Thursday, October 19, 2017

Lesson 4-5: The Perpendicular Bisector Theorem (Day 45)

This is officially my 600th post. Earlier, I'd announced another post as my 600th, but I've actually deleted a few posts and miscounted. The deleted posts were summer posts in which I was unsure whether I'd be hired at a new school or not. I erased all those posts in order to avoid confusion.

This is what Theoni Pappas writes on page 292 of her Magic of Mathematics:

"The Arabs used different arrangements of the 3 * 3 magic square to signify the signs of fire, earth, water, and air."

This is the second page of the subsection on magic squares. Pappas continues to tell us how early Islamic Arabs believed that this arrangement of squares really was "magic."

Pappas returns to the magic square from earlier:

124 19   94
49   79   109
64   139 34

Pappas points out that the two numbers in the upper-right corner are 19 and 94 -- as in 1994, the year that she writes this book. This is a nod to the magic square of Albrecht Durer, who created a magic square of order 4 in the year 1514. Two of the numbers that appear in this square are 15 and 14:

Pappas explains how she managed to get 19 and 94 to appear in her magic square -- she begins with the usual 3 * 3 square:

8 1 6
3 5 7
4 9 2

multiplies each element by 15, and then adds 4. In fact, all 3 * 3 magic squares are equivalent in that any such square can be obtained from another via either an algebraic transformation (linear) or a geometric transformation (reflection, rotation).

In fact, let's try to make the current year, 2017, appear in a magic square. We begin by rotating the square 90 degrees clockwise to make the numbers 4 and 3 appear in the upper-left corner. Then we multiply by 3 to convert these numbers into 12 and 9. Finally, adding 8 gives 20 and 17 as desired:

20 17 32
35 23 11
14 29 26

Pappas informs us some properties shared by all magic squares:

1) Each row, column, and corner-to-corner diagonal total the same number. Each magic square has a magic constant. Pappas provides a formula for the constant -- (1/2)(n(n^2 + 1)) if the entries are just the natural numbers 1 to n. If the entries instead form an arithmetic sequence with initial term a and common difference d, then the formula is an + d(n/2)(n^2 - 1).

2) Any two numbers (in a row, column, or diagonal) that are equidistant from its center are complements (2 numbers whose sum equals the sum of the largest and smallest numbers of the square.)

This last statement on page 292 is obviously true for the order 3 square, since the center number is 5 and so the other pairs must add up to 1 + 9 = 10 in order for the magic constant to be 15. But now we learn that this is true for magic squares of any order.

Today you may notice that there's a Google Doodle for a scientist, Subrahmanyan Chandrasekhar, who was born in India on today's date in 1910. As it turns out, Ian Stewart, in his book Calculating the Cosmos, mentions Chandrasekhar in one of his chapters. When I first bought Stewart's book, I didn't make it a side-along reading book but instead covered it in parts -- often when a Google Doodle of a famous scientist appears.

And so today I want to write about Chapter 14 of Stewart's book. But oops -- I'm already scheduled to cover Chapter 5 of Paul Hoffman's book today! And so once again, I'm writing about two chapters -- albeit in different books -- meaning that I'll be rushing through them yet again.

Chapter 14 of Ian Stewart's Calculating the Cosmos is called "Dark Stars." Just like Hoffman and many other authors, Stewart opens with a quote:

HOLLY: Well, the thing about a black hole, its main distinguishing feature, is -- it's black. And the thing about space, the colour of space, your basic space colour, is -- it's black. So how are you supposed to see them?
-- Red Dwarf, Series 3 episode 2, 'Marooned'

Stewart begins by telling us that traveling to the moon has been a fascination of science fiction writers for nearly 2000 years, dating back to the second century playwright Lucian of Samosata. But the author informs us that the escape velocity of the earth is 11.2 km/s, so we must travel at least that fast to reach the moon. The sun has a larger escape velocity -- 42.1 km/s -- and so this is how fast we must go to escape the solar system. But there are some objects that are so massive that their escape velocity is faster than c, the speed of light. As the title and opening quote already give away, these objects are called "dark stars" or "black holes."

We ought to skip directly to the Chandrasekhar part of this chapter, but I can't help but notice what Stewart writes about black hole geometry -- since this is, first and foremost, a geometry blog. And so Stewart writes:

"In Euclid's geometry, the natural transformations are rigid motions, and these preserve distances between points. The analogues in special relativity are Lorentz transformations, and these preserve a quantity called the interval."

Recall that "rigid motions" appear in the Common Core Standards, and the U of Chicago gives these transformations another name -- "isometries." So once again, we see isometries appearing in contexts outside of the Core, even though traditionalists make it sound as if the Core invented them.

Stewart proceeds to give us a distance formula for intervals, which is slightly different from the usual Euclidean-Pythagorean distance formula:

"The square of the interval is the square of the space coordinate minus that of the time coordinate. This difference is zero along the 45-degree lines, and positive inside the light cone."

In other words, time is essentially an imaginary -- as in sqrt(-1) -- form of space.

OK, I admit that I found much of this chapter fascinating. Stewart goes on to explain why people speculate that black holes lead to "wormholes" and potential other universes using light cones and Minkowski diagrams. (The previous Chapter 13 is equally interesting, as Stewart speculates on what aliens might actually look like.) But this is taking us further away from Chandrasekhar.

So let's skip to what Stewart actually says about Chandra:

"In 1931 Subrahmanyan Chandresekhar used relativistic calculations to predict that a sufficiently massive body, composed of electron degenerate matter, must collapse under its own gravitational field to form a neutron star, composed almost entirely of neutrons. A typical neutron star manages to compress twice the mass of the Sun into a sphere with a 12 kilometer radius. If the mass is less than 1.44 times that of the Sun, a figure called the Chandrasekhar limit, it forms a white dwarf instead of a neutron star."

This is why the Google Doodle depicts the scientist measuring out 1.44 solar masses. On the other hand, stars of about five solar masses or more become black holes.

Chapter 5 of Paul Hoffman's Man Who Loved Only Numbers is called "God Made the Integers." As usual, we begin with a quote -- another limerick:

A graduate student at Trinity
Computed the square of infinity
But it gave him the fidgits
To put down the digits,
So he dropped math and took up divinity.
-- Anonymous

The title of the chapter quotes Leopold Kronecker, an 19th century German mathematician. But it definitely reminds us of the SF and His infinite math book. At the start of this chapter, Hoffman points out that over 100,000 new theorems are proved every year. If each theorem is a page of the SF's book, then that's 100,000 newly discovered pages every year. Yet there are still infinitely many undiscovered SF pages to go!

I actually don't mind doing a rush job on this chapter, since much of its contents are ideas that we've visited already, in either Ogilvy or earlier posts. Some of these topics include:

-- Carl Gauss and his early life, including adding up 1 to 100 at age ten
-- Fibonacci numbers and the golden ratio
-- pi^2/6 as the sum of the reciprocals of the squares
-- Prime Number Theorem
-- standard algorithms vs. Roman numerals
-- divergence of the harmonic series
-- Georg Cantor, countability of the rationals, and uncountability of the reals

One thing that's new to us -- and the only idea related to Erdos -- is his probabilistic method. This method proves the existence of objects in Ramsey theory by making random choices. The example Hoffman gives is his usual example of random friends and acquaintances at a party. The author points out that this is another example of an existence proof -- we know that an answer exists, but we don't know what it is.

There are hundreds of thousands of new theorems every year, and Hoffman points out that some theorems are more important than others. In Geometry, one important theorem appears in our very next lesson.

Lesson 4-5 of the U of Chicago text is "The Perpendicular Bisector Theorem." (This theorem is part of Lesson 5-5 of the modern edition.) This is what I wrote two years ago about today's lesson:

Lesson 4-5 of the U of Chicago text covers the Perpendicular Bisector Theorem. This theorem is specifically mentioned in the Common Core Standards, so it's important that we prove it.

There are many ways to prove the Perpendicular Bisector Theorem. The usual methods involve showing that two right triangles are congruent. But that's not how the U of Chicago proves it. Once we have reflections -- and we're expected to use reflections in Common Core, the proof becomes very nearly a triviality.

Here's the proof, based on the U of Chicago text but rewritten so that the column "Conclusions" and "Justifications" become "Statements" and "Reasons," and with "Given" as the first step (as I explained in one of the posts last week):

Given: P is on the perpendicular bisector m of segment AB.
Prove: PA = PB

Statements                        Reasons
1. m is the perp. bis. of AB 1. Given
2. m reflects A to B            2. Definition of reflection
3. P is on m                       3. Given
4. m reflects P to P            4. Definition of reflection
5. PA = PB                       5. Reflections preserve distance.

So the line m becomes our reflecting line -- that is, our mirror. Since m is the perpendicular bisector of AB, the mirror image of A is exactly B. After all, that was exactly our definition of reflection! And since P is on the mirror, its image must be itself. Then the last step is the D of our ABCD properties that are preserved by reflections. The tricky part for teachers is that we're not used to thinking about the definition or properties of reflections as reasons in a proof. Well, it's time for us to start thinking that way!

Now the text writes:

"The Perpendicular Bisector Theorem has a surprising application. It can help locate the center of a circle."

This is the circumcenter of the triangle, one of our concurrency proofs. I mentioned last week that this is the easiest of the concurrency proofs. At first, it appears to be a straightforward application of the Perpendicular Bisector Theorem plus the Transitive Property of Equality. But there's a catch:

"If m and n intersect, it can be proven that this construction works."

Here m and n are the perpendicular bisectors of AB and BC, respectively. But the text doesn't state how to prove that these two lines must intersect. As it turns out, the necessary and sufficient conditions for the lines to intersect is for the three points AB, and C to be noncollinear. Well, that's no problem since right at the top of the page, it's stated that the three points are noncollinear -- and besides, we don't expect there to be a circle through three collinear points anyway. (And if this is part of a concurrency proof, then the three points are the vertices of a triangle, so they are clearly noncollinear.)

The Two Perpendiculars Theorem doesn't require Playfair, but the Perpendicular to Parallels Theorem does, so the above proof requires Playfair. As it turns out, in hyperbolic geometry, it's possible for three points to be noncollinear and yet no circle passes through them -- and so there's a triangle with neither a circumcenter nor a circumcircle!

And, if one has any lingering doubts that the existence of a circle through the three noncollinear points requires a Parallel Postulate, here's a link to Cut the Knot, one of the oldest mathematical websites still in existence. It was first created the year after I passed high school geometry as a student, and it has recently been redesigned:

Of the statements that require Euclid's Fifth Postulate to prove, we see listed at the above link:

"3. For any three noncollinear points, there exists a circle passing through them."

Now I didn't plan on giving Playfair's Postulate until Chapter 5. But here's something I noticed -- Playfair is used to prove the Parallel Consequences -- that is, the theorems that if parallel lines are cut by a transversal, then corresponding (or alternate interior) angles are congruent -- of which the Perpendicular to Parallels Theorem is a special case. But only that special case is needed to prove our circumcenter theorem. Indeed, I saw that the proof that the orthocenters are concurrent needs only that special case, and so does the Two Reflection Theorem for Translations.

But the Perpendicular to Parallels Theorem is tricky to prove. And we've already seen what other texts do when a theorem is tricky to prove, yet useful to prove medium- or higher-level theorems later on. We just simply declare the theorem to be a postulate! So instead of giving Playfair in Chapter 5, I state the Perpendicular to Parallels Postulate. (Note -- this is what I did last year. I have yet to decide whether I will change this for this year or not.)

Notice that the Perpendicular to Parallels Postulate can then be used to prove full Playfair. A proof of Playfair is given in the text at Lesson 13-6. The only changes we need to make to that proof is making sure that angles 1, 2, and 3 are all right angles. The new Step 2 can read:

2. The blue line is the line passing through P perpendicular to line l, which uniquely exists by the Uniqueness of Perpendiculars Theorem (which we proved on this blog last week). So angle 1 measures 90 degrees. So, by the new Perpendicular to Parallels Postulate, since the blue line is perpendicular to l, it must be perpendicular to both x and y, as both are parallel to l. So angles 2 and 3 both measure 90 degrees.

Then Playfair is used to prove the full Parallel Consequences. Notice that like Dr. Franklin Mason, we plan on adding a new postulate. But unlike his Triangle Exterior Angle Inequality Postulate, my postulate can replace Playfair, while Dr. M's TEAI Postulate must be used in addition to Playfair.

The final thing I want to say about a possible Perpendicular to Parallels Postulate is that this postulate is worth adding if it will make things easier for the students. I believe that it will. Notice that the full Parallel Consequences require identifying corresponding angles, alternate interior angles, same-side interior angles, and so on, and students may have trouble remembering which is which. But the Perpendicular to Parallels Postulate simply states that in a plane, if m and n are parallel and l is perpendicular to m, then l is perpendicular to n -- no need to remember what alternate or same-side interior angles are! So, in the name of making things easier for the students, I just might include this postulate in Chapter 5.

But I won't include it right now. And so I'll skip that part of the lesson -- and throw out the questions like 4 and 5 that require it.

That makes this lesson rather thin. So we ask, is there anything else that can be included? Let's look at the Common Core Standard that requires the Perpendicular Bisector Theorem again:

Prove theorems about lines and angles. Theorems include [...] points on a perpendicular bisector of a line segment are exactly those equidistant from the segment's endpoints.

We notice a key word there -- exactly. It means that if we rewrote this statement as an if-then statement, then it would have to be written as a biconditional:

A point is on the perpendicular bisector of a segment if and only if it is equidistant from its endpoints.

That is -- we need the converse of the Perpendicular Bisector Theorem. For some strange reason, the U of Chicago text makes zero mention of the converse! As it turns out, we can prove the converse quite easily, but it requires a theorem that's still two sections away. Once we reach Lesson 4-7, then we can finally prove the Converse of the Perpendicular Bisector Theorem.

And so there's not much left in this section -- but one could say that the Perpendicular Bisector Theorem is so important that it nonetheless merits its own section. Notice that I kept Question 6, which is similar to the construction of the circumcircle, except it's given that lines e and f intersect at point C. So we don't need a Parallel Postulate to prove that they intersect.

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