Today I subbed in a high school math class. It's another one of those classes where the regular teacher is teaching from home, so all I need to do is watch the class. Since I don't do any real teaching today, I've decided not to do "A Day in the Life Today" -- but as usual, I will describe the class.
This school is in my new district, and as it turns out, it's the high school for which my long-term middle school is a feeder. There are some schedule differences between this and the other high school I've subbed for in the district. First of all, it has a block schedule, except that Tuesday/Thursday is even classes, rather than the expected odds.
Also, for some reason, Cohort A attends class on Tuesdays and Fridays, rather than the expected Tuesdays and Wednesdays. I've written about Hybrid Plan I and II before, but I never considered the combination where one cohort is Tues./Fri. while the other cohort is Wed./Thurs. instead. Just don't ask me why the schedule goes this way.
This teacher has a zero period class, which is followed by second period since today is even. Both classes are Honors Algebra II/Trig, and the classes are learning about conic sections. (Conics were briefly mentioned last week in Lesson 9-4 of the U of Chicago Geometry text, but they are usually covered in more detail in Algebra II.) But then like most teachers with a zero period, she has no sixth period -- and as it turns out, her conference period is fourth. So my short day of subbing ends at 10:15, after tutorial.
Of the two classes that meet today, zero period has six in-person students and second period has ten. And here's how many total students there are in each grade:
Period 0: 11 freshmen, 12 sophomores, 9 juniors
Period 2: 9 freshmen, 20 sophomores, 5 juniors
And in fact, since younger students are more likely to attend in-person than older students, it happens that freshmen were in the majority of in-person students in both classes. These students are working at the advanced Bruce William Smith level, while the sophomores are at SteveH level.
In second period, I tell the students that I was a long-term sub at their feeder middle school, and so it's possible I met some of their younger siblings there. They don't appear to recognize the name of the regular teacher I covered for. Then again, the freshmen in this class had Geometry last year -- and the eighth grade Geometry this year (and presumably last year as well) wasn't my regular teacher. Actually, it's the "lead Math 8 teacher" who also has a section of Geometry. (I never mentioned this on the blog, since there was no reason to -- my focus was on my own classes, not eighth grade Geometry.)
The second period students stay for tutorial. As usual, I solve a few problems on ellipses. And I decide that there's a large enough audience for a rendition of "Palindrome Song" in this period.
By the way, the zero period students are preparing for a quiz on circles and parabolas tomorrow, but the second period students take this quiz today. That's because zero period doesn't meet on Mondays (which are Late Start days, just like my long-term school):
Period 0: 60+ minutes each Tuesday-Friday (yes, this means that there's no pure distance day for per. 0)
Block Periods: 40+ minutes on Monday, 105 minutes twice per week
It turns out that both classes about the same number of minutes per week, but they are distributed very much differently. Notice that if we can easily cover the Monday block lesson in zero period on Tuesday, but then there's 20 extra minutes. This is confusing -- if I were a zero period teacher, I'd have to plan zero period separately from the blocks, almost as if I had an extra prep. So instead of two preps (Algebra II and AP Stats) she actually has three (block Algebra II, zero period Algebra II, and Stats).
With my subbing day ending at 10:15, I was strongly tempted to drive down to my long-term middle school (obviously just a few blocks away) for a visit. But I don't -- there's no reason for me to interrupt their classes. After school, I do see one eighth grade girl from my long-term -- she was riding in the back of her car, most likely to pick up her older sibling at the high school.
Lecture 10 of Prof. Arthur Benjamin's The Mathematics of Games and Puzzles: From Cards to Sudoku is called "Solving Sudoku." Here is a summary of the lecture:
- Without a doubt, Rubik's Cube has been the most physical puzzle ever invented, but in the last 10 years [by now about 20], the world's most popular puzzle, requiring less physical dexterity and more logical reasoning is sudoku.
- The challenge can be described in a single sentence: Enter the numbers 1 through 9 in such a way that each number appears once in each row, column, and 3 * 3 box.
- How many different Sudoku puzzles are there? To answer this, we first count how many Shidoku puzzles there are -- a Shidoku contains only numbers 1 to 4.
- There are 4! or 24 different ways to fill in the upper left 2 * 2 box. This leaves 2 ways to finish the first row and 2 ways to complete the first column. Then there are only 3 ways to fill in the rest of the puzzle. So the total Shidoku is 24 * 2 * 2 * 3 = 288.
- It was shown in 2006 that there are over 6 sextillion different completed Sudoku grids -- about 100 times as many Sudokus as Rubik's cube positions. Every Sudoku needs a unique solution.
- It's possible to be given all but four numbers and still admit two different solutions -- for example, if there are two missing in the upper right corner and two in the lower right. On the other hand, it's possible to be given as few as 17 clues and have a unique solution.
- A certain Sudoku in the book Taking Sudoku Seriously has only 18 clues. Not only does it have rotational symmetry, but the given digits are the first 18 digits of pi. (It helps, of course, that there is no zero among the first 30 digits of that constant.)
- Palmer Mebane and Thomas Snyder -- two Sudoku world champions -- taught the professor the following strategy.
- But first, we must define the key terms -- a Sudoku puzzle consists of 81 squares with some given clues. It has nine rows, nine columns, and nine boxes. Numbers that may possibly fit in a square are called candidates for that square.
- Don't start by listing the candidates for each square. Instead, ask yourself where a certain number can possibly go. For example, in the Pi Sudoku mentioned above, the digit 3 appears as a clue four times, so we might start by asking, where else can a 3 go?
- The tic-tac-toe method considers the nine boxes in three "big rows" and "big columns." In the Pi Puzzle, big column two has a pair of 3's, in little column five and six. So there's a 3 that needs to be in little column four. But it can't go in the top or middle boxes, leaving only the bottom row.
- The 3 that we just placed in the bottom row, fourth column is called a hidden single. It's the only legal place it can go in its column. But if there are two possible places in a row where a 3 can go, then we can write this number tiny in both boxes. This is the only time we write tiny numbers.
- The cross hatching method is used to eliminate places for 3's -- if there's a column with a 3 and a row with a 3, then we can eliminate all other squares in that row and column. Often this leaves only one place in a box for the 3 -- and that helps us with the tiny 3's from earlier.
- All the 3's have been placed. we can now look for structures, such as right angles. For example, the lower box contains three numbers in a right angle, leaving the top three squares and the left three squares empty. This may be susceptible to crosshatching and a chain reaction
- With the mini-box structure, we often need only one number to make progress. For example, there's a mini-box in the top left box and a 1 in the third column, leaving two possibilities for the 1 in the third row. This is a pointing pair -- eliminating the possibility for other 1's in that row.
- A hidden pair occurs when there are two possibilities for two squares -- we know which two numbers go in the squares but not in which order. It's also possible to have a pointing triple that also eliminates some possibilities. There's only one place for the 4 -- a foregone conclusion.
Lesson 9-6 of the U of Chicago text is called "Views of Solids and Surfaces." In the modern Third Edition of the text, views of solids and surfaces appear in Lesson 9-5. (Recall that Lesson 9-4 of the new edition is "Drawing in Perspective," which is Lesson 1-5 of the old edition.)
This lesson is perfect for an architect. This is what I wrote last year about today's lesson:
Only two words are defined in this lesson -- views and elevations:
"Underneath the picture of the house are views of the front and right sides. These views are called elevations."
There's not much more for me to say about this lesson, except to add that the modern Third Edition includes some examples using isometric graph paper. This reminds me a little of the (mis)adventures my middle school students had with isometric graph paper four years ago (at the old charter school). And last year, I subbed in an Engineering Graphing Design class where the students dealt with front, side, and top views.
And so we return to Euclid. Of course, none of his definitions or propositions have anything to do with views of solids and surfaces, so we just look at the next proposition in order:
Proposition 9
This is the three-dimensional analog of yet another theorem, previously studied in Lesson 3-4:
Transitivity of Parallelism Theorem:
In a plane, if l | | m and m | | n, then l | | n.
Some texts omit "in a plane," (after all, as Euclid is about to prove, it holds in three dimensions as well) but only prove it for a plane. The U of Chicago text is more honest and admits that it is only proving it for the plane case. Indeed, the text gives an informal argument in terms of slope -- and of course, we only define slope for lines in a (coordinate) plane.
Euclid's proof is interesting in that unlike those of the previous propositions, he does not use plane Transitivity of Parallelism to prove the spatial case.
Given: AB | | EF, CD | | EF (noncoplanar)
Prove: AB | | CD
Statements Reasons
1. bla, bla, bla 1. Given
2. Choose G on EF, 2. Point-Line-Plane, part b (Ruler/Protractor Postulates)
H on AB with GH perp. EF,
K on CD with GK perp. EF
3. EF perp. plane(GH, GK) 3. Proposition 4 from last week
(call it plane P)
4. AB perp. plane P, 4. Proposition 8 from yesterday
CD perp. plane P
5. AB | | CD 5. Proposition 6 from last week
But in this case, Euclid's original proof is simple enough to show to high school students as it is:
Perhaps the only change we might make is replace "plane through HG and GK" with "plane P" in order to avoid repeating that cumbersome phrase.
David Joyce mentions another proof at the above link -- midpoint quadrilaterals, which are also known as Varignon quadrilaterals. The U of Chicago text only mentions midpoint quadrilaterals in an Exploration exercise (likewise in the new Third Edition, the only difference being that the new text actually mentions Varignon's name). But it's easy to prove that they are parallelograms -- just divide the quadrilateral into two triangles and apply the Midsegment Theorem of Lesson 11-5 to each.
But now Joyce points out that the theorem is true even for "space quadrilaterals." Our definition of polygon (and hence quadrilateral) states that the vertices must be coplanar. But even if we relax this requirement and allow for space quadrilaterals, Varignon's Theorem still holds. After all, even space quadrilaterals can be divided into two triangles (and each triangle lies in a plane), and so we can still apply the Midsegment Theorem to each one. The difference is that today's Proposition 9 is used to prove that opposite midsegments are parallel -- each is parallel to the the diagonal of the quadrilateral but the three lines aren't coplanar.
Notice that the final parallelogram always lies in a single plane -- this follows from last Friday's Proposition 7 that two parallel lines and a transversal are always coplanar.
Meanwhile, I've been continuing to think about a proof of Proposition 4 that is based on rotations, since this might be simpler than Euclid's proof that requires seven pairs of congruent triangles. Let's recall what we're supposed to prove -- given that line l is perpendicular to both lines m and n, prove that l must be perpendicular to the entire plane containing m and n.
Euclid begins by defining points A, B on m, C, D on n, and E, F on l. (Actually, E is where all three lines intersect.) These points have the additional property that AE = BE = CE = DE. My goal was to demonstrate that a rotation of 180 degrees about axis EF maps A and B to each other, as well as C and D to each other. Of course, E and F are fixed points of this rotation.
Let's first find A', the rotation image of A. We first notice that rotations, like all isometries, preserve angle measure, and so Angle A'EF = AEF, which is known to be 90. Also, since rotations preserve distance, AE = A'E. Finally, since the magnitude of the rotation is 180, Angle AEA' = 180. The only point satisfying all these requirements is B, so A' = B. Similarly, we have B' = A, C' = D, and D' = C.
Of course, this requires a more rigorous definition of rotation about an axis. For plane rotations about a center O, we expect A' to be a point such that AO = A'O and Angle AOA' equals the magnitude. But for rotations about an axis, we can't use the same definition unless we know where O is along the entire axis. Otherwise we won't know where to measure AO, A'O, or Angle AOA'.
The correct answer is that O is chosen along the axis such that AO is perpendicular to the axis. Since AE is perpendicular to the axis EF, E is the correct point to choose. So that's why AE = A'E and Angle AEA' must be 180. If AE weren't perpendicular to the axis, choose O to be a different point instead.
Once this is complete, then we choose line o with G and H the points of intersection. It then follows that H = G' since rotations preserve collinearity, and the proof is complete.
Or is it? I'm wondering whether I slipped and accidentally assumed what we're trying to prove. For example, can we be sure that the rotation maps o to itself unless we already know that o is perpendicular to l? Also, if we define a rotation as the composite of two reflections in intersecting planes (similar to the definition for plane rotations), then we might not be able to prove that rotations work the way I said they do unless we've already assumed that Proposition 4 is true! (Why, for example, do we choose O such that AO is perpendicular to the axis?)
Making the proof work requires much more thought than I'm willing to take now, considering that we're not actually trying to teach Proposition 4 in a classroom.
Here is the worksheet for today: Students can fill out the front of the worksheet with examples of both a 3D figure (such as a pyramid) and a block-building (similar to #2 on the back) with the respective front, right, and top views of each.
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