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.