Revised 2011

Taxi Cab Geometry with Technology:

Some Exploration Materials

by

Jim Wilson

Overview of Taxi Cab Geometry

Taxi Cab Geometry has the following distance function between points A(x1,y1) and B(x2,y2):

D = |x2 - x1| + |y2 - y1|

The claim is made that all of axioms and theorems in Neutral Geometry (Chapter 1) up to the SAS congruence will hold.

Construct Graphing Calculator and/or GSP files (or GeoGebra files) for

Taxi Cab Circle

Graphing Calculator 3.5 File for center A and radius d.

|x - a| + |y - b| = d

Graphing Calculator 3.5 File for center A through B

|x - a| + |y - b| = |g - a| + |h - b|

GSP File for center A through B

If A(a,b) is the origin (0,0), the the equation of the taxicab circle is |x| + |y| = d.

In particular the equation of the Taxicab Unit Circle is |x| + |y| = 1. Graph it.

In the Euclid Plane we use the Unit Circle to define the cosine, sine, and tangent ratios. Is there an analogous way to define trigonometric ratios for the Taxicab plane?

The ratio of the circumference to the diameter of a circle in the Euclidean plane is a constant -- pi.

Is the ratio of the circumference to the diameter of a TC circle a constant? What is it?

Taxi Cab Ellipse

A GCF file

Using the TC distance metric, and the definition of an ellipse as the set points where the sum of the distance from two fixed points is a constant d, we can write an equation for the ellipse with foci at A(a,b) and B(g,h) as

(|x - a| + |y - b|) + (|x - g| + |y - h|) = d

Create your own GCF file, or use the one in the link above, to explore the shapes and loci that result from this equation.

Here are six examples for different choices of A, B, and d. Each TC ellipse in each of the first 5 is made up of six or eight segments. These segments are either parallel to the x-axis or y-axis (not shown here) or segments at a slope of 1 or slope of -1.

VERIFY BY COUNTING THE GRID LINES THAT EVERY POINT ON THE DEPICTED SEGMENTS ARE PART OF THE TCEllipse.

The last one occurs when AB = d. If we include it as a TCEllipse it is a degenerate case. Will the region always be a square? Why? Verify that for every point in the square, the sum of the TC distances from A and B is a constant.

FOR YOUR GSP CONSTRUCTIONS you want to trace the intersection of the two TC circles you generate.

First of all there are 4 intersection points to trace, not 2. You only see 2 at at time.

Next, there are 4 times at which the sides of the TCcircles coincide on the same segment. At those times, all the points on the segment that coincides satisfy the equation. Thus the ellipse, if it exists and is not degenerate, is a closed figure with 8 or 6 segments.

Verify.

From this GSP file you can readily move the focal points A and B. I have not figured out a way to trace the segments of slope 1 or -1 from this animation.

Taxi Cab Hyperbola

GCF file for Taxi Cab Hyperbola

The definition of a hyperbola is the set of points where the difference in the distance from two fixed points is a constant. Try this equation:

|(x - a) + (y - b)| - |(x - g) + (y - h)| = d

for fixed points A(a,b) and B(h,g). It seems the another set of points that satisfy the constant difference would be determined by this equation:

|(x - g) + (y - h)| - |(x - a) + (y - b)| = d

Therefore, a single equation that gets both sets of points would be:

||(x - a) + (y - b)| - |(x - g) + (y - h)|| = d

Like with the TCEllipse, construction of the GSP version may require some attention to segments of overlapping squares and not just points of intersection.

Verify that the figure at the right is a TCHyperbola with A(6,7), B(14,14), and d = 3

Explore other foci and other values of d.

What happens with d > AB? What happens when d = AB?

There is other craziness here. Explore these three case for d = 2:

1. A(5,5) and B(11, 14)

2. A(5,5) and B(12,14)

3. A(5,5) and B(13, 14)

EXPLAIN

Taxi Cab Parabola

The usual Euclidean definition of a Parabola is the locus of points equidistant from a line and a fixed point.

Determine a way to express the distance from a line and use that to write an equation for a parabola that can be graphed with Graphing Calculator 3.5.

First of all, we need to recognize that distance from a point to a line in Taxicab Geometry has the following definition:

Definition: Let l be a line and P a point. Then the distance from P to l is the minimum TC distance PQ where Q is a point on l.

Discussion:

Three cases are presented. Taxi cab distance is the sum along a horizontal plus a vertical. If the point Q is on the horizontal with P then h would be the taxicab distance from P to Q. If the point Q is on the vertical with P then v is the taxicab distance from P to Q. The TC distance from P to the line would be the minimum for all points Q on the line.

Thus

if the slope of l is more than 1, the h is the minimum distance.

if the slope of l is less than 1, the v is the minimum distance.

if the slope of l is equal to 1, the h = v and all points between BC have the same TC length.

Similar figures could be examine for negative slopes.

The definition of a Parabola is the locus of points such that a point on the Parabola is equidistant from a line called the directrix and a point called the focus. In Euclidean Geometry, the distance of a point from the line is taken along the perpendicular from a point on the directrix. It makes no difference what the slope of the line is. There should be a caution flag waving to warn that something a little different will be done with Taxicab Geometry.

GRAPHING CALCULATOR 3.5 for the TC Parabola

Case1. |m| > 1.

|If the line is y = mx + c and we have point F(a,b), then we need to set the two TC distances equal to one another. |x - a| + |y - b| gives us the distance from F. The distance to the line would be along a horizontal from any point (x,y) on the locus. Therefore the coordinates of a point Q that is minimum distance from (x,y) would be ( (y-c)/m, y). Thus the distance to the line would be |x - (y-c)/m| + |y - y| = |x - (y-c)/m|. Thus the equation for the TC Parabola is

|x - a| + |y - b| = |x - (y-c)/m|

Here is a GCF file for exploring this graph.

Case2. |m| < 1

As in case one, we have y = mx + c and point F(a,b). The distance to the line would be along the vertical from any point (x,y) on the locus. Therefore the coordinates of a point Q that is minimum distance from (x,y) would be (x, mx+b). Thus the distance to the line would be |x - x| + |y - (mx+b)| or, that is, |y - (mx+b)|. Thus, the equation for the TC Parabola is

|x - a| + |y - b| = |y - (mx+b)|

Here is a GCF file for exploring this graph

Case 3. m = 1

Use either of the two previous GCF files for exploring this case.

Construction of the TC Parabola with GSP.

A key point to consider with the GSP constructions is that the TC distance of a point from a line would be along a segment parallel to one of the axes. Determine a TC distance in this way and use it to develop a TC circle about the point F. We then look for the intersection of a line moving parallel to the directrix and the TC circle about F. All of the cases are implemented in the GSP file (four pages) below. Also, GSP script tools have been built into the file.

GSP file for TC Parabolas

Perpendicular Bisector of a Line Segment

The perpendicular bisector of a line segment is the set of points equidistant from the two ends of the segment. So the equation should be:

|(x - a) + (y - b)| = |(x - g) + (y - h)|

for A(a,b) and B(g, h). Click Here for a GCF file implementing this equation.

Verify this graph of a TC Perpendicular Bisector for A(1,1) and B(8,7).

Examine a similar graph for A(1,1) and B(7,7)

Examine a similar graph for A(1,1) and B(6,7). Contrast and explain the 3 graphs.

Examine the graph with A(1,1) and B(1, 7)

Examine the graph with A(1,1) and B(7, 1)

Is the midpoint of AB always on the graph? Prove or disprove.

How is the graph of A(1, 8) B(8,3) different or similar to the ones you have examined so far?

Triangles

Construct several Taxi Cab equilateral triangles. What do you observe?

Consider a segment AB and construct a TC equilateral triangle on it. The red TC circle has center at A and the green TC circle has center at B. The two circles share points C and C' as well as all the points on the segment CC'.

Therefore triangle ABC is TC equilateral, triangle ABC' is TC equilateral, AND, if C'' is any point on CC', then triangle ABC'' is TC equilateral.

VERIFY these or disprove them. Does this prove that SSS congruency does not hold?

Are there Euclidean equilateral triangles that are TC equilateral?

Are there Euclidean isosceles triangles that are TC isosceles?

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