Thursday, April 27, 2017

Lesson 14-2: Lengths in Right Triangles (Day 142)

This is what Theoni Pappas writes on page 117 of her Magic of Mathematics:

"Every math enthusiast at one time or another has discovered or been delighted by tricks or oddities involving numbers. Here are two for you to explore and hopefully enjoy."

Pappas is "Playing with Numbers" in this section. She begins with summing and squaring:

1 + 2 + 1 = 2^2
1 + 2 + 3 + 2 + 1 = 3^2
1 + 2 + 3 + 4 + 3 + 2 + 1 = 4^2
1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5^2

This reminds us of the sum of odd numbers pattern from two weeks ago. Notice that this and the odd number sums are related -- take a look at that last sum again:

1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5^2

The first term is 1, which is also the first odd number. If we cross it out, the new first and last terms are 2 + 1 = 3, the second odd number. If we cross them out, the new first and last terms are 3 + 2 = 5, the third odd number. If we cross them out, the new first and last terms are 4 + 3 = 7 -- by now, we get the pattern. We obtain:

1 + 3 + 5 + 7 + 9 = 5^2

which is the pattern from page 108.

Another way to think about this pattern is to rotate a 5 x 5 square of dots 45 degrees to form a diamond -- hey, wasn't I writing something about how a diamond is just a square yesterday? We can view this diamond as containing nine rows -- one in the first row, two dots in the second row increasing up to five dots in the fifth row, then four dots in the sixth row decreasing down to one dot in the ninth row. This is a visual justification for:

1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5^2

Here is the second pattern Pappas mentions today -- the 1's pyramid:

1^2 = 1
11^2 = 121
111^2 = 12321
1111^2 = 1234321
11111^2 = 123454321
111111^2 = 12345654321
1111111^2 = 1234567654321
11111111^2 = 123456787654321

Pappas asks, "When does it stop?" Well, let's try a few more:

111111111^2 = 12345678987654321
1111111111^2 = 1234567900987654321

So that's when the pattern ends. But at this point, we notice that most of the digits of this number still follow the pattern. Indeed, what seems to prevent the pattern is that there's no digit for ten -- and in fact, look at what happens in base 11, where we can use X to represent ten:

1111111111^2 = 123456789X987654321

And in dozenal, we can use E to represent eleven. (I mentioned "dozenal," or base 12, several times on this blog, most recently in my September 22nd post.)

11111111111^2 = 123456789XEX987654321

So it appears that the pattern is limited only by the digits available in our base. To see why it works, we let b be the base of our numeration system. Then we obtain:

11 = b + 1
111 = b^2 + b + 1
1111 = b^3 + b^2 + b + 1
11111 = b^4 + b^3 + b^2 + b + 1

So squaring 11, 111, and so on amounts to squaring polynomials in b:

(b + 1)^2 = b^2 + 2b + 1
(b^2 + b + 1)^2 = b^4 + 2b^3 + 3b^2 + 2b + 1
(b^3 + b^2 + b + 1)^2 = b^6 + 2b^5 + 3b^4 + 4b^3 + 3b^2 + 2b + 1

So now we can see where the coefficients come from. The b^3 coefficient of this last squared polynomial is 4 because there are four b^3 terms (b^3)(1), (b^2)(b), (b)(b^2), (1)(b^3), while the b coefficient is 2 because there are only two b terms (b)(1), (1)(b).

An interesting case occurs when b = 1. Then we obtain:

(1 + 1)^2 = 1 + 2 + 1
(1 + 1 + 1)^2 = 1 + 2 + 3 + 2 + 1
(1 + 1 + 1 + 1)^2 = 1 + 2 + 3 + 4 + 3 + 2 + 1
(1 + 1 + 1 + 1 + 1)^2 = 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1

which should look familiar. That's right -- when b = 1, the second pattern that Pappas gives today reduces to the first pattern!

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

Lesson 14-2 of the U of Chicago text is on lengths in right triangles -- specifically, those lengths that are related to the altitude and involve the geometric mean.

Geometric Mean Theorem:
The geometric mean of the positive numbers a and b is sqrt(ab).
(Note: This may sound like a definition, but actually the U of Chicago defines geometric mean to be the number x such that a/x = x/b, so we need a theorem to get the geometric mean as sqrt(ab).)

Right Triangle Altitude Theorem:
In a right triangle:
a. The altitude of the hypotenuse is the geometric mean of the segments dividing the hypotenuse.
b. Each leg is the geometric mean of the hypotenuse and the segment adjacent to the leg.

In this lesson, I give the proof of the Pythagorean Theorem based on similarity, but this time I gave the proof in the book, which mentions the geometric mean. Let's look at the proof -- as usual, with an extra step for the Given:

Given: Right triangle
Prove: a^2 + b^2 = c^2

Proof:
Statements                                Reasons
1. Right triangle                       1. Given
2. a geometric mean of c & x,  2. Right Triangle Altitude Theorem
b geometric mean of c & y
3. a = sqrt(cx), b = sqrt(cy)       3. Geometric Mean Theorem
4. a^2 = cxb^2 = cy                 4. Multiplication Property of Equality
5. a^2 + b^2 = cx + cy               5. Addition Property of Equality
6. a^2 + b^2 = c(x + y)              6. Distributive Property
7. x + y = c                                7. Betweenness Theorem (Segment Addition Postulate)
8. a^2 + b^2 = c^2                    8. Substitution (step 6 into step 7)

It is uncertain whether this is the proof that Common Core intends the students to learn, or whether my earlier proof that avoids geometric means suffices.

Actually, since posting this last year, I've decided to check both the PARCC and SBAC released test questions for those related to the proof of the Pythagorean Theorem. There were a few questions that required use of Pythagoras, but none directly related to the proof. Of course, some people lament that there aren't very many proofs on the Common Core tests.

Speaking of the Pythagoran Theorem, I've been thinking last week about how I presented it to my eighth graders earlier this year:

Monday: Coding
Tuesday: Illinois State project
Wednesday: Traditional lesson on the theorem
Thursday: "Learning Centers" (actually the traditional lesson continued)
Friday: Test

which too many students failed. But as I think about it more, we can see how the students were set up for failure with only two days between introduction and assessment. (As it turned out, even the original project failed because it was on the day that the Hidden Figures movie extra credit was due, and many students focused on finishing that rather than the project.)

It might have been better to teach the theorem on Tuesday, and then the students can apply that knowledge to the project on Wednesday. (And it would have been even better to have the traditional lesson on Monday, but I had no control over the coding teacher's schedule.) The problem is that Illinois State, on its pacing guide, insists that the project be given before the traditional lesson. This in itself leads back to the traditionalist debate, which I'm not getting into in today's post. Another problem is that this might work for sixth and eighth grades, but not for seventh grade, as I had less time with them.

(Speaking of seventh grade, today I taught the volumes of rectangular and triangular prisms to seventh graders.)