Tracing a Point, Spinning a Vertex: How Circles are
Made
Emily Banks, Seth Cooper, Laura Hudson, and Elizabeth Miller
Department of Mathematics and Computer Science
Carleton College, Northfield, Minnesota 55057
[email protected], [email protected], [email protected], [email protected]
April 5, 2005
Abstract
This paper proves several theorems involving how special points of a triangle
trace circles as one of the vertices of the triangle is moved around the circumcircle. We investigate the larger issue of why this happens using constant distance
ratios, and finally, we prove some theorems about strophoids and use a “lighthouse”
construction to show that another special point of a triangle traces a strophoid.
1
The following question was the starting point of our research [5].
Given a circle, what is the locus of all possible incenters of the triangles for which that
circle is a circumcircle?
Put more simply, this question asks to:
1) Construct three points on a circle.
2) Connect each point to form a triangle.
3) Bisect each angle of the triangle to create the incenter of that triangle.
4) Move the vertices of the triangle around the circle and look at the paths
traced by the incenter.
1
The following is an outline of the construction we used to answer our initial question.
The notation used in this construction will apply throughout the paper.
C
X0
c1
B
A
Incenter Construction
Construct circle c1 , and select 3 points, A, B, and C along its circumference. Join AB,
BC, and AC to form a triangle circumscribed by c1 .
We refer to circle c1 as the circumcircle.
Now bisect 6 ABC, 6 BAC and 6 ACB. It is a common Euclidean theorem that these
bisectors are concurrent. The point of intersection is called the incenter, and we will refer
to it as X0 .
1
It is a well-known theorem that each of the angle bisectors of a triangle intersect at a common point
known as the incenter. The incenter of a triangle is the center of its incircle, the unique circle that is
mutually tangent to each side of that triangle.
2
When we animated this construction on Geometer0 sSketchpad, we immediately noticed
that when vertices A and B remain fixed and C moves along the circumcircle, Xo traces
out a football-like shape that appears to be sections of two different circles. In the diagram
below we have included these two arcs. We already knew a fair amount about circles, and
as a result decided to tackle this problem first.
PROOF THAT THE INCENTER LIES ON A CIRCLE
O2
C
ε δ
c1
β
X0
γ
α
A
α
B
O1
l
c2
Construction
Construct circle c1 and choose two points, A and B on c1 . Join AB and construct the
perpendicular bisector, l, of AB. Label the intersections of l with circle c1 , O1 and O2 ,
where O1 lies below AB and O2 lies above AB.
Construct circle c2 , with center O1 and radius O1 A. Choose a point X0 on circle c2 such
that X0 lies within circle c1 . Join AX0 and BX0 . Let 6 ABX0 = α and 6 BAX0 = γ.
3
Double 6 ABX0 and call the intersection of the ray from B with c1 , C. By construction,
X0 BC = α.
6
Join AC and CX0 . Let 6 CAX0 = β, 6 ACX0 = ² and 6 BCX0 = δ.
Theorem
The incenter of 4ABC is point X0 and will always lie on circle c2 .
Proof
We will show that point X0 is the incenter of 4ABC by proving that AX0 is the angle
bisector of 6 BAC. Thus, we wish to show that γ = β.
We must first show that points O1 , X0 and C are collinear. Recall that
6
AO1 X0 =
26 ABX0 = 2α since both angles cut the same arc of circle c2 and one is from the center
of the circle and the other is from a point on the circle. As a result, 6 AO1 X0 = 6 ABC.
Since O1 and B both lie on circle c1 , they must therefore cut the same arc of c1 . Thus, if
we extend O1 X0 , it must intersect circle c1 at point C. Therefore, points O1 , X0 and C
are collinear.
We now note that 6 BO1 X0 = 26 X0 AB = 2γ since both angles cut the same arc of circle
c2 . Further, 6 BCO1 = 6 BAO1 = δ because they cut the same arc of circle c1 and 6 O1 CA
= 6 ABO1 = ² because they cut the same arc of circle c1 .
Looking at 4ACB, the sum of the angles is equal to
6
ABX0 + 6 X0 BC + 6 BCX0 + 6 X0 CA + 6 CAX0 + 6 X0 AB = π
α+α+δ+²+β+γ=π
2α + δ + ² + β + γ = π.
Now looking at 4BO1 A, the sum of the angles is equal to
6
O1 BA + 6 BAO1 + 6 AO1 C + 6 CO1 B = π
² + δ + 2α + 2γ = π.
4
Setting these equations equal, we get
γ = β.
Since γ = β, we can conclude that X0 A is the angle bisector of 6 BAC. Therefore, point
X0 is the incenter of 4ABC and given that X0 was constructed to lie on circle c2 , we
have proved that the incenter always lies on circle c2 .
Note that we can follow the same proof by constructing circle c3 with center O2 and radius
O2 A, allowing our proof to account for all cases.
Returning to the original question:
The original question asked: What is the locus of all possible incenters of the triangles
for which c1 is a circumcircle? The observant reader will note that by fixing points A and
B we have not accounted for all possible triangles. However, given any point, P , that lies
within circle c1 , it is possible to find a triangle for which P is the incenter.
EXCENTERS
Having proved that the incenter of 4ABC traces sections of circles as vertex C moves
around its circumcircle, we set out to discover where the rest of these two circles lay.
Some experimentation on Geometer0 sSketchpad quickly revealed that the answer lay in
4ABC’s excenters.
There are three excenters for any given triangle. The excenters of a triangle can be
constructed by extending the sides of that triangle to form lines and constructing all
of the circles that are mutually tangent to the sides of 4ABC. The three new circles
formed under this construction are referred to as excircles and their centers are known as
excenters.
5
The diagram below is an illustration of this construction:
We can also construct the excenters of a triangle by finding the points of intersection of
its interior and exterior angle bisectors. Exterior angle bisectors lie perpendicular to their
interior angle bisectors. This particular construction gives us more to work with than the
excircle construction, and as a result this will be the one we use in the ExcenterP roof s
that follow. The following is an outline of the construction of the excenters of circle, c1 .
Again, the notation used in this construction will apply throughout the paper.
Excenter Construction
Building from the incenter construction, we construct 3 more special points, X1 , X2 , and
X3 . These points are known as excenters and are located at the following intersections:
X1 : The interior
bisector of B and the exterior
6
X2 : The interior
6
bisector of A and the exterior
X3 : The interior
6
bisector of C and the exterior
6
bisectors of A and C.
6
6
6
bisectors of B and C.
bisectors of A and B.
As we move vertex C along the path of our circumcircle c1 , all 4 of our special points
move as well. We wish to prove that each of the 4 special points traces part of 2 circles,
c2 and c3 .
X1
C
X2
X0
c1
A
B
X3
While examining the preceding theorems, we ran into an interesting problem with Geometer’s Sketchpad. When C passes through A or B the angle effectively turns inside out and
the angle bisector there goes from interior to exterior. Geometer’s Sketchpad has trouble
following a single line through this switch and would change focus from the old interior
bisector to the new one every time this happened. This sent our excenters jumping from
one side of the circle to the other. We referred to this phenomenon as “jumping spots.”
We wanted the motion of the excenters to be consistent with the continuous motion of C.
To alleviate the problem, we began to consider our angle bisectors as pairs of perpendicular lines spinning around some fixed point. By shifting our focus to these spinning pairs,
the animation became continuous, like it should be. This idea of spinning perpendiculars
will be addressed again later in our paper.
7
EXCENTERS TRACE CIRCLES
In order to prove this, we have to break our proof into two different parts:
1) Proof that excenters X1 and X2 lie on circle c3 where c3 is the lower traced circle.
2) Proof that excenter X3 lies on circle c2 where c2 is the the upper traced
circle.
Proof that excenters X1 and X2 lie on circle c3
Although this proof only deals with excenter, X2 . It can be applied to excenter X1 as
well.
In the following proof we will show:
1) The interior angle bisector of vertex A and the exterior angle bisector at
vertex C always intersect.
2) By constructing excenter, X2 , we will show that the segment joining X2
and vertex A is the interior angle bisector of 6 CAB.
O2
c1
2γ
C
c3
X2
β
γ
B
A
O1
8
Construction
Given circle c1 , choose points A, B, and C on c1 . Join AB, BC, and AC. Draw the ⊥
bisector of AB and label its intersections with c1 , O1 and O2 . Join O2 B. Construct circle
c3 with center O2 and radius O2 B. Construct the interior and exterior angle bisectors of
ACB. Join AX2 . Let 6 BAX2 = γ and 6 CAX2 = β.
6
Lemma
The points O2 , C and X2 are collinear.
Proof
It has already been shown that the interior angle bisector of 6 ACB intersects the perpendicular bisector of AB on c1 at point O1 , given that the interior angle bisector of 6 ACB
and the perpendicular bisector of AB are two different ways to bisect the arc subtended
by 6 ACB.
Given O2 and O1 are on opposite ends of a diameter, 6 O1 CO2 = π2 .
Since CX2 and CO1 are exterior and interior angle bisectors of 6 ACB,
6
O1 CX2 = π2 .
Then,
6
O1 CO2 + 6 O1 CO2 = π.
Therefore, O2 , C, and X2 are collinear.
Theorem
The excenter, X2 , always lies on circle c3 .
Proof
By constructing the exterior angle bisector of 6 ACB and labeling its intersection with c3 ,
X2 , we will prove that AX2 is the interior angle bisector of 6 BAC and therefore X2 is an
excenter and always lies on circle c3 .
9
Given that 6 BAX2 and 6 BO2 X2 cut out the same arc of circle, c3 and 6 BO2 X2 is at the
center of circle c3 and 6 BAX2 is at a point on c3
6
On circle c1 ,
6
BO2 X2 = 26 BAX2 = 2γ.
BO2 C and 6 BAC cut out the same arc. We have already shown that
BO2 C = 2γ, and by construction, 6 BAC = γ + β.
6
Setting these two equations equal,
2γ = γ + β
γ=β
Given
6
BAX2 = γ
We have thus shown that 6 BAX2 = 6 CAX2 .
We have therefore proved that the excenter, X2 lies on a circle, c3 as vertex C moves
around the circumcircle c1 . As noted before, this proof also applies to the excenter, X1 .
Having proved that excenters, X1 and X2 trace sections of the circles also traced by
incenter, X0 , we now look to our third and final excenter, X3 . We will prove that excenter,
X3 lies on the lower circle traced by X0 . This circle is denoted as c2 .
Proof that excenter X3 lies on circle c2
We know that the perpendicular bisector of AB and the interior angle bisector of 6 ACB
intersect on circle c1 at point O1 . We therefore know that O1 lies on the interior angle
bisector of 6 ACB. In order to prove that X3 lies on circle c2 , it is sufficient to show that
points C, O1 and X3 are collinear. We already know that O1 C intersects c2 at X0 . If XO ,
O1 and X3 are collinear, then X0 and X3 are on opposite ends of a diameter of circle c2
and therefore X3 always lies on circle c2 .
10
c1
C
A
B
α
F
O1
c2
X3
Construction
Given circle c1 , choose points A, B, and C on c1 . Join AB, BC, and AC. Construct the
angle bisector of 6 ACB. Construct the ⊥ bisector AB and label its intersection with c1
O1 . Join O1 B and O1 A.
Construct circle c2 with center O1 and radius O1 B. Construct ray CA and choose a point,
F , along this ray. Bisect 6 F AB. Denote the intersection of this bisector with circle c2 as
X3 . Join O1 X3 . Let 6 F AX3 = 6 X3 AB = α.
Proof
Join CO1 and O1 X3 . Given 6 CO1 B cuts the same arc of circle c1 as does 6 CAB, then
CO1 B = 6 CAB = β.
6
The angle X3 AB cuts the same arc of circle c2 as 6 X3 O1 B. Given 6 X3 O1 B is at the
center of circle c3 , we can conclude that 6 X3 O1 B = 2α.
Given that points C, A, and F are collinear,
6
CAF = β + 2α = π.
11
However, we now know that
6
CO1 X3 = β + 2α.
Thus, C, O1 and X3 are collinear. This proves that excenter X3 lies on circle c2 .
We have thus proved that our four special points: incenter XO and excenters X1 , X2 and
X3 of 4ABC trace circles, c2 and c3 , as vertex C moves around its circumcircle, c1 .
EXAMINING OTHER POINTS
After examining the incenter and excenters, we started looking at other special points
of triangles. Clark Kimberling has gathered a list of special points of triangles, and we
began by looking at one of the most common points on this list: the orthocenter.
The Orthocenter
The orthocenter is the point of intersection of the altitudes (the perpendicular dropped
from a vertex to the opposite side) of a given triangle. It is a common Euclidean theorem
that all three altitudes intersect at a single point. Furthermore, the orthocenter lies on
the Euler line. Given the geometric importance of this point, it seems a relevant point
to study. So, we set out to determine the locus of all orthocenters for triangles inscribed
within a given circle. In order to take a systematic approach, we again fixed points A and
B and moved point C along the circumcircle.
12
C
c1
α β
Q
P
D
A
R
B
c2
Construction
Construct circle c1 with center O and choose points A, B, and C that lie on c1 to form
4ABC. From vertex B, drop a perpendicular to AC and label the point of intersection
Q. Repeat this from vertices A and C and label the points of intersection with BC and
AB P and R respectively. Label the point of intersection D.
Construct circle c2 by constructing the circumcircle to 4ABD. The center of this circle
will be the point at which the perpendicular bisectors of 4ABD intersect.
Let 6 ACD = α and let 6 BCD = β.
Theorem
The orthocenter of 4ABC always lies on circle c2 .
Proof
In this proof we will only consider the case where the orthocenter lies within 4ABC;
however, the proof can be applied to the case where the orthocenter lies outside of the
triangle as well.
In order to prove that D traces a circle, it is sufficient to show 6 ADB is constant.
Since 6 ACB subtends a constant arc, 6 ACB remains constant.
13
Let α + β = 6 ACB = γ, where γ is constant.
6
Consider 4CQD. Given that this is a right triangle, we can say that 6 CDQ =
BDR, by vertical angles. By the same logic, 6 CDP =
π
2
π
2
-α=
- β = 6 ADR.
Thus,
6
ADB = 6 BDR + 6 ADR
= ( π2 - α) + ( π2 - β)
= π - (β + α)
= π - γ, which is constant.
Therefore, 6 ADB is constant and D lies on a circle with A and B.
Apples and Oranges
Eager to discover something that did not trace circles, we began to examine other points.
One such point that we believed would trace something other than a circle was a case we
began to call Apples and Oranges. This example involves the point of intersection of the
perpendicular bisector of one side and the angle bisector of the adjacent angle. Unlike
the other cases that we examined, this case is not symmetric. The construction does not
involve doing the same thing to each angle or side of the triangle, so we expected some
interesting results. The construction is as follows.
14
l
m
C
E
α α
α
M
D
O
γ
α
A
B
Construction
Construct circle c1 with center O, and select three points, A, B, and C along its circumference. Join AB, BC, and AC to form three sides of a triangle circumscribed by
c1 .
Now bisect 6 ACB. Call this line l. Construct the perpendicular to l at point C and call
it m. Call each of the angles formed by l α. Next, construct the perpendicular bisector
of AC and call the midpoint of AC M . This line intersects line l at point D, and m at
E. Join AD and AE.
We will next prove that circles c2 and c3 are traced by points D and E respectively. Before
beginning this proof, however, we must note where the centers of circles c1 and c2 lie.
If C is fixed at any arbitrary point on c1 (causing D and E to be fixed as well) c2 is the
circumcircle of 4ADO and c3 is the circumcircle of 4AEO. Thus the center of c2 is the
intersection of the perpendicular bisectors of the sides of 4ADO and the center of c3 is
the intersection of the perpendicular bisectors of the sides of 4AEO.
15
l
C
c1
E
D
M
O
c3
B
A
c2
The following proof will apply directly to the case where D lies between O, the center of
the circumcircle, and our special point M , the midpoint of AC; however, the proofs for
the other cases are quite similar.
Theorem
Points D and E trace circles c2 and c3 respectively as point C moves around circle c1 .
A Brief Aside
Shortly after asserting our belief that these points trace circles as C moves around circle
c1 , John Conway came to visit Carleton College. During the course of his visit, we were
lucky enough to have the opportunity to sit with him and discuss our research. Shortly
after presenting our Apples and Oranges Conjecture to him, he stood up and wrote a
proof that D traces a circle on the board. Thus, the first proof that follows must be
credited to the great John Conway [1].
16
Proof
We will prove that D lies on circle c2 , by proving that 6 ADO remains constant regardless
of where C lies on circle c1 .
By SAS, 4AM D ∼
= 4CM D. Thus, 6 M AD = 6 M CD = α. Note that α is constant
16
2
given 6 ACB is constant and α =
ACB.
Begin by looking at 4AM D.
6
M AD + 6 AM D + 6 M DA = π
α+
π
2
+ 6 M DA = π
So,
6
M DA =
π
2
− α = γ, which is constant.
Given that 6 M DA and 6 ADO are supplementary angles. We know that
6
M DA + 6 ADO = π
γ + 6 ADO = π
6
ADO = π − γ
Given that γ is constant, 6 ADO is constant and our proof is complete. The point D lies
on a circle.
Theorem
Point E always lies on circle c3 .
Proof
We will prove that E lies on c3 by proving that 6 AEO is constant.
Let 6 ECA = β.
Given that l is the interior angle bisector of 6 ACB and m is the exterior angle bisector
of the same angle, 6 ECA + 6 ACD = 6 ECD = π2 .
17
In 4CM E,
6
Given 6 EM C =
π
2
ECA + 6 CEM + 6 EM C = π
and 6 ECA = β
6
CEM =
π
2
- β = α.
By Side-Angle-Side 4EMC ∼
= 4EM A and therefore 6 CEM = 6 AEM = 6 AEO = α,
which is constant.
Therefore, E lies on circle c3 .
So, circles continue to appear in surprising and unpredictable places.
SEARCHING FOR GENERAL CONCEPTS
After proving that several special points of a triangle do in fact trace circles and observing
instances of points that do not trace circles, we wanted to get a more general sense of what
was happening. We wanted some insight into how many points traced circles, and the
reason why some special points traced circles and others did not. We were also looking for
an explicit way to tell whether or not a given point would trace a circle in our construction.
It appeared that all of the points we found that traced circles were in fact at the intersection of two sets of rotating axes. Thus, we conjectured that a point at the intersection of
two rotating axes would in fact trace a circle in our construction. However, as we tested
this theory on Geometers Sketchpad we found some counterexamples. For instance, the
intersection of the axes created by the interior and exterior angle bisector at A and the
axes created by taking the perpendicular bisector of AC its intersections with circle c1
and the perpendiculars at those points clearly does not trace a circle.
18
This is shown in the following figure (the path of the special points are shown in red):
Non-circle intersection of
rotating axes example
C
Perpendicular bisector of
AC and perpendiculars at
intersection with C1
B
Angle bisectors at A
A
Still, the intersection of rotating axes did seem to have something to do with special
points tracing circles. However, there was some other characteristic that the intersections
that traced circles had and the intersections that did not trace circles did not have. We
could not figure out what this characteristic was so we moved on.
CONSTANT-DISTANCE RATIOS
Our next idea had to do with constant-distance ratios. A constant-distance ratio is just
that: a ratio of distances between points which remains constant. While searching for
insight in papers on special points in triangles we found “Central Points and Central Lines
in the Plane of a Triangle” by Clark Kimberling [3]. One of the things that Professor
Kimberling discusses in his paper is constant-distance ratios among special points of
triangles. It occured to us that if there were a relationship that maintained a constantdistance ratio between point C (which is known to trace a circle in our construction) and
some other special points then it would make sense that these points would trace a circle
in our construction as well. Applying this idea to points other than C would lead to a
19
kind of cascading effect of points that will trace circles. Thus, we set out to prove that
this was the case.
Construction
Given three points, Z, F , and Z 0 , where
ZF
Z0F
6
= c, F is a fixed point, c is a constant, and
ZF Z 0 = γ, where γ is also a constant.
Theorem
If Z traces a circle, c1 , centered at o with radius r, Z 0 also traces a circle with radius cr.
S'
O'
Z'
Z
O
F
S
Proof
Since Z traces a c1 at some point it is collinear with F and O at this point stop the ZF Z 0
“system.” Then extend F Z 0 , if need be, and mark point O0 such that
ZO
ZO0
= c. Let S be
the rotation of Z and S’ be the rotation of Z’ about fixed point F. To show that Z’ is on
a circle we will show that Z 0 O0 = S 0 O0 .
20
Since
6
SF S 0 = 6 SF Z + 6 ZF S
and
6
ZF Z 0 = 6 SF Z + 6 S 0 F Z 0
S 0 F Z 0 = 6 SF Z.
6
By construction
ZF
Z0F
=
SF
S0F
= c.
Thus,
4F ZS ∼ 4F Z 0 S 0 .
By definition of similar triangles,
6
SZF = 6 S 0 Z 0 F
and
6
SZO = 6 S 0 Z 0 O0 .
Thus, (note: SZ = cS 0 Z 0 and OZ = cO0 Z 0 from before)
4ZSO ∼ 4Z 0 S 0 O0 .
Therefore, Z 0 O0 = S 0 O0 and Z 0 traces a circle with center O0 and radius cr.
Where does this get us in our research?
There are many fixed points in our construction: points A and B, the center of circle
c1 , any point on AB and any point on circle c1 . Thus, if we can find points that have
the same initial conditions as the previous proof we can show that more points will trace
circles. In fact, there are plenty of special points of triangles that do satisfy our initial
21
conditions. Take for example point A as the fixed point and consider the midpoint, M ,
of AC: since AC : AM = 2 : 1, 6 CAM is constant (these points are collinear) and C
traces a circle, then M will trace a circle. Also it is a well known Euclidean fact that the
centroid (the intersection of the medians) lies 2/3 of the distance from the vertex to the
midpoint of the opposite side, thus the centroid will trace a circle. It is also a well known
Euclidean fact that the centroid is on the Euler line and that the other special points on
the Euler line are at constant distance ratios. Since one of the points on the Euler line is
the circumcenter, and thus fixed in our construction, then all of these special points on
the Euler line (the orthocenter, nine-point center, De Longchamps point) will also trace
circles as C moves along the circumcircle (and the Euler line rotates). This cascading
effect easily gives us many other points that will trace circles.
LIGHTHOUSES
After our sojourn into constant distance ratios, our advisor, Dr. Stephen Kennedy,
brought us a problem that related to our earlier ideas about spinning pairs of lines. On
the way home from a conference, Dr. Kennedy ran into Dr. Richard Guy in the airport.
They got to talking and Dr. Guy mentioned that he had been working on the following
problem:
Given two lighthouses, each projecting a beam of light and each spinning at some rate,
what is the path of the intersection of those beams?
This can best be envisioned by finding two friends to extend their arms and spin around.
In the absence of friends, however, this picture will suffice.
22
Lighthouse A
Lighthouse B
Dr. Guy had already derived an equation for the path of the intersection when both
beams spin at the same speed as well as when one spins twice as fast as the other. His
parameterization reduces to a circle when the beams are spinning at the same speed and
to a cubic curve when their speeds differ [2].
Experimenting with this idea, we produced some of these cubic curves on Mathematica.
As expected, they were distinctly un-circular. The following is an example of the shape
that is traced by the intersection of these beams when one lighthouse beams spins twice
as fast as the other.
23
STROPHOID
After thumbing through geometry books, we finally found a curve that fit the bill, the
strophoid [4]. This curve is constructed by first choosing two fixed points A and B, and
constructing any line through A. On that line, choose a point C and construct circle
c1 with center C and radius AC. Finally, draw the line through C and B and label its
intersections with c1 , E and F . The path that E and F take as C slides along the line
through A is called a strophoid.
B
E
C
A
F
Because saying “Gee, that sure looks like a strophoid”is not mathematically satisfying,
we came up with the following proof.
This proof shows that as the lighthouses spin, one twice as fast as the other, the intersection of the beams of light remains on the strophoid.
24
l
D
m
B
t
A
θ
w
y
3π
-t-θ
2
w
x
F
C
E
Construction
Construct line AB. At A, construct a line l at 6 θ to AB. Choose points C and D on l
such that A is between D and C. Construct circle c1 centered at C with radius CA.
Construct line BC, and call it m. Label the intersections of m with circle c1 , E and F .
Join AE and AF . Also, note that 6 EAF is right since E and F lie on opposite ends of
a diameter.
Proof
In this proof we will derive a relationship between t and y and we will use this to show
that a path of intersection of lighthouse beams will follow the strophoid curve when one
beam is moving twice as fast as the other.
25
Let
AF B = x
6
and
ABF = y
6
Given CF and CA are radii of circle c1 ,
6
CF A = 6 CAF = w
Let
6
DAE = t.
Given
t + θ + 6 BAF +
6
BAF =
3π
2
π
2
= 2π
−t−θ
At this point, it is helpful to think about which is the strophoid and which is the lighthouse construction. The lines l and m are exactly as they were in the above strophoid
construction. In this case, however, the line m also represents the beam of light from a
lighthouse at B, and the line AE represents a lighthouse beam from A.
w = π − θ − 6 BAF
=π−θ−
3π
2
= - π2 + t
We also know
x + w = π.
Combining the last two equations yields
26
+θ+t
x=
The angle sum of 4ABF =
Letting x =
3π
2
3π
2
3π
2
−t
− θ − t + x + y = π.
− t in the above equation, we get
2t = 2π − θ + y.
This is, in fact, sufficient to show that our lighthouses, when one is moving twice as fast
as the other, trace a strophoid. To see this, begin by noting that the lighthouse beam
from A began at time 0 when it was along line l. Since we are assigning the relative speed
of lighthouse A to be t radians per unit time, we see that after spinning for 1 unit time,
the beam from A is exactly t radians from where it started. We now ask where the beam
from B began. If we rewind this beam back 2t, using the relation we just found, we find
that B started parallel to where A started. This means that whatever t is, if B is spinning
twice as fast as A, the two beams of light will always intersect on the strophoid.
By using the relationship in a different way, we can further clarify the situation. The
question is centered around what happens when C moves along l and the beams from
the lighthouses spin. Thus, consider the same situation along with what happens when
C slides to a new position C 0 , E goes to E 0 , and F goes to F 0 accordingly. We want to
show that the angle traveled by the beam from B is twice the angle that the beam from
A has traveled. In other words,
6
E 0 BE = 26 E 0 AE
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This idea is outlined in the following diagram.
A
A
B
B
C
C
E
E
C'
E'
Now take a closer look at our translated construction.
r
A
t
z
y
B
C
E
C'
E'
It is helpful to note that the red lines correspond to the construction from the previous
proof, while the blue lines represent the new position along the strophoid line.
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Using the equation we found previously, we know that
2t = 2π − θ + y
2r = 2π − θ + z.
Subtracting the two equations we get
2r − 2t = z − y
and
2(r − t) = z − y.
But we know that
6
E 0 AE = (r − t) and 6 E 0 BE = (z − y).
Therefore
6
E 0 BE = 26 E 0 AE.
Interestingly, this proof ties in nicely with a puzzle we encountered long before we took
up the issue of lighthouses and strophoids. In our Apples and Oranges construction, E
traces a circle when C moves around c1 . Unlike all of our other constructions, however,
this one is not symmetric. What happens when we move the other vertices?
When B moves around the circle, AC stays fixed and as a result, E and F merely slide
up and down the perpendicular bisector of AC. When we move A, however, something
more interesting happens: E and F trace a curve that looks like (and is) a strophoid.
The key to this proof is “finding the lighthouses” in the construction.
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E
Lighthouse A (C)
F
Lighthouse B (O)
A
B
The lighthouse construction requires two fixed points, and two sets of spinning lines anchored at those points. In Apples and Oranges we have exactly that. Points C (relabeled
Lighthouse A) and O, the center of the circle, (relabeled Lighthouse B) stay fixed and
both have “beams” coming from them, the angle bisector of 6 ACB and the perpendicular
bisector of AC respectively.
Knowing that we can look at Apples and Oranges in terms of lighthouses, leaves us only to
prove that the beam from Lighthouse B spins twice as fast as the beam from Lighthouse
A.
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We can show that this is true by examining the angles that these beams sweep out as A
makes a full circuit.
Lighthouse A (C)
Lighthouse B (O)
A
B
D
Proof
Consider the segment AO, this segment moves around the circle at a speed corresponding
to the speed that A moves around the circle. Since the beam from O bisects 6 AOC, we
know that that beam will spin half as fast as the segment AO.
Now, if we consider the segment DC, the beam from C, we can see that point D bisects
the arc AB. Thus, D will move around the circle half as fast as point A. However, since
the segment DC is anchored at the edge of the circle, we can see that DC moves
1
4
as
fast as AO. So AO moves twice as fast as the beam from O and four times as fast as the
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beam from C. Therefore, the beam from O moves twice as fast as the beam from C.
Thus we have shown that Apples and Oranges is in fact identical to the lighthouse construction. We therefore know that Apples and Oranges also traces a strophoid when A
moves around the circle.
Conclusion
In the course of our research we proved several theorems about special points of a triangle
and how they trace circles. This lead us to a greater understanding of why these points
trace circles based on their distance from fixed points. Finally our quest to understand the
shape of things led us to investigate another curve, the strophoid. Using our lighthouse
construction, we identified other special points of a triangle that trace strophoids.
Further Research
We are still left with several open questions:
Are there other shapes traced by special points of a triangle that we have not yet
encountered? We chose to study the circle and the strophoid, but there are likely other
interesting shapes to study.
Is there a way to parameterize our findings? From the beginning of our research we
wished to develop some sort of “algebraic machine” that would explain our findings.
Would other geometric coordinate systems assist us in looking at this problem? Barycentric and trilinear coordinates are the standard way of discussing special points of a triangle,
using these could be helpful.
*********************************
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References
[1] John Conway, personal communication
[2] Richard Guy, The Lighthouse Theorem - A Budget of Paradoxes, to appear, and
personal communication
[3] Clark Kimberling, Central Points and Central Points in Triangles, Mathematics Magazine Vol. 67, No. 3 (June 1994) 163–187.
[4] J. D. Lawrence, A Catalog of Special Plane Curves, New York: Dover, 1972.
[5] William Mueller, Centers of Triangles of Fixed Center: Adventures in Undergraduate
Research, Mathematics Magazine Vol. 70, No. 4 (Oct., 1997) 252–262.
A Special thanks to:
Stephen Kennedy, Advisor
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