Recall that in this district, spring break has nothing to do with Easter. Instead, spring break is the first full week after the start of the fourth quarter (that is, the last quarter starts just before the break). On the other hand, Easter isn't ignored completely -- school is closed Good Friday and Easter Monday. In some years (such as 2017 and 2019), spring break is several weeks before Easter, and so students have both a full week off and then a four-day weekend in the spring.

On the other hand, in years such as 2018, the spring break week might include Easter Monday (rare) or, in this case, Good Friday. In such years, students and teachers would be cheated a day off, since Good Friday is already part of the spring break week.

So instead, another is taken off in lieu of Good Friday. It's known as a "floating day." I'm not sure why May 4th was chosen for the floating day. (No, it's not for Star Wars Day.) I suspect it's because it's the last day before AP exams. Notice that the first AP exam (Chemistry) is always on the first Monday in May -- the district has no control over the date of this national test. Thus a good day for the floating day is the Friday before the the first Monday in May (tomorrow), since it's the latest possible day off that doesn't affect testing.

Notice that this floating day occurs right around the end of the seventh quaver. Well, it's the mathematical 7/8-point of the year, but the second semester isn't mathematically 1/2 of the year, so I don't know if this is the ceremonial end of the quaver or not. In my new district, today is Day 151, so it's the start of the sixth hexter instead.

Meanwhile, today on her

*Mathematics Calendar 2018*, Theoni Pappas writes:

3 distinct lines can intersect in at most _____ distinct points.

The relevant theorem clearly is the Line Intersection Theorem of Lesson 1-7:

Two different lines intersect in at most one point.

But here we have three lines, not just two lines, so what effect does the third point have? We're trying to maximize intersection points, so we avoid parallel and concurrent lines -- neither of which increase the number of intersections. As long as the three lines are neither parallel nor concurrent, then there will be three intersections -- the three vertices of a triangle. So the answer is three -- and of course, today's date is the third.

Notice that this is really an Algebra II problem masquerading as Geometry. We can show that if there are

*n*lines, then there are

*n*choose 2 (that is,

*n*nCr 2 on the TI) possible intersections, since there are the same number of possible pairs of lines. All pairs of lines produce distinct intersection points, provided no two lines are parallel and no three lines are concurrent. These are easy to avoid if we draw the lines on a coordinate plane. To avoid parallel lines, choose a different slope for each line. To avoid concurrent lines is trickier, but notice that at each step, only finitely many

*y*-intercepts lead to concurrent lines. So just choose another

*y*-intercept among the infinitely many that are left.

Okay, let's start out traditionalists post -- and now I hear you saying,

*traditionalists again?*Well, if you're tired of traditionalists posts, tell the traditionalists to stop posting long posts that draw lots of comments from other traditionalists:

http://www.joannejacobs.com/2018/04/rethinking-high-school-math/

This is the Joanne Jacobs website, so you know that traditionalist Bill posts a comment here. But there's another major traditionalist -- one I haven't quoted in some time -- who posts here as well.

But let's start with the original link from Jacobs:

Math literacy — using numbers to understand real-world problems — should be the goal of high school math classes, concludes a report from the National Council of Teachers of Mathematics.

Don't forget that many of the Common Core Standards come from NCTM -- and in fact, the reason that many of the U of Chicago lessons fit the Core is that the text was also influenced by NCTM. We keep this in mind as we read the comment thread.

Our first comment comes from someone with the initials C T:

C T:

Even if they think they’re going to increase math ability, time will prove their costly mistake. For the result will be a giant slide down greased by constructivist delusions and refusals to use time-tested, efficient algorithms. The homeschoolers, a few charter schools, and the rich (who can afford tutors) will the only ones to escape the slide into math ignorance.

This, of course, is the usual traditionalist line. Common Core math is based on constructivism rather than the standard algorithm, and only by homeschooling, attending a certain charter, or hiring a private tutor can one escape the Common Core.

Our next comment comes from Darren. This Darren is the same Northern Californian who posts at the Right on the Left Coast website. Thus his posts are usually more about political conservatism than mathematical traditionalism, although he does comment on the latter from time to time:

Darren:

NCTM describes a system that both walks and quacks like tracking, but says they don’t want that particular duck. I’m forced to agree with CT that the goal seems to be to dumb down the high math achievers in order to have “equity”, not to help the lower end of the math spectrum.

I suspect that NCTM opposes traditionalism for demographic reasons. I wonder what Darren would consider to be legitimately helping the lower end of the math spectrum -- in a way that the parents of such students recognize as truly helping. We know that parents of students placed on lower tracks do

*not*view tracking as helping (otherwise there'd be no opposition to tracking).

Deirdre Mundy:

Also, even kids NOT on the STEM path need Algebra II/Trig. Machinists, Plumbers, Electricians, Carpenters… they use this stuff. So by putting kids in ‘consumer math’ ‘math for citizens’ or whatever, you’re preparing them for a life of low-end service jobs. ALL OF THE TRADES NEED MATH.

And yet kids actually sitting in those Algebra II classes ask "When will we use this in real life?" -- even the ones who

*want*to become machinists and plumbers. Also, by disparaging "Consumer Math,"

Mundy ignores those adults who say "I wish I learned how to do my taxes/balance my checking account instead of all this useless Algebra II."

Notice that Mundy says that she's a (former) high school teacher. She writes about her algebra (but which, Algebra I or II?) students who struggle with fractions. I have no problem giving students more practice with fractions, but that doesn't make "When will I ever need Algebra II?" or "I wish I learned Consumer Math instead of Algebra II so many years ago" disappear.

lgm:

Anyone here that paid attention enough to be ‘on grade level’ or higher must pay for math courses via the Dual Enrollment option with the local Community College after Regents Algebra 2…there is nothing free after that point, as its not required for the Regent’s Advanced Diploma, and we mustn’t steal money needed the remedial students.

I know that "Regents" means New York State exams. This is the first time I've ever heard that some schools don't offer anything higher than Algebra II -- the closest complaint I've heard is that the

*Common Core*doesn't require anything higher than Algebra II.

But notice that even full-blown tracking does not make the "money needed for remedial students" issue disappear, since both tracks would spend money. The only way to make the issue disappear is for

*remedial students*not to be offered any math class (as the Pre-Calc students are now). Still, I partly agree with lgm -- we shouldn't

*force*students to take Calc, but we should at least

*offer*it.

In a later post, lgm continues:

lgm:

Singapore Math and living math teach everything much more quickly and with less confusion than the classroom, since the classroom now is full inclusion, spending massive amounts of time on rote memorization with things like skip counting and the nines finger trick rather than the part/whole concepts and the properties.

I've never heard of "living math," but I know that traditionalists like Singapore Math. I assume that "skip counting" means counting by twos or fives, while "the nines finger trick" is a trick for multiplying by nine (taking advantage of the omega rules). So lgm wants to see less of this and more of "part/whole concepts" (fractions, I assume) and the "properties" (of fractions???).

I've mentioned before that some multiplication facts are easier to learn than others -- and twos, fives, and nines are easier because of the skip counting and omega finger tricks. One doesn't memorize the whole table at once -- some facts are learned before others.

It's hard for me to tell what lgm is trying to say here. The phrase "more quickly" is a hint -- the idea is to get to where we can answer 9 * 5 = 45 in one second. Supposedly, skip counting and omega finger become crutches where it takes several months to learn the whole table, whereas the "Singapore method" (which I don't know, since I don't have a Singapore text) might get students to 9 * 5 = 45 in one second after only, say, one month of learning.

A poster named "President Sun and Moon" makes a side remark about the two Koreas becoming unified under a single -- um, time zone. Two months ago I made several posts about clocks and time zones, so I find this mildly interesting. But let's just get to Bill:

Bill:

You can’t get admitted to electricians school without a solid working knowledge of algebra, period…The first thing covered is Ohm’s Law which is E = IR or V = IR (depending on how old you are) .

Kumon and Singapore Math should be taught in our schools…but the educational leadership won’t stand for it…

So here's another ringing endorsement for both tutoring ("Kumon") and Singapore Math -- and Bill says that both should be taught in schools (instead of Common Core). I wonder whether Bill knows that Singapore high school math is integrated, so there is no separate course called "algebra."

And now here's a sight for sore eyes -- the traditionalist Ze'ev Wurman:

Ze'ev Wurman:

I find it funny that the NCTM complains about “too much” symbolic manipulations. Fluency with symbolic manipulation is *precisely* one of the key skills for later use of using math in engineering, physics, chemistry, bio … also known as STEM (smile).

Recall what they say about symbolic manipulations -- every appearance of an equation in a book cuts its sales in half. Symbolic manipulations scare students away and make them hate math

*more.*

*Ze'ev Wurman:*

Interestingly, the fraction of California students taking Algebra 1 in grade 8 dropped from 58% in 2013 to 19%(!!) in 2017. I guess Mike Kirst and Jerry Brown call it “progress.”

I usually think of eighth grade Algebra I as a class that only above average math students should take, not the majority. Of course, here he's referring to the old pre-Core California Standards, which used to encourage eighth grade Algebra I. Actually, I'd call it progress if we could have a smaller percentage of students who

*hate*math, and have a smaller percentage of students ask questions like "When will we use fractions or algebra in real life?"

There's one more Bill post, but he mainly repeats a Calculus limit from an earlier post, so there's no need to repeat the quote.

Due to my erasing several posts last year (the Post Purge of 2017) after my employment situation was in flux, there is no record of Lesson 15-8 from last year. Therefore, this is what I wrote two years ago about today's lesson:

Today we are covering Lesson 15-8 of the U of Chicago text, on the Isoperimetric Inequality. It's a brand new lesson that I didn't cover last year. In fact, I was inspired by some of the side-along books that I've read so far this year to include this lesson.

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

[2018 Update: I dropped the posted proofs from two years ago. 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.

By the way, I'm surprised that all the figures from Cut the Knot actually copied into this post! I have a few comments to make about this proof:-- 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.

According to the MAA link, Steiner himself didn't bother to prove

*H*. A later German mathematician Oskar Perron would criticize Steiner's omission by giving the following fallacious proof:

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.

Because of the floating day (no school) tomorrow, my next post will be on Monday.

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