Friday, May 29, 2015

PARCC Practice Test Question 32 (Day 177)

Today, for the second day in a row, I subbed in a middle-school math class. This time, I was in a sixth grade classroom, and the lesson was on integer operations.

Some people may notice that the word "integers" doesn't appear in the Common Core Standards. Instead, the standards refer to the set of "rational numbers":

Apply and extend previous understandings of numbers to the system of rational numbers.

Understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.g., temperature above/below zero, elevation above/below sea level, credits/debits, positive/negative electric charge); use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation.

If we think about it, we notice that the students first learn about natural numbers, then about fractions (of which the natural numbers are a subset), then we suddenly want to jump to integers -- of which the set of fractions is not a subset. The students already know about fractions, so as soon as they learn about positive and negative signs, they technically already have the entire field Q of rational numbers. Of course, the simplest examples -- including all the examples that they saw today -- were integers.

Students learned about adding and subtracting integers today -- next week they will learn how to multiply
and divide them. Today's lesson is actually a seventh grade topic:

Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram.

Then again, this is the end of the year. Most sixth graders have already completed the SBAC (yet some students are still taking it), and so this lesson is a preview of next year.

With all these visits to middle school math classes this week, angles and parallels are still on my mind. Recall that I'm still trying to come up with a plan to prove the Corresponding Angles Test as a theorem without using Playfair's Parallel Postulate.

Of course, you may ask, why can't I just accept the Corresponding Angles Test as a postulate, just as Dr. Franklin Mason is now doing? I guess that the words of Dr. David Joyce still haunt me:

Chapter 7 [of the Prentice-Hall text -- dw] is on the theory of parallel lines. It begins by postulating that corresponding angles made by a transversal cutting two parallel lines are equal. One postulate is enough, but for some reason two others are also given: the converse to the first postulate, and Euclid's parallel postulate (actually Playfair's postulate). A proliferation of unnecessary postulates is not a good thing. One postulate should be selected, and the others made into theorems.

Dr. M's original presentation did select only one postulate -- Playfair's. Then he used Playfair and the Parallel Tests to prove the Parallel Consequences. The Parallel Tests are themselves proved using the Triangle Exterior Angle Inequality, which in turn follows from SAS. It was only later on when Dr. M added a Corresponding Angles Test Postulate to avoid the TEAI proof -- which may be a bit tricky for high school students to understand.

My concern is mainly that I want Corresponding Angles to be first, and then derive the theorems for Alternate Interior and Same-Side Interior from Corresponding Angles. It's possible to modify the TEAI proof so that it refers to corresponding rather than alternate interior angles, but a natural reading of other proofs -- such as Dr. Hung-Hsi Wu's rotation proof -- uses alternate interior angles.

I was just reading the geometry text of the mathematician Adrien Legendre -- which, if you recall, was the first text to challenge the dominance of Euclid's Elements. Now Legendre's Proposition 60 appears to give priority to a Same-Side Interior Angle Test Theorem. Here's is Legendre's proof, converted to two-column format (and replacing the Euclidean phrase "sum to two right angles" with "are supplementary angles"):

Given: Angles CAB, ABD are supplementary
Prove: Lines AC | | BD

Statements                              Reasons
1. CABABD supplementary    1. Given
2. Let G be the midpoint of AB 2. Point-Line-Plane (Ruler) Postulate
3. Draw EG perpendicular to     3. Through point there's line perp. a given line
AC intersecting BD at F
4. GBF, GBD supplementary     4. Linear Pair Theorem
5. GAE, GBD supplementary     5. GAE, GBD are just CAB, ABD renamed
6. Angle GAE = GBF               6. Angles suppl. to same angle are congruent
7. AG = BG                             7. Definition of midpoint
8. Angle AGE = BGF               8. Vertical Angles Theorem
9. Triangle AGE = BGF           9. ASA Congruence Postulate
10. Angle AEG = BFG             10. CPCTC
11. BD, EG perpendicular         11. All right angles are congruent.
12. AC | | BD                           12. Two Perpendiculars Theorem

Notice that just like Legendre, I on the blog have already proved the Two Perpendiculars Theorem independently of any previous theorem. So we both can use the Two Perpendiculars Theorem to prove this Same-Side Interior Angle Test.

But then we notice that this proof works better as an Alternate Interior Angles Test -- we notice that Step 6 of this proof ultimately shows that the alternate interior angles are congruent, so we might as well have this be given and drop Steps 4-5 altogether. I mentioned earlier that Wu's rotation proof can be converted to a proof using triangle congruence. Well, this is exactly the proof -- notice that the two triangles proved congruent (AGE and BGF) can be mapped to each other via exactly the rotation that appears in the Wu proof.

So if Legendre could change a natural Alternate Interior Angles Test into a Same-Side Interior Angles Test, then I can change it into a Corresponding Angles Test instead. We simply give that the corresponding angles are congruent and use the Vertical Angle Theorem for the alternate interior angles necessary to produce the congruent right triangles.

Then again, I wonder whether we can just take the original Wu rotation proof and attempt to adjust it so that it gives a Corresponding Angles Test instead. Notice that for every pair of corresponding angles, one is exterior and the other is interior. So we simply rotate the exterior angle to produce what one could call its "alternate exterior angle." We then use Vertical Angle Theorem to obtain the angle that is corresponding with the original preimage angle, without anyone wondering why we don't apply Vertical Angles to the preimage to give alternate interior angles.

Speaking of corresponding angles, they appear in the final PARCC question. Question 32 of the PARCC Practice Test is about constructions:

Part A:

The figure shows line r, points P and T on line r, and point Q not on line r. Also shown is ray PQ.

(A standard construction for parallel lines appears. Point W is constructed on line r, and point S is constructed not on line r.)

Consider the partial construction of a line parallel to r through point Q. What would be the final step in the construction?

(A) draw a line through P and S
(B) draw a line through Q and S
(C) draw a line through T and S
(D) draw a line through W and S

Part B:

Once the construction is complete, which of the reasons listed contribute to proving the validity of the construction?

(A) When two lines are cut by a transversal and corresponding angles are congruent, the lines are parallel.
(B) When two lines are cut by a transversal and vertical angles are congruent, the lines are parallel.
(C) definition of segment bisector
(D) definition of angle bisector

For Part A, even if one isn't familiar with this particular construction, we notice that the question asks for the final step. Since we are trying to construct a line parallel to r through Q, the final step ought to be -- to draw the line parallel to r through Q, of course! Thus drawing a line through P, T, or W make no sense as all of these points lie on r, so any line passing through these points must intersect r, not be parallel to r. The only possible line parallel to r is line QS. So choice (B) is correct.

For Part B, we want to prove that lines r and QS are parallel. We are copying the angle QPW, and this angle and the copied angle are corresponding angles. So by the Corresponding Angles Test, we conclude that the lines are parallel. So choice (A) is correct.

Notice that the U of Chicago text doesn't even give the construction of copying an angle -- much less the construction of a line parallel to a given line through a point not on the line, which depends on the former construction. For all its faults, the Prentice-Hall text mentioned by Joyce gives copying an angle as the second construction in its Section 1-1, and we've already mentioned that the Pearson Integrated Math II test gives the same construction in its own first section as well. The U of Chicago text focuses mainly on constructing perpendicular bisectors -- even its angle bisector construction is really just a form of its perpendicular bisector construction.

Indeed, on the blog, I gave a parallel line construction based on the perpendicular bisector construction (which is probably how the U of Chicago would present it if it were to do so). This construction is justified by the Two Perpendiculars Theorem.

But the construction on the PARCC exam is justified by the Corresponding Angles Test. So once again, we want Corresponding Angles to have priority -- there's nothing wrong with performing the construction to use alternate interior rather than corresponding angles, but it's just not done. I wonder whether a student who saw Alternate Interior Angles given priority and Corresponding Angles proved using Alternate Interior and Vertical Angles might even be tricked into selecting choice (B) rather than (A) for Part B above.

Today is an activity day. I decided that the students need more practice with constructions, and so I have them draw a simple design that requires them to construct parallel lines. This is similar to the tic-tac-toe design that I had them draw earlier -- except there are no perpendicular lines. So students will have to use the parallel line construction based on corresponding angles that is covered on today's worksheet.

This is the last of the PARCC Practice Test questions. According to the school district whose calendar I'm using on the blog, there are three days of school left -- finals week. And so in my next post, I will post a final exam based on both U of Chicago and PARCC practice questions. I don't provide a review worksheet since all 32 PARCC questions actually are review questions. On Monday I post only the test itself.

PARCC Practice EOY Exam Question 32
U of Chicago Correspondence: Section 3-4. Parallel Lines
Key Theorem: Corresponding Angles Postulate:

If two coplanar lines are cut by a transversal so that two corresponding angles have the same measure, then the lines are parallel.

Common Core Standard:
Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.).Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.

Commentary: The construction of parallel lines doesn't appear anywhere in the U of Chicago text at all. One can use the perpendicular bisector twice to construct parallel lines, but this is never directly stated in the text.

Thursday, May 28, 2015

PARCC Practice Test Question 31 (Day 176)

Last night I tutored my geometry student. Section 9-5 of the Glencoe text is on dilations, which, as readers of this blog already know, are the gateway to similarity.

One major difference that I noticed between the Glencoe and the U of Chicago texts is that Glencoe actually allows the scale factor to be negative. The U of Chicago text makes it clear that its focus is on positive scale factors:

"Any number but zero can be the magnitude, but in this book k > 0 unless otherwise stated."

The only problem that I could find with a negative scale factor is in the Exploration part of Section 12-1 in the text. Question 18 reads:

Use Triangle JKL of Question 5.
a. Multiply all the coordinates of the vertices by -3.
b. Describe the figure that results.
c. Generalize parts a and b.

I've briefly mentioned what a dilation of scale factor -1 would look like. Such a transformation would actually be a rotation of 180 degrees centered at the same point as the dilation. I've stated how rotations of 180 degrees and dilations of positive scale factors have several properties in common -- most notably, lines and their images under both types of transformations are parallel.

Dilations with a negative scale factor other than -1 can be thought of as the composite of a dilation with the corresponding positive scale factor followed by the 180-degree rotation. So the dilation in Question 18 above is the composite of a dilation of scale factor 3 (the dilation mentioned in Question 5) and a 180-degree rotation. The dilation and rotation commute, so it doesn't matter which one of them we perform first. The set of all dilations with nonzero scale factor, along with the translations, form a group, with the set of all dilations with positive scale factor and translations a subgroup.

Dilations with negative scale factor, like all rotations, preserve orientation. In all dimensions, the dilation of scale factor -1 is a point inversion (or point "reflection"). It preserves orientation if there are an even number of dimensions, but if there are an odd number of dimensions, then the dilation of scale factor -1 reverses orientation. This includes the 1D inversion (which is merely a reflection) and the 3D inversion (a roto-reflection).

Today I also subbed in a middle school math class. As it turned out, it was a class of seventh graders who were getting ready to take a test on angles. Printed on each test was the Common Core Standard being tested, which is:

Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure.

During the seventh grade class, an eighth grader was sent to my room to make up her own test. I glanced at her test and saw that it was also about angles -- except on parallel lines. In some ways, both of these tests fit better with yesterday's featured PARCC question that today's -- even though I'm comparing middle school tests to the high school PARCC question. Of course, the basic angle relationships of Chapter 3 of the U of Chicago text (the Linear Pair and Vertical Angle Theorems) come up over and over again, first in middle school and again in high school.

That's enough about yesterday's question -- now for today's. Question 31 of the PARCC Practice Test has four parts. It involves the coordinate plane and the perimeter and area of various quadrilaterals.

Luke purchased a warehouse on a plot of land for his business. The figure represents a plan of the land showing the location of the warehouse and parking area. The coordinates represent points on a rectangular grid with units in feet.

In the figure, here are the vertices:
Plot of land: (0, 0), (50, 0), (70, 60), (0, 60)
Warehouse: (4, 20), (39, 20), (39, 50), (4, 50)
Parking area: (4, 0), (50, 0), (39, 20), (4, 20)

Part A:

What is the perimeter of the plot of land?

Express your answer to the nearest tenth of a foot.

Part B:

What is the area, in square feet, of the plot of land that does not include the warehouse and the parking area?

Part C:

Luke is planning to put a fence along two interior sides of the parking area. The sides are represented in the plan by the legs of the trapezoid. What is the total length of fence needed?

Express your answer to the nearest tenth of a foot.

Part D:

In the future, Luke has plans to construct a circular storage bin centered at coordinates (50, 40) on the plan. Which of the listed measurements could be the diameter of a bin that will fit on the plot and be at least 2 feet away from the warehouse?

Select all that apply.

(A) 10 feet
(B) 15 feet
(C) 18 feet
(D) 22 feet
(E) 25 feet

For Part A, three of the sides clearly have length 50, 60, and 70 feet. The remaining side requires the Distance Formula: sqrt((70 - 50)^2 + (60 - 0)^2) = sqrt(4,000) = 63.246 feet. So the total perimeter is 50 + 60 + 70 + 63.246 = 243.246 feet, or 243.2 feet to the nearest tenth.

For Part B, it's easier to find the area of the total plot of land and then subtract. We use the Trapezoid Area Formula as all three quadrilaterals are trapezoids (including the rectangle, inclusive definition):

Plot of land: 1/2 (60)(50 + 70) = 3,600 square feet
Warehouse: 1/2 (30)(25 + 25) = 750 square feet
Parking area: 1/2 (20)(46 + 25) = 710 square feet

So the net area is 3,600 - 750 - 710 = 2,140 square feet.

For Part C, one leg is clearly 20 feet, and the other requires the Distance Formula. We find the other leg as follows: sqrt((39 - 50)^2 + (20 - 0)^2) = sqrt(521) = 22.83 feet. So the total fence has length 20 + 22.83 = 42.83 feet, or 42.8 feet to the nearest tenth.

Finally, Part D is very tricky. The closest point from (50, 40) to the warehouse is clearly (39, 40), which is clearly 11 feet away. But we need a 2-foot gap, so the maximum radius is 9 feet. But now we must make sure that the other side will fit on the plot of land. The shortest distance between a point and a line is the perpendicular distance. The right side of the plot has slope 3, and so if we drop down a perpendicular it would have slope -1/3. The equation of the line passing through (50, 0) with slope 3 is found to be:

y = 3(x - 50)

and the equation of the line passing through (50, 40) with slope -1/3 is:

(y - 40) = -1/3 (x - 50)

To solve this system, we substituting, we obtain:

3x - 150 - 40 = -1/3 x + 50/3
9x - 570 = -x + 50
10 x = 620
x = 62
y = 3(62 - 50)
y = 36

The distance formula gives the distance from (50, 40) to (62, 36) as sqrt((62- 50)^2 + (36 - 40)^2), which equals sqrt(148) or 12.2 feet, which is clearly greater than 9 feet. So only the distance to the warehouse matters the most. The radius must be at most 9 feet, and the diameter at most 18 feet, so the correct choices are (A), (B), and (C).

I admit that if I were a student taking this test, I wouldn't want to do all of that work for Part D just to find out that the bin is closer to the warehouse than the edge of the plot anyway -- especially after having already worked on 30 other PARCC questions. I'm more likely just to ignore the edge and choose (A), (B), and (C) anyway -- either that, or guess that the edge is slightly closer and choose only (A) and (B).

In addition to today's worksheet -- especially since neither yesterday nor today did I have much room to include extra practice problems -- I present the worksheet on linear pairs and vertical angles on which the seventh graders practiced to prepare for today's test.

PARCC Practice EOY Exam Question 31
U of Chicago Correspondence: Section 11-2, The Distance Formula
Key Theorem: Distance Formula:

The distance between two points (x_1, y_1) and (x_2, y_2) in the coordinate plane is:
sqrt((x_2 - x_1)^2) + (y_2 - y_1)^2).

Common Core Standard:
Use coordinates to compute perimeters of polygons and areas of triangles and rectangles, e.g., using the distance formula.

Commentary: The Distance Formula is in this section, although there is not much on area in this chapter at all. Section 8-6 of the text gives some good area problems involving trapezoids on the coordinate plane.

Wednesday, May 27, 2015

PARCC Practice Test Question 30 (Day 175)

Question 30 of the PARCC Practice Test is on angles. Students are to complete two simple short proofs involving angles:

In the figure shown, CF intersects AD and EH at points B and F, respectively.

Part A:

Given: Angle CBD = BFE
Prove: Angle ABF = BFE

Statements                    Reasons
1. Angle CBD = BFE    1. Given
2. Angle CBD = ABF    2.
3. Angle ABF = BFE    3.

Which two of the given reasons could be used to correctly complete the proof?

(A) Definition of congruent angles
(B) Congruence of angles is reflexive
(C) Congruence of angles is symmetric
(D) Congruence of angles is transitive
(E) Vertical angles are congruent

Part B:
Given: Angle CBD = BFE
Prove: Angle BFE + DBF = 180 degrees

Statements                                      Reasons
1. Angle CBD = BFE                      1. Given
2. Angle CBD + DBF = 180 degrees 2.
3. Angle BFE + DBF = 180 degrees 3.

Which two of the given reasons could be used to correctly complete the proof?

(A) Adjacent angles are congruent
(B) Adjacent angles are supplementary
(C) Linear pairs of angles are supplementary
(D) Reflexive property of equality
(E) Substitution property of equality
(F) Transitive property of equality

This question shouldn't be too difficult to answer. In the proof for Part A, Angles CBD and ABF are vertical angles and therefore congruent, Since CBD is congruent to both BFE and ABF, therefore ABF is congruent to BFE by the Transitive Property. So the correct answers are (D) and (E).

Now in the proof for Part B, this time we have that CBD and DBF form a linear pair, and so these angles are supplementary. Then we can just substitute BFE in for CBD since these angles are given as having equal measures. So the correct answers are (C) and (E).

There are a few issues in this problem, though. Notice that a correct answer for Part A is "Congruence of angles is transitive," whereas most texts write it as, "Transitive Property of Congruence." I once told a student in an Algebra I class where I was student teaching, after she asked which number to multiply first, I told her that multiplication is commutative, instead of use the Commutative Property of Multiplication -- which may have confused her. Then again, notice that Part B calls it "Transitive property of equality."

Then again, one wonders whether we actually use transitivity, whatever we call it. Notice that the Transitive Property usually states that if x = y and y = z, then x = z. But Part A is actually in a different form -- if x = y and x = z, then z = y. Transitivity, in its pure form, requires that the RHS of an equation or congruence be the same as the LHS of another -- not that the LHS of two equations are the same (or the RHS of two equations). An argument can be made that we should rewrite x = z first as z = x, using the Symmetric Property. Now the RHS of one is the LHS of the other, so that we can use the Transitive Property. This is not insignificant, considering that the Symmetric Property is one of the other choices in Part A.

In practice, people call it the Transitive Property even when two LHS or two RHS are the same, even though there's a Symmetric Property. The only time that I've explicitly seen the Symmetric Property in a proof is using the Pappus proof of the Isosceles Triangle Theorem -- in triangle ABC where sides AB and AC are congruent, we are given one S, and using the Symmetric Property, AC = AB is in fact the other S. The A is given by the Reflexive Property as Angle A is congruent to itself, so that Triangles ABC and ACB are congruent by SAS.

But I still haven't addressed the big elephant in the room for this problem. Notice that, believe it or not, the word "parallel" appears nowhere in this problem -- even though the figure associated with this problem looks very much like two parallel lines, AD and EH, cut by a transversal. In both parts, we are given that the corresponding angles are congruent. In Part A, we are asked to show that the alternate interior angles are congruent, and in Part B, we are asked to that the same-side interior angles are supplementary.

Notice that the lines AD and EH don't merely appear to be parallel -- they are provably parallel because we are given that the corresponding angles are congruent. This is what the U of Chicago calls the Corresponding Angles Postulate. But if we were instead given that the lines are parallel, then we would be, in some ways, assuming the converse of the Corresponding Angles Postulate (called the Parallel Lines Postulate in the U of Chicago) and using that postulate to prove two theorems -- one for alternate interior angles and the other for same-side interior angles. So of course we shouldn't use those theorems in the proofs, because we're trying to prove those theorems. Instead, we show how those theorems go back to the Vertical Angle and Linear Pair Theorems.

At any rate, this is a good time to re-evaluate how parallel lines are taught in the U of Chicago and other texts as well as here on the blog, and whether my way is satisfactory or I should find a new way to teach parallel lines, in light of today's featured PARCC question.

Let's begin our review with the U of Chicago text. In many ways, the U of Chicago's presentation of parallel lines is weak. The two postulates mentioned above are in Section 3-4, but neither alternate interior nor same-side interior angles appear until Chapter 5. Moreover, alternate interior angles appear in Section 5-6, but same-side interior angles don't appear at all except in the context of trapezoids in Section 5-5.

Now Dr. Hung-Hsi Wu uses rotations to develop his properties of parallel lines. Theorem 10 states that if lines are parallel then alternate interior and corresponding angles are congruent, and Theorem 12 is the converse of Theorem 10. Notice that rotations lead directly to alternate interior angles, since if we use the diagram for PARCC Question 30, the rotation image of Angle BFE 180 degrees centered at the midpoint of BF is ABF (alternate interior angle), not CBD (corresponding angle). So vertical angles are then used to establish the corresponding angle theorems. Like the U of Chicago, same-side interior angles are just an afterthought for Wu.

We notice that whereas the U of Chicago begins with corresponding angles and Wu begins with alternate interior angles, the Pearson Integrated Math II text that I mentioned last week begins with same-side interior angles! So far, we've seen three different starting points for the teaching of the parallel line theorems.

Dr. Franklin Mason is the one who distinguishes between the Parallel Tests -- statements of the form "if ... then the lines are parallel" and the Parallel Consequences -- "if the lines are parallel then ..." He has altered his parallel lessons several times. Originally Dr. M proved the Triangle Exterior Angle Inequality and used it to prove the Parallel Tests, beginning with the Alternate Interior Angles Test, just like Wu. Then Dr. M made the TEAI a postulate. Finally, he dropped the TEAI Postulate and just made the Corresponding Angles Test a postulate, just like the U of Chicago.

Also, there is the classic Parallel Postulate -- through a point not on a line, there is exactly one line parallel to that line. This is the Parallel Postulate of Playfair. Originally Playfair was the only postulate about parallels mentioned in Dr. M, and he used Playfair to derive the Parallel Consequences but no parallel postulate was needed to derive the Parallel Tests. This is similar to how Euclid proved these theorems many, many centuries ago. To this day Dr. M still uses Playfair only to prove the Consequences -- the Tests go back to the new Corresponding Angles Postulate.

Now which approach best prepares us for the PARCC exam, in particular today's question? Today's featured question shows us how to use corresponding angles to prove the Alternate Interior Angles and Same-Side Interior Angles Consequences. So it's best that we begin with corresponding angles, just like the U of Chicago or the new Dr. M, but not like Pearson, Wu, or the old Dr. M. Again, this is probably why Dr. M switched to making Corresponding Angles a postulate.

On this blog, I started out by using the Wu approach -- which we now consider unacceptable because it leads to Alternate Interior instead of Corresponding Angles. Wu uses Alternate Interior Angles because these naturally flow from his proof based on 180-degree rotations.

I mentioned last summer and fall that what would be more appealing is if instead of rotations, we could use translations to derive the properties of parallel lines. After all, if we take the figure in today's featured PARCC problem and translate Angle BFE by the vector FB, we obtain CBD -- which is indeed the corresponding angle for BFE. And this would justify the use of the name "corresponding angles" for both parallel lines and congruent triangles, since both of them are the preimages and images of certain isometries.

But when we tried this, we ran into a major problem. The properties of translations were proved using parallel lines -- since after all, a translation is the composite of reflections in parallel lines. So we would have circularity if we used translations to prove the parallel line theorems and the parallel line theorems to prove the translation theorems. Instead, I ended up assuming the weakest possible parallel postulate that will allow us to the properties of translations. That turned out to be the Perpendicular to Parallels Theorem, which I took to be our new postulate. This postulate also allowed us to prove Playfair, and then Playfair can be used to derive the Parallel Consequences, just as in Dr. M.

But that's assuming that we've already proved the Parallel Tests. Let's recall how Dr. M proves the Parallel Consequences. He uses a common trick to prove the converses of statements that are much easier to prove -- by combining the forward statement with a uniqueness condition. Thus the Converse of the Pythagorean Theorem is proved using the forward Pythagorean Theorem combined with the uniqueness condition of SSS Congruence (i.e., given three side lengths, there is at most one triangle up to isometry with those lengths). Likewise, Dr. M proves the Corresponding Angles Consequence using the forward statement (the Corresponding Angles Test), with Playfair as the uniqueness condition (through a point and a line, there is at most one line parallel to the given line). This explains why Dr. M chose the Corresponding Angles Test and Playfair as his two postulates.

Now I can only think of two ways to maintain the priority of Corresponding Angles. One is simply to introduce a Corresponding Angles Postulate, as Dr. M does now. The other is to go back to Dr. M's old TEAI method -- a slightly modification of the TEAI proof yields corresponding instead of alternate interior angles. Now that SAS (as well as SSS and ASA) has been moved up to Unit 2, the proof of TEAI itself can now be given.

Notice that we've already given a special case of the Corresponding Angle Postulate here on the blog without any need for TEAI -- the Two Perpendiculars Theorem. We already know that Wu's rotation proofs for parallelograms can be written using triangle congruence instead. It's possible to give his rotation proof of Theorem 10 using congruent triangles as well. If we are given that alternate interior angles such as BFE and ABF above are congruent, we may drop a perpendicular from the midpoint of BF (i.e., the center of the rotation in the original Wu proof) down to line AD. This produces two triangles that are congruent by ASA, and then CPCTC allows us to conclude that newly-drawn perpendicular to line AD must also be perpendicular to line EH. Then the Two Perpendiculars Theorem tells us that AD and EH are parallel lines. But once again, this theorem goes back to alternate interior angles, not corresponding angles.

And so we're left with either TEAI (which can be written to give priority to either corresponding or alternate interior angles) or to use the Wu proof or its ASA modification (which prioritizes alternate interior angles only). Since we want corresponding angles to have priority, we're left with TEAI.

PARCC Practice EOY Exam Question 30
U of Chicago Correspondence: Section 3-2, Types of Angles
Key Theorem: Linear Pair and Vertical Angle Theorems

If two angles form a linear pair, then they are supplementary.
If two angles are vertical angles, then they have equal measures.

Common Core Standard:
Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment's endpoints.

Commentary: I give the theorems for linear pairs and vertical angles as these are the only ones needed to answer Parts A and B. But the question hints at the theorems for parallel lines -- and we notice that the Common Core Standard quoted above mentions both the Vertical Angle Theorem itself and the Parallel Line Consequences. Proofs of these are given in Sections 5-5 and 5-6 of the U of Chicago Text. Question 11 of Section 5-6 asks students to prove the Alternate Interior Angles Test.

Tuesday, May 26, 2015

PARCC Practice Test Question 29 (Day 174)

I hope that everyone enjoyed the Memorial Day weekend -- regardless of whether your school took a three- or four-day weekend. My situation was a little more complicated, because I'm a substitute teacher in two different districts -- one of them had a three-day weekend, while the other observed a four-day weekend. And so while one of my districts was closed on Friday, I was actually subbing in the other district. A few interesting things happened on Friday, but I couldn't post them here on the blog because I'd already written here that it was a four-day weekend.

Anyway, I began my day in an Integrated Math I class for freshmen, and I ended up in an Algebra I class for sophomores grandfathered on the traditionalist pathway. In my last post I mentioned some integrated math texts, and so of course I have more to say about that today. The freshmen were studying out of a text published by Core Connections. Here are its contents:

Volume 1:
Chapter 1: Functions
Chapter 2: Linear Functions
Chapter 3: Transformations and Solving
Chapter 4: Modeling Two-Variable Data
Chapter 5: Sequences

Volume 2:
Chapter 6: Systems of Equations
Chapter 7: Congruence and Coordinate Geometry
Chapter 8: Exponential Functions
Chapter 9: Inequalities
Chapter 10: Functions and Data
Chapter 11: Constructions and Closure

The students in this class were in Chapter 6, on Systems of Equations. Students had to determine whether to use substitution or elimination to solve certain systems, and then solve them.

Even though the other class was traditionalist Algebra I, there just happened to be some integrated materials in the classroom. One was a stack of orange MathLinks packets -- the ones that are labeled eighth grade yet are being used in freshmen classes. This packet, numbered 8-15, was titled, "Geometry Discoveries." Here are its contents of its four sections, with the contents of the first section in greater detail:

15.1 Similar Triangles
-- Establish the angle-angle criterion for similarity of triangles.
-- Apply the angle-angle criterion to solve problems.
-- Link concepts of parallel lines and similar triangles to slopes of lines.
-- Prove a famous theorem [the Pythagorean Theorem] using similar triangles.
15.2 Volume of Cylinders
15.3 Volume of Cones and Spheres
15.4 Skill Builders, Vocabulary and Review

There was also another integrated math text in this classroom, published by Carnegie Learning. I've already posted chapter lists for several texts in the past week, so I won't write out the chapter list for the Carnegie Learning text. The important thing to note is that Carnegie Learning, just like Core Connections and Pearson, has a paperback Integrated Math I text divided into two volumes. For some reason, this is apparently common among integrated texts.

Recall that the gold standards for integrated texts are usually considered to be Saxon and Singapore Math texts. I'm starting to change my mind about the Singapore Math text, only because the series that I discussed on the blog a few months ago -- the New Elementary Math text -- has four volumes for grades 7-10, but the ninth and tenth grade texts are out-of-print. It's awkward to keep discussing texts that no one has access to over and over again. Often when traditionalists recommend Singapore Math, they are referring to grades K-6, occasionally K-8, but never any grade higher than eighth.

So this leaves the Saxon texts as our sole remaining gold standard for integrated high school math. I have already mentioned that traditionalists want there to be a path for seniors to take Calculus. The Saxon texts are usually denoted by two grade levels, as in Saxon 87, which is for below-average eighth graders and above-average seventh graders. Obviously, if we want there to be a path to Calculus, we must assign this text to the lower grade, so Saxon 87 is for seventh graders. Recalling how the Saxon texts are labeled, we give the following integrated path from Saxon 87 to Calculus:

Grade 7: Math 87
Grade 8: Algebra 1/2 (half)
Grade 9: Algebra 1 (3rd edition, integrated)
Grade 10: Algebra 2 (3rd edition, integrated)
Grade 11: Advanced Math
Grade 12: Calculus

So in our discussion of the various integrated texts, we must see how the text assigned to eighth graders compares to Algebra 1/2, how the freshman text compares to Algebra 1, and so on.

I already concede that in order to get students to Calculus, the Common Core math classes labeled Integrated Math I, II, and III must be taken in grades 8-10. Then juniors can take Precalculus, and this allows the seniors to take Calculus. Notice that we are comparing three different course numbering systems -- the course to be taken by freshmen on the Calculus path is called Singapore New Elementary Math 3, Common Core Integrated Math 2, and Saxon Algebra 1.

I now want to discuss my ideal Integrated Math 1 course. Originally, I wanted to divide my class into twelve units. This was ideal because I've noticed that it takes about three weeks to cover one chapter of a math text. Since a typical school year is 38 weeks long (including partial weeks like Memorial Day week, but not vacation weeks like winter break), this allows each semester to span six chapters plus a finals week.

But now I'm considering having just ten units rather than twelve. When I was student teaching in an Algebra I class, one of the texts we used was Holt Explorations in Core Math. This text was divided into ten chapters, and I noticed that we covered Chapter 1 in September, Chapter 2 in October, Chapter 3 in November, all the way up to Chapter 10 in June. So the ten chapters roughly correspond to the months of the school year. This makes pacing much easier. Instead of counting out intervals of three weeks and trying to remember whether the current week was the first, second, or third week of the chapter, I could just look at the calendar. If it was around the 15th of the month, then the class should be halfway through the chapter, and when it's the last week of the month, the class should be getting reading for the test.

So now here's my ideal Integrated Math I course. My ideal course starts out with geometry -- and indeed, I'd start out my ideal Geometry course the same way:

Unit 1: Geometry Basics
Unit 2: Reflections
Unit 3: Rotations
Unit 4: Translations
Unit 5: Glide Reflections
Unit 6: Dilations

At this point Integrated Math I and Geometry diverge. Integrated Math I now goes into algebra:

Unit 7: Slopes and Lines
Unit 8: Inequalities
Unit 9: Systems
Unit 10: Functions

For Geometry, the remaining four units are:

Unit 7: Area
Unit 8: Volume
Unit 9: Trigonometry
Unit 10: Circles

Let's look at these units in detail. I would like the first six units of the course to be similar to what I posted here on the blog. In particular, Unit 1 corresponds roughly to the first three chapters of the U of Chicago Geometry text, while Units 2-5 correspond respectively to Chapters 4-7 of the text.

Now it may seem weird to spend roughly a semester just on transformations. But most of the other material in geometry ultimately go back to transformations -- this is why transformations are the cornerstone of Common Core Geometry. But I definitely want to make some changes to the way that I covered the material.

In particular, the congruence theorems SSS, SAS, and ASA are covered much too late in the U of Chicago text. The text uses transformations -- translations, reflections, and rotations -- to prove the three theorems SSS, SAS, and ASA, and since translations don't appear until Chapter 6, triangle congruence can't appear until Chapter 7. But this approach makes SSS, SAS, and ASA merely afterthoughts -- when we've seen several questions requiring SSS, SAS, or ASA on the PARCC.

Instead, we move SSS, SAS, and ASA up into Unit 2 with reflections. Notice that we can still prove SSS, SAS, and ASA using reflections only -- translations aren't needed at all. So we can cover much of the first half of Chapter 7 (U of Chicago) with Chapter 4 and make them into my Unit 2. I make these changes just as Dr. Franklin Mason has changed his curriculum after seeing what appears on the PARCC and SBAC exams.

Unit 3 can remain very much how it's posted here. It corresponds roughly to Chapter 5 of the U of Chicago, as we discuss polygons. Rotations will still be used to develop the properties of parallel lines, just as Dr. Hung-Hsi Wu does. A parallel postulate can appear right at the start of the second quarter of the course.

I was thinking about how I wanted to deal with the end of the first semester. I mentioned earlier that I wanted similarity to start the second semester, rather than end the first semester. My Unit 6 will be a tough unit, and I don't want to tantalize students into having a good grade for most of the semester, only to have it drop after covering the similarity chapter. Indeed, my idea is that the last unit before the end of the semester should be something that's easy to do, but hard to remember. So grades won't drop because the unit is easy, and though it's difficult to remember, the final occurs right afterward, before the students forget the material.

Furthermore, we consider the fact that if Unit 1 is in September, then Unit 4 is in December -- that is, just before winter break. On the Early Start Calendar, Unit 1 may be in August, which means that Unit 4 would be in November -- just before Thanksgiving break. So Unit 4 is likely to precede a holiday week no matter what. This is, of course, not necessarily the best time to introduce difficult material either. Instead, I want to put something more enjoyable here. Since many students enjoy drawing, this may be a good time (after teaching translations from Chapter 6) to introduce some of those constructions that appear on the PARCC but not in the U of Chicago text.

Unit 5 will contain glide reflections, as these are difficult to remember. Properties of parallelograms are also hard to remember, and so I can end the unit the same way I ended the semester here on the blog, with the last part of Chapter 7.

Then we reach Unit 6 -- the big chapter on dilations and similarity. This will cover material from Chapters 11 and 12 of the U of Chicago -- just as I began second semester here on the blog. Notice that there is so much to cover here -- the Dilation Distance Postulate, similarity using AA/SAS/SSS, the Pythagorean Theorem, the coordinate plane, the distance/midpoint formulas. transformations on the coordinate plane (barely covered in U of Chicago, strongly covered on the PARCC), and finally the slope formula.

By now you can surely figure out what I'm trying to accomplish here -- we teach the slope formula in Unit 6, and then segue right into slope and the linear equations of algebra in Unit 7. After all, this is the whole point of taking an integrated course -- to show the connections between the various branches of mathematics. But this won't work if Unit 6 becomes so unwieldy, with so much information packed in that it becomes unteachable.

Notice that we could begin Unit 6 with the Dilation Distance Postulate, and then move into similarity the same way that MathLinks packet 8-15 does it. Notice that we could get away with teaching only AA Similarity and not SAS or SSS. After all, the heavy-duty results of the similarity chapter -- the slope formula, Pythagorean Theorem, and eventually trigonometry all use AA Similarity only. The only theorem I saw in the U of Chicago text whose proof required SAS Similarity was the Converse to the Side-Splitter Theorem -- but of course the forward theorem used AA, and no theorem required SSS at all. By teaching AA only, there would be time to teach more of the bigger results.

As I mentioned in my last post, one thing that I like to do is have some lesson involving the number pi close to Pi Day. On a Labor Day Start calendar, March is Unit 7, while on an Early Start calendar, March would be Unit 8. In my traditionalist Geometry course, I made sure that Units 7 and 8 are on area and volume, so that both units teach formulas involving pi.

Now on my Integrated Math pathway, I delay area to Math II and volume to Math III. By doing so, both courses have lessons on pi that can be taught near Pi Day. Indeed, notice how we can continue this pattern into Precalculus, when trigonometry using radians is taught. The common angles whose trig values can be memorized are those of simple ratios of pi. And so that particular constant can be taught in Precalc near Pi Day as well.

But Integrated Math I, unfortunately, doesn't mention pi much at all. After all, most of the second semester is on Algebra I, and that class generally avoids mention of pi. The closest I can come is to mention the properties of real numbers during Units 7 and 8 -- and to explain what "real numbers" are, we can mention irrational numbers such as pi (and sqrt(2), since we just taught a lesson on the Distance Formula, which involves square roots). This is why I made Unit 8 on Inequalities -- this is another good place to sneak in pi, since we have the inequality 3 < pi < 22/7 involving pi.

To close out my Integrated Math I class, Unit 9 will be on systems -- in a Labor Day start class, Unit 9 will be in May, and the class I subbed in last week was learning systems. Unit 10, like Unit 5, is at the end of a semester, and June is certainly not the time when students want to see a very difficult math lesson! Here I decided to put in a short unit on functions. I've noticed that in many Algebra I texts, there is a chapter on functions just before the chapter on linear functions (and slope). The teacher often spends a day or two on the functions chapter, then goes straight into linear functions. I put my lesson on functions at the end, so that teachers can spend as little or as much time as is left at the end of the year before reviewing for the final.

By the way, since I'm so eager to show the connection between similarity and slope, one might wonder why I teach functions in Unit 10 rather than tie it to geometry transformations -- since transformations actually are functions. My fear is that in Unit 10, I wish to teach that the domain of a function y = f (x) consists of the input, or the set of x-values, while the range of a function consists of the output, or the set of y-values. But the input of a geometry transformation like T(x, y) has both an x- and a y-value -- and so does the output. So I avoid this confusion by placing functions far away from transformations -- although Math II or III could mention that transformations are functions.

Let's compare my Integrated Math I course to that from the Core Connections text -- which is supposed to be based on the PARCC Integrated Math I exam. One key omission in my text is the simple solving of linear equations. Here I'm assuming that solving at least one-step (and possibly two-step) equations was taught in the sixth- or seventh-grade courses, and unlike fractions, this is one thing that students do typically remember. Simple equations can also appear in some of the first semester geometry lessons. Yes, I know that I've said that I don't want to see too much algebra during the geometry lessons, but that was because I was worrying about students coming off of a C- or D- grade in Algebra I getting to Geometry and having so see so much algebra again -- but this is not a concern in an Integrated Math I class. And besides, my concern was more about having, say similarity questions that lead to quadratic equations. Of course, proportions will appear in the similarity unit, and some equation manipulation (such as solving for y) appears in the slope unit. All other types of linear equations (including equations with x on both sides, which often confuse students) can be taught or reviewed during the inequalities unit -- another reason that I placed that unit there.

But again, the gold standard for Integrated Math isn't Core Connections -- it's Saxon. So let's compare my Integrated Math I course to Saxon Algebra 1/2. The algebra side of my Integrated Math I course is on par with Algebra 1/2 -- mine is probably a little ahead, since I cover systems that Saxon doesn't cover until Algebra 1. Notice that both Core Connections and Saxon include lessons on collecting and graphing data, exponential functions, and other applied word problems. These can be incorporated into my Units 7-10 wherever they belong -- including exponential functions, which can be sneaked into Unit 10 if desired.

On the other hand, my geometry lessons differ from all of the other integrated texts. Similarity appears in both Core Connections and Saxon later than when I'd teach it -- but I want to show the connection between similarity and slope. In exchange, area and volume must appear later in my lessons than in Core Connections or Saxon.

There's one more thing that I must mention about my Integrated Math I course. My placement of Unit 5 was highly dependent on how students would feel about the semester final. But in order for students to reach calculus by senior year, Math I must be taken in the eighth grade -- a grade which, at many middle schools, has neither semesters nor finals. At middle schools where there are trimesters, it may be convenient to teach only nine units, as this would be three units per trimester. With three units per trimester, the first two trimesters would be geometry with only one trimester of algebra. There would be a tough Unit 6 at the end of the second trimester -- but once again, there are no finals, and the students already received a first trimester grade. (Of course, this is also an argument to go back to 12 units, as 12 is convenient for both semesters and trimesters.)

I owe the development of my Geometry and Integrated Math I courses to the mathematicians whom I regularly quote on the blog. The idea to teach SSS, SAS, and ASA early goes back to Dr. M, who changed his own course to reflect what appears on the PARCC and SBAC exams. From Dr. Wu, I keep the idea to use rotations to teach parallel lines. From Dr. David Joyce, I keep the idea to use similar triangles to teach slope. And one could argue that I first got the idea to teach only AA and no other similarity theorem from Danica McKellar, who gives only AA in her geometry for girls text.

Of course, neither my ideal Geometry nor my Integrated Math I course can become reality now, because both classes are bound by what appears on their respective PARCC exams. Furthermore, the fact that the PARCC exam is given so early in the year forces me to accelerate Units 7-9 so that they are completed before the PARCC PBA test. Unit 10 of my Geometry course, on circles, must remain the last chapter to be taught since questions from it appear on the PARCC EOY but not the PBA.

And one of those questions is today's featured PARCC problem. Question 29 of the PARCC practice exam is indeed on circles:

Point B is the center of a circle, and AC is the diameter of the circle. Point D is a point on the circle different from A and C.

Part A

Indicate which statements must be true.

Select all that apply.

(A) AD > CD
(B) Angle CBD =1/2 (Angle CAD)
(C) Angle ADC = 90 degrees
(D) Angle ABD = 2 (Angle ACD)
(E) Angle ABD = Angle DBC

Part B

If Angle BDA = 20 degrees, what is Angle CBD?

(A) 20 degrees
(B) 40 degrees
(C) 70 degrees
(D) 140 degrees

Like most other PARCC questions involving circles and angles, Section 15-3, the Inscribed Angle Theorem, is the only section of the U of Chicago text that really matters. For Part A, there are two correct answers. Angle ABD = 2 (Angle ACD) is correct because it gives the central angle as equal to double the inscribed angle, and Angle ADC = 90 degrees because it's inscribed in semicircle ADC. So the answers are (C) and (D).

For Part B, we must go back to isosceles triangles again, just like many other PARCC questions on inscribed angles. Triangle BDA is isosceles because two of its sides are radii. So Angle BAD, the same as CAD, must also be 20 degrees. Now CBD is the central angle for the inscribed angle CAD, so its measure must be double that of CAD, so it's 40 degrees. The correct answer is (B).

A common error for Part A is to include (B) as a correct choice. But we notice how (B) claims that the CBD is half of CAD, when it's actually double. Notice that if (B) had been correct, the answer to Part B would have been 10, not 40, degrees. Indeed, I'm actually surprised that 10 degrees wasn't one of the wrong choices for Part B.

There's nothing much more I can about this question that wouldn't be a repeat of my comments about the other circle-angle problems.

PARCC Practice EOY Exam Question 29
U of Chicago Correspondence: Section 15-3, The Inscribed Angle Theorem
Key Theorem: Inscribed Angle Theorem

In a circle, the measure of an inscribed angle is one-half the measure of the intercepted arc.

Common Core Standard:
Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.

Commentary: I repeat my comments from Questions 8 and 24. Most of the inscribed angle problems in the U of Chicago text give the intercepted arc measure, but a few do require students to calculate the arc measure first before finding the inscribed angle measure. In the SPUR section, Objective B, including Questions 7 through 11, are good questions to consider.

Thursday, May 21, 2015

PARCC Practice Test Question 28 (Day 173)

The school whose calendar I'm following here on the blog observes a four-day weekend for Memorial Day, with school closed Friday and Monday. Therefore today is the last day of the Long April/May, which is actually the longest stretch without a day off on my calendar.

Today I subbed in a math class at a continuation school -- which means that the students were at many different levels. One thing I noticed near the front of the classroom were textbooks for Integrated Math II and III. These texts were published by Pearson -- and one reason that many people oppose Common Core is that they feel it's less about helping students and more about making money for Pearson.

Unfortunately, I didn't see a copy of their Integrated Math I text anywhere. But let's review the contents of the two texts that I do see.

Pearson Mathematics II
Volume 1
Chapter 1: Reasoning and Proof
Chapter 2: Proving Theorems About Lines and Angles
Chapter 3: Congruent Triangles
Chapter 4: Proving Theorems About Triangles
Chapter 5: Proving Theorems About Quadrilaterals
Chapter 6: Similarity
Chapter 7: Right Triangles and Trigonometry
Chapter 8: Circles

Volume 2
Chapter 9: Surface Area and Volume
Chapter 10: Properties of Exponents With Rational Exponents
Chapter 11: Polynomials and Factoring
Chapter 12: Quadratic Functions
Chapter 13: Probability
Chapter 14: Other Types of Functions
Chapter 15: Sequences and Series

Pearson Mathematics III
Volume 1
Chapter 1: Drawing Conclusions From Data
Chapter 2: Linear Equations, Inequalities, and Functions
Chapter 3: Quadratic Functions and Equations
Chapter 4: Polynomials and Polynomial Functions
Chapter 5: Rational Expressions and Functions
Chapter 6: Radical Expressions and Functions

Volume 2
Chapter 7: Exponential and Logarithmic Functions
Chapter 8: Trigonometric Functions
Chapter 9: Sequences and Series
Chapter 10: Applying Geometric Concepts
Chapter 11: Connecting Algebra and Geometry
Chapter 12: Circles

Since I don't have a copy of the Math I text, I can't make any conclusions. But I've always thought that on the Integrated Pathway, geometry would be split among all three years of math. Yet we notice that almost the entire geometry curriculum appears in Math II, Volume 1, with only surface area and volume appearing in the second volume.

I assume that Pearson intended for the two volumes into which the company divides both Math II and III to correspond to the semesters. So we see that the first semester of Math II is all Geometry, while the second semester is a mixture of Algebra I and II. The first semester of Math III starts the same way that an Algebra II course would, while the second semester ends with a little bit of Geometry (as Chapter 11 implies, it connects the two subjects).

Let's focus on Math II Volume 1, as this text is the most relevant to our geometry blog. Section 1-1 of the text is on basic constructions -- and I already see something that I don't like about the text. We recall that the Prentice-Hall text that David Joyce criticizes on his website also begins with constructions, and Joyce doesn't like that one bit. Joyce writes:

On pages 40 through 42 four constructions are given: 1) to cut a line segment equal to a given line segment, 2) to construct an angle equal to a given angle, 3) to construct a perpendicular bisector of a line segment, and 4) to bisect an angle. Later in the book, these constructions are used to prove theorems, yet they are not proved here, nor are they proved later in the book. There is no indication whether they are to be taken as postulates (they should not, since they can be proved), or as theorems. At the very least, it should be stated that they are theorems which will be proved later. In summary, the constructions should be postponed until they can be justified, and then they should be justified.

Notice that the four constructions in the Pearson text are exactly the same as the four in Prentice-Hall, even numbered in the same order! I checked the list of authors for the Pearson text and saw that there are several, since both algebra and geometry writers are needed. But I notice that one of the geometry authors is Laurie Bass, who is also, according to Joyce, the author for Prentice-Hall -- small wonder, then, that the same four constructions are here. The fact that Bass doesn't prove any of her constructions is especially galling considering that on last Monday's featured PARCC question, students had to justify the angle bisector construction (the last construction of Section 1-1)! Moreover this lesson also requires students to purchase compasses close to the first day of school.

Section 1-2, on Patterns and Inductive Reasoning, is a much better lesson. I've mentioned that many texts have an early lesson on patterns, but the U of Chicago doesn't. If I were teaching a class using the Pearson text, I'd be tempted to skip Section 1-1 entirely and start the course with 1-2. In short, Chapter 1 of the Pearson text corresponds mostly to Chapter 2 of the U of Chicago text, with some material from later chapters. Section 1-5, on Deductive Reasoning, contains some of the logic from U of Chicago's Chapter 13, and Section 1-7, on Proving Angles Congruent, contains some of the material from Chapter 3.

Pearson's Chapter 2 is on Lines and Angles. Section 2-2 introduces a postulate on parallels -- and it happens to be the Same-Side Interior Angles Postulate. This is highly unusual, as most texts choose either Corresponding Angles (U of Chicago) or Alternate Interior Angles as a postulate. Section 2-3 contains the Parallel Tests -- and all of the are listed as theorems. The first test is the Converse of the Corresponding Angles Theorem, and its proof is said to be given in Section 13-5 -- but that chapter is on Probability. I once saw another text that does give the proof -- it's an indirect proof, and so Pearson probably intended to wait until Chapter 4 to give the proof, but it mistakenly omits it. In summary, most of this chapter corresponds to Chapter 3 of the U of Chicago text, except for Section 2-5 on Parallel Lines and Triangles, which contains another Parallel Postulate (Playfair's), which isn't given in the U of Chicago until Chapter 13 (where it's actually proved as a theorem).

Pearson's Chapter 3 is on Congruent Triangles. We notice that in Sections 3-2 and 3-3, SSS, SAS, and ASA are given as postulates. But then in true Common Core fashion, congruence transformations (isometries) are used to verify these postulates in Section 3-8. I notice that as important as the isometries are to Common Core, translations, reflections, and rotations aren't defined anywhere in this section. I suspect that since these isometries form the foundation of Common Core Geometry, Pearson actually defines them in its Math I text. There's no way for me to know for sure since I don't have access to the Math I text. In short this corresponds to the first part of Chapter 7 in U of Chicago, except for isosceles triangles, which appear at the start of Chapter 5.

Pearson's Chapter 4 is on Proving Theorems About Triangles. Here is where the concurrency theorems are shown. These appear in Chapter 10 of Dr. M but nowhere in the U of Chicago. In Section 4-5, indirect proofs appear for the first time, and the rest of the chapter is devoted to proving the inequalities associated with triangles, including Triangle Inequality and Hinge Theorem. These appear in Chapter 5 of Dr. M but are scattered in the U of Chicago, as I discussed here last week.

Pearson's Chapter 5 is on Proving Theorems About Quadrilaterals. Most of this material appears in the same-numbered chapter of the U of Chicago, except some of the theorems on parallelograms don't appear until Chapter 7. Of course I had to check to see that Section 5-7 of Pearson gives the exclusive definition of trapezoid. As I said earlier, the inclusive definition was spotted in a PARCC question, but I have yet to see it. Then the rest of the chapter gets into coordinate geometry -- which I suppose makes sense as many coordinate proofs involve quadrilaterals. The harder concurrency theorems from Chapter 4 are given coordinate proofs here as well. Many of these coordinate proofs require the Distance or Midpoint Formulas, which are never introduced. Once again, I suspect that these formulas are taught in the Math I text and students are expected to remember them.

Pearson's Chapter 6 is on Similarity. This corresponds to Chapter 12 of the U of Chicago. Right triangle similarity appears in Section 6-4, but not until Chapter 14 in the U of Chicago. The chapter ends by defining dilations and similarity transformations -- and just as in Chapter 3, these are used to verify AA, SAS, and SSS Similarity,

Pearson's Chapter 7 is on Right Triangles and Trigonometry. This corresponds to Chapter 14 of the U of Chicago. Angles of elevation/depression, inverse trig, and areas of regular polygons using the apothem all appear in this chapter, but are only mentioned briefly if at all in the U of Chicago.

Pearson's Chapter 8 is on circles. In some ways, this is more like Dr. M's Chapter 9 than the U of Chicago's Chapter 15, since all of the circle material is right here in one chapter. Section 8-2 is on the areas of a circle. Notice that the area of no other basic figure appears in any chapter of this text. In some ways, this is alarming because the text defines the area of a segment of a circle as the area of a sector minus the area of a triangle -- except that the area of a triangle isn't defined in the text. Once again, we can only assume that any gaps must be covered in the Math I text.

As I mentioned earlier, Chapter 9 is the first chapter of Volume 2, and this chapter is on surface area and volume. Whenever I look at the curriculum for any text, I like to see whether I can time it so that the lessons mentioning pi can occur around Pi Day. In this text, Chapters 8 and 9 mention pi. But it would be tough to squeeze in all of Chapters 10-15 between mid-March and the PARCC exam (recall that PARCC does have an Integrated Math test), especially considering that these contain some tough algebra lessons.

For Volume 2, I will only mention the remaining chapters that are relevant to geometry. Chapter 12 is on quadratic equations, including completing the square and the quadratic formula -- but then it ends up with the equation of a circle. This is convenient since we know that equations of a circle that require completing the square appear on the PARCC.

As I mentioned, Integrated Math III contains some geometry. But notice that Chapter 8 of Math II and Chapter 12 of Math III are identical chapters. Every exercise in both chapters is the same, and Section 12-1 of Math III even mentions Postulate 19 and Theorem 72, even though the first 18 postulates and 71 theorems don't appear in the Math III text! I know that much of the material from Algebra I is repeated in Algebra II and then again in Precalculus, but this is the first time that I've seen a textbook publisher try to get away with including identical material in different courses!

As for the other geometry chapters in Math III, Chapter 10 includes some key problems that appear on the PARCC exam. These include some more constructions in Section 10-1 (most notably the inscription of a hexagon in a circle, specifically mention in Common Core), density in Section 10-2 (which we've already seen on the PARCC), areas of similar figures in Sections 10-3 and 10-6.

Chapter 11 contains some area formulas -- including the ones that I thought were missing from the Math II text. I wouldn't be surprised to find out that Chapter 11, like 12, is the repetition of a topic from an earlier text.

I spent much of today's blog discussing the Pearson Integrated Math texts, but notice that none of the students at the continuation school where I was subbing actually studied from any of them. Some of the students were reading the McDougal-Littell Algebra I text -- recall that continuation students are behind in credits, so these students are one of the last cohorts to take an Algebra I course. The other students were reading Mathematics: A Human Endeavor by Harold R. Jacobs. This text appears to teach a stripped-down version of Algebra II, as many Algebra II topics (sequences, quadratic equations, logarithms, conic sections) appear at a basic level. Section 10-2 of this text is on the Seven Bridges of Konigsberg -- my first day of school activity. The students who were reading the Jacobs text were officially enrolled in "Coordinated Math." Most of the students were only on Chapters 2 or 3 of their respective texts. Of course, the students who were classified as juniors still had to take the SBAC -- and they were pulled out of classes today to do so.

I've been thinking about integrated math courses. We've spent time looking at the Pearson Integrated Math texts, and we've already discussed the Saxon and Singapore Integrated texts on the blog. And so, if I'm going to spend an entire post criticizing the Pearson text, I should ask myself, what would my ideal Integrated Math classes look like?

I notice that much of the high school material naturally falls into three dimensions. Think about it -- my Integrated Math I course can begin with geometry, from lines, angles, and transformations all the way up to similarity (Chapters 1 through 7 and 12 of U of Chicago). Then we go from similarity to the coordinate plane and the constancy of a line's slope -- and that's a great point to segue into linear (one-dimensional) equations and algebra. We could probably cover the first semester of an Algebra I course before the end of the first year.

Integrated Math II stands for the second dimension. So on the geometry side, this class should focus on measuring two-dimensional figures -- perimeter and area. Many area questions lend themselves to quadratic equations, and so the algebra side of the class should work on quadratic functions, like the second half of an Algebra I class.

Integrated Math III stands for the third dimension. By now, you've figured out that the geometry side of the class should have surface area and volume. For algebra, we can focus on cubic polynomials, and some non-polynomials functions studied in Algebra II, like exponential functions. So far, this division is incomplete, but if I do it right, I can cover all of the algebra and geometry needed for Common Core and make connections between the topics better than Pearson can, which is what Integrated Math is all about. For example, Chapter 10 in Pearson's Math III ends with the concept of locus -- a set of all points satisfying a certain condition. This would be a great place to segue into conic sections, which are defined as sets of points satisfying certain conditions -- in a word, loci.

Before I leave the classroom where I subbed, I saw that there were a few worksheets left over from an activity that this class did last week. I noticed that this activity was divided into parts, "Act One," "Act Two," "Act Three," "Sequels," and "Title." And of course, as soon as we see three acts, the readers of this blog and I should immediately recognize it as a Dan Meyers Three-Act activity.

The question in Act One is, "Where will the stacks of cups tie?" So I went directly to the dy/dan website and tried to find this Three-Act activity, but I couldn't find it anywhere. The following link was the closest activity that I could find:

How many Styrofoam cups would you have to stack to reach the top of your math teacher’s head?

So I suppose that "tie" here must mean "tied score." If the stack of cups is shorter than the teacher, then the stack "loses," if the stack of cups is taller, then the stack "wins," and so if the stack of cups is the same height as the teacher, then the stack "ties." Yes, this is an entertaining activity that could fit in Geometry, Algebra I, or Integrated Math I (the one-dimensional course).

So let's get back to PARCC now. Question 28 of the PARCC Practice Test is still another transformations question. There are some tricky reflections and rotations mentioned in this one:

Quadrilaterals ABCD and EFGH are shown in the coordinate plane.

(Here are the vertices: A(-4, 4), B(-3, 7), C(1, 6), D(-2, 2), E(4, -4), F(3, -7), G(-1, -6), H(2, -2).)

Part A:

Quadrilateral EFGH is the image of ABCD after a transformation or sequence of transformations.

Which could be the transformation or sequence of transformations?

Select all that apply.

(A) a translation of 3 units to the right, followed by a reflection across the x-axis
(B) a rotation of 180 degrees about the origin
(C) a translation of 12 units downward, followed by a reflection across the y-axis
(D) a reflection across the y-axis, followed by a reflection across the x-axis
(E) a reflection across the line with equation y = x

Part B:

Quadrilateral ABCD will be reflected across the x-axis and then rotated 90 degrees clockwise about the origin to create quadrilateral A'B'C'D'. What will be the y-coordinate of B'?

The answer to Part A may be obvious just by looking at the coordinates above, but I only post them because there's no other way to convey to the readers of this blog in ASCII what the quadrilaterals look like. But the student will see only the quadrilaterals, not the list of coordinates. So let's approach this from a student's perspective.

The easiest thing for the student to check is not the coordinates of the vertices, but the orientation. So we notice that both ABCD and EFGH have a clockwise orientation. This means that we can eliminate three of the choices -- the reflection and the two glide reflections. As for the two remaining choices, one might recall that the composite of a y-axis reflection and an x-axis reflection actually is a rotation of 180 degrees about the origin. After a quick check of corresponding coordinates -- for example, we count off the graph to see the coordinates B(-3, 7) and F(3, -7) -- we see that the this transformation really is a point inversion about the origin. So both of the remaining choices, (B) and (D), are correct.

For Part B, we start with point B(-3, 7). The x-axis reflection switches the sign of the y-coordinate only, so that gives us (-3, -7). The 90-degree rotation clockwise about the origin rotates the point from the third back into the second quadrant, to B'(-7, 3). So the y-coordinate of B is 3.

Common mistakes for Part A include choosing transformations that describe some of the points, but not all of them. For example, if we reflect A(-4, 4) about the line y = x, we do get E(4, -4), and likewise the reflection of D(-2, 2) is H(2, -2). But the mirror images of B and C are not F and G, so choice (E) is wrong. In a similar manner, performing the glide reflection in choice (C) maps C to F, but none of the other points are mapped correctly. Only (B) and (D) map all of the points on the graph to their proper images.

There are a few more comments to make about this problem. First of all, we notice that the transformations in choices (A) and (C) really are glide reflections. The phrase "glide reflection" doesn't appear on the PARCC, but they are listed by the information that matters most for a glide reflection, namely the translation and the reflection. We see that the PARCC will perform the translation first -- as opposed to the U of Chicago which always performs the reflection first. It doesn't really matter as the translation and reflection will always commute in this case, but it's good to know how glide reflections will be written on the PARCC.

I wrote earlier that there are some transformations that are easy enough to appear on the PARCC, and some that shouldn't appear at all. All translations are easy, but only certain reflections and rotations are easy enough for the students. (By this measure, a glide reflection is easy provided the reflection part of it is.) The easiest reflections and rotations are the ones that appear in choices (B) and (D) -- reflections across the axes, rotation 180 about the origin. The next best reflections and rotations are the ones that appear in choice (E) and Part B: reflection across the line y = x (or y = -x), and rotation of 90 degrees in either direction about the origin.

Of these, reflection in y = x is the easiest -- the image of (x, y) is just (y, x). Reflection in the other line y = -x maps (x, y) to (-y, -x), so it switches the coordinates along with their signs. Rotations of 90 degrees are important because they appear in the proof of the perpendicular slope formula. Rotating (x, y) 90 degrees clockwise gives (y, -x), and rotating it 90 degrees counterclockwise gives (-y, x). In either case, a rotation of 90 degrees maps the line passing through the origin and (x, y) (notice that the slope is y/x) to the line passing through the origin and (y, -x) (notice that the slope is -x/y). This constitutes a coordinate proof of the slopes of perpendicular lines as opposite reciprocals. QED

All four of the transformations switch the coordinates of their preimages, but they each have a different effect on the signs of the coordinates. Since tomorrow is Memorial Day weekend, I can make today an activity day. The following is a short activity -- one of my favorites that I like to show during tutoring -- that helps students remember how to perform these four transformations.

We take a small coordinate plane and label its axes x and y -- actually, let's mark the signs as well, so we have all four directions +x (east), +y (north), -x (west), and -y (south). We should label these on the front and back (or just use marker, so the directions bleed through the paper). We cut it out and place it on a larger coordinate plane, also with the four directions marked.

Now we can perform all of the various transformations. Let's say we want to do the transformations listed in Part B. To reflect over the x-axis, we fold and ultimately flip the smaller plane. Now the +y of the larger plane lines up with the -y of the larger plane. This tells us that to reflect over the x-axis, we must switch +y with -y (i.e., change the sign of y). Restoring the smaller plane to its original position, we now fold and flip over the diagonal line y = x. Now the +y of the larger plane lines up with the +x of the smaller plane, indicating that we switch x and y. If we want to try a rotation, let's put the plane in its original position and try a 90-degree rotation clockwise. Now the +x of the larger plane lines up with +y of the smaller and +y of the larger plane lines up with -x of the smaller. Thus (x, y) of the larger (original) plane maps to (y, -x) of the smaller (rotated) plane.

To perform a composite, we perform one transformation right after the other. First let's reflect across the x-axis, so +y original is now -y transformed, but +x didn't change. Now we rotate 90 degrees in the clockwise direction. So now we have +x original lined up with -y transformed and +y original lines up with -x transformed, so (x, y) now maps to (-y, -x). Thus B(-3, 7) maps to B'(-7, 3) and so the answer to Part B is 3.

Notice that (x, y) maps to (-y, -x) in Part B. This is the same as a reflection across y = -x. The set of all eight rotations and reflections mentioned today form a group -- the symmetry group of a square centered at the origin and sides parallel to the axes.

Oh, and here's the answer to yesterday's question -- find some (discontinous) transformation T such that T o T is reflection in the x-axis:

Divide the plane into horizontal strips. Each strip covers a different section of the y-axis. One strip is for y = 0 (i.e., the x-axis itself). The other strips cover the interval (n - 1, n] on the y-axis for each natural number n. Then T acts on each horizontal strip as follows:

n = 0: T maps the x-axis to itself
n > 0 and odd: T translates one unit up
n > 0 and even: T reflects across the line y = 1/2
n < 0 and odd: T translates one unit down
n < 0 and even: T reflects across the line y = -1/2

I was going to make this challenge part of the activity today, but I realized that this will be too difficult even for the brightest high school students. So I threw it out and replaced it with Dan Meyer's cup activity instead.

Thus concludes this post. My next post will be Tuesday, May 26th. Happy Memorial Day!

PARCC Practice EOY Exam Question 28
U of Chicago Correspondence: Section 6-6, Isometries
Key Theorem: Definition of glide reflection

Let r_m be a reflection and T a translation with nonzero magnitude and direction parallel to m. Then G = T o r_m is a glide reflection.

Common Core Standard:
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: The U of Chicago defines glide reflections as T o r_m, but on the PARCC, what appears is r_m o T. Reflections over y = x and 90-degree rotations don't appear at all.