Tuesday, October 20, 2015

More on Rotations (Day 38)

Part X of Benoit B. Mandelbrot's The Fractal Geometry of Nature is called "Random Tremas; Texture" and contains Chapters 31 through 35. Mandelbrot writes that in Chapters 32, 33, and 35, he is continuing to model the galaxies in the universe using random fractals.

In Chapter 34, meanwhile, Mandelbrot comes up with a new concept -- texture. He writes that texture "is an elusive notion that mathematicians and scientists tend to avoid because they cannot grasp it," and so he comes up with the aspect of "lacunarity." A fractal is said to be lacunar if it contains many gaps, such as the Cantor dust. At the end of the chapter, he gives an example of two fractals -- Sierpinski carpets (just like the famous triangle, but based on squares instead) that have the same dimension (log 8/log 3 or 1.8928), but different sized gaps and thus different lacunarity.

Here is a link to the Sierpinski carpet -- albeit just the original version of the fractal. You'll have to find Mandelbrot's text to see how to create it with a different lacunarity:


Also in this chapter, Mandelbrot laments that the Cantor set is not translation-invariant -- that is, no translation maps it to itself. He writes that straight lines are of course translation-invariant, and we already know that infinite tessellations are translation-invariant.

Of course, here's an easy way to make the Cantor set translation-invariant -- simply copy the fractal from [0, 1] to every integer interval. Notice that this is not the same as making it dilation-invariant (that is, self-similar on all scales) -- to do that, we only copy the fractal to certain intervals, such as [2, 3], [6, 7], [8, 9], [18, 19], and in general to all real numbers whose ternary representations, both to the left and to the right, can be written without 1's. But what Mandelbrot finds lacking is a fractal that is both translation- and dilation-invariant.

Now let's get to today's lesson -- but in discussing today's lesson, I must refer to changes that I'm making to my curriculum, both earlier this year and upcoming this week.

Last year at this time, I figured that the U of Chicago Lesson 6-3 isn't sufficient as a full introduction to rotations. And so I added an extra rotations lesson from another source -- the well-known Engage New York Common Core curriculum. I wrote then that the EngageNY curriculum is ultimately based on Dr. Hung-Hsi Wu's writings. But the lesson that I'm posting today has nothing to do with parallel lines or 180-degree rotations, so it's safe for me to post.

But I was looking ahead a major change that I'm making later this week -- namely to replace the Wu lessons with the SSS, SAS, and ASA Congruence Theorems. But because I'm changing the order of the lessons around, I must amend the proofs that I'm writing in order to avoid circularity.

As it turns out, I'll be able to keep the U of Chicago proof of ASA intact. But the U of Chicago proofs of both SAS and SSS have problems. In particular, they use the Isosceles Triangle Theorem -- but I want eventually to use SAS to prove the Isosceles Triangle Theorem (as it is in most pre-Core texts), so this would be circular.

The SAS proof will be easy to fix, but SSS will be trickier. As it turns out, I have an easy proof of SSS -- instead of the Isosceles Triangle Theorem, it will depend on the closely related Converse of the Perpendicular Bisector Theorem.

But here's the problem -- we skipped the Converse of the Perpendicular Bisector Theorem! This was because the week I was supposed to cover it, I subbed in a Geometry class instead, and so I wanted to drop a lesson in order to discuss what I covered in class that week. The fact that the ink on my old Perpendicular Bisector Converse page was fading made it a no-brainer to drop that lesson -- and besides, I could always pick it up when covering the Isosceles Triangle Theorem.

I know -- I find it frustrating myself when I change my mind regarding the curriculum, only to change it back later on. At first I wanted to avoid the Perpendicular Bisector Converse, but now I want to bring it back. Notice that I could do what Dr. Franklin Mason does -- begin with SAS in one lesson, then move on to the Isosceles Triangle Theorem in the next lesson, and then proceed with SSS in still a third lesson. But I don't necessarily want to break up SAS and SSS into separate lessons like Dr. M.

But now I go back to the lesson that I'm posting today on rotations -- in particular, students are given the preimage and image of a rotation and are asked to construct the center of the rotation. As it turns out, the proof that the construction works requires -- the Perpendicular Bisector Converse (as students must construct two perpendicular bisectors)!

And so I will add a new worksheet to today's lesson. On this worksheet, I will give the proofs of both the Perpendicular Bisector Converse and the construction of the center of rotation.

I will still keep the old proof of the Perpendicular Bisector Converse, though. Recall from last year that it's really the proof of the Isosceles Triangle Symmetry Theorem from U of Chicago Lesson 5-1, but I'll come up with a new proof for isosceles triangles later.

By the way, the fact that all of this switching around is just to set up the proof of SSS reminds me of a major concern regarding proofs in Geometry courses. Some statements are much easier to use than they are to prove. Of the three major congruence theorems SAS, ASA, and SSS, we see here that SSS is the hardest to prove. But of all the congruence theorems, SSS is the easiest to use. After all, in order to use SAS or ASA we must make sure that we have the included angle or side. But if we have two triangles, each with side lengths 4, 5, and 6, then we know that these two triangles are congruent by SSS, and we are done without having to worry about the "included" anything.

Likewise, consider the three Common Core transformations -- reflections, rotations, translations. Of these three, the easiest for students to understand is arguably the translation -- especially if these transformations are occurring on a coordinate plane. All one has to do to translate a figure on a plane is add coordinates, whereas there's more to remember for reflections and rotations. Yet the properties of translations are the hardest to prove. Translation proofs require an extra postulate (the Parallel Postulate) that the others don't -- and besides, we wish to define translations in terms of reflections.

It may be easier to teach SSS and translations before SAS and reflections -- and indeed, this is exactly what happens in Michael Serra's Discovering Geometry, where the order in which the material is taught is determined by what's easiest to use (and understand), not what's easiest to prove. And so in Lesson 5.4, SSS comes before SAS, and in Lesson 8.1, translations come before reflections. And the circle properties, which are difficult to prove, appear in Serra's Chapter 7, which is earlier than in most other texts. (Recall that these chapter notations are for my old version of the Serra text -- these might occur one chapter earlier in the new versions.)

But in a proof-based course -- and this course on the blog is somewhat proof-based as it partly derives from Dr. M's ideas -- we are bound to respect the order in which the results are proved. And so students have to struggle through the more difficult reflections and rotations before they can see the easier translations, and they have to struggle with recognizing "included" angles and sides before they can see the easier SSS.

This is what I wrote last year for the Perpendicular Bisector Converse:

Converse of the Perpendicular Bisector Theorem:
If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

Given: PA = PB
Prove: P is on the perpendicular bisector m of segment AB.

Let m be the line containing the angle bisector of angle APB. First, since m is an angle bisector, because of the Side-Switching Theorem, when ray PA is reflected over m, its image is PB. Thus A', the reflection image of A, is on ray PB. Second, P is on the reflecting line m, so P' = P. Hence, since reflections preserve distance, PA' = PA. Third, it is given that PA = PB. Now put all of these conclusions together. By the Transitive Property of Equality, PA' = PB. So A' and B are points on ray PB at the same distance from P, and so A' = B. That is, the reflection image of A over m is B.

But, by definition of reflection, that makes m the perpendicular bisector of AB -- and we already know that P is on it. Therefore P is on the perpendicular bisector m of segment AB. QED

And this is what I wrote last year for today's rotation lesson -- including the derivation of the rotation center construction from the Perpendicular Bisector Converse:

So far, our work on rotations focuses the composite of two reflections in intersecting lines, for that's how the U of Chicago text defines them. But ordinarily, when we think about rotations, we want to think about its center and magnitude -- not the two reflections. Yet the U of Chicago text doesn't spend enough time on rotations do accomplish this. So we must go to another source.

Recall that for Wu, rotations are of primary importance. But the Wu link that I gave yesterday doesn't give exercises or problem sets, since it mainly focuses on definitions and theorems. Luckily, here's a link to the Common Core Geometry curriculum developed for New York State.


I suspect that much of the Empire State's curriculum is based on Wu. We see that transformations are covered in Topic C, Lessons 12-21. Rotations come first (Lesson 13) -- just like Wu. Reflections come next (Lesson 14) and then translations (Lesson 16). But the giveaway that Topic C is based on Wu is Lesson 18 -- which uses 180-degree rotations to construct parallel lines and ultimately prove the Alternate Interior Angles Theorem!

As it turns out, Lesson 15 covers the relationship between a reflection and a rotation -- so it's just yesterday's lesson. And Lesson 17 is essentially the Perpendicular Bisector Theorem, which we've already covered last week. Our focus today will be on Lesson 13, on rotations.


Notice that unlike this blog, this New York lesson prescribes what questions are meant to be the Opening Activity, Exercises, Exit Ticket, and Problem Se, as well as how many minutes are to be devoted to each learning task. I don't do that -- but of course, I did sort of hint that the folding tasks at the beginning of my reflection lessons would make good Opening Activities. Actually, yesterday's task of reflecting the letter F in the y-axis and then the x-axis would also make a good Opening Activity or Anticipatory Set -- I considered posting a separate page to perform the reflections, but I want to conserve paper and avoid excessive visits to the Xerox machine. Of course, nothing is stopping the teacher from having the students take a blank sheet of paper, draw a coordinate plane with a large F in the first quadrant, and then perform the reflections, possibly even by folding.

We observe that the New York lesson also uses positive angles to denote counterclockwise and negative angles to denote clockwise. But it also uses the abbreviation CW for clockwise. Also, the New York lesson uses the letter R to denote a rotation, with two subscripts -- the first being the center and the second being the magnitude (that is, the angle). The U of Chicago uses the letter r for reflection, but no letter for translation. This blog doesn't use subscript function notation at all.

Also, I found it interesting that the New York lesson uses the Greek letter theta to denote the measure of an angle. I rarely see the theta symbol used in math texts until Precalculus, or perhaps an Honors Algebra II class with Trigonometry.

For all our talk about how the Wu and New York lessons don't define rotations as a composition of reflections, it does define rotations as a composition of rotations. In particular, a rotation of 180 degrees is the composition of two 90-degree rotations, and rotations with magnitudes greater than 180 are also the composition of rotations whose magnitudes add up to the given magnitude. In the U of Chicago text, 180 degrees is not a special case -- it's included as part of the Two Reflection Theorem for Rotations (where a "non-obtuse" angle could be a right angle, so twice its measure would be a straight angle, or 180 degrees). For larger angles, the text states that we can add or subtract multiples of 360 until the magnitude is between -180 and 180 degrees.

The New York lesson begins with its Opening Task, where the students cut out an angle and then use it to perform the rotation of a given figure. The first exercise asks students to use a protractor to measure the magnitude of a rotation. The second exercise is basically the same task as the first.

But at Exercise 3, things begin to get interesting. This exercise is similar to the first two, where we're rotating the letter M. But there's a problem -- we don't know the center of rotation! So the lesson gives the following algorithm for finding the center of rotation:

a. Draw a segment connecting points A and A'.
b. Using a compass and straightedge, find the perpendicular bisector of this segment.
c. Draw a segment connecting points B and B'.
d. Find the perpendicular bisector of this segment.
e. The point of intersection of the two perpendicular bisectors is the center of rotation. Label this point P.

Can we prove that this algorithm works? It looks a bit familiar -- because it is similar to the construction of an circle through three noncollinear points. But I threw that construction out when we did Section 4-5 -- because the proof that the construction works requires a Parallel Postulate. In this case, even if we had a Parallel Postulate, it's not obvious how to prove that the construction from a-e above works -- we have four points, not just three, and there's nothing saying that three of the points, or even all four of them, aren't collinear. It could be that the perpendicular bisectors of AA' and BB' do not intersect at all, much less intersect at the center of the rotation.

As it turns out, we don't have to worry about that. We already know -- that is, we are given -- that the points A' and B' are rotation images of A and B respectively. That is, we are given that there exists a point P such that P is the center of said rotation. And so we can prove that P lies on both of the perpendicular bisectors -- that is, we can prove that the bisectors must intersect!

And so here is our proof, in paragraph form:

Given: A'B' are the images of AB under a rotation centered at C
Prove: C lies on the perpendicular bisectors of AA'BB'.

The result follows directly from the proof of the Two Reflections Theorem for Rotations. Indeed, here we label the center C rather than P for emphasis -- result (3) of yesterday's proof tells us that AC = A'C (and similarly, BC = B'C). By definition of equidistant, C is equidistant from A and A' (and likewise, C is equidistant from B and B'). Therefore, by the Converse to the Perpendicular Bisector Theorem, C lies on the perpendicular bisector of AA' (and similarly that of BB' as well). QED (The proof sketch given in the New York lesson mentions chords of a circle, which the U of Chicago text doesn't cover until Chapter 15. But we just used that Converse instead.)

Notice that this lesson is clearly a straightedge and compass lesson. But we can find the perpendicular bisector of a segment by folding -- in this case, we fold A over so that it lies on top of A'. The crease of the fold is the perpendicular bisector of AA'. This may seem a bit tricky -- we used folds earlier in order to reflect points, and yes, the reflection image ofA over the first fold line is A', yet A' is supposed to be a rotation image of A, not a reflection image. As it turns out, A' can be the image of A via either a reflection or a rotation. The folding is simply to find the perpendicular bisector, not to perform a reflection.

We notice that finding the center rotation is more difficult than finding a line of reflection. If we are given a point A and its reflection image A' -- as long as A' is not A itself, we immediately know the reflecting line to be the perpendicular bisector of AA'. But to find the center of rotation, it's not enough to know a single point and its image A' (unless A' is A itself) -- we saw that a second point B and its image B' were necessary. Because of this, we say that a reflection has one degree of freedom, while a rotation has two degrees of freedom. (A translation, like a reflection, also has only one degree of freedom.)

For the problem set, I decided to omit the part about a straightedge and compass. The reason for this is that the New York lesson assumes that the students can construct 45-, 60-, and 120-degree angles using only a straightedge and compass. Technically, the students in our course can do this -- we know how to draw perpendicular lines with their 90-degree angles and bisect them to get 45 degrees, and the First Theorem of Euclid's Elements lesson gives us equilateral triangles, hence 60 degrees -- but we can't prove the angle measures of an equilateral triangle until Chapter 5. So we expect the students just to use a protractor. Problem Set 3 discusses vertical angles. Interestingly enough, this is how Wu proves the Vertical Angles Theorem -- but we already proved it in Chapter 3 another way. The answers to Problem Set 4 and 5 are "rectangle" and "rhombus" respectively, but we don't define either of these terms until Chapter 5 as well. So I threw these out.

The New York lesson also mentions Geogebra as a possible tool to perform the rotations -- I suggested that earlier this month as well.

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