## Monday, January 29, 2018

### Lesson 9-8: The Four-Color Problem (Day 98)

Lesson 9-8 of the U of Chicago text is called "The Four-Color Problem." This lesson doesn't appear anywhere in the modern Third Edition, because this is one of those "extra" lessons that we include mainly for fun.

In the past, I've mentioned several books and lectures which discuss the Four-Color Conjecture. One of these was David Kung's lectures. Two years ago, his lectures inspired me to create a worksheet for Lesson 9-8, even though we never formally covered Chapter 9. (Thus Lessons 9-5 and 9-8 were the only ones in this chapter for which I'd made a worksheet -- until this year.) I most recently mentioned the prover of Four Colors -- Wolfgang Haken -- two months ago, in my November 27th post.

And so today I post the worksheet from two years ago -- exactly. Yes, the actual date I posted the original worksheet was January 29th, 2016. That day, I also posted a worksheet on reflections on the coordinate plane (not three-dimensional space), and so I'll repost them both today. I admit that it may be awkward to include planar reflections during the same unit that spatial reflections are taught (part of Lesson 9-5). But again, I want to prepare students for the PARCC and SBAC, and these Common Core tests are likely to ask questions about planar reflections over an axis. Unfortunately, this topic isn't fully covered in the U of Chicago text (at least not my old Second Edition, anyway).

This is what I wrote exactly two years ago about today's lesson:

Then Kung moves on to a famous theorem -- the Four-Color Theorem. Just like Lesson 12-7 "Can There Be Giants" from yesterday, the Four-Color Theorem appears in the U of Chicago text -- in fact it's Lesson 9-8. We skipped over it only because we omitted Chapter 9 entirely (as the rest of the material, 3D figures, are covered more thoroughly along with volume in Chapter 10).

The Four-Color Theorem states that any map can be colored with at most four colors. The U of Chicago tells us that the theorem holds on either a plane or a sphere. Kung points out that on other surfaces, different numbers of colors are required. On a Mobius strip, six colors are needed. Both Kung and the U of Chicago text mention that for a special doughnut shape (or is it a coffee cup shape?) called a torus, seven colors are necessary.

CCSS.MATH.CONTENT.8.G.A.3
Describe the effect of dilations, translations, rotations, and reflections on two-dimensional figures using coordinates.

Notice that eighth grade is the first in which transformations appear.

Here are the two lessons that I'm posting instead. One is based on the reflections worksheet that the students worked on in class, and the other is based on the Four-Color Theorem -- another lesson inspired by Kung's lecture. Next week, Kung will continue with topology.

Before we return to Euclid, I want to make a few MTBoS-related links. First, the Queen of the MTBoS -- Fawn Nguyen, that is -- has spoken:

http://fawnnguyen.com/how-i-use-between-two-numbers/

Hmm, I could have sworn that Nguyen already explained what "Between 2 Numbers" is and then I repeated her explanation here on my own blog, but I guess neither of us ever did it. So anyway, Nguyen writes:

It has become one of our regular warm-up routines. (We do visual patterns and math talks, too. Duh.) Before I launched the site, I was just referring to this routine with my kiddos as a “tidbit.”
Take tidbit #5, for example. I ask three questions on Google Form, the first two are identical.
Question 1 is asking for a guess, an estimate, a gut check, a what-do-you-think.
I notice that all three of Nguyen's Warm-Ups are links. And in this example, students don't merely estimate some real-world value, but submit it online. I've mentioned the Warm-Ups that Illinois State required us to give. These are given online, and I project them onto the front board, but they aren't interactive the ways Nguyen's Warm-Ups are.

The other link I wish to give here is to a member of MTBoS who actually teaches the Four-Color Theorem in class:

http://eatplaymath.blogspot.com/2015/10/the-four-color-theorem-and-pumpkin-time.html

Lisa Winer is the author of this post that is over two years old. She doesn't specify in what state she lives, nor does she make it easy for me to figure out what grade or class this is.

Anyway, in Winer's class, she uses the term "chromatic number" to describe the fewest number of colors required to fill in a map. The Four-Color Theorem, therefore, states that the chromatic number of any planar map is four. On a Mobius strip the maximum chromatic number is six, and of course on a torus the maximum is seven.

It's time to return to Euclid. Of course, he writes nothing about Four Colors or reflections across an axis, and so we proceed with the next proposition instead:

Proposition 11.
To draw a straight line perpendicular to a given plane from a given elevated point.

Propositions 11 and 12 are both constructions. Many of Euclid's propositions are constructions -- indeed, "The First Theorem in Euclid's Elements" (that is, Proposition I.1) featured in Lesson 4-4 is actually a construction.

Classical constructions are performed with a straightedge and compass, and David Joyce writes about the importance of actually proving constructions as theorems. But it's awkward to ask our students to perform a construction in three dimensions.

In this construction, we have a point A and a plane P, and we wish to construct the line perpendicular to P through A. How can our students do this? Is P the flat plane of the paper and A a point floating up in space?

It might be interesting to attempt Euclid's construction in the classroom. Here's how: We choose A to be a point on the ceiling and P is the plane of the floor. Thus our goal is to draw a point on the floor directly below A.

The key to this construction is to hang a rope from point A -- a rope that should be longer than the room is high. We can pull the rope at any angle and double-mark the points where the rope is touching the floor. I say "double-mark" because the point on the floor (where the rope touches) is marked (say with chalk), and then the point on the rope (where the floor touches) is marked (say with a piece of tape). The rope now can serve as a compass -- the point of the compass is at A, and the opening of the compass is set to the distance between A and the tape. The locus of all points on the floor that are the same distance from A as the point marked on the floor is a circle, and the locus of all points on a given line on the floor that are the same distance from A is a pair of points. So if we have a point (say B) drawn on a line on the floor, then we could find the unique point C on that line such that AB and AC are congruent.

All the lines on the floor can be drawn in chalk. There will be some plane constructions drawn on the floor as well, so we could use a large compass where the pencil has been replaced with chalk.

OK, so let's begin the construction. We start by drawing any line on the floor, and then we label any point on that line B. We now find C on this line exactly as given above -- we double-mark B on both the rope and floor, and then swing the rope to find C such that AB = AC.

Now we use the chalk compass to find the perpendicular bisector of BC. The midpoint is D.

Then we double-mark D with a second piece of tape, and then find the point on the last line we drew (that is, the perp. bisector of BC), to be labeled E, such that AD = AE. The second piece of tape must be higher up than the first since AD < AB, and so there's no danger of confusing which piece of tape is which.

Finally, we find the perpendicular bisector of DE. The midpoint is F. Euclid's G and H are any points on this last line -- their location doesn't matter. Only F is relevant here. AF is the desired line through A that is perpendicular to the plane of the floor, and F is directly below A.

Of course, this whole construction seems silly because of gravity. We can just hang a rope freely from A, label the point where the rope touches the ground F, and then we're done! The difference, of course, is that Euclid's three-dimensional space isn't physical space, and so there's no direction that's "favored" because of gravity or any physical force.

And so I'm not quite sure how David Joyce has in mind when he says he wants "the basics of solid geometry" to be taught better. Does he include Euclid's spatial constructions -- does he really want students to perform them? Or maybe he merely desires that students visualize the proofs in their minds while looking at the proof.

(Do you remember Euclid the Game, which is played on computers? Maybe in higher levels, players can make three-dimensional constructions that are difficult to perform in the real world!)

By the way, we can still modernize Euclid's proof:

Given: the segments and angles in the above construction.
Prove: AF perp. plane P

Proof:
Statements                              Reasons
1. bla, bla, bla                         1. Given
2. BC perp. plane (ED, DA)   2. Proposition 4 from two weeks ago
(Call it plane Q. In the classroom, Q is an invisible plane parallel to a wall.)
3. GH | | BC                            3. Two Perpendiculars Theorem (planar version)
4. GH perp. plane Q               4. Perpendicular to Parallels (spatial, last week's Prop 8)
5. AF in plane Q                     5. Point-Line-Plane, part f (A, D, F all in plane Q)
6. GH perp. AF                       6. Definition of line perpendicular to plane
7. AF perp. plane (GH, DE)   7. Prop 4 (AF perp. DE is part of "Given")
8. AF perp. plane P                 8. From construction (both lines were drawn in plane P)

It might be tricky to reconcile this proof with the "rope" construction from above. In Euclid's construction, AD is designed to be perpendicular to BC, likewise AF is perp. to DE. Both of these perpendicular constructions technically occur in planes other than the floor -- yet earlier I direct you to perform perpendicular constructions on the floor -- which is the wrong plane.

But think about it -- given a point A and a line, how do we construct a line through A perpendicular to the given line? The answer is that, using the compass, we find points B and C on that line that are equidistant from A, and then find the perpendicular bisector (in that plane) of BC.

But technically, all we really need is D, the midpoint of BC. Then the line through points A and D is automatically the perpendicular bisector of BC in the correct plane. It doesn't matter how we obtain the midpoint D -- all that matters is that we find it. This includes finding the perpendicular bisector of BC in the wrong plane (that is P, the plane of the floor). This is why Euclid is able to assert and use statements like AD perp. BC in his proof, even though this isn't obvious from our ropes. (And as it happens, the perpendiculars in plane P appear later in the proof anyway, so we might as well construct these.)

In the end, let's just stick to the Four-Color Theorem and two-dimensional reflections. Here are the worksheets for today: