*Magic of Mathematics*:

"Albrecht Durer (1471-1528) was an artist of many talents. He felt the study of mathematics enhanced art, in particular geometry, perspective and ideas of projective geometry."

On this first page of the Durer section, Pappas shows a woodcut from Durer's

*Treatise on Geometry*. I post another Cut the Knot link on some of Durer's geometric ideas:

http://www.cut-the-knot.org/pythagoras/DurerPentagon.shtml

This is a link to the construction of a "regular" pentagon with straightedge and compass. The regular pentagon construction goes back to Euclid, but it's a bit complicated -- which explains why we have high school students construct squares and regular hexagons but not regular pentagons.

As it turns out, Durer doesn't construct a regular pentagon either. His much simpler construction does produce a pentagon that is equilateral, but not quite equiangular. It may be interesting to have high school students try to construct Durer's pentagon.

Now let's get on with today's lesson. Today is supposed to be Lesson 12-9 of the U of Chicago text, on the AA and SAS Similarity Theorems, since today is Day 129. But there are two problems here.

First, Chapter 12 is the only chapter of the U of Chicago text with a full ten lessons. This causes a wrinkle in our digit-based pacing plan. Chapter 13, for example, has only eight lessons, so we can cover Lessons 13-1 to 13-8 on Days 131 to 138, then use Day 139 to review for the Chapter 13 Test to be given on Day 140. Then Lesson 14-1 can begin the next day, Day 141. For some shorter chapters, such as Chapter 11, this is even easier. The last lesson of the chapter is 11-6 on Day 116, then this leaves four days to review for and give the Chapter 11 Test.

But with ten lessons in Chapter 12, the pattern would have us cover Lesson 12-9 on Day 129, 12-10 on Day 130, and then 13-1 on Day 131 with no time for the Chapter 12 Test. The best thing to do seems to be to squeeze in the Chapter 12 Test on the same day as Lesson 12-10 -- at least it would be were it not for the second problem.

And that problem is -- Cesar Chavez Day. I first wrote about Chavez Day two years ago on the blog as a day celebrated in the LAUSD, but not most other school districts. Well, not only does the blog calendar observe all LAUSD holidays, but as I mentioned in February, the blog calendar also includes an extra PD day. The PD day is tomorrow to create a four-day weekend for the students.

This is what I wrote two years ago about Chavez Day:

Some schools -- and most notably universities -- here in California observe Chavez Day. As I mentioned earlier, there's a trend to sever the link between spring break and Easter, in order to avoid having the school holidays be tied to a holiday that can vary by over a month like Easter. In the University of California and California State University systems, spring break is tied not to Easter, but to Chavez Day.

Notice that for schools whose the first day of school is after Labor Day, that end-of-summer holiday is the most important holiday on the calendar. The first day of school is either early or late depending on whether Labor Day is early or late, which means that the last day of every quarter and trimester, including the last day of school, is early or late depending on the previous Labor Day date. In 2014, Labor Day fell on its earliest possible date, September 1st, while [in 2015], Labor Day will fall on its latest possible date, September 7th. This means that dates for the end of each semester will fall later during the 2015-6 school year than in the 2014-5 school year. Schools on an Early Start calendar may choose a different holiday to be the most important holiday on the calendar. For example, some Early Start schools are set up so that the last day of school is before Memorial Day, so that holiday is the most important. Others set up the year so that a fixed number of weeks occur before winter break, thereby making Christmas the most important holiday. The LAUSD Early Start calendar appears to have a fixed number of weeks before winter break -- but winter break itself is defined to begin three weeks after Thanksgiving. I won't know for sure until the holiday reaches its earliest possible date again on November 22nd, 2019, but it appears that Thanksgiving is the most important holiday on the LAUSD academic calendar.

Well, the most important holiday for the UC and Cal State college is Chavez Day. For example, at my alma mater, UCLA, the school year is divided into three quarters -- fall, winter, and spring. Spring break contains the observed Chavez Day and separates winter quarter from spring quarter. Since each quarter has a fixed length (ten weeks plus a finals week), and the length of the breaks is also fixed, all the dates in the UCLA academic calendar can be determined starting from Chavez Day.

There is a slight difference between the Cal State and UC calculations of the Chavez Day date. In the Cal State system, Chavez Day is on the labor leader's actual birthday, March 31st, and so spring break is the week containing the last day of March. Since most Cal States operate on a semester system, the spring break week is actually halfway during the spring semester, so that midterms can be given the week before the holiday. But at UC's -- well, at least at UCLA -- Chavez Day is defined to be the last

*Friday*in March -- so that [in 2015], March 27th was Chavez Day Observed. This means that at the Cal States, this week is spring break, while at the UC's, spring break was last week (just as it was on the blog) and spring quarter has already begun. Notice this means that there are often classes held at UC on the labor leader's actual birthday on the 31st, while there is never school this day at Cal State. I point out that in either case, it's more convenient, especially at UC, to tie spring break to Chavez Day rather than to Easter, in order to avoid quarters of differing lengths in years when the Christian holiday is exceptionally early or late.

LAUSD also observes a Chavez Day holiday. It is observed on the actual date, March 31st, in some years, while in others it's moved to the nearest Monday or Friday. [Of course in 2017, March 31st is a Friday, so the LAUSD observes Chavez Day on the actual date this year -- dw]

[Four] years ago, Easter actually fell on Chavez Day. There was a Google Doodle that day celebrating Cesar Chavez -- which angered some Christians expecting to see a Google Doodle for Easter. I point out that Chavez was a devout Catholic, and so one could argue that Chavez

*himself*would have rather seen Google celebrate Easter than himself.

Returning to 2017, here are a few more things I want to say about Chavez Day. The UCLA calendar is relevant to my classroom this year due to the presence of the Bruin Corps students. We see that the university observes Chavez Day on the last Friday in March, which is also the 31st. Thus this entire week is the UCLA spring break, and so the Bruin Corps students weren't in my classroom. It also means that last week was winter quarter finals -- and with so much study time needed, Bruin Corps wasn't in my room last week either. One of them should be back in the classroom next week for the first week of spring quarter -- but for the other student, winter quarter was his final quarter at UCLA, and he's now graduating. I wish him congratulations.

As I wrote two years ago, it's strange that UCLA would select the last

*Friday*in March to observe as Chavez Day -- this means that the holiday could occur as early as the 25th (as it did last year). The students always return to school the following Monday -- which means that students usually return on or before the 31st. To me, this seems to be

*dishonoring*the labor leader -- ironically, students are more likely to attend school on the 31st now than before Chavez Day was implemented (which was my final year at UCLA), when a different formula was used to determine spring break.

It would have been more logical for UCLA to observe the last

*Monday*in March as Chavez Day rather than the last Friday. Then the entire following week would be spring break -- and this week always contains the 31st. Students start the spring quarter the following Monday -- the first Monday in April, which would always be after the labor leader's actual birthday. The county library also closed Monday for the holiday -- when I was at the book sale on Saturday, I saw signs announcing the holiday closure.

I was curious to see whether any other California K-12 districts close for Chavez Day. Here in Southern California, Lynwood Unified also observes the holiday on Friday. I also found a few Northern California districts that close Friday as well.

So this means that with the PD day tomorrow, the Big March actually ends today -- by definition, the Big March extends from President's Day to the next school holiday, which for me is Chavez Day, not Easter or spring break. It does mean that there is only a short time left until our spring break. Notice that the Big March is thus almost never the longest stretch without a break. Even when Easter is late, the existence of Chavez Day means that the stretch from President's Day to spring break is broken up by Chavez Day, while the stretch from Easter to Memorial Day never is. Still, I consider the Big March to be the toughest time of the year. At least in April and May we can see summer just beyond the horizon, while in March summer is still far away.

But now that there's a four-day weekend, you can see why we can't have the Chapter 12 Test on Day 130 as the digit pattern suggests. Day 130 falls on a Monday after a four-day weekend, and that's the last day we'd ever want to give a test.

So my idea is to give the test today, just before the long weekend. We squeeze in Lesson 12-9, which is on AA~ and SAS~, before the test. Fortunately, I'm using last year's worksheets, and on the worksheets I created last year, I actually combined AA~ with SSS~ on the 12-8 worksheet, and so today's worksheet has only SAS~. This means that the students had an extra day to study AA~ before the test. (The reason I did this last year has nothing to do with Chavez Day -- it wasn't even March when I posted Chapter 12 last year!)

Here I will cut-and-paste what I wrote last year on SAS~ followed by the test. Notice that I'm using last year's Chapter 12 test -- but which I actually posted last year was a Chapter 11-12 Test! (It would be more logical to wait to have a Chapter 12-13 Test, as that solves the Day 129-130 problem.) So I only post the second part of the test, where the questions are numbered 10-20. Of these eleven questions, only one would have to be thrown out, #15, as it strongly references Lesson 12-10 -- this leaves ten questions, which is convenient for grading out of 100. Of the remaining questions, three questions use AA~ (including two proofs), which the students learned yesterday. One question mentions SAS~, which is today's lesson -- but that question only requires students to identify two triangles that are similar by SAS~, not give a proof.

As for the SAS~ material from last year, it's rather long -- but hey, it's a four-day weekend and besides, it's all cut-and-paste anyway. Scroll down past the SAS~ material for test answers!

I've referred to the American mathematician George David Birkhoff several times. Birkhoff was the mathematician who first came up with the Ruler and Protractor Postulates, which are often two of the first postulates to appear in a modern Geometry text. Actually, Birkhoff showed that only four postulates are required to derive all of Euclidean geometry:

-- Through any two points, there is exactly one line.

-- The Ruler Postulate

-- The Protractor Postulate

-- SAS~

This is astounding -- only

*four*postulates are required? Even Euclid himself had five postulates, and now Birkhoff claimed that he can derive Euclid's geometry in only four? And as Birkhoff's first postulate is the same as Euclid's, the American's other three postulates should somehow be equivalent to the Greek sage's other four. But how can this be?

Let's think about what theorems can be proved from Birkhoff's postulates. One should immediately jump out at us -- from SAS Similarity, we should be able to prove SAS

*Congruence*. In fact, the proof is almost trivial -- two figures are congruent iff they are similar with scale factor 1. Thus if we assume any similarity statement as a postulate, we can immediately prove the corresponding congruence statement.

We can also see how to derive some of Euclid's other postulates. That a line can be extended indefinitely goes back to the Ruler Postulate -- every point on a line corresponds to a real number, so just as the real numbers go on indefinitely, so do the points on a line. That we may draw a circle with any center and radius also comes from the Ruler Postulate, though this is less obvious. But think about it -- starting from the center, we can imagine placing a ruler in any direction. If the center is marked with the real number 0, we consider the point marked with the real number

*r*, the radius. The locus of all points found in this manner is the desired circle. That all right angles are equal obviously comes from the Protractor Postulate, as all right angles measure 90 degrees.

This leaves us with just one of Euclid's postulates -- but it's the one that caused the most trouble for millennia, the Fifth Postulate. It's possible to derive the Fifth Postulate from Birkhoff's axioms, but this is very complicated. But as we try to work it out, we'll learn much about Birkoff's geometry.

But first, let's think about what we've proved so far. We have Euclid's first four postulates and we also have SAS Congruence. This means that we can prove Euclid's first 28 propositions -- the ones that don't require a Parallel Postulate.

In particular, we can prove his Proposition 26, which is ASA. Recall that Dr. Franklin Mason also reproduces Euclid's proof, where he uses SAS to prove ASA.

Now the we have ASA Congruence, we can derive another similarity theorem. We can do it the same way that it's done in many pre-Core Geometry texts -- one similarity result is assumed as a postulate, and we use that postulate and the corresponding triangle congruence statements to derive each similarity theorem. In such texts, AA~ is usually the postulate, and so these texts use SAS to prove SAS~ and SSS to prove SSS~. We are given two triangles satisfying, say, the SSS~ condition and we wish to prove them similar -- the trick is to come up with a third triangle that is both similar to one of the given triangle via the AA~ Postulate and congruent to the other via SSS. This proves that the two given triangles must also be similar.

Birkhoff can use the same trick, except SAS~ is the postulate this time. We can now use SAS~ and ASA to prove a similarity theorem -- but which one? It's the one that corresponds to ASA. A moment's reflection should convince you that the similarity corresponding to ASA is in fact AA~! It can't be ASA~ -- what does it mean for

*one*pair of sides to be proportional? Today's Lesson 12-9 of the U of Chicago text points out that both ASA and AAS correspond to AA~. (Recall that the text proves all three of AA~, SAS~, SSS~ as theorems by using a dilation to produce similar triangles.)

So now we have proved AA~. The next result is often called the Third Angle Theorem (which Dr. M abbreviates as TAT) -- if two angles of one triangle are congruent to two angles of another triangle, then the third pair of angles is also congruent. This easily follows from AA~ as follows -- since the two triangles have two congruent pairs of angles, they are already similar by AA~. And if two polygons are similar, then

*all*of the corresponding pairs are congruent.

Here's a note of warning -- notice that we have

*not*yet proved a Triangle Sum Theorem. If two angles of a triangle have measure

*x*and

*y*, then all we know is that there exists some function

*f*such that the third angle has measure

*f*(

*x*,

*y*). We don't know that

*f*(

*x*,

*y*) =

*a*-

*x*-

*y*for some constant

*a*, much less that

*a*= 180 degrees.

So the Third Angle Theorem is a much weaker result than the full Triangle Sum. But we can use TAT to prove an interesting result:

-- Every Lambert quadrilateral is a rectangle.

And now you're thinking -- what the...? You thought that we were done with all this Lambert and Saccheri nonsense, and here I go talking about Lambert quadrilaterals again!

But notice that we've already proved the first four Euclid postulates thus far. In other words, so far we're in

*neutral geometry*. And so I'm going to use terms that a neutral geometer would use, such as Lambert quadrilateral. And besides, a Lambert quadrilateral only means a quadrilateral with three right angles -- and of course, a rectangle is a quadrilateral with four right angles. We could have said:

-- If three angles of a quadrilateral are right angles, then so is the fourth.

Also, my claim that we can use TAT to prove this theorem is strange. TAT is all about the

*third*angle of a

*triangle*, and now we'll use it to prove something about the

*fourth*angle of a

*quadrilateral*?

Well, here's the proof. Let's call our Lambert quadrilateral

*ABCD*, and declare that

*A*,

*B*,

*C*are the right angles. So our goal is to prove that angle

*D*is also a right angle.

Let's extend sides

*and*~~AB~~

*a little. That is, we choose point*~~CD~~

*E*on ray

*AB*beyond point

*B*, and point

*F*on ray

*CD*beyond point

*D*. Notice that rays

*AB*and

*CD*point in opposite directions. So when we connect

*E*and

*F*, we have that

*G*and

*H*respectively.

Now we look at triangles

*AEH*and

*BEG*. In these triangles, angles

*A*and

*EBG*are congruent as they are both right angles (with

*EBG*forming a linear pair with the given right angle

*B*), and the triangles have angle

*E*in common. Thus

*AEH*and

*BEG*are similar, and by TAT the third angles,

*AHE*and

*BGE*, must be congruent.

Notice we have vertical angles --

*AHE*and

*DHF*are congruent, as are

*BGE*and

*CGF*. Since

*AHE*and

*BGE*are congruent, we conclude that

*CGF*and

*DHF*are congruent.

Now we consider triangles

*GFC*and

*HDF*. We just proved that

*CGF*and

*DHF*are congruent, and the triangles have angle

*F*in common. Thus

*CGF*and

*DHF*are similar, and by TAT the third angles,

*C*and

*FDH*, must be congruent.

But we are given that

*C*is a right angle. Thus

*FDH*is also a right angle, and (with

*FDH*forming a linear pair with

*CDH*, the same as the given angle

*D*)

*D*is a right angle. Therefore the quadrilateral

*ABCD*is a rectangle. QED

So we just proved that every Lambert quadrilateral is a rectangle. Now as it turns out, it's known in neutral geometry that the statement "every Lambert quadrilateral is a rectangle" is one of many statements equivalent to Euclid's Fifth Postulate. The proof is not simple -- here's a link to a neutral geometry course where this is proved:

http://personal.bgsu.edu/~warrenb/Courses/09STheorems.pdf

The two relevant proofs are listed as 9->6, which takes us from "every Lambert quadrilateral is a rectangle" to Triangle Sum, and then 6->2, which takes us from Triangle Sum to Playfair. This is sufficient, since the most commonly mentioned Parallel Postulate is Playfair. So this concludes how we get from Birkhoff's four axioms to a Parallel Postulate.

Both David Joyce and David Kung say that there should be as few postulates as possible, and you can count me as a third David who agrees. If the fewness of the postulates were all that mattered, Birkhoff's axiomatization would be the winner.

But that's

*not*all that matters. Imagine a high school Geometry course trying to get from Birkhoff's axioms to Playfair. Some of the steps are reasonable for high school students -- SAS to ASA is not commonly done, but the indirect proof does appear on Dr. M's website. The phrase "Lambert quadrilateral" should

*never*be spoken in a high school course, but if we write it as "if three angles of a quadrilateral are right angles, then so is the fourth," the proof isn't terrible.

Of course, the proof that takes us from the rectangle theorem to Triangle Sum is very inappropriate for a high school class. And besides -- this proof sequence is the opposite from what we normally want to do in high school Geometry. We want to use Playfair to prove Triangle Sum, not vice versa, and we want to use Triangle Sum in our proof about the angles of a rectangle, not vice versa.

But actually, this isn't even where the problems with Birkhoff's axioms begin. They actually start with the very first proof -- using SAS Similarity to prove SAS Congruence. Yet of all the proofs, this is by far the simplest, almost trivial: two triangles are congruent iff they are similar with scale factor 1.

An argument can be made that similarity is much more important than congruence. Think about it:

-- What can we use to set up scale models and maps? (similarity, not congruence)

-- What can we use to prove the Pythagorean Theorem? (similarity, not congruence)

-- What can we use to derive the trig ratios? (similarity, not congruence)

-- What can we use to derive the slope formula? (similarity, not congruence)

...and so on. Congruence matters very little outside of Geometry class, but similarity matters throughout subsequent high school class, college, and careers.

If we were to introduce an SAS~ Postulate, we instantly have a proof of SAS Congruence. Of course, we already proved SAS Congruence, but we can imagine a class where similarity is taught before congruence is. (I actually once saw a text that teaches similarity before congruence.) Then we can get to the Pythagorean Theorem and the slope formula, and we can still teach congruence as a special case of similarity with scale factor 1. And we can just throw in Playfair as an extra postulate, so we don't run into the same problems as Birkhoff's axioms did.

So what is the problem with this approach? Similarity is a more

*difficult concept*than congruence -- and students don't fully understand similarity until they know what congruence is. This is why I tried so hard to avoid teaching similarity until the second semester, in hopes that students can get a good first semester grade without being confused by similarity. So minimizing the number of axioms is secondary to maximizing

*student understanding*. And so in today's lesson, the first of the second semester, we introduce SAS~, from Lesson 12-9 of the U of Chicago text

There's one more thing I want to say about SAS~. Last week, I implied that I would have to assume SAS~ as a postulate first, then use SAS~ to prove the properties of dilations, and finally use dilations to define

*similar*and prove the other similarity theorems.

But this is hardly logical. To see why, let's look at a statement of SAS~:

SAS Similarity Theorem:

If, in two triangles, the ratios of two pairs of corresponding sides are equal and the included angles are congruent, then the triangles are ______________.

That blank there is intentional, to remind you that we

*haven't defined "similar" yet*! And it's circular to use SAS~ to prove that triangles are "similar," use it to prove the properties of dilations, and

*only then*use dilations to define "similar."

All of this stems from that PARCC question that I mentioned last week. The Common Core Standards tell us that we should use dilations to define similarity:

CCSS.MATH.CONTENT.HSG.SRT.A.2

Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar; explain using similarity transformations the meaning of similarity for triangles as the equality of all corresponding pairs of angles and the proportionality of all corresponding pairs of sides.

yet that PARCC question does the opposite -- we have to use SAS~ (and therefore know what "similar" means) in order to prove the properties of dilations.

If all I had was the Common Core standard above, I would have introduced a Dilation Postulate that assumes the properties of dilations, then used dilations to derive SAS~ and the other postulates. If all I had was the PARCC question, I would have followed the classical pre-Core definition of similarity, proved SAS~, and then used it to prove the properties of dilations at the very end of the unit.

I decided to go back to Dr. M's website to see how he teaches similarity, and I see that his method is a compromise of the Common Core method and the PARCC method. He teaches similarity in his Chapter 7, and in his Lesson 7.3 he introduces a postulate:

The Polygon Similarity Postulate:

Given a polygon

*P*and a positive quantity

*k*, we may construct a second polygon

*Q*such that

*P*~

*Q*, with scale factor

*k*.

In this case,

*similarity*has already been defined classically, while

*dilation*isn't defined yet. Yet this postulate, while it doesn't say specifically that a

*dilation*(with scale factor

*k*, of course) maps

*P*to

*Q*but only that such a

*Q*exists, serves the same purpose as my Dilation Postulate.

We can then prove all three similarity statements -- SAS~, SSS~, and AA~ -- as theorems by using the corresponding congruence theorem plus the Polygon Similarity Postulate. In particular, we use the postulate to produce a triangle similar to the first given triangle, with the correct scale factor that makes it congruent to the second triangle via the corresponding theorem.

This solves most of our problems. Today we use SAS and Polygon Similarity to prove SAS~, with the idea that SAS~ can be used to prove the properties of dilations as in the PARCC question. Yet it still has the same idea expressed in the Common Core Standards -- we don't just assume any of the three major similarity statements, but prove all of them by producing similar triangles.

Test Answers:

10. b.

11. Yes, by AA Similarity. (The angles of a triangle add up to 180 degrees.)

12. Yes, by SAS Similarity. (The two sides of length 4 don't correspond to each other.)

13. Hint: Use Corresponding Angles Consequence and AA Similarity.

14. Hint: Use Reflexive Angles Property and AA Similarity.

15. [deleted in 2017]

16. 9 in. (No, not 4 in. 6 in. is the shorter dimension, not the longer.)

17. 2.6 m, to the nearest tenth. (No, not 1.5 m. 2 m is the height, not the length.)

18. 10 m. (No, not 40 m. 20 m is the height, not the length.)

19. $3.60. (No, not $2.50. $3 is for five pounds, not six.)

20. 32 in. (No, not 24.5 in. 28 in. is the width, not the diagonal. I had to change this question because HD TV's didn't exist when the U of Chicago text was written. My own TV is a 32 in. model!)

Chavez Day is on Friday, so the next post will be Monday, April 3rd. On that day we'll finish Lesson 12-10 and then begin Chapter 13, Logic and Indirect Reasoning.

## No comments:

## Post a Comment