## Wednesday, October 4, 2017

### Lesson 3-5: Perpendicular Lines (Day 35)

This is what Theoni Pappas writes on page 277 of her Magic of Mathematics:

"Using only a straightedge on which distances can be indicated, can you bisect the given angle, and prove why it is bisected?"

The angle in this "Is It Possible?" puzzle is arbitrary, so you can try this out with any angle. As usual, I'll give the answer in tomorrow's post.

Meanwhile, Pappas actually fits a second puzzle on this page, "Can you make a Latin Square?":

"If no number is repeated in any row or column of a square of 18 tiles, it is called a Latin square. The square below is not a Latin square, since some numbers repeat in rows or columns. For example, the top row has two 6s and two 4s and the right side has two blanks and two 4s. Either rearrange this, or start a new one to make a Latin square."

Well, for this one I can show you how the spots on the dominoes are arranged (with 0=blank):

644615
103534
366510
531624
243203
052520

But this doesn't indicate how these 36 numbers are divided into 18 dominoes -- that is, whether the dominoes are horizontal or vertical. Thus I will inform you that all the dominoes in the first two columns on the left are horizontal, while all the tiles in the third and sixth columns are vertical. In the fourth and fifth columns, there are two horizontal dominoes in the middle two rows, and all other tiles in these columns are vertical. So now you can try to solve this puzzle. The solution will be tomorrow.

Meanwhile, here's the answer to yesterday's Milkman's Puzzle:

10 qt.     10 qt.     5 qt.     4 qt.
10          10          0           0 -- starting
5            10          5           0
5            10          1           4
9            10          1           0
9            10          0           1
4            10          5           1
4            10          2           4
8            10          2           0
8            6            2           4
10          6            2           2 -- ending

Notice where the Diophantine steps are -- "pour from river to 5" and "pour from 5 to 4" are the first two steps, and these two steps repeat near the middle of the solution.

Lesson 3-5 of the U of Chicago text is called "Perpendicular Lines." This corresponds to Lesson 3-8 in the modern Third Edition of the text. Meanwhile, Lesson 3-7 of the new edition is actually an introduction to "size transformations" (or dilations). This introduction is basically the same as Lesson 12-1 in my old version, and dilations are studied in more detail in Chapter 12 of both editions.

Two years ago, we didn't really cover Lesson 3-5. The main reason was that the two main theorems of this lesson, Two Perpendiculars and Perpendicular to Parallels, are proved directly from the two postulates on corresponding angles from yesterday. Since I didn't cover those postulates two years ago, I couldn't cover the theorems either.

Three years ago, we did officially cover Lesson 3-5. That year, I rewrote the lesson in order to avoid mentioning the postulates. I've decided to reblog all of the worksheets I posted that day, even though I'm teaching it differently this year.

In these worksheets, I mentioned Two Perpendiculars but not Perpendicular to Parallels. It would be easy to add that theorem in. On the other hand, I didn't mention slopes of perpendiculars either. In fact, I like the idea of using rotations to teach perpendicular slopes, but that's not how they're taught in the text.

This is some of what I wrote three years ago about today's lesson. After yesterday's long explanation, I don't need to compare yet again my old way of teaching this lesson to the text's. But first, let me post a YouTube video from Square One TV on perpendicular lines.

Officially, I'm doing Section 3-5 now, but then again, not really. Let's consider the contents of this particular section:

-- The definition of perpendicular has already been covered. I'm moved it to Section 3-2 when I defined right angles, because I wanted to get it in before jumping to Chapter 4 on reflections, since reflections are defined in terms of the perpendicular bisector.

Now as I imply in this post, most of what I write on this blog is derived from mathematicians like Dr. M and Dr. Wu, who have written extensively about Common Core Geometry. But my plan to include Perpendicular to Parallels as a postulate appears to be original to me. I've searched and I have yet to see any text or website who will derive all the results of parallel lines from a Perpendicular to Parallels Postulate. Then again, what I'm doing here is, in some ways, as old as Euclid.

Let's look at Playfair's Parallel Postulate again:

Through a point not on a line, there is at most one line parallel to the given line.

This is a straightforward, easy to understand rendering of a Parallel Postulate, and Dr. M uses this postulate to derive his Parallel Consequences. But let's look at Euclid's Fifth Postulate, as written on David Joyce's website:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/post5.html

That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

No modern geometry text would word its Parallel Postulate in this manner. For one thing, even though the use of degrees to measure angles dates back to the ancient Babylonians, Euclid never uses degrees in his Elements. So the phrase "less than two right angles" is really just Euclid's way of writing "less than 180 degrees." Indeed, in Section 13-6, the U of Chicago text phrases Euclid's Fifth Postulate as:

If two lines are cut by a transversal, and the interior angles on the same side of the transversal have a total measure of less than 180, then the lines will intersect on that side of the transversal.

But let's go back to the Perpendicular to Parallels Theorem as stated in Section 3-5:

In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.

Now count the number of right angles mentioned in this theorem. The transversal being perpendicular to the first line gives us our first right angle, and the conclusion that the transversal is perpendicular to the second line gives us our second right angle. So we have two right angles -- just like Euclid! So in some ways, making the Perpendicular to Parallels Theorem into our Parallel Postulate is making our postulate more like Euclid's Fifth Postulate, not less.

Of course, if we wanted to make our postulate even more like Euclid's, we could write:

If a plane, if a transversal is perpendicular to one line and form an acute angle (that is, less than right) with another, then those two lines intersect on the same side of the transversal as the acute angle.

But this would set us up for many indirect proofs, which I want to avoid. So our Perpendicular to Parallels Postulate is the closest we can get to Euclid without confusing students with indirect proofs.

So this is exactly what I plan on doing. Since we're in Section 3-5, the section that has Perpendicular to Parallels given, I could include it here -- we don't need to worry about how to prove it since I want to make it a postulate. But I already said that I want to wait until Chapter 5 before including any sort of Parallel Postulate. And so the new postulate will be given in that chapter.

Returning to Lesson 3-5:

-- The Perpendicular Lines and Slopes Theorem must also wait. Common Core Geometry gives an interesting way to prove this theorem, but the proof depends on similar triangles, which I don't plan on covering until second semester.

So that leaves us with only one result to be covered in 3-5: the Two Perpendiculars Theorem:

If two coplanar lines l and m are each perpendicular to the same line, then they are perpendicular to each other.

This theorem doesn't require any Parallel Postulate to prove. Indeed, even though I just wrote that I don't want to use indirect proof, in some ways this theorem is just begging for an indirect proof:

Indirect Proof:
Assume that lines l and m are both perpendicular to line n, yet aren't parallel. Then the lines must intersect (as they can't be skew, since we said "coplanar") at some point P. So l and m are two lines passing through point P perpendicular to n. But the Uniqueness of Perpendiculars Theorem (stated on this about two weeks ago) states that there is only one line passing through point P perpendicular to n, a blatant contradiction. Therefore l and m must be parallel. QED

But this would be a very light lesson indeed if all I included is this one theorem. Because of this, I decided to include a theorem that I mentioned back in July -- the Line Parallel to Mirror Theorem (companion to the Line Perpendicular to Mirror Theorem mentioned a few weeks ago):

Line Parallel to Mirror Theorem:
If a line l is reflected over a parallel line m, then l is parallel to its image l'.

Lemma:
Suppose T is a transformation with the following properties:
-- The image of a line is a line.
-- Through every point P in the plane, there exists a line L passing through P such that L is invariant with respect to T -- that is, T maps L to itself.
Then any line that doesn't contain a fixed point of T must be parallel to its image.

Now all Common Core transformations satisfy the first property -- that the image of a line is itself some line. (Forget about that Geogebra "circle reflection" where the image of a line can be a circle, since that's not a Common Core transformation.) As it turns out, there are five types of Common Core transformations that satisfy the second property:

-- Any reflection
-- Any translation
-- Any glide reflection
-- Any dilation
-- A rotation of 180 degrees

So notice that the only Common Core transformations not satisfying the second property are rotations of angles other than 180 degrees.

Here is a proof of the lemma. (By the way, "lemma" means a short theorem that is mainly used to prove another theorem.) Let l be the original line and l' its image, and let P be any point on l. Since l doesn't contain any fixed points, the image of P can't be P itself -- so instead, it must be some point distinct from P, which we'll call P'. So of course P' lies on l'. Now the point P lies on some invariant line L -- and by invariant, we mean that P' lies on it. Now through the two points P and P', there is exactly one line, and that line is L, not l. Since P lies on l, it means that P' can't lie on l. But this is true for every point P on l. For every point P on lP' is not on l. So l can't intersect its image l', since every point on l fails to have an image on l. In other words, l and its image l' are parallel. QED

OK, that's enough from three years ago -- we don't need to go any further because we haven't actually learned reflections this year. So here are all the old worksheets from that year -- except that I replaced Line Parallel to Mirror with a worksheet that is more nearly aligned with Lesson 3-5. There are now questions referring to Perpendicular to Parallels, and Perpendicular Lines and Slopes.