## Monday, October 19, 2015

### Lesson 6-3: Rotations (Day 37)

Part IX of Benoit B. Mandelbrot's The Fractal Geometry of Nature is called "Fractional Brown Fractals" and covers Chapters 27 through 30.

In these chapters, Mandelbrot uses randomness as in the Brownian motion of the previous chapter, but applies it to more examples from earlier in the text. For example, in Chapter 28, he revisits the coastline problem, but describes what he calls "Brown relief" to generate new fractals that look realistically like coastlines. Once again, the dimension is the key -- although it's easy to generatea Brown relief fractal with dimension 1.5, the dimension needs to be closer to 1.2 to look realistic.

Because the world is round, Mandelbrot must take this shape into consideration. Here we can think back to the spherical geometry that we considered over the summer -- in particular, we recall how points directly opposite each other are called antipodes, or antipodal points. Mandelbrot writes that for his simple Brown relief fractal, "Thus, a big hill at the point P corresponds to a big hole at the antipodal point P'." Later on in the chapter, he points out that this invokes non-Euclidean geometry -- in particular, Riemann's non-Euclidean (elliptic or spherical) geometry.

Between Chapters 29 and 30, Mandelbrot places "a book-within-the-book." It's the only section of the book that's in color. Indeed, tt's nice to see some of the more well-known fractals in color -- including the self-squared fractal dragon, the Apollonius gasket, and several pictures of fictitious planets generated using Brown relief.

We are currently in the Rotations unit, and so we actually get to rotations today. In the new versions of the U of Chicago text, all of the transformations are in Chapter 4, but since I have the old version of the text, I must skip up to Lesson 6-3 in order to reach rotations.

When I covered rotations last year, much of the lesson was to set up for Dr. Hung-Hsi Wu's lessons on 180-degree rotations and parallel lines. This year I'm not following Dr. Wu's curriculum exactly, so I'm only posting the part of last year's post that has nothing to do with Wu:

Since I already started the U of Chicago order by giving reflections first, we must remind ourselves why I don't continue the U of Chicago order and give translations next. The problem is that the U of Chicago's definition of translation is not valid in non-Euclidean geometry and requires a Parallel Postulate to be valid -- so we end up with circularity if we try to derive the properties of parallel lines from translations (but we will try to find a way around this soon). On the other hand, the definition of rotation is valid in neutral geometry, so we can give it without worrying about a Parallel Postulate.

So reflections and rotations, unlike translations, have the common property that they work in all types of geometry. Another common property is that reflections and rotations -- but not translations (unless it is the identity translation) -- have fixed points. The fixed point of a rotation is its center, while a reflection has an entire line (its axis) full of fixed points. In higher math, we find out that one can perform certain reflections and rotations in the coordinate plane using matrix multiplication (which require that the origin be a fixed point), but translations require addition instead.

Also, the fixed points of reflections and rotations allow polygons to have either reflectional or rotational symmetry -- but not translational symmetry. We expect a symmetry of a regular polygon to have at least fixed point -- namely the center of the polygon. As it turns out, only infinite figures (tessellations) can have translational symmetry.

And so I present Lesson 6-3 of my U of Chicago text, rotations. We start with a definition:

A rotation is the composite of two reflections over intersecting lines.

Some texts define rotations the way you'd expect them to be defined -- in terms of a center and angle of rotation -- and then prove that the composition of two reflections is a rotation. But here, we define rotations to be that composite, then prove that it has the properties you expect. That theorem is called the Two Reflection Theorem for Rotations:

If m intersects l, then the composite of the reflection over m following the reflection over l "turns" figures twice the non-obtuse angle between l and m, measured from l to m, about the point of intersection of the lines.

The text provides the following proof. We are reflecting over lines l and m, and C is the point where they intersect. A is an arbitrary point that we're reflecting. We first reflect A over l to obtain A*, then reflect A* over m to obtain A'D and E are arbitrary points on lines l and m, respectively, and are used to form angles.

Proof:
(1) C is on both reflecting lines. So C' = C* = C.
(2) By the Figure Reflection Theorem, angle ACD reflected over l is angle A*CD and A*CE reflected over m is angle A'CE. Since reflections preserve angle measure, angles ACD and A*CD have the same measure (call it x) and angles A*CE and A'CE have the same measure (call it y). From the Angle Addition Postulate, the measure of angle DCE is x + y and the measure of angle ACA' is 2(x + y). By substitution, angle ACA' has twice the measure of angle DCE.
(3) Since AC = A*C and A*C = A'C (reflections preserve distance), AC = A'C. QED

In my proof, I prefer to use A' and A" to represent the two images, to emphasize that we are performing two reflections. As in many texts, positive is counterclockwise and negative is clockwise, so we can simply used signed numbers to represent the angle of rotation.

For these exercises, I include many exercises from Lesson 6-3, but I throw out most of the review Questions, since these are review from mostly Chapter 5 and the first part of Chapter 6, which we obviously haven't covered yet. I only include Question 23 -- but it's a tricky question about the hands of a clock (yet it's relevant here as the clock's hands are actually rotating). I also include the bonus (Exploration) Question 24. Students will have to figure out that if the rotation image of a point A is the point B, then any point equidistant from A and B -- that is, any point on the perpendicular bisector of AB -- can serve as the center of the rotation.