Lesson 4-6: Reflecting Polygons (Day 46)

This is what Theoni Pappas writes on page 293 of her Magic of Mathematics:

3) There are many ways to transform an existing magic square into a new magic square.

This is the third page of the magic square subsection. On this page Pappas gives Example (a), which is to transform each term linearly. We're already familiar with this, since Pappas uses such linear transformations to make "19 94" appear in the magic square of order 3 -- and then we did the same thing to make "20 17" appear. Examples (b), (c), and (d) don't appear until page 294 -- but since tomorrow's the weekend and a non-posting day, let me post these examples today:

(b) If two rows or two columns, equidistant from the center are interchanged, the resulting square is a magic square.
(c) Interchange quadrants in an even order magic square.
(d) Interchange partial quadrants in an odd order square.

Most of page 293 is taken up by Ben Franklin's huge 16 * 16 magic square. Of course I won't write out all 256 entries here, so let me just provide a link instead:

http://www.math.wichita.edu/~richardson/franklin.html

Some of the properties of this magic square are mentioned at the link above. Nonetheless, I'll rewrite the entire caption from Pappas anyway:

"This is Benjamin Franklin's super duper 16 * 16 magic square. It has all the properties of the regular magic square except its corner to corner diagonals do not total to its magic number, 2056. But at the diagram illustrates its magic number pops up in so many ways, such as -- broken 8-diagonals, broken 8-parallel rows, any 4 * 4 square; and perhaps you can find more. In the 1952 Journal of the Franklin Institute, Albert Chandler contends that this magic square is not Franklin's original, but one that was set incorrectly by a printer."

Chapter 6 of Paul Hoffman's The Man Who Loved Only Numbers is called "Getting the Goat." Since we're covering only one chapter today, we should be able to cover it completely. As usual, the chapter begins with an opening quote:

"My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home."
-- Monty Hall

Wait, surely that's sounds familiar -- Monty Hall! Yes, the game show legend passed away three weeks ago, and I spent my first two posts afterward writing about the Monty Hall problem. It goes without saying that Hoffman is mentioning Hall in this post not because he's into game shows, but because he wants to write about the game shows, but about the Monty Hall problem (actually, he will use the word "dilemma" instead of problem. So yes, this means that this will my third post which I'm devoting to the Monty Hall dilemma.

In this chapter, Hoffman writes about a famous solver of the Monty Hall dilemma. Marilyn vos Savant writes a weekly column, described by Hoffman as "Hints from Heloise" for the mind. Notice that vos Savant is still alive, at age 71 this year. Indeed, she's still a regular contributor to Parade, and here is her latest post, dated today:

https://parade.com/609547/marilynvossavant/our-english-accents/

Anyway, vos Savant writes about Monty Hall dilemma in her colum in 1990. She explains that the car probability of staying is 1/3. And here is the response she receives:

"You blew it, and you blew it big! I'll explain: After the host reveals a goat, you now have a one-in-two chance of being correct. Whether you change your answer or not, the odds are the same. There is enough mathematical illiteracy in the country, and we don't need the world's highest IQ propagating more. Shame!"
-- Scott Smith, Ph.D., University of Florida

Here Smith is referring to vos Savant's measured IQ of around 200. Well, if you recall the description of the problem from earlier this month, we can see what's going on here.

Dr. Smith writes about "mathematical illiteracy in the country." Recall that this was in 1990, a few years after the infamous "third pounder" burger incident:

http://mentalfloss.com/article/76144/why-no-one-wanted-aws-third-pound-burger

A mathematician like Smith would've been upset that so many people were tricked into believing that 1/3 < 1/4, and now here comes vos Savant claiming that 1/3 = 1/2. Indeed, after she receives many more letters, another mathematician, E. Ray Bobo, asks:

"How many irate mathematicians are needed to get you to change your mind?"

In reality, we know that all these mathematicians are really claiming that 1 = 2, while vos Savant is telling them that 1 < 2. And so no matter how many hundreds, thousands, or even more are making the claim that 1 = 2, it doesn't make the claim any more true.

Some of the letters attack vos Savant for her gender -- almost 200 years after Sophie Germain was criticized for her gender. But I'd argue that gender in this case is a red herring. If vos Savant were instead a male making the 1/3 car probability claim, she (well, he) would still receive just as many letters questioning the claim. The letters would instead omit any reference to gender.

Indeed, this reminds me of NBA players who complained about fouls charged to them by Lauren Holtkamp, the league's third female referee. The players make sexist comments against Holtkamp after receiving the fouls, saying that she "doesn't belong" on the court. But does this mean that the players would have meekly accepted the foul calls had a male ref made them? I doubt it.

I suspect the same is true of race as well. If a driver of another race cuts you off on the road, you might angrily tweet later on, "This [racial expletive] cut me off today!" But if a driver of the same race cuts you off, you'd be just as angry, except you'd replace the expletive with "moron." (Think back to when race was mentioned by a student during my last subbing assignment.)

Nonetheless, here's a YouTube video discussing how so many "mansplainers" were stumped by the Monty Hall dilemma. (And yes, this is the second video that begins with the words "I hate math" that I post to the blog.)

Finally, let's go back to vos Savant's website, where she keeps a record of the complaints:

http://marilynvossavant.com/game-show-problem/

Vos Savant, according to Hoffman, doesn't make it any easier when she criticizes both the Wiles proof of FLT and Einstein's Theory of Relativity -- but then again, this is in 1993, three years after the Monty Hall column. (And besides, the Wiles proof of FLT really was wrong in 1993 -- the correct proof isn't written until 1994.)

By now, you may be wondering what do vos Savant and Monty Hall have to do with Paul Erdos? As it turns out, one of the mathematicians skeptical of the 1/3 car probability is the great Erdos. He is told this problem by his childhood friend Vazsonyi. In the end, Vazsonyi shows him a Monte Carlo situation in order to help him see the answer. A Monte Carlo simulation is similar to the activity that the students do at the end of the video, except it's randomized by a computer.

After the similation, Erdos grudgingly accepts the result, but not the proof. To him, this is similar to the proof of the Four Color Theorem (Lesson 9-8 in the U of Chicago text), which is also proved by trial and error on a computer. Erdos would prefer an elegant proof worthy of the SF's book to a proof by computer.

The chapter ends with the death of the great Paul Erdos. His passing is on September 20th, 1996 at the age of 83. But there are still two chapters remaining in this book, so his story isn't over yet. But let's get back to Geometry.

Lesson 4-6 of the U of Chicago text is called "Reflecting Polygons." This lesson doesn't appear in the new Third Edition -- instead, its material is incorporated into Lesson 4-2.

Two years ago, I didn't cover Lesson 4-6 as its scheduled day was blocked by a subbing day. So instead, we must go back three years to find the lesson:

Section 4-6 of the U of Chicago text considers what happens when we reflect an entire polygon -- not just individual points or even a segment or angle.

Still, the section begins with a theorem on what happens when we reflect a single point twice. Suppose we have two points, F and G and a reflecting line m. Now suppose I told you that the mirror image of F is G. So where do you think the mirror image of G is? If we drew this out and showed it to a student, chances are the student will say that the mirror image of G is F. The book gives a proof of this fact -- by the definition of reflection, G as the mirror image of F means that m is the perpendicular bisector of FG. But FG is the same segment as GF, so its perpendicular bisector is still m. And so, by the definition of reflection again, this would make F the mirror image of G. QED

The text calls this the Flip-Flop Theorem:

If F and F' are points or figures and r(F) = F', then r(F') = F.

Recall that the text often uses the function notation r(F) to denote the reflection image of F. So the theorem can be written as:

If F and F' are points or figures and the mirror image of F is F', then the mirror image of F' is F.

And one can use even more function notation than the text and write the theorem as:

If F is a point or figure, then r(r(F)) = F.

So here's a two-column proof of the Flip-Flop Theorem:

Given: r(F) = F'
Prove: r(F') = F

Proof:
Statements                        Reasons
1. r(F) = F'                       1. Given
2. m is the perp. bis. of FF' 2. Definition of reflection (meaning)
3. FF' = F'F                      3. Reflexive Property of Equality
4. m is the perp. bis. of F'F 4. Substitution Property of Equality
5. r(F') = F                       5. Definition of reflection (sufficient condition)

Notice that this proof uses both the meaning and the sufficient condition parts of the definition of reflection -- this occurs in other proofs as well. For example, a proof of the theorem "all right angles are congruent" (Euclid's Fourth Postulate) uses both the meaning and the sufficient condition parts of the definition of right angle.

But the above proof is a little strange. We explained earlier the significance of Statement 3 in the above proof -- but the problem is that we need a reason as to why FF' and F'F are the same segment. There is no actual definition, postulate, or theorem that states this directly. The reason I wrote "Reflexive Property" above is that this often occurs in other proofs -- especially triangle congruence proofs that are used to prove that certain quadrilaterals are parallelograms. For example, in Section 7-7, we wish to prove that quadrilaterals with opposite sides congruent are parallelograms. The proof at the beginning of that lesson divides quadrilateral ABCD into two triangles, ABD and CDB, which the text then proves are congruent by SSS. But Step 2 of that proof reads:

2. BD is congruent to DB   2. Reflexive Property of Congruence

And so I did the same in the above proof. Of course, it's awkward to follow a statement that uses the "Reflexive Property" (that some object equals itself) with one that uses the "Substitution Property." (So we're substituting an object for itself?)

Some people may point out that now we're being overly formalistic here. The Flip-Flop Theorem is obviously true -- the two-column proof only serves to confuse the students. Perhaps if even I, as a teacher, have trouble filling in all the steps in the "Reasons" column (like Step 3 above), it means that the proof is so simple that it's better written as a paragraph proof (as the U of Chicago text has done) and not as a two-column proof.

Here's one final way to state the Flip-Flop Theorem:

A reflection is an involution.

An involution is simply a function or translation such that performing it twice on a point or figure gives the original point or figure. Therefore composition of an involution with itself is the identity. In function notation, f(f(x)) = x.

Now the other concept introduced in this chapter is orientation. The important concept, added to the Reflection Postulate as part f, is that reflections switch orientation.

But what exactly is the "orientation" of a polygon? The text explains that, in naming the vertices of a polygon, we can move either clockwise or counterclockwise around the polygon. The important idea here is that if pentagon ABCDE is clockwise and we reflect it, then A'B'C'D'E' is counterclockwise.

Then the book proceeds to tell us that "orientation" is undefined -- just like point, line, and plane. As we mentioned earlier, we only discover what an undefined term is by using postulate. So we have the Point-Line-Plane Postulate to tell us what points, lines, and planes are, and we have part f of the Reflection Postulate to tell us what orientation is. We may not know what orientation actually is, but we do know that whatever it is, reflections switch it.

The idea that reflections switch orientation shows up later on. In particular, translations and rotations preserve orientation, because they are the compositions of two reflections -- so the first reflection switches it, and the second switches it back.

Also, a question that often comes up is, if translations and rotations are the compositions of two reflections, maybe reflections are the composition of two rotations, or two of something else. As it turns out, this is impossible. Reflections can't be the composition of two of the same type of transformation, because of orientation. Either the orientation is switched and switched back, or it isn't switched at all. (If you want a reflection to be some transformation composed with itself, you must do something complicated, such as cut the plane into strips, then translate some of the strips and reflect the others.)

Is it possible to define "orientation"? We think back to Chapter 1, where the term "point," although undefined, can be modeled with an ordered pair. If we know all of the x- and y-coordinates of the vertices of the polygon, then we can plug it into a complicated formula such that if the answer is positive, then the orientation must be counterclockwise, and if the answer is negative, then the orientation must be clockwise. (If it's zero, then the points are collinear, which means that they don't form a polygon at all.) What's cool about the formula is that the number -- not just the sign -- actually means something. In particular, if we divide the number by two, we get the area of the polygon! But I won't give the formula here.

There's also a simpler version of the formula, but it only works if the polygon is convex. Notice the picture of octagon FGHIJKLM in the text. The book points out that determining its orientation is more difficult because it's nonconvex.

A much more intuitive way of thinking about orientation is if the preimage and image aren't figures, but words. If we hold up words to a mirror, then unless we're lucky and choose a word like MOM, the image will be illegible, since reflections reverse orientation. But if we translated the words instead, then we can still read the words (unless by "translation" we mean translation into another language).

One final note about orientation: A well-known math teacher blogger named Kate Nowak -- she calls her blog "Function of Time" or f(t) in function notation -- recently gave an Opening Task to her geometry classes:

http://function-of-time.blogspot.com/2014/08/arguing-about-shapes.html

Now Nowak gave her classes pairs of figures, and the students had to identify whether the two figures are "the same" or "not the same." As it turned out, the students easily reached a consensus if the two figures have the same orientation, but they disagreed if the orientations were different:

One group: "We said set C is not the same because you have to flip it."
Me [Nowak -- dw]: "Great."
Other group: "Wait a minute, we said set C is the same because we thought flipping was okay."
Me: "Also great."
Yet another group: "So which is it? We said they are the same."
Me: "...   ...   ...  because... ?" 

Okay, let's return to 2017. Kate Nowak's website still exists, but she hasn't posted since the first day of school, and apparently she's creating a new middle school curriculum.

Today's an activity day. Of course, we could give the activity that Nowak describes in her 2014 post above, or even the Monty/Monte (that is, Hall/Carlo) simulation from above. But instead, let's go back two years when I wrote about Euclid the Game:

A few weeks ago, I mentioned the math teacher Lisa Bejarano, who had posted something called "Euclid: The Game" in one of her recent posts. And when I saw that part of the game reminded me of ancient geometer's Proposition 1 from Lesson 4-4, I couldn't resist checking the game out.

First, here's a link to Euclid: The Game:

http://euclidthegame.com/

Apparently, this is a one-player game. The goal is, on each level, to construct the figure in the diagram at the top of each page. The possible moves are the same as those allowed in classical Greek construction -- drawing an arbitrary point, drawing a point at an intersection, drawing a segment given two endpoints, drawing a ray given the endpoint and another point, and drawing a circle given the center and a point on the circle.

Now Level 1 is indeed Euclid's first proposition -- to draw an equilateral triangle given a side. This one, despite being Level 1, may be tough for students seeing this for the first time -- but of course, our students who remember yesterday's Lesson 4-4 should have no trouble with this one. Notice that according to Kasper Peulen, the creator, this game is powered by Geogebra -- and we were just talking about John Golden and his Geogebra lessons this week. Yes, I'm definitely going to keep going back to Bejarano, Golden, and other teachers when looking for good geometry activities.

Level 2 requires students to construct midpoints. The usual way to perform this construction is to construct the perpendicular bisector -- it intersects the original segment at its midpoint. For our students, this will be a preview of next week's Lesson 4-5 on perpendicular bisectors.

Level 3 requires students to construct angle bisectors. As we've already seen here on the blog, angle bisectors appear on the Common Core tests, yet are given short shrift in the U of Chicago text. The construction is buried in a Question in Lesson 4-7. Here's how to bisect Angle AOB:

Step 1. Circle O containing A
Step 2. Circle O intersects Ray OB at C.
Step 3. Subroutine: Line PQ, the perpendicular bisector of AC

As it turns out, the Euclid game has an equivalent of a "subroutine" -- like many computer and video games, passing a level unlocks a new "tool." In Level 2, I had already unlocked the midpoint tool. So I decided to follow the U of Chicago suggestion -- I drew a circle A (to label points of intersection B and C), found the midpoint D of BC, and then drew Ray AD. I passed the level with a minimum number of moves, three.

Level 4 requires students to find the perpendicular to a line through a point on the line. In the U of Chicago text, this is Example 2 of Lesson 3-6. This time, following the U of Chicago construction doesn't give me the minimum number of moves -- I needed four, but the minimum is three.

Level 5 requires students to find the perpendicular to a line through a point not on the line. It is the line given in this week's Uniqueness of Perpendiculars Theorem.

Of course, not every classroom has access to a computer -- then again, Euclid obviously didn't have a computer in ancient Greece either. So I decided to create worksheets for the first six levels of Euclid: the Game, and students will have to solve them the way that Euclid would have.