Friday, February 16, 2018

Lesson 11-1: Proofs with Coordinates (Day 111)

Chapter 11 of the U of Chicago text is called "Coordinate Geometry." And today, I have plenty of things to say about this new chapter -- but first things first.

Today on her Mathematics Calendar 2018, Theoni Pappas writes:

The area of the large concentric circle is 25pi sq. units. Find the area of the square circumscribing the smaller circle.

(No, the square circumscribed about the small circle is not inscribed in the large circle. Instead, a difference of 3 is given between the radii of the circles.)

To solve this, we notice that since the area of the large circle is given, finding its radius is easy:

A = pi r^2
25pi = pi r^2
r = 5

The large radius is 5, and the small radius is 3 units shorter than this, so the small radius is 2. This is the radius of the inscribed circle -- that is, it's the apothem of the square.

Now just yesterday, I wrote that the apothem of a square is half of its side length -- that is, the diameter of the inscribed circle is equal to the side length. Since this has come up again today, let's look at this statement in more detail.

The statement that the diameter of the circle equals the side of the square appears "obvious," but technically speaking, we haven't proved it yet. Indeed, a full proof actually requires us to wait until Lesson 13-5, when we learn that a tangent to a circle is perpendicular to the radius of the circle drawn at that point. Because of these right angles, the diameter of the circle really divides the square into two rectangles. Rectangles (like all other parallelograms) have congruent opposite sides. Each rectangle as a diameter as one side, and the opposite side is also a side of the square. Thus the diameter of the circle equals the side of the square, and the radius of the circle (the apothem of the square) is half of the side of the square. QED

So the square has an apothem of 2, hence a side length of 4. Its area is 4^2 or 16. Therefore the answer is 16 square units -- and of course, today's date is the 16th.

Lesson 11-1 of the U of Chicago text is called "Proofs with Coordinates." In the modern Third Edition of the text, proofs with coordinates appear in Lesson 11-4.

Coordinate proofs are mentioned in the Common Core Standards:

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).

Despite this, I've never covered Lesson 11-1 on the blog before. The reason for this omission is, as usual, a long story.

In Lesson 11-1, we are given the coordinates of the vertices of a polygon, and we are asked to prove that the polygon is a parallelogram, right triangle, or rectangle. The key to these coordinate proofs is to find and compare the slopes of the sides.

But here's another Common Core Standard:

Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.

So students are supposed to use similarity to prove the properties of slope. David Joyce, whom we mentioned throughout Chapter 9, also endorses the use of similarity to prove slope -- and indeed, he has harsh words to say about the treatment of coordinate geometry in most Geometry texts:

In a return to coordinate geometry it is implicitly assumed that a linear equation is the equation of a straight line. A proof would depend on the theory of similar triangles in chapter 10. At this point it is suggested that one can conclude that parallel lines have equal slope, and that the product the slopes of perpendicular lines is -1. The only justification given is by experiment. (A proof would require the theory of parallels.)

And in our text, similar triangles don't appear until Chapter 12. Thus to follow the Common Core and David Joyce, we should wait to teach Chapter 11 until after Chapter 12. Students need to have mastered similar triangles before they can begin learning about slope.

This has been the source of many headaches in my blog posts over the years. First of all, I'd start with Chapter 12, then go back to teach slope and some other Chapter 11 topics -- but I'd never actually reach Lesson 11-1.

The problem, of course, is that slope is an Algebra I topic. High school students are thus going to see slope well before they ever see similarity, because they take Algebra I before Geometry.

I also wrote extensively about Integrated Math courses. But even Integrated Math usually covers slope before similarity -- indeed, slope is a Math I topic, while similarity is a Math II topic. I once tried to devise my own Integrated Math courses that teach similarity before slope, but I failed. It's difficult to justify teaching similarity (from the second half of Geometry) before slope (which is from the first half of Algebra I).

In fact, we notice that the Common Core Standard requiring students to use similarity to prove the slope properties is an eighth grade standard, not a high school Geometry standard. This now makes sense -- students are introduced to slope in eighth grade in order to prepare to study it in more detail in Algebra I.

I think back to last year's eighth grade class. Of course, student behavior and classroom management were issues. But another problem was that I began teaching translations, reflections, and rotations -- and rotations, understandably, confused some students. The extra time spent on isometries meant less time on dilations -- and dilations are the bridge to similarity and slope.

I now sometimes wonder whether it's better to teach only one of the isometries -- perhaps reflections, since they generate all isometries (i.e., all isometries are the composite of one or more reflections) -- and then skip directly into dilations. But this contradicts the Common Core Standards that explicitly mention translations and rotations -- and these might appear in PARCC or SBAC questions.

At any rate, if the connection between similarity and slope is covered in eighth grade, then it doesn't need to be introduced in high school Geometry. And so we can write about slope in Chapter 3 without having to prove anything about similarity first. As I mentioned before, Chapter 3 is a great time to teach slope, since it's a review topic from Algebra I, and Chapter 3 is often taught right around the time of the PSAT (where slope questions will appear).

When David Joyce wrote about slope and similarity, he forgot that there's a class called "Algebra I" where students learn many things about slope and coordinates without proving everything. In the end, I did say that this year I'd adhere to, not Joyce's suggestions, but the order of the U of Chicago text.

And by the order of the text, I mean the order of the old Second Edition. Earlier, I wrote that Lesson 11-1 appears as Lesson 11-4 of the new Third Edition. So what exactly appears in the first three lessons of the modern version?

Well, Lessons 11-1 through 11-3 of the new text correspond to Lessons 13-1 through 13-4 of the old version of the text. Indeed, the new Chapter 11 is called "Indirect and Coordinate Proofs." You might recall that Chapter 13 of the old text has been destroyed, and its lessons are now included as parts of different chapters. And so the first half of the old Chapter 13 now forms the first part of 11. (There are now only three lessons instead of four because the old Lesson 13-2, "Negations," has now been incorporated into the other three lessons.)

Otherwise Chapter 11 remains intact in moving from Second to Third Edition. Chapter 11 of the old edition has six lessons, and these correspond roughly to Lessons 11-4 through 11-9 of the new text.

Let's finally take a look at the new Lesson 11-1 worksheet. We begin with the two examples from the text -- the first problem lists four ordered pairs and asks us to prove that they are the vertices of a parallelogram, while the second lists three pairs that may be the coordinates of a right triangle. In each case, students are to calculate the slopes of the sides formed by adjoining vertices, and show that these slopes are either equal or opposite reciprocals.

Today is an activity day. This section has only one Exploration Question #22. But I decided to include Question #9, because students are asked to prove that EFGH is a rectangle -- and rectangles, unlike parallelograms or right triangles, are explicitly mentioned in the Common Core Standards:

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).

The second part of that standard, on circles, will have to wait until later in the chapter. To complete the rectangle question, students must calculate the four slopes, and show that slopes of opposite sides are equal, while slopes of adjacent sides are opposite reciprocals.

For the activity question, students are given three vertices of a parallelogram and asked to find the location of the fourth vertex. They are told that there are at least two possibilities for the fourth vertex and asked whether there are any others.

The answer is that there are three possibilities for the fourth vertex. If you think about it, the three given order pairs are already the vertices of a triangle. To make this triangle into a parallelogram, we perform a 180-degree rotation, centered at the midpoint of one side of this triangle. There are three sides to choose from, hence three possible parallelograms. Each rotation maps the endpoints of the chosen side to each other and the remaining vertex to the fourth vertex of the parallelogram.

To make this easier to understand, label the three given vertices A, B, and C. The three possible rotations (all of them are 180 degrees) are:

  • Rotate A about the midpoint of BC to obtain A'.
  • Rotate B about the midpoint of AC to obtain B'.
  • Rotate C about the midpoint of AB to obtain C'.
Then the possible parallelograms are ABA'C, BAB'C, and CAC'B.

But students don't need to perform a rotation to find the fourth vertex -- nor do they even need to use the Slope Formula. Instead, translations suffice. We start with the one of the three given vertices and count how many horizontal and vertical steps it takes to reach one of the other vertices. Then we begin at the third vertex and count out the same number of steps in each direction -- the final location is where the fourth vertex is. The reason this works is that we are actually finding the composite of two translations, which is tantamount to adding their vectors. And the sum of two vectors can be found using a parallelogram rule. (For example, vector AB + vector AC = vector AA')

There are only three possibilities because only one choice matters -- which point to pick first. If we choose C as the first point, it doesn't matter whether A or B is the second point. Either choice will wind up at the same fourth point, namely C'. (Again, this is easy to see if we use vectors -- the second choice doesn't matter because vector addition is commutative.)

This concludes a whirlwind week of holidays and celebrations -- Lincoln's Birthday on Monday, Mardi Gras on Tuesday, Valentine's Day on Wednesday, the partial solar eclipse yesterday, and Chinese New Year today. This upcoming Monday is one more holiday -- President's Day, when all schools are closed. Therefore my next post will be on Tuesday.

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