Thursday, July 26, 2018

Semester 2 Review and Summer Preview (Day 180)

Today on her Mathematics Calendar 2018, Theoni Pappas writes:



How many lines of symmetry does a regular hexagon have?



In the U of Chicago text, regular polygons are introduced in Lesson 7-6, but that lesson has little to say on its lines of symmetry.



Fortunately, the modern Third Edition of the U of Chicago text discusses the symmetries of a regular polygon in much more detail. The relevant theorem, which appears in Lesson 6-8, is:



Regular Polygon Reflection Symmetry Theorem:

Every regular polygon has reflection symmetry about:



  1. each line containing its center and a vertex.
  2. each perpendicular bisector of its sides.
And that text adds:


"You may have noticed...that when a regular n-gon has an odd number of sides, the perpendicular bisector of one side contains a vertex on the opposite side of the polygon. When n is even, each symmetry line either bisects two angles of the n-gon or is the perpendicular bisector of two opposite sides. Thus, every n-gon has n lines of symmetry."


Today's Pappas question asks about the hexagon (n = 6). Therefore the regular hexagon has six lines of symmetry -- and of course, today's date is the sixth.


It's a shame that the symmetries of the regular polygon don't appear in the old version of the text, but I'm glad they appear in the new edition. We notice the following Common Core Standard:


CCSS.MATH.CONTENT.HSG.CO.A.3

Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.




And yes, the modern text writes about the rotations that map the regular polygon to itself as well. We notice that the n lines of symmetry form 2n rays emanating from the center of the polygon, and so the angle between two consecutive rays is 360/(2n). Then the magnitude of the smallest rotation mapping the polygon to itself must be exactly twice this value, or 360/n (according to the Two Reflection Theorem for Rotations).



But actually, the new U of Chicago text proves that a regular polygon has rotation symmetry first, and then we can prove that it has reflection symmetry. That a regular polygon has rotation symmetry follows from the existence of its center -- the Center of a Regular Polygon Theorem. And this is the one theorem that is actually proved in the old Second Edition of the text. Perhaps Lesson 6-8 of the Third Edition is a good lesson to emphasize for the SBAC -- but then again, we didn't see any symmetry problem during our recent look of the practice SBAC questions.



Today is the last day of school in my old district. It isn't the last day of school in my new district, where it is only Day 174. But the blog is following the old calendar.



Usually, today is when I post a preview of the upcoming school year, but of course, summer school is on my mind right now. In my new district, today is the last summer school training meeting.



As of now, I still don't know whether enough students will sign up for summer school. Remember that even if my summer class is canceled, there won't be a Great Post Purge of 2018 -- that is, I won't go back and delete every post that mentions summer school. Even when I first found out about summer school, I knew that it was dependent on there being enough students to sign up for the class.



Assuming that the summer school class happens, then what are my plans? Of course, the class doesn't start next week (since this is only Day 174), but the week after. The classes are three weeks, four days per week, and I'm scheduled to teach two classes, each a little less than two hours.



When I see my fellow summer Algebra I teacher today (yes, she's the current student teacher at one of the district high schools), we confirm the following plans. She continues to think in terms of the Glencoe text that she uses during the school year, even though we're actually using Edgenuity, an online curriculum:



First Week: Chapters 1-2

Second Week: Chapters 3-4

Third Week: Chapter 5 and District Final



In the name of purity, I should use the names of the actual units in Edgenuity:

  • Solving Linear Equations
  • Introduction to Functions
  • Analyzing Functions
  • Linear Functions
  • Point-Slope Form and Linear Equations
  • Solving Equations and Inequalities
The first unit, Solving Linear Equations, actually corresponds to Chapter 2 of Glencoe. Again, Chapter 1 of Glencoe doesn't actually appear in Edgenuity (since Chapter 1 is based on middle school standards in Common Core), and so we'll teach them Chapter 1 material without a computer -- mostly Order of Operations and the Distributive Property.



In fact, it's possible that my supplemental lessons could come the U of Chicago Algebra I text. The Order of Operations is taught well in Lesson 1-4 of the text. Unfortunately, the Distributive Property is spread out among several lessons in Chapter 6. The distributive property is first introduced in Lesson 6-3, but this is mostly about combining like terms -- as in 2x + 2x = (2 + 2)x = 4x -- as well as discount and markup questions -- as in x - 0.25x = (1 - 0.25)x = 0.75x for 25% off.



More general examples of the Distributive Property appear in Lesson 6-8, whose title is "Why the Distributive Property Is So Named." Examples of distributing a negative value appear in the next lesson, "Subtracting Quantities." All three of these Chapter 6 lessons contain equations to solve, since the U of Chicago text introduces solving equations before the Distributive Property.



Along with us Algebra 1A teachers, the Algebra 1B and Algebra II teachers continue to be concerned with how Edgenuity presents some of the lessons. In fact, when I explored some Edgenuity lessons, I observe that in the videos, fraction and decimal equations are solved directly without clearing the fractions or decimals first. But then the ensuing quiz asks students to identify the number by which they must multiply both sides to clear the fractions or decimals -- and there are two or three such questions on a ten-question quiz! The Algebra II teachers notice similar problems on their respective quizzes -- to the extent that they're seriously considering foregoing the Edgenuity quizzes and just creating and printing their own tests.



This just goes to show us that we teachers shouldn't blindly assume that Edgenuity is teaching all of the students properly. Throughout the entire process, we must continue to monitor the students to see whether they are actually learning -- or are the videos too confusing.



Especially since I still don't know whether I'll have an actual class this summer, I do want to look ahead to the upcoming school year. Of course, it's always possible that I'll be hired as a regular teacher in the fall. But assuming that my employment situation doesn't change, I'll be preparing the fifth year of this blog and the fourth year of posting U of Chicago Geometry.



This year, the digit pattern for pacing served me well. I liked how the digit pattern allows me to schedule lessons -- Lesson 2-1 on Day 21, Lesson 3-2 on Day 32, Lesson 3-4 on Day 34, Lesson 13-3 on Day 133, and so on -- without needing to think about it. If I were teaching a real class with real students, then I might want to be more careful with my pacing. And as my partner summer teacher informs me, the district encourages her to cover the chapters in a different order anyway. But on my blog, it's easiest for me just to post the lessons in order. So I'll continue the digit pattern next year.



Meanwhile, I'll continue to follow the calendar for my old district where I do still get occasional subbing calls, even though the lion's share of calls will be for my new district. I prefer letting the blog follow an Early Start Calendar (that is, where first semester finals are before winter break) rather than a near-Labor Day Calendar (that is, where the first day of school is within a week of that holiday).



On the Early Start Calendar, the first semester is usually slightly less than 90 days -- so that only seven full chapters are covered (with Chapter 7 on Days 71-80) rather than eight chapters. I find that the end of Chapter 7 is a natural semester dividing point. On the other hand, the end of the semester on near-Labor Day Calendars are unconstrained by winter break, and so that first semester ends on the actual Day 90 (which would force eight chapters into the first semester).



But there are a few differences between this year's district calendar and next year's. This year, the first semester contained 83 days, so we finished Chapter 7 (Lesson 7-8) on Day 78, reviewed two days for the final, and then we had the three official finals days.



This year, the first semester contains 85 days. This means that according to the digit pattern, Lessons 8-1 and 8-2 are before the final, Lessons 8-3 through 8-5 are blocked by the final (just as Lessons 8-1 through 8-3 were blocked this year), and the new semester begins with Lesson 8-6. I don't mind squeezing in Lessons 8-1 and 8-2 into the first semester, as these are easy lessons (which are on perimeter and tessellations). But Lesson 8-5, on the areas of triangles, is important. I definitely don't want to skip this lesson and force the students to begin with Lesson 8-6, on the areas of trapezoids, right after returning from winter break. An interesting way to combine Lessons 8-5 and 8-6 is to think of a triangle as a trapezoid with one of its bases having length zero -- then the trapezoid area formula reduces to the triangle area formula.



In this district, students always return from winter break on a Tuesday, since Monday is a PD day for teachers only. So Days 86-89 are on Tuesday through Friday, and then Day 90, the Chapter 8 Test, lands on a Monday. I know that Monday tests are tough since students forget stuff over the weekend, but it beats the alternative of squeezing Chapter 8 even shorter. I have plenty of time to decide how exactly I'll teach Chapter 8 this year. Once again, I could just switch to the other district calendar, where there's plenty of time to teach Chapter 8 in the first semester. (After seeing my partner teacher delay Glencoe Chapter 6 to the second semester of Algebra I, this reminds me that I don't wish to end the semester with a tough chapter like 8).



Let's look at how other tests fit the pacing plan, besides the first semester final. This year, the PSAT occurred near the end of Chapter 3, which fits since Chapter 3 has only six sections. Also, I like the idea of not having so much algebra at the start of the Geometry course -- but Chapter 3 is a great time to teaching equations of lines (Lesson 3-4 on parallel lines) since it's right before the PSAT.



We see that in this district, the SBAC is given around Chapter 14 -- which isn't good because Chapter 14 material and Lesson 15-3 appear on the SBAC. There's not much I can do about this -- but I will move the Chapter 15 Test up to Day 160 and start the SBAC review on Day 161. This will allow me to cover all 34 questions on the practice SBAC (two a day on Days 161-177) before finals week begins with Day 178. Unfortunately, Day 160 is on a Monday next year, so this will be yet another Monday test. (Both Chapters 8 and 15 contain nine sections, so neither test can be moved up to the previous Friday.)



Oh, and there's one more problem to watch out for. This year, Pi Day is on Day 130, so that it's the date of the Chapter 12 Test. Giving a test interferes with the celebratory nature of Pi Day, but there's no alternative since Chapter 12 has ten sections, and then we're running up against the day that third quarter grades are due. (Then again, last year in my new district, some eighth graders had to take a test on Pi Day -- but that matters less since eighth grade isn't the year students learn about pi.)



It might be better to use the new Third Edition of the U of Chicago text next year rather than my trusty rusty old Second Edition. It fits the SBAC better, since there are only 14 chapters, and no Chapter 14 material appears on the state standardized test. Chapter 12 is shorter so we can avoid a test on Pi Day, but Chapters 3 (PSAT) and 8 (first semester finals) are worse in the new edition than in the old one. I'll probably stick to the Second Edition, since that's the text I've been using to blog lessons for years now.



I've been continuing to reflect upon yesterday's subbing, especially the sixth period incident. No, I really didn't break the third resolution on bring up past incidents, but now I worry that I might have broken the second resolution on arguing/yelling in class.



You see, before the boys started throwing rulers, I asked one girl to answer some short questions about the Quadratic Formula (such as how to find -b when b = -3). But she didn't want to answer -- she claimed that she already answered all the problems on the worksheet in her head. (The Quadratic Formula in your head -- really?)



I eventually write her name on the Bad List of names to give the regular teacher -- but I continued to talk to her about the incident anyway. Even though I don't think I yelled at her, I must admit that anything I said to her after writing her name down counts as arguing. As soon as I write down the name, the incident is over.



I suspect that by continuing the argument, it made me appear to lose control of the class -- at least in the eyes of the students. And since I was no longer in control, the kids felt that they could do whatever they wanted, including throwing rulers.



Also, I could have given the girl an incentive to do the work -- singing the song. I remember that she in particular wanted me to sing "Pop Goes the Weasel." I could have told her that I'd sing the song if she completed the work. Again, this naturally fit the lesson -- each time we started a new problem, I must set up the Quadratic Formula, and so I could sing the song. So if the class gets through more problems, then I could sing more renditions of the song. I embarrass myself in the eyes of blog readers -- I devote so many posts to music, specifically music in the classroom. But then when I have a chance to sing something, I don't use the songs to the fullest extent. A powerful use of my songs is as incentives for the kids to work.



The week between the end of regular school (in my old district) and the start of summer school (in my new district) is similar to spring break. My plan is to post twice during this break. One of these posts will probably be on Friday, June 15th -- the day we have access our summer classrooms. This assumes that the class isn't cancelled -- in which case I'll use those posts to announce the cancellation.

Music Post: More on 20EDL and Other Scales

Table of Contents



1. Introduction

2. Yet Another Traditionalists Post

3. Composing Music in 20EDL

4. Converting 20EDL to EDO Scales

5. Exploring 29EDO

6. What About 28EDO?

7. Converting EDO Scales to EDL Scales

8. Eighth and Sixteenth Notes



Introduction



This is officially my first summer post. I already wrote that my second summer post will be Friday, June 15th -- just ahead of the start of summer school. So I might as well make my first summer post today, so that both posts are on Fridays.



No announcement has been made regarding whether I'll actually have a summer class or not. And so I'll just use today's post to continue our discussion of various musical scales. Again, all of this is to prepare songs to play during music break in a possible summer Algebra I class.



Yet Another Traditionalists Post



You knew it was too good to be true. Just days after I give up on Barry Garelick and SteveH and blog about some other traditionalists, guess who makes a surprise post today!



https://traditionalmath.wordpress.com/2018/06/08/stop-me-if-youve-heard-this-dept/



It's almost as if Barry Garelick read this blog and suddenly remembered that he hadn't posted anything in a while. Anyway, let's see what Barry Garelick has to say today:



This article talks about a book titled “Systems for Instructional Improvement”   coauthored by the dean of the University of Southern California ed school.  It is described as  “dedicated to improving math instruction in the U.S.”
Why is it that just about every book, article, tweet, and Linked-In polemic that purports to put math education back on track starts from the following assumption:
“For the past 25 years or so, there’s been a growing recognition that students at the middle-school level, in particular, aren’t developing a deep understanding of mathematics,” said Thomas Smith, dean of UC Riverside’s Graduate School of Education. “A big piece of that is because of the way students in the U.S. are taught; current math instruction tends to be highly procedural — as in ‘use these steps to solve these types of problems’ — instead of allowing students to investigate real-life problems and experiment with different types of solution strategies.”


Implied in the quoted paragraph is that traditional math leads to students wondering "Why do we have to learn this?" and "When will we use this in real life?" -- so by switching to reform math and real-life examples, those questions would disappear. Well, let's see what Garelick says about that:



Singapore has boasted high scores on international tests for years, but the problems that students solve there may be held in disdain by math reform types. Perhaps they think they are not relevant to students concerns. If students are not given proper instruction on how to solve such problems, and are expected to discover “strategies” for solving, they will tend to ask “When am I ever going to use this in real life?”   If given proper instruction with scaffolded problems that are variations on the initial worked example, students generally will tackle such problems. The “When will I ever use this” question is generally an expression of frustration.



Hmm. On one hand, I must give Garelick credit for being a middle school teacher in an actual class with real students. If he can really make the "When will I ever use this?" disappear in his classroom by using traditional direct instruction, then I must accept this as fact.



On the other hand, Garelick himself would probably say that much of the "When will I ever use this?" frustration is due to lack of traditional math in elementary school, so that by the time the students get to him, they're already struggling to understand the new material. This implies that he could reduce "When will I ever use this?" even more if the students had better elementary texts (such as Singapore, mentioned in the above paragraph). This, of course, can't be proved, since Garelick has no power to change the elementary texts.



So far, Garelick's post has drawn two comments. Both of these posts are written by, well, the one you know who's been itching to comment for over a month now:



SteveH:

They NEVER explain what this magic “deep understanding” means. In middle school, the process culminates with Algebra I, where “deep” understanding has to do with individual mastery of the problem sets in a good textbook. I never hear them talk about success in those concrete terms because they think that homework and tests are not “authentic.” Real life is not authentic?



Homework and tests are real life?



SteveH:

A person complimented my son last week on his college graduation (one degree was in math) and said it was based on my fatherly philosophy of “the process is the product.” I don’t know where that came from. I never think that way and I don’t even know what it means. This caused me to look up the phrase and I found that I completely disagree – even though you can make it mean whatever you want. A good process may be statistically better, but parents and teachers should individualize the process to achieve better product results – results drive the process. It’s never the case where the product doesn’t matter.



Traditionalists like to claim that they want to "individualize the process" -- if some individuals learn better using traditional math, then they should be taught traditional math, not reform math. But once again, suppose the tables were turned, and traditional math was the dominant paradigm. How would the traditionalists treat those individuals who don't want to learn math traditionally? I reckon the traditionalists would just ignore them -- they'd be the ones saying "the process is the product."



I congratulate SteveH's son for graduating with his BS in math. Of course, SteveH will be the first to say that his son's success was due to his pushing the youngster to learn traditional math beyond what his secondary schools taught him. And SteveH would suggest that if only the schools would push traditional math, eighth grade Algebra I, and senior-year Calculus, then maybe four or five more of his middle/high school classmates would have earned their math degrees last week as well.



In fact, in his second comment SteveH proceeds to write more about his son's very early education:



SteveH:

Note that my involvement in the “Math Wars” started when my son was in preschool and the teacher told me that our school used MathLand, a product so bad it was wiped off the face of the internet. It was replaced in our schools by Everyday Math [U of Chicago elementary texts -- dw], which tells teachers to keep moving and to “trust the spiral” (process). If the product is not achieved, then blame students, parents, peers, society, not enough engagement, or whatever. Meanwhile, they never ask us parents of their best students what we had to do at home. This foolishness has been going on for decades now. It’s what they have been pushing. They own it. It disappears in traditionally-taught high school AP Calculus tracks – the only path to a STEM career. So let’s redefine it as STEAM so that everyone can be successful and they can ignore all individual lost educational opportunities because the process is the product.



SteveH tells us that his son actually enjoyed the traditional problem sets (p-sets) that he completed as a teenager. But once again, how many students sitting in his son's classes agreed that these traditional p-sets are enjoyable? For every "SteveH's son," there are four or five students who don't enjoy these p-sets, and one student who, when assigned a p-set, doesn't even look at #1 or #2. The students don't care about "individual lost educational opportunities" to earn a math or STEM degree -- they just to see something different in secondary math other than endless p-sets.



My upcoming summer school class consists of students who failed Algebra I during the year. Most likely, these students don't enjoy math and probably didn't do many of the p-sets -- otherwise they probably wouldn't have failed the class. And these are high school students -- not eighth graders. If they failed Algebra I as freshmen, imagine how they would have fared if they'd been forced onto SteveH's vaunted "AP Calculus track" and took eighth grade Algebra I.



Traditionalists will be happy that there's no time for projects in my class. But they hold online courses like Edgenuity in disdain. I'm the first to admit that Edgenuity isn't a perfect program, but I reckon that students are more likely to answer questions on the computer (even if they're just guessing) than they are to write down #1 or #2 from the p-set in the text.



By the way, since I'll need to supplement Edgenuity with some written material anyway, some traditionalists might wonder, why can't I use a text that they approve of, such as Dolciani? Well, that the particular Dolciani text that I purchased at the book sale two months ago isn't an Algebra I text or even Pre-Algebra, but more like pre-Pre-Algebra. The Order of Operations doesn't appear, and the Distributive Property is listed only as one of six "Properties of Multiplication" in Lesson 3-3, so there's not enough coverage there either.



Composing Music in 20EDL



Let's review the 20EDL scale -- currently the scale of the week:



The 20EDL scale:

Degree     Ratio     Note

20            1/1         green C

19            20/19     khaki C#

18            10/9       white D

17            20/17     umber D#

16            5/4         white E

15            4/3         green F

14            10/7       red F#

13            20/13     ocher G

12            5/3         white A

11            20/11     amber B

10            2/1         green C



http://www.haplessgenius.com/mocha/



10 CLS

20 N=8

30 FOR A=0 TO 6

40 B=4

50 X=A-INT(A/2)*2

60 IF X=0 THEN D=20 ELSE D=19

70 PRINT D;

80 L=RND(B)

90 SOUND 261-N*D,4*L

100 IF L>1 THEN FOR I=1 TO L-1:PRINT "   ";:NEXT I

110 B=B-L

120 IF B>0 THEN D=21-RND(11):GOTO 70

130 PRINT

140 NEXT A

150 PRINT 20

160 SOUND 261-N*20,16



Don't forget to click on "Sound" to turn it on!



This randomizer is based on C and C# chords. I've suggested 20:16:13 as a possible C major chord, but what chord can be built on C#? We probably should at least use note 16 so that the fawn minor third 19/16 can be played. But neither 20 nor 19 has a perfect fifth above it. At this point we might prefer to use a D minor chord (18:15:12), or even F major (15:12:10), as these are chords that at least sound more compatible with C major. (Of these, Dm is considered easier to play on the guitar.)



As for actual songs in 20EDL, I keep saying that I want to stay away from just converting ordinary (12EDO) songs to 20EDL, but this week I sang the Quadratic Formula song ("Pop Goes the Weasel") and it's tempting to convert this to the new scale. This song is a good song to convert to 20EDL -- the scale's lack of perfect fifth is less important since the note G isn't stressed in the song:



NEW

10 CLS

20 N=8

30 FOR X=1 TO 27

40 READ D,T

50 SOUND 261-N*D,T

60 NEXT X

70 DATA 20,8,20,4,18,8,18,4

80 DATA 16,4,13,4,16,4,20,12

90 DATA 20,8,20,4,18,8,18,4

100 DATA 16,12,20,12

110 DATA 20,8,20,4,18,8,18,4

120 DATA 16,4,13,4,16,4,20,12

130 DATA 12,12,18,8,15,4

140 DATA 16,12,20,12



Notice that this song is in 3/4 time, and so a measure consists of three quarter notes. This contrasts with the 4/4 songs that the randomizer produces.



By the way, if you want the randomizer to produce 3/4 music, then change the following lines:



40 B=3

160 SOUND 261-N*20,12



Recall that I prefer minor scales for most classroom music. Therefore I'm more likely to use 12EDL or 14EDL (or possibly the neutral 16EDL) than 18EDL or 20EDL.



Converting 20EDL to EDO Scales



Recall that we've been using the following Mocha program to convert EDL scales to EDO scales:



NEW

10 INPUT "EDL"; L

20 A=0:B=L

30 A=A+1: E=0

40 FOR X=L/2+1 TO L-1

50 C=LOG(L/X)/LOG(2)*A

60 D=INT(C+.5)

70 E=E+ABS(C-D)

80 NEXT X

90 IF E<B THEN B=E:PRINT A;

100 GOTO 30



If we input 20 into this program for 20EDL, then the following EDO's are produced:



1, 4, 5, 7, 8, 12, 13, 27, 29, 53, 72, 217, ...



Since 20EDL contains ten notes, nothing less than 10EDO should even be considered. Thus the first nontrivial EDO's in the list are 12EDO and 13EDO.



The presence of 12EDO in this list might be surprising. After all, 20EDL contains both 11 and 13, and both of these are poorly approximated in 12EDO. On the other hand, we see that the first few notes of 12EDO approximate the 20EDL scale. In fact, 20EDL begins gC, kC#, wD, uD#, wE, gF, rF#, and so playing C to F# in 12EDL sounds good enough,



In fact, every EDL has an EDO that approximates its first few notes. For 14EDL we have 9EDO, for 16EDL we have 10EDO, and for 18EDL we have 11EDO. We know that in terms of cents, the notes of an EDL are more spread apart as we ascend the scale. Thus for 14EDO, the first step 14/13 is 128.3 cents, a little less than 9EDO's step size of 133.3 cents. For 16EDO, the first step 16/15 is 111.7 cents, a little less than 10EDO's step size of 120 cents, and so on.



In the case of 20EDL, the first step 20/19 is 88.8 cents, the second step 20/18 (= 10/9) is 182.4 cents, and the third step 20/17 is 281.4 cents. Each of these is fairly close to the steps of 12EDO. We notice that 13EDO also appears in the list, since its step size (92.3 cents) is even closer to 20/19.



But since the step sizes of an EDL increases in cents as we ascend the scale, the higher notes of the 20EDL scale no longer match 12EDO or 13EDO. Thus 20/15 (= 4/3) is far from the corresponding size of 13EDO. Of course, the perfect fourth still sounds accurate in 12EDO, but two steps later is 20/13, which is no longer accurate in 12EDO. Thus 12EDO and 13EDO are accurate enough for the lower parts of the 20EDL scale, but not the entire octave.



So the first reasonable EDO's for 20EDL are 27EDO and 29EDO. We've already seen 27EDO as a good approximation for 14EDL. This tells us that 27EDO is reasonable enough for 13-limit, even though 17 and 19 aren't quite as good as 13 in 27EDO. Therefore we will now take a closer look at the next scale, 29EDO.



Exploring 29EDO



Here is a link to 29EDO on the Xenharmonic website. (All links will be dead after July.)



http://xenharmonic.wikispaces.com/29edo



29edo divides the 2:1 octave into 29 equal steps of approximately 41.37931 cents. It is the 10th prime edo, following 23edo and coming before 31edo.



29 is the lowest edo which approximates the 3:2 just fifth more accurately than 12edo: 3/2 = 701.955... cents; 17 degrees of 29edo = 703.448... cents. Since the fifth is slightly sharp, 29edo is a positive temperament -- a Superpythagorean instead of a Meantone system.



Let's see the link between 29EDO and Kite's color notation:



Combining ups and downs notation with color notation, qualities can be loosely associated with colors:

qualitycolormonzo formatexamples
downminorblue{a, b, 0, 1}7/6, 7/4
minorfourthward white{a, b}, b < -132/27, 16/9
upminorgreen{a, b, -1}6/5, 9/5
"jade{a, b, 0, 0, 1}11/9, 11/6
downmajoramber{a, b, 0, 0, -1}12/11, 18/11
"yellow{a, b, 1}5/4, 5/3
majorfifthward white{a, b}, b > 19/8, 27/16
upmajorred{a, b, 0, -1}9/7, 12/7
All 29edo chords can be named using ups and downs. Here are the blue, green, yellow and red triads:

color of the 3rdJI chordnotes as edostepsnotes of C chordwritten namespoken name
blue6:7:90-6-17C Ebv GC.vmC downminor
green10:12:150-8-17C Eb^ GC.^mC upminor
yellow4:5:60-9-17C Ev GC.vC downmajor or C dot down
red14:18:270-11-17C E^ GC.^C upmajor or C dot up


Once again, 29EDO isn't meantone, and so the major third isn't C-E (instead, it's C-Ev). We keep this in mind as we convert 20EDL to 29EDO:



The 20EDL scale (converted to 29EDO, starting on C):

Degree     Ratio     Note

20            1/1         C

19            20/19     Db

18            10/9       Dv (D-down)

17            20/17     D^ (D-up)

16            5/4         Ev (E-down)

15            4/3         F

14            10/7       F#

13            20/13     G^ (G-up)

12            5/3         Av (A-down)

11            20/11     A#

10            2/1         C



Here's a video of a song in 29EDO. Just as I don't want to do (but did anyway), the author just took a song in 12EDO and converted it to 29EDO:





What About 28EDO?



You might remember that two months ago, I posted an Easter song in 28EDO. The notes of this Easter song are based on the dates of Easter.



When we look at the list of good EDO's for 20EDL, we notice that both 27EDO and 29EDO make the list, but not 28EDO. Indeed, 28EDO hasn't made the list for any EDL at all. The reason for this is that 28EDO is inferior to both 27EDO and 29EDO for representing just intonation.



There is one interval that is approximated extremely well in 28EDO -- 5/4, the major third. A just 5/4 works out to be 386.3 cents, while nine steps of 28EDO make 385.7 cents. Indeed it's approximated better in 28EDO than in any lower EDO. Since 5/4 sounds good in 28EDO, its inversion 8/5, the minor sixth, must also play well in 28EDO. Furthermore, since nine steps makes up the major third, it follows that three steps are a trienthird (1/3 third), 14/13.



The reason that 27EDO and 29EDO are superior to 28EDO is that the first two EDO's have much better perfect fifths -- which are more important than thirds. This may seem strange when considering 20EDL since this EDL has a major third but no perfect fifth. Yet 20EDL does have a perfect fourth -- the inversion of a perfect fifth. And so 20EDL's perfect fourth sounds better in 27EDO and 29EDO than in 28EDO.



Moreover, all EDO's contain several copies of the simplest intervals -- when we consider all of the intervals of scale, not just the ones on the base. In fact, we see that 20EDL contains three perfect fourths/fifths -- Degrees 20/15, Degrees 18/12 (perfect 5th), and Degrees 16/12 (perfect 4th). On the other hand, the scale contains only two major thirds -- Degrees 20/16 and Degrees 15/12. Thus even in 20EDL, perfect fourths/fifths are more prevalent than major thirds.



In the end, this means that EDO's that approximate the perfect fifth well are more likely to show up in lists for approximating EDL's, even if they fail at other intervals. We see that 29EDO has an excellent perfect fifth -- its fifth is more accurate than any lower EDO, including 12EDO. On the other hand, 28EDO has a comparatively poor fifth.



The fifth of 28EDO is the same as that of 7EDO (that is, 28 is contorted in the 3-limit) -- 686.7 cents, much flatter than a just 3/2 of 702 cents. A circle of fifths contains only seven notes -- in other words, B-F is a fifth, and the difference between F-F# or Bb-B is tempered out. So in 28EDO, we must use ups and downs, since sharps and flats are tempered to the unison.



The fifth of 28EDO is so flat that the interval C-E (which means, by definition, four fifths reduced by two octaves) is flatter than a major third (even though 81/64 is sharper than M3). In other words, 28EDO is not meantone since it doesn't temper out the syntonic comma -- instead, the syntonic comma is mapped to -1 (negative one) step. This is one of the weird things that happen when we use an EDO with an inaccurate perfect fifth.



So why, then, did I choose 28EDO for the Easter song? The truth is, 28 = 4 * 7, and both 4 and 7 are important in calculating the Easter date. (The significance of seven is obvious -- the seven days of a week from Sunday to Sunday. I explained why four is important back in my Easter post).



In fact, suppose we define an Easter interval to be the difference between consecutive Easters. (This year Easter was on April 1st and next year it's on April 21st, so 20 is an Easter interval.) Then 28EDO is the only scale in which the inversion of an Easter interval is itself an Easter interval. For example, in 28EDO, the inversion of 20 is 8, which is an Easter interval. (Compare the consecutive Easters April 12th, 2020 and April 4th, 2021). But in 27EDO, the inversion of 20 is 7, which is not an Easter interval. (No consecutive Easters can be seven days apart, since both dates obviously can't land on Sundays in both years.)



Thus 28EDO is the best EDO for the Easter song, even though as a musical scale, it's inferior to its neighbors 27EDO and 29EDO.



Here's what Xenharmonic has to say about 28EDO. It's not as extensive as 27EDO and 29EDO, for reasons that I've made clear:



http://xenharmonic.wikispaces.com/28edo



Basic properties

28edo, a multiple of both 7edo and 14edo (and of course 2edo and 4edo), has a step size of 42.857 cents. It shares three intervals with 12edo: the 300 cent minor third, the 600 cent tritone, and the 900 cent major sixth. Thus it tempers out the greater diesis 648:625. It does not however temper out the 128:125 lesser diesis, as its major third is less than 1 cent flat (and its inversion the minor sixth less than 1 cent sharp). It has the same perfect fourth and fifth as 7edo. It also has decent approximations of several septimal intervals, of which 9/7 and its inversion 14/9 are also found in 14edo.



Converting EDO Scales to EDL Scales



When I wrote the Easter song, I played it in Mocha. But the Easter song is written in an EDO scale (28EDO), while Mocha can play only EDL scales.



Just as it's possible to convert EDL's to EDO's, the reverse conversion is possible. And we can write the program for finding the best EDL scale for a given EDO in Mocha itself:



NEW

10 INPUT "EDO"; L

20 A=0:B=L

30 A=A+2: E=0

40 FOR X=1 TO L-1

50 C=2^(-X/L)*A

60 D=INT(C+.5)

70 E=E+ABS(C-D)

80 NEXT X

90 IF E<B THEN B=E:PRINT A;

100 GOTO 30



Once again, we press the up-arrow for the ^ symbol in Line 50. This gives us the powers of two used in the given EDO. The list of EDL's begins with the trivial 2EDL and then counts up by two, since we're assuming that octaves are available.



Just like the EDL->EDO program, this EDO->EDL program runs forever, so we must press the Esc key in order to stop the program. It's also possible to have the program stop at 260EDL (since Mocha only goes up to Degree 260). Then we can actually have Mocha play the winning EDL scale, which we'll keep track of in variable W:



90 IF E<B THEN B=E:W=A:PRINT W;

100 IF A<260 THEN 30

110 PRINT

120 FOR X=0 TO L

130 C=2^(-X/L)*W

140 D=INT(C+.5)

150 PRINT D;

160 SOUND 261-D,4

170 NEXT X



Here I have the computer print the Degrees of each note in the winning scale. If you'd rather the computer print the Sounds of each note instead, then change Line 150 to print 261 - D rather than D.



For the simplest nontrivial EDO (2EDO), Mocha prints out the following list:



2EDO: 2, 4, 10, 24, 58, 140, ...



This means that 140EDL is the best scale in which to play the 600-cent tritone -- the only nontrivial interval of 2EDO. The Degrees played are 140-99-70 -- which means that the ratios 140/99 and 99/70 are being used to approximate this tritone. Notice that these are both rational approximations of sqrt(2) (the exact ratio of the tritone).



Here are the EDL lists for a few selected EDO's:



4EDO: 2, 4, 6, 10, 20, 44, 106, 232, ...

5EDO: 2, 4, 6, 8, 12, 24, 38, 54, 94, 132, 256, ...

8EDO: 2, 4, 8, 12, 14, 20, 44, 62, 74, 88, 170, 202, ...

10EDO: 2, 10, 14, 16, 26, 54, 164, 256, ...

12EDO: 2, 4, 6, 8, 16, 18, 20, 24, 64, 126, 232, ...

16EDO: 2, 8, 22, 24, 26, 42, 74, 214, ...

18EDO: 2, 16, 24, 26, 28, 84, 98, 246, ...

20EDO: 2, 10, 26, 28, 30, 32, 54, 94, 108, 112, 116, 164, 224, ...

23EDO: 2, 12, 32, 34, 36, 102, 122, 126, 188, 256, ...

28EDO: 2, 24, 38, 40, 42, 64, 74, 128, 210, ...

30EDO: 2, 16, 18, 40, 42, 44, 46, 54, 138, 148, 170, 228, 256, ...

31EDO: 2, 16, 40, 44, 46, 48, 60, 82, 136, 152, 170, 234, ...

32EDO: 2, 28, 46, 48, 50, 74, 156, 232, ...



In this list, we see that both 5EDO and 10EDO are approximated by 256EDL. This is logical, since 5EDO is a subset of 10EDO. Likewise 30EDL (another multiple of 5EDO) has as 256EDL as the best approximation. But 20EDL isn't approximated by 256EDL despite being a multiple of 5EDL, while 23EDL is approximated by 256EDL despite not being a multiple.



As for 12EDO, we see that it's approximated by 232EDL, as is 4EDO. Another multiple of 4EDO has its best approximation in 232EDL, namely 32EDO. But the other multiples of 4EDO aren't closely approximated by the 232EDL scale.



Here's what 12EDO looks like in 232EDL. Degree 232 itself is closest to F#/Gb, and so we simply fill in the other notes of 12EDO:



12EDO as approximated by 232EDL:

Degree     Note of 12EDO

232          F#/Gb

219          G

207          G#/Ab

195          A

184          A#/Bb

174          B

164          C

155          C#/Db

146          D

138          D#/Eb

130          E

123          F

116          F#/Gb



But only estimating macrotonal EDO's (that is, those less than 12EDO, where each step is more than 100 cents) is recommended. As the number of steps increases, the EDL's are less accurate. This is because the step sizes in EDL's (in cents) increase as we ascend the scale, while the step sizes in EDO's are all equal.



Since step sizes are smaller near the low parts of the scale, the assumption is that approximating an EDO by EDL is more accurate if the degree is large. The above list of EDO's (from 4EDO to 32EDO) shows only those where an EDL close to 256 appears in the list. Interestingly enough, more even EDO's appear than odd EDO's. (On the other hand, another way to determine the accuracy is to look at the variables E, for error, and B, for minimum errors. As it turns out, odd EDO's have smaller values of B/L -- the average minimum error for each step -- than even EDO's, thus implying that it's the odd EDO's that are more closely approximated.)



If a microtonal EDO (past 12EDO) must be played, then only 16EDO or 18EDO are recommended, since the accuracy quickly drops off from there. (If we wish to play 12EDO itself, then there's already a PLAY command available for this scale.)



This means that 28EDO is not a recommended scale to play, as the accuracy is not good. Yet I play the Easter song in this scale -- only because I have no instrument that can play 28EDO and is more accurate than Mocha. Again, the computer is designed to play EDL scales, not EDO scales. And according to the list, the best approximation to 28EDO is 210EDL. Therefore, the Easter song in Mocha is actually written in 210EDL.



Eighth and Sixteenth Notes



The Mocha randomizer that I've been using so far chooses among four different lengths for the notes, namely quarter, half, dotted half, and whole notes. But real songs -- ones which fit lyrics -- often need faster notes such as eighth and sixteenth notes.



In fact, sometimes I've used the randomizer, and was disappointed because Mocha randomly chose to play three whole notes in a row. It's difficult to sing lyrics over long whole notes. The song "Pop Goes the Weasel" above uses quarter, half, and dotted half notes (since it's in 3/4 time), but the other songs I coded in the past month (the Dren and Packet songs) used mainly quarter notes.



Well, here's a program that adds eighth and sixteenth notes to the arsenal:



NEW

10 CLS

20 N=8

30 FOR A=0 TO 6

40 B=4

50 L=RND(B)

60 BB=4

70 LL=RND(BB)

80 D=21-RND(11)

90 P=A*32+(4-B)*8+(4-BB)*L*2

100 D$=RIGHT$(STR$(D),2)

110 PRINT @ P,D$

120 SOUND 261-N*D,L*LL

130 BB=BB-LL

140 IF BB>0 THEN 70

150 B=B-L

160 IF B>0 THEN 50

170 NEXT A

180 PRINT @ 224,"20"

190 SOUND 261-N*20,16



Here's how this program works -- no, it doesn't randomly choose a length from 1 (the sixteenth note) to 16 (the whole note). This is because some note lengths, such as 5 (a quarter note plus a sixteenth note) or 7 (a double-dotted quarter note) are rare.



Instead, it chooses a note length L, from 1 to 4. But then whichever note length is chosen is itself divided into four parts, LL, which also ranges from 1 to 4. For example if L = 2, then this denotes a half note. Then LL = 1 indicates an eighth note (1/4 of a half note), LL = 2 is a quarter note, LL = 3 is a dotted quarter note, and LL = 4 is a half note. If L = 1, then sixteenth notes and dotted eighth notes are possible. If L = 4, then it runs like the old program, where LL = 1 is a quarter note and LL = 4 is a whole note.



The only rare note length that this produces is when L and LL are both 3. Then the total note length is 9/16 (a half note plus a sixteenth note). Hopefully, L = LL = 3 will be rare enough that this won't make too much difference. (Such a note will always be preceded or followed by a dotted eighth note to add up to three whole beats.) But if it causes to many problems, then we can add extra lines to make sure that L and LL can't both be 3. For example, you might change Lines 30-40 to:



30 FOR A=0 TO 13

40 B=2

180 PRINT @ 448, "";



This effectively changes the song to 2/4 time, and by doing so, this will also block all dotted half notes and whole notes (except for the very last note, which is always a whole note). Then again, this might be desirable if we have long lines of lyrics and we want to be sure to avoid long notes. (We must also change Line 180 -- the final note no longer prints at the wrong place, but fortunately we already know what the final note is so we don't need to print it.)



Lines 90-110 are there to print the Degrees on the screen in columns to imply the note lengths. The idea is that once we create a random song that we like, we write it down on paper and then write a new program that plays the song we created. Every two columns corresponds to a sixteenth note, and the string commands (using $, the dollar sign) are there to make sure that the Degrees print correctly.



The program is currently set up for 20EDL. We can change these to other EDL's as follows:



80 D=13-RND(7)

80 D=15-RND(8)

80 D=17-RND(9)

80 D=19-RND(10)



for 12EDL, 14EDL, 16EDL, 18EDL respectively. In the first line, RND(7) is a random number from 1 to 7, so 13-RND(7) is a random Degree from 12 to 6 -- the Degrees of 12EDL. As usual, N in Line 20 tells us which key the song is in. Larger values of N are lower keys -- the maximum value of N is chosen to that N times the size of the EDL is no larger than 260. Once again, for 12EDL, N can be as large as 21, while for 20EDL, the max value of N is 13. Oh, and if we change the program from 20EDL, then the 20 in lines 180 and 190 (for the final note) must be replaced with the proper value.



For this program, I dropped the rule that the notes alternate between chords (for example in 12EDL, the first measure starts with 12, 10, or 8 while the second measure starts with 11, 9, or 7). The program is hard enough to write without keeping track of the chords. We'll just have to figure out which chords to play on the guitar after the song is written.



Oh, and the following change again allows us to write songs in 3/4 time instead of 4/4:



40 B=3



Also, change the final 16 in the last line to 12 to make the last note sound as a dotted half note.



Now I believe I have all the code I need to create my own songs. My next post will be on Friday, June 15th -- the day I hope I'll get in my summer school classroom. If that happens, then in that post I'll write about one final EDL scale, as well as my plans for classroom management.

Van Brummelen Chapter 4: The Medieval Approach

Table of Contents



1. Summer School Announcement

2. Review: What Is Spherical Geometry?

3. Review: What Is Spherical Trig?

4. Van Brummelen Chapter 4: The Medieval Approach

5. Exercise 4.2

6. Exercise 4.3

7. Exercise 4.9

8. A Resolution of the Cliffhanger?



Summer School Announcement



Well, this is exactly what I feared would happen -- not enough students signed up for first semester Algebra I in summer school, and so I no longer have a class to teach this summer. And so let the Great Post Purge 2018 begin...



...just kidding! I already wrote that I won't delete any posts this year. And so all those posts I wrote over the past month about both Algebra I that I'd teach and the music that I'd play will remain posted, even though I don't have a class.



I'd actually run one of the Mocha programs to create a song just minutes before I received the email message that summer school was cancelled. The song would have been a redo of "Solve It," one of the last songs I played at the charter school -- since after all, one of the earliest lessons in the Algebra I class is solving equations.



Meanwhile, I was originally planning on writing about one last scale this week, 22EDL. But since I'm not teaching a class, there's no reason to write about 22EDL any more. And my original original plan (that is, from back in April before the possibility of summer school even arose) was to write about EDL scales throughout the summer. I decided that I've already written enough on EDL scales over the last month that I don't need to spend the summer writing about music some more.



And so instead, today I'll begin my other other plan for this summer -- returning to spherical trig. Last summer, if you recall, we began Glen Van Brummelen's Heavenly Mathematics: The Forgotten Art of Spherical Trigonometry. Last year we covered the first three chapters, and so this year my hope is to finish the book with Chapters 4 through 9.



Spherical trig might not be compatible with Euclidean geometry -- but hey, at least it is Geometry, as opposed to the Algebra I and music topics that have dominated this blog for a month now.



Today's chapter is on the medieval approach to spherical trig. During this time, Europe was stuck in the Dark Ages, and so many of the mathematicians who were active this time were Muslim. It's only fitting because depending on the moon sighting, today is Eid al-Fitr, one of the most important holidays in the Islamic calendar. In fact, some school districts (such as New York, observing the holiday for the second time in 2018) are closed today for Eid al-Fitr, just as many districts are closed on Jewish holidays.



Review: What Is Spherical Geometry?



This is what I wrote last summer about spherical geometry and spherical trig. Today I begin by quoting Chapter 2 of Van Brummelen, which is the chapter in which he introduces a little bit of spherical geometry. I wrote:



Van Brummelen assumes that we're already familiar with earth's surface, and so he devotes part of this chapter to describing the celestial sphere. He writes:



"We've already seen the most obvious feature of the celestial sphere, namely its daily rotation around us. Given the sphere's unfathomably large size, rendering the Earth as an infinitesimal pin prick at its center, one can only imagine how quickly it is actually moving."



The author defines several key terms here:



-- The celestial equator rises from the east point of the horizon and sets in the west.

-- The ecliptic is the path the sun takes as it makes a complete circuit around the celestial sphere.

-- The obliquity of the ecliptic, symbolized by the Greek letter "epsilon," is the tilt between the celestial equator and the ecliptic. Its current value is 23.44 degrees.

-- The equinoxes are the two points where the celestial equator and the ecliptic intersect.

-- The summer solstice is the most northerly point on the ecliptic, halfway between the equinoxes.



I'll quote -- but not prove -- the some key theorems of spherical geometry. Again, you can go back to last year's posts to relearn these theorems. Better yet, you can go back to posts from previous summers, when I first started writing about spherical geometry. (Oh, and by the way, those old posts from previous years were originally based on the writings of 19th-century French mathematician Adrien Legendre.)



Theorem:

Every cross-section of the sphere by a plane is a circle.



Lemma (Triangle Inequality):

The third side of any spherical triangle cannot exceed the sum of the other two.



Theorem:

The sum of sides in a spherical triangle cannot exceed 360 degrees.



Theorem:

The polar triangle of a polar triangle is the original triangle.



Polar Duality Theorem:

The sides of a polar triangle are the supplements of the angles of the original triangle, and the angles of a polar triangle are the supplements of the sides of the original.



Theorem:

The angle sum of a triangle must exceed 180 degrees.



Review: What Is Spherical Trig?



And in the following chapter, Van Brummelen introduces spherical trig:



Van Brummelen begins by writing about Hipparchus of Rhodes, the founder of trigonometry. But unfortunately, not much of this Greek scholar's work survives today. Van Brummelen proceeds:



"We must therefore move more than two centuries ahead, to a figure almost as elusive as Hipparchus. We are aware that Menelaus of Alexandria lived in Rome in the late first century AD because Ptolemy tells us he made some observations there, but that is all we know."



At this point Van Brummelen states and proves the key theorems of the book -- the theorems of ancient Greek mathematician Menelaus, which have both plane and spherical versions. Again, let me only state the main theorem and all lemmas used to prove it -- I won't prove the theorem again:



Menelaus's Plane Theorem: In figure 3.2, AK/KB = AT/TD * DL/LB.

Given (I insert a given section here, since you can't see figure 3.2):

A-D-T (that is, D is between A and T), A-K-BDB and TK intersect at L



Lemma A: In figure 3.5, AB/BC = sin alpha/sin beta.

Given: A-B-C, with A and C on the surface of the sphere (so B is interior to the sphere). A vertical diameter is drawn through B. The arcs drawn from A to the diameter and C to the diameter are labeled alpha and beta, respectively.



Lemma B: In figure 3.6, AC/AB = sin alpha/sin beta.

Given: A-B-C, with B and C on the surface of the sphere (so A is interior to the sphere). A horizontal diameter is drawn through A. The arcs drawn from C to the diameter and B to the diameter are labeled alpha and beta, respectively.



Menelaus's Theorem A: sin AZ/sin BZ = sin AG/sin GD * sin DE/sin EB.

By the way, figure 3.4 is the same as 3.2, except there are additional arcs on the sphere. Point Z is now on Arc AB, and G is chosen so that D is on Arc AG, Arcs DB and GZ intersect at point E. (Once again, this is much easier seen than described.)



Menelaus's Theorem B: sin AB/sin AZ = sin BD/sin DE * sin GE/sin GZ.



Van Brummelen then leaves Menelaus and describes al-Kumi, an Iraqi mathematician. (Yes, here we begin with our Islamic mathematicians.) I don't wish to repeat everything I wrote last year, so again I'll state only the givens and the goal as he applies the Menelaus result to the celestial sphere:



  • The equator and the ecliptic intersect, as always, at Aries.
  • The equator intersects the horizon (another great circle) at E. The arc from Aries to E is measured as theta.
  • The Sun is always along the ecliptic, and it's rising on the horizon.
  • We drop a perpendicular from Sun to horizon at point M. As always, the arc from Sun to M is given as delta (declination).
  • The arc from E to Sun is given as eta.
  • The arc from E to M is given as n. As always, the arc from Aries to M is given as alpha (right ascension), therefore theta + n = alpha.
  • The arc from Sun to M can also be extended upward to N, the North Pole.
  • Arc NGZ is the "equator" to the "pole" at Aries. (Recall that on the sphere, all great circles have "poles," and all points have "equators.") Point G lies on the ecliptic (so the arc from Aries to G must be 90) and Z lies on the equator (so the arc from Aries to Z must be 90), and as before, the arc from G to Z must be epsilon (obliquity of the ecliptic).
  • Arc NHCQ is the "equator" to the "pole" at E. Point H lies on the horizon (so the arc from E to H must be 90), C lies on the ecliptic, and Q lies on the equator (so the arc from E to Q is 90). The arc from N to H is labeled as phi (which equals our latitude).


And here is the final result:



sin 90/sin MZ = sin 90/sin(90 - lambda) * sin(90 - delta)/sin 90



or sin MZ = cos lambda/cos delta. Van Brummelen explains that theta = Arc ZQ = Arc MQ - MZ.



Again, you can return to last year's posts for further review. For now, let's start the new chapter.



Van Brummelen Chapter 4: The Medieval Approach



This is what Van Brummelen writes in Chapter 4 of his book, "The Medieval Approach":



"Reading al-Kuhi's statement defending the advantages of Menelaus's Theorem in the previous chapter is a bit like eavesdropping on someone holding a telephone conversation. We have a rough idea of what was said, but important parts of the debate are a blank to us. We are never told the name of the advocate of the new theorem, nor even what the new theorem was."



So in this chapter, the author writes about some of the theorems of al-Kuhi's contemporaries. We begin with Abu Nasr, the discoverer of the polar triangle. He proposes two theorems in his Book of the Azimuth (which unfortunately no longer exists):



Rule of Four Quantities: sin BD/sin CE = sin AD/sin AE.

Abu Nasr's Second Theorem: sin DF/sin EF = sin AD/sin AB.



Here are the givens (in Figure 4.1): both BD and CE are perpendiculars to ABC that intersect at F. It's been a while since we've thought about spherical geometry, so now's a good time to remember that perpendiculars to the same line aren't parallel on the sphere -- indeed, no lines are parallel in spherical geometry. So this is how both BD and CE can be perpendicular to the same line yet intersect at F. In fact, we call point F the pole of line ABC.



Van Brummelen states that the proof of this theorem is easy:



"At first it appears that these theorems are nothing more than corollaries to Menelaus, and in a mathematical sense they are."



Proof of the Rule of Four Quantities: Apply Menelaus's conjunction theorem to figure 4.1: we get 1/sin CE = 1/sin BD * sin AD/sin AE. QED



At this point the author gives us a trick to help remember some of these spherical theorems:



"The Rule of Four Quantities is also our first example of the principle of locality. Imagine a spherical triangle shrinking in size until it almost vanishes. As it gets smaller it begins to resemble a plane triangle; and when it is very small, it almost becomes one. Therefore any statement about a spherical triangle, applied to a triangle shrinking to nothingness, becomes a statement about a plane triangle. In our case, imagine the configuration of figure 4.1 shrinking until it is so small that the sides are almost straight. In radian measure, as x -> 0, the value of sin x essentially becomes x itself. So replacing the sines of the arcs in the Rule of Four Quantities with the arcs themselves, we find that for two nested right triangles, the ratios of the altitudes to the hypotenuses are equal. It's similar triangles."



This explains why, as we proceed through this chapter and the book, many of the spherical theorems look just like Euclidean theorems with lengths replaced with sines. Again, very small triangles on the sphere look like Euclidean triangles -- which is why if we draw a small triangle on the spherical earth, the triangle looks Euclidean, with the sum of its angles being essentially 180.



At this point, Van Brummelen shows how the Rule of Four Quantities can be used to prove al-Kumi's rising times problem -- the problem I quoted again in the last section of this blog. Indeed, figure 4.2 is a repeat of the figure I mentioned earlier (the equator and ecliptic intersecting at Aries, etc.), and so I don't need to repeat the givens. Let's just jump into the proof:



Solution of al-Kuhi's rising times problem, using the Rule of Four Quantities:

(1) We begin with figure AriesSunGZM, from which we find sin delta/sin lambda = sin epsilon/1, so sin delta = sin lambda sin epsilon.

(2) Use figure ESunHQM, from which we have sin delta/sin eta = sin(90 - phi)/1, or sin eta = sin delta/cos phi.

(3) Use figure NSunMQH, which gives sin(90 - eta)/sin(90 - delta) = sin MQ, or sin MQ = cos eta/cos delta.

(4) Finally, use figure NGZMSun to get sin(90 - lambda)/sin(90 - delta) = sin MZ/1, or sin MZ = cos lambda/cos delta.



"Just as before, theta = MQ - MZ, and we are done. There is no doubt about it: the Rule of Four Quantities is much easier to apply, and we get results much more quickly. Menelaus and al-Kuhi didn't stand a chance."



Van Brummelen now proceeds to derive a better-known theorem -- the Spherical Law of Sines. The following proof is due to Abu 'l-Wafa, the author of the Almagest (not to be confused with Ptolemy):



Derivation of Spherical Law of Sines, given Triangle ABC (figure 4.3):

Choose C to be one of the vertices, so that its perpendicular projection onto the opposite side AB lands between A and B, at D. [In other words, let C be the largest angle -- dw]. Let EZ be the equator corresponding to pole A, and let HT be the equator for pole B; extend the sides of the original triangle as shown. [Recall that just as every line has two poles, every point is the pole of an equator. I point out that by definition of "equator" of a point, several right angles are formed -- AEZ, AZE, BHT, as well as BTH. Also, sides AE, AZ, BH, BT are quadrants.] Then apply the Rule of Four Quantities to two configurations, both involving CD. Firstly, on ACZED we get



sin CD/sin B = sin EZ/sin AZ, or sin CD = sin A * sin b.



Secondly, on BCTHD we get



sin CD/sin a = sin TH/sin TB, or sin CD = sin B * sin a.



Combine the two equations and eliminate the shared term sin CD. A little juggling results in



sin a/sin A = sin b/sin B.



But we could have started the argument equally well with any of the three vertices, not just C. If we had applied it to A, for instance, we would have ended up with sin b/sin B = sin c/sin C. Combining these two results, we are left with the breathtakingly simple



Spherical Law of Sines: sin a/sin A = sin b /sin B = sin c/sin C.



It goes without saying that Van Brummelen compares this to the Euclidean Law of Sines:



Planar Law of Sines: a/sin A = b/sin B = c/sin C.



Again, the author uses the principle of locality to explain why the spherical law looks like the planar law with sines of the sides rather than the lengths of the sides.



At this point, Van Brummelen takes an aside. He informs us that even though most mathematicians working during this time were Islamic, there were a few working on spherical trig in India as well.



The author gives one example of Indian trig here -- finding declinations of arcs of the ecliptic. Most of the problem is in stating all the givens from the diagram (figure 4.4):

  • The equator and the ecliptic intersect, as always, at Aries.
  • The Sun lies on the ecliptic. The foot of perpendicular dropped from Sun to equator is A. Arc lengths are as follows: AriesSun = lambda, AriesA = alpha, SunA = delta.
  • B lies on equator, and C lies on ecliptic, each a quadrant away from Aries. Thus BC is epsilon, the angle of the ecliptic (about 23.4 degrees).
  • O is the center of the sphere.
  • Other points are in the interior of the sphere, the plane through the center and equator. The foot of perpendicular from Sun to this plane is D. The foot of perpendicular from D to OAries is E, and the foot of perpendicular from A to OAries is F. And finally, the foot of perpendicular from C to OB is K.
Van Brummelen proceeds as follows:


The two right triangles SunED and COK, called the "kranki-setras" or "declination triangles," are similar since they share the angle epsilon between the planes of the equator and the ecliptic. Therefore


SunD/SunE = CK/CO.


But SunD = sin delta (to see why, consider the vertical circular segment ODASun) and similarly SunE = sin lambda and CK = sin epsilon. CO is the radius, so it is equal to 1, and the standard formula sin delta = sin lambda sin epsilon follows. (He omits the derivation of alpha here.)


Now the author returns to Arabic culture. He points out that two of the Five Pillars of Islam require knowledge of astronomy -- the Ramadan fast, and the five daily prayers. Since today, after all, is Eid al-Fitr (that is, the end of fasting), let's look more closely at Ramadan. Van Brummelen writes:


"Consider the monthly fast. The Arabic calendar is lunar, so each month begins when the lunar crescent reappears from behind the Sun after New Moon. Miss the crescent on a particular day, and you may end up violating the fasting requirement unawares. Muslim scientists worked hard attempting to predict the first appearance of the lunar crescent, with varying degrees of success."



Indeed, missing today's Shawwal new moon is just as bad as missing the Ramadan new moon, since just as fasting is required on 1st Ramadan, fasting is forbidden on 1st Shawwal. So this shows how important it was for medieval Islamic mathematicians to predict the crescent dates.



But Van Brummelen now turns to the final pillar -- the daily prayers. As he explains:



"When the moment occurs, worshipers are enjoined to face the Ka'ba, the most sacred site of Islam. The Ka'ba, a cubical building (figure 4.5) that houses the Black Stone, is the destination of the pilgrimage that Muslims are asked to embark upon once in their lives. The direction of the Ka'ba -- the qibla -- serves several purposes besides the daily prayers, including determining the direction in which Muslims should face when they are buried."



So let's begin the calculations:



"On the face of it the qibla does not seem difficult to calculate. Since the positions of both Mecca and the worshiper are given, we know the local latitude phi_L, phi_Mecca = 21.67 degrees, and the difference in longitude. So we would seem to have a right triangle on the Earth's surface with values of the two sides adjacent to the right angle (figure 4.6)."



But, as the author points out, the side of this "triangle" isn't a line (that is, a great circle), since parallels of latitude aren't great circles. By the way, notice what effect this fact has on the qibla -- it means that Muslims on the same latitude as Mecca (that is, 21.67N) do not face due east or west during prayer, even though the holy city lies due east or west of them. (The shortest path between two points directly east or west of each other is in general not due east or west.) And it's even possible that a worshiper north of 21.67N might even have to face slightly north of east (or west) to Mecca.



Anyway, Van Brummelen performs the calculations for Ghazna, Afghanistan. Let's start with his description of the givens in the diagram:



"To give a reader a taste of ancient and medieval diagrams, we have reproduced al-Biruni's diagram in figure 4.7. Although it looks two-dimensional, appearances are deceiving. Imagine that you are looking directly down on Ghazna from above the celestial sphere. All the curves on the figure (even the two straight lines) are great circle arcs on the celestial sphere seen from above, so G is the zenith directly above Ghazna. The line connecting north and south through G, actually a great circle called the meridian of Ghazna, passes through the north pole P; the outer circle is Ghazna's horizon. M is the point on the celestial sphere that an observer at Mecca would perceive as the zenith. WM connects the west point on the horizon to M, and extends to A on the meridian. PMB is the meridian of Mecca."



And now let's start the calculation of the qibla:



Al-Biruni's geographical coordinates for Ghazna and Mecca were phi_L = 33.58 degrees, phi_M = 21.67 degrees, and a longitude difference of lambda = 27.37 degrees. Now phi_L is the altitude NP of the North Pole, the northernmost segment of Ghazna's meridian; but both NG and PC are 90, so GC = phi_L = 33.58. [Van Brummelen doesn't say this explicitly, but C is where Ghazna's meridian meets the equator -- dw] So the arc from the worshiper's zenith perpendicularly down to the equator is equal to the local latitude. This fact must also apply to the zenith of Mecca, so MB = phi_M = 21.67. Finally, the difference in longitude is equal to the angle at the North Pole between the two zeniths, so Angle MPG = BC = 27.37. Now that we have transferred all the data onto arcs in the diagram, we are ready to begin the actual mathematics.



We shall use nothing but the Rule of Four Quantities. Starting with configuration CAPMB we have



sin PM/sin MA = sin PB/sin BC, or sin(90 - phi_M)/sin MA = 1/sin lambda,



so sin MA = cos phi_M sin lambda, which gives the "modified longitude" MA = 25.29. Our second configuration is WMACB, from which we get



sin WM/sin MB = sin WA/sin AC or sin(90 - MA)/sin phi_M = 1/sin AC,



so sin AC = sin phi_M /cos MA, and we have the "modified latitude" AC = 24.11. Then GA = GC - AC = phi_L - 24.11 = 9.47.



With the modified longitude and latitude in hand, we turn out attention to the outer horizon circle for Ghazna, which is where the qibla resides. It will take two steps. Firstly, from WMASD [where S is the southern horizon, a quadrant south of Ghazna -- dw]:



sin WM /sin MD = sin WA/sin AS or sin(90 - MA)/sin MD = 1/sin(90 - GA),



so sin MD = cos MA cos GA, which gives MD = 63.10. Our final step applies the Rule of Four Quantities to figure GMDSA:



sin GM /sin MA = sin GD/sin DS or sin(90 - MD)/sin MA = 1/sin DS,



so sin DS = sin MA/cos MD. This gives us the qibla, because DS = 70.79 is the number of degrees west of south that we must turn to face Mecca.



Van Brummelen wraps up the chapter by describing the medieval qibla tables:



"The best of these tables was a set composed by Shams al-Din al-Khalili, an astronomical timekeeper employed by the Umayyad mosque in Damascus. Its sixteen pages contain almost 3000 entries of the qibla for every degree of latitude and difference in longitude for all Earthly locations that mattered. The effort involved must have been Herculean."



Exercise 4.2



Let's try some of the exercises from this chapter. I choose Exercise 4.2 -- and you'll find out why I selected it in a moment:



Repeat question 2 of chapter 3, but use only the Rule of Four Quantities.



Oh, now my choice makes sense. Last last we did Exercise 3.2, so this year we do Exercise 4.2. This is what I wrote last year about Exercise 3.2 (in preparation for today's 4.2):



Choose a particular date (say May 20), and a particular latitude (say 49.3 degrees N). Use Menelaus's Theorem to calculate the following quantities:



(a) the Sun's declination delta

(b) the ortive amplitude eta

(c) the equation of daylight n

(d) the rising time theta



First of all, I don't choose a latitude of 49.3N. Instead, I choose my home latitude, which is 34N.



For today's exercise, of course I'm changing the date to June 15th, today's date. (Notice that today is just a few days before the summer solstice.) And now we may begin our solution using the Rule of Four Quantities. Again, we refer to the description of figure 4.2 from earlier in today's post for the givens from the relevant diagram.



(a) We begin with figure AriesSunGZM, from which we find sin delta/sin lambda = sin epsilon/1. For this solution, I'll plug the values directly into the equation -- lambda = 82.9 degrees from the chart (it becomes 90 degrees around June 22nd-23rd, the summer solstice) and epsilon is always 23.4 degrees:



sin delta/sin lambda = sin epsilon/1

sin delta/sin 82.9 = sin 23.4

sin delta/0.99233 = 0.39714

sin delta = 0.3941

delta = 23.2 degrees



(b) Use figure ESunHQM, from which we have sin delta/sin eta = sin(90 - phi)/1.



sin delta/sin eta = sin(90 - phi)/1

0.3941/sin eta = sin (90 - 34)

0.3941/sin eta = 0.82904

sin eta = 0.47537

eta = 28.4 degrees



(c) Use figure NSunMQH, which gives sin(90 - eta)/sin(90 - delta) = sin MQ.



sin(90 - eta)/sin(90 - delta) = sin MQ

sin (90 - 28.4)/sin(90 - 23.2) = sin MQ

0.87978/0.91907 = sin MQ

0.95726 = sin MQ

MQ = 73.2 degrees

n = 90 - MQ (from last year)

n = 16.8 degrees



(d) Finally, use figure NGZMSun to get sin(90 - lambda)/sin(90 - delta) = sin MZ/1.



sin(90 - lambda)/sin(90 - delta) = sin MZ

sin (90 - 34)/sin(90 - 23.2) = sin MZ

0.82903/0.91907 = sin MZ

0.90204 = sin MZ

MZ = 64.4 degrees



theta = MQ - MZ

theta = 73.2 - 64.4

theta = 8.8 degrees

theta = about 35 minutes



Exercise 4.3



Prove Abu Nasr's second theorem using Menelaus.



To do this proof, let's repeat the givens and the goal from earlier:



Abu Nasr's Second Theorem: sin DF/sin EF = sin AD/sin AB.



Here are the givens (in Figure 4.1): both BD and CE are perpendiculars to ABC that intersect at F. It's been a while since we've thought about spherical geometry, so now's a good time to remember that perpendiculars to the same line aren't parallel on the sphere -- indeed, no lines are parallel in spherical geometry. So this is how both BD and CE can be perpendicular to the same line yet intersect at F. In fact, we call point F the pole of line ABC.



Van Brummelen makes this problem sound so simple. We take the theorem of Menelaus, plug in the given values, simplify the equation, and voila! The whole proof should take less than five minutes.



But I struggled with this problem for more than five hours. And no matter how many times I try it, I can never obtain sin DF/sin EF = sin AD/sin AB as a result.



Let's review how to get from Menelaus to the Rule of Four Quantities, since Van Brummelen implies that the proof here is similar. The Menelaus Conjunction Theorem for this problem is:



sin CF/sin CE = sin BF/sin BD * sin AD/sin AE



Since we're going to be finding the sine of everything, let's agree to abbreviate sine as "s," so we don't have to keep writing "sin" over and over:



sCF/sCE = sBF/sBD * sAD/sAE



Now since F is the pole of BC, both BF and CF are 90 degrees, and s90 (= sin 90) = 1. So we have:



1/sCE = 1/sBD * sAD/sAE



We multiply both sides by sBD to obtain the desired result:



sBD/sCE = sAD/sAE



Now we attempt to prove Abu Nasr 2. We recall that there's a second Menelaus theorem -- often known as the Disjunction Theorem:



sCE/sEF = sAC/sAB * sBD/sDF



Since the desired theorem has sDF/sEF on the left side, we multiply both sides by sDF and then divide both sides by sCE:



sDF/sEF = sAC/sAB * sBD/sCE



And we have sAB on the right side, so all we need to make appear is sAD. If we could somehow show that sAD = sAC * sBD/sCE, then we're done. Let's return to the Four Quantities and solve for sAD:



sBD/sCE = sAD/sAE

sAD = sAE * sBD/sCE



Oops -- we needed sAC there, not sAE. In fact, we can multiply the previous formula by sAE/sAE:



sDF/sEF = sAC/sAB * sBD/sCE * sAE/sAE

sDF/sEF = (sAE * sBD/sCE) * 1/sAB * sAC/sAE

sDF/sEF = sAD/sAB * sAC/sAE



And so we have an extra factor of sAC/sAE that we can't get rid of.



Our second attempt is to relabel the Menelaus diagram. We can interchange A with F and B with E to obtain a new Menelaus diagram. The two Menelaus formulas become:



Conjunction: sAC/sBC = sAE/sDE * sDF/sBF

Disjunction: sBC/sAB = sCF/sEF * sDE/sAD



Let's take the disjunction formula. We know that sCF = 1, so we have:



sBC/sAB = 1/sEF * sDE/sAD

sAD/sAB = 1/sEF * sDE/sBC



We now start working on the conjunction formula, aware that sBF = 1:



sAC/sBC = sAE/sDE * sDF/1

sDE/sBC = sAE/sAC * sDF



Let's plug this into the earlier equation:



sAD/sAB = 1/sEF * sDE/sBC

sAD/sAB = 1/sEF * sAE/sAC * sDF
sAD/sAB = sDF/sEF * sAE/sAC
sDF/sEF = sAD/sAB * sAC/sAE



And so once again, we have this extra factor of sAC/sAE that we can't eliminate.



By the way, you might wonder under what situation would we have sAC/sAE = 1. Well, if sAC = sAE, and AC = AE, then Triangle ACE would be isosceles. The Isosceles Triangle Theorem is valid in spherical geometry, and so Angle AEC = ACE, which is given to be 90. This would then force A to be the pole of CEF, so both AC and AE would be quadrants. In other words, sAC can't equal sAE unless both equal 1. And we know for a fact that no, angle AEC isn't intended to be a right angle, otherwise the Rule of Four Quantities would transform into the Rule of Three Quantities (sBD/sCE = sAD, as sAE would equal 1).



My next attempt involves the Spherical Law of Sines (even though we're not asked to use this to prove the theorem). In Triangle ABD, we have:



sAD/sABD = sAB/sADB



Now Angle ABD = 90, and so sABD = 1:



sAD/1 = sAB/sADB

sAD/sAB = 1/sADB



OK, so there's the right side of Abu Nasr 2. We notice that Angle ADB = EDF as these are vertical angles, and so let's try to apply Spherical Law of Sines to Triangle DEF:



sEF/sEDF = sDF/sDEF

1/sEDF = sDF/sEF * 1/sDEF



If we set 1/sADB = 1/sEDF, then we obtain:



sAD/sAB = sDF/sEF * 1/sDEF

sDF/sEF = sAD/sAB * sDEF



So this time, the extra factor we can't eliminate is sDEF. Notice that the only way to make this extra factor equal 1 is for Angle DEF = 90. Once again, it all works out only when A is the pole of CEF.



If we think about, we notice that our goal is to take an equation with six values (Menelaus) and wind up with an equation with four quantities (Abu Nasr 2). The only two values that are known to disappear to 1 are sBF and sCF. This implies that we need a Menelaus equation that connects the four values in Abu Nasr 2 (DF, EF, AD, AB) to BF and CF. Yet there's no way to rearrange either form of Menelaus so that it uses only these six values.



At this point, I give up. Instead, I attempted to find a counterexample. To do so, I wanted to choose values of each arc so that both Menelaus and Four Quantities are satisfied, yet Abu Nasr 2 isn't. Of course, I tried to use as many measures of 30, 45, 60, and 90 degrees as possible, since we know the sines of all of these angles.



I actually found values that satisfy both Menelaus Theorems and the Rule of Four Quantities and not Abu Nasr 2. This implies that Abu Nasr 2 can't be algebraically derived from Menelaus. But recall that we found two more Menelaus equations (the ones we found after we switched A-F and B-E), and the values I chose do not satisfy those equations. This means that, while I've found an algebraic counterexample, I don't have a geometric counterexample.



Of course, even if I stumble upon a proof of Abu Bakr 2, we still have a valid proof of the same equation with the sAC/sAE factor thrown in -- a factor which is never 1 except in a special case. And since Van Brummelen doesn't use Abu Nasr 2 at all in the rest of the chapter, it almost makes me believe that the author made a mistake -- that the correct Abu Nasr equation is:



sDF/sEF = sAD/sAB * sAC/sAE



Exercise 4.9



In this question we shall work through another of al-Biruni's methods for finding the qibla of Ghazna in The Determination of the Coordinates of Cities. The diagram is identical to figure 4.7, except that AMW is omitted, and GFH is drawn from G perpendicular to PM. BC = lambda = 27.37 degrees,

GC = phi_L = 33.58 degrees, and MB = phi_M = 21.67 degrees.



[By the way, I was considering changing this from Ghazna to my hometown to find my own qibla, but I don't know whether these methods work if Mecca is more than a quadrant away.]



(a) Use the Rule of Four Quantities on figure PGCBF to find FG.

sin FG/sin BC = sin PG/sin PC

sin FG/sin 27.37 = sin 56.42/1 (as PG is 90 - 33.58)

sin FG = sin 56.42 sin 27.37

sin FG = 0.38301

FG = 22.52 degrees



(b) Use the Rule of Four Quantities on figure EFPNH to find PE, and from it find ME.

sin FH/sin PN = sin EF/sin PE

sin 67.48/sin 33.58 = 1/sin PE (as E is the pole of HFG, implied but not stated)

sin PE = 0.59876

PE = 143.22 degrees (also implied by the diagram is PE > 90)

ME = PE - PM

ME = 143.22 - 68.33 (as PM is 90 - 21.67)

ME = 74.89 degrees



(c) Use the Rule of Four Quantities on figure EMFHD to find MD.

sin MD/sin FH = sin ME/sin EF

sin MD/sin 67.48 = sin 74.89/1

sin MD = sin 67.48 sin 74.89

sin MD = 0.8918

MD = 63.1 degrees



(d) Finally, use the spherical Law of Sines on Triangle PGM to find PGM, the direction of the qibla.

sin p/sin P = sin g/sin G

sin (90 - 63.1)/sin lambda = sin(90 - 21.67)/sin PGM

sin 26.9/sin 27.37 = sin 68.33 /sin PGM

sin PGM = sin 27.37 sin 68.33/sin 26.9

sin PGM = 0.94434

Angle PGM = 70.79 or 109.21 degrees



Technically speaking, PGM is 109.21 degrees. From the graph, this is the number of degrees west of north (since P is the North Pole). The other value of the arcsine, 70.79 degrees, is the number of degrees west of south. This matches the qibla value we found earlier in this post.



A Resolution of the Cliffhanger?



Last summer, at the end of my last Van Brummelen post, I wrote the following:



In fact, I believe I now have a proof of the following:



Theorem:

A line and its translation image are parallel.



Last year on the blog, I believed that I had a proof of this statement. But that proof is invalid -- and I know it because nowhere in the proof did I use the Fifth Postulate. So the proof would work in hyperbolic geometry -- yet the result is false in hyperbolic geometry. Once again, I use the third geometry to check the validity of a proof.



This post is getting long enough, and I definitely want to double- and triple-check this proof before I attempt to post it. But the following appears to be a preliminary lemma needed for the proof:



Lemma:

Suppose line l is the image of a reflection and l' is its image.

(a) If l and l' are parallel (and not identical), then there is exactly one possible position for the mirror.

(b) If l and l' intersect, then there are exactly two possible positions for the mirror.

(c) If l and l' are identical, then there are infinitely many possible positions for the mirror.



This lemma holds in both Euclidean and spherical geometry. Of course, case (a) doesn't apply in spherical geometry, so only (b) and (c) matter here.



Well, I'm now wondering whether I ever had a valid proof. But let me at least say a little about what I was thinking at the time.



The theorem to prove is:



A line and its translation image are parallel.



Now a translation, if you recall, is the composite of reflections in parallel mirrors. So we can rewrite the theorem as:



Given a composite of two reflections, if the mirrors are parallel, then so are the line and its image.



And this is equivalent, via the contrapositive, to:



Given a composite of two reflections, if the line and its image are nonparallel, then so are the mirrors.



But now let's recall what "nonparallel" is. Here I'm using the U of Chicago definition of "parallel," which allows a line to be parallel to itself. (This is significant because it's easy to map a line to itself via a translation -- just slide the line in the direction of that line.) Thus, we must show that if the line and its image are distinct and intersecting, then the mirrors are also distinct and intersecting.



Here's the idea behind the lemma -- we let l be the original line and l" the image of the composite. We can now let l' be any line. Now we must find mirrors m and m' so that a reflection over m maps l to l' , and a reflection over m' maps l' to l". The lemma tells us how many possibilities for m and m' exist.



But this is where I'm stuck. I don't remember exactly what I was thinking last year -- and even if I did, that proof might never have been valid (which is why I didn't just post it right then last year).



Last year, I wrote that I like thinking about spherical trig because it reminds me how it feels to be a student in our math classes. I worked on Question 4.3, struggled, and gave up -- just as many students do when searching for proofs in our Euclidean geometry classes.



Of course, this last problem applies to Euclidean geometry, not (just) spherical geometry. I don't expect students to figure it out on their own -- the idea was for me to find the proof just once and use it as a new way to teach transformations and parallel lines in Geometry. But as of now, I don't have a valid proof. I will definitely continue to think about this problem.



I don't know when my next post will be yet -- it won't be Monday, since I'm no longer teaching a summer school class that day. So I return to the randomness of summer posting.



To any and all Muslim readers, have a blessed Eid al-Fitr!