Thursday, April 27, 2017

Lesson 14-2: Lengths in Right Triangles (Day 142)

This is what Theoni Pappas writes on page 117 of her Magic of Mathematics:

"Every math enthusiast at one time or another has discovered or been delighted by tricks or oddities involving numbers. Here are two for you to explore and hopefully enjoy."

Pappas is "Playing with Numbers" in this section. She begins with summing and squaring:

1 + 2 + 1 = 2^2
1 + 2 + 3 + 2 + 1 = 3^2
1 + 2 + 3 + 4 + 3 + 2 + 1 = 4^2
1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5^2

This reminds us of the sum of odd numbers pattern from two weeks ago. Notice that this and the odd number sums are related -- take a look at that last sum again:

1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5^2

The first term is 1, which is also the first odd number. If we cross it out, the new first and last terms are 2 + 1 = 3, the second odd number. If we cross them out, the new first and last terms are 3 + 2 = 5, the third odd number. If we cross them out, the new first and last terms are 4 + 3 = 7 -- by now, we get the pattern. We obtain:

1 + 3 + 5 + 7 + 9 = 5^2

which is the pattern from page 108.

Another way to think about this pattern is to rotate a 5 x 5 square of dots 45 degrees to form a diamond -- hey, wasn't I writing something about how a diamond is just a square yesterday? We can view this diamond as containing nine rows -- one in the first row, two dots in the second row increasing up to five dots in the fifth row, then four dots in the sixth row decreasing down to one dot in the ninth row. This is a visual justification for:

1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5^2

Here is the second pattern Pappas mentions today -- the 1's pyramid:

1^2 = 1
11^2 = 121
111^2 = 12321
1111^2 = 1234321
11111^2 = 123454321
111111^2 = 12345654321
1111111^2 = 1234567654321
11111111^2 = 123456787654321

Pappas asks, "When does it stop?" Well, let's try a few more:

111111111^2 = 12345678987654321
1111111111^2 = 1234567900987654321

So that's when the pattern ends. But at this point, we notice that most of the digits of this number still follow the pattern. Indeed, what seems to prevent the pattern is that there's no digit for ten -- and in fact, look at what happens in base 11, where we can use X to represent ten:

1111111111^2 = 123456789X987654321

And in dozenal, we can use E to represent eleven. (I mentioned "dozenal," or base 12, several times on this blog, most recently in my September 22nd post.)

11111111111^2 = 123456789XEX987654321

So it appears that the pattern is limited only by the digits available in our base. To see why it works, we let b be the base of our numeration system. Then we obtain:

11 = b + 1
111 = b^2 + b + 1
1111 = b^3 + b^2 + b + 1
11111 = b^4 + b^3 + b^2 + b + 1

So squaring 11, 111, and so on amounts to squaring polynomials in b:

(b + 1)^2 = b^2 + 2b + 1
(b^2 + b + 1)^2 = b^4 + 2b^3 + 3b^2 + 2b + 1
(b^3 + b^2 + b + 1)^2 = b^6 + 2b^5 + 3b^4 + 4b^3 + 3b^2 + 2b + 1

So now we can see where the coefficients come from. The b^3 coefficient of this last squared polynomial is 4 because there are four b^3 terms (b^3)(1), (b^2)(b), (b)(b^2), (1)(b^3), while the b coefficient is 2 because there are only two b terms (b)(1), (1)(b).

An interesting case occurs when b = 1. Then we obtain:

(1 + 1)^2 = 1 + 2 + 1
(1 + 1 + 1)^2 = 1 + 2 + 3 + 2 + 1
(1 + 1 + 1 + 1)^2 = 1 + 2 + 3 + 4 + 3 + 2 + 1
(1 + 1 + 1 + 1 + 1)^2 = 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1

which should look familiar. That's right -- when b = 1, the second pattern that Pappas gives today reduces to the first pattern!

This is what I wrote last year about today's lesson:

Lesson 14-2 of the U of Chicago text is on lengths in right triangles -- specifically, those lengths that are related to the altitude and involve the geometric mean.

Geometric Mean Theorem:
The geometric mean of the positive numbers a and b is sqrt(ab).
(Note: This may sound like a definition, but actually the U of Chicago defines geometric mean to be the number x such that a/x = x/b, so we need a theorem to get the geometric mean as sqrt(ab).)

Right Triangle Altitude Theorem:
In a right triangle:
a. The altitude of the hypotenuse is the geometric mean of the segments dividing the hypotenuse.
b. Each leg is the geometric mean of the hypotenuse and the segment adjacent to the leg.

In this lesson, I give the proof of the Pythagorean Theorem based on similarity, but this time I gave the proof in the book, which mentions the geometric mean. Let's look at the proof -- as usual, with an extra step for the Given:

Given: Right triangle
Prove: a^2 + b^2 = c^2

Statements                                Reasons
1. Right triangle                       1. Given
2. a geometric mean of c & x,  2. Right Triangle Altitude Theorem
    b geometric mean of c & y
3. a = sqrt(cx), b = sqrt(cy)       3. Geometric Mean Theorem
4. a^2 = cxb^2 = cy                 4. Multiplication Property of Equality
5. a^2 + b^2 = cx + cy               5. Addition Property of Equality
6. a^2 + b^2 = c(x + y)              6. Distributive Property
7. x + y = c                                7. Betweenness Theorem (Segment Addition Postulate)
8. a^2 + b^2 = c^2                    8. Substitution (step 6 into step 7)

It is uncertain whether this is the proof that Common Core intends the students to learn, or whether my earlier proof that avoids geometric means suffices.

Actually, since posting this last year, I've decided to check both the PARCC and SBAC released test questions for those related to the proof of the Pythagorean Theorem. There were a few questions that required use of Pythagoras, but none directly related to the proof. Of course, some people lament that there aren't very many proofs on the Common Core tests.

Speaking of the Pythagoran Theorem, I've been thinking last week about how I presented it to my eighth graders earlier this year:

Monday: Coding
Tuesday: Illinois State project
Wednesday: Traditional lesson on the theorem
Thursday: "Learning Centers" (actually the traditional lesson continued)
Friday: Test

which too many students failed. But as I think about it more, we can see how the students were set up for failure with only two days between introduction and assessment. (As it turned out, even the original project failed because it was on the day that the Hidden Figures movie extra credit was due, and many students focused on finishing that rather than the project.)

It might have been better to teach the theorem on Tuesday, and then the students can apply that knowledge to the project on Wednesday. (And it would have been even better to have the traditional lesson on Monday, but I had no control over the coding teacher's schedule.) The problem is that Illinois State, on its pacing guide, insists that the project be given before the traditional lesson. This in itself leads back to the traditionalist debate, which I'm not getting into in today's post. Another problem is that this might work for sixth and eighth grades, but not for seventh grade, as I had less time with them.

(Speaking of seventh grade, today I taught the volumes of rectangular and triangular prisms to seventh graders.)

Wednesday, April 26, 2017

Lesson 14-1: Special Right Triangles (Day 141)

This is what Theoni Pappas writes on page 116 of her Magic of Mathematics:

"Take the year you were born. To this add the year of an important event in your life. To this sum add the age you will be at the end of 1994. Finally, add to this sum the number of years ago that the important event took place."

Here Pappas is showing us some "Number Magic." Notice that 1994 is the year in which she first wrote this book. The following is the example that she gives:

1953 -- Year born
1980 -- Year Pappas (claimed to have) traveled into outer space. (Really, Pappas?)
    41 -- Age as of 1994
    14 -- Years elapsed since 1994 of important event
3988 -- Total

According to Pappas, the answer is always 3988. Of course, this isn't really "magic" at all -- notice that there are two pairs that add up to 1994. The year of her birth and her age add up to 1994, as do the important event and the number of years since that event. So it's not surprising that the numbers always add up to 1994 * 2 = 3988.

Let me try another example. This time, I'll change it to the current year, 2017, so my four numbers should add up to 2017 * 2 = 4034. The year of my birth is 1980 -- hey, that's the same year that Pappas (supposedly) traveled into space. And as for the year of my important event -- I just think I'll choose the year 1994. After all, it's the year I took my favorite math class, Geometry. Oh, and a lady with the initials TP wrote one of my favorite books in that year.

1980 -- Year born
1994 -- Year of important event
    37 -- Age at the end of this year
    23 -- Years elapsed since important event
4034 -- Total

Unlike Pappas, here's something that really did travel into space -- Cassini. There was a special Google Doodle marking the space probe's arrival at the rings of Saturn. Again, I'm a science teacher, so I like to point out important scientific events. In this case, researchers are excited over the possibility that there is life on at least one of the ringed planet's moons. Cassini will orbit Saturn 22 times to gather data from the Cronian moons.

This is what I wrote last year about today's lesson:

Chapter 14 of the U of Chicago text is on Trigonometry and Vectors. Here's the plan:

Today, April 26th -- Lesson 14-1: Special Right Triangles
Tomorrow, April 27th -- Lesson 14-2: Lengths in Right Triangles
Friday, April 28th -- Activity (includes Lesson 14-3: The Tangent Ratio)
Monday, May 1st -- Lesson 14-4: The Sine and Cosine Ratios
Tuesday, May 2nd -- Lesson 14-5: Vectors
Wednesday, May 3rd -- Lesson 14-6: Properties of Vectors
Thursday, May 4th -- Lesson 14-7: Adding Vectors Using Trigonometry
Friday, May 5th -- ActivityMonday, May 8th -- Review for Chapter 14 Test
Tuesday, May 9th -- Chapter 14 Test

So the plan for this chapter is straightforward. The one thing to note is how the day that Lesson 14-4 would have occurred, there is a planned activity day. I've noticed how many texts, including the U of Chicago, discuss the tangent ratio in a separate lesson from sine and cosine. I suppose that in many ways, sine and cosine are alike in a way that tangent isn't. The sine or cosine of any real number is between -1 and 1, while the tangent can be any real number. Therefore the graphs of sine and cosine resemble each other. The tangent ratio involves two legs, while the sine and cosine ratios involve one leg and the hypotenuse. Even the name "cosine" includes the word "sine," while the name "tangent" doesn't include "sine."

Yet I will end up covering sine, cosine, and tangent all on the same day. In the past, I've seen many teachers simply teach SOH-CAH-TOA all in the same lesson, and then when they come to me for tutoring, they look at each triangle in the homework to determine whether sine, cosine, or tangent is needed to solve the problem. But as it turns out, all of the questions require tangent because the student is actually reading the tangent lesson in the text! If the student is going through all of that, then we might as well have all three trig ratios in the same lesson.

And so this is exactly what I'll do. This will then free a day for an activity. My planned activity will actually be one that I found off of another teacher's website. (Actually, I'm still debating whether to do the activity on Friday or on Thursday, since this teacher presents this activity before teaching the students about sine, cosine, and tangent.)

[2017 Update: The activity day is now on Friday and the trig ratios are on Monday. Thus the other teacher's original intent has been restored.]
But that's for later this week -- how about today's lesson? Lesson 14-1 of the U of Chicago text is on Special Right Triangles -- that is, the 45-45-90 and 30-60-90 triangles. The text emphasizes how these triangles are related to the regular polygons. In particular, the 45-45-90 and 30-60-90 triangles are half of the square and the equilateral triangle, respectively. We can obtain these regular polygons, in true Common Core fashion, by reflecting each right triangle over one of its legs. The regular hexagon is also closely related to the 30-60-90 triangle.

The questions that I selected from the text refers to these regular polygons and using the triangles to measure lengths related to the regular polygons. I mentioned today how I like to watch baseball over summer break -- well, a baseball "diamond" (really a square) appears on the worksheet. Also, a honeycomb, with its hexagonal bee cells, also appears.

The review questions that I selected are also preview questions. Two of the questions involve similar right triangles in preparation for geometric means in Lesson 14-2, and the other one is about how to simplify radicals, so we can explain in Lesson 14-4 why the sine and cosine of 45 degrees are usually written as sqrt(2)/2.

Tuesday, April 25, 2017

Chapter 13 Test (Day 140)

This is what Theoni Pappas writes on page 115 of her Magic of Mathematics:

"It seems the further out one ventures in the field of whole numbers, the primes become rarer and rarer. One might think that because they appear less and less frequently, that perhaps they end somewhere. As early as about 300 B.C. Euclid provided the first proof that the prime numbers are infinite."

Euclid -- hey, that name sounds familiar to readers of this Geometry blog! Yes, the same Euclid after whom Euclidean geometry is named is also the one who proved the infinitude of primes.

Pappas continues, "He used indirect reasoning..." -- that's right, as in indirect proof. It's interesting that just as we cover the U of Chicago's Chapter 13 on indirect reasoning, we read about so many indirect proofs in Pappas.

Notice the similarity between Cantor's Diagonal Argument and Euclid's infinitude of primes. Cantor assumes that the reals are countable and derives a contradiction, so the reals are uncountable. And so Euclid assumes that the primes are finite and derives a contradiction, so the primes are infinite.

Pappas assumes that the finite list 2, 3, 5, 7, and 11 contains all the primes. She then multiplies these primes and adds 1 to obtain 2311. But this number can't be divisible by 2, 3, 5, 7, or 11 since each leaves a remainder of 1. Therefore 2311 must be a new prime! Pappas then generalizes this to the case where n is the largest prime. In each case we generate a new prime not on the list, and so n is not the largest prime -- a contradiction, therefore there is no largest prime. The primes are infinite. QED

By the way, tonight's the first episode of Genius, about Albert Einstein, on National Geographic. I'm a science teacher, so I'll watch the miniseries, but I may or may not write about it on the blog.

Notice that Euclid's proof generates a much larger prime that's not on the list, in this case 2311. It doesn't necessarily give the actual next prime, which is 13. Oh, and speaking of 13...:

This is what I wrote two years ago about today's test. As it turned out, I originally posted this test in 2015 on Friday the 13th, and I mentioned this fact throughout the test. Unfortunately, this year today is neither Friday nor the 13th. But if it's any consolation, this is my 13th post blog post in April:

Here's an answer key for the test:

1. a. 90 degrees. I could have made this one more difficult by choosing a heptagon, or even a triskaidecagon, but I just stuck with the easy square.

b. Here is the Logo program:

Notice that the side length is 13. I'll still find a way to sneak 13, if possible, into each problem.

2. a. If a person is not a Rhode Islander, then that person doesn't live in the U.S.
b. If a person doesn't live in the U.S., then that person isn't a Rhode Islander.
c. The inverse is false, while the contrapositive is true.

Notice that Rhode Island is the thirteenth state.

3. y = 10.

4. There is a line MN. (M is the thirteenth letter of the alphabet.)

5. Every name in this list is melodious.

6. The equation has no solution. (This question references 13, as 13x appears in the expansion.)

7. a. 13, 11, 9 (descending odds).
b. 13, 17, 19 (increasing primes -- of course, Euclid proved that this sequence is infinite).

8. a = 2, b = 1, c = 3.

9. kite.

10. I discussed this problem earlier this week. It is the same as the problem from the Glencoe text, except that I only drew half of the figure -- the part where a contradiction appears.

Assume that the figure is possible. Then ABC is isosceles, therefore angles A and C are each 40 degrees (as the third angle of the triangle is 100). Then ABO is isosceles (as it has two 40 degree angles), so AO = BO = 3. Then by the Triangle Inequality, 3 + 3 > 8, a contradiction.

11. Through any two points, there is exactly one line. (This is part of the Point-Line-Plane Postulate.)

12. a. KML measures 13 degrees.
b. K measures less than 167 degrees.
c. L measures less than 167 degrees. (This is the TEAI, Exterior Angle Inequality.)

13. a. Law of Ruling Out Possibilities.
b. You forgot to rule out another possibility -- that nothing bad will happen to you today. Hopefully, this will be true for you.

Monday, April 24, 2017

Review for Chapter 13 Test (Day 139)

This is what Theoni Pappas writes on page 114 of her Magic of Mathematics:

"Then [Cantor] looked at the numbers along the diagonal of this list. He formed a number by taking the digits landing on the diagonal, and changed each digit on this diagonal to another digit (e.g. he may have changed them by simply adding 1 to each digit). Then by placing a decimal point in front of it, the number had to be between 0 and 1."

As it turns out, we're right in the middle of a section that began on page 113 -- "Cantor & the Uncountable Real Numbers." Here "Cantor" refers to the great 19th century German mathematician, Georg Cantor.

We don't need to back up to page 113 to figure out what Pappas is talking about here. I actually wrote about this topic last year, when I was watching David Kung's lectures. And so it's time for a little more cut and paste. This is what I wrote last year about today's Pappas topic, as given by Kung:

We begin as Cantor did -- let's imagine what a solution would look like. We would have every decimal in a room -- and we'd be able to prove it by giving a list of all of the decimals. So all we have to do is check the list -- if all the decimals are listed then we have a 1-1 correspondence, and if there's a decimal missing then we don't have a 1-1 correspondence.

But how can we tell whether there's a decimal missing from the list? Kung suggests playing another game -- a game of Dodgeball. So let's modify our guessing game a little. Instead of having you try to guess my number, I will instead try to dodge your guesses. Each time you give a guess, I'll give a digit that will avoid all of your guesses. If I can avoid all of your guesses, then I have a number that isn't on the list and I win -- if I can't, then you win. Of course, if you have a complete list of decimals (a 1-1 correspondence) then there would be no way for me to dodge all of them.

To make the game simpler, Kung suggests having only five digits, and letting each digit be either zero or one. So on your turn, you post five binary digits, and on my turn, I add one more binary digit to my one number. The game ends after five turns. If I was forced to choose one of your numbers then you win, and if I dodged all of your numbers, then I win. So let's play:

Your Turn: 01000
My   Turn: 1
Your Turn: 11101
My   Turn: 10
Your Turn: 10100
My   Turn: 100
Your Turn: 10010
My   Turn: 1000
Your Turn: 10000
My   Turn: 10001

Let's see who won: your numbers are 01000, 11101, 10100, 10010, and 10000, while my number turns out to be 10001. I dodged all your numbers, so I won!

Of course, now you may be saying that it's easy for me to win. Each time you wrote a zero in a given place, I wrote a one, and each time you wrote a one, I wrote a zero. So it's easy for me to win. But that is precisely the point.

Now let's play with actual real numbers:

Your Turn: .9293020214...
My   Turn: .6
Your Turn: .8569378772...
My   Turn: .66
Your Turn: .3578110034...
My   Turn: .666
Your Turn: .1769761505...
My   Turn: .6666
Your Turn: .4552619933...
My   Turn: .66668
Your Turn: .6835320759...
My   Turn: .666686
Your Turn: .9302868685...
My   Turn: .6666866
Your Turn: .2668818128...
My   Turn: .66668666
Your Turn: .7235519668...
My   Turn: .666686668
Your Turn: .7981213176...
My   Turn: .6666866688...

Notice that I always choose to write a six -- unless you wrote a six, when I write eight instead. After my first turn, we know that my number won't be in Room 1. After my second turn, we know that my number won't be in Room 2. After my third turn, we know that my number won't be in Room 3. So after infinitely many turns, we know that my number won't be in any of the rooms. This is called Cantor's Diagonal Argument, since the digits we focus on fall on a diagonal of the list. Using Cantor's Diagonal trick, I win every single time you give me a list.

This trick works for any list that you can give me. In particular, it works for a list that you claim is a 1-1 correspondence between N and the set of decimals. But how can it, when by definition, every decimal is on the list, so that there's no decimal left for me to win?

The answer is -- you only assumed that every decimal is on the list. I just showed that the assumption that every decimal is on the list leads to a contradiction. And so, we actually just gave an indirect proof of something -- but of what? We assumed that there exists a 1-1 correspondence between N and the set of decimals, so we actually just proved that no such 1-1 correspondence exists. QED

The set of all decimals has a special name -- the set of real numbers, symbolized R. So what we just proved is that N(R) > N(N). The set of reals is uncountable. And so the answer to our original problem is that there is no algorithm that will allow you to win the game with real numbers (whether as the original Pascal guessing game or as the Dodgeball game). As the title of this episode implies, there is more than one type of infinity.

So my old intuition was right -- there are more real numbers than integers. But my way of thinking about it was completely wrong -- it has nothing to do with there being infinitely many reals between the integers (since there are infinitely many rationals between them, yet these are countable). In particular, it is misleading to say that there are "infinity" integers and "infinity^2" reals. Instead, Cantor shows us how to use 1-1 correspondences to determine set size, or cardinality.

Here are a few more things I want to say about cardinality. My old set theory book also shows that there is another special set that has Q as a subset yet is still countable -- the algebraic numbers. These are the numbers x for which there exists a polynomial P, with integer (or rational) coefficients, such that P(x) equals 0. The number sqrt(2) is algebraic, and we can even count complex numbers such as sqrt(-1), which is i. But numbers like pi or e still aren't algebraic -- they are transcendental.

Many people have trouble realizing that Q is countable and R is uncountable, yet I had no problem with either of these. Instead, I couldn't believe that the set of algebraic numbers -- a set that contains irrationals (and even some complex numbers) yet is still countable. My set theory book explained that there is a trick similar to infinitely many buses -- the bus number can be the degree of the polynomials and the seating can be arranged by the sum of the absolute values of the coefficients.

Now think about this -- R is uncountable, while there are countably many rationals. So what about the set of all irrationals? It's easy to come up with an indirect proof to show that the set of irrationals is uncountable: first assume that the set of irrationals is countable. Then it would be easy to show that the set of all reals would also be countable, using the integer or zipper tricks -- just place the rationals on one side and the irrationals on the other, and pull up the zipper. But this is a contradiction, since the set of all reals is uncountable. So the assumption that the set of the set of irrationals is countable leads to a contradiction -- therefore the set of irrationals is uncountable. QED

So we now have two uncountable sets -- R and the set of irrationals (often written R \ Q). One of my favorite questions is, is there a 1-1 correspondence from R to R \ Q? As it turns out, there does exist such a 1-1 correspondence, and we can prove it. Let's think of in terms of a Hilbert's Hotel -- except it's an uncountable hotel with a room for every irrational. Now a bus arrives containing all of the rational numbers. How can we fit them into the hotel?

Well, let's take our favorite irrational, pi, and so we add pi to every rational. We see that the sum of a rational and an irrational is irrational, and so we can ask each rational to go to the room where that irrational is presently sitting. And where do those irrationals already in those rooms go? By now we should know the Hilbert's Hotel pattern -- they add pi to themselves to get new irrationals, and then go to the room where those new irrationals are. In other words:

The number x + n * pi (x, rational, n a whole number) goes to room x + (n + 1)pi.

Of course, we must be careful that x + (n + 1)pi isn't accidentally rational. But there is an indirect proof that it's always irrational: assume that x + (n + 1)pi = y is rational. Then rearranging the equation, we obtain pi = (y - x) / (n + 1), the quotient of two rationals, so is itself rational. So we get that pi is rational -- a contradiction since pi is irrational. Therefore x + (n + 1)pi is irrational. QED

We notice that in this case, there are uncountably many rooms, yet only countably many occupants need to change rooms. Most irrationals don't have to move -- in particular, sqrt(2) doesn't move, neither does e, and neither does -pi.

Here's another thing I've noticed -- sets of numbers that are special, or fit some sort of pattern, tend to be countable, while sets of numbers that don't fit patterns tend to be uncountable. The natural numbers fit the ultimate pattern -- 1, 2, 3, so they are countable. The rational numbers are countable (they fit the pattern a / b), while the irrationals are countable. The algebraic numbers are countable (they fit into polynomials), while the transcendental numbers are uncountable.

What about the set of all numbers consisting of the digits 0 and 1 only -- is it countable? Even though this may sound like a pattern, it really isn't. After all, in binary, all reals consist of only two digits, so consisting of two digits isn't special enough to make the set countable. The set of reals with only two digits is uncountable, regardless of which digits or the base. (In the special case of 0 and 1, we can play Kung's original Dodgeball game to see why.) If the digits are 0 and 2, and the base is changed to ternary (base 3), we obtain a famous uncountable set -- the Cantor fractal set!

Returning to 2017, notice that Cantor's Diagonal Argument is essentially an indirect proof -- he assumed that the set of real numbers is countable and derived a contradiction, therefore the set of reals is uncountable. This indirect proof fits right in with our current Chapter 13.

And so we now reach today's lesson, which is review for the Chapter 13 Test. Recall that last year, I didn't actually give a Chapter 13 Test. So this is what I wrote two years ago about today's worksheet:

Let's begin reviewing for the Chapter 13 test. As usual, here I will discuss the rationale for including the questions that I included.

My test ended up including most of the logic from Chapter 13, and consequently not as much of the inequalities from Chapter 5. The first question is based on yesterday's lesson. I have separated it into an (a) and (b) part. The (a) part simply asks for the exterior angle of a regular hexagon, while the (b) part asks the student to write a Logo program to draw a regular hexagon using that value of the exterior angle. This way, if the teacher doesn't want to do Logo, the (b) part can be skipped, and only the (a) part will be required.

Question 2 is a simple question on converse, inverse, and contrapositive.

Questions 3-6 ask the students to make conclusions based on logic. I was going to ask the students to name the logical rules that they used (Law of Detachment, etc.), but I decided against it.

Question 7 was inspired by something that I once saw in another text (not Glencoe), for a different student I was tutoring in geometry at least [three years] ago. [2017 update: I gave my middle school students sequences like these during the first week of school. See my August 19th post for more information -- dw] One of them is mathematical and straightforward. The second one was a fun one for my students to figure out the answer -- it's the first names of the presidents in order:

George (Washington), John (Adams), Thomas (Jefferson), James (Madison), ...?

and so the next three answers will be the next three presidents:

James (Monroe), John (Quincy Adams), Andrew (Jackson)

I've seen this question modified so that it gives the order of the presidents that appear on money:

George (Washington), Thomas (Jefferson), Abe (Lincoln), Alexander (Hamilton), ...?

with the answer:

Andrew (Jackson) [Harriet (Tubman) in a few years -- dw], Ulysses (Grant), Benjamin (Franklin)

And yet another variation gives the first ladies:

Martha (Washington), Abigail (Adams), Martha (Jefferson), Dolley (Madison), ...?

with the answer:

Elizabeth (Monroe), Louisa (Adams), Rachel (Jackson)

Question 8 is a logic problem. This one comes directly from the SPUR section of U of Chicago. No, I won't include any of Fireball's so-called "easy" logic problems, as these would not be appropriate for a math test.

Question 9 is on tangents to circles. By the way, even though I had to squeeze in Section 13-5 right in between the indirect proof and inequalities lessons, there is a benefit to including this lesson. I'm expecting that by the time we finally reach Chapter 15 of U of Chicago (on circles), we'll be rushing in order to finish it before the PARCC and SBAC exams. I'm not sure how much of Chapter 15 might appear on the PARCC or SBAC, but at least one topic that's likely to appear -- the fact that tangents to circles are perpendicular to their corresponding radii -- has already appeared right now.

Question 10 is an indirect proof. I would've included the U of Chicago indirect proofs, except that I got tired of the "prove that the square root of 9800 isn't 99" questions. I did notice that one of the questions in the U of Chicago was "prove that the square root of 2 isn't 577/408." In some ways, this sort of question can be said to lead up to one of the most famous indirect proofs -- namely that the square root of 2 is irrational. Here is a link to a common indirect proof that sqrt(2) is irrational:

Neither the U of Chicago nor Glencoe gives the proof outright. But both hint at it -- I just mentioned the U of Chicago's square root proofs. The Glencoe text asks the students to prove that if the square of a number is even, then it is divisible by four. As we can see at the above link, this fact is directly mentioned in the irrationality proof.

I remember once reading the proof of the irrationality of sqrt(2) in my textbook back when I was an Algebra I student. Until then, I had always heard that sqrt(2) was irrational, but I never realized that it was something that could be proved. So I was fascinated by the proof. Naturally, the text only included this as an extra page between the main sections, so it was something that the teacher skipped and most students probably ignored.

The irrationality of sqrt(2) has an interesting history. It goes back to Pythagoras -- he was one of the first mathematicians to use sqrt(2), since his famous Theorem could be used to show that the diagonal of a square has length sqrt(2). The website Cut the Knot, which has many proofs of the Pythagorean Theorem, also contains many proofs of the irrationality of sqrt(2):

Now there is a famous story regarding sqrt(2) and Pythagoras. At the following link, we see that Pythagoras was the leader of a secret society, or Brotherhood:

Now Pythagoras and his followers believed that only natural numbers were truly numbers. Not even fractions were considered to be numbers, but simply the ratios of numbers -- numberhood itself was reserved only for the natural numbers. In some ways, this attitude resembles that of algebra students today -- when the solution of an equation is a fraction, they often don't consider it to be a real answer, even though modern mathematics considers fractions to be numbers. (The phrases real number and imaginary number reflect a similar attitude about 2000 years after Pythagoras -- that some numbers aren't really numbers.) So of course, the idea that there were "numbers" that weren't the ratio of natural numbers at all was just unthinkable.

Pythagoras and his followers must have spent years searching for the correct fraction whose square is 2, but to no avail. Finally, one of his followers, Hippasus, discovered the reason that they were having such bad luck finding the correct fraction -- because there is no such fraction! And, as the story goes, Pythagoras was so distraught, afraid that the secret that sqrt(2) was irrational would be revealed, that he ordered to have poor Hippasus drowned at sea!

But as I said, nowadays students simply complain when they have a fractional, or worse irrational, answer to a problem. No one has to drown any more just because of irrational numbers.

Question 10 on my test review, therefore, is actually the final step of that proof, since that's the step where the contradiction occurs. They are given a triangle with sides of length 3 and 8, and two angles each 40 degrees (one of which is opposite the side of length 3). The students are to use the Converse of the Isosceles Triangle Theorem to show that the missing side must also be of length 3, and then the Triangle Inequality to show that 3 + 3 must be greater than 8, a contradiction.

When I wrote this problem, I had trouble deciding how difficult I wanted my indirect proof to be. For example, I considered giving 100 as the measure of the angle opposite the side of length 8, and give only one 40-degree angle instead. Then the students would have to use the Triangle Angle-Sum Theorem to find the missing angle as 40 degrees before applying the Isosceles Converse.

Or, to go even further, we can derive a contradiction without making the angle isosceles at all. For example, we could make the angle opposite the 8 side to be, say, 90 degrees instead of 100. Then the missing angle would be 50 instead of 40. If the triangle is drawn so that 50 degrees is opposite the 3 side, then by the Unequal Angles Theorem, the missing side would be less than 3, so the sum of the two legs would still be less than the longest side.

But this might confuse the students even more -- especially if the 90-degree angle is marked with a box (to indicate right angle) rather than "90." A right triangle might lead a student to use the Pythagorean Theorem to find the missing leg. Although this still eventually leads to contradiction -- the missing side would be sqrt(55), which isn't less than 3 -- that irrational side length might still cause some students to drown.

And so I wrote my Question 10 on the review so that it will actually help the students prepare for the corresponding question on the test. I balance out this tough question with some easier questions about logic (converse, inverse, etc.). Hopefully the test won't be too hard for the students.