Friday, October 12, 2018

Lesson 4-2: Reflecting Figures (Day 42)

In the district whose calendar the blog is following, there is no school for students on Monday. It appears that this will be a PD day for teachers, but a day off for students and subs.

Notice that this off-day has nothing to do with Columbus Day. In previous years, I used the term "Columbus Day" to refer to certain days in October when schools were closed, but those were in fact different districts. And indeed, Columbus Day was observed on Monday, October 8th, but my current district will be closed a week later, on Monday, October 15th.

We also notice that the first day of the semester was of course Day 1, and I already gave last day of the semester as Day 83. And so Day 42 is exactly halfway between 1 and 83. This shows us a more likely reason as to why Monday is a no-school day -- it represents the start of the second quarter.

(Oh, and why did I post last Wednesday -- another day with no math class -- yet I won't this Monday? It's because Wednesday was Day 40, but Monday will remain an unnumbered day, with Day 43 on Tuesday.)

Meanwhile, today I subbed in a seventh grade history class. It is, in fact, the very same class that I subbed in back on Tuesday. Of course, this is in my new district, where today is Day 33. Thus it's not the end of the first quarter, but more like the end of the first hexter (in other words, the midpoint of the first trimester).

As I often do for repeated and multi-day sub visits, the focus resolution defaults to the first one, on classroom management. So let's do another "Day in the Life":

Today I subbed in a seventh grade history class. It's at the same middle school that I've been subbing a lot lately, including on Friday. Since this is at a middle school, I wish to focus more deeply on classroom management, and thus I'm doing "A Day in the Life" right now.

Oh, and it appears that this teacher will be out on Monday as well. But because there's no school that day in my old district, I won't post anything at all on Monday, much less "A Day in the Life."

7:05 -- That's right, this teacher has zero period. And if you remember from Tuesday, for some reason, it's P.E. instead. Once again the students have free play -- they can either walk laps or play with a ball or Frisbee. Oh, and like Tuesday, I actually have to supervise the P.E. class today, especially considering that I must cover for the only zero period P.E. teacher.

In the morning, I do see the regular teacher on his way to a meeting. He's frantically entering grades into the computer -- after all, what did I just say about it being the end of the first hexter, with progress reports about to come out?

I ask some of the students why they must have a zero period P.E. class. They tell me it's so that they can have a second elective class.

8:15 -- This is homeroom. I show the class the announcements.

8:20 -- As is usual for middle schools, there is a rotation today, and it starts with second period. At this particular school, students report to third period for homeroom.

Second period is the first history class. All of this teacher's history classes are honors, and so there are very few behavior problems. Here in California, seventh grade is Medieval World History, and one major unit is the origins of Islam. This time, instead of a video, the students are working on an interactive chart on eight different aspects of Islamic culture

9:15 -- Second period leaves. Third period is this teacher's other P.E. class. Because it's Friday, the students are doing something different. This time, all P.E. classes at that time join together. After one lap on the track, they can have free play outside or enter the gym to play "Castleball." I never see what this game is like, since I decide to walk an extra lap with some of the students, and by then the door to the gym is closed.

10:10 -- Third period leaves, and the short break begins. This leads into the fourth period conference, and unlike Tuesday, I'm not given another class to cover at this time.

11:15 -- Fifth period arrives. This is another honors class.

12:05 -- Fifth period leaves and lunch begins. This leads into the sixth period conference -- again, this teacher has two conference periods because he has a zero period.

2:00 -- First period arrives. This is another honors class.

2:55 -- My school day ends, and I go home to type this blog entry.

There are two main management issues today. The first is in zero period P.E., where I learn of the fallout due to the kids running out to the blacktop on Tuesday. When the regular teacher returned on Wednesday, he assigned them an essay to write.

Of course, punishment essays remind me of my charter class from two years ago. Oops, ixnay on the arterchay -- wait a minute, side-along reading is over, so I can bring up my old charter class again!

I've been thinking about why many of my punishments from that year -- including punishment essays for the whole class -- were ineffective. Since then, I've realized that if I show that I won't punish bad behavior in time, then I'll lose control of the class. But once I lose control of the class, nothing -- not even punishing bad behavior -- will make me regain control of the class easily.

When this teacher made his P.E. students write lines on Wednesday, the whole class probably became silent quickly as they realized the seriousness of their misbehavior. The students acted up, but he still has control of his class.

But when I made my students write lines two years ago, they continued to talk loudly even while they were writing them. I've lost control of the class, and not even punishing the students allowed me to regain control. Also, some special ed students that year had trouble writing the paragraphs. At today's school, I doubt that these zero period kids (who are taking seven classes) are special ed students.

Eventually, two years ago the director (principal) told me that the essay-writing was taking too much time away from teaching math. Today's teacher didn't take too much time away from P.E. -- just one day of punishment is enough to get the kids back on track.

In fact, I end up naming zero period the best P.E. class of the day today. Meanwhile, once again the best history class is the one that meets just before lunch (in this case fifth period).

The other problem occurs in first period. In the sub plans, the regular teacher specifies that students are not to go to the restroom unless it's an emergency. I know better than to ask the students who want to go whether it's an emergency, since they'll always claim that it is indeed an emergency, and then I'll have ten or more "emergencies" in one day. My rule of thumb is that if more than one student in a period claims an emergency, it's likely not a real emergency. The probability of two true emergencies in one period approaches 0.

To discourage students from claiming an emergency when it isn't an emergency, I write down the name of any student who claims such an emergency for the regular teacher. When the teacher wants as few students to leave the room as possible, he should know exactly who's leaving the room.

On Tuesday, only one student asks for a restroom pass. Today, two students in fifth period ask to go, but fortunately the second student asks less than ten minutes before the end of class, so it was easy for me to convince the second student to wait.

But then two students ask to leave near the midpoint of first period as well. Based on the way the second student asks, I believe that he has a genuine emergency. (There's no reason for me to post any details on a blog, since he wouldn't want me to tell the whole world.)

As for the first student, let me describe his behavior before he asks to go to the restroom. He starts disturbing other students, including a girl who sits two seats away. It takes him a while even to take the chart he's supposed to be working on out of his backpack. It appears that he doesn't feel like working and just wants to take a quick breather by claiming he needs to go to the restroom. And I let him go, since at the time he was the first student who asks.

I think back to another day I had problems with students asking for restroom passes when the teacher specifies a strict restroom policy in the lesson plans -- March 20th. There are a few differences between March 20th and today, though.

Days like this make me think about how I'd like to come up with my own restroom policy if I ever return to teaching full-time someday. I wrote earlier than in any particular period, the probability of two real emergencies in one period approaches 0. But as a regular teacher, five periods a day for 180 days would be 900 periods. The probability of two real emergencies in one of the 900 periods in a school year actually approaches 1, because there are so many periods. So I fear that if I came up with my own "one emergency per period," I'd be burned the day that there are two genuine emergencies in the same period.

Another fear I have is that one day, the first person comes up with some lame excuse of an emergency while the second person has a genuine emergency. I might let the first person go and not the second only because he asked a minute too late, even though his emergency is genuine. It's possible that this is exactly what happens today. In this case, he tells me that his previous period teacher (also a sub) didn't let him use the restroom, and so he waited until my class to go.

When the first student returns, I complain that his emergency wasn't real. I write his name on the bad (detention) list -- not just for using the restroom, but for disturbing other students earlier. This is another issue of line-crossing -- even though I'm punishing him for disturbing others, I make it appear that he "crossed the line" by asking for the restroom pass.

Two years ago, I wanted to avoid the "first person false, second person genuine" problem. So in the middle of the year, if one person in a week asks for a restroom pass, there's no punishment. If a second person asks for a pass in a week, then I'd give a short detention to both students who ask for the passes. If a third person asks in a week, then I'd give a harsher punishment to all three students who ask for the passes. The idea is to discourage someone from asking for a pass on Tuesday and thereby block someone with a genuine emergency from asking on Wednesday or Thursday.

But in the end, the students are confused. They continue to ask for passes early in the week and then get angry when they suddenly get punishments later in the week. Sometimes it's the same student who asks early and late in the week. In fact, it now appears that any restroom policy that depends on how many previous students have asked for passes is doomed to fail -- it doesn't discourage students from coming up with flimsy excuses early in the period/day/week.

In the end, the student agrees to be good for the rest of the period, and so I take his name off of the bad (detention) list. I calmly explain why I'm writing the note to the teacher. I tell him that the sub plans for Monday include that same emergency rule -- and on that day, his class will rotate into the period right after lunch. Hopefully, next week he'll go during lunch.

Meanwhile, today on her Mathematics Calendar 2018, Theoni Pappas writes:

The area of the 6 regular congruent hexagons is 108 sq. units. What is the area of the black triangles?

(The only relevant given info from the diagram is that there are four black triangles, and every side of every triangle is also a side of one of the hexagons.)

First, simple division tells us that each regular hexagon has an area of 18. The key to this problem is knowing that a regular hexagon can be divided into six congruent equilateral triangles, each with the same side length as the hexagon. (Think in terms of its angles -- each angle of an equilateral triangle is 60 degrees, which is 360/6. Thus six triangles can be put together to form the hexagon.)

So the area of each triangle is 3, and since there are four such triangles, the desired area of all of them is 12 square units -- and of course, today's date is the twelfth.

OK, let's finally get to the U of Chicago text!

Lesson 4-2 of the U of Chicago text is called "Reflecting Figures." This is what I wrote two years ago about today's lesson:

Let me state the Reflection Postulate, because it is so important, directly from the U of Chicago text:

Reflection Postulate:
Under a reflection:
a. There is a 1-1 correspondence between points and their images.
This means that each preimage has exactly one image, and each image comes from exactly one preimage.

b. If three points are collinear, then their images are collinear.
Reflections preserve collinearity. The image of a line is a line.

c. If B is between A and C, then the image of B is between the images of A and C.
Reflections preserve betweenness. The image of a line segment is a line segment.

d. The distance between two preimages equals the distance between their images.
Reflections preserve distance.

e. The image of an angle is an angle of the same measure.
Reflections preserve angle measure.

This postulate corresponds to Dr. Franklin Mason's "Rigid Motion Postulate," in the old version of his Lesson 3.1 last year. Since then, Dr. M has completely changed his Chapter 3 -- this is almost certainly because isometries ("rigid motions") aren't emphasized on the Common Core texts nearly as much as either of us thought they would when we first read the standards. Nowadays, Dr. M uses the classical definition of congruent polygons (i.e., equality of corresponding measures). He assumes SAS as a postulate (just as the mathematician Hilbert did a century ago), and uses Euclid's ancient proof to derive ASA. But for SSS, Dr. M still uses rigid motions to move one of the triangles into place (similar to the start of the U of Chicago proof) before using SAS and Isosceles Triangle Theorem to prove SSS.

Part a is a very important part of the Reflection Postulate. Without it, a point A could have two reflection images -- there could be two distinct points B and C such that the reflecting line m is the perpendicular bisector of both AB and AC. (I believe that the Ruler and Protractor Postulates, or their U of Chicago equivalents, are sufficient to prove the existence of at least one reflection image, but to prove that at most one reflection image exists requires the Reflection Postulate.) This was a problem for Dr. Hung-Hsi Wu, who decided to define rotation before reflection and then use the properties of rotations to prove that every point has at most one reflection image. But since I want to use reflections to define rotation, I am forced to assume that reflection images uniquely exist as part of the postulate to avoid circularity.

According to the text, reflections preserve:
Angle measure
Betweenness
Collinearity
Distance

a nice little mnemonic for the students.

The first theorem of this chapter is the Figure Reflection Theorem:
If a figure is determined by certain points, then its reflection image is the corresponding figure determined by the reflection images of those points.

This theorem is used to conclude, for example, that if A' is the image of A, and B' is the image of B, then A'B' is the image of AB. An informal proof is given. A formal proof is a little bit tricky -- sure, we know that if C is between A and B, then C' is between A' and B'. But the problem is the converse -- if there's a point D between A' and B', how do we know there's a point C between A and B such that C' is exactly D? The best way is probably to use the Distance part of the Reflection Postulate -- we choose C to be the point on AB such that AC = A'D (which exists by Ruler Postulate). Then since reflections preserve distance, C' must be exactly D.

But that is the sort of proof that I don't want to confuse students with. It's best just to use the informal proof given in the text and save formal proofs for later.

The text states that when a figure intersects the reflecting line, the image must intersect the reflecting line in the same point or points. This follows immediately from the fact that the image of a point on the reflecting line is the point itself.

But what I find interesting is the related statement -- if a figure intersects its reflection image, then it must intersect the reflecting line in the same point or points. This statement is false in general, but it's true if the figure to be reflected is itself a line. This fact helps us greatly -- for example, consider Question 21 from the text:

The reflection image of Triangle ABC is Triangle XYZ. Now Lines AC and XZ intersect at a point -- which we now know must lie on the reflecting line. And Lines BC and YZ intersect at a point -- which we now know must lie on the reflecting line as well. And those two points determine exactly one line -- the reflecting line! So all the student has to do is draw the line through the two points of intersection.

And as a corollary, it follows that if a line is parallel to the reflecting line, it must be parallel to its reflection image, Last year, I called this the "Line Parallel to Mirror Theorem" (where "mirror" refers to the reflecting line) But I will wait a few days before introducing that theorem to students. (Notice that the Common Core Standards state that a line must be parallel to its dilation image, so why not give the conditions when a line is parallel to its reflection, rotation, or translation images?)

Today is an activity day. Two years ago, I posted several activities after Lesson 4-4, and so I might as well reblog those activities today, in addition to today's Reflection Postulate worksheet. This is what I wrote two years ago about the activities:

Meanwhile, I created worksheets for the proof of reflection over the axes. This is all in addition to the original worksheet for today, the Centauri challenge.

The following activity is another one from Michael Serra's text -- it appears in his Chapter 15, since this one often goes with two-column proofs (and proofs appear in his book late). In this activity, strings consisting of the letters P, Q, R, and S are converted into others using four rules (that end up being our postulates):

Rule 1. Any two adjacent letters in a string can change places with each other. (PQ=>QP)
Rule 2. If a string ends in the same two letters, then you may substitute a Q for those two letters. (RSS=>RQ)
Rule 3. If a string begins in the same two letters, then you may add an S in front of those two letters. (PPR=>SPPR)
Rule 4. If a string of letters starts and finishes with the same letter, then you may substitute an R for all the letters between the first and last letters. (PQRSP=>PRP).

Then the text gives the following theorem, PQQRSS=>QRQ. (Notice that I chose to write "=>" where the text writes ">>" since, after all, we've already used the former to denote the hypothesis and conclusion of a conditional.)

Proof:
Statements            Reasons
1. PQQRSS          1. Given
2. PQQRQ            2. By Rule 2
3. QPQRQ            3. By Rule 1
4. QRQ                 4. By Rule 4

So for the students, this is a puzzle which gets them thinking about the logical structure of proofs without having to think about geometry.

The text calls this the "Centauri challenge," which I assume refers to Alpha Centauri, the closest star system to the sun. Notice that many of the Cooperative Problem Solving challenges in Serra are said to take place in a futuristic lunar colony. For this one, the inhabitants of this colony are trying to communicate with aliens from (Alpha) Centauri, but apparently, the Centaurian alphabet consists of only four letters.

My worksheet contains all of Challenge 1, then adds Challenge 2 as a Bonus. In case you're curious, here are my answers to Challenge 2.

1. Can you produce a string of five or more letters that cannot be reduced to RQ?

My answer is that I can't -- but now I must prove it. Here is my proof, in paragraph form:

Proof:
Our string has five letters, but there are only four letters available. So one of those letters must appear at least twice! (This is called the Pigeonhole Principle.) Let's call the letter that appears twice X. (I know, it's actually P, Q, R, or S, not X, but here I'm using X as a variable to stand for one of the letters P, Q, R, or S, since I want this proof to be as general as possible.)

Using Rule 1, we take the first appearance of X and change it with the letter on its left. Now take that X and change it with the new letter on its left. Keep doing this until the X is the first letter. Now take the last appearance of X and change it with the letter on its left. Keep doing this until this other X is the last letter.

Now our string begins with X and ends with X. So by Rule 4, it becomes XRX. Using Rule 1, we can change the first X and the R to obtain RXX. Finally, this string ends in the same two letters (whatever letter the X stands for), so by Rule 2, it becomes RQ. QED

After doing this, the second question becomes obvious.

2. One of the rules of Centauri can be removed without losing any of the first five theorems proved in Challenge 1. Which of the rules can be removed?

Look at which rules have been used in the post so far, and notice which one's missing. That's the rule that can be removed. I like the similarity between determining which theorems are provable with or without a certain rule, and finding out which theorems in geometry require a Parallel Postulate. [2017 update -- this is what Russell, Whitehead, Godel, Erdos etc. did in general -- try to figure out what proofs require which axioms, or postulates, or rules.]

Here is the activity:






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