This is what I wrote last year about today's lesson:
We already discussed translations and rotations for Common Core Geometry, but those transformations applied to the two-dimensional plane. The physical world has three dimensions, and so it's the 3-D transformations that actually apply to physics.
Lesson 9-5 of the U of Chicago text is on reflections in space. I find this to be an interesting topic, but since it doesn't appear on the PARCC, some might consider it a waste of time to cover it in class. But it's a regular lesson now, and so there will be a worksheet for this lesson after all!
The text begins by defining what it means for a plane in 3-D, rather than a line in 2-D, to be a perpendicular bisector:
"In general, a plane M is the perpendicular bisector of a segment
Now we can definition 3-D reflections almost exactly the same way we define 2-D reflections:
"For a point A which is not on a plane M, the reflection image of A over M is the point B if and only if M is the perpendicular bisector of
If you think about it, when you (a 3-D figure!) look at yourself in a mirror, the mirror itself isn't a line, but rather a plane. Mirrors in 2-D are lines, while mirrors in 3-D are planes. So now we can define what it means for two 3-D figures to be congruent:
"Two figures F and G in space are congruent figures if and only if G is the image of F under a reflection or composite of reflections."
Most texts don't actually define what it means for two 3-D figures to be congruent. We know that the traditional textbook definition, that congruent figures have corresponding sides and angles congruent, only applies to polygons. It doesn't even apply to circles, much less 3-D figures. But we can simply use the Common Core definition -- two figures are congruent if and only if there exists an isometry (i.e., a composite of reflections) mapping one to the other -- and it instantly applies to all figures, polygons, circles, and 3-D figures.
In 2-D there are only four isometries -- reflections, translations, rotations, and glide reflections. An interesting question is, how many isometries are there in 3-D?
Well, for starters, translations and rotations exist in 3-D. We can define both of these exactly the same way that we do in 2-D -- a translation is the composite of two reflections in parallel planes, while a rotation is the composite of two reflections in intersecting planes.
Notice that every 2-D rotation has a center -- the point of intersection of the reflecting lines. The same thing happens in 3-D, except that the intersection of two planes isn't a point, but a line. Thus, a 3-D rotation has an entire line as its center -- every point on this line is a fixed point of the rotation. But usually, instead of calling the line the center of the rotation, we call it the axis of the rotation. One 3-D object that famously rotates is the earth, and this rotation has an axis -- the line that passes through the North and South Poles. Confusingly, the mirror of a 2-D reflection is often called an axis -- but in some ways, these two definitions are related. One can perform a 2-D reflection by taking a 2-D figure and rotating it 180 degrees about the axis in 3-D (and recall that A Cube has reflected A Square in exactly this manner).
Glide reflections also exist in 3-D -- although these are often called glide planes in 3-D. A glide reflection is the composite of a reflection and a nontrivial translation parallel to the mirror. Notice that there are infinitely many directions to choose from for our translation in 3-D, but if this were a 2-D glide reflection there are only two possible directions for a translation that's parallel to the mirror.
But are there any other isometries in 3-D? Well, we notice that a glide reflection is the composite of two other known isometries, a reflection and a translation. So the next natural possibility to consider is, what if we find the composite of the other two combinations? What is the composite of a reflection and a rotation, or a translation and a rotation?
In 2-D, the composite of a translation and a nontrivial rotation is another rotation. This is possible to prove, as follows: let T be a translation and R be a nontrivial rotation, and G be the composite of T following R. Both translations and rotations preserve orientation, and so their composite G must preserve orientation as well. In 2-D, three mirrors suffice -- that is, every isometry is the composite of at most three reflections. Since G preserves orientation, it must be the composite of an even number of rotations. Therefore G is the composite of two reflections (or the identity transformation -- the transformation that maps every point to itself), and so G is either a translation or a rotation.
Let's try an indirect proof -- assume that G is a translation. That is:
T o R = G
Since T is a translation, it has a translation vector t, and as we're assuming that G is a translation, it must also have a translation vector g. Now let U be the inverse translation of T -- that is, the translation whose vector is -t, the additive inverse of t. We now compose U on both sides:
U o T o R = U o G
Now since U and T are inverses, U o T must be I, the identity transformation:
I o R = U o G
Since I is the identity transformation, I o R must be R:
R = U o G
Notice that G and U are both translations, whose vectors are g and -t, respectively. Then their composite must be another translation V whose vector is g - t. So now we have:
R = V
that is, a rotation equals a translation. Now no rotation can equal a translation (as the former has a fixed point, while the latter has no fixed point) unless both are the identity -- which contradicts the assumption that R is a nontrivial rotation (i.e., not the identity). Therefore, the composite of a translation and a nontrivial rotation isn't a translation, so it must be a rotation. QED
But this proof is invalid in 3-D. This is because the proof uses a step that only works in 2-D -- namely that three mirrors suffice. We must show how many mirrors suffice in 3-D.
Let's recall why mirrors suffice in 2-D. Let G be any 2-D isometry, and let A, B, and C be three noncollinear points whose images under G are A', B', and C'. The first mirror maps A to A', the second mirror fixes A' and maps B to B'. It could be that the image of C under both mirrors is already C', otherwise a third mirror maps it to C'. Notice that the proof of the existence of these mirrors is nontrivial and depends on theorems such as the Perpendicular Bisector Theorem, since reflections are defined using perpendicular bisectors.
As it turns out, four mirrors suffice in 3-D. To prove this, we let G be any 3-D isometry, and let A, B, C, and D be four noncoplanar points. The first mirror maps A to A', the second mirror fixes A' and maps B to B', the third mirror fixes A' and B' and maps C to C', and the fourth, if necessary, fixes A', B', and C' and maps D to D'.
And so this opens the door for there to be a new transformation in 3-D, one that is the composite of a translation and a rotation as well as the composite of four reflections. We can imagine twisting an object like a screw. A screwdriver rotates the screw about its axis, but then it's being translated into the wall in the same direction as that axis. And because of this, this new transformation is often called a screw motion.
We still have one last combination, the composite of a reflection and a rotation. It is subtle why the composite of a reflection and a rotation in 2-D is usually a glide reflection -- why should the composite of a reflection and a rotation equal the composite of a (different) reflection and a translation? And it's even subtler why the composite of a reflection and a rotation may be a new transformation in 3-D.
But the simplest example of this roto-reflection is the inversion map. In 3-D coordinates, we map the point (x, y, z) to its opposite point (-x, -y, -z). This map is the composite of three reflections -- the mirrors are the three coordinate planes (xy, xz, and yz). As the composite of an odd number of reflections, it must reverse orientation. Yet it can't be a reflection, since it has only a single fixed point (0, 0, 0) and not an entire plane. Similarly, it can't be a glide plane because glide planes don't have any fixed points at all.
(Note: Some sources refer to "roto-reflections" as "improper rotations.")
Roto-reflections are formed when the axis of the rotation intersects the reflecting plane in a single point -- and of course, this single point is the only fixed point of the roto-reflection.
So now we have six isometries in 3-D -- reflections, translations, rotations, glide planes, screw motions, and roto-reflections. Are there any others? As it turns out, these six are all of them -- and the proof depends on the fact that four mirrors suffice in 3-D.
Returning to the U of Chicago text, we have the definition of a reflection-symmetric figure:
"A space figure is F is a reflection-symmetric figure if and only if there is a plane M such that the reflection of F in M is F."
Similarly, a figure can be rotation-symmetric, as well as roto-reflection-symmetric. In the text, figures such as the right cylinder have reflection, rotation, and roto-reflection symmetry.
But a figure can't have translation symmetry unless it's infinite, as translations lack fixed points. So likewise, figures that have glide reflection or screw symmetry must also be infinite, as these transformations are based on translations.
Of course, there exist dilations in 3-D space as well. There is little discussion of similarity in 3-D, except to compare the surface areas and volumes of 3-D figures.
Now that we are reading about the fourth dimension, the next thing we wonder is, what would a reflection in 4-D look like? We follow the same pattern that we followed for 2-D and 3-D -- we begin with the perpendicular bisector of a segment. In 2-D the perpendicular bisector is a 1-D line, in 3-D it's a 2-D plane, and so in 4-D it's a 3-D hyperplane. This hyperplane M is the mirror of a 4-D reflection, mapping each point A to its image B if and only if M is the perpendicular bisector of
We know that the 3-D isometries include all of the 2-D isometries (translations, rotations, reflections, and glide reflections) plus two new isometries (screws and roto-reflections). So we expect the 4-D isometries to include all the 3-D isometries plus two new ones.
One of these new isometries is called a double rotation, as it is the composite of two rotations. Notice that the center of a 2-D rotation is a 0-D point, and the center (axis) of a 3-D rotation is a 1-D line, and so the axis of a 4-D rotation is a 2-D plane. The double rotation is the composite of two rotations whose respective axes (planes) intersect in a single point. A point inversion is a double rotation.
The other new 4-D isometry is a roto-glide reflection. It is the composite of a roto-reflection (which requires three dimensions) and a translation in the fourth dimension. Just as three mirrors suffice in 2-D and four mirrors suffice in 3-D, we see that five mirrors suffice in 4-D.
OK, let's return to 2019 and continue our reading of Euclid. Of course, Euclid doesn't write about our isometries, and so we proceed directly to the next proposition:
According to David Joyce, this is essentially a converse to Proposition 6 -- which, as we found out last Thursday, is the analog of the Two Perpendiculars Theorem. Therefore Proposition 8 is the analog of the Perpendicular to Parallels Theorem of Lesson 3-5. Again, the old Perpendicular to Parallels Theorem will appear as a step in today's proof.
Given:
Prove:
Proof:
Statements Reasons
1. bla, bla, bla 1. Given
2.
(call it plane Q)
3. Choose E in plane P s.t. 3. Point-Line-Plane, part a (Ruler/Protractor Postulates)
4.
5.
6. BD = BD 6. Reflexive Property of Congruence
7. Triangle ABD = EDB 7. SAS Congruence Theorem [steps 3,4,6]
8. AD = BE 8. CPCTC
9. AE = AE 9. Reflexive Property of Congruence
10. Triangle ABE = EDA 10. SSS Congruence Theorems [steps 3,8,9]
11.Angle ABE = Angle EDA 11. CPCTC
12.
13.
14.
15.
16.
17.
This proof is a bit complex. But since we just learned about isometries today, we might wonder whether isometries can be used to make the proof easier (just as the U of Chicago text uses symmetry to prove the Isosceles Triangle Theorem, for example).
We notice that in step 7, Triangles ABD and EDB are congruent, and in step 10, triangles ABE and EDA are congruent. The isometry that maps both ABD to EDB and ABE to EDA must switch A with E and B with D. Thus the desired isometry is in fact a rotation of 180 degrees. Its axis passes through the midpoints of
But I'm not quite sure whether this knowledge actually helps to simplify the Proposition 7 proof. So instead, I'm wondering whether our Proposition 4 proof can be simplified using a rotation. In that proof, there are seven pairs of congruent triangles. Every single pair switches A with B and C with D, and G with H, and fixes E and F. Thus we can use a rotation of 180 degrees about axis
There are other things going on in the world that I wish to discuss in this post. First, a lunar eclipse occurred a few days ago. Unfortunately, it had to occur over a three-day weekend with consecutive non-blogging days. This is why I couldn't discuss the eclipse until today.
Like all lunar eclipses, this one occurred at the full moon. (It was the middle of the month on most lunar calendars, including the Yerm and David Lunar Calendars. On the Hebrew Calendar, it was the full moon holiday of Tu B'Shevat, a tree holiday. But it was the start of a new month on the Nirvana Lunar Calendar, based on our assumption that his/her lunar months begin on full moons.)
And like all lunar eclipses, this one was visible from any location in the world where it was nighttime at the moment of the full moon. This was Sunday night at about a quarter after 9 PM Pacific Time, making the eclipse visible in most of the Americas.
At the time of the eclipse, I happened to be watching a DVD movie, The House with a Clock in Its Walls -- a film that had been recommended to me. Little did I know that the climax of this movie actually features a lunar eclipse! As I watched the movie, I saw the moon turn red on the screen -- and then I ran outside to watch the real moon turn red, just before totality.
Once the movie ended, I observed the moon every few minutes, during and after totality. (It was partly cloudy but very windy, and so the clouds moved quickly across the moon.) Since this is a Geometry blog, I must comment on the shape of the moon as it emerged after the eclipse.
Recall that a lune is the region of a sphere's surface bounded by two great circles. (Even though we discussed spheres in my last post on Friday, I didn't mention lunes, since the emphasis there was on plane cross sections. Instead, we return back to my old spherical geometry posts for lunes.) The key is that during an ordinary phase, the visible portion of the moon is a lune.
When the moon slowly reappeared after the eclipse, a near-crescent was visible. If I had taken a photo of the moon at the time, it would be difficult to discern from a crescent moon -- and someone just then looking up at the sky -- unaware of the eclipse -- might assume that the moon was currently in its crescent phase.
But when the moon was about 50% illuminated, it definitely didn't resemble the semicircle we associate with the first or last quarter phases. Instead, it appeared to be a circle with another circle cut out of it. In other words, the illuminated portion was not a lune. This makes sense when we think about what causes lunar eclipses -- the earth's shadow. Since the earth is approximately a sphere, its shadow is always a circle. (Now this does go back to Friday's Lesson 9-4, since shadows are more or less cross-sections.) Thus as the eclipse ends and the moon moves slowly out of the earth's shadow, the illuminated portion must be a circle cut out of a circle -- not a lune.
In the movie, I saw the protagonist's uncle consult a chart that shows circles cut from circles. This clearly indicates the eclipse that's about to occur in the film, since the moon is never shaped like that except before or after an eclipse.
(By the way, as the movie opens, I saw some store depicted as having a "Lunar Eclipse Sale." I didn't notice any stores in the real world attempt to hold a Lunar Eclipse Sale over the weekend -- but I could have easily missed it.)
The lunar eclipse was billed as a "Super Blood Wolf Moon." Here "Super" refers to perigee -- when the moon, in its elliptical orbit, is at its nearest point to the earth. This and the next two full moons (in February and March) are close enough to perigee to be called super moons. "Blood" refers to the red color of the lunar eclipse, and the "Wolf" Moon is the January full moon.
In the movie, I remember mention of a "Harvest Moon." Unlike the other traditional moon names, "Harvest" refers to the fall equinox, so the Harvest Moon -- a full moon near the equinox -- could be in September or October. (Depending on how the Nirvana Lunar Calendar is rigorously defined, New Year's Day in that calendar could coincide with the Harvest Moon.) On the other hand, the Wolf Moon is always in the Gregorian month of January. (Notice that it's the "Harvest Moon" -- and not the jack o'lanterns in the front yard -- that indicates what season House with a Clock takes place.)
Lunar eclipses always occur within a fortnight of a solar eclipse. The solar eclipse associated with this lunar eclipse occurred the evening of January 5th. But it was visible mostly in Asia, rather than the Americas.
In fact, one solar and one lunar eclipse occurs during each eclipse season -- about six months. And these of course are lunar months. Because 12 lunar months are short of a solar year, it's possible to have three solar eclipses in a calendar year -- one the first week in January, the second around midyear, and the third the last week in December. Indeed, 2019 is such a year -- the second eclipse will be July 2nd, and the last eclipse will be December 26th. Only the July solar eclipse is total, and none of them are visible from North America.
Returning to lunar eclipses, the last total eclipse was 12 lunar months ago. I wrote about nearly a year ago, back in my January 31st post. That moon wasn't a Wolf Moon, but a blue moon, as it was the second full moon in January. That morning (unlike this weekend), as the moon slowly returned I could only see a crescent, since the moon set (and the sun rose) before it returned completely (so I never saw the special eclipse shapes). In other North American time zones east of Pacific, the eclipse was fully after sunrise, hence not visible.
The next total lunar eclipse won't be until May 26th, 2021. This eclipse is similar to last year's as far as visibility is concerned -- in California, the eclipse won't end fully until after sunrise (though most of the eclipse is before sunrise) and on the East Coast, nearly all the eclipse will be after sunrise. As this is a Wednesday, it's possible that I'll be awake to get ready for school, but note that sunrise is earlier in May than in January (so even teachers might not wake up until after sunrise).
The other issue in the news right now is the LAUSD teacher strike. A deal has finally been reached, and teachers will return to class tomorrow.
Here are some details of the agreement. Perhaps most relevant to us as math teachers, all class sizes will be reduced in high school English and math. In the 2019-20 school year the maximum class size will be 39 (down from a mind-boggling 46), and it will continue to drop by one or two students the next three years.
Also relevant to us math teachers, there is a plan to reduce district assessments by 50%. Teacher salaries will increase by 6%. And more librarians and counselors will be hired.
As for charters, the union will gain more power to stop co-located charters. (The old charter I worked at two years ago was co-located at the time, but has since moved to its own campus.) And while neither the union nor the district has the power to stop new charters, the board will vote on a resolution to encourage the state to establish a charter cap.
Recall that I've worked for both charter and traditional public schools, and so I recognize that both sides have merit. But once again, as a former teacher, I stand in solidarity with teachers. And so when I drove past an LAUSD middle school on Wednesday, I honked my horn in support of teachers.
I also spotted a striking teacher today, in a 7-Eleven around late afternoon. She was wearing a "Red for Ed" T-shirt that many teachers around the country wore during their strikes. I introduced myself as a former teacher, and congratulated her and her union for what was accomplished today.
By the way, last week I wrote about how I'd run my class if I were an LAUSD teacher knowing that a strike was imminent (namely give a full week of activities). This means that I'd treat tomorrow as the actual first day of the semester, and begin with the first traditional lesson -- on coordinate geometry, according to the LAUSD pacing plan. (As it happens tomorrow according to the LAUSD calendar is Day 91, the mathematical first day of the second semester anyway.)
Here is the worksheet:
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