Just as at my old charter from two years ago, Wednesdays are Common Planning Days. School is out about an hour earlier today. (I've obviously subbed at this school several times before, but never before on a Wednesday.)
For today's "Day in the Life" we might as well revisit the resolutions in order:
2. Keep a calm voice instead of yelling at students.
7:15 -- Zero period, the first Algebra I class, arrives. Yesterday the students learned about solving systems by substitution, so it goes without saying that today they learn about the elimination method in Lesson 4-3 of the Big Ideas text. This time, the homework assignment is posted correctly on the students' Chromebooks.
8:15 -- Zero period leaves and homeroom arrives, which leads directly into first period, the second Algebra I class. This time homeroom is the usual ten minutes instead of the extended homeroom due to yesterday's digital citizenship lesson.
9:10 -- Recall that at this school, period rotation is in effect after first period. Since it's Wednesday, the rotation begins with fourth period, which is Math Support. Students continue to work on either ALEKS or homework. Only two girls choose ALEKS today.
9:55 -- Fourth period leaves and snack begins. This leads directly into fifth period conference.
10:55 -- Sixth period, the third Algebra I class, arrives.
11:40 -- Sixth period leaves and lunch begins.
12:25 -- Second period, the final Algebra I class, arrives.
1:10 -- Second period leaves and third period arrives. This is the ASB class, and so students continue to work on hanging a few last minute election posters.
1:50 -- Third period ends, thus concluding my day.
Whenever I teach a lesson on the elimination method, I think back to Christie Bradshaw, the teacher who once proposed using dry erase packets in the classroom. She actually posted her dry erase packet for her lesson on the elimination method. I actually created my own version of this packet, but then I never posted it on the blog -- since it has nothing to do with Geometry -- nor did I ever use it in an actual class. I never had the opportunity to do so -- until today, that is.
But I didn't bring the dry erase packets with me, nor did I bring any markers. (Perhaps I should have already known to do so -- even though I didn't receive a lesson plan until midway through zero period, I already knew that I'd at least be subbing today, and it's logical to assume that the next lesson after substitution would be elimination.) I won't be able to bring the packets tomorrow, but I just might have them in time for Friday. (But I might not have enough -- the teacher's largest class is second period, which has 35 students.)
Notice that subs really shouldn't deviate from the regular teacher's lesson plan. But technically, using dry erase packets wouldn't have been deviating from the lesson plan. I was directed only to go over the systems in the text -- using dry erase packets would still be "going over" the systems.
In first period, it's obvious that some students have already learned the elimination method. When I asked the students how to solve a particular system, one girl says to multiply one equation by -2 to eliminate x when the system is already set up to eliminate y without needing to multiply -- but this is before I give any examples that require multiplication. Then at the end of class, another girl asks me "Aren't you going to give any where you multiply both equations?" when there are no such systems given in the examples (yet they do appear in the homework).
Clearly no one who's seeing elimination for the first time would make such remarks. As it turns out, the first girl was studying for the SSAT (a private school admissions test), and she tells me that she scored at the 98th percentile -- wow! I neglect to ask her what school she's trying to get into -- among the private schools in Southern California, most of them use the ISEE, while the Catholic schools only require a placement test, HSPE. So I wonder who required this girl to take the SSAT.
Meanwhile, the other girl previously studied at Kumon. Hmm, the traditionalists have something to say about this eighth grade Algebra I student. They'd say that her success in math is due to Kumon, no thanks to Common Core -- had it not been for the tutoring, she'd probably be stuck in Math 8.
As far as behavior is concerned, sixth period is the best class, followed closely by second period. I'm able to avoid yelling today, though I was tempted yesterday. One student asked for a restroom pass, but I didn't let him go because another student had already left. And of course, this had to be the class that was right after snack (fourth period).
Two years ago at the old charter school, some students gave me a stress ball. Recall the reason for this second resolution -- one problem at my old school was that I yelled too much, and so these students wanted to help me out with the stress ball. Back then, what angered me was when I told students to do something and they wouldn't listen.
Since then, I realized that other actions tend to annoy me as well, such as when students ask for restroom passes right after snack or lunch -- especially when it's multiple students. (It seems as if the classes when they ask for passes the most lately are the class after snack and the class after lunch!) So when this happens, I get out the ball and start squeezing. It has really helped me release stress so that I don't yell at students, thus fulfilling the second resolution.
OK, that's enough Algebra I for today -- let's get back to Geometry. Today on her Mathematics Calendar 2018, Theoni Pappas writes:
If the circle's diameter is 16 units, then AB = ?.
[Here is some given info from the diagram -- the distance from the center to chord AB is sqrt(15).]
Notice that the radius of the circle is 8. Let O be the center and let C be the foot of the perpendicular from O to
So now we can use the Pythagorean Theorem:
AC^2 + OC^2 = OA^2
AC^2 + sqrt(15)^2 = 8^2
AC^2 + 15 = 64
AC^2 = 49
AC = 7
And of course AB = 2 AC. (To see why, just repeat Pythagoras on Triangle OBC.) This tells us that the length of the chord is 14 units -- and of course, today's date is the fourteenth.
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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