Taxicab Geometry: Not the Shortest Ride Across Town (Exploring Conics with a Non-Euclidean Metric)

Creative Component Christina Janssen Iowa State University July 2007

1 Introduction Pg 3 Taxicab geometry: What is it and where did come from? What is a metric? Proof Euclidean/Taxicab are metrics Proof Taxicab distance Euclidean distance Circles Pg13 Circumference Area Equations (two dimensions) Sphere (three dimensions) Formula Summary Ellipses Pg 28 Circumference Area Equations Formula Summary Parabolas Pg 42 Equations Formula Summary Hyperbolas Pg 53 Equations Formula Summary Conclusion Pg 59 Jaunts Pg 61 1 Metrics 2 Triangle Inequality Theorem 3 Right Triangles 4 Congruency 5 Rigid Motion 6 Quadrilaterals 7 Area Limits 8 High School 9 Activities Charts Pg 84 Bibliography Pg 88

2 Introduction

This paper journals my investigation through the conic sections with the Taxicab metric. I am a high school teacher and graduate student and attempted to focus not only on my personal learning experience, but also on what I could incorporate into my classroom for the good of my students. As most teachers will attest, learning concepts that you believe will motivate your students is an exciting prospect. My goals in this creative component are multifold:

1. An opportunity to practice and hone my newly acquired problem solving

skills in an individual and in-depth investigation.

2. A chance to delve deeper into a topic (Taxicab geometry) that intrigues me.

3. To acquire a deeper understanding of the structure of mathematics.

4. To become more knowledgeable about conics.

5. To discover tools and ideas to bring back to my students.

6. Completion of my Masters in School Mathematics degree.

The main text of the paper is broken into four sections, each highlighting a conic section. It begins with Taxicab circles, and continues with Taxicab ellipses, Taxicab parabolas and Taxicab hyperbolas. Each section has various topics and observations. The content, other than definitions and historical information is a strict representation of my own work. Work was performed from scratch, with rudimentary sketches with decisions on future directions being made along the way. The investigation has been empowering. I do not by any stretch know all there is to know, and I am possibly left with more questions then when I started. But, my mathematical growth and increased confidence in my ability to problem solve, has been remarkable. I hope you enjoy the paper.

3 Taxicab geometry: What is it and where did come from?

“The usual way to describe a (plane) geometry is to tell what its points are, what its lines are, how distance is measured, and how angle measure is determined.” (Krause

2) Taxicab geometry will use points and lines as defined in Euclidean geometry. “A point is a location.” and “A line is made up of points and has no thickness or width.” (Glencoe

6) Additionally, we shall define an angle to be “the union of two rays that share a common endpoint. The point is called the vertex.” The measure of an angle shall be “the amount of rotation about the vertex needed for one side to overlap the other.”(Geometry to Go 062) The difference in Taxicab geometry shall be that distance measured can be only horizontal and /or vertical. The well-known and loved Euclidian distance formula for points P(x , y ) and Q(x , y ) defined d P, Q = (x ! x )2 + ( y ! y )2 is not P P Q Q E ( ) p Q P Q the metric for your typical cab driver (unless his or her taxi happens to be an air cab).

Taxicab geometry is built on the metric where distance is measured d P,Q = x ! x + y ! y and will continue to be measured as the shortest distance T ( ) P Q P Q possible.

Taxicab geometry was proposed as a metric long before it was labeled Taxicab. A

Russian by the name of Hermann Minkowski wrote and published an entire work of various metrics including what is now known as the taxicab metric. In 1952 an exhibit was displayed at the Museum of Science and Industry of Chicago, which highlighted geometry. A small pamphlet was distributed entitled, “You Will Like Geometry.” It was in the pages of this booklet that the Minkowski’s geometry was coined Taxicab geometry. (Reinhardt 38)

4 What is a metric?

A metric is a mathematical function that measures distance. It is important to note that both the Euclidean distance formula and the Taxicab distance formula fulfill the requirements of being a metric. The three axioms for metric space are as follows.

Let P, Q, and R be points, and let d(P,Q) denote the distance from P to Q.

Metric Axiom 1 d(P,Q) ! 0 and d(P,Q) = 0 if and only if P = Q

Metric Axiom 2 d(P,Q) = d(Q, P)

Metric Axiom 3 d(P,Q) + d(Q, R) ! d( P, R) (Reynolds 124)

In simple terms this means that the distance between two points is always greater than zero and only equal to zero if the two points are actually the same point. The distance between two points is the same despite which point you begin your measure.

The distance from a first point to an intermediate point and then from the intermediate point to a final point must be farther than or equal to the distance you would travel if you went directly from the first to a final point. Both distance formulas, Euclidean and

Taxicab, satisfy these axioms. Proof is shown below. (See Jaunt 1 for additional metric examples.)

Proof Euclidean distance formula is a metric:

Let P(x , y ) , Q(x , y ) and R(x , y ) !"2 . P P Q Q R R

Metric Axiom 1: d(P,Q) ! 0 and d(P,Q) = 0 if and only if P = Q .

If P = Q then d(P,Q) = d(P, P)

2 2 = x ! x + y ! y ( P P ) ( P P )

5 2 2 = 0 + 0 ( ) ( )

= 0 .

If P ! Q then either x ! x or y ! y . P Q P Q

2 2 And either x ! x > 0 or y ! y > 0 . ( P Q ) ( P Q )

Suppose x ! x P Q

2 2 Then x ! x + y ! y ( P Q ) ( P Q )

2 ! x " x ( P Q )

= x ! x P Q

> 0

2 2 Therefore, x ! x + y ! y > 0 ( P Q ) ( P Q )

Metric Axiom 2: d(P,Q) = d(Q, P)

2 2 d(P,Q) = x ! x + y ! y ( P Q ) ( P Q )

2 2 = 1 x x 1 y y (! ) ! P + Q + (! ) ! P + Q ( ( )) ( ( ))

2 2 = 1 x x 1 y y (! ) Q ! P + (! ) Q ! P ( ( )) ( ( ))

2 2 2 2 = !1 x ! x + !1 y ! y ( ) ( Q P ) ( ) ( Q P )

2 2 = x ! x + y ! y ( Q P ) ( Q P )

6 = d(Q, P)

Metric Axiom 3: d(P,Q) + d(Q, R) ! d( P, R)

Let P , Q , and R be non-collinear, thus forming a triangle.

2 2 2 2 d(P,Q) + d(Q, R) = x ! x + y ! y + x ! x + y ! y ( P Q ) ( P Q ) ( Q R ) ( Q R )

2 2 d(P, R) = x ! x + y ! y ( P R ) ( P R )

According to the Triangle Inequality Theorem: The sum of the lengths of any two sides of a triangle is greater the length of the third side.” Proof of the theorem is provided in

Jaunt 2.(Glencoe 261)

2 2 2 2 2 2 x ! x + y ! y + x ! x + y ! y " x ! x + y ! y ( P Q ) ( P Q ) ( Q R ) ( Q R ) ( P R ) ( P R )

and d(P,Q) + d(Q, R) ! d( P, R) .

Let P , Q , and R be collinear. For three points collinear, one point must be between the other two. Betweenness of points requires the points be collinear, and if Q is between P and R then PQ + QR = PR .

Proof Taxicab distance formula is a metric:

Let P(x , y ) , Q(x , y ) and R(x , y ) !"2 . P P Q Q R R

Metric Axiom 1: d(P,Q) ! 0 and d(P,Q) = 0 if and only if P = Q

If P = Q then d(P,Q) = d(P, P)

= x ! x + y ! y P P P P

= 0 + 0

7 = 0 .

If P ! Q then either x ! x or y ! y . P Q P Q

and x ! x > 0 or y ! y > 0 . P Q P Q

Suppose x ! x P Q

Then x ! x + y ! y P Q P Q

! x " x P Q

> 0

Therefore, x ! x + y ! y " 0 . P Q P Q

Metric Axiom 2: d(P,Q) = d(Q, P)

d(P,Q) = x ! x + y ! y P Q P Q

= !1 !x + x + !1 !y + y ( )( P Q ) ( )( P Q )

= !1 x ! x + !1 y ! y ( )( Q P ) ( )( Q P )

= !1 x ! x + !1 y ! y ( ) ( Q P ) ( ) ( Q P )

= x ! x + y ! y Q P Q P

= d(Q, P)

Metric Axiom 3: d(P,Q) + d(Q, R) ! d( P, R)

Let P , Q , and R be non-collinear, thus forming a triangle.

d(P,Q) + d(Q, R) = x ! x + y ! y + x ! x + y ! y P Q P Q Q R Q R

8 = x ! x + x ! x + y ! y + y ! y ( P Q Q R ) ( P Q Q R )

! x " x + x " x + y " y + y " y ( P Q Q R ) ( P Q Q R )

= x ! x + y ! y P R P R

= d( P, R)

hence, d(P,Q) + d(Q, R) ! d( P, R) (Reynolds 125)

Note that the Triangle Inequality was not used in the proof of axiom 3, as it was in the Euclidean proof. Instead only the fact that A + B ! A + B . A + B = A + B

occurs when A or B or both are zero, or A and B are both positive or negative.

A + B > A + B when A and B have opposite signs.

The Triangle Inequality Theorem, which is strictly greater than, does not hold true in Taxicab geometry. figure 1shows a counterexample of two sides of a triangle not being greater than the third side as measured with the Taxicab metric.

4 d T(B,C)=7 d T(A.B)=4

2 A C

d T(A.C)=3

5 10

figure 1

Fortunately our metric axiom requires the sum of two distance to be ! the third direct distance and certainly the Taxicab metric does fulfill that axiom.

9 Let P , Q , and R be collinear. Allowing Q to be between P and R , it is again true that d(P,Q) + d(Q, R) = d(P, R) .

Before we drive on, let’s look at how the Taxicab metric could be applied. On the following graph which is closer to A, point B or point C?

C 4

d(A,C)=5

d(A,B)=6 A B

2 4 6 8

-1 figure 2

With the measurements given in the Cartesian plane above one would simply state that since d(A,C) = 5 and d(A, B) = 6 that d(A,C) < d(A, B) . Hence, C is closer to A than

B . Euclid and Pythagoras would be proud.

Now consider the points (oriented in a similar fashion) in the following map. If you are driving and plan to keep your license, your distances are measured in a very different fashion.

10 C

A B

figure 3

It now seems that the distance from A to C is 7 blocks, while the distance from A to B is 6 blocks. Unless we choose to go off-road, B is now closer to A than C. Euclidean distance formula has its place, but on the city streets is not the most useful location.

Taxicab distance is sometimes equal to Euclidean distance, but otherwise it is greater than Euclidean distance.

Euclidean distance ! Taxicab distance Proof:

Absolute values guarantee non-negative value

0 ! 2 x " x y " y P Q P Q

Addition property of inequality

2 2 2 2 x ! x + y ! y + 0 " x ! x + y ! y + 2 x ! x y ! y P Q P Q P Q P Q P Q P Q

11 Absolute value may be dropped because squaring guarantees the positive

value of the expression

2 2 2 2 x ! x + y ! y + 0 " x ! x + y ! y + 2 x ! x y ! y ( P Q ) ( P Q ) P Q P Q P Q P Q

Factoring the right hand side

2 2 2 x ! x + y ! y + 0 " x ! x + y ! y ( P Q ) ( P Q ) ( P Q P Q )

Take the square root of both sides

2 2 2 x ! x + y ! y " x ! x + y ! y ( P Q ) ( P Q ) ( P Q P Q )

Simplify to

2 2 x ! x + y ! y " x ! x + y ! y ( P Q ) ( P Q ) P Q P Q

d ! d E T

Only when x = x or y = y (they cannot both be true simultaneously or else P Q P Q we do not have two distinct points) will the Taxicab metric be equal to the Euclidean metric, or in simple terms our points (x , y ) and (x , y ) are either vertically or P P Q Q horizontally, oriented from one another. Otherwise, if x ! x and y ! y then Taxicab P Q P Q distance is greater than Euclidean distance. It is worth mentioning, the ease at which the

following proof can be made with a simple substitution. For instance, let A = xP ! xQ and

B = yP ! yQ . The entire proof can be written

0 ! 2 A B

2 2 2 2 A + B ! A + 2 A B + B

12 2 A2 + B2 + 0 ! A + B ( )

A2 + B2 ! A + B

(This simplification may assist high school students from getting bogged down in the notation of the distance formulas.)

Climb in and let’s go for a ride. The driver will attempt the most direct route possible, but as is the case with Taxicab geometry ( d ! d ), it is not always possible. E T

We have a long way to travel and the meter is running, let the adventure begin.

DESTINATION CIRCLES

“A circle is the locus of all points in a plane which are all equidistant from a given point called the center of the circle.” (Glencoe 522) With the Euclidean metric, d P, Q = (x ! x )2 + ( y ! y )2 the graph of a circle with radius three appears so. E ( ) p Q P Q

-5 5

-2

-4

figure 4

13 A circle with the Taxicab metric d P,Q = x ! x + y ! y , and radius three T ( ) P Q P Q appears so.

-5 5

-2

-4

figure 5

The simple, yet stunning difference makes clear how very important the metric is. Further complicating the situation, the following is not a circle.

-5 5

-2

-4

figure 6

14 Upon inspection, it becomes apparent that the Taxicab distance from the center, at the origin for this situation, (0,0) to the point (3,0) is three, while the distance from the center to the point (3,3) is six. Therefore, by definition that all points are equidistant from the center, a square with a side oriented horizontally in the plane is no longer a circle. A circleT appears as a square with four sides and four vertices (which is how we will refer to the segments of a circleT). Considering our plane with the positive y-axis being oriented north and the positive x-axis oriented east, the vertices of a circleT must be oriented directly east, north, west and south.

Conjecture: Formulas can be written for circumference of a circleT with respect to its radius.

“The circumference of a circle is the distance around the circle.” (Glencoe 523)

The Euclidean metric formula for circumference is C= 2!r with r being the radius. With some investigation, we see that the following is true when using the Taxicab metric.

Radius r Circumference C

1 (2)(4)=8

2 (4)(4)=16

3 (6)(4)=24

chart 1

The distance from a vertex to a consecutive vertex of the circle is twice the radius.

o o o (Note: This means that the length of a hypotenuse of a 45 ! 45 ! 90 triangle no longer has length 2 times the leg, but instead two times the leg. (A fact that would thrill every high school student whoever incorrectly simplified a 2 + b2 = c2 to a + b = c ) This occurs

15 when a leg is oriented horizontally in the plane. (See Jaunt 3 for additional work on right triangles in Taxicab geometry.) An equation for circumference with respect to side is

C = 4s

Substituting the radius for a length of twice one side results in the following equation for circumference with respect to radius.

(1.1) C = 8r

An interesting question arises as we consider the nature of circlesT . They have four sides with the same Taxicab measurement. Can it be claimed that the sides are congruent? Congruency indicates that two things have equal size and shape. By this definition the sides of our Taxicab circle are indeed congruent, but beware, unlike

Euclidean geometry, not all Taxicab segments with equal measures are congruent. (More on congruency can be found in Jaunt 4)

Conjecture: Formulas can be written for area of a circleT with respect to its radius.

2 Area of an Euclidean circle has the formula A = !r . When considering the area of a Taxicab circle one must consider the side measured with the Taxicab metric. Recall, in the Taxicab metric the length of one side of our circle with radius three is six. It no longer seems reasonable to say that the area of what appears as a square has area equal to the side squared. (See jaunt 5 for work in rigid motion and jaunt 6 for a trip through

Taxicab quadratics.) This would produce an area of thirty-six units squared. Upon inspection and simple counting of the squares on the graph, we see this is unreasonable.

Again, the driver is reminded that although a circleT appears as a square, it cannot be treated as such in the classical Euclidean way.

16 Let us consider area measured with strictly horizontal and vertical lengths. Area is defined to be “the number of unit squares equal in measure to the surface.” (Webster’s

101) The Taxicab circle can split into four congruent isosceles triangles each with a base and height of r . It is important to notice that the formula for area holds true based on the fact that the base and height are measured horizontally and vertically and therefore the lengths of the segments are the same for the Taxicab and Euclidean metric.

2 h

-5 b 5

-2

-4 figure 7

1 1 1 The area of each triangle is (base)(height ) or (r)(r) = r 2 . Multiplying by 2 2 2

1 four, to account for the number of congruent triangles and we have A = (4) r 2 , or 2 equivalently,

2 (1.2) A = 2r

It is interesting that the conclusion is the same as if the measurements were based upon the Euclidean metric with sides measuring r 2 and area of a square being side

2 2 squared resulting in r 2 = 2r . ( )

17 Additionally, a formula can be written using the length of the four congruent sides of a circleT. The length of a side is twice the radius of a circleT. Using our previous

2 s formula A = 2r we may use the substitution s = 2r solve for radius, r = . The result is 2

2 ! s $ a formula for area with respect to a side, A = 2 . Simpified, "# 2 %&

1 A = s2 2

For more detail on why area is the same in Euclidean and Taxicab geometry see jaunt 7.

Major Conjecture: There is a connection between geometry and algebra in the Taxicab

Geometry. Therefore, we should be able to write equations for circles.

Any two circlesT will have the same angle measurements and corresponding sides will be proportional. By definition, this makes any two circlesT similar. Hence, all circlesT will have four sides with slopes of ±1. Slope is used with the traditional

meaning. The slope of a line containing two points with coordinates (xP ,yP ) and

yP ! yQ (xQ , yQ ) is where xP ! xQ . In Euclidean as well as Taxicab geometry slope is xP ! xQ measured horizontally and vertically and are therefore equivalent. Graphically, each vertex will occur horizontally or vertically from the center. Let the center of any circleT have a center (x , y ) . The northeast side of a circle has a slope of negative one and c c T endpoints of (x , y + r) and (x + r, y ) , r being radius. c c c c

18 (xc,yc+r) N W E

(xc,yc) S (xc-r,yc) (xc+r,yc)

(xc,yc-r)

figure 8

In search of a formula for the side the following calculation takes place

y = mx + b General slope y-intercept equation of a line.

y = !1x + b Slope of the line equals negative one.

y = !1(x + r) + b Substitute one known point in for x and y. c c

b = y + x + r Solve for b. c c

y = !x + y + x + r Rewrite equation c c

The preceding equation is only appropriate on a domain from x to x + r . c c

Calculating the other sides of the circle in similar fashion a formula for a circleT is created. Working counterclockwise through the four standard quadrants from the positive x-axis we have,

#(sideNE )y = !x + yc + xc + r,xc " x " xc + r ' % % %(sideNW )y = x + yc ! xc + r,xc ! r < x < xc % Circle (x)$ ( T (side )y x y x r,x r x x % SW = ! + c + c ! c ! " " c % &%(sideSE )y = x + yc ! xc ! r, xc < x < xc + r )%

19 With careful scrutiny, it seems that our equations for Taxicab circles have domains that overlap. This is due to the fact that, similar to Euclidean circles, Taxicab circles are not functions. (Note: Graphically it can be observed that both types of circles fail to pass the vertical line test at all but two points. An equation for the circleE can be written as

(x ! x )2 + ( y ! y )2 = r 2 c c

or y = ± r 2 ! (x ! x )2 + y c c

Taking into consideration order of operations, the equation now reads: start the circle at the origin, then translated horizontally, then the plus or minus results in a reflection over the x-axis. These changes occur prior to the vertical translation. Using the same concepts, the following simplification will consolidate our Taxicab equations to a simpler form. First, consider any circle to have a center oriented on the x-axis and T vertically translated after a reflection has occurred to determine both the northern and southern segments. Therefore, y is now zero and the following changes take place in our c graph.

(xc,r)

(xc,0) (xc-r,o) (xc+r,0)

(xc,-r)

figure 9

20 With vertical translation considered after our reflection, our formulas can now be written in a more concise form.

y = mx + b General slope y-intercept equation of a line.

y = !1x + b Slope of the line equals negative one.

0 = !1(x + r) + b Substitute one known point in for x and y. c

b = r Solve for b.

y = !x + r Rewrite equation

Now taking into consideration the reflection and vertical translation we have,

%#(sideNW sideSW ) ± (x ! xc + r) + yc ,xc ! r " x " xc %' (1.3) Circle (x) = $ ( . T (side side ) ± (!x + x + r) + y , x < x " x + r &% NE SE c c c c )%

An example using the formula is a circleT with center (5, -3) and radius 2. The equation would therefore be

#±(x !5 + 2) + (!3),5! 2 " x " 5 & CircleT (x) = $ ' %±(!x + 5+ 2) + (!3),5 < x " 5 + 2(

simplified,

#±(x ! 3) + (!3),3" x " 5 & CircleT (x) = $ ' %±(!x + 7) + (!3),5 < x " 7(

21 5

-2

-4

-6 figure 10

Euclidean circles can be rotated about a diameter to form a three-dimensional sphere. Similarly, we can rotate our Taxicab circle around a diameter to create Taxicab spheres. The first consideration we must take is where is a diameter on a Taxicab circle?

“In a circle, a chord that passes through the center of a circle” is considered a diameter.

With a chord defined as “a segment with endpoints that are on the circle.” (Glencoe R13)

Euclidean circles are symmetric in all diameters. Our Taxicab circles are not so simple.

Consider the following three diameters.

-5 5

-2

-4

figure 11

22 The various diameters will result in different three-dimensional objects. We will take each in turn.

Case 1: Diameter from midpoint of a side (sketched in green)

figure 12

The resultant figure is equivalent to a right circular cylinder albeit tipped at a 45 angle from having a horizontal base. The radius of the cylinder is equivalent to the radius of the circleT and the height is a diameter and therefore has a length equal to two radii.

Volume for three-dimensional objects are the same in Euclidean and Taxicab geometry, but their equation are not. We will begin our alteration with the equationE for a cylinder.

V = !r 2h Euclidean volume of a right circular cylinder

V = 2!r 3 height always equals two radii

(1.4) V = 2!r 3

Similar manipulation can be performed to obtain a formula for surface area

S = 2!rh + !r 2 Euclidean surface area of a right circular cylinder

S = 4!r 2 + !r2 height always equals two radii

23 (1.5) S = 5!r 2

Case 2: Diameter from a vertex (sketched in red)

figure 13

The sphere is equivalent to two cones connected at their bases with each height and radius of the cones equal to the radius of the circleT.

1 V = !r2 h Euclidean volume of a cone 3

1 V = !r2 r height always equals radius 3

1 V = 2 !r 3 multiply by two 3

2 (1.6) V = !r 3 3

S = !r r 2 + h 2 Euclidean lateral surface area of a cone

S = !r r 2 + r 2 height equals radius

24 S = 2 2!r2 multiply by two

(1.7) S = 2 2!r2

Case 3: Diameter from points between vertex and midpoint of a side (sketched in blue)

This diameter is perhaps the most intriguing of the Taxicab spheres. First notice that the diameter itself can be placed in an infinite variety of slopes of which create an infinite variety of three-dimensional spheres. My biggest initial roadblock was in being able to visualize what object would be created if I rotated around the given diameter. (It is for that reason I have withheld the figure at this stage to allow the reader to conjecture for themselves. I must admit my lack of ability to do so, left me less than impressed. So I began resorting to manipulatives. I cut out a Taxicab circle and folded it on the desired diameter. The result looked like figure 14.

figure 14

I must admit, (and likely you’ll agree) it didn’t look like much. I used the very rudimentary strategy of taping the diameter to a pencil and spinning it. As primitive as this may sound, it seemed to help. Going back to paper and pencil I sketched the outline of my paper cutout and added its other corresponding other half.

figure 15

The picture was coming clearer. This is my current conjecture of the appearance of the sphere.

figure 16

Although I have not yet finalized formulas for volume and surface area, I would like to share a few interesting observations I have discovered on that path. I honestly admit it took quite some time before I noticed that figure 15 is simply two circlesT overlaid.

figure 17

With this revelation the investigation began to continue forward with the following findings about figure 17.

1. The triangles formed all have the same angle measurements.

2. The triangles are right and congruent using the metricE, no matter what the

slope of the diameter is that is use to create the sphere.

3. With the triangles removed the middle is itself an octagon formed out of 8

congruent triangles with the metricE..

4. Each side of one of the overlaid circlesT is the sum of the three different sides

lengths of the triangle with the metricE.

There is much more to discover. Circles have been a fascinating leg of our trip.Let us continue onto new, undiscovered paths with other conic sections.

Circle Formula Summary:

Circumference with respect to the radius

(1.1) C = 8r

Circumference with respect to the radius

2 (1.2) A = 2r

27 Circle with center (xc , yc ) and radius r .

%#(sideNW sideSW ) ± (x ! xc + r) + yc ,xc ! r " x " xc %' (1.3) (Circle (x) = $ ( T (side side ) ± (!x + x + r) + y , x < x " x + r &% NE SE c c c c )%

Volume of sphere with midpoint diameter with respect to radius r .

(1.4) V = 2!r 3

Surface area of a sphere with midpoint diameter with respect to radius r .

(1.5) S = 5!r 2

Volume of a sphere with vertex diameter with respect to radius r.

2 (1.6) V = !r 3 3

Surface area of a sphere with vertex diameter with respect to radius r .

(1.7) S = 2 2!r2

DESTINATION ELLIPSES

With the Euclidean metric the standard definition of an ellipse with center (h, k)

(x ! h) (y ! k) and major and minor axes of lengths 2a and 2b , where a > b is 2 + 2 = 1 a b

(x ! h) (y ! k) with a horizontal major axis, or 2 + 2 = 1 with a vertical major axis. The foci b a

2 2 2 lie on the major axis, c units from the center, with c = a ! b . (Larson 640)

An example of an ellipseE appears so, as graphed by the TI-89 Titanium graphing calculator.

figure 18

Through investigation and many manually constructed graphs, a Taxicab ellipse appears as a hexagon,

figure 19 or appear as an octagon.

figure 20

The distinguishing factor is whether the foci are collinear on a horizontal or vertical line.

If so a hexagon results, otherwise an octagon results. We shall begin by defining terms of the Taxicab ellipse.

29 6

Street4 N

Alley Alley 2 W E S Street Center foci

-5 5 Street 10 foci

-2 Alley Alley

Street -4

figure 21

The center and foci will remain as defined with Euclidean ellipses. The horizontal and vertical segments shall be named streets, while non-horizontal and non-vertical segments will be named alleys. Each segment shall be distinguished according to its direction. The sum of the distances from one focus to any point on the ellipse and from that point to the other focus shall, in keeping in concert with Euclidean geometry when possible, be a distance 2a . For ellipses and the remainder of this paper if

!1 < slope < 1and not equal to zero, we shall consider this a non-steep slope. If the slope is < !1or > 1it shall be considered steep. Undefined, zero and slopes of ±1shall be considered special cases. Finally, in ellipses as in the rest of our travels we shall often be interested in the individual components of the Taxicab metric d P,Q = x ! x + y ! y . We shall define the vertical component of distance T ( ) P Q P Q y ! y = m and the horizontal component of distance x ! x = n . P Q P Q

30 Conjecture: Formulas can be written for circumference of an ellipseT based on the horizontal and vertical distance between the foci, and distance 2a.

Through the construction of many ellipses the following patterns (memorialized in a chart at the end of the paper) were discovered. There is one north street and one south street equal in length to each other and the horizontal distance n between foci. ( )

Similarly, there is one east street and one west street, also equal in length to each other and the length of the vertical distance m between foci. Each alley is the hypotenuse of ( )

! ! ! a 45 ! 45 ! 90 triangle with a horizontal leg. Therefore, the length of the hypotenuse is double the length of either leg. To find the length of one leg, consider the following in relation to the graph.

leg

m f1 n

figure 22

One vertices of the triangle is oriented directly horizontal from the focus and one vertical from the focus. The sum of the distance from any point of the ellipse to each focus is 2a. Let the endpoint of the hypotenuse be considered point P . A focus f2 is the vertex of the triangle at the right angle. Therefore, the distance between P and f2 is the length of the leg. To determine the distance from point P to the other focus f1we need to find the sum of the length of the leg with mand n .

31 By the definition of an ellipse, this total distance is 2a . The following calculation can therefore take place.

d (P, f ) + d (P, f ) = 2a T 1 T 2

(leg) + (leg + m + n) = 2a

2leg = 2a ! m ! n

1 leg = (2a ! m ! n) 2

The four alleys are all congruent with a length twice that of a leg and therefore have length 2a ! m! n . Hence, the following calculation for circumference C takes place.

circumference=north street + south street + east street + west street + 4 alleys

C=north street + north street + east street + east street + 4 alleys

C=2 north street + 2 east street + 4 alleys

C = 2n + 2m+ 4(2a ! m ! n)

C = 2n + 2m+ 8a ! 4m! 4n

For a final formula of

(2.1) C = 8a ! 2n ! 2m

It is important to note that this formula works for all ellipses including those with a horizontal or vertical major axis. When this occurs either n is equal to zero or m is equal to zero, eliminating the term from the formula and eliminating the lengths of two sides leaving a hexagon.

Note that the circumference of a Taxicab ellipse is equal to the area of a rectangle circumscribed around the ellipseT.

figure 23

This is due to the fact that alley length can be calculated as the Taxicab distance measured inside along the aforementioned triangle or along the outside with the legs of the triangle being part of the rectangles sides. One must always be cognizant that unlike

Euclidean distance, there are many paths in which Taxicab distance can be measured.

Conjecture: Formulas can be written for area of an ellipseT based on the horizontal and vertical distance between the foci, and distance 2a.

Area of an ellipse can be broken down into sections

B D D 2

C A C

-10 -5 5 10

D B D -2

-4

-6

figure 24

33 The following calculation can take place

! ! ! Area=1rectangle A+ 2rectangle B+ 2rectangle C + 4 ( 45 ! 45 ! 90 ) triangle D

2 " 1% " 1 % A = mn + n(2a ! m ! n) + m(2a ! m ! n) + 4 (2a ! m ! n) #$ 2&' #$ 2 &'

" 1% " 1% A = mn + (2a ! m ! n)(n + m)+ 4 (2a ! m! n)2 #$ 2&' #$ 4&'

For a final area formula of

1 (2.2) A = mn + (2a ! m ! n)(n + m)+ (2a ! m ! n)2 2

As with circumference, the formula for area is good for all varieties of ellipses.

Major Conjecture: There is a connection between geometry and algebra in the Taxicab

Geometry. Therefore, we should be able to write equations for ellipses.

For the sake of our investigation, we will take a variety of ellipseT in turn.

Case 1: Major axis horizontal, center at origin.

f1(xf1,yf1) c(0,0) f2(xf2,yf2)

figure 25

Let the investigation begin with the center at the origin. As the major axis is the x- axis, and horizontal, a hexagon results. The ellipse is comprised of a north and south street and four alleys (a NE, NW, SW and SE). The graph of the ellipse is dependent upon three factors: The position of the center C , the horizontal and vertical distance

34 between the foci m and n and lastly the sum distance from any point on the ellipse to each focus 2a .After creating a chart of findings from various graphs of ellipsesT, the following equation has been derived.

# alley y x a,x x a ' ( NE ) = ! + f < < % 2 % % 1 % street y 2a n , x x x %( N ) = ( ! ) f " " f % 2 1 2 % % 1 % alley y x a, a x x % %( NW ) = + ! < < f % 2 1 EllipseT = $ ( % alley y x a, a x x % ( SW ) = ! ! ! " < f % 1 % % 1 % % street S y = ! 2a ! n ,x f " x " xf % ( ) 2 ( ) 1 2 % % alley y x a, x x a %( SE ) = ! f < " % & 2 )

Since the ellipse with horizontal major axis is symmetrical we can rewrite in a fashion similar to that of Euclidean ellipse formula. An Euclidean ellipse centered at the origin has formula

x2 y2 + = 1 a 2 b2 solved for y

2 " x % y = ± b $1! 2 ' # a &

(This type of investigation may aide students in not only their understanding of ellipses and metrics, but a deeper understanding of reflections, translations, why a graph is or is not a function and issues of domain to name a few. “Ms. Janssen, I solved for y but my calculator will only graph half a circle!”)

35 The Euclidean ellipse formula separates the formula into two functions using a plus and minus, one part a reflection over the major axis of the other. We will use a similar technique to simplify our taxicab equation.

# alley y x a , x x a ' ( NE / SE ) = ±(! + ) f < " % 2 % % 1 % (2.3) Ellipse street y 2a n , x x x T = $( N / S ) = ± ( ! ) f " " f ( % 2 1 2 % alley y x a , a x x %( NW / SW ) = ±( + ) ! " < f % & 1 )

Case 2: Major axis horizontal, center not at origin.

f2 xf ,yf f x ,y ( 1 1) c(xc,yc) 2( f 2 f 2)

figure 26

An ellipse with horizontal major axis but with the center moved away from the origin is a simple translation from our last equation. Formulas can be derived assuming horizontal shift initially and vertical shift after reflection across the major axis. We shall use the center point, but formulas could be acquired also using the foci, as we know the foci are related to the center by length m and n .

# 1 ' alley y = ± !x + x + a + y ,x + n < x " x + a %( NE / SE ) ( c ) c c 2 c % % % % 1 1 1 % (2.3) Ellipse street y (2a n) y ,x n x x n T = $( N / S ) = ± ! + c c ! " " c + ( % 2 2 2 % % 1 % alley y = ± x ! x + a + y ,x ! a " x < x ! n %( NW / SW ) ( c ) c c c 2 % & )

36 Case 3: Major axis vertical

f x ,y 1( f 1 f 1)

c(xc,yc)

f x ,y 2( f 2 f 2)

figure 27

Using similar techniques, written in terms of y, so that a reflection over the major axis

may still be utilized and with f1 being the northern most point the formula is such.

) 1 - alley x = ± !y + y + a + x , y + m < y " y + a +( NE / NW ) ( c ) c c 2 c + + + + # 1 & 1 1 + (2.4) Ellipse street x a m x , y m y y m T = *( E/W ) = ±% ! ( + c c ! " " c + . + $ 2 ' 2 2 + + 1 1 + alley x y y a x , y a y y m +( SE/ SW ) = ±( ! c + ) + c c ! " < c ! + , 2 2 /

Case 4: Major axis with non-steep slope (!1 < slope < 1,slope " 0 )

f x ,y 2( f 2 f 2) c(xc,yc) f x ,y 1( f 1 f 1)

figure 28

37 The ellipse with non-horizontal slopes becomes a scenic route. The shortest

1 distance to a point on the ellipse horizontally or vertically from a focus is (2a ! m ! n) . 2

This fact will be used in forming equations for ellipseT. For the north street we need the equation y equal to the height. y=the center y-coordinate plus half the vertical length m

1 plus the distance (2a ! m ! n) 2

1 1 y = y + m + (2a ! m ! n) c 2 2

1 1 1 y = y + m + a ! m ! n c 2 2 2

1 y = y + a ! n c 2

To find the northeast alley the following can be calculated.

y = mx + b General equation of a line.

y = 1x + b Substitute slope of alley, equal to one.

" 1 1 % " 1 % $ yc + m + (2a ! m ! n)' = !1$ xc + n' + b Substitute in one known point. # 2 2 & # 2 &

1 1 1 1 y + m + a ! m ! n + x + n = b Solve for b, the y-intercept c 2 2 2 c 2

yc + a + xc = b Simplify

y = !x + xc + yc + a Rewrite equation of a line.

Using similar calculations and the previous idea that the southern half of the ellipse is a reflection of the northern half, the following formula has been found.

38 ) " 1 % 1 1 - street x = ± a ! m + x , y ! m ( y ( y + m +( E/W ) $ ' c c c + + # 2 & 2 2 + + 1 1 + +(alley )y = ± !x + x + a + y , y + m < y < y + a ! n+ + NE / NW ( c ) c c 2 c 2 + (2.5) Ellipse * . T " 1 % 1 1 + street y = ± a ! n + y , x ! n ( x ( x + n + +( N / S ) #$ 2 &' c c 2 c 2 + + + + 1 1 + (alley )y = ± x ! x ! a + y , y ! a + n < y < y ! m + SE / SW ( c ) c c c + , 2 2 /

Example:

f2(3,2 5 ) 4

c(2,1)

-5 5 10 15

-2 f1(1,2-2 5 )

-4

-6 figure 29

When an ellipseT has a steep major axis the graph has more height than width, but the previous equations hold strong. Considering the above foci 1,2 ! 2 5 and 3,2 5 , the ( ) ( ) ellipse would contain center (2,1), and slope m = 4 5 ! 2 and n = 2 , we shall chose 2a for example to be 16 and our generic equation results in the following

39 ) " 1 % 1 1 - street x = ± 8 ! 4 5 ! 2 + 2,1! 4 5 ! 2 ( y ( 1+ 4 5 ! 2 +( E/W ) $ ( )' ( ) ( )+ + # 2 & 2 2 + + 1 1 + +(alley )y = ± x ! 2 + 8 +1,1 + 4 5 ! 2 < y < 1+ 8 ! 2 + + NE / NW ( ) 2( ) 2( ) + Ellipse * . T " 1 % 1 1 + street y = ± 8 ! 2 +1,2 ! 2 ( x ( 2 + 2 + +( N / S ) #$ 2 ( )&' 2( ) 2( ) + + + + 1 1 + (alley )y = ± x ! 2 ! 8 +1,1 !8 + 2 < y < 1! 4 5 ! 2 + SE / SW ( ) ( ) + , 2 2 ( ) /

# Street x = ± 9 ! 2 5 + 2,2 ! 2 5 " y " 2 5' %( E /W ) ( ) % %(Alley )y x 2 9,2 5 y 8 % % NE / NW = ±( ! ) + < < % EllipseT $ ( Street y 7 1,1 x 3 %( N / S ) = ± + " " % % % %(Alley ) y = ± x ! 2 ! 7,!6 < y < 2 ! 2 5 % & SE/ SW ( ) )

Resulting in the following graph of the ellipse.

(1,8) (3,8) 8

(1+2( 5 -4),2 5 ) (3-2( 5 -4),2 5 ) f2(3,2 5 ) 4

c(2,1)

-5 5 10 15

-2 (1+2( 5 -4),2-2 5 ) f1(1,2-2 5 ) (3-2( 5 -4),2-2 5 )

-4

(1,-6) (3,-6) -6

figure 30

There are more questions to be answered about ellipsesT. What would an ellipsoid look like? Its volume? Its surface area? Are there everyday objects that an ellipsoid

40 would mimic? Can the formulas be condensed into one formula that encompasses all ellipsesT? These and many other questions are yet to be answered.

Ellipse Formula Summary:

Circumference with respect to vertical distance m, horizontal distance n and length 2a.

(2.1) C = 8a ! 2n ! 2m

Area with respect to vertical distance m, horizontal distance n and length 2a.

1 (2.2) A = mn + (2a ! m ! n)(n + m)+ (2a ! m ! n)2 2

Ellipse with horizontal major axis with respect to vertical distance m, horizontal distance n,

length 2a and center (xc , yc ) .

Ellipse with vertical major axis with respect to vertical distance m, horizontal distance n,

length 2a and center (xc , yc ) .

Ellipse with non-steep or steep major axis with respect to vertical distance m, horizontal

distance n, length 2a and center (xc , yc ) .

41 ) " 1 % 1 1 - street x = ± a ! m + x , y ! m ( y ( y + m +( E/W ) $ ' c c c + + # 2 & 2 2 + + 1 1 + +(alley )y = ± !x + x + a + y , y + m < y < y + a ! n+ + NE / NW ( c ) c c 2 c 2 + (2.5) Ellipse * . T " 1 % 1 1 + street y = ± a ! n + y , x ! n ( x ( x + n + +( N / S ) #$ 2 &' c c 2 c 2 + + + + 1 1 + (alley )y = ± x ! x ! a + y , y ! a + n < y < y ! m + SE / SW ( c ) c c c + , 2 2 /

DESTINATION PARABOLAS

“ A parabola is the set of all points in a plane equidistant from a particular line

(the directrix) and a particular point (the focus) in the plane.” (Demana 634) We know how this appears measured in Euclidean geometry and we have equations for those parabolas however stretched, translated, reflected, etc.

(Encarta) figure 31

42 Since parabolas have neither finite circumference nor finite interior (Although the graph could be bound by an axis and/or other graphs) with which to calculate area the driver shall steer directly to equations.

Major Conjecture: There is a connection between geometry and algebra in the Taxicab

Geometry. Therefore, we should be able to write equations for parabolas.

Let us consider parabolas with the Taxicab metric. To begin the investigation various types of parabolas were sketched. These included parabolasT with directrices that are horizontal, vertical, have slopes equal to ±1or any of the infinite non-steep or steep slopes. These options can be further compounded by which side of the parabolaT the focus is located and how far the focus is located from the directrix. (This has potential to be an interesting classroom exercise in categorization.)

Since our graphs can be oriented many ways in the plane, the following notation will be used.

N Street Street W E

Focus Intersection Intersection S f Alley Alley Vertex

Directrix

figure 32

For all purposes of this trip, and ease to the passenger, the plane will continue to be oriented with north being in the direction of the positive y-axis (vertical), and east

43 being in the direction of the positive x-axis (horizontal). Horizontal and vertical lines will continue to be called streets and others alleys. The vertex will be where the parabola intersects the line the line that contains both the focus and the point on the directrix closest to the focus. Some initial observations include: The streets extend continuously to positive or negative infinity either horizontally or vertically. Some, but not all of the parabolas are symmetric with respect to the line through the focus vertex. Intersections and vertices are always horizontal or vertical from the focus. Many of the variations of parabolas are a simple rotation or reflection of others. It is these initial observations that will assist in mapping our trip through parabolas.

Although the number of parabolasT seems huge, the similarities are stronger than the differences. Let’s begin by taking a look at some parabolas categorized by their directrix’s slope. There is a graph and initial observations of each case to help get an overview of the situation.

Case 1: Horizontal directrix,

-5 5

-2

-4

-6

-8 figure 33

44 When the focus is plotted to the north the parabola opens up. The streets are parallel and go forever vertically to the north. The alleys are congruent and have slopes of

1 ± . When the focus is below the directrix a reflection over the directrix results. 2

Case 2: Vertical directrix

figure 34

The focus is plotted to the east, and the parabola opens easterly. Streets are again parallel, but horizontally. Slope of the alleys is ±2 , the reciprocal of the horizontal case.

This parabola is a rotation or reflection over the line y = x of case 1. The focus could also be plotted to the west and result in a reflection over the vertical directrix.

Case 3: Diagonal directrix

-5 5

-2

-4

45 figure 35 Graphed above with a slope of negative one and a northern focus an interesting change takes place. The streets are no longer parallel but perpendicular to one another.

Additionally, the arrows have merged into one straight segment, with a slope equal to that of the directrix. Reflections or 90° rotations would result in variations with southern foci and/or directrices with slope of ±1.

Case 4: Non-horizontal, non-vertical, non-diagonal directrix

2 f

5 10 15

-2

figure 36 Graphed is a parabola with a negative non-steep directrix. Once again positive, and/or steep slopes are only a reflection away. The streets have returned to their parallel state. The alleys now become a major point of interest as each has a different slope and length than its counterpart.

The cases 1 through 3 are the basis for understanding parabolas with the Taxicab metric. (Those three cases were where my investigation began and would be an excellent place to begin investigation at the high school level. Those three cases contain a variety of difficulties in the parabola’s forms. They lend themselves to problem solving for a

46 varied knowledge base. The difficulty can then be raised to case 4) For the purposes of this paper we will narrow our scope to an in-depth investigation of parabolas with directrices being non-steep slope and the focus above that given line.

Investigation of case 4: The symmetry has fled the plane.

First, lets consider a formula for the vertex. Let the focus be x , y and the ( f f ) distance from the focus to the directrix be called d. In Euclidean geometry we had the steadfast rule to find the distance which is to be the shortest possible: “The distance from a line to a point not on the line is the length of the segment perpendicular to the line from the point.” (Glencoe 159) In Taxicab geometry the shortest distance from a line to a point not on the line is the length of the segment from the point to the line vertically if the line is non-steep, horizontally if the line is steep and either horizontal or vertical, since the distance will be equal, when the line has a slope of ±1. For special cases of finding distance to a horizontal line from a point not on the line, we use vertical distance, and in the case of a vertical line we horizontal distance. We have limited our scope to non-steep directrices, Therefore the vertex is half the distance d south of the focus.

Conjecture: For parabolas where -1directrix slope<1, and focus is above the directrix,

! 1 $ the coordinates of the vertex are x , y - d . "# f f 2 %&

Second, lets consider a formula for slope of the alleys. During the investigation into slope, it became apparent that no matter how the construction of the parabola proceeded, the triangle formed by the alleys and the segment between the intersections were similar for the same directrix. The movement of the focus dilated the triangle (larger triangle for greater d), but as long as the slope of the directrix does not change, neither do

47 the angles of the triangle, the orientation of the angles in the plane or the slope of the alleys.

Conjecture: The triangle formed by the vertex and intersection of streets and alleys are similar for all directrices with equal slope.

This observation allowed the driver to look at various cases of directrix slope and

!1 !1 generalize. (full chart at end of paper) The investigation began with the slopes , , 2 3

!1 !1 , and . Once the numbers were investigated and written in a similar fashion, a 4 5 pattern emerged. A generalization has been written in the bottom row.

Directrix slope Alley mw Alley me

!1 !3 1

2 4 4 !1 !2 1

3 3 3 !1 !5 3

4 8 8 !1 !3 2

5 5 5 m x y

n n n chart 2

Conjecture: For parabolas with directix slope ! ±1, the slopes of the alleys sum to the slope of the directrix.

48 x y m The following equations have been formulated + = , and x + y = m . It also n n n appeared that x+y=n. With patterns emerging and further investigation, the following equations for slope emerged.

m ! n (3.1) M w = 2n

n + m (3.2) Me = 2n

These slope equations worked well for directrices with rational slopes. For a final step and confidence in the work thus far, an additional case shall be considered. What if we have irrational slope?

2 Irrational example: Directrix with slope 4

2 12 12 ,0 ,0 ( 2-4 ) ( 4+ 2 ) f

-10 -5 5 10

-3 0, -2 ( 2 )

2 m= -4 4

figure 37

49 1 m ! n 1 m ! n Alleyw = (slopew )x + y f ! d ! (slopew)x f = ( )x + y f ! d ! ( )xf = 2 2n 2 2n

2 ! 4 1 2 ! 4 2 ! 4 3 ( )x + 0 ! (3) ! ( )(0) =( )x ! 2(4) 2 2(4) 8 2

similarly,

1 n + m 1 n + m Alleye = (slopee )x + y f ! d ! (slopee )x f = ( )x + y f ! d ! ( )x f = 2 2n 2 2n

4 + 2 1 4 + 2 4 + 2 3 ( )x + 0 ! (3) ! ( )(0) =( )x ! 2(4) 2 2(4) 8 2

SUCCESS! Slope has been found! The driver is happy to not be driving the wrong direction down a one-way street.

Third, find the intersections of streets and alleys. Intersections are horizontal from the focus, the investigation continued into the distance measured from the focus to the intersections,

Directrix Distance Distance Distance Distance slope d focuswest focuseast wfe !1 1 2 2 8

2 1 3 1 3 !1 4 8 8 32

2 1 3 1 3 !1 1 3 3 9

3 1 4 2 4 !1 4 12 12 36

3 1 4 2 4 !2 1 4 4 32

8 1 5 3 15 !1 2 10 10 25

5 1d 6 4 6 m d dn dn dn2

n n +1 n !1 n +1 chart 3

50 It is important to note, as this is a journal of investigation, that the charts have been written in a form conducive to seeing the desired pattern. The original charts were neither neat and simple, nor free from errors resulting in often great struggles to find patterns. (One of the things I liked best was the graphs of Taxicab conics strongly lead one to believe that there must be a pattern. If students think there is no relationship, they will discontinue the search.) It never crossed my mind that a pattern was not hidden in the problem.

Intersections have the following formulas

" dn % intersectionW = $ xf ! , yf ' # n +1 &

" dn % intersectionE = $ xf + , yf ' # n !1 &

We know have everything required to complete the task at hand.

Fourth, write formulas for the parabola

1 Alleys = (slope)x + yf ! d ! (slope)xf 2

Note that since it accounts for the varied slope, the above equation should hold for both west and east alleys. In addition, this generalization holds for the horizontal directrix, allowing it to be good for slopes between ±1.

Using our discovery of intersections to determine the valid domains of the alleys and equations for the streets. The final formula for a Taxicab parabola with a non-steep slope and focus above the directrix is as follows.

51 * " m! n% 1 " m! n% dn . Alley y = x + y ! d ! x , x ! ( x ( x ,( w ) $ ' f $ ' f f f , , # 2n & 2 # 2n & n +1 , , " m + n% 1 " m + n% dn , , Alley y = x + y ! d ! x , x ( x ( x + , ,( e ) #$ 2n &' f 2 #$ 2n &' f f f n !1 , (3.3) ParabolaT = + / , dn , , Streetw x = x f ! , y f ( y < ) , ( ) n +1 , , , dn , Street x = x + , y ( y < ) ,( E ) f f , - n !1 0

Though the trip could continue with multiple variations of more specific equations, we shall brake here. Perhaps a more interesting question is, can a formula be derived that encompasses all parabolas? We shall leave that question in the rear view mirror and turn the corner to our next destination.

Formula Summary:

Slope of the west alley of a non-steep parabola with vertex to the north.

m ! n (3.1) M w = 2n

Slope of the east alley of a non-steep parabola with vertex to the north.

n + m (3.2) Me = 2n

m Parabola with directrix with non-steep slope ± , distance d from the directrix to the focus n

(x f ,y f ) .

52 * " m! n% 1 " m! n% dn . Alley y = x + y ! d ! x , x ! ( x ( x ,( w ) $ ' f $ ' f f f , , # 2n & 2 # 2n & n +1 , , " m + n% 1 " m + n% dn , , Alley y = x + y ! d ! x , x ( x ( x + , ,( e ) #$ 2n &' f 2 #$ 2n &' f f f n !1 , (3.3) ParabolaT = + / , dn , , Streetw x = x f ! , y f ( y < ) , ( ) n +1 , , , dn , Street x = x + , y ( y < ) ,( E ) f f , - n !1 0

DESTINATION HYPERBOLAS

“ A hyperbola is the set of all points in a plane whose distances from two fixed points in a plane have a constant difference. The fixed points are called the foci of the hyperbola.

The line through the foci is the focal axis. The point on the focal axis midway between the foci is the center. The points where the hyperbola intersects its focal axis are the vertices of the hyperbola.” (Demana 656) With the Euclidean distance formula the graph of an ellipse looks so.

figure 38

53 With the Taxicab distance formula the following graph is created.

N street

W E

focus S street center alley alley street focus

street

figure 39

Again, for the sake of ease we shall call horizontal and vertical rays of the hyperbola, streets, while non-horizontal and non-vertical segments shall be called alleys.

Intersections occur at turns. The difference in the distance from any point on the hyperbola to each focus shall be a > 0 .

Case 1: Center the origin. Foci horizontal

xf ,yf xf ,yf ( W W) ( W W)

figure 40

54 Visually, and at this point in the trip, it makes sense that the hyperbola has become two vertical lines. The desired difference from focus to focus occurs somewhere between the foci (the vertex). Since the foci are horizontal, any movement vertically then adds to both distances. Focus one –focus two = difference becomes focus one +constant –

(focus two +constant) =difference. With the difference and m=0 Equations for the situation with W representing the west line and E representing the east are as follows

" 1 & $xw = x f + (n ! a)$ $ W 2 $ (4.1) HyperbolaT # ' $ 1 $ xE = x f ! (n ! a) %$ E 2 ($

Case 2: focal axis vertical

(x fN,yfN)

(x fS,yfS)

figure 41

Similarly, the equations for a vertical focal axis are as follows

" 1 & $yN = y f ! (m ! a)$ $ N 2 $ (4.2) HyperbolaT # ' $ 1 $ yS = y f + (m ! a) %$ S 2 ($

55 Case 3: focal axis with slope of ±1

(x fN ,yfN )

(x fS ,yfS )

figure 42

The slopes of the alleys are –1 the negative reciprocal of the focal axis slope. The intersections occur horizontally or vertically from the foci. From a point of intersection to

! m + n + d $ the nearest focus, the distance # & . With slope and the intersections found the " 2 % equation for a hyperbola with focal axis slope of –1, and difference in distances d is as follows.

* ! m + n + d $ ! m + n + d $ . (Alley )y = x + y + x + , x ' x ' x + S fS fS # & fS fS # & , " 2 % " 2 % , , , , ! m + n + d $ ! m + n + d $ , (AlleyN )y = x + yf N + xfN ( # &, x fN ( # & ' x ' x fN , " 2 % " 2 % , , , , ! m + n + d $ , (Street )y = y ( ,x < x < ) , E fE # & fN , , " 2 % , (4.3) HyperbolaT + / ! m + n + d $ ,(Street )x = x ( ,y < y < ) , , N f N "# 2 %& fN , , , ! m + n + d $ ,(Street )y = y + ,) < x < x , W fS # & fS , " 2 % , , , , ! m + n + d $ , (Street S )x = yfS + # & ,) < y < yfS -, " 2 % 0,

56 Case 4: non-steep focal axis

xf ,yf ( 2 2)

(xc,yc)

xf ,yf ( 1 1)

figure 43

The alleys of the hyperbola have slopes of ±1. Additionally the intersections remain horizontal or vertical from their respective foci. Can the previous formulas be used to find a new equation for this specific hyperbola? Will the previous equation work as it is written? What happens if the hyperbola is rotated around the focal axis? These and many other questions remain to be answered.

Though the road trip in conics may never be complete, this leg of the journey is.

How many more trips need be taken to truly understand Taxicab conics has yet to be seen. Though this is the passenger’s stop, the driver will continue.

Formula Summary:

Hyperbola with horizontal focal axis with respect to horizontal distance n between foci

x , y and x , y ( fW fW ) ( fE fE )

57 " 1 & $xw = x f + (n ! a)$ $ W 2 $ (4.1) HyperbolaT # ' $ 1 $ xE = x f ! (n ! a) %$ E 2 ($

Hyperbola with vertical focal axis with respect to horizontal distance n between foci

x , y and x , y ( fW fW ) ( fE fE )

" 1 & $yN = y f ! (m ! a)$ $ N 2 $ (4.2) HyperbolaT # ' $ 1 $ yS = y f + (m ! a) %$ S 2 ($

Hyperbola with a focal axis of –1 with respect to difference d, horizontal distance n and vertical distance m between the foci x , y and x ,y . ( fS fS ) ( fN fN )

58 Conclusion

As I read back through my journey, I must admit, a lot of the glow has worn off.

Showing a friend pictures of a vacation, does capture the excitement of the trip.

I wish the reader could have been present when I found the slope for non-steep alleys of a parabola. I almost laugh now thinking about it. I jumped up from my chair,

(where I likely had been mumbling to myself for hours.) and began to do a bit of a dance down the hallway, into my son’s room, singing a little song that could have been entitled

“I got the sloooooo-ope.” My son looked at me like I was crazy and yet I continue my jig.

(Fortunately for him he did not have company over.)

If a student of mine feels this elation in solving a problem in my class, I feel the problem has been successful. One feels the most pride and success from a struggle that has been overcome. Too often students want us to “just tell them how to do it.” Although this is an option to solve a problem it is not “problem solving.”

Students (and teachers) often do not have the skills, experience and confidence it takes to be great problem solvers. It takes an understanding of various strategies that can be employed in diverse situations. It requires running into roadblocks and having the confidence and foresight to continue. Problem solving requires the opportunity to problem solve. Teachers have the ability to offer that opportunity in a safe environment with support. It is with that opportunity that skills and confidence will emerge.

Many students have spent the majority of their mathematical education being taught rules and then practicing said rules. I was one of those students all the way through my undergraduate degree. It was not until the opportunities given to me in my graduate studies that I can truly say I began to problem solve mathematically. Taxicab geometry

59 has been a chance for me personally to feel what it is like to create math. I knew I could memorize it, repeat it, learn it, etc. But, it took a different part of me to discover all that I did in this paper. As I mentioned in the introduction, this work is mine. You see the path that succeeded written here, what you don’t see is all the wrong ideas, and paths I went down. It was on those wrong roads that I improved my problem solving skills. It was also on those wrong paths that I began to understand mathematics so much better.

I can honestly say that my understanding of the basic structure of geometry was sadly very weak when I was introduced to the Taxicab metric. I was stunned by my first sketch of a Taxicab circle. That’s not right at all! Was perhaps my first gut instinct. Even a bit of fear. Don’t tell me everything I know is wrong! My love with mathematics was based in part on my impression of its concrete nature. My first Taxicab parabola was a disaster because I was still under the belief that the shortest distance from a line to a point not on the line was perpendicular. That’s not right at all! As I have traveled through

Taxicab geometry I have had to stop and reevaluate everything. How can you not understand Euclidean Geometry better with that type of comparing and contrasting. My geometric understanding has grown beyond belief and the great thing is, now that the seed is planted, it is still growing.

I promised one of my geometry classes that I would show them some of what I had been working on for this paper before the end of the year. When the time came about a month before the end of the school year, I was very anxious that showing them this new metric, for fear that it would confuse them. Not only did it not confuse them, the class ended up having an incredible discussion on congruency. This was not the direction that I had expected. It began with kids coming up with all sorts of ideas. Soon a student came

60 up and sketched a triangle that had all sides with equal length as measured with the

Taxicab metric. Then the argument came to whether or not that was an equilateral triangle. Soon kids were quickly looking up definitions in their books, trying to defend their position. They quickly found the sides needed to be congruent. Back to the books they went. We ran out of time that day and I agreed that we would come back to the discussion the next day. So, they were told, they should be prepared to defend their positions. Defend their positions, they did. The conversation continued the next day and got brought up at various times during the rest of the school year. Taxicab geometry has this awesome property of being catchy. To my surprise, it never seemed to confuse them at all, just the opposite, it seemed to clarify.

It takes a certain level of mathematical strength and confidence to be comfortable in taking a new path or in allowing students to travel down their own path and take you on their trip rather than the other way around. Taxicab geometry has the ability to encourage problem solving. Create a deeper understanding of the structure of mathematics (geometry). Allow students to use the language the have acquired in mathematics to communicate their personal ideas. Broaden their view of what math is and what it can do..

In closing let me just say, for yourself and your students,“Take the road less traveled.”

JAUNT 1: METRICS

The Taxicab metric is only one metric in a sea of metrics. If the Taxicab metric can be defined to measure only horizontally and vertically, and the Euclidean metric is as the crow flies, there must be other metrics. (Students may not realize that we often use other

61 metrics everyday. For example, the distance d(P, Q) = Q ! P for real numbers on the number line. Yes this could be considered a subset of the Euclidean metric, but it is its own metric, in its own space.) Indeed there are many others. Including one tied to a game, and one with connections to computer science.

Consider the game of chess. Each piece moves according to a specific rule. The rook for example moves somewhat like the taxicab metric, forwards or backwards, right or left, essentially only at right angles. Bishops can only move diagonally. With this movement there are actually spaces on the board that a bishop cannot reach. For example a bishop on black squares can’t occupy a white square. The king can move one space per turn onto any square adjacent his position. It is the moves of the king that that are related to a metric called the Chebyshev distance, or chessboard distance. “ In mathematics the

Chebyshev distance is a metric defined on vector space where the distance between two vectors is the greatest of their differences along any coordinate dimension.” Interestingly,

Chebyshev circles are also squares, similar to Taxicab circles, but the sides lay parallel to the axes.

The Levenshtein (character edit) distance, measures the dissimilarity of two strings P and Q. distance is considered to be the number of deletions, substitutions, or insertions that area needed to transform on string into the other. (Gilleland)

An example of this distance would be

if P = (word ) and Q = (word) , Then d(P, Q) = 0 .

This is due to the fact that they are the same word . Therefore no deletions, substitutions or insertions are necessary. While

if P = (word ) and Q = (work) , Then d(P, Q) = 1

62 It will take the one substitution of the letter d for the letter k to make the words identical.

As the distance in strings increase the possibilities also increase. For example, consider the following. Let P and Q be actual words in the English dictionary.

If P = ( pot ) and, d(P, Q) = 1 , then what is Q ?

Option: Let the first letter be incorrect. Some possible answers could be tot, lot, or dot.

Option: Let the middle letter be incorrect. Some possible answers could be put, pit or pat.

Option: Let the last letter be incorrect. Some possible answers could be pod, pog, or pop.

This is not all the options available, you must also consider the options of adding or deleting a letter. For example, Q could be post or spot.

This seems like a lot of options but manageable. Now consider the following.

If P = ( pot ) and, d(P, Q) = 2 , then what is Q ?

The possibilities just became enormous.

(I chose this metric due to my belief that it has the possibility of being very strong for the classroom. It is easily understood. Can be made simple or challenging for advanced students. It could be used in combination with matrices as a way of organizing data. Additionally it has multiple real life applications to interest students in our technological based world.)

This metric has been used in applications such as spell checking, speech recognition software, DNA analysis and in the software used to detect plagiarism.

(Gilleland)

(There are many metric mentioned in mathematics literature. Consider finding ones that will interest your students or fits with your curriculum.)

63 JAUNT 2: THE TRIANGLE INEQUALITY THEOREM

(Euclidean metric)

C B

Given: !ABC , Prove: The Triangle Inequality Theorem AB < AC + BC

To begin we shall construct CD so that AC ! CD . Consider D to be a point collinear to points B and C and C shall be between B and D . AD can also be constructed because any two points determines a line.

C D B

!ACD is an isosceles triangle, therefore < CAD !< CDA (base angles) and their measures are equal. m < BAC + m < CAD = m < BAD , by the angle addition postulate. Substituting the equivalent angle measure of < ADC for that of < CAD , results in m < BAC + m < ADC = m < BAD . Because it takes the positive m < BAC to make m < ADC equal to m < BAD it follows that m < ADC < m < BAD (definition of inequality). Longer sides of triangle are across from larger angles therefore, AB < BD .

Using the segment addition postulate BC + CD = BD and therefore AB < BC + CD . A final substitution and AB < BC + AC .

64 JAUNT 3: RIGHT TRIANGLES

Any right triangle with a leg oriented horizontally using the Taxicab distance formula will satisfy the following equation.

a+b=c

This is based on the fact that the legs are both horizontally and vertically oriented and so the measurement of the hypotenuse can be measured along the path of the legs.

A right triangle with a horizontal hypotenuse has the following properties

i) A short leg with vertical distance m and horizontal distance n has

a length m + n .

ii) The long leg with respect to the short leg has a length

2 ! m$ ! m$ m + mn # & m + # & n or equivalently " n % " n % n

iii) The hypotenuse with respect to the short leg has a length

2 2 ! m$ n + m n + # & m or equivalently " n % n

Similar to special Euclidean triangles, the length of the other two legs can be determined by the short leg. These formulas could also be rewritten in terms of any side of a right triangle with the hypotenuse horizontal, or be rewritten in terms of the hypotenuse being vertically oriented.

JAUNT 4: CONGRUENCY

Congruent (!) : Having exactly the same size and shape. (Geometry to go 450) In

Euclidean geometry it is enough to know that segments have the same length.

65 AB = CD iff AB ! CD . This does not hold true in Taxicab geometry. The following are examples of segments with a Taxicab distance of ten.

-10 -5 5 10

-2

figure 45

Although they all have equal lengths, the shape of each segment is not the same. Hence, in Taxicab geometry the size and shape must both be taken into consideration. For segmentsT to be congruent the following must be true

x ! x and y ! y are equal. P Q P Q

This satisfies that the segmentsT are the same shape and also that the addition of the two for d P,Q = x ! x + y ! y will also be true resulting in equal distance. (This was T ( ) P Q P Q a great revelation to my students as to why congruency and equal lengths are not the same thing.)

JAUNT 5: RIGID MOTION

Area does not change with rigid motion. “The rigid motions (congruent transformations) are transformations in space or in a plane, in which the original figure and its image are congruent.” (Newman 374)

66 For Euclidean geometry in a plane these include

1. Parallel displacement or translation

2. Rotation about a point

3. Transformations obtained by successive translations and rotations

Adding in rigid motion is space

1. Parallel displacement or translation

2. Rotation about a straight line

3. Transformations obtained by successive applications of transformations of

types 1, 2. (Newman 375)

(In more k-12 classroom friendly terms: a slide, flip or turn or similarly a

translation, reflection or rotation.) How do these isometries apply in Taxicab

geometry?

A Translation or slide preserves congruency. Since the segments in a

figure slopes remain fixed with respect to the axes. Below is an example of a

translation seven horizontal and four vertical.

6 A'

2 C' A B'

-5 5

C -2

-4

figure 46

67 A reflection in Taxicab geometry is another story

2 A A'

-5 5

C C' -2

B B'

-4 figure 47

It is true that the image of a figure reflected across a horizontal or vertical line remains congruent to the original figure (figure 47). While the reflection of an image across a non-horizontal or non-vertical line may not be congruent. When the angle of segments in

Taxicab geometry are changed we have the possibility of changing the distance between points. The length of a segment is dependent upon its slope and placement in the plane.

4 A'

2 A B'

-5 5 C' C -2

B -4

figure 48

A rotation does not preserve congruency (unless though a multiple of right angles). Although a rotation preserves the shape of segments, when the slope of a segment changes the length also changes.

68 JAUNT 6: QUADRATICS

Euclidean Quadratics can be looked at in the following flow chart.

Quadralateral

Parallelograms Trapezoids

Rectangles Rhombi Isosceles Trapezoid

Square

figure 49

Let us consider each in turn.

“Quadrilateral: A polygon with four sides. (Geometry to go 156)

The definition of a polygon is “ A closed figure formed by a finite number of coplanar segments called sides such that the sides that have a common endpoint are non- collinear and each side intersects exactly two other sided, but only at their endpoints.”

(Glencoe R22). There does not appear to be any discrepancies in the definition of a polygon to cause difficulties. Distance is not a factor and therefore quadrilaterals are the same in Euclidean and Taxicab geometry.

“Trapezoid: A quadrilateral with exactly one pair of parallel sides” (Geometry to go 165)

What is parallel? Parallel lines are coplanar and never intersect. This is dependent upon slope and slope is calculated with horizontal and vertical components and is equivalent in both of our geometries.

“Isosceles Trapezoid: A trapezoid with congruent legs.” (Geometry to go 134)

69 Once again the important distinction of congruency has saved the day. If the definition of isosceles trapezoid stated that the length of the legs were equal, we would have a difference (figure 50) , but fortunately our traditional notion holds strong.

figure 50

“Parallelogram: A quadrilateral in which both pairs of opposite sides are parallel”

(Geometry to go 157)

Again the definition of parallel has to do with placement in the plane which holds true for Taxicab metric.

“Rectangle: A quadrilateral with four right angles.” (Glencoe R22)

In the original definition of Taxicab geometry, we stated that angles are measured the same in both Euclidean and Taxicab geometry.

Rhombi: A quadrilateral with four congruent sides.” (Glencoe R24)

Congruency requires a same size and shape. Our definition and original concept of rhombi holds true.

“Square: A quadrilateral with four right angles and four congruent sides.” (Glencoe R26)

Since both rectangle and rhombi hold strong the combination of the two also does.

We must add an addition to our list of quadrilaterals. A circle.

A circle is a quadrilateral with four right angles and four congruent sides with vertices horizontal and vertical from the center. Our flow chart now appears thus.

70 Quadralateral

Parallelograms Trapezoids

Rectangles Rhombi Isosceles Trapezoid

Square

Circle

figure 51

JAUNT 7: AREA LIMITS

Partitioning and limits allow the area of various figures to be determined. A simple figure such as a triangle in the first quadrant can be partitioned into rectangles of equal width and varied height.

figure 52

71 Note that although the triangle’s side is neither horizontal nor vertical the rectangles sides are. The area for the sum of rectangles can be written as

!(height)(width)

n " f (mi )!x i=1

With a and b the endpoints of width and n the number of partitions,

(b ! a)(i !1) th m = a + is the right endpoint of the i rectangle, i n

" (b ! a)(i !1)% f a + is the height of each rectangle and, #$ n &'

b " a !x = is the width of each rectangle. n

Our formula now becomes

n " (b ! a)(i !1)% b ! a ( f $a + ' i=1 # n & n

!3 For our specific example with f (x) = x + 6 that sum would be 2

n " !3" (4 ! 0)(i !1)% % ($ $ 0 + ' + 6' (1) i=1 # 2 # 4 & &

n " !3 % ($ (i !1) + 6' = i=1 # 2 &

" !3 % " !3 % " !3 % " !3 % $ (1! 1) + 6' + $ (2 ! 1) + 6' + $ (3 !1) + 6' + $ (4 !1) + 6' = # 2 & # 2 & # 2 & # 2 &

This can be adjusted to account for rectangles of smaller and smaller width. For example allowing the partitions to increase to sixteen changes the figure to

72 6

figure 53

The area of the rectangles approaches the area of the figure itself. As the number of partitions increases to infinity, the area of the rectangles becomes equal to that of the triangle.

Area of congruent figures does not change between Euclidean distance formula and Taxicab distance as the limit of the area is equal.

JAUNT 8: HIGHSCHOOL

The high school classroom is filled with individuals with varying mathematical strengths. The topics touched on in a simple idea such as the parabola has far reaching notions. This type of activity has the ability to reach students of differing abilities and strengths. Based on the level of students this could be an open-ended activity, as is being performed here, or it could be carefully directed for the individual needs of the student.

This has the potential to be a strong group activity. If each student creates one or two parabolas, the charts for an in-depth search could come together quickly with everyone’s participation. Other topics that could be investigated:

73 Conics are results of a plane intersecting with a cone. Is conics still a reasonable

name with the Taxicab metric?

Is there an object that the Taxicab conics (if they may be called that) are sliced out

of with a plane rather than a cone?

What if we applied the Taxicab metric on sphere? (driving on the earth)

Consider what arc length would mean on a Taxicab circle or ellipse.

What would a Taxicab ellipsoid, look like? What about its volume? What about

Surface area?

What if you can’t cut between blocks? (In other words you are restricted to unit

roads. Where should road blocks be set up in a 5 mile perimeter?)

Could a tool be written in Geometers sketchpad for Taxicab distance?

Can we find a strategy for finding the area under a Taxicab parabola?

(Integration?)

Could we convert our ideas/equations into polar coordinates?

Do the Congruency Theorems for triangles still hold with the Taxicab metric?

(SSS, SAS?)

The questions and expansion seems endless. These and numerous other ideas could be addressed in the secondary classroom. Though the driver herself does not have the answers to all of these questions, as was explained in the beginning, it is an investigation and a journey, as it should be for the students. We can learn as much from our students as they do from us, if we all just open our minds.

JAUNT 9: ACTIVITIES

74 Metrics Name______

1. Sketch points P(5, 4) and Q(1,2) . 6

-10 -5 5 10

-2

-4

2. Find the distance (Euclidean) betw-6een points.

3. Find the distance between the points if you can only move horizontally and/or vertically.

4. Give an example/describe a situation when measurement would be more accurate using problem three’s measurement?

5. Give an example of when measurement would be more accurate using problem two’s measurement?

6. If one point is (x1,y1 ) and the other is (x2 , y2 ) . Write a formula for your new distance formula measuring only horizontally and/or vertically.

7. Use your formula to find the distance between points R(-2,4) and T(6,-5). Verify your answer graphically.

75 8. Seth and Thad are looking for an apartment within walking distance of Seth’s job and Thad’s college. Let the coordinate for Seth’s job be J(6,2) and the coordinate for Thad’s college be C(-2,-4). They would like to find an apartment where both walk an equal distance that is as short as possible. Graph the possibilities (Using horiz/vert distance).

-5 5 10

-2

-4

9. After a week of not finding a place to live, Seth and Thad decide to extend their search. They still want to have equal walking distance for each of them, but will not require a minimum distance to each destination. Graph the possibilities.

-5 5 10

-2

-4

76 10. Since Thad has classes morning and evening he will have to walk from home to classes more than once a day. Seth says they should still get an apartment as close as possible to both work and school, but that he would be willing to walk further than Thad, as he only walks to work once a day. Graph the possibilities.

-5 5 10

-2

-4

11. Assume for this final situation, that Seth and Thad can walk directly to work or school, without sticking to horizontal or vertical paths. How would this change problem 10? Graph the possibilities.

-5 5 10

-2

-4

77 Taxicab metric Name______

Let all work use the Taxicab distance formula dT = x2 ! x1 + y2 ! y1

1. Plot G(-4, 1). Plot all points that are a distance of 4 units from G. What type of figure is it?

-10 -5 5 10

-2

-4

-6 2. Plot the points A(1,1), B(5,1), C(5,5) and D(1,5), What kind of figure is it?

3. Write an equation for perimeter of problem 2 with respect to its side length.

4. Write two equation for perimeter of problem 1, one with respect to its side and one w.r.t. its radius.

5. Write an equation for area of problem 2 with respect to its side.

6. Write two equation for area of problem 1, one with respect to its side and one w.r.t. its radius.

78 7. A builder wants to construct a gas station within six blocks of the school and within four blocks of the grocery store. If the school is at S(-3,0) and the grocery store is at G(2,2). Graph the area in which the builder should locate the station.

-10 -5 5 10

-2

-4

-6

8. There are three schools in a large city. Larson school is located at L(-4,3), Clark School at C(2,1) and Swanson school at S(-1,-5). Graph in school boundaries such that each student is attending the closest school to their home. 6

-10 -5 5 10

-2

-4

-6

79 9. An additional school (Good School) is built at G(2,5). Redraw the school boundaries.

-10 -5 5 10

-2

-4

-6

10. What type of 3 dimensional figure is created if you spin a circle on its diameter? graph the possibilities.

80 Taxicab Parabolas Name______

2 2 d E = (x2 ! x1) + (y2 ! y1 ) and dT = x2 ! x1 + y2 ! y1

1. Prove d E ! dT analytically. Explain when it is less than, and when it is equal.

2. A parabola is created graphically according to distance. A parabola is all points equidistant from a line called the directrix and a point called the focus. Graph A

parabola using d E given directrix y=-4 and focus f(0,0).

-5 5 10

-2

-4

3. Write an equation for the above parabola.

81 4. Graph A parabola using dT given directrix y=-4 and focus f(0,0).

-5 5 10

-2

-4

5. Write an equation for the parabola.

6. Graph A parabola using dT given directrix y=-x and focus f(2,2).

-5 5 10

-2

-4

7. Write an equation for the parabola

82 8. Investigate triangles below. Choose at least two topics. Some topics may be, but are not limited to, equilateral, equiangular, isosceles, and/or right triangles, perimeter, area, special formulas…

-5 5 10

-2

-4

9. Results? Equations? Further ideas for investigation?

Homework adapted and aided by Taxicab Geometry by Eugene F Krause.

83 CHARTS

2a=sum distance from any point on an ellipse to each focus. d=distanceT between foci. m=the vertical distance between foci. n=the horizontal distance between foci.. l=the distanceT between the foci and the nearest point P on the ellipse. (also may

referred to as the length of the leg of isosceles triangle)

! = the angle to the first vertex counterclockwise from the positive x-axis

! = the angle to the second vertex counterclockwise from the positive x-axis

Ellipse chart

2a d m n m/n l ! !

4 2 0 2 0 1 NA NA

6 4 0 4 0 1 NA NA

8 6 0 6 0 1 NA NA

6 2 0 1 0 2 NA NA

84 6 4 0 4 0 1 NA NA

8 4 2 2 2 2 " 1% " 3% tan!1 tan!1 2 #$ 3&' #$ 1&'

12 9 3 3 3 2 " 2% " 4% tan!1 tan!1 3 #$ 4&' #$ 2&'

16 12 6 6 6 4 " 3% " 5% tan!1 tan!1 6 #$ 5&' #$ 3&'

11 7 3 4 3 2 " 1% " 2% tan!1 tan!1 4 #$ 2&' #$ 1&'

9 5 4 1 4 2 " 2 % " 4 % tan!1 tan!1 1 #$ 2.5&' #$ .5&'

15 7 6 1 6 4 " .5% " 4.5% tan!1 tan!1 1 #$ 7 &' #$ 3 &'

10 6 2 4 2 2 " 1% " 3% tan!1 tan!1 4 #$ 4&' #$ 2&'

12 8 2 6 2 2 " 1% " 3% tan!1 tan!1 6 #$ 5&' #$ 3&'

85 18 10 2 8 2 4 " 1% " 5% tan!1 tan!1 8 #$ 8&' #$ 4&'

24 16 8 8 8 4 " 4% " 8% tan!1 tan!1 8 #$ 8&' #$ 4&'

20 12 4 8 4 4 " 2% " 6% tan!1 tan!1 8 #$ 8&' #$ 4&'

Parabola chart

Directrix Distance Alley Alley Distance Distance Distance slope fdx mw me fw fe wfe !1 1 !3 1 2 2 8

2 1 4 4 3 1 3 !1 2 !3 1 4 4 16

2 1 4 4 3 1 3 !1 3 !3 1 6 6 24

2 1 4 4 3 1 3 !1 4 !3 1 8 8 32

2 1 4 4 3 1 3 !1 1 !2 1 3 3 9

3 1 3 3 4 2 4 !1 2 !2 1 6 6 18

3 1 3 3 4 2 4 !1 3 !2 1 9 9 27

3 1 3 3 4 2 4 !1 4 !2 1 12 12 36

3 1 3 3 4 2 4 !2 1 !5 3 4 4 32

8 1 8 8 5 3 15 !2 2 !5 3 8 8 64

8 1 8 8 5 3 15

86 !2 3 !5 3 24 24 32

8 1 8 8 10 6 5 !1 1 !3 2 5 5 25

5 1 5 5 6 4 12 !1 2 !3 2 10 10 25

5 d 1 5 5 6 4 6 m x y dn dn dn2

n n n n +1 n !1 n +1

87 BIBLIOGRAPHY

Demana, Franklin D. Precalculus: Graphical, Numerical, Algebraic, Addison Wesley, Boston. 2007.

Encarta. Microsoft Corportation. Redmond, WA 1993-2003.

Geometry. Glencoe Mathematics. New York, New York. 2005.

Geometry To Go. Great Source Education Group. Houghton Mifflin Company. Wilmington MA. 2001.

Gilleland, Michael. Levenshtein Distance, in Three Flavors. Merriam Park Software.

Krause, Eugene F. TAXICAB GEOMETRY, An adventure n Non-Euclidean Geometry. Dover Publications, Inc. New York. 1975.

Larson Roland E. Calculus with Analytic Geometry. Sixth edition. Houghton Mifflin Co. New York. 1998.

Newman, James R. The Universal Encyclopedia of Mathematics. Simon and Schuster. New York. 1964.

Reinhardt, Chip. Taxi Cab Geometry: History and applications. The Montana.

Webster,s New World Dictionary. Third College Edition. Prentice Hall. New York. 1991.

Wikipedia, The free encyclopedia. Metric Space. May 30, 2007.

88 What people are saying about Taxicab Geometry

“Spectacular! The ride of a lifetime!”

Tara Clark S.P.H., O.H.S.

The immortal Lou Erickson once said, “Life is like a taxi. The meter just keeps a ticking

whether you are getting somewhere or just standing still.”

Michael Good, A.O.K., O.H.S.

“What is said in Calculus, stays in Calculus!”

“You don’t want to miss this ride.”

Stephanie Haberer

“ The road to hell is thick with Taxicabs.”

Don Herald (U.S. Cartoonist and humorist 1889-1966)