§18.1 Taxicab Geometry The student will learn about: a geometric model that conforms to many of the Euclidean axioms but not all. 1 1

Introduction to Taxicab Geometry The mathematician Hermann Minkowski (1864-1909) developed a rather unique geometry just by changing how distance between two points is measured. Minkowski grew up in Konigsberg. In 1879, he won his first prize in mathematics. He was a teacher at Gottingen University and colleague of David Hilbert from 1902 until his untimely death of appendicitis at the age of 45.

Hermann Minkowski Besides his monumental work in the theory of numbers, he is also known for his mathematical foundation for the Theory of Relativity, which Einstein – a student of Minkowski – used extensively in his work. The physicist Max Born, also a student of Minkowski, one said: “Minkowski laid out the whole arsenal of relativity mathematics…as it has been used ever since by every theoretical physicist.”

Introduction We shall introduce this rather unusual geometry. This geometry satisfies many of the axioms of Euclidean geometry but will not satisfy many of them including the SAS postulate. We will develop this geometry in two dimensions only although it may be extended to three.

Definitions Points will be defined as for the Euclidean plane. A point is an ordered pair of real numbers – A(x, y). Plot the following points on a piece of graph paper. A (3, 1) B (7, 7) C (7, 1) D (-2, 5)

Definitions Lines will be defined as for the Euclidean plane. A line is all the points that satisfy and equation of the form Ax + By = C or y = mx + b. Plot the line AB as defined above. Plot the line BC as defined above.

Distance The Euclidean definition for distance Will be replaced by the following definition - B (6,7) A (2, 1) Find AB * NOTE: AB* = |Δx| + |Δy|. Δx Δy

To explain taxi cab distance consider the following map of a section of a city. A taxicab wants to go from the corner of Ash and Apple Streets to the corner of Dogwood and Elderberry Streets. How far does the taxi cab drive? Does the route taken make a difference? Elderberry Street Dewberry Street Banana Street Apple Street Cherry Street Ash Street Birch Street Cyprus Street Dogwood Street Do you see the significance of the name “Taxicab” geometry?

Distance Note that there are many ways to travel from point A to point B in this geometry and that ALL have the same distance. Find the following distances: AB* = B F AC* = = 10 BC* = D E AD* = DE* = EF* = FB* = = 10 A C

Axiomatics Because points and lines are the equivalent to Euclidean points and lines the incidence axioms all hold. Because angle measure is the same as Euclidean the angle measure axioms all hold. The distance postulates (uniqueness, equality and zero distance) are obvious so we need only establish the “Ruler Postulate”.

Axiomatics – Ruler Postulate AB* = | a - b | We need to establish that AB* = | a - b | Start with our distance formula. And since P and Q are both on y = mx + b. y 1 = mx 1 + b and y 2 = mx 2 + b

Axiomatics – Ruler Postulate AB* = | a - b | To set up the “ruler” let P (x, y) ↔ (1 + |m|)x where m is the slope of the ruler. Then A (x 1, y 1) ↔ (1 + |m| ) x 1 = a and B (x 2, y 2) ↔ (1 + |m| ) x 2 = b and AB* = ( 1 + |m| ) | x 1 – x 2 | from 3 above. And = | ( 1 + |m| ) x 1 – ( 1 + |m| ) x 2 | = | a - b | Establishing the ruler postulate.

Ruler Postulate - Examples Let A (2, 1), B (5, 6) and C (7, -3) Then A (x 1, y 1) ↔ (1 + |m| ) x 1 = a For AB*; |m| = 5/3 so A(2, 1) ↔ (1 + |5/3| ) 2 = 16/3 = a and B(5, 6) ↔ (1 + |5/3| ) 5 = 40/3 = b and AB* = | a - b | = | 16/3 - 40/3 | = 8 Find AC* m = 4/5, and AC* = | 9/5 - 56/5 | = 9 Find BC* m = 9/2, and BC* = | 55/2 - 77/2 | = 11 Establishing the ruler postulate. Yes, of course it is easier to use the other distance formula, but we have established a ruler.

Axiomatics Does the betweenness postulate hold? Note that from a previous slide betweenness is established. C B A F E D The proof will be left for homework.

Additional Equivalencies Because of the previously established postulates we can now use the Euclidean definitions for segments, rays, and angles.

Taxicab Triangles The definition of a triangle in Euclidean geometry will carry over into Taxicab geometry. That is we need three vertices and three sides.

Let’s Play From the warm up we know C B A Calculate the lengths of the three sides using the taxicab metric. AB* = 3 + 1 = 4 BC* = 2 + 2 = 4 AC* = 1 + 3 = 4. An equilateral triangle. Is it equiangular?

More on Triangles Consider ∆ ABC with A (0, 6), B (0, 0), and C (6, 0) and ∆ XYZ with X (6, 3), Y (9, 6), and Z ( 12, 3). Both of these triangles are right triangles. With right angles at B and Y respectively. Find the taxicab lengths of the sides of ∆ ABC. AB* = 6, BC* = 6, AC* = 12 Find the taxicab lengths of the sides of ∆ XYZ. XY* = 6, YZ* = 6, XZ* = 6 Under the correspondence ABC  XYZ are corresponding angles congruent? Under the correspondence ABC  XYZ are corresponding sides congruent? Is SAS satisfied? Are the triangles congruent?

Definitions Sketch the set of points that are equidistant from the segment ends defined by the two points A (0, 0) and B (0, 6). Looks normal! What is the name given to this set of points? 19

Definitions Sketch the set of points that are equidistant from the segment ends defined by the two points A (0, 0) and B (4, 6). Looks normal! Now that’s interesting! What is the name given to this set of points?

Definitions Sketch the set of points that are equidistant from the segment ends defined by the two points A (0, 0) and B (4, 4). Looks normal! Now that’s really interesting! What is the name given to this set of points?

Euclidean Distance from a Point to a Line.

Taxicab Distance from a Point to a Line. 5 5 6 So the circle method is the one that makes the most sense and hence our definition.

Vertical Lines Find the line parallel to the y – axis through the point (4, 1). 24

Lines with Slope ± 1 Sketch the set of points that are equidistant from the line y = - x and through the point (5, 5). What is the distance between these two lines? 4 25

In General Given two parallel lines y = mx + b 1 and y = mx + b 2 the distance between them is given by d 26

PARALLEL LINES From the previous slides it is evident that through a point not on a line there exist exactly one parallel.

Summary. We learned about “taxicab” geometry We learned about a new distance metric. We learned that many of the Euclidean axioms are satisfied by taxicab geometry but not all of them. We played around with triangles. We learned about parallel lines.

With the remaining class time lets work on our homework assignment 18 29

Assignment: §18.1