Friday, October 31, 2014

Chapter 5 Test (Day 56)

Some people may question the wisdom of posting a test on Halloween. My thinking is that it's better for the students to take a test on Halloween and then have a homework-free night, when they will want to celebrate the holiday. Today I heard someone suggest that districts have a staff development day on Halloween -- students are simply in no mood to study at all on Halloween, in a way that can't be said of Columbus or Veterans Day -- the federal holidays for which schools actually close.

Meanwhile, I continued to look at the Glencoe Geometry text. I noticed that as bad as Section 1-6, on Two-Dimensional Figures is, Section 1-7, on Three-Dimensional Figures, is even worse. Students are now expected to cram in several surface area and volume formulas, all in the first chapter!

If I were teaching a class with the Glencoe text, I would cover Sections 1-1 and 1-2 (undefined terms), skip up to Sections 1-4 and 1-5 (on angles -- this is the one thing that makes Glencoe better than the U of Chicago, where angles don't appear until Chapter 3), and then just jump up and begin Chapter 2. Maybe I'd cover the names of the polygons in 1-6 and possibly even a polyhedron in 1-7, but certainly no formulas for area, volume, or distance. After all, the last four chapters (10 through 13) are already there to cover area, volume, and right triangles anyway.

So now I wonder, what is the fascination with the Distance Formula that both Prentice-Hall and Glencoe feel a need to place it in Chapter 1? Well, the first chapter of any geometry text should cover basic concepts, and while the Distance Formula isn't a basic concept, distance itself is. Perhaps the textbook writers feel that in order to cover distance fully, the Distance Formula is required.

But I don't want to torture my students with the Distance Formula right at the beginning of the year -- I agree with David Joyce that it should wait until after the Pythagorean Theorem. The U of Chicago's Section 1-8, on One-Dimensional Figures, contains all the information on length and distance that I'd want to cover at the beginning of the year. By the way, the student I tutored last night had no trouble with the area questions on his homework assignment, just those on perimeter of a triangle where the Distance Formula is required to find the length of one or two of the sides. (I wonder how he'll fare on the surface area and volume questions, though.)

As I mentioned earlier, today I subbed for a P.E. class. On the back of many of the students' P.E. uniforms were printed vocabulary words for English and formulas for math -- probably so that the students are still learning academics even during P.E. class. And as it turned out, one of the formulas was -- you guessed it -- the Distance Formula. Other formulas I saw include the Area of a Circle, Midpoint Formula, Pythagorean Theorem, and Quadratic Formula.

Now here are the answers to my scary Halloween test.

1-2. constructions (or drawings). Notice that a construction for #2 is halfway to constructing a square inscribed in a circle for Common Core.

3. 95 degrees.

4. 152.5 degrees.

5. 87 degrees.

6. 86 degrees.

7. x degrees. This is almost like part of Euclid's proof of the Isosceles Triangle Theorem (except I think that his proof focused on the linear pairs, not the vertical angles).

8. 27 degrees.

9a. x = 60

9b. 61, 62, 58 degrees.

10. 25, 69, 128, 138 degrees.

11. polygon, quadrilateral, parallelogram, rectangle, square.

12. kite.

13. rectangle.

14. false. A counterexample is found easily.

15. Yes, the perpendicular bisector of the bases.

16. 46. Yes, after seeing my student's assignment last night, I decided that even though I don't want to cover area so soon, perimeter is a concept that could be developed more in these early lessons. My question actually defines perimeter since my lessons haven't stressed the concept yet. This is the simplest possible perimeter problem that I could have covered, where only the definition of kite is needed to find the two missing lengths. I could have given an isosceles trapezoid instead, where the Isosceles Trapezoid Theorem is needed to find a missing side length. Or since I squeezed in the Properties of a Parallelogram Theorem in our Section 5-6, I could have even put a parallelogram here with only two consecutive side lengths given.

17. The conjecture is true, and is a key part of the proof of Centroid Concurrency Theorem.

18. Statements                     Reasons
      1. angle G = angle FHI 1. Given
      2. EG | | FH                   2. Corresponding Angles Test
      3. EFHG is a trapezoid 3. Definition of trapezoid (inclusive def. -- it could be a parallelogram)

19. Statements                   Reasons
      1. O and P are circles  1. Given
      2. OQ = OR, PQ = PR 2. Definition of circle
      3. OQPR is a kite         3. Definition of kite (inclusive def. -- it could be a rhombus)

20. Figure is at the top, then below it is quadrilateral. Branching out from it are kite, trapezoid. Then below trapezoid is parallelogram. Kite and parallelogram rejoin to have rhombus below. (Once again, these are inclusive definitions!)
I hope you have a wonderful Halloween.

Thursday, October 30, 2014

Review for Chapter 5 Test (Day 55)

Today was first day as a substitute teacher in my second district. But today I subbed for a sixth grade English class and tomorrow it will actually be a P.E. class, so once again there will be no math to post.

But I did finally tutor my geometry student today, and once again, I showed him some worksheets from earlier. Today, his school covered Section 1-6 of a text published by Glencoe, a section called "Two-Dimensional Figures." This section covers the perimeter and area of rectangles, circles, and triangles, as well as the Distance Formula and coordinate geometry.

Recall that as David Joyce points out with the Prentice Hall text, giving the Distance Formula, well before its proof using the Pythagorean Theorem, is a terrible idea. I find it fascinating that so many texts would give the Distance Formula in Chapter 1. And I've mentioned many times that I don't want to scare students just coming off of algebra with so many heavy-duty formulas in the first chapter -- so I'm glad that the U of Chicago waits until the second half of the book before teaching area and the distance formula (Chapters 8 and 11).

I showed my student my worksheets from Sections 2-1 and 2-7 of the U of Chicago text -- about convexity and polygons. My student did well on them -- to him, they were a breath of fresh air after seeing so many hard triangle perimeter problems requiring the Distance Formula.

And today, we now see why my plan to post the Fifth Postulate Lesson on Day 55 -- all those fives coming together -- backfired. Aligning my schedule to the school calendar at my first district meant that I needed to finish Chapter 5 by Day 56 in order to finish Chapters 6 and 7 by the end of the semester, so today, Day 55, needs to be test review, not a new lesson.

There are many important concepts covered on this test -- isosceles triangles, quadrilaterals, and parallel lines. Here are the types of questions that appear on this test -- all of these coming from the SPUR section of Chapter 5 of the U of Chicago text.

  • The first two questions direct the student to draw an example of the figure using ruler, compass, or protractor. I don't expect most classrooms to have a compass -- and if the students don't have a ruler and protractor, close enough is good enough. For students with a straightedge and compass, the necessary constructions are perpendicular lines (for the scalene right triangle) and congruent line segments (for the isosceles acute triangle -- a single arc is sufficient).
  • Question 3 requires use of the Trapezoid Angle Theorem and algebra. Once again, this test will require the most algebra of any first semester chapter.
  • Questions 4-5 require use of the Parallel Consequences and algebra. The first question is the Alternate Interior Angles Consequence while technically, the second one is actually Same-Side Exterior Angles. Savvy students should be able to figure it out by using, for example, the Linear Pair Theorem followed by Corresponding Angles. One way I remember is that any two angles formed by two parallel lines and a transversal are either equal or supplementary -- and one can eyeball it to see whether the angles are acute or obtuse to tell which ones are which.
  • Questions 6-7 require use of the Isosceles Triangle Theorem -- neither one is straightforward, though, as either the Linear Pair or Vertical Angles Theorems are necessary. Once again, the students can look for acute and obtuse angles to find which ones are equal or supplementary.
  • Questions 8-9 require use of the Triangle-Sum Theorem. The first one only needs arithmetic, while the second needs algebra.
  • Question 10 requires use of the Quadrilateral-Sum Theorem. Once again, linear pairs, vertical angles, and algebra are needed.
  • Question 11 requires use of the Quadrilateral Hierarchy Theorem. Students must arrange the various shapes from most general to most specific.
  • In Questions 12-13, students must identify the quadrilaterals. The second one is a rhombus, not necessarily a square, since nowhere is it stated that the angles are right angles. Some might note that here I appear to contradict myself -- didn't I just say that in Question 5, students are supposed to assume that angle 1 is obtuse and angle 7 is acute in order to determine that they are supplementary, but here in Question 13 students should not assume that any of the angles are right angles? The difference is that in Question 5, identifying acute and obtuse angles is used as a mnemonic to remember actual theorems such as Linear Pair or Corresponding Angles Consequence, but in Question 13 nothing warrants knowing that there are any right angles.
  • Question 14 is a true-or-false question about the Quadrilateral Hierarchy. If the statement is false, the students should draw a square that isn't a parallelogram. But they shouldn't -- because the statement happens to be true.
  • In Question 15, students must determine whether the quadrilateral has any symmetry lines. As it turns out, the figure is a kite, so by the Kite Symmetry Theorem, it has a symmetry line.
  • Question 16 is another true-or-false question about the Quadrilateral Hierarchy, but this time students are not directed to draw a counterexample. As it turns out, the statement is true -- that the bases have a common bisector is mentioned in the Isosceles Trapezoid Symmetry Theorem.
  • In Question 17, students are to evaluate a conjecture. The U of Chicago text asks students to rank on an A-E scale whether they believe the conjecture to be true or false -- but I don't do this on a test, or otherwise every student would just choose C ("I'm not sure!"). So the students' only choices are true and false. As it turns out, the conjecture is true, and it can proved simply by knowing that half of 180 is 90. But students don't have to prove it -- proofs can wait until...
  • ...Questions 18-19. The keys to the first question are the Alternate Interior Angles Test and the definition of trapezoid. The keys to the second question are two definitions -- those of circle and kite. Notice that the first proof underscores the preference for inclusive rather than exclusive definitions. Under the exclusive definition, EFHG could be a parallelogram -- and the way it is drawn, it almost looks like a parallelogram. We can't prove that it's a parallelogram -- but we can't prove that it's not a parallelogram either, which means that under the exclusive definition, we can't prove that it's a trapezoid either! The second question isn't a problem for the exclusive definition -- OQPR is clearly nonconvex while all parallelograms are convex. But even if point O had been drawn outside of circle P, we could still prove that it's an inclusive kite, but it could be a parallelogram (hence a rhombus) and not an exclusive kite.
  • Question 20 has the students draw part of the Quadrilateral Hierarchy. It goes without saying that inclusive definitions should be used, so a rectangle is an isosceles trapezoid.

Wednesday, October 29, 2014

Section 5-7: Sums of Angle Measures in Polygons (Day 54)

Section 5-7 of the U of Chicago text discusses the sum of the angle measures in polygons, including triangles, quadrilaterals, and higher polygons. To me, this is the most arithmetic- and algebra-intensive lesson in all of the first semester.

The lesson begins with a discussion of Euclidean and non-Euclidean geometry. The 19th-century mathematician Karl Friedrich Gauss wanted to determine whether Euclidean geometry was true -- that is, that it accurately described the measure of the earth -- by experiment. The text shows a photo of three mountaintops that Gauss used as the vertices of a triangle, and the mathematician found that the sum of the angle measures of the triangle was, to within experimental error, 180 degrees.

Later on, the text states that if Gauss could have used a larger triangle -- say with one vertex at the North Pole and two vertices on the equator -- the angle-sum would have been greater than 180. The geometry of a sphere is not Euclidean, but is a special type of non-Euclidean geometry -- often called spherical geometry. As stated in the text:

"In a plane, two perpendiculars to the same line cannot intersect to form a triangle, but this can happen on a sphere. The surface of the earth can be approximated as a sphere. A triangle formed by two longitudes (north-south lines) an the equator is isosceles with two right base angles! Since there is a third angle at the North Pole, the measures add to more than 180 degrees. Thus neither the Two Perpendiculars Theorem nor the Triangle-Sum Theorem works on the surface of the earth."

But hold on a minute. It's obvious that the Triangle-Sum Theorem only holds in Euclidean geometry, as its proof uses the Alternate Interior Angles Consequence that depends on the Fifth Postulate. But we were able to prove the Two Perpendiculars Theorem on this blog, without using any sort of Parallel Postulate at all! So the Two Perpendiculars Theorem ought to hold for all types of geometry, both Euclidean and non-Euclidean -- yet it clearly doesn't hold for spherical geometry.

The truth is that spherical geometry differs from Euclidean geometry much more strongly than hyperbolic geometry differs from Euclidean.  We can obtain hyperbolic geometry from Euclidean simply by dropping the Fifth Postulate and replacing it with an axiom stating that there are many parallels through a point not on the line. But we can't obtain spherical geometry in a similar way.

First of all, what exactly is a line in spherical geometry? (Recall that line is one of the undefined terms, so we can't rely on its definition.) Any figure that we think is a "line" on earth goes all the way around the world, and so is actually a circle. What we want is for a "line" to be the shortest distance between two points. Notice that smaller circles on the globe clearly look curved, but larger circles that go around the world look like straight lines to a traveler. Therefore the most "linear" circle is the largest possible circle -- one that shares a center with the earth. This is called a great circle -- and this is why the example in the text mentions two longitudes and the equator -- these are great circles. But the so-called "parallels of latitude" are not great circles and so are not "lines" (geodesics).

Now what postulates does this spherical geometry violate? Notice that there are no parallel lines on the sphere, because any two great circles intersect. (Once again, note that "parallels of latitude" are not great circles.) Any two longitudes meet at the poles, and so the Unique Line Assumption part of the Point-Line-Plane postulate is violated -- through the poles there are infinitely many lines rather than just one.

But any two great circles that intersect at the North Pole must intersect at the South Pole. And any two great circles that intersect at one point intersect at the point directly opposite that point -- often called the antipodes, or antipodal point. So one way to avoid this problem is to declare that two antipodal points are actually one point. The resulting geometry is called elliptic geometry.

Yet elliptic geometry still violates the postulates. Here I link to David Joyce's website for more discussion of elliptic geometry:

Notice that this is a link to the first theorem of Euclid that fails in elliptic geometry. It is the Triangle Exterior Angle Inequality Theorem, or TEAI. Originally, Dr. Franklin Mason followed Euclid and used the TEAI to derive the Parallel Tests. The Parallel Tests do not hold in elliptic geometry (of course not, since they prove lines parallel and there are no parallel lines).

In discussing which of Euclid's five postulates that fail in elliptic geometry, the link above writes:

Elliptic geometry satisfies some of the postulates of Euclidean geometry, but not all of them under all interpretations. Usually, I.Post.1, to draw a straight line from any point to any point, is interpreted to include the uniqueness of that line. But in elliptic geometry a completed “straight line” is topologically a circle so that any pair of points on it divide it into two arcs. Therefore, in elliptic geometry exactly two “straight lines” join any two given “points.”
Also, I.Post.2, to produce a finite straight line continuously in a straight line, is sometimes interpreted to include the condition that its ends don’t meet when extended. Under that interpretation, elliptic geometry fails Postulate 2.

Both of these are essentially part of our Point-Line-Plane Postulate. So this is the postulate that we'd have to rewrite if we want elliptic or spherical geometry. Our other postulates still hold -- we can still measure angles, we can still perform reflections, and we still have plane separation (of course, we'd call these halves "hemispheres").

Notice that ironically, our Fifth Postulate still holds in spherical geometry. Of course, it's vacuously  true -- there are no parallel lines, so any statement of the form "if lines are parallel, then..." or something about a line intersecting two parallel lines, is vacuously true. The Parallel Consequences are also vacuously true in spherical geometry. Playfair also holds, provided that we write it the way that Dr. M writes it on his site:

"Through a point not on a given line, there’s at most one line parallel to the given line."
(emphasis mine)

"At most one" allows for the possibility of zero parallel lines. Technically, this is the form of Playfair that we proved earlier this week -- we only showed that at most one parallel line exists. The proof that at least one parallel line exists uses rotations and is not valid in spherical geometry.

Some teachers believe that we should briefly introduce high school students to non-Euclidean geometry -- and usually spherical geometry is suggested as it describes the earth. This is opposite what a college non-Euclidean geometry class would do -- in college, the emphasis is usually on hyperbolic geometry because its theorems are more similar to those of Euclidean geometry.

But it's often interesting to discuss with students how spherical geometry affects the earth. A classic brainteaser often goes as follows:

  • A bear hunter sets out from camp and walks one mile south.
  • He sees a bear and is about to shoot it.
  • The bear grabs his gun and eats it.
  • The hunter runs away one mile east.
  • He then walks one mile north and gets back to his camp and changes his underwear.
  • What colour was the bear?
The answer is that the "colour" (sorry -- this is obviously from a British website) of the bear is white, since the puzzle describes a polar bear at the North Pole. Technically, this is not a spherical triangle, since the "one mile east" is along a parallel of latitude, not a great circle. It's not even close to being a great circle -- if the hunter ran approximately six miles east he would have walked in a complete circle around the pole.

Here's another puzzle related to spherical geometry. I've tutored students who've taken a long transoceanic flight, from California to Seoul, South Korea. Along the way, the plane ends up flying very close to Alaska. The question is, why does it fly so close to Alaska, rather than take a more sensible route closer to, say, Hawaii? The answer is that the flight near Alaska is actually shorter -- the flight follows a great circle, and the great circle through California and Korea passes near Alaska.

One final related question -- any two great circles meet at two antipodal points. Where exactly is the point on the globe that is antipodal to where we are standing now? Despite all the talk about "digging a hole to China," that country is not antipodal to the United States. As it turns out, most of the Lower 48 United States are not antipodal to land at all. If one dug a straight hole through the center of the earth starting anywhere in California, we'd end up in the Indian Ocean. But Hawaii is antipodal to parts of Botswana and Namibia in Africa, and of course Alaska is antipodal to Antarctica.

Here is a link to a map that calculates antipodes:

Returning to Euclidean geometry, here's the proof of the Triangle-Sum Theorem given in the U of Chicago text. Since the book gives a two-column proof, I'll convert it to a paragraph proof:

Triangle-Sum Theorem:
The sum of the measures of the angles of a triangle is 180 degrees.

Given: Triangle ABC
Prove: angle A + angle B + angle C = 180

Draw line BD with the measure of angle 1 (ABD) equal to angle A. By the Alternate Interior Angles Test, lines BD and AC are parallel. Then angle 3 (the angle on the other side of BC -- the text doesn't name it, but we can call it CBE if E is a point such that BE and BD are opposite rays) has the same measure as angle C, by the Alternate Interior Angles Consequence. By the Angle Addition Postulate, angles 1, 2 (ABC), and 3 add up to 180 degrees. Substituting, we get that angles AABC, and C add up to 180 degrees. QED

Right now, I am a substitute teacher, but recently I interviewed for a position as a regular teacher, and one of the things I was asked to prove was the Triangle-Sum Theorem. (I also had to derive the Quadratic Formula.) I gave a two-column proof similar to the one given in the text, and the principal told me that it was satisfactory, but that he might have preferred something like this:

Statements                                              Reasons
1. Draw line BD parallel to line AC        1. Uniqueness of Parallels (Playfair)
2. angle 1 = angle A, angle 3 = angle C    2. Alternate Interior Angles Consequence
3. angle 2 = angle ABC                            3. Reflexive Property of Equality
4. angle 1 + angle 2 + angle 3 = 180        4. Angle Addition Postulate
5. angle A + angle ABC + angle C = 180  5. Substitution (steps 2 and 3 into step 4)

So we include step 3, to show students that we are making three substitutions. Calling the same angle by two different names -- angle 2 and angle ABC -- emphasizes the need for a Reflexive Property to show that this angle equals itself. The U of Chicago just changes angle 2 to angle ABC without any explanation whatsoever. On the other hand, the U of Chicago distinguishes between the Angle Addition Postulate and the Linear Pair Theorem (which is the just the Angle Addition Postulate in the case that the angles add up to 180). The hope is that this form of the proof is the best for students to understand, which is our goal.

The Quadrilateral- and Polygon-Sum Theorems are just corollaries of the Triangle-Sum Theorem, as we expect. As I mentioned earlier, calculating (n - 2)180 (and dividing by n to find each angle of a regular polygon) is the most complicated algebra that I want students to have to do in first semester of the geometry class.

Tuesday, October 28, 2014

Section 5-6 (and other sections): Properties of Trapezoids (Day 53)

Once again, we are jumping around the text. The students will definitely need more practice on the Parallel Consequences from Section 5-6, plus there are theorems on isosceles trapezoids left in Section 5-5 that we need to cover.

The first theorem that we have is the Isosceles Trapezoid Symmetry Theorem:
The perpendicular bisector of one base of an isosceles trapezoid is the perpendicular bisector of the other base and is a symmetry line for the trapezoid.

Notice that this theorem is similar to the Kite Symmetry Theorem. In many ways, there is a sort of dualism between the kite and the isosceles trapezoid. A kite is defined by having consecutive equal sides, while an isosceles trapezoid has consecutive equal angles. The symmetry line for the kite bisects two of the angles, and the symmetry line for the isosceles trapezoid bisects two of the sides.

In fact, this dualism in some ways justifies our definition of isosceles trapezoid. Defining an isosceles trapezoid to be a trapezoid having a pair of equal opposite sides -- which works for the exclusive definition of trapezoid but not the inclusive definition -- would be just like defining a kite in terms of having a pair of equal opposite angles.

Here is the proof of the Isosceles Trapezoid Symmetry Theorem as given by the U of Chicago. This time, since Section 5-5 gives a paragraph proof, let me post a two-column proof here on the blog.

Given: ZOID is an isosceles trapezoid with angles I and D equal in measure.
           m is the perpendicular bisector of ID
Prove: m is the perpendicular bisector of ZO
           m is a symmetry line for ZOID

Statements                           Reasons
1. m is the perp. bis. of ID  1. Given
2. D' = I, I' = D                    2. Definition of reflection
3. angle I = angle D               3. Given
4. ray DZ reflected is IO      4. Reflections preserve angle measure.
5. Z' lies on ray IO               5. Figure Reflection Theorem
6. ZOID is a trapezoid         6. Given
7. ZO | | DI                           7. Definition of trapezoid
8. ZO perpendicular to m     8. Fifth Postulate
9. Z' lies on line ZO              9. Definition of reflection
10. Z' = O                             10. Line Intersection Theorem
11. O' = Z                             11. Flip-Flop Theorem
12. ZOID reflected is OZDI 12. Figure Reflection Theorem
13. m symm. line of ZOID   13. Definition of symmetry line
14. m is the perp. bis. of OZ 14. Definition of reflection

This proof is quite long and can be intimidating for students. A teacher can break it down by actually folding the isosceles trapezoid along line m. The teacher can ask, "Where does D fold to?" The students will probably answer I, only to have the teacher ask "Why?"

Then the tougher part is to show that Z folds to O. We do it one step at a time -- first we show that since reflections preserve angle measure, Z' must be somewhere on ray OI -- although not necessarily O (but of course students will want to jump to that conclusion). Notice that we then have to the Fifth Postulate to show that side ZO is also perpendicular to m, so that Z' also lies on ZO -- and this is why I stated our Fifth Postulate yesterday. We didn't use Playfair or any other Parallel Postulate, but we used the Perpendicular to Parallels Postulate directly. This is one reason why I let Perpendicular to Parallels be our Fifth Postulate.

Now we have another theorem, the Isosceles Trapezoid Theorem:
In an isosceles trapezoid, the non-base sides are equal in measure.

In other words, our definition of isosceles trapezoid implies what we usually think of when we hear the word isosceles. The text states that this is merely a corollary of the previous theorem -- since the reflection of one of the non-base sides is the other.

For readers of this blog, let's go back to what I wrote about non-Euclidean geometry. I stated that in hyperbolic geometry, there's something called a Saccheri quadrilateral -- defined as a quadrilateral with the base angles both right angles and the non-base sides equal. As it turns out, a Saccheri quadrilateral is also an (inclusive) isosceles trapezoid and it also has a line of symmetry. Cutting the Saccheri quadrilateral along this line of symmetry gives two new figures called Lambert quadrilaterals. But the above Isosceles Trapezoid Symmetry Theorem, as stated above, can't be used to prove that a Saccheri quadrilateral is symmetrical, because its proof uses the Fifth Postulate.

Instead, we can prove that a Saccheri quadrilateral is symmetrical, as follows: the first five steps can remain as they are. But we can easily get that Z' is O, because the fact that DZ = IO is now part of the given information (rather than proved as a corollary later). So Z' is the point on ray IO that is the correct distance from I -- and that point is exactly O. Then we pick up the proof from step 11 of the proof above. (Now we have a new corollary -- namely that ZO and ID are parallel. This follows from the fact that m is the perpendicular bisector of both ZO and ID, and so by the Two Perpendiculars Theorem -- which still holds in hyperbolic geometry -- these two sides are parallel.)

Returning as a text, another corollary given is the Rectangle Symmetry Theorem:
Every rectangle has two symmetry lines, the perpendicular bisectors of its sides.

This corollary follows from the Isosceles Trapezoid Symmetry Theorem in the same way that the Rhombus Symmetry Theorem (which shows that a rhombus also has two symmetry lines) follows from the Kite Symmetry Theorem. It shows again the beauty of using inclusive definitions.

So far in this section, we've mentioned the symmetries of the kite, rhombus, isosceles trapezoid, as well as the rectangle. A general trapezoid, as we've mentioned before, has no symmetry. But there's one special quadrilateral that's missing -- the parallelogram. So does a parallelogram have symmetry?

Also, so far in this section we've mentioned reflections and reflection symmetry, but there's another transformation that we've learned about -- rotations and rotational symmetry. So do any of our special quadrilaterals have rotational symmetry.

We can answer both of these questions at once: A parallelogram has rotational symmetry!

Notice that the U of Chicago text doesn't discuss rotational symmetry of any figure, much less that of the parallelogram. Yet we see in the Common Core Standards:

Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.
(emphasis mine)

And so we should discuss rotational symmetry in general and that of the parallelogram in particular.

But let's begin with the rhombus and rectangle before going on to the general parallelogram. We recall that our definition of rotation is the composite of reflections in intersecting lines -- and if those lines are symmetry lines of a figure, each one maps the figure to itself. Therefore, the entire rotation must map the figure to itself. It follows that any figure that contains two lines of symmetry must automatically have rotational symmetry as well! And the Two Reflection Theorem for Rotations tells us exactly what the center and magnitude of the rotation are. The center is where the two symmetry lines intersect, and the magnitude is double the angle between the intersecting lines.

Corollary to Two Reflection Theorem for Rotations:
If l and m are intersecting symmetry lines of a figure, then it also has rotational symmetry, where the magnitude is twice the non-obtuse angle between l and m and the center is the point of intersection of l and m.

Applying this to the rhombus, we see that both its diagonals are symmetry lines. Because a rhombus is a kite, its diagonals are perpendicular. So the magnitude of the rotation is twice 90, or 180 degrees, about the point where the diagonals intersect.

And now let's look at the rectangle. Its symmetry lines are the perpendicular bisectors of its sides, and we can use the Two Perpendiculars Theorem and the Fifth Postulate to show that these symmetry lines are also perpendicular. So the magnitude of the rotation is twice 90, or 180 degrees, about the point where the diagonals intersect.

But as it turns out, all parallelograms -- not just the rhombuses and rectangles -- can be shown to have 180-degree rotational symmetry. The U of Chicago doesn't prove this, but Hung-Hsi Wu does. In his Theorem 4, he proves that the opposite sides of a parallelogram are equal, by showing that the parallelogram has rotational symmetry.

Here is Wu's proof of his Theorem 4, in paragraph form (as Wu himself gives it):

Given: ABCD is a parallelogram
Prove: AD = BC

Let M be the midpoint of the diagonal AC and we will use Theorem 1 -- Wu's First Theorem (that a 180-degree rotation maps a line to a parallel line, already proved on this blog) to explore the implications of the 180-degree rotation R around M.

Because MA = MC and rotations preserve distance, we have C" = A, so that R maps line BC to a line passing through A and parallel to line BC. Since the line AD has exactly the same two properties by assumption, Playfair implies that R maps line BC to line AD. Similarly, R maps line AB to CD. Thus since B lies on both AD and CD, its image B" is a point that lies on both AD and CD. By the Line Intersection Theorem, this point is exactly D.

Recall we also have C" = A. Therefore R maps the segment BC to the segment joining D (which is B") to A (which is C") by the property that a rotation (like a reflection) maps segments to segments. The latter segment has to be the segment DA, by the Unique Line Assumption. Thus by the Figure Reflection Theorem, R maps ABCD to CDAB, so ABCD has rotational symmetry. Since rotations preserve distance, BC = AD, as desired. QED

Notice that Playfair is used in the proof. A Saccheri quadrilateral, despite being a parallelogram, doesn't have rotational symmetry.

This proof is very similar to the Isosceles Trapezoid Symmetry Theorem. Both theorems depend on what Wu calls "Lemma 6" -- identifying the reflection or rotation image of a point by finding two lines that intersect at the preimage point and noting that the images of these two lines must intersect at the image of the point.

Like the Isosceles Trapezoid Symmetry Theorem and its corollary the Isosceles Trapezoid Theorem, we should call part of the proof the Parallelogram (Rotational) Symmetry Theorem and then derive that opposite sites are equal as a corollary. Wu also derives as a corollary that opposite angles are equal -- but we can also derive this by applying the Trapezoid Angle Theorem twice -- since a parallelogram is a trapezoid. Even the third major property of parallelograms -- that their diagonals bisect each other -- can be derived as a corollary (since rotations preserve distance, we must also have MB = MD in the above proof).

So we can prove all the major properties of parallelograms using only rotations and not using triangle congruence at all. The U of Chicago derives the properties of other figures using reflection symmetry but reverts to triangle congruence for the parallelogram properties (in Chapter 7).

Finally, notice that since rhombuses and rectangles are parallelograms, we can show that they have 180-degree rotational symmetry without looking at their lines of symmetry. But I still like using the Corollary to the Two Reflection Theorem for Rotations because it can be used to determine the rotational symmetry for figures other than quadrilaterals. We earlier proved that an equilateral triangle has three lines of symmetry -- so it also has rotational symmetry.

Tomorrow I will be tutoring my geometry student, so I'll have another chance to test out some of my geometry material posted to this blog.

Monday, October 27, 2014

Section 5-5 (and other sections): The Fifth Postulate (Day 52)

Today we finally reach a Parallel Postulate -- and this is what makes geometry Euclidean. Here I make some key decisions regarding the postulate as well as cover many key theorems that depend on the postulate for their proofs.

Let's start with the postulate itself. Earlier, I stated that the most convenient postulate for proofs is the Perpendicular to Parallels Postulate (a theorem in the U of Chicago and most other texts). I've decided that I will give this postulate a new name:

The Fifth Postulate:
In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.

The name "Fifth Postulate" has several levels of significance. First of all, it obviously refers to Euclid's Fifth Postulate, and we are using this postulate just as Euclid used his -- namely, as the main postulate for parallels. As I mentioned earlier, both my Fifth Postulate and Euclid's actually refer to two right angles. By calling my postulate the "Fifth Postulate," I can still use the phrase "the proof of this theorem depends on the Fifth Postulate" and any reader will know that the theorem is provable in Euclidean geometry as opposed to non-Euclidean geometry.

Here's a link to the page on David Joyce's website regarding Euclid's Fifth Postulate:

Furthermore, one significance of this postulate is that it can be directly applied in a few other theorems, including our translation theorems (for Chapter 6) and the concurrency theorems. These proofs involve a line that are perpendicular to one of two parallel lines and showing that it is perpendicular to the other. So our Fifth Postulate is convenient for these proofs.

Finally, notice that this is actually the fifth named postulate on this blog. (The other four postulates are the Point-Line-Plane Postulate, the Angle Measure Postulate, the Reflection Postulate, and the Plane Separation Postulate.) Technically speaking, we've used more than five postulates as the properties of arithmetic are actually postulates -- but I think of these as being more like Euclid's "Common Notions" than his postulates. Notice that there is more connection to the number five -- we are now in Chapter 5, and in that chapter, the first section requiring a Parallel Postulate is 5-5. I was actually considering waiting until Day 55 before posting this postulate, but ran out of time (due to last week being skipped), but at least I posted this in the 50's.

Indeed, my skipping last week has major consequences for the pacing. I want to be out of Chapter 5 by the end of October, as I want to devote November and December to Chapters 6 and 7 -- these cover translations and triangle congruence which are vital for Common Core. Originally, I wanted to slow down to prove the many consequences of our Fifth Postulate last week and this week, but instead we must finish Chapter 5 by this week. Then again, my pacing rush resembles the problems faced in many actual classrooms -- we must rush Chapter 5 if we want to finish Chapter 7 by the end of the semester.

And so today's lesson will cover more topics than I originally wanted to in one lesson. Notice that even though this lesson is numbered 5-5, much of the material comes from other sections.

First, we stated the Perpendicular to Parallels Theorem, found in Section 3-5 of the U of Chicago text, as our Fifth Postulate.

Second, we prove the Uniqueness of Parallels Theorem, found in Section 13-6. Notice that this is actually Playfair's Parallel Postulate -- John Playfair being a 19th century Scottish mathematician. It's best for us, just as the U of Chicago does, to call it the Uniqueness of Parallels Theorem, since both of us actually prove it (so it's not a postulate). But sometimes on this blog, I just call this statement "Playfair" for short.

Our proof is similar to the U of Chicago's. The first difference is that since our Fifth Postulate requires perpendicular lines, we just draw the blue line in the picture so that it's perpendicular to line l -- that is, make angle 1 90 degrees. Then by the Fifth Postulate, angles 2 and 3 are also 90. The other difference is that we want to avoid indirect proof. Notice that step 3 in the text concludes that if x and y are both lines through P parallel to l, then they are identical -- so there is really only one line through P parallel to l. This isn't actually an indirect proof at all -- many proofs in higher math do this. To show that a number x with a certain property is unique, we simply let x and y be two numbers with that property and then prove that x = y.

The resulting proof looks something like this:

Given: x contains P, y contains P, x || l, y || l
Prove: x and y are identical

Statements                                        Reasons
1. x, y contain P                                 1. Given
2. Let b be line through P perp. to l  2. Uniqueness of Perpendiculars Theorem
3. b perp. to x, b perp. to y               3. Fifth Postulate
4. intersects x, y at 90                     4. Definition of perpendicular
5. x intersects y at 0 degrees              5. Angle Addition Property
6. x and y are identical                       6. Zero Angle Property

Another, more interesting proof of Playfair depends on a theorem given in Section 3-4:

Transitivity of Parallelism Theorem:
In a plane, if l || m and m || n, then l || n.

When we covered Lesson 3-4, we skipped this theorem because it was proved using slope, and we don't want to discuss slope until second semester. But we can use our Fifth Postulate to prove the Transitivity of Parallelism Theorem, as follows:

Draw any line b perpendicular to m. Then by our Fifth Postulate, b must be perpendicular to both l and n as well. Therefore by the Two Perpendiculars Theorem, l and n are parallel. QED

And now notice that Playfair immediately follows from this. To show that there can be only one line parallel to m through a point P, we let l and n be two lines parallel to m through a point P. By the Transitivity of Parallelism Theorem, l and n are parallel to each other -- yet the lines both contain the same point P. Then, by our definition of parallel, the only way for two parallel lines to have a point in common is for them to be identical, so l and n are the same line. I like this proof better.

Once we have Playfair, we can now derive the Parallel Consequences -- that is, statements of the form "if two parallel lines are cut by a transversal, then ...." Now I was considering deriving Corresponding Angles Consequence the same way that Dr. Franklin Mason does this -- but unfortunately the proof that he gives is indirect. So again we must go back to the trick used for Lesson 13-6 earlier:

Given: l || m with transversal t, intersecting m and P
Prove: Corresponding angles 1 and 2 are equal in measure.

Statements                     Reasons
1. l || m                            1. Given
2. Let n intersect t at P,  2. Angle Measure Postulate
at corr. angle 3 = angle 1
3. l || n                             3. Corresponding Angles Test
4. m and n are identical,  4. Uniqueness of Parallels Test (Playfair)
so angle 2 = angle 3
5. angle 1 = angle 2         5. Transitive Property of Equality

This may be easier for students to understand if we think of angle 1 having a specific measure, something like x. We want to prove that angle 2 also has measure x. To prove it, we let n be the line through P so that the angle between n and t already has the measure we want -- x. Then by the Corresponding Angles Test, n is a line through P parallel to l. But we already have a line through P parallel to l -- namely m. So by Playfair n is just another name for m. And since n intersects t at x degrees and n really means m, m intersects t at x degrees, which is what we want to prove.

As I mentioned before, this is a trick for proving converses. The Corresponding Angles Consequence is just the converse of the Corresponding Angles Test. The trick to proving converses is to combine the forward statement with a uniqueness criterion. And this is exactly what we did here -- we proved the Corresponding Angles Consequence by combining the Corresponding Angles Test with the Uniqueness of Parallels Theorem (Playfair).

The remaining statements are proved quickly. We prove the Alternate Interior Angles Consequence using the U of Chicago proof from Section 5-6 -- I included this in the exercises.

Then I prove the Same-Side Interior Angles Consequence -- this isn't proved in the U of Chicago text per se, except as part of the Trapezoid Angle Theorem in Section 5-5. I decided to include both this theorem and the theorem about both pairs of base angles of an isosceles trapezoid being equal. But the next theorem, on the symmetry of an isosceles trapezoid, is saved for the next section.

The proofs in this theorem occur in Euclid: Proposition 29 being the Parallel Consequences and Proposition 30 being the Transitivity of Parallels Theorem. These are the first two theorems of Euclid requiring his Fifth Postulate.

With this new image, the worksheet I posted back in July where the Perpendicular to Parallels is given as a theorem is now obsolete.

Friday, October 24, 2014

The Common Core Debate, Grade 3 Continued: Proposed Standards

Today was my second day as a substitute teacher. I ended up subbing for an English teacher at a continuation school. But I do have something to say that's related to math. For the classes were held in the library, and while the students were watching a video, I found on the shelf an old favorite book on traditionalist math from the 1980's -- Quick Arithmetic (2nd ed.) by Carman and Carman. I always get a laugh at the Peanuts and other comics interspersed throughout the text.

Now that I've stated that I oppose the Common Core Standards for grades K-3, the next question is, what would I replace them with? For this, let's focus on the third grade, since if we can fix the third grade -- and that's both the standards and the standardized tests like PARCC/SBAC based on them -- then that will automatically fix all the grades below that grade.

The goal is to make the standards more traditionalist in the lower grades. Now, if we were to ask any traditionalist what a third grader should learn in math, the answer is obvious -- the multiplication table.

But hold on a minute. The Common Core already seems to have a standard for third grade multiplication. Here is the relevant standard:

Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.

So why would a traditionalist object to this standard? Well, one problem is that the third grade standards fall into five domains -- Operations & Algebraic Thinking, Number & Operations in Base Ten, Number & Operations -- Fractions, Measurement & Data, and Geometry. And this above standard is just the seventh standard within one of the domains -- that is, it's relatively buried within.

But a traditionalist whose third grade child mastered four of the five domains, yet didn't know the multiplication tables, would consider the year a failure and seek to blame the teacher, the school, and the Common Core Standards. By contrast, a traditionalist whose third grader knew very little about four of the five domains, yet can correctly answer what six times seven is and all the other single-digit multiplication problems in one second or less, would consider it a success. And a traditionalist who decides to homeschool a third grader in order to avoid the Common Core would begin by teaching that child the multiplication tables.

And so our first step is to give this standard priority over all other standards:

New Third Grade Standard #1:
By the end of Grade 3, know from memory all products of two one-digit numbers.

Notice that I dropped the first part about division. Division is an admirable goal, but until the multiplication table has been memorized, there can be no division.

But it's not enough just to write this standard. In order to emphasize its importance, we must make sure that it has a prominent place in on the PARCC/SBAC.

Recall that both PARCC and Smarter Balanced are computer-based exams. So there should be no problem with imposing a time limit for each question. So we can have the PARCC and SBAC begin by asking the student to answer a simple multiplication problem, such as "6 x 7 =," and imposing a time limit.

How much time should we give the students? We want to make sure that it's impossible to get the right answer without having memorized the table. Earlier, I stated that the student should be able to answer 42 in one second or less. But I wouldn't want to give a one-second time limit -- I myself had already memorized the entire multiplication table by the end of kindergarten, yet I'd be so nervous with such a short time limit that I'd press the wrong keys on the keyboard. So let's give a more reasonable time limit, say ten seconds.

The first 100 problems on the PARCC/SBAC exams can be the 100 single-digit multiplication facts. This will take no more than 1000 seconds, or almost 17 minutes, to complete, leaving time for the other domains.

And now here's where things really get interesting. If the student doesn't earn a satisfactory score on these 100 questions -- if we want to be tough, we can set 90 to be the passing score -- then the test immediately ends and a failing score is reported for the student, without any question from any other domain being asked of the student.

This would immediately drive home the importance of memorizing the times tables to schools that would be judged by their PARCC/SBAC scores. The way that I have proposed structuring the tests, nothing else at all matters in third grade math more than memorizing the multiplication tables. Therefore, third grade teachers should spend however many weeks it takes for the students to know their times tables -- whether it be a trimester, a semester, or even the entire year up to the window for the tests.

Now let's move on to our next traditionalist standard:

New Third Grade Standard #2:
Fluently add and subtract multi-digit whole numbers using the standard algorithm.

This standard is certainly traditionalist, since it emphasizes use of the standard algorithm for addition and subtraction. The problem is that it doesn't appear in the current Common Core third grade standards -- instead, it appears in the fourth grade standards. There is a gap between when the students learn how to add and subtract and when they are to learn the standard algorithm -- and many schools fill in this gap with progressive algorithms instead. We fill in the gap by dropping this fourth grade standard to third grade to avoid this temptation -- and even that may be too late, since nonstandard algorithms might be taught even in second grade. But for now I keep the standard under third grade, since that is the year that I'm emphasizing right now.

Notice that there's nothing wrong with nonstandard algorithms per se -- after all, even Carman's text, the one I mentioned at the start of this post, mentions nonstandard algorithms in order to motivate the use of the standard algorithm as a kind of shortcut. The problem is when the standard algorithm doesn't appear until a year or two later, as in the current Common Core standards.

Here is the nonstandard algorithm mentioned by Carman:

+  95

That is, it adds up one column at a time. But as I mentioned earlier, the standard algorithm is introduced quickly after this nonstandard algorithm is given.

But a progressive algorithm might look like this: to add 132 + 95, we perform an easier addition, 132 + 100, to obtain 232. Then, since we added 5 too much, we must subtract that 5 to obtain 227 as the answer. I admit that often times, I perform mental math in my head using this method, often called the "Plus-Minus Method", but the problem is that the Plus-Minus method is too confusing for a student in the third grade to perform and understand. This is why the emphasis should be on the standard algorithm.

I see nothing wrong with the rest of the third grade standards in Common Core except that they take time away from the most important two. I decided to list only the standards that are the most important for students to learn. The first two standards listed above should form the bulk of the third-grade math course, and so all current standards that take time away from these should be dropped.

Here are the remaining standards that I've deemed the most important:

New Third Grade Standard #3:
Use place value understanding to round whole numbers to the nearest 10 or 100.

Traditionalists have no problems with students learning to round -- the problem is when students only add or subtract the rounded numbers and never learn how to add the unrounded numbers.

New Third Grade Standard #4:
Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b.

New Third Grade Standard #5:
Represent a fraction 1/b on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. Recognize that each part has size 1/b and that the endpoint of the part based at 0 locates the number 1/b on the number line.

New Third Grade Standard #6:
Represent a fraction a/b on a number line diagram by marking off a lengths 1/b from 0. Recognize that the resulting interval has size a/b and that its endpoint locates the number a/b on the number line.

New Third Grade Standard #7:
Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.

Notice that the current third grade standards include fractions. Using the number line to motivate fractions is a strong traditionalist method. I specifically mention Standard #7 in reaction to a story I once read where McDonalds customers didn't know that a Third Pounder would be larger than a Quarter Pounder -- because they couldn't compare two fractions with the same numerator, 1/3 and 1/4.

New Third Grade Standard #8:
Partition shapes into parts with equal areas. Express the area of each part as a unit fraction of the whole. For example, partition a shape into 4 parts with equal area, and describe the area of each part as 1/4 of the area of the shape.

The other main traditionalist method of teaching fractions is by dividing areas, so this standards follows naturally from the fraction standards. As much as I like geometry, the other current third grade standard on quadrilaterals is considered too distracting to be included in a traditionalist curriculum.

New Third Grade Standard #9:
Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step "how many more" and "how many less" problems using information presented in scaled bar graphs. For example, draw a bar graph in which each square in the bar graph might represent 5 pets.

New Third Grade Standard #10:
Solve real world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters.

These standards follow naturally from the others.

My goal is to change attitudes about math. Currently, people who are great at math, say algebra and above, are derided as nerds. Instead, I propose a word to describe those who aren't proficient at basic math, such as the math included in the above standards. Until I come up with a better word, I will spell the word nerd backwards, and call such an adult a dren.

Thus concludes this post. My next post will be on Monday, when I will proceed with Day 52 of the geometry course.

Wednesday, October 22, 2014

The Common Core Debate, Grades K-3

Today was my first day as a substitute teacher. As it turned out, I didn’t sub today for any math class, much less geometry. Instead, today the class that I covered was a science class, AP Physics 1 -- but this class is closely related to algebra. The students took a test where they had to graph position and velocity vs. time, using y = mx + b.

But I also tutor students on the side, and one student I tutor is just starting a geometry class. And so I showed him pages posted on this blog -- Lessons 1-6 and 1-7, plus the Opening Activity on Euler’s bridges, and he succeeded on these first assignments -- and he quickly thought to add an extra bridge to make the impossible network into a traversable one. He also enjoyed the story of why students should learn math.

As promised, today I discuss the pros and cons of the Common Core Standards in Math, from kindergarten through grade three. The plan is to address many concerns that teachers and parents have regarding the standards, and suggest how I would change the standards in order to solve the problems caused by the current Common Core. I will only discuss Common Core on its mathematical merits. In particular, I will not discuss:

Constitutionality: Many people believe that anything other than 50 different sets of standards, one for each state, is unconstitutional and violates the Tenth Amendment. One good thing about national standards is that they make it easy for a family that does not want to change standards be able to avoid changing standards (that is, a family forced, because of a job, to move to another state). On the other hand, one bad thing about national standards is that they make it hard for a family that wants to change standards to be able to change standards.

Politics: As I mentioned earlier, much of the opposition to Common Core is political. Many of those who oppose the standards are Republicans, as it was a Democratic administration that proposed them. 
The ELA standards: The ELA standards for Common Core have their own pros and cons. One major point of contention is their emphasis on informational nonfiction.

AP US History and other subjects: Earlier on this blog, I mentioned that some people are mentioning changes to the way AP US History is taught in their arguments against Common Core. And recently, I found a counterargument to my statement that the Common Core Standards have nothing to do with US History -- someone claimed that the Common Core and the AP US History standards have a common author. But of course, this all goes back to politics again -- the claim is that because of Common Core, AP US History will now be taught with a liberal slant. 

The Next Generation Science Standards: Once again, my focus will be on math, not science -- but I may mention science only to the extent that math and science are closely related. After all, I just mentioned that much of what one learns in Algebra I shows up again in AP Physics 1.

And now enough about what I won’t discuss, and on to what I will discuss. Many Common Core horror stories begin with a photo of a math worksheet that a young student brings home. As I mentioned earlier, I was drawn into a debate on Facebook, where the following photo was posted:

Now this second grade worksheet begins by noting that the whole number 227 stands for 2 hundreds, 2 tens, and 7 ones. But the problem is that later on, it states that 227 is 1 hundred, 12 tens, and 7 ones. So, the second-grader and her parent are left to wonder, what’s going on here?

We begin by asking, what should a second-grade worksheet look like? Well, one thing we might expect to see is a subtraction problem -- for example, what is 227 - 132? And we begin by subtracting 2 from 7 in the ones place, leaving 5. Now we move to the tens place, but we can't take 3 away from 2. So we borrow 1 from the hundreds place and make the 2 a 12, so now we can subtract 3 from 12 to give us 9. Now look at the top -- after borrowing, the top row now reads 1, 12, 7 -- 1 hundred, 12 tens, and 7 ones!

So now we see the problem -- the student and her mother alike were confused when they saw 1, 12, 7 on this worksheet, but I bet there would have been no problem if the paper had simply asked for the answer to 227 - 132. So why wasn’t this question on the worksheet instead?

The so-called Math Wars have been fought for much of the 20th century, and they have spilled into the 21st with the advent of Common Core. There are two main schools of thought -- the traditionalists and the progressives. Here are some of the differences between these two philosophies:

A progressive teacher is often described as “a guide on the side.”
A traditionalist teacher is often described as “the sage on the stage.”
A progressive teacher abhors worksheets full of math facts as “drill and kill.”
A traditionalist teacher adores worksheets full of math facts as “drill and skill.”
A progressive teacher focuses on the end result -- applying math to the real world.
A traditionalist teacher focuses on the beginning -- mastering basic skills.

Around the year 2000, there were two websites that corresponded to these philosophies -- Mathematically Correct took the traditionalist side of the Math Wars, and Mathematically Sane took the progressive side.

Now undoubtedly, Common Core leans toward progressivism. Some Common Core defenders point out that nowhere in the standards is progressivism recommended. But even if the Common Core Standards don’t themselves recommend progressivism, many progressive teaching methods are being implemented in the name of Common Core. For example, if the mother whose daughter brought home the above worksheet were to ask why so many strange questions were being asked, the teacher would certainly reply that it was because of Common Core.

Second, it’s not as much what the standards say, but what material appears in the PARCC and SBAC assessments. If the standards mention traditional math but the test questions are progressive, then progressive math is what will be taught.

Finally, although the Mathematically Correct website no longer exists (and disappeared before the Common Core debate), the Mathematically Sane site still does (though it’s rarely update). About Common Core, Mathematically Sane writes:

“The release of the Common Core State Standards (CCSS) is a welcome milestone in the standards movement that began more than 20 years ago when the National Council of Teachers of Mathematics published Curriculum and Evaluation Standards for School Mathematics.”

And so this clinches it -- a known progressive website calls the Common Core a “welcome milestone.” Even if Common Core doesn’t directly promote progressivism, it is the de facto cause of the sharp increase in progressivism in the classroom and progressive worksheets for the students to take home.

Now I am sympathetic to both the traditionalist and progressive philosophies -- both of them have a purpose. In particular, many of the arguments for progressivism apply more to older students. For example, oppositional teenagers may resent adult authority. They may not want a “sage on the stage” forcing them to do “drill and kill” worksheets. They may ask, “When are we ever going to do this in real life?” -- implying that they will do no more than the bare minimum of math required to survive in the real world. Unless it is obvious from the beginning the applications of the math they are learning, they won’t even begin to do the work at all. And this is where progressivism shines -- it focuses on the real-world applications of the math that the students are supposed to learn.

But that’s just it -- this argument focuses on oppositional teenagers. Younger students, on the other hand, are often eager to learn. They want to please the adults in their lives and prove to them how smart they are. And one place where little children can prove how smart they are -- to demonstrate their skills -- is on “drill and skill” worksheets! And so we actually want “drill and skill” for these students.

And so I prefer traditionalism in the lower grades. But now I must define what exactly the “lower grades” are. Certainly, the grades that are too young to take the PARCC or SBAC assessments are too young for progressivism. If we make the third grade -- the youngest PARCC/SBAC year -- traditionalist, then that would automatically make all grades younger than that grade traditionalist as well, since the primary teachers would have to prepare the students for the traditionalist third grade.

I’ve mentioned in passing that my opinion of Common Core is mixed. For the youngest students, my opinion isn’t so mixed -- but instead is crystal clear. The official position of this blog is to support traditionalism from kindergarten to third grade. And so, I am opposed to the Common Core Standards in math for grades K-3. 

Friday, October 17, 2014

Activity: Kites (Day 51)

Go fly a kite! Well, this short activity won't have the students actually fly a kite -- but perhaps the students can at least construct one using straightedge and compass.

What's the easiest way to construct a kite? Let's consider the properties of a kite -- in particular, the Kite Diagonal Theorem again:

The symmetry diagonal of a kite is the perpendicular bisector of the other diagonal and bisects the two angles at the ends of the kite.

Aha, there's the perpendicular bisector! So to construct a kite, we can let any segment be one of the diagonals, and then we construct its perpendicular bisector to find the other (symmetry) diagonal. So after that we join the vertices of these two segments to form the sides of the kite.

If we were to do this using the U of Chicago's perpendicular bisector construction, the two diagonals end up bisecting each other and the kite becomes a rhombus. If one wants a non-rhombus kite, we can use the straightedge to extend the symmetry diagonal in one direction before joining the vertices.

In a classroom without compasses, we can use paper folding instead. We begin by drawing any triangle, and then reflect it with one of its sides as the mirror. The result will be a kite. Notice that in a  pre-Common Core class, we use SSS to prove that the two halves of the kite are congruent -- that is, we use SSS to prove the kite properties. But in the U of Chicago's Chapter 7, we end up using the kite properties to prove SSS! There's that circularity again, that we must be careful to avoid.

This activity does not require me to scan and post an image, and so I won't.

So now I go into a week off from this blog, in order to align the school calendar on this blog with the Early Start calendar at the school where I will become a sub here in Southern California. (Note: With all the talk about the leadership at LAUSD in the news, let me state that the district where I'll be a sub is not LAUSD. Also, I'm considering subbing in a second school district with a traditional Labor Day Start calendar, and so further adjustments may need to be made to may day counts.)

See you on Day 52, Monday, October 27th!

Thursday, October 16, 2014

Who Am I? And Section 5-4: Properties of Kites (Day 50)

In yesterday's post, I began revealing some personal information about myself. I mentioned that I am 33 years old, and my weight is -- did I actually state my weight? Well, recently I went to the doctor who weighed me, and I measured between 190 and 200 pounds. (In honor of last week's Metric Week, let me also give my mass as between 85 and 90 kilograms.)

But as much as the students are fascinated by this information, readers of this blog aren't? Instead, you readers want to know such information as, am I a teacher? Where and whom do I teach?

Well, let me give answer that question. Today I was hired to work as a substitute teacher right here in Southern California. That's right -- for the first fifty school days of this blog's existence, I wasn't working in a school at all. Much of what I posted here is based on my knowledge of the Common Core Geometry Standards, as well as what I observed in the classroom in the past. I have completed my California Clear Credential in Single Subject math, but times are tough for newly credentialed teachers. I'm hoping to sub this year and try again to be hired as a full teacher in time for next year.

Now I know that math teachers who read blogs aren't interested in lesson plans and worksheets -- they can get those out of textbooks and other sources. No -- actually, readers of math teacher blogs want to see how teachers interact with students, and how well lessons go with students. But as long as I had no students, there was no way to gauge how well my lessons were written.

So now things are going to change. I'm hoping that I'll be able to sub for a geometry teacher, see how the students react to the new Common Core Geometry lessons, and report the results on this blog.

This means several noticeable changes on this blog. Recall that I've been numbering the school days -- today is Day 50. Technically, this was based on a hypothetical Early Start calendar in which I labeled the last day before winter break as Day 90 -- the end of the semester -- and counted backwards. But now that I'm working for an actual district, I will use the actual calendar for the district.

School in my district started about a week later than Day 1 on this blog. And so to make up for it, I will take an entire week off from posting. I already mentioned that today's lesson, Section 5-4 of the U of Chicago text, is a natural dividing point for the chapter (because I want to state a Parallel Postulate before covering Section 5-5). So I'll post Section 5-4 today, then a short activity tomorrow (Day 51), and then I won't post any school days the following week. This schedule makes out Day 52 to be Monday, October 27th, which aligns better with my district calendar. Counting backwards, the school year in my district began on Tuesday, August 12th, which I should count as Day 0 because only freshmen attended school that day. Day 1 for all students was the next day, which was Wednesday, August 13th. (Oh, and by the way, my district actually took Columbus Day off!)

Starting on Day 52, I will be posting in the late afternoon or early evening -- that is, after school. This way, if I sub for a geometry teacher, I can post on this blog how the students reacted. I plan on beginning with the Parallel Postulate, but that will change depending on what I teach that day. I don't necessarily like jumping around in the text, but it will be much valuable to you readers to let the actual classroom, not a textbook, determine the order of the lessons posted to this blog.

Finally, as a substitute teacher, I already know that students are generally less cooperative for subs than they are for full-time teachers. I've seen students who, the instant they see a sub in the class, give excuses such as "Nobody's going to do this assignment!" Notice it's not I won't do the work, but nobody will -- the implication being that for a sub to do anything other than give the students a free period is highly unreasonable.

To fight this, my strategy will be to give a lesson similar to yesterday's activity. The first question I ask the students to answer will be "What is my age?" This will immediately disarm the student who was planning on saying "Nobody's going to do this assignment!" -- especially when some groups are already shouting out guesses! Only after giving the first two questions will I ask any mathematical question -- even if the full-time teacher has left an assignment, I plan on asking my two questions first before anything from the assignment. Of course, I'll keep tally of the points. I want the students to be behind in points and eager to answer the next question -- and then they find out that they have to complete the teacher's assignment in order to earn the points! The winning students or groups will be the students whose names I leave the regular teacher as having performed and behaved well that day.

And now let's get on with today's lesson. Section 5-4 of the U of Chicago text covers kites. The kite is a relatively new quadrilateral classification. Not only did Euclid never define kite, but many texts made no mention of kites -- including my class geometry textbook from 20 years ago. Nowadays most texts define kite, but some include kites only in bonus questions, not in the main text.

Here's what John Conway wrote about the kite -- nearly 20 years ago, right around the time that I was taking my geometry class:

In fact it's not quite true, either, because "kite" is not
a very traditional name - it was obviously inserted because
this was a type of quadrilateral that SHOULD have received a
traditional name, but didn't, until recently.

Why do we include the kite - plainly because it represents
the one type of symmetry not otherwise mentioned. But this
reason suggests we should also EXCLUDE the non-isosceles

David Joyce, meanwhile, doesn't find kites to be necessary at all:

Too much is included in this chapter. The sections on rhombuses, trapezoids, and kites are not important and should be omitted.

Notice that both Conway and Joyce want to exclude trapezoids -- but Joyce is the only writer I know who wants to omit rhombuses. Of course, for Joyce, the emphasis should be on triangles and parallel lines, not quadrilaterals like rhombuses. Also, notice that kites are not specifically mentioned in the Common Core Geometry Standards. But I find that students can identify a kite more readily than a rhombus or trapezoid.

Now every kite contains two special vertices, known as its "ends." As defined by the U of Chicago:

"The common vertices of the equal sides of a kite are the ends of the kite."

Notice that Michael Serra doesn't define ends, but does give a name to the two angles whose vertices are the ends of the kite -- the vertex angles, in analogy with the vertex angle of an isosceles triangle. I see that from a proof standpoint, this makes sense, since the first thing that we do in the proof of our main theorem is divide the kite into two isosceles triangles.

And now here is our main theorem: the Kite Symmetry Theorem. As I mentioned back in the lesson on isosceles triangles, we use symmetry to determine the properties of kites. (A pre-Common Core proof might divide the kite into two triangles to be proved congruent by SSS.) I will post a proof here of the Kite Symmetry Theorem. It is taken directly from the U of Chicago, except that I, as always, add a Given step to the beginning of the proof. Since the U of Chicago's proof has eight steps, mine has nine:

Kite Symmetry Theorem:
The line containing the ends of a kite is a symmetry line for the kite.

Given: ABCD is a kite with ends B and D.
Prove: Line BD is a symmetry line for ABCD.

Statements                                           Reasons
1. ABCD is a kite with ends B and  1. Given
2. AB = BC, AD = DC                         2. Definition of ends of kite (meaning)
3. Tri. ABC and ADC are isosceles      3. Definition of isosceles triangle (sufficient)
4. Let m be the perp. bis. of AC          4. A segment has exactly one perp. bisector
5. A' = C, C' = A                                  5. Definition of reflection (sufficient)
6. m contains B and D                         6. The perp. bis. of the base = angle bis. of the vertex angle
                                                                 (so it contains the vertex)
7. B' = B, D' = D                                 7. Definition of reflection (sufficient)
8. ABCD reflected over m is CBAD     8. Figure Reflection Theorem
9. m (Line BD) is a symmetry line      9. Definition of symmetry line (sufficient)

Notice that more than half of the reasons in this proof are definitions. This underlines how important definitions are to the study of quadrilaterals.

As for the other theorems in this lesson, the Kite Diagonal Theorem follows directly from lines 4 and 6 of the above proof. It makes the symmetry diagonal the perpendicular bisector of the other diagonal and so the diagonals of a kite are perpendicular, and the symmetry diagonal bisects the other one.

Finally, we have the Rhombus Symmetry Theorem. It states that a rhombus has two symmetry lines, as both of the lines containing its diagonals are symmetry lines. This follows directly from the classification of a rhombus as a kite. In texts that define kite exclusively, the theorems "The diagonals of a kite are perpendicular" and "the diagonals of a rhombus are perpendicular" are two separate theorems, often in two separate sections. But here we can easily see why the diagonals of both the kite and the rhombus are perpendicular -- because the rhombus is a kite! Also, since we are defining kite inclusively, we don't need any extra steps in the proof to ensure that our kite isn't a rhombus.

This is our last section for now that doesn't require a Parallel Postulate. After all, a kite doesn't have parallel sides (unless it is a rhombus), and so the Parallel Postulate can't appear in the proofs.