## Monday, April 30, 2018

### Lesson 15-5: Angles Formed by Secants or Tangents (Day 155)

Today on her Mathematics Calendar 2018, Theoni Pappas writes:

In Triangle ABC, C is a right angle, AC = sqrt(27), AB = 6, angle A = x. What is x?

(Actually, today's Pappas problem is completely diagrammatic. Here I convert it from diagrammatic to symbolic form.)

We notice that in this right triangle, the hypotenuse and a leg adjacent to the angle are known. Thus it's possible to use the cosine function:

cos A = sqrt(27)/6
A = arccos(sqrt(27)/6)

But the U of Chicago text doesn't use the inverse cosine function. Instead, only the inverse tangent function appears in Lesson 14-7. So let's use the Pythagorean Theorem to find the other leg:

a^2 + b^2 = c^2
a^2 + sqrt(27)^2 = 6^2
a^2 + 27 = 36
a^2 = 9
a = 3

So the other leg is 3. And as soon as we see this, we can do away with inverse tangent altogether -- the opposite leg is half of the hypotenuse. Thus this is a special right triangle, 30-60-90 -- a fact that would have been made more obvious had the original problem stated the length of the longer leg as 3sqrt(3), not sqrt(27). (Once again, consider how many special triangles have appeared on the Pappas this month.) Therefore the desired angle is 30 degrees -- and of course, today's date is the thirtieth.

April 30th, 1777 -- the birthday of Carl Friedrich Gauss, a very famous mathematician. There's a Google Doodle for "The Prince of Mathematicians" -- and so of course I'm going to write about him on the blog today.

I've referred to the story of Gauss as a young child so many times on the blog. Most recently, I mentioned him back in my September 15th post, as part of reading Stanley Ogilvy's book. I know that our side-along reading book today is Wickelgren, but let's go back to Ogilvy for one day so we can celebrate the young prodigy:

"Carl Friedrich Gauss, possibly the greatest mathematician of all time, showed his arithmetical skill at an early age. When he was ten years old his class at school was given what was intended to be a long routine drill exercise by a tyrannical schoolmaster: 'Find the sum of the first 100 positive integers.' This was easy for the schoolmaster, who knew how to sum arithmetic progressions, but the formula was unknown to the boys. Young Gauss did not know how to do it either, but he invented a way, instantly and in his head. Writing the answer on his slate, he handed it in at once. When the rest of the students' calculations were collected an hour later, all were found to be incorrect except Gauss's. We are told that we did it by pairing the terms and then multiplying the value of each pair by the number of pairs. If the pairs could each total 100, so much the easier: 100 + 0, 99 + 1, etc. This would make 50 pairs of 100 each for 5000, plus 50 left over (the middle number), for a total of 5050."

A few years ago, I was student teaching an Algebra I class. My master teacher also taught Algebra II, and in late April that year, she was about to reach the lesson on arithmetic series. She decided to look up the biography of Gauss at the following link:

http://www-groups.dcs.st-and.ac.uk/history/Biographies/Gauss.html

And much to her surprise, his birthday was April 30th -- the exact date when she was planning on giving the arithmetic series lesson! So just like e Day (and Pi Day for Geometry), Gauss Day is a day that Algebra II teachers might wish to celebrate in class. The arithmetic series lesson is given in Lesson 13-1 of the U of Chicago Advanced Algebra text, and so it's plausible that we might reach this lesson on the final day in April (though not if we use the digit pattern).

Gauss is also mentioned in the U of Chicago Geometry text. This is what Lesson 5-7 has to say about the great mathematician (copied from my November 6th post):

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.

Today, I decided to search for any teachers giving a Gauss lesson on Twitter. Instead, I found only a tweet by a group of teachers in Hernando County, Florida:

Google celebrates a mathematician & we think THAT is worth celebrating!

Meanwhile, today is my third day of five in the high school Economics class. Unfortunately, the class doesn't go well. In first period, I have all the passwords I need except the wi-fi password to access the Internet, on a day when the regular teacher wants me to play a YouTube video. A student allows me to use his wi-fi account, but there's not enough time to play the whole video -- which mentions the answers to tonight's homework.

By second period, I figure out how to use my own email address as a wi-fi password -- except that for a whole hour, no password works. This wi-fi blackout starts a few minutes after the start of second period and lasts into third period. (Today is a Late Start Monday, so there is no tutorial. Had today not been Monday, only third period would have been affected by the blackout.) As luck would have it, the Internet is working perfectly on another computer -- one that isn't connected to the projector.

Only fifth period is able to watch the entire video and can do the homework tonight. Fortunately, fourth period World History is starting a research project that doesn't require me to play a video.

By the way, during the classes when I couldn't play the video, one girl starts working on her math assignment -- I couldn't tell whether the class was Stats or IB Math Studies, but there was clearly a bell curve drawn on the worksheet. I tell her that the bell curve is due to Gauss (which is why it's part of the Google Doodle) -- and then I jump right into the story of the ten-year old mathematician.

In case you're curious, here's the video I try to play in the Econ classes today:

John Stossel, the narrator of this 1998 video, is a well-known conservative journalist. Thus, I've just jumped into politics by linking to his video. But then again, these days have been very politically charged for education. In more states, including nearby Arizona, teachers are walking out. And here in California, the two major gubernatorial candidates are on opposite sides of the district/charter school debate, with Gavin Newsom winning the support of major teacher unions and Antonio Villaraigosa winning the support of major charter backers. (Stossel himself has recently written articles supporting charters. But Villaraigosa is a Democrat -- I haven't heard what any Republican candidates are saying about education yet.) Here on the blog, I don't take either side of the debate, since I've worked for both district schools (this year as a sub) and charters (last year as a teacher).

It's obvious that the regular teacher I'm subbing for is politically conservative -- he has a picture of the current president posted on the wall. And on his desk is a 100-page book, Why Socialism Works, by Harrison Lievesley. Every single page contains only two words -- "It Doesn't."

I make no comment on the blog about whether "socialism works" or not. But I do wish to comment on Stossel's video in more detail -- skipping to the parts of the video relevant to education. This is in the interest of keeping the blog politically balanced -- last month I swung to the left by writing about the March 14th gun control walkout, and so this month I go rightward with Stossel.

At one point (17:50), economist Walter Williams, when interviewed, says that education is one industry that based more on "kindness" (publicly-funded schools) than on "greed" (the profit motive), but the main mention of schools appears at the 21:33 mark. This example involves the traditionalist debate, so here's yet another post labeled "traditionalists." I also add the "subbing" label, to emphasize that the reference to politics in the post is directly related to the video I'm assigned to play.

At 21:33, Stossel says:

"Still, I don't like to admit that self-interest or greed is the motivator. And we certainly don't like to teach that to our kids. But what if we did?"

The scene moves to a high school in New York. According to the narrator, the school is so tough that:

"One teacher's hair was set on fire."

Teacher Steve Mariotti is seen showing the students how to add 40 + 20. The students greeted him by putting him in a headlock. So he asked them, "Why are you doing this to me?" And the response he received was, "You are boring."

Mariotti pressed further. "Was there ever a time when I was a good teacher?"

And so Mariotti began not only to tell more stories about his firm, but how his students could strive to start their own businesses. The transformation in his students was nothing short of amazing: "Here's a guy that had been defined as brain damaged and emotionally upset...and here he re-created, in total, a Harvard Business School stat sheet."

For example, here's one of Mariotti's success stories: "Howard Johnson grew up in a poor section of the Bronx. Mariotti counseled him to rent a hot dog stand. Now he owns five and he makes as much as \$3,000 a week."

A certain CEO summarizes the whole class as follows: "Capitalism opens up opportunities to climb up that economic ladder."

But let's focus on the educational, not economic, issues here. Nowhere in the video does it say that Mariotti is a math teacher. Perhaps this is an Econ class, and 40 + 20 just happened to appear as the calculation of some dollar figure.

Yet it's also likely that this is a math class after all. The students are far below grade level, and so Mariotti must teach them basic arithmetic. A traditionalist would agree that students must master the basics before they can learn anything like Algebra I or high school math.

If this is true, notice that the students are rejecting these traditional lessons. Traditionalists act as if students enjoy traditional lessons, and that the only people getting between traditional teachers and their traditionalist-loving students are progressive reformers and "educationists." But those weren't progressive reformers putting Mariotti in a headlock -- those were the students. And Mariotti drops his traditional lessons in favor of "engaging" activities -- not because progressive reformers tell him to, but because his students tell him that's what they want.

Traditionalists will point out that anyone who doesn't know what 40 + 20 is won't be able to know enough math to become a successful businessman (or woman). And of course, they like to brand anyone not on the "AP Calculus track" (which these kids clearly aren't on) as losers. But these students weren't motivated to learn 40 + 20 until they saw how it would help them make money. So now they know what 40 + 20 is, and a whole lot more.

There are a few things that traditionalists might respond with that I'd agree with. Of course, the students should have learned 40 + 20 in elementary school. Even the Common Core, with its delaying of standard algorithms, lists addition of multiples of ten as a first grade standard:

CCSS.MATH.CONTENT.1.NBT.C.4
Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10...Understand that in adding two-digit numbers, one adds tens and tens,

And in first grade, the students aren't big enough to put the teachers in a headlock. Students should learn as much arithmetic as possible before the age at which they start to question (and possibly harm) authority.

Meanwhile, traditionalists might also appreciate that this math teacher (if he really is a math teacher) used to work in industry. We know that some traditionalists believe that second career teachers are the most effective (and least "brainwashed" by the pedagogues).

Chapter 11 of the Wayne Wicklegren's How to Solve Problems is "Problems from Mathematics, Science, and Engineering." It is the final chapter of the book -- and it's the longest, since as its title implies, it contains many problems. (If this were a side-along reading of Harrison Lievesley's book, I'd be done, since I've already summarized the entire book!)

"This chapter is designed to establish the generality of the problem-solving methods discussed throughout the book. In previous chapters, the problems used to illustrate the methods were deliberately selected so that they could be solved by the reader with no more background than a high school student with one year of algebra and one year of plane geometry."

In other words -- the audience for the first ten chapters is our students. Since many of these examples from STEM fields require knowledge beyond our Geometry class, I'll cover only some of these.

The first example involves systems of equations from Algebra II:

"The solution of systems of simultaneous linear equations provides a simple example of the use of evaluation functions, hill climbing, and subgoals."

Here is the system:

2x + y - 3z = 1
x + 2y + 5z = 9
3x - 3y - 10z = 4

Wickelgren lists three allowable operations:

(a) multiply both sides of an equation by the same number
(b) add equals to equals (or subtract equals from equals)
(c) substitute equals for equals

The goal is to derive three expressions of the form x = ____, y = ____, and z = ____, where specific numbers appear in the blanks. Now stop reading and try to solve the problem.

OK, I won't waste your time -- if you're a high school math teacher, then you already know how to solve a system of three equations in three variables. Wickelgren does state the steps in terms of goals and subgoals. The first subgoal is to obtain two equations in two variables. The second subgoal is to solve this system to derive a single equation involving one unknown. The third subgoal is to derive another single equation involving a single unknown. The final subgoal is, of course, to solve for whichever variable is left. By the way, the solution is (5, -3, 2).

Let's skip the next example (solving an exponential equation using logs) and go straight into trig, which at least is closely related to Geometry:

"Determine the altitude, h, of a general scalene triangle, given the length of one side (its base b) and the angles made by the two other sides with the base (the two base angles, alpha and gamma), as illustrated in [the figure]."

The figure shows Triangle ABC, with angles A, B, C measuring their respective Greek letters (alpha, beta, gamma). As usual, b is opposite angle B, and so the altitude h is drawn from vertex B. Stop reading and try to solve the problem.

Let's try defining a subgoal first. For example, why is the altitude h of a triangle significant -- and why do we use the letter h for "altitude." That's right -- h really stands for height, and its significance is that it appears in the formula for the area of a triangle, A = (1/2)bh.

Thus Wickelgren lists this as our subgoal -- find the area of the triangle. We can use the area formula to find the height. Stop reading and try again to solve the problem -- if you did not do so before.

So we wish to find the area of a triangle given ASA. At this point, the author assumes that we have access to a trig text that gives the proper formula:

A = (b^2 sin alpha sin gamma)/(2 sin beta)

But notice that beta isn't part of the given information. Then again, beta is easy to find, because the three angles of a triangle add up to 180 degrees. This allows us to solve the problem.

Wickelgren's next example, from analytic geometry, is highly relevant to state testing:

Determine the location and geometric properties of the figure specified by the equation:

x^2 + y^2 - 5x + 7y = 3

This, of course, is the equation of a circle. We know that circle equation questions appear on the PARCC and SBAC exams. Stop reading and try to solve the problem.

We'll follow Wickelgren's steps in more detail:

"The given expression is x^2 + y^2 - 5x + 7y = 3, and the goal expression is of the form..." -- well, the author uses a, b, c, but we'll use h, k, r as in Lesson 11-3 -- (x - h)^2 + (y - k)^2 = r^2. "Once again, we might define a subgoal by means of working backward from the goal expression."

Starting from the goal:

(x - h)^2 + (y - k)^2 = r^2
(x^2 - 2hx + h^2) + (y^2 - 2ky + k^2) = r^2

"Stop reading and try to solve the problem, if you did not do it before."

2h = 5
h = 5/2

2k = 7
k = -7/2

x^2 - 5x + 25/4 + y^2 + 7y + 49/4 = 3 + 74/4
(x - 5/2)^2 + (y + 7/2)^2 = 86/4
(x - 5/2)^2 + (y + 7/2)^2 = sqrt(86)/2

"Therefore, the coordinates of the center of the circle are (5/2, -7/2), and the radius of the circle is sqrt(86)/2, and the problem is solved." Of course, hopefully the questions we'll see on the actual PARCC or SBAC won't have such messy numbers.

Here's an interesting-sounding problem:

"Determine equations for the new coordinates of a point in a plane when the new coordinate system is obtained by translation and rotation from an old coordinate system. Translation of a coordinate system means the origin is changed to a new point, and rotation of a coordinate system means both axes are turned through the same angle in the same direction, pivoting about the origin."

Hmm, Common Core Geometry is all about translations and rotations. But notice that here we're actually transforming the axes, not a figure on the axes, or even a function. Stop reading and try to solve the problem.

"If x and y are the original coordinates, and the coordinates of the new origin in terms of the original (x, y) axes are (x_0, y_0), then let the new coordinates obtained by the translation be represented by" x', y' -- here I changed Wickelgren's x_1, y_1. "The formulas for simple translation of coordinates are as follows:"

x = x' + x_0
x' = x - x_0

y = y' + y_0
y' = y - y_0

Actually, let's change Wickelgren's (x_0, y_0) to (h, k)

x = x' + h
x' = x - h

y = y' + k
y' = y - k

This finally explains why, when we translate f(x) right h units, it's f(x - h), and when we translate the original function left h units, it's f(x + h). We're actually translating the axes, not the function -- and so the sign of h or k indicates the direction in which we translate the axes! Stop reading and try to solve the rest of the problem (the rotation part), if you haven't already.

Well, rotations of axes aren't usually covered in Algebra II classes, but here are the formulas relating (x_2, y_2), the rotation image of (x_1, y_1) counterclockwise by magnitude alpha:

x_1 = x_2 cos alpha - y_2 sin alpha
y_1 = x_2 sin alpha + y_2 cos alpha

Wickelgren uses this formula to give another rotation of axes problem, but I wish to end on a Calculus note:

Prove that, within the set of triangles having a constant base and constant perimeter, the isosceles triangle has the maximum area.

Well, we haven't quite reached Lesson 15-8 yet, but this is another one of those isometric problems in the same line as "among all plane figures with a constant perimeter, the circle has the largest area" and "among all rectangles with a constant perimeter, the square has the largest area." It seems to be generalized -- among all figures of a particular type, the most symmetrical figure possible tends to have the largest area.

I won't post the Calculus proof, although Wickelgren also adds another proof, using Heron's area formula instead of Calculus.

Let's conclude the book -- a highly enjoyable read. I wish I could have posted more problems from the book on the blog:

"The method of special case is sometimes equivalent to the subgoal method."

It's something to think about whenever we solve problems from now on.

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

Lesson 15-5 of the U of Chicago text is on "Angles Formed by Chords or Secants." There is one vocabulary term as well as two theorems to learn.

The vocabulary word to learn is secant. The U of Chicago defines a secant as a line that intersects a circle in two points. This is in contrast with a tangent, a line that intersects the circle in one point.

At this point, I often wonder why we have tangent and secant lines as well as tangent and secant functions in trig. Well, here's an old (nearly 20 years!) Dr. Math post with the explanation:

http://mathforum.org/library/drmath/view/54053.html

```Now, the tangent and the secant trigonometric functions are related to
the tangent and secant of a circle in the following way.

Consider a UNIT circle centered at point O, and a point Q outside the
unit circle. Construct a line tangent to the circle from point Q and
call the intersection of the tangent line and the circle point P. Also
construct a secant line that goes through the center O of the circle
from point Q. The line segment OQ will intersect the circle at some
point A. Next draw a line segment from the center O to point P. You
should now have a right triangle OPQ.

A little thought will reveal that the length of line segment QP on the
tangent line is nothing more but the tangent (trig function) of angle
POQ (or POA, same thing). Also, the length of the line segment QO on
the secant line is, not surprisingly, the secant (trig function) of
angle POA.```

And now let's look at the theorems:

Angle-Chord Theorem:
The measure of an angle formed by two intersecting chords is one-half the sum of the measures of the arcs intercepted by it and its vertical angle.

Given: Chords AB and CD intersect at E.
Prove: Angle CEB = (Arc AD + Arc BC)/2

Proof:
Statements                                            Reasons
1. Draw AC.                                          1. Through any two points there is exactly one segment.
2. Angle C = Arc AD/2,                        2. Inscribed Angle Theorem
Angle A = Arc BC/2
3. Angle CEB = Angle C + Angle A    3. Exterior Angle Theorem
4. Angle CEB = Arc AD/2 + Arc BC/2 4. Substitution

Angle-Secant Theorem:
The measure of an angle formed by two secants intersecting outside the circle is half the difference of the arcs intercepted by it.

Given: Secants AB and CD intersect at E
Prove: Angle E = (Arc AC - Arc BD)/2

Proof:
Statements                                            Reasons
1. Draw AD.                                          1. Through any two points there is exactly one segment.
2. Angle ADC = Arc AC/2,                   2. Inscribed Angle Theorem
Angle A = Arc BD/2
3. Angle A + Angle E = Angle ADC     3, Exterior Angle Theorem
4. Angle E = Angle ADC - Angle A      4. Subtraction Property of Equality
5. Angle E = Arc AC/2 - Arc BD/2       5. Substitution

In the end, I must admit that of all the theorems in the text, I have trouble recalling circle theorems the most.

I decided to include another Exploration Question as a bonus:

The sides of an inscribed pentagon ABCDE are extended to form a pentagram, or five-pointed star.
a. What is the sum of the measures of angles, FGHI, and J, if the pentagon is regular?

Notice that each angle satisfies the Angle-Secant Theorem. So Angle F is half the difference between CE (which is two-fifths of the circle, Arc CD + Arc DE = Arc CE = 144) and AB (which is one-fifth of the circle, Arc AB = 72). So Angle F = (144 - 72)/2 = 36 degrees. All five angles are measured the same way, so their sum is 36(5) = 180 degrees.

b. What is the largest and smallest this sum can be if the inscribed polygon is not regular.

Well, let's write out the Angle-Secant Theorem in full:

Angle F + G + H + I + J
= Arc (CD + DE - AB + DE + EA - BC + EA + AB - BC + AB + BC - DE + BC + CD - EA)/2
= Arc (CD + DE + EA + AB + BC)/2
= (360)/2 (since the five arcs comprise the entire circle)
= 180

So the largest and smallest this sum can be is 180. The sum of the five angles is a constant.

## Friday, April 27, 2018

### Lesson 15-4: Locating the Center of a Circle (Day 154)

Today on her Mathematics Calendar 2018, Theoni Pappas writes:

If this sphere's radius is 3cbrt(6pi^2)/(2pi), what is its volume?

Well, Lesson 10-8 of the U of Chicago text gives us the formula for the volume of a sphere:

V = (4/3)pi r^3
V = (4/3)pi (3cbrt(6pi^2)/(2pi))^3
V = (4/3)pi 27(6pi^2)/(8pi^3)
V = (4/3)(6/8)(pi^3/pi^3)27
V = 27

Therefore the volume is 27 cubic units -- and of course, today's date is the 27th. This is another problem where the radius is some crazy irrational number just to make the volume work out to be a whole number.

Today, my old school hosts a Math Day, sponsored by the California Math Council. Here is a link describing such a festival in more detail:

http://cmc-math.org/math-festival-programs/

Of course, while this is going on, I'm subbing in my new district, the second of five days in the Economics classroom. Today the seniors watch Boiler Room, a movie from the year 2000 whose main character becomes a stock broker. Meanwhile, the sophomore World History students watch a documentary, produced by the History Channel, about aircraft carriers during World War II.

The main classroom management issue involves a girl who asks for a restroom pass right at the start of third period. This is not the class right after snack (which would be fourth period), and so I allow her to go. But then she never returns. During the remaining three days of this assignment, I must watch out for this student. If she asks for another restroom pass, I might have to tell her that she can't go until the last 15 minutes of class, or just wait for snack.

Chapter 10 of Wickelgren's How to Solve Problems is "Topics in Mathematical Representation":

"As stated in Chapter 2, problems contain information concerning givens, actions, and goals. The first and most basic step in problem solving is to represent this information in either symbolic or diagrammatic form."

Wickelgren points out that many problems contain both forms -- for example, today's Pappas problem contains both symbolic and diagrammatic notation. I don't post the diagram part on the blog -- instead, only the symbolic form is given.

"The first step in solving such a problem is to translate from the representation given explicitly or implicitly in the original statement of the problem to a more adequate representation. This chapter is concerned with selected topics in the mathematical or precise representation of information in problems."

The author warns us that many of these problems are related to set theory, modern algebra, and combinatorial mathematics:

"Such students should consult regular mathematics books concerned with these topics, rather than try to master the material."

Even though I've written about these topics on the blog, they aren't the focus of this blog. Except for one example from Geometry (diagrammatic, of course), there isn't much of our favorite math course in this chapter. So let's keep this description mercifully brief -- a respite after yesterday's super-long post (where examples from Geometry abound).

Wickelgren lists several reasons why it's better to represent information about a problem on paper rather than in the head:

"Problems that involve tables or matrices of information are especially difficult to retain as a visual image purely in the mind."

Now let's get to the Geometry example. Wickelgren gives a diagram that we might see at the start of a Geometry problem. Since I can't post the diagram here, let's follow the author as he converts the given diagram to symbolic form:

"For example, the spatial information represented in [the figure] can be represented symbolically as follows: lines a, b, and h meet at common vertex B, lines a and d meet at vertex A, lines d, h, and c meet at vertex D, lines b and c meet at vertex C, lines c and d are collinear and line h is perpendicular to lines d and c."

I've had practice converting Pappas problems to symbolic form here on the blog. (By the way, Wickelgren's c and d are more like line segments than lines. Segment d can be also written  AD while segment c is DC.)

The authors says of postulates and other basic theorems:

"Since such prior knowledge is often assumed implicitly to be part of the givens in a problem, it is clearly important that you have access to your memory for such information."

Let's skip a few pages here. Wickelgren tells us that when converting a word problem to algebra, it's better to use as few variables as possible:

"However, it must be recognized that you are performing two steps at once: representing the concepts in the problem and expressing some relatively simple relations between concepts."

The author tells us there may be situations when subscripts are needed -- for example, if we're finding the volume of A, a complex figure consisting of a cylinder, two cubes, and a box:

"There are also complex containers B and C, each composed of subcontainers. How should you represent the volume of each subcontainer in container A, for example?"

Here's how I'd write subscripts in ASCII: V_c^A is the volume of the cylinder part of A. This means that c is a subscript and A is a superscript. It's also possible to have multiple subscripts:

"Such cases arise frequently in statistics, where x_ijk might represent the wheat yield on the kth plot of land, subjected to the ith value of one treatment dimension (for example, the amount of some kind of fertilizer), and subjected to the jth value on another treatment dimension (for example, the amount of water)."

Let's get to an example problem:

Tom, Dick, and Harry mow lawns in the summer to earn money. They each have a lawn mower, and one Saturday they decide to mow a 5,900 square foot lawn together, using all three lawn mowers. Tom mows 70 square feet per minute, Dick 50, and Harry 40. Dick and Harry start mowing the lawn at the same time, but Tom has trouble starting his mower and is delayed for 30 minutes. All three boys stop mowing at the same time, when the lawn is finished. How long does Tom mow?

Here Wickelgren lets t represent the time that Dick and Harry mow and t - 30 represent the time that Tom mows. This is better than using t_0, t_1, and t_2 for the times, or t_T, t_D, and t_H. Then he writes an equation, 70(t - 30) + 50t + 40t = 5900. I'd personally would prefer to let t represent the time that Tom mows and t + 30 for the others. Not only does this avoid negative values, but then I'm done as soon as I find t, whereas I might forget to subtract 30 from t = Dick/Harry's time. In either case, we find that Tom mows for 20 minutes.

A gym teacher wishes to put on a balancing demonstration in which one of the students will be to have four boys stand on each others' shoulders in a single tower. Out of the class of 20 boys, the gym teacher wishes to select the most stable tower of four boys once and time how long they are able to balance successfully on each others' shoulders without falling over. How many such towers of four boys must the gym teacher investigate? Stop reading and try to solve the problem.

The answer, found on a TI calculator, is 20 nPr 4 = 20!/(20 - 4)! = 20!/16!. Notice that it's nPr, not nCr, since the order matters -- one boy is on the bottom and other is on the top.

Wickelgren ends the chapter by discussing y = f(x) for function notation. One way to save ourselves a variable is to write y = y(x) -- so y is the name of the function as well as the dependent variable. (My first college math professor at UCLA wrote y = y(x) all the time.)

"This is a useful mnemonic trick in simplifying notation in problems where there is no possibility of confusing the concept of the function (f) with the concept of the dependent variable (y). However, when such confusion is possible, this trick should be avoided."

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

Lesson 15-4 of the U of Chicago text is "Locating the Center of a Circle." According to the text, if we are given a circle, there are two ways to locate its center. The first is the perpendicular bisector method, which first appears in Lesson 3-6. (Recall that the perpendicular bisectors of a triangle are a concurrency required by Common Core.) This section gives the right angle method:

1. Draw a right angle at P (a point on the circle). AB (where angle sides touch circle) is a diameter.
2. Draw a right angle at Q (another point on the circle). CD is a diameter.
3. The diameters AB and CD intersect at the center of the circle.

This method is based on the fact that a chord subtending a right angle is a diameter -- a fact learned in the previous lesson. Indeed, "an angle inscribed in a semicircle is a right angle" is a corollary of the Inscribed Angle Theorem.

Notice that unlike the perpendicular bisector method, this is not a classical construction. That's because the easiest way to construct a right angle is to construct -- a perpendicular bisector, which means that if we have a straightedge and compass, we might as well use the first method. The text writes that drafters might use a T-square or ell to produce the right angles, while students can just use the corner of a sheet of paper.

[2018 Update: Today is an activity day. I've decided to expand the Exploration Question from this lesson into a full activity page.]

"Each of the three circles below intersects the other two. The three chords common to each pair of circles are drawn. They seem to have a point in common. Experiment to decide whether this is always true."

As it turns out, these three chords are indeed concurrent, except for a few degenerate cases such as if the circles have the same center or if the centers are collinear. (The concurrency of perpendicular bisectors has the same exceptions.) The students are asked to experiment rather than attempt to prove the theorem that these three lines (called radical lines) intersect at a common point (radical center, or power center). The name "power center" refers to "power of a point" -- a dead giveaway that we must wait until Lesson 15-7 before we can attempt to prove the theorem.

## Thursday, April 26, 2018

### Lesson 15-3: The Inscribed Angle Theorem (Day 153)

Today on her Mathematics Calendar 2018, Theoni Pappas writes:

Find the volume of this right triangle pyramid with height 13' and AB = 2sqrt(6).

(AB is the hypotenuse of the base, a triangle with a 45-degree angle.)

So the base is clearly another special triangle -- 45-45-90. We determine that each of its legs must have length 2sqrt(6)/sqrt(2) = 2sqrt(3). The area of the right triangle is half the product of its legs, and so it's 2sqrt(3)2sqrt(3)/2 = 6 square feet.

The volume of the pyramid is one-third the product of its base and height, and so it's (6)(13)/3, which works out to be 26. Therefore the volume is 26 cubic feet -- and of course, today's date is the 26th.

Today I subbed in a high school economics class. It's the first of a five-day assignment, since the teacher is on an unexpected trip out of the country. In fact, one reason I couldn't have stayed in the middle school math class after Tuesday is that I was already assigned this class.

The original plan is no "Day in the Life" for non-math classes (unless it's middle school), but I do wish to have one for the first day of a multi-day assignment. The focus resolution is the first one:

1. Implement classroom management based on how students actually think.

And once again, it's because if I don't manage well on the first day, I'll have lost control of the class by the fifth day. So let's begin:

7:05 -- Yes, it's yet another zero period class -- officially called "first period" in high school. Here in California, Econ is generally a one-semester class for seniors (with Government usually as the other semester class).

Today, the students are watching a DVD episode of Frontline from about twenty years ago. The episode is on the history of the stock market. Students are to take notes and answer fifteen questions about the video. The assignment is worth one point per question.

It takes a while for me to figure out where the DVD player even is, and with the episode being just under an hour in length, I need to have started the video as soon as the class begins in order to reach Question #15 by the end of class. The video is in the middle of Question #13 when I realize that I must collect a homework assignment that's due today. So in my notes for the teacher, I inform him that students should only be responsible for Questions #1-12, since he's grading it a point a question.

As far as classroom management is concerned, the key rules to enforce are "no cell phones" and "no gum," since these rules are explicitly written on the side board.

8:00 -- First period leaves and second period arrives. In this Econ class, I know to start playing the video right away.

8:55 -- This school has a tutorial period. Only four students, all girls, attend it. As often happens in tutorial, I see one girl taking out her math assignment for an IB Math Studies class. The question she's on happens to be a stats question -- but unfortunately, stats is my weakest math class. I let her know this, and she replies that she'll just ask her teacher for help. I'm hoping that she'll move on to the next question, which is more geared towards algebra. But she succumbs to temptation -- she starts talking to the other three girls, who simply put their own (non-math) assignments away. In the end, the quartet accomplishes nothing academic during the tutorial period.

Here's what I should have done -- I should have insisted on helping the first student with her math assignment, even if the question is stats. I know enough about stats that I could assist her with parts of that question, and we could have struggled through the rest together. And by sitting next to the first girl, the others would be more encouraged to work on their own assignments.

9:30 -- Third period is slightly longer than the other classes, because this is like homeroom. The video announcements are shown at the beginning, and at the end is a senior survey where they vote for their favorite teacher. So there's just as much time for the Econ video as there is in second period.

10:30 -- Third period leaves for snack.

10:50 -- Fourth period arrives. This is the only non-Econ class -- instead, it's World History, a sophomore class in California. The students are labeling maps of Europe -- latitude and longitude, bodies of water, countries and their capitals, and so on. I tell the students that they must at least fill in all the countries today -- the rest is a homework assignment due tomorrow.

This is the period where I'm strict about students who go to the restroom right after a snack break. As it happens, two girls ask to leave -- one to go to her locker to get her history textbook (even though the teacher yesterday told the class to bring their texts the rest of the week), and the other to correct an issue with yesterday's attendance (but the regular teacher isn't here to verify that attendance).

11:45 -- Fifth period is the last Econ class. Each class runs more smoothly than the previous Econ class, since by now I know what to expect. I start the video as soon as the tardy bell rings. Question #1 isn't answered until eight minutes into the DVD -- so I have these eight minutes talk over the video, in order to take attendance and collect last night's homework. Question #15, meanwhile, is answered at the 48-minute mark -- just two minutes before the end of class. I have the students leave their worksheets on their desks, since there's no time to collect it before the dismissal bell rings.

Meanwhile, as far as enforcing the two main rules is concerned, phone use and gum chewing seem to increase as the day goes on. I point to the side board where the rules are written -- and most of all, I try to avoid arguments and yelling. Also, in one of the classes, one guy lays his head on the desk during the DVD. I remind him that many of the students are just four months away from their first college class, and napping in class won't set them up for success in college. (I don't remember which class this is, but I know it's not first period -- the class you'd expect student to enter half-asleep.)

12:35 -- Fifth period leaves for lunch, and so I collect the DVD worksheets. As it turns out, sixth period is the teacher's conference period, and most teachers who have a first period don't have a seventh (unless they're an athletic coach). Thus my day actually ends at lunch!

As it turns out, most of what Econ's doing this week is on video -- including Monday's video which will be on YouTube. On the other hand, the sophomores get only one video tomorrow -- next week, they'll start a multi-day project.

I hope I will have a successful week in the classroom, even though the class isn't math. But don't expect any more "Day in the Life" posts until I'm done with this assignment.

Chapter 9 of Wayne Wickelgren's How to Solve Problems is called "Relations Between Problems":

"When mankind has a satisfactory theory of problems, it will be possible to state many deep and detailed relations between different types of problems. But even without such a theory, we can still state certain basic types of relations between different problems."

Here are the five basic relations Wicklegren defines in this chapter:

1. Problem a is unrelated to problem b.
2. Problem a is equivalent to problem b.
3. Problem a is similar to problem b.
4. Problem a is a special case of problem b.
5. Problem a is a generalization of problem b.
He continues:

"In determining whether any of these five relations holds between two problems, it is important to note that the critical problem elements concern the types of operations and the relations that can obtain between different expressions or things, not the specific expressions or things themselves."

For example, we can compare Wickelgren's five-disk Tower of Hanoi problem to Brian Harvey's tower of six disks:

"Any two special cases of the general Tower of Hanoi problem are similar problems, though I would hesitate to call them equivalent problems. Similarly, in the nim problem...."

...well, I already linked to a similar, yet not equivalent, problem involving flags.

The author's first problem in this chapter is the fox, goose, corn problem:

A man went on a trip with a fox, a goose and a sack of corn. He came upon a stream which he had to cross and found a tiny boat to use to cross the stream. He could only take himself and one other - the fox, the goose, or the corn - at a time. He could not leave the fox alone with the goose or the goose alone with the corn. How does he get all safely over the stream?

"Stop reading and try to solve the problem by recalling the methods used to solve a similar (partly analogous) problem."

The similar problem Wickelgren has in mind is the missionaries-and-cannibals problem from all the way back in Chapter 5. In both problems, a detour is necessary to obtain the solution:

The answer is: Take the goose over first and come back. Then take the fox over and bring the goose back. Now take the corn over and come back alone to get the goose. Take the goose over and the job is done!

In this case, "bring the goose back" to the original side of the river is the detour.

Yesterday's second Square One TV song tells us to solve "simpler but similar problems," and yes, Wickelgren would agree, but he warns us that a "simpler" problem might turn out to be harder:

(Yes, there's more on the Cut the Knot site besides 100+ Pythagorean proofs!)

There are 10 stacks of 10 identical-looking coins. All of the coins in one of these stacks are counterfeit, and all the coins in the other stacks are genuine. Every genuine coin weighs 10 grams, and every fake weighs 11 grams. You have an analytical scale that can determine the exact weight of any number of coins. What is the minimum number of weighings needed to identify the stack with the fake coins?

Stop reading and try to solve the problem.

Now the author tells us that we might try to simplify the problem by having only one coin in each stack instead of ten. But in the end, that would increase the number of weighings. On the other hand, if we reduce the number of stacks, say to three, then we can reach a solution in just one weighing:

Number the coin stacks from 1 to 10. Take 1 coin from the first stack, 2 coins from the second, and so on, until all 10 coins are taken from the last stack. Weigh all these coins together. The difference between this weight and 550, the weight of (1 + 2 + ... + 10) = 55 genuine coins, indicates the number of the fake coins weighted, which is equal to the number of the stack with the fake coins. For example, if the selected coins weigh 553 grams, 3 coins are fake and hence it is the third stack that contains the fake coins.

The next problem is a Geometry example. It's not mentioned in the U of Chicago text, but it ought to be (and it does appear in other texts -- possibly the Glencoe text):

A ray of light travels from point A to B in (the figure) by bouncing off a mirror represented by the line CD. Determine the point X on the mirror such that the distance traveled from point A to point B is a minimum. What is the relationship between the angles alpha and beta?

Wicklegren tells us that the easier problem to solve is to replace B with E, where E is the reflection image of B over line CD. Then it's trivial to find the shortest distance from A to E through CD -- the straight-line distance AE. Since reflections preserve distance, X, the intersection of AE and CD, must be the desired point in the original problem. And angle alpha must equal angle gamma (the reflection image of angle beta) as they are vertical angles. Since reflections preserve angle, alpha = beta -- that is, the angle of incidence equals the angle of reflection.

This idea isn't directly stated in the U of Chicago text -- it's only hinted at in Lesson 6-4, the lesson on miniature golf and billiards.

There's one more Geometry problem in this chapter -- the walking-fly problem:

In a rectangular room (a cuboid) with dimensions 30' * 12' * 12', a spider is located in the middle of one 12' * 12' wall one foot away from the ceiling. A fly is in the middle of the opposite wall one foot away from the floor. If the fly remains stationary, what is the shortest total distance (i.e., the geodesic) the spider must crawl along the walls, ceiling, and floor in order to capture the fly?

Stop reading and try to solve the problem. (In Wickelgren's example, the fly is the hunter, not the target -- instead, a lollipop is the target.)

As it turns out, the solution is to make a room into net, using the ideas of Lesson 9-7. Click on the link above, since I won't reproduce the net here.

You are given the following: (a) A straight line equals an angle of 180 degrees, (b) A right angle equals 90 degrees. (c) If two parallel lines are cut by a transversal, the alternate interior angles are equal. Prove that the sum of the angles of any triangle equals 180 degrees. Stop reading and try to prove this theorem, making use of the method of special case.

This is, of course, the Triangle-Sum Theorem of Lesson 5-7. We already know how to prove this theorem from the U of Chicago text. But that proof doesn't involve a special case. Since one of the givens, namely (b), mentions a right angle, perhaps the special case is a right triangle.

And this is exactly what the author does. There's not much difference between the right angle case and the usual proof of the general case. But perhaps for those who don't already know the proof, the right angle case might be easier to prove first.

Here's another Geometry example given later in this chapter:

A cylindrical hole 10 inches long is drilled through the center of a solid sphere, as shown in [the figure]. What volume remains in the sphere? Stop reading and try to solve the problem, using the method of special case.

Wickelgren tells us that this problem makes an unstated assumption -- that the volume remaining in the sphere is independent of the diameter of the hole (or of the sphere). And so it's possible to consider a special degenerate case -- when the diameter of the hole is zero.

Then the length of the hole equals the diameter of the sphere -- 10 inches. A hole of zero diameter has no volume, and so the volume remaining in the sphere must its entire volume. Using the sphere volume formula, (4/3)pi r^3 = (4/3)pi 5^3 = 500pi/3. Therefore, provided that a well-defined answer exists, it must necessarily be equal to 500pi/3.

Triangle ABC is formed by three tangents to a circle, as shown in [the figure]. Angle DAE = 26 degrees. Solve for angle COB.

Let me describe the diagram, which contains important given information. The entire circle, except for the points of tangency, lies in the interior only of angle A, not the other two angles of the original given circle O. Thus Triangle ABC lies completely in the exterior of circle O, except for a single point of tangency. The points of tangency on segments AB, AC, and BC are D, E, and F respectively. Stop reading and try to solve the problem.

The statement of the problem implies that the measure of angle COB depends only on the measure of angle DAE and not on the relative location of the points. A special case to consider is to let the three points O, F, and A be collinear.

The first step the author uses is a tricky one. He writes:

"Since angle AED and angle ADE intersect the same arc of the circle, these two angles are equal."

This is tricky, since this is implied by a theorem that appears later in Chapter 15. But the subgoal here is to prove that Triangles AOE and AOD are congruent. We can do this by using HL instead -- angles AEO and ADO are both right angles (Radius-Tangent Theorem, Lesson 13-5), they each have a radius for one leg and a common hypotenuse OA. Let's pick up the rest of Wickelgren's solution from here:

• Angle EAO = DAO (CPCTC) = 13 (since angle DAE = 26)
• Angle AOE = AOD = 90 - 13 = 77 (Triangle-Sum Theorem)
• Triangles OEC and OFC are congruent (They are right triangles by Radius-Tangent, each with a radius for a leg and sharing a hypotenuse. It's interesting that Wickelgren uses HL here, but not to prove AOE and AOD congruent earlier.)
• Triangles ODB and OFB are congruent (same reason)
• Angle COF = 1/2 EOF, angle FOB = 1/2 FOD (implied by CPCTC)
• Angle COB = 1/2 EOD = 1/2 (77 + 77) = 77 degrees
and the problem is solved.

The final idea mentioned in this chapter is generalization. An example from Geometry is given by -- don't tell SteveH -- Polya:

The problem is to find a plane that passes through a given line and bisects the volume of a given octahedron.

The solution is to consider the center of symmetry of the octahedron. Then since a line and a point not on the line determine a plane (asserted by Euclid but only implied in the U of Chicago), this plane is the solution. (If the center lies on the given line, then any plane containing that line works.) This solution generalizes to any solid with a center of symmetry -- which we'll define here loosely as the intersection of all planes of symmetry, if it exists.

Wickelgren concludes the chapter:

"Thus, once again we see that the role of generalization and the role of representation of information (as discussed in Chapters 3 and 10) are very closely linked and perhaps identical."

In [the figure], angle BAD = 20, AB = AC, and AD = AE. Solve for angle CDE.

In the diagram, we have Triangle ABC with D between B and C, and E between A and C. Stop reading and try to solve the problem, making use of the method of a special case.

Wickelgren points out that not only are we not given any side lengths, but it's possible to draw many non-similar triangles satisfying these conditions --in other words, angle DAE can take on a wide range of values. In the special case, he lets DAE = 20 (that is, equal to BAD). Then he obtains:
• Angle BAC = 40 (Angle Addition)
• Angle ABD = ACD (Isosceles Triangle Theorem) = 70 (Triangle-Sum Theorem)
• Angle ADE = AED = 80 (same reason)
• Angle DEC = 100 (Linear Pair Theorem)
• Angle CDE = 180 - 100 - 70 (Triangle-Sum) = 10
The author tells us that we can choose DAE to be many other values, or even DAE = y. But it's easier just to plug in one specific value and find a solution

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

Lesson 15-3 of the U of Chicago text is on the Inscribed Angle Theorem. I admit that I often have trouble remembering all of the circle theorems myself, but this one is the most important:

Inscribed Angle Theorem:
In a circle, the measure of an inscribed angle is one-half the measure of its intercepted arc.

2018 Update: By a fortuitous coincidence, this just happens to be one of the problems given in the current Wickelgren chapter! So if you thought we were done with Wickelgren today, guess again!

"Stop reading and try to solve the problem by first considering a special case."

Well, the special case is when one of the given sides of the angle contains a diameter. This is given as Case I in the U of Chicago text:

Given: Angle ABC inscribed in Circle O
Prove: Angle ABC = 1/2 * Arc AC

Proof:
Case I: The auxiliary segment OA is required. Since Triangle AOB is isosceles [both OA and OB are radii of the circle -- dw], Angle B = Angle A. Call this measure x. By the Exterior Angle Theorem, Angle AOC = 2x. Because the measure of an arc equals the measure of its central angle, Arc AC = 2x = 2 * Angle B. Solving for Angle B, Angle B = 1/2 * Arc AC. QED Case I.

Notice that the trick here was that between the central angle (whose measure equals that of the arc) and the inscribed angle is an isosceles triangle. We saw the same thing happen in yesterday's proof of the Angle Bisector Theorem -- the angle bisector of a triangle is a side-splitter of a larger triangle, and cutting out the smaller triangle from the larger leaves an isosceles triangle behind.

Wickelgren admits that the general case divides into two subcases -- where the diameter is in the interior of the angle and whether is not in the interior. The U of Chicago calls these Cases II and III:

Let's move onto Case II. Well, the U of Chicago almost gives us a two-column proof here, so why don't we complete it into a full two-column proof. For Case II, O is in the interior of Angle ABC.

Statements                                                     Reasons
1. O interior ABC                                           1. Given
2. Draw ray BO inside ABC                            2. Definition of interior of angle
3. Angle ABC = Angle ABD + Angle DBC       3. Angle Addition Postulate
4. Angle ABC = 1/2 * Arc AD + 1/2 * Arc DC 4. Case I and Substitution
5. Angle ABC = 1/2(Arc AD + Arc DC)           5. Distributive Property
6. Angle ABC = 1/2 * Arc AC                         6. Arc Addition Property and Substitution

The proof of Case III isn't fully given, but it's hinted that we use subtraction rather than addition as we did in Case II. Once again, I bring up the Triangle Area Proof -- the case of the obtuse triangle involved subtracting the areas of two right triangles, whereas in the case where that same angle were acute, we'd be adding the areas of two right triangles.

Wickelgren writes:

"In proving the theorem for each of these two more general special cases, the truth of the theorem for the special case was used in the proof."

Here's the rest of what I wrote last year:

The text mentions a simple corollary of the Inscribed Angle Theorem:

Theorem:
An angle inscribed in a semicircle is a right angle.

The text motivates the study of inscribed angles by considering camera angles and lenses. According to the text, a normal camera lens has a picture angle of 46 degrees, a wide-camera lens has an angle of 118 degrees, and a telephoto lens has an angle of 18. I briefly mention this on my worksheet. But a full consideration of camera angles doesn't occur until the next section of the text, Lesson 15-4 -- but we're only really doing Lesson 15-3 today.