PARCC Practice Test Question 14 (Day 157)

Today is the second of five days that I'm subbing in an Algebra II and Integrated Math II class. It is also the midpoint of the fourth quarter of the year -- the end of the seventh quaver.

In Algebra II, the students begin working on a Pizzazz packet for basic trig. Today's worksheet is page 165, "Greek Decoder." Students are given two sides of a right triangle and they must use the Pythagorean Theorem to find the length of the third side. Then they match up the letter for each problem with the Greek letters in the answer key to decode the message. (I assume that the Greek part is because Pythagoras was Greek.) The letters don't actually correspond to the real alphabet -- for example, "nu" is S and "pi" is O. The final message is, "Pythagoras was a famous Greek who knew all the right angles!" I believe most students figured out how to complete the worksheet.

During my third period conference, I am assigned to an AP Spanish class. Apparently, it is a class tradition to watch a movie the week after the AP Spanish exam. And the movie they watch today is Stand and Deliver, about the world's most famous math teacher -- Jaime Escalante, who successfully taught AP Calculus AB to students at a school right here in Southern California. (Well, maybe he's the most famous math teacher other than Pythagoras, who taught the members of his brotherhood.)

In Integrated Math I, students continue working on the Geometric Constructions Project. Here are a few things I want to say about the project:

-- First of all, the teacher suggests that the students watch some YouTube videos in order to learn how to perform the constructions. Here are the first two videos mentioned on this project worksheet:

The teacher provides some more links, but I don't post them here to the blog. This is because three of them show how to construct the regular octagon and pentagon, but constructing these challenging figures is not necessary for the project. The last link (Julie Nowak) is actually to a playlist of all of the same videos listed under David Nicholls, so there's no need for me to link to the playlist.

-- On this blog, I'm focusing on the Math I classes because Math I is almost identical to Common Core Math 8 -- a class I will be teaching next year. But as we found out last week, constructions don't appear in Math 8 (to the dismay of the traditionalists who want to see them in middle school), but they do appear in Integrated Math I -- well, Honors, at least. Actually, I spoke to one of the other teachers, and she told me that this is definitely an Honors-only assignment -- the regular students were having trouble just trying to bisect a segment! Again, we apologize to the traditionalists who want to push this down into middle school.

-- I recommend to the students that they start by constructing a hexagon -- even if the final figure contains an equilateral triangle. Let's say the students are drawing Figure #1 (from the project pages I linked to on Friday), which is two interlocking triangles (Star of David). It's much better to draw the hexagon first and connect every other vertex than to draw an equilateral triangle. It's also easier to remember how to construct the hexagon -- I might be tempted to use the hexagon to draw Figure #28, even though it's just an equilateral triangle with no hexagon necessary.

-- I decide to create a completed project of my own to show the students. Since students are to choose one design of their own, I also include my own design. I choose two interlocking circles and two equilateral triangles, as constructed in Lesson 4-4 of the U of Chicago text (Euclid's first theorem) or Level 1 of Euclid the Game.

That's right -- Euclid the Game! By now, many readers were probably wondering how long it would take me to mention Euclid the Game in this post, since every time I mentioned constructions this year on the blog, I inevitably bring up Euclid the Game.

Well, since the students are looking at the two YouTube videos listed above anyway, I want to mention the Euclid the Game website. But as it turns out, Euclid the Game is blocked on the district computers, as are all game websites! You'd think that the district would make an exception for an obviously educational game like Euclid, but oh well!

Actually, the author of the Euclid the Game website is in the process of changing his website and adding five new levels, to bring the total to 25. I tell the students that I've completed about 7-9 levels of Euclid so far, but of course this means nothing to them unless I can show them the game.

-- A major source of confusion is whether the students can complete this assignment in groups, or must they do it as individuals. Despite the teacher directing the class to divide into groups for the assignment, he wants the students to complete the assignment as individuals -- which he clarifies by sending a text.

-- So far, Figures #16 and #17 are the most popular -- probably because they are easy. To me, it's much better to do an easy figure correctly than a more complex figure incorrectly. Recall that constructions are all about exactness!

-- Many of the students insist on using a protractor as a straightedge, instead of a ruler. I don't know why they refuse to use a ruler for a straightedge. We know that protractors are forbidden in classical constructions, and the teacher specifically tells them to avoid protractors on the worksheet. I must be careful to make sure that they're not trying to sneak angle measure into their constructions.

-- Finally, I think I know why I'm unable to draw a circle with the teacher's whiteboard compass. My problem is that, as I hold down my finger to represent the center, the ribbon is wrapping around my finger, so that the radius is shorter after I wrap it around. Again, constructions are all about exactness, so it's not good for me to draw a whiteboard circle that's visibly imperfect and then insist that the students' circles be drawn correctly (even if they have the advantage of using a real compass).

I'll continue to write about the Math I students' progress as they work their way through the Geometric Constructions Project.

Chapter 14 of Morris Kline's Mathematics and the Physical World is "The Motion of Projectiles." In this chapter, Kline shows us the paths of objects that have been launched.

"I value the discovery of a single even insignificant truth more highly than all the argumentation on the highest questions which fails to reach a truth." -- Galileo

So we're still looking at Galileo's research on mechanics. Kline writes:

"By analyzing motion along a straight line, whether it be of bodies falling straight down or sliding down inclined planes, Galileo discovered some fundamental physical principles, namely, the first two laws of motion. These laws are in themselves quantitative statements about velocity, force, mass, and acceleration."

Now Kline considers an object (possibly a bomb) dropped from an airplane flying horizontally at 100 feet per second. Once again, the key to this idea goes back to my high school physics teacher and his line that "vectors operating at right angles are independent." So he writes equations for horizontal:

x = 100t

and vertical:

y = 16t^2

components of the motion. Then Kline uses substitution to eliminate t:

y = x^2 / 625

And this is the equation of a parabola. Later on in the chapter, Kline considers a projectile launched from a gun pointing at an angle A above the ground at velocity V. In order to consider the horizontal and vertical components of the velocity, he must use trig -- just as my Algebra II students are getting ready to learn:

v_x = V cos A
x = (V cos A)t = Vt cos A

v_y = V sin A - 32t
y = Vt sin A - 16t^2

Again, Kline eliminates t to obtain:

y = (tan A)x - 16 / (V^2 cos^2 A)x^2

which is yet another parabola. So Algebra students who wonder why they have to study parabolas only need to look at the equations of projectile motion to learn of their application.

An interesting question Kline asks is, what value of A maximizes the horizontal distance the projectile can travel? He points out that when A is 0, the projectile doesn't move from the ground, and when A is 90 degrees, it just goes straight up and down, not horizontally. So the maximum likely lies somewhere in between 0 and 90 degrees. To determine the answer, Kline sets the vertical height to 0 (that is, when the projectile lands on the ground), solves for t, then plugs this value into the equation for the horizontal distance, to obtain:

x = (V^2/16) sin A cos A

It's possible to use Calculus to find the maximum value of x. It's also possible to use trig to find the maximum -- consider the double-angle formula:

sin 2A = 2 sin A cos A

to obtain:

x = (V^2/32) sin 2A

Then the maximum occurs when sin 2A = 1 -- that is, when 2A = 90, so A = 45 degrees. But Kline uses neither Calculus nor the double-angle formula -- instead, he uses a clever trick.

He reminds us of the definitions of sine and cosine -- in a right triangle with legs a and b and hypotenuse c, sin A is just a/c and cos A is just b/c. So he writes:

sin A cos A = (a/c)(b/c) = ab/(c^2)

Kline then decides to fix c^2, and to maximize ab, we can maximize its square a^2 b^2 instead. So his goal is to maximize a^2 b^2 under the constraint that c^2 (which is a^2 + b^2) is fixed. And now he has reduced this to a previously solved problem:

"We can regard a^2 and b^2 as sides of a rectangle of semiperimeter c^2, which is fixed. The quantity a^2 b^2 is the area of such a rectangle. Of all such rectangles, the area is largest when the rectangle is a square, that is, when a^2 = b^2. Hence the product ab is also a maximum when a = b. In this case, the right triangle above is isosceles and angle A is 45 degrees. That is, for any given muzzle velocity, the maximum range is obtained by firing at an angle of 45 degrees. This result, a famous one, is Galileo's."

So we can see how powerful this trig stuff and the Pythagorean Theorem -- on which my Algebra II students are working today -- can be. It also shows us how remarkable Galileo was to have discovered so many properties of moving objects. But Kline's coverage of the famous Italian scientist is at an end, for Kline closes Chapter 14 with the following:

"But Galileo had already done more than one man's share and in 1642 his infirm body refused to function any longer."

By the way, I've posted the Geometric Constructions Project as if it were a one-day lesson, for last Friday's activity day. But we see that in the actual class, the students have a full week. I don't post week-long projects on the blog and treat everything like a one-day lesson or activity. And so I will post PARCC problems here on the blog as scheduled. But for teachers who are reading this blog, keep in mind that it's more realistic for the students to take a full week to finish the activity.

Question 14 of the PARCC Practice Exam is on rotations and reflections on a coordinate plane:

14. The right triangle in the coordinate plane is rotated 270 degrees clockwise about the point (2, 1) and then reflected across the y-axis to form triangle A'B'C'.

Drag and drop the correct orientation for triangle A'B'C' into the correct orientation on the coordinate plane.

(Here are the coordinates of triangle ABC: A(2, 1), B(5, 6), C(5, 1).)

We've seen before when rotating triangles that when the center of rotation is not the origin, it's always placed at one of the vertices of the triangle -- in this case A. So the rotation image of A is again A. At this point, we could try to find the rotation images of B and C -- but since the question asks us to choose the correct orientation and place it on the plane, it's far better to keep track of the orientation of a single vertex and the orientation of the triangle. Notice that 270 degrees clockwise is equivalent to 90 degrees counterclockwise, so our image would be blue triangle in the upper-left corner, placed so that the image of A is itself.

But that's just the rotation image. Our transformation is the composite of a rotation and a reflection, so we must still perform the reflection across the y-axis. So the final image A'B'C' looks like the blue triangle in the upper-right corner, and the mirror image of A(2, 1) is A'(-2, 1), so this is where we place the image A'.

PARCC Practice EOY Question 14
U of Chicago Correspondence: Lesson 6-3, Rotations
Key Theorem: Two Reflection Theorem for Rotations

The rotation r_m o r_l, where m intersects l, "turns" figures twice the non-obtuse angle between l and m, measured from l to m, about the point of intersection of the lines.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.CO.A.5
Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, e.g., graph paper, tracing paper, or geometry software. Specify a sequence of transformations that will carry a given figure onto another.

Commentary: Even though the Integrated Math I classes are studying constructions now, in the classroom I found several old worksheets scattered around -- including these Kuta worksheets that I saw. Notice that Kuta's rotations are always centered at the origin, unlike the PARCC's which are often centered at a vertex. On the other hand, Kuta's reflections sometimes have mirrors that are parallel to the axes, not just at the axes themselves.

PARCC Practice Test Question 13 (Day 156)

Today I subbed in a high school math class -- in fact, this teacher is scheduled to be out for a full week, up to and including this upcoming Thursday. So I will have plenty of things to say about this class here on the blog.

This teacher has three sections of grandfathered Algebra II (1st, 4th, and 5th periods), as well as two sections of Honors Integrated Math I (2nd and 6th periods).

The Algebra II students are taking a quiz on statistics. But according to the plans for next week, the students will be working on trigonometry. I'm actually surprised that I'm seeing trig in a regular Algebra II class (as opposed to Honors). Then again, these worksheets appear to be the simpler trig of Chapter 14 of the U of Chicago Geometry text. We'll see how these students do on these Pizzazz worksheets next week.

By the way, during the teacher's 3rd period conference, I cover another math teacher, who had an Integrated Math II class. But this teacher tried to leave a video for the students to watch -- the video didn't work, and so all of the subs providing period coverage just make it a free period. It appeared, though, that the Math II students are also learning about trig --it's also Chapter 14 Trig.

Now the class I want to focus on here on the blog is Integrated Math I. The students in that class are starting a project that involves geometric constructions. Because today is an activity day anyway, I will post this assignment as today's activity. (Yes, I know that this project has nothing to do with PARCC Question 13, but activities that occur in the classroom always take priority over other plans for the blog.)

Here is the information from the worksheet:

Geometric Constructions Project
Overview: Your job is to construct any two of the attached figures using only a straightedge and a compass and then to create one more of your own design. Each of the three figures must be neat and decorated sufficiently to hang on the wall. For one of the three figures you will be required to turn in at least three partially completed constructions (illustrating the progression of steps required to create it) and a fully completed, but undecorated version (all construction marks remain, but it isn't colored). All of the work is due at the end of class on 5-13-16 [the day that the teacher returns -- dw].

This project is somewhat tricky. The students tell me that the they've seen their teacher perform a few basic constructions such as drawing a circle using a compass, but they haven't seen anything major yet such as constructing an equilateral triangle or hexagon.

Of course, these appear in the Common Core Standards:

CCSS.MATH.CONTENT.HSG.CO.D.13
Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle

Today, I show the students how to perform these constructions, so that they can create the more difficult designs next week.

Before we get to our PARCC question, let me point out that this is a traditionalists-labeled post. But now that I'm getting ready to work at a school in the fall, I want to tone it down from having all these weekly posts where I quote comments posted on articles from various newspapers. The focus of any teacher blog should be what's happening in the classroom. It will still be a few more months before I get my own classroom, but I want to start thinking about my future classroom here on the blog.

(By the way, I have a special traditionalist post planned for this upcoming Wednesday. It will refer to something that will happen both in the class I'm subbing this week and the class that I will be teaching this fall.)

Still, I have several loose ends to tie up regarding traditionalists today:

-- Minnesota has adopted the pre-Core Massachusetts standards. Earlier, I'd written that they were considering the change -- since then that change has been made official.

-- The game show Jeopardy! made fun of Common Core again. On last night's episode, there was a category called Common "Core" -- the word "core" in quotes implies that this four-letter word will be a part of every correct question. One of the contestants, a history teacher, used this opportunity to criticize the Common Core Standards. She wasn't specific -- she just implied that the Common Core was bad, period.

Of course, we know that there are no Common Core Standards for history, so this teacher shouldn't be affected by the Core. Then again, there's a link between AP US History and Common Core in that David Coleman is considered the architect of both. Some people believe that Coleman introduced a liberal bias to the AP US History test (which students are taking today, in fact), and so many people link this bias to the Common Core itself.

-- First Daughter Malia has made her college decision. She'll be going to Harvard in fall 2017. I'd been writing about Malia because of Presidential Consistency -- many Common Core opponents don't like it when politicians, especially the president, promote Common Core yet insulate their own children from the ill effects of the Core.

In particular, I wasn't sure which of the three Sidwell Friends tracks Malia was placed. But we must assume that she was placed on the highest track since she's been admitted to an Ivy League school. Of course, this track leads to AP Calculus (presumably, Malia took her AP test yesterday).

This will draw the ire of the traditionalists and other Common Core opponents. President Obama promotes the Common Core, which doesn't encourage eighth grade Algebra I or senior year Calculus, yet he makes sure that his daughter is on the Calculus track at Sidwell. His own daughter will get into Harvard, but it will be harder for Common Core students to get into the Ivy League, which will be looking for Calculus on the transcript.

-- And traditionalist Dr. Katharine Beals is at it again. This question comes from the 8th grade Common Core test given in her home state of Pennsylvania:

http://oilf.blogspot.com/2016/05/math-problems-of-week-common-core.html

Kelsey draws a series of right triangles with sides that have the lengths shown in the table below:

Lengths of Sides of Kelsey's Right Triangles (inches)
Triangle    First Leg    Second Leg    Hypotenuse
       A                1                   1                sqrt(2)
       B                1                   2                sqrt(5)
       C                1                   3                sqrt(10)
       D                1                   4                sqrt(17)
       E                 1                  5                sqrt(26)

Kelsey continues making right triangles following the same pattern she used to make the first five right triangles.

C. [Beals omits Parts A and B here, likely because she only has a problem with Part C -- dw] Explain why none of the right triangles Kelsey makes will have a hypotenuse with a rational number length.

Beals then provides an example of "a complete explanation," according to the scorer:

The only square roots that produce rationals are those that come out even or give an exact value. Like the square root of 9 or 16 or 25, unlike any of those that are listed. Kelsey's triangles all have lengths of hypotenuses that are not rational and produce numbers that do not come out even or give exact values. Like sqrt(17) which starts coming out as 4.12310562562... and never stops.

As usual with her "Math problems of the week" series, Beals asks several "extra credit questions":

Extra Credit: 1. In what sense is this a complete explanation? Hint: Consider the terms "even," "exact value," and "never stops." Consider as well the difference between answering a question and rephrasing the question as a statement. 2. Discuss the human element involved in scoring responses to open-ended questions on tests taken by millions of students. 3. Discuss the human element involved in scoring the verbal explanations submitted by tens of thousands of English-impaired students.

So let's look at each of these in turn:

1. I wonder what sort of explanation Beals would prefer to see. Let's attempt to rewrite the student response without using those three terms she wants us to avoid:

The only whole numbers whose square roots are rational are perfect squares, like 9, 16, or 25, unlike any of those that are listed. Kelsey's triangles all have lengths of hypotenuses that are irrational and whose decimals neither terminate nor repeat, like sqrt(17), which is approximately 4.12310562562.

Of course, stopping the decimal expansion at that point makes it appear that "562" might repeat, but in reality they don't. (In fact the last "2" is a rounded value -- it's actually a "1," but the next digit is "7," so "1" is rounded up to "2.")

But wait a minute -- we're avoiding phrases like "comes out even" and replacing them with more mathematically precise terms as "perfect square." Yet Beals often refers to mathematically precise terms as "labels" that should be avoided (for example, "number sentence," which is more precise than incorrectly calling an inequality an "equation"). So what does Beals really want here?

Let's save Question 2 for last and move on to Question 3:

3. When Beals worries about English learners, we've seen what she's really saying before -- she wants there to be a more symbolic answer. But we know that in reality, most students -- and probably most people -- dislike reliance on symbols.

If we use a, b, and c for the sides of a right triangle, then we can let a = 1 be the first leg of the triangle, to obtain c = sqrt(b^2 + 1). The real problem is that b^2 + 1 is never a perfect square, as b^2 obviously is a perfect square. The only two consecutive whole numbers that are perfect squares are 0 and 1, and 0 can't be the side length of a triangle. So c is always irrational.

The problem with an answer such as this one is eighth graders are unlikely to give it, because it requires algebraic expressions such as sqrt(b^2 + 1). Yet this is the sort of symbolism that Beals likes to see more of.

2. To some extent, I agree with Beals here. The real solution to this conundrum is simply not to include this sort of question on the test. I believe that if our Common Core tests are going to be given on the computer, then they should be able to report the scores instantly. So any question that can't be scored immediately, such as this Part C question here, should be thrown out. Presumably there's no problem with Parts A or B, so these can remain on the test.

By the way, in second period some students wonder why they have to learn classical constructions. I've stated before that the age-old question of "Why should we learn this?" or "When will we use this?" isn't asked when the lesson is easy, fun, or high-status even if they'll never use it in life. Though the drawing and coloring parts of our project are fun, the constructions aren't. And I obviously fail to show that constructions are easy when I have trouble figuring out how to use the teacher's "compass" (really a small ribbon tied to a marker) to draw a circle. (And we know that it's next-to-impossible for anything taught in math to lead to a high status.)

But since this is still a traditionalists-labeled topic, let's put it on them. Why, according to the traditionalists, should students learn classical constructions? How would they answer the question "Why should we learn this?" to students who will ask that question when any lesson isn't easy, fun, or of a high status?

I performed a Google search at Beals's website, but I couldn't find any mention of constructions, straightedges, or compasses. I tried another search for classical constructions using a different name, the known traditionalist, Barry Garelick. All I found were complaints that Common Core students must wait until high school to study constructions, rather than learn them in middle school (probably by 7th grade at the latest, since they want 8th graders to learn Algebra I).

So let's try to reconcile that -- traditionalists think these students should have learned constructions two years ago, while the students think they shouldn't have to learn constructions at all.
So for sixth period, I tried to prepare an answer the best I could. And here's what I wrote:

Why should we study classical constructions?

-- The ancient Greeks were sticklers for exactness. If it isn't exact, it isn't mathematics.
-- If done correctly (and that's a big if), straightedges and compasses allow you to construct circles, equilateral triangles, and regular hexagons exactly.
-- To the Greek mathematicians, the challenge of using only compasses and straightedges to construct exact geometric figures was fun, like a game or puzzle!
-- Doubling the cube, squaring the circle, and trisecting an angle were three ancient construction problems that even the Greeks found to be too difficult. After almost 2000 years, someone finally proved that these three problems are impossible.

At the end of the period, one student said that he enjoyed the lesson. So hopefully the rest of this activity will go well.

By the way, I saw two references to Theoni Pappas and her Mathematical Calendar today. First, in the Math II class I covered during conference period, apparently this teacher likes setting up calendars full of her own problems, with the date as the correct answer. All twelve months were posted around the classroom. (This is also something I might do in my own classroom next year.) The other reference is, today's question asks to give the number of lines of a certain figure -- which turns out to be a six-pointed star similar to the first question on today's project. Of course, that figure has six lines of symmetry -- and today is the 6th.

With everything else going on today at school, I hardly have time for Chapter 13 of Morris Kline's Mathematics and the Physical World, which is called "Motion on an Inclined Plane."

"But where the senses fail us reason must step in." -- Galileo

Yes, it was Galileo who first used an inclined plane for experiments. According to Kline, Galileo proved that neglecting friction, the amount of time it takes an object to roll down an inclined plane is the same as the time it takes the same object dropped from the same height to fall. My own high school physics teacher simplified this as:

Vectors operating at right angles are independent.

In this case, the vertical vectors are independent of the horizontal vectors -- which is why only the vertical vectors determine the time it takes the object to roll down.

Sorry, but we need to get into the PARCC question pronto:

13. Part A

The number of people who live in a unit of area is called the population density of the area. It is usually given as people "per square mile" or "per square kilometer."

A map of the Orchard Hill neighborhood is shown. The population of Orchard Hill is 360 people. The length of each block is the same and the length of 20 blocks is one mile.

What is the area in square miles of Orchard Hill?
A. 0.03 square mile
B. 0.15 square mile
C. 0.35 square mile
D. 0.60 square mile

Part B

What is the population density of the Orchard Hill neighborhood, given as the number of people per square mile?

For Part A, each block is 0.05 mile. The area of the neighborhood is 4 blocks by 3 blocks, or 0.2 mile by 0.15 mile, or 0.03 square mile, which is choice (A). For Part B, the density is 360 people/0.03 square mile, which equals 12,000 people per square mile. This may sound high, but actually most urban areas have a large population density -- the density of Washington DC is over 10,000 people per square mile.

PARCC Practice EOY Exam Question 13

U of Chicago Correspondence: Section 8-3, Fundamental Properties of Area

Key Theorem: Area Postulate

b. Rectangle Formula: The area of a rectangle with dimensions l and w is lw.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.MG.A.2
Apply concepts of density based on area and volume in modeling situations (e.g., persons per square mile, BTUs per cubic foot).

Commentary: The U of Chicago text discusses area, but not population density. The activity for today is not about population density, but about constructions.

PARCC Practice Test Question 12 (Day 155)

Chapter 12 of Morris Kline's Mathematics and the Physical World is called "Vertical Motion." In this chapter, we find out how fast objects fall.

"To give us the science of motion God and Nature have joined hands and created the intellect of Galileo." -- Fra Paolo Sarpi, contemporary of Galileo.

Kline begins:

"Galileo had formulated his program and he proceeded to put it into effect. Whereas scientists of Greek and medieval times had tried to embrace the whole of man and nature, Galileo, 'with the restraint that shows the master,' decided to select...the subject of motion."

So as we can see, Kline is still writing about Galileo's work. But now we've finally reached the great Italian scientist's most famous experiment:

"Hence, if air resistance is neglected, all bodies take the same time to fall a given distance. This is the lesson Galileo is supposed to have learned by dropping objects from the leaning tower of Pisa."

Galileo showed that all falling objects accelerate at 32 feet per second squared. From this, Kline is able to derive the equation d = 16t^2 for the distance an object falls in a given time. He does this without Calculus, but simply by finding the average velocity and multiplying by the time. This is the source of those -16t^2 problems that often appear in Algebra I courses (usually in the chapter on solving quadratics).

Question 12 of the PARCC Practice Exam is about segments and partitions:

The diagram shows MN graphed on a coordinate plane.

(Here are the endpoints: M(-4, 4), N(6, -2).)

Point P lies on MN and is 3/4 of the way from M to N. What are the coordinates of point P?

As it turns out, the correct answer is (7/2, -1/2) or (3.5, -0.5). Just by looking at the answer, we can tell that students will have trouble with this problem since the answer is not a lattice point. I hope that they wouldn't have trouble entering the answer in -- I assume that simply pressing the period key produces the decimal point and pressing the dash key produces the negative sign, and those are the only keys they need in addition to the digits.

But of course, we must figure out how the students are supposed to solve this problem. I wrote extensively about this type of question last year. This is what I wrote last year -- but there's one major change that I need to point out. (I also changed the discussion so that it refers to this year's problem, not last year's.):

We've discussed this type of problem here on the blog before. I pointed out that, while Lesson 11-4 of the U of Chicago text is on the Midpoint Formula, this sort of question where we are finding a point that divides the segment into a ratio other than 1:1 doesn't appear in the U of Chicago. Dr. Franklin Mason includes some questions like this in his text (his Lesson 13.3), and I added a quick activity on this blog, because both of us knew that this is mentioned in the Common Core Standards.

There are three ways to solve this problem. One of them is to start by using the Midpoint Formula to find O, the midpoint of MN, as ((-4 + 6)/2, (4 - 2)/2) = (1, 1). Then we use the Midpoint Formula to find the midpoint of ON, ((1 + 6)/2, ((1 - 2)/2) = (3.5, -0.5).

But the problem with this method is that it only works if we're dividing the segment into fourths or eighths (to find ratios such as 1:7 or 3:5), but not thirds (1:2) or fifths (1:4, 2:3). So we must find another method that works for general divisions of a segment. This is a lot of work, but dividing segments into ratios is important -- for example, in acoustics and music.

Here's one way I might teach this lesson. We note that the U of Chicago text introduces midpoints by discussing the center of gravity. We learned that the center of gravity of a set of points is the point whose x- and y-coordinates are the average, or mean, of the all of the respective coordinates of the points in that set. The center of gravity of two points is just the midpoint of the segment joining them.

(By the way, speaking of gravity, no -- Kline hasn't mentioned gravity yet. Galileo only calculated how fast objects fall, and not what causes objects to fall. We're still a few chapters away from gravity and Newton.)

Now as it turns out, we can divide the segment into other ratios by taking the mean of a list of coordinates with one of the points repeated as many times as indicated by the ratio. Since we want a 1:3 ratio here, we repeat one of the points three times:

Mean of MMMN: ((-4 + 6 + 6 + 6)/4, (4 - 2 - 2 - 2)/4) = (3.5, -0.5)

Notice that the resulting point P is closer to M or N depending on which point is repeated. In general, P will be closer to whichever point is repeated more often.

The U of Chicago text doesn't discuss finding the center of gravity when one or more points are repeated -- but then again, it doesn't cover the division of a segment into ratios adequately at all. Still, this is the best way to get from what appears on the U of Chicago to what appears on the PARCC.

The tricky part is making sure that the chosen points divide the segment in the correct ratio. A huge problem will be that the points that divide the segment into the ratio 1:3 are actually a quarter of the way on the segment. We see that when MP = 3 PN, we have MN = 4 PN. This sort of confusion often occurs in similarity problems as well -- the dilation mapping PN to MN has scale factor 4 (and if there are similar triangles with sides PN and MN, the sides of the latter would be 4 times those of the former), even though MP is only 3 times PN.

Depending on the wording of the problem, my center of gravity method minimizes this sort of error. To find the points dividing the segment into 1:3, we list one point once and the other point thrice. So the numbers in the ratio tell us how many times to list each point. Of course, the number 4 is still involved, as we must divide by 4 to find the mean, but at least we see where the 1 and 3 come from.

Another method that often appears is a vector method. We consider the coordinates of M(-4, 4) and N(6, -2) to be the vectors m and n, and we find the vector n + 1/4 (m - n)  -- that is, we start at one point and add 1/4 of the vector that gets us from one point to the other. But this would be very confusing -- first of all the number 3 doesn't appear at all (unless we change 1/4 to 3/4 to show 3/4 of the way from one to the other, which gives us m + 3/4 (n - m), and also this is prone to sign errors as it's not as obvious when we want m - n and when we want n - m. If we use the wrong vector difference, then our point will still end up on line MN, but it won't be between M and N.

The vector method might be preferable if the question was stated as "1/4 of the way from J to K." But since the PARCC test uses the ratio 1:3, I like my center of gravity method better.

...that is, I liked the center of gravity method better last year. That method made sense last year because last year's question was worded as "divides JK into two parts with in a ratio 1:3." But this year's question is worded differently -- "3/4 of the way from M to N." For this wording, the vector method makes more sense, since the fraction 3/4 (or 1/4) actually appears in the calculation.

But actually, as I think about it more and more, I'm wondering whether the best method to solve both last year's and this year's questions is just to find O as the midpoint of MN, and then P as the midpoint of ON. Yes, I said that this method won't work for thirds or fifths -- but notice that so far, the only division we actually see on the PARCC are quarters. The two-step midpoint process works whether we're asked to "divide a segment in a 1:3 ratio" or "find the point 3/4 of the way," so students should be less confused.

Not only that, but think about the following. Last year, we were asked to find M, the point that divides JK in a 1:3 ratio. Note that the letters J, K, and M are in alphabetical order, but skipping L. It's almost as if the PARCC wants us to find L as the midpoint of JK and M as the midpoint of LK. And in the same way, this year, we are asked to find P, the point that is 3/4 of the way from M to N. Note that the letters M, N, and P are in alphabetical order, but skipping O. It's almost as if the PARCC wants us to find O as the midpoint of MN and P as the midpoint of ON.

That settles it! On today's worksheet, students will repeatedly use the Midpoint Formula to divide the segment into common ratios like quarters, since this will be the easiest way for the students to get the PARCC questions correct. (But with our luck, the PARCC will probably throw in a question where we must divide a segment into thirds or fifths.)

PARCC Practice Test Question 12

U of Chicago Correspondence: Lesson 11-4, The Midpoint Formula

Key Theorem: Midpoint Formula

If a segment has endpoints (a, b) and (c, d), its midpoint is ((a + c)/2, (b + d)/2).

Common Core Standard:

CCSS.MATH.CONTENT.HSG.GPE.B.6
Find the point on a directed line segment between two given points that partitions the segment in a given ratio.

Commentary: The U of Chicago only focuses on midpoints -- that is, points that divide the segment into a 1:1 ratio. But we can apply this formula repeatedly to divide a segment into quarters and eighths.

PARCC Practice Test Question 11 (Day 154)

Chapter 11 of Morris Kline's Mathematics and the Physical World, "Explanation Versus Description," introduces the reader to the birth of modern science.

"Nor should it be considered rash not to be satisfied with those opinions which have become common. No one should be scorned in physical disputes for not holding on to the opinions which happen to please other people best." -- Galileo

And there's yet another name that needs no further introduction. As we see, Kline credits the 17th century Italian physicist Galileo Galilei as the father of modern science:

"Descartes and Fermat [Pierre de Fermat, mentioned several times on the blog before -- dw] created one of the key mathematical toolds for the develoment of modern science. Galileo Galilei created modern science."

Kline describes how Galileo radically changed the way people thought about the world. Ever since the days of Aristotle, educated Europeans assumed that there were only four elements -- air, fire, earth, and water -- and that objects fell because each element was seeking out its natural place, which for solid objects (earth) was the ground.

But Galileo challenged these notions. Kline writes:

"Galileo decided that he would seek descriptions of how things worked rather than explanations of why they worked or what purpose they served."

Although Kline introduces us to some of Galileo's early experiments, his most famous experiment of all will have to wait for tomorrow's chapter.

Today is Day 154 according to the blog calendar, which is based on one of my districts. According to the other district, today is Day 160. At the school where I will teach in the fall, today is Day 152.

Question 11 of the PARCC Practice Exam is on rotations -- but a different sort of rotation:

11. Each of the two-dimensional figures shown will be rotated 360 degrees about the respective line, creating a three-dimensional figure.

Drag the appropriate two-dimensional figure to identify the correct representation of the resulting three-dimensional figure.

(Here are the 2D figures: a circle not intersecting the axis of rotation, a semicircle whose diameter lies on the axis, a rectangle whose length is on the axis, and a right triangle whose leg is on the axis. Here are the 3D figures: a torus, a cone, a cylinder, and a sphere.)

There was a question like this on last year's practice PARCC. Here are the correct matches -- the circle creates the torus, the triangle generates the cone, the rectangle creates the cylinder, and the semicircle generates the sphere.

There are several things going on in this question. First of all, this is the only problem on the PARCC where transformations are performed in 3D. Recall that just as the mirror of a reflection is a line in 2D and a plane in 3D, the center of a rotation is a point in 2D and a line (axis) in 3D. This is why we are given a line to rotate about, rather than a point.

Second, our rotations are actually producing a new solid. This is called a solid of revolution. Notice that tomorrow, many Calculus BC students will be struggling to calculate the volumes of solids of revolution as they take their AP test. And as we found out last year, identifying solids of revolution appears on the PARCC Geometry test.

Only one solid of revolution appears in the U of Chicago text, and I mentioned it a month and a half ago -- Lesson 10-7, Question 15. In this case, a right triangle is rotated -- its axis is the line containing one of its legs -- to form a cone. But this problem tells us that the solid is a cone -- the students' task is to find its volume, if the length of the other leg and the the hypotenuse are given. If the given figure had been a rectangle, then the solid would have been a cylinder rather than a cone. I wonder whether Question 11 on the PARCC would be more palatable to traditionalists if the dimensions of the rectangle were given and students had to find the volume of the cylinder.

PARCC Practice Test Question 11

U of Chicago Correspondence: Lesson 10-7, Volumes of Pyramids and Cones

Key Theorem: none

Common Core Standard:

CCSS.MATH.CONTENT.HSG.GMD.B.4
Identify the shapes of two-dimensional cross-sections of three-dimensional objects, and identify three-dimensional objects generated by rotations of two-dimensional objects.

Commentary: The only question relevant to solids of revolution is Question 15, and it rotates a triangle to form a cone, rather than a rectangle to form a cylinder. Another U of Chicago section that may be relevant to the first part of the above standard is Section 9-4, which is on "Plane Sections" (cross-sections). On this blog, we jumped around Chapters 9 and 10, so Section 9-4 may not have been adequately covered.

PARCC Practice Test Question 10 (Day 153)

Chapter 10 of Morris Kline's Mathematics and the Physical World is called "The Wedding of Curve and Equation." In this chapter, we graph equations.

"I have resolved to quit only abstract geometry, that is to say, the consideration of questions which serve only to exercise the mind, and this, in order to study another kind of geometry, which has for its object the exploration of the phenomena of nature." -- Descartes

And of course the author of this quote needs no introduction. When it's time to graph equations, the first thing we think about is the coordinate or Cartesian plane, which is named after Rene Descartes, the 17th century French mathematician.

There are many very interesting topics in this chapter. First Kline looks at the equations of circles, derived from the Pythagorean Theorem, just as in Lesson 11-3 of the U of Chicago text. Then he writes about the derivation of the equation of a line:

"Had we considered a straight line inclined more steeply to the horizontal, for example, one that rises 2 units for each horizontal difference of 1 unit, then from similar triangles OQ'P' and OQP we might have argued that y/x = 2/1, or that y = 2x is the equation of the line."

So Kline is deriving the equation of a line given its slope using similar triangles -- which is also expected in the eighth grade Common Core Standards.

Then Kline writes about a line like y = 2x + 3, which is obtained from the graph of y = 2x simply by moving each point up three units -- that is, by performing a translation. And of course a line and its translation image are parallel, so the graphs of y = 2x and y = 2x + 3 are parallel.

Then Kline moves on to parabolas, including their derivation from the definition of parabola as the locus of all points equidistant from a point (the focus) and a line (the directrix).

Question 10 of the PARCC Practice Exam is on dilations:

10. The figure shows line segment JK and a point P that is not collinear with points J and K.

Suppose that line segment J'K' is the image of line segment JK after a dilation with scale factor 0.5 that is centered at point P. What statement best describes the position of line segment J'K'?

A. Line segment J'K' is parallel to line segment JK.
B. Line segment J'K' is perpendicular to line segment JK.
C. Line segment J'K' intersects line segment JK at one point, but it is not perpendicular to line segment JK.
D. Line segment J'K' lies on the same line as line segment JK.

This question is simple if you know the theorem -- a line is parallel to its dilation image. This means that the correct answer is A.

That a line is parallel to its dilation image is one of the two big properties of dilations proved in the first three lessons of Chapter 12 -- the other being Dilation Distance.

Recall that I don't like the U of Chicago proof of Dilation Distance because, according to the Common Core Standards, it's circular. We should be using dilations to prove the properties of the coordinate plane, not using coordinates to prove the properties of dilations!

I also point out that it's good to look at all of the transformations we have learned, and see which ones have lines that are parallel to their images, as well as lines that are their own images (invariant lines):

Reflections:
Invariant lines -- the mirror itself, any line perpendicular to mirror
Lines parallel to images -- any line parallel to mirror

180-Degree Rotations:
Invariant lines -- any line through the center
Lines parallel to images -- any other line

Translations:
Invariant lines -- any line parallel to direction of slide
Lines parallel to images -- any other line (as Kline does in Chapter 10)

Glide Reflections:
Invariant lines -- the mirror itself
Lines parallel to images -- any line parallel or perpendicular to mirror

Dilations:
Invariant lines -- any line through the center
Lines parallel to images -- any other line

Recall that a fixed point is a point that is actually mapped to itself. Every line on the mirror of a reflection is a fixed point, as is the center of a rotation or dilation. Translations and glide reflections have no fixed points.

Notice that 180-degree rotations (also called inversions), translations, and dilations all have a property in common -- every line is either invariant or parallel to its image. As it turns out, we can prove a theorem about all three transformations simultaneously. (In some ways, a 180-degree rotation or inversion is also a dilation with scale factor -1.)

Theorem:
Let T be a transformation preserving betweenness and collinearity, and such that through any point, there is an invariant line, and such that no fixed point lies on a line that is not invariant. Then any line not invariant is parallel to its image.

Indirect Proof:
Let l be any line that is not invariant -- that is, the image of l is not l. We are to show that l must be parallel to its image l'. So assume the contrary, that l and l' are not parallel -- that is, that they intersect at some point P. Every point on l' has an preimage on l. In particular, the preimage of P is on l -- call that point Q (that is, Q' is P). Since by hypothesis no fixed point can lie on l, so Q is not P.

By hypothesis, every point lies on some invariant line. In particular, Q lies on some invariant line -- call that invariant line q. Since l is not invariant, l is not q. By the definition of invariant, the image of every point on q is another point on q. In particular, the image of Q is on q. But the image of Q is P, so P is on q.

So notice that Q and P are both on q -- but notice that Q and P are both on l. In other words, we have two distinct lines, q and l, intersecting at two distinct points Q and P. This is a contradiction, since two lines can intersect at most one point. Therefore l | | l'. QED

PARCC Practice EOY Question 10
U of Chicago Correspondence: Lesson 12-3, Properties of Size Changes
Key Theorem:

A line and its image under a size transformation are parallel.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.SRT.A.1.A
A dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged.

Commentary: The U of Chicago text doesn't give a satisfactory proof of the this theorem, nor many problems based on the theorem, but it does mention the theorem. The Common Core Standards only emphasize that a line and its dilation image are parallel, but there's no reason why we can't look at other lines and their images -- recall the Line Parallel to Mirror and Line Perpendicular to Mirror Theorems here on the blog.

PARCC Practice Test Question 9 (Day 152)

Chapter 9 of Morris Kline's Mathematics and the Physical World is "The Scientific Revolution." In this chapter, Kline writes about how people began to think about science during the 16th, and even more in the 17th, century.

"Besides the mathematical arts there is no infallible knowledge except it be borrowed from them." -- Robert Recorde, 16th century Welsh mathematician (famous for the "equals" sign =)

Kline begins:

"The mathematics that has been created since the year 1600 is enormously greater than that which even the Greek geniuses produced. Moreover, this newer mathematics, in intimate collaboration with science, has affected, one might say molded, the character of our modern civilization so markedly that we now recognize that we live in a scientific age."

There actually isn't much math or science in this chapter. Instead, Kline mainly writes about how attitudes in Europe changed as history entered the Renaissance. As you can see with the quote above, this is when math began to look like, well, math, with the introduction of the equals and plus signs.

Oh, and speaking of the Renaissance, I did go to that Renaissance Fair over the weekend -- but to my disappointment, I didn't see that "Earth is the center of the universe" bulls-eye game at all. Oh well -- at least I had my own personal Renaissance Fair by reading Chapters 8 and 9 of Kline. I read about some of the leading scientists (like Galileo and da Vinci), as well as some important religious leaders (like Luther and Calvin) who lived during, and shaped, the Renaissance.

In my last post, I mentioned that I plan on teaching full-time at a middle school next year. So now you may be wondering, how will this affect my plans for next year on the blog -- considering that I intended this to be a high school Geometry blog.

Well, I'm expecting to have three preps next year -- Math 6, Math 7, and Math 8. I've decided that the focus of this blog will be on the Math 8 course. After all, even though the Common Core Standards for all three middle school years contain some geometry, it's Math 8 where the Common Core focus on transformations begins. Therefore I'll be writing about my Math 8 class here on the blog. I will write about the class the entire year, not just when I'm covering a geometry lesson.

Lately, I had been starting to write my plans for next year on the blog -- indeed, just a few posts back, I wrote about the school calendars at the districts where I'm subbing, as well as how I'm setting up the blog to line up days of school with U of Chicago lessons (such as Lesson 15-2 on Day 152). Well, never mind all of that -- things change when I least expect it. The focus will be on my new school, new school calendar, and new text. More details are to come regarding this.

I'm almost tempted just to restart Chapter 15 right now (since I'd said so often that I'd do this), but I've already committed myself to the PARCC. That being said, Question 9 of the PARCC Practice Exam, the first in the calculator section, is on similar triangles:

9. Given the two triangles shown, find the value of x.

(Here is the information shown in the diagram: in triangle ABC, Angle A = 68, Angle C = 38, AB = 6,  and BC = 10. In triangle DEF, Angle D = 38, Angle E = 74, EF = 15, and DE = x.)

The value of x is (Choose...4, 11, 12, 19, 20, 25).

To solve this, we notice that the sum of the angle measures mentioned is 68 + 38 + 74 = 180. Thus Triangles ABC and FED are similar by AA Similarity. It remains only to set up the proportion:

DE/CB = EF/BA
x/10 = 15/6
x = 150/6 = 25

Therefore the correct answer is the largest choice, 25. Common errors include setting up the proportion incorrectly -- the smallest choice, 4, is the solution to x/10 = 6/15, for example. Indeed, I'm very surprised that 9 isn't one of the answer choices -- especially as the way the triangles are drawn, students are likely to assume that ABC is similar to DEF rather than FED. I'm not sure where PARCC gets the other wrong answer choices from, but note that 19 is the solution to x - 10 = 15 - 6, so perhaps some students might try to subtract instead of divide in their proportion.

I actually don't have much to complain about with this problem. It's an excellent similarity question that requires students to think about both the Triangle Sum and Angle Similarity Theorems.

I do admit that students might be tricked by this sort of question. I remember once when I was teaching or tutoring a student who was in the similarity chapter of the text. Upon seeing the two triangles in a question much like this one, he immediately started writing a proportion. I told him that a proportion can be set up only if the triangles are similar, and so I asked him, how did he know that the triangles are similar? His response was, of course the triangles were similar because we were in the similarity chapter of the text!

Yes, my student cleverly figured out that in the similarity chapter, nearly every problem would have a pair of similar triangles. But this problem illustrates what's wrong with his thinking. First of all, this question is on the PARCC test -- not in the similarity chapter of any textbook -- so he has to be able to recognize similar triangles without using a book for clues. Second, even if this problem were in a book, he has to know something about similarity or else he might assume that ABC is similar to DEF, not FED as is the case -- and set up the proportion wrong.

Indeed, here are two problems that I wrote on today's worksheet:

Find the value of x. Assume triangles ABC and DEF:
8. Angle A = D = 27, Angle C = F = 52, AC = 5, DF = 15, DE = 12, AB = x
9. Angle A = D = 37, Angle C = E = 90, AC = 60, DF = 15, DE = 12, AB = x

These two questions may look alike, but they aren't. Notice that in Question 9, Angle C is congruent to Angle E, not F. Therefore triangle ABC is similar to DFE, not DEF. This makes a big difference -- the value of x is now 75, not 48 as one might expect.

Sometimes I wonder whether this problem from my worksheet is an unfair problem, like (y + k)^2 from last week. I'd argue that when a student is first learning about similarity, it would indeed be unfair to build up a student's confidence, and then knock it down with "Wrong! Triangle ABC is similar to DFE, not DEF! Can't you see Angle C is congruent to Angle E?" But on a review worksheet, it is a fair question to ask -- especially when preparing for a test like the PARCC where such tricky questions might appear. (It isn't a convention that ABC must be similar to DEF. On the other hand, (y - k)^2 and ax^2 + bx + c = 0 are conventions, so I call questions that violate them unfair, even if they appear on the PARCC.)

Meanwhile, here's something about this problem that I must point out. Going back to the original problem, we have:

In triangle ABC, Angle A = 68, Angle C = 38, c= 6,  and a = 10.

Notice that now I'm using a to denote the side opposite A, as in the Law of Sines:

a/sin A = c/sin C

In this problem, all four values are known:

10/sin 68 = 6/sin 38
10.7853... = 9.7456...

So now suddenly, ten point something is equal to nine point something? The left hand side of the equation exceeds the right by more than one whole unit!

This is actually a fairly common problem with similarity questions -- most of the time, the triangles in such questions are overdetermined -- that is, we provide more than enough information to solve the whole triangle. And if we try to solve it, most of the time the values contradict each other. This is mostly because trig values are inexact, but I'm surprised that it would be this far off.

When I set up my worksheet, I choose whole numbers for the sides, and then find the angles using the Law of Cosines, sometimes rounding up and sometimes down to the nearest degree. Then I know that the error is no more than one degree in each angle. Rounding, say, 36.9 up to 37 degrees (as occurs in the 3-4-5 right triangle), or even down to 36 degrees, is more accurate than rounding either 9.04 up to 10, or 6.64 down to 6 (which must have occurred when setting up this PARCC problem). Actually, 9.04 down to 9 wouldn't have been that bad, but that's not what PARCC did.

PARCC Practice EOY Question 9
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: The U of Chicago text contains only two questions where students need to use Triangle Sum before they can apply AA~ -- Question 5 of Lesson 12-9, and Question 30 of the SPUR Review section. In neither question do the students then have to set up a proportion to find the length of a side. This question is therefore more sophisticated than any that appears in the U of Chicago text.