SBAC Practice Test Questions 5-6 (Day 164)

This is what Theoni Pappas writes on page 150 of her Magic of Mathematics:

"Namely, the diagonal numbers 42; 25, 35 convert to 42 + (25/60) + (35/3600) = 42.42638889 while 30sqrt(2) = 42.42640687."

Oops -- well, we're on the last page of the first section, "Babylonians & Square Roots: What Were They Doing with Accurate Square Root Approximations?" We missed the rest of this section due to the Memorial Day weekend.

So let's just summarize the entire section. Pappas is describing the mathematics of the ancient Babylonians, who wrote on cuneiform tablets between 3000 and 200 BC.

The first key is the Babylonian numeral system itself. The number 1 was written as a down arrow, and then 2 through 9 were written as the same number of down arrows. Then the number 10 was written as a left arrow, and multiple left arrows denote 20, 30, 40, 50.

What makes Babylonian numerals unique is that they are based on sixty (sexagesimal), not ten. We still use the Babylonian system today when we consider minutes (1/60) and seconds (1/3600). Both Babylonian and modern minute-second systems are best described as six-on-ten rather than pure sixty, so, for example, the number 71 would be written as down, left, down. Pappas writes this with a comma as 1, 11, for one in the sixties place and eleven in the ones place. But I actually prefer 1:11, since the colon reminds us of the minute-second system. In any rate, the semicolon represents the "decimal" (or sexagesimal) point, so 1; 11 is actually 1 + 11/60.

And it's these fractions that are the focus of this chapter. Pappas writes that the Babylonians were able to estimate irrational square roots in the Babylonian system. They were able to estimate sqrt(2) to three sexagesimal places as 1; 24:51:10. This value is accurate to one part in 60^3 = 216000. Pappas calculates the decimal value as:

1 + 24/60 + 51/60^2 + 10/60^3 = 1 + 2/5 + 51/3600 + 1/21600 = 1.4142129+

which compares to sqrt(2) = 1.414213562.... She writes that the Babylonians might have found this value using a repetitive approximation and division method often used by the Greeks.

Last week's SBAC questions and our discussion of approximating square roots are on my mind. Of course we don't teach the approximation and division method any more because it's cumbersome to use during the current calculator age. But it was a great way to find square roots back before calculators existed. Notice that the Illinois State method gives sqrt(2) as 1 + 1/3, while the derivative method gives 1 + 1/2. Both of these are inaccurate, but notice that halfway between these is 1 + 5/12, which is fairly good. In sexagesimal, 1 + 5/12 is 1; 25, which is the same as the Babylonian rounded off to one place. (As 51 is more than half of 60, 1; 24:51 rounds up to 1; 25.)

The Pappas approximation of 30sqrt(2) would have been a cinch for the Babylonians to find once they had sqrt(2), since 30 is half of the base 60. They only needed to multiply by 60 (one-place shift) and then divide by two.

By the way, the numbers sqrt(2) and 30sqrt(2) were found drawn on clay tablets near the diagonal of a square. This implies that the Babylonians knew that the diagonal of a square is sqrt(2) times the length -- and this might have been centuries before Pythagoras.

Speaking of irrationals, Question 5 of the SBAC Practice Exam is on irrational numbers:

5. Determine for each number whether it is a rational or an irrational number.

-- 1/sqrt(9)
-- sqrt(17)
-- -1 2/3
-- 0.423

Well, -1 2/3 and 0.423 are clearly rational. For 1/sqrt(9), we must recall that sqrt(9) = 3, so that this equals the rational number 1/3. Only sqrt(17) is irrational, since 17 is not a perfect square. So the answer must be:

rational - irrational - rational - rational

By the way, a good Babylonian approximation to sqrt(17) is 4; 7:23:11, to three sexagesimal places.

Question 6 of the SBAC Practice Exam is on the equation of a line:

6. Consider the line shown on the graph.

Enter the equation of the line in the form y = mx where m is the slope.

Naturally, the line passes through the origin since there is no b in the equation. This is another graph where the slope appears to be 1 until we look at the scale on each axis. We observe that the y-scale is 5 while the x-scale is 10. So the slope is 5/10 = 1/2, and so the equation is y = (1/2)x.

SBAC Practice Exam Question 3
Common Core Standard:
Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number.

SBAC Practice Exam Question 4
Common Core Standard:
Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.

Commentary: The first question confused my eighth graders, until they are reminded first what an irrational number is and second what sqrt(9) is. For the second question, slope should have been still familiar for the students after having seen last week's worksheet. For my students, entering y = (1/2)x on the computer is a little tricky, but fortunately, they've had practice entering equations with fractions on IXL.

SBAC Practice Test Questions 3-4 (Day 163)

This is what Theoni Pappas writes on page 146 of her Magic of Mathematics:

"Only a few of the many thousands of mathematicians and ideas are mentioned in this chapter. Those I discuss in no way reflect their superiority or stature over those not appearing. I urge you to use these sections as spring boards for further study and to enhance your understanding of the magic of the past."
[emphasis the author's]

This is last page of the chapter introduction. Pappas provides a long list of six dozen famous mathematicians (and scientists) on this page. Here are some of them:

Kepler, Galois, Einstein, Zeno, Jacobi, Godel, Lobachevsky, Laplace, Legendre, Bolyai, Eudoxus, Poincare, Thales, Riemann, Kronecker, etc.

I chose these names from her list at random. So her disclaimer applies -- the names I chose don't reflect their superiority either. As far as further study is concerned, notice that I've already devoted several posts to some of these mathematicians:

-- I mentioned Kepler and Thales while reading Morris Kline's Mathematics and the Physical World.
-- I mentioned Galois while reading Mario Livio's The Equation That Couldn't Be Solved.
-- I'm currently watching Einstein on National Geographic's Genius.
-- I mentioned Zeno's and Godel's paradoxes while watching David Kung's Mind Bending Math.
-- Lobachevsky, Bolyai, and Poincare are the fathers of hyperbolic geometry. (I mentioned Poincare.)
-- Legendre and Riemann are the fathers of spherical geometry. (I did mention these two of course.)
-- Kronecker was quoted by Stephen Hawking, the subject of the movie The Theory of Everything.

Well, Pappas, it appears that my readers and I already have a spring board for further study of the famous mathematicians -- namely this blog.

Question 3 of the SBAC Practice Exam is on irrational numbers:

3. Select True or False to indicate whether each comparison is true:

-- sqrt(37) < 5 1/4
-- 3pi > 3sqrt(3)
-- sqrt(5) > 5/7
-- 15/sqrt(10) > 8.38

This is tricky. In theory, there's supposed to be an online calculator, but I haven't been able to figure out how to access it.

Fortunately, there's a trick to answering this without a calculator. Notice that every single radicand in this question is either one more or one less than a perfect square. This makes it easy to estimate the values of the square roots -- for example sqrt(37) is a little more than 6. So rounding off gives us:

sqrt(37) < 5 1/4
6 < 5 1/4

which is clearly false. And notice that sqrt(37) is even greater than 6, so if 6 isn't less than 5 1/4, sqrt(37) can't be less than 5 1/4 either. Notice that had this been sqrt(35) rather than sqrt(37), we couldn't easily compare it to 5 1/4, since sqrt(35) is less than 6, but we don't know how much less than 6 it is, so it could be less than 5 1/4. (In reality, we must drop all the way to sqrt(27) before it becomes less than 5 1/4.) So the first answer is False.

Fortunately, all of the rounding is in the correct direction to answer all of these problems. We see that sqrt(5) is more than 2, which is definitely greater than 5/7, so sqrt(5) > 5/7 is True. We also see that sqrt(10) is greater than 3, so 15/sqrt(10) must be less than 15/3 or 5 (as increasing the denominator makes the value smaller). Since 5 isn't greater than 8.38, neither can 15/sqrt(10) be greater than 8.38, so the answer to the last part is False.

This leaves us with 3pi > 3sqrt(3), which contains pi as well as sqrt(3). We see that sqrt(3) is less than 2, so the right hand side is less than 6. We estimate pi as 3, so the left hand side is more than 9. So we have that something more than 9 is greater than something less than 6, which is clearly True. Thus the correct answer is False, True, True, False.

Question 4 of the SBAC Practice Exam is also on irrational numbers:

Select all possible values of x in the equation x^3 = 375.

-- 5cbrt(3)
-- cbrt(375)
-- 75cbrt(5)
-- 125cbrt(3)

The obvious answer is cbrt(375). It turns out that this can be simplified since 375 is thrice 125, the cube of 5. So the other correct answer is 5cbrt(3). Notice that both wrong answers involve removing factors from the radicand without taking their cube root -- both would be correct if they were given as cbrt(75)cbrt(5) and cbrt(125)cbrt(3).

SBAC Practice Exam Question 3
Common Core Standard:
Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., π2). For example, by truncating the decimal expansion of √2, show that √2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations.

SBAC Practice Exam Question 4
Common Core Standard:
Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that √2 is irrational.

Commentary: These questions frustrated my eighth graders. Earlier in the year, I talked a little about how sqrt(37) is slightly more than six, but only briefly. Notice that the Illinois State square root arts project might have helped, as could a Square Root Approximation Day. As for the second, I'd only made brief mention of how to simplify roots, so sqrt(27) = 3sqrt(3) -- and no mention of simplifying cube roots at all. Now I know what to emphasize more in the future.

Monday is Memorial Day, so my next post will be on Tuesday.

SBAC Practice Test Questions 1-2 (Day 162)

This is what Theoni Pappas writes on page 145 of her Magic of Mathematics:

"History shows that mathematical creativity is not privy to any particular culture, time period, civilization, or gender. The wealth of amazing ideas and contributions that were developed over the centuries is truly incredible and exciting to explore."

We're in the middle of the chapter introduction. Pappas is describing how so many different groups of people contributed to mathematics throughout history. She repeats the chart from page 142 showing how to count in several different numeration systems -- except for some reason, this chart counts up to 12 rather than 11.

Pappas also mentions how the concept of zero developed, independently by the Babylonians and Mayans before used in our Hindu-Arabic system.

We have reached the end of the U of Chicago text. Typically, in my first blog post after the last chapter test, I write about my most popular post of the past year. This year, it's no contest -- my number one post is my review of the movie Hidden Figures. This post has over 300 hits, nearly four times its closest competitor. Most likely, many people were searching for information about the movie and stumbled across my post.

In a way, Hidden Figures represents the truth of the Pappas quote at the start of this post. Creativity in math is not limited to any culture or gender, which is why Katherine Johnson was able to find success in math.

In the past, I would devote an entire post to the worksheet I created from my most popular entry. But of course, there is no worksheet associated with that post -- unless you count the extra credit page that I made for my students. I don't post it though, since I didn't save any copies of it.

Instead, we go straight to the next thing I usually do after completing the book -- test prep. In the past, I'd post worksheets based on the PARCC Practice Exam. But recall that I live in California, which administers the SBAC, not the PARCC. It's silly to continue posting PARCC questions when I live in one of the SBAC states -- with my students actually taking the SBAC.

In fact, I'm going to post eighth grade SBAC test prep questions this year, since I did choose after all to focus on eighth grade this year. And besides, some of the eighth grade questions are ultimately related to geometry anyway.

The reason I favored PARCC over SBAC is because the former is easier to access. Even though both tests are given on the computer, the PARCC website has a written practice test -- which I used the first two years of this blog. The SBAC website forces you to take the practice test at once -- and it's impossible to reach Question 33 (the last practice question) without answering the first 32. This is very inconvenient for setting up worksheets based on one or two questions at a time.

In the past, I reviewed only one question each day. There is less time available, so this time I'll review two questions per day. And yes, I know that my students finished the SBAC last week, and most teachers reading this have probably finished state testing as well, but I'm posting these worksheets now anyway.

Question 1 of the SBAC Practice Exam is on exponents:

1. Enter the value of n for the equation 5^n = 5^11 * 5^3.

This question is straightforward for those who remember the laws of exponents. To multiply powers we just add exponents, so the correct answer is 11 + 3 = 14.

Question 2 of the SBAC Practice Exam requires a graph:

2. The distance (d) in meters a car travels in t seconds is shown in the table.

d          t
10        1
20        2
30        3
40        4
50        5

Use the Add Arrow tool to graph the proportional relationship between the distance (d) traveled by a car and the time (t).

This question is tricky because the distance is shown on the y-axis even though the table lists distance in the left column, where the x-values usually are given. The scale on the distance axis is 10 while the scale on the time axis is 1, so the graph ends up looking like the identity function even though its slope is officially 10, not 1.

SBAC Practice Exam Question 1
Common Core Standard:
Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 32 × 3-5 = 3-3 = 1/33 = 1/27.


SBAC Practice Exam Question 2
Common Core Standard:
Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed.

Commentary: In this section, I'll write about how my own eighth graders dealt with these questions as they faced them. Naturally, many of them forgot the laws of exponents. This might have been a good time to sing my exponents song -- the eighth grade verse of my parody of the UCLA fight song. The graphing was a little tricky because of the x-y reversal.

Chapter 15 Test (Day 161)

This is what Theoni Pappas writes on page 144 of her Magic of Mathematics:

"In most sciences one generation tears down what another has built and what one has established another undoes. In mathematics alone each generation adds a new story to the old structure." -- Hermann Hankel, 19th century German mathematician

"In these days of conflicts between ancient and modern studies, there must surely be something to be said for a study which did not begin with Pythagoras and will not end with Einstein, but is the oldest and youngest of all." -- G.H. Hardy, 20th century British mathematician (who helped Ramanujan!)

Pappas begins Chapter 6 with a few pages of introductory material. Unfortunately, we won't get past these intro pages and start the first section of Chapter 6 until next week.

Here are the Chapter 15 Test answers:

1. 144pi - 288 square units.
2. 178 degrees.
3. Arc DE = 65 degrees.
4. Many answers are possible, for example Angle A = 47.5 degrees.
5. 14 degrees.
6-7. These are visual, so I can't put the answers here.
8. 21.
9. a. (-2, 9). b. 7. c. Many answers are possible. To find lattice points on the circle, we go right, left, up, and down seven units, to obtain (5, 9), (-9, 9), (-2, 16), and (-2, 2).
10. a. (0, 0). b. sqrt(72). c. This time, sqrt(72) = 6sqrt(2), so we can go diagonally to find lattice points on the circle, to obtain (6, 6), (-6, 6), (6, -6), and (-6, -6).
11. This is the complete the square question -- included because such problems are on PARCC!
x^2 + y^2 - 8y = 9
x^2 + y^2 - 8y + 16 = 25
x^2 + (y - 4)^2 = 25
So this gives us:
11. a. (0, 4). b. 5. c. (5, 4), (-5, 4), (5, 9), and (5, -1).
12. a. A circle with radius 20 feet. b. 40pi feet.
13. Draw any circle.
14. About 1.68%. I set this question in Brazil because the Olympics are about to start!
15. 15.
16. Cavalieri's Principle. Take that, traditionalists!
17. a. When the line and circle intersect in a point. b. When the line is perpendicular to the radius at the point of tangency. PARCC contains a few tangent problems, and all of them appear to involve angle measures, so that right angle is important.
18. a. 36 degrees. b. 18 degrees. PARCC also contains problems on inscribed angle measure -- possibly in the same question as tangents.
19. a. About 2.5 or 2.6 cm. b. The ratio to the circumference to the diameter is -- what else -- pi. We see that we estimate pi as either 3.08 or 3.2 using this measurement. Interestingly enough, 3.14 is almost exactly halfway between these two estimates.
20. a. 24 square units. (The height is 4, using the Pythagorean Theorem) b. About 38.5 square units.

Review for Chapter 15 Test (Day 160)

This is what Theoni Pappas writes on page 143 of her Magic of Mathematics -- a table of contents for the new Chapter 6, "Mathematical Magic From the Past":

-- Babylonians & Square Roots
-- The Chinese Method of Piling Squares
-- The Ladder That Inches on the sqrt(2)
-- Early Random Number Generators
-- Egyptian Multiplication
-- The First Scientific Laboratory
-- Plato Doubles the Square
-- The Romans & the Area of a Circle
-- How the Gnomon Trisects an Angle
-- Unsolved Mathematical Mysteries
-- Fermat's Last Theorem
-- Galileo & Proportion
-- Mathematics & Containers
-- Geometries -- Old & New
-- What's in a Name?
-- Euler's Magic Formula -- F + V - E = 2

There are many fascinating things for Pappas to discuss in this chapter -- and I definitely look forward to writing about them.

Let's get to today's Geometry lesson. Notice that Chapter 15 contains nine sections, so normally we'd have to squeeze in the Chapter 15 test today (Day 160) so we could begin Lesson 16-1 tomorrow (which is Day 161). But notice that Chapter 15 is the last of the U of Chicago text. Since there is no Chapter 16, we can take our time and review today for the test tomorrow.

Last year I wrote a "Chapter 15 Test." But last year I didn't cover Chapter 15 as a whole -- instead it was a general "circles and spheres" test with questions from five different chapters. This year we covered Chapter 15 straight through, so I could post a genuine Chapter 15 test today.

I decided to take last year's test and create a new first page, since all of the Chapter 15 questions are included on the second page. My new first page includes two questions each from Lessons 15-3 and 15-4 and one each from 15-1, 15-5, 15-6, and 15-7.

Last year's Page 2 contains some questions from Lessons 15-3 and 15-8. But there are also some questions from different sections.

Questions 9 through 11 are on the equation of a circle, Questions 14 and 16 are on the volume/surface area of a sphere, Question 15 is on the Pythagorean Theorem, and Question 17 is on the relationship between radius and tangent.

The new version of the test still contains these questions from four other chapters, but now most of the questions are from Chapter 15.

Lesson 15-9: The Isoperimetric Theorems in Space (Day 159)

This is what Theoni Pappas writes on page 142 of her Magic of Mathematics:

(nothing)

This is another introduction page. Chapter 6 of the Pappas book is "Mathematical Magic From the Past," and the actual title page for this chapter is on page 143.

But as usual, Pappas doesn't simply leave page 142 blank. Instead, she fills this page by counting from zero to eleven in several different numeration systems:

-- Hindu-Arabic
-- Babylonian
-- Greek
-- Egyptian Hieroglyphic
-- Chinese Script
-- Hebrew
-- Chinese Rod Numerals
-- Roman
-- Egyptian Hieratic
-- Mayan
-- Binary Numerals

Not all of these numbers can be rendered easily in ASCII though. The Hindu-Arabic numerals from zero to eleven are very simply rendered in ASCII:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11...

That's right -- the standard numerals we use today are actually Hindu-Arabic. The Roman numerals are obviously easy to type in ASCII as well:

I, II, III, IV, V, VI, VII, VIII, IX, X, XI...

Notice that the Roman, as well as some of the others, begin with one rather than zero, as these systems lack a zero. The final set of numbers easy to type in ASCII are the binary numerals:

0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011...

With the Greek numerals, the number 1 is A and 2 is B, but we can't do 3 in ASCII. That's because the A is actually "alpha" and the B is "beta," but the 3 is "gamma," which doesn't look like any letter that we can type in ASCII. Number 9 is "theta," a letter which we associate with angles.

As it happens, both Greek and Hebrew simply use the first ten letters of the alphabet for 1-10. The symbol for 1 in Hebrew is "aleph," which Cantor used to denote infinity. The fact that every letter in Hebrew and Greek is a number is used in the "Bible Code," where people try to find hidden messages in the scriptures by converting the words in the original language to numbers.

Lesson 15-9 of the U of Chicago text is on "The Isoperimetric Theorems in Space." These are the 3D analogs of the theorems we discussed on Friday.

Isoperimetric Theorem (space version):
Of all solids with the same surface area, the sphere has the most volume.
Of all solids with the same volume, the sphere has the least surface area.

We don't even bother trying to prove these theorems. As we've seen, the 2D proofs are very difficult, so imagine how much more so the 3D proofs would be.

This is the final lesson in the U of Chicago text. Here is how the U of Chicago closes the text:

"The Isoperimetric Theorems involve square and cube roots, pi, polygons, circles, polyhedra, and spheres. They explain properties of fences, soap bubbles, and sponges. They demonstrate the broad applicability of geometry and the unity of mathematics. Many people enjoy mathematics due to the way it connects diverse topics. Others like mathematics for its uses. Still others like the logical way mathematics fits together and grows. We have tried to provide all these kinds of experiences in this book and hope that you have enjoyed it."

Well I for one have certainly enjoyed this text, and I hope you, the readers of this blog, have as well.

One of the bonus Exploration questions mentions the ancient Carthaginian queen Dido. I wrote about her last year as well:

"According to one of the legends of history, Dido, of the Phoenician city of Tyre, ran away from her family to settle on the Mediterranean coast of North Africa. There she bargained for some land and agreed to pay a fixed sum for as much land as could be encompassed by a bull's hide."

"Her second bright idea was to use this length to bound an area along the sea. Because no hide would be needed along the seashore she could thereby enclose more area."

We know that the solution to the Isometric Problem is the circle -- the curve that encloses the most area for its length. We've also seen questions in which we are to maximize area by building a fence along a river to enclose a rectangular area -- the answer is a rectangle whose width is exactly half of its length. Combining these two ideas, we can solve the Dido problem:

"According to the legend, Dido thought about the problem and discovered that the length of hide should form a semicircle."

So we see that without water, the largest area is a circle -- with water, it's a semicircle. If we restrict to rectangles, without water the largest area is a square -- with water, it's a semi-square (that is, half of a square, or a rectangle whose width is half of its length).

"[Dido's new lover] Aeneas was a man on a mission, and he soon departed to found a new civilization in Rome. Dejected and distraught, Dido could do no more for Aeneas than to throw herself on a blazing pyre so as to help light his way to Italy...Rome made no contributions to mathematics whereas Dido might have."

By the way, let's tie this back to Pappas and ask, what numerals would Dido have used? Carthage is actually derived from the Phoenician culture. Notice that the Square One TV video "The Mathematics of Love" seems to imply that the Phoenicians used our (current) numerals (in contrast with the Roman), even though Phoenician has nothing to do with Hindu-Arabic. So in the end, I don't know what numerals Carthaginians might have used.

Lesson 15-7: Lengths of Chords, Secants, and Tangents (Day 157)

This is what Theoni Pappas writes on page 138 of her Magic of Mathematics:

"If you're a surfer, you know it's sometimes difficult to know in advance when the surf will be up. Sometimes the surf appears perfect from the beach, but when you get in the water it's died down, so you wait for that perfect wave for what sometimes seems like hours."

This is a new section, "Mathematics Rides the Crest of the Wave." In this section, Pappas writes about many of the factors influencing wave size -- the wind, an earthquake, the wake of a boat, and the moon and sun controlling the tides. But she doesn't reach the math part until the next page.

Before we get to today's lesson, let me point out that our two weeks of testing are almost over. If I had been still participating in Tina Cardone's "Day in the Life" project, today would have been my post for the special day "A Testing Day," because it would have also doubled as my monthly posting date of the 18th.

Lesson 15-7 of the U of Chicago text is on "Lengths of Chords, Secants, and Tangents." Here are the two big theorems of this lesson:

Secant Length Theorem:
Suppose one secant intersects a circle at A and B, and a second secant intersects the circle at C and D. If the secants intersect at P, then AP * BP = CP * DP.

Given: Circle O, secants AB and CD intersect at P.
Prove: AP * BP = CP * DP.

There are two figures, depending on whether P is inside or outside the circle, but proofs are the same.

Proof:
Statements                    Reasons
1. Draw DA and BC.     1. Two points determine a line.
2. Angle BAD = BCD,  2. In a circle, inscribed angles intercepting
    Angle ADC = ABC       the same arc are congruent.
3. Triangle DPA ~ BPC 3. AA~ Theorem (steps 2 and 3)
4. AP / CP = DP / BP    4. Corresponding sides of similar
                                           figures are proportional.
5. AP * BP = CP * DP   5. Means-Extremes Property

This leads, of course, to the definition of power of a point.

Tangent Square Theorem:
The power of point P for Circle O is the square of the length of a segment tangent to Circle O from P.

Given: Point P outside Circle O and Line PX tangent to Circle O at T.
Prove: The power of point P for Circle O is PT^2.

Proof:
Statements                    Reasons
1. Draw Ray TO which 1. Two points determine a line.
intersects Circle O at B.
2. Let PB intersect         2. A line and a circle intersect in at most
Circle O at A and B.           two points.
3. PT perpendicular TB 3. Radius-Tangent Theorem and def, of semicircle
   and TAB in semicircle    
4. PTB right triangle     4. Definitions of right angle, right triangle,
with altitude TA,                and altitude
5. PT^2 = PA * PB        5. Right Triangle Altitude Theorem
6. The power of point P 6. Definition of power of a point
for Circle O is PT^2

By the way, we can now finally prove the Bonus Question from Lesson 15-4. I think I'll dispense with two-column proofs here and give a paragraph proof. We begin with a lemma:

Lemma:
Suppose two circles intersect in two points. Then for each point on their common secant line, the power of that point for first circle equals the power of that point for the second circle.

Given: Circles A and B intersect at C and D, Point P on secant CD
Prove: The power of point P for Circle A equals the power of point P for Circle B

Proof:
For both circles, the power of P is CP * DP, no matter whether P is inside or outside the circle. This common secant has a name -- the radical (or power) axis of the two circles. QED

Theorem:
Suppose each of three circles, with noncollinear centers, overlaps the other two. Then the three chords common to each pair of circles are concurrent.

Proof:
The proof of this is similar to that of the concurrency of perpendicular bisectors of a triangle (which I'll compare to this proof in parentheses). Let A, B, and C be the three circle centers. Every point on the radical axis of A and B has the same power for both circles. (Compare how every point on the perpendicular bisector of XY is equidistant from the points X and Y.) Every point on the radical axis of B and C has the same power for both circles. (Each point on the perpendicular bisector of YZ is equidistant from Y and Z.) So the point where these chords intersect has the same power for all three circles, and thus must lie on the radical axis of Circles A and C. (So the point where the perpendicular bisectors of XY and YZ must be equidistant from all three points, and so must lie on the perpendicular bisector of XZ.) The point where all three radical axes intersect is called the radical center (for perpendicular bisectors, it's called the circumcenter.) QED

Notice that if three centers are collinear, then the three radical axes are parallel (just as if three points are collinear, then the perpendicular bisectors are parallel).

I decided to make this the activity day for this week, since the Exploration section includes two questions rather than one. This is more like a puzzle, since the key to both questions isn't Geometry but arithmetic (or Algebra) and the properties of multiplication!

Lesson 15-6: Angles Formed by Tangents (Day 156)

This is what Theoni Pappas writes on page 137 of her Magic of Mathematics:

"As nature puts forth its wonders in the garden, most people are oblivious to the massive calculations and mathematical work that have become so routine in nature."

Here Pappas wraps up the story about of the gardener who is surprised by all of the mathematics that appears in her garden. On this page she writes about a triple junction -- a point where three line segments meet and the angles at the intersection are each 120 degrees. According to Pappas, triple junctions are found in soap bubbles, as well as the cracking of earth or stone and on kernels on a cob of corn (bringing this back to the garden).

Lesson 15-6 of the U of Chicago text is on "Angles Formed by Tangents." The theorems in this lesson are similar to those in yesterday's lesson.

Tangent-Chord Theorem:
The measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc.

Given: AB chord of Circle O, Line BC tangent to Circle O
Prove: Angle ABC = Arc AB/2

Proof:
Statements                                     Reasons
1. Draw diameter BD.                    1. Through any two points (B, O) there is exactly one line.
2. Arc AD = 180 - AB                    2. Arc Addition Postulate
3. CB perpendicular BD                 3. Radius-Tangent Theorem
4. Angle ABC = 90 - ABD             4. Angle Addition Postulate
5. Angle ABC = 180/2 - Arc AD/2 5. Inscribed Angle Theorem
6. Angle ABC = (180 - Arc AD)/2 6. Distributive Property
7. Angle ABC = Arc AB/2              7. Substitution Property of Equality

Tangent-Secant Theorem:
The measure of the angle between two tangents, or between a tangent and a secant, is half the difference of the intercepted arcs.

Given: Line AB secant, Ray EC tangent at point C, forming Angle E,
Arc AC = x, Arc BC = y
Prove: Angle E = (x - y)/2

Proof ("between a tangent and a secant"):
Statements                                     Reasons
1. Draw AC.                                   1. Through any two points there is exactly one line.
2. Angle DCA = x/2, EAC = y/2     2. Inscribed Angle Theorem
3. Angle DCA = EAC + E              3. Exterior Angle Theorem
4. Angle E = DCA - EAC               4. Subtraction Property of Equality
5. Angle E = x/2 - y/2                     5. Substitution Property of Equality
6. Angle E = (x - y)/2                     6. Distributive Property

In the text, the "between two tangents" is given as an exercise. The Given part of this proof with the way the points are labeled is completely different from the first part.

Given: Ray PV tangent at Q, Ray PU tangent at R
S on Circle O (same side of QR as P), T on Circle O (opposite side of QR as P)
Prove: Angle P = (Arc QTR - QSR)/2

Proof ("between two tangents"):
Statements                                     Reasons
1. Draw QR.                                   1. Through any two points (B, O) there is exactly one line.
2. Angle VQR = Arc QTR/2,          2. Inscribed Angle Theorem
    Angle PQR = Arc QSR/2
3. Angle VQR = PQR + P              3. Exterior Angle Theorem
4. Angle P = VQR - PQR               4. Subtraction Property of Equality
5. Angle P = Arc QTR/2 - QSR/2  5. Substitution Property of Equality
6. Angle P = (Arc QTR - QSR)/2   6. Distributive Property

In some ways, the Tangent-Chord Theorem is just like yesterday's Angle-Chord Theorem, except that one of the intercepted arcs is 0 degrees. The bonus question concerns a solar eclipse.

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

This is what Theoni Pappas writes on page 136 of her Magic of Mathematics:

"She [yesterday's gardener] had no idea that the garden abounded with equiangular spirals. They were in the seed-heads of the daisies and various flowers."

So Pappas writes about an equilateral spiral, also known as a logarithmic spiral. Equilateral spirals are so-called because at each point on the spiral, the angle formed by segment joining that point to the center of the spiral and the tangent (in the Calculus sense) at that point is constant. Circles also have that property -- the constant angle equals 90 degrees. Equiangular spirals occur when the constant angle is not 90.

Equiangular spirals are self-similar in that there exists a dilation mapping the spiral to itself. The center of the dilation is, of course, the center of the spiral. The magnitude k of this dilation depends only on the constant angle theta -- k = e^(2pi/tan theta). If k is known, then we can solve this for theta to obtain theta = arctan(2pi/ln k) -- and there's the "logarithm" in the name. It turns out that a dilation of other scale factors is actually equivalent to a rotation of the spiral!

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, 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.

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

This is what Theoni Pappas writes on page 135 of her Magic of Mathematics:

"How well the entire garden was shaping up and exploding with new growth! Admiring the new green leaves..."

This page is right in the middle of a new section, "The Mathematically Annotated Garden." Pappas writes this section as a story told from the perspective of a gardener admiring all of the various plants in her backyard. On this page, she notices the symmetry of the maple leaf, the phyllotaxis nature uses to arrange the leaves on branches in stems, and the tessellation of carrots in the patch.

Symmetry and tessellations are both Common Core topics -- they refer to figures that are invariant with respect to reflections and translations respectively. Phyllotaxis is a new topic, though. It refers to the observation that the number of leaves on many plants is a member of the Fibonacci sequence -- 1, 1, 2, 3, 5, 8, and so on.

Before I get to today's lesson, I want to mention the news that First Son Barron Trump's school for the fall has been announced. Of course, I try to avoid politics during school year posts, but I've written so much about Presidential Consistency that I want to write about this now. Recall that Presidential Consistency stems from the frequent accusation that Common Core is knowingly terrible, yet politicians promote the Core because they know that they'll send their children to private schools that don't use the Core.

Therefore, Presidential Consistency is the notion that whenever a new president with school aged children is elected, the curriculum at the children's new school is automatically defined to be the new Common Core.

Barron's new school is St. Andrew's Episcopal. Here's a link to its curriculum:

https://bbk12e1-cdn.myschoolcdn.com/ftpimages/658/misc/misc_138116.pdf

Many people have noted that unlike the children of previous presidents, Barron waited until a brand new school year before moving to the DC area. Since he's eleven years old, this means that he'll be a sixth grader -- so waiting for summer makes it a clean transition from elementary to middle school.

As we look over the curriculum at St. Andrew's, there's much for traditionalists to like:

-- Unlike Sidwell Friends, there is no integrated pathway. All students must take the traditional classes Algebra I, Geometry, and Algebra II to graduate from St. Andrew's.
-- The highest class offered is Multivariable Calculus. Therefore students can take Algebra I as early as seventh grade in order to make it to Multivariable Calc.

I notice that there's a class called Functions and Trig -- I assume that this corresponds to Algebra III at Sidwell and other schools. There's also a non-AP Calculus class in addition to both AB and BC. I find it interesting that even though a B+ in Algebra II is required for Honors Pre-Calc, where a B is required to advance to AP Calc, any Calc class (including non-AP Calc) may be used to advance to Multivariable Calc!

Now let's look at what traditionalists aren't going to like -- the middle school curriculum:

-- The sixth grade class is Connected Math. As described on the St. Andrew's website, "CMP is a problem-centered curriculum where concepts are sequenced and developed to allow students to explore them in depth."

The traditionalist Barry Garelick doesn't have much to say about Connected Math, but Katharine Beals was outspoken in her opposition to this curriculum. Here's a link to her old website, where at this time four years ago, she featured Connected Math in her "Math problems of the week" series:

http://oilf.blogspot.com/2013/05/math-problems-of-week-6th-grade_16.html

In many ways, CMP 6 and CMP 7 correspond to Common Core 6 and 7, respectively, but the eighth grade course is called Pre-Algebra. So there's more algebraic content ("operations with polynomials") and a little less geometric content than in Common Core 8. Unfortunately for traditionalists, the word "project" appears three times in the description of the Pre-Algebra course.

All students take Pre-Algebra before Algebra I. Therefore students on the Multivariable Calc track take Pre-Algebra in sixth grade and those on the AP Calc track take Pre-Algebra in seventh -- and can take the CMP 7 class in sixth grade.

We can't be sure which class Barron will take next year. We note that his father is an outspoken opponent of Common Core, but even if Barron takes Pre-Algebra in sixth grade he'll still have to complete those three darned "projects" that Core opponents dislike.

Suppose we wanted to implement Presidential Consistency and adjust the Common Core so that it matches the St. Andrew's curriculum. The best way to do so would be to abolish the Common Core 8 standards and replace it with St. Andrew's Pre-Algebra. The resulting PARCC and SBAC tests for Pre-Algebra can be administered to students in any middle school grade. Students who score sufficiently high on the Pre-Algebra PARCC can take the Algebra I End of Course PARCC the following year rather than any middle school test. Those in SBAC states who score high enough in Pre-Algebra can be exempt from taking any more SBAC math tests until high school.

Notice that transformational geometry isn't mentioned in normal Geometry -- but it is mentioned under Honors Geometry. The final unit of Honors Geometry is on spherical geometry. Perhaps in the name of Presidential Consistency, transformations should be dropped from all Geometry courses unless they are Honors courses.

St. Andrews has a Labor Day Start Calendar, so Barron will start there in September. The school uses trimesters for all grades -- including high school, where trimesters are a rarity. Usually, trimester high schools are on a 5 x 3 block schedule. but this doesn't seem to be the case at St. Andrews. According to the website, high school students do indeed take five classes -- which means five academic classes, each of which is a full year. The bell schedule has eight periods, but one of these is lunch. The remaining seven periods are for the five academic classes, and then two periods for non-academic courses like art, P.E., and religion, which are the only trimester-length classes.

By the way, note that my own middle school doesn't offer any sports. St. Andrew's is the exact opposite -- the only P.E. offered to middle school students is competitive sports! So Barron will have to choose sports for at least the first and third trimesters -- the second trimester, which is in the winter, is optional. Some people speculate that Barron will choose golf and soccer -- but these are both fall sports, so he'll need to choose another sport for the third trimester.

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.

There are a few things I want to say about today's worksheet. One of the questions from the text mentions a fictional school, Emmy Noether High School. This isn't just a random name -- it's in fact the name of a famous mathematician. I wrote about her on the blog two years ago:

"At Göttingen, after 1919, Noether moved away from invariant theory to work on ideal theory, producing an abstract theory which helped develop ring theory into a major mathematical topic. Idealtheorie in Ringbereichen (1921) was of fundamental importance in the development of modern algebra."

The other thing I want to mention is the bonus question. This is an Exploration question in the text:

"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.

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

This is what Theoni Pappas writes on page 132 of her Magic of Mathematics:

"Select a continent. Let's use South America as an example. Suppose we want to measure the distance of its coastline."

This section is called "Fractalizing the Surface of the Earth." Hey -- we're still talking about fractals, so let's continue with what I wrote about Mandelbrot last year:

Part III of Mandelbrot's The Fractal Geometry of Nature is called "Galaxies and Eddies," and consists of Chapters 9 through 11. In this chapter, Mandelbrot argues that many problems that appear in science can be solved by using fractal geometry.

For example, in Chapter 9, Mandelbrot discusses the distribution of stars in the universe. He writes about Olbers paradox, which poses the question, if the universe is uniformly full of stars -- that is, if there are stars everywhere -- then why is the sky dark at night? Mandelbrot -- who prefers to call this the "Blazing Sky Effect" rather than Olbers paradox -- proposes that it could be because the universe is not uniformly full of stars. Instead, the universe could be a fractal with dimension less than 3, and so the distribution of stars follows a fractal pattern.

In the ensuing chapters, Mandelbrot mentions other scientific ideas -- turbulence and fluid motion -- that could also be modeled using fractals. So we see that these fractals are hardly some abstract mathematical idea that has no practical significance, but instead have some scientific value.

Returning to 2017, let me say that Mandelbrot does write about coastlines in his Part VII. So coastlines are another scientific problem that is solved using fractal geometry. Pappas writes about how the length of a coastline seems to increase when measured with a shorter ruler. The reason is that the coastline doesn't have a finite one-dimensional length. The dimension of the coastline is between 1 and 2 somewhere.

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.

The text divides the proof into three cases -- depending on whether the center of the circle is inside, outside, or on the inscribed angle. The easiest case occurs when the center is on the angle, and this case is used to prove the other two cases. The exact same thing occurred when we proved the Triangle Area Formula back in Chapter 8. Then, the three cases were whether the altitude was inside, outside, or aside of the triangle -- and the case when the altitude was aside of the triangle (i.e., when the triangle was a right triangle) was used to prove the other two.

I will reproduce the paragraph proof of the Inscribed Angle Theorem from the U of Chicago:

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.

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.

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.