A 36-sided regular polygon has exterior angles of measure _____ degrees.
The key theorem is in Lesson 13-8 of the U of Chicago text:
Exterior Angles of a Polygon Sum Theorem:
In any convex polygon, the sum of the measures of the exterior angles, one at each vertex, is 360.
Thus in a regular 36-gon, each angle must measure 360/36 = 10. Therefore the measure of each angle is ten degrees -- and of course, today's date is the tenth. Let's draw a regular 36-gon in Logo:
repeat 36 [forward 10 right 10]
Today I subbed at the same middle school as Monday and Tuesday, but not the same class. Instead, it was an "Intervention" class -- which really means in-school suspension. The subbing assignment doesn't even begin until 11:00. One guy is sent there to do his work, mostly science and history. At lunch time, several more students enter. Two seventh graders come to do their math homework. They are working on measurement questions. The two of them have trouble with the area of a trapezoid -- a formula I can't find in their Big Ideas text, even though trapezoids appear in the exercises.
For the last class of the day (fourth period, according to the Thursday rotation at this school). I ended up covering an eighth grade Spanish class. (Today is Mother's Day in Mexico, Dia de las Madres.) I won't post "Day in the Life" since there's little true math and little true classroom management today.
We have reached the last chapter of the U of Chicago text. Typically, in my first post after the last chapter test, I would write about my most popular post of the previous year. This year, the top three posts by hit count are all summer posts. I've decided this year to highlight my most popular Geometry post, which is from Friday, October 13th (yes, Friday the 13th):
http://commoncoregeometry.blogspot.com/2017/10/lesson-4-2-reflecting-figures-day-42.html
This is Lesson 4-2 of the U of Chicago text, on reflecting figures. We've seen that many teachers --especially eighth grade Algebra I teachers -- are reviewing transformations ahead of SBAC. And so I'll reblog this lesson today.
Usually, I follow this up with a study of some released SBAC questions. This year, I want to give questions from the practice SBAC eleventh grade test -- even though this test will mix Geometry questions with Algebra I question. But that's OK -- creating review worksheets for Algebra I will help prepare me for my summer Algebra I class. I don't want to begin our SBAC review in the middle of the week, so I'll start it next week.
And so let me give today's Mocha music post before I reblog the reflections activity. The goal today is for me to compose some actual music in the scale of the week, 12EDL.
One thing that's unfortunate about Mocha is that it can't play more than one note at a time (unless we bypass BASIC using machine language). Today I'm going to write about the possible chords suggested by the notes, but harmony actually isn't possible with the SOUND command. That's a bummer, since EDL's just intonation is well-suited for harmony.
The 12EDL scale:
Degree Ratio Note
12 1/1 white A
11 12/11 amber B
10 6/5 green C
9 4/3 white D
8 3/2 white E
7 12/7 red F#
6 2/1 white A
The primary triad of this scale is the minor triad 12:10:8, which reduces to 6:5:4. The secondary triad of this scale consists of the remaining three notes 11:9:7.
By the way, notice that I'm writing these triads utonally. We usually write the ratios of triads otonally, so that 4:5:6 is a major triad and 10:12:15 is a minor triad. In utonal notation, these are reversed, so that 6:5:4 is minor while 15:12:10 is major. The utonal triad 11:9:7 becomes 63:77:99 when we rewrite it in otonal notation.
You might wonder why I'm so eager to code music in Mocha when it's so limited. Well, I think back to last year, when some of my music break songs flopped. I often chose random notes to make up the tunes, which made them sound bad when I tried to sing words to them. It's obviously much easier to sing parodies -- the work of creating a tune has already been completed, and so all I have to do is invent the lyrics. Thus the students often enjoyed the parodies more than my original songs. My goal is to have Mocha play the tune, so I can practice singing the lyrics. Once I've memorized my lyrics, then I can take out the guitar and practice playing the tune.
The 12EDL scale is a good one to compose music in. Since it's primary triad is minor, its sad sounds will fit in with the general mood of the class -- sad that they are forced to take a summer math class while all of their friends are sleeping in or enjoying the beach.
The general idea is that the guitar can alternate between the two chords 6:5:4 and 11:9:7. But this raises a question -- what does it mean to play 11:9:7 on a guitar fretted to the usual 12EDO scale rather than 12EDL?
If we just naively remove the colors, then the 11:9:7 chord simply becomes B-D-F# -- which, as you might notice, is itself a B minor chord. As I wrote yesterday, the amber B is Bd (B quarter-flat), which is about halfway between B and Bb. Perhaps it might be better to play Bb than B -- then the secondary chord becomes Bb-D-F#, a Bb augmented chord. We know that the augmented chord is dissonant (remember the train whistle?) and so it seeks to resolve to the consonant A minor. On the other hand, B minor is consonant in its own right.
I've decided to keep the note as B anyway. The note B is easier to play on the guitar than Bb, mainly because there's a B string, not a Bb string. For the B minor chord, we place fingers at the second fret of the A string (to produce B) and the high E string (to produce F#). With the D and B strings open, this leaves only the G string to finger. An interesting choice is to stop this string at the second fret as well, to produce A (a Bm7 chord). We avoid the index finger and use the other three fingers on the second fret in order to ease the transition between the Bm7 and Am chords.
If we prefer a pure Bm chord, we can use a bar chord. This time, we avoid the index finger when playing Am, and then slide these fingers up two frets each before placing the bar finger for Bm. In either case, Bm is easier to play than Bbaug. On a piano, I might play Bb instead so that the Bb can serve as a descending leading tone to A. (It might sound good to play B to form B minor but then switch to Bb just before A. But this is cheating -- we're using seven notes to play 12EDL, which is a six-note scale.)
Here's a program that creates a short random song using these parameters:
http://www.haplessgenius.com/mocha/
10 CLS
20 N=16
30 FOR A=0 TO 6
40 B=4
50 X=A-INT(A/2)*2
60 D=14-RND(3)*2-X
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=13-RND(7):GOTO 70
130 PRINT
140 NEXT A
150 PRINT 12
160 SOUND 261-N*12,16
Here's how this works -- there are seven measures, labeled from 0 to 6. The reason I number the first measure 0 rather than 1 is so that the even measures are associated with the even Degrees 12, 10, 8 (Am) while the odd measures are associated with the odd Degrees 11, 9, 7 (Bm). The first note of each measure is randomly chosen from the correct triad for the measure, then all remaining notes are chosen from the entire scale. The note lengths are chosen at random -- either a quarter note, half note, dotted half note, or a whole note. The final note isn't random -- it's the root note of the scale, played as a whole note. The Degrees (12 is low, 6 is high) are printed so that if I like a tune, I can record it and then write lyrics for it (since running the code again creates a different tune each time).
In practice, I can modify the tunes generated by the computer. For example, in certain tunes, the last note might sound better as Degree 6 rather than Degree 12. I might also wish to run the code twice and use the first tune for a refrain and the second tune as a bridge. The final note of the bridge doesn't necessarily need to be the root of the scale.
I'll play around with this program. There's still time before the summer class begins and I need to have a song prepared.
OK, here is my most popular lesson reblogged, including the worksheets:
Reflection Postulate:
Under a reflection:
a. There is a 1-1 correspondence between points and their images.
This means that each preimage has exactly one image, and each image comes from exactly one preimage.
b. If three points are collinear, then their images are collinear.
Reflections preserve collinearity. The image of a line is a line.
c. If B is between A and C, then the image of B is between the images of A and C.
Reflections preserve betweenness. The image of a line segment is a line segment.
d. The distance between two preimages equals the distance between their images.
Reflections preserve distance.
e. The image of an angle is an angle of the same measure.
Reflections preserve angle measure.
This postulate corresponds to Dr. Franklin Mason's "Rigid Motion Postulate," in the old version of his Lesson 3.1 last year. Since then, Dr. M has completely changed his Chapter 3 -- this is almost certainly because isometries ("rigid motions") aren't emphasized on the Common Core texts nearly as much as either of us thought they would when we first read the standards. Nowadays, Dr. M uses the classical definition of congruent polygons (i.e., equality of corresponding measures). He assumes SAS as a postulate (just as the mathematician Hilbert did a century ago), and uses Euclid's ancient proof to derive ASA. But for SSS, Dr. M still uses rigid motions to move one of the triangles into place (similar to the start of the U of Chicago proof) before using SAS and Isosceles Triangle Theorem to prove SSS.
Part a is a very important part of the Reflection Postulate. Without it, a point A could have two reflection images -- there could be two distinct points B and C such that the reflecting line m is the perpendicular bisector of both
According to the text, reflections preserve:
Angle measure
Betweenness
Collinearity
Distance
a nice little mnemonic for the students.
The first theorem of this chapter is the Figure Reflection Theorem:
If a figure is determined by certain points, then its reflection image is the corresponding figure determined by the reflection images of those points.
This theorem is used to conclude, for example, that if A' is the image of A, and B' is the image of B, then
But that is the sort of proof that I don't want to confuse students with. It's best just to use the informal proof given in the text and save formal proofs for later.
The text states that when a figure intersects the reflecting line, the image must intersect the reflecting line in the same point or points. This follows immediately from the fact that the image of a point on the reflecting line is the point itself.
But what I find interesting is the related statement -- if a figure intersects its reflection image, then it must intersect the reflecting line in the same point or points. This statement is false in general, but it's true if the figure to be reflected is itself a line. This fact helps us greatly -- for example, consider Question 21 from the text:
The reflection image of Triangle ABC is Triangle XYZ. Now Lines AC and XZ intersect at a point -- which we now know must lie on the reflecting line. And Lines BC and YZ intersect at a point -- which we now know must lie on the reflecting line as well. And those two points determine exactly one line -- the reflecting line! So all the student has to do is draw the line through the two points of intersection.
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