7:15 -- That's right -- this teacher has a zero period class. This is an Algebra I class. Hmm, since this is an eighth grade math class, you know what that means -- it's the second day of the Math Diagnostic Placement Exam.
You might wonder why these students would need a placement exam, since it's unlikely that any Algebra I eighth grader would need a Math Support or Algebra 1A class as a freshman. Well, there are still several possible pathways for such students. High scorers are placed in Honors Geometry, middle scorers are placed in regular Geometry, and low scorers will have to repeat Algebra I. So if the students wish to remain on SteveH's "AP Calculus track," they need to prove they belong there on today's test.
The regular teacher is actually here to help set up the computer exam. He tells me that he tried to give this test to zero period yesterday (since after all, zero period isn't part of the block schedule -- it still meets everyday), but the computers weren't working. So he delayed it until today. Fortunately, the test goes without a hitch in zero period. One or two students fail to finish -- but before he leaves, he tells the students that with his admin password, he can give them more time next week upon his return.
8:15 -- Zero period leaves and homeroom arrives. And now here's another part of today's confusing schedule due to the block schedule -- recall that in yesterday's post, I explained that students took the test in periods 1-3 that day and periods 4-6 today. So that all testing takes place before lunch, the rotation is set up so that periods 4-6 meet in the morning and periods 1-3 meet in the afternoon.
Now at other middle schools in the district, the rotation 4-5-6-1-2-3 is quite common -- but this period order is quite unusual at this school. That's because normally, first period is always in the morning after homeroom -- in fact, they're the same kids in both classes. All the other classes rotate -- and since five classes rotate, there's a 1-1 correspondence between rotations and days of the week. In fact, the ordinary Tuesday schedule is 1-3-4-5-6-2.
Yesterday, homeroom was ten minutes, immediately followed by 70 minutes of first period. Today, passing period is necessarily for the students to get from homeroom to fourth period. Therefore, homeroom is only five minutes long.
As usual, we play the morning announcements video in class. Today's video was 4 minutes and 58 seconds in length, to be played during the five-minute homeroom. Hence there literally wasn't time to do anything else in homeroom -- not even take attendance. The computer system wasn't set up to take homeroom attendance anyway, since homeroom normally blends into first period.
8:20 -- The five-minute homeroom ends. Like yesterday, fourth period happens to be the teacher's conference period. I walk down to the classroom of the teacher I co-taught with yesterday (in periods 2 and 6), just to say hello. (Her prep period is also fourth.) I tell her that I'm subbing for another math teacher and that I was able to give the Geometry Placement Exam to zero period with no problems.
9:40 -- Fifth period, another Algebra I class, begins. I set up the placement exam for the students.
10:00 -- Suddenly, students ask about the error messages that appear on their Chromebooks. The word "Error" shows up whenever they select an answer, and the "Submit" button doesn't work. It turns out that for some unknown reason, the "testing window" was programmed to close at 10:00, even though the block schedule allows for testing in both fifth and sixth period until lunch. No one finishes the test by 10:00, but some students say that they are tantalizingly close to answering all thirty-five questions. I say the same thing I tell the one guy in zero period -- that the regular teacher can give them more testing time using his admin password.
I think to myself, what would SteveH say about this situation? (No, this won't turn into yet another traditionalists post -- thank goodness!) I reckon he'd agree with trying to get as many students into freshman Geometry (part of the AP Calculus track) as possible, but he'd disagree with using a computer test to do it, since problems such as today's can happen. I'd counter by pointing out that paper-and-pencil tests aren't without their problems (such as the tests getting lost in transit between here and the high school). Then again, for computer problems such as today's to be worth it, there should be some benefits, such as instant (or within a few days) test scores. This is mainly an issue not with placement exams, of course, but the computerized SBAC (and PARCC) exams.
Today, I end up handing out the extra credit worksheet left by the regular teacher left for students who finish the test early -- technically, everyone "finishes" the test early today. But this worksheet confuses the students a little -- they're asked to solve a system of two equations -- one linear and the other quadratic. This fits the following Common Core Standard:
CCSS.MATH.CONTENT.HSA.REI.C.7
Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. For example, find the points of intersection between the line y = -3x and the circle x2 + y2 = 3.
But this seems a bit too advanced for Algebra I, even with all of the quadratic equations being parabolas (rather than the circle given in the above example). So of course, the students struggle with these systems. I point out that they've probably seen systems of linear equations months ago, and they just saw quadratic equations in the last unit. Now they are combining the two ideas.
By the way, I wonder whether this might have been a good opportunity to play the Conjectures/"Who Am I?" game that I mention during subbing assignments in past years. I haven't played the game since starting in the new district, since my focus is on following the seven resolutions (and I fear that the game might have become a "crutch" to avoid true classroom management). But if any time's a good time to play the game, it's during an unplanned, difficult lesson after a computer failure.
10:50 -- Fifth period finally ends and it's time for snack. During the break, I walk to the classroom of the special ed class where I sub for yesterday, and I tell the aide about the testing problem. She and the regular teacher for that class have just given their students the Algebra I Placement Exam without any issues whatsoever (and some students have even remained in snack to finish the test). Apparently, only this Geometry Placement Exam has a testing window that ends at 10:00. (The two tests have different sources -- the Algebra I placement test comes from a nearby university, while the Geometry placement test is from the district.)
As I head back to my room, I see a fellow sub whom I met while we both subbed at a high school in the district. She tells me that her daughter is in the fifth period class where I just subbed, and I inform her that we're unable to finish the test. She's concerned -- and understandably so. After all, imagine if her daughter gets placed in the wrong class next year just because the computers aren't working. And if this one parent is worried about the test, imagine how the others would feel if they knew that their children might not be able to complete the placement test.
11:05 -- Sixth period begins. Fortunately, this is a Math Support class for eighth graders who have already taken a placement test in their main math classes. Thus I don't have to worry about the exam failure, as these students are unaffected.
These students are still working on Chromebooks, but on websites that are working. This includes ALEKS and Prodigy software.
12:15 -- Sixth period ends and it's time for lunch.
1:00 -- First period is one of two Algebra I classes who took the placement exam yesterday. And so this class is working on a new lesson -- the Pythagorean Theorem.
These students aren't working from a Glencoe text -- instead, they use the Big Ideas text. (Sorry to bring up traditionalists again, but Big Ideas is the text that Barry Garelick mentions as the "official" text for his eighth grade Algebra I text -- yet he uses this text only to supplement Dolciani 1962.)
I cover all of the examples in the text -- solving for a, b, and c, as well as a few word problems on the coordinate plane to motivate the Distance Formula (to be covered in tomorrow's lesson). I notice that according to the lesson plans left for the rest of the week (for another sub), the students will have a quiz on these two lessons on Thursday. After last year's failure to teach the Pythagorean Theorem, this year I wish I could sub here the whole week so that I can see how well I'm teaching it now. This quiz, unlike the one I gave last year, is open-book -- probably because of the short preparation time, the short period today, and the unexpected week with the regular teacher out.
1:35 -- First period leaves and second period arrives. This is another Algebra I class.
I'm impressed when one student starts doing all the problems in his head, even the square roots -- for example, he sees a right triangle with legs 8 and 15 already drawn on the board and quickly calls out "17" as the hypotenuse. It's good that I don't play "Who Am I?" in this class, or his group would have scored all of the race points and dominated the game without giving anyone else a chance.
Then again, I use the student's mental speed to crush out "dren" behavior. I ask the students to find the length of the hypotenuse if the legs are 7 and 10 -- without using a calculator to square 7 and 10 or add 100 and 49.
2:15 -- Second period leaves. Third period is an ASB Leadership class. The students meet next door, where they work on videos to advertise Spring Spirit Week next week (which will likely be played after the morning announcements). The class is divided into four groups, and each group works on a different day's video. One group uses an iPad while the others record on cell phones. All four groups finish filming today -- all they need to do the rest of this week is edit.
2:55 -- Third period leaves, and thus ends a long day of subbing.
So what should I choose for my focus resolution today? Well, almost all the problems I had today go back to the fifth period computer failure. I suppose I can repeat the seventh resolution:
7. If there is an official assignment to review for state testing, then implement it fully.
But once again, today's computer exam has nothing to do with the SBAC.
Still, I wonder whether I could have handled the computer failure differently. For example, should I have called someone to try to fix the computers? I'm not sure whether anyone would have been able to do anything. This is a software problem, not a hardware problem -- once again, for some reason the testing window is scheduled to close at 10. I don't know what a technician could have done to make the 10:00 deadline disappear. I do admit that I probably would have been more aggressive in asking for help if sixth period also needs to take a test today. Since fifth period is the final testing period, I feel that asking for help would just be closing the barn door after the horse has escaped.
The regular teacher doesn't leave the lesson plans on a single sheet of paper -- instead, each class is written on a different sheet. On the lesson plan for third period, he leaves his phone number -- probably because he fears something will go wrong with the ASB filming. But I don't read the third period lesson plan until second period, long after the test is over. While the students are filming, I do dial the number, but I get only voicemail. Once again, I figure that there's no real point in leaving a message about the computer failure, considering how late it will be once he gets the message. I wonder whether he would have answered his phone had I known to call him during fifth period.
So the only things I could have done differently is call the office immediately during fifth period in the hope that someone can extend the testing window, or play the "Who Am I game?" (or perhaps even both -- play the game while waiting for a tech guy to show up). Also, I could have had the students work on the other side of the extra credit worksheet, which contains quadratic problems that have nothing to do with systems. (I'd chosen the systems side because, while it wasn't a Pizzazz worksheet, had a Pizzazz-type riddle. This is often a double-edged sword -- the riddle makes it easier for both students and teachers to check answers. But it also makes it easy for students to guess the answers with no work once they figure out the riddle.)
In the end, I wish that today's schedule could have been the rotation starting with fifth period rather than fourth. Then all of my testing could have been completed before the 10:00 window closure.
Chapter 7 of Wayne Wickelgren's How to Solve Problems is called "Contradictions." It begins:
"As mentioned in Chapter 3, amateur problem solvers often do not pay enough attention to the goal or the set of possible goals as part of the information in a problem. They apply operations to the givens in an attempt to get to the goal, but they frequently do not consider applying operations to the possible goals in order to get to the givens or to meet the givens halfway."
Wickelgren tells us it's possible, in problem solving, to show that a goal can contradict the givens:
"This method of contradiction is appropriate for several types of problems in which you must decide which of two or more goals could be derived from the givens."
The author tells us that there are four ways contradictions can be used in problem solving. As Geometry teachers, we are very familiar with the first one -- indirect proof, introduced in Lesson 13-4 of the U of Chicago text. (Indeed, the term contradiction itself is defined in this lesson.) Another type, classificatory contradiction, is used when there is a large search space rather than just one or two alternative goals:
"It is necessary to devise some more efficient search procedure that contradicts large classes of alternative goals simultaneously."
But our first examples will involve indirect proof:
- Given that you have already proved the theorem that all squares of nonzero integers are positive, prove that the equation x^2 + 1 = 0 has no integer solution. Stop reading and attempt to prove the theorem, using the method of contradiction (indirect proof).
This one is simple -- subtract 1 from both sides, obtaining x^2 = -1. The contradiction is that x can't be equal to any integer c, otherwise c^2 = -1, contradicting the theorem that the square of an integer can't be negative. (Obviously, a similar proof shows us that x^2 + 1 = 0 has no real solution.)
The next example is the proof that sqrt(2) is irrational. This fits with the Pythagorean Theorem that I taught today -- and indeed, I told the students that sqrt(2) is irrational (mainly because the school's vocabulary word of the day today is "rational").
- Given an isosceles right triangle with sides of unit length, the Pythagorean Theorem asserts that the length of the hypotenuse equals the square root of the sum of the sqaures of the lengths of the sides; namely, c = sqrt(1^2 + 1^2) = sqrt(2). Prove that the length of the hypotenuse of this triangle -- namely sqrt(2) -- is irrational. Stop reading and try to solve the problem, using the method of contradiction.
I might as well repeat the proof that I posted on the blog two years ago:
Indirect Proof:
Assume that sqrt(2) is rational -- that is, sqrt(2) = a/b for some whole numbers a and b. Just as in all the other square root indirect proofs, we square both sides to get 2 = a^2 / b^2. Multiplying gives us the equation a^2 = 2b^2. So a must be even as its square is even -- that is, a = 2c for some other whole number c. Substituting gives us (2c)^2 = 2b^2, or 4c^2 = 2b^2, or 2c^2 = b^2. So b must be even as its square is even. As a and b must both be even, a/b can't be written in lowest terms -- but this is a contradiction as every fraction has a lowest-terms form. Therefore sqrt(2) is irrational. QED
Assume that sqrt(2) is rational -- that is, sqrt(2) = a/b for some whole numbers a and b. Just as in all the other square root indirect proofs, we square both sides to get 2 = a^2 / b^2. Multiplying gives us the equation a^2 = 2b^2. So a must be even as its square is even -- that is, a = 2c for some other whole number c. Substituting gives us (2c)^2 = 2b^2, or 4c^2 = 2b^2, or 2c^2 = b^2. So b must be even as its square is even. As a and b must both be even, a/b can't be written in lowest terms -- but this is a contradiction as every fraction has a lowest-terms form. Therefore sqrt(2) is irrational. QED
Here is Wickelgren's next example:
"You are given three assumptions or previously proved theorems. (1) If c > 0 and a = b, then ac = bc. (2) If c > 0 and a > b, then ac > bc. (3) The law of trichotomy obtains: for any a and b, one and only one of three alternatives holds: a < b, a = b, or a > b. Using these givens, prove that, if c > 0 and ac < bc, then a < b. Stop reading and try to prove the theorem, using the method of contradiction."
Notice that (1) and (2) are very similar to the Postulates from Arithmetic of Lesson 1-7. But (3) isn't mentioned in the text, though it's strongly implied. Indeed, trichotomy is used in the indirect proof of the Unequal Angles Theorem of Lesson 13-7.
This proof is very easy. We are given c > 0 and ac < bc, and there are only two possibilities to rule out, a = b and a > b. But both easily lead to contradiction -- assume a = b, then by (1) ac = bc, and if we assume a > b, then by (2) ac > bc. Either case contradicts ac < bc. Therefore a < b. QED
Wickelgren now turns to some of my favorite examples -- that is, those from Geometry:
Given the assumption that two distinct points determine one and only one straight line, prove that two lines can intersect at no more than one point.
This indirect proof is also given in Lesson 1-7, though it's not stated as an indirect proof. Two lines can't intersect in both A and B, since by the First Postulate there is only one line through A and B. In fact, this is the first theorem of the U of Chicago text, the Line Intersection Theorem. Recall that David Joyce makes a big deal about this theorem:
Chapter 1 introduces postulates on page 14 as accepted statements of facts. The four postulates stated there involve points, lines, and planes. Unfortunately, the first two are redundant. Postulate 1-1 says 'through any two points there is exactly one line,' and postulate 1-2 says 'if two lines intersect, then they intersect in exactly one point.' The second one should not be a postulate, but a theorem, since it easily follows from the first. And what better time to introduce logic than at the beginning of the course. No statement should be taken as a postulate when it can be proved, especially when it can be easily proved.
Here is our next example:
You are given two assumptions or previously proved theorems. (1) A straight angle is a 180-degree angle. (2) Two lines are perpendicular, if they make a 90-degree angle when they intersect. From these assumptions, prove that from a point on a line, only one perpendicular line can be erected.
In our text, (1) is part of the Angle Measure Postulate of Lesson 3-1, while (2) is considered to be the definition of perpendicular in Lesson 13-5. But this "uniqueness of perpendiculars" (through a point on a line) is implied in the text though never explicitly proved. Wickelgren's indirect proof assumes that there are two such lines, with angle alpha between them. But then the three lines form three angles that add up to 180 -- yet two of them are 90 (definition of perpendicular) and the third is given as alpha. Since 90 + alpha + 90 > 180, this is a contradiction. Therefore uniqueness of perpendiculars (though a point on a line) holds. QED
The other three sections of the this chapter in Wickelgren contain no formal Geometry, so let's get through these quickly in this, yet another super-long text. In "Multiple Choice -- Small Search Space," the author starts with a quick example:
- The solution of sqrt(7x - 3) + sqrt(x - 1) = 2 is: (A) x = 3, (B) x = 3/7, (C) x = 2, (D) x = 1, (E) x = 0. Stop reading and try to solve the problem, using the method of contradiction.
If this were an SAT problem, the easy way to find the solution is just to plug each number in for x -- the correct answer is (D).
In this section, Wickelgren returns to liars and "truars" (knights and knaves). He also gives a logic problem in the spirit of Lesson 13-3, the famous Smith, Jones, and Robinson problem:
On a train, Smith, Robinson, and Jones are the fireman, brakeman, and the engineer, but NOT respectively. Also aboard the train are three businessmen who have the same names: a Mr. Smith, a Mr. Robinson, and a Mr. Jones. Using the clues below, can you determine the identity of the Engineer?
1. Mr. Robinson lives in Detroit.
2. The brakeman lives exactly halfway between Chicago and Detroit
3. Mr. Jones earns exactly $20,000 per year.
4. The brakeman's nearest neighbor, one of the passengers, earns exactly three times as much as the brakeman.
5. Smith beats the fireman in billiards.
6. The passenger whose name is the same as the brakeman's lives in Chicago.
1. Mr. Robinson lives in Detroit.
2. The brakeman lives exactly halfway between Chicago and Detroit
3. Mr. Jones earns exactly $20,000 per year.
4. The brakeman's nearest neighbor, one of the passengers, earns exactly three times as much as the brakeman.
5. Smith beats the fireman in billiards.
6. The passenger whose name is the same as the brakeman's lives in Chicago.
Stop reading and try to solve the problem, using the method of contradiction.
The solution, of course, is given at the above link. But the author uses a table, similar to the ones appearing in Lesson 13-3.
In the section "Classificatory Contradiction -- Large Search Space," Wickelgren gives the following verbal arithmetic problem:
DONALD + GERALD = ROBERT
where D = 5 and every digit corresponds to a different letter. (The most famous verbal arithmetic problem is likely SEND + MORE = MONEY.) Stop reading and try to solve the problem.
Wickelgren tells us that some digits can be found without contradiction -- for example, since D is 5 and D + D = T in the last column, T must be 0 (with a carry of 1). But the digits N, B, and O can't be found without trying different digits and seeing whether they lead to a contradiction.
The last section is "Iterative Contradiction in Infinite Search Spaces." Well, the term "infinite" is used loosely here:
In numbering the pages of a book, a printer used 3,289 digits. How many pages were in the book, assuming that the first page in the book was numbered 1?
Stop reading and try to solve the problem, using the method of classificatory contradiction to rule out all but one of the infinity of positive integer answers.
Wicklegren's final example involves solving a polynomial equation
Determine the roots (permissible values of x) that satisfy the equation:
x^6 - 4x^5 + 2x^4 + 3x^3 - 7x^2 + 13x - 30 = 0
Stop reading and attempt to specify an iterative method of contradiction, by which one might determine each root (permissible value of x) for the preceding equation.
The author ultimately assumes that the roots lie in some interval like [-1000, 1000] (otherwise the x^6 term becomes too big), and then we start searching the interval for sign changes in the value of the polynomial. Notice that this is exactly how graphing calculators like the TI-83 and TI-84 find zeros (when you press the key sequence 2nd TRACE 2).
Wickelgren concludes by warning us about this last method:
"However, it clearly gets more and more time consuming and more and more complicated, the greater the number of unknowns or the more complex the functions."
This is what I wrote last year about today's lesson:
We now reach the final chapter of the U of Chicago text, "Further Work with Circles." I didn't cover this chapter in great detail the first two years of this blog, since by the time we reached it we were pressed up against the PARCC and SBAC tests. And besides, the only section that's really tested on PARCC is Lesson 15-3, the Inscribed Angle Theorem.
This year, I follow the digit pattern and so we cover Chapter 15 on Days 151-159. But for those who start PARCC next week and need to complete Lesson 15-3, we'll reach it on Thursday.
Lesson 15-1 of the U of Chicago text is on Chord Length and Arc Measure. The key theorem of this lesson is:
Chord-Center Theorem:
a. The line containing the center of the circle perpendicular to a chord bisects the chord.
b. The line containing the center of the circle and the midpoint of a chord bisects the central angle determined by the chord.
c. The bisector of the central angle of a chord is perpendicular to the chord and bisects the chord.
d. The perpendicular bisector of a chord of a circle contains the center of the circle.
Proof:
Each part is only a restatement of a property of isosceles triangles.
a. This says the altitude to the base is also a median.
b. This says the median to the base is also an angle bisector.
c. This says the angle bisector is also an altitude and a median.
d. This says the median to the base is also an altitude. QED
Why didn't the text just say "diameter" instead of "the line containing the center of a circle"? I assume it's because a diameter is a segment, but bisectors are lines. Here is the other key theorem:
Arc-Chord Congruence Theorem:
In a circle or in congruent circles:
a. If two arcs have the same measure, they are congruent and their chords are congruent.
b. If two chords have the same length, their minor arcs have the same measure.
The U of Chicago text points out that we can't use the terms "are congruent" and "have the same measure" interchangeably. Two angles are congruent if and only if they have the same measure, but two arcs with the same measure aren't necessary congruent. A 50-degree arc of a tiny circle is nowhere near congruent to a 50-degree arc of a large circle. The theorem tells us that an additional condition is needed -- the circles must be congruent also.
But what does it mean for two circles to be congruent? The U of Chicago text proves that two circles are congruent if and only if their radii are equal. Recall that in Common Core Geometry, we can only show two figures congruent by showing that some isometry maps one to the other:
Lemma:
Two circles are congruent if and only if they have equal radii.
Proof of Lemma:
If two circles X and Y have equal radii, then one can be mapped onto the other by the translation mapping X to Y. So they are congruent. Of course, if they do not have equal radii, since isometries preserve distance, no isometry will map one to the other. QED
Proof of Part a of Arc-Chord Congruence:
In circle O, you can rotate Arc AB about O by the measure of Angle AOC to the position of CD. Then the chordAB rotates to CD also, and AB is congruent to CD, Thus in a circle, arcs of the same measure are congruent and have congruent chords. QED
Part b is left in the text as an exercise. A hint is given -- the measure of an arc equals the measure of its central angle. This suggests that we could use a traditional two-column proof via SSS:
Given: AB = CD in Circle O
Prove: measure Arc AB = measure Arc CD
Proof:
1. AB = CD 1. Given
2. AO = CO, BO = DO 2. All radii of a circle are congruent.
3. Triangle AOB = COD 3. SSS Congruence Theorem
4. Angle AOB = COD 4. CPCTC
5. Arc AB = CD 5. Definition of arc measure
The text warns us that in circles with different radii, arcs of the same measure are not congruent -- they are similar. This isn't proved in the text, but notice that one of the Common Core Standards directs students to "prove that all circles are similar." So let's do so right here:
Theorem:
All circles are similar.
Proof:
If the two circles are concentric, then their common center is also the center of a dilation, with the scale factor obviously R/r, with r the smaller radius and R the larger radius. If the two circles have different centers, then we can translate one of the circles so that its center matches the other, then perform the dilation. QED
Actually, a single dilation will work if the centers aren't the same, but it's difficult to locate the center of this dilation, so it's easier just to translate first.
Here is my first newly created worksheet for Chapter 15:
This year, I follow the digit pattern and so we cover Chapter 15 on Days 151-159. But for those who start PARCC next week and need to complete Lesson 15-3, we'll reach it on Thursday.
Lesson 15-1 of the U of Chicago text is on Chord Length and Arc Measure. The key theorem of this lesson is:
Chord-Center Theorem:
a. The line containing the center of the circle perpendicular to a chord bisects the chord.
b. The line containing the center of the circle and the midpoint of a chord bisects the central angle determined by the chord.
c. The bisector of the central angle of a chord is perpendicular to the chord and bisects the chord.
d. The perpendicular bisector of a chord of a circle contains the center of the circle.
Proof:
Each part is only a restatement of a property of isosceles triangles.
a. This says the altitude to the base is also a median.
b. This says the median to the base is also an angle bisector.
c. This says the angle bisector is also an altitude and a median.
d. This says the median to the base is also an altitude. QED
Why didn't the text just say "diameter" instead of "the line containing the center of a circle"? I assume it's because a diameter is a segment, but bisectors are lines. Here is the other key theorem:
Arc-Chord Congruence Theorem:
In a circle or in congruent circles:
a. If two arcs have the same measure, they are congruent and their chords are congruent.
b. If two chords have the same length, their minor arcs have the same measure.
The U of Chicago text points out that we can't use the terms "are congruent" and "have the same measure" interchangeably. Two angles are congruent if and only if they have the same measure, but two arcs with the same measure aren't necessary congruent. A 50-degree arc of a tiny circle is nowhere near congruent to a 50-degree arc of a large circle. The theorem tells us that an additional condition is needed -- the circles must be congruent also.
But what does it mean for two circles to be congruent? The U of Chicago text proves that two circles are congruent if and only if their radii are equal. Recall that in Common Core Geometry, we can only show two figures congruent by showing that some isometry maps one to the other:
Lemma:
Two circles are congruent if and only if they have equal radii.
Proof of Lemma:
If two circles X and Y have equal radii, then one can be mapped onto the other by the translation mapping X to Y. So they are congruent. Of course, if they do not have equal radii, since isometries preserve distance, no isometry will map one to the other. QED
Proof of Part a of Arc-Chord Congruence:
In circle O, you can rotate Arc AB about O by the measure of Angle AOC to the position of CD. Then the chord
Part b is left in the text as an exercise. A hint is given -- the measure of an arc equals the measure of its central angle. This suggests that we could use a traditional two-column proof via SSS:
Given: AB = CD in Circle O
Prove: measure Arc AB = measure Arc CD
Proof:
1. AB = CD 1. Given
2. AO = CO, BO = DO 2. All radii of a circle are congruent.
3. Triangle AOB = COD 3. SSS Congruence Theorem
4. Angle AOB = COD 4. CPCTC
5. Arc AB = CD 5. Definition of arc measure
The text warns us that in circles with different radii, arcs of the same measure are not congruent -- they are similar. This isn't proved in the text, but notice that one of the Common Core Standards directs students to "prove that all circles are similar." So let's do so right here:
Theorem:
All circles are similar.
Proof:
If the two circles are concentric, then their common center is also the center of a dilation, with the scale factor obviously R/r, with r the smaller radius and R the larger radius. If the two circles have different centers, then we can translate one of the circles so that its center matches the other, then perform the dilation. QED
Actually, a single dilation will work if the centers aren't the same, but it's difficult to locate the center of this dilation, so it's easier just to translate first.
Here is my first newly created worksheet for Chapter 15:
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