Friday, April 29, 2016

PARCC Practice Test Question 8 (Day 151)

Chapter 8 of Morris Kline's Mathematics and the Physical World is called "The Revolutions of the Heavenly Spheres." As the title implies, this chapter is all about the orbits of the earth and moon.

"This terror then and darkness of mind must be dispelled not by the rays of the sun and glittering shafts of light, but by the aspect and law of nature." -- Lucretius

This chapter is mostly about Copernicus and Kepler and their heliocentric theory. These two 17th century astronomers transformed the universe from one centered at the earth, with the sun and planets moving around it in "epicycles," to one with a focus at the sun and the earth and other planets moving around it in ellipses. An "epicycle" is like a circle going around another circle, such as the moon going around the sun (except that this is more like an ellipse upon an ellipse."

By the way, this weekend I plan on attending a local Renaissance Faire. One of the carnival games involves shooting water at a bulls-eye. The targets are the planets and sun, and the bulls-eye is the center of the universe -- the earth. Copernicus, of course, lived late in the Renaissance, so the geocentric universe was the prevailing opinion for most of the Renaissance.

I don't have much to write about this chapter, because I have an announcement to make. My employment situation is about to change. I am on the verge of accepting a teaching position for the 2016-7 school year at a charter middle school in Los Angeles. I'll have more to say about that on the blog as I receive more information.

Meanwhile, let's back to the traditionalists. Last week, I mentioned how some states are considering dropping Common Core, including Michigan. Well, here's yet another heated Common Core debate thread from that particular state.

One poster in particular definitely appears to be a traditionalist -- Kelley. In response to a comment that the Common Core doesn't dictate teaching methods, Kelley writes:

Then why are all the public schools in the U.S. having the exact same problems with math, English, etc?  I challenge you to find one public school kid, in 9th grade or below that can do a long division problem, or one who knows his times tables.  Common Core is federally run education, and that is unconstitutional. 
So we see that in addition to a traditionalist, Kelley is a strong Tenth Amendment supporter (her comments about the Core being unconstitutional). Remember that I'm torn between whether there should be national standards or 50 different state standards.

Continuing with her traditionalist criticism of the curriculum, she writes:

Disastrous is the fact that kids are going through elementary and middle school without knowing so much as their times tables.  Forget about long division.  What you fail to see is that CC standards were written by test company employees, were never field tested, and when sent to validation, were stricken down by the only two experts in their fields on that committee.  Now those two experts, Dr. James Milgram from Stanford, and Dr. Sandra Stotsky travel the U.S. speaking on the dangers of CC.  They were not state led.  As a matter of fact, most states signed on before they were even written.  I have been studying CCSS for the past four years and I've learned a ton.  I think everyone should because it's the single biggest change in education in history, it was shoved down our throats with no vote of any kind.   Get educated on what CC really is, because I don't have enough room here to explain all of the tentacles involved with CC.  It's not about education, at all.
Later on, when another poster asked to look at the new proposed standards first:

Why?  You didn't approve of CC standards.  Now you want to approve standards that have been PROVEN in MA?  They had the best in the U.S. before CC.  Look it up. 
So Kelley likes the pre-Core Massachusetts standards, which Michigan is considering adopting. But some Tenth Amendment supporters oppose even two states having the same standards -- they refer to it as an illegal interstate compact.

As for myself, I consider regional standards to be a good compromise between national standards and 50 separate state standards. But before Michigan adopts the Bay State's pre-Core standards, let's see whether Massachusetts itself will adopt the old standards. Then I'd like to see the standards expand to the other nearby New England states, as many of these have small populations. My own state of California, meanwhile, is so large that our state standards would a priori be regional standards!

Meanwhile, a pro-Core writer brings up an idea that I've mentioned before -- Presidential (or Gubernatorial, in this case) Consistency:

YEAH the government officials do not care if the kids are up to speed. Why? Simple they send their own kids to private school.
They just want to use a special test that will make them look good but in reality because will think we are up to international standards when in reality we are not. We finally got it right with common core and now going backwards again.

Question 8 of the PARCC Practice Test is about the equation of a circle:

8. Part A
A circle in the xy-coordinate plane has the equation x^2 + y^2 + 6y - 4 = 0. If the equation of the circle is written in the form x^2 + (y + k)^2 = c, where k and c are constants, what is the value of k?

Part B
What is the radius of the circle?
A. 2
B. 4
C. sqrt(13)
D. 13

And so this is our second circle equation with completing the square this week! Here is the answer:

x^2 + y^2 + 6y - 4 = 0
x^2 + y^2 + 6y = 4
x^2 + y^2 + 6y + 9 = 13
x^2 + (y + 3)^2 = 13.

This gives us the answers k = 3 and sqrt(13), or choice (C) as the answer to Part B. Again, the wrong answer choices for Part B give away the most common student errors. Choice (D) occurs when students give the value of r^2 rather than r. Choice (A) occurs if students stop after the second step, see 4 on the right side alone, and assume that this is r^2 without completing the square. Choice (B) occurs if students make both errors.

But my concern with this question is Part A. Notice that the center of the circle is (0, -3). But the question does not ask for the y-coordinate of the center -- it asks for the value of k in the special equation provided, x^2 + (y + k)^2 = c. Since this equation includes (y + k)^2 and not (y - k)^2, this means that k = +3, not -3.

Let me be honest -- I don't like this part of the question one bit. Students are expected to know the equation for the area of a circle, (x - h)^2 + (y - k)^2 = r^2. Using (y + k)^2 only ends up confusing students more than necessary.

To see what's wrong with this question, suppose Part B had been written like this:

Part B?
If the equation of the circle is written in the form x^2 + (y + k)^2 = c, where k and c are constants, what is the value of c?
A. 2
B. 4
C. sqrt(13)
D. 13

Now suddenly (D) is the correct answer, even though (C) is the radius of the circle. And the following version of Part B is even worse:

Part B?
If the equation of the circle is written in the form x^2 + (y + k)^2 = r, where k and r are constants, what is the value of r?
A. 2
B. 4
C. sqrt(13)
D. 13

Here the answer is again (D), even though (C) is still the radius. That is, r is not the radius. But almost every text is going to use r for the radius in this situation. So to me, using r to denote the square of the radius, rather than the radius itself, isn't merely confusing -- it's downright deceitful!

At this point, someone might point out that students might make the mistake I mentioned earlier, and think that the right hand side of the equation is the radius rather than its square. In this case, two wrongs actually make a right -- the students don't know that they need to take the square root to find the radius, and they don't know that the question is asking for the square of the radius!

Likewise, the original Part A as given in the text also has two wrongs making a right. A student could think that +3 is the y-coordinate of the center and not realize that the question is actually asking for the opposite of the y-coordinate.

In fact, some might argue that the purpose of this question is to protect student from having to give a negative sign. But then this defeats the purpose of giving this problem -- in such a problem, it's possible for a student to make 0, 1, or 2 errors. It's set up so that a student will make 0 errors -- that is, a student will likely make either 1 or 2 errors. Since two wrongs here make a right, a student gets credit for making 2 errors, but not for 1 error. A student is rewarded for making more mistakes! This is surely not the message we want to send our students!

Furthermore, suppose the following question were to appear on a PARCC Algebra I exam:

-- Give the general solution of the equation ax^2 + bx - c = 0.

The correct answer is x = (-b +/- sqrt(b^2 + 4ac))/2a. The negative sign on c in the original problem forces the discriminant to be b^2 + 4ac. Sure, the Quadratic Formula has b^2 - 4ac, but the question asks for a general solution to the given equation, not for the Quadratic Formula itself. And the following question is even worse:

-- Give the general solution of the equation ax^2 + bx = c

where the negative sign on c isn't as apparent. Here's yet another bad question:

-- Give the general solution of the equation cx^2 + bx + a = 0

where the correct answer is x = (-b +/- sqrt(b^2 - 4ac))/2c. The following problem isn't as bad:

-- Give the general solution of the equation ax^2 + bx = d

where at least the use of the letter d is a tip-off that something is different. We can solve this by using the Quadratic Formula with a = a, b = b, and c = -d. There's no reason to assume that c = d, while there's a great reason to assume that c = c!

One might argue that questions like ax^2 + bx = c force students to think, rather than allow them to write the Quadratic Formula without any thought at all. But once again, there's a fine line between challenging and deceiving, and the question ax^2 + bx = c does the latter.

There's no inherent reason that a must be the quadratic coefficient, b the linear coefficient, and c the constant in the equation ax^2 + bx + c = 0. But this is a convention -- virtually every single Algebra I text that teaches the Quadratic Formula uses these variables. Conventions are used not only in mathematics, where a, b, and c are coefficients, (h, k) is the center of the circle with radius r, but also in the sciences (a for acceleration, c for the speed of light), computer science, and so on.

Using variables for a purpose that directly oppose their conventional use not only deceives students, but also reinforces the notion that algebra is more about manipulating symbols rather than actually solving problems. This goes back to Kline and the true purpose of mathematics.

So I'd like to discard this sort of problem and throw it away. But of course, we can't because it appears on the PARCC. Therefore questions such as these fuel the assertions of Kelley and other traditionalists that yes, the Common Core really does dictate the curriculum.

Actually, this is a bit tricky. I've once said before that the only way to have a set of standards that does not dictate curriculum is for the test to ask higher-level questions, which are more likely to be made curriculum-independent. (Deceptive questions like today's shouldn't be on any sort of test no matter what the standards are.)

Questions that ask about reflections, rotations, and translations obviously require that congruence be developed using transformations. This is why I wonder whether it's possible to use area arguments in place of transformations, since area is something that the students should be learning anyway.

PARCC Practice EOY Question 8
U of Chicago Correspondence: Lesson 11-3, Equations for Circles
Key Theorem: Theorem (Equation for a Circle)

The circle with center (hk) and radius r is the set of points (xy) satisfying

(x - h)^2 + (y - k)^2 = r^2.

Common Core Standard:
Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation.

Commentary: It's the way this question is worded that makes it so tricky, so I just added in questions from last year's review worksheet. Today is an activity day, so the rest of the worksheet is based on yesterday's question, as it lends itself to activity more naturally.

Thursday, April 28, 2016

PARCC Practice Test Question 7 (Day 150)

Chapter 7 of Morris Kline's Mathematics and the Physical World is called "The Dimensions of the Heavenly Spheres." In this chapter, Kline measures the size of the earth, moon, and sun.

"On he passed, far beyond the flaming walls of the world, and traversed in mind and spirit the immeasurable universe." -- Lucretius, 1st century BC Roman poet

Kline begins,

"According to history, Euclidean geometry arose from man's effort to measure areas of land, and the literal meaning of the word geometry, namely, earth-measure, supports history."

And so in this chapter, Kline truly performs geometry, or earth-measure. He introduces the three main trig ratios of sine, cosine, and tangent. Then he proceeds to measure the sizes of the three main bodies of the earth, moon, and sun.

This sounds familiar -- and it should, because we just did it earlier this month, in Lesson 13-5. The U of Chicago text provides us with the earth-moon distance, 240,000 miles, the earth-sun distance of 93,000,000 miles, and the moon's radius of 1080 miles, and then similar triangles are set up to determine the sun's radius.

Kline does likewise. But let's imagine this problem from the perspective of the ancient Greeks, who weren't given the three values (1080, 240,000, and 93,000,000) found in the U of Chicago text. How were they able to estimate the radius of the sun? The answer is -- using trig. Kline writes:

"...the trigonometric methods of obtaining such seemingly inaccessible facts as the distance to the sun and the size of the sun are far simpler than the geometric proof that the area of a circle is pi r^2."

Kline writes about Hipparchus, the second century BC Rhodes mathematician. But the story begins with Erathosthenes, the 3rd century BC Cyrenian who first proved that the earth is round. By measuring shadows at two different locations, he obtained a value of about 24,000 miles for the circumference of the earth, and thus a value of about 4000 miles for the radius of the earth.

So Hipparchus used this value to find the distance to the moon. Here E is the center of the earth, M is the center of the moon, P is a point on earth's surface where the moon is on the horizon, and Q is another point on earth's surface where the moon is directly overhead. This makes EPM a triangle, and just as in Lesson 13-5, MP is tangent to the earth and so angle P is a right angle. Since the moon is directly overhead at Q, Q is on EM. Hipparchus measured the value of angle E using the system of longitude that he invented, and obtained 89 degrees, 4 minutes, 12 seconds, according to Kline (that is, 89 + 4/60 + 12/3600 degrees, or abour 89.07 degrees). So we obtain:

cos E = PE/EM = 4000/EM
.0163 = 4000/EM
.0163EM = 4000
EM = 245,000

According to Kline, Hipparchus obtained a value 280,000 miles. This value can be used to find the radius of the moon. This time, Q is a point on the moon's surface such that PQ is tangent to the moon, so now PMQ is a right triangle with the right angle at Q. Angle MPQ (which Kline writes as angle A can be measured on the earth to be about 15 minutes (1/4 of a degree). So we write:

sin A = QM/PM
sin 15' = QM/240,000
QM = 240,000 sin 15'
QM = 1056

The distance to the sun can be estimated the same way as the distance to the moon, but the angles involved are much harder to measure. According to Kline, Hipparchus obtained 10,000,000 miles, which is off by a full order of magnitude. Once these values are obtained, now we can find the radius of the sun using the method mentioned in the U of Chicago text.

Today is Day 150 on the blog calendar. It marks the midpoint of the third trimester -- in some districts, this is known as the end of the fifth hexter.

Today's also a special day because there's a featured Google Doodle. Hertha Marks Ayrton was a 19th century British engineer -- but she was also a mathematician! The following website describes Ayrton in more detail:

Let's read a little more about Ayrton's mathematical work:

From 1881 to 1883, Marks worked as a private mathematics tutor, as well as tutoring other subjects. In 1884 she invented a draftsman's device that could be used for dividing up a line into equal parts...

Dividing up a line into equal parts -- hey, doesn't that sound familiar?

Find the point on a directed line segment between two given points that partitions the segment in a given ratio.

Let's get back to her biography:

as well as for enlarging and reducing figures.

That is, as well as for performing dilations on figures! So we see that two of Ayrton's mathematical accomplishments are directly related to the Common Core Geometry standards.

Now Kline tells us that mathematics matters only when it can be applied to science. So let's look to see how Ayrton applied her mathematics:

Marks began her scientific studies by attending evening classes in physics at Finsbury Technical College given by Professor William Ayrton, whom she married in 1885. She assisted her husband with his experiments in physics and electricity, becoming an acknowledged expert on the subject of the electric arc. She published several papers from her own research in electric arcs in the Proceedings of the Royal Society of London and The Electrician, and published the book The Electric Arc in 1902.

An "electric arc" occurs when electrons travel between two electrodes in an arclike direction. And an electric field propagates from an electric charge in round waves in all directions. So everything goes back to ratios, lines, circles, and spheres.

Question 7 of the PARCC Practice Test is on -- what else -- circles. (It's too bad that we couldn't have a dividing segments question or a dilation question in Ayrton's honor, but this is the next best thing.)

7. The figure illustrates an informal argument for the area of a circle. The circle is divided into congruent sectors, and the sectors are rearranged to form a shape that resembles a parallelogram, as shown. As the number of sectors increases, the rearranged shape more closely resembles a parallelogram with area A, given by the formula A = bh, where b is the base and h is the height of the parallelogram.

Select the correct value for b and h to develop the area of the circle in terms of r, the radius of the circle.

(Both b and h lead to drop-down menus. The menu for b provides choices pi, r, pi r, 2pi, and 2pi r, while the menu for r provides choices pi, r, r * r, 2r, and 2pi.)

This problem demonstrates a very well-known proof of the circle area formula -- the answer is that the base is pi r and the height is r. It appears in many Geometry texts, including the U of Chicago text in Lesson 8-9. Let's recall what Kline writes about the derivation of the circle area formula again:

"...the trigonometric methods of obtaining such seemingly inaccessible facts as the distance to the sun and the size of the sun are far simpler than the geometric proof that the area of a circle is pi r^2."

And yet the PARCC expects students to provide the geometric proof that the area of a circle is pi r^2.

Notice that this is not how other authors derive the circle formulas. Dr. Hung-Hsi Wu actually starts with the area of a circle and then uses the formulas for perimeter/area of a regular polygon (which approaches the circle in the limit) to obtain the circumference of a circle. Dr. Franklin Mason goes the other way -- he starts with the circumference and uses regular polygons to derive the area.

Unfortunately, the way I taught the pi lessons on the blog ended up as a hodgepodge. My original intention was to follow Wu's lesson, but then I gave the circumference formula first. I ended up reteaching circumference as part of a seventh grade lesson I saw as a sub earlier this month.

By the way, a week later I ended up speaking to that seventh grade teacher again. He told me that he actually taught the seventh graders the circle area derivation given by U of Chicago and PARCC. He said that it was a bit confusing for them -- but again, remember what Kline said about the formula.

Notice it's also possible to begin with the area of a circle as in Wu, then divide it into sectors to form the parallelogram. This time, the area and height of the parallelogram are known, and the students find the height and therefore the circumference.

PARCC Practice EOY Question 7
U of Chicago Correspondence: Lesson 8-9, The Area of a Circle
Key Theorem: Circle Area Formula

The area A of a circle with radius r is A = pi r^2.

Common Core Standard:
Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri's principle, and informal limit arguments.

Commentary: Lesson 8-9 of the U of Chicago text gives the desired proof. In fact, the first exercise on the worksheet is the same as the first question from the text.

Wednesday, April 27, 2016

PARCC Practice Test Question 6 (Day 149)

Chapter 6 of Morris Kline's Mathematics and the Physical World is "The Laws of Space and Forms," and this chapter is all about our favorite branch of mathematics -- geometry!

" the science that it hath pleased God hitherto to bestow on mankind." -- Thomas Hobbes, 17th century English philosopher

Kline begins:

"The story is told that the Greek philosopher Aristippus and some friends were shipwrecked on what appeared to be a deserted island near Rhodes. The company was downcast at its ill fortune when Aristippus noticed some geometric diagrams drawn on the beach sand. 'Be of good cheer,' he told his companions, 'I see traces of civilized men.'"

In other words, wherever there is civilization, there is geometry. Kline discusses the development of geometry, beginning with the Ancient Egyptians and Babylonians, and moving to the Greeks and -- who else? -- Euclid.

Kline writes that even Euclid's contemporaries wondered why the ancient sage spent so much time with intuitively obvious theorems, such as SAS, the Triangle Inequality, and the theorem that the shortest distance between a point and a line is perpendicular to the line. Kline answers:

"Many geometrical theorems indeed seem obvious, but if they can be deduced from axioms, then the theorems are utterly beyond question."

He then gives the Triangle Midsegment Theorem as an example of a theorem that isn't obvious and requires proof. More to the point, if we relied only on intuition, rather than proof, we may end up accepting results that appear to be true, but are actually false.

Kline devotes the rest of this chapter to two main results. One of which is a continuation of the Isometric Theorem from yesterday, and the other involves reflections and light.

Yesterday, we looked at Kline's algebraic proof that among all rectangles with the same perimeter, the square has the most area. Well today, he adds a geometric proof of the same result. The following is his geometric proof:

Let ABCD be any rectangle with perimeter p. If the rectangle is not a square then one side, AB, is larger than the other. On the larger side let us construct a square, EBHG, with perimeter p. [That is, E is on AB and C is on BH, according to the diagram in the book -- dw] Now AE + EB + BC is half the perimeter of the rectangle. Hence

(1) AE + EB + BC = p/2.

But since the square has the same perimeter as the rectangle, half of its perimeter is also p/2; in symbols,

(2) EB + BC + CH = p/2.

If we subtract equation (2) from equation (1) we obtain

(3) AE - CH = 0.

Addition of CH to both sides of this equation yields

(4) AE = CH.

Now the smaller side, AD, of the rectangle must be less than any side of the square because the square has the same perimeter as the rectangle. Hence

(5) AD < GH.

The area of rectangle II [ADFE -- dw] is AE * AD and the area of rectangle III [CFGH -- dw] is CH * GH. But equation (4) tells us that AE = CH, while equation (5) tells us that AD < GH. Hence the area of rectangle II is less than the area of rectangle III.

Since the area of the entire rectangle ABCD is the sum of the areas of I [BCFE -- dw] and II and the area of the square is the sum of areas I and III, the square has greater area than the rectangle. QED

Kline ends his proof by writing, "We have proved precisely the same result that was established in the preceding chapter. It is a matter of taste whether one prefers the geometric proof to the algebraic one."

Now Kline considers the case where one of the sides of the rectangle whose area is to be maximized lies along a river. The maximum area occurs when the side along the river is twice the length of the other side -- and he proves this simply by reflecting the rectangle across the river!

Speaking of reflections, Kline moves on to the reflection of light. He writes that the first person to write about the properties of light was Euclid himself, in his work simply titled Optics. Euclid writes that the angle of incidence equals the angle of reflection.

Another theorem was proved by the mathematician Heron in the first century A.D. -- this is the same Heron whose famous formula produces the area of a triangle given its three sides. This time, Heron proves that the light actually follows the shortest path from a point to the mirror to a second point. I have seen this fact mentioned in other texts (but not the U of Chicago text), and it shows us how, for example, to get from a starting point to a river (where we can fill a bucket with water) and from there to a burning house in the shortest distance. Kline writes that this same idea shows up with billiard balls (and this does appear in the U of Chicago text, Lesson 6-4).

I'd like to give Heron's proof here, but this post is getting long enough, and we still need to get to today's PARCC question.

Before I get to the PARCC question, I want to make a comment about the blog calendar. Today is Day 149 in one of the two districts where I work. In the other district, today is Day 155. That second district has an Early Start Calendar, where the first semester ended before Christmas. As it turns out, the second district has already completed its SBAC testing, before the AP.

The blog calendar is based on my first district, where the SBAC is after the AP. It is on what I call a Middle Start Calendar -- but next year it will complete its transition to an Early Start Calendar. In fact, the first day of school will be exactly the same in both districts. I'm curious as to when the first district will give the SBAC next year, now that it's also on an Early Start Calendar.

Question 6 of the PARCC Practice Test is about rectangles and triangles:

6. In the diagram, quadrilaterals FBAG and CDEF are rectangles.

(Here is some other info from the diagram: A on CE, B on FC, BC = 3, CD = 12, G on EF, EG = 7.)

How long is DE rounded to the nearest tenth?

There are two things we need to know in order to solve this problem. First, we need to know that the opposite sides of a rectangle are congruent. We can obtain several lengths this way -- since CD = 12, its opposite side EF is also 12. This means that GF = 5, and so its opposite side AB = 5.

The other thing we need is that the opposite sides of rectangles are parallel. Since DE | | FC, the alternate interior angles CED and ECF are congruent.

This gives us two pairs of congruent angles in triangles ABC and CDE -- the right angles ABC and CDE are congruent, and the angles ACB (same as ECF) and CED. Thus by AA Similarity, triangles ABC and CDE are similar. This allows us to set up a proportion:

DE / BC = CD / AB
x/3 = 12/5
x = 36/5 = 7.2 exactly

We are told to round off the answer to the nearest tenth, but the answer DE = 7.2 is exact.

This is an interesting problem. It requires looking at similar triangles, but it's not obvious which triangles in this problem are similar. This is more complicated than anything that appears in Chapter 12 of the U of Chicago text, where it's always obvious which triangles are similar. This is the type of problem that traditionalists want to see more of, instead of dilation questions or whatnot. It requires students to think -- which is why many students will have trouble with this problem.

Today's question also brings up the idea of slope and its relationship to similar triangles -- which, as we know, is an eighth grade Common Core Standard. We wish to show that the slope along line CE is constant -- that is, the slope along CA equals the slope along AE.

But how do we find the slope along CA? Well, that equals the rise CB (which is negative) divided by the run BA. And the slope along AE equals the rise AG divided by the run GE. In other words, the triangles CBA and AGE are slope triangles.

As usual, we can prove that the slope triangles are congruent. They already have one pair of congruent angles -- the right angles CBA and AGE. And the angles ACB and EAG are congruent because they are corresponding angles, as CB | | AG. So triangles CBA and AGE are similar slope triangles, and so the slopes CB / BA and AG / GE are equal.

Here's an interesting question -- is it possible to solve this problem without similar triangles? As it turns out, it is possible. Instead of similar triangles, we use area. Let h be the value of DE that we are trying to find -- h standing for height (of the rectangle CDEF, of course).

Area(CBA) + Area(AGE) + Area (FBAG) = Area (CFE)
15/2 + 7(h - 3)/2 + 5(h - 3) = 12h/2
15/2 + (7h - 21)/2 + 5h - 15 = 12h/2
15 + 7h - 21 + 10h - 30 = 12h
17h - 36 = 12h
5h = 36
h = 7.2

This is clearly not the method intended by PARCC. After all, we look at the alignment document (provided at the same link where I'm getting the questions), and we see the standard for this question listed as G-SRT.5, where SRT stands for similarity and right triangles. But we could have sneaked this question into Chapter 8 (area) of the U of Chicago text instead of Chapter 12 (similarity).

Notice that this proof is somewhat more algebraic than the similarity proof -- that is, the students' mental burden is placed more on calculating the areas and solving the equation than on figuring out which triangles are similar. This goes back to what Kline writes about how it's a matter of taste whether an algebraic proof or a geometric proof of the same result is better.

But I've mentioned before that in many ways, similarity and area are equivalent. Many questions that are solved via similarity can be answered using area instead, and this is one of them. I wonder whether any student who sees this problem on the PARCC, but fails to notice the similar triangles, would try to find the area of the triangles and rectangles instead.

Sometimes I reflect on Common Core Geometry and how the course is organized. Many people don't like the reliance of Common Core Geometry on transformations. I like the idea that many statements that are postulates in pre-Core Geometry, such as SAS (both similarity and congruence), the slope formula, and so on, are provable in Common Core Geometry. But many, especially traditionalists, don't consider the extra level of rigor to be worth confusing students with reflections and dilations.

One idea that has been forming in my mind is replacing similarity and dilations in many Common Core proofs with area. This is already often done with the Pythagorean Theorem, but we can do it with the slope formula as well, as suggested in today's post. In some ways, this makes sense -- students have been learning about the area of a rectangle since third grade, years before they learn anything about similarity or dilations.

But there's one major problem with this idea -- beginning with the Area Postulate of Chapter 8 requires knowing that the opposite sides of a rectangle are parallel and congruent. After all, how we can say that the area of a rectangle is length times width if we don't know that the length of the top of the rectangle equals the length of the bottom of the rectangle? And I don't see how we can prove these rectangle properties without some sort of postulates about congruence. Well, it was just a thought.

PARCC Practice EOY Question 6
U of Chicago Correspondence: Lesson 12-9, The AA and SAS Similarity Theorems
Key Theorem: AA Similarity Theorem

If two triangles have two angles of one congruent to two angles of the other, then the triangles are similar.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.SRT.B.5Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.

Commentary: This question is a bit more sophisticated than any of the problems in the U of Chicago text for this lesson. Naturally, I include several more exercises similar to this PARCC problem, where students must search for similar triangles, on the worksheet.

Tuesday, April 26, 2016

PARCC Practice Test Question 5 (Day 148)

Chapter 5 of Morris Kline's Mathematics and the Physical World has the title of "Numbers, Known and Unknown." In this chapter, Kline writes all about unknown numbers -- that is, algebra.

"Algebra begins with the known and ends with the unknowable." -- Anonymous

And of course, the author of quote about the unknown is, well, unknown! But Kline begins his chapter writing about a mathematician whose very well-known indeed:

"When Karl Friedrich Gauss, one of the greatest mathematicians of all time, was in elementary school, his teacher, harried by endless clerical work, assigned to the class the problem of finding the sum of all the numbers from 1 to 100."

We've talked about Gauss before on the blog, but not in the context of this problem. But then again, the story that Kline tells here is so famous that many Algebra II teachers still tell it today in their classrooms today. In short, Gauss was able to add the numbers from 1 to 100 quickly, not using arithmetic, but algebra. The sum of the naturals from 1 to n is n(n + 1)/2, and this formula is given to students to show how useful algebra is.

If "math" is a subject that many students hate, then "algebra" is the reason that they hate it. Kline talks a little about the origin of the word algebra:

"The historical associations of the word algebra almost substantiate the sordid character of the subject. The word comes from the title of a book written by the ninth-century Arabian mathematician Al-Khowarizmi.... Figuratively, al-jebr meant restoring the balance of an equation in transposing terms.... [In Moorish Spain], in the form algebrista, it came to mean a bonesetter, or a restorer of broken bones.... Thus it might be said that there is a good historical basis for the fact that the word algebra stirs up disagreeable thoughts."

So we have this word "algebra," a foreign-sounding Arabic word, used to describe this mixture of numbers and letters that might as well be a foreign language which, as Kline writes, "many people complain about the need to learn...."

J.K. Rowling, the author of the Harry Potter series, writes that "the fear of a name increases fear of the thing itself." This is why, as I wrote a month and a half ago, I'm not sure whether I want to use the name "Algebra III" for the governor's proposed senior math course in California. In fact, I once read about a college algebra course where the professor was able to motivate the students to learn algebra simply by avoiding the word "algebra" until just before the final!

I also mentioned the following back when our state still had a high school exit exam, the CAHSEE:

-- The CAHSEE tested up to eighth grade math.
-- The CAHSEE tested up to Algebra I.

The first sentence sounds as if the CAHSEE requirements were too low -- why should students have to master only eighth grade math in order to graduate from the twelfth grade? And indeed, this argument was used by those arguing that the CAHSEE should have been more challenging. On the other hand, the second sentence sounds as if the CAHSEE requirements were too high -- why should students have to master a subject that hardly any adults use in order to graduate? And indeed, this argument was used by those arguing that the CAHSEE should have been easier.

But at the time the CAHSEE existed, our state standards recommended eighth grade Algebra I. And so the two statements above were actually equivalent! Yet it was easy to sway opinion from "CAHSEE is too easy" to "CAHSEE is too hard" simply by using a seven-letter Arabic-derived word.

Let's get back to Kline. He writes about how the ancient Greek mathematician Archimedes used algebra to prove that King Hiero's gold crown was a fake. I wrote about Archimedes earlier this month regarding his discovery of the sphere volume formula -- that day, I posted a YouTube video containing a Square One TV song about Archimedes. One line from the song mentions "the king's gold crown."

Kline also shows how algebra is used to solve the Isoperimetric Problem. That day, I stated that among all rectangles with a given perimeter, the square has the greatest area -- but I didn't provide a proof of that statement. Well, Kline shows us how to prove this. Here is his proof: let x be the side of a rectangle whose perimeter is 100, but isn't a square, so x is not 25. Then Kline writes:

(25 - x)^2 > 0
625 - 50x + x^2 > 0
625 - x(50 - x) > 0

That is, the area of the square of perimeter 100 (that is, 625) minus the area of the rectangle with side x and perimeter 100 is greater than 0 -- so the square is larger. Afterwards, Kline proves that the area of a circle with circumference is still greater than this -- hinting at the full Isoperimetric Theorem.

Since "algebra" is an Arabic word, and here in California seventh graders first learn about that part of the world, it makes since to introduce seventh graders to algebra.

Question 5 of the PARCC Practice Test is on reflections and rotations in the coordinate plane:

5. Triangle ABC is graphed in the xy-coordinate plane, as shown.

(Here are the coordinates of ABC: A(3, 2), B(4, 6), C(6, 4).)

Part A

Triangle ABC is reflected across the x-axis to form triangle A'B'C'. What are the coordinates of C' after the reflection?

A. (-6, 4)
B. (3, -2)
C. (4, -6)
D. (6, -4)

Part B

Triangle ABC in the xy-coordinate plane will be rotated 90 degrees counterclockwise about point A to form triangle A"B"C". Which graph represents A"B"C"?

A. (A"(3, 2), B"(7, 1), C"(5, -1))
B. (A"(-4, 2), B"(-5, 6), C"(-7, 4))
C. (A"(3, 2), B"(-1, 3), C"(1, 5))
D. (A"(-3, 4), B"(-2, 0), C"(0, 2))

Part A should be straightforward. To reflect over the x-axis, we switch the sign of y -- that is, the point (x, y) reflected over the x-axis is (x, -y). Since the preimage C has coordinates (6, 4), the image C' has coordinates (6, -4), which is (D). We can see what the common student errors for Part A will be just by looking at the wrong choices. Choice (A) occurs if students switch the sign of x instead of y, choice (B) occurs if the students find A' instead of C', and choice (C) occurs if the students find B' instead of C'.

Part B is trickier, because the center of the rotation is not the origin. One way to eliminate choices is to note that since A is the center of rotation, the image of A must be A itself, so that enables us to eliminate some choices right off the bat. Of the remaining choices, we see that (A) appears to be rotated clockwise while only (C) is rotated counterclockwise, so choice (C) is correct.

The U of Chicago text covers reflections in Chapter 4 and rotations in Lesson 6-3. But we do know of one deficiency with the way transformations are covered in the text -- coordinates are emphasized on the PARCC, but not in the text. I consider this to be a wasted opportunity by the authors of the U of Chicago text -- at least the simplest reflections (those where the mirror is either axis) could be completed on the coordinate plane.

Rotations centered at points other than the origin are usually not taught in any text -- that is, we aren't given a simple formula as (x, y) -> (x, -y) is for x-axis reflections. It is possible to use algebra -- going back to the topic of today's Kline chapter -- to find the answer. We first notice that the rotation 90 degrees counterclockwise, but centered at the origin, is:

(x, y) -> (-y, x)

Then the same rotation, only centered at (a, b), has the form:

(a + x, b + y) -> (a - y, b + x)

Replacing x with x - a and y with y - b gives us:

(x, y) -> (a + b - y, b - a + x)

But this is error-prone and difficult to remember. It also depends on the evident, yet unproved, assertion that a rotation centered at the point (a, b) is the composite of a translation mapping (a, b) to the origin, followed by the same rotation centered at the origin, followed by a translation mapping the origin back to (a, b).

Instead, I once noticed that on the PARCC, in every problem where we are asked to rotate a triangle, the center is either the origin or one of the vertices of the triangle. This helps us greatly, since the rotation always maps its center to itself. To rotate the other two vertices, it's probably better just to look at the vectors from the center to each of the other two vertices (which are sides of the triangle we're trying to rotate) and just rotate them by counting squares in the new direction (just as we do to find the slope of perpendicular lines). This is assuming, of course, that the correct image triangle isn't evident from looking at the graphs (for example, in a problem where graphs aren't provided).

PARCC Practice EOY Question 5
U of Chicago Correspondence: Lesson 6-3, Rotations
Key Theorem: Formulas for Reflection and Rotation

The reflection over the x-axis maps (x, y) to (x, -y), and the rotation 90 degrees clockwise around the origin maps (x, y) to (-y, x).

Common Core Standard:
Represent transformations in the plane using, e.g., transparencies and geometry software; describe transformations as functions that take points in the plane as inputs and give other points as outputs.

Commentary: The U of Chicago text covers transformations, but doesn't emphasize the coordinate plane enough. Today's worksheet provides the students with some much-needed practice with transformations on the coordinate plane.

Monday, April 25, 2016

PARCC Practice Test Question 4 (Day 147)

Chapter 4 of Morris Kline's Mathematics and the Physical World has the title of "The Deeper Waters of Arithmetic." In this chapter Kline describes the discovery of irrational and negative numbers.

"...where ignorance is bliss, 'Tis folly to be wise." -- Thomas Gray, 18th century English poet

Kline begins Chapter 4 as follows:

"Mathematics could have been a rather simple and innocuous subject if the concept of number had been limited to the whole numbers and fractions."

This chapter reminds us that for centuries after their discoveries, many mathematicians would rather not learn about irrationals or negatives at all, rather than accept them as numbers. And in many ways, our students in the modern middle school classroom feel the same way.

Kline starts his chapter out with some Geometry -- the Pythagorean Theorem, in fact. But he only brings up Pythagoras in order to introduce his discovery that sqrt(2) is irrational. He mentions the proof of the irrationality of sqrt(2) -- the same one I mentioned on the blog a few weeks ago right after Square Root Day:

"There is a legend that the discovery of sqrt(2) was made by a member while the entire group of Pythagoreans was at sea. The member was thrown overboard and the rest of the group pledged to secrecy."

While Pythagoras and his followers were struggling to accept irrational numbers, negative numbers were being discovered, not in classical Greece, but in India, 3000 miles away and 1000 years later. As Kline writes:

"The Hindus saw that when the usual, positive numbers were used to represent assets, it was helpful to have other numbers represent debts."

But, as Kline continues, it took centuries for numberhood to be granted to both of these new, strange types of number. He writes that it was not until the 17th century when mathematicians fully embraced irrationals, and not until the 19th century before they accepted negative numbers! To many mathematicians, the idea that there could be numbers that aren't ratios or are less than zero was too undesirable for irrationals and negatives to be granted numberhood.

What did it take for mathematicians to accept these new numbers at long last? Kline writes:

"These new numbers, suggested by physical uses such as the representation of lengths in the case of irrational numbers or debts in the case of negative numbers, were as legitimate as the whole numbers and fractions. They recognized further that the axioms about number -- that is, the premises on which reasoning about number is based -- applied as well to the new numbers as to the old ones."

Kline goes on to describe the commutative, associative, and distributive axioms -- that is, the postulates, or properties, from Lesson 1-7 of the U of Chicago text. The new numbers, the irrationals and negatives, satisfy these properties as well as the whole numbers and fractions. But it's additional properties that ultimately forced the acceptance of irrationals and negatives.

In particular, the Additive Inverse Property tells us that every number has an opposite. The idea that there can be numbers without additive inverses is more undesirable than the idea that some numbers can be less than zero. And more advanced properties from Calculus, such as the least upper bound property, force the existence of irrationals. The idea that there can be a bounded sequence without a least upper bound is even more undesirable than the idea that some numbers aren't irrational.

Kline compares the distinction between definitions and axioms of mathematical terms to that between the definition and axioms of citizenship. The definition of citizenship tells us exactly who is a citizen (sufficient condition), but the axioms tell us what rights it bestows. Just as every real number has the right to an additive inverse, citizens have the right to vote and defend self and country.

As I reflect upon this chapter, I think about the five named sets of numbers -- natural numbers, integers, rational numbers, real numbers, and complex numbers. These five sets are so important that they are labeled with bold letters:

-- N, Z, Q, R, and C.

This chain approximately represents the order in which these sets are discovered -- people were working with natural numbers for many millennia before they considered real or complex numbers.

Likewise, students learn about natural numbers in kindergarten or even earlier, but they don't see complex numbers until Algebra II. Even if we ignore the complex numbers (since we haven't reached them in Kline yet), the real numbers aren't fully studied until eighth grade under the Common Core.

But this isn't exactly the order in which the set were discovered. As Kline tells us, not only did the Egyptians and Babylonians use fractions well before the use of Z, but the Greeks even discovered irrational numbers well before the Hindus started using integers. So the order in which the sets were discovered looks more like:

-- N, Q+ (fractions), R+ (unsigned reals), R, C

The proof website Metamath builds up complex numbers from sets, and this is the order in which the sets are built up -- naturals, fractions, unsigned reals, signed reals, complex numbers.

On the other hand, this is the order in which students learn the sets under Common Core:

-- N, Q+, Q, R, C

This explains why the seventh grade Common Core Standards mention "rational numbers" so often -- this is the year when students learn about positive and negative numbers, but since they already learned about fractions, they can now work with all of Q.

I've also mentioned the Classical Curriculum here on the blog. It is based on the idea that students, in four-year intervals -- learn all of history, and align English and science to the historical period about which they are learning.

Math is left out of this Classical Curriculum pattern, but there's no reason we can't include it. Here in California, for example, both Ancient Greece and India are studied in sixth grade history. So we could introduce both irrationals and negatives to sixth graders, right around the same time of year that they are learning about the cultures. Then as irrationals and negatives weren't fully accepted until around the time of the American Revolution, these concepts can be revealed until eighth grade, when the students are taking U.S. history.

As interesting as this idea is, it may be troublesome because it would require linking Common Core national standards to history state standards. And as controversial as Common Core is, the idea of national history standards are somewhat dangerous, since it's too easy to teach history with a specific ideological bias -- and we wouldn't want biases to be enshrined in national standards.

Kline ends the chapter thusly:

"One might expect that the power of mathematics to tackle and explore deeper and more significant physical problems would be considerably increased thereby. We shall soon see whether the game is worth the candle."

Question 4 of the PARCC Practice Test is on the equation of a circle:

4. The equation x^2 - 10x + 17 = -y^2 - 2y describes a circle in the coordinate plane. Find the radius of the circle and the coordinates of the center.

This is one of those questions that requires completing the square. We begin by setting up the equation for completing the square, with the variable terms on the left and the constant on the right:

x^2 - 10x + 17 = -y^2 - 2y
x^2 - 10x + y^2 + 2y = -17
x^2 - 10x + 25 + y^2 + 2y + 1 = -17 + 25 + 1
(x - 5)^2 + (y + 1)^2 = 9

And so the radius is 3 and the center is (5, -1). There are all sorts of points in this problem where students might make mistakes:

-- They might complete the square on -y^2 - 2y, instead of transposing these terms to the left side.
-- They might forget to take half of -10 and 2 before squaring.
-- They might miss the negative sign on 17 and obtain a wrong value on the right side.
-- They might forget that the right side is r^2, not r, and claim a radius of 9.
-- They might forget (x - h) and (y - k), not (x + h) or (y + k), and claim a center of (5, -1).

To me, this problem belongs in Algebra II, not Geometry. The U of Chicago text has a lesson on Equations for Circles, Lesson 11-3. In this lesson, all the equations are already in the correct form, and so only the last two student errors in the above list are possible. But PARCC requires the students to complete the square to put the equation in the correct form. Now we see that there are five common mistakes students might make, rather than just two.

Furthermore, we know that completing the square isn't taught in many Algebra I classes anymore. The U of Chicago Algebra I text doesn't mention completing the square at all. And in many Algebra I classes whose texts do contain completing the square, teachers end up skipping the lesson in the rush to finish the unit on Quadratic Equations before the PARCC or other end-of-year test.

In some ways, the U of Chicago text represents a compromise between the idea that students should see circle equations in Geometry and the opposite idea that they should wait until Algebra II. In our text, the students at least see that there's a connection between algebra and geometry as they are exposed to the derivation of the circle equation. They learn that not only do lines have equations -- as they've seen in Algebra I -- but so do circles. And they are now aware that they must be careful not to confuse r with r^2 or h and k with -h and -k. This will prepare them for completing the square, which they will see the following year in Algebra II. But this compromise is impossible. As long as equations of circles and completing the square appear in the Common Core Geometry Standards, we are obligated to teach it in Geometry, period.

By the way, I decided to look up the conic section unit in an actual Algebra II text. In the text I chose, the first lesson reviews the Distance and Midpoint Formulas. This is followed by one lesson on each of the major conic sections (parabolas first, then circles, ellipses, and hyperbolas), but in these lessons, the vertex (of the parabola) or the center is always at the origin. Only in the following section are the conics translated and completing the square required. So surely if we are to each it in Geometry, we ought to do the same as the Algebra II text, and devote the first day to circles centered at the origin and the second to translated circles and completing the square. But of course, I didn't have time to teach circle equations that way on the blog.

PARCC Practice EOY Question 4
U of Chicago Correspondence: Lesson 11-3, Equations for Circles
Key Theorem: Theorem (Equation for a Circle)

The circle with center (h, k) and radius r is the set of points (x, y) satisfying

(x - h)^2 + (y - k)^2 = r^2.

Common Core Standard:
Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation.

Commentary: This standard is a bit difficult for students at this level, nevertheless the students are expected to learn it. When we covered circle equations a few weeks ago on the blog, I made sure to include completing the square as I was aware of the presence of this sort of question on the PARCC. The worksheet gradually introduces equations with more and more algebraic manipulation required to put them in center-radius form.

Friday, April 22, 2016

PARCC Practice Test Question 3 (Day 146)

Chapter 3 of Morris Kline's Mathematics and the Physical World is "The Science of Arithmetic." In this chapter, Kline, as the title implies, applies the simplest mathematics to nature.

As usual, Kline begins every chapter with a quote:

I had been to school...and could say the multiplication table up to 6 x 7 = 35, and I don't reckon I could ever get any further than that if I was to live forever. I don't take stock in no mathematics, anyway. -- Huckleberry Finn

I've seen this Mark Twain quote before. It is often mentioned in progressive math books to emphasize how many students don't work hard to learn math as they see no point in it. Kline continues by describing why many students exhibit Huck Finn's attitude towards math:

"The Greeks had a word for it. They called the boring manipulations of arithmetic that are used in commerce, trade, calendar-reckoning, and military problems logistica. Most young people would endorse this Greek deprecation of arithmetic for many reasons. But the reason that is pertinent is that we are obliged to learn the subject before we can really understand it. Because the knowledge of counting, adding, subtracting, and the like is regarded as a preparation for 'life' we are taught it mechanically from early childhood. The practice takes precedence over the principles. No doubt this introduction to life is not especially cheering. It enables us to handle money efficiently and perhaps serves to arouse our cupidity, but it hardly inspires us."

And of course, this sounds exactly like the traditionalist philosophy of teaching. So here Kline is directly attacking traditionalism -- and because of this, I am labeling this post as "traditionalists."

Kline writes that as opposed to boring logistica, the Greeks were interested in the concept of number and its applications to nature, which they called arithmetica. And so Kline devotes the rest of this chapter with using the natural numbers to describe the natural world.

Again, let's recall my definition of dren -- a reverse-nerd who can't even do third grade math. By this definition, Huck Finn is undoubtedly a dren, as 6 x 7 -- which equals 42, not 35 -- counts as a third grade math problem. But let's say that Huck at least knows the tables to 5 x 5. Arguably, this would allow him to satisfy the following second grade Common Core standard:

Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns; write an equation to express the total as a sum of equal addends.

Again, Huck says that even if he were immortal, he could never master third grade math -- much less Algebra, let alone the Calculus that traditionalists want students to take. Instead, he feels that this second grade standard represents all the math he'll ever know.

There have been many stories in the news lately about states that are dropping Common Core, including Huck Finn's home state of Missouri. I've mentioned before that the Show-Me State was considering ending Common Core -- well, it's official:

Traditionalists may appreciate the fact that the Missouri standards explicitly divide the high school math standards into classes Algebra I, Geometry, and Algebra II.

Meanwhile, New York State isn't dropping Common Core yet, but the opt-out movement is as strong as ever in the Empire State:

Finally, Massachusetts is also considering dropping Common Core Standards and replacing them with the old pre-Core standards -- and Michigan may adopt the Bay State's old standards too:

This last link leads to a heated 2000+ comment debate thread. Many of the posts are off-topic, but here is one from an apparent traditionalist:

I've dealt with this very issue as a result of a move and was smart enough to marry a math wiz who's primary area of study was aeronautical engineering. Our primary concern was our oldest child being challenged enough in the new school, which was properly addressed with advanced math class options. 
If the issue was left up to me, I would consult with the right people on a local level and make the proper decision. The last thing I would need is the federal government being involved.  

Now try looking at this post from Huck Finn's perspective -- Huck Finn, of course, would avoid the advanced classes mentioned in this post.

(The rest of that comment turned into the federal vs. state vs. local control debate. As I've mentioned in previous posts, I'm torn between favoring federal and state control as many other countries have national standards while others have the equivalent of state standards. Purely local standards, on the other hand, deviate too far from what other countries do.)

Question 3 of the PARCC Practice Test is on dilations:

3. A dilation with a center at P(0, 0) and a scale factor k is applied to MN. Let M'N' represent the image of MN after the dilation.

Select each correct statement.
A. If k > 0, then M'N' > MN.
B. If k > 1, then M'N' > MN.
C. If 0 < k < 1, then M'N' < MN.
D. If 0.5 < k < 1.5, then M'N' < MN.
E. If k = 1, then M'N' = MN.
F. If k = 0.5, then M'N' = 0.5(MN).

This question is a simple dilation problem. It can be solved using the following theorem from Lesson 12-1 of the U of Chicago text:

Let S_k be the transformation mapping (x, y) onto (kx, ky). Let P' = S_k(P) and Q' = S_k(Q). Then
(1) Line P'Q' | | PQ,
(2) P'Q' = k * PQ.

This theorem is stated again in Lesson 12-3 as "Size Change Distance Theorem." We know that "size change" means dilation, so we could call it the Dilation Distance Theorem. On the blog, we actually call it the Dilation Postulate in order to avoid circularity down the road.

Whatever we call this result, this is what we use to answer the question. If the scale factor k is more than 1, then the dilation is an expansion, and M'N' > MN, so (B) is correct. If the scale factor k is between 0 and 1, then the dilation is a contraction, and M'N' < MN, so (C) is correct. If the scale factor k is exactly 1, then the dilation is the identity, and M'N' = MN, so (E) is correct. And choice (F) is exactly the Dilation Postulate in the case k = 0.5, so (F) is correct.

Choice (A) is wrong, because it is directly refuted by (C). And choice (D) is wrong, because it is directly refuted by (B). So the correct answers are (B), (C), (E), and (F). For the second day in a row, we have four correct answers, and so for the second day in a row, the most common student error will be the failure to mark all four answers.

Today is an activity day. Recall that we had an entire link of lessons to try out back in February. Well, here is a lesson on dilations:

The author of this blog is Julia Finneyfrock, a North Carolina private high school teacher. Finneyfrock refers to herself as the "Designated Deriver."

Like many of the lessons I linked to back in February, Finneyfrock's is highly computerized. She mentions three different apps -- PearDeck, Desmos, and polygraph -- in her lesson (and she even mentions a fourth app, Haiku, just to check homework). Here is an excerpt of the lesson:

When they are finished checking their homework, my students then enter our class code in onPeardDeck . PearDeck creates interactive lectures. I embedded a Desmos Activity into the first part of my lesson for every transformation. As soon as my students log into PearDeck they are taken directly to this activity and are able to start at their own pace.

Next, students were able to practice dilations on PearDeck together. I gave my students a shapes and the dilation and they drew the dilation on their slide (I let them draw pictures if they finish early). I was then able to overlay all of the drawings to see if any of the students were off, and I could show individual students work as well.

We practiced 4 dilations together, then they were taken to another Desmos Activity. This activity is a polygraph game that I created for students to practice describing all of the transformations they learned so far (reflections, translations, dilations). Students played this game for about 10-15 minutes at the end of class. Students LOVE polygraph. I literally had to kick students out of my class because they didn’t want to leave.

I have nothing against computerized lessons -- after all, the PARCC and SBAC are on computers, so students must be familiar with them in order to be successful. And of course, one application of dilations is computer graphics -- contracting objects so that they fit on the screen.

The problem I have is that we can't have a computerized lesson unless we have the software. All the PearDeck this and Desmos that in the world means nothing unless we have PearDeck and Desmos installed on enough computers in the classroom for the students to enjoy them. As a sub, I've seen a few (mostly eighth grade) math classes with computers. But these were used mostly as online texts, not for anything described in Finneyfrock's post. The best I can do is convert her lesson into a more paper-based lesson and post it onto the blog.

How did Finneyfrock's students react to this assignment? Let's find out:

“Class is over already? I don’t want to go to my next class!”
“I love this! Can we do this more often?”
“This is so much fun”

Unfortunately, I doubt that my paper version of this assignment will draw the same response. Any interactive computer assignment will be more fun than a paper version. Even Huck Finn, if he lived in the computer age, might actually enjoy this math assignment for once! On the other hand, let's see whether any assignment endorsed by traditionalists can elicit the same response from students in a Geometry class.

It's clear that Finneyfrock has completely embraced the use of computers and other forms of technology in her classroom. If you have access to some of the same software that she does, you may wish to read about the assignment directly from her blog at the link above.

Otherwise, I am posting my paper version below. One thing I noticed about most of Finneyfrock's transformations is that she often scales by a different factor in the x- and y-directions. I've mentioned before that while such a transformation is a linear transformation, it is not a true dilation. The Dilation Postulate/Dilation Distance Theorem, the main result of this lesson, only applies to true dilations with the same scale factor in all directions.

PARCC Practice EOY Question3
U of Chicago Correspondence: Lesson 12-3, Properties of Size Changes
Key Theorem: Size Change Distance Theorem (Dilation Postulate)

Under a size change with magnitude k > 0, the distance between any two image points is k times the distance between their preimages.

Common Core Standard:
The dilation of a line segment is longer or shorter in the ratio given by the scale factor.

Commentary: The first three lessons of Chapter 12 all cover basic properties of dilations. In the posted activity, I focus on true dilations, not mere horizontal/vertical stretches. But I maintain the spirit of Finneyfrock's lesson by providing an answer sheet. If this answer sheet is printed directly onto a transparency, then it can be placed directly over a student sheet in order to check the answers, just like Finneyfrock's lesson does, except ours is comparatively low-tech.

Thursday, April 21, 2016

PARCC Practice Test Question 2 (Day 145)

Today I subbed in an English class, but this teacher also has an AVID class. Today is a tutorial day, and I've said before that more often than not, the subject the students need help in is math.

These students are sophomores, so I assume they're enrolled in Integrated Math II. The students appeared to need help on a hodgepodge of problems, including many geometry problems:

-- Exterior Angles of a Triangle
-- Slopes of Parallel and Perpendicular Lines
-- Circumcenter of a Triangle
-- Triangle Inequality
-- Vertical Angles
-- Complementary/Supplementary Angles
-- Unequal Sides/Angles
-- Translations on a Coordinate Plane

These topics are all over the place -- it makes me wonder, are the students preparing for a unit test, or a semester final? We know, of course, that geometry is squeezed into the final modules of the Common Core Integrated Math texts, but still, why must so many topics appear on one test?

Well, I guess I can't complain. Last week's "Chapter 15 Test" posted to the blog arguably contains so many topics as well. The problem was that I wanted to cover three lessons on circles that are important for PARCC, but they appear in three different chapters in the text (11-3, 13-5, 15-3). That plus returning to Chapter 8 to celebrate Pi Day and Square Root Day made for a crowded test indeed.

Back in October, I gave the contents of an Integrated Math II text, so I can only assume that the students are working out of that text. That text contains 24 module, with the second volume having Modules 14-24, mostly on geometry and probability.

Notice that most of these topics appeared during the first semester here on the blog. Actually, most first semester Geometry topics appear during Integrated Math I, with Math II consisting of the second semester Geometry topics like similarity and basic trig. In this text, Modules 14 and 15 are a review of Math I geometry, with Modules 16-21 containing the new material for Math II.

Some readers may point out that this is an argument against Integrated Math and in favor of the more traditionalist Algebra I, Geometry, Algebra II sequence. Most texts contain review lessons because we can't expect students to remember everything from previous years -- but this text reviews an entire semester (first semester of Geometry or second semester of Math I) in just two modules! On the other hand, with the Algebra I, Geometry, Algebra II sequence, there's no need to review Geometry because it's all contained in a single course. Of course, there will still be much review of Algebra I during Algebra II, but this can be accomplished more gradually since the entire year of Algebra II is devoted to algebra topics.

Today I help out one student on the translation problem. He was given coordinates the three vertices of a triangle and had to translate the triangle five units right. I notice that he draws the triangle first and then moves the triangle, when he could have added five to the x-coordinates first. I can't be sure, though, that this is how the teacher prefers him to do it.

The questions that trouble me the most are the circumcenter questions. This is because this topic was not emphasized as much before the Common Core. I must double-check to make sure that my answers are correct.

Unlike the last time I covered an AVID class, this time I'm more careful to avoid simply blurting out the answers. One student is working on a distance formula problem and writes 4^2 = 8 midway through the calculation. So I just asked "What's 4^2?" The student soon corrects herself. At the end, I say that I did a much better job of helping out today than I did last time.

Chapter 2 of Morris Kline's Mathematics and the Physical World is called "Discovery and Proof." I must emphasize that word "proof" that comes up so often in Geometry class.

Once again, Kline starts out the chapter with a quote:

Since then reason is divine in comparison with man's whole nature, the life according to reason must be divine in comparison with usual human life. -- Aristotle

Kline begins:

"Mathematical reasoning is commonly regarded as different from and even superior to the reasoning other studies utilize."

And of course, many Geometry students lament the differences between mathematical and other reasoning when it's time to write proofs.

Kline then distinguishes between inductive and deductive reasoning. Inductive reasoning is when we extrapolate from a few occurrences of an event that the event will always occur. His example is the assumption that all substances heat when expanded, since most do. But inductive reasoning doesn't always work -- a counterexample to that assumption above is water. Mathematicians therefore employ deductive reasoning in their proofs. I notice that some Geometry texts make a big deal about inductive and deductive reasoning, often in their chapters on logic (usually Chapter 2). But the U of Chicago text does not mention inductive or deductive reasoning.

Well, I do know of one book that mentions inductive and deductive reasoning in its second chapter --namely Kline's. And much of this chapter is indeed all about Geometry.

Kline's main example is the sum of the angles of a triangle. He explains how at first consider this question inductively by looking at a few triangles and measuring their angles. Then he gives the usual deductive proof that their sum is 180 degrees -- the same proof that we find in Lesson 5-7 of the U of Chicago text. He points out that this theorem can be generalized to find the angle sum of all polygons, and indeed Lesson 5-7 does exactly that. But then Kline gives a less familiar example:

"AB, CD, EF, and GH are parallel chords of a circle. The centers of these chords lie on the one line. A circle, however, is a special case of an ellipse. Hence it is likely that a similar theorem holds for the ellipse. This generalization can be proved and yields another theorem."

This theorem is not proved anywhere in the U of Chicago text, not even for the circle. There is a theorem in Lesson 15-1 of the text that may help us out:

Chord-Center Theorem
a. The line containing the center of a circle perpendicular to a chord bisects the chord.
b. The line containing the center of a circle and the midpoint of a chord bisects the central angle determined by the chord.
c. The bisector of the central angle of a chord is perpendicular to the chord and bisects the chord.
d. The perpendicular bisector of a chord of a circle contains the center of the circle.

Using this theorem, we can prove Kline's claim about circles. For lack of a better name, let's call it:

Kline's Theorem:
If three or more chords of a circle are parallel, then their midpoints are collinear.

Our plan is to use part d of the Chord-Center Theorem, so we begin by looking at the perpendicular bisectors of our chords (which, by definition, pass through their respective midpoints). We use the same trick that the U of Chicago text uses to prove the Two Reflection Theorem for Translations -- since the chords are parallel to each other, their respective perpendicular bisectors must also be parallel to each other, using the Two Perpendiculars Theorem and Perpendicular to Parallels (which we call the "Fifth Postulate" here on the blog). But all of those "parallel lines" contain the same point, namely the center of the circle (by part d). So those lines are in fact identical. (The U of Chicago text includes identical lines as parallel -- if we wish to avoid this, we replace "parallel" with "parallel or identical" above.) That single line is the common perpendicular bisector of all the parallel chords -- it contains all the midpoints, so they are collinear. QED

Of course, we don't bother to prove the theorem for ellipses. This also goes back to a debate as to whether circles counts as ellipses. Some people say that a circle isn't an ellipse, but rather the limiting position of an ellipse (where the distance between the foci is zero), just as a parabola is the limiting position of an ellipse (where the distance between the foci is infinite).

Here Kline does count a circle as an ellipse, and states a theorem about ellipses that automatically applies to circles as well. So in the spirit of the U of Chicago's inclusive definitions, we should definitely count a circle as an ellipse. But the theorem about ellipses is not worth attempting to prove here on the blog (since ellipses don't come up in Geometry classes). I notice that based on the pictures in his book, the line containing the common midpoint in an ellipse isn't necessarily perpendicular to all of the chords (though of course it is in a circle, as we just proved it).

Kline ends the chapter with a quote from Alexander Pope:

"Who could not win the mistress, must be content to win the maid."

Question 2 of the PARCC Practice Test is about the symmetries of a regular polygon:

2. Octagon PQRSTVWZ is a regular polygon with its center at point C.
Which transformations will map octagon PQRSTVWZ onto itself?
Select each correct transformation.
A. reflecting over QV
B. reflecting over RW
C. reflecting over TZ
D. rotating 45 degrees clockwise around point Z
E. rotating 135 degrees clockwise around point C
F. rotating 90 degrees counterclockwise around point C

This question has multiple answers. For the three reflections, the valid mirrors are any line that bisects two angles, two sides, or one angle and one side. In particular, the mirror must pass through the center of the polygon, so only (A) and (B) are lines of symmetry. For the three rotations, any rotation whose magnitude is a multiple of 360/n degrees, in either direction, maps a regular n-gon to itself -- here n = 8 and so we need multiples of 45 degrees. Once again, though, the center of the rotation must be that of the polygon, so only (E) and (F) map our octagon to itself.

So there are four correct answers (A), (B), (E), and (F). As with any problem with multiple answers, the most common student error will be to omit some of the right answers.

I think this is a great problem to include on the PARCC test. My only problem is that the coverage of this type of problem in the U of Chicago text -- a text which emphasizes transformations more than most -- is highly disappointing.

One of the Common Core Standards states that the students should know of the transformations that map a figure to itself. The existence of an isometry that maps a figure to itself has a special name in math -- symmetry. So this question asks for the symmetries of the regular octagon -- that is, the ways in which the regular octagon is reflection-symmetric or rotation-symmetric.

Lesson 4-6 of the U of Chicago text is on reflecting polygons, and the following lesson introduces reflection-symmetric figures. But reflection-symmetric polygons aren't covered until Chapter 5 -- and for the most part, only isosceles triangles and trapezoids appear (as in the Isosceles Triangle Symmetry Theorem of Lesson 5-1, the Isosceles Trapezoid Symmetry Theorem of Lesson 5-5). In particular, the symmetry of polygons with five or more sides is not mentioned. And what's worse is that the rotational symmetry of any polygon fails to appear anywhere in the U of Chicago text!

The term regular polygon is defined in Lesson 7-6 of the U of Chicago text. In this lesson, the only theorem involving regular polygons is stated and proved:

Center of a Regular Polygon Theorem:
In any regular polygon there is a point (its center) which is equidistant from all its vertices.

Given: regular polygon ABCD...
Prove: There is a point O equidistant from A, B, C, D, ...

I won't restate the proof here, but in the course of the proof, it's stated that OABC is a kite with symmetry diagonal OB. Also, we know that OAB is an isosceles triangle, and the perpendicular bisector of AB is the symmetry line for that triangle. But we still have yet to prove that the symmetry lines of the kite or triangle are also symmetry lines for the entire regular polygon.

The text proves the Center of a Regular Polygon Theorem using a technique called induction, which I explained back in December during the Lesson 7-6 post. Induction is not the same as inductive reasoning, even though some authors (like Kline, for example) use the term "induction" when they mean "inductive reasoning." Indeed, mathematical induction is considered deductive reasoning!

After this proof, the U of Chicago text proceeds:

"The Center of a Regular Polygon Theorem implies that there is a circle which contains all the vertices of a regular polygon. This enables regular polygons to be drawn quite easily. Draw the circle first and equally space the vertices of the polygon around the circle."

This is tantalizingly close to a description of the symmetries of the regular polygon. Indeed, the symmetries of the polygon are the exactly the symmetries of the circle that preserve that set of vertices spaced equally around the circle.

But believe it or not, there is actually one more lesson in the text that mentions regular polygons -- Lesson 15-2, "Regular Polygons and Schedules." This lesson teaches us how we can schedule a round-robin tournament  -- one in which every team plays every other team. The opening round of the World Cup of Soccer is round-robin in that every team in the group plays every other team. But we rarely seen round-robin tournaments with more than four teams.

I skipped Lesson 15-2 since I doubt that round-robin tournaments appear on the PARCC. But let's look at how the text schedules a round-robin tourney for seven teams:

Step 1. Let the 7 teams be the vertices of an inscribed regular 7-gon (heptagon).
Step 2. (the first week) Draw a chord and all chords parallel to it. [But how do we know that any of the chords joining vertices are parallel? dw] Because the polygon has an odd number of sides, no two chords have the same length. This is the first week's schedule.
Step 3. (the second week) Rotate the chords 1/7 of a revolution.

And now we have the text rotating the regular heptagon to itself without even stating, much less proving, any sort of symmetry result.

So let's finally state and prove our theorem:

Regular Polygon Symmetry Theorem:
a. Any perpendicular bisector of a regular polygon is a symmetry line for the polygon.
b. The line containing any angle bisector of a regular polygon is a symmetry line for the polygon.

So how should we prove our theorem? We see Step 2 above about parallel chords -- and we're immediately reminded of Kline's Theorem that we proved earlier in this post! By the way, my intention in reading Kline and covering the practice PARCC at the same time was not to apply anything mentioned in Kline to the PARCC questions. But if that opportunity presents itself, then of course we're going to seize that opportunity.

We need one more theorem, also mentioned in Lesson 15-1, before we can prove our main result:

Arc-Chord Congruence Theorem:
In a circle or in congruent circles:
a. If two arcs have the same measure, then they are congruent and their chords are congruent.
b. If two chords have the same length, then their minor arcs have the same measure.

By the way, we skipped Lessons 15-1 and 15-2 because they don't appear on the PARCC. Notice that we're using material from those lessons to prove the Regular Polygon Symmetry Theorem which does appear on the PARCC, but not the results from 15-1 and 15-2 directly.

So let's begin the proof of the Regular Polygon Symmetry Theorem.

To prove part a, we let AB be any side of a regular polygon. We want to show that the perpendicular bisector of AB is a symmetry line for the polygon.

Let C be the next vertex after B on the opposite side of A, and D be the next vertex after A on the other side of B. In other words, D, A, B, C are four consecutive vertices of the polygon, and we have DA, AB, BC as three consecutive sides of the polygon.

Now DA = BC as these are sides of a regular polygon. So by part b of the Arc-Chord Congruence Theorem, minor arcs DA and BC are congruent. Notice that the longer arcs ADC and BCD also ahve the same measure by the Arc Addition Postulate. Now the angles ABC and BAD are inscribed in the arcs ADC, BCD respectively, so by the Inscribed Angle Theorem the measures of the two angles are half those of the arcs. Since the arcs have equal measure, so do the angles ABC and BAD.

We also can use Arc Addition and add the arc AB to the congruent arcs DA and BC. This tells us that the arcs DAB and ABC are congruent, and so their inscribed angles BCD and ADC are congruent.

Now we look at the quadrilateral ABCD. We see that it has two pairs of congruent angles -- ABC and BAD, then BCD and ADC. Notice that this is sufficient information to conclude that ABCD is in fact an isosceles trapezoid. If we use the blog definition of isosceles trapezoid, then a pair of congruent adjacent angles and a pair of congruent opposite sides (DA = BC, if you recall) is enough to prove that ABCD is an isosceles trapezoid, and since an isosceles trapezoid is a trapezoid, we conclude that sides AB and CD are parallel.

At this point, you may be wondering why we dealt with arc measure at all -- we already know that the angles ABC and BAD are congruent because they're angles of a regular polygon, without any need to look at arc measure. The reason is that angles BCD and ADC are not angles of a regular polygon, so we need the argument above to prove that they are congruent. Moreover, the next step is to take E as the next vertex after D and F as the next vertex after C. The goal is to prove that CDEF is also an isosceles trapezoid -- and none of the angles of CDEF are angles of the regular polygon.

But I admit that it is possible to avoid the Arc-Chord and Inscribed Angle Theorems by using induction, similar to the Center of a Regular Polygon Theorem. The initial step is that ABC and BAD are congruent (as they are angles of a regular polygon). The inductive step is that since ABC and BAD are congruent, ABCD is an isosceles trapezoid, and so ADC and BCD are congruent. These congruent angles are subtracted from the angles ADE and BCF (which are angles of the regular polygon, hence congruent) to conclude that CDE and DCF are congruent. This is enough to conclude that CDEF is an isosceles trapezoid, and so on down the regular polygon.

So we have divided the regular polygon into isosceles trapezoids. Each vertex is connected to another vertex via one of several parallel chords. So now by Kline's Theorem, these midpoints of all these chords are collinear -- indeed, we see from the proof of Kline's Theorem above that this line passes through the center of the circle (i.e., it's a diameter) and is the perpendicular bisector of all of them.

But recall that this line -- the perpendicular bisector of all the chords, including the original AB -- is exactly the mirror of our desired reflection! Therefore, by the definition of reflection, the mirror image of C must be D, the mirror image of E must be F, and so on down the polygon. Therefore the perpendicular bisector of AB is a symmetry line of the polygon. QED part a

Now for part b, we look at the bisector of angle B, which we take to be between the vertices A and C, in that order. Again AB = AC as these are the sides of a regular polygon, so ABC is, in fact, an isosceles triangle. In Lesson 5-1, we know that the bisector of the vertex angle is the perpendicular bisector of the base, so our mirror is the perpendicular bisector of AC. Then we proceed the same way as part a -- ACFD is an isosceles trapezoid because it has congruent base angles CAD and ACF, found using either arcs or subtracting the angles BAC and BCA (congruent base angles of the isosceles triangle ABC) from the angles BAD and BCF (congruent angles of the regular polygon).

In either case a or b, the final figure at the bottom of the regular polygon may be either an isosceles triangle or an isosceles trapezoid (depending on odd or even number of sides). In both cases, the entire polygon has been divided into isosceles triangles and trapezoids via a number of parallel chords all with the same diameter, the mirror, as a perpendicular bisector. Therefore the mirror, the angle bisector of B, is a symmetry line of the polygon. QED part b

This completes the proof of the Regular Polygon Symmetry Theorem. It tells us that every regular polygon has many lines of symmetry. But though an n-gon has n angle bisectors and n perpendicular bisectors, it has only n total lines of symmetry. This is because every line of symmetry is a diameter that touches the polygon twice. So every angle bisector must also bisect a second angle, or else is the perpendicular bisector of a side, and the same is true of every perpendicular bisector. This double counting tells us that there are only n symmetry lines, not 2n.

All that's left now is rotational symmetry. We see that the composite of two reflections in intersecting lines is a rotation, and all of our symmetry lines intersect at O, the circumcenter of the polygon. Any figure with intersecting symmetry lines also has rotational symmetry, with the center of the rotation the same as that of the polygon. But what is the magnitude of the rotation?

We can draw n radii to divide a regular n-gon into n triangles. These triangles are all congruent isosceles triangles by SSS, and the radii each bisect an angle of the polygon. Since the angles at O all are congruent and add up to 360 degrees, the central angle between two angle bisectors is 360/n. The perpendicular bisectors also bisect these central angles (Lesson 5-1), and so the central angle between an angle bisector and a perpendicular bisector is 360/2n. That is, the symmetry lines of the polygon are spaced 360/2n degrees around O (but again, there are only n lines, not 2n, since each line is a diameter, not just a radius).

The Two Reflection Theorem for Rotations tells us that the magnitude of the rotation is exactly twice the angle between the mirrors -- and the angles are multiples of 360/2n. So the magnitude of the rotation must be multiples of 720/2n, or 360/n. So a regular n-gon has n-fold rotational symmetry -- rotating it by any multiple of 360/n degrees maps the polygon onto itself.

The set of all symmetries of a figure is called the symmetry group of that figure. The Regular Polygon Symmetry Theorem tells us that an n-gon has a symmetry group with 2n elements -- n of these are reflections and the other n are rotations.

PARCC Practice EOY Question 2
U of Chicago Correspondence: Lesson 7-6, Properties of Special Figures
Key Theorem: Center of a Regular Polygon Theorem

In any regular polygon there is a point (its center) which is equidistant from all its vertices.

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

Commentary: Lesson 7-6 of the U of Chicago text only mentions that a regular polygon has a center -- not that the polygon has any symmetry. We come up with a Regular Polygon Symmetry Theorem to help us answer this question. The theorem tells us that reflecting our regular octagon in any diagonal containing the center C, or rotating it around the center C in any multiple of 360/8 = 45 degrees, maps the octagon to itself.