I first found out about this new challenge by reading the blog of Fawn Nguyen, a fellow California middle school math teacher whose blog I've linked to in the past:
http://fawnnguyen.com/hashtag-mtbos30/
I’ve written two posts in the last 8 months. Naturally that means I’ll [try to] write 30 posts in the next 30 days. Anne Schwartz @sophgermain is entirely behind this. Hashtag MTBoS30.
So Hashtag MTBoS30 (i.e., #MTBoS30) is the name of the new MTBoS challenge. Apparently, all we math teacher have to do to participate is to post once a day on the blog for 30 days -- there's no weekly topic or anything like that (or requirements to comment on the three blogs above you) as there was for the real challenge. So although I'm a few weeks behind -- busy with my week-long subbing assignment, I had little time to check out other blogs -- all I have to do is declare my posts for the last 30 days of school (Days 151-180) to be my MTBoS posts and voila! There are my thirty posts!
Day 151 is a great post to begin my MTBoS challenge, because that's the day I found out I was hired to work at my new middle school. It was the day that the tone of my posts began to change, gradually away from a hypothetical Geometry class that I'm not teaching and towards an actual middle school class that I will be teaching. Indeed, many of my posts prior to that day were just cut-and-paste posts from last year. It seems cheating to cut-and-paste thirty posts from last year and call it an MTBoS "challenge" to post them again -- that wouldn't be very challenging at all. But from Day 151, most of my posts are completely new content (although ironically, I'm cutting and pasting something I wrote about music last year into today's post).
Besides the start of the transition from substitute to regular teacher, other topics I have discussed and will continue to discuss during the thirty MTBoS posts include:
-- Morris Kline's 1959 book Mathematics and the Physical World
-- PARCC questions released earlier this school year
-- The week (May 6th-13th) I spent subbing in an Algebra II/Integrated Math II class
Of course, it may be argued that these 30 posts still don't fulfill the challenge completely. I'm calling my Day 151 post on April 29th to be the first MTBoS post, but Fawn Nguyen didn't declare the challenge until the next day, April 30th. At the time she wrote about the challenge, my district had six weeks of school left and 6 * 5 = 30, but because there is no school on Memorial Day, there were only 29 days of school left at the time.
Moreover, it appears that the 30 days actually refer to 30 calendar days, not school days. In fact, the 30 days probably refer to the month of May. But May has 31 days, not 30. Most likely, the teachers didn't want the challenge to begin on a Sunday, but weekends are nonetheless included in the 30 days from May 2nd-31st.
In the calendar month of May, I will post only 21 times (Days 152-172). This, frankly, may be more than most of the other MTBoS participants -- I've seen so far that several posters (including Nguyen herself) are already a little behind on their 30 posts. (But of course, now we're sending the message to students that only working hard enough to earn 21/30 = 70%, the lowest possible C, is acceptable.) I could just start posting on weekends (I did, after all, post on a Saturday in January in order to fufill MTBoS requirements), but I won't.
Still, using my own counting, today is Day 162, the twelfth (out of 30) MTBoS posts. I could go back and add the label MTBoS to all my posts starting with Day 151, but I choose not to. Instead, I will use that label only for today's post and my usual MTBoS post during the last week of school
Chapter 19 of Morris Kline's Mathematics an the Physical World is called "Oscillations of the Air." I wrote about the oscillations of a spring, but now we will look at how air vibrates:
"I have never been able to fully understand why some combinations of tones are more pleasing than others, or why certain combinations of tones not only fail to please but are even highly offensive." -- Galileo
Even though it's Galileo who is quoted here, the Italian scientist has nothing to do with the solution to this musical mystery. Kline begins:
"Since the phenomenon of sound almost constantly engages out attention, it is not surprising that mathematicians and scientists include the subject in their investigations."
In fact, the first to investigate this were Pythagoras and his followers. Kline writes:
"They found further that the sounds emitted by two equally taut strings are harmonious when the lengths of these two strings bear simple numerical ratios to each other. Thus two strings, one of which is one-half the length of the other, give forth a harmonious combination of sounds. The musical interval between these sounds is called an octave."
In fact, this is one of my favorite topics, and I've written about it before. This is what I wrote last year about musical intervals (except that I changed a few references to a TV show that I watched last year to the book I'm now reading, Kline).
The next simple ratio mentioned by Kline is 3:2 -- that is, the longer string is half again as long as the shorter string. This interval is called a perfect fifth. The note a perfect fifth above C is G. So notice that we already have two of the three notes of the C chord, the C and the G.
The next simple ratio mentioned by Kline is 4:3 -- that is, the longer string is a third again as long as the shorter string. This interval is called a perfect fourth. The note a perfect fourth about C is F. This gives us the three Pythagorean intervals -- the octave, the perfect fifth, and the perfect fourth.
Musicians are familiar with the circle of fifths. We begin at C and move up a perfect fifth to G. Then we move up another perfect fifth to D, and then to A, then to E, and so on. After twelve steps on the circle of fifths, we return back to C again -- or so we are told. That is, twelve perfect fifths are declared to be equal to a whole number of octaves -- seven, as it turns out.
But what exactly are seven octaves? Notice that one octave is 2:1, two octaves are 4:1, three octaves are 8:1, and so on. Since these are ratios, they must be multiplied, not added, just as performing a dilation with scale factor 2 thrice gives a figure that is eight times the original. So seven octaves are the ratio 128:1.
Now let's try twelve perfect fifths. Each fifth is 3:2, so twelve of them would be 531441:4096, which is not exactly 128:1. Cross multiplying these ratios would give 531441 = 524288, which is false. And we shouldn't be surprised that twelve fifths don't actually equal seven octaves. Since combining musical intervals amounts to multiplying, breaking them up amounts to factoring -- and we can't factor the same interval in two different ways (i.e., as seven octaves and twelve fifths) for the same reason that we can't factor the same number in two different ways -- this would violate the Fundamental Theorem of Arithmetic. The "equation" 531441 = 524288 states that a power of three equals a power of two, which is impossible.
But in music, we declare 531441 and 524288 to be equal. So the ratio 531441:524288 is considered to be equal to 1. In honor of the ancient Greek mathematician, we refer to the ratio 531441:524288 as the Pythagorean comma. And it's because it's the twelfth power of three that is approximately a power of two that our octaves consist of twelve notes -- and why it's the twelfth fret on the guitar that gives us the octave. (Notice that computer scientists have their own "comma." They declare 1024 = 1000 -- that is, a power of two equals a power of ten -- in order to justify the names "kilobyte," "megabyte," etc.)
Kline mentions only the octave, perfect fourth, and perfect fifth, since these were the only intervals with which Pythagoras was concerned. But notice that these give us the notes C, F, G, while the C chord is actually C, E, G. So where does the note E come from?
This was actually discovered about 500 years or so after Pythagoras, by the musician Didymus. We notice that Pythagoras stopped at the ratio 4:3. The next natural ratio to consider is 5:4. This ratio is now known as a major third, and represents the interval from C to E. Then the full C major chord is the extended ratio 4:5:6.
Notice that there's already an E on the circle of fifths. Declaring E to be a major third above C is the same as setting four perfect fifths to equal the ratio 5:1 -- which, as we know, is impossible. Four fifths would actually be 81:16, so cross-multiplying gives us 81 = 80. The ratio of 81:80 is considered to be another comma -- Didymus's comma.
We are now ready to produce the full C major scale. We start with the three intervals of Pythagoras, the octave, perfect fourth, and perfect fifth -- C, F, G. On each of these three notes, we build a major chord: the C chord (C, E, G), F chord (F, A, C), and G chord (G, B, D). Putting these notes in order gives us the full major scale -- C, D, E, F, G, A, B, C.
I remember taking a piano class the summer after kindergarten. As it was only a beginners' course, the only major scale taught was the C major scale, which was played on the white keys. But I wondered to myself why one couldn't play a major scale beginning on notes other than C. The following December (either for my birthday or Christmas) I received a small electronic keyboard as a gift, and naturally I tried playing other scales. beginning on D, then E -- but none of them sounded like the proper major scale Do, Re, Mi, etc.
But one of these scales sounded almost right -- the scale beginning on G. The scale G, A, B, C, D, E, F, G sounded correct except for the last few notes. But I didn't know how to make the last part sound like the major scale. Disappointed, I started playing around with chromatic scales -- where I included the black keys as well as the white keys. But as I was still tantalized by the G scale, I sometimes started out by playing the part of the G scale that sounded right -- G, A, B, C, D, E -- and then switched to the chromatic scale. The black key between F and G is called F sharp, or F#, so what I played was G, A, B, C, D, E, F, F#, G, and I often played around with this "scale" for awhile.
Then one day, I accidentally skipped the F note, so what I ended up playing was G, A, B, C, D, E, F#, G. And what I played sounded exactly like the major scale that I had been seeking! And I still remember to this day how excited my seven-year-old self was to "discover" the G major scale! After this, I quickly realized that I could make all the other scales (the D scale, E scale, and so on) sound right by including some of the black keys as well as the white keys on my keyboard.
The next ratio to consider is 6:5. This interval is called a minor third. To produce a minor chord, we take the major chord C, E, G and replace the E with the black key just below E, called E flat, often rendered in ASCII as Eb. So the C minor chord is C, Eb, G.
Why do minor chords sound "sadder" than major chords? Recall that a major chord has its notes in the ratio 4:5:6. Well, a minor chord has its notes in the ratio 10:12:15 -- that is, 5:6 is now the lower ratio and 4:5 is now the higher ratio. Since 4:5:6 is simpler than 10:12:15, the 4:5:6 major chord sounds brighter than the 10:12:15 minor chord. Many popular songs on the radio nowadays tend to be about one of two topics -- falling in love and breaking up. The former songs tend to be in major keys, while the latter songs tend to be in minor keys. Over the past few decades, popular hits in minor keys have increased.
The second keyboard that I owned, when I was in the second grade, had several songs built in, including the classic Greensleeves. This song was in the key of A, but I noticed that it was based on A minor rather than A major, as it contained the chord A, C, E (whereas A major would be A, C#, E). I soon learned that there were minor scales as well as major scales, and tried to deduce what the A minor scale was based on the song. Unfortunately, the version of Greensleeves that was built into my keyboard wasn't truly in A minor, but a scale called the Dorian mode. Instead, I was under the misconception that A, B, C, D, E, F#, G, A was the A minor scale -- that is, A minor was just like G major in that it contained F# rather than F natural.
It was not until the third grade until I took private piano lessons. My third grade classroom had a piano, and the teacher was a pianist. So she taught me piano one a week after school until she left on maternity midway through the year, after which she referred me to her mother to continue my study of the piano. This was when I finally learned that the A natural minor scale was actually the notes A, B, C, D, E, F, G, A -- that is, it has no sharps or flats, just like C major. This minor scale is formed by taking the three Pythagorean intervals (octave, fourth, and fifth) and building a minor chord on each note, rather than a major chord.
Here are a few more links discussing major and minor scales:
http://music.stackexchange.com/questions/22236/why-am-i-always-sticking-to-minor-keys
which in turn links to:
http://www.nme.com/blogs/nme-blogs/the-science-of-music-why-do-songs-in-a-minor-key-sound-sad
OK, I could talk about music forever, but that's enough for now. Let's get back to PARCC. Question 19 of the PARCC Practice Test is on dividing a segment in 1-D:
19. Points X and Z are on a number line, and point Y partitions
This is the second PARCC question dealing with dividing a segment into parts. But there are several differences between Question 12 and today's question:
-- Question 12 was in 2-D, but today's question is in 1-D.
-- Question 12 used a fraction (3/4 of the way), but today's uses a ratio (5:7).
-- Question 12 used a simple division (quarters), but today's is more complicated (twelfths).
-- Question 12 used whole numbers for endpoints, but today's uses decimals.
-- Question 12 gives both endpoints, while this question gives the midpoint and the dividing point.
These last three points make today's question more complicated than Question 12, but at least we have the mitigating factor of the question being only on a 1-D number line, not a 2-D coordinate plane.
I have several things to say about this problem. First of all, we know that dividing a segment into parts doesn't appear in the U of Chicago text. I pointed out earlier that dividing a segment on the 1-D number line into parts could actually fit in Lesson 1-8 of the text (on 1-D figures), but this topic doesn't appear there.
Next, we see that after I discussed so much about how the PARCC questions tend to be about dividing segments into halves, quarters, and eighths, suddenly we have a question requiring twelfths! This means that the midpoint method I mentioned in that post doesn't work for this question. Moreover, the vector method is awkward when we are given a ratio (5:7) rather than a fraction (5/12 of the way).
So this leaves us with the center of gravity method of Lesson 11-4. We can calculate the necessary coordinate via a weighted average:
(1.3 + 1.3 + 1.3 + 1.3 + 1.3 + 1.3 + 1.3 + z + z + z + z + z)/12 = 3.8
(9.1 + 5z)/12 = 3.8
9.1 + 5z = 45.6
5z = 36.5
z = 7.3
As you can see, this problem is not easy. Even I was somewhat confused when reading this problem, with the ratio 5:7 looking so much like the decimals 1.3 and 3.8, plus it was so tempting to treat 3.8 as one of the endpoints instead of the dividing points. If this confused me, imagine how the students feel about this question!
Also, it's not obvious why the coordinate of X must appear seven times in the weighted average and the unknown coordinate of Z (which I simply labeled as lowercase z) only five times in order to indicate the ratio XY : YZ = 5:7. The best way to explain it is that since XY < YZ, Y must be closer to X than to Z, and it's the closer point that must appear more in the weighted average.
Here's another method that works when one endpoint and a dividing point are given, and we are to find the other endpoint: we want the ratio XY : YZ = 5:7 -- that is, if it's five steps from X to Y, it must be seven steps from Y to Z. Since XY = 3.8 - 1.3 = 2.5, each step must be 2.5/5 = 0.5. Thus seven steps are 7(0.5) = 3.5, and as the coordinate of Y is 3.8, Z must be at 3.8 + 3.5 = 7.3.
This method certainly appears easier for this particular question. The real problem is that for some questions this method is the easiest, but for others, such as Question 12, it isn't. To apply this method to Question 12, we must note that if we want P to be 3/4 of the way from M to N, this is telling us that if it's four steps from M to N, it is three steps from M to P. But we must still figure out how to determine how far one step is -- and the only straightforward way to do this is to consider the x- and y-coordinates of the endpoints separately. It might be easier to subtract one step from N than it is to add three points to M -- maybe.
By the way, the ratio 7:5 into which we are dividing the segment in this PARCC problem can also be seen as a musical interval. It is the next natural interval to consider after the minor third 6:5 -- 7:5 is known as the septimal tritone. The word "septimal" means seven, referring to the fact that this is a ratio involving the number seven. The word "tritone" refers to an interval comprising three whole tones, such as F-B (the first tone is F-G, the second is G-A, and the third is A-B). Ordinarily a tritone is considered to be dissonant -- the devil's interval -- because it is not one of the simple intervals that sound harmonious or consonant. But once we involve the number seven, the septimal tritone is sometimes considered to be consonant. (The septimal tritone is also known as Huygens's tritone -- Christiaan Huygens being a 17th century Dutch physicist Kline alludes to in today's chapter.)
Oh, that's right -- I never finished discussing Kline's chapter. Well, Kline explains that air vibrates in sine waves, and when the frequencies are in a simple ratio, these sine waves often reach a maximum or minimum at the same time, which leads to harmony. Then he writes that any periodic function can be written as a sum of sine waves via Fourier analysis, named for the 19th century French mathematician Joseph Fourier.
PARCC Practice EOY Question 19
U of Chicago Correspondence: Lesson 1-8, One-Dimensional Figures
Key Theorem: Betweenness Theorem
If B is between A and C, then AB + BC = AC.
Common Core Standard:
CCSS.MATH.CONTENT.HSG.GPE.B.6
Find the point on a directed line segment between two given points that partitions the segment in a given ratio.
Commentary: The U of Chicago text contains only one question that is anything like this PARCC problem -- Question 13 from Lesson 1-8. In this problem, students are given the coordinates of A and B and that AB = BC (that is, B divides
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