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.

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