## Tuesday, May 15, 2018

### SBAC Practice Test Questions 3-4 (Day 165)

Today on her Mathematics Calendar 2018, Theoni Pappas writes:

The shaded regions's area is 10 sq. units. Find the area of Triangle ABC.

Here are the diagrammatic given and goal, converted to symbolic form: the region whose area we're given is a hexagon. It's formed by trisecting each side of Triangle ABC. If we label the points of trisection (using dashes to mean "between") as A-D-E-B, B-F-G-C, and C-H-I-A, then the hexagon with an area of 10 is DEFGHI.

It turns out that the key to solving this problem is to draw in the diagonals DG, EH, and FI. These three diagonals are concurrent -- indeed, they intersect at O, the centroid of the triangle. These three diagonals divide the hexagon into six small triangles (DEO, FOE, OFG, GHO, IOH, OID) -- and notice that there are three small triangles (ADI, EBF, HCG) that are part of the large triangle's area, but not part of the hexagon's area.

All nine of these triangles are congruent. This is proved first using SAS Similarity to show that the three triangles ADI, EBF, HCG are all similar to ABC with scale factor 1/3. Thus the three triangles are all similar to each other with scale factor 1 -- that is, they're congruent to each other. Then the Side-Splitting Converse is used to show that opposite sides of the hexagon are parallel. This allows us (using Alternate Interior Angles) to find many congruent angles -- including enough congruent angles to allow us to prove the other small triangles congruent by ASA.

So Triangle ABC has been divided into nine congruent nonoverlapping small triangles -- six of which form the hexagon whose area is known to be 10. All that remains is to solve a proportion:

6/9 = 10/x

Cross multiplying,

6x = 90
x = 90/6 = 15

Therefore the area is 15 sq. units -- and of course, today's date is the fifteenth.

Here's the thing about questions like today's Pappas question -- can we really be sure that the three diagonals DG, EH, FI are concurrent? If we know that they are, then the problem becomes easy because the hexagon's area is six-ninths of the triangle's. But we can't just draw a point O unless we prove that the point exists -- and if it doesn't, then the whole proof is invalid. I end up pulling my hair out just to prove that O exists, all to justify a proportion that takes 15 seconds to solve.

Notice that earlier, I referred to O as the centroid of Triangle ABC. So here's a different approach -- instead of assuming that DGEHFI are concurrent and intersect at a point O (which may or may not exist), just let O be the centroid of Triangle ABC. Here we assume a previously proved result -- namely that a centroid O exists, and that it lies 2/3 of the way along each median. So our subgoal now is to show that DG (and likewise EH, FI) must contain point O.

We're given that D lies 2/3 of the way from B to A, and G lies 2/3 of the way from B to C. So by Converse Side-Splitting, DG | | AC. Now let's draw in the median to Triangle ABC from B. This intersects side AC at the midpoint of AC, some point M (which is neither H nor I, by the way).

Now we apply forward (i.e., not the converse) Side-Splitting to Triangle ABM (or BCM). We have a line DG that is parallel to side AM, and so it must divide sides BA and BM proportionally. Thus the intersection of DG and BM must lie 2/3 of the way along median BM. But of course, the point 2/3 of the way along median BM is exactly point O. Therefore O lies on DG -- and a similar argument shows us that O lies on EH and FI. (It turns out that not only does O lie on all three diagonals, but it's in fact the midpoint of all three diagonals. In fact, a 180-degree rotation centered at O maps the hexagon to itself.)

It's time to start our second week of music posts. This week, we'll be exploring our next EDL scale, which is 14EDL.

The 14EDL scale:
Degree     Ratio     Note
14            1/1         red F#
13            14/13     ocher G
12            7/6         white A
11            14/11     amber B
10            7/5         green C
9              14/9       white D
8              7/4         white E
7              2/1         red F#

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

10 INPUT N
20 FOR D=7 TO 14
30 SOUND 261-N*D,4
40 NEXT D

This is a descending scale. To make the scale ascend, use:

20 FOR D=14 TO 7 STEP -1

This is our first EDL that contains the 13-limit. New Kite colors are need for the 13-limit -- emerald (otonal) and ocher (utonal). Since EDL's are utonal, the scale uses the note "ocher G."

We know that 12-EDL is now "The New 11-Limit Scale," but what about 14-EDL? Typically, limits are named for the largest prime they contain. Thus both 14-EDL and 16-EDL would both count as new 13-Limit scales. The term "odd-limit" refers to the largest odd number they contain. And thus 14EDL is 13 odd-limit, while 16EDL would be a 15 odd-limit scale.

Degree 13 is an ocher G, while Degree 26 is also ocher G an octave lower. Meanwhile, Degree 27 happens to be a white G. Thus the ratio between white G and ocher G is 27/26 -- about 65.3 cents. It's an interval that's about 2/3 of a semitone or 1/3 of a tone -- hence it's called a "tridecimal third-tone."

The Helmholtz-Ellis accidental for the tridecimal third-tone is 2/3 of a sharp sign, since it's about 2/3 of the semitone that an ordinary sharp represents. This symbol resembles a capital "H" in ASCII, but on the blog I'll use a lowercase "h." So Degrees 13 and 26 can both be written as "Gh." As I wrote earlier, it's redundant to write "ocher Gh" -- either "Gh" or "ocher G" is acceptable.

But for some, calling the note "ocher G" is awkward. Since the tridecimal third-tone is more than half of a semitone, ocher G is closer to white G# than white G. Thus some might argue that this note should be called "ocher G#." Kite himself leaves it open whether Gh should be "ocher G" or "ocher G#," just as he leaves it open whether Bd is "amber B" or "amber Bb."

I've decided to call it "ocher G," since my rule is to use the color to represent all Helmoltz-Ellis accidentals, leaving only classical accidentals in the name. You might wonder why H-E then calls this note Gh rather than G#* (with * replaced by some symbol for a partial 13-limit flat).

Well, here's the reason -- the symbol "h" denotes the 27/26 interval. The note white G# must be seven perfect fifths about G. Thus G# is (3/2)^7 = 2187/128 above G, or 2187/2048 octave-reduced. This must be lowered to the note that is 27/26 above G -- the ratio works out to be 1053/1024. Even though this is only 48.3 cents (as opposed to the 65.3-cent third-tone), 1053/1024 is undoubtedly a much more complex interval than 27/26. Therefore H-E defines an interval for 27/26, not 1053/1024, and so the note is "Gh" or "ocher G."

Unfortunately, the conventions for both 11 and 13 will cause problems down the line. We'll see what these problems are as we proceed in this post.

Let's look at the 14EDL scale again:

The 14EDL scale:
Degree     Ratio     Note
14            1/1         red F#
13            14/13     ocher G
12            7/6         white A
11            14/11     amber B
10            7/5         green C
9              14/9       white D
8              7/4         white E
7              2/1         red F#

Notice that when Kite defines a scale, he usually begins the scale with 1/1, the "white unison." But here our tonic is red F#, not a white note. We can force the tonic to be white by adding the opposite color of red -- namely blue -- to the color of every note:

The 14EDL scale (Kite colored intervals):
Degree     Ratio     Note
14            1/1         white unison
13            14/13     ocher-blue second
12            7/6         blue third
11            14/11     amber-blue fourth
10            7/5         bluish fifth
9              14/9       blue sixth
8              7/4         blue seventh
7              2/1         white octave

We see here that "green" plus "blue" makes "bluish." Here are more traditional interval names:

The 14EDL scale (traditional intervals):
Degree     Ratio     Note
14            1/1         tonic
13            14/13     small tridecimal neutral second, "trienthird"
12            7/6         subminor third
11            14/11     undecimal diminished fourth
10            7/5         small septimal tritone
9              14/9       subminor sixth
8              7/4         subminor seventh, "harmonic seventh"
7              2/1         octave

The name "trienthird" for 14/13 is a strange name -- it actually means "1/3 of a (major) third." Let's work it out in cents:

14/13 = 128.3 cents
5/4 = 386.3 cents

And three times 128.3 cents is 384.9 cents, which is close enough to 5/4. Another name for the trienthird is "2/3 tone," since it's about twice as large as the third-tone we defined earlier.

Notice that even though Gh is closer to G# than G, the interval 14/13 (F#> to Gh) is closer to 100 cents (a 12EDO semitone) than to 200 cents. The fact that F# is red rather than white mitigates some of the error in naming Gh "ocher G." A rule of thumb is that red, jade, and ocher intervals are all higher than the corresponding white intervals, while blue, amber, and emerald intervals are all lower than the corresponding white intervals. The interval 14/13 is "ocher-blue second," and so "blue" lowers the interval about half as much as "ocher" raises it.

The real problem is trying to name the interval 14/11. This interval is 417.5 cents, which is much closer to 400 cents (a 12EDO major third) than to 500 cents. Yet the names "red F#" and "amber B" make it appear that this interval should sound more like a perfect fourth than a major third -- to the extent that some people call 14/11 an "undecimal major third" rather than any sort of "fourth."

The Kite name for 14/11 is "amber-blue fourth." Both "amber" and "blue" are colors that lower the note, which is why this interval is much smaller than a perfect (white) fourth. Kite's names require us to call this interval a "fourth," though not necessarily a perfect fourth (as using Kite rather than H-E, we could have called the note "amber Bb," and F#-Bb is a diminished fourth).

Notice that both 12/11 (150.6 cents) and 11/10 (165 cents) are both "undecimal neutral seconds." But I've seen some sources distinguish between these:

12/11 = undecimal superminor second
11/10 = undecimal submajor second

We use these to find a name for 14/11:

(14/11) = (7/6)(12/11) = (14/12)(12/11)
= subminor third + superminor second
= minor third + minor second (since "sub" and "super" cancel)
= diminished fourth

Another problematic interval is 13/11. The interval Gh-Bd appears to be a major third (G-B), but at 289.2 cents, it's narrower even than a minor third. In Kite this is an "emerald-amber third." If we follow the same pattern for tridecimal neutral thirds:

14/13 = tridecimal superminor second
13/12 = tridecimal submajor second

(13/11) = (13/12)(12/11)
= submajor second + superminor second
= major second + minor second
= minor third

Thus 13/11 is a tridecimal minor third. (This is why Kite came up with colors in the first place, so he doesn't have to figure out what "undecimal diminished fourth" and "tridecimal minor third" are!) And Kite himself suggests rounding 13 and 11 both up to G# and B, or both down to G and Bb, so that the interval between them is spelled like a minor third. But we're stuck with Gh and Bd.

Finally, notice that 12/11 is superminor and 11/10 is submajor. Yet 12/11 is A-Bd and 11/10 is Bd-C, which makes the 150.6-cent interval appear wider than the 165-cent interval. But I justified the name "amber B" by looking at thirds, not seconds -- G-Bd is 27/22, 354.5 cents, Bd-D is 11/9, 347 cents. I avoided taking 11/10 because it uses the prime 5 while the jade/amber thirds use only 2, 3, and 11.

Let's look at the 14EDL scale once more:

The 14EDL scale:
Degree     Ratio     Note
14            1/1         red F#
13            14/13     ocher G
12            7/6         white A
11            14/11     amber B
10            7/5         green C
9              14/9       white D
8              7/4         white E
7              2/1         red F#

The 14EDL scale is unique in that it's the only EDL that contains the same number of tones as our usual major or minor modes -- seven notes (plus the octave). But 14EDL doesn't sound like either the major or minor scale. So which classical mode does it sound like?

The major scale ends with a leading tone, as does the harmonic minor scale. But EDL's don't end with leading tones -- instead, the number of cents between the notes increases throughout the scale, from 14/13, 128.3 cents, to 8/7, 231.2 cents. This suggests a mode which has semitones near the beginning and whole tones near the end of the scale, such as the Locrian mode.

In fact, if we ignore the colors, the resulting scale F#-G-A-B-C-D-E-F# is indeed F# Locrian. It's not quite F# Locrian, of course, since 14/11 sounds more like a major third than a perfect fourth. Both scales contain a diminished fifth, but no mode has a diminished fourth the way 14EDL does. In many ways, 14EDL is the subminor EDL, since it has a subminor third (plus a subminor sixth and seventh), just as 12EDL is the minor EDL.

Finally, we might be tempted to make a 13EDL scale -- if we ignore the colors, then it certainly looks like a major scale (G-A-B-C-D-E-F#). But the interval from ocher G to amber B is more like a minor third, so the resulting scale would sound more like G minor than G major. And since 13 is an odd number, it has no octave available. So there is no leading tone F#-G at the end of this scale.

Question 3 of the SBAC Practice Exam is on solution sets:

Click the table to indicate whether each equation has no real solution, one real solution, or infinitely many real solutions.
• 5/(20x) = 1/(4x)
• 3x = 4 + 5x
• sqrt(2x + 3) + 6 = 0
Of these three equations, only one is undoubtedly a first semester Algebra I problem. It's easy to solve this linear equation:

3x = 4 + 5x
-2x = 4
x = -2 (one real solution)

The first equation could be taught in first semester Algebra I. Even though it's not linear, it's easily transformable into a linear equation using cross multiplying, which may be studied in first semester:

5/(20x) = 1/(4x)
20x = 20x (infinitely many solutions)

Notice that the solution isn't "all real numbers" -- x = 0 can't be a solution, since x appears in the denominators in the original equation. But the SBAC question indicates "infinitely many solutions," which is not the same as saying that all real numbers are solutions.

The last equation is definitely not a first semester Algebra I problem. Notice that just last week, I subbed in an eighth grade Algebra I class that was studying radical equations -- and last week was clearly in the second semester:

sqrt(2x + 3) + 6 = 0
sqrt(2x + 3) = -6
2x + 3 = 36
2x = 33
x = 16.5

But let's check this answer:

sqrt(2 * 16.5 + 3) + 6 = 0
sqrt(36) + 6 = 0
6 + 6 = 0
12 = 0 (no real solution)

This is an extraneous root. In fact, it should have been obvious at the step sqrt(2x + 3) - 6 that the principal square root of (2x + 3) can never be -6.

Since my goal is to prepare for the summer Algebra I class, let me write out how exactly I'd teach solving the second equation to my students:

3x = 4 + 5x

Let's get all the x's on the left side and all the numbers on the right side. We want to get rid of the 5x on the right side, so instead of adding 5x, let's subtract 5x:

3x - 5x = 4 + 5x - 5x
-2x = 4

So this is -2 times x. But I don't want -2 times x -- I want just x. So instead of multiplying by -2, let's divide by -2:

-2x/-2 = 4/-2
x = -2

Question 4 of the SBAC Practice Exam is on adding and subtracting polynomials:

Enter an expression equivalent to (3x^2 + 2y^2 - 3x) + (2x^2 + y^2 - 2x) - (x^2 + 3y^2 + x) using the fewest number of possible terms.

This is definitely a second semester Algebra I problem. Let's take care of the subtraction first:

(3x^2 + 2y^2 - 3x) + (2x^2 + y^2 - 2x) - (x^2 + 3y^2 + x)
= 3x^2 + 2y^2 - 3x + 2x^2 + y^2 - 2x - x^2 - 3y^2 - x
= 3x^2 + 2x^2 - x^2 + 2y^2 + y^2 - 3y^2 - 3x - 2x - x
= 4x^2 - 6x

This is a straightforward combining like terms problem, except that all y^2 terms are eliminated.

SBAC Practice Exam Question 3
Common Core Standard:
Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.

SBAC Practice Exam Question 4
Common Core Standard:
Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials.

Commentary: Equations with variables on both sides appear in Lesson 6-6 of the U of Chicago Algebra I text, while proportions appear in Lesson 5-7. For the square root equation, my eighth graders last week pointed out that -6 is also a square root of 36, yet x = 16.5 isn't a solution because the sqrt() symbols indicate the principal square root only. It's also possible for two wrongs to make a right here -- they claim that 2x = 33 has no solution because it's impossible to divide an odd number by 2!