*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:

https://twitter.com/HernandoCCTM

Google celebrates a mathematician & we think THAT is worth celebrating! https://www.google.com/amp/amp.timeinc.net/time/5259372/johann-carl-friedrich-gauss-google-doodle … #mtbos #iteachmath

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?"

"The only time you ever had value to me, was when you told us about your import/export business."

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:

2

*x*+

*y*- 3

*z*= 1

*x*+ 2

*y*+ 5

*z*= 9

3

*x*- 3

*y*- 10

*z*= 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 - 5

*x*+ 7

*y*= 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 - 5

*x*+ 7

*y*= 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 - 2

*hx*+

*h*^2) + (

*y*^2 - 2

*ky*+

*k*^2) =

*r*^2

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

2

*h*= 5

*h*= 5/2

2

*k*= 7

*k*= -7/2

*x*^2 - 5

*x*+ 25/4 +

*y*^2 + 7

*y*+ 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

*E*.

Prove: Angle

*CEB*= (Arc

*AD*+ Arc

*BC*)/2

Proof:

Statements Reasons

1. Draw

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

*E*

Prove: Angle

*E*= (Arc

*AC*- Arc

*BD*)/2

Proof:

Statements Reasons

1. Draw

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,

*F*,

*G*,

*H*,

*I*, 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.

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