| Transcript
for the Math
Jam "AoPS Classes Math Jam"
on Jun 4. |
| Math Jam hosted by rrusczyk
(Richard Rusczyk ). |
rrusczyk19:33:05
Hello, and welcome to an Art of Problem Solving Math Jam. Today we'll be discussing the Intermediate Algebra and Intermediate Counting & Probability courses. We will go through a couple example problems from each class, and discuss both what these classes cover and how they work.
rrusczyk19:33:30
For those of you who are enrolled in the MATHCOUNTS, Introduction to Algebra, or Introduction to Counting classes, most of the math we'll discuss tonight will be well beyond you (for now -- your time will come!). I won't be able to give all the background necessary to bring you up to speed, so you'll mostly be using this Math Jam as an opportunity to try out the classroom.
rrusczyk19:33:53
My name is Richard Rusczyk. I founded Art of Problem Solving and have written several Art of Problem Solving textbooks.
rrusczyk19:34:17
The classroom is moderated: students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. 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.
rrusczyk19:34:41
Note that it is not possible for the instructor to personally respond to every comment that you submit -- 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.
rrusczyk19:35:00
The virtual classroom is LaTeX enabled. LaTeX allows users to make nice equations and other math expressions. If you would like to learn how to write in LaTeX, click on the tab on the left side panel of our site and there is a tutorial and reference guide there.
rrusczyk19:35:29
You do not need to learn LaTeX to use our classes or our classroom!
rrusczyk19:35:39
Using LaTeX in the virtual classroom is slightly different than using it on the message board or in a LaTeX editor. If anything you type up in a post uses LaTeX, then you must use a semicolon (;) to begin your post. For example, if you type
rrusczyk19:35:45
rrusczyk19:35:50
This message will look like this when posted in the classroom:
rrusczyk19:35:54
rrusczyk19:36:13
Just remember, if your post uses LaTeX, use the semicolon (;) to begin your post!
rrusczyk19:36:24
In this Math Jam, I will briefly describe a course, then go through a few example problems. Then, I will hold a question-and-answer session about that class.
rrusczyk19:36:41
We'll start with:
rrusczyk19:36:42
Intermediate Algebra
rrusczyk19:37:10
The Intermediate Algebra class will be held on Wednesday evenings (7:30 - 9:00 PM ET, 4:30 - 6:00 PM PT), from June 18 to Dec 3. It will be taught by Mathlinks website founder Valentin Vornicu, who twice represented Romania at the International Math Olympiad (IMO) and has three times been asked to be a judge at the IMO.
rrusczyk19:37:27
Our Intermediate Algebra 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.
rrusczyk19:37:45
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.
rrusczyk19:38:18
The textbook for the course is our new Intermediate Algebra text, by Richard Rusczyk and Mathew Crawford. The text is required for the course.
rrusczyk19:38:35
I'll now proceed with a couple more 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.
rrusczyk19:39:35
Again, a reminder, if you are enrolled in MATHCOUNTS or the Introduction classes, then these problems will likely not be at your level. I won't have time to cover all of Introduction to Algebra tonight to get you up to the Intermediate level! So, you'll have to let the Intermediate folks do most of the heavy lifting on the problems.
rrusczyk19:39:51
Here's our first problem:
rrusczyk19:39:52
rrusczyk19:40:09
Yikes. Where might we start with those two equations?
FibonacciFan19:40:33
difference of cubes?
FantasyLover19:40:33
x^3-y^3?
rrusczyk19:40:48
Interesting. Why might we think of that?
FantasyLover19:41:19
factor it.
gh625_219:41:19
It's a difference of cubes, which has a nice factorization.
rrusczyk19:41:24
We know how to factor a difference of cubes. So, let's try subtracting one equation from the other.
rrusczyk19:41:34
rrusczyk19:41:41
What do we do with that?
virtuoso19:42:03
nikki19:42:03
factor the 11
rrusczyk19:42:08
What else can we factor?
FantasyLover19:42:28
x^3-y^3=(x-y)(x^2-xy+y^2)
e^i_penguin19:42:28
x^3-y^3=(x-y)(x^2+xy+y^2)
rrusczyk19:42:34
rrusczyk19:42:39
Now what?
virtuoso19:42:55
Divide both sides by x-y
nikki19:42:55
once you expand the diff of cubes, you can cancel the (x-y)
e^i_penguin19:42:55
rrusczyk19:42:59
virtuoso19:43:05
Since lxl =/ lyl, x-y will not be 0
rrusczyk19:43:15
Exactly. That's why we can divide by x-y.
rrusczyk19:43:24
But, what do we do with the equation we have now?
rrusczyk19:43:52
Hmmm . . .
rrusczyk19:44:14
We got somewhere with the difference of cubes, but not all the way.
rrusczyk19:44:18
What else might we consider?
FibonacciFan19:44:42
sum of cubes
rrusczyk19:44:52
And how might we introduce sum of cubes into this problem?
FibonacciFan19:45:35
add the two equations and factor out x+y like before
rrusczyk19:46:00
Instead of subtracting the original two equations, we L:
istos19:46:10
add the second two together
MathMathMath...19:46:10
How about adding them together?
rrusczyk19:46:17
rrusczyk19:46:22
And what does this do for us?
virtuoso19:46:45
Factor out 19 on the right, and sum of perfect cubes on left.
Kevin Zhang19:46:48
Factor x^3 + y^3
FibonacciFan19:46:53
x^2-xy+y^2=19
e^i_penguin19:46:53
FantasyLover19:46:53
x^2-xy+y^2=19
rrusczyk19:46:58
qwertythecucumber19:47:03
two equations to work with
rrusczyk19:47:15
2 equations. Hopefully that will be better than 1:
rrusczyk19:47:17
rrusczyk19:47:21
Now what?
FloodFilter:19:47:37
add the two together
FantasyLover19:47:37
add two equations
virtuoso19:47:37
Add these.
istos19:47:37
we could add them together
MathMathMath...19:47:46
2 xsquared plus 2 ysquared equals 30
e^i_penguin19:47:48
rrusczyk19:47:51
rrusczyk19:47:57
Um. . . Now what?
BOBCA19:48:01
subtrack both equations?
smartazn19:48:01
subtract 2nd equation from first
FibonacciFan19:48:01
and then subtract them
rrusczyk19:48:07
OK:
rrusczyk19:48:11
rrusczyk19:48:20
virtuoso19:48:36
Yes!
e^i_penguin19:48:36
yes
rrusczyk19:48:39
Show me!
Kevin Zhang19:49:02
Expand the expression
rrusczyk19:49:08
rrusczyk19:49:17
OK. How does that help?
Quickster9419:49:26
rrusczyk19:49:35
Ooh. Nice.
rrusczyk19:49:52
virtuoso19:50:15
225-4(16) = 161
e^i_penguin19:50:15
225-64=161
rrusczyk19:50:18
Now we just substitute the values we already found, and we get
rrusczyk19:50:27
rrusczyk19:51:01
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.
rrusczyk19:51:32
That problem is a little easier than the average problem in the Intermediate Algebra class.
rrusczyk19:51:39
Let's look at a somewhat tougher problem:
rrusczyk19:51:42
rrusczyk19:51:55
(The link is just a copy of the problem.)
rrusczyk19:51:59
I'll give you a minute to think about it.
UnendingPi19:53:04
whats relatively prime?
rrusczyk19:53:14
No common divisors except 1
prithviramesh19:53:22
Where do we start?
rrusczyk19:53:30
Good question. Where do we start?
rrusczyk19:53:44
First, what does |a+bi| = 8 tell us?
e^i_penguin19:54:03
a^2+b^2=64?
FibonacciFan19:54:03
a^2+b^2=64
rrusczyk19:54:56
rrusczyk19:55:16
(To those of you asking what i is, you'll have to wait until you have more of an algebra background to tackle this problem!)
rrusczyk19:55:30
We have a very wordy problem. We'll have to convert those words to equations.
rrusczyk19:55:44
One 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.
rrusczyk19:56:06
In terms of f and z, what is "the image of the point z"?
gh625_219:56:47
f(z)=z(a+bi)
rrusczyk19:56:58
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."
rrusczyk19:57:13
Now, what is the distance between f(z) and z?
virtuoso19:57:27
The magnitude?
rrusczyk19:57:31
The magnitude of what?
MathMathMath...19:58:20
What's magnitude?
rrusczyk19:58:33
|a+bi| is referred to as the magnitude of the number a+bi.
UnendingPi19:58:41
length
e^i_penguin19:58:41
distance
istos19:58:41
it's the length of a vector, basically
rrusczyk19:58:46
The length of what?
rrusczyk19:59:04
We have the points f(z) and z in the complex plane. What is the distance between them as an algebraic expression?
isabella229619:59:40
f(z) - z
smartazn19:59:42
f(z)-z?
rrusczyk19:59:53
The distance between f(z) and z is |f(z) - z|.
rrusczyk20:00:05
And what is the distance between f(z) and the origin?
gh625_220:00:24
|f(z)|
FloodFilter:20:00:24
|f(z)|
rrusczyk20:00:34
This distance, by definition, is simply |f(z)|. So, what is our equation?
RollandWu20:00:55
|f(z)|=|f(z)-z|
FantasyLover20:00:55
|f(z)|=|f(z)-z|
rrusczyk20:01:01
FibonacciFan20:01:18
|(a+bi)z|=|(a+bi)z-z|
virtuoso20:01:18
Try solving this?
rrusczyk20:01:32
We can now use the definition of f in the problem to write
rrusczyk20:01:36
rrusczyk20:01:50
Now what might we do?
FibonacciFan20:02:16
factor out z?
gh625_220:02:19
|z(a+bi-1)|=|z(a+bi)|
yu.wang20:02:29
|(a+bi-1)z|=|(a+bi)z|
rrusczyk20:02:42
Why does this help?
FibonacciFan20:02:52
use the fact that |wz|=|w||z|
MathMathMath...20:02:55
divide both by z
rrusczyk20:03:02
rrusczyk20:03:17
Ah, now what?
istos20:03:35
use the definition of absolute value?
virtuoso20:03:35
Remove absolute value bars.
rrusczyk20:03:53
How? (Also, in complex numbers, we refer to this as "magnitude" instead of "absolute value")
FibonacciFan20:04:15
(a-1)^2+b^2=a^2+b^2 implies that a-1=pm a which implies that a=1/2
RollandWu20:04:15
(a-1)^2+b^2=a^2+b^2
rrusczyk20:04:34
We use the definition we stated at the beginning for the magnitude of a complex number.
rrusczyk20:04:44
(This is NOT the same as what you do with a real number!)
rrusczyk20:04:58
rrusczyk20:05:09
Applying this to the equation |a-1 + bi | = |a+bi |
rrusczyk20:05:11
gives us:
rrusczyk20:05:29
gh625_220:05:48
(a-1)^2=a^2
rrusczyk20:05:56
And if we keep going, we get:
happyof20:06:00
a=1/2?
yu.wang20:06:00
a=0.5
e^i_penguin20:06:00
a=1/2
rrusczyk20:06:11
rrusczyk20:06:25
However the problem asked us to find the value of b^2. Is there any piece of information in the problem that we haven't used yet? (Whenever you're stuck on a long, complicated problem, this is a good question to ask yourself: What haven't I used yet?)
KaneHsu20:06:42
Use |a+bi|=8
rrusczyk20:07:06
gh625_220:07:42
BOBCA20:07:42
a^2+b^2=64?
Kevin Zhang20:07:42
63.75 = b^2
FloodFilter:20:07:42
a^2+b^2=64
rrusczyk20:08:03
rrusczyk20:08:12
What is our final answer?
virtuoso20:08:49
259
istos20:08:57
259
yu.wang20:08:57
259
prithviramesh20:08:57
259
MathMathMath...20:08:57
259
rrusczyk20:09:04
rrusczyk20:09:37
This problem is a bit harder than the average problem in the Intermediate Algebra class (and we cover the necessary prerequisite knowledge of complex numbers in the class).
rrusczyk20:09:55
The Intermediate Algebra 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:
rrusczyk20:10:00
rrusczyk20:10:05
rrusczyk20:10:10
rrusczyk20:10:18
rrusczyk20:10:20
rrusczyk20:10:32
Are there any questions about the course?
MathMathMath...20:10:55
Yes. How long is the course?
rrusczyk20:11:00
The course is 24 weeks.
smartazn20:11:09
looks way harder than the intermediate algebra course at stanford online...
rrusczyk20:11:26
It may well be. (Our average student is probably stronger than theirs.)
e^i_penguin20:11:30
june13-december 3 or something like that?
rrusczyk20:11:42
June 18.
MathMathMath...20:11:46
just try it! you'll probably learn more too!
rrusczyk20:11:48
:)
qwertythecucumber20:11:52
So, this would be about the school equivalent of Algebra II (but with more stuff)?
rrusczyk20:12:15
Algebra II + all the non-trig pre-calculus, plus a lot of more challenging material you won't see in a regular classroom.
The Distribuator20:12:25
how many kidsare in one course?
rrusczyk20:12:41
We expect 40-50. If the class gets too large, we split it and run two sessions.
UnendingPi20:12:43
how long is this jam gonna be?
rrusczyk20:12:49
Probably 30 more minutes.
rrusczyk20:13:26
There are extra instructors in each class, so don't be daunted by the class size. Any questions that need be answered will be answered in class or privately by one of the instructors in the room.
Quickster9420:13:31
mr v is teaching right?
rrusczyk20:13:41
That's correct - Valentin Vornicu.
Quickster9420:13:43
are matrices covered?
rrusczyk20:13:54
No, we will be adding that to our Trig course when I write the book.
The Distribuator20:13:56
Is the class just problems, or is their basic instruction first?
rrusczyk20:14:27
Both instruction and problems, but the "instruction" will be through a series of fundamental problems. You'll always be thinking about something rather than just reading what the instructor writes.
yu.wang20:14:44
Are we going to discuss anyother courses tonight?
rrusczyk20:15:02
Yes, we will cover the Intermediate Counting class after Q&A.
BOBCA20:15:04
can we try another problem?
rrusczyk20:15:15
We'll do a counting problem after the question/answer session.
BOBCA20:15:17
Does the class remind us formulas or concepts just in case we don't know them?
rrusczyk20:15:25
Yes.
Pericles20:15:27
what happens if we miss a few classes
rrusczyk20:15:53
There is a full transcript of every class available right after class, so you can catch up on the course homepage.
Quickster9420:16:02
does this class also have challenge sets?
rrusczyk20:16:25
Yes. 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, however you may post them to the message board if you like. The class also has 4 longer problem sets for which you should write up your full solutions and submit them. These solutions will be read, and you will receive detailed feedback.
Kevin Zhang20:16:36
Will we cover modulos?
rrusczyk20:16:42
That is in the Introduction to Number Theory course.
MathMathMath...20:16:49
Are you going to have a q&a session after each class?
rrusczyk20:17:15
The instructor will invite questions during and after the class. And you can always use the class message board.
FibonacciFan20:17:16
how do you calculate the cost of a class? Why is this class more expensive than the AIME class?
rrusczyk20:17:42
This class is twice as long as the AIME class, and it has assignments that we evaluate and give feedback on (the AIME class does not).
RollandWu20:18:02
Does this course have some time overlap with AIME, in Dec?
rrusczyk20:18:12
They overlap one week, but they are on different days.
MathMathMath...20:18:14
Class message board?
rrusczyk20:18:20
There is a board in the Forum for the class.
qwertythecucumber20:18:43
Is there a basic format of the classes?
Kevin Zhang20:18:43
Where does this class take place?
rrusczyk20:18:47
The class is online.
rrusczyk20:19:00
The format is similar to the Math Jam, in that most of the class is a discussion of problems.
rrusczyk20:19:14
Some of these problems will be basic, to introduce concepts. Others will be more advanced.
BOBCA20:19:16
How many questions are we going to do per class?
rrusczyk20:19:25
It varies a great deal from class to class.
BOBCA20:19:27
What if we can't solve some of the problems?
rrusczyk20:19:43
That's OK! No one is able to solve all of them! You ask questions then!
prithviramesh20:20:16
What if you still don't get them?
rrusczyk20:20:21
Then you ask again!
MathMathMath...20:21:01
Is there an explanation for each problem?
rrusczyk20:21:05
In class, yes.
BOBCA20:21:38
are the problems arranged form easy to hard?
rrusczyk20:21:40
Yes.
hello12320:21:41
about what level are most of the problems AMC AIME etc
rrusczyk20:21:59
They range from very basic to introduce an idea, up through AIME and even a few beginning Olympiad level problems.
Kevin Zhang20:22:03
Will you give us homework to practice with?
rrusczyk20:22:16
There is homework after each class in the course, but there won't be homework tonight!
rrusczyk20:22:25
Let's look at the other class we have to discuss tonight:
rrusczyk20:22:30
Intermediate Counting & Probability
rrusczyk20:22:39
The Intermediate Counting & Probability class will cover a variety of powerful counting and probability tools. The topics in the course will include discrete mathematics, including clever one-to-one correspondences, principle of inclusion-exclusion, generating functions, distributions, pigeonhole principle, induction, constructive counting and expectation, combinatorics, recursion, and conditional probability.
rrusczyk20:22:54
The course will be taught by Dave Patrick, who is also the author of the Introduction to Counting & Probability and Intermediate Counting & Probability textbooks (the latter of which will be used in this class).
rrusczyk20:23:39
We'll do a sample problem (one of my favorites!), and then I'll give a few more details about the course.
rrusczyk20:23:48
rrusczyk20:24:14
We're going to do this problem a couple different ways. Are there any suggestions about how to approach it?
e^i_penguin20:24:29
casework
rrusczyk20:24:31
We can try cases. What cases should we consider?
MathMathMath...20:25:11
rows and columns, then diagonals.
Kevin Zhang20:25:11
columns, rows, and diagonals
UnendingPi20:25:11
diagonals, columns, rows, and 3Ds
rrusczyk20:25:17
Our cases are those sets which are not diagonals, sets which are '2D' diagonals (such as those which are the diagonal of a square), and '3D' diagonals.
rrusczyk20:25:25
rrusczyk20:25:30
rrusczyk20:25:38
rrusczyk20:25:42
So, we start with our first case:
rrusczyk20:25:46
rrusczyk20:25:52
How many are there?
rrusczyk20:26:21
Give a justification!
rrusczyk20:27:28
I see some right answers with wrong justifications (or incomplete ones), and wrong answers with wrong justifications. . .
KaneHsu20:27:53
16*3=48 because each dimension has 16 and there are the x,y,z dimensions
Pericles20:27:53
48 because on each face there are four possibilities vertically and four horizontally and there are 6 faces so 8 x 6
rrusczyk20:28:05
In the direction shown, there are 4 on each level. Therefore, there are 4^2 = 16 with that direction since there are 4 levels.
rrusczyk20:28:11
rrusczyk20:28:14
There are two other directions we could have chosen, examples of which are here in red and blue:
rrusczyk20:28:22
rrusczyk20:28:29
There are 4^2 in each of these directions, as well. So our total for the 'not diagonal' case is 3(4^2) = 48.
rrusczyk20:28:34
How about the '2D Diagonal' case?
rrusczyk20:28:41
virtuoso20:29:06
2 per plane.
rrusczyk20:29:12
rrusczyk20:29:17
We note that in any given square, there are two of these 2D diagonals:
rrusczyk20:29:21
How many such squares are there?
yu.wang20:29:33
12
FibonacciFan20:29:33
12
happyof20:29:43
4 on each diagonall plane, so there ate 12
rrusczyk20:29:45
There are 4 of these 'squares' facing in each direction, and there are 3 possible orientations of these groups of squares, so there are 3(4) squares. Thus, for the '2D diagonal' case we have 2*3*4 = 24 4-in-a-row configurations.
rrusczyk20:29:53
What about the 3D case:
rrusczyk20:29:59
KaneHsu20:30:21
4 total
thepianistalex20:30:21
four total
BOBCA20:30:21
4
rrusczyk20:30:30
There are 4 interior diagonals of the cube, so here we have 4.
rrusczyk20:30:34
So what is the total number of winning sets?
MathMathMath...20:31:30
76 4s!
KaneHsu20:31:30
76?
virtuoso20:31:36
76
rrusczyk20:31:39
rrusczyk20:31:44
That's the orderly casework approach - we teach that general approach in Introduction to Counting. In Intermediate Counting, we learn a slick way to do it.
rrusczyk20:31:53
Does anyone see a slick way to do this?
rrusczyk20:32:13
Hmmm . . . Let's look at that expression again:
rrusczyk20:33:14
3*4^2 + 6*4 + 4 = 3*4^2 + 3*4*2 + 2^2
rrusczyk20:33:30
Does that expression on the right vaguely remind you of anything in math?
virtuoso20:34:00
Pascal's Triangle?
rrusczyk20:34:17
Pascal's triangle. . . powers of numbers. Where do these two go together?
virtuoso20:34:31
Binomial Theorem?
Quickster9420:34:36
binomial theorem?
e^i_penguin20:34:36
binomial theorem?
rrusczyk20:34:49
And where do we get a couple 3's in a row in the Binomial Theorem?
virtuoso20:35:13
1 3 3 1
e^i_penguin20:35:15
cubes
qwertythecucumber20:35:15
1,3,3,1?
rrusczyk20:35:32
(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
rrusczyk20:35:37
Binomial Theorem.
rrusczyk20:35:49
3*4^2 + 3*4*2 + 2^2...
rrusczyk20:36:01
What might we stick into our Binomial theorem for x and y?
rrusczyk20:37:00
We have a 4^2 and a 2^2.. . .
yu.wang20:37:13
4 and 2?
happyof20:37:20
4 and 2?
rrusczyk20:37:51
rrusczyk20:38:12
Sorry, that should be
rrusczyk20:38:37
rrusczyk20:39:17
We're trying to produce 3*4^2 +3*4*2 + 2^2. What might we do next to isolate that on the right?
RollandWu20:40:01
-4^3 and divide by 2
KaneHsu20:40:01
Subtract 4^3
Kevin Zhang_220:40:01
subtract 4^3 then divide by 2
rrusczyk20:40:04
rrusczyk20:40:12
Then, we divide by 2:
rrusczyk20:40:39
rrusczyk20:40:50
We know that expression on the right counts what we want.
rrusczyk20:41:15
That expression on the left is very simple!! It's just (6^3 - 4^3)/2.
rrusczyk20:41:27
Simple expressions in counting problems usually mean there are simple explanations.
rrusczyk20:41:39
So, let's try to find a simple explanation for why the answer comes out to this.
rrusczyk20:42:00
Now, it's clear where we might come up with a 4^3 in this problem -- there are 4^3 dots.
rrusczyk20:42:07
But where does the 6^3 come from?
rrusczyk20:42:53
4x4x4 cube gives us 4^3 = 64 dots. . .
rrusczyk20:43:07
What should we be looking for to find 6^3?
yu.wang20:43:20
6x6x6
Quickster9420:43:20
a 6x6x6 cube?
rrusczyk20:43:34
And how might we relate such a cube to a 4x4x4 cube?
MathMathMath..._220:44:24
difference of cubes should have something to do with it.
rrusczyk20:44:45
OK - we have a 6x6x6 cube and think we want to do something about subtracting a 4x4x4 cube . . .
rrusczyk20:45:01
What 4x4x4 cube should we be thinking about once we have a 6x6x6 cube?
happyof20:45:26
the inside one?
rrusczyk20:45:34
In the Intermediate counting class we will spend a couple days on 1-1 correspondences. What this basically means is to find a relationship between what we want to count and something that's easy to this count.
rrusczyk20:45:40
Sometimes this feels almost like magic. In the course we will talk about how to see these 1-1 correspondences. Here, we have to see something that isn't there. Suppose we take a slightly larger cube:
rrusczyk20:45:48
rrusczyk20:45:57
Our 4x4x4 cube is embedded in a 6x6x6 cube. How can we relate our 4-in-a-rows to the 6x6x6 cube?
rrusczyk20:47:28
First of all, how is 6^3 - 4^3 related to this diagram?
arb9520:47:57
it's the outer region
virtuoso20:47:59
All of the dots not in the smaller cube
rrusczyk20:48:02
We have 4^3 dots in the small cube, and 6^3 in the big one, so there are 6^3 - 4^3 on the outside.
rrusczyk20:48:18
And we think we want to divide this by 2 to count the 4-in-a-rows inside the little cube.
rrusczyk20:48:36
So, how do we relate any one 4-in-a-row to these outside dots?
happyof20:49:04
contineus to 2 of them
rrusczyk20:49:10
Any 4 in the row when extended hits our big cube in two places:
rrusczyk20:49:16
rrusczyk20:49:20
There are four examples in that diagram: notice how when we extend the red 4-in-a-row it hits our big cube in 2 places. And when we extend the blue 4-in-a row it hits the big cube in 2 places, and so on.
rrusczyk20:49:51
Conversely, if we go the other way, start from a point on the big cube and find a 4-in-a-row in the little cube, there's only one way to do it:
rrusczyk20:49:56
rrusczyk20:50:01
Look at the red point. There's only one way to put a line through it to get a 4-in-a-row in the smaller cube. Same for the green point in the upper left corner. Same for the purple point; same for the blue.
rrusczyk20:50:11
So? How does this give us a way to count our 4-in-a-rows?
Kevin Zhang_220:50:47
6^3 - 4^3 divided by 2!!!!!!!!!!!!
yu.wang20:50:47
every dot has a line and each line connects two dots so you have to divide by 2
FloodFilter:20:50:55
there are 152 dots on the outside, 2 dots make a line, 152/2 = 76 linew
rrusczyk20:51:01
For every 4-in-a-row, there's a pair of points on the outside of the big cube. For every pair of points on the outside of the big cube, there's a four in a row. Thus, we have what we call a 'one-to-one correspondence' between opposite pairs of points on the outside of the cube and 4-in-a-rows.
rrusczyk20:51:05
Therefore, to count our 4-in-a-rows, we just count the points on the outside of the cube and divide by two.
rrusczyk20:51:10
rrusczyk20:51:31
Notice that we've completed the problem in two different ways and found the same answer both times, so we're probably right.
rrusczyk20:51:40
(This is my favorite way to check counting problems.)
FibonacciFan20:51:45
that is so clever!
BOBCA20:51:45
ooohhhhhhhhhhhhh
Kevin Zhang_220:51:45
Wow...
rrusczyk20:51:53
Indeed, it's one of my favorite problems.
rrusczyk20:51:57
Here are a couple more problems that can be solved with clever correspondences:
rrusczyk20:52:03
Define the alternating sum of a set as follows:
List the elements of the set in decreasing order. Take the first number in the list, subtract the second, add the third, subtract the fourth, and so on. Thus, the alternating sum of the set {3,5,11,7} is 11-7+5-3=6.
Find the sum of the alternating sums of all of the subsets of {1,2,3,4,5,6,7,8,9,10}.
rrusczyk20:52:08
And:
rrusczyk20:52:12
Our club has 8 members. We are going to form two committees.
Each person must serve on at least one committee (and may serve on both), and each committee must have at least one member. The two committees are indistinguishable, so the pair of committees {Ajai, Bob}, {Mary} is the same as the pair {Mary}, {Ajai, Bob}. How many ways can we form the committees?
rrusczyk20:52:35
These problems are roughly middle-level difficulty for the course. The example we did was slightly easier than the average problem in the course. (The first solution is slightly easier that is.)
FibonacciFan20:52:57
sorry i don't see why there is only one way to put a line through, the red point e.g. and get a 4-in-a-row...should that just be obvious?
rrusczyk20:53:02
This is a subtle point!
rrusczyk20:53:11
Kevin Zhang_220:53:59
Just extend the line through to the other side( to FibonacciFan)
rrusczyk20:54:09
But why isn't there more than one way to do it?
rrusczyk20:54:33
You can think of it this way: if you start at a point on one side, then go in a straight line by the dots to the other side, then in each dimension, you must either stay the same all the way, or change by 1 every time.
rrusczyk20:54:51
I'll let you sort out the rest from there :)
rrusczyk20:55:00
You can find more questions like those we cover in the course by checking out the Post Test for the course here:
rrusczyk20:55:07
Students should have a complete mastery of basic counting (at the level of Introduction to Counting & Probability) before taking this course. Students should also have a solid algebra background through at least algebra II. Students who have completed the Art of Problem Solving Intermediate Algebra and Introduction to Counting & Probability classes should feel comfortable taking this class. (However, students are not required to take these classes before taking Intermediate Counting & Probability.)
rrusczyk20:55:19
The course will meet for 18 weeks on Thursdays, starting June 19, at 7:30 PM Eastern / 4:30 PM Pacific. The last class date is October 16. Each class is 90 minutes.
rrusczyk20:55:28
This course will use a textbook in conjunction with the course: our new Intermediate Counting & Probability 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 an excerpt from the book here:
rrusczyk20:56:08
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 are recommending 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. We also expect to spend some class time answering students' questions about problems from the textbook.
rrusczyk20:56:20
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, however you may post them to the message board if you like. The class also has three longer problem sets -- one for each 6-week period of the course -- for which you should write up your full solutions and submit them. These solutions will be read, and you will receive detailed feedback.
rrusczyk20:56:37
That's it for the math -- are there any questions about the course?
UnendingPi20:57:10
is it fun, in your opinion?
rrusczyk20:57:32
Absolutely. The Intermediate Counting class is probably the most fun one we offer in my opinion.
yu.wang20:57:37
Do you know Jeremy Hahn?
rrusczyk20:57:59
Yes - I taught him back when he only knew a lot of math, instead of the terrifying amount he knows now.
MathMathMath..._220:58:03
how many courses are there for the subject?
rrusczyk20:58:11
I don't understand the question.
happyof20:58:32
he meant how many counting courses are there
rrusczyk20:58:43
There are two courses: one intro, one intermediate.
Harrison20:58:56
I have never seen anything like this course on a high school level. What would be the college class equivalent for this course?
rrusczyk20:59:29
Good question. The reason you don't see this in high school is that most middle and high school classes are not designed for future mathematicians and scientists, and they also have not caught up to the digital age.
rrusczyk21:00:04
This course would be called discrete math in college, and any computer scientist would tell you biggest problem with middle and high school math is that discrete math is not taught enough.
Kevin Zhang_221:00:06
What about advanced/expert?
rrusczyk21:00:13
There's not a higher level.
FibonacciFan21:00:20
do you know of any sites similar to aops that offer college level classes? If not I suggest you start one or move AOPS in that direction.
rrusczyk21:00:30
We will be expanding in that direction over the next couple years.
BOBCA21:00:33
umm im in junior high? So is the contents of this ok for me?
rrusczyk21:00:40
I recommend you start with the Intro classes.
Kevin Zhang_221:00:43
Do you want us to do anything else besides practice?
rrusczyk21:00:46
Read the book!
FloodFilter:21:01:21
is there a class in fall?
rrusczyk21:01:39
The Interm Algebra will be repeated in the fall. The Interm Counting in the spring.
RollandWu21:01:44
I mean, there is a intermediate algebra course immediately after this probability course, Right?
rrusczyk21:01:55
We already discussed the Algebra course (we did that first)
mfranzs21:01:58
I am enrolled in a mathcounts class that starts next week. I missed the other math jams. Is there anything I need to do to prepare?
rrusczyk21:02:04
You're all set! See you in class next week.
Kevin Zhang_221:02:06
Are there any strategies you have for us?
rrusczyk21:02:21
Look at the Articles in the Resources section. There are a number of strategies there.
