And so you know -- without my saying it -- the name of the movie I watched over the weekend. I usually don't discuss movies on the blog unless there is some relevance to math or Common Core.
Well, believe it or not, there actually is some relevance to the blog. In the first scene that appears in a school, the main characters are asked to take a standardized test -- perhaps one similar to the Common Core tests. I was wondering whether the movie writers would choose to make fun of Common Core at this point -- especially since we've seen other pop culture sources ridiculing the Core. But the only test question actually mentioned in the movie begins, "Suppose you have six red tomatoes..." -- at which point, the main character Charlie Brown gets distracted. The word red reminds him of his crush, the Little Red Haired Girl. He ends up filling in bubbles at random just before the time limit.
Another student, Peppermint Patty, also appears not to take the test seriously. When the test is over, we find that she has drawn a happy face over the bubbles on her answer document. This is a frequent criticism of both pre-Core and Common Core standardized tests -- many students, particularly older teenagers, don't take the test seriously because the tests don't affect them directly. Notice that it is never directly stated what grade levels the characters are in, although it's implied later on that Charlie Brown's younger sister Sally is in kindergarten.
Now what I'm about to say is a movie spoiler. Don't worry -- I only spoil the standardized testing plot and not the main plot of the movie (which is about Charlie Brown trying to win the heart of his favorite redhead). Still, I will label the following paragraph as spoilers, so those who want to skip the following paragraph may do so:
As it turns out, Peppermint Patty -- despite drawing pictures on her answer sheet -- actually earned a perfect score on her test. But due to a mix-up, Charlie Brown is given credit for the high score. The other students treat Charlie Brown as a hero until he admits that it wasn't really his test. This is interesting, since students rarely think of their classmates with perfect scores as heroes.
The only other part of the movie that is somewhat related to the blog content is that Charlie Brown spends parts of the movie trying to fly a kite -- and of course, last week one of the quadrilaterals in the hierarchy we discussed was the kite.
By the way, I recognized a few references to earlier episodes. Many casual fans know only about the Christmas (with its 50th anniversary this year) and maybe the Great Pumpkin episodes, but the movie referred to many lines and names mentioned in the lesser-known episodes or even the comics only. As a longtime Peanuts fan, I can point out many of these references for you:
A Charlie Brown Christmas: Obvious.
A Charlie Brown Thanksgiving: "I only know how to fix toast!"
Happy New Year, Charlie Brown: The name of the book Charlie Brown must read.
A Charlie Brown Valentine: The name of their teacher.
Life Is a Circus, Charlie Brown: The name of Snoopy's love interest.
You're in Love, Charlie Brown: Much of the main plot, a few lines, and the final scene.
Snoopy's Reunion: The appearance of Snoopy's family during the credits.
When the moving van shows up, I think the company name was something Melendez. Bill Melendez was the longtime voice of Snoopy.
Finally, there are a few characters who rarely appear on television (but appeared often in the comics) whose names are on the list showing everyone's test scores.
There is actually a Peanuts episode which discusses math tests the first five minutes. This episode is called "There's No Time for Love, Charlie Brown." In one scene, Sally is reading a text which appears to introduce arithmetic using set theory, when she exclaims, "All I want to know is how much is two and two!"
This may sound like a criticism of Common Core math, but this special was produced in 1973 -- decades before the Core. The episode was most likely mocking several pre-Core reforms often known as the New Math. Considering that Sally is merely a kindergartner, I must agree with the criticisms, and that the earliest years should be devoted to ensuring that the students know simple math facts, such as 2 + 2, as simply and quickly as possible. Here's a link to more information about that episode:
I could talk about the Peanuts all day, but it's time to get back to Geometry. We've reached an interesting sequence of lessons, where we are going to use translations to develop the properties of parallel lines.
The goal is to begin by defining translation, and then we show that the translation image of a line is a parallel line. This leads to the Parallel Tests. By the end of the week (Friday), we will introduce a Parallel Postulate. Once we introduce that postulate, we'll have made Euclidean geometry. Until then, our geometry is still neutral.
Last week, I was discussing black boxes, and if we don't want to see how to use translations to derive the Parallel Tests, we can just skip over it and treat the Parallel Tests as postulates. In other words, just treat the entire week as movie spoiler and teach only Friday's lesson on the Parallel Postulate.
But we ordinarily don't want to give all three Parallel Tests as postulates. Instead, of Corresponding Angles, Alternate Interior Angles, and Same-Side Supplementary Angles, we want to use one of them to use the other two. It's difficult to maintain a black box because we've seen different texts that assume each of these three as postulates to use to prove the other two.
We've seen a question on the PARCC, though, that uses Corresponding Angles to prove Alternate Interior Angles. Indeed, this is why we're using translations to prove Corresponding Angles, because translations directly lead to Corresponding Angles. Other methods, such as Dr. Hung-Hsi Wu and his 180-degree rotations, lead to Alternate Interior Angles rather than Corresponding Angles.
So let's get to our lesson on translations. But here's the other thing about this lesson -- we must find a way to keep the lesson neutral and non-circular. Let's look again at the definition of translation given in Lesson 6-2 of the U of Chicago text:
A translation (or slide) is the composite of two reflections over parallel lines.
But here's the correct, neutral definition of translation:
A translation (or slide) is the composite of two reflections over ultraparallel lines.
where we have this strange word "ultraparallel." This word is easy to define -- two lines are said to be ultraparallel if they have a common perpendicular. This concept is strange to someone accustomed to Euclidean geometry, because all parallel lines are ultraparallel -- and moreover, any line perpendicular to one of the parallel lines must be perpendicular to the other. But in neutral geometry, this isn't the case -- only ultraparallel lines have a common perpendicular, and even then, they have only one such common perpendicular.
Just like "Saccheri quadrilateral" and similar terms, the word "ultraparallel" should never see the light of day in a high school Geometry text. So here's how I will define a translation:
A translation (or slide) is the composite of two reflections over parallel lines with both mirrors perpendicular to the same line, called the direction of the translation.
And so just as we can't really define circle without mentioning a special point or special distance, we won't define translation without mentioning a special line -- the common perpendicular. And just as the special point and distance for a circle are called its center and radius, the special line for a translation is called its direction. And the name "direction" for this line isn't arbitrary -- it refers to the actual direction of the slide.
Many, but not all, of the properties of translations in Euclidean geometry are still available to us in neutral geometry. First of all, translations are still isometries because they are always the composite of two reflections -- both of which are themselves isometries, so the translation must preserve everything that each of the reflections does. Translations definitely preserve orientation, since there is nothing in the U of Chicago's line "the first reflection switches orientation, and the second switches it back" that violates neutrality.
If we look at the Two Reflection Theorem for Translations given in the U of Chicago text, we see that the line AA" is perpendicular to both mirrors j and k. Therefore, the results of that theorem only apply to points on that common perpendicular -- that is, every point on the line is mapped to another point on that line, and the distance between the preimage and the image is the same for every single point on that line. Notice that in every single point on a Euclidean plane lies on a line that's perpendicular to both mirrors, and so the theorem applies to every point on the plane. But in neutral geometry, not every point lies on such a common perpendicular.
But I can prove that every point really does move in the same direction even in neutral geometry. As I pointed out, this is implied by something I once found at the website of teacher Jennifer Silverman:
Recall that even though Silverman -- just like Dr. Wu -- uses 180-degree rotations to derive the properties of parallel lines, I once pointed out something that she wrote at the above link:
These statements are valid only in Euclidean geometry. But when we combine the lines, we get something that is valid in neutral geometry:
-- Lines equidistant at 2 places are parallel.
To prove this, notice the distance between a point and a line is defined as the perpendicular distance between the point and line. So if two points P and P" are equidistant from a line l, what we're really saying is to drop perpendiculars from both P and P" to l. Let's call the points where these two perpendiculars meet the line A and A", respectively. Then to say that P and P" are equidistant from the line l is to say that AP = A"P".
We then consider the quadrilateral AA"P"P. By design, angles A"AP and AA"P" are right angles, and we know that AP = A"P". So AA"P"P is a Saccheri quadrilateral, and in every Saccheri quadrilateral, the base and summit are parallel. So AA" | | PP", which is what we wanted to prove. QED
In our classes, we don't use the term "Saccheri quadrilateral." So instead, we say "isosceles trapezoid" (which is why I made such a big deal defining that term). Then just as we proved last week, every isosceles trapezoid is a trapezoid -- so the sides AA" and PP" are parallel.
Let's bring this back to translations. So l is a common perpendicular of the mirrors (the directing line of the translation) and P is the point that we're trying to translate. To do this, we translate the segment
But as it turns out, even though our translation moves every point in the same direction, it doesn't move every point the same distance unless the geometry is Euclidean -- that is,
The next thing to show is that a line and its translation image are parallel. Actually, if l is a common perpendicular of the mirrors, then any line parallel to l must be invariant. We've basically already shown this -- if P is any point not on l, then line PP" is parallel to l (same as AA" above).
So suppose we wish to translate a line k that's not parallel to l. Let's assume towards a contradiction that k and its image k" intersect at some point Q. Since Q is on k", this means that there exists some point P on k whose image is Q -- that is, Q = P". So we've identified two points on k, both P and P", and so line k is the same as line PP". But we've shown that line PP" is parallel to l, and so we conclude that k both is and is not parallel to l, a contradiction. Therefore if k is a line that's not parallel to the directing line l, then k | | k". QED
Now finally we will prove the Corresponding Angles Test, which is the whole point of this. Say we have two lines j and k cut by a transversal l such that corresponding angles are congruent. Let's call the points where j and k meet this transversal J and K, respectively.
Now we consider a translation mapping K to J. (There are several suitable pairs of mirrors that can produce this translation -- for example, let the first mirror be a line perpendicular to l through K and the second the perpendicular bisector of
Now we consider the angle where l and k intersect, and we wish to find its translation image. Of course, the line l is invariant. Since K is on k and K" = J, we see that J must be on k". Now translations preserve angle measures, so the angle where l and k intersect must map to a congruent angle with vertex J and with one of its sides along l. And conveniently, there's already an angle of the correct size where l and j intersect -- because corresponding angles are congruent.
This implies that the translation image of k is exactly j. And we know that lines and their translation images are parallel. Therefore k | | j, whjch is exactly what we wanted to prove. QED
I've mentioned before that the inspiration for this comes from the following site:
We can see an animation where the translation is performed. It is clear from the animation that corresponding angles and congruent and the lines are parallel. This is why I prefer this proof to the ones using 180-degree rotations -- an angle under this rotation maps to its alternate interior (or exterior) angles, while translations map to corresponding angles.
But that's another thing. How do we know that translations maps angles to their corresponding angles, and not, say, same-side interior angles? This would be problematic, as we don't want to prove accidentally that if same-side interior angles are congruent (rather than supplementary), then the lines are parallel.
Last year, we took care of this with Dr. Wu's Plane Separation Postulate. This tells us that every line divides the plane into two half-planes. We found out that any line entirely contained in one of these half-planes must be parallel to the boundary line. And we also discovered what happens when we reflect a half-plane -- if the mirror is perpendicular to the boundary line, then the half-plane maps to itself (that is, it's invariant). But if the mirror is the boundary, then it maps each half-plane to the opposite half-plane.
We can observe what affect this has on rays. Even though a line is invariant with respect to reflection in any mirror perpendicular to that line (Line Perpendicular to Mirror Theorem), a ray is not invariant for instead, such a reflection maps a ray to one pointing in the opposite direction. But a ray that's parallel to its mirror is mapped to a ray pointing in the same direction. In either case, after two such reflections, the image ray is pointing in the same direction as the original ray.
Therefore, a ray and its translation image are pointing in the same direction. And so an angle and its translation image consist of rays pointing in the same direction -- and these are what we refer to as corresponding angles. We might not want to write a formal proof of this in the classroom, but we may wish to point it out, especially when it's time to define corresponding angles and the other angles formed when two lines are cut by a transversal.
But we can save that for tomorrow's lesson. Today we want to focus on the translations themselves, up to the proof that a line and its translation image are parallel.
I decided to start with a new worksheet to show this year's versions of the proofs, but then I included the exercises from last year's worksheet. Some of questions from that worksheet include graphing translations on a coordinate plane -- which I don't want to omit, since I know how important such transformations are to the Common Core tests.
We can't prove yet that the transformation (x, y) -> (x + h, y + k) is a translation. I believe that we can prove that (x, y) -> (x + h, y) or (x, y + k) are translations -- the common perpendicular to the mirrors are the x- and y-axis respectively. I suspect that a proof that the full translation (x, y) -> (x + h, y + k) not only requires a non-neutral proof, but is proved most easily using second-semester coordinate formulas -- we show that each point is moved sqrt(h^2 + k^2) units along a line of slope k / h.
It might be simpler to show that the composite of two translations is a translation. I suspect that this can be proved only using congruence, not similarity -- if A and B are any two points, A' and B' their images under the first translation, and A" and B" their images under the second translation (and the translations aren't along the same line, for which the proof would be trivial), then we can show that the triangles AA'A" and BB'B" are congruent via SAS. But this proof would hardly be neutral.