| Transcript
for the Math
Jam "AoPS Classes Math Jam"
on Jun 2. |
| Math Jam hosted by rrusczyk
(Richard Rusczyk ). |
rrusczyk19:36:04
Hello, and welcome to an Art of Problem Solving Math Jam. Today we'll be discussing the Algebra 1, Algebra 3, and Intermediate Number Theory courses.
rrusczyk19:36:14
My name is Richard Rusczyk. I founded Art of Problem Solving and have written several Art of Problem Solving textbooks.
rrusczyk19:36:28
Before we get started I would like to take a moment to explain our Virtual Classroom to those who have not previously participated in a Math Jam or one of our online classes.
rrusczyk19:36:48
The classroom is moderated: students can type into the classroom, but only the moderators can choose a comment to drop into the classroom for everyone to see. So, when you send a message, it will not appear immediately, and may not appear at all. This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read. Also, only moderators can enter into private chats with other people in the classroom.
rrusczyk19:37:30
Note that it is not possible for the instructor to personally respond to every comment that you submit during the Math Jam -- please do not take it personally if your comment is not posted or responded to! I will try to respond to all questions to the extent that I can. I will let you know when to start asking questions about the classes.
rrusczyk19:37:39
A couple of you have asked about sound. There is no sound in the classroom. This webpage explains why:
rrusczyk19:37:53
You can read that later; you don't have to check it out now. There will be a full transcript of this class available on the Math Jam pages about an hour after we finish.
rrusczyk19:38:06
In this Math Jam, I will briefly describe the course, then go through an example problem. Then, I will hold a question-and-answer session about the class.
rrusczyk19:38:18
Before we get started, I'd like to note that the mathematics we will discuss today cover a *very* wide range of difficulty. Moreover, I know that many of you are here just to check out the classroom before your classes start this summer.
rrusczyk19:38:48
Please understand that if you are enrolled in one of our introductory classes, or haven't much experience yet with advanced problem solving, then much of the material we cover when discussing the Algebra 3 and Intermediate Number Theory classes will be well beyond you right now. We won't be able to teach you all the math you need to understand this material in one night! So, don't be frustrated if you don't understand the problems we discuss for those classes -- your time will come!
rrusczyk19:39:15
We do have two assistants to help you out tonight, Feiqi Jiang (beta) and Billy Dorminy (solafidefarms). Sometimes they will answer your questions my whispering answers to you, or by opening windows with you to work one-on-one.
rrusczyk19:39:27
Feiqi has been an Art of Problem Solving community member for many years. In high school, he participated in various math competitions, including qualifying several times for the USAMO. He has just completed his first year studying mathematics at the University of Michigan.
rrusczyk19:39:48
Billy was a student of many AoPS classes before joining us as an assistant. As part of one of our classes, he started a project that he refined over the following years into research that would eventually win $10,000 through the Davidson Fellows Program. He attended MathCamp and will head to MIT in the fall. (He also has a wicked vocabulary.)
rrusczyk19:40:20
Let's get to the classes.
rrusczyk19:40:23
Algebra 1
rrusczyk19:40:34
The Art of Problem Solving Algebra 1 covers the fundamental concepts of algebra, including exponents and radicals, linear equations and inequalities, ratio and proportion, systems of linear equations, factoring quadratics, complex numbers, completing the square, and the quadratic formula. (Note: This class is equivalent to the first half of the Introduction to Algebra course we used to offer.)
rrusczyk19:40:59
Our algebra classes do not align precisely with the standard curriculum. We have divided them up and named them the way we have in order to provide some guidance for choosing which of our classes to take.
rrusczyk19:41:05
If you have never taken an algebra course, then our Algebra 1 class is probably the right place to start.
rrusczyk19:41:12
If you have taken an algebra 1 class in school, then much of our Algebra 1 course would be review (although there are some more challenging problems in our Algebra 1 class than you would see in school). You may wish to skip our Algebra and move on to Introduction to Counting & Probability, Introduction to Number Theory, and/or Algebra 2.
rrusczyk19:41:45
If you have taken Algebra 2 in school, then Algebra 3 is probably the correct algebra class to take with us (though other non-algebra classes will likely be appropriate as well).
rrusczyk19:42:20
Our algebra classes, emphasize problem solving and conceptual understanding rather than rote memorization. So in addition to teaching students how to manipulate equations, we teach them why the techniques are logically sound, and we talk about general problem solving strategies. Our class meetings are largely interactive, meaning that most of the time is spent solving problems. As much as possible, the students do the solving; the teacher only guides them along and provides useful hints.
rrusczyk19:42:59
(In other words, the classes aren't a lot like what this has been so far -- teacher going on and on!)
rrusczyk19:43:07
There are a couple major goals of Algebra 1:
rrusczyk19:43:11
Students should develop the ability to translate a situation (which might be a real-life situation, a puzzle, etc.) into the abstract language of equations, and then manipulate the equations to develop insight into the original problem. This skill is the key that opens the door to numerous other fields of study, like higher math, chemistry, physics, and engineering.
rrusczyk19:43:35
The real world is the taking-off point which motivates many mathematical concepts. Once one begins to study those concepts, many new questions arise: Do all equations have solutions? Is every number a fraction? Can we invent a number whose square is negative? We delve into some of these questions so that students gain an appreciation for the structure of mathematics, the ability to think abstractly, and the confidence to tackle very difficult questions.
rrusczyk19:43:55
We are now going to go through a sample problem, and then I'll take questions about the course.
rrusczyk19:44:08
What's the solution to the following equation?
rrusczyk19:44:12
rrusczyk19:44:38
Where might we start?
Yankzgodzilla5519:44:51
start with the square root
davidgu19:44:51
get rid of the radical
rrusczyk19:44:56
And how might we do that?
rrusczyk19:45:08
People have suggested this:
rrusczyk19:45:11
rrusczyk19:45:15
After all, the square root of the square of a number seems to be the original number. For example,
rrusczyk19:45:18
rrusczyk19:45:22
Let's try this and see where it goes. What's our new equation?
TheAnswerIs4219:45:36
x-2x=1
draops19:45:36
x-2x=1
remy114019:45:36
x-2x=1
psherman4219:45:36
x-2x = 1
Jackrabbit Oswald19:45:39
x-2x=1
rrusczyk19:45:42
rrusczyk19:45:46
Now what?
Yankzgodzilla5519:46:01
simplify
uap00119:46:02
-x=1
psherman4219:46:02
-x = 1
Jackrabbit Oswald19:46:02
-x=1
Yingaxia19:46:02
combine like terms
rrusczyk19:46:06
We can combine like terms to give
rrusczyk19:46:10
rrusczyk19:46:13
So, what is x?
raycat19:46:26
x=-1
xiaofanpan19:46:26
x=-1
ktothethird19:46:26
-1
cacadodokk19:46:26
-1
Yingaxia19:46:26
-1
davidgu19:46:26
-1
CantonMathGuy19:46:26
-1
rrusczyk19:46:28
x = -1
rrusczyk19:46:35
Alright, now, it's always a good idea to check your work by plugging your solution into the original equation. Let's do that. If we plug x = -1 into
rrusczyk19:46:37
rrusczyk19:46:40
what do we get on the left hand side?
bubbles4819:46:59
3
ThinkFlow19:46:59
1 + 2 = 3
Mousepool19:46:59
1 - 2(-1) = 1 + 2 = 3
rrusczyk19:47:07
rrusczyk19:47:17
Uhoh! Is x = -1 a solution?
Mousepool19:47:29
no
ktothethird19:47:29
no
bargello19:47:29
no
nuggetfan19:47:29
no
froggy19:47:29
no
karebear24819:47:29
No
rrusczyk19:47:40
Hurm...
rrusczyk19:47:45
Where did we go wrong?
froggy19:48:12
sqrt (x^2) = plus or minus x
mathemagician172919:48:12
the square root must be negative
CantonMathGuy19:48:12
sqrtx^2=x or -x
Kiyoshi19:48:12
we didn't take both the possitive and the negative of the square root
ktothethird19:48:15
sqrtx^2=x,-x
ThinkFlow19:48:21
rrusczyk19:48:27
It was wrong to assume that
rrusczyk19:48:29
rrusczyk19:48:40
We can only make this assumption if we know that x is positive. If x is negative, what does
rrusczyk19:48:47
rrusczyk19:48:50
come out to?
nuggetfan19:49:16
-x
ThePurloinedLetter19:49:16
-x
qwertythecucumber19:49:16
-x
bargello19:49:16
-x
Jackrabbit Oswald19:49:16
-x
rrusczyk19:49:27
rrusczyk19:49:33
Let's try solving the equation again, this time assuming x is negative. What does the equation become if x is negative and we simplify the square root?
Mousepool19:49:58
-3x = 1
davidgu19:49:58
-x-2x = 1
CantonMathGuy19:49:58
-3x=1
rrusczyk19:50:03
rrusczyk19:50:13
And what is x?
nuggetfan19:50:22
x=-1/3
Yingaxia19:50:22
x=-1/3
Mousepool19:50:22
x = -1/3
raycat19:50:22
-1/3
rrusczyk19:50:27
rrusczyk19:50:30
qwertythecucumber19:50:34
now we check by plugging in again
rrusczyk19:50:41
Good idea. Does it work?
remy114019:51:11
Yes. 1/3-(-2/3)=1
davidgu19:51:11
1/3 +2/3 = 1
rrusczyk19:51:49
strikerJ19:51:51
yes,1/3+2/3=1
rrusczyk19:51:56
Now we've found the solution.
rrusczyk19:52:01
This problem illustrates the importance of not making any hidden assumptions when you attempt to solve an equation. You must keep *all* possible solutions under consideration rather than assuming the solution is positive.
rrusczyk19:52:34
In the Algebra 1 class, we will see a lot of different types of equations that we will learn to solve.
rrusczyk19:52:51
We'll also learn how to apply all these skills to a variety of word problems.
rrusczyk19:53:06
The course will meet for 15 weeks on Wednesdays, starting June 10, at 7:30 PM Eastern / 4:30 PM Pacific. Each class is 90 minutes, and each is 7:30 - 9 PM ET (4:30 - 6 PM PT). The last day of the class is September 16.
rrusczyk19:53:16
This course will use a textbook in conjunction with the course: our own Introduction to Algebra book. The material covered in the textbook is roughly equivalent to the material covered in the course. You can see the table of contents and some excerpts from the book here:
rrusczyk19:53:43
The book is required for the course. Students will be able to read additional material that complements the lectures, and will have access to a large number of practice problems at varying levels of difficulty. We recommend that students read the corresponding chapter(s) in the book before each lecture, and attempt some of that chapter's Review and Challenge Problems after each lecture.
rrusczyk19:54:06
I am the instructor for the course. As background, I wrote the textbook, along with 4 other AoPS texts, and created the Art of Problem Solving website. I learned most of my math by studying for contests like MATHCOUNTS and the AMC series of tests, and now spend most of my time building resources I wish I'd had when I was a student. You can find a more detailed bio here:
rrusczyk19:54:29
The homework for the class consists of weekly problems that will be posted to the class message board -- for these problems, you do not turn your solutions in, but you may post them to the message board if you like. The class also has 5 Challenge Sets for which you should write up your full solutions and submit them. You will receive thorough feedback for your work on these Challenge Sets that will comment both on your mathematical accuracy and how well you write solutions.
rrusczyk19:54:46
Are there any questions about this class or the textbook?
uap00119:55:39
Does this class fully cover the high school Algebra 1 course content?
Yankzgodzilla5519:56:43
how much algebra 2 is covered in the textbook
rrusczyk19:56:46
This course together with our Algebra 2 course covers all of what a typical school does in Algebra 1 plus most of Algebra 2. Our two algebra classes together are one year; so we get through all of a typical school Algebra 1 and Algebra 2 in one year.
rrusczyk19:57:27
The rest of a typical Algebra 2 is covered in our Algebra 3 course, along with a lot more. ("Algebra 1" and "Algebra 2" don't mean to us the same thing it means to most schools.)
mathematics139619:57:29
so in the class, the teacher will only use the message board, not talk
rrusczyk19:57:40
The class will be very much like what we just did. There is no audio.
rrusczyk19:57:47
Here is why:
soccerplayer343219:57:52
I might miss two weeks in August. How do I make up the classes?
skylord581619:58:18
There are complete transcripts online.
rrusczyk19:58:23
There are full class transcripts available for every class, so you can review everything you miss, and you can use the class message board to ask questions if you have any.
sinj00719:58:28
does this class cover fractions
rrusczyk19:58:59
It covers fractions with variables and square roots, but not things like 2/6 + 7/10. We will assume that you know how to do that.
AGAK19:59:02
What is the average time for homework?
rrusczyk19:59:26
We expect students to spend 5-8 hours/week to get the most out of class.
davidgu19:59:28
are there basic graphs discussed, like absolute value and lines
rrusczyk19:59:56
Graphs of linear equations and linear inequalities are covered in Algebra 1.
benb20:00:00
are there problems in the text book that we will do?
rrusczyk20:00:11
Yes; many of the weekly message board problems come from the text.
Ronnicus20:00:14
If I didn't see that twist in the problem and can't do all of the problems on the post-test, should I take the class? (I've taken Alg I in school).
rrusczyk20:00:23
You should use the Post-test to decide that:
rrusczyk20:00:33
It is here:
Krrish20:00:46
can we do another practice problem, so i feel comfortable with what we are going to do?
rrusczyk20:01:18
We won't be doing any more Algebra 1 problems tonight; we will be doing the more advanced classes in about 10-15 minutes. The first class will be more gradual than that first problem was.
monkeygirl1220:01:24
Are there any open spots
rrusczyk20:01:28
Yes
davidgu20:01:30
is there a grade limit for this class?
rrusczyk20:01:32
No
cocoarules20:01:34
do we cover logarithms?
rrusczyk20:01:43
In the Algebra 2 and Algebra 3 courses.
hyperspace31120:01:45
do we get a grade for the class?
rrusczyk20:01:57
If you need one, we can assign one. Most students do not need one.
cocoarules20:02:01
How are we going to discuss the graphs?
rrusczyk20:02:07
I can post pictures in the room :)
psherman4220:02:18
What class should one take if he/she took Pre-Calc and is preparing for AMC 12 and AIME
rrusczyk20:02:41
Email me your math background at classes@artofproblemsolving.com, including any contest scores, and I'll give guidance.
cacadodokk20:02:44
are there tests?
rrusczyk20:03:01
No. There are Challenge Sets, on which we will give you thorough feedback.
remy114020:03:03
About how much of Algebra 2(in school) is covered in the Algebra 2 course in AoPS?
rrusczyk20:03:45
Varies a lot by school. Anywhere from 50-90% of a typical Algebra 2 course is covered in our Algebra 1 + Algebra 2 (which together are one year)
monkeygirl1220:03:49
Will there be assistdents
rrusczyk20:03:50
yes
xiaofanpan20:03:53
What ages should take this class
rrusczyk20:04:01
Anyone who is ready to start algebra.
Jackrabbit Oswald20:04:05
Are graphs like parabolas covered in Algebra 1, 2, or 3?
rrusczyk20:04:10
In Algebra 2 and 3.
cocoarules20:04:15
How are you going to show us the graphs; do you enter a picture in the class?
rrusczyk20:04:17
Exactly.
monkeygirl1220:04:22
Wil you be the only one teachingthe class
rrusczyk20:04:32
There may be a sub if I am sick or traveling.
gggggr20:04:48
how advanced is the class
rrusczyk20:05:00
You can use the Post-test to gauge that:
skylord581620:05:12
Can I transfer credit to a middle school math course?
rrusczyk20:05:18
Only if your middle school says OK.
uap00120:05:23
how many problems do each of the challange test have?
rrusczyk20:05:26
Around a dozen.
hyperspace31120:05:28
it seems like the problems are harder than a typical alg 1 class in school. is this true?
rrusczyk20:05:50
*Very much so*. Typical algebra classes are far, far too easy for AoPS students.
rrusczyk20:05:57
Our classes are much more challenging.
AGAK20:06:02
Mr.rusczyk do we have any breaks during lectures?
rrusczyk20:06:09
Not in the classes we will discuss tonight.
mathematics139620:06:21
has the advanced mathcounts class been successful for past students who were hoping to reach the national level
rrusczyk20:06:43
Yes. Many of the top 12 have taken the class. All of the top 5 at Nationals have taken classes at Art of Problem Solving.
davidgu20:06:47
is trigonometry covered in these classes?
rrusczyk20:07:17
We introduce trig in Intro to Geometry and do a thorough treatment of it in our Precalculus class (much more thorough than what you'll see in school). It is not in the algebra classes.
Ronnicus20:07:42
how many classes should one take over the summer?
rrusczyk20:07:47
Depends on how much time you have.
rrusczyk20:08:15
Budget 5-8 hours for subject classes without text, and 3-4 for the Problem Series classes.
ThePurloinedLetter20:08:17
Can you share any advice for younger kids who may not be quite ready for Algebra 1?
rrusczyk20:08:35
Students who have gone through Singapore Math tend to do well when they enter our material.
rrusczyk20:08:54
Also, MOEMS is a great source of problems for students who are not yet ready for algebra.
rrusczyk20:09:01
There are 3 books of theirs in our bookstore.
candyrox20:09:05
what is the best thing to do to make the class easier
rrusczyk20:09:28
Read the book before coming to class; this is how I made college *very* easy for me (and I studied chemical engineering at Princeton).
Krrish20:09:50
will we cover bisectors and stuff like that?
rrusczyk20:09:55
That is in the geometry class.
uap00120:09:57
is it too late to join classes that has already started (e.g. mathcounts)?
rrusczyk20:10:05
The MATHCOUNTS Basics class is full.
candyrox20:10:08
will this help us in the mathcounts competion
rrusczyk20:10:10
Yes.
monkeygirl1220:10:12
How much homework would there be for both the number theory and algrebra 1
rrusczyk20:10:20
5-8 hours per week per clas.
Jackrabbit Oswald20:10:22
Do we have to know Latex before taking the class?
rrusczyk20:10:24
No.
rrusczyk20:10:28
Most students don't use it.
spatel20:10:31
what about if someone has not gone through Singapore Math
rrusczyk20:10:47
You'll be fine -- we have lots of students who don't use Singapore.
candyrox20:11:04
r we doing any more questions or not
rrusczyk20:11:11
Yep. We are going to now move on to
rrusczyk20:11:20
Algebra 3
rrusczyk20:11:25
Our Algebra 3 class contains much of the algebra of a typical Algebra II class, all of the non-trig, non-matrices algebra of a typical precalculus class, plus a number of advanced topics that are excluded from the standard curriculum.
rrusczyk20:11:36
The course starts with a review of linear and quadratic equations, functions, and complex numbers, then goes on to cover conics, polynomials, advanced factoring techniques, classical inequalities, techniques for solving hard systems of equations, symmetric polynomial sums, sequences and series, identities and induction, greatest and least integer functions, advanced methods for dealing with logarithms, functional equations, and much more.
rrusczyk20:11:53
The textbook for the course is our new Intermediate Algebra text, by Richard Rusczyk and Mathew Crawford. The text is required for the course.
rrusczyk20:12:11
I'll now proceed with a couple challenging problems from the course. If you find these problems very, very easy, then you might be too advanced for the course. If you find them challenging but not completely impossible, then the class is probably a good fit for you.
rrusczyk20:12:40
Warning: those of you who are here to test out the classroom before an Introduction course or Algebra 1 will likely find these problems nearly impossible.
rrusczyk20:12:50
rrusczyk20:13:09
Where might we start?
draops20:13:45
subtracting the equations?
rrusczyk20:13:50
Why would we do that?
bargello20:14:05
Could we add the equations?
rrusczyk20:14:10
Why would we do that?
bargello20:14:40
I notice we'll get 19x+19y on one side so that might be nice.
davidgu20:14:40
um....try to cancel terms (elimination)
rrusczyk20:14:59
OK, let's give adding the equations a try. We're not sure it will help, but we'll try it.
rrusczyk20:15:07
One reason we might be inspired to try this is:
tornado.adv420:15:08
difference of cubes or addition of cubes
rrusczyk20:15:15
rrusczyk20:15:25
Why does this help? What can we do with that?
davidgu20:15:53
factor...
remy114020:15:53
x^3+y^3=(x+y)(x^2-xy+y^2)
bargello20:15:53
We can factor the sum of cubes: x^3+y^3=(x+y)(x^2-xy+y^2)
mathemagician172920:15:53
factor into (x+y)(x^2-xy+y^2)
rrusczyk20:16:02
Why does factoring like that make us happy?
qwertythecucumber20:16:11
the question doesnt ask for x or y. if you try adding or subtracting the equations, some components we see in the result might show up in the expansion of what we're trying to find
rrusczyk20:16:33
That's a good insight; we might be able to get to the answer without finding x and y!
Mousepool20:16:40
x^3 + y^3 = (x + y)(x^2 - xy + y^2) = 19(x + y), so we know that (x^2 - xy + y^2) = 19
benjamin7xx20:16:40
x^2 - xy + y^2 = 19 (?)
CantonMathGuy20:16:40
19x+19y=19(x+y)
remy114020:16:40
So (x^2-xy+y^2)=19.
karebear24820:16:44
divide by (x+y)
bargello20:16:44
And from the original condition, we can divide both sides by x+y.
rrusczyk20:16:53
Exactly; we can factor x+y out of both sides.
rrusczyk20:16:58
rrusczyk20:17:12
That seems useful. But it's not quite enough to finish.
rrusczyk20:17:15
What might we try next?
qwertythecucumber20:17:53
subtracting equations...?
rrusczyk20:17:58
Subtracting what equations?
benjamin7xx20:18:02
(x^2 - y^2)^2 = ((x+y)(x-y))^2
rrusczyk20:18:17
Interesting insight; we'll remember that, just in case we find x+y and x-y.
CantonMathGuy20:18:54
diff of cubes
RoFlLoLcOpT20:18:54
rrusczyk20:19:19
Adding the initial two equations seemed to get us somewhere. Let's try subtracting them.
Mousepool20:19:25
x^3 - y^3 = (x - y)(x^2 + xy + y^2) = (15x + 4y) - (4x + 15y) = 11(x - y), so that (x^2 + xy + y^2) = 11
ThinkFlow20:19:32
rrusczyk20:19:51
Interesting (I think you want x^3 - y^3 there). Let's investigate.
rrusczyk20:20:02
If we subtract the second given equation from the third, we get:
rrusczyk20:20:06
rrusczyk20:20:13
And what do we do with this?
draops20:20:42
the difference of cubes?
applepie3141520:20:43
factor
rrusczyk20:20:46
Keep going...
Caelestor20:20:50
divide by (x-y)
rrusczyk20:20:55
And what are we left with?
Caelestor20:21:10
(x^2+xy+y^2)=11
Mousepool20:21:10
x^2 + xy + y^2 = 11
Jackrabbit Oswald20:21:15
x^2 + xy + y^2 = 11
rrusczyk20:21:18
rrusczyk20:21:34
rrusczyk20:21:41
What do we do with those?
skylord581620:21:52
subtract!
Caelestor20:21:52
add together and cancel the xy terms
XxAwesomexX20:21:52
add them
qwertythecucumber20:21:52
now add or subtract them
rrusczyk20:22:00
Which should we do? Add or subtract?
CantonMathGuy20:22:20
+ and -
CantonMathGuy20:22:20
both
Jackrabbit Oswald20:22:20
Can we do both?
rrusczyk20:22:31
Why make decisions when you can have it both ways.
rrusczyk20:22:39
What does adding give us?
Jackrabbit Oswald20:23:04
2(x^2 + y^2) = 30 after addition.
alexk20:23:04
x^2 + y^2 = 15
Caelestor20:23:04
2x^2+2y^2=30
Caelestor20:23:04
or x^2+y^2=15
raycat20:23:04
x^2+Y^2=15
rrusczyk20:23:07
rrusczyk20:23:14
And what does subtracting give us?
Mousepool20:23:34
xy = -4
Caelestor20:23:35
-2xy=8?
CantonMathGuy20:23:35
xy=-4
draops20:23:35
-2xy=8
qwertythecucumber20:23:35
rrusczyk20:23:39
rrusczyk20:24:09
So. . . Is this enough to solve the problem?
rrusczyk20:24:17
Remember, we don't have to find x and y!
davidgu20:24:48
yes
blueshark20:24:48
yes
XxAwesomexX20:24:48
YES!
rrusczyk20:24:54
Indeed we can finish. How?
LilyBreeze20:25:22
(x-y)^2=x^2+y^2-2xy= 15+8=23
Caelestor20:25:22
add these two up to (x-y)^2
Jackrabbit Oswald20:25:22
yes...using (x+y)^2 will give us the answer
Caelestor20:25:22
x^2+y^2-2xy=(x-y)^2
rrusczyk20:25:54
There's one way -- you can compute (x-y)^2 and (x+y)^2 separately with our values of x^2 + y^2 and xy. What do we get?
Mousepool20:26:36
(x^2 - y^2)^2 = (x + y)^2(x - y)^2 = (x^2 + 2xy + y^2)(x^2 - 2xy + y^2) = (15 + 2(-4))*(15 - 2(-4)) = 7*23 = 161
Mousepool20:26:36
7*23 = 161
topofmath20:26:51
rrusczyk20:27:02
We also could have written (x^2 - y^2)^2 in terms of x^2 + y^2 and xy like this:
rrusczyk20:27:09
rrusczyk20:27:23
(We know (x^2 + y^2) = 15, so (x^2 + y^2)^2 = 225. We know that xy = -4, so 4x^2y^2 = 4(-4)^2 = 64.)
rrusczyk20:27:29
The key to this problem was simply working with the information that we were given long enough to see the factorizations that we know how to perform. A couple of times we cleverly added and subtracted equations to create new equations that we could manipulate in ways we know how.
skylord581620:27:32
And we have our answer!
rrusczyk20:27:35
Exactly.
rrusczyk20:27:48
Are there any questions about this problem?
rrusczyk20:28:07
(Also, I should note that if you're just here for Algebra 1 or one of the Intro classes, you can safely leave now if you want.)
Ronnicus20:28:10
are all algebra 3 problems that hard?
rrusczyk20:28:18
This is a touch above average difficulty.
rrusczyk20:28:34
We would work up to this with some easier problems that would give you some insight what to do on this one.
Caelestor20:28:35
would this be an AMC 12 question?
rrusczyk20:28:43
I believe this is an old AIME problem.
topofmath20:28:45
Will we do any more problems for Algebra 3?
rrusczyk20:28:49
1 more.
ThinkFlow20:28:52
This would be on the easy side, right?
rrusczyk20:29:18
I think it might be a touch harder than average, but I'm not sure. It would be about average for a Challenge Set (which will naturally be more challenging than average!)
tornado.adv420:29:21
will there be problems like this in the book or only in the class?
rrusczyk20:29:27
There are many problems like this in both.
topofmath20:29:29
What competitions does Algebra 3 help with?
rrusczyk20:29:44
AMC 12, AIME, ARML, Mandelbrot, even very beginning olympiad.
ThinkFlow20:29:46
What's the hardest type of Algebra 3 problem? Does the class sort of follow the order of Intermediate Algebra?
rrusczyk20:30:12
After the next problem I'll show some sample problems that are harder. The class is close to the book in topic coverage.
davidgu20:30:14
does Algebra 3 equal a hard Precalc course; or does it equal a easy Calc course; or both?
rrusczyk20:30:19
It's not an easy anything.
rrusczyk20:30:53
It covers all of the non-trig precalculus and the harder stuff you'd see in a good Algebra 2 course, plus a great deal that you won't see in any typical classroom.
rrusczyk20:31:00
Let's look at another problem as an example.
rrusczyk20:31:07
Caelestor20:32:07
this is definitely an AIME problem
rrusczyk20:32:18
Yes, you know the lion by its paw, so to speak :)
davidgu20:32:23
what does relatively prime mean?
rrusczyk20:32:30
No factors in common besides 1
rrusczyk20:32:36
Where do we start?
rrusczyk20:33:31
It's a very wordy problem. What do we usually want to do with a wordy problem?
ThinkFlow20:33:42
make it into equations
davidgu20:33:42
translate it into math
rrusczyk20:33:47
We have a very wordy problem. We'll have to convert those words to equations.
rrusczyk20:33:58
The sentence that we have to focus on is, "This function has the property that the image of each point in the complex plane is equidistant from that point and the origin." We have to turn this into an equation somehow. We'll do it step-by-step.
rrusczyk20:34:03
In terms of f and z, what is "the image of the point z"?
ThinkFlow20:34:35
f(z)
topofmath20:34:35
f(z) is the image
rrusczyk20:34:46
The image of the point z under the function f is simply f(z). So, we can replace "the image of each point" with f(z) and "the point" with z. This leaves us the sentence, "This function has the property that f(z) is equidistant from z and the origin."
rrusczyk20:34:53
Now, what is an expression for the distance between f(z) and z in the complex plane?
rrusczyk20:35:55
Let's take a step back for a minute: what is an expression for the distance between any two complex numbers w and z in the complex plane?
ThinkFlow20:36:26
|w-z|
topofmath20:36:26
rrusczyk20:37:02
rrusczyk20:37:07
(This fact will be taught in the course.)
qwertythecucumber20:37:09
|f(z)-z| is then the distance
rrusczyk20:37:26
Exactly. We now have an expression for the distance between f(z) and z.
rrusczyk20:37:29
The distance between f(z) and z is |f(z) - z|.
rrusczyk20:37:32
And the distance between f(z) and the origin?
ThinkFlow20:38:00
|f(z)|
Mousepool20:38:00
lf(z)l
topofmath20:38:00
tacoman20620:38:00
abs(f(z))
rrusczyk20:38:05
This is simply |f(z)|. So, what is our equation?
rrusczyk20:38:41
(Our goal was to turn all those words into an equation.)
skylord581620:38:46
|f(z)-z| = |f(z)-0|
skylord581620:38:46
|f(z)-z|=|f(z)|
topofmath20:38:46
Mousepool20:38:46
lf(z) - zl = lf(z)l
RoFlLoLcOpT20:38:46
tacoman20620:38:46
|f(z)| = |f(z)-z|
rrusczyk20:38:54
rrusczyk20:39:08
What we just did is the single biggest step in a great many wordy AIME and USAMO problems.
rrusczyk20:39:22
Turn the words into math. We did it with small steps, but we've taken a big leap.
rrusczyk20:39:35
Now, what can we do with this equation? What is our next step?
rrusczyk20:39:58
What is difficult about this equation?
topofmath20:40:21
the absolute value signs
Caelestor20:40:21
absolute signs
davidgu20:40:21
it has absolute value and functions in it?
tacoman20620:40:21
the absolute values
rrusczyk20:40:27
The absolute values and the f(z).
rrusczyk20:41:09
I see many of you saying that the quantities are equal or opposites. That doesn't work in complex numbers. For example, |3+4i| = |5i|. We'll talk about this in the course.
davidgu20:41:11
so we would try to get rid of the f(z) by substituting
tacoman20620:41:25
can't you substitute (a+bi)z for f(z)
rrusczyk20:41:29
OK, we know what f(z) is, so we can substitute. Let's see what taht does:
rrusczyk20:41:35
rrusczyk20:41:42
What can we do with that?
Mousepool20:42:29
factor out z
RoFlLoLcOpT20:42:51
bargello20:42:59
Can we write the left as |(a-1+bi)z|?
rrusczyk20:43:07
Indeed, and then what might we do?
ThinkFlow20:43:22
Cancel out z
tacoman20620:43:22
can't we divide out z?
Caelestor20:43:22
then we can divide by z
rrusczyk20:43:27
rrusczyk20:43:46
(We also could have noted that it has to be true for z=1, so we could let z=1 to get to this point.)
rrusczyk20:43:49
Now what do we do?
rrusczyk20:45:37
Again, I see many of you removing the absolute value signs and making four equations. Doesn't work that way in complex numbers (and we'll explain why in the course.)
ThinkFlow20:45:41
RoFlLoLcOpT20:45:41
ThinkFlow20:45:41
skylord581620:45:41
substitute |a+bi| for 8
rrusczyk20:45:56
Indeed (when stuck, ask what you haven't used yet).
rrusczyk20:46:16
If we use |a+bi| = 8, we have |a-1 + bi| = 8. Now what?
topofmath20:47:18
RoFlLoLcOpT20:47:18
rrusczyk20:47:27
We now have two equations for a and b:
rrusczyk20:47:32
|a+bi| = 8
rrusczyk20:47:41
|a-1 + bi| = 8.
rrusczyk20:48:05
Note that we cannot use the first to say that a+bi is +8 or -8. That is not how absolute value works in complex numbers.
rrusczyk20:48:17
This is the definition of "absolute value" in complex numbers:
CantonMathGuy20:48:20
|x+yi|=sqrt(x^2+y^2)
rrusczyk20:48:30
So, using these two equations, we have
rrusczyk20:48:40
rrusczyk20:48:52
topofmath20:48:54
davidgu20:48:55
so a^2 +b^2 = 64
rrusczyk20:49:08
Squaring them both gets rid of the radicals and gives us nice equations.
rrusczyk20:49:10
Now what?
ThinkFlow20:49:32
Mousepool20:49:32
subtracting gets us -2a + 1 = 0
rrusczyk20:49:54
Subtracing one for the other eliminates b and gives a^2 - (a-1)^2 = 0. Solving this equation gives a=1/2.
CantonMathGuy20:49:58
a=0.5
topofmath20:50:23
rrusczyk20:50:35
And then we just substitute and do some arithmetic to find b^2. What is b^2?
Mousepool20:51:04
b^2 = 255/4
topofmath20:51:05
davidgu20:51:05
63.75
RoFlLoLcOpT20:51:05
Mousepool20:51:05
a^2 + b^2 = (1/2)^2 + b^2 = 1/4 + b^2 = 64, so b^2 = 255/4
rrusczyk20:51:09
rrusczyk20:51:23
What is our final answer? (Read the question last. Always read the question again last.)
ThinkFlow20:51:27
Mousepool20:51:27
m + n = 255 + 4 = 259
topofmath20:51:27
255+4=259
qwertythecucumber20:51:27
since b^2 = m/n, m+n=259
rrusczyk20:51:34
rrusczyk20:51:39
This problem is a bit harder than the average problem in the Algebra 3 class (and we cover the necessary prerequisite knowledge of complex numbers in the class).
rrusczyk20:51:44
The Algebra 3 class also involves a variety of other algebraic topics including methods of substitution, functions, polynomials, sequences and series (including the use of difference equations), binomial expansion, logarithms, advanced systems of equations, and greatest/least integer functions. Here are a few harder problems we will tackle in the course:
rrusczyk20:51:49
rrusczyk20:51:54
rrusczyk20:51:57
rrusczyk20:52:03
rrusczyk20:52:10
topofmath20:52:26
That's scary
rrusczyk20:52:28
:)
rrusczyk20:52:36
The course will meet for 24 weeks on Thursdays, starting June 11. Each class starts at 7:30 PM Eastern / 4:30 PM Pacific, and is 90 minutes long. In addition to message board problems, there are 8 Challenge Sets in this course, one every 3 weeks.
rrusczyk20:52:46
The course is taught by Dan Zaharopol. Dan has taught math at numerous programs across the country. He has been an instructor at Canada/USA Mathcamp, at the Boston Math Circle, and with the Splash and HSSP programs at the MIT Educational Studies Program. Dan participated in numerous local and national math competitions in high school and was a finalist in the 1999 USA Computing Olympiad. He holds an SB from MIT in mathematics, and two MS's from the University of Illinois in mathematics and teaching mathematics. He currently is working on launching MIT's Splash program as a nation-wide effort.
davidgu20:52:59
is there probability?
rrusczyk20:53:08
That is in the Counting & Probability courses.
skylord581620:53:18
What does the big E do?
rrusczyk20:53:27
That's for summations; you'll learn that in the course.
topofmath20:53:33
Is this a USAMO problem or USAMO-level problem?
rrusczyk20:53:51
The last problem we did (and most of the ones I listed above) are toughish AIME problems.
rrusczyk20:54:37
Are there any questions about the Algebra 3 class?
froggy20:55:13
Don't you need more than algebra 2 (in school) to solve the AIME problems?
rrusczyk20:55:14
Absolutely. That's why people come to us.
solafidefarms20:55:24
I've taken classes from Dan in person at Mathcamp; I can vouch that he's a very, very good teacher. Although all AoPS instructors are very good teachers :).
remy114020:55:33
Is this course going to cover precalc and more?
rrusczyk20:55:38
This course covers non-trig precalc, plus much, much more.
mathmongoose20:56:30
About how many hours of work outside of the class does it normally require?
rrusczyk20:56:44
To get the most out of class, 5-8 hours per week, including class time.
bubbles4820:57:01
About how many students are in the class?
rrusczyk20:57:08
I expect 40-50 in this class.
froggy20:57:26
What about in the others?
rrusczyk20:57:48
Anywhere from 40-70.
rrusczyk20:58:36
I'll do the Intermediate Number Theory class and then take more questions.
rrusczyk20:58:39
Intermediate Number Theory
rrusczyk20:58:46
The Intermediate Number Theory Seminar is an 8 week course for students who have both a strong foundation in basic number theory and solid algebraic skills. The course starts June 10 and meets on Wednesdays, 7:30 PM - 9 PM ET, until July 29. It is recommended that students take the Introductory Number Theory course and Algebra 3 course before enrolling, or be confident with the material covered in those classes.
rrusczyk20:58:54
The instructor for the course in Sean Markan. He participated in numerous math and science programs in high school, including the Math Olympiad Summer Program in 2001 and the US Physics Team in 2000 and 2002. He also won the Mandelbrot Competition in 2002. He graduated from MIT with a degree in Physics in 2006.
rrusczyk20:59:06
Topics covered in the Intermediate Number Theory Seminar include algebraic methods of problem solving in number theory, base number problems involving algebra, counting, and qualitative problem solving techniques, divisibility and divisor problems involving algebra, Diophantine equations, modular arithmetic with an emphasis on algebraic applications, perfect squares, an introduction to Pell's equations, Fermat's Little Theorem, Euler's Phi Function, and Euler's Theorem.
rrusczyk20:59:26
As with all AoPS classes, the Intermediate Number Theory Seminar will have its own private message board in the AoPS Forum. The Intermediate Number Theory class will include more message board problems per week of class than other AoPS courses to be sure that students get ample practice with the concepts discussed in each lesson. There are no Challenge Sets in the Intermediate Number Theory course.
rrusczyk20:59:40
We will now work on some problems from the class. These problems are a little easier than average in the course. Unfortunately, we cannot sample some of the most important topics covered in class because it would take too much time developing the background to discuss them here. This background will be developed during the course, so you'll get a crack at those advanced concepts there.
rrusczyk20:59:49
rrusczyk20:59:57
Where do we start?
tacoman20621:00:32
factor f(x) to (x+2)(x+1)
benjamin7xx21:00:32
factor
ThinkFlow21:00:32
Factor?
rrusczyk21:00:35
Algebra and number theory together almost always means we should consider looking for ways to factor.
rrusczyk21:00:44
rrusczyk21:00:50
How does this help?
remy114021:01:59
either one of the expressions should be a mutliple of 6, one of them can be a mult. of 2 and the other a mult. of 3.
Mousepool21:02:05
f(x) needs to be divisible by 2 and 3
rrusczyk21:02:13
Is either of these particularly easy to deal with?
Ari21:02:35
one number must be even, one must be odd, that means that we only have to look out for 3 instead of 6
rrusczyk21:02:41
We can now see that f(x) will always be even because either x + 1 or x + 2 must be even. This examination of the evenness or oddness of the integers is called a "parity" argument.
rrusczyk21:02:50
So now, what's all that's left to be done?
tacoman20621:03:25
find out which numbers make either x+2 or x+1 divisible by 3
davidgu21:03:28
look at all the numbers from 1 to 25 for the divisible by 3 property
bargello21:03:28
Find out when we get multiples of 3
rrusczyk21:03:28
Our goal now is to find out exactly when f(s) will be a multiple of 3.
rrusczyk21:03:34
When will f(s) be a multiple of 3?
Mousepool21:04:28
for f(x) to be divisible by 3, x == 2(mod 3) or x==1(mod 3), s just has to not be a multiple of 3
Caelestor21:04:28
if x is not divisible by 3
alligator11221:04:29
1,2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,25
qwertythecucumber21:04:29
when s isnt
rrusczyk21:04:32
f(s) = (s+1)(s+2) will be a multiple of 3 when either s + 1 or s + 2 is a multiple of 3. This is the same thing as saying that f(s) will be a multiple of 3 when s is NOT a multiple of 3.
rrusczyk21:04:41
So, how many members of S are not multiples of 3?
alligator11221:05:17
17
benjamin7xx21:05:17
So, 0 is the first value of s that doesn't work. From there, 3, 6, 9, ..., 24, so there are 9 possibilites that don't work. Then, 26 - 9 = 17.
Caelestor21:05:17
17
CantonMathGuy21:05:17
17
rrusczyk21:05:39
We now count the members of S that aren't multiples of 3 and we count a total of 17 of them. (I like the way you wen't about it the counting benjamin!)
rrusczyk21:05:49
The key to this problem was factoring the function and then examining the result in terms of the prime factors of 6. Once we did that, the actual divisibility arguments were relatively simple. But notice how the factorization and focusing on primes made the problem much easier (many of you gave me all sorts of more complicated casework).
rrusczyk21:05:58
One more problem!
rrusczyk21:06:03
rrusczyk21:06:18
Where do we start?
rrusczyk21:06:25
(Hint: wordy problem . . . . )
davidgu21:06:44
translate into math!
topofmath21:06:47
Easier to see factors
ThinkFlow21:06:47
EQUATIONS!!!
bargello21:06:47
translate into equations
rrusczyk21:06:54
We must find a way to write this in mathematical terms!
rrusczyk21:06:59
So, what will we do?
Caelestor21:07:01
put n into base 2
rrusczyk21:07:08
rrusczyk21:07:18
OK, and how else might we write that?
rrusczyk21:07:33
(The a_i are the digits in base 2, which is what the little 2 tells us.)
tacoman20621:08:00
a0 + 2a1 + ... + (2^k)ak
RoFlLoLcOpT21:08:19
rrusczyk21:08:27
rrusczyk21:08:28
And what do we know about all those a_i?
qwertythecucumber21:09:04
theyre either 0 or 1
Caelestor21:09:04
equal to 1 or 0
bargello21:09:04
They must be 0s or 1s.
ThinkFlow21:09:04
They are 0 or 1
rrusczyk21:09:18
They're 0s and 1s... OK.
rrusczyk21:09:20
What next?
bargello21:09:44
write the other number in expanded form
ThinkFlow21:09:44
Same thing, but in base 3 is equal to 2n
rrusczyk21:09:47
rrusczyk21:09:52
Now we're getting somewhere!
rrusczyk21:09:57
Er...where would that be?
benjamin7xx21:10:38
two times the first equation equals the second.
rrusczyk21:10:48
We can now apply what I like to call the Fundamental Principle of Algebraic Problem Solving -- find two ways to express the same thing and set them equal!
rrusczyk21:10:56
rrusczyk21:11:05
All we're doing here is wiping n out of our equations.
rrusczyk21:11:22
How might we organize these equations a bit better?
davidgu21:11:51
group
qwertythecucumber21:11:51
combine like terms?
ThinkFlow21:11:51
Write them as coeefficients of the a_i
rrusczyk21:12:00
rrusczyk21:12:05
How does this help?
Caelestor21:13:15
well the ao and ai terms can only be 0 or 1
rrusczyk21:13:28
True, the a_i terms can only be 0 or 1.
rrusczyk21:13:31
What else?
ThinkFlow21:13:47
The coeficcients of a_2 to a_k are negative
rrusczyk21:14:10
Caelestor21:14:49
a_0+a_1-a_2=0
Caelestor21:14:49
and a_3 and greater have to be 0, or it'll get too negative
rrusczyk21:15:01
To see why this is the case, we can do this:
ThinkFlow21:15:08
put a_2 to a_k on the other side
rrusczyk21:15:13
benjamin7xx21:15:42
Coefficient of a_2 is -1, but the coefficient of a_>2 becomes much higher negative numbers. Thus, in order for the starting assumption to be true, either a_0, a_1, a_2 are all 0, or a_2 is 1 and either a_0 or a_1 is 1, in order to set the equation to 0. The a_>2 are all 0's.
rrusczyk21:15:48
Good explanation.
rrusczyk21:16:07
rrusczyk21:16:21
So, once we wipe out the terms after a_2, we have a much simpler equation:
rrusczyk21:16:42
rrusczyk21:16:51
And how do we finish?
alligator11221:17:23
0,101,110
ThinkFlow21:17:23
We see that n is 000, 110, 101, so our answer is 3.
benjamin7xx21:17:23
0 + 0 = 0
1 + 0 = 1
0 + 1 = 1
TheMan199821:17:26
test values for n
alligator11221:17:31
either they are all 0 or a_2 is 1 and a_0 is 1 or a_2 is 1 and a_1 is 1: 0,101,110
rrusczyk21:17:34
Checking values in our range we find that n = 0, 5, or 6, so our answer is 3.
rrusczyk21:17:40
The key to this problem was using a sequence of variables to represent the digits of n so that we could use these variables to compare the two different ways we had to express the value of n. Once we did that, we could fall back on our algebraic skills using equations and inequalities.
rrusczyk21:17:49
Words -> equations. Very important step.
davidgu21:17:53
wow, was that an AMIE problem?
rrusczyk21:18:00
I don't know where that problem comes from.
rrusczyk21:18:30
That's it for the math. You can see harder problems like those in the course here:
topofmath21:18:34
Is that above, below, or above average?
rrusczyk21:18:46
Those two problems are easier than average.
qwertythecucumber21:18:48
how well does aops vol. 2 cover number theory?
rrusczyk21:18:54
Not nearly as well as this course.
davidgu21:18:56
so...about what level math does this class use?
rrusczyk21:19:19
It assumes you know what modular arithmetic is, and uses all your algebra skills through algebra 2 at least.
ThinkFlow21:19:21
Was it AIME difficulty?
rrusczyk21:19:27
On the easy end of that.
benjamin7xx21:19:29
This seemed easier than Algebra 3 (?)
rrusczyk21:19:46
Yes, these sample problems are easier than the samples in the Algebra 3 class.
rrusczyk21:20:05
That's because we can't assume people in the Math Jam know much advanced number theory that we'll teach in the class.
smokewisps21:20:08
will there be an intermediate number theory book?
rrusczyk21:20:10
Someday.
ThinkFlow21:20:13
How much overlap is there with this course and Introductions to Number Theory and Inequalities? (written by C J Bradley and offered on the AoPS bookstore)
rrusczyk21:20:26
I'm not sure; I haven't spent much time with that book. I imagine there is some.
davidgu21:20:28
what's modular arithmetic???
rrusczyk21:20:39
Look on the AoPS Wiki (or the Intro Number Theory book)
benjamin7xx21:20:54
Can you link an Algebra 3 practice test? I don't remember if you did.
davidgu21:21:03
can any of these problems be solved with calculus...by any chance?
rrusczyk21:21:12
Probably, but you don't need it.
rrusczyk21:21:42
Are there any more questions?
davidgu21:22:07
when is the course?
rrusczyk21:22:12
Details there.
ThinkFlow21:22:45
Woud the techniques taught in this course be applicable to easy AIME? Hard AIME? USAMO?
rrusczyk21:22:57
Easy and hard AIME, and even beginning USAMO.
davidgu21:23:29
this couse does not cover AMC stuff?
rrusczyk21:23:38
It will help with the harder problems on the AMC.
rrusczyk21:23:51
All right -- thanks for coming to the Math Jam. If you have any more questions, you can write me at classes@artofproblemsolving.com.
