α

2

x=

sin π2 − θ

2 + α2

r. ∴ g = r

sin π2 − θ

2 − α2

sin π2 − θ

2 + α2

. Replacing α with sin−1 12 sin θ and

realizing that sin(π2 − φ) = cosφ provides the desired result,

g < rcos[12(θ + sin−1(1

2 sin θ))]cos[12(θ − sin−1(1

2 sin θ))].

12

Now that we have g bounded above, we consider the lower bound. Clearly we need to includeg = 0 because we need to include equilateral triangles. We have always looked at our triangles with Gon the right side of O . If we look at the triangles we get with G on the left side of O we realize theyare the same as the ones with G on the right side. They can be mapped to each other by reflectingacross the line perpendicular to the Euler line at O . Thus,

0 ≤ g < rcos[12(θ + sin−1(1

2 sin θ))]cos[12(θ − sin−1(1

2 sin θ))].

The bounds for θ are a lot harder to sort out because they come in cases which depend on g Forthe following propositions, we will denote the θ value for an isosceles triangle with a vertex and Glying on the Euler line on the same side of O as θR . Also, we will denote the θ value for an isoscelestriangle with a vertex and G lying on the Euler line on the opposite sides of O as θL .

Proposition 15. If g < r3 , then cos−1(3g+r

2r ) ≤ θ ≤ cos−1(3g−r2r ) .

θmax θmin

C'

A'

C

A

B' BHO G

Figure 13: The bounds are created by an isoceles triangle on either end.

Proof. Since g < r3 , every triangle with a vertex on the Euler line is an isosceles triangle. Consider

the isosceles triangle with vertex B on the Euler line on the same side of O that G is on. So,θR = m∠GOA . Let Mb be the midpoint of AC . Since this is an isosceles triangle, Mb lies on theEuler line and m∠AMbO = π

2 . Also, r−g = BG = 2(GMb) . Therefore, OM = r−3g2r . We also know

m∠AOM = π − θR . So cos(π − θR) = − cos θR = r−3g2r . Therefore,

θR = cos−1(3g − r

2r) .

A consequence of this isosceles triangle case is that ∠GOA ∼= ∠AOB . Also, ∠GOA ∼= ∠GOC ,and both measure θ , so we can think of ∠GOA as θ . By the law of cosines, we know that

∠AOB = cos−1(2r2 − (AB)2

2r2) .

13

In the appendix, we prove that

AB =

√3r2 − 3rg cos θ − 3rg sin θ

√3r2 − 9g2 + 6rg cos θ√r2 + 9g2 − 6rg cos θ

.

When θ = cos−1(3g−r2r ) , (after some simplification) we see

∂(θ −m∠AOB)∂θ

= 1− 3g

r.

So, since g < r3 ,

∂(θ −m∠AOB)∂θ

> 0.

This means that as we increase m∠GOA , the difference m∠GOA−m∠BOA increases. So,

m∠GOA−m∠BOA > 0.

Therefore, A and B must be on the same side of the Euler line. This means that θ is made by∠GOC , and

∠GOC < cos−1(3g − r

2r),

otherwise C and B would be on the same side of the Euler line (and B cannot be on both sides ofthe Euler line). This proves that θR is the upper bound for θ when g < r

3 .Likewise, the lower bound for θ is found in the other isosceles triangle case. That is, when we

have an isosceles triangle with a vertex (without loss of generality, C ) on the opposite side of O fromG . Let θL be made by ∠GOA . Again, let Mc be the midpoint of AB . Since this is an isoscelestriangle, Mc lies on the Euler line and m∠AMcO = π

2 . Also, g + r = CG = 2(GMc) . Therefore,OMc = 3g+r

2r . We also know m∠AOG = θL . So,

θL = cos−1(3g + r

2r) .

The consequence of this isosceles triangle case is that ∠GOA + ∠AOC = π . Also, ∠GOA ∼=∠GOB , and both measure θ , so we can think of ∠GOA as θ . By the law of cosines, we know that

∠AOC = cos−1(2r2 − (AC)2

2r2.

In the appendix, we prove that

AC =

√3r2 − 3rg cos θ + 3rg sin θ

√3r2 − 9g2 + 6rg cos θ√r2 + 9g2 − 6rg cos θ

.

When θ = cos−1(3g+r2r ) , (after some simplification) we see

−∂(θ + m∠AOC − π)∂θ

= −(1 +3g

r) .

So,

−∂(θ + m∠AOC − π)∂θ

< 0 .

This means that as we decrease m∠GOA , the m∠GOA + m∠BOA− π decreases. So,

m∠GOA + m∠COA < π.

14

Therefore, A and C must be on the same side of the Euler line. This means that θ is made by∠GOB , and

∠GOB > cos−1(3g + r

2r),

otherwise C and B would be on the same side of the Euler line (and C cannot be on both sides ofthe Euler line). This proves that θL is the lower bound for θ when g < r

3 .Combining the upper and lower bounds we get,

cos−1(3g + r

2r) ≤ θ ≤ cos−1(

3g − r

2r) .

Note that if we consider the g = 0 case, we have defined θ = π2 which falls between the bounds.

Proposition 16. If g = r3 , then 0 < θ ≤ π

2 .

Proof. To find the upper bound for θ when g = r3 we again look at the isosceles triangle with vertex

B on the same side of O as G . We know that in this case, O and Mb are the same point, andthat m∠AOG = π

2 . We also know that A , O , and C are collinear, so if we increase ∠AOG wenecessarily decrease ∠GOC , and the two angles must sum to π . The right triangle case is the reasonwe had to add the requirement that θ be the smaller of the two angles if we have two “one-vertexsides.” In this case, the largest value that the smaller angle can take is π

2 . When we go to look atthe isosceles triangle with vertex C on the opposite side of O from G , we realize that we cannotmake this triangle because two vertices would have to fall on on the Euler line, which is impossible.So if we keep B at the right angle, we can move vertex C arbitrarily close to the Euler line on theopposite side of the circle from B . This means that we can make m∠GOC arbitrarily close to π ,so m∠GOA can be made arbitrarily close to 0 . So when g = r

3

0 < θ ≤ π

2.

Proposition 17. If g > r3 , then cos−1(3g−r

2r ) ≤ θ ≤ cos−1(3g2−r2

2rg ) .

Proof. Again we consider the isosceles triangle with vertex B on the Euler line on the same side ofO that G is on. So,

m∠GOA = θR = cos−1(3g − r

2r) .

As before we have ∠GOA ∼= ∠AOB . Also, ∠GOA ∼= ∠GOC , and both measure θ , so we canthink of ∠GOA as θ . Following the same procedure for the isosceles triangle with θ = θR (from theg < r

3 case), we see,∂(θ −m∠AOB)

∂θ= 1− 3g

r.

But this time, since g > r3 ,

∂(θ −m∠AOB)∂θ

< 0.

This means that as we increase m∠GOA , the difference m∠GOA−m∠BOA decreases! So,

m∠GOA−m∠BOA < 0.

15

θmax

θmin

B,C

C'

A'

B' HO G

A

Figure 14: The bounds are created by an isoceles triangle and a line.

Therefore, A and B must be on opposite sides of the Euler line. This proves that θR is the lowerbound for θ when g > r

3 .When g > r

3 there is no other isosceles triangle to look at. This is because BMb = 32(r + g) >

32(4r

3 ) = 2r , so the midpoint of the side opposite B and therefore the side itself is not contained inthe circumcircle. Even though this means that our usual means of finding the other bound is notpossible, it gives us another possibility. Now we set out to find where the triangle disappears. To dothis, we find the angle that forces Ma to land on the circumcircle (also forcing A, MA, and G tobe collinear. Therefore, we are interested in the case where OA = r = OMa . This means 4AOMa

is an isosceles triangle, and ∠OAMa∼= ∠AMaO . If we define l = AG , then AMa = 3l

2 . By the lawof cosines, r2 = r2 + 9l2

4 − 3rl cos∠AMaO . After some simple algebra we discover,

2r(cos∠AMO) =3l

2.

By the Law of Cosines in 4AOG we know g2 = r2 + l2 − 2rl cos∠OAMa . Substituting andsolving for l2 , we get l2 = 2(r2 − g2) . We again use the law of cosines in 4AOG and get l2 =r2 + g2 − 2rg cos θ . Substituting for l2 and solving for θ we get

θ = cos−1(3g2 − r2

2rg) .

Thus,

cos−1(3g − r

2r) ≤ θ ≤ cos−1(

3g2 − r2

2rg) .

The nice part about the last upper bound for θ is that it is the inverse function to our upperbound for g . With that in mind, here is a cross-section of our space for r = 1 :

As you see in the picture, we have built in a flipping effect that happens when g crosses the r3 line.

This stems directly from our definition of θ as the angle formed to the one-vertex side. It is a problemwe can live with and will deal with shortly, but we do so with the belief that our definition of θ is the

16

Figure 15: A cross-section of the space when r = 1 .

most natural way to eliminate overcounting. First, we must quickly return to the situation where wefix θ and allow g to vary. If we do this and allow g to cross the r

3 threshold, it is not immediatelyclear what happens. Upon reflection of the g = r

3 case, we realize that (θ, g, r) ∼= (π − θ, g, r) .Now that we know where all of our triangles are and what triangle space looks like, it is natural to

define a metric to determine how close two triangles are to being congruent. We would like the metricto have two additional properties:

i) Account for the fact that all triangles with g = 0 and the same r are congruent, andii) Account for the flipping effect when g crosses r

3 .First, we will account for the flipping by defining a new function,

θ =

θi giri3

π−θi gi>ri3

.

Let s and t be triangles such that s = (θ1, g1, r1) and t = (θ2, g2, r2) . Define

D(s, t) =√

(g1θ1 − g2θ2)2 + (g1 − g2)2 + (r1 − r2)2 .

Attaching the g ’s onto the θ terms accounts for how alike triangles with small g values are. Italso leads to two nice propositions dealing with similar triangles, but first we will show a result abouttriangles who have the same r and g values.

Definition 8. A gr -family of triangles is a family of triangles with the same values of g and r .

Proposition 18. If triangles s and t are in the same gr -family, with s = (θ1, g, r) and t =(θ2, g, r) , then D(s, t) = g|θ1 − θ2| .

Proof.

D(s, t) =√

(gθ1 − gθ2)2 + (g − g)2 + (r − r)2 =√

g2(θ1 − θ2)2 = g|θ1 − θ2| .

17

Proposition 19. Let s and t be triangles such that s = (θ, g, r) and t = (θ, kg, kr) . Then

D(s, t) = |k − 1|√

(gθ)2 + g2 + r2

Proof. Begin with the definition of

D(s, t) =√

(kgθ − gθ)2 + (kg − g)2 + (kr − r)2.

Then factor a (k − 1)2 from each term, and bring it outside the square root as a |k − 1| .Proposition 20. Given triangles p = (θ1, g, r) , p′ = (θ1, kg, kr) , q = (θ2, g, r) , q′ = (θ2, kg, kr) ,then

D(p′, q′) = kD(p, q).

Proof.

D(p′, q′) =√

(kgθ1 − kgθ2)2 + (kg − kg)2 + (kr − kr)2 = kg|θ1 − θ2| = kD(p, q).

5 Tracing Theorems

Now that the foundation of our construction has been laid out, we turn our attention to the behaviorof triangle families. Several interesting properties were found with great assistance from the animationfeature of Geometer’s Sketchpad, which allows us to study the behavior of two families of triangles.We first study the family of triangles obtained through varying g , which we call the rθ -family.

B'

C'

M'aMa

B

C

O G

A

G'

Figure 16: rθ -family: the parallel line.

Proposition 21. Let 4ABC be given with circumcenter O and centroid G . If we fix r , θ andmove G along the Euler line, we get a family of triangles denoted as the rθ -family. Then the locusof midpoints Ma for side BC forms a line segment parallel to the Euler line.

18

Proof. For a 4ABC determined by θ , g and r in our triangle construction, let 4AB′C ′ be thetriangle formed by moving G away from O along the Euler ine. 4AB′C ′ is determined by θ ,g′ and r . Let M ′

a be the midpoint of B′C ′ and G′ be the centroid of 4AB′C ′ . Then, consider4AMaM

′a . By construction, G lies on AMa , and G′ lies on AM ′

a . Now, we know that AG = 2GMa

and AG′ = 2G′M ′ . Thus, with ∠MaAM ′a in common, by SAS similarity, 4MaAM ′

a ∼ 4GAG′ .This implies that ∠AGG′ ∼= ∠AMaM

′a . Thus, GG′ ‖ MaM ′

a . Therefore, as G moves away from Oon the Euler line, Ma moves along a line parallel to OG .

Mp

M''

Mc

B

Mb

C

O G

A

Figure 17: rθ -family: the circle.

Proposition 22. In our rθ -family of triangles, the locus of midpoints Mb for side AC and Mc forside AB forms part of a circle centered at the midpoint M ′′ of OA .

Proof. Let O be the circumcenter of 4ABC and let Mb and Mc be the midpoints of sides B andC respectively. Also, let Mp be the midpoint of segment AMc . Now, construct OA and call itsmidpoint M ′′ . By SAS similarity, 4OAMc ∼ 4M ′′AMP . Thus, M ′′MP is parallel to OMC , whichis perpendicular to AB since MC is the midpoint of AB and O is the circumcenter.

Now, we know that M ′′MP is the perpendicular bisector of the segment AMC . From this, weknow that AM ′′ ∼= MCM ′′ . By the same reasoning, we know that AM ′′ ∼= MBM ′′ .

So, M ′′ is the circumcenter of 4AMBMC .

Definition 9. For any 4ABC , the circle determined by the midpoints of each side also passes throughthe feet of its altitudes and the midpoints of the segments joining each vertex with the orthocenter.This circle is known as the nine-point circle, or the Euler circle.

Proposition 23. Recall that the collection of triangles sharing a circumcircle O and centroid G isdefined as the gr -family. This family of triangles shares the same nine-point circle, which is formedby the locus of the midpoints of their sides.

Proof. Consider 4ABC with Ma , Mb and Mc as the midpoints of BC , AC , and AB respectively.Extend OG to O′ such that GO′ = OG

2 . Because GO′ = OG2 , GMC = CG

2 and ∠O′GMC∼= ∠OGC ,

4O′GMC ∼ 4OGC . Therefore, O′MC = OC2 = r

2 . By similar logic, O′ is equidistant to Ma , Mb

and Mc . Thus, as θ changes, Ma , Mb , and Mc trace the nine-point circle. Another way to lookat this is that since the gr -family of triangles share the same circumcircle and centroid, they shouldalso share the collection of complementary points for the circumcircle, i.e., the nine-point circle.

19

O'

Ma

Mc

B

Mb

C

H

O

G

A

Figure 18: gr -family: the nine-point circle.

6 The Symmedian Point

Having examined the loci of certain well-known points in our construction, we now turn our atten-tion to a more obscure triangle center–the symmedian point. In the collection of triangle centers,few could be considered well-known to geometers and even fewer to the average geometry student.Occasionally, however, centers which should be a part of mathematicians’ base knowledge disappearfrom the contemporary consciousness. The symmedian point is one such center. Well explored manyyears ago, the symmedian point has a plethora of useful and fascinating properties. In his work 19thand 20th Century Euclidean Geometry, Ross Honsberger calls it “one of the jewels of modern geom-etry” [Honsberger, pg. 53]. In order to begin our brief study of this geometric gem, we first needto understand some established definitions and theorems. Thus, let us define the concept of isogonalconjugates.

Definition 10. Let ∠A be given. Then, the isogonal conjugate of AP , where P is any point inthe plane, is the reflection of AP over the angle bisector of ∠A . AP and its reflection are calledisogonal conjugate lines or simply isogonal conjugates.

A

P

Figure 19: Isogonal conjugate lines.

In Figure 19 above, the middle line is the angle bisector of ∠A , and the two thick black lines are

20

isogonal conjugates. Moreover, as we can see above, one direct consequence of the definition of anisogonal conjugate is that the gray angles are congruent, and the black angles are congruent.

Q

P

B

C

A

Figure 20: Isogonal conjugate points.

Since isogonal conjugates inherently involve angles, one question which naturally arises is howisogonal conjugates relate to triangles. As Theorem 2 below states, they have at least one fascinatingproperty.

Theorem 2. Let P be a point in the plane, and let 4ABC be given. Then the lines isogonal toAP , BP , and CP , meet at a point Q . Points P and Q are called isogonal conjugate points, orisogonal conjugates. [Honsberger, pg. 53]

A proof of this theorem can be found in Ross Honsberger’s work, Episodes in Nineteenth andTwentieth Century Euclidean Geometry. We will omit it here. However, Theorem 2 is an importantpart of this section since it states that isogonal conjugate points, like isogonal conjugate lines, occurin pairs. Surprisingly, O and H , the circumcenter and orthocenter of a triangle, are one such pair.Although this fact is well-known, our proof of it is included below. Before we begin, however, we muststate a property of isogonal conjugates which we will use in the proof.

F

E

QP

A

B

C

D

Figure 21: Proposition 24.

21

Proposition 24. Let AP and AQ be isogonal conjugate lines through vertex A , and let D be apoint on AP . Then the line EF connecting the feet of the perpendiculars from D to AB and ACis perpendicular to AQ . [Honsberger, pgs. 64—65]

A proof of Proposition 24 can be found in Honsberger. We will omit it here and move directly toour proof that the circumcenter and the orthocenter are isogonal conjugates.

Proposition 25. The circumcenter and the orthocenter of a triangle are isogonal conjugates.

D

Ha

P Mb

Pa

A

B

C

Figure 22: Isogonal conjugates: the circumcenter and the orthocenter.

Proof. Let 4ABC be given. Construct AHa , the altitude from vertex A to BC . Construct theisogonal conjugate of AHa . It will meet BC at a point Pa . Now, construct the perpendicularbisector of AC . It will meet APa at a point P . Drop a perpendicular from P to AB , and call theintersection of the two D . Construct DMb , where Mb is the midpoint of AC . By Proposition 24,DMb must be perpendicular to AHa . However, by construction, AHa is perpendicular to BC .Thus, DMb ‖ BC . By AAA similarity, 4ADMb ∼ 4ABC . So, AD

AB = AMbAC = 1

2 . Therefore, Dis the midpoint of AB , and P must be the circumcenter of 4ABC . Thus, P lies on the isogonalconjugate to the altitude from vertex A . By a similar argument, P will also lie on the isogonalconjugate to the altitude from vertices B and C . Therefore, the circumcenter and the orthocenterof 4ABC are isogonal conjugates.

Pa

Pb

Pc

P

C

B

A

Figure 23: A pedal triangle.

22

Returning to more general isogonal conjugates, another known and interesting property which wewill reference later relates the circumcircles of the pedal triangles of two isogonal conjugate points.Before we state and prove this property, however, we need to define the concepts of a pedal triangleand a cyclic quadrilateral, and prove a useful fact about cyclic quadrilaterals.

Definition 11. Let P be a point inside a given triangle 4ABC . Construct the perpendiculars fromP to AB , BC , and AC . Label the points of intersection Pc , Pa , and Pb , respectively. Then,4PaPbPc is the pedal triangle of P with respect to 4ABC .

The next definition and lemma deal with certain types of quadrilaterals.

Definition 12. A quadrilateral ABCD is called cyclic if all of its vertices lie on a circle.

Lemma 1. Let ABCD be a quadrilateral with right angles at vertices A and C . Then, quadrilateralABCD is cyclic.

B

A

M

B

C

A

D

D

C

Figure 24: A generic cyclic quadrilateral and a cyclic quadrilateral with right angles.

Proof. Connect B to D . Then, the midpoint M of BD will be the circumcenter of ∆ABD andof ∆CBD . Thus, MA ∼= MD ∼= MB ∼= MC . Thus, ABCD lies on the circle centered at M , andABCD is cyclic.

Now, let us state and prove the property of isogonal conjugates which relates the circumcirclesof the pedal triangles of two isogonal conjugate points. Note that although this fact is known, thefollowing proof is our own.

Proposition 26. Let triangle 4ABC be given with isogonal conjugate points P and Q . Then,the midpoint O of the segment PQ is the circumcenter of the pedal triangles of both P and Q .Moreover, both pedal triangles share the same circumcircle. [Honsberger, pgs. 67–69]

Proof. The following proof utilizes Figure 25.

Let P and Q be isogonal conjugate points, and let O be the midpoint of segment PQ . Let Mbe the midpoint of segment AP , and let triangle 4PaPbPc be the pedal triangle determined by thepoint P . Let I be the intersection of segment PbPc with OM , and let J be the intersection ofsegment PbPc with AQ .

We first show that OM ‖ AQ . By construction, MP = 12AP and OP = 1

2PQ . Since ∠MPOis common, it follows that 4MPO ∼ 4APQ . Thus, ∠PMO ∼ ∠PAQ , which indicates thatOM ‖ AQ .

23

LJ

IM

Pa

Pb

Pc

OP

Q

B

A

C

Figure 25: Proof that the pedal triangles of points P and Q share the same circumcenter.

Now consider triangles 4APcP and 4APbL , where L is the intersection of PbP with AQ .By construction, m∠PPcA = π

2 = m∠LPbA . Moreover, since P and Q are isogonal conjugates,∠PcAP ∼= ∠PbAL . By angle subtraction, ∠PcPA ∼= ∠PbLA . Thus, by AAA Similarity, 4APcP ∼4APbL .

Now, since m∠LPbA = π2 = m∠PPcA , quadrilateral APbPPc is cyclic by Lemma 1. Thus,

∠PcAP ∼= ∠PcPbP . Since m∠PcAP + m∠PcPA = ∠PcPbP + m∠PbLA = π2 , it follows that

m∠PbJL = π2 . This implies that m∠PcIO = π

2 since OM ‖ AQ .Now, since quadrilateral APbPPc is cyclic, the points A , Pb , P, and Pc lie on a circle centered at

M, the midpoint of AP . Thus, MPc∼= MPb . Moreover, since segment MI is common, 4MPcI and

4MPbI both have right angles, congruent hypotenuses, and one congruent leg. They are consequentlycongruent. This then indicates that IPc

∼= IPb , making I the midpoint of PbPc .Therefore, OM is the perpendicular bisector of PbPc . A similar argument shows that O also lies

on the perpendicular bisectors to segments PaPb and PaPc , making O the circumcenter of 4PaPbPc .Repeat the argument above for Q . Then, the midpoint O of segment PQ is the circumcenter of

the pedal triangles constructed from P and Q .To show the two pedal triangles share not only the same circumcenter, but also the same circum-

circle, consider the following argument, which utilizes Figure 26 below.Let points P and Q be isogonal conjugates, and let O be the midpoint of segment PQ . Drop

perpendiculars from P , O , and Q to AC , and let Pb , Ob , and Qb be their intersections withAC , respectively. Since m∠PPbQb = m∠OObQb = m∠QQbOb = π

2 , it follows that PPb ‖ OOb ‖QQb . Construct PQb and let N be the intersection of PQb and OOb . Then, ∠PNO ∼= ∠PQbQand ∠PON ∼= ∠PQQb . Since ∠NPO is common, 4PNO ∼ 4PQbQ . Thus, PO

PQ = PNPQb

= 12 ,

and N is the midpoint of segment PQb . By a similar argument, 4QbObN ∼ 4QbPbP . Thus,PNPQb

= PbObPbQb

= 12 . This implies that Ob is the midpoint of segment PbQb . Thus, since PbOb

∼= PbQb ,∠PbObO ∼= ∠QbObO , and OOb is common, by SAS congruence, 4OObPb

∼= 4OObQb . This indicatesthat OPb

∼= OQb . However, OPb is the radius of the pedal triangle of P , and OQb is the radius ofthe pedal triangle of Q . Therefore, the pedal triangles of P and Q share the same circumcircle.

Now that we have a basic understanding of isogonal conjugates and some of their properties, weare ready to define the symmedian point.

24

N

Ob

Qb

Pb

O

P

Q

C

A

B

Figure 26: Proof that the pedal triangles of points P and Q share the same circumcircle.

Definition 13. The symmedian point K is the isogonal conjugate of the centroid.

Interesting properties of the symmedian include such statements as Proposition 27 below.

Proposition 27. The symmedian point is the centroid of its own pedal triangle. [Honsberger, pgs.72–73]

This proposition is well known. A proof of it can be found in Honsberger. Moreover, utilizing itand Proposition 26, we can immediately prove the following corollary.

Corollary 2. The centroid of a 4ABC always lies on the Euler line of the pedal triangle of itssymmedian point K .

Proof. Let 4ABC be given. Since the centroid and the circumcenter of a triangle determine its Eulerline, it follows from Propositions 26 and 27 that GK , where G is the centroid of 4ABC and K isits symmedian point, is the Euler line of the pedal triangle of K .

Corollary 2 leads to an interesting observation. The locus of symmedian points in a gr -family oftriangles form either an entire circle or an arc of a circle centered on the Euler line. Let us state thismore formally.

Theorem 3. Let Ω be the family of triangles produced by fixing a circumcenter O , a centroid G , acircumradius r , and varying ∠GOA from 0 to 2π . Let g denote the distance between O and G .Let E ⊆ [0, 2π] such that for all φ ∈ E , there exists a triangle 4ABCφ ∈ Ω with m∠GOA = φ .Let Kφ be the symmedian point of 4ABCφ .

Let Kn be the point on the ray−−→OG such that OKn = 2gr2

r2−g2 . Then, for all φ ∈ E , Kφ lies on

the circle with radius 2g2rr2−g2 centered at Kn . We call this circle the Carleton Circle and its center

Kn the Knights’ Point.If g < r

3 , every point on the Carleton Circle is the symmedian point of a triangle in Ω . If g = r3 ,

then every point except (r, 0) is the symmedian point of a triangle in Ω . If g > r3 , then every point

P on the Carleton Circle such that P is strictly contained in the interior of the disc enclosed by thecircumcircle of Ω is the symmedian point of a triangle in Ω .

25

While in the specific case in which g = r3 a geometric proof of Theorem 3 is apparent, in the

general case, such a proof is not obvious. Thus, we will approach the general case from an analyticperspective. However, in order to more clearly understand precisely what Theorem 3 states, let usbegin our exploration with the geometric proof of the case in which g = r

3 .

Lemma 2. Let Ω be a gr –family of triangles in which g = r3 . Let Kn lie on segment OH such

that OKn = 3KnH . Let the circle centered at Kn with radius KnH be called the Carleton Circle.Then, for every 4ABCφ ∈ Ω , Kφ , the symmedian point of 4ABCφ , lies on the Carleton Circle.Moreover, every point except H on the Carleton Circle is the symmedian point of a triangle in thegiven Ω .

Note that Lemma 2’s geometric description of the location of Kn and the radius of the CarletonCircle correspond to the analytic description in Theorem 3. This is because by Proposition 7 andCorollary 1, when g = r

3 , H will coincide with one vertex of 4ABCφ , making OH a radius ofthe circumcircle. Thus, Kn will lie the following distance along the Euler line of 4ABC : 2gr2

r2−g2 =2g(3g)2

(3g)2−g2 = 18g3

8g2 = 9g4 = 3r

4 = 34OH . Moreover, the radius of the Carleton Circle will be 2g2r

r2−g2 =2g2(3g)

(3g)2−g2 = 6g3

8g2 = 3g4 = r

4 = 14OH .

Now, to prove Lemma 2, we will use the following proposition.

Proposition 28. Let 4ABC be a right triangle with the right angle at vertex B . Then, the sym-median point K is the midpoint of the symmedian from vertex B . [Honsberger, pgs. 59–60]

This fact is commonly known; a proof of it can be found in Honsberger. Let us now prove Lemma 2.

KnM

KqKb

J

C

BO

A

Figure 27: Proof that the symmedian point lies on a circle when 4ABC is a right triangle.

Proof. By Corollary 1 and Proposition 6, one vertex of 4ABCφ must coincide with H . Withoutloss of generality, assume that B is that vertex. Construct the symmedian from B , and let Kb bethe intersection of that symmedian with AC . We will first show that Kb lies on the circle centeredat M, the midpoint of segment OH .

Construct the angle bisector of ∠ABC and label it JB , where J is its intersection with AC .Now, since the symmedian is the isogonal conjugate of the median, ∠OBJ ∼= ∠JBKb and ∠KbBA ∼=∠OBC . Since O is the circumcenter of 4ABC , it follows that 4OBC is isosceles with OB ∼= OC .

26

Thus, ∠ACB ∼= ∠OBC . Now, m∠ABC = π2 . This implies that m∠ACB + m∠CAB = π

2 whichin turn implies that m∠KbBA + m∠CAB = π

2 . Thus, m∠AKbB = π2 , and Kb lies on the circle

centered at M with radius MO . Note that in this shows that Kb is the altitude from vertex B ,and that the circle it lies on is the circumcircle of the pedal triangles of O and H .

Now, by Proposition 28, Kφ is the midpoint of KbB . Construct KφM . By SAS similarity,4KφBM ∼ 4KbBO . Thus, m∠MKφB = m∠OKbB = π

2 . Therefore, Kφ lies on the circlecentered at the midpoint Kn of MB , 3

4 of the way from O to H. The radius of the circle will be12MB = 1

4OH .To show that every point except B on the Carleton Circle is a symmedian point of a triangle in

the given gr – family, let J be a point on the Carleton Circle. Since H is on the circumcircle, byProposition 6, it must coincide with one vertex of every triangle in the given gr – family. Withoutloss of generality, let B be that vertex. Construct BJ . Then, extend BJ past J to a point Jb suchthat JJb

∼= BJ . By the argument above, Jb is the foot of the altitude from B . Construct the lineperpendicular to JbB through Jb . This line will intersect the circumcircle at points A and C . Byconstruction, 4ABCφ has symmedian point J . Thus, every point except B on the Carleton Circlecentered at Kn is a symmedian point of a triangle in the given gr –family.1

To show the general case of Theorem 3, we will first think about placing the gr− family of trianglesin a coordinate system. We will find the x - and y -coordinates of the Knights’ Point Kn and of thesymmedian point Kφ of any triangle in that family. Then, we will use the distance formula to findthe distance between these two points. If that distance is not dependent on φ , then, since Kn isfixed in a gr− family, Kφ will always lie on a circle centered at Kn .

The proof below will make use to two propositions, both of which are stated below. We will notinclude proofs of either here.

Proposition 29. Let vectors ~a and ~b be given. Then, ~c = |~b|~a + |~a|~b bisects the angle created byvectors ~a and ~b . [Stewart, pg. 850]

Proposition 30. Let triangle 4ABC with altitude AHa be given. Then, the symmedian point K of4ABC lies on the line connecting the midpoint Mh of AHa to the midpoint Ma of BC . [Honsberger,pgs. 65–67]

Now, let us prove Theorem 3.

Proof. Let Ω be the gr -family of triangles with circumcenter O , centroid G , and circumradius rgiven. As we have previously done, let g denote the distance between the circumcenter and thecentroid. Let φ denote m∠GOA where A is the chosen vertex in the triangle construction. ChooseO to be the origin of the coordinate system and the Euler line OG to be the x -axis. We will firstfind expressions for the coordinates of the vertices A , B , and C of any triangle in the gr− family.Our method of finding them will algebraically mimic the geometric construction described earlier.

By construction, O has coordinates (0, 0) , G has coordinates (g, 0) , and H has has coordinates(3g, 0) . Since A lies on the circle with radius r , and since angle φ is the angle between

−→OA and

1The reason that B cannot be a symmedian point is a result of the fact that the symmedian point K of a 4ABCcan never fall on the circumcircle of 4ABC . Suppose that K does lie on the circumcircle. Then, either K is coincidentwith one vertex, say vertex A , or K is not coincident with any vertex of 4ABC . Suppose K is coincident with vertexA . Then, AB and AC are the symmedian lines from vertices B and C . Consequently, BC must be the medianfrom both vertices B and C . However, that would force C to be the midpoint of AC and B to be the midpointof AB . This would imply that A , B , and C are coincident. This is a contradiction. Now, suppose that K is notcoincident with any vertex of 4ABC . Then, K lies outside 4ABC . However, that would force the median of at leastone vertex to lie outside of 4ABC . This is a contradiction. Thus, K must always be strictly inside the circumcircleof 4ABC .

27

the x -axis, it follows that A has coordinates (r cosφ, r sinφ) . Thus, we can find the equation of theline AG by computing its slope and its y -intercept.

The slope of the AG isr sinφ− 0r cosφ− g

=r sinφ

r cosφ− g.

To find the y -intercept b , we use the point (0, g) and the slope:

0 = gr sinφ

r cosφ− g+ b

b = −gr sinφ

r cosφ− g

Thus, the equation of AG is y =r sinφ

r cosφ− gx− g

r sinφ

r cosφ− gor equivalently y = r sin φ(x−g)

r cos φ−g .

Next, we find the coordinates of the midpoint Ma of BC . Since the y -coordinate of G is 0 , and

since Ma lies on the ray−→AG such that AMa = 3

2AG , the y -coordinate of Ma must be−r sinφ

2.

We solve for the x -coordinate using the equation of AG :

r sinφ

r cosφ− gx− g

r sinφ

r cosφ− g=−r sinφ

2r sin θ

r cosφ− gx =

−r sinφ

2+ g

r sinφ

r cosφ− g

r sin θ

r cosφ− gx =

−r sinφ(r cosφ− g) + 2gr sinφ

2(r cosφ− g)

x = (−r sinφ(r cosφ− g) + 2gr sinφ

2(r cosφ− g))(

r cosφ− g

r sin θ)

x =g − r cosφ + 2g

2

x =3g − r cosφ

2.

Thus, the coordinates of Ma are (3g−r cos φ2 , −r sin φ

2 ) . Let us now find the equation of AH .The slope of the AH is

r sinφ− 0r cosφ− 3g

=r sinφ

r cosφ− 3g.

Thus, we now only need its y -intercept d , which we find using the point (0, 3g) and the slope:

0 = 3gr sinφ

r cosφ− 3g+ d

d = −3gr sinφ

r cosφ− 3g

The equation of AH is y =r sinφ

r cosφ− 3gx− 3g

r sinφ

r cosφ− 3gor equivalently y =

r sinφ(x− 3g)r cosφ− 3g

.

To find BC , we use the fact that BC is perpendicular to AH . From this, it follows that the slope

of BC is the negative reciprocal of the slope of AH . Thus, the slope of BC must ber cosφ− 3g

r sinφ.

We now find the y -intercept i of BC using its slope and the coordinates of Ma :

28

(3g − r cosφ

r sinφ)(

3g − r cosφ

2) + i =

−r sinφ

2(3g − r cosφ)2

2r sinφ+ i =

−r sinφ

2

i =−r sinφ

2− (3g − r cosφ)2

2r sinφ

i =−r2 sin2 φ− (3g − r cosφ)2

2r sinφ.

Thus, the equation of BC is as follows:

y =3g − r cosφ

r sinφx +

−r2 sin2 φ− (3g − r cosφ)2

2r sinφ.

To find the coordinates of B and C , we solve the system of equations

y =3g − r cosφ

r sinφx +

−r2 sin2 φ− (3g − r cosφ)2

2r sinφ.

andy2 + x2 = r2.

Using Mathematica to solve this complicated system, we see that the two solutions are

xB =54g3 + 12gr2 − 2r(27g2 + r2) cos φ + 6gr2 cos 2φ

4(9g2 + r2 − 6gr cosφ)

+r√

3 csc φ(cos 2φ− 1)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

4(9g2 + r2 − 6gr cosφ).

and

xC =54g3 + 12gr2 − 2r(27g2 + r2) cosφ + 6gr2 cos 2φ

4(9g2 + r2 − 6gr cosφ)

− r√

3 csc φ(cos 2φ− 1)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

4(9g2 + r2 − 6gr cosφ).

To find the y -coordinates of B and C , we plug their x -values into the equation of BC . Doingso, we find that

yB = − sinφ(9g2r2 + r4 − 6gr3 cosφ

2r(9g2 + r2 − 6gr cosφ)

+r√

3 cscφ(3g − r cosφ)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

2r(9g2 + r2 − 6gr cosφ))

= − sinφ(r2(9g2 + r2 − 6gr cosφ)2r(9g2 + r2 − 6gr cosφ)

+√

3 csc φ(3g − r cosφ)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

2(9g2 + r2 − 6gr cosφ))

= − sinφ(r

2+√

3 csc φ(3g − r cosφ)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

2(9g2 + r2 − 6gr cosφ)) .

29

and

yC = − sinφ(9g2r2 + r4 − 6gr3 cosφ

2r(9g2 + r2 − 6gr cosφ)

− r√

3 cscφ(3g − r cosφ)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

2r(9g2 + r2 − 6gr cosφ))

= − sinφ(r2(9g2 + r2 − 6gr cosφ)2r(9g2 + r2 − 6gr cosφ)

−√

3 csc φ(3g − r cosφ)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

2(9g2 + r2 − 6gr cosφ))

= − sinφ(r

2−√

3 csc φ(3g − r cosφ)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

2(9g2 + r2 − 6gr cosφ)) .

Thus, the coordinates of A , B , and C are as follows:

A = (r cosφ, r sinφ) .

B =(

54g3 + 12gr2 − 2r(27g2 + r2) cosφ + 6gr2 cos 2φ

4(9g2 + r2 − 6gr cosφ)

+r√

3 cscφ(cos 2φ− 1)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

4(9g2 + r2 − 6gr cosφ),

− sinφ(r

2+√

3 cscφ(3g − r cosφ)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

2(9g2 + r2 − 6gr cosφ))

).

C =(

54g3 + 12gr2 − 2r(27g2 + r2) cos φ + 6gr2 cos 2φ

4(9g2 + r2 − 6gr cosφ)

− r√

3 cscφ(cos 2φ− 1)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

4(9g2 + r2 − 6gr cosφ),

− sinφ(r

2−√

3 cscφ(3g − r cosφ)√−27g4 + r4 + 36g3r cosφ− 4gr3 cosφ− 6g2r2 cos 2φ

2(9g2 + r2 − 6gr cosφ))

).

With the coordinates of the vertices of a generic triangle in Ω , we can now find the coordinatesof the symmedian point.

We will begin by finding the equation of the symmedian line from vertex A . In order to do so,we need to bisect ∠BAC . However, unlike in non-analytic geometry, this is not a simple matteralgebraically. We thus make use of Proposition 29 above.

Let us consider segments AB and AC to be vectors with their heads at B and C , respectively. Touse Proposition 29 to bisect them, we first need to find their components. This is easily accomplishedby subtracting the coordinates of their tails from the coordinates of their heads. We next find themagnitude of each vector. We will write neither the components nor the magnitudes of vectors

−−→AB

and−→AC here as they are algebraically intensive. However, they are in the Mathematica document in

the Appendix B .

Our next task is to multiply the components of−−→AB by the magnitude of

−→AC and to multiply the

components of−→AC by the magnitude of

−−→AB . Adding these new scaled vectors by together, we find

30

by Proposition 29 a vector which bisects vectors−−→AB and

−→AC . Again, as this would simply be an

exercise in precise copying, we have only included the resulting components in the Appendix B .Now, to find the equation of the angle bisector of ∠BAC , we first find its slope—the y -coordinate

over the x -coordinate of the angle bisector. Then, using this slope and the coordinates of vertex A ,we can find the y -intercept. Thus, we have the equation of the angle bisector of ∠BAC .

As discussed previously, transformations such as reflections are significantly more difficult alge-braically than they are geometrically. Thus, to find a point on the symmedian line from vertex A,we first find the equation of the line through G , ( (g, 0) in our coordinate system) perpendicular tothe angle bisector of ∠BAC . This appears in Appendix B . Then, we find the x -coordinate of theintersection J of this new perpendicular line (which we will now refer to as GJ ) with the anglebisector. This value also appears in Appendix B .

Next, we find the point L on GJ such that LJ = GJ . We do this by subtracting g from thex -value of J then adding that difference to the x -value of J . To find the y -value of L , we evaluatethe equation of GJ at the x -value we just computed. Thus, we now have the coordinates of L . Lwill be a point on the symmedian line from vertex A by the following geometric argument:

LJE

D

C

BHG

O

A

Figure 28: Constructing a point on the symmedian line from vertex A .

Let 4ABC be given with centroid G . Construct the angle bisector of ∠BAC . Then, constructthe line through G perpendicular to the angle bisector of ∠BAC . We have two cases:

Case 1. Suppose that AG is not the angle bisector of ∠BAC . Then, the perpendicular linethrough G will meet the angle bisector at a point J . Now, construct the point L on GJ such thatLJ = GJ . Then, since m∠LJA = m∠GJA = π

2 and segment AJ is common, it follows from SAScongruence that 4AJG ∼= 4AJL . Thus, ∠GAJ ∼= ∠LAJ , which implies that L is a point on thesymmedian line from vertex A .

Case 2. Suppose that AG is the angle bisector of ∠BAC . Then, the line through G perpendicularto AG will intersect AG at G . Then, G , J , and L will be coincident, and G will trivially be apoint on the symmedian line from vertex A .

Now, using the coordinates of vertex L and vertex A , we can find the equation of the symmedianline from vertex A . Its equation simplifies nicely to the following.

y =(−2gr2 + 3g2x + r2x− 2grx cosφ) sinφ

(3g2 + r2) cos φ− gr(3 + cos 2φ).

31

Shortly after setting out to apply the method we employed to find the symmedian from A to findthe symmedian from B , one realizes that given the coordinates of vertices B and C , the algebra ispractically impossible to do by hand and takes a long time for even a program such as Mathematicato complete. Thus, we will use Proposition 30 above and find the equation of the line connecting themidpoint of side BC to the midpoint of segment AHa , where Ha is the intersection of the altitudefrom A and BC .

Thus, we first find where AH intersects BC . Recall that the equation of AH is

y =r sinφ(x− 3g)r cosφ− 3g

,

and that the equation of BC is

y =3g − r cosφ

r sinφx +

−r2 sin2 φ− (3g − r cosφ)2

2r sinφ.

Thus, we solve the system of equations above for x .

r sinφ(x− 3g)r cosφ− 3g

=3g − r cosφ

r sinφx +

−r2 sin2 φ− (3g − r cosφ)2

2r sinφ

−r2 sin2 φ(x− 3g) = (3g − r cosφ)2x− r2 sin2 φ

2(3g − r cosφ)− (3g − r cosφ)3

2... = ...

x =9g(3g2 + r2)− r(27g2 + r2) cosφ

2(9g2 + r2 − 6gr cosφ).

To find the y -coordinate of this point, we evaluate the equation of AH at the given value ofx above. Doing this, we find that the y -coordinate of HA is r(9g2−r2) sin φ

2(9g2+r2−6gr cos φ). We now find the

midpoint of AHa by adding the x -coordinates of A and Ha and dividing by 2 .

12(9g(3g2 + r2)− r(27g2 + r2) cos φ

2(9g2 + r2 − 6gr cosφ)+ r cosφ)

=27g3 + 3gr2 − 9g2r cosφ + r3 cosφ− 6gr2 cos2 φ + 6gr2 sin2 φ

4(9g2 + r2 − 6gr cosφ).

We find its y -coordinate by plugging this value into the equation of AH . Doing this, we find thatthe y -value of the midpoint of AHa is as follows:

r(27g2 + r2 − 12gr cosφ) sinφ

4(9g2 + r2 − 6gr cosφ).

Thus, we now have all the information we need to find the equation of the line MaHa . We firstfind its slope.

r(27g2+r2−12gr cos φ) sin φ4(9g2+r2−6gr cos φ)

− (− r sin φ2 )

27g3+3gr2−9g2r cos φ+r3 cos φ−6gr2 cos2 φ+6gr2 sin2 φ4(9g2+r2−6gr cos φ)

− (3g−r cos φ2 )

=r(15g2 + r2 − 8gr cosφ) sinφ

r(15g2 + r2) cosφ− g(9g2 + 3r2 + 4r2 cos(2φ)).

32

Next we find j , the y -intercept of MaHa , by using its slope and the point Ma with coordinates(3g−r cos φ

2 , −r sin φ2 ) .

−12r sinφ = (

r(15g2 + r2 − 8gr cosφ) sin φ

r(15g2 + r2) cosφ− g(9g2 + 3r2 + 4r2 cos(2φ)))(

3g − r cosφ

2) + j

j = −12r sinφ− (

r(15g2 + r2 − 8gr cosφ) sinφ

r(15g2 + r2) cos φ− g(9g2 + 3r2 + 4r2 cos(2φ)))(

3g − r cosφ

2)

j =2gr(9g2 + r2 − 6gr cosφ) sinφ

−r(15g2 + r2) cos φ + g(9g2 + 3r2 + 4r2 cos 2φ).

Thus, we see that the equation of MaHa , the line joining the midpoint of AHA to the midpointof BC is as follows:

y =r(15g2 + r2 − 8gr cosφ) sinφ

r(15g2 + r2) cos φ− g(9g2 + 3r2 + 4r2 cos(2φ))x

+2gr(9g2 + r2 − 6gr cosφ) sinφ

−r(15g2 + r2) cos φ + g(9g2 + 3r2 + 4r2 cos 2φ)

=r(−18g3 − 2gr2 + 15g2x + r2x + 4gr(3g − 2x) cosφ) sin φ

−9g3 − 3gr2 + 15g2r cosφ + r3 cosφ− 4gr2 cos2 φ + 4gr2 sin2 φ.

By Proposition 30, the symmedian point Kφ of 4ABCφ lies on MaHa . Since the symmedianpoint also lies on the symmedian line from vertex A , it follows that these two lines intersect at Kφ .To find the coordinates of Kφ , we solve the system of these two equations for x .

(−2gr2 + 3g2x + r2x− 2grx cosφ) sinφ

(3g2 + r2) cos φ− gr(3 + cos 2φ)

=r(−18g3 − 2gr2 + 15g2x + r2x + 4gr(3g − 2x) cosφ) sin φ

−9g3 − 3gr2 + 15g2r cosφ + r3 cosφ− 4gr2 cos2 φ + 4gr2 sin2 φ.

Thus,

x =2gr((9g3 + 6gr2) cos φ− r(9g2 + r2 + 6g2 cos 2φ− gr cos 3φ))

(g2 − r2)(9g2 + r2 − 6gr cosφ).

To find the y -coordinate of the symmedian point, we substitute this value of x into the equationof the symmedian from vertex A . Doing this, we see that

y =2g2r(9g2 + r2 − 12gr cosφ + 2r2 cos 2φ) sin φ

(g2 − r2)(9g2 + r2 − 6gr cosφ).

Thus, the coordinates of the symmedian point are as follows:

Kφ = (2gr((9g3 + 6gr2) cos φ− r(9g2 + r2 + 6g2 cos 2φ− gr cos 3φ))

(g2 − r2)(9g2 + r2 − 6gr cosφ),

2g2r(9g2 + r2 − 12gr cosφ + 2r2 cos 2φ) sin φ

(g2 − r2)(9g2 + r2 − 6gr cosφ)) .

33

In order to prove that the symmedian point is always a fixed distance from the point Kn lyingon ray

−−→OG such that OKn = 2gr2

r2−g2 , we first recognize that−−→OH spans the Euler line, which, in our

construction, is the x -axis. Since O is defined to be the origin, and H lies on the positive x -axis,Kn must have coordinates ( 2gr2

r2−g2 , 0) . We use the distance formula to find the distance between thesetwo points.

((2gr((9g3 + 6gr2) cos φ− r(9g2 + r2 + 6g2 cos 2φ− gr cos 3φ))

(g2 − r2)(9g2 + r2 − 6gr cosφ)− 2gr2

r2 − g2

)2

+(

2g2r(9g2 + r2 − 12gr cosφ + 2r2 cos 2φ) sin φ

(g2 − r2)(9g2 + r2 − 6gr cosφ)− 0

)2) 1

2

=

√4g4r2

(g2 − r2)2

=

√4g4r2

(−1)2(r2 − g2)2

=2g2r

r2 − g2.

Thus, the distance between Kn and Kφ is not dependent on φ , which implies that Kφ willalways lie on the circle of radius 2g2r

r2−g2 centered at Kn . Additionally, it is nice to note that since rand g are distances, they are always strictly greater than zero. Moreover, since g cannot equal r , itfollows that 2g2r

r2−g2 > 0 and is always defined.Let us now turn our attention to the second half of Theorem 3. We have three cases: g < r

3 ,g = r

3 , and g > r3 .

Case 1. Let g < r3 . To show that every point on the Carleton Circle is the symmedian point of

a triangle in Ω , we first reexamine the coordinates of the symmedian point. They are copied belowfor easy reference.

Kφ = (2gr((9g3 + 6gr2) cos φ− r(9g2 + r2 + 6g2 cos 2φ− gr cos 3φ))

(g2 − r2)(9g2 + r2 − 6gr cosφ),

2g2r(9g2 + r2 − 12gr cosφ + 2r2 cos 2φ) sin φ

(g2 − r2)(9g2 + r2 − 6gr cosφ)) .

Let

A(φ) =2gr((9g3 + 6gr2) cos φ− r(9g2 + r2 + 6g2 cos 2φ− gr cos 3φ))

(g2 − r2)(9g2 + r2 − 6gr cosφ),

and let

B(φ) =2g2r(9g2 + r2 − 12gr cosφ + 2r2 cos 2φ) sin φ

(g2 − r2)(9g2 + r2 − 6gr cosφ).

Now, by definition, K(φ) = (A(φ), B(φ)) . We will show that K(φ) is continuous. Notice thatg2 − r2 6= 0 since g 6= r .2 Moreover, 9g2 + r2 − 6gr cosφ ≥ 9g2 + r2 − 6gr = (3g − r)2 > 0 sinceg 6= r

3 . Thus, (g2 − r2)(9g2 + r2 − 6gr cosφ) 6= 0 . Since the numerators of both A(φ) and B(φ) are

2If g were equal to r , then this would force all three of the vertices of a 4ABC to be coincident with G . However,a point is not a triangle.

34

combinations of continuous functions, it follows that both A(φ) and B(φ) are continuous, makingK(φ) = (A(φ), B(φ)) continuous.

Now, in Proposition 15, we saw that every triangle in a given gr -family (up to congruence) fallswithin certain bounds on φ . We will use these bounds for φ , so they are copied below for easyreference.

cos−1(3g + r

2r) ≤ φ ≤ cos−1(

3g − r

2r) .

Setting φ = cos−1(3g+r2r ) , we find using Mathematica that K(cos−1(3g+r

2r )) = ( 2grr−g , 0) . Similarly,

setting φ = cos−1(3g−r2r ) , we find that K(cos−1(3g−r

2r )) = ( 2grr+g , 0) .

Since the interval [cos−1(3g+r2r ), cos−1(3g−r

2r )] is a connected set and K(φ) is continuous, it fol-lows from the fact that continuous functions map connected sets to connected sets that the setK([cos−1(3g+r

2r ), cos−1(3g−r2r )]) is connected.3 Moreover, K(cos−1(3g+r

2r )) = ( 2grr−g , 0) and

K(cos−1(3g−r2r )) = ( 2gr

r+g , 0) . These are the two points of intersection of the Carleton Circle withthe Euler line. Since in the first part of the proof we showed that for all φ , K(φ) lies on the Car-leton Circle, it follows that the only way K([cos−1(3g+r

2r ), cos−1(3g−r2r )]) can be a connected set is if

it encompasses minimally either the upper or lower half of the Carleton Circle (including the inter-sections of the Carleton Circle with the Euler line). In other words, K([cos−1(3g+r

2r ), cos−1(3g−r2r )])

must be at least (x, y) | (x − 2gr2

r2−g2 )2 + y2 = 4g4r2

(r2−g2)2∧ either y ≥ 0 or y ≤ 0 . To show that

K([cos−1(3g+r2r ), cos−1(3g−r

2r )]) encompasses only one half of the Carleton Circle, we use Mathematicato take the derivative of A(φ) .

A′(φ) =6g2r(27g4 + 36g2r2 + r4 − 4gr(18g2 + 5r2) cos φ + 2r2(15g2 + r2) cos 2φ− 4gr3 cos 3φ) sin φ

(r2 − g2)(9g2 + r2 − 6gr cosφ)2.

Setting A′(φ) equal to 0 , we find the following critical points of A(φ) in [0, 2π] :

φ = 0φ = π

φ = ± cos−1(3g + r

2r)

φ = ± cos−1(3g − r

2r)

φ = ± cos−1(3g2 + r2

4gr)

Certainly ± cos−1(3g±r2r ) 6∈ (cos−1(3g+r

2r ), cos−1(3g−r2r )) . Moreover, for ± cos−1(3g2+r2

4gr ) to exist,

−1 ≤ 3g2+r2

4gr ≤ 1 . Thus,

3See Munkres, pg 150.

35

3g2 + r2

4gr≤ 1

3g2 + r2 ≤ 4gr

3g2 + r2 − 4gr ≤ 0(3g − r)(g − r) ≤ 0

This implies that either

1. 3g − r ≤ 0 ∧ g − r ≥ 0 or

2. 3g − r ≥ 0 ∧ g − r ≤ 0 .

However, g − r ≥ 0 → g ≥ r , which cannot occur. Moreover, 3g − r ≥ 0 → 3g ≥ r , whichcontradicts our assumption that g < r

3 . Thus, ± cos−1(3g2+r2

4gr ) is not a valid critical number forg < r

3 .

Finally, recall that cos−1(3g+r2r ) and cos−1(3g−r

2r ) were found by constructing the two isoscelestriangles in a gr -family and measuring the positive ∠GOP where P is one of the two vertices of4ABC not on the Euler line of the gr -family. Since one vertex of 4ABC on the Euler line alwaysforces the remaining two vertices to lie on opposite sides of the Euler line, it follows that m∠GOP ∈(0, π) . Thus, [cos−1(3g+r

2r ), cos−1(3g−r2r )] ⊂ (0, π) . This implies that 0, π 6∈ [cos−1(3g+r

2r ), cos−1(3g−r2r )] .

Therefore, A(φ) is monotone on [cos−1(3g+r2r ), cos−1(3g−r

2r )] . This in turn indicates thatK([cos−1(3g+r

2r ), cos−1(3g−r2r )]) can only be one half of the Carleton Circle.

To determine whether that half is the upper of lower half, we find the derivative of B(φ) .

B′(φ) = 6g2r(3g2(9g2+7r2) cos φ+r(−4g(9g2+2r2) cos 2φ+r(15g2+r2) cos 3φ−2g(9g2+r2 cos 4φ)))(g2−r2)(9g2+r2−6gr cos φ)2

.

Evaluating B′ at φ = cos−1(3g+r2r ) , we find that B′(cos−1(3g+r

2r )) = 6g2

r−g > 0 . Thus, at φ =cos−1(3g+r

2r ) , B(φ) is increasing, which implies that K([cos−1(3g+r2r ), cos−1(3g−r

2r )]) is the upper halfof the Carleton Circle. In other words, K([cos−1(3g+r

2r ), cos−1(3g−r2r )]) = (x, y) | (x− 2gr2

r2−g2 )2 + y2 =4g4r2

(r2−g2)2∧ either y ≥ 0 = C1 .

Thus, if (x, y) ∈ C1 , it follows that there exists a 4ABCφ in Ω , φ ∈ [cos−1(3g+r2r ), cos−1(3g−r

2r )] ,with symmedian point (x, y) . If (x, y) is a point on the lower half of the Carleton Circle, then bysymmetry it will be the symmedian point of 4ABC−φ where (x,−y) is the symmedian point of4ABCφ , φ ∈ [cos−1(3g+r

2r ), cos−1(3g−r2r )] .

Therefore, every point on the Carleton Circle is the symmedian point of a triangle in Ω .

Case 2. Let g = r3 . This case was covered in Lemma 2.

Case 3. Let g > r3 . To show that for all (x, y) ∈ (x, y) | (x− 2gr2

r2−g2 )2+y2 = 4g4r2

(r2−g2)2∧ x2+y2 <

r2 , there exists a triangle in Ω with symmedian point (x, y) , we follow a method similar to that ofCase 1.

Now, for the case in which g > r3 , Proposition 17 states that every triangle in a given gr -family

(up to congruence) falls within the following bounds on φ :

cos−1(3g − r

2r) ≤ φ < cos−1(

3g2 − r2

2rg) .

36

As in Case 1, we will use these bounds. We saw earlier that K(cos−1(3g−r2r )) = ( 2gr

r+g , 0) . UsingMathematica, we find that

K(cos−1(3g2 − r2

2rg)) =

(3g2 + r2

4g,−1

4r

√10− 9g2

r2− r2

g2

).

Since the y -coordinate of K(cos−1(3g2−r2

2gr )) involves a square root, we must verify that

cos−1(3g2−r2

2gr ) is actually in the domain of K(φ) . To do this, we note that

−14r

√10− 9g2

r2− r2

g2= −1

4r

√10− 9g4 + r4

g2r2

= −14r

√10− 9g4 + r4 + 6r2g2 − 6r2g2

r2g2

= −14r

√10− (3g2 + r2)2 − 6r2g2

r2g2

= −14r

√10− (

(3g2 + r2)2

r2g2− 6r2g2

r2g2)

= −14r

√16− (3g2 + r2)2

r2g2.

We need to show that(3g2 + r2)2

r2g2≤ 16 . Thus, we examine the conditions under which this

inequality holds.

(3g2 + r2)2

r2g2≤ 16

(3g2 + r2)2 ≤ 16r2g2

3g2 + r2 ≤ 4rg

3g2 − 4gr + r2 ≤ 0(3g − r)(g − r) ≤ 0

This implies that either

1. 3g − r ≤ 0 ∧ g − r ≥ 0 or

2. 3g − r ≥ 0 ∧ g − r ≤ 0 .

In the first case, g − r ≥ 0 → g ≥ r , which is a contradiction. In the second, 3g − r ≥ 0 →g ≥ r

3 , which holds by assumption, and g − r ≤ 0 → g ≤ r , which is true. Thus, we see that

−14r

√10− 9g2

r2 − r2

g2 will be a real number for all g ≥ r3 .

37

Now, again by the preservation of connectedness under continuous functions, since[cos−1(3g−r

2r ), cos−1(3g2−r2

2rg )] is connected, K([cos−1(3g−r2r ), cos−1(3g2−r2

2rg )]) is connected. This implies

that K([cos−1(3g−r2r ), cos−1(3g2−r2

2rg )]) minimally contains (x, y) | (x− 2gr2

r2−g2 )2 + y2 = 4g4r2

(r2−g2)2∧ y ≤

0 . However, from Case 1, we know that A(φ) has the following critical points in [0, 2π] :

φ = 0φ = π

φ = ± cos−1(3g + r

2r)

φ = ± cos−1(3g − r

2r)

φ = ± cos−1(3g2 + r2

4gr)

Since 3g+r2r > r+r

2r = 1 , we see that ± cos−1(3g+r2r ) are not real numbers, and thus are not valid

critical points. By construction [cos−1(3g−r2r ), cos−1(3g2−r2

2rg )) ⊆ (0, π) . This implies that 0, π 6∈[cos−1(3g−r

2r ), cos−1(3g2−r2

2rg )] . Moreover, by the following argument, cos−1(3g2+r2

4gr ) ∈ (0, π) :

A

O G

Ma

Figure 29: The angles cos−1(3g2−r2

2rg ) and cos−1(3g2+r2

4gr ) .

This proof utilizes Figure 29 above. Let Ω∗ be a gr -family of triangles in which g > r3 . Then,

it follows that there exists an ∠GOA such that the midpoint Ma of BC lies on the circumcircle ofΩ∗ . As discussed in Proposition 17, the measure of this angle is cos−1(3g2−r2

2gr ) . We will show that

m∠GOMa = cos−1(3g2+r2

4gr ) .

To begin, construct the segment AMa where m∠GOA = cos−1(3g2−r2

2gr ) . Then, by the law ofcosines,

38

AG2 = r2 + g2 − 2gr cos (cos−1(3g2 − r2

2gr))

= r2 + g2 − 2gr3g2 − r2

2gr

= r2 + g2 − 3g2 + r2

= 2r2 − 2g2 .

So, AG =√

2(r2 − g2) . Now, by Proposition 4, AG = 2GMa . Thus, GMa =√

2(r2−g2)

2 . To findm∠GOMa , we again use the law of cosines.

GM2 = g2 + r2 − 2gr cos(m∠GOMa)(√2(r2 − g2)

2

)2

= g2 + r2 − 2gr cos(m∠GOMa)

r2 − g2

2= g2 + r2 − 2gr cos(m∠GOMa)

r2 − g2 = 2g2 + 2r2 − 4gr cos(m∠GOMa)

−r2 − 3g2 = −4gr cos(m∠GOMa)

3g2 + r2

4gr= cos(m∠GOMa)

cos−1(3g2 + r2

4gr) = m∠GOMa .

Now, notice that the only instance in which m∠GOA can equal π occurs when g = r3 . Thus,

since g > r3 , it follows that m∠GOA < π . However, m∠GOA < π indicates that Ma and A are

on opposite sides of the Euler line of Ω∗ . Consequently, ∠GOMa ∈ (0, π) .

Now that we have established that both cos−1(3g2+r2

4gr ) and cos−1(3g−r2r ) lie in the interval (0, π) ,

we can state the following:

(3g + r)(g − r) < 0

3g2 − 2gr − r2 < 0

3g2 − 2gr < r2

6g2 − 2gr < 3g2 + r2

2g(3g − r) < 3g2 + r2

3g − r <3g2 + r2

2g

3g − r

2r<

3g2 + r2

4gr

cos−1(3g − r

2r) > cos−1(

3g2 + r2

4gr) .

39

Since cos−1(3g−r2r ) ≤ cos−1(3g2−r2

2rg ) , we have now established that there are no critical points

of A(φ) on (cos−1(3g−r2r ), cos−1(3g2−r2

2rg )) . Thus, A(φ) is monotone on [cos−1(3g−r2r ), cos−1(3g2−r2

2rg )] .

Since B(cos−1(3g2−r2

2rg )) = −14r

√16− (3g2+r2)2

r2g2 , this implies that K([cos−1(3g−r2r ), cos−1(3g2−r2

2rg )]) is

an arc of the Carleton Circle lying below the Euler line. Thus, K([cos−1(3g−r2r ), cos−1(3g2−r2

2rg )]) =

(x, y) | (x− 2gr2

r2−g2 )2 + y2 = 4g4r2

(r2−g2)2∧ y ≤ 0 .

Now, as discussed in a footnote in the proof of Lemma 2, the symmedian point of a given 4ABCcan never fall on the circumcircle of 4ABC . Using Mathematica, we find the intersections ofthe circumcircle of Ω with the Carleton Circle to be the points

(3g2+r2

4g , 14r

√10− 9g2

r2 − r2

g2

)and

(3g2+r2

4g ,−14r

√10− 9g2

r2 − r2

g2

). Thus, K(cos−1(3g2−r2

2rg )) =(

3g2+r2

4g ,−14r

√10− 9g2

r2 − r2

g2

)is not ac-

tually a valid symmedian point even though it is in the domain of K . With this in mind, werestrict our original interval to [cos−1(3g−r

2r ), cos−1(3g2−r2

2rg )) . Since [cos−1(3g−r2r ), cos−1(3g2−r2

2rg )) ⊆[cos−1(3g−r

2r ), cos−1(3g2−r2

2rg )] , it follows that K([cos−1(3g−r2r ), cos−1(3g2−r2

2rg ))) ⊆K([cos−1(3g−r

2r ), cos−1(3g2−r2

2rg )]) . Therefore, K([cos−1(3g−r2r ), cos−1(3g2−r2

2rg ))) will be the connected

set (x, y) | (x− 2gr2

r2−g2 )2 + y2 = 4g4r2

(r2−g2)2∧ 2gr

g+r ≤ x < 3g2+r2

4g ∧ y ≤ 0 = C2 .

Thus, if (x, y) ∈ C2 , it follows that there exists a 4ABCφ in Ω , φ ∈ [cos−1(3g−r2r ), cos−1(3g2−r2

2rg )) ,with symmedian point (x, y) . If (x, y) is a point on the reflection of C2 over the Euler line of Ω ,then by symmetry it will be the symmedian point of 4ABC−φ where (x,−y) is the symmedian pointof 4ABCφ , φ ∈ [cos−1(3g−r

2r ), cos−1(3g2−r2

2rg )) .

Therefore, every point on the arc of the Carleton Circle (x, y) | (x − 2gr2

r2−g2 )2 + y2 = 4g4r2

(r2−g2)2∧

2grg+r ≤ x < 3g2+r2

4g is the symmedian point of a triangle in Ω .Finally, to ensure that we have not missed any points on the Carleton Circle which fall within

the circumcircle, we recall from above that the intersections of the two circles occur at x = 3g2+r2

4g .

This implies that for any x > 3g2+r2

4g , if (x, y) is a point on the Carleton Circle, then (x, y) will lieoutside the circumcircle of Ω . Thus, we have discussed all points on the Carleton Circle which couldbe symmedian points. For all points (x, y) on the Carleton Circle and inside the circumcircle of Ω ,there exists a 4ABC ∈ Ω with symmedian point (x, y) .

It is indeed interesting that the symmedian points of a gr –family of triangles form a completecircle or a segment of a circle. To extend on this idea, we have the next corollary.

Corollary 3. Every point on GK traces a circle as φ varies.

Proof. Let triangle 4ABC be given with centroid G , circumcenter O , symmedian point K , andKnights’ Point Kn . Consider the vector

−−→GK . Vector

−−→GK spans the line GK . Thus, for all

points P on GK ,−−→GP = a

−−→GK , where a ∈ R . Let K1 and K2 be the intersections of the

Carleton Circle with the Euler line of 4ABC . Then, let P1 and P2 be two points on the Eulerline such that

−−→GP1 = a

−−→GK1 and

−−→GP2 = a

−−→GK2 . Now, since a = GP

GK = GP1GK1

, and angle ∠KGK1 iscommon, it follows that 4KGK1 ∼ 4PGP1 . By the same argument, 4KGK2 ∼ 4PGP2 . Thus,PP1KK1

= GPGK = PP2

KK2. Moreover, P1P2 = GP2 − GP1 = aGK2 − aGK1 = a(GK2 − GK1) = aK1K2 .

This implies that P1P2K1K2

= a . Thus, by SSS similarity, 4PP1P2 ∼ 4KK1K2 . Since K lies on thecircle with diameter K1K2 , m∠K1KK2 = π

2 . Thus, m∠P1PP2 = π2 . This implies that P lies on

the circle with diameter P1P2 . We will call this circle the Carleton a –Circle.

40

Pn

P2P1

P

K

C

BKn

K2

K1

O G

A

Figure 30: Every point on GK lies on a circle centered on the Euler line of a given gr –family oftriangles.

Now, the center Pn of the Carleton a Circle is the point located halfway between P1 and P2 .So, GPn = GP1 + P1Pn = aGK1 + P1P2

2 = aGK1 + aK1K22 = aGK1 + aK1Kn = aGKn . If we consider

the Euler line of triangle 4ABC to be the x -axis and O to be the origin, then OPn = g + GPn =g + aGKn = g + a( 2gr2

r2−g2 − g) = g + a 2gr2

r2−g2 − ag = (1− a)g + a 2gr2

r2−g2 .Now, by the proof of Theorem 3, we know that every point on the Carleton Circle which lies inside

the circumcircle of 4ABC corresponds to a symmedian point of some triangle in the gr –family of4ABC . Define the function h :

−−−→GKφ ⇒ −−→

GPφ such that h(−−−→GKφ) = a

−−−→GKφ =

−−→GPφ . Now, K(φ) is a

continuous function, or, simply put, the coordinates of Kφ are continuous. Moreover, the componentsof−−−→GKφ are found through subtraction of g and 0 from

2gr((9g3 + 6gr2) cosφ− r(9g2 + r2 + 6g2 cos 2φ− gr cos 3φ))(g2 − r2)(9g2 + r2 − 6gr cosφ)

and2g2r(9g2 + r2 − 12gr cosφ + 2r2 cos 2φ) sin φ

(g2 − r2)(9g2 + r2 − 6gr cosφ),

respectively. Thus, the components of−−−→GKφ are continuous. Since the function h just multiplies the

vector−−−→GKφ by a real number, it follows that h must be continuous. Thus, since continuous functions

map connected sets to connected sets, by an argument similar to that in the proof of Theorem 3, ifg < r

3 , then P will trace the entire Carleton a –Circle. If g = r3 , P will trace the entire Carleton

a –Circle except for the farthest intersection of the Carleton a –Circle with the Euler line of the gr –family. If g > r

3 , then P will trace an arc whose endpoints correspond to the places at which theCarleton Circle meets the circumcircle of the gr –family.

Finally, the center Pn of the Carleton a –Circle is the point located halfway between P1 andP2 . So, GPn = GP1 + P1Pn = aGK1 + P1P2

2 = aGK1 + aK1K22 = aGK1 + aK1Kn = aGKn .

If we consider the Euler line of triangle 4ABC to be the x -axis and O to be the origin, thenOPn = g + GPn = g + aGKn = g + a( 2gr2

r2−g2 − g) = g + a 2gr2

r2−g2 − ag = (1− a)g + a 2gr2

r2−g2 .

41

7 The Sum of Squares

Now that we have examined what happens to certain points when we vary ∠GOA in a gr –family,let us turn to another interesting property that naturally emerges from our construction.

Proposition 31. For any point P on 4ABC ’s plane, define function D = PA2 + PB2 + PC2 .Then D has constant values for the gr -family of triangles, and the locus of points sharing constantD -values forms circles centered at the centroid G .

Proof. To show this is acutally true, we build upon the coordinate system we placed 4ABC in.Following our previous definitions, point A ’s polar coordinates are ( r cosφ , r sinφ ). Set point P ’spolar coordinates to be (x, y) = (s cosβ, s sinβ) . Next we just need to prove D is independent of φ .Recall in the previous section we have found the coordinates for the three vertices of 4ABC . Here,we place them into function D . After simplification, we get the following results:

D = 3(r2 + s2 − 2gs cosβ) .

Notice in this case the function D is indeed independent of φ . If we convert it into Cartesiancoordinates and treat D as a constant D , we get an equation of circles centered at G(g, 0) for anyD > 3r2 + 3g2 :

(x− g)2 + y2 =D

3− r2 − g2,

which completes our proof.

8 Conclusion

We have achieved our objective of finding a natural way to relate triangles. We created a constructionbased on the Euler line, an element intrinsic to a triangle. Using that construction, we parameterizedtriangles and grouped them into families. Then, we bounded the construction, enabling us to visualizetriangle space. In order to algebraically conceive of distances between triangles and to account forcertain peculiarities in our space, we created a metric. Then, we looked at the loci of certain trianglecenters and special points in our triangle families, discovering for example that a gr –family has aconstant Nine-Point Circle. We also looked at the symmedian point in a gr –family and proved thatit will always lie on a circle (the Carleton Circle) centered at the Knights’ Point on the Euler line ofthat family. Finally, we noticed that given a point in the plane, the sum of squares of distances fromthat point to the vertices of any triangle in a gr -family is constant.

While the work here is a basis for this way of conceiving triangles, it is certainly not exhaustive.Euclidean Geometry has been a branch of mathematics for thousands of years. Despite this, there isclearly still much to discover.

42

Bibliography

[Altshiller Court] Altshiller Court, Nathan. College Geometry: An Introduction to the Modern Ge-ometry of the Triangle and the Circle. New York: Barnes & Noble, Inc., 1952.

[Honsberger] Honsberger, Ross. Episodes in Nineteenth and Twentieth Century Euclidean Ge-ometry. Washington: Mathematical Association of America, 1995.

[Kimberling] Kimberling, Clark. “Triangle Centers.” Encyclopedia of Trian-gle Centers - ETC. University of Evansville. 12 Mar. 2007.http://faculty.evansville.edu/ck6/tcenters/index.html.

[Mueller] Mueller, William. “Centers of Triangles of Fixed Center: Adventures in Under-graduate Research.” Mathematics Magazine. Vol. 70, No. 4 (Oct., 1997), pp.252-262.

[Munkres] Munkres, James R. Topology. 2nd ed. Upper Saddle River, NJ: Prentice Hall,Inc., 2000.

[Stewart] Stewart, James. Calculus. 5th ed. Australia: Brooks/Cole, 2003.

43

9 Appendix A

In this appendix, we find explicit representations for the lengths of the sides of the triangle in ourconstruction.

r

g

θθθθ

M

C

BHO G

A

Figure 31: Finding Side Lengths.

First, we will find the length of BC . Using Law of Cosines on 4AOG :

AG2 = r2 + g2 − 2rg cos θ ⇒ AM =32

√r2 + g2 − 2rg cos θ

Using the Law of Sines on 4AOG :

sin∠OAG

g=

sin θ√r2 + g2 − 2rg cos θ

⇒ sin∠OAG =gsinθ√

r2 + g2 − 2rg cos θ

⇒ cos∠OAG =r − g cos θ√

r2 + g2 − 2rg cos θ

Now:OM2 = AM2 + AO2 − 2(AM)(AO) cos ∠OAG

⇒ OM2 =94(r2 + g2 − 2rg cos θ) + r2 − 3r

√r2 + g2 − 2rg cos θ · r − g cos θ

r2 + g2 − 2rg cos θ

⇒ OM2 =14(r2 + 9g2 − 6rg cosβ)

⇒ OM =12

√r2 + 9g2 − 6rg cos θ

⇒ BM2 = r2 − (14r2 − 9g2 + 6rg cos θ)

⇒ BM2 =34(r2 − 3g2 + 2rg cos θ)

44

⇒ BM =12

√3(r2 − 3g2 + 2rg cos θ

BC = 2BM

∴ BC =√

3(r2 − 3g2 + 2rg cos θ)

Next, we will find the length of AB :

OG2 = OM2 + MG2 − 2(OM)(MG) cos∠OMG

g2 =r2

4+

9g2

4−6rg cos θ

4+

r2

4+

g2

4−2rg cos θ

4−1

2

√r2 + g2 − 2rg cos θ

√r2 + 9g2 − 6rg cos θ·cos∠OMG

⇒ cos∠OMG =r2 + 3g2 − 4rg cos θ√

r2 + g2 − 2rg cos θ)(r2 + 9g2 − 6rg cos θ)

⇒ ∠OMG = cos−1

(r2 + 3g2 − 4rg cos θ√

r2 + g2 − 2rg cos θ)(r2 + 9g2 − 6rg cos θ)

)

⇒ ∠AMB =π

2− cos−1

(r2 + 3g2 − 4rg cos θ√

r2 + g2 − 2rg cos θ)(r2 + 9g2 − 6rg cos θ)

)

⇒ cos∠AMB = sin

[cos−1

(r2 + 3g2 − 4rg cos θ√

r2 + g2 − 2rg cos θ)(r2 + 9g2 − 6rg cos θ)

)]

Using a right triangle:

x22 = (r2 + g2 − 2rg cos θ)(r2 + 9g2 − 6rg cos θ)− (r2 + 3g2 − 4rg cos θ)2

(x2)2 = 4r2g2 sin2 θ

⇒ x2 = 2rg sin θ

⇒ cos∠AMB =2rg sin θ√

(r2 + g2 − 2rg cos θ)(r2 + 9g2 − 6rg cos θ)

⇒ AB2 = AM2 + BM2 − 2(AM)(BM) cos∠AMB

⇒ AB2 =94r2 − 9

2rg cos θ +

34r2 +

32rg cos θ − 3

√3(r2 − 3g2 + 2rg cos θ) ·

(rg sin θ

r2 + 9g2 − 6rg cos θ

)

⇒ AB2 = 3r2 − 3rg cos θ − 3rg sin θ ·√

3r2 − 9g2 + 6rg cos θ√r2 + 9g2 − 6rg cos θ

∴ AB =

√3r2 − 3rg cos θ − 3rg sin θ ·

√3r2 − 9g2 + 6rg cos θ√r2 + 9g2 − 6rg cos θ

Finally, we will find the length of AC :

∠AMC =π

2+ ∠OMG

cos∠AMC = sin

[cos−1 r2 + 3g2 − 4rg cos θ√

(r2 + g2 − 2rg cos θ)(r2 + 9g2 − 6rg cos θ)

]

⇒ cos∠AMC =2rg sin θ√

(r2 + g2 − 2rg cos θ)(r2 + g2 − 6rg cos θ)

∴ AC =

√3r2 − 3rg cos θ + 3rg sin θ ·

√3r2 − 9g2 + 6rg cos θ√r2 + 9g2 − 6rg cos θ

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