## Friday, November 3, 2017

### Lesson 5-6: Alternate Interior Angles (Day 56)

This is what Theoni Pappas writes on page 307 of her Magic of Mathematics:

"Sam Loyd's Battle of the Four Oaks Puzzle...."

This is the second page of the subsection on "a few oldies" from Sam Loyd. There are two puzzles on this page, and the first is visual, so as usual I provide a link:

http://www.puzzles.com/PuzzlePlayground/FourOaks/FourOaks.htm

No, this isn't technically a checkerboard problem, but I like how the link uses the board to indicate the locations of the four oaks. We must divide the board into four pieces of the same size and shape -- in other words, four congruent pieces (this is a Geometry blog after all) -- such that each piece contains one oak. Pappas doesn't use a board in presenting the problem -- but her solution is just like the one given at the link above, where an 8 * 8 board suddenly appears. I won't post the solution, since it's at the link above.

There is one more puzzle on this page, which doesn't require a picture:

"Sam Loyd's Weight of a Brick Puzzle...."

If a brick balances with three quarters of a brick and three quarters of a pound, then how much does the brick weigh?

Hey, this isn't a puzzle -- this is algebra! I won't bother to post a solution here, since readers of a Geometry blog should be able to solve what's basically an Algebra I question. Then again, notice that this is a way to "trick" Algebra I students into solving an equation by labeling it as a "puzzle."

That's leaves us with one puzzle for which you're still awaiting a solution -- Sam Loyd's Get Off the Earth Puzzle. How are we able to rotate 12 people into 13 people?

Here's the solution -- each member of the thirteen is only 12/13 as tall as a member of the dozen! The dozen are oriented so that different fractions of each member lie on the part that rotates (the earth) and the part that doesn't rotate -- the top 12/13 with the bottom 1/13, the top 11/13 with the bottom 2/13, and so on until the top 1/13 with the bottom 12/13. After the rotation, the top 11/13 now lines up with the bottom 1/13, the top 10/13 with the bottom 2/13, and so on until the top 1/13 is now with the bottom 11/13. So there are eleven members who add up to 12/13. The top 12/13 has no bottom, and the bottom 12/13 has no top. So there are a total of thirteen members who are really 12/13. The missing 1/13 is different in each person (some have no top, no bottom, no middle, etc.), but this missing part is so tiny that they appear to be 13 whole people!

This is the last page that we're covering in Pappas. The next two pages, 308 and 309, are blocked by the weekend, and these are the last two pages of this final chapter. Then all that's left in the book are the answers (which we've already given) and the index. Therefore, we have actually reached the end of the Pappas book.

But let's recall why we're reading this book in the first place -- it's supposed to tide us over until January 1st, 2018, when the new Pappas Mathematical Calendar begins. There are two months left, and so I'll continue to post Pappas until December 31st.

And I didn't actually purchase this old Pappas book until March -- which means that we missed pages from the start of the book. So starting Monday, we'll return to Chapter 1 -- since Monday's date is the sixth, we'll begin on page 6. We'll read Chapter 1 this month and Chapter 2 in December, and then that will take us to Chapter 3 (which we read back in March). This way, we'll have covered every chapter of her book and then be ready to return to her actual calendar in 2018.

Lesson 5-6 of the U of Chicago text is called "Alternate Interior Angles." The modern Third Edition of the text covers alternate interior angles slightly earlier, in Lesson 5-4 (along with Same Side Interior Angles, which aren't emphasized in the Second Edition).

I wrote yesterday that I've changed Lessons 5-5 and 5-6 two to three years ago. In attempting to restore the original order, I'll post one side of a worksheet that contains Lesson 5-5 material. The other page will be a new activity (albeit based on an old activity), since this is an activity day.

This is some of what I wrote two years ago about today's lesson. I'll post some of this anyway even though it doesn't refer to how I'm teaching the lesson any longer (but in the end, I prefer what I wrote two years ago to what I wrote three years ago):

Today's lesson focuses on proving the Parallel Consequences -- that is, statements of the form, "if two lines are cut by a transversal, then ..."

This lesson will be set up almost exactly like Dr. Franklin Mason's Lesson 4.4. We wish to prove the converses of the Parallel Tests. We do so by using my favorite trick for proving converses -- we use the forward theorems along with a uniqueness statement. The uniqueness statement we need is the Uniqueness of Parallels Theorem -- in other words, Playfair.

It's possible to prove all of the Parallel Consequences by using the respective test plus Playfair -- so we'd prove the Corresponding Angles (CA) Consequence using CA Test plus Playfair, the Alternate Interior Angles (AIA) Consequence using AIA Test plus Playfair, and then Same-Side Interior Angles Consequence using that test plus Playfair.

But Dr. M only proves one of the consequences using Playfair -- he then uses vertical angles and linear pairs to derive the other consequences, as is traditionally done. We've seen that students should definitely be familiar with using one of the consequences to prove the others. Dr. M uses the Alternate Interior Angles Consequence to prove the others, but as we've discussed before, I'm changing this to the Corresponding Angles Consequence instead.

I looked back at last year's lesson and compared it to what I'm doing this year. In my first lesson after posting the Fifth Postulate last year, I posted some properties of two types of quadrilaterals, isosceles trapezoids and parallelograms. This year, we've already proved the isosceles trapezoid properties.

And now here's the activity I created two years ago -- a little something on spherical geometry (which is alluded to in Lesson 5-7):

Since I cut off the second part of the worksheet, I wish to replace it with something. I notice that in the U of Chicago and many other texts, students are given a taste of what would happen if we didn't have a parallel postulate. Yes, we'd have non-Euclidean geometry.

We've spent so much time on the blog discussing what's possible in neutral geometry and what requires a parallel postulate. The U of Chicago text introduces the students to spherical geometry -- as did I over the summer on the blog. But note that spherical geometry is not neutral -- neutral geometry includes only Euclidean and hyperbolic geometry. Technically, if we want to show students what impact the parallel postulate has on geometry, we should be showing them hyperbolic geometry, not spherical geometry. But since we live on a globe, spherical geometry is far easier for students at this age to understand.

Indeed, I often point out that the Fifth Postulate and its equivalents are, believe it or not, true in spherical geometry! This is because most of these statements begin with, "if two parallel lines..." or "if two lines are parallel." But in spherical geometry there are no parallel lines, so all Parallel Consequences are vacuously true! We might as well replace every occurrence of "parallel lines" with "unicorns," and all of the statements are still vacuously true:

Perpendicular to Parallels: If a line is perpendicular to one unicorn, it's perpendicular to the other.
Transitivity of Parallels: If a line is parallel to one unicorn, it's parallel to the other.
Parallel Consequences: If two unicorns are cut by a tranversal, corresponding angles are congruent.
Playfair: Using the version of Playfair given in Lesson 13-6 of the U of Chicago text doesn't work, but it does if we use Dr. Franklin Mason's version: "Through a point not on a given line, there's at most one line parallel to the given line." The phrase at most one allows for the possibility of zero.
Euclid's Fifth Postulate: If the two same-interior angles add up to less than two right angles, the lines will intersect on that side. This is true in spherical geometry because all lines (great circles) will intersect on both sides no matter what the angles add up to. Euclid's Fifth says nothing about what happens when the angles do add up to two right angles, only when they don't.

But still, I will mention spherical geometry as an example of a non-Euclidean geometry in which parallel lines don't work the way we expect them to.