Friday, May 8, 2015

PARCC Practice Test Question 19 (Day 164)

Today I wrapped up Mario Livio's The Equation That Couldn't Be Solved by reading the appendices. I point out some of the things that appear in the some of these appendices:

Appendix 3 and 4 are about the Greek mathematician Diophantus. A Diophantine equation is one whose solutions must be whole numbers. The example given in Appendix 4 is 29x + 4 = 8y. This is a linear equation with infinitely many solutions, and it's easy to plug in any real number for x and obtain a real number for y. But a Diophantine equation only admits whole number solutions, and merely plugging in a whole number for x doesn't guarantee that y will be a whole number. So Livio describes how to find the whole number solution (4, 15) for this Diophantine equation. Then he moves on to another problem: find two numbers whose sum is 20 and whose sum of squares is 208. Diophantus came up with the trick of letting 10 + x and 10 - x be the two numbers, which makes the resulting quadratic equation easier to solve. The two numbers turn out to be 12 and 8.

In Appendix 7 discusses the roots of quadratic equations. He calls the two solutions x_1 and x_2 -- once again, these are subscripts that I'm writing in ASCII. He shows that the sum of the roots must be -b / a and the product of the roots must be c / a. Then he writes this admittedly strange expression, 1/2[(x_1 + x_2) +- sqrt((x_1 + x_2)^2 - 4x_1 x_2)]. As it turns out, we can simplify this to equal x_1 when the + sign is chosen and x_2 when the - sign is chosen. So we replace x_1 + x_2 with -b / a and x_1 x_2 with c / a, and we obtain the Quadratic Formula. But what's the point of this? In the expression 1/2[(x_1 + x_2) +- sqrt((x_1 + x_2)^2 - 4x_1 x_2)], it doesn't matter which root is x_1 and which one is x_2 -- that is, the formula is symmetric in x_1 and x_2. So the point is to show a relationship between symmetry and the solutions of equations -- and this is how Galois proved that there are Quadratic, Cubic, and Quartic Formulas, but no Quintic Formula.

All in all, Mario Livio's The Equation That Couldn't Be Solved is an enjoyable read, and I recommend it, especially to math teachers -- whether Algebra or Geometry. Let me conclude my discussion of the book with another YouTube video -- the song "Do You Hear the People Sing" from Les Miserables, sung at the rebellion that almost occurred at Galois's death:



Question 19 of the PARCC Practice Exam is on reflections and rotations:

Triangle ABC has vertices at A(1, 2), B(4, 6), and C(4, 2) in the coordinate plane. The triangle will be reflected over the x-axis and then rotated 180 degrees about the origin to form Triangle A'B'C'. What are the vertices of Triangle A'B'C'?

(A) A'(1, -2), B'(4, -6), C'(4, -2)
(B) A'(-1, -2), B'(-4, -6), C'(-4, -2)
(C) A'(-1, 2), B'(-4, 6), C'(-4, 2)
(D) A'(1, 2), B'(4, 6), C'(4, 2)

So we are asked to find the composite of a reflection and a rotation. But notice that the rotation is itself a composite of reflections according to the Two Reflection Theorem for Rotations, mentioned in Section 6-3 of the U of Chicago text. To find the two mirrors that produce our desired rotation, we see that these two mirrors must intersect at the center of the rotation -- namely the origin -- and the angle between them must be half of the magnitude of the rotation. Since half of 180 is 90, the two mirrors must be perpendicular and intersect at the origin. But since we already have a reflection over the x-axis, we might as well let the x-axis be one of the mirrors for our rotation -- then the other mirror will end up being the y-axis.

So our final isometry is the composite of a reflection over the x-axis, then another reflection over the x-axis, and then a reflection over the y-axis. The two reflections over the x-axis cancel (this is known as the Flip-Flop Theorem), so we are left with a single reflection over the y-axis. Reflecting the three vertices A, B, and C over the y-axis produces the images in choice (C), which is the correct answer.

We notice how the composite of a reflection over the x-axis and a 180-degree rotation about the origin is a reflection over the y-axis -- while the composite of a reflection over the x-axis and a reflection over the y-axis is a 180-degree rotation about the origin. Finally the composite of a reflection over the x-axis and itself is the identity. We are beginning to suspect that the set of four isometries -- the identity, x-axis reflection, y-axis reflection, and origin-reflection -- is a group. And not only is this set a group, but this group is isomorphic to -- you guessed it -- the Klein four-group that Livio mentions in his book.

If we have a rectangle centered at the origin with its sides parallel to the axes, or a rhombus centered at the origin with its diagonals on the axes, then this group of four isometries is the symmetry group of our quadrilateral.

The easiest way to solve our PARCC problem is if we know how to perform these isometries on ordered pairs by manipulating the signs of the coordinates. A reflection in the x-axis keeps the signs of the x-coordinate, but changes the sign of the y-coordinate. Likewise, a reflection in the y-axis keeps the sign of the y-coordinate, but changes the sign of the x-coordinate. A 180-degree rotation about the origin changes the signs of both coordinates -- and of course, the identity keeps both signs.

So the reflection over the x-axis switched the y-coordinates, while the rotation switched both signs, restoring the original x signs but switching the y signs. This is a much easier way to obtain (C) as the correct answer.

Once again, we notice that the U of Chicago text only hints at a relationship between transformations and coordinates. And once again, the problem is proof and circularity -- we don't know that reflection in the x-axis maps (x, y) to (x, -y) until we define slope and distance in the plane (since reflections are defined using the perpendicular bisector). And we can't define slope and distance and prove that they work unless we've already defined dilations and similar triangles. (Even though U of Chicago doesn't use the Dr. Hung-Hsi Wu proofs of slope and distance using similar triangles, it doesn't focus on coordinates until Chapter 11 when transformations are in Chapter 6.)

If we stick to the principle that nothing should be taught in Geometry until it can be proved, then we must develop translations, reflections, rotations, and dilations completely without coordinates. Then we use dilations to define similarity, similarity to prove the Pythagorean Theorem, and then prove the formulas for slope and distance. Only then can we finally prove that transformations in the coordinate plane work as they should -- so that the reflection in the x-axis does map (x, y) to (x, -y).

But in this case, the insistence that proof precede use hinders student learning. Indeed, I reckon that it would be easier for students to understand transformations by graphing them. So we can declare to students that the rotation of 180 degrees maps (x, y) to (-x, -y) by fiat so that they understand what such rotations look like, with the promise that it can be formally proved later on after coordinates have been developed.

The geometry student I tutor just finished Section 9-2 of the Glencoe text, and so either yesterday or today he is covering Section 9-3 on rotations. And we can be sure that his text is teaching him that the 180-degree rotation about the origin maps (x, y) to (-x, -y). Knowing what effect these four transformations mentioned today have on coordinates will help students answer questions on the PARCC test -- including today's question.

Today is an activity day. My activity focuses on two ideas -- the first is that the four transformations (line reflections over axes, point reflections over origin, identity) can be accomplished simply by manipulating the signs, and second, that these four transformations are related (i.e., that they end up forming a group).

PARCC Practice EOY Exam Question 19
U of Chicago Correspondence: Section 6-3, Rotations
Key Theorem: Two Reflection Theorem for Rotations

If m intersects l, then the composite of the reflection over m following the reflection over l "turns" figures twice the non-obtuse angle between l and m, measured from l to m, about the point of intersection of the lines.

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

Commentary: There is only one question in the text that requires students to perform the rotation of 180 degrees about the origin. Much more emphasis on transformations and the coordinate plane than is provided in the U of Chicago text is needed here.



No comments:

Post a Comment