Lesson 15-8: The Isoperimetric Inequality (Day 158)

This past weekend was an important one for my well-being. That's because I finally receive my first dose of the coronavirus vaccine.

The decision to take the vaccine was an easy one for me to make. I mentioned on the blog that a few years ago, a former coworker died of the flu. I'd never taken flu shots before his passing, but I've taken the shots every year since. Thus, while I know that many people oppose flu and coronavirus vaccinations, I will continue to take both as necessary to stay healthy.

It took me a bit longer than expected to receive the shot, though. Two months ago, Orange County educators were offered the opportunity to get it. But it was ambiguous whether we subs were included in with these educators. I was told that subs could get it if we presented a letter from the superintendent showing that we regularly worked in the district.

Yet when I tried to enroll for an appointment through the county's Othena system, I was rejected. This is what I suspect -- if we typed in the single word "teacher" when asked for our occupation, then we would get an appointment, but if we added anything else, such as "substitute," "sub," or "long-term," then Othena would reject us. The superintendent letter was something for us to show human eyes at the vaccination supersite, but that wasn't enough to get through Othena.

Instead, I actually received my appointment through my LA County district. We were given a code, and one of the available vaccination sites was at a Kaiser in the county. Interestingly enough, I was still in Kaiser's system as an inactive member -- I'd been enrolled through my old charter school. The only reason I was able to get an appointment was because Kaiser allowed me to book it well ahead -- the earliest weekend appointment was Saturday, May 1st. And so on Saturday, I finally received the first dose of the Moderna vaccine. My second appointment will be in four weeks.

While I received the vaccine was little later than I thought two months ago, it's ahead of schedule from what I was thinking at this time last year. On the blog, I mentioned late June as a possible target for getting the vaccine. Instead, I should have received both doses and be fully inoculated two weeks after the second dose, which takes us to mid-June. The timeliness of the vaccine production is a tribute to the hard working scientists (and mathematicians) at Moderna and the other vaccine companies.

The hope, of course, is that enough of us will be vaccinated in order to return to mostly normal in time for the first day of school this fall. Oh, and by coincidence, there was a vaccination Google Doodle on the day that I received my first shot.

Speaking of Google Doodles, there's another one today. This week is Teacher Appreciation Week, celebrated the first full week in May. It's definitely been emphasized the past few years. Sometimes special meals are served for us in the lounge, although I'm not sure how much I'll see that during this pandemic as opposed to past years. Well, I'll find out soon enough.

Of the five inspirational stories in that Doodle, my favorite has to be Jose and Carlos. The teacher Carlos encouraged Jose to join the running team, which reminds me of McFarland USA. Notice that Jose, just like the McFarland runners, is a Mexican immigrant living here in California.

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. 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.

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:

1. Among all triangles with the same perimeter, the equilateral one has the largest area
2. Among all quadrilaterals with the same perimeter, the square has the largest area
3. In particular, among all rectangles with the same perimeter, the square has the largest area
4. This latter fact is equivalent to √ab ≤ (a + b)/2, a particular case of the inequality between the geometric an arithmetic means.
5. ...
6. Among any finite number of regular polygons with the same perimeter, the one with the largest number of sides has the largest area.
7. 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].
8. Each of the statements above has an equivalent where the area is given.
9. 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.
10. 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.