Friday, January 30, 2015

Chapter 11 Test (Day 103)

Earlier this week I subbed in a high school special education classroom. Therefore, the students were working well below grade level. Written on the board were objectives that were clearly based on the Common Core Standards. One that I recognized was:

Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction.

And so, even though I only sub at middle and high schools, I still come in contact with elementary math standards. The above standard is one that traditionalist Common Core opponents dread. As I've already stated on the blog, this is one of those standards that mentions "strategies" and "algorithms" other than the standard algorithm. Once again, this is a third grade standard, and the corresponding standard that requires the traditionalist algorithm appears a year later:

Fluently add and subtract multi-digit whole numbers using the standard algorithm.

My proposed standards placed the traditionalist algorithm one year earlier, in third grade, and dropped the other standard about "strategies" altogether. But the fact that I found this in a special ed classroom brings up another issue: My proposal endorses traditionalism up to the third grade, and then a mixture of traditionalist and progressive methods for fourth grade and above -- the thought that preteens and teens don't respond well to direct instruction from a "sage on the stage."

Now this is a third grade standard, so it's intended for students young enough for traditionalism, but it's being taught to teenagers in a special ed class, students too old for pure traditionalism. So which philosophy rules, the intended age of the audience or the actual age of the audience?

My rule of thumb is, when students stop learning a subject just because a teacher -- "the sage on the stage" -- tells them to, and starts questioning what they are learning, as in "Why are we learning this?" and "When will we ever need this?" -- that is when pure traditionalism should end. I suspect that this aligns more with chronological age than with the level of material. This means that a young student who is well above grade level can learn more advanced material via direct instruction than my proposals recommend, only because they can reach such advanced material before reaching the age of questioning authority. Conversely, students who are well below grade level are still learning basic material well past the age of questioning authority. They may benefit from extra "strategies" that are suggested in the Common Core Standards.

This appears to be consistent with my observation. Clearly this group of students that I saw this week would not learn just because I told them to. I had to cover a science lesson where I played for the students a video about the 4.5-billion year history of the earth. It was a simple assignment where all the students had to do was take notes, but it was difficult to get the students to show any interest in the material at all, answering the questions I asked them with silence. This is consistent with their chronological age and shows the problems with purely direct instruction at that age.

On the other hand, I read anecdotes from traditionalist homeschooling parents about how their second grader learned fifth grade math effectively via direct instruction. I don't consider this to be a valid argument that the fifth grade math standards should be written from a purely traditionalist view -- because fifth graders, more than second graders, will start questioning why they have to learn how to compute with fractions.

But this does mean that the traditionalists' favored standard algorithms and memorization of basic math facts are to be taught as soon as possible, and not delayed a year as in Common Core. One common complaint among traditionalists is that students are never made to memorize basic multiplication facts. Questions such as six times nine or seven times eight should be considered very easy questions that take no more than a second to answer. But not only do many people consider such problems to be difficult, but it has become fashionable to consider those who have difficulties with such problems to be normal and those who find such problems easy to be outliers -- nerds.

It's often pointed out that people would feel deeply ashamed to admit that that can't read at a third grade level, yet are proud to admit that they can't do third grade math. Since I've stated that third grade math is something that students should have learned traditionally -- that is, have memorized -- I should do something about it in my classes.

The thought is that, rather than have those who find single-digit multiplication to be easy be outliers who get the label nerd, it's those who can't multiply by the time they reach middle and high school who should be considered outliers -- just as someone who can't read at a third grade level is taken to be an outlier. But of course, it's improper for me, a teacher, to start calling my students derogatory names such as idiot, no matter how low their understanding of math is.

So I need a word that criticizes the student, yet is proper for me to use in a classroom. Well, since I want my word to have the opposite effect of the word nerd, I briefly mentioned at the end of one of my posts a few months back that I made up my own word, by spelling the word nerd backwards, to obtain "dren."

My plan is to use my new word "dren" in such a way to make it sound as if a "dren" is not what a student wants to be. For example, when we reach the unit on area, students will need to multiply the length and width to find the area of a rectangle. So I might say something like, "A dren will have trouble multiplying six inches by nine inches. Luckily you guys are too smart to be drens, so you already know that the area is ...," and so on. Similarly, if a student, say, starts to reach for a calculator to perform the single-digit multiplication. I can say, "You're not a dren. You know how to multiply six times nine ...," and so on.

Notice that in these examples, I don't call anyone a dren directly. But every time I say the word "dren," I want to be annoying enough so that the students will want to do what it takes to avoid my having to say that word.

I coined the word "dren" to be the word nerd spelled backwards. But ironically -- according to my new Simpsons book -- the word nerd is already spelled backwards! Originally, the word was "knurd," which is drunk spelled backwards. The net result is that my word "dren" is basically just an abbreviation of drunk. Of course, the word drunk isn't a word that I should use in the classroom!

Here are the answers to today's test.

1. Using the distance formula, two of the sides have the same length, namely sqrt(170). This is how we write the square root of 170 in ASCII. To the nearest hundredth, it is 13.04.

2. The slopes of the four sides are opposite reciprocals, 2 and -1/2. Yes, I included this question as it is specifically mentioned in the Common Core Standards!

3. Using the distance formula, all four sides have length sqrt(a^2 + b^2).

4. Using the distance formula, two of the medians have length sqrt(9a^2 + b^2).

5. 60.

6. From the Midpoint Connector Theorem, ZV | | YW. The result follows from the Corresponding Angles Parallel Consequence.

7. From the Midpoint Connector Theorem, BD | | EF. The result follows by definition of trapezoid.

8. 4.5.

9. (0.6, -0.6). Notice that four of the coordinates add up to zero, so only (3, -3) matters.

10. At its midpoint.

11. 49.5 cm. The new meter stick goes from 2 to 97 cm and we want the midpoint.

12. Using the distance formula, it is sqrt(4.5), or 2.12 km to the nearest hundredth.

13. sqrt(10), or 3.16 to the nearest hundredth.

14. 1 + sqrt(113) + sqrt (130), or 23.03 to the nearest hundredth.

15. sqrt(3925), or 62.65 to the nearest hundredth. (I said length, not slope!)

16. -1/2. (I said slope, not length!)

17. (2a, 2b), (-2a, 2b), (-2a, -2b), (2a, -2b). Hint: look at Question 5 from U of Chicago!

18. (0, 5).

Thursday, January 29, 2015

Review for Chapter 11 Test (Day 102)

It's time to prepare for the next test. I'm labeling this as the "Chapter 11 Test," even though we didn't cover all of that chapter in the U of Chicago, and instead included sections from other chapters.

In particular, this test is based on the SPUR objectives for Chapter 11. As usual, I will discuss which items that I have decided to include and exclude, and the rationale for each:

Naturally, I had to exclude Objective G: equations for circles, which I take to be an Algebra II topic, not a Geometry topic. (If this had been an Integrated Math course, I would have delved more into graphing linear equations, as we covered this week.) Also, I left out Objective J, three-dimensional coordinates, as we haven't covered Chapters 9 or 10 on 3D geometry yet.

One major topic that I had to include is coordinate proof, as this appears in Common Core. I did squeeze in some coordinate proofs involving the Distance or Midpoint Formulas, but not slope. So therefore, the coordinate proofs included on this review worksheet all involve either distance or midpoint, not slope. The only proofs involving parallel lines had these lines either both vertical or both horizontal. Once again, a good coordinate proof would often set it up so that the parallel lines that matter are either horizontal or vertical.

What good are coordinate proofs, anyway? Well, a coordinate proof transforms a geometry problem into an algebra problem. Sometimes I can't see how to begin a synthetic geometry proof, so instead I just start labeling the points with coordinates and see what develops.

So coordinate geometry reduces an unknown problem (in geometry) to one whose answer is solved (in algebra, in this case). Mathematicians reduce problems to previously-solved ones all the time -- enough that some people make jokes about it:

I ended up including six straight problems -- Questions 8 through 13 from U of Chicago. Most of these questions are from Objective C -- the Midpoint Connector Theorem. The text covers this here in Chapter 11, but we actually covered it early, in our Similarity Unit, because we actually used the Midpoint Connector Theorem to start the proof of the basic properties of similarity. Still, this was recent enough to justify including it on the test.

Next are a few center of gravity problems. This is straightforward, since all we have to do is average the coordinates. Afterwards are a few midpoint problems, including two-step questions where one must calculate the distance or slope from one point to the midpoint of another segment.

Then there are a few more coordinate proofs where one has to set up the vertices -- notice that some hints are given in earlier questions. Finally, one must find the point where two lines intersect. This involves either substitution or elimination (from Algebra I), but these problems are both simple.

Wednesday, January 28, 2015

Equations of Lines (Day 101)

Equations of lines don't appear in the U of Chicago geometry text at all. Instead, we go back to U of Chicago's Algebra (that is, Algebra I) text. There, equations appear in the following sections in Chapter 8, "Slopes and Lines":

Section 8-4: Slope-Intercept Equations for Lines
Section 8-5: Equations for Lines with a Given Point and Slope
Section 8-6: Equations for Lines Through Two Points
Section 8-7: Fitting a Line to Data
Section 8-8: Equations for All Lines

Obviously, in Lesson 8-4, the equation y = mx + b appears. But as it turns out, the point-slope form doesn't appear in the U of Chicago at all. Instead, the slope-intercept form is used to find equations in both 8-5 and 8-6. Then again, I notice that most students have trouble remembering point-slope, so we might as well teach only the slope-intercept form. In Lesson 8-8, the standard form Ax + By = C of a linear equation appears.

A question that is often asked during this lesson is, why so we use m to denote slope? One urban legend is that it refers to the French word monter, meaning to climb. An English cognate of this word is "mountain," and of course mountains have slopes. The problem with this theory is that there is no evidence that the mathematician Rene Descartes ever used the letter m. One would think that if m had a French origin, Descartes -- you know, the French creator of the Cartesian plane on which slope is usually measured -- would have been the first to use it. A discussion appears at this thread:

Notice that John Conway -- the mathematician I previously mentioned as an advocate of the inclusive definition of trapezoid -- is a participant in this 20-year old thread. (That's right -- when Conway wrote in this thread, I was myself a young geometry student!) Conway suggests that m may stand for "modulus of slope." One teacher tells his students that m stands for "move" and b stands for "begin," since this is how students learn to graph lines in slope-intercept form.

The origin of b is little more well-known. It refers to the idea that a is the x-intercept of a graph, and b is the y-intercept. The equation of an ellipse centered at the origin uses a and b in this way -- but if the ellipse is translated so that its center isn't at the origin, a and b no longer stand for the intercepts.

Finally, back when I was student teaching, my students came up with their own mnemonic for the slope-intercept formula. The letters y = mx + b stand for "your mom's ex-boyfriend."

Slope-Intercept Property:
The line with equation y = mx + b has slope m and y-intercept b.

Once we have the slope-intercept formula, we can algebraically prove that its slope is m and that its y-intercept is b. The y-intercept is the point that lies on the y-axis -- and as we saw yesterday, this is the point whose x-intercept is 0. Setting x = 0 in the slope-intercept formula gives y = b -- that is, the y-intercept is b. To find the slope, we let (x_1, y_1) and (x_2, y_2) be two arbitrary points on the line:

(y_2 - y_1) / (x_2 - x_1)
= (mx_2 + b - (mx_1 + b)) / (x_2 - x_1)
= (mx_2 + b - mx_1 - b) / (x_2 - x_1)
= (mx_2 - mx_1) / (x_2 - x_1)
= m(x_2 - x_1) / (x_2 - x_1)
= m. QED

Finally, we can show how the standard form Ax + By = C is also the equation of a line. We notice that if B = 0, we have the horizontal line x = C / A. Otherwise, we may divide by B:

Ax + By = C
        By = -Ax + C
          y = (-A/ B)x + C / B

and so we have a line with slope -A / B and y-intercept C / B. QED

Tuesday, January 27, 2015

Parallel and Perpendicular Lines: Parts of Sections 3-4 and 3-5 (Day 100)

According to the blog calendar, today is the 100th day of school -- often celebrated with a big party in first grade classrooms. This is mostly because 100 is traditionally the number to which students are expected to learn how to count. But we see a change with the advent of the Common Core Standards:

Count to 100 by ones and by tens.

Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.

So now counting to 100 is a kindergarten, rather than first grade standard. For some strange reason, the counting target for first graders is now 120. Considering that the counting target for second graders is 1000, both before and under Common Core, I wonder why 200 wasn't chosen as a target between 100 and 1000, rather than 120.

Here is an article from about a year ago which discusses the change in the counting to 100 standard, from first grade under California standards to kindergarten under Common Core:

There's a long comment thread at the end of this article. Many of the issues discussed on this blog appear in this thread, including traditionalism vs. progressive constructivism (and since the topic is early elementary math, I agree with the traditionalist), the delay by a year of the memorization of basic math facts, and even high school Integrated Math. (The commenter appears mainly concerned with the transition from a traditionalist to an integrated pathway.)

To maintain consistency with the Common Core, kindergarten teachers would now celebrate the 100th day of school, while first grade teachers should wait until the 120th day of school. The only problem with this is that the 120th day of school marks the end of the second trimester -- so it's often the day of a (district-mandated) test, not a party.

Now let's return to geometry. It's now time for the slopes of parallel and perpendicular lines. In Section 3-4 of the U of Chicago text, we see the following theorem:

Parallel Lines and Slopes Theorem:
Two nonvertical lines are parallel if and only if they have the same slope.

Of course, Common Core now expects us to prove this theorem. The text writes:

"A proof requires quite a bit of algebra, and is omitted."

Actually, I don't see that much algebra needed for the proof at all. In fact, the proof of the theorem is very similar to yesterday's proof of the well-definition of slope. This time, instead of the four points (x_1, y_1), (x_2, y_2), (x_3, y_3), and (x_4, y_4) all being on the same line, we place the first two  points on the first of two parallel lines and the other two on the other line.

But now we're given the two parallel lines, not the points so we are allowed to choose any points on the line that we wish. Since we want to set up corresponding angles, we begin by selecting a line to be the transversal and then let (x_1, y_1) and (x_3, y_3) both lie on this transversal. Any line may be chosen as the transversal as long as it intersects both given parallel lines.

Well, since we're given that the two parallel lines aren't vertical (as vertical lines have no slope), we can let any vertical line be the transversal. In fact, why don't we choose the easiest vertical line to work with -- the y-axis? These means that our two points (x_1, y_1) and (x_3, y_3) are now the two points (x_1, 0) and (x_3, 0). Notice that these two chosen points are where the two lines intersect the y-axis -- that is, they are the y-intercepts. So let's change those x's to b's -- and so the two points in our proof will be labeled (b_1, 0) on the first line and (b_2, 0) on the second line.

Notice that by letting the transversal be the y-axis and two of our points have 0 as their second coordinate, we're demonstrating the hallmarks of a good coordinate proof. We want to place as many key lines and segments along the coordinate axes as possible.

Now let's consider our second points on each line. Since our first points have 0 as the y-coordinate, let's let the second points have 1 as the y-coordinate. So these points are (x_1, 1) and (x_2, 1).

Now just as we did for the Distance Formula and Slope Well-Definition proofs, we add the points E(x_1, 0) and F(x_2, 0) to form two right triangles. Notice that these two right triangles aren't merely similar, as they were in the Slope Well-Definition proof -- they are in fact congruent.

Given: k | | l, A(b_1, 0), B(x_1, 1) on k, C(b_2, 0), D(x_2, 1) on l
Prove: slope of k = slope of l

Statements                     Reasons
1. k | | l, etc.                    1. Given
2. Angle BAE = DCF     2. Corresponding Angles Consequence
3. Angle AEB = CFD     3. All right angles are congruent.
4. BE = DF                     4. All segments of unit length are congruent.
5. Tri. AEB, CFD cong. 5. AAS Congruence Theorem
6. AE = CF                     6. CPCTC
7. slope k = AE/1 = AE,  7. Definition of slope
    slope l = CF/1 = CF
8. slope k = slope l          8. Substitution

We see that there are several cop-outs in the above proof to avoid excessive formalism. Technically speaking, "all segments of unit length (that is, of length 1) are congruent" is not a valid reason in a formal proof. We actually need four steps here -- the first to prove that BE = 1, the second to prove that DF = 1, the third to prove that BE = DF, and final to prove that these equal lengths imply that the segments are congruent. Arguably, we'd need to do the same with the right angles. Also, several steps are skipped in finding the slopes of both lines -- it may be instructive to calculate the slope by plugging the values into the formula to show that the slope of k is x_1 - b_1, while the slope of l is x_2 - b_2, and these are the values proved equal in the CPCTC step. (Notice that the theorem to be proved is stated as "if and only if," so we need to prove the converse -- that is, if the slopes are equal, then the lines are parallel. A proof of the converse requires SAS rather than AAS. For the converse, our given lines have the same slope -- so we can choose the familiar letter m to denote that slope!)

But there is one more subtlety that we need to discuss. We assumed that AE = x_1 - b_1 in the proof above, but actually by the definition of 1D distance, AE = |x_1 - b_1|, not x_1 - b_1. That is, we've ignored the sign of the slope in the above proof! Without this, it could be the fact that parallel lines have slopes with opposite sign. Even the proof of the well-definition of slope from yesterday isn't really complete since there was no mention of any sign.

What we actually need here is that extra postulate that Dr. Wu gave us -- Plane Separation. Our lines have constant slope because we showed that, if we start at a point on the line and can move v units up and w units right to arrive back on the line, then the same is true no matter where we start. There is no point where we can move w units right, but we have to go v units down instead of up in order to get back on the line.

The reason is that set of all points w units right of the line is a translation image of the line, and as a translation image, it is parallel to its pre-image. So it can't cross from one half-plane defined by the original line to the other. In particular, it can't cross from the half-plane containing the line v units above the starting line to the half-plane containing the line v units below the starting line. And so it's the Plane Separation Postulate that ultimately guarantees that slope is well-defined and that parallel lines have equal slope.

As Wu often points out, this is a subtle point that is ignored in most math classes. The problem, of course, is that we're about to jump to the case where the sign of the slope makes a huge difference -- the slope of perpendicular lines.

Now here's a mini-activity that I like to begin with to illustrate the slopes of parallel and perpendicular lines. I choose two students and ask them to draw a line that is parallel to my given line, which has a predetermined slope of, let's say, 2/3. I inspect the two lines and declare the student whose slope more clearly represents going up 2 and right 3 units to be the winner.

Then I ask the students to draw a line that is perpendicular to my given line. This will be harder for someone who hasn't learned the formula yet, and so it's likely that both lines will be far away from a truly perpendicular line. And so I come in and draw my own perpendicular line. To do this, I point out that since perpendicular means 90 degrees, I rotate my paper 90 degrees and then redraw the original line on the rotated paper. I make sure that my new line also has rise of 2 and run of 3. Then I rotate the paper -90 degrees to retain the original position, and notice that the rise of 2 rotates to a run of 2, and the run of 3 rotates to a rise of -3. So the new slope is -3/2 -- the opposite reciprocal of the original slope. And then I declare myself to be the winner!

I don't include a formal proof of the Perpendicular Lines and Slopes Theorem on the worksheet, but we can figure it out -- instead of a translation, we use a rotation.  We rotate the given line AB 90 degrees about some point (say the origin) to obtain a new line A'B', and we see that its slope is the opposite reciprocal of the original slope, since rise and run become run and negative rise. Then, by the Two Perpendiculars Theorem, any other line CD that is perpendicular to AB must be parallel to A'B', and so has that same slope as A'B' (which is the opposite reciprocal of that of AB). It may be possible to skip a step here and simply rotate about the point where the two given lines AB and CD intersect -- then the image of line AB would be line CD itself (but A doesn't necessarily rotate into either C or D).

Monday, January 26, 2015

Slope: Based in part on Section 3-4 (Day 99)

Section 3-4 of the U of Chicago text is where slope -- that very important concept -- is defined. The definition in the text is simple enough and is typical for most high school math texts. Here it is, rendered in ASCII:

The slope of the line through (x_1, y_1) and (x_2, y_2), with x_1 != x_2, is (y_2 - y_1) / (x_2 - x_1).

But this isn't good enough for Common Core. Let's look at the relevant standard:

Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point).
[emphasis mine]

So we have to prove that slope works. It isn't good enough just to say that slope formula works just because we defined it to work. To see why, consider the following definition:

The favorite food of the class containing the student S is the food that S likes to eat the most.

And now you instantly see the problem with this "definition." Naturally, not every student in the class likes the same food. The food that student S loves to eat may be a food to which student T is indifferent and a food that absolutely disgusts student U. The "favorite food" of a class depends on which student we ask to name a favorite food. Simply writing the word "Definition:" and setting the phrase "favorite food" in bold doesn't magically force everyone to have the same favorite food.

Likewise, writing the word "Definition:" and setting the word "slope" in bold doesn't magically force the slope to be the same no matter which two points we choose. It could be that if we choose points (x_1, y_1) and (x_2, y_2), we get a different slope from if we choose  (x_3, y_3) and (x_4, y_4), just as if we choose student S we get a different favorite food (or favorite song) from if we choose student T instead of S.

In mathematics, we would say that "favorite food of a class" is not well-defined. In order for slope to be well-defined, we must prove that the slope is independent of which two points we choose to plug into the formula. We can't prove that the favorite food (or song) is independent of which student we choose -- indeed, it's trivial to find a counterexample: simply declare one student's favorite to be the class's favorite, and watch all the counterexample students call out how much they don't like the declared favorite!

And so the main theorem for today is to prove that slope is well-defined. We want to prove that slope of a line is independent of which two points we choose to plug into the formula. The trick for doing this will remind us of the proof of the Distance Formula.

Friday, January 23, 2015

Activity: The Pythagorean Theorem and Coordinate Proofs (Day 98)

There has been much discussion this week about the president's State of the Union address -- in particular, the proposal that community college be free for qualifying students.

Obviously, it's impossible to escape politics when discussing this proposal. Those who are of the same party as the president tend to support the plan, whereas members of the opposing party tend to reject the plan.

My blog, of course, is all about K-12 education, but there is so much discussion about this proposal that it's hard for me not to comment on it. Naturally, I'm of two minds on this issue. I'd like to compare this proposal to the college systems in other countries -- especially those that are considered superior to that of the United States. But most other countries don't have an exact equivalent of our community colleges.

On one hand, I support the idea of making sure that all people, regardless of demographics, are able to attain a comfortable middle-class lifestyle. To the extent that many students can't afford the college necessary to earn the degrees that make them attractive to employers, the proposal might become a good idea.

On the other hand, last week I proposed a Dickens age, namely 15 years of age. The proposal, where college is made free up to the normative age of 20, contradicts the concept of Dickens age. Recall that I defined the Dickens age to be the age at which a youngster can be encouraged to have a job -- that is, employment. A supporter of this Dickens age wouldn't be in favor of free schooling up to five years past that age. Instead of a plan to make college free, they would applaud a proposal to reduce the unemployment rate of youngsters from the Dickens age and above, to around the desirable rate for adults -- which, as I mentioned, is 5%

Reducing the unemployment rate for 15-year-old youngsters to 5% is politically complex. From an educational -- and especially a mathematical -- perspective, the idea is that a business that doesn't require its workers to know 12th grade math (calculus) should have no problem hiring a 15-year-old with only a ninth grade education.

Perhaps here's an idea that could combine the president's proposal to make school more affordable with the concept of a Dickens age. Recall that the normative student completes 7th grade math, which is my graduation requirement, at age 13 -- two years before the Dickens age. Then those intervening years could serve the same purpose as community college. Opportunities available at the junior college level -- including, most importantly, vocational training -- could be made available to those students age 13-15, so that when they leave school at the Dickens age, they will be more attractive to some employers.

I'd argue that as far as math is concerned, the normative student would keep on taking math classes at the 8th and 9th grade standards mentioned earlier this week (based on Singapore Secondary Two and Three), but it would not be terrible if they don't pass these classes, especially if they plan on applying to a job not requiring that much math.

Today is an activity day. As I mentioned earlier this week, I just can't resist presenting Pythagorean Proof #9 (from the Cut the Knot webpage) as an activity. To do this, we cut out the four right triangles (with sides of length a, b, and c) and the three squares (one square each with sides of length a, b, and c).

Then the students try to fit the four triangles and the squares of length a and b inside of the larger target square. Afterward, the students try to fit the four triangles, and now only the square of length c, inside of the same target square. I've found that this second puzzle is harder for the students, since they usually don't think to tilt the square of area c to make it fit with the four triangles.

The Cut the Knot webpage states that this proof doesn't require the students to know the formula for the area of the triangle. All that's needed is the area of a square -- and that's trivial, since we just take the side length and square it. So we can simply call the area of each triangle T. Notice that the area of the target square is both 4T + a^2 + b^2 (four triangles, a and b squares) and 4T + c^2. We write:

4T + a^2 + b^2 = 4T + c^2
a^2 + b^2 = c^2

therefore completing the proof of the Pythagorean Theorem. QED

This activity likely won't take the whole period. And so I provide another activity. Once again, the Exploration Questions at the end of the sections often lend themselves to good activities. Since we just finished Section 11-4 on the Midpoint Formula, we consider Question 26:

The center of gravity of a polygonal region with more than three sides is not generally the same point as the center of gravity of its vertices. Cut a nonsymmetric convex polygonal region out of cardboard.

a. By trial and error, find the point where the region will balance on the tip of a pin.
b. By putting the region on graph paper, find the center of gravity of its vertices and determine how close your point in part a is to this point.

Let's look at this question in more detail. First of all, the question mentions the tip of a pin, but the photo just below the question clearly shows a pencil. Actually, I'm not sure whether either method is truly feasible for finding the center of mass. Instead, one can take the polygon and place it right on the edge of the desk. Then one slides (translates) the polygon slightly, and keeps doing so until the polygon falls off the table. There should be a segment corresponding to the cross-over line between balancing on the desk and falling -- this should be marked. Then one turns (rotates) the polygon and repeats the process. Where the two marked segments intersect is the center of gravity. If done right, this point should be the unique point of the region such that the polygon balances if and only if this point is still on the desk.

Notice that if there is sufficient cardboard, one can do both of these activities -- the Pythagorean proof and this center of mass -- on cardboard. In fact, one can even choose one of the Pythagorean right triangles as the first polygon whose center of gravity we are to find. But for a triangle, the vertices and the entire region do have the same center of gravity -- and this is the centroid of the triangle, a discussion of which we squeezed into the final week of the first semester (as it's required under Common Core). This is why the instructions mention "more than three sides." And these same instructions also say "nonsymmetric," so we shouldn't use a square as our second polygon -- we could probably guess its center by eyeballing it rather than doing the activity. Let the students draw their own quadrilateral or pentagon. Doing more than five sides isn't recommended, as the students are more likely to make a mistake in finding the average of the vertices.

There's one more activity that can occur today -- coordinate proofs. Let's look at the Common Core standards again more closely:

Use coordinates to prove simple geometric theorems algebraically. For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point (1, √3) lies on the circle centered at the origin and containing the point (0, 2).

So we see that coordinate proofs are required under Common Core. In the U of Chicago text, Sections 11-1 and 11-5 both contain coordinate proofs. But most of these proofs require slope to be taught first -- and my coverage of slope next week squeezes out time to do most of the needed coordinate proofs. Indeed, there's even a question right in Lesson 11-1 that specifically addresses the standard "prove or disprove that a figure ... is a rectangle" -- Question 9. But we have to cover slope first, in order to show that the sides are perpendicular.

The other part of that standard involves equations of circles. Although the U of Chicago text gives this in Section 11-3, this is not typically covered in a geometry class. Recall that the Common Core standards aren't divided into traditionalist Algebra I, Geometry, and Algebra II courses. Although most of the geometry standards correspond, naturally, to the Geometry course, this and a few others on conic sections properly belong in Algebra II. Since circle equations appear in the U of Chicago, I might have considered including it here, but we don't have the time.

A few more standards from this section here:

Find the point on a directed line segment between two given points that partitions the segment in a given ratio.

This standard naturally fits with today's activity. To partition a segment into a given ratio, say 1:2, we count one of the endpoints once and the other twice, and then average the points, just as we averaged the vertices of a polygon to find the center of mass. Likewise, to partition a segment 2:3, count one of the points twice and the other thrice. The resulting point will be closer to whichever point was weighed more heavily, in the correct ratio.

So to find the point one-third of the way from (1, 2) to (10, 5), average (1, 2), (1, 2), and (10, 5). The result will be (4, 3), which is indeed one-third of the way from (1, 2) to (10, 5).

Finally, we have this standard:

Use coordinates to compute perimeters of polygons and areas of triangles and rectangles, e.g., using the distance formula.

A question addressing this standard was already included in the lesson earlier this week on the Distance Formula -- at least for perimeter. More may be included as we go along. Of course, area will have to wait until Chapter 8.

Thursday, January 22, 2015

Section 11-4: The Midpoint Formula (Day 97)

Last night I tutored my geometry student. Section 4-5 of the Glencoe text is on ASA and AAS Congruence, so this clearly corresponds to Chapter 7 of the U of Chicago text.

I wanted to show him my worksheet from this blog -- the one where I proved the ASA Congruence Theorem using Common Core principles. At first he only took one glance at my proof outline and was intimidated when he saw that it was a ten-step proof. But then he had to finish his homework assignment, which required what turned out to be a nine-step proof.

Notice that the congruence chapters in the two texts are organized differently. In the U of Chicago, SSS, SAS, ASA, and AAS are all theorems that need to be proved. And so Section 7-2 contains these proofs, so that Section 7-3 has students prove the higher-level theorems that use these medium-level Congruence Theorems, and then Section 7-4 contains more complex proofs, the ones with overlapping triangles.

But Glencoe, like most pre-Common Core texts, gives SSS, SAS, and ASA as postulates. Because of this, the lessons go straight into having the students make higher-level proofs. And some of these proofs had overlapping triangles. So unlike the U of Chicago, with separate sections for the simpler and harder proofs, Sections 4-4 (SSS, SAS) and 4-5 (ASA, AAS) contained both simpler and harder proofs.

We look at Section 7-4 of the U of Chicago, and see Example 1 of a proof with overlapping triangles. Here, B lies on AD and C lies on AE.

Given: AC = ABAD = AE
Prove: Angle D = E

Statements                    Reasons
1. AC = AB, AD = AE  1. Given
2. Angle A = A              2. Reflexive Property of Congruence
3. ABE, ACD cong. tri.  3. SAS Congruence (steps 1, 2, 1)
4. Angle D = E              4. CPCTC

Now Glencoe contains a similar problem, except that the given is different. F is where BE and CD meet.

Given: AB, AC cong. seg., BDF, CDE cong. tri.
Prove: D, E cong. angles

Statements                      Reasons
1. BDF, CDE cong. tri.   1. Given
2. BD cong. CE              2. CPCTC
3. AB cong. AC              3. Given
4. BD = CE, AB = AC    4. Segment Congruence Theorem
5. AB + BD = AC + CE  5. Addition Property of Equality
6. AD = AE                    6. Betweenness Theorem
7. Angle A cong. A          7. Reflexive Property of Congruence
8. ABE, ACD cong. tri.    8. SAS Congruence (steps 3, 7, 6)
9. Angle D cong. E          9. CPCTC

The reason that this proof is so long is that in many texts, including Glencoe, the statements "segment AB is congruent to AC" and "length AB is equal to AC," while equivalent, are not identical. In the U of Chicago, the Segment Congruence Theorem of Chapter 6 tells us that segments of equal lengths are congruent -- that is, that there is an isometry mapping one segment to the other. But after introducing the theorem, the book freely interchanges the congruence of the segments and the equality of the lengths, without mentioning it as a separate step in the proof.

Here the problem is that we need the Betweenness Theorem -- what Glencoe calls the Segment Addition Postulate -- and that only numbers, not figures, can be added. A length is a number, but a segment is a figure -- so we must convert the given segment congruence statement into an equal length statement before we can add anything, and then convert it back to segment congruence before we can invoke SAS. Most geometry teachers just handwave over the distinction between segment congruence and length equality -- as does the U of Chicago text, as do I on this blog, since the congruence symbol is hard to type in ASCII.

But the Glencoe text painfully includes every distinction between segment congruence and length equality -- as became apparent when I looked at one of the answers (a flow proof) in the back of the text. Moreover, my student's math teacher has been known to deduct half-points for leaving out such steps!

Even in the proof I posted above, notice that Step 6 is AD = AE, but I failed to convert this back to the congruence of segments AD and AE. Not only that, but technically speaking, the Segment Addition Postulate tells us that AB + BD = AD and AC + CE = AE. This should be the step that immediately follows step 5, and then the next step should derive AD = AE by substituting the above two equations into step 5.

So a fully complete proof would actually have eleven steps, not nine. The student and I came to the agreement that these nine steps should be sufficient for the teacher not to deduct any points -- but we won't know until today.

To me, having to make the painful distinction between segment congruence and length equality, as well as adding the substitution step, changes this problem to an exercise in formality, rather than help the student to see how Segment Addition can be used in an overlapping triangles proof. Even Dr. Wu admits that he skips steps in proofs when they don't enhance student understanding -- which is the primary goal of teaching proofs.

After this proof, I hardly had time to show my student the full proof of the ASA Congruence Theorem, as planned. What I decided to do was cut out two triangles satisfying the ASA condition, and then showed how I could slide, flip, and turn one triangle so that it coincided with the other. In particular, I showed him exactly how the Side-Switching Theorem was used to show that the reflection image of C' is F.

Afterward, my student noticed another proof in the Glencoe text and wondered whether he could use reflections, as I had to in the proof of the ASA Congruence Theorem, to prove the triangles in that problem congruent. Actually, he had a point there. Notice that in general, in a congruence proof, the first given triangle isn't necessarily a mere reflection image of the second. But in overlapping triangles, the triangles actually are mirror images of each other. Indeed, we look at the diagrams in U of Chicago Lesson 7-4, and notice that every single diagram in the lesson has reflectional symmetry!

In the diagram for this problem, point E lies on AB.

Given: Angles BCE, DCE congruent, Angles BEC, DEC congruent
Prove: AB, AD congruent

Let's compare the traditional Glencoe proof with a Common Core/U of Chicago proof.

Glencoe Proof:
Statements                        Reasons
1. Angles BCE, bla, bla     1. Given
2. CE cong. CE                 2. Reflexive Property of Congruence
3. BCE, DCE cong. tri.     3. ASA Congruence (steps 1, 2, 1)
4. BE cong. DE                 4. CPCTC
5. AEB, CEB linear pair,   5. Definition of linear pair
    AED, CED linear pair
6. Angle AEB cong. AED  6. Congruent Supplements Theorem
7. AE cong. AE                  7. Reflexive Property of Congruence
8. AEB, AED cong. tri.      8. SAS Congruence (steps 4, 5, 6)
9. AB cong. AD                 9. CPCTC

For the Common Core/U of Chicago proof, we'll start out just like the ASA Congruence Theorem proof and proceed from there:

Common Core/U of Chicago Proof:
Statements                                       Reasons
1. Angles BCE, bla, bla                    1. Given
2. Ray EB refl. over line AC is ED  2. Side-Switching Theorem
3. Ray CB refl. over line AC is CD  3. Side-Switching Theorem
4. The image of B lies on ED, CD   4. Figure Reflection Theorem
5. The image of B is D                     5. Line Intersection Theorem
6. The image of A is A                      6. Definition of reflection (point on mirror refl. to itself)
7. AB refl. over AC is AD                 7. Figure Reflection Theorem
8. AB cong. AD                                 8. Definition of congruence (since the refl. is the isometry)

The Common Core proof is slightly shorter -- mainly because of the wasted step in the Glencoe about showing that the angles form a linear pair. And I forget whether Glencoe would make the student add yet another step, that the angles in a linear pair are supplementary! It's because of these extra formalistic steps that make the reflectional proof shorter.

Also, this week I've been subbing at the same continuation school for the second week in a row. One student is preparing to take the national test known as the Armed Services Vocational National Battery -- or ASVAB, for short. He is having trouble passing the ASVAB because it covers much material from the second semester of Algebra I, whereas he has just barely finished the first semester of algebra.

Notice that the ASVAB has two test formats -- computerized and written -- and it is given by the federal government -- specifically the military.

Section 11-4 of the U of Chicago text covers the other important formula of coordinate geometry -- the Midpoint Formula. As the text states, this is one of the more difficult theorems to prove.

In fact, the way we prove the Midpoint Formula is to use the Distance Formula to prove that, if M is the proposed midpoint of PQ, then both PM and MQ are equal to half of PQ. The rest of the proof is just messy algebra to find the three distances. The U of Chicago proof uses slope to prove that M actually lies on PQ. Since we don't cover slope until next week, instead I just use the Distance Formula again, to show that PM + MQ = PQ, so that M is between P and Q. The algebraic manipulation here is one that's not usually used -- notice that instead of taking out the four in the square root of 4x^2 to get 2x (as is done in the last exercise, the review question), but instead we take the 2 backwards inside the radical to get 4, and then distribute that 4 so that it cancels the 2 squared in the denominator.

I don't have nearly as much to say about the Midpoint Formula as the Pythagorean Theorem and its corollary, the Distance Formula. To me, it's a shame that I had to bury the Pythagorean Theorem in the middle of this Coordinate Geometry unit. The main theorem named for a mathematician really deserves its own lesson, but due to time constraints I had to combine it with the Distance Formula the way I just did it in yesterday's lesson.

Wednesday, January 21, 2015

The Pythagorean Theorem and the Distance Formula (Day 96)

Section 8-7 of the U of Chicago text is on the Pythagorean Theorem, and Section 11-2 of the same text is on the Distance Formula. I explained yesterday that I will cover these two related theorems in this lesson.

The Pythagorean Theorem is, of course, one of the most famous mathematical theorems. It is usually the first theorem that a student learns that is named for a person -- the famous Greek mathematician Pythagoras, who lived about 2500 years ago -- a few centuries before Euclid. I believe that the only other named theorem in the text is the Cavalieri Principle -- named after an Italian mathematician from 400 years ago. Perhaps the best known named theorem is Fermat's Last Theorem -- named after the same mathematician Fermat mentioned in yesterday's worksheet. (We discuss some other mathematicians such as Euclid and Descartes, but not their theorems.)

It's known that Pythagoras was not the only person who knew of his named theorem. The ancient Babylonians and Chinese knew of the theorem, and it's possible that the Egyptians at least knew about the 3-4-5 case.

We begin with the proof of the Pythagorean Theorem -- but which one? One of my favorite math websites, Cut the Knot (previously mentioned on this blog), gives over a hundred proofs of Pythagoras:

The only other theorem with many known proofs is Gauss's Law of Quadratic Reciprocity. Here is a discussion of some of the first few proofs:

Proof #1 is Euclid's own proof, his Proposition I.47. Proof #2 is simple enough, but rarely seen. Proofs #3 and #4 both appear in the U of Chicago, Section 8-7 -- one is given as the main proof and the other appears in the exercises. Proof #5 is the presidential proof -- it was first proposed by James Garfield, the twentieth President of the United States. I've once seen a text where the high school students were expected to reproduce Garfield's proof.

So far, the first five proofs all involve area. My favorite area-based proof is actually Proof #9. I've tutored students where I've shown them this version of the proof. Just as the Cut the Knot page points out, Proof #9 "makes the algebraic part of proof #4 completely redundant" -- and because it doesn't require the students to know any area formulas at all (save that of the square), I could give this proof right now. In fact, I was considering including Proof #9 on today's worksheet. Instead, I will wait until our next activity day on Friday to post it.

But it's the proof by similarity, Proof #6, that's endorsed by Common Core. This proof has its own page:

Here is Proof #6 below. The only difference between my proof and #6 from the Cut the Knot webpage is that I switched points A and C, so that the right angle is at C. This fits the usual notation that c, the side opposite C, is the hypotenuse.

Given: ACB and ADC are right angles.
Prove: BC * BC + AC * AC = ABAB (that is, a^2 + b^2 = c^2)

Statements                                Reasons
1. ADC, ACBCDB rt. angles   1. Given
2. Angle A = A, Angle B = B     2. Reflexive Property of Congruence
3. ADC, ACB, CDB sim. tri.     3. AA Similarity Theorem
4. AC/AB = AD/AC,                 4. Corresponding sides are in proportion.
    BC/AB = BD/BC
5. AC * AC = AB * AD,           5. Multiplication Property of Equality
    BC * BC = AB * BD
6. BC * BC + AC * AC =         6. Addition Property of Equality
    AB * BD + AB * AD
7. BC * BC + AC * AC =         7. Distributive Property
    AB * (BD + AD)
8. BC * BC + AC * AC =         8. Betweenness Theorem (Segment Addition)
    AB * AB

I mentioned before that, like many converses, the Converse of the Pythagorean Theorem is proved using the forward theorem plus a uniqueness theorem -- and the correct uniqueness theorem happens to be the SSS Congruence Theorem (i.e., up to isometry, there is at most one triangle given three sidelengths). To prove this, given a triangle with lengths a^2 + b^2 = c^2 we take another triangle with legs a and b, and we're given a right angle between a and b. By the forward Pythagorean Theorem, if the hypotenuse of the new triangle is z, then a^2 + b^2 = z^2. (I chose z following the U of Chicago proof.) Then z^2 = c^2 by transitivity -- that is, z = c. So all three pairs of both triangles are congruent -- SSS. Then by CPCTC, the original triangle has an angle congruent to the given right angle -- so it's a right triangle. QED

Interestingly enough, there's yet another link at Proof #6 at Cut the Knot, "Lipogrammatic Proof of the Pythagorean Theorem." At that link, not only is Proof #6 remodified so that it's also an area proof (just like Proofs #1-5), but, as its author points out, slope is well-defined without referring to similar triangles!

The Common Core Standards only require that the Pythagorean Theorem be proved using similarity -- not the concept of slope (which we'll cover here on the blog next week). So now I'm wondering whether it might be easier for the students to understand this derivation of slope.

It was David Joyce who pointed out that slope requires similarity to prove. He also criticized the area-based proof of the Pythagorean Theorem given in the Prentice-Hall text -- but this is because he wanted the Parallel Tests and Consequences to be taught first. (According to Cut the Knot, the Pythagorean Theorem is equivalent to the Parallel Postulate.)

For now, I think I'll stick to my plan to use similarity to prove slope -- but my proof may still be based on the one given at the above link.

Tuesday, January 20, 2015

Section 1-3: Ordered Pairs as Points (Day 95)

No, that's not a typo. This week we begin coordinate geometry, and in order to do that, we need to review coordinates. This is given in Section 1-3 of the U of Chicago text. When we started the school year, we began with Section 1-4, so we haven't actually done Section 1-3 yet.

But first, there's a few more things I wanted to say about how I would change Common Core, testing, and tracking if I were in charge. First of all, my computer adaptive replacement to Smarter Balanced has the computer keep track of the students' answers not only as they are taking the test, but also from year to year. This would encourage both acceleration and remediation.

But one major criticism of Common Core testing is that it requires the computer to keep track of student information, which may be seen as a violation of privacy. Even traditionalists who favor tracking might be opposed to it if government computers are the ones doing the tracking. If asked, they'd probably say that they'd prefer the teacher whose been with the student the whole year to be the one to determine on which track to place the student.

My problem is that this is where the bias seeps in -- the bias that doomed the old form of tracking. It is only human nature to have even subconscious biases, and such biases may result in students placed on the tracks based on demographics rather than what's best for the students. And so, to satisfy the MLK Jr. dream ideal, I'd much prefer that the computer be the one to place students on tracks, with as few humans with their subconscious biases involved as possible.

Now, here's the other thing I want to address here. I've set the Dickens age, the age at which students can leave school, to be 16. But now, some people are wondering, how can I claim that my plan satisfies the MLK ideal, when MLK himself graduated school at 15?

Actually, age 15 was my first choice for the Dickens age. I changed it to 16 because this fits the Singapore standards better. But now, on second thought, I'm going to set the Dickens age back to my first choice of 15. A bright student like MLK may complete all the high school requirements by this age, and the normative student will have completed ninth grade by then.

For now, I'll still set the graduation standard to 800 -- that is, the completion of the seventh grade (Singapore Secondary One) standards. This means that a standardized test is given every year until the score 800 is reached, and then again at the Dickens age to show what has been achieved when the student is ready to leave school. The normative student will have ended up taking a test each year from grades 3-7 and then 9.

So now I'd have to reorganize my ninth grade standards. Once again, it's too bad that I don't have access to Singapore New Elementary 3. The standards link that I gave in my last post shows that the Secondary Three/Four standards are combined. All I know is that all this material is taught in three semesters -- corresponding to all of 9th grade and the first half of 10th grade -- and the second semester of 10th grade is for review.

I could change it so that the second half of 9th grade -- the last semester before the Dickens age -- is now the review half. So now all I have to do is choose which of the Singapore standards to fit into the first semester of 9th grade. Since there are three major topics -- Numbers and Algebra, Geometry and Measurement, and Statistics and Probability -- it would make sense just to include the first topic, Numbers and Algebra. This topic includes the sub-topic "Applications of mathematics in practical situations," including utilities bills, hire-purchase, and simple/compound interest. This is a great topic to include the final year before the Dickens age. But it also includes "Matrices" -- and this seems a bit advanced for this level.

Or perhaps one can keep up the Integrated Math spirit of my standards, and perhaps include a little of all three topics. "Applications" is great from the Numbers strand, "Congruence and similarity" is good from the Geometry strand (notice that the eighth grade course introduces congruence, but SSS, SAS, and ASA don't appear until the ninth grade course), and I'd include everything from the Statistics and Probability strand.

Notice that if we wanted to divide the grades into paths, now the division K-1, 2-3, 4-5, 6-7, 8-9 makes a little more sense. From 4-5 to 6-7 is the jump from elementary to middle school, 6-7 to 8-9 is the jump from Pre-algebra to Algebra, and 8-9 to 10-12 is the jump across the Dickens age.

OK, that's enough about MLK and my tracking plan. Let's get into the new chapter. We are now beginning coordinate geometry. Unlike similarity, where my unit is based completely on Dr. Wu, coordinate geometry will be based completely on Dr. Franklin Mason's geometry course.

Dr. M named his final Chapter 13 "Analytic Geometry." His lessons are organized as follows:

Section 13.1 -- The Coordinate Plane
Section 13.2 -- Distance
Section 13.3 -- The Midpoint Formula
Section 13.4 Pt. 1 -- Slope
Section 13.4 Pt. 2 -- Parallels and Perpendiculars
Section 13.5 -- Equations of Lines

Let's convert these into U of Chicago sections, and this will give my plan for the next two weeks:

Today, January 20th -- Ordered Pairs as Points (Section 1-3)
Tomorrow, January 21st -- The Distance Formula (Section 11-2)
Thursday, January 22nd -- The Midpoint Formula (Section 11-4)
Friday, January 23rd -- Activity
Monday, January 26th -- Slope (Section 3-4)
Tuesday, January 27th -- Parallel Lines and Perpendicular Lines (Sections 3-4 cont., 3-5)
Wednesday, January 28th -- Equations of Lines (no exact correspondence in geometry)
Thursday, January 29th -- Review for Chapter 11 Test
Friday, January 30th -- Chapter 11 Test

Notice that officially, I stated that I would be covering "Chapter 11" after Chapter 12. But as we see above, we jump around among Chapters 1, 3, and 11 of the U of Chicago text. Indeed, only two sections -- the ones on the Distance and Midpoint Formulas -- actually come from Chapter 11.

Today's lesson is straightforward. I simply remind students about the coordinate plane that they surely learned in a previous course -- Algebra I, if not earlier.

The main concern is about tomorrow's lesson. Tomorrow I am scheduled to teach the Distance Formula -- but this is supposed to be proved using the Pythagorean Theorem, which I haven't taught on this blog yet. If I teach the Distance Formula now, I'm no better than the Prentice-Hall and Glencoe texts that try to teach the Distance Formula in Chapter 1.

The problem is that with all of my jumping around back and forth, the Pythagorean Theorem has been orphaned without a chapter. The U of Chicago gives the Pythagorean Theorem in Chapter 8, on Measurement Formulas. It appears after area, since the text uses area in its proof of the theorem. But not only are we delaying Chapter 8 until March, but an area proof isn't even how the Common Core wants us to prove the theorem anyway:

Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity.

(We've already proved the first two in Section 12-10, as the Side-Splitting Theorem.) So we see that the Pythagorean Theorem is to be proved using triangle similarity, not area.

Dr. M proves the Pythagorean Theorem using similarity in his Chapter 8. Now then, his Chapter 8 corresponds roughly to the U of Chicago's Chapter 14. In Section 14-2 of the U of Chicago, Question 11 asks the students to give a proof of the Pythagorean Theorem using Similarity. As it turns out, this proof, buried in an exercise in the U of Chicago, is the primary proof that Dr. M and Common Core want the students to learn.

I could give that proof right now, since we have already covered similarity. The problem is that the proof uses geometric means, and that's covered earlier in Section 14-2. Dr. M devotes a whole section, 8.1, to the geometric means and then 8.2 is the Pythagorean Theorem. I could simply let tomorrow's lesson be Section 14-2 and teach both geometric means and Pythagoras. But then the primary topic of tomorrow -- the Distance Formula -- would be buried.

The only other choice would be to reshuffle the chapters once again -- just teach U of Chicago Chapter 14 right now and let Chapter 11 be last, just as Dr. M's Chapter 13 is last. But I've already said that I would do Chapter 11 now, and I feel that I've already delayed the Distance and Midpoint Formulas long enough.

So the plan for tomorrow is to start with a proof of the Pythagorean Theorem. Since we haven't given geometric means yet, it will just set up the necessary proportions directly without mentioning geometric means at all. Then it ends with the Distance Formula -- the emphasis on the fact that the Distance Formula is simply the Pythagorean Theorem applied to the coordinate plane. My hope is that students will understand both Pythagoras and the Distance Formula better if they are taught in the very same lesson.

As for today's lesson, it's always tough when I need to create a worksheet on graph paper. I chose to get the grid from another site:

I hope that this will show up on the page when printing.

Friday, January 16, 2015

Chapter 12 Test (Day 94)

Yesterday, I mentioned Darren Miller's post about Integrated Math. While I don't agree with everything that he wrote, there's one thing I must point out. He is also opposed to his district's new requirement that students take three years of math in order to graduate. He fears that the Integrated Math classes would be watered-down in order to get non-mathematically inclined students to pass them and graduate, making it even harder for any students to reach AP Calculus.

With this, I agree. I certainly don't like the trend of having many school districts -- including not only Miller's district, but also the LAUSD right here in Southern California -- requiring the equivalent of Algebra II just to graduate high school. Yes, Algebra II is required to enter college, but one shouldn't have to be college-bound just to graduate high school. According to California's master plan, the UC system is for the upper eighth, or 12.5%, of high school graduates, while the Cal State system is for the upper third, or 33%, of high school grads. So I agree with Miller, and disagree with the districts that require 100% of students to take what only 33% of them need.

But then this raises the question, which students should take the college-bound classes? This leads naturally to that highly controversial topic -- tracking.

I wrote that I wanted to devote an entire post to tracking, and this post is it. The idea of tracking sounds good in theory, but in practice, it often leads to segregation among demographic groups, and the leaders might show that they only care about the outcomes of those students with whom they share a demongraphic group.

So before stating a tracking plan, one should emphasize that one cares about the outcomes of all students, including those on higher and those on lower tracks, and regardless of the demographics such as income and -- the big one -- race. I don't want to have to make this about race, but any discussion of tracking inevitably leads to race. On this Martin Luther King Jr. weekend, let's maintain the MLK ideal and provide opportunities for all students, independent of race, to attain a comfortable middle-class life.

Earlier on this blog, I proposed part of a tracking plan -- the Path Plan -- based on a scheme that my own elementary school had for a few years. Before I return to discussing that plan, let me put my money where my mouth is and state what I just said that anyone should state before attempting to propose a tracking plan:

I care about the outcomes of all students, including those on higher and those on lower tracks, and regardless of the demographics such as income and race. I want to provide opportunities for all students, independent of race, to attain a comfortable middle-class life.

Now, let's review the plan. Students are not divided into grade levels based strictly on age, but instead into various paths. The normative span of each track is two years, and they correspond roughly to grades K, 1-2, 3-4, 5-6, 7-8, 9-10, and 11-12. Students are placed into paths based on their reading or ELA skills, but may take other subjects, most notably math, on a different path. Students have more than one teacher per day -- the exact number depends on the path. Here are the classes that I proposed to be taught by a teacher other than the homeroom teacher (defined here as the ELA teacher):

Path K: none
Path 1-2: Math
Path 3-4: Math, Elective
Path 5-6: Math, Elective, Science
Path 7-8: Math, Elective, Science, PE
Path 9-10: All except ELA

The closest known plan to this Path Plan is the Joplin Plan. The difference between the Joplin Plan and the Path Plan is that with Joplin, ELA and math are the two classes taught by another teacher for all grades, and so all other subjects are given in the homeroom.

Both the Joplin and Path Plans should be less prone to demographic profiling than a pure tracking plan where within each grade level, there is a high-track and a low-track class. With the Joplin and Path Plans, an above-average second grader and a below-average fifth grader may end up in the same 3-4 math class. The teacher can't simply give the below-average fifth grader an inferior education -- as a fifth grader on a low-track might receive -- because doing so would hurt the above-average second grader in the same class.

The tradeoff, of course, is that students of vastly different ages are in the same class. The above scenario places students three years apart in age in the same class. I consider three years to be the borderline between acceptable and unacceptable. Above a three-year difference, we run the risk of having the older students in the class beat up the younger kids -- especially since an older student in such a class is likely a below-average "bully" and the younger an above-average "nerd."

Some people who propose tracking plans say that, for the higher grades, all the below-average students can do in class is disrupt it, since they are hopelessly behind the other students. Instead, they say that such students should simply be expelled from school. If attending school were no longer mandatory for students beyond a certain age, school can be made more rigorous for those who remain in the school. One can have full Algebra I in eighth grade and AP Calculus in 12th, because the students who can't pass those classes unless watered-down would no longer even be in school. And instead, those children can get jobs and gain work experience (for in such a world, lack of a high school diploma would no longer be an impediment to getting a full-time job).

The problem with this plan is Charles Dickens. That is, his 19th-century novels describe a world in which businesses would exploit their youngest employees. Our language has a word, Dickensian, to describe the bleak working conditions these children face. And so I don't like the idea of children working 100 hours a week for very low pay in some factory at such a young age just because they don't do well in Algebra I.

I define the Dickens age to be the age at which a person can be an employee at a business for his or her own benefit, rather be exploited by a business as in a Dickens novel. Then school should be mandatory until a student reaches the Dickens age. For my blog, I set the Dickens age to be around 16 years old, the conclusion of sophomore year. I wouldn't mind setting it a little lower, to 15 years, but I'm basing this on the Singapore standards, where students attend school up until they finish their O-level exams, at the end of the equivalent of 10th grade.

Some traditionalists say that the high school diploma has become devalued, because it has been made easier to obtain in order to allow more to obtain it. To make a diploma worth something, they say, one has to accept the fact that some people aren['t qualified to get it -- but then these people would be unable to get jobs or a comfotable middle-class lifestyle.

So how easy should it be to obtain the 10th-grade "diploma" that I'm proposing here? Well, during presidential elections, pundits discuss how hard it is for an incumbent (or his party) to be reelected if the unemployment rate is high. Usually, an acceptable unemployment rate is around 5% -- that is, about 95% of the people have jobs. And so this is what I want for my 10th-grade diploma -- it should be attainable by around 95% of the population.

We can fit this into our proposed math standards as follows -- an exit exam can be devised to that a certain amount of math is required to earn this diploma. In particular, the level of math required should be such that 95% of the population can learn the material by the Dickens age. If we line this up with the proposed Singapore math standards, then I'd set it to be the equivalent of the Secondary One standards -- the equivalent of seventh-grade pre-algebra. The Secondary Two eighth grade standards contain too much Algebra I to expect 95% to master them by the Dickens age. (Under the old California standards, I might have said up to the first semester of eighth grade Algebra I, but Secondary Two contains many difficult topics right at the start of the year.)

Now let's think about how this test would be given. There are already two existing tests, the PARCC exam, given to grades 3-8 and at the end of every math course, and the Smarter Balanced or SBAC exam, given to grades 3-8 and 11. I live here in California, an SBAC state. If the equivalent of the SBAC were given, then passing the seventh grade test would be sufficient for graduation.

But recall that the Smarter Balanced exam is supposed to be computer-adaptive -- that is, questions are tailored to each student based on answers to earlier questions. But the way the SBAC is currently set up is not designed to take full advantage of its computer adaptability! There's actually a way to set up a computer-adaptive test so that it supports grade-level (or path-level, using the paths as described above) acceleration. Simply put, if a student answers all of, say, the third grade questions correctly, then the student should start answering fourth grade questions, and if all of these are answered, then the computer should move on to fifth grade questions, and so on.

Then the scores can be determined so, say 300 represents the attainment of an average third-grader, 400 the attainment of an average fourth-grader, 500 a fifth-grader, and so on. Actually, I'd would add 150 points to each of those levels, so that 450 is the average third-grader. This way, any score within fifty points of this -- in the four hundreds -- should be the range of the entering fourth graders. The score therefore can represent placement of a student into a particular grade level or path. Then the following year, the computer-adaptive test can begin by asking grade-level questions based on the student's score the previous year -- so a student scoring in the five hundreds the previous year should
begin receiving fifth grade questions.

A student scoring 800 or above should be placed into eighth-grade math, which, as we stated, contains much Algebra I. So this would end the debate over who should take Algebra I. Since seventh grade math is the highest required for the diploma, the score of 800 would be the score required for receiving our "diploma." (The score of 800 is already a magical number -- it represents a perfect score on a section of the SAT and a proficient score on our old California State Tests.)

Students who have earned an 800 shouldn't have to take an annual SBAC, but instead only the comprehensive high school SBAC exam. This should be given at the Dickens age -- so that means the end of sophomore, not junior, year. Therefore students should keep on taking higher and higher math courses until reaching the Dickens age. The type of student who reaches 800 early is the same type of student who should be taking higher and higher math. The test given at the Dickens age would be integrated, containing both algebra and geometry. But a school or district could still have traditionial pathway classes (Algebra I, Geometry, Algebra II), since such a student can still get the Algebra I classes even if he or she doesn't reach Geometry by the Dickens age.

This was a problem when California first adopted the Common Core. Under the Common Core, states were allowed to augment the standards by 15%. Naturally, California wanted to add Algebra I to the eighth grade math standards. The problem was that the eighth graders would have been still required to take the SBAC for Common Core Grade 8 -- so they would've been in a class that covers all of the Algebra I standards missing in Common Core Grade 8 (such as polynomials and factoring) plus all of the geometry standards from Common Core Grade 8 (such as dilations). The class would no longer have been reasonable for eighth graders to take. So California dropped the idea of adding Algebra I to the Common Core Standards.

The simple solution would have been to say that eighth graders in Algebra I don't have to take the SBAC, just as ninth graders in Algebra I don't have to take the SBAC. But unfortunately, No Child Left Behind requires a test every year from third to eighth grade. Furthermore, I can easy see schools putting their lowest eighth graders into Algebra I, so that they wouldn't have to take the SBAC and so wouldn't lower the school score. This is why the plan that I'm discussing describes how things would be if I ruled the world -- where I can change SBAC to support acceleration and ignore the NCLB requirements however I please.

We should also set the lowest age at which a student could take our computer-adaptive exam. The current SBAC begins at the third grade. Recall that for me, third grade marks the end of pure traditionalism, and pure traditionalism emphasizes the multiplication tables. And so the SBAC should first be given as soon as a student has memorized the times tables. I even argued earlier that if a student attempts to take the test but fails to answer the simple one-digit multiplication problems correctly, the test could immediately end and a failing score be given.

What would the lowest score on the exam be? I once read about someone who argued that even the SAT supports grade/score inflation, because the lowest score on each section is 200, not zero. The score of 200 was chosen because it represents three standard deviations (each s.d. being 100 points) below the theoretical mean (500 points), and only a statistically insignificant number of students would score beyond three s.d. above or below the mean -- hence the scale is 200 to 800. On my scale, a score of 0 would represent the lowest kindergartner. If one wishes, instead of having the test end after a student fails to multiply, several addition and subtraction questions can be given so that scores between 0 and 400 can now be distinguished.

Sometimes, I wonder whether I should actually fit the standards into the path system, so that a given set of standards corresponds to two grade levels. The Singapore Secondary Math Standards and the Common Core ELA Standards already combine grades 9-10, so one can simply work the way down and combine grades 7-8, 5-6, 3-4, and 1-2 into paths.

But this would be a bad idea, for several reasons. First of all, we definitely want to separate seventh and eighth grades and not combine them into a single set of standards, because we want to require seventh, but not eighth, grade math for our "diploma." Second, these paths were based on my own elementary school, which was a K-6 school. Many elementary schools are now K-5, so combining fifth and sixth grades into a single path would be awkward at such schools -- even moreso in my current district, which has a few upper elementary schools for grades 4-5 only.

A third reason is that it's in our human nature to oppose something that we would have originally supported if we feel that we're being forced to support it, especially if it's the federal government doing the forcing. Many traditionalists say that they oppose the Common Core Standards because they don't like its standards being divided into grades and forcing all students in the same grade -- that is, age -- to learn the same thing. If my proposed standards were divided into my proposed paths instead of, then it would separate these standards from being tied down to a specific age -- which is exactly what the traditionalists want. But then, I suspect, they would criticize my standards for forcing this new "path plan" on everyone -- especially if I started using my old school's unfamiliar names like "Primary Path" and "Transition Path." Human nature would cause them to defend the grade plan that they formerly opposed -- going right from Why should all ten-year-olds be forced into the fifth grade? to I know what the third grade is, but what exactly is this "Preparatory Path" supposed to mean?

So my plan is to keep the standards tied down to age, but take advantage of the SBAC's computer adaptibility to allow students to test at different ages.

Here are the answers to the test that I am posting:

1.-3. These are drawings. (Hopefully you can figure out what the dilations should look like!)

4. 22. (No, not 17.6. 88 is the length of NR, not OR!)

5. XY = 336. (No, not 21. The scale factor is 4, not 1/4. This was hard for me to draw accurately while still getting it to fit on the page.)

6. 5.4, to the nearest tenth. (Happy MLK Day!)

7. d.

8. a. This is a drawing.
b. 3.
c. SU = 9, TU = 12.

9. Answers may vary. One possibility is t/m = h/a. (More complex answers look bad here in ASCII.)

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. 3000 ft. (No, not 5250 ft. It's 7000 ft. from Euclid to Menelaus, not Euclid to Pythagoras.)

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!)