Today on her Daily Epsilon on Math 2021, Rebecca Rapoport writes:
If the smaller circle rotates without slipping within the larger circle, what is the length of the path of P?
(Here are some givens from the diagram: The two circles are internally tangent at P. The diameter of the smaller circle equals the radius of the larger circle, namely 7.)
We can use Common Core transformations -- specifically rotations to solve this problem. There are two rotations involved -- the revolution of the smaller circle as it moves inside the larger circle, and the rotation of the smaller circle around its own center.
This is similar to our planet's path around the sun -- it both revolves around the sun every year and rotates around its axis each day. But a more apt comparison is to two interlocking gears. If we turn one gear, the other turns in the opposite direction. And if the radius of one gear is half that of the other, then it will rotate twice as fast. Even though one circle is inside the other, the same analogy applies -- the two rotations are in opposite directions, with one twice the magnitude of the other.
OK, so this tells us what the two rotations are. If the center of the larger circle is N, then the rotation centered at N has magnitude theta. And if the center of the smaller circle is O, then the rotation centered at O has magnitude -2theta. Notice that N, O, P are collinear -- indeed, PN is the diameter of Circle O.
Our goal is to see what affect these rotations have on point P:
(R_O,-2theta o R_N,theta)(P)
But what is the composite of these two rotations? By the Two Reflection Theorem for Rotations, this is equivalent to the composite of four reflections. And we can use the theorem to find suitable positions for the four mirrors. For the second rotation centered at O, the mirrors must intersect at O and have an angle of -2theta/2 = -theta between them. So we can let the line through points N, O, P be the first mirror and line l, where l intersects this line at O at angle -theta, be the second mirror:
(r_l o r_PN o R_N,theta)(P)
As for the first rotation centered at N, we need two lines intersecting at N with angle theta/2. We could let PN be the first mirror and the second mirror intersect it at theta. But we want the second-first mirror (second mirror for first rotation) match the first-second mirror (first mirror for second rotation). So instead, PN is the second mirror and the first mirror m intersects this at N with angle -theta/2 (since going from -theta/2 to 0 is like going from 0 to +theta/2 -- both are going +theta/2).
(r_l o r_PN o r_PN o r_m)(P)
The reason we want the two inner mirrors to match is because of the Flip-Flop Theorem -- the composite of any reflection with itself is the identity:
(r_l o I o r_m)(P)
(r_l o r_m)(P)
So we are left with the composite of two reflections. These two mirrors intersect PN at different angles, and so they can't be parallel. Thus this composite is another rotation, centered at the point where the two mirrors intersect, which we'll call Q. And the magnitude of this rotation must be -theta, since the angle between the mirrors is -theta/2. The first mirror has an orientation of -theta/2 and the second has orientation -theta, so we're rotating in the negative direction.
R_Q,-theta(P)
Now let's find some angles. From the way we set up the mirrors, we have Angle PNQ = theta/2 and Angle POQ = theta. POQ is the sum of the remote interior angles, so NOQ = theta/2. Since Triangle NOQ has two angles of measure theta/2, it must be isosceles -- OQ = ON.
But we already know that OP = ON since O is the center of a circle with P and N on the circle. And so we have OP = OQ, and so Triangle OPQ is isosceles, with vertex angle POQ = theta. And so its base angles OPQ = OQP = 90 - theta/2.
So Angle NQP = NQO + OQP = theta/2 + 90 - theta/2 = 90 -- that is, NQP is a right angle. And if we drop a perpendicular from Q to PN and label the foot R, then we just divided right triangle NQP into two similar right triangles. So Angle PQR = theta/2.
Now let P' be the reflection image of P over line QR. Then P' lies on line PN since this line is perpendicular to the mirror. We now claim that:
R_Q,-theta(P) = P'
This is because P' lies on the correct ray -- PQR = theta/2, so PQP' = theta -- and P' is the correct distance from Q -- QP = QP'. Thus the image of P under a rotation centered at Q, of magnitude -theta, is exactly P'.
What matters is that P' lies on line PN -- so the path whose length we're asked to find is merely a line segment on PN. To find its length, we only need to figure out how far the image can get from P.
Notice that if we replace theta with -theta, PQN is reflected over line PN, but P' works out to be the same point for both angles. This suggests that as theta goes from 0 to 360, the image of P will slide from 0 to 180 and then return from 180 to 360. The farthest the image goes must be when theta = 180:
(R_O,-2theta o R_N,theta)(P)
(R_O,-360 o R_N,180)(P)
(I o R_N,180)(P)
R_N,180(P)
So we rotate P around N 180 degrees, which makes P' be the point diametrically opposite from P on the other side of Circle N. Thus PP' is just the diameter of N. And since its radius is given to be 7, its diameter is 14. Therefore the desired length is 14 -- and of course, today's date is the fourteenth.
The fact that P moves in a straight line as the circle rotates is exploited in a Numberphile video:
Today is the third day of my four-day assignment in the high school math class. Thursdays are even periods, and Cohort B meets in person. As we saw on Tuesday, first period is Geometry, third period is Algebra I, and fifth period is conference.
Of the in-person classes, third period Algebra I has the most students -- ten in-person students, out of a dozen who signed up for hybrid, out of 19 total students in Cohort B. This is somewhat expected, as we know that younger students are more likely to attend in-person than older students.
These students are learning how to multiply polynomials. While the regular teacher did show them FOIL earlier for multiplying two binomials, today they multiply a binomial by a trinomial. For these, the teacher shows them the box method -- another method introduced by the famous teacher-blogger Sarah Carter:
https://mathequalslove.net/multiplying-polynomials-using-box-method/
With ten students in the classroom today, I decide that this is enough to justify a song. As I promised earlier, it's the "GCF Song," which anticipates their upcoming unit on factoring polynomials. I discussed this song during the summer -- I won't repeat that discussion, but I post the lyrics and the final tune that I came up with:
GCF!
Greatest Common Factor (A-highC-lowC-E-G-A)
GCF!
List every factor (A-lowC-E-G-A)
GCF! (D-E-A)
Circle the ones in common (lowC-C-D-G-low-C-C-D)
GCF! (D-E-A)
Choose the biggest one (lowC-D-lowC-D-lowC)
LCM! (G-lowC-F)
Least Common Multiple (A-highC-lowC-E-G-A)
LCM! (G-lowC-F)
List some multiples (A-lowC-E-G-A)
LCM! (D-E-A)
Circle the ones in common (lowC-C-D-G-low-C-C-D)
LCM! (D-E-A)
Choose the smallest one (lowC-D-lowC-D-lowC)
After I finished teaching at my middle school, I had my guitar fixed and then kept it at home. So I only perform vocally today, so even though I'm adding the "music" link today, I won't make any discussion of guitar chords. I will make another guitar post soon, but not now.
Lecture 3 of Prof. Arthur Benjamin's The Mathematics of Games and Puzzles: From Cards to Sudoku is called "Optimal Blackjack and Simple Card Counting." Here is a summary of the lecture:
- Among all the games that you may play against the casino, Blackjack is the game that gives you the best chance of winning if you play it properly. On the other hand, if you don't play the rihgt strategy, Blackjack can be a very expensive game.
- The name Blackjack refers to getting an ace and a 10 (including face cards), which pays 3:2 unless the dealer also has a Blackjack, in which case it's a push (tie).
- If the dealer has an ace showing, you can buy insurance -- you win if the dealer has a 10 or face card for Blackjack. It's a bad idea to buy insurance except when card counting.
- According to basic strategy, you should hit on hard 12-16 if the dealer has a 7 or better showing, stand on 12-16 if the dealer has a 5 or 6, and stand on 13-16 if the dealer has 2, 3, or 4.
- If you play basic strategy, your expected value is -0.5 cents per $1 bet. This is better than Craps with an EV of -1.4 cents, and much better than Roulette with an EV of -5.3 cents.
- Basic strategy is derived from dynamic programming. For example, when the dealer has a 7 showing, you should hit on 16. To prove this, we see that B(16) = 8/13 (probability that the dealer will bust), or about 0.615. And B(7) = 0.262.
- If you have 17 against the dealer's 7, you will win 26.2% and tie 36.9% of the time, for a effective win probability of 44.7%. But if you stand on 16, you can't win unless the dealer busts, which happens only 26% of the time. If you hit, you win about 29% of the time.
- In card counting, when the count gets high, low cards have been played, and there are plenty of high cards (10, face, ace) left. You can take insurance and make larger bets.
- If the count is sufficiently positive, the player has a slight advantage. This occurs if the count is twice as high as the number of decks remaining. Even so, you should never bet more than 1% of your bankroll.
- The dealer must always follow a set strategy, and so is predictable. In other games, your opponent has options and isn't as predictable.
Lesson 8-9 of the U of Chicago text is called "The Area of a Circle." We all know the famous formula that appears in this lesson.
Last year, my Pi Day activity was more geared towards the area. Therefore, I'm posting that Pi Day worksheet today for Lesson 8-9.
Meanwhile, many chapters in the second half of the book are longer than those in the first half -- and this causes a problem in setting up the chapter review and chapter test. Tuesday is Day 91, which is when Lesson 9-1 will be taught, and tomorrow is the Chapter 8 Test. This means that today needs to be the Chapter 8 Review as well as Lesson 8-9. Get used to this, since there are several more long chapters coming up in the text.
Earlier I wrote that I wasn't sure whether to have a full Chapter 8 Test or just make it a Chapter 8 "Quiz," considering the timing of this assessment. Hmm -- perhaps after what I see in Geometry today, I should make it a Chapter 8 "Quest" instead!
This is what I wrote last year about Lesson 8-9:
I visited several other teacher blogs for ideas on lessons. One of these blogs has a lesson that's perfect for Pi Day:
https://theinfinitelee.wordpress.com/2016/02/08/lesson-area-of-a-circle-or-how-i-got-students-hungry-for-the-formula/
Laura Lee is a middle school math teacher from Minnesota. Here is how she teaches her seventh graders about pi:
Notice that last year, I posted a lesson that actually covered area before circumference. Lee's lesson restores the order from the U of Chicago text, with circumference (Lesson 8-8) before area (8-9).
Let's just skip to the part where, as Lee writes, a pizza makes an appearance:
Method 2 (I had been talking about pizza all morning with my first 2 classes, so last hour I decided I should just order a pizza and use that to derive the area formula)
- Order a pizza (Domino’s large cheese worked great!)
- Reveal pizza to class, watch them go insane!
- Have students gather around your front table
- Slice pizza into 16 slices,
- talk about circumference of 8 of the slices or half of the pizza: πr, record this on the pizza box
- then start arranging pizza slices into a rectangle, listen to student “Ahas!” and “No ways!” when they see it is clearly starting to form a rectangle
- Talk about dimensions of rectangle and then the area
The U of Chicago text does something similar in its Lesson 8-9. The difference, of course, is that the text doesn't use an actual pizza.
Lee writes that for her, the key is proportionality. This fits perfectly with the Common Core:
CCSS.MATH.CONTENT.HSG.C.A.1
Prove that all circles are similar.
Then again, notice that Common Core seems to expect a proof here. How does Common Core expect students to prove the similarity of all circles without Calculus?
Unfortunately, none of our sources actually prove that all circles are similar. What I'm expecting is something like this -- to prove that two circles are similar, we prove that there exists a dilation mapping one to the other. For simplicity, let's assume the circles are concentric, and the radii of the two circles are r and s. So we let D be the dilation of scale factor s/r whose center is -- where else -- the common center O of the two circles. If R is a point on the circle of radius r, then OR = r, and so its image R' must be a point whose distance from O is r * s/r = s, and so it must lie on the other circle of radius s. Likewise, if R' is a point on the circle of radius s, its preimage must be a point whose distance from O is s / (s/r) = r, and so it must like on the circle of radius r. Therefore the image of the circle of radius r is exactly the circle of radius s.
Of course, this only works if the circles are concentric. If the circles aren't concentric, then it's probably easiest just to compose the dilation with an isometry -- here a translation is easiest -- mapping the center of one circle to that of the other. Therefore there exists a similarity transformation mapping any circle to any other circle. Therefore all circles are similar. QED
To get from the area of the unit disk (pi) to the area of any disk (pi * r^2), we are basically using the Fundamental Theorem of Similarity from Section 12-6 of the U of Chicago. This time, though, we are using part (b) of that theorem:
Fundamental Theorem of Similarity:
If G ~ G' and k is the ratio of similitude [the scale factor -- dw], then
(b) Area(G') = k^2 * Area(G) or Area(G') / Area(G) = k^2.
Although Wu attempts to prove a special case of the Fundamental Theorem of Similarity using triangles, it's much easier to do it using squares, as the U of Chicago does. If G can be divided into A unit squares, then G' can be divided into A squares each of length k. And the area of a square of length k is clearly k^2, so the area of G' must be Ak^2. For the circle problem k is the radius r, and A is the area of the unit circle or pi, so the area of a circle is pi * r^2. We can do this right on the same worksheet -- there's already a circle drawn of radius 10 times the length of a square, so instead of the length of each square being 1/10, let it be 1 instead. Then the area of the circle of radius 10 is equal to the number of shaded boxes, or 314, since the old unit square has been divided into 100 unit squares.
Next, I'll add some of what I wrote a few years ago (on a subbing day, just before I started at the old charter school) about the circumference activity:
CCSS.MATH.CONTENT.7.G.B.4
Know the formulas for the area and circumference of a circle and use them to solve problems; give an informal derivation of the relationship between the circumference and area of a circle.
And we all know what this means -- today was the day the students begin learning about pi!
Of course I am posting today's worksheet on the blog. For this activity, the students are given four round objects and a tape measure, and they are to measure the circumference and diameter of each of the objects. For example, one of the objects is a heart tin -- its circumference is about 47 cm and its diameter is about 15 cm.
You may notice that there's room to measure five objects, not just four. Well, the fifth object is the circle painted on the outdoor basketball court. This is convenient because its diameter is already marked (the free-throw line). But the students, instead of bringing the tape measure outside, use a nonstandard unit to measure the circle -- their own feet. With basketball on the mind of so many Californians today -- here in the south we celebrate Kobe Bryant's final game, while those in the north hope the Warriors win their 73rd game today -- it's great to incorporate the sport into today's lesson.
Notice that students are not to fill out the column "What relationship do you see?" yet. But some students try to come up with a relationship anyway. One student tries subtracting the diameter from the circumference, to write something like, "The circumference is 32 cm more than the diameter." I argue that this student is actually on the right track, if you think about it.
Meanwhile, a few students have already heard of pi, so they already know the relationship. One student cheats by measuring the diameters and simply multiplying each one by 3.14. The regular teacher will probably reveal the relationship between the circumference and diameter tomorrow.
Most of the students enjoy the lesson, but a few wonder why we are doing this activity. But most likely, these students are upset because they finish measuring the basketball court before any other group and is hoping for a reward. Instead, they are caught by another teacher for attempting to return to the classroom and fool around while I'm still out watching the other students.
Let's think about where this lesson fits in the seventh grade curriculum. Last week I wrote that if I were teaching the class, I'd try to reach Chapter 8 by Pi Day. As we see, this class came close -- certainly much closer than last week's Chapter 2 class.
But it can be argued that today is actually a "Pi Day" of sorts. You see, instead of 3/14, today is April 13th, which is 4/13. As the digits of pi appear in reverse, we can think of this as "Opposite Pi Day."
And now you're thinking -- here we go grasping at straws to come up with another math holiday. We already have Pi Day on March 14th, Pi Approximation Day on July 22nd, and Pumpkin Pi Day on the 314th day of the year in November. We had Square Root Day of the Decade on 4/4/16, Square Root Day of the Century on 4/5/2025, and several Square Root Days of the Month -- including yesterday, April 12th, which can serve as sqrt(17) day. And now I insist on adding yet another Pi Day on April 13th just because 3/14 reversed is 4/13! Do I really think that anyone is actually going to celebrate any of these extra so-called "Pi Days"?
And this is what I wrote last year about Review for Chapter 8 Test:
As I mentioned earlier, the Chapter 8 Test is tomorrow, which means that the review for the Chapter 8 Test must be today.
For example, most students learn about area at some point in their geometry texts, but only the U of Chicago text includes tessellations in the area chapter. Yet the very first question on this area test is about -- tessellations. So a teacher who assigns this worksheet to the class will then have the students confused on the very first question!
Let's review the purpose of this blog and the reason why I post worksheets here. The purpose of this blog is to inform teachers about the transformations (isometries, similarity transformations) and other ideas that are unique to Common Core method of teaching geometry. The worksheets don't make up a complete course, but instead are intended to be used with a non-Common Core text -- the one that teachers already use in the classroom, in order to supplement the non-Common Core text with Common Core ideas. Another intent is for those teachers who do have Common Core texts, but are unfamiliar with Common Core, to understand what Common Core Geometry is all about. My worksheets are based mainly on the U of Chicago text because both this old text and the Common Core Standards were influenced by NCTM, National Council of Teachers of Mathematics.
So this means that a teacher interested in Common Core Geometry may read this blog, see this worksheet, decide to assign it to the class, and then have all the students complain after seeing the first question because their own text doesn't mention tessellations at all.
I decided to include the tessellation question because it appear in the U of Chicago text. But as of now, it's uncertain that tessellations even appear on the PARCC or SBAC exams. So it would be OK, and preferable, for a teacher to cross out the question or even change it. There's a quadrilateral, a kite, that's already given in the question, so the question could be changed to, say, find the area of the kite, especially if the school's text highlights, instead of tessellations, the formula for the area of a kite.
I admit that it's tricky to accommodate all the various texts on a single worksheet. I included tessellations since this is a drawing assignment that is fun, and I'd try to include them if I were teaching a class of my own. But I also want to include questions that may be similar to those that may appear on the PARCC or SBAC exams.
This year, I restored the tessellations lesson, but that was all the way before winter break began. Meanwhile, I couldn't really skip the triangle area questions, even though we only barely discussed Lesson 8-5 this year.
For example, Questions 2 and 9 are exactly the type of "explain how the..." questions that many people say will appear on those Common Core exams. And so it was an easy decision for me to include those questions.
Then there is a question where students derive the area of a parallelogram from that of a trapezoid. I point out that in other texts -- especially those where trapezoid is defined inclusively -- this isn't how one derives the area of a parallelogram. In the U of Chicago, the chain of area derivations is:
rectangle --> triangle --> trapezoid --> parallelogram
But in other texts, it may be different, such as:
square --> rectangle --> parallelogram --> triangle --> trapezoid
This year, of course, we've followed a different pattern:
trapezoid --> parallelogram --> rectangle --> triangle
Thus parallelogram still comes after trapezoid, hence this test question is still valid.
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