Thursday, April 30, 2015

PARCC Practice Test Question 13 (Day 158)

Chapter 4 of Mario Livio's The Equation That Couldn't be Solved, "The Poverty-Stricken Mathematician," is about one of the two mathematicians who independently discovered that there is no Quintic Formula. Just as dal Ferro and Tartaglia independently co-discovered the Cubic Formula, Abel and Galois independently co-discovered the non-existence of the Quintic Formula.

Niels Henrik Abel was a Norwegian mathematician born at the turn of the nineteenth century. As the title of the chapter implies, Abel was poor his entire life. According to Livio, Abel was the child of two parents who were often in trouble with the law. The family was so poor that they had to eat horse meat. But despite this difficult family life, the pastor was able to homeschool Abel until he became a teenager, at which time he was sent to the Cathedral School in Norway's capital.

As a teacher myself, I am intrigued by the way Livio contrasts two of Abel's math teachers at the Cathedral School. His first teacher, Hans Peter Bader, was a terrible teacher. Corporal punishment was common back then, and Bader frequently applied it to his students. In fact, Bader was fired after being accused of beating one of his students to death. Abel was deeply discouraged and his grades dropped dramatically.

Abel's other math teacher, after the firing of Bader, was Bernt Michael Holmboe. Holmboe was certainly a much better teacher than his predecessor. Abel excelled under his new teacher, who encouraged the youngster to go beyond the standard curriculum and study the great mathematicians like Euler, Newton, and Gauss. Holmboe wrote of his student, "With the most incredible genius he unites an insatiable interest in and ardor for mathematics, such that quite probably, if he lives, he will become one of the great mathematicians."

Of these two, Bader and Holmboe, which one do I want to emulate more? This is a no-brainer -- of course I wish to be more like Holmboe. But sometimes, I feel that my teaching style is more like Bader's than Holmboe's. Of course, I don't mean that I would implement corporal punishment, but sometimes I wonder whether I'm verbally attacking students if they say the wrong answer. I want to encourage my weaker students to study math -- not discourage the stronger students from learning, as Bader unwittingly did to Abel.

Livio writes that Abel was "whizzing through the standard curriculum," which is why Holmboe encouraged Abel to go above and beyond that curriculum. But many people believe that this would be impossible today, especially if "standard curriculum" means "Common Core." This is not an easy question to answer. The ideal answer would be that the test should be one that contains items that a prodigy like Abel could study on his own and omit those that require special Common Core training to answer. In the lower grades, it's obvious what should be included (i.e., traditionalist arithmetic) but in higher grades it's not as clear-cut. I also favor changing the test in such a way that rewards above grade-level students. But still, even with Common Core and all that it entails, I owe it to my above grade-level students to make them feel that they are learning in my class, and to inspire the other students to work harder to they can become more like the top students.

We can see many similarities between Abel's world in 1815 -- the year he entered Cathedral School -- and the world of many of our poorer students two centuries later. Abel's parents both lived unsavory lives -- just like many of our students' parents -- and he was often depressed whenever he was separated from his friends -- again, just like kids today. Yet Abel never used his poverty as an excuse not to study -- and neither should our students. Abel's story is one that I would like to tell my students someday in my own classroom.

Abel thought that he had discovered the Quintic Formula. It was a few years later, in 1823. when Abel proved the opposite -- that there is no Quintic Formula. Livio writes that Abel's proof is far too technical to include in his book. But he points out the basic idea -- Abel began by assuming that there is a Quintic Formula and shows that this assumption leads to a contradiction. In other words, Abel's proof is an indirect proof -- the sort of proof that we learn about in Section 13-4 of the U of Chicago math text. Livio himself uses the Latin phrase reductio ad absurdum -- "reduction to the absurd" -- which is formal name for indirect proof.

When one thinks about it, it makes perfect sense for Abel's proof to be indirect. How in math do we prove that something is impossible? The only real way is to assume that it's possible and show that this leads to a contradiction. Many of our proofs that lines are parallel were indirect, because we showed that it's impossible for the lines to intersect. Abel's proof is now often referred to as Abel's Impossibility Theorem. It's also often called the Abel-Ruffini Theorem -- after Paolo Ruffini, the mathematician who first mentioned the possibility that a Quintic Formula is impossible.

Compared to Abel's Impossibility Theorem, today's question is very simple. Question 13 on the PARCC Practice Test is on dilations:

The figure shows line AC and line PQ intersecting at point B. Lines A'C' and P'Q' will be the images of lines AC and PQ, respectively, under a dilation with center P and scale factor 2.

Which statement about the image of lines AC and PQ would be true under the dilation?

(A) Line A'C' will be parallel to line AC, and line P'Q' will be parallel to PQ.
(B) Line A'C' will be parallel to line AC, and line P'Q' will be the same line as PQ.
(C) Line A'C' will be perpendicular to line AC, and line P'Q' will be parallel to PQ.
(D) Line A'C' will be perpendicular to line AC, and line P'Q' will be the same line as PQ.

The necessary theorem is repeated several times through the first three sections of Chapter 12. A line and its image under a size transformation (dilation) are parallel. So we can already eliminate the two choices that mention "perpendicular." As it turns out, because the center of the dilation is P, that point P is a fixed point of the dilation, and any line passing through P must be an invariant line of the dilation -- that is, its image under the dilation is the line itself. Therefore the line P'Q' will be the same line as PQ, and so the correct answer is (B).

The idea that a line is parallel to its dilation image is emphasized in the Common Core Standards. It appears in the U of Chicago and a few questions in the first three sections of Chapter 12 mention it, but it is not as strongly highlighted as in the Common Core. In particular, the idea that a line passing through the center of the dilation is invariant doesn't appear in the U of Chicago text at all.

Some readers might point out that the U of Chicago text is therefore wrong. The theorem states that a line and its dilation and its image are parallel. Nowhere in the theorem does it state that the line does not pass through the center of the dilation -- so we can't rule out the possibility that the line passes through the center. Since such a line is mapped to itself while the theorem claims that the line is mapped to a parallel line, we've found a contradiction (there's that indirect proof again!) that proves that the theorem is incorrect.

As it turns out, this is not really a contradiction, because in the U of Chicago text, a line is considered to be parallel to itself. So a line passing through the center of the dilation actually is mapped to a parallel line -- namely the line itself.

Not everyone, of course, accepts this unusual definition of "parallel." In some ways, this is actually an inclusive definition of "parallel" in the same way that considering a parallelogram to be a trapezoid entails an inclusive definition of "trapezoid." From one perspective, it makes sense to define a line to be parallel to itself -- after all, parallel lines have the same slope, and a line clearly has the same slope as itself, so therefore a line is parallel to itself. Notice that the proof of this theorem that a line is parallel to its dilation image is a coordinate proof that merely shows that the line and its image have the same slope.

Indeed, we can prove the special case that a line passing through the center of the dilation is invariant as follows: we already know that the line and its image are parallel by the above theorem. Since the center of the dilation is a fixed point, the image must also contain the center. If two parallel lines have a point in common, then they must be identical. So the line and its image -- two parallel lines having the center in common -- must be identical. Therefore the line is an invariant line. QED

On the other hand, calling the U of Chicago's definition of parallel an "inclusive definition" seems a bit strange. Having two pairs of parallel lines is a special case of having one pair of parallel lines, so it makes sense to consider the parallelogram a trapezoid. But in what world can having every point in common be a special case of having no points in common?

If we use the so-called "inclusive" definition of parallel, then choice (B) is correct, but choice (A) remains correct as line PQ is parallel to itself. The correctness of choice (B) doesn't automatically eliminate choice (A), as we've seen PARCC questions with more than one correct answer. But the intended answer is for choice (B) to be correct, not (A) (otherwise, we would have seen disclaimers such as "select all that apply").

So the U of Chicago definition fails to distinguish between a correct answer (B) and an incorrect answer (A) on a very specific PARCC problem. Therefore, any course that claims to be a Common Core Geometry course must reject the U of Chicago definition. I will be rewriting any worksheet that uses the U of Chicago definition of parallel in time for next year.

But now it remains -- how do we prove that a line that doesn't pass through the center of the dilation is parallel to its image using the most common definition of parallel?

In Algebra I classes, teachers often avoid merely saying "Parallel lines have the same slope." Instead, they say that parallel lines have the same slope and different y-intercepts. We can use this to convert the proof from Section 12-1 of the U of Chicago into one that uses the common definition of parallel.

Let S_k be a dilation centered at the origin with scale factor k, and (0, b) be the y-intercept of a line not passing through the center. Since this line doesn't pass through the origin, b can't be 0. So the image of the point (0, b) must be (0, kb), which is a different point. So the lines do not have the same y-intercept, but the U of Chicago text already proves that they have the slope. The line and its image have the same slope and different y-intercepts, therefore they are parallel. QED

There are a few things to point out about this proof. First of all, we assumed that k can't be 1 -- otherwise b and kb would be equal after all. But a dilation with scale factor 1 is just the identity -- and we could argue that every point is the center of a dilation with scale factor 1. So there's no such thing as a line that doesn't pass through the center.

Another point to make is that Section 12-1 assumes that the center of the dilation is the origin. Does this proof still work if the center is the origin? Recall that this is a coordinate proof. In coordinate proofs we typically place the origin and axes at convenient locations. Well, the most convenient location for the origin is the center of the rotation. We do the same with coordinate proofs involving, say, quadrilaterals. Let's place the coordinates of a rectangle at (0, 0), (a, 0), (0, b), and (a, b) -- that is, we placed the origin at one of the vertices and the axes along two of the sides. Then we can prove that both diagonals have length sqrt(a^2 + b^2) and thus are congruent. This coordinate proof doesn't work only for rectangles with a vertex at the origin -- it works for all rectangles. And so it is with our dilation proof.

The real problem, though, is that we can't use a coordinate proof to properties of dilations because in Common Core, we use dilations to prove the properties of coordinates! We discussed this problem all throughout our study of Chapter 12 back in January. Let's briefly recall that proof -- which ultimately goes back to Dr. Hung-Hsi Wu.

We actually begin with the fact that a line passing through the origin is invariant. This follows basically from the definition of dilation -- if O is the center, then any point A is mapped to a particular point on ray OA. Then the image of any point on line OA is on ray OA -- which is part of line OA. So any point on a line passing through O is mapped to point on the same line. The image of line OA is the line itself.

Now let's show that a line not passing through O must be parallel to its image. Our earlier proof from January was a direct proof, but we can do it more elegantly with an indirect proof. Assume that the line and its image are not parallel -- that is, that they have a point in common. That is, there exists a point A on the line such that its image A' is also on the line (that's where the line and its image intersect). So the original line can now be written as line AA'. Notice that O is also on line AA' -- once again, it follows from first case above that the points O, A, and A' are collinear. But this contradicts the assumption that O is not on the original line. Since the assumption that the line and its image intersect leads to a contradiction, we conclude that the line and its image are parallel. QED

Speaking of indirect proofs, what about Abel's indirect proof? Livio writes down that Abel was finally rewarded with a professorship in Berlin -- a few days after he had died of tuberculosis. He was only 26 years old when he died. Today there is a major prize -- the Abel prize -- that is offered every year in Abel's home country of Norway to a top mathematician, similar to the Nobel Peace Prize (and is sometimes considered to be the analog of a Nobel Prize in mathematics). Just as Holmboe had written just a few years earlier, who knows how great a mathematician Abel would have been if he had lived.

PARCC Practice EOY Exam Question 13
U of Chicago Correspondence: Section 12-2, Size Changes Without Coordinates
Key Theorem: Size Change Theorem

Under a size change:
(d) lines and their images 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: There are a few questions about lines being parallel to their images throughout the first three sections of Chapter 12 and in the SPUR section under Properties. The idea that a line passing through the center is unchanged doesn't appear in the U of Chicago at all.


Wednesday, April 29, 2015

PARCC Practice Test Question 12 (Day 157)

Chapter 3 of Mario Livio's The Equation That Couldn't Be Solved is, "Never Forget This in the Midst of Your Equations." At this point, Livio makes a sudden jump from geometry and symmetry to algebra and equations. But in describing this third chapter, I want to make a point about both algebra and geometry in the Common Core Standards, including which topics should be studied and which shouldn't be studied by high school students.

Livio begins by describing the history of polynomial equations. First, we have linear equations, which appear extensively in the ancient Egyptian Rhind Papyrus. Second, we have quadratic equations, and a form of the Quadratic Formula was used by the ancient Babylonians in order to relationship between the perimeter and area of a rectangular plot of land.

We all know the Quadratic Formula from Algebra I. Here I state both the general quadratic equation and the Quadratic Formula in order to demonstrate how I will render polynomials in ASCII in order to discuss them on the blog.

Quadratic Equation: ax^2 + bx + c = 0
Quadratic Formula: (-b +- sqrt(b^2 - 4ac))/2a

The Quadratic Formula is notoriously difficult to remember, and so a common way to get students to remember it is to make it into a song, such as "Pop Goes the Weasel":


I remember back when I was first learning about the Quadratic Formula, I often wondered whether there was such a thing as a Cubic Formula -- a formula that could solve any cubic equation of the form ax^3 + bx^2 + cx + d = 0. As it turns out, mathematicians in early Renaissance Italy also sought out a Cubic Formula.

Livio devotes much of Chapter 3 to the story of four Italian mathematicians who worked on solving the cubic equation. As we've seen before on this blog, the stories behind mathematicians can be highly intriguing. (Last week, I didn't get to sub in any math courses but I was in a eighth grade physical science class, and they were studying buoyancy, including the Principle of Archimedes. So of course I couldn't resist telling the story of how Archimedes discovered his Principle while taking a bath -- as I mentioned in my April Fool's post.)

Livio credits Scipione dal Ferro with discovering the solution of a simple case of the cubic equation, namely ax^3 + bx = c, in the year 1515. He did not publish his formula, but instead told it directly to his student Antonio Maria Fiore. Now Fiore kept the formula a secret as well, but he wanted to prove to everyone how smart he was, so he challenged the mathematician Niccolo Tartaglia -- which means Niccolo the Stammerer -- to a duel, a mathematical duel. The two of them would give each other 30 cubic equations to solve. Unfortunately for Fiore, not only had Tartaglia independently discovered dal Ferro's formula, but he had also found out how to generalize it to all cubics.

In between all of this, another mathematician, Ludovico Ferrari, not only generalized all of these results on cubic equations, but also discovered a Quartic Formula -- one that would solve equations of degree four, ax^4 + bx^3 + cx^2 + dx + e = 0. Livio tells the story much more dramatically than I have here on the blog, but by the end of the 16th century, the Quadratic, Cubic, and Quartic Formulas were all known. So this raises the question -- why do we expect high school students to learn only the Quadratic Formulas and not the Cubic or Quartic Formulas? After all, if the Babylonians developed the Quadratic Formula to solve perimeter and area problems, then why can't high school students learn the Cubic Formula to solve volume problems?

Well, let's take a look at this Cubic Formula. Livio states the dal Ferro formula in Appendix 5:

Cubic Equation: x^3 + px = q
Cubic Formula: cbrt(q/2 + sqrt(p^3/27 + q^2/4)) + cbrt(q^2 - sqrt(p^3/27 + q^2/4))

So we see that even for this simpler form of the cubic ax^3 + bx^2 + cx + d = 0, where a = 1, b = 0, c = p, and d = -q, this is already more complicated than the full Quadratic Formula. Since the Quadratic Formula involves square roots, it's not surprising that the Cubic Formula uses cube roots, abbreviated "cbrt" above.

Many Algebra II texts claim that imaginary numbers were created because mathematicians wanted to find solutions for quadratic equations such as x^2 + 1 = 0. But this is false -- mathematicians around fifteenth century Italy would have simply declared x^2 + 1 = 0 to have no solution. Complex numbers weren't created to solve quadratic equations -- they were created to solve cubic equations. The cubic equation mentioned earlier, x^3 - 15x = 4, requires imaginary numbers that end up canceling out to produce a real solution. As it turns out, if a cubic has just one real root, the dal Ferro formula gives the root without imaginary numbers appearing, but if it has three real roots, then the formula doesn't give any of the roots without requiring imaginary parts that eventually cancel out.

Finally, we notice that we have solved the full cubic ax^3 + bx^2 + cx + d = 0 yet. Well, we can easily eliminate the a coefficient simply by dividing by a, just as we do when completing the square. As it turns out, we can eliminate the term bx^2 by doing something like "completing the cube" -- we ultimately perform the transformation x' = x - b/3a. Notice that this transformation is, in fact, a horizontal translation. So we see that solving cubic equations is much more complicated than solving quadratic equations.

But then we could ask the opposite question. If the Cubic Formula is deemed too hard for high school students to learn, then why do we make them learn the Quadratic Formula? After all, quadratics can be solved using numerical methods on a computer instead as well. Any argument against teaching the Cubic Formula can be used to argue against teaching the Quadratic Formula too.

The status quo in high school math is that students should learn all of the units that help prepare them for calculus, and within each unit, learn as much as they reasonably can at that stage. If the Common Core Standards required students to learn the Cubic Formula, then opponents would point out that the Cubic Formula is too difficult and would prevent students from succeeding and making it to Calculus, but if the Common Core Standards didn't require students to learn the Quadratic Formula, then opponents would say that the Common Core is watering down Algebra I. Only the current path, where students learn the Quadratic but not the Cubic Formula, would be acceptable to most traditionalists.

And so I return to Geometry. Many people discuss the Core's emphasis on transformations and wonder whether students should have to learn translations, reflections, and rotations. Are the Common Core's transformations like the Quadratic Formula -- not completely necessary, but still within the range of student understanding? Or are they like the Cubic Formula -- completing frustrating to students and would be a huge barrier to advancement in math.

My answer is that only time will tell. We have to wait and see how well students are able to learn about transformations on the Common Core tests. If it can be shown that students don't learn about congruence using transformations well, but succeeded when SSS, SAS, and ASA were all considered postulates, then transformations should be dropped in favor of the old path.

With that, we look at today's PARCC Question. Question 12 on the PARCC Practice Test is on tangents to circles:

The figure shows two semicircles with centers P and M. The semicircles are tangent to each other at point B, and Ray DE is tangent to both semicircles at F and E.

(In the figure, Ray DA passes through P and M, with AB and BC diameters.)

If PB = BC = 6, what is ED?

(A) 6
(B) sqrt(48)
(C) 8
(D) sqrt(72)

Here's how we solve this problem. Section 13-5 of the U of Chicago text tells us that the tangent of a circle is perpendicular to the radius, so both DFP and DEM are right angles. Both Triangles DFP and DEM have angle D in common, so by AA Similarity (Section 12-9) these are similar triangles.

To find the scale factor, we notice that PF is a radius of circle P, so its length must be equal to that of PB, or 6. This side corresponds to ME, a radius of circle M. A diameter of circle M is BC, or 6, so the radius ME must be half of this, or 3. So the scale factor must be 1/2.

So DM must be half of DP -- and notice that MP must be the other half of DP. Now MP is the sum of the radii MB and PB, so it must be 3 + 6 = 9. So DP = 18 and DM = 9. Now DM is the hypotenuse of right triangle DEM, with known leg ME = 3, and we want to find the other leg ED. So all that remains is to use the Pythagorean Theorem (Section 8-7):

ME^2 + ED^2 = DM^2
3^2 + ED^2 = 9^2
9 + ED^2 = 81
ED^2 = 72
ED = sqrt(72), which is choice (D).

This is a very tricky problem. As we can see, we needed information from three Sections, 13-5, 12-9, and 8-7, in order to get an answer. Students might know the Pythagorean Theorem, but it's not obvious that Pythagoras must be used especially since no triangles at all -- much less right triangles -- appear in the drawing.

This is just like those similarity problems where one ends up needing to use the Quadratic Formula in order to find the desired length. Students may remember the Quadratic Formula from the previous year, but not realize that they need to use it in the problem.

Some people believe that students need to be challenged. They need to know how to use the Pythagorean Theorem and Quadratic Formula even in situations where it's not obvious that those are the tools to be used. But in questions like this one, it may end up frustrating students, who are likely to get this question wrong, especially with the correct answer being the last choice (D).

Here's an analogy -- suppose a third-grade teacher, being a strong traditionalist, wanted to test the students' knowledge of the multiplication tables from 0 * 0 to 9 * 9, but she doesn't want to grade 100 questions for every student. So instead, she decides to have the students multiply two 10-digit pandigital numbers -- that is, numbers containing all of the digits:

    4019782536
*  4092578361

Since the numbers are pandigital, students must know all of 0 * 0 to 9 * 9 to get the answer, and as the correct answer has 20 digits, it's easier for the teacher to grade than 100 problems. If all 20 digits are correct, the student gets a score of 100%, if one digit is wrong, the student gets 95%, and so on.

Obviously, this is a lousy test. It's very easy for the students to make a mistake, since there are so many digits to keep track of and so much carrying and adding. Notice that if students use the much maligned lattice method, the lattice will contain a complete multiplication table before the students have to add and carry -- where they will probably make a mistake!

If we decide that the best questions are content-based skill questions, then notice that a question that gives a right triangle or a quadratic equation and asks for students to use the Pythagorean Theorem or Quadratic Formula to solve it is indeed a skill-based question. Notice that there's a question in the Chapter 13 SPUR section similar to this PARCC question -- and it's a Properties question, not a Skills question. After all, it's a property of the tangent line that produces the right angle needed for the Pythagorean Theorem. Yesterday's question was more obviously a Properties question since it asks for the properties of rotations, but today's question is no less a Properties question.

To finish, let me point out that after Ferrari discovered the Quartic Formula so quickly after finding the Cubic Formula, the race was on to find the Quintic Formula -- a formula that would solve the fifth-degree equation ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0. But the solution to this quintic wasn't discovered as quickly as the Italian mathematicians had hoped. According to Livio, it wasn't until the end of the 18th century when Paolo Ruffini suspected that the reason that no one had discovered the Quintic Formula yet was because there is no Quintic Formula. In other words, the quintic equation is, as mentioned in the title of the book, The Equation That Couldn't Be Solved.


PARCC Practice EOY Exam Question 12
U of Chicago Correspondence: Section 13-5, Tangents to Circles and Spheres
Key Theorem: Radius-Tangent Theorem

A line is tangent to a circle if and only if it is perpendicular to a radius at the radius's endpoint on the circle.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.C.A.2
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: There is a question in this section that looks much like the PARCC problem, namely Question 7. This question is written below:
7. Line CA below is a common tangent to circles P and Q at points A and B. If CB = 5, AB = 10 and the radius of circle Q is 3, what is the radius of circle P?
Notice that this question does not require the Pythagorean Theorem to solve. A question that would be more analogous to the PARCC problem would be to ask for PQ, as this would require the Pythagorean Theorem. Also, PQ cannot be determined simply by adding the radii since, unlike the PARCC Question, the circles are not tangent to each other. In fact, we calculate that PA = 9, yet PQ = 8. The drawing is not to scale as point Q should actually be inside circle P. We see that there are more questions like this in the SPUR section, under Properties -- Questions 24 and 25 under Objective F.



Tuesday, April 28, 2015

PARCC Practice Test Question 11 (Day 156)

Today marks the midpoint of the fourth quarter -- that is, it marks the end of the seventh quaver -- according to the blog calendar.

Chapter 2 of Mario Livio's The Equation That Couldn't Be Solved, "eyE s'dniM eht ni yrtemmyS" -- spell it backwards to decipher the title -- continues by discussing symmetry. In this chapter Livio describes the object known as a group.

A group is a simpler object than a ring. (Don't forget that Common Core does not -- repeat, not -- expect high school students to learn group or ring theory.) Livio defines a group as a set with an operation that satisfies four properties -- closure, associativity, identity, and inverse. (A ring, meanwhile, also satisfies commutativity, plus it has has a second operation -- called "multiplication" while the first is called "addition" -- and the multiplication also closed and associative.)

The important thing Livio emphasizes is that the set of all symmetries on a figure is a group. For example, the set of all symmetries on a human being (or an isosceles trapezoid) is a group with two elements, the identity and a reflection about a vertical plane. Even though students don't have to know what a symmetry group is, we note the following Common Core Standard:

CCSS.MATH.CONTENT.HSG.CO.A.3
Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.

So students do have to know what transformations maps a figure to itself -- that is, the symmetry transformations -- even if they don't have to think of it as a group. We already saw about two weeks ago what such a question will look like on the PARCC. That day, we unwittingly determined the set of symmetries -- that is, the symmetry groups -- of all of the named quadrilaterals:

Isosceles Trapezoid: {identity, reflection over perpendicular bisector}
Kite: {identity, reflection over angle bisector}
Parallelogram: {identity, rotation 180 degrees}
Rectangle: {identity, reflection over perp. bis, reflection over other perp. bis., rotation 180 degrees}
Rhombus: {identity, reflection over angle bis., reflection over other angle bis., rotation 180 degrees}
Square: {identity, reflect 1st perp. bis., reflect 2nd perp. bis., reflect 1st angle bis., reflect 2nd angle bis., rotate 90 degrees, rotate 180 degrees, rotate 270 degrees}
Trapezoid: {identity}

Indeed, when the British mathematician John Conway was defending the inclusive definition of trapezoid, he mentioned symmetry groups in his defense:

http://mathforum.org/kb/message.jspa?messageID=1074038

On the other hand, the question of just which symmetry groups
N-gons can have actually interested me even as a professional
mathematician, although for any particular N the answer can be
found by a schoolchild.

He then proceeds to name the groups for each type of quadrilateral. Once again, we don't expect high school students to know the names of groups, but Conway assigns the group D2e to the isosceles trapezoid, D2v to the kite, and D4v to the rhombus. In each case, the number (2 or 4) tells how many members the group has. The letter D means "dihedral," which means "two sides" -- in other words, reflection symmetry. The "e" stands for edges, since isosceles trapezoids has opposite edges congruent, while "v" stands for vertices, since kites has congruent sides that meet at a vertex.

There are a few other groups that are of interest here. Given any point on the plane, the set of all rotations about that point (of any magnitude) and reflections in any line passing through that point is a group -- the symmetry group of any circle centered at that point.

The set of all isometries on a plane is a group -- the symmetry group of the entire plane. If we only consider the translations, then this set is also a group, since it is closed (the composite of two translations is another translation), associative, has an identity, and has inverses.

For any figure, the set of isometries that preserve orientation is also a group. For the entire plane, this is the set of all translations and rotations. For a square, this is a set with four elements -- rotations by 0 (identity), 90, 180, and 270 degrees. For a kite, this is a set with one element (identity).

Even though dilations aren't isometries, they are elements of groups as well. The set of all similarity transformations on the plane is a group. The set of all similarity transformations that preserve orientation is a group. The set of just the dilations is not a group, because it's not closed -- if we dilate a figure with scale factor 2 and then dilate it with scale factor 1/2 with a different center, then the composite is not a dilation, but a translation. So the set of all translations and dilations is a group. (In fact, some sources use the word "dilation" to mean "dilation or translation," and use another word, "homothety," to mean dilation that isn't a translation.)

Livio writes about structural information theory -- the idea that if a figure is symmetrical, less information is needed to describe it. He writes that because a rectangle is symmetrical, we can denote a rectangle of dimensions a and b by either:

a 90 b 90 a 90 b 90
2 * (a 90 b 90)

This should immediately remind us of the LOGO programming language:

to rectangle :a :b
forward :a left 90
forward :b left 90
forward :a left 90
forward :a left 90
end

to rectangle :a :b
repeat 2 [forward :a left 90 forward :b left 90]
end

The last thing I want to write about the Chapter 2 in Livio is the concept of invariant. He writes, "Symmetry plays an important role in the recognition of similarity because it represents a true invariant -- an immunity to change." The isometries and similarity transformations preserve key properties of the figure, and we call these "invariants."

This is a good place to segue into the next PARCC problem. Question 11 of the PARCC Practice Test is about the invariants of a certain transformation -- in this case, a rotation:

Triangle ABC is shown in the xy-coordinate plane.

(The figure shows the vertices A(-5, 2), B(-2, 5), C(3, 4).)

The triangle will be rotated 180 degrees around the point (3, 4) to create Triangle A'B'C'. Which characteristics of Triangle A'B'C' will be the same for the corresponding characteristic of ABC?

Select all that apply.

(A) the coordinates of A'
(B) the coordinates of B'
(C) the perimeter of Triangle A'B'C'
(D) the area of Triangle A'B'C'
(E) the measure of Angle B'
(F) the length of segment A'B'

To answer this question, we must consult three sections of the U of Chicago text. The first is Section 6-3, since this is on rotations. We learn that the only fixed point (i.e., invariant point) of a rotation is its center. Notice that the center of the rotation is (3, 4) -- which is exactly point C. Therefore the coordinates of C' will be the same as those of C -- and had "the coordinates of C'" been one of the choices, we'd be able to select it. But neither A nor B can be fixed points as C is the only one -- so we cannot mark either of the first two choices.

The next place that we check out will be Section 6-5, on congruent figures. We go here since Triangles ABC and A'B'C' must be congruent, since the latter is the rotation image of the former. We see that every isometry -- including a rotation -- preserves Angle measure, Betweenness, Collinearity, and Distance -- A-B-C-D. So we can mark choice (E), as angle measure is invariant, and we can mark choice (F), as distance is invariant. Choice (C) is also correct, since a perimeter is by definition just the sum of distances.

So we are left with choice (D), area. We must go to Section 8-3, Fundamental Properties of Area:

Area Postulate:
c. Congruence Property. Congruent figures have the same area.

So this implies that area is also invariant under any isometry. So this gives us four correct answers -- and these are (C), (D), (E), and (F).

I find this to be a much better question than the dilation question from last week. Today's question would be much worse if students had to find the coordinates of A' and B'. In general, rotations of 180 degrees (or 90 or 270 -- that's right, the symmetries of a square) aren't terrible to determine their coordinates, but it's much easier when the center is the origin. If we reflect the point (x, y) 180 degrees about the origin, we obtain the point (-x, -y) -- so in some ways, a rotation of 180 degrees is just a dilation of magnitude -1.

One way to find the coordinates of a point rotated 180 degrees about a point other than the origin is to note that the center C of the rotation must be the midpoint of segments AA' and BB'. We've seen questions before where we were given the midpoint and one endpoint of a segment and we had to find the other endpoint. It's because of this that the rotation of 180 degrees is also known by a special name -- point inversion (and if a point inversion maps a figure to itself, as it does for a parallelogram, then the figure is said to have point symmetry).

Point inversions -- just like translations and dilations -- also have the property of mapping lines to parallel lines. (Indeed, the set of all translations and point inversions is a group, and the set of all translations, point inversions, dilations, and inversion-dilations is yet another group.) I pointed out that the U of Chicago develops transformational geometry by using reflections to define rotations, but Dr. Hung-Hsi Wu uses rotations to prove that reflections are well-defined. One can combine these ideas without circularity, because full rotations aren't needed to complete the Wu proofs that reflections are well-defined. Only point inversions are needed.

I've never seen point inversions mentioned in any high school text -- although point symmetry appears in some texts (but it's point inversions that justify the name "point symmetry"). I would agree that performing a point inversion -- like the coordinates of A' and B' in today's question -- should not appear on any Common Core test unless the center is the origin. Fortunately, they don't appear in today's question. That means that we've now had three straight questions in the calculator section that don't require any calculations!

To summarize, these are the only transformations T which I think it's fair for the PARCC to give the coordinates (x, y) and ask the student to give the image T(x, y)

-- any translation
-- a reflection (or possibly a glide reflection) whose mirror is the x-axis, y-axis, or possibly the lines y = x or y = -x
-- a rotation centered at the origin of magnitude 180, or possibly 90 or 270 degrees
-- any dilation centered at the origin

To repeat, my permissible rotations and reflections are the symmetries of a square centered at the origin -- the easier ones are the symmetries of a rectangle centered at the origin -- with sides parallel to the axes.

PARCC Practice EOY Exam Question 11
U of Chicago Correspondence: Section 6-5, Congruent Figures
Key Theorem: A-B-C-D Theorem

Every isometry preserves Angle measure, Betweenness, Collinearity (lines) and Distance (lengths of segments).

Common Core Standard:
CCSS.MATH.CONTENT.HSG.CO.B.6
Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent.

Commentary: Question 14 in this section is an interesting one to ask here. Here it is in full:
14. Multiple choice. If r(Tri. ABC) = Tri. DEF, then according to the definition of congruence:
(a) AB is congruent to DE
(b) m Angle C = m Angle F
(c) Tri. ABC is congruent to Tri. DEF
Notice that all three of these statements are true, but the intended answer is (c), since this is the only statement that follows from the definition of congruence (i.e., two figures are congruent if some isometry maps one to the other) rather than the A-B-C-D Theorem. If we drop the phrase "according to...," then all three answers will be correct -- so it will be just like the PARCC question where there is more than one correct answer.




Monday, April 27, 2015

PARCC Practice Test Question 10 (Day 155)

Over the weekend, I finally got those Mario Livio books that I was discussing on the blog a couple of weeks ago. Though I didn't get the book that was mentioned in the NOVA episode -- Is God a Mathematician? -- I did buy the one that I wanted to mention here on the blog -- The Equation That Couldn't Be Solved: How Mathematical Genius Discovered the Language of Symmetry. I also got checked out from the library another Livio book, an audiobook this time -- Brilliant Blunders, From Darwin to Einstein: Colossal Mistakes by Great Scientists That Changed Our Understanding of Life and the Universe.

In Brilliant Blunders, I listened to the first two chapters and a short part of the third. Chapter 1 is introduction, and Chapter 2 is the first scientist Livio describes in his book, British naturalist Charles Darwin, and his theory of evolution. Chapter 3, based on the short part that I heard, will reveal that Darwin's error is that, while he knew that natural selection occurred, he didn't know why. That would have to wait until the discovery of genetics, by the Czech scientist Gregor Mendel.

So far, I have read Chapter 1 of The Equation That Couldn't Be Solved. This first chapter is simply titled "Symmetry." In this chapter, Livio describes four types of symmetry -- mirror-reflection symmetry (his example is the bilateral symmetry of animals), rotational symmetry (snowflakes), translational symmetry (19th century British artist William Morris and 18th century Austrian musician Wolfgang Amadeus Mozart) and glide-reflection symmetry (more art plus the motion of humans walking and snakes slithering). And these are, of course, the four isometries that appear on the Common Core tests. Livio even mentions a fifth symmetry that occurs in 3D -- screw symmetry (leaves of certain flowers). I mentioned the 3D screw isometry back during spring break.

Livio also mentions the 20th century American mathematician George David Birkhoff. I mentioned Birkhoff earlier on this blog as the originator of the Ruler Postulate, the first postulate in many geometry texts (and part of the Point-Line-Postulate in the U of Chicago). Here Livio credits Birkhoff with creating a formula, M = O/C, where M is the aesthetic measure of a work of art, O is the order, or symmetry, present in the work, and C is the complexity of the work.

Question 10 from the PARCC Practice test is on trigonometry. Fortunately, this will be one of the simpler trig questions if students know what they are doing.

In right triangle ABC (and with the right angle at A), Angles B and C are not equal. Let sin B = r and cos B = s. What is sin C - cos C?

(A) r + s
(B) r - s
(C) s - r
(D) r/s

We think back to Section 14-4 of the U of Chicago text, since this is on sines and cosines. When we discussed 14-4 on the blog, I mentioned how I often tell my students where the name "cosine" actually comes from. Let me repeat the explanation given in the U of Chicago text -- this refers to Example 1 from the text, where the right angle is C and A is an acute angle. Since in the PARCC question the right angle is A instead, I will change the letter A to C in order to avoid confusion:

...notice that sin C = cos B and sin B = cos C. This is because the leg opposite either angle is the leg adjacent to the other. Angles C and B are also complementary. This is the origin of the term "cosine"; cosine is short for complement's sine.
[emphasis U of Chicago's]

(As an aside, often Precalculus students wonder why, once they learn about the cotangent, we do have cot B = 1/tan B, but not cos B = 1/sin B. This is because, if C and B are complementary, we do have tan C = 1/tan B, but not sin C = 1/sin B. The "co" in cosine, cotangent, and cosecant stands for "complement," not "reciprocal.")

Once we remember this, the question becomes trivial. So sin C = cos B = s, and cos C = sin B = r, and so sin C - cos C = s - r, which is choice (C). The trick is that if the students think that the question is asking for sine and cosine of B instead of C, they might choose (B) instead of (C) -- an easy mistake to make since choice (B) precedes choice (C).

So far, of the three questions we've seen from the calculator section of the PARCC Practice Test, two of the questions don't require any calculations at all! Thursday's question involved two angles that add up to 90 degrees, and we had to calculate one of the measures. Friday's question involved two angles that add up to 90 degrees, but we only had to figure out which angles those were, not any of their individual measures. Today's question involves two angles that add up to 90 degrees, but we only have to find the relationship between their sines and cosines, not calculate those values.

One more thing I want to point out is that extra mention in the problem that Angles B and C do not have equal measure. I suspect the reason for this mention is that if Angle B = C, then both (B) and (C) would be correct answers as r - s = s - r = 0. So that line is just to avoid the loophole that someone may choose (B), then claim that we can't rule it out in case that B = C.

I haven't had the opportunity yet to mention this relationship between sines and cosines to the geometry student I tutor. We had to jump into the Law of Sines so fast that I didn't have time to talk about cosine as the complement's sine. Hopefully I'll be able to mention it by the end of this week.

PARCC Practice EOY Exam Question 10
U of Chicago Correspondence: Section 14-4, The Sine and Cosine Ratios
Key Theorem: Definition of sine and cosine

In right triangle ABC with right angle C,
the sine of Angle A, written sin A, is leg opposite Angle A / hypotenuse
the cosine of Angle A, written cos A, is leg adjacent to A / hypotenuse

Common Core Standard:
CCSS.MATH.CONTENT.HSG.SRT.C.7
Explain and use the relationship between the sine and cosine of complementary angles.

Commentary: That "cosine" means "complement's sine" is mentioned in U of Chicago, but it is not emphasized there as strongly as in the Common Core Standards. We see that Question 3 asks for cos F and sin G in triangle FGH, and Question 6 asks for sin 48 and cos 42. And technically speaking, Question 5 asks for sin 45 and cos 45 -- and these are equal because 45 degrees is its own complement. In the SPUR section, Questions 13-14 ask for sin A and cos B, as part of Objective C in the Skills section.




Friday, April 24, 2015

PARCC Practice Test Question 9 (Day 154)

Question 9 of the PARCC Practice Test is on perpendicular lines -- specifically, the conditions which make two lines perpendicular:

The figure shows lines r, n, and p intersecting to form angles numbered 1, 2, 3, 4, 5, and 6. All three lines lie in the same plane.

(In the figure, the three lines are concurrent, and they are lettered r, n, and p in clockwise order. The angle numbered 1 is formed between lines r and n, and the other angles are numbered clockwise.)

Based on the figure, which of the individual statements would provide enough information to conclude that line r is perpendicular to line p?

Select all that apply.

(A) Angle 2 = 90
(B) Angle 6 = 90
(C) Angle 3 = Angle 6
(D) Angle 1 + Angle 6 = 90
(E) Angle 3 + Angle 4 = 90
(F) Angle 4 + Angle 5 = 90

This question is straightforward. If we draw in boxes to denote right angles between r and p, we see that the following combinations of angles would have to be 90 degrees:

Angle 1 + Angle 2
Angle 3
Angle 4 + Angle 5 (choice F)
Angle 6 (choice B)

We eliminate choices (A), (D), and (E) as these would make p perpendicular to n rather than r. And we can eliminate choice (C) as Angles 3 and 6 are vertical angles, so they would have equal measure regardless of whether any of the lines were perpendicular. The correct answers are (B) and (F).

The question itself is not difficult. If we had to show which section of the U of Chicago text the question corresponds to, then I'd have to say Section 3-5, since this is on Perpendicular Lines. But the problem the students will have most likely wouldn't be that they don't know which angle is 90, but that they won't select more than one answer. Choice (B) is correct, since it correct states that if Angle 6 is a right angle then lines r and p are perpendicular. So then students will just mark (B) and not check the rest of the answers to see whether a second answer is correct.

Another possible issue could be confusing a statement with its converse. The question asks students to tell which statements imply that r is perpendicular to p -- not which statements the perpendicularity of r and p imply. If it were the latter, (C) would be correct, since if r were perpendicular to p, then Angles 3 and 6 would indeed be equal as both would be 90 degrees. The problem is that we can have Angles 3 and 6 equal without r and p being perpendicular. But I expect the far more common mistake for students would be to omit (F) as a correct answer, rather than include (C).

I'm in the mood to post some Square One TV videos. Let's begin with a song that's definitely relevant to today's PARCC, "Perpendicular Lines":



Now today is an activity day. I've decided that I want this activity to focus not on perpendicular lines, but on the idea that sometimes more than one answer can be correct.

I rarely post entire episodes of Square One TV, but I couldn't find this as an isolated clip -- but fortunately, what I want is at the beginning of the show, right after the theme song. This is a game show called "Square One Challenge" -- a parody of To Tell the Truth. The difference here is that sometimes they're both telling the truth -- and other times they're both bluffing:



My activity page is based on this game. Notice even before clicking to start the video that the first question is based on folding paper and reflections, a Common Core transformation.

PARCC Practice EOY Exam Question 9
U of Chicago Correspondence: Section 3-5, Perpendicular Lines
Key Theorem: Definition of perpendicular

Two segments, rays, or lines are perpendicular if and only if the lines containing them form a 90-degree angle.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.CO.A.1
Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, distance along a line, and distance around a circular arc.

Commentary: Because perpendicular lines is such a basic concept, I argue that this question is more about logic and knowing how to choose more than one correct answer than about perpendicular lines per se. There are no logic questions like this one anywhere in the U of Chicago text.



Thursday, April 23, 2015

PARCC Practice Test Question 8 (Day 153)

Last night I tutored my geometry student. Section 8-6 of the Glencoe text is on the Law of Sines. This topic is not mentioned in the U of Chicago text, and is traditionally a Precalculus topic, yet it's included in the Glencoe text. Of course, we can't help but notice that the Law of Sines is mentioned in the Common Core Standards:

CCSS.MATH.CONTENT.HSG.SRT.D.11
(+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces).

Once again, we have the problem that the Common Core Standards aren't divided into courses. So we don't know whether the Law of Sines is intended to be a Geometry topic or for a later course except by seeing whether there is such a question on the PARCC Geometry Tests. We haven't seen a Law of Sines question yet, but then all of the PARCC questions we've done are non-calculator questions. To remind everyone what the Law of Sines is:

Law of Sines:
In any triangle ABC, the equation a/sin A = b/sin B = c/sin C holds.

My student and I appear to have found an error in one of the Glencoe questions. This is the Glencoe text, not the PARCC exam, but this sounds like the sort of question that could appear on PARCC:

5. SURVEYING. To find the distance between two points A and B that are on opposite sides of a river, a surveyor measures the distance to point C on the same side of the river as point A. The distance from A to C is 240 feet. He then measures the angle across from A to B as 62 degrees and measures the angle across from C to B as 55 degrees. Find the distance from A to B.

Here is the correct answer that my student found:

240/sin 63 = x/sin 55
x = 240 sin 55/sin 63
x = 220.6 ft.

But, as this is an odd-numbered question, here is the answer as given by Glencoe:

240/sin 63 = x/sin 62
x = 240 sin 62/sin 63
x = 237.8 ft.

There are two ways to explain how Glencoe obtained this erroneous answer. The text may have given the length of CB rather than AB. Or the text may have given the length of AB after all, but intended the phrase "angle across from A to B" to refer to angle ACB instead of the natural reading BAC,

One fear Common Core opponents have about the PARCC exam is that there may be poorly worded questions that are ambiguous. Of course, so far the only poorly worded questions I've found have been in the Glencoe text, not on PARCC. But then again, if there is a badly worded question on the PARCC Practice Test we'll never know, because the practice test doesn't provide any answers.

There are a few SSA questions in Glencoe as well as the easier AAS and ASA. Fortunately, all of the SSA questions have the given angle opposite the longer of the two given sides. This is what the U of Chicago calls the SsA case (in the congruence chapter, not the trig chapter). The SsA case never leads to the ambiguous case, so we never have to worry about there being two triangles.

Dr. Franklin Mason also includes a Law of Sines in his text -- in his text that section is numbered 8.7, just one section off from the Glencoe numbering. This implies the possibility that we may be seeing the Law of Sines on the PARCC. Dr. M avoids SSA and therefore the ambiguous case -- yes, the ambiguous case can definitely wait for Precalculus.

But the next question for us to cover is not on the Law of Sines. Question 8 of the PARCC Practice Test is on inscribed angles of a circle. This is the first question of the calculator section of the exam:

The figure shows Triangle ABC inscribed in circle D. If Angle CBD = 44 degrees, find Angle BAC.

Today's question is nowhere near as bad as yesterday's. Since BAC is an inscribed angle, we expect the Inscribed Angle Theorem, included in the U of Chicago's Section 15-3, to be used here. It's because of questions like these that I included 15-3 on the blog -- the only section of the text's final chapter so included.

Notice that CBD is, strictly speaking, an inscribed angle as well. But D is the center of the circle, not a point on the circle, and the point diametrically opposite B on the circle isn't given (so in particular, it's not A). So yes, the arc subtended by CBD would be 88 degrees, but that arc has nothing to do with this problem.

Instead. we must look at Triangle BCD. This looks just like the first part of the proof of the Inscribed Angle Theorem -- the key fact is that Triangle BCD is isosceles, because both DB and DC are radii of the same circle. Then by the Isosceles Triangle Theorem, Angles CBD and BCD are congruent, and so the measure of BCD must also be 44 degrees. Since the measures of the angles of a triangle add up to 180, BDC must measure 92 degrees. This is the central angle that subtends arc BC, which must also measure 92 degrees. Finally, the inscribed angle that intercepts arc BC, angle BAC, must measure half of this, or 46 degrees.

Some people may notice that the given angle is 44 degrees and the correct answer is 46 degrees -- and these two angles add up to 90 degrees. So some people might wonder whether we can prove that Angles CBD and BAC must be complementary. As it turns out, this is the case. Here is the proof of this statement:

Given: Triangle ABC inscribed in circle D
Prove: Angles CBD, BAC are complementary

Statements                              Reasons
1. ABC inscribed in circle D     1. Given
2. DB = DC                            2. Definition of circle
3. Angles CBD = BCD            3. Isosceles Triangle Theorem
4. CBD + BCD + BDC = 180   4. Triangle Angle-Sum Theorem
5. CBD + CBD + BDC = 180   5. Substitution (4 into 5)
6. BDC = 180 - 2m CBD         6. Subtraction Property of Equality
7. Arc BC = Angle BDC          7. Definition of arc measure
8. Angle BAC = 1/2 Arc BC     8. Inscribed Angle Theorem
9. BAC = 1/2 (180 - 2m CBD)  9. Substitution (6 and 7 into 8)
10. BAC = 90 - CBD               10. Distributive Property
11. CBD + BAC = 90              11. Addition Property of Equality
12. CBD, BAC complementary 12. Definition of complementary

This means that all a student has to do is subtract 90 - 44 and enter the number 46 into the test. But it's not at all obvious that angles CBD and BAC should be complementary. The calculation that we performed earlier in this post will be typical for most students.

Unlike most of the problems that we've covered so far this week, traditionalists should have no problem with today's question. Calculating measures of inscribed angles from measures of intercepted arcs is definitely a content-based skill.

PARCC Practice EOY Exam Question 8
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:
CCSS.MATH.CONTENT.HSG.C.A.2
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: 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.



Wednesday, April 22, 2015

PARCC Practice Test Question 7 (Day 152)

Question 7 of the PARCC Practice Test is on dilations on a coordinate plane. It is the last question in the non-calculator section:

In the xy-coordinate plane, Triangle ABC has vertices at A(1, -2), B(1, 0.5), and C(2, 1) and Triangle DEF has vertices at D(4, -3), E(4, 2), and F(6, 3).

The triangles are similar because Triangle DEF is the image of Triangle ABC under a dilation. What is the center and scale factor of this dilation?

Select the two true statements.

(A) The center of the dilation is at (-2, -1).
(B) The center of the dilation is at (-1, -2).
(C) The center of the dilation is at (0, 0).
(D) The scale factor is 1/2.
(E) The scale factor is 2.
(F) The scale factor is 4.

Since we are dealing with dilations on a coordinate plane, Section 12-1 of the U of Chicago text, "Size Changes on a Coordinate Plane," seems appropriate here. So let's attempt to find the scale factor based on what is taught in Section 12-1. We see that point B has coordinates (1, 0.5), and we notice that its image E has coordinates (4, 2). Since the coordinates of E are exactly four times those of B, we are tempted to say that the scale factor is 4 -- which happens to be one of the choices.

But then point C has coordinates (2, 1), and its image F has coordinates (6, 3), so the coordinates of the image are only thrice those of the preimage. Not only does the scale factor 3 not match the earlier calculated scale factor 4, but 3 isn't even one of the choices. And moreover, we can't multiply the coordinates of A, (1, -2), by anything to obtain the coordinates of its image D, (4, -3) So what gives?

The problem is that the formula given in Section 12-1 works only when the center of the dilation is the origin, that is (0, 0). As it turns out, we just proved that the center of the dilation must not be the origin as soon as we obtained different scale factors for different points! The next section, 12-2, discusses how to find the center and scale factor of a dilation given a preimage and image. But none of these questions appear on the coordinate plane.

In short, the center of every dilation that appears in the U of Chicago text has either coordinates (0, 0) or no coordinates at all. But this PARCC question involves a dilation whose center has coordinates other than (0, 0). So this problem has no analog in the U of Chicago text.

Thus once again, we have another PARCC problem that I don't necessarily like. Many traditionalists don't like the Common Core transformations at all. I have no problem with dilations, but notice that this is another thinking backwards problem. Instead of giving the center, scale factor, and preimage and asking students to produce the image, this question provides the preimage and image and asks for the center and scale factor. Yesterday, I wrote that thinking backwards questions should be given only when the concept being tested is easy. I wouldn't mind this question had the center been at the origin, but I disagree with the presence on the PARCC of a thinking backwards question involving a dilation centered at a point other than the origin.

So how exactly would we answer this problem and complete this PARCC question? We look at Section 12-2 and the directions for Questions 14 through 16:

In 14-16, trace each figure. Use a ruler to determine the center and the scale factor k for each size transformation represented. (The image is blue.)

But just "use a ruler" is no help to solve the PARCC question. We're given no graph, but just a list of the coordinates. So we back up to the examples and find out how to draw a dilation image:

Step 1. Measure OA.
Step 2. On Ray OA, locate A' so that OA' = 2.5 * OA. That is, A' is 2.5 times [in this problem, the scale factor is 2.5 -- dw] as far from the center as A. Point A' is the size change [dilation] image of A.

In this example, the center O, scale factor 2.5, and preimage A are given. In our problem, we are given the preimage A and image D and wish to find the center O and the scale factor. We can convert this algorithm into one to find the center O as follows:

Step 1. Choose any point A and its image A' (or D), and draw line AA'.
Step 2. Choose another point B and its image B' (or E), and draw line BB'.
Step 3. Then O must be the point where lines AA' and BB' intersect.

At this point, one may ask, how do we know that lines AA' and BB' intersect? Well, because of the definition of dilation, we can prove that O must lie on both lines. Therefore, lines AA' and BB' are provably not parallel, unless they are identical. So all we have to do is choose B to be any point on the preimage that's not on line AA'. Since in general we're transforming figures like triangles and rectangles rather than lines, it should be easy to find a point B that's not on line AA'. For the best results, B should be as far away from line AA' as possible. (If we're doing Question 14 from the U of Chicago's 12-2, and we choose A to be upper-left corner of the rectangle, then choosing B to be the lower-right corner is a bad idea since this point would be too close to line AA', even if it's not exactly on line AA'. Choosing B to be the upper-right corner is a much better strategy.)

Step 4. Measure OA' and OA, and divide to find the scale factor.

And that's it! If we are given pictures as in Questions 14 through 16 of the U of Chicago, then we can use a ruler, as directed, to perform Steps 1 through 3 above. But to answer PARCC Question 7, we can perform steps 1 through 3 algebraically.

Step 1. Point A has coordinates (1, -2) and its image D has coordinates (4, -3). This is a classic Algebra I problem, to find the equation of a line given two points. We must first find the slope:

slope of line AD = (-3 + 2)/(4 - 1) = -1/3

Then we use the Point-Slope Formula for the equation of a line:

y + 2 = -1/3(x - 1)
y = (-1/3)x - 5/3

Step 2. Point B has coordinates (1, 0.5) and point E has coordinates (4, 2). So we have:

slope of line BE = (2 - 0.5)/(4 - 1) = 1.5/3 = 1/2
y - 1/2 = 1/2(x - 1)
y = (1/2)x

We could have chosen points A and C instead of A and B, if we want to avoid the decimal 0.5. But we cannot choose points B and C, because point C lies on line BE.

Step 3. To find out where these lines intersect, we solve the system of equations:

y = (-1/3)x - 5/3
y = (1/2)x

(1/2)x = (-1/3)x - 5/3

3x = -2x - 10
5x = -10
x = -2

y = (1/2)(-2)
y = -1

And so we conclude that (-2, -1) is the center of the dilation, which is choice (A). All that remains is to find the scale factor of the dilation.

Step 4. The scale factor is equal to OD/OA. To find this distance, we use the Distance Formula:

OA = sqrt((-2 - 1)^2 + (-1 + 2)^2) = sqrt(10)
OD = sqrt((-2 - 4)^2 + (-1 + 3)^2) = sqrt(40) = 2sqrt(10)
OD/OA = 2sqrt(10)/sqrt(10) = 2

So 2 is the scale factor of the dilation, which is choice (E). Notice that one advantage to using dilations for this problem is that there's a definite preimage ABC and image DEF. So we know that the scale factor must be OD/OA, not OA/OD. This is important since we see that OA/OD = 1/2 is one of the wrong choices. If we were simply given ABC ~ DEF, then there would be no way to tell whether (D) or (E) is the correct answer. This was the problem that my geometry student had when completing the similarity questions in Chapter 7 of the Glencoe text.

This method gives the correct answer, but it is definitely a lot of work. Not only that, but it involves too much Algebra I. Many Geometry students won't remember how to solve all of these problems -- and even if they do, they are not likely to see this problem on the PARCC and reason, "To solve this problem, I have to use Point-Slope to write two equations, solve the system of equations to find the center, and use the Distance Formula and division to find the scale factor."

Fortunately, there are many ways to simplify some of the steps. For starters, to find the scale factor in Step 4, we can divide DE/AB rather than OD/OA. This follows from a theorem in Section 12-3:

Size Change Distance Theorem:
Under a size change [dilation] with magnitude [scale factor] k > 0, the distance between any two image points is k times the distance between their preimages.

And it's easier to find AB and DE than OA or OD, because AB and DE are vertical. When solving any problem that involves finding distance and there is a choice of which distance to find, we should always choose a distance that is vertical or horizontal -- that is, segments whose endpoints have matching x- or y-coordinates -- whenever possible. So we have:

A(1, -2), B(1, 0.5), AB = 0.5 - (-2) = 2.5
D(4, -3), E(4, 2), DE = 2 - (-3) = 5
DE/AB = 5/2.5 = 2

This avoids the Distance Formula and the need for square roots -- though unfortunately, the division 5/2.5 may be tricky since this is the non-calculator section.

But even if this is easier, we still have Steps 1 through 3, with the Point-Slope Formula to remember and a system of equations to solve. As it turns out, there's a way to avoid this as well.

We begin by finding the scale factor first. Our DE/AB trick shows is that it's possible to find the scale factor to be 2 without having to know what point O is. Then we have to find what the point O must be in order to have OD = 2 * OA. To do this, we use vectors. If we think of all the points as vectors, so point A is the vector a, D is the vector d, and O is the unknown vector o, then we are trying to solve:

o - d = 2(o - a)
o - d = 2o - 2a
2a - d = o

Since o equals 2a - d, we only need to substitute:

o = 2a - d
o = 2(1, -2) - (4, -3)
o = (2, -4) + (-4, 3)
o = (-2, -1)

I mentioned something similar at the end of Monday's lesson, when we were trying to divide a segment into a given ratio. For today's problem, using vectors turns out to be easier, unlike Monday.

But even this method may still be confusing to students. Notice that for this particular problem, there's still another way to find the center O, using a process of elimination. Recall that we began today's post by noting that the center of dilation can't be the origin, since the scale factors for BE and CF aren't equal. So we've already eliminated choice (C). Then we look at the coordinates B(1, 0.5) and E(4, 2), and note that in each case, the y-coordinate is exactly half of the x-coordinate. So the line passing through these points has equation y = (1/2)x -- and we know that O must lie on this line (as it must lie on AD and CF as well). Choice (B) doesn't satisfy y = (1/2)x, and choice (C) has already been eliminated, leaving choice (A) as the correct center. But this trick only works for certain problems of this type -- it doesn't work in general.

Once again, I don't like this PARCC question one bit. There are better ways to determine whether students really understand what dilations are. We could keep the thinking backwards format, but make sure the center of dilation is the origin. Then to find the center, our simple division at the beginning of this post will work. The other way to do it is to have the center be a point other than the origin, but give the scale factor and preimage and ask for the image. In this case, vectors will probably be the best way to do it. It might even be good to have a scale factor less than 1 in this case -- then this problem could be done almost the same way as PARCC Question 5 from Monday's post, so students can see how Question 5 and this question are related. This is how I designed today's worksheet -- I included a thinking backwards dilation problem centered at the origin, a straightforward dilation problem centered elsewhere, and finally a question at the PARCC level.

Notice that Dr. Frankin Mason not only dropped dilations completely, but the coordinate plane does not appear any more in his chapter on similarity (which is his Chapter 7, just like Glencoe). We know that anything Dr. M drops from his text is because he lacked the time to teach it and decided that it wasn't as important. That makes this PARCC question look even worse.

PARCC Practice EOY Exam Question 7
U of Chicago Correspondence: Section 12-2, Size Changes Without Coordinates
Key Theorem: Definition of Size Change (Dilation)

Let O be a point and k be a positive real number. For any point P, let S(P) = P' be the point on Ray OP with OP' = k * OP. Then S is the size change [dilation] with center O and magnitude [scale factor] k.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.SRT.A.1
Verify experimentally the properties of dilations given by a center and a scale factor.

Commentary: Even though this problem includes coordinates, we state 12-2 to be the corresponding section rather than 12-1 since all of the dilations from 12-1 are centered at the origin, while today's dilation is not centered at the origin. There are no dilations in the U of Chicago text where the center is given coordinates other than (0, 0). Because of its difficulty, today's question is another horrible question to include on the PARCC.



Tuesday, April 21, 2015

PARCC Practice Test Question 6 (Day 151)

Last night I tutored my geometry student. Section 8-5 of the Glencoe text is specifically on Angles of Elevation and Depression.

I mentioned that the U of Chicago text doesn't have a separate section for angles of elevation. The term "angle of elevation" is mentioned in Section 14-3 of the U of Chicago, on the Tangent Ratio:

Example 2. At a location 50 m from the base of a tree, the angle of elevation of the tree is 33 degrees. Determine the height of the tree.

We notice in the Questions that there are several problems where the phrase "angle of elevation" could appear, but the U of Chicago avoids this term. For example, we look at Question 4:

4. When the sun is 32 degrees up from the horizon, the wall of a store casts a shadow 25 meters long. How high is the wall?

This question could be rewritten as:

4'. When the angle of elevation of the sun is 32 degrees, the wall of a store casts a shadow 25 meters long. How high is the wall?

Similarly, we notice Question 10:

10. From eye level 5' off the ground and 20' away from a flagpole, a person has to look up at a 40 degree angle to see the top of the pole. How high is the pole?

Rewritten, this becomes:

10'. From eye level 5' off the ground and 20' away from a flagpole, the angle of elevation is 40 degrees to the top of the pole. How high is the pole?

It makes sense for angles of elevation to appear in the lesson on tangents rather than sines and cosines since, as I pointed out, most angle of elevation problems involve vertical height and horizontal distance, so the hypotenuse of the triangle doesn't appear. In the Glencoe text, most of the problems in Section 8-5 involve the tangent ratio, but a few use sine instead. I must also point out that several of the problems involve inverse sine or tangent, while the U of Chicago text mentions only the inverse tangent among the arcfunctions. I consider this to be a deficiency of the U of Chicago text.

In the U of Chicago, the phrase angle of depression appears in Section 14-5 on sines and cosines:

14. From the top of a building, you look down at an object on the ground. If your eyes are 55 feet above the ground, and the angle of sight, called the angle of depression, is 50 degrees below the horizontal, how far is the object from you?

Notice that this question is written so that the cosine is needed, rather than sine or tangent.

In the SPUR review section at the end of Chapter 14, angles of elevation and depression are certainly under the section U for USES -- and yes, my student groaned slightly when I told him that all of the questions in this section are word problems The four questions under Objective I (the ninth letter of the alphabet), namely Questions 52 through 55, could be written use the phrase angle of elevation (depression) yet none of them do.

I will have more to say on angles of elevation and depression, and trigonometry in general, when we get to the PARCC trig questions. But today's PARCC question isn't on trig. Question 6 of the PARCC Practice Test is on completing the square:

Use the information provided to answer Part A and Part B for question 6.

The equation x^2 + y^2 - 4x + 2y = b describes a circle.

Part A
Determine the y-coordinate of the center of the circle.

Part B
The radius of the circle is 7 units. What is the value of b in the equation?

So far on this PARCC Practice Test, I pointed out that some traditionalists may object to some of the questions, but I myself have no problem with them -- even though I acknowledge that some students may be tricked on questions where may need to choose more than one answer. But all of this changes with this problem. Question 6 is the first question that I believe doesn't belong on a Geometry test.

To solve this question, we must complete the square:

x^2 + y^2 - 4x + 2y = b
x^2 - 4x + y^2 + 2y = b
x^2 - 4x + 4 + y^2 + 2y + 1 = b + 4 + 1
(x - 2)^2 + (y + 1)^2 = b + 5

Comparing this with the equation (x - h)^2 + (y - k)^2 = r^2, we see that the center of the circle must be the point (2, -1). So we get k = -1 as the answer to Part A.

For Part B, we notice that the right-hand side of the question, b + 5, must equal r^2. Since r = 7, we have r^2 = 49, so b + 5 = 49. Subtracting 5 gives us b = 44 as the answer to Part B.

The first objection I have is the need to complete the square. A few weeks ago, I saw that equations for circles appeared on the PARCC and so I quickly included Section 11-3 from the U of Chicago on a worksheet. But Section 11-3 doesn't contain any completing the square questions. When I created my worksheet for 11-3, I squeezed in completing the square questions because I knew that they were coming on the PARCC.

Many students haven't seen completing the square since Algebra I -- and many haven't seen it at all since many Algebra I teachers skip it. This is why I'd prefer this question to be on an Algebra II test, not a Geometry test.

Of course, a common error to be made on this sort of question, even by Algebra II students, would be to mix up the sign. Had this been a multiple choice question, I can almost guarantee that one of the wrong choices for Part A would be 1. And the other two wrong choices may be 2 and -2 -- the test writers banking on students confusing the x- and y-coordinates of the center.

Part B just makes this problem even worse. This question provides the radius of the circle and students are asked to find the value of b. It would have been much more straightforward to give the value of b -- that is, begin with the equation x^2 + y^2 - 4x + 2y = 44 -- and then ask the students for the radius. I have no problem with having questions with a Part B on the test, and this would've made a better Part B than what is actually included here. Notice that I will organize today's worksheet differently for multi-part questions -- there will be just one or two questions divided into parts similarly to the PARCC question.

Such questions that force the students to think backwards appear several times on the PARCC. I have no problem with this thinking backwards questions, and would've had no problem with it here in Question 6 had this not been an Algebra II question masquerading as a Geometry question. If we are going to have thinking backwards questions on the test, then they should be saved for questions when the underlying concepts are simple. If we are going to have completing the square to find the equation for a circle on a Geometry test, then there should only be straightforward questions, such as just to identify the center and radius of the circle.

Before we leave this question, let me point out that if this had been an Integrated Math course, this might have been an excellent question to ask. Integrated Math courses ought to focus on the connection between algebra and geometry. I could envision an Integrated Math II or III course in which one covers circles -- their circumferences, areas, tangent lines, inscribed angles, and equations, and then segue from there to the equations of the other conic sections.

But as a question on a traditionalist Geometry test, this question is terrible. Notice that Dr. Franklin Mason steers away from equations for circles completely in his text. Unfortunately, this is only the first of several terrible questions on the PARCC.

PARCC Practice EOY Exam Question 6
U of Chicago Correspondence: Section 11-3, Equations for Circles
Key Theorem: Equation for a Circle

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

(x - h)^2 + (y - k)^2 = r^2.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.GPE.A.1
Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation.

Commentary: The U of Chicago gives equations of circles, but never equations where students have to complete the square to find the center and radius. The fact that this Common Core standard is immediately followed by standard involving the equations of conic sections should have been a red flag that this standard belongs in Algebra II, not Geometry. But this all goes back to the fact that the Common Core doesn't divide its standards into courses.