These students are sophomores, so I assume they're enrolled in Integrated Math II. The students appeared to need help on a hodgepodge of problems, including many geometry problems:
-- Exterior Angles of a Triangle
-- Slopes of Parallel and Perpendicular Lines
-- Circumcenter of a Triangle
-- Triangle Inequality
-- Vertical Angles
-- Complementary/Supplementary Angles
-- Unequal Sides/Angles
-- Translations on a Coordinate Plane
These topics are all over the place -- it makes me wonder, are the students preparing for a unit test, or a semester final? We know, of course, that geometry is squeezed into the final modules of the Common Core Integrated Math texts, but still, why must so many topics appear on one test?
Well, I guess I can't complain. Last week's "Chapter 15 Test" posted to the blog arguably contains so many topics as well. The problem was that I wanted to cover three lessons on circles that are important for PARCC, but they appear in three different chapters in the text (11-3, 13-5, 15-3). That plus returning to Chapter 8 to celebrate Pi Day and Square Root Day made for a crowded test indeed.
Back in October, I gave the contents of an Integrated Math II text, so I can only assume that the students are working out of that text. That text contains 24 module, with the second volume having Modules 14-24, mostly on geometry and probability.
Notice that most of these topics appeared during the first semester here on the blog. Actually, most first semester Geometry topics appear during Integrated Math I, with Math II consisting of the second semester Geometry topics like similarity and basic trig. In this text, Modules 14 and 15 are a review of Math I geometry, with Modules 16-21 containing the new material for Math II.
Some readers may point out that this is an argument against Integrated Math and in favor of the more traditionalist Algebra I, Geometry, Algebra II sequence. Most texts contain review lessons because we can't expect students to remember everything from previous years -- but this text reviews an entire semester (first semester of Geometry or second semester of Math I) in just two modules! On the other hand, with the Algebra I, Geometry, Algebra II sequence, there's no need to review Geometry because it's all contained in a single course. Of course, there will still be much review of Algebra I during Algebra II, but this can be accomplished more gradually since the entire year of Algebra II is devoted to algebra topics.
Today I help out one student on the translation problem. He was given coordinates the three vertices of a triangle and had to translate the triangle five units right. I notice that he draws the triangle first and then moves the triangle, when he could have added five to the x-coordinates first. I can't be sure, though, that this is how the teacher prefers him to do it.
The questions that trouble me the most are the circumcenter questions. This is because this topic was not emphasized as much before the Common Core. I must double-check to make sure that my answers are correct.
Unlike the last time I covered an AVID class, this time I'm more careful to avoid simply blurting out the answers. One student is working on a distance formula problem and writes 4^2 = 8 midway through the calculation. So I just asked "What's 4^2?" The student soon corrects herself. At the end, I say that I did a much better job of helping out today than I did last time.
Chapter 2 of Morris Kline's Mathematics and the Physical World is called "Discovery and Proof." I must emphasize that word "proof" that comes up so often in Geometry class.
Once again, Kline starts out the chapter with a quote:
Since then reason is divine in comparison with man's whole nature, the life according to reason must be divine in comparison with usual human life. -- Aristotle
"Mathematical reasoning is commonly regarded as different from and even superior to the reasoning other studies utilize."
And of course, many Geometry students lament the differences between mathematical and other reasoning when it's time to write proofs.
Kline then distinguishes between inductive and deductive reasoning. Inductive reasoning is when we extrapolate from a few occurrences of an event that the event will always occur. His example is the assumption that all substances heat when expanded, since most do. But inductive reasoning doesn't always work -- a counterexample to that assumption above is water. Mathematicians therefore employ deductive reasoning in their proofs. I notice that some Geometry texts make a big deal about inductive and deductive reasoning, often in their chapters on logic (usually Chapter 2). But the U of Chicago text does not mention inductive or deductive reasoning.
Well, I do know of one book that mentions inductive and deductive reasoning in its second chapter --namely Kline's. And much of this chapter is indeed all about Geometry.
Kline's main example is the sum of the angles of a triangle. He explains how at first consider this question inductively by looking at a few triangles and measuring their angles. Then he gives the usual deductive proof that their sum is 180 degrees -- the same proof that we find in Lesson 5-7 of the U of Chicago text. He points out that this theorem can be generalized to find the angle sum of all polygons, and indeed Lesson 5-7 does exactly that. But then Kline gives a less familiar example:
"AB, CD, EF, and GH are parallel chords of a circle. The centers of these chords lie on the one line. A circle, however, is a special case of an ellipse. Hence it is likely that a similar theorem holds for the ellipse. This generalization can be proved and yields another theorem."
This theorem is not proved anywhere in the U of Chicago text, not even for the circle. There is a theorem in Lesson 15-1 of the text that may help us out:
a. The line containing the center of a circle perpendicular to a chord bisects the chord.
b. The line containing the center of a circle and the midpoint of a chord bisects the central angle determined by the chord.
c. The bisector of the central angle of a chord is perpendicular to the chord and bisects the chord.
d. The perpendicular bisector of a chord of a circle contains the center of the circle.
Using this theorem, we can prove Kline's claim about circles. For lack of a better name, let's call it:
If three or more chords of a circle are parallel, then their midpoints are collinear.
Our plan is to use part d of the Chord-Center Theorem, so we begin by looking at the perpendicular bisectors of our chords (which, by definition, pass through their respective midpoints). We use the same trick that the U of Chicago text uses to prove the Two Reflection Theorem for Translations -- since the chords are parallel to each other, their respective perpendicular bisectors must also be parallel to each other, using the Two Perpendiculars Theorem and Perpendicular to Parallels (which we call the "Fifth Postulate" here on the blog). But all of those "parallel lines" contain the same point, namely the center of the circle (by part d). So those lines are in fact identical. (The U of Chicago text includes identical lines as parallel -- if we wish to avoid this, we replace "parallel" with "parallel or identical" above.) That single line is the common perpendicular bisector of all the parallel chords -- it contains all the midpoints, so they are collinear. QED
Of course, we don't bother to prove the theorem for ellipses. This also goes back to a debate as to whether circles counts as ellipses. Some people say that a circle isn't an ellipse, but rather the limiting position of an ellipse (where the distance between the foci is zero), just as a parabola is the limiting position of an ellipse (where the distance between the foci is infinite).
Here Kline does count a circle as an ellipse, and states a theorem about ellipses that automatically applies to circles as well. So in the spirit of the U of Chicago's inclusive definitions, we should definitely count a circle as an ellipse. But the theorem about ellipses is not worth attempting to prove here on the blog (since ellipses don't come up in Geometry classes). I notice that based on the pictures in his book, the line containing the common midpoint in an ellipse isn't necessarily perpendicular to all of the chords (though of course it is in a circle, as we just proved it).
Kline ends the chapter with a quote from Alexander Pope:
"Who could not win the mistress, must be content to win the maid."
Question 2 of the PARCC Practice Test is about the symmetries of a regular polygon:
2. Octagon PQRSTVWZ is a regular polygon with its center at point C.
Which transformations will map octagon PQRSTVWZ onto itself?
Select each correct transformation.
A. reflecting over
B. reflecting over
C. reflecting over
D. rotating 45 degrees clockwise around point Z
E. rotating 135 degrees clockwise around point C
F. rotating 90 degrees counterclockwise around point C
This question has multiple answers. For the three reflections, the valid mirrors are any line that bisects two angles, two sides, or one angle and one side. In particular, the mirror must pass through the center of the polygon, so only (A) and (B) are lines of symmetry. For the three rotations, any rotation whose magnitude is a multiple of 360/n degrees, in either direction, maps a regular n-gon to itself -- here n = 8 and so we need multiples of 45 degrees. Once again, though, the center of the rotation must be that of the polygon, so only (E) and (F) map our octagon to itself.
So there are four correct answers (A), (B), (E), and (F). As with any problem with multiple answers, the most common student error will be to omit some of the right answers.
I think this is a great problem to include on the PARCC test. My only problem is that the coverage of this type of problem in the U of Chicago text -- a text which emphasizes transformations more than most -- is highly disappointing.
One of the Common Core Standards states that the students should know of the transformations that map a figure to itself. The existence of an isometry that maps a figure to itself has a special name in math -- symmetry. So this question asks for the symmetries of the regular octagon -- that is, the ways in which the regular octagon is reflection-symmetric or rotation-symmetric.
Lesson 4-6 of the U of Chicago text is on reflecting polygons, and the following lesson introduces reflection-symmetric figures. But reflection-symmetric polygons aren't covered until Chapter 5 -- and for the most part, only isosceles triangles and trapezoids appear (as in the Isosceles Triangle Symmetry Theorem of Lesson 5-1, the Isosceles Trapezoid Symmetry Theorem of Lesson 5-5). In particular, the symmetry of polygons with five or more sides is not mentioned. And what's worse is that the rotational symmetry of any polygon fails to appear anywhere in the U of Chicago text!
The term regular polygon is defined in Lesson 7-6 of the U of Chicago text. In this lesson, the only theorem involving regular polygons is stated and proved:
Center of a Regular Polygon Theorem:
In any regular polygon there is a point (its center) which is equidistant from all its vertices.
Given: regular polygon ABCD...
Prove: There is a point O equidistant from A, B, C, D, ...
I won't restate the proof here, but in the course of the proof, it's stated that OABC is a kite with symmetry diagonal
The text proves the Center of a Regular Polygon Theorem using a technique called induction, which I explained back in December during the Lesson 7-6 post. Induction is not the same as inductive reasoning, even though some authors (like Kline, for example) use the term "induction" when they mean "inductive reasoning." Indeed, mathematical induction is considered deductive reasoning!
After this proof, the U of Chicago text proceeds:
"The Center of a Regular Polygon Theorem implies that there is a circle which contains all the vertices of a regular polygon. This enables regular polygons to be drawn quite easily. Draw the circle first and equally space the vertices of the polygon around the circle."
This is tantalizingly close to a description of the symmetries of the regular polygon. Indeed, the symmetries of the polygon are the exactly the symmetries of the circle that preserve that set of vertices spaced equally around the circle.
But believe it or not, there is actually one more lesson in the text that mentions regular polygons -- Lesson 15-2, "Regular Polygons and Schedules." This lesson teaches us how we can schedule a round-robin tournament -- one in which every team plays every other team. The opening round of the World Cup of Soccer is round-robin in that every team in the group plays every other team. But we rarely seen round-robin tournaments with more than four teams.
I skipped Lesson 15-2 since I doubt that round-robin tournaments appear on the PARCC. But let's look at how the text schedules a round-robin tourney for seven teams:
Step 1. Let the 7 teams be the vertices of an inscribed regular 7-gon (heptagon).
Step 2. (the first week) Draw a chord and all chords parallel to it. [But how do we know that any of the chords joining vertices are parallel? dw] Because the polygon has an odd number of sides, no two chords have the same length. This is the first week's schedule.
Step 3. (the second week) Rotate the chords 1/7 of a revolution.
And now we have the text rotating the regular heptagon to itself without even stating, much less proving, any sort of symmetry result.
So let's finally state and prove our theorem:
Regular Polygon Symmetry Theorem:
a. Any perpendicular bisector of a regular polygon is a symmetry line for the polygon.
b. The line containing any angle bisector of a regular polygon is a symmetry line for the polygon.
So how should we prove our theorem? We see Step 2 above about parallel chords -- and we're immediately reminded of Kline's Theorem that we proved earlier in this post! By the way, my intention in reading Kline and covering the practice PARCC at the same time was not to apply anything mentioned in Kline to the PARCC questions. But if that opportunity presents itself, then of course we're going to seize that opportunity.
We need one more theorem, also mentioned in Lesson 15-1, before we can prove our main result:
Arc-Chord Congruence Theorem:
In a circle or in congruent circles:
a. If two arcs have the same measure, then they are congruent and their chords are congruent.
b. If two chords have the same length, then their minor arcs have the same measure.
By the way, we skipped Lessons 15-1 and 15-2 because they don't appear on the PARCC. Notice that we're using material from those lessons to prove the Regular Polygon Symmetry Theorem which does appear on the PARCC, but not the results from 15-1 and 15-2 directly.
So let's begin the proof of the Regular Polygon Symmetry Theorem.
To prove part a, we let
Let C be the next vertex after B on the opposite side of A, and D be the next vertex after A on the other side of B. In other words, D, A, B, C are four consecutive vertices of the polygon, and we have
Now DA = BC as these are sides of a regular polygon. So by part b of the Arc-Chord Congruence Theorem, minor arcs DA and BC are congruent. Notice that the longer arcs ADC and BCD also ahve the same measure by the Arc Addition Postulate. Now the angles ABC and BAD are inscribed in the arcs ADC, BCD respectively, so by the Inscribed Angle Theorem the measures of the two angles are half those of the arcs. Since the arcs have equal measure, so do the angles ABC and BAD.
We also can use Arc Addition and add the arc AB to the congruent arcs DA and BC. This tells us that the arcs DAB and ABC are congruent, and so their inscribed angles BCD and ADC are congruent.
Now we look at the quadrilateral ABCD. We see that it has two pairs of congruent angles -- ABC and BAD, then BCD and ADC. Notice that this is sufficient information to conclude that ABCD is in fact an isosceles trapezoid. If we use the blog definition of isosceles trapezoid, then a pair of congruent adjacent angles and a pair of congruent opposite sides (DA = BC, if you recall) is enough to prove that ABCD is an isosceles trapezoid, and since an isosceles trapezoid is a trapezoid, we conclude that sides
At this point, you may be wondering why we dealt with arc measure at all -- we already know that the angles ABC and BAD are congruent because they're angles of a regular polygon, without any need to look at arc measure. The reason is that angles BCD and ADC are not angles of a regular polygon, so we need the argument above to prove that they are congruent. Moreover, the next step is to take E as the next vertex after D and F as the next vertex after C. The goal is to prove that CDEF is also an isosceles trapezoid -- and none of the angles of CDEF are angles of the regular polygon.
But I admit that it is possible to avoid the Arc-Chord and Inscribed Angle Theorems by using induction, similar to the Center of a Regular Polygon Theorem. The initial step is that ABC and BAD are congruent (as they are angles of a regular polygon). The inductive step is that since ABC and BAD are congruent, ABCD is an isosceles trapezoid, and so ADC and BCD are congruent. These congruent angles are subtracted from the angles ADE and BCF (which are angles of the regular polygon, hence congruent) to conclude that CDE and DCF are congruent. This is enough to conclude that CDEF is an isosceles trapezoid, and so on down the regular polygon.
So we have divided the regular polygon into isosceles trapezoids. Each vertex is connected to another vertex via one of several parallel chords. So now by Kline's Theorem, these midpoints of all these chords are collinear -- indeed, we see from the proof of Kline's Theorem above that this line passes through the center of the circle (i.e., it's a diameter) and is the perpendicular bisector of all of them.
But recall that this line -- the perpendicular bisector of all the chords, including the original
Now for part b, we look at the bisector of angle B, which we take to be between the vertices A and C, in that order. Again AB = AC as these are the sides of a regular polygon, so ABC is, in fact, an isosceles triangle. In Lesson 5-1, we know that the bisector of the vertex angle is the perpendicular bisector of the base, so our mirror is the perpendicular bisector of
In either case a or b, the final figure at the bottom of the regular polygon may be either an isosceles triangle or an isosceles trapezoid (depending on odd or even number of sides). In both cases, the entire polygon has been divided into isosceles triangles and trapezoids via a number of parallel chords all with the same diameter, the mirror, as a perpendicular bisector. Therefore the mirror, the angle bisector of B, is a symmetry line of the polygon. QED part b
This completes the proof of the Regular Polygon Symmetry Theorem. It tells us that every regular polygon has many lines of symmetry. But though an n-gon has n angle bisectors and n perpendicular bisectors, it has only n total lines of symmetry. This is because every line of symmetry is a diameter that touches the polygon twice. So every angle bisector must also bisect a second angle, or else is the perpendicular bisector of a side, and the same is true of every perpendicular bisector. This double counting tells us that there are only n symmetry lines, not 2n.
All that's left now is rotational symmetry. We see that the composite of two reflections in intersecting lines is a rotation, and all of our symmetry lines intersect at O, the circumcenter of the polygon. Any figure with intersecting symmetry lines also has rotational symmetry, with the center of the rotation the same as that of the polygon. But what is the magnitude of the rotation?
We can draw n radii to divide a regular n-gon into n triangles. These triangles are all congruent isosceles triangles by SSS, and the radii each bisect an angle of the polygon. Since the angles at O all are congruent and add up to 360 degrees, the central angle between two angle bisectors is 360/n. The perpendicular bisectors also bisect these central angles (Lesson 5-1), and so the central angle between an angle bisector and a perpendicular bisector is 360/2n. That is, the symmetry lines of the polygon are spaced 360/2n degrees around O (but again, there are only n lines, not 2n, since each line is a diameter, not just a radius).
The Two Reflection Theorem for Rotations tells us that the magnitude of the rotation is exactly twice the angle between the mirrors -- and the angles are multiples of 360/2n. So the magnitude of the rotation must be multiples of 720/2n, or 360/n. So a regular n-gon has n-fold rotational symmetry -- rotating it by any multiple of 360/n degrees maps the polygon onto itself.
The set of all symmetries of a figure is called the symmetry group of that figure. The Regular Polygon Symmetry Theorem tells us that an n-gon has a symmetry group with 2n elements -- n of these are reflections and the other n are rotations.
PARCC Practice EOY Question 2
U of Chicago Correspondence: Lesson 7-6, Properties of Special Figures
Key Theorem: Center of a Regular Polygon Theorem
In any regular polygon there is a point (its center) which is equidistant from all its vertices.
Common Core Standard:
Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.
Commentary: Lesson 7-6 of the U of Chicago text only mentions that a regular polygon has a center -- not that the polygon has any symmetry. We come up with a Regular Polygon Symmetry Theorem to help us answer this question. The theorem tells us that reflecting our regular octagon in any diagonal containing the center C, or rotating it around the center C in any multiple of 360/8 = 45 degrees, maps the octagon to itself.