"Complexity is an emerging science which may hold answers or at least explanations to such questions as...."
This is the last page of the subsection on complexity and the present. Pappas gives a full list on many questions that have to do with complexity. I'll only give part of her list below:
How is it that
- the universe emerged out of the void?
- on January 17, 1994 Los Angeles suffered an earthquake of unexpected magnitude and destruction?
- Yugoslavia was thrown suddenly into severe internal wars?
- for no apparent reason the stock market surges upward or plunges downward?
By the way, as a Southern Californian, I lived through the 1994 Northridge earthquake that occurred a few months before Pappas wrote her book. I was a seventh grader at the time. Notice that January 17th that year was Martin Luther King Day, and so all schools were closed for the holiday. I woke up with a jolt early that morning, but otherwise I was too far away from the epicenter to suffer damage.
Let me give excerpts from the rest of this page, so we can hurry up and get into Szpiro's book:
"The underlying common factor of these events is that each represents a very complex system. The factors which act on such a system are ever growing and changing. There seems to be a continual tug of war between order and chaos. It is the means by which the system regains equilibrium by changing and adapting itself to constantly changing factors/circumstances. These scientists and mathematicians fell that today's mathematics, along with other tools and high tech innovations, are capable of creating a complexity framework that can impact major aspects of our global world, especially economics, the environment, and politics."
In short, complex math is used to study complex systems.
Chapter 3 of George Szpiro's Poincare's Prize is called "The Forensic Engineer." It begins at the end, which a newspaper article reporting a mathematician's passing:
"Monsieur Henri Poincare, Professor of Mathematical Astronomy in the University of Paris, Member of the French Academy and the Academie des Sciences, died suddenly this morning of an embolism of the heart. The death of Henri Poincare at the comparatively early age of 58 deprives the world of one of its most eminent mathematicians and thinkers."
This chapter and the next are the story of Poincare, who first formulated the conjecture that would haunt mathematicians for over a century. Szpiro now begins with his birth:
"Jules-Henri Poincare was born in Nancy, a town in the northeast of France, on April 29, 1854. His father, Leon, was a doctor and professor of medicine at the University of Nancy; his uncle, Antoni, inspector of public works and an acknowledged authority on meteorology."
The first half of this chapter is all about the young Henri's education -- and as a teacher, I like to write about the educations of the subjects whose biographies I read. Because he is diagnosed with diphtheria at the age of five, his mother homeschools him for the first three years of his education. But they are worried when it's time for him to start at a public school:
"His very first essay was described as a small masterpiece by his teacher, who ranked the boy first in his class. Henri would keep that rank in all subjects throughout his school years."
But, as Szpiro points out, math isn't always Poincare's best subject:
"His favorite subjects were history and geography, and his aptitude for literary and philosophical essays aroused the interest of his educators. Henri's interest in mathematics only become apparent when he was fourteen years old."
After surviving the Franco-Prussian war, he prepares to take some big tests:
"In August 1871, Poincare sat for the literary baccalaureat, the French high school exams. He did extremely well in the Latin and French essays and excelled in all other subjects."
In high school, I myself took French, and as part of learning the language, we were taught all about French culture, including the big tests that our overseas counterparts must take. Furthermore, some traditionalists believe that the French bac and similar tests are superior to the PARCC, SBAC, and other state tests.
For one thing, notice that Poincare takes the tests in 1871, when he is 17 years old -- in other words, a high school senior. Standardized testing in our own country ends in junior year -- for some reason, seniors don't take tests like PARCC or SBAC. In France, the entire senior year is devoted to taking the bac. In addition, we see that the literary bac -- the equivalent of the ELA exam (or even the verbal section of SAT or ACT) -- contains a Latin section. We also see that Poincare sits his literary exam early in the school year -- in August -- and the math bac is three months later:
"Amazingly, Henri nearly failed the written test. The exam question asked for a proof of a formula to compute the sum of a geometric series. Henri had arrived late for the examination."
Fortunately, Poincare passes his math test on the strength of the oral part of the exam. Of course, American math standardized tests don't have an oral section, and of course high school students wouldn't be ask to prove (rather than merely use) the geometric series sum on a test.
"Following high school, young Frenchmen with exceptional aptitude for mathematics underwent, and still undergo today, a two-year grueling course in mathematics in preparation for entry to the grandes ecoles, the prestigious engineering schools."
Again, there is no equivalent in the American education system. Imagine if students here had to study for two extra years just to get in to an Ivy League school, MIT, or Caltech. During these two years, Poincare does so well that he passes his math courses without taking many notes, and even wins several national math competitions.
"Henri's next challenge was to enter the competition to gain admission to one of France's top engineering schools. Of all the grandes ecoles the most prestigious was, and still is, the Ecole Polytechnique in Paris."
The Polytechnique was the math and science school in France. I've mentioned the famous university twice on the blog in connection with other mathematicians (Evariste Galois and Sophie Germain -- the latter being unable to matriculate there due to her gender, but, as Szpiro points out, women can go there now).
Eventually, Poincare does attend the Polytechnique, and moves on to a special grad school for prospective mining engineers. Before we leave, Szpiro does tell us about the grading scale in France. Officially, there is a 20-point scale, but I already knew that earning a 20 is basically impossible, according to Numberphile:
The Numberphile video implies that earning 19 out of 20 is possible, but according to Szpiro, even 18 isn't achievable -- and scores from 16 to 18 are "reserved for the professor himself." Most of the grades earned by Poincare are between 15 and 16 -- except for Geometry (our subject!) and a closely related subject, drawing.
After he graduates, he works in the coal mines for a year. At this point, Szpiro spends the rest of the chapter telling an fascinating story about an explosion at the mines, where 16 people were killed. And Poincare uses mathematics to solve the mystery of what causes the explosion -- which I choose not to write about here, since we must hurry up to get to Geometry. (Again, you can read Szpiro's book for yourself to get the whole story.) His solution is similar to that of one of our recent logic puzzles -- indeed, the mine is a very complex system, just as we read about in Pappas today,
At the end of the chapter, Poincare leaves the mines and starts thinking about math again, as he works to become a math professor:
"Poincare was a formidable, even heroic engineer. His real calling, however, lay elsewhere."
Lesson 6-3 of the U of Chicago text is called "Rotations." In the modern Third Edition, we must backtrack to Lesson 4-5 to learn about rotations.
"Monsieur Henri Poincare, Professor of Mathematical Astronomy in the University of Paris, Member of the French Academy and the Academie des Sciences, died suddenly this morning of an embolism of the heart. The death of Henri Poincare at the comparatively early age of 58 deprives the world of one of its most eminent mathematicians and thinkers."
This chapter and the next are the story of Poincare, who first formulated the conjecture that would haunt mathematicians for over a century. Szpiro now begins with his birth:
"Jules-Henri Poincare was born in Nancy, a town in the northeast of France, on April 29, 1854. His father, Leon, was a doctor and professor of medicine at the University of Nancy; his uncle, Antoni, inspector of public works and an acknowledged authority on meteorology."
The first half of this chapter is all about the young Henri's education -- and as a teacher, I like to write about the educations of the subjects whose biographies I read. Because he is diagnosed with diphtheria at the age of five, his mother homeschools him for the first three years of his education. But they are worried when it's time for him to start at a public school:
"His very first essay was described as a small masterpiece by his teacher, who ranked the boy first in his class. Henri would keep that rank in all subjects throughout his school years."
But, as Szpiro points out, math isn't always Poincare's best subject:
"His favorite subjects were history and geography, and his aptitude for literary and philosophical essays aroused the interest of his educators. Henri's interest in mathematics only become apparent when he was fourteen years old."
After surviving the Franco-Prussian war, he prepares to take some big tests:
"In August 1871, Poincare sat for the literary baccalaureat, the French high school exams. He did extremely well in the Latin and French essays and excelled in all other subjects."
In high school, I myself took French, and as part of learning the language, we were taught all about French culture, including the big tests that our overseas counterparts must take. Furthermore, some traditionalists believe that the French bac and similar tests are superior to the PARCC, SBAC, and other state tests.
For one thing, notice that Poincare takes the tests in 1871, when he is 17 years old -- in other words, a high school senior. Standardized testing in our own country ends in junior year -- for some reason, seniors don't take tests like PARCC or SBAC. In France, the entire senior year is devoted to taking the bac. In addition, we see that the literary bac -- the equivalent of the ELA exam (or even the verbal section of SAT or ACT) -- contains a Latin section. We also see that Poincare sits his literary exam early in the school year -- in August -- and the math bac is three months later:
"Amazingly, Henri nearly failed the written test. The exam question asked for a proof of a formula to compute the sum of a geometric series. Henri had arrived late for the examination."
Fortunately, Poincare passes his math test on the strength of the oral part of the exam. Of course, American math standardized tests don't have an oral section, and of course high school students wouldn't be ask to prove (rather than merely use) the geometric series sum on a test.
"Following high school, young Frenchmen with exceptional aptitude for mathematics underwent, and still undergo today, a two-year grueling course in mathematics in preparation for entry to the grandes ecoles, the prestigious engineering schools."
Again, there is no equivalent in the American education system. Imagine if students here had to study for two extra years just to get in to an Ivy League school, MIT, or Caltech. During these two years, Poincare does so well that he passes his math courses without taking many notes, and even wins several national math competitions.
"Henri's next challenge was to enter the competition to gain admission to one of France's top engineering schools. Of all the grandes ecoles the most prestigious was, and still is, the Ecole Polytechnique in Paris."
The Polytechnique was the math and science school in France. I've mentioned the famous university twice on the blog in connection with other mathematicians (Evariste Galois and Sophie Germain -- the latter being unable to matriculate there due to her gender, but, as Szpiro points out, women can go there now).
Eventually, Poincare does attend the Polytechnique, and moves on to a special grad school for prospective mining engineers. Before we leave, Szpiro does tell us about the grading scale in France. Officially, there is a 20-point scale, but I already knew that earning a 20 is basically impossible, according to Numberphile:
The Numberphile video implies that earning 19 out of 20 is possible, but according to Szpiro, even 18 isn't achievable -- and scores from 16 to 18 are "reserved for the professor himself." Most of the grades earned by Poincare are between 15 and 16 -- except for Geometry (our subject!) and a closely related subject, drawing.
After he graduates, he works in the coal mines for a year. At this point, Szpiro spends the rest of the chapter telling an fascinating story about an explosion at the mines, where 16 people were killed. And Poincare uses mathematics to solve the mystery of what causes the explosion -- which I choose not to write about here, since we must hurry up to get to Geometry. (Again, you can read Szpiro's book for yourself to get the whole story.) His solution is similar to that of one of our recent logic puzzles -- indeed, the mine is a very complex system, just as we read about in Pappas today,
At the end of the chapter, Poincare leaves the mines and starts thinking about math again, as he works to become a math professor:
"Poincare was a formidable, even heroic engineer. His real calling, however, lay elsewhere."
Lesson 6-3 of the U of Chicago text is called "Rotations." In the modern Third Edition, we must backtrack to Lesson 4-5 to learn about rotations.
This is what I wrote last year about today's lesson:
Yesterday, we discussed translations on the coordinate plane, and so now we move on to rotations. I point out that we learned how to perform translations of the form (x, y) -> (x + h, y + k) -- which turns out to be every translation in the plane.
But with rotations, we only perform a precious few of them. The only rotations that appear on the PARCC and other Common Core tests are those of magnitude 90, 180, or 270. Yet we've seen a few of these rotations centered at points other than the origin on the PARCC.
We'll begin with rotations that are centered at the origin, though. Just as we used the Two Reflections Theorem for Translations yesterday, today we'll use the Two Reflections Theorem for Rotations. So to perform the rotation of 180 degrees about the origin, we compose two reflections in mirrors that intersect at the origin, at an angle of half of 180, or 90 degrees. The obvious choices for mirrors are the x- and y-axes. We've already proved that the reflection image of (x, y) in the x-axis is (x, -y) and the reflection image of (x, y) in the y-axis is (-x, y). It doesn't matter in which order we compose these as reflections in perpendicular mirrors always commute. So we prove that the rotation image of (x, y) centered at the origin and of magnitude 180 degrees is (-x, -y).
Now our other common rotation magnitude is 90 degrees -- and this time, it will make a difference whether it's clockwise or counterclockwise. The angle between the mirrors will now have to be half of 90, or 45 degrees. There's one mirror to consider that will help us with a 45-degree angle -- the line whose equation is y = x.
We've hinted at several proofs involving reflection over the line y = x. Let's look at the quadrilateral whose vertices are (0, 0), (a, 0), (a, a), and (0, a). We can show that this figure is a kite.
So now we can apply the properties of a kite -- the Kite Symmetry Theorem. The diagonal of our kite running from (0, 0) to (a, a) bisects the angle between the x- and y-axes -- and since we know that the angle between the axes is 90 degrees, the diagonal must form a 45-degree angle with each axis. And reflecting across this symmetry diagonal must map the axes to each other and x = a to y = a.
Recall that at this point, we don't know the equations of lines, so we aren't yet certain that the graph of y = x is even a line (which we'd better figure out before trying to use it as a mirror). But we see that the value of a in the above proof is arbitrary -- it's true for every single real number a (although in case a is negative, we should probably say that the kite has sides of length |a|, not a). Therefore every single point of the graph of y = x lies on the bisector of the angle between the axes -- that is, the graph of y = x is exactly that line. And reflecting in that line maps x = a to y = a and vice versa -- that is, it switches x and y. Therefore the image of (a, b) must be (b, a).
Now that we know how to reflect in the line y = x, let's use it to perform a 90-degree rotation. It's probably easiest just to start with the reflection in y = x first, so (x, y) maps to (y, x). As for the second mirror, it depends on whether we want to go clockwise or counterclockwise. To go clockwise, the second mirror must be 45 degrees clockwise of the first mirror, y = x. That is the x-axis, and to reflect in it, we change the sign of the second coordinate. So (y, x) reflected in the second mirror is the point (y, -x), so mapping (x, y) to (y, -x) rotates points 90 degrees clockwise. To go counterclockwise, the second mirror must be 45 degrees counterclockwise of the first mirror, y = x. That is the y-axis, and to reflect in it, we change the sign of the first coordinate. So (y, x) reflected in the second mirror is the point (-y, x), so mapping (x, y) to (-y, x) rotates points 90 degrees counterclockwise.
Notice that some of the PARCC questions mention 270-degree rotations -- for example, there was a released question that mentions a 270-degree clockwise rotation. Of course, a 270-degree clockwise rotation is equivalent to a 90-degree counterclockwise rotation, so it maps (x, y) to (-y, x). If students forget this, they can still take half of 270 degrees to get 135 degrees clockwise, and they can see that 135 degrees clockwise from the line y = x is still the y-axis, just as it would have been if they'd gone 45 degrees counterclockwise instead.
Last year, I created a quick worksheet to help students perform any of the reflections and rotations mentioned in this post. (This was late in the year when we were covering PARCC questions, but now I'm giving this lesson much earlier.) It takes the coordinate plane labeled with the positive x-, negative x-, positive y-, and negative y-axes. Students can then perform the rotations on the axes to see what happens. For example, let's try our 270-degree clockwise rotation. After we rotate the paper 270 degrees clockwise, we see that where the x ought to be, we see -y instead, and where the y ought to be, it's +x. Thus the image must be (-y, x).
Okay, so we've taken care of all the rotations centered at the origin, But on the PARCC, there are questions with rotations centered at other points. These questions that I've seen direct the students to take a triangle ABC and rotate it around one of its vertices -- let's say C. Well that makes things a little easier, since then the rotation image of C is C itself. So then there are only two points that we need to find, A' and B'.
It's possible, in principle, to find formulas to determine the image of (x, y) under reflections in mirrors parallel to the axes and rotations centered at points other than the origin. We've seen, for example, that the point (x, y) reflected in the line x = a is (2a - x, y). An interesting question is, where exactly does the 2 in 2a - x come from?
To find out, we notice that if we were reflecting in the y-axis (which is parallel to x = a), then the point (x, y) is mapped to (-x, y). Now that extra 2a term looks just like a horizontal translation of exactly 2a units.
So somehow, our reflection in the line x = a appears to be the composite of a reflection in the y-axis and a horizontal translation. (This is not a glide reflection between the mirror is perpendicular to the direction of translation -- we found out last year that such a composite yields a simple reflection, not a glide reflection.)
Using symbols, let's call the composite transformation T. It is the composite of a y-axis reflection, which we'll call r_y, and a horizontal translation of 2a units, H_2a:
T = H_2a o r_y
But the horizontal translation is itself the composite of two reflections. The two mirrors here must be vertical mirrors spaced exactly half of 2a, or a units apart. We might as well let the two mirrors be the y-axis itself and the line x = a.
T = H_2a o r_y
= r_(x = a) o r_y o r_y
= r_(x = a) o I
= r_(x = a)
which is exactly what we wanted -- a reflection in the line x = a.
Likewise, we see that the reflection in the line y = b maps (x, y) to (x, 2b - y). The composite of both reflections is a 180-degree rotation about the point (a, b), which maps (x, y) to (2a - x, 2b - y) -- and that's also the composite of a 180-degree rotation about the origin and yet another translation.
Now 90-degree rotations about points other than the origin are even trickier, because now we'd have to reflect about mirrors with equations like y = x + b -- and we don't even know that's a line yet. The algebra involved in this reflection gets very messy.
Of course, if we try to visualize the rotation, another composite transformation jumps at us. To perform a 90-degree rotation (either clockwise or counterclockwise) about the point (a, b), it appears that we can first perform the translation that maps (a, b) to (0, 0), then perform the rotation centered at the origin, and finally translate (0, 0) back to (a, b).
This seems to work, but is there any reason why it should? Let's use symbols again -- in order to remember what the symbols stand for, we let "rot" stand for the rotation and "trans" stand for the translation mapping (0, 0) to (a, b). Then trans^-1 can stand for the inverse translation -- the one mapping (a, b) to (0, 0). This gives us:
T = trans o rot o trans^-1
This composite has a name in classes like linear algebra and above -- conjugation. That is, we are conjugating the rotation by the translation.
We now want to rewrite both the translation and origin-rotation with two mirrors each. And as usual, we want to choose the mirrors strategically so that some of the reflections cancel out. For the translation, we'll let k be the line joining the points (0, 0) and (a, b). Then l will be the line perpendicular to k passing through the origin, m will be the line perpendicular to k passing through the midpoint of (0, 0) and (a, b), and n will be the line perpendicular to k passing through (a, b). Then the rotation can be written as r_l o r_k, and the translation can be written as r_m o r_l. Notice that the inverse translation can be written r_l o r_m -- but it can also be written as r_m o r_n (as either l and m, or m and n, are the correct distance apart). So we write it:
T = trans o rot o trans^-1
= r_m o r_l o r_l o r_k o r_m o r_n
= r_m o I o r_k o r_m o r_n
= r_m o r_k o r_m o r_n
= r_k o r_m o r_m o r_n (as reflections in perpendicular mirrors commute)
= r_k o I o r_n
= r_k o r_n
which is the composite of reflections in perpendicular mirrors intersecting at (a, b). And so T is in fact the rotation centered at (a, b), which is what we were expecting.
Once again, though, this is not the sort of symbolic manipulation I'd want my students to see. But then, what should we expect students to do when faced with a PARCC question where they have to rotate around a point other than the origin?
Most likely, this is something that can wait until we discuss the Slope Formula -- especially since it's this rotation that leads to the slopes of perpendicular lines. For now, one can consider such rotations only informally -- after all, the PARCC questions usually include graphs, so students might be able to perform the rotations just by counting units on the graph, rather than using an algebraic formula or manipulating mirrors and symbols.
But with rotations, we only perform a precious few of them. The only rotations that appear on the PARCC and other Common Core tests are those of magnitude 90, 180, or 270. Yet we've seen a few of these rotations centered at points other than the origin on the PARCC.
We'll begin with rotations that are centered at the origin, though. Just as we used the Two Reflections Theorem for Translations yesterday, today we'll use the Two Reflections Theorem for Rotations. So to perform the rotation of 180 degrees about the origin, we compose two reflections in mirrors that intersect at the origin, at an angle of half of 180, or 90 degrees. The obvious choices for mirrors are the x- and y-axes. We've already proved that the reflection image of (x, y) in the x-axis is (x, -y) and the reflection image of (x, y) in the y-axis is (-x, y). It doesn't matter in which order we compose these as reflections in perpendicular mirrors always commute. So we prove that the rotation image of (x, y) centered at the origin and of magnitude 180 degrees is (-x, -y).
Now our other common rotation magnitude is 90 degrees -- and this time, it will make a difference whether it's clockwise or counterclockwise. The angle between the mirrors will now have to be half of 90, or 45 degrees. There's one mirror to consider that will help us with a 45-degree angle -- the line whose equation is y = x.
We've hinted at several proofs involving reflection over the line y = x. Let's look at the quadrilateral whose vertices are (0, 0), (a, 0), (a, a), and (0, a). We can show that this figure is a kite.
So now we can apply the properties of a kite -- the Kite Symmetry Theorem. The diagonal of our kite running from (0, 0) to (a, a) bisects the angle between the x- and y-axes -- and since we know that the angle between the axes is 90 degrees, the diagonal must form a 45-degree angle with each axis. And reflecting across this symmetry diagonal must map the axes to each other and x = a to y = a.
Recall that at this point, we don't know the equations of lines, so we aren't yet certain that the graph of y = x is even a line (which we'd better figure out before trying to use it as a mirror). But we see that the value of a in the above proof is arbitrary -- it's true for every single real number a (although in case a is negative, we should probably say that the kite has sides of length |a|, not a). Therefore every single point of the graph of y = x lies on the bisector of the angle between the axes -- that is, the graph of y = x is exactly that line. And reflecting in that line maps x = a to y = a and vice versa -- that is, it switches x and y. Therefore the image of (a, b) must be (b, a).
Now that we know how to reflect in the line y = x, let's use it to perform a 90-degree rotation. It's probably easiest just to start with the reflection in y = x first, so (x, y) maps to (y, x). As for the second mirror, it depends on whether we want to go clockwise or counterclockwise. To go clockwise, the second mirror must be 45 degrees clockwise of the first mirror, y = x. That is the x-axis, and to reflect in it, we change the sign of the second coordinate. So (y, x) reflected in the second mirror is the point (y, -x), so mapping (x, y) to (y, -x) rotates points 90 degrees clockwise. To go counterclockwise, the second mirror must be 45 degrees counterclockwise of the first mirror, y = x. That is the y-axis, and to reflect in it, we change the sign of the first coordinate. So (y, x) reflected in the second mirror is the point (-y, x), so mapping (x, y) to (-y, x) rotates points 90 degrees counterclockwise.
Notice that some of the PARCC questions mention 270-degree rotations -- for example, there was a released question that mentions a 270-degree clockwise rotation. Of course, a 270-degree clockwise rotation is equivalent to a 90-degree counterclockwise rotation, so it maps (x, y) to (-y, x). If students forget this, they can still take half of 270 degrees to get 135 degrees clockwise, and they can see that 135 degrees clockwise from the line y = x is still the y-axis, just as it would have been if they'd gone 45 degrees counterclockwise instead.
Last year, I created a quick worksheet to help students perform any of the reflections and rotations mentioned in this post. (This was late in the year when we were covering PARCC questions, but now I'm giving this lesson much earlier.) It takes the coordinate plane labeled with the positive x-, negative x-, positive y-, and negative y-axes. Students can then perform the rotations on the axes to see what happens. For example, let's try our 270-degree clockwise rotation. After we rotate the paper 270 degrees clockwise, we see that where the x ought to be, we see -y instead, and where the y ought to be, it's +x. Thus the image must be (-y, x).
Okay, so we've taken care of all the rotations centered at the origin, But on the PARCC, there are questions with rotations centered at other points. These questions that I've seen direct the students to take a triangle ABC and rotate it around one of its vertices -- let's say C. Well that makes things a little easier, since then the rotation image of C is C itself. So then there are only two points that we need to find, A' and B'.
It's possible, in principle, to find formulas to determine the image of (x, y) under reflections in mirrors parallel to the axes and rotations centered at points other than the origin. We've seen, for example, that the point (x, y) reflected in the line x = a is (2a - x, y). An interesting question is, where exactly does the 2 in 2a - x come from?
To find out, we notice that if we were reflecting in the y-axis (which is parallel to x = a), then the point (x, y) is mapped to (-x, y). Now that extra 2a term looks just like a horizontal translation of exactly 2a units.
So somehow, our reflection in the line x = a appears to be the composite of a reflection in the y-axis and a horizontal translation. (This is not a glide reflection between the mirror is perpendicular to the direction of translation -- we found out last year that such a composite yields a simple reflection, not a glide reflection.)
Using symbols, let's call the composite transformation T. It is the composite of a y-axis reflection, which we'll call r_y, and a horizontal translation of 2a units, H_2a:
T = H_2a o r_y
But the horizontal translation is itself the composite of two reflections. The two mirrors here must be vertical mirrors spaced exactly half of 2a, or a units apart. We might as well let the two mirrors be the y-axis itself and the line x = a.
T = H_2a o r_y
= r_(x = a) o r_y o r_y
= r_(x = a) o I
= r_(x = a)
which is exactly what we wanted -- a reflection in the line x = a.
Likewise, we see that the reflection in the line y = b maps (x, y) to (x, 2b - y). The composite of both reflections is a 180-degree rotation about the point (a, b), which maps (x, y) to (2a - x, 2b - y) -- and that's also the composite of a 180-degree rotation about the origin and yet another translation.
Now 90-degree rotations about points other than the origin are even trickier, because now we'd have to reflect about mirrors with equations like y = x + b -- and we don't even know that's a line yet. The algebra involved in this reflection gets very messy.
Of course, if we try to visualize the rotation, another composite transformation jumps at us. To perform a 90-degree rotation (either clockwise or counterclockwise) about the point (a, b), it appears that we can first perform the translation that maps (a, b) to (0, 0), then perform the rotation centered at the origin, and finally translate (0, 0) back to (a, b).
This seems to work, but is there any reason why it should? Let's use symbols again -- in order to remember what the symbols stand for, we let "rot" stand for the rotation and "trans" stand for the translation mapping (0, 0) to (a, b). Then trans^-1 can stand for the inverse translation -- the one mapping (a, b) to (0, 0). This gives us:
T = trans o rot o trans^-1
This composite has a name in classes like linear algebra and above -- conjugation. That is, we are conjugating the rotation by the translation.
We now want to rewrite both the translation and origin-rotation with two mirrors each. And as usual, we want to choose the mirrors strategically so that some of the reflections cancel out. For the translation, we'll let k be the line joining the points (0, 0) and (a, b). Then l will be the line perpendicular to k passing through the origin, m will be the line perpendicular to k passing through the midpoint of (0, 0) and (a, b), and n will be the line perpendicular to k passing through (a, b). Then the rotation can be written as r_l o r_k, and the translation can be written as r_m o r_l. Notice that the inverse translation can be written r_l o r_m -- but it can also be written as r_m o r_n (as either l and m, or m and n, are the correct distance apart). So we write it:
T = trans o rot o trans^-1
= r_m o r_l o r_l o r_k o r_m o r_n
= r_m o I o r_k o r_m o r_n
= r_m o r_k o r_m o r_n
= r_k o r_m o r_m o r_n (as reflections in perpendicular mirrors commute)
= r_k o I o r_n
= r_k o r_n
which is the composite of reflections in perpendicular mirrors intersecting at (a, b). And so T is in fact the rotation centered at (a, b), which is what we were expecting.
Once again, though, this is not the sort of symbolic manipulation I'd want my students to see. But then, what should we expect students to do when faced with a PARCC question where they have to rotate around a point other than the origin?
Most likely, this is something that can wait until we discuss the Slope Formula -- especially since it's this rotation that leads to the slopes of perpendicular lines. For now, one can consider such rotations only informally -- after all, the PARCC questions usually include graphs, so students might be able to perform the rotations just by counting units on the graph, rather than using an algebraic formula or manipulating mirrors and symbols.
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