Monday, December 10, 2018

Lesson 7-6: Properties of Special Figures (Day 76)

Today I subbed in a high school special ed class. Like many special ed teachers, this regular teacher has one period of co-teaching (a sophomore World History class). His own classes are one section each of Government and Econ (the two one-semester senior social science class) and two sections of Integrated Science.

(Recall that in this district, "Integrated Science" is for special ed students only -- even though I myself took Integrated Sci as a young high school student. There are students of all grade levels in these classes, but the majority are juniors.)

All four of the regular teacher's classes have the same assignment on Mondays -- a current event, which they look up on Chromebooks. This is after the students watch "CNN 10," a daily ten-minute news video intended for school students. (From my experience, special ed teachers are apparently more likely to show "CNN 10" than gen ed teachers.) For the two senior classes, a representative from Jostens gives a presentation on caps, gowns, and class rings. (He even shows the students an actual Super Bowl ring, produced by Jostens.)

There will be no "Day in the Life" since most of the classes have a special aide. This includes the class I must cover during my conference period -- which turns out to be yet another special ed class (this time a self-contained class). During the period I'm in there, three students in this class are giving a presentation on "How I Spent My Weekend."

But seventh period, the last class of the day, has no aide. It's an Integrated Science class with mostly juniors (so there's no Jostens this period either).

The students are supposed to complete the current event assignments on forms. Earlier in the day, the aide notices that there aren't any forms, so she makes more copies -- but unfortunately, she neglects to make enough copies to last all the way through seventh period. I have one of the students pass out the CE forms, and he realizes that there's only enough for about half of the 18 students in this class. (By contrast, the other Integrated Science class has only four students -- and two of them are absent!)

I also point out that sometimes, the students complete the CE forms in Google Classroom rather than on paper forms -- but there's no place for this week's CE forms on Google. (Keep in mind that the regular teacher probably went home Friday thinking he'd be here today, so he had no reason to set up Monday's CE assignment in advance, either on paper or online.)

This sets up a domino effect. One guy refuses to work on his current event -- he just starts playing nonacademic videos on YouTube instead of searching for an article. I tell him that he's required to answer the first three questions on the CE, and he says that he will -- just to get me off his back. He has no intention of stopping the video. Of course, if I try to force him to do the current event, all he has to do is point to the side of the classroom without CE forms and insist on not working, just as those students aren't.

And then some students start packing up with about six or seven minutes to go! Part of it is because a student with a walker is allowed to leave early with her assistant. But it's more because I check to see whether the other guy has answered his three questions, and so many students associate checking work with time to go. Students are now lined up at the door wondering what's taking so long for the bell to ring (answer: they lined up too soon). Three students leave early -- I catch one girl. She insists that she should be allowed to wait outside for the bell -- after all, two others are already gone.

In the past, I've asked, is it a good idea to ask students to answer a certain number of questions to avoid being written on the bad list. We've seen before that, especially in special ed classes, I might set the bar too high and hard-working students are unable to reach it. But in this case, I only require the students to answer the following three questions:

  1. What is the title of the article?
  2. What is the source of the article?
  3. What is the date of the article?
Students should be able to answer these trivial questions in under ten minutes, if not five. Any student who fails is most likely doing something nonacademic.

But the main issue here is, what should I do if I'm passing out papers and run out of copies. It's possible that had I not lost my conference period, I might have noticed the lack of papers during the break -- enough time to find a Xerox machine. (This has happened recently to me while subbing at another school.)

A simple solution might have been to drop current events completely to play the "Conjectures"/"Who Am I?" game. (I don't want to wait until my next birthday to play it -- next year, my birthday will fall on Saturday, and I'll have to wait two years, until my 40th in 2020, for it to fall on Monday.) Of course I'd inform the regular teacher of the reason to deviate from his plan (the lack of copies). It might have been tricky to come up with Integrated Science questions -- I see a General Science text in the classroom but I don't know what page they're on. I can always stall for time with the "What's my age/weight?" questions with "Let me give you time to guess," while I search the teacher's desk for evidence of a science lesson.

Lesson 7-6 of the U of Chicago text is called "Properties of Special Figures." (In the new Third Edition of the U of Chicago text, this is Lesson 7-7. Oh, and the title of the new Lesson 7-7 makes it clear that the "special figures" referred to here are parallelograms.)

I've made many changes to this lesson over the past four years. (The new Third Edition also blows up the old Chapter 13 by moving all of its topics to other chapters. Ironically, Chapter 7 isn't one of them -- Lesson 7-6 of the new edition, on tessellations, appears in Chapter 8 of the old edition, and the other two extra lessons in the new edition don't appear in the old edition at all.)

Anyway, this is what I wrote last year about today's lesson:

The unit test will not be given until just before winter break, leaving us with an extra week to fill. So I asked myself, what else can I fit into this unit?

And so I decided just to post the parallelogram properties after all. This week we cover Lesson 7-6, on the Parallelogram Consequences, and Lesson 7-7, on the Parallelogram Tests.

We've already seen how useful the Parallelogram Consequences really are. The reason that they are delayed until 7-6 in the U of Chicago text is that they are best proved using triangle congruence, but triangle congruence doesn't appear until Chapter 7.

It would probably make more sense to cover parallelograms along with the other quadrilaterals in Chapter 5, but it's too late now. Fortunately, I have this opening in the schedule now to cover both Lessons 7-6 and 7-7.

And today's Lesson 7-6 also includes the Center of a Regular Polygon Theorem, which the U of Chicago text proves using induction. This fits with the Putnam-based lesson that I posted last week.

Oh, and by the way, I found the following page about the Fibonacci sequence and generalizations:

(That's right -- this is my second straight week on the same Putnam problem.) It shows why numbers of the Fibonacci sequence have other such numbers as factors -- it's because the Fibonacci numbers can be written as polynomials that can be factored to get other Fibonacci polynomials.

So how does Dr. Merryfield solve Putnam problems like 2014 B1 or 2015 A2? As it turns out, he uses a proof technique called mathematical induction. Most proofs in a high school geometry course aren't proved used mathematical induction. Indeed, only one proof in the U of Chicago text is proved this way -- and it just happens to be the Center of a Regular Polygon Theorem! Furthermore, Dr. Wu uses induction in his proofs on similar triangles, so this is a powerful proof technique indeed. (On the other hand, Wu simply defines a regular polygon as an equilateral polygon with a circle through its vertices, so he would have no need to prove the following theorem at all.)

[2018 update: I retain this discussion from 3-4 years ago about old Putnam problems in order to demonstrate mathematical induction. In a way, this year's 2018 A6 is sort of like induction in the way we prove certain numbers are rational, but not quite. Last year, I wrote about 2017 A1 where we proved that certain numbers are "sexy," and we can do the same with 2018 and replace "rational" with "sexy." So the sum of two sexy numbers is sexy, as is their difference, product, and so on. Oh, and RIP Dr. Merryfield yet again.]

So here is the proof of the Center of a Regular Polygon Theorem as given by the U of Chicago -- in paragraph form, just as printed in the text (rather than converted to two columns as I usually do):

Analyze: Since the theorem is known to be true for regular polygons of 3 and 4 sides, the cases that need to be dealt with have 5 or more sides. What is done is to show that the circle through three consecutive vertices of the regular polygon contains the next vertex. Then that fourth vertex can be used with two others to obtain the fifth, and so on, as many times as needed.

Given: regular polygon ABCD...
Prove: There is a point O equidistant from ABCD, ...

Draw: ABCD...

Write: Let O be the center of the circle containing AB, and C. Then OA = OB = OC. Since AB = BC by the definition of regular polygon, OABC is a kite with symmetry diagonal OB. Thus ray BO bisects angle ABC. Let x be the common measure of angles ABO and OBC. Since triangle OBC is isosceles, angle OCB must have the same measure as angle OBC, namely x. Now the measure of the angles of the regular polygon are equal to 2x, so angle OCD has measure x also. Then triangles OCB and OCD are congruent by the SAS Congruence Theorem, and so by CPCTC, OC = OD. QED

Now the "Analyze" part of this proof contains the induction. If the first three vertices lie on the circle, then so does the fourth. If the fourth vertex lies on the circle, then so does the fifth. If the fifth vertex lies on the circle, then so does the sixth. If the nth vertex lies on the circle, then so does the (n+1)st. I point out that this is induction -- from n to n+1.

Every induction proof begins with an initial step, or "base case." In this proof, the base case is that the first three points lie on a circle. This is true because any three noncollinear points lie on a circle -- mentioned in Section 4-5 of the U of Chicago. The induction step allows us to prove that one more point at a time is on the circle, until all of the vertices of the regular polygon are on the circle.

We have seen how powerful a proof by induction can be. We have proved the Center of a Regular Polygon Theorem. But now we wish to prove another related theorem -- one that is specifically mentioned in the Common Core Standards:

Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.

What we wish to derive from the Center of a Regular Polygon Theorem is that we can rotate this polygon a certain number of degrees -- about that aforementioned center, of course -- or reflect it over any angle bisector or perpendicular bisector. But the U of Chicago text, unfortunately, doesn't give us a Regular Polygon Symmetry Theorem or anything like that.

I'm of two minds on this issue. One way would be to take this theorem and use it to prove that when rotating about O, the image of one of those isosceles triangles with vertex O and base one side of the polygon is another such triangle. The other way is to do Dr. Wu's trick -- he defines regular polygon so that it's vertices are already on the circle. Then we can perform rotations on the entire circle. (Rotations are easier to see, but it's preferable to do reflections because a rotation is the composite of two reflections.) Notice that the number of degrees of the rotation depends on the number of sides. In particular, for a regular n-gon we must rotate it 360/n degrees, or any multiple thereof.

The modern Third Edition of the text actually mentions rotation symmetry. This section tells us that a parallelogram has 2-fold rotation symmetry, and the statements about regular polygons actually appear in Lesson 6-8 of the new edition.

Before we end this post, let me add the "traditionalists" label. Our main traditionalists have posted over the weekend:

I wrote this 4 years ago regarding a column in USA Today. I commented and got into an argument with Linda Gojak, former president of NCTM. She presents the usual obfuscation and claims as evidence that students lack ‘understanding” because they cannot apply procedures in a variety of different problem solving situations. Well, if you ignore the novice-expert spectrum and put an expectation of expert thinking on novices, then yes, there’s your evidence I guess.

Now here Garelick quotes himself from four years ago. But who am I to complain, considering that I just quoted myself from four years ago in our Lesson 7-6 discussion.

Linda Gojak, former president of NCTM, decides to answer my comment on a comment she made in response to someone else and … Where was I? Well, it was a USA Today article proclaiming that Common Core math is not fuzzy.
Here’s what I said: “Linda Gojak Some understanding is critical, but not all. Sometimes procedural fluency leads to that understanding. It works in tandem. “
Recall that NCTM is strongly Pro-Core -- in fact, this is why the U of Chicago text resembles the Common Core. Both were influenced by NCTM.
We might as well skip to SteveH's comment, since you know that's where we're going:
Linda Gojak is a past president of NCTM with a, BSEd, Elementary Education and Teaching, from Miami University.
Kids struggle in math because schools don’t enforce mastery of basic skills in K-6, no matter how much talk of understanding and in-class group work they offer – which has now been going on in K-6 for 2+ decades! All proper skills require and ensure some understanding. After that, students still might have understanding/flexibility issues. The solution is not top down, but bottom up from mastery of scaffolded skills, and Ms. Gojak has no clue what we mathematicians, scientists, and engineers went through and really know about the many levels of understanding. We don’t expect kids to be able to add and subtract in binary or octal, even though they already know something about dozens and 60 minutes in an hour. Do we expect them to know why months have different numbers of days?
I assume SteveH means "dozenal or sexagesimal," since those bases fit better with his later references to dozens and 60's.
No, SteveH, we don't need to know other number bases or why months are different lengths to use a calendar. Nor do they need to know how arithmetic works in order to use arithmetic. But if they view arithmetic as a set of unrelated procedures, then they're likely to get confused at to which procedure at to use and at what time.
They need to first ensure individual mastery of homework P-sets in K-6 (which they supposedly believe in) and then we can talk about higher levels of understanding that might be missing.
P-sets that are left blank lead to neither learning nor understanding. Just ask that student whose class I subbed in today. How much does he learn about current events by leaving his CE sheet blank today?

Friday, December 7, 2018

Lesson 7-5: The SSA Condition and HL Congruence (Day 75)

Today is a special day, for those of you familiar with my blog over the years. It is my 38th birthday.

Also, today I subbed in a high school English class (different school from yesterday). There are two freshman English classes and an English as a Second Language (ESL) class -- a single class that lasts for two periods. There are students of all high school grades in this class.

I won't do "A Day in the Life" today, but I will describe the classes. All classes have a vocabulary quiz today -- ten words for ESL, twenty for the freshmen. All the words on each quiz start with the same letter -- "D" for ESL, "F" for the freshmen. (D and F -- those are the last letters I want to think about at quiz time!) In the second hour, the students work on Chromebooks (ESL, Studio 44). The last period of the day is team football. As I often do, I actually lift some weights with the athletes.

One thing I enjoyed back on my 36th birthday at the old charter school is the Conjectures game, also known as "Who Am I?" (I'd established this tradition the previous year, my 35th, when I was subbing in a math class.) And so I can't resist playing the game today as a review -- even though I only rarely played it in classes other than math. It also gives the students something to do, rather than just say "review your vocabulary for 20-25 minutes" (after which the students would talk or play games on their phones the whole time).

In second period ESL, the game doesn't work very well, probably because most of the students have very low English skills. Most of the students will apparently fail the quiz anyway. In the two freshman classes the game works out much better. The classes are a bit loud, but this game encourages it. This time I begin with "guess my college" (easy since I'm wearing a UCLA shirt). Then I do the usual "guess my weight" and of course, "guess my age" (since it's my birthday).

I name third period as the best class of the day, since it isn't quite as loud as fifth period. Third period is also more engaged with the quiz review. Recall that some questions in this game are answered as a "race" (first group to answer gets the point) while others require a response from every group. One of the latter type asks students to give the definition of "fawn." Some groups in third period answer "a spotted deer" while others say "yellowish-brown in color." Well, the original vocab worksheet lists "fawn" as an adjective, and so I only accept the color.

(Readers of this blog should already know that "fawn" is a color, since it's one of Kite's old colors in his notation for 19-limit music. And of course "Fawn" is also the name of one of my favorite math teacher bloggers, Fawn Nguyen.)

In the end, I like how my "Who Am I?" game often leads to a great discussion among the students -- even if it is a bit loud. Once again, this is why my game is much better suited to introducing new concepts rather than reviewing them before a test.

As it turns out, one student in this class is also celebrating his birthday today. He's a freshman, so this is his fifteenth birthday (just 23 years younger than I am). He's in the group that wins the "Who Am I?" game and so I write his name on my good list for the teacher -- and once again, he's in the best-behaved period of the day. But the birthday boy doesn't do well on his test, answering only three of the 20 vocab questions.

Some students believe that they're entitled to a non-academic free day on their birthday. I was never that student -- I always worked as hard as a young student on December 7th as I did any other day. But I recall one year when I was student teaching an Algebra II class. The master teacher and I decided to give the test the Tuesday before Thanksgiving (at the time, most schools in the area had classes through Wednesday of Thanksgiving week), since the next day was a minimum day and we figured that the students wouldn't concentrate on the test. But one girl refused to take the test because it was her birthday. In fact, she was perfectly willing to take the test on Wednesday, the day before the holiday, because it was no loner her birthday. And I think she passed the test with flying colors!

I don't ever remember having a major test on my birthday. But I do recall that in college, I had to take two major tests on December 8th (the GRE to get into grad school at UCLA, and two years later, the final exam for a set theory class). I have a funny story about the GRE. I knew that the GRE had math and verbal sections, but I didn't even know about the logic section until the night before the test (that is, my birthday night)! So I had to cram for an entire section on my birthday. Still, I know I must have passed that section because I obviously got into grad school at UCLA. (Note that the GRE logic section was eliminated the year after I took the test.)

Today, my 38th, is a milestone birthday. It is my second Metonic birthday, referring to the 19-year cycle of the moon. This means that the phase of the moon on my 19th, 38th, 57th, etc., birthdays should match the lunar phase on the day I was born. Today is a new moon, and so we conclude that I must have been born under a new moon as well.

(Notice that in my own Pacific Time Zone, the new moon occurred at 11:20 last night, but in every other time zone except Pacific, the new moon is today. If Proposition 7 were in effect, then the new moon would have been 12:20 AM on the 7th in California as well.)

New moons around this time of year occur during the Jewish holiday of Hanukkah. Observant Jews light the sixth Hanukkah candle tonight -- and typically, the new moon occurs around the sixth or seventh night of the holiday. Thus we conclude that the night I was born was probably also the sixth or seventh night of Hanukkah that year.

I actually remember learning about the Metonic cycle during my senior year in high school. I had obtained a free discarded book from our school library. It was a recreational math book that contained puzzles, including charts to help calculate the date of Easter. It was then that I learned about the 19-year cycle. I looked ahead on the calendar to the following December and noticed that my birthday would fall on a new moon. And this was my nineteenth birthday -- my first Metonic birthday. And so I realized that I must have been born under a new moon.

There's one more thing I want to say about 38th birthday. In one Simpsons episode, Homer remarks that since he is 38.1 years old and the average male lifespan is 76.2 years, he's exactly halfway through his life -- the definition of middle-aged. A Google search reveals that the average lifespan of an American male is now 78.6 years old, so I won't be middle-aged until next year, at age 39.3.

Lesson 7-5 of the U of Chicago text is called "The SSA Condition and HL Congruence." This lesson introduces the final congruence theorems.

This is what I wrote last year about today's lesson:

Lesson 7-5 of the U of Chicago text is on SSA and HL. I've already mentioned how I'll be able to prove HL without using AAS, since we have to wait before I can give an AAS proof.

Meanwhile, we know that SSA is invalid, but the U of Chicago text provides us with an SsA Congruence Theorem, where the size of the S's implies that it must be the longer of the congruent sides that is opposite the congruent angle. The text doesn't provide a proof of SsA because the proof is quite difficult -- certainly too difficult for high school Geometry students.

But you know how I am on this blog. I'm still curious as to what a proof of SsA entails, even if we don't ask high school students to prove it. Last year, I mentioned how SsA leads to the ambiguous case of the Law of Sines.

I've noticed that when using the Law of Sines to solve "both" triangles for the SsA case, the "second" triangle ends up having an angle sum of greater than 180 degrees. We can use the Unequal Sides Theorem to see why this always occurs -- that theorem tells us that the angle opposite the "s" must be smaller than the angle opposite the "S" (the known angle). We know that if two distinct angles between 0 and 180 have the same sine, then they are supplementary. So the second triangle would have angle sum of at least 180 minus the smaller angle plus the larger angle -- which always must be greater than 180. (This is often called the Saccheri-Legendre Theorem, named for two mathematicians with whom we're already familiar and associate with non-Euclidean geometry. Of course spherical geometry is not neutral, and Saccheri-Legendre fails in spherical geometry.)

That the sum of the angles of a triangle can never be greater than 180 (but we don't necessarily know that it's exactly 180) is neutral, but the Law of Sines is not neutral. Nonetheless, it is known that SsA is a neutral theorem. So the Law of Sines can't be behind the secret proof that U of Chicago text doesn't print in its text.

But one of the questions from the U of Chicago text hints at how to prove SsA -- and it's one that I included on my HL/SsA worksheet last year. Here is the question:

Follow the steps to make a single drawing of a triangle given the SSA condition:
a. Draw a ray XY.
b. Draw angle ZXY with measure 50 and XZ = 11 cm.
c. Draw circle Z with radius 9 cm. Let W be a point where circle Z and ray XY intersect.
d. Consider triangle XZW. Will everyone else's triangle be congruent to yours?

The answer is that they most likely won't. In step (c) we are to let W be a point -- not the point -- where the circle and the ray intersect. This implies that there could be more than one point where they intersect -- and in fact, there are two such points. But there can never be a third such point, because a circle and a ray (or line) intersect in at most two points. We can prove this indirectly using the Converse of the Perpendicular Bisector Theorem:

A line and a circle intersect in at most two points.

Indirect Proof:
Assume towards a contradiction that there exists a circle O that intersects line l in at least three points AB, and C, and without loss of generality, let's say that B is between A and C. By the definition of circle, O is equidistant from AB, and C. From the Converse of the Perpendicular Bisector Theorem, since O is equidistant from A and BO lies on m, the perpendicular bisector of AB -- and again using the converse, since O is equidistant from B and CO lies on n, the perpendicular bisector of BC. We know that m and n are distinct lines because m intersects l at the midpoint of AB and n intersects l at the midpoint of BC -- and those midpoints are distinct because B lies between them.

Now both m and n are said to be perpendicular to l, since each is the perpendicular bisector of a segment of l. So by the Two Perpendiculars Theorem, m and n must be parallel -- and yet O is known to lie on both lines, a blatant contradiction. Therefore a line and a circle can't intersect in three points, so the most number of points of intersection is two. QED

So let's prove SsA now using this lemma and the construction from the problem above. Once again, the proof must be indirect. Even though most of our congruence theorem proofs call the two triangles ABC and DEF, I will continue to use the letters XZW so that it matches the above question.

Given: AB = XZ < BC = ZW, Angle A = Angle X
Prove: Triangles ABC and XZW are congruent.

We begin by performing the usual isometry that maps AB to XZ. As usual, we wish to show that the final reflection over line XZ must map C to W -- that is, C' must be W.

So assume towards a contradiction that C' is not W. As usual, the given pair of congruent angles allows us to use the Flip-Flop Theorem to map ray AC to ray XY. Just as in the problem from the text, we know that C' must be a point on ray XY (since ray AC maps to ray XY), and it must be the correct distance from Z (since reflections preserve distance). Now the set of all points that are the correct distance from Z is the circle mentioned in the above problem. We know that the intersection of a circle and a ray is at most two points, and so the assumption that C' is not W implies that W must be one of the two points of intersection, and C' must be the other point.

We must show that this leads to a contradiction in the SsA case -- that is, when AB and XZ are longer than the sides BC and ZW. We see that if C' and W are distinct points equidistant from Z, then the triangle ZC'W must be isosceles, and so its base angles ZC'W and ZWC' must be congruent.

Now we look at triangle ZC'X. It contains an angle, ZC'X, which forms a linear pair with ZC'W, so its measure must be 180 - m/ZC'W -- that is, m/ZC'X = 180 - m/ZWC' by substitution. It also contains an angle ZXC' -- which (renamed as ZXW), we see must be larger than ZWC' (renamed as ZWX) by the Unequal Sides Theorem -- ZXW is opposite the longer side ZW (in triangle ZXW), so it must be the bigger angle.

So now we add up the measures of two of the three angles in triangle ZC'W -- the angle ZC'X has measure 180 - m/ZWC', and angle ZXC' is known to be greater than ZWC'. So the sum of the two angles is greater than 180 - m/ZWC' + m/ZWC' -- that is, it is greater than 180. That is, the sum of the angles of triangle ZC'W is greater than 180 -- which is a contradiction, since by the Triangle-Sum (actually Saccheri-Legendre) Theorem, the sum must be at most 180.

Therefore the assumption that C' and W are different points is false. So C' must be exactly W. QED

Of course we wouldn't want to torture our students with this proof. It depends on three theorems -- Isosceles Triangle, Unequal Sides, and Triangle-Sum -- that we have yet to prove. On my worksheet, I added an extra note to this problem explain how it leads to SsA.

Again, I retain references to non-Euclidean geometry from the old post. The new Third Edition of the U of Chicago text includes an actual SsA proof. It's similar, but not quite like, the proof given here. I dropped the mention of SsA on the worksheet since today's the second day of the SSASS activity. Oh, and since SsA works, so does SsAsS. Two short sides are adjacent to the angle, which implies a convex quadrilateral.

Oh, and since the Spherical Geometry label is on this post, it makes me wonder about the Quadrilateral Congruence Theorems in Spherical Geometry. I'm not quite sure which theorems are spherically valid, since very few authors write about quadrilateral congruence on the spherical. I suspect that at least SASAS and ASASA would be valid on the sphere.