## Monday, May 4, 2015

### PARCC Practice Test Question 15 (Day 160)

Chapter 6 of Mario Livio's The Equation That Couldn't Be Solved is all about "Groups." So far, the first two chapters of Livio's book is about symmetry and the next three are about equations. In this chapter, the author reveals the connection between symmetry and equations to be groups.

Livio begins by discussing permutations. A permutation is a one-to-one function whose domain and range are the same set. He gives a simple example of a permutation -- the mapping from GALOIS to AGLISO -- and then describes two popular puzzles based on permutations. One of them is the 14-15 puzzle of Sam Loyd from the 1870's. Loyd offered \$1000 to anyone who could solve it. Of course, the puzzlist already knew that it was impossible to solve. Here's a link to a Cut the Knot page, which gives a group theoretic proof as to why it's impossible:

http://www.cut-the-knot.org/Curriculum/Combinatorics/Fifteen.shtml

http://mathworld.wolfram.com/RubiksCube.html

Livio proceeds by stating that the set of all permutations on a base set is a group, which he denotes by the symbol S_n (which is actually the subscript n, but this is how I write it in ASCII). So the group of all permutations of the three letters A, B, and C is S_3. This set has six elements:

ABC, ACB, BAC, BCA, CAB, CBA

He points out that if we think of these not at letters but as points A, B, and C, then these six elements correspond to the symmetries of an equilateral triangle. If ABC is the original triangle (the identity), then ACB is a reflection over the angle bisector of A, BAC is a reflection over the angle bisector of B, BCA is a 120-degree rotation, and so on. So the group of permutations on three letters is essentially the same as the symmetry group of an equilateral triangle. According to Livio, there's a name for two groups that are essentially the same -- isomorphic.

Notice there is a connection between the symmetric group of an equilateral triangle and the even and odd permutations of Loyd's 14-15 puzzle. The odd permutations -- the ones where one makes an odd number of swaps, such as ABC to ACB -- correspond to isometries that reverse orientation, while the even permutations correspond to isometries that preserve orientation. So in a way, Loyd's 14-15 puzzle is impossible to solve for the same reason that you can't wear your left shoe on your right foot.

The symmetry group of an equilateral triangle is isomorphic to S_3, while the symmetry group of an isosceles triangle is isomorphic to S_2. But notice that there is no quadrilateral whose symmetry group is isomorphic to S_4 -- not even the square. To see why, we notice that if ABCD is a square, then ABDC is not a square -- nor are ACBDACDB, or ADBC. But ADCB is a square -- it is the original square reflected about the diagonal AC.

Livio then describes another group, the Klein four-group. He states that there are many groups isomorphic to the Klein four-group, including the four symmetries of a pair of jeans (the identity, turning it back to front, turning it inside-out, and doing both of these). Another group involves four symmetries of a box centered at the origin -- the identity and three 180-degree rotations, one about each axis. This is not the full symmetry group of a box as it omits reflections, but it is the set of symmetries of a box that preserve orientation. But the full symmetry group of a rectangle or rhombus in 2D is indeed isomorphic to the Klein four-group, as it includes an identity, two reflections, and a rotation of 180-degrees that is the composite of the two reflections.

I won't repeat Livio's full explanation of what these symmetries have to do with Galois's proof of the impossibility of solving the quintic, but I will summarize it. Basically, just as every figure (such as an equilateral triangle, square, rectangle, etc.) has a symmetry group, every equation has a symmetry group, called the Galois group. The Galois group of a quadratic equation is S_2, as it clearly has reflection symmetry -- after all, Algebra I teachers talk about the axis of symmetry of a parabola. As it turns out, some quintic equations have S_5 as their Galois group. If S_5 is the Galois group, then the equation is not solvable, but if it's a different group. it could be solvable. I'll leave anyone interested in a more detailed explanation to read Livio directly.

The next-to-last section in this chapter is definitely on-topic for this blog -- "What Is Geometry?" We read that the German mathematician Felix Klein -- for whom the Klein four-group is named -- claimed that geometry is group theory! This section begins by describing Euclid and his famous Fifth Postulate, then moves on to non-Euclidean geometry. Livio writes about the rigid motions -- that is, the isometries of the plane. (I use the term "isometries" as this is the term that appears in the U of Chicago text, but technically speaking I should be calling them "rigid motions," as this what the Common Core standards calls them.) He writes:

"Take for instance the group of rigid motions -- the motions that preserve distances and angles, and consequently shapes. Since such motions are the bread and butter of Euclidean geometry, the latter can be defined as the geometry that remains invariant under all the transformations that are in the group of rigid motions. A circle of a given radius remains the same circle no matter how you turn it. Two triangles that overlap precisely (and are the subject of so many theorems in Euclidean geometry and are a constant source of headache to high school students) stay congruent even if you translate, rotate, or reflect them. Klein's radical idea, however, allowed for a much wider variety of geometries to exist."

So when traditionalists oppose Common Core Geometry on the grounds that they include translations, reflections, and so on, notice that they are actually opposing what Livio calls "the bread and butter of Euclidean geometry"! And if one decides to take the more traditionalist path of accepting SSS, SAS, and ASA as postulates, then note that SSS is true only because there is always an isometry mapping one triangle to another triangle whose sides are the same length as those of the first. So traditionalists who begin with SSS, SAS, and ASA are, deep down, invoking translations, reflections, and rotations whether they like it or not!

Livio writes that we can define the Euclidean plane as the plane whose symmetry group is the set of all translations, reflections, rotations, and glide reflections. It was Klein who realized that we can define a non-Euclidean plane as a plane with a different symmetry group. I've mentioned on the blog that in spherical geometry there are no translations, while in hyperbolic geometry there are two different types of translation.

With that, let's move on to the PARCC Question for the day. Question 15 of the PARCC Practice Test is on the volumes of pyramids:

The table shows the approximate measurements of the Great Pyramid of Giza in Egypt and the Pyramid of Kukulcan in Mexico.

Great Pyramid of Giza:
Height: 147 meters
Area of Base: 52,900 square meters

Pyramid of Kukulcan:
Height: 30 meters
Area of Base: 3,025 square meters

Approximately what is the difference between the volume of the Great Pyramid of Giza and the volume of the Pyramid of Kukulcan?

(A) 1,945,000 cubic meters
(B) 2,562,000 cubic meters
(C) 5,835,000 cubic meters
(D) 7,686,000 cubic meters

Oops! You know how I like to time my math questions to the holidays, and I came so close to timing it so that a question mentioning a pyramid in Mexico occurs on Cinco de Mayo. Oh well!

This question is straightforward provided one knows the volume of a pyramid, V = 1/3 Bh. Unlike most of last week's problems, our calculator will come in handy here! We calculate the volume of the Great Pyramid to be 1/3(147)(52,900), or 2,592,100 cubic meters. Notice that it's easy to find the volume of the Pyramid of Kukulcan without a calculator, since 1/3 of 30 is obviously 10 and 10 times 3,025 is obviously 30,250. Subtracting these gives 2,561,860, and since the question mentions the key word "approximately," we round this off to obtain one of the choices, 2,562,000, which is choice (B).

A common mistake will be to forget the factor of 1/3. We see that choice (D) is exactly thrice as much as choice (B), and so those who forget 1/3 will end up choosing (D). Traditionalists should have no problem with this question.

PARCC Practice EOY Exam Question 15
U of Chicago Correspondence: Section 10-7, Volumes of Pyramids and Cones
Key Theorem: Pyramid-Cone Volume Formula

The volume V of any pyramid or cone equals 1/3 the product of its height h and its base area B.

V = 1/3 Bh

Common Core Standard:
CCSS.MATH.CONTENT.HSG.GMD.A.3
Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.

Commentary: Example 1 actually asks students to find the volume of the Great Pyramid. The slight difference in the volume given by U of Chicago and that given by PARCC is because the former gives 231 m as the length of a side, while the latter gives 230 m instead. Question 11 mentions a pyramid in Mexico, but it's not the same as the one given on the PARCC test. No question in the U of Chicago asks for the difference between the volumes of two pyramids.