Even though art clearly isn't representative of the class that I want to teach someday, I'll still do "A Day in the Life" today. It's a middle school class, and so I need to take another look at my classroom management.
8:15 -- This is the middle school where homeroom leads directly into first period. This is the first of three seventh grade art classes.
As I wrote in January, seventh grade art is a trimester class. That day, the students sculpted a whistle out of clay. As it turns out, now the third tri students are supposed to work on the same whistle project -- but instead, the regular teacher decides to assign a simple worksheet on drawing a bird.
9:20 -- And once again, we know that this school has a period rotation after first period. The rotation for Thursday goes 1-5-6-2-3-4.
Fifth period is the only eighth grade art class. These students are working on applying three layers of glaze to the mugs that they sculpted.
10:15 -- Fifth period leaves for snack.
10:30 -- Sixth period arrives. This is the second of the seventh grade art classes.
11:25 -- Sixth period leaves and second period arrives. As it turns out, only four of the five classes this teacher has is art -- the other class is P.E., believe it or not! This eighth grade class plays badminton in the gym.
12:20 -- Second period leaves for lunch.
1:05 -- Third period is conference period -- except I have another P.E. class to cover. (I believe that most of the P.E. teachers have CPR training, including the art/P.E. teacher.) This seventh grade class plays Ultimate Frisbee outside.
2:00 -- Third period leaves and fourth period arrives. This is the last of the three seventh grade art classes.
2:55 -- Fourth period leaves, thus ending my day.
By the way, notice that seventh period art lasts only for a trimester. It's part of what we often call an "exploratory wheel" of electives. Eighth grade art is for the entire year -- so presumably only committed artists would sign up for this class.
Back in January I posted a focus resolution, so let me repeat it for today's post:
3. Move on from past incidents instead of bringing them up with students or flouncing around.
In fifth period, three students are not following the seating chart, even though the lesson plan specifically states to enforce the seating chart. My basic assumption is that any student who claims that the teacher just forgot to change the chart is lying, unless I see some evidence. But just in case the teacher really forgot, I take the liberty of updating the seating chart for him. If the students are lying, then he'll find out who lied and where they tried to sit instead. Recall that eighth grade art is yearlong -- and one of the students isn't listed on the seating chart at all. This counts as some evidence in favor of her seat story.
But then in sixth period, there are three more students not following the chart. I tell the students that it's understandable that eighth graders who have the yearlong class might have changed seats without being marked, but why should the chart not be current for the seventh grade tri classes? (In this district today is Day 151, the start of the sixth hexter. In the old district this is the end of the seventh quaver.) Then again, just as in the eighth grade class, there is one seventh grader not on the seating chart (a special ed student).
Unfortunately, this might count as bringing up past incidents -- mentioning the eighth grade seating arrangement during the seventh grade class. Once again, I must find a way to figure out what's happening with the seventh grade seating chart without mentioning eighth grade.
In sixth period, some of the students have earned the right for me to play music in class. As I wrote in January, this is something that this regular teacher offers them. In order to avoid an argument over what music to play on YouTube similar to the first period argument, I choose the song. I start to tell them about the first period argument -- then stop myself just in time.
By the way, the sixth period students want to hear the song "Old Town Road" by Lil Nas X. Instead, I play another country song with a black lead -- "Nine, Nine, Nine" from Square One TV. Actually, the sound is much stronger for the late Reg E. Cathey's other country hit, "Round It Off." (This is the first time I've played that song in class.) I play a cleaned-up version of "Old Town Road" for classes that earn it (no more than one student on the bad list).
Other issues that occur are in second period P.E., when two teams have only three players each (instead of 5-6) because some students are fooling around. In second period, I finally say that phrase that I'm keeping track off -- "Because I said so!" -- when one guy asks why he can't switch seats.
With so much in this post devoted to subbing and the U of Chicago lesson, I don't even have time to write much about Dialogue and Chapter 17 of Douglas Hofstadter's Godel, Escher, Bach. Yes, Hofstadter's Law strikes again!
Dialogue 17 of Douglas Hofstadter's Godel, Escher, Bach is called "The Magnificrab, Indeed." In this Dialogue, the Crab tells Achilles and Tortoise about an Indian mathematician named Najunamar who has proved several famous theorems. One of them is the 1729-Color Theorem (as opposed to the Four-Color Theorem of Lesson 9-8) -- and speaking of 1729, Achilles remarks that he took taxicab 1729 over to Tortoise's house that day. Najunamar also proved that every even prime is the sum of two odd numbers (in other words, 1 + 1 = 2) and Fermat's Last Theorem for exponent n = 0 (in other words, 1 + 1 = 2 again).
Then Crab plays some theorems of TNT on his flute. This is similar to the Metamath music page:
http://us.metamath.org/mpegif/mmmusic.html
As it turns out, Metamath has set the proof of the second Peano postulate to music while Crab does the same with the third Peano (or "Piano") postulate.
Chapter 17 of Douglas Hofstadter's Godel, Escher, Bach is "Church, Turing, Tarski, and others." In this chapter, he describes several mathematicians -- including Church, Turing, Tarski, and others.
Here are some key ideas from this chapter:
Church's Theorem. There is no infallible method for telling theorems of TNT from nontheorems.
Church-Turing Thesis, Tautological Version: Mathematics problems can be solved only by doing mathematics. Hofstadter gives several concrete versions of this theorem in the Chapter, including:
Church-Turing Thesis, Reductionist Version: All brain processes are derived from a computable substrate.
Church-Turing Thesis, Theodore Roszak Version: Computers are ridiculous. So is science in general.
Tarski's Theorem: There is no TNT-formula with a single variable a which can be translated thus:
"The formula whose Godel number is a expresses a truth."
Here "Turing" refers to Alan Turing, the subject of the movie Imitation Game. The author also writes about the Indian mathematician Ramanujan, the subject of the movie The Man Who Knew Infinity. (I have written about Turing and Ramanujan in past blog posts.) Hofstadter jokes about Ramanujan in the Dialogue -- the mathematician Crab meets has Ramanujan's name spelled backwards, and 1729 is a famous taxicab number that the Indian resolves into the sum of cubes.
Today we are covering Lesson 15-8 of the U of Chicago text, on the Isoperimetric Inequality. Here are the key theorems of this lesson:
Isoperimetric Theorem:
Of all plane figures with the same perimeter, the circle has the most area.
Equivalently, of all plane figures with the same area, the circle has the least perimeter.
Isoperimetric Inequality:
If a plane figure has area A and perimeter p, then A < p^2/(4pi).
The U of Chicago text writes:
The proof of this theorem requires advanced calculus, a subject usually not studied until college. The reason the proof is so difficult is that it requires discussing all sorts of curves.
Readers of this blog should know by now that of course I'm not just going to leave it at that! I'm curious about the proof, and so I had to do some research.
Here is a link to a proof of the Isometric Inequality:
https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/blasjo526.pdf
According to the link, the 19th century Swiss mathematician Jakob Steiner was the first to prove the Isometric Inequality -- indeed, he gave five different proofs of the theorem! And, just as the U of Chicago tells us, the proofs all involve Calculus that is well beyond AP Calculus BC.
But the link also tells us about the ancient Greek mathematician Zenodorus, who most likely lived a little after Archimedes (and thus well after Euclid). The text tells us that the full proof is so difficult because "it requires discussing all sorts of curves," but Zenodorus is able to give a simple proof of the case that a circle has a greater area than any polygon with the same perimeter.
Zenodorus's polygon proof requires demonstrating three theorems. I will provide the proofs of all three theorems as given at the link above -- with commentary, as usual.
Theorem. For regular polygons with the same perimeter, more sides implies greater area.
Proof. Consider the apothem, the radius-like perpendicular drawn from the center to a side. Half the product of the apothem by the fixed perimeter yields the area of the polygon. The apothem is the height of the triangle [one of the n congruent triangles into which a regular n-gon is divided -- dw].
If we increase the number of sides, the base of the triangle is shortened and the angle is decreased. It is clear that the height increases. We would prove this by trigonometry; Zenodorus had to rely on the usual pretrig bag of tricks. It is routine for us, and it probably was for Zenodorus as well. QED
Theorem. A circle has greater area than any regular polygon with the same perimeter.
Proof. Archimedes proved that the cut-and-roll area formula also holds for the circle. [We actually discussed this back on Pi Day. Although the formulas C = 2pi r and A = pi r^2 are difficult to derive on their own, it's easy to derive one from the other using cut-and-roll. Last year we started with area and used cut-and-roll to derive the circumference, and this year we did the opposite direction. -- dw]
So we must show that the apothem of any regular polygon is shorter than the radius of the circle with the same perimeter. Rescale the perimeter so that it circumscribes the circle [a dilation! -- dw].
The perimeter is now greater than the perimeter of the circle, and therefore greater than before the scaling. Thus the scaling was a magnification [an expansion, using U of Chicago terminology -- dw], with the apothem magnified to the size of the radius of the circle. QED
Theorem. A regular n-gon has greater area than all other n-gons with the same perimeter.
Proof: Among isoperimetric triangles with the same area, the isosceles triangle covers the greatest area. so the maximal n-gon must be equilateral. Otherwise we could improve on it by making it equilateral.
We now know that the maximal n-gon must be equilateral. Suppose that it is not equiangular [an indirect proof -- dw]. Consider two dissimilar triangles [dividing the polygon -- dw]. Now make them similar [congruent -- dw] by redistributing perimeter from the pointy to the blunt angle until the two angles are the same. This increases the area. Accordingly the maximal n-gon must be equiangular: if not, we could improve on it. QED
Combining these three proofs, we conclude that the circle has a greater area than any other polygon with the same perimeter. Now we can see why not even Zenodorus's proof isn't given in the U of Chicago text -- it depends on so many theorems (as in how fixing the perimeter of a regular polygon and increasing the number of sides must lengthen the apothem) that are difficult to prove.
The following link leads to Cut the Knot, one of my favorite websites. It gives a proof of the Isoperimetric Inequality that is essentially the first Steiner proof given at the MAA link above:
http://www.cut-the-knot.org/do_you_know/isoperimetric.shtml
I dropped the posted proofs from past years. Just follow the link if you're really eager to see them again.
Elegant as it is this proof of Statement 1 contains a flaw. On each of the three steps we assumed that the shape answering conclusions of the steps existed and the Lemmas have only been proved under this assumption. Ultimately, we assumed that there exists a figure having a maximum area among all the shapes with the same perimeter. Under this assumption we proved that such a shape is bound to be a circle. Denote the existence hypothesis as H. What we have actually shown is an implication H ⇒ A. In order to prove A we still have to demonstrate that H holds true.
Existence of the optimal shape in the sense of Statement 1 is not at all obvious. For example, if Statement 1 required us to determine a shape with the smallest area for a given perimeter, such shape would not exist at all. Once we understood this point it's less important to actually complete the proof. H is proven with a limiting procedure which is quite simple but requires some basic elements of Calculus.
- Notice that reflections appear twice in this proof -- first in proving that the optimal shape is convex and then again in the proof of Lemma 1. In fact, we see that dilations also make an appearance -- the CTK proof that the two forms of the Isoperimetric Theorem are equivalent (Statements 1 and 2) also uses dilations. Indeed, it's a bit surprising that the U of Chicago text doesn't prove this equivalence using dilations -- instead the text solves the Isoperimetric Inequality for p. But it does show us how the Common Core transformations keep showing up in proofs.
- The proofs of these lemmas are all indirect. ("Assume it's not" convex... "Assume, on the contrary, that the area" is larger... Assume that SPT is not a right angle....)
- In yesterday's Lesson 15-3 we proved that an angle inscribed in a semicircle is a right angle. In this post we use the converse -- if every angle inscribed in an arc is right, then the arc is a semicircle.
- Just like the Zenodorus proof from earlier, we have a non-obvious statement whose proof is not given: "Among all triangles with two given sides, the one whose sides enclose a right angle has the largest area." But this is easy to prove using trig -- in fact, many Geometry texts (but not the U of Chicago) give the following formula: A = ab sin(theta)/2. The area is maximum when sin(theta) = 1 -- that is, when theta = 90 degrees.
- CTK states that this proof is incomplete because there's a statement H that has yet to be proved -- that there even exists a shape with a maximum area. It's the proof of H that requires Calculus -- otherwise we could almost include this proof in the U of Chicago text.
Theorem. Among all positive integers, 1 is the largest.
Proof. For any integer that is not 1, there is a method (to take the square) by which one finds a larger positive integer. Therefore 1 is the largest integer. QED?
(This "theorem" is also known as Perron's Paradox.) Here's what Perron actually proved: "either 1 is the largest integer or there is no largest integer." And of course, it's the latter statement that's true, but if Perron's proof is invalid, then so is Steiner's, unless H can be proved. It would take nearly a century after Steiner before H was finally proved.
But notice that all of these theorems can be generalized. Given a class of figures with the same perimeter, notice that the one with the largest area tends to be the most symmetrical. Cut the Knot generalizes this observation:
- Among all triangles with the same perimeter, the equilateral one has the largest area
- Among all quadrilaterals with the same perimeter, the square has the largest area
- In particular, among all rectangles with the same perimeter, the square has the largest area
- This latter fact is equivalent to √ab ≤ (a + b)/2, a particular case of the inequality between the geometric an arithmetic means.
- ...
- Among any finite number of regular polygons with the same perimeter, the one with the largest number of sides has the largest area.
- Among all n-gons (n fixed) with the same perimeter the regular one has the largest area. (This is known as Zenodorus Theorem, see [Tikhomirov, pp 11-15].
- Each of the statements above has an equivalent where the area is given.
- As in Lemma 2, among all plane curves of fixed length with fixed endpoints, a circular arc encloses a maximum area between it and the line joining its endpoints.
- Of all polygons with n sides inscribed in a given circle, the regular one has the largest area.
The three Zenodorus proofs help us out with the bonus question on the worksheet as well. It asks for the dimensions of a polygon with perimeter 100 feet and area greater than 625 square feet. We see that a square with perimeter 100 ft. has sides of length 25 ft. and area 625 sq. ft. -- and we know that the square has the greatest area of all quadrilaterals with perimeter 100 feet (generalization #2 from the above list). So the answer to the bonus question can't be a quadrilateral.
But we also proved that as the number of sides of a regular polygon increases, the area of the polygon increases as well (generalization #6 from the above list). So any regular polygon with more than four sides will work. For example, a regular pentagon with sides of length 20 ft. will have perimeter 100 ft. and area more than 625 square feet, and a regular decagon with sides of length 10 ft. will have an even larger area.
According to one of the review questions, "if we finish the next lesson, our class will have done every lesson in the book." Note that Lesson 15-8 is the penultimate lesson of the U of Chicago text.
No comments:
Post a Comment