Activity: Jeopardy! (Day 41)

I have finished reading Benoit B. Mandelbrot's The Fractal Geometry of Nature. Let me tell you a joke about Mandelbrot:

Q: What does the B. in Benoit B. Mandelbrot stand for?
A: Benoit B. Mandelbrot.

Notice that this makes Mandelbrot's name self-similar, just like the fractals that he studied. Actually, as it turns out, this no joke. Mandelbrot had no middle name, and so he intentionally show the initial so that his name could be self-similar.

The readers of this blog may wonder, why did I read and discuss Mandelbrot's book anyway? It's because in learning something new like fractal geometry (or category theory, like that book that I read over the summer), I can gain a better perspective of my students when they learn new things.

I admit that Mandelbrot's book was a tough read -- especially the second half. There were so many equations that I couldn't understand -- for example, Calculus students learn about differentiation and integration, but Mandelbrot describes something called "integro-differentiation."

But as I have trouble with fractal geometry, my students have trouble with regular Geometry. As I struggle with 2.5-dimensional objects, my students struggle with 2- and 3-dimensional objects. I don't know everything there is to know about geometry, and so I can't expect my students to know everything there is to know about geometry either. And so I can better sympathize with my students when don't understand a Geometry topic the first time I teach it. All in all, I highly recommend Mandelbrot's book to those who want to expand their horizons.

Today is an activity day. Now last year I didn't plan on having any activities between Lessons 7-2 and 7-3, but with the changes to my pacing plan, this year I want an activity at this point.

So what activity should I post? Well, all this week (from last Monday, Day 37, up to and including this Monday, Day 42), I've been subbing in the same continuation school that I was in last month (back on Day 18). And as I wrote in September, every Friday that class plays a game of Jeopardy.

And so I decided to devote today activity to yet another game of Jeopardy -- except that instead of an afterthought to the already scheduled Daffynition Game that I posted in September, this time Jeopardy will be the main activity.

In the actual classroom today, math was one of the four categories. This time, the topic for math is percents and discounts. Unfortunately, the Jeopardy game was canceled because I couldn't come up with good questions for the other categories.

Now for the Jeopardy game that I'm posting for this Geometry blog -- which of course won't be cancelled -- the four categories refer to the four lessons that I posted this week -- 6-3, the extra rotations lesson, 7-1, and 7-2. Of course, the first two lessons are so closely related, as are the last two lessons, that it may be difficult to distinguish the lessons among the categories.

Here's what I came up with for the names of the categories:
1. Rotations = 2 Reflections
2. Properties of Rotations
3. Triangle Congruence (general theorems like SAS, SSS, etc.)
4. Can We Prove Congruence? (specific numbers and shapes are given)

I've decided that I won't post a worksheet for this, because it's best done not on worksheets, but rather an index card for each question. So let me type in the questions right here in this post:

Rotations = 2 Reflections
100. A rotation centered at O maps P to P'. Why is OP = OP'? (Rotations preserve distance.)
200. A rotation maps Angle N to Angle T. Why are they equal? (Rotations preserve angle measure.)
300. The angle between two mirrors is 37. What is the angle of the clockwise rotation? (74)
400. The angle of a rotation is 172. What is the angle between the mirrors? (86)
500. The angle of a rotation is -160. What is the angle between the mirrors? (80)

Properties of Rotations
100. What is the sign of a clockwise rotation? (negative)
200. What is the sign of a counterclockwise rotation? (positive)
300. Rotate a 30-60-90 triangle about the midpoint of its longest side. What do you get? (rectangle)
400. Rotate a 45-45-90 triangle about the midpoint of its longest side. What do you get? (square)
500. Rotate a 45-degree angle about its vertex. What type of angles are 45 degrees? (vertical angles)

Is This a Valid Triangle Congruence Theorem?
100. SSS (yes)
200. AAA (no)
300. ASA (yes)
400. SAS (yes)
500. SSA (no)

Can We Prove Congruence?
100. Triangle ABC has sides 2, 8, 7 cm. Triangle DEF has sides 7, 2, 8 cm. (yes, SSS)
200. Triangle ABC has angles 30-60-90. Triangle DEF has angles 30-60-90. (no)
300. AB = DE = 20, Angle A = D = 60, Angle B = E = 30. (yes, ASA)
400. AB = DE = 20, Angle A = D = 60, AC = DF = 10. (yes, SAS)
500. AB = DE = 20, Angle A = D = 60, BC = DF = 10. (no)

Final Jeopardy
(Draw a kaleidoscope image with hexagonal symmetry. Label the center O, then label three points A, B, and C equally spaced around the center at 12 o'clock, 4 o'clock, and 8 o'clock, respectively.)
What is the angle of the rotation centered at O mapping B to C? (-120)

Now since today was an activity day rather than a test day, I really don't need to have another topic about traditionalists today. But the traditionalist Dr. Katharine Beals is still posting her Math Problem of the Week series comparing Common Core to pre-Core Geometry, and so I will address it.

Last week, Dr. Beals posted one of the two proofs that appear on the PARCC Practice Performance Based Assessment (PBA) exam. Today she posted the other one:

Marcella drew each step of a construction of an angle bisector.
9. Part A
Angle Z is given in Step 1. Describe the instructions for Steps 2 to 5 of the construction. Enter your description in the space provided.
Part B
Marcella wants to explain why the construction produces an angle bisector. She makes a new step with line segments AC and BC added to the proof, as shown.
Using the figure, prove that ray ZC bisects angle AZB. Be sure to justify each statement of your proof.

Now in each post in this series, Beals asks a few "Extra Credit Questions":

1. In how many steps can this proof be completed?
2. Does practice constructing lengthy proofs play any role in preparing for college and career in the 21st century?

We can tell by the way Beals asks the questions what answers she's expecting. Indeed, the second question contains the intended answer to the first -- she wrote "lengthy proofs," so the answer to the first is something like "too many." And we can easily figure out the intended answer to her second question as well -- "No, the long proof doesn't prepare anyone for college or career, so completing it is a complete waste of time." In other words, Beals, who laments that there aren't enough proofs in Common Core Geometry, is now implying that a Geometry proof is a waste of time!

So what's going on here? The only logical explanation is that Beals considers some proofs to be a waste of time and others to be essential to a rigorous Geometry class. So how can we distinguish between a useless proof and an essential proof?

The obvious guess -- that Beals views proofs based on reflections and other transformations to be useless and those based on triangle congruence as useful -- is invalid because we know that PARCC is expecting a proof based on triangle congruence for Part B. This is because a similar problem appears as Question 25 of the Practice End-of-Year (EOY) Exam. That question asks about the angle bisector construction, and it specifically mentions congruent triangles. Nonetheless, Beals doesn't like this proof appearing on the standardized test.

We think back to Dr. David Joyce, who also emphasizes a strong proof-based course. Here's what he writes in particular about the proof of constructions such as the angle bisector:

The book does not properly treat constructions. Constructions can be either postulates or theorems, depending on whether they're assumed or proved. For instance, postulate 1-1 above is actually a construction. On pages 40 through 42 four constructions are given: 1) to cut a line segment equal to a given line segment, 2) to construct an angle equal to a given angle, 3) to construct a perpendicular bisector of a line segment, and 4) to bisect an angle. Later in the book, these constructions are used to prove theorems, yet they are not proved here, nor are they proved later in the book. There is no indication whether they are to be taken as postulates (they should not, since they can be proved), or as theorems. At the very least, it should be stated that they are theorems which will be proved later.

So we see that Joyce -- unlike Beals -- believes that constructions such as 4) to bisect an angle should indeed be proved in class.

Beals asks her readers to count the number of steps required in the proof. Let's try writing a proof to find out:

Given: Angle Z
Prove: Ray ZC, as constructed, bisects Angle AZB.

Proof:
Statements                                         Reasons
1. Angle Z                                          1. Given
2. Construct points A, B on Angle Z 2. Compass rule
such that ZA = ZB
3. Construct point C inside Angle Z 3. Compass rule
such that AC = BC
4. ZC = ZC                                         4. Reflexive Property of Equality
5. Triangles AZC, BZC congruent     5. SSS Congruence Theorem (steps 2, 3, 4)
6. Angle AZC = BZC                         6. CPCTC
7. Ray ZC bisects Angle AZB            7. Definition of angle bisector

Now one might argue that the "compass rule" is intended to refer to the collapsible compass, which the compass that appears in the problem is not. Well, if we insist on a collapsible compass, then we can just use Euclid's proof of his Proposition I.9.

Let the angle BAC be the given rectilinear angle.

It is required to bisect it.

I.3
I.Post.1, I.1

Take an arbitrary point D on AB. Cut off AE from AC equal to AD, and join DE. Construct the equilateral triangle DEF on DE, and join AF.

I say that the angle BAC is bisected by the straight line AF.

Since AD equals AE, and AF is common, therefore the two sides AD and AF equal the two sides EA and AF respectively.

I.Def.20, I.8

And the base DF equals the base EF, therefore the angle DAF equals the angle EAF.

Therefore the given rectilinear angle BAC is bisected by the straight line AF.

Q.E.F.

Now Euclid's proof clearly also uses seven steps -- perhaps only six, since "it is required to bisect it" isn't really a step at all.

We can shorten this proof even more if we just put the steps ZA = ZB and AC = BC right into the Given information, since it's given by the use of the compass that these two statements are true.

Given: ZA = ZB and AC = BC
Prove: Ray ZC, as constructed, bisects Angle AZB.

Proof:
Statements                                         Reasons
1. ZA = ZB and AC = BC                   1. Given
2. ZC = ZC                                         2. Reflexive Property of Equality
3. Triangles AZC, BZC congruent     3. SSS Congruence Theorem (steps 1, 1, 2)
4. Angle AZC = BZC                         4. CPCTC
5. Ray ZC bisects Angle AZB            5. Definition of angle bisector

By this point, the proof looks more like a classic Lesson 7-3 proof -- which we'd think is the type of proof of which Beals approves.

Of course, we know that Beals holds the Weeks & Adkins Geometry text in high regard. I can't help but wonder how Weeks & Adkins presents the angle bisector construction -- since unfortunately, Beals doesn't provide a Weeks & Adkins text today for her usual comparison. (I'm also curious about what sort of proof Joyce expects to see for the angle bisector construction.)

So instead, let's look at the Weeks & Adkins page that Beals posted last week, and find out how many steps it takes to solve that problem:

Given: P is a point on the side AB of Triangle ABC such that AP : AB = 1 : 3. Q is the point on BC such that CQ : CB = 1 : 3. AQ and CP intersect at X.
Prove: AX : AQ = 3 : 5.

Proof:
Statements                                         Reasons
1. AP : AB = 1 : 3, CQ : CB = 1 : 3   1. Given
2. BP : AB = 2 : 3, BQ : CB = 2 : 3    2. Subtraction Property of Proportions
3. Angle ABC = ABC                         3. Reflexive Property of Equality
4. Triangle PBQ ~ ABC                     4. SAS Similarity Theorem (steps 2, 3, 2)
5. PQ : AC = 2 : 3                              5. Corresponding sides of similar triangles are proportional.
6. Angle BPQ = BAC                         6. Corresponding angles of similar triangles are congruent.
7. Line PQ | | AC                                7. Corresponding Angles Test
8. Angle QPX = ACX, PQX = CAX   8. Alternate Interior Angles Consequence
9. Triangle QPX ~ ACX                     9. AA Similarity Theorem (steps 8, 8)
10. QX : AX = 2 : 3                            10. Corresponding sides of similar triangles are proportional.
11. AX : AQ = 3 : 5                            11. Addition Property of Proportions

It might be possible to shorten this proof by using the Side-Splitting Theorem (found in Lesson 12-10 in the U of Chicago text). In particular, the converse of that theorem would allow us to jump directly from Step 2 directly to Step 7, a savings of four steps. But the problem is that we need Step 5 in order to find ratios in Steps 10 and 11 -- 2 : 3 is the bridge between the given 1 : 3 and the desired 3 : 5. So I had to stick to this 11-step proof, with both sets of similar triangles.

Beals complains about the seven-step proof of the angle bisector construction, yet has no problem with this 11-step similarity proof. Therefore, I must disagree with Beals on this one.

Now Beals might have had a good point here. One difference between the derivation of SAS from transformations (which she opposes) and the derivation of the Isosceles Triangle Theorem from SAS (which she approves of) is that the latter proof will be much shorter than the former. Our proof of the latter will contain only five steps (much shorter than in many texts), but just look at our proofs of SAS, SSS, and ASA yesterday -- seven, eight, and ten steps respectively. (I do point out that I gave different levels in detail in each proof -- SSS ought to have the longest proof, not ASA.) So we can let students avoid those long proofs by making SAS, SSS, and ASA all postulates. But then she weakened her argument by endorsing the similarity question with its 11-step proof.

It's all about what will engage the students to learn. In another post, Beals writes about how it's students with disabilities who prove that the progressive "guide on the side" classroom is ineffective, and that students will thrive in a traditionalist "sage on the stage" environment. Well, this week, I see how it's the continuation school students who prove that the traditionalist classroom isn't always the best -- the students are the least likely to listen to a teacher just because he/she's a sage on a stage.

As I said about Levy yesterday, the trick is to hook the students with an interesting result. Then we can move on to the formalism of proving the basic theorems of Geometry.