Because this exercise simultaneously exercises your understanding of a planar proof and your understanding of the differences between spherical and plane geometry, there is likely to be something similar to it on the final.

**Lemma:** *On the sphere*, the perpendicular bisector of
segment AB is the set of points equidistant from points A and B. (In
other words, |PA| = |PB| if and only if P is a point on the
perpendicular bisector of segment AB.)

The proof of this in the plane goes something like:

**Planar Proof:** If P is on the perpendicular bisector and M is
the midpoint of AB, then by side-angle-side triangle AMP is congruent
to triangle BMP. Therefore |AP| = |BP|.

On the other hand, if |AP| = |BP| and M is the midpoint of AB then by side-side-side triangle AMP is congruent to triangle BMP. Angles PMA and PMB are congruent supplementary angles, so must be right angles. Therefore line PM is perpendicular to line AB, and so line P lies on the perpendicular bisector of segment AB.//

Does this proof work on the surface of a sphere?

**Discussion of spherical proof:**
Since we've defined distance in terms of angle measure (see page 223)
and we've described how to measure angles, we can define the
perpendicular bisector of a segment to be a great circle that passes
through the midpoint of a segment at right angles to that segment. So
at least everything in the statement of the lemma makes sense!

Looking at the proof of the lemma, the first thing we see that we're not sure of is "by side-angle-side". Is side-angle-side true on the sphere? A quick Google search says it is, but we'd prefer a less ephemeral reference than a web site. Page 214 of the text says side-angle-side is true on the sphere, and that's a good enough reference for a homework assignment.

Continuing through the proof, we see that we need side-side-side to be true on the sphere; page 214 says it is. Are congruent supplementary angles still guaranteed to be right angles? "Congruent" means that α = β and "supplementary" means that α + β = 180°, so supplementary angles on the sphere must be right angles. It looks like every step of the proof on the plane is also true on the sphere, so we can use an identical proof to the planar one.

Suppose also that you only pick up customers at intersections. Then
your map of customer pick-up and drop-off locations looks like the set
of points in the plane with integer coordinates. The geometry we are
about to describe is called a *discrete geometry* because each
point in the space is isolated, not part of a continuum.

All of the streets in our imaginary city are two way streets, so we can always move horizontally or vertically from any intersection to the next. However, we cannot cut across the lots or cut through the buildings on the blocks; we must drive on the streets to reach our destination.

What would a segment look like in this "taxicab geometry"? Is there always a "segment" connecting any pair of points? Is our first axiom still true?

What do we mean by "segment"? On the sphere and hyperbolic plane we talked about geodesics -- shortest paths between two points. As taxi drivers we will want to drive along the shortest route between two locations, so our segments might consist the points (intersections) we go through as we move from one point (origin) to another (destination).

What about the second axiom? Given a segment, is there a way to
extend it to a line? Is there a *unique* way to extend it to a
line?

What is a circle in this geometry, and can we draw a circle of
*any* center and radius? What does "any radius" mean in this
geometry?

Does the concept of angle make sense in this geometry? How might you define an angle? If you can define an angle, would two right angles always be congruent?

What about the parallel postulate: Given a line L and a point P not on the line, there exists a unique line through P that is parallel to L. Is this true in taxicab geometry? If not, what goes wrong? What does "parallel" mean in this context? (What does "line" mean in this context?)

Now think about axioms 6 through 8. What is the definition of an isometry in this space? Are any of these axioms true in taxicab geometry? Are any of these axioms false?

**Names:**

Working in groups of two or three, discuss whether and how the following axioms apply to taxicab geometry. Record your observations and conclusions on this worksheet; you will hand in one copy of this worksheet per group.

- We can draw a unique line segment between any two points.
- Any line segment can be continued indefinitely.
- A circle of any radius and center can be drawn.
- Any two right angles are congruent.

Given the difficulty associated with defining a line in taxicab geometry, we won't discuss the parallel postulate. However, we can discuss axioms 6 through 8.

How would we define an isometry of taxicab space?

Is it true that given any two points in taxicab space, we can find an isometry that takes one to the other?

Given a point P and two points Q and R so that |PQ| = |QR|, is it true that there's an isometry of taxicab space that fixes P and moves Q to R?

Are there any reflections that are isometries of taxicab space? If so, what are they?

Using linear algebra, we can describe all of the isometries of taxicab space that fix a specified point.

Think of the point that stays fixed as the origin, (0,0). Let your x-axis run east/west from the origin, and your y-axis run north/south. Then you can label the intersections in your street grid with Cartesian coordinates: (0,1) is one block north of (0,0); (1,0) is one block east of (0,0); and so on.

If an isometry has to leave (0,0) where it is, what can it do with the rest of the points in taxicab space? It can move (1,0) to: (1,0), (0,1), (-1,0) or (0,-1). Any other point is too far away from (0,0); isometries have to preserve distance.

It can move (0,1) to (1,0), (0,1), (-1,0) or (0,-1), but whichever of those it moves to it must be the next poitn either clockwise or counterclockwise from the image of (1,0).

Linear algebra lets us represent all of these possibilites using matrices. To find out where the isometry corresponding to a matrix sends a point, we multiply on the left by a 2 by 2 matrix. The matrix shown below fixes all the points whose x and y coordinates are equal, and swaps the x and y coordinates of all the other points in taxicab space.

[0 1] [a] [b] [ ]*[ ] = [ ] [1 0] [b] [a]What are the other isometries that fix (0,0) and preserve taxicab distance? Take some time and make a list of them.

We could use vector addition to describe some isometries that do not fix (0,0). Can we describe all the isometries of taxicab space as combinations of some isometry that fixes (0,0) and a vector addition?

[1 0] [1 0] [0 1] [ 0 1] [-1 0] [-1 0] [0 -1] [0 -1] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [0 1] [0 -1] [1 0] [-1 0] [0 1] [0 -1] [1 0] [-1 0]