| Transcript
for the Math
Jam "AoPS Classes Math Jam"
on Oct 1. |
| Math Jam hosted by rrusczyk
(Richard Rusczyk ). |
rrusczyk19:16:50
Hello. This is the Math Jam classroom. The Math Jam will be starting at 7:30 PM ET (4:30 PM PT).
This is not the classroom for the Algebra 3 class or the AMC 10 Problem Series . This is the AoPS Classes Math Jam. If you are enrolled in Algebra 3 or the AMC 10 Problem Series, then you should leave the classroom now, click Classroom on the sidebar again, and choose the appropriate link on the next page.
rrusczyk19:17:29
The classroom is moderated, so your comments will not appear right after you type them. They come to the instructors.
rrusczyk19:27:11
We'll get started in just a few minutes.
rrusczyk19:27:20
This is not the classroom for the Algebra 3 class or the AMC 10 Problem Series . This is the AoPS Classes Math Jam. If you are enrolled in Algebra 3 or the AMC 10 Problem Series, then you should leave the classroom now, click Classroom on the sidebar again, and choose the appropriate link on the next page.
rrusczyk19:31:23
Hello, and welcome to an Art of Problem Solving Math Jam. Today we'll be discussing the following three classes: Algebra 3, Intermediate Counting & Probability, and Calculus.
rrusczyk19:31:35
My name is Richard Rusczyk. I founded Art of Problem Solving and have written several Art of Problem Solving textbooks.
rrusczyk19:31:41
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.
rrusczyk19:31:53
The classroom is moderated: students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. 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.
rrusczyk19:32:16
As for questions about the classes, we will try to answer all of those tonight. I will let you know when to start asking questions about specific classes.
helagha19:32:30
not too many people today...
rrusczyk19:32:56
True, this is a pretty small Math Jam, so you better speak up when we start doing problems, so we can solve them!
rrusczyk19:33:36
In this Math Jam, I will briefly describe a course, then go through an example problem (except for calculus -- we won't be doing sample problems for the calculus class). Then, I will hold a question-and-answer session about the class.
rrusczyk19:33:53
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 fall.
rrusczyk19:34:15
Please understand that if you are enrolled in one of our introductory classes, or haven't much experience yet with advanced problem solving, then some of the problems we discuss tonight will be very hard for you. 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!
rrusczyk19:34:40
Algebra 3
rrusczyk19:34:50
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.
rrusczyk19:34:56
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.
rrusczyk19:35:25
The textbook for the course is our new Intermediate Algebra text, by Richard Rusczyk and Mathew Crawford. The text is required for the course.
rrusczyk19:35:41
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. After we work through the problems, I'll take questions about the course.
rrusczyk19:35:59
rrusczyk19:36:11
Where should we start?
number.sense19:36:26
subtract the two equations
helagha19:36:33
subtract the two
rrusczyk19:36:35
Why do we think to do that?
number.sense19:37:18
that way we get x^3-y^3 on one side which is divisible by x-y and 11x-11y on the other
rrusczyk19:37:22
Let's see!
rrusczyk19:37:25
rrusczyk19:37:32
What can we do with that?
helagha19:37:45
so we can get to the x^2 and the y^2 (after factoring x^3-y^3)
aunch19:37:45
factor
Iggy Iguana19:37:45
factor the right
limac19:37:45
Factor both the sides.
rrusczyk19:37:50
What does factoring give us?
number.sense19:38:18
factor the right side as (x-y)(x^2+xy+y^2)
rrusczyk19:38:22
Keep going!
limac19:38:35
(x-y)(x^2+xy+y^2)=11(x-y)
helagha19:38:35
left is 11(x-y)
rrusczyk19:38:44
aunch19:38:52
(x-y)(x^2+xy+y^2) = 11(x-y)
helagha19:39:01
we now know that x^2+xy+y^2 = 11
limac19:39:01
so the (x-y)'s cancel out leaving us with: x^2+xy+y^2=11
Iggy Iguana19:39:01
divide by x-y
rrusczyk19:39:05
number.sense19:39:10
since |x|-|y| isn't = 0, we can divide by x-y
rrusczyk19:39:24
Good point: it's important to note why we can divide by x-y.
rrusczyk19:39:33
Well. Now what?
Iggy Iguana19:39:57
(x+y)^2 =11
number.sense19:39:57
we can add the original equations to get a different ending equation
CantonMathGuy19:39:57
add
number.sense19:40:11
now we can add both equations
rrusczyk19:40:32
Subtracting the two equations got us somewhere. But it didn't finish the problem.
rrusczyk19:40:39
Let's try adding them.
rrusczyk19:40:43
rrusczyk19:40:46
How does this help?
number.sense19:41:14
The lhs=(x+y)(x^2-xy+y^2) and the rhs=19(x+y) after adding
CantonMathGuy19:41:14
factor
aunch19:41:14
factor out (x+y)
shanmug100019:41:14
factor!
limac19:41:14
so, we are left with: x^2-xy+y^2=19 after canceling out the (x+y)
rrusczyk19:41:22
We do the same thing we did after subtracting!!!
rrusczyk19:41:27
helagha19:41:30
we get x^2 - xy +y^2 = 19 using the same method
rrusczyk19:41:34
rrusczyk19:41:42
rrusczyk19:41:50
Now what?
aunch19:41:59
add the two new equations to cancel out xy
$LaTeX$.19:41:59
cancle the xy's
limac19:41:59
Finally, we add both the equations above, which cancesl out the xy terms.
Iggy Iguana19:41:59
add
rrusczyk19:42:08
And what does that give?
amywu19:42:24
cancle the xy's
$LaTeX$.19:42:24
2x^2+2y^2
number.sense19:42:24
x^2+y^2=15
CantonMathGuy19:42:24
x^2+y^2=15
rrusczyk19:42:28
limac19:42:29
2(x^2+y^2)=30 => x^2+y^2=15
rrusczyk19:42:38
Is that enough to solve the problem?
aunch19:42:47
no
helagha19:42:47
no
untaik19:42:51
no
CantonMathGuy19:42:51
no
$LaTeX$.19:42:51
no
amywu19:42:54
no
rrusczyk19:43:00
Nope. What else can we do?
shanmug100019:43:16
subtract the 2 equations
aunch19:43:16
subtract
rrusczyk19:43:49
What does that give?
number.sense19:44:15
xy=-4
aunch19:44:15
xy=-4
countingrocks19:44:15
to get 2xy = -8
untaik19:44:15
-2xy=8
limac19:44:15
-2xy=8
number.sense19:44:15
xy=-4
CantonMathGuy19:44:15
xy=-4
rrusczyk19:44:21
rrusczyk19:44:38
Now, before we continue, I must introduce tonight's assistant, matt276eagles.
rrusczyk19:44:56
His real name is Matt Superdock, and he's currently a student at Princeton. He was a 2-time USAMO Honorable Mention (to 24 in the country), and is one of the best Guitar Hero and Rock Band players in the world.
rrusczyk19:45:19
That means he's a lot cooler than I am. He will answer some of your math questions tonight, if you have trouble with anything we are working on.
rrusczyk19:45:29
So, speak up if you don't understand something!!
matt276eagles19:45:32
Hi guys, and I highly doubt I'm cooler than you, Richard
rrusczyk19:45:33
Now, back to our story.
rrusczyk19:45:53
We had two equations: x^2 + y^2 = 15, xy = -4.
rrusczyk19:45:58
CantonMathGuy19:46:15
yes
number.sense19:46:15
yes, it is
helagha19:46:23
yes
arbala19:46:23
yes
shanmug100019:46:23
yes...
rrusczyk19:46:27
How?
limac19:46:56
Expand it out
rrusczyk19:47:06
What do we get when we expand?
helagha19:47:44
limac19:47:49
x^4-2x^2y^2+y^4
rrusczyk19:48:01
How can we use what we know to evaluate that?
countingrocks19:48:42
(x^2+y^2)^2 - 4xy
number.sense19:48:42
which is (x^2+y^2)^2 - 4x^2y^2
CantonMathGuy19:48:42
(x^2-y^2)^2=(x^2+y^2)^2-4(xy)^2=15^2-4(-4)^2=225-64=161
rrusczyk19:48:52
Exactly!
rrusczyk19:48:55
rrusczyk19:49:19
Any questions about what we did there?
rrusczyk19:49:39
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.
rrusczyk19:50:18
Let's try one more algebra problem.
rrusczyk19:50:23
rrusczyk19:51:08
That's a pretty wordy problem. What would we like to do with all those words?
limac19:51:28
turn them into math!
number.sense19:51:28
make some equations, etc
shanmug100019:51:28
put them into equations!
CantonMathGuy19:51:28
turn to math
rrusczyk19:51:32
We'll have to convert those words to equations.
rrusczyk19:51:37
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
rrusczyk19:51:42
In terms of f and z, what is "the image of the point z"?
CantonMathGuy19:52:13
f(z)
rrusczyk19:52:16
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."
rrusczyk19:52:39
Now, what is an expression for the distance between f(z) and z in the complex plane?
number.sense19:53:11
that means that f(z) lies on the perpendicular bisector of the segment connecting z and 0
rrusczyk19:53:30
Interesting -- see if you can come up with a geometric approach to this problem (on your own, matt276 may have some ideas to help :) )
rrusczyk19:53:40
Meanwhile, we'll continue with our algebraic approach:
limac19:53:43
|f(z) - z|
rrusczyk19:53:50
The distance between f(z) and z is |f(z) - z|.
rrusczyk19:53:59
And what's the distance between f(z) and the origin?
number.sense19:54:23
|f(z)| = |z-f(z)|
limac19:54:23
|f(z)|
rrusczyk19:54:29
This is simply |f(z)|.
rrusczyk19:54:42
rrusczyk19:55:44
I realize some of you may not have much background in complex numbers; we start from the basics in class. You may not be familiar with |a+bi| --- what does that equal?
limac19:56:25
countingrocks19:56:35
(a^2+b^2)^0.5
rrusczyk19:56:38
rrusczyk19:56:57
CantonMathGuy19:56:59
a^2+b^2=8^2
number.sense19:56:59
a^2+b^2=64
rrusczyk19:57:15
Now that we understand what the magnitude of a complex number is, let's return to:
rrusczyk19:57:21
rrusczyk19:57:27
What can we do with this?
aunch19:57:31
solving time!!!
rrusczyk19:57:37
What do we do as our first step?
rrusczyk19:58:33
Uh-oh. We're stuck staring!
rrusczyk19:58:44
We have an equation involving f(z). Do we know anything about f(z)?
CantonMathGuy19:59:01
f(z)=(a+bi)z
limac19:59:01
It is (a+bi)z
rrusczyk19:59:22
OK, so, let's not just stare --- let's use that! What does that make our equation?
limac20:00:10
rrusczyk20:00:15
We substitute (a+bi)z in for f(z), and we have |(a+bi)z - z| = |(a+bi)z |
rrusczyk20:00:22
All we're doing here is substitution.
rrusczyk20:00:27
Nothing magic (yet!!)
rrusczyk20:00:46
Hmmm. What will we do with that equation?
Iggy Iguana20:01:05
factor left
rrusczyk20:01:36
OK, what does that give?
rrusczyk20:02:50
That is, what do we get when we factor the expression on the left side?
aunch20:03:03
z(a+bi-1)
rrusczyk20:03:11
OK, that gives |z(a-1 + bi)| = |(a+bi)z|
rrusczyk20:03:39
(My answer is the same as aunch's; I've just grouped the real parts together in a-1+bi.)
rrusczyk20:03:41
Now what?
aunch20:03:43
take out z!
rrusczyk20:03:50
What do you mean by "take out z"?
$LaTeX$.20:03:56
l(a-1+ bil=l(a+bi)l
limac20:03:56
That is the same as |z||a-1 + bi| = |(a+bi)||z|
rrusczyk20:04:04
Exactly, we can factor the |z| out:
rrusczyk20:04:12
smartalec1720:04:16
divide by z on both sides
rrusczyk20:04:21
Are we allowed to do that?
Iggy Iguana20:04:51
yes
aunch20:05:08
why?
rrusczyk20:05:12
Good question.
rrusczyk20:05:20
Why can we just "cancel out" the |z| terms?
helagha20:05:49
they could be zero
rrusczyk20:06:00
Ah, good point. |z| could be 0.
rrusczyk20:06:09
Could we divide by |z| if |z| is nonzero?
limac20:06:39
yes
CantonMathGuy20:06:39
yes
shanmug100020:06:39
i think.
rrusczyk20:06:56
Sure! If |z| is nonzero, then |z| is some real nonzero number.
rrusczyk20:07:01
We can divide by that :)
rrusczyk20:07:04
That would give us
rrusczyk20:07:14
|a-1 + bi | = |a+bi |
rrusczyk20:07:46
The equation
rrusczyk20:07:54
|z||(a+bi) -1| = |a+bi| |z|
rrusczyk20:08:22
must hold for all z, so it must hold when z is 1.
rrusczyk20:08:32
We can also think of what we just did as rearranging the equation to
rrusczyk20:08:54
|z| ( |(a+bi)-1| - |a+bi|) = 0.
rrusczyk20:09:31
This must hold for *any* z, so the coefficient of |z| here must be 0.
rrusczyk20:09:38
Now, what do we do with:
rrusczyk20:09:46
aunch20:10:15
somehow take away the absolute value signs
rrusczyk20:10:36
Key strategy in algebra problems: find ways to get rid of annoying notation. How do we do that here?
Cheepskape20:11:47
(a-1)^2+b^2 = a^2+b^2
limac20:11:47
That is the same as: | (a-1)^2 + b^2 | = | a^2+b^2 |
limac20:11:51
$(a-1)^2 + b^2 = a^2+b^2$
rrusczyk20:11:59
rrusczyk20:12:01
Now what?
limac20:12:31
that just means that (a-1)^2 = a^2
rrusczyk20:12:35
Keep going
shanmug100020:12:43
expand the equation
Nlakkamraju20:12:59
that just means that (a-1)^2 = a^2
rrusczyk20:13:04
So?
limac20:13:07
or a^2-2a+1 = a^2 => 1=2a => a=1/2.
Cheepskape20:13:07
a=1/2
rrusczyk20:13:18
Nlakkamraju20:13:22
a=1/2
rrusczyk20:13:39
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?)
helagha20:14:00
CantonMathGuy20:14:00
|a+bi|=8
rrusczyk20:14:09
rrusczyk20:14:30
What do we get for b?
CantonMathGuy20:14:32
substitute a=1/2
aunch20:14:34
plug in a
rrusczyk20:14:40
And what comes out for b?
Cheepskape20:15:09
b^2 = 255/4
CantonMathGuy20:15:09
b^2=255/4 ans=255+4=259
rrusczyk20:15:21
How do we get this from |(1/2) + bi| = 8?
CantonMathGuy20:16:33
use definition then square
rrusczyk20:16:44
We use the definition of |a+bi|.
rrusczyk20:16:52
rrusczyk20:17:00
rrusczyk20:17:46
rrusczyk20:17:53
Here's a concise version of our solution:
CantonMathGuy20:18:00
|(a+bi-1)z|=|(a+bi)z| => a-1=-a => a=0.5 => |0.5+bi|=8 => b^2=63.75=255/4 => answer=255+4=259
rrusczyk20:18:11
matt276 has a nice geometric solution:
matt276eagles20:18:14
Here's a geometric solution: Let's figure out what f(1) maps to, since f(1) = a + bi. Since f(1) must be equidistant from 1 and the origin, it lies on the vertical line through (1/2, 0). Also, |f(1)| = |a + bi| * |1| = 8. Now f(1) can be on either side of the x-axis, but either way we have a right triangle with one leg 1/2 and hypotenuse 8. Then b^2 = 64 - 1/4 = 255/4, so our answer is 259.
aunch20:18:42
that's very clever
rrusczyk20:18:45
You can think about that on your own later. It requires some understanding of the geometry of complex numbers.
rrusczyk20:18:54
This problem is harder than the average problem in the Algebra 3 class (and we cover the necessary prerequisite knowledge of complex numbers in the class).
rrusczyk20:19:18
So, if you that second problem was pretty scary, don't worry too much. You'll get the requisite knowledge you need before tackling problems like that.
rrusczyk20:19:24
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:
rrusczyk20:19:29
rrusczyk20:19:33
rrusczyk20:19:37
rrusczyk20:19:41
rrusczyk20:19:45
rrusczyk20:19:51
rrusczyk20:19:57
The course will meet for 24 weeks on Wednesdays, starting October 7. 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.
rrusczyk20:20:05
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.
rrusczyk20:20:39
Are there any questions about the course?
aunch20:21:03
how many people will be in a class?
rrusczyk20:21:23
I'm guessing there will be 30-50.
limac20:21:50
The course weeks contents doesn't state conics being, covered. But earlier you said that they will be covered, which week will that be?
rrusczyk20:22:10
I think they'll be discussed at a few points throughout the course, and there is a thorough chapter in the text on them.
shanmug100020:22:21
do we have grades?
rrusczyk20:22:52
No, but we can provide one if you need one for school. We provide detailed feedback on your work on the Challenge Sets.
aunch20:22:55
which book do we need?
rrusczyk20:23:01
Our Intermediate Algebra text
aunch20:23:30
will we have homework?
rrusczyk20:23:32
Yes; there will be practice message board problems after each class. You don't turn these in for evaluation. There are 8 Challenge Sets that you turn in for evaluation.
$LaTeX$.20:24:10
what course would a.g 3 come under in a conventional school curriculum?
rrusczyk20:24:11
This class covers much of Algebra 2, all the non-trig/vector precalc, and lots of stuff that just isn't covered at all in most schools.
Cheepskape20:24:29
I am using your "the Art of Problem Solving: vlumes 1 and 2. do the books cover Algebra 1 and 2?
rrusczyk20:24:31
Yes, they do, and a lot more.
rrusczyk20:24:42
Any more questions?
shanmug100020:25:26
if i am doing al 2, which book should i get?
helagha20:25:44
all of them :)
rrusczyk20:25:49
That's a good answer :)
rrusczyk20:26:05
The Intro Algebra book covers some Algebra 2, and our Intermediate Algebra book covers the rest.
rrusczyk20:26:36
Tough to say which would be a better fit if you're just starting with algebra 2. Probably the latter half of the Intro book.
helagha20:26:39
when is the calc book coming out?
rrusczyk20:26:45
Around year-end, we hope.
rrusczyk20:26:49
Same for the Precalc book.
helagha20:27:07
before or afdter the couse starts?
rrusczyk20:27:21
Calculus, after the course starts for sure. Precalculus, not sure. It'll be close.
rrusczyk20:27:54
Let's do some counting, and then we'll discuss the counting and calculus classes.
rrusczyk20:27:58
Intermediate Counting & Probability
rrusczyk20:28:01
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.
rrusczyk20:28:06
Here is a sample problem:
rrusczyk20:28:10
shanmug100020:29:37
casework will be needed, right?
aunch20:29:37
we have to break this problem into cases
rrusczyk20:29:43
We're going to do this problem a couple different ways.
rrusczyk20:29:49
First, let's do some casework.
rrusczyk20:29:54
What cases should we consider?
rrusczyk20:29:59
(It's very important when doing casework to set up your cases in an organized fashion, so you can be sure that you count every possibility once, and only once.)
shanmug100020:30:19
one cse for column, row, internal diagonal, and reg. diagonal.
rrusczyk20:30:26
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.
rrusczyk20:30:30
rrusczyk20:30:36
rrusczyk20:30:40
rrusczyk20:30:45
So, we start with our first case:
rrusczyk20:30:49
rrusczyk20:30:56
How many are there that are not diagonal?
Iggy Iguana20:31:33
48
rrusczyk20:31:35
Why?
rrusczyk20:32:55
Let's take this step-by-step.
rrusczyk20:33:00
How many are in the direction shown?
rrusczyk20:33:22
(When doing casework, the key is to be organized!)
Iggy Iguana20:33:37
16
rrusczyk20:33:41
In the direction shown, there are 4 on each level. Therefore, there are 4^2 = 16 with that direction since there are 4 levels.
rrusczyk20:33:57
So, how many 4-in-a-rows are "not diagonal"?
aunch20:34:10
48
rrusczyk20:34:12
Why?
shanmug100020:34:19
16*3= 48
rrusczyk20:34:25
Where does the *3 come from?
Iggy Iguana20:34:39
3 dimensions
shanmug100020:34:39
for 3 dimensions.
rrusczyk20:34:43
There are two other directions we could have chosen, examples of which are here in red and blue:
rrusczyk20:34:47
rrusczyk20:34:50
There are 4^2 in each of these directions, as well. So our total for the 'not diagonal' case is 3(4^2) = 48.
rrusczyk20:34:53
How about the '2D Diagonal' case?
rrusczyk20:34:59
rrusczyk20:35:13
How can we go about counting these?
Iggy Iguana20:35:34
each side has 2 diagonals
rrusczyk20:35:52
Each square has 2 diagonals. How many squares are there?
mathq20:36:14
12 squares
aunch20:36:14
there are 12
rrusczyk20:36:23
We note that in any given square, there are two of these 2D diagonals:
rrusczyk20:36:27
rrusczyk20:36:31
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.
rrusczyk20:36:36
What about the 3D case:
rrusczyk20:36:45
rrusczyk20:36:49
How many such diagonals are there?
CantonMathGuy20:37:01
4
shanmug100020:37:02
4
aunch20:37:06
there are only 4
Iggy Iguana20:37:06
4
$LaTeX$.20:37:06
4
rrusczyk20:37:08
There are 4 interior diagonals of the cube (one for each vertex on top), so here we have 4.
rrusczyk20:37:13
So what is the total number of winning sets?
aunch20:37:34
76
Iggy Iguana20:37:34
76
rrusczyk20:37:37
shanmug100020:37:42
48+24+4=76
arbala20:37:46
76
rrusczyk20:37:48
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.
rrusczyk20:37:52
Now, does anyone see a slick way to do it?
rrusczyk20:38:30
(There's no reason this should come easily -- it takes some practice and guidance to learn how to find these sorts of magical solutions.)
rrusczyk20:38:34
Here's a hint:
rrusczyk20:38:40
Think "outside the box".
rrusczyk20:39:32
We'd like to relate our 4-in-a-rows to something that is easy to count.
rrusczyk20:39:43
What might we relate our 4x4x4 cube?
rrusczyk20:39:58
*What might we relate it to?
rrusczyk20:40:15
We don't have much else in the problem, just this 4x4x4 cube of dots.
rrusczyk20:40:31
It's not obvious how we might constructively relate it to a smaller cube.
rrusczyk20:40:39
But might we relate it to a larger one?
rrusczyk20:40:45
Let's see:
rrusczyk20:40:48
rrusczyk20:40:59
Our 4x4x4 cube is embedded in a 6x6x6 cube. How can we relate our 4-in-a-rows to the 6x6x6 cube?
rrusczyk20:41:20
Here's a link if you want to open the pic in another window.
shanmug100020:41:22
6x6x6? why?
rrusczyk20:41:37
Well, let's see! Start with a 4-in-a-row in the little cube.
rrusczyk20:41:45
How would you relate it to the big cube?
CantonMathGuy20:42:15
extend
rrusczyk20:42:17
We extend a 4-in-a-row, we hit the big cube in exactly 2 places.
rrusczyk20:42:23
rrusczyk20:42:27
img=http://www.artofproblemsolving.com/Community/MJImages/bigcubeext.gif
rrusczyk20:42:33
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.
rrusczyk20:42:58
Now, what should we wonder once we think of this?
aunch20:43:01
how does that help???
rrusczyk20:43:22
Good question -- we extend the 4-in-a-rows, we hit the big cube twice. What should we wonder?
rrusczyk20:44:09
We should wonder if we can run this the other way---can we start on the outside cube and always produce a 4-in-a-row.
rrusczyk20:44:35
This is the heart of the idea of "1-to-1 correspondence", which we'll discuss a great deal in the Intermediate Counting course.
rrusczyk20:44:52
For this problem, part of discovering this is:
CantonMathGuy20:44:54
does every 4-in-a-row hit a different pair
rrusczyk20:45:17
Does each 4-in-a-row hit a different pair of dots on the outside cube?
Iggy Iguana20:45:29
yes
rrusczyk20:45:59
Yes, if we have a 4-in-a-row through two outside spots, it's the only one that goes through both of the outside spots it hits.
rrusczyk20:46:11
How does that help?
aunch20:46:34
1 to 1 correspondence
rrusczyk20:46:37
How?
rrusczyk20:46:38
Can we run it backwards? Can we start from any two outside points and get a 4-in-a-row?
Iggy Iguana20:46:55
no
rrusczyk20:46:58
Why not?
aunch20:47:27
take two points on an edge of the cube
rrusczyk20:47:42
Right. We can't take any 2 points on the outside and find a corresponding 4-in-a-row.
rrusczyk20:47:44
But. . .
rrusczyk20:47:52
Can we start from any *one* outside point and get a 4-in-a-row?
aunch20:48:40
there is always at least one way from a vertex, edge, center of a face
rrusczyk20:49:04
*at least* What would we like *at least* to be?
rrusczyk20:49:08
rrusczyk20:49:15
Here's a cube to help you think about it.
aunch20:49:37
1
rrusczyk20:50:06
Exactly -- we want to be able to say that starting from any point on the outside of the cube, there is exactly 1 4-in-a-row that hits it.
aunch20:50:09
from a vertex, there is only one way
rrusczyk20:50:21
Clearly, that's true for the green point in the upper left.
rrusczyk20:50:45
Is it true for the purple point on the edge and the red point on the interior of a face?
aunch20:51:26
yes
Iggy Iguana20:51:26
yes
rrusczyk20:51:31
Yes, it's true for these points too -- start with any point on the outside, and there is only one way to find a 4-in-a-row that we can extend to hit that point.
aunch20:51:54
wow there is exactly one way for each of them!!!
rrusczyk20:52:00
Precisely.
rrusczyk20:52:03
So? How does this give us a way to count our 4-in-a-rows?
mathq20:52:29
so count the points at the cover, and divide by two so you get the all number of lines
aunch20:52:32
count all of the outside points
rrusczyk20:52:39
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.
rrusczyk20:52:54
Therefore, to count our 4-in-a-rows, we just count the points on the outside of the cube and divide by two. How do we quickly count the dots on the outside of the cube?
CantonMathGuy20:53:24
6^3-4^3
rrusczyk20:53:31
mathq20:53:33
number of points covering the cube=152 so 152/2=76
rrusczyk20:53:39
Notice that we've completed the problem in two different ways and found the same answer both times, so we're probably right. (This is the best way to avoid mistakes on the AIME in counting problems, by the way.)
rrusczyk20:53:43
But wait, there's more.
rrusczyk20:53:53
Take a look at our two answers:
rrusczyk20:53:56
rrusczyk20:54:01
That these are the same is a result of the Binomial Theorem.
rrusczyk20:54:05
rrusczyk20:54:09
We'll use the Binomial Theorem and many other tools to develop all sorts of neat relationships when we talk about combinatorial identities.
aunch20:54:11
could you take more than a 6x6x6 cube?
rrusczyk20:54:24
See if you can generalize this problem by thinking about starting with a larger cube.
rrusczyk20:54:34
(Starting from the very beginning)
rrusczyk20:54:38
Here are a couple more problems that can be solved with clever correspondences:
rrusczyk20:54:43
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}.
rrusczyk20:54:47
And:
rrusczyk20:54:50
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?
rrusczyk20:54:56
These problems are roughly middle-level difficulty for the course. The example we did was slightly easier than the average problem in the course.
rrusczyk20:55:11
(We will spend a lot of time talking about how to discover these sorts of correspondences.
rrusczyk20:55:15
You can find more questions like those we cover in the course by checking out the Post Test for the course here:
rrusczyk20:55:29
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 Algebra 3 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.)
rrusczyk20:55:33
The course will meet for 18 weeks on Tuesdays, starting October 6, at 7:30 PM Eastern / 4:30 PM Pacific. The last class date is February 23. Each class is 90 minutes.
rrusczyk20:55:42
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.
rrusczyk20:55:55
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:
rrusczyk20:56:03
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.
rrusczyk20:56:15
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 3-week period of the course -- for which you should write up your full solutions and submit them. These solutions will be evaluated by AoPS graders, and you will receive detailed feedback.
rrusczyk20:57:25
OK; I'll quickly go over the Calculus class -- there are no problems to cover.
rrusczyk20:57:29
The AoPS Calculus class is a course in single-variable calculus. This course covers all of the standard single-variable calculus topics: limits, continuity, derivatives and their applications, definite and indefinite integrals, infinite sequences and series, plane curves, polar coordinates, and basic differential equations.
rrusczyk20:57:34
At the conclusion of the course, students should have sufficient preparation to take the AP Calculus BC exam; however, "AP exam preparation" is not the main focus of the course. (Note: the AP exam is not offered by Art of Problem Solving—you will have to privately arrange to take the exam at your local school if you are interested.)
rrusczyk20:57:57
This class will differ from the "typical" high-school calculus class in two ways.
rrusczyk20:58:01
First, our primary focus will be to understand the concepts of calculus, and not just to learn a bunch of algorithms in order to compute things. In particular, we will not spend a large amount of time doing "real-world applications" of calculus; rather, we will be focusing on the mathematical content of calculus. Nor will we spend a lot of time playing with graphing calculators, as the class will focus more on concepts than on calculation.
rrusczyk20:58:19
Second, we will be tackling much more difficult problems than those in a typical calculus course or on the AP exam. We will routinely discuss Putnam-level calculus problems (the Putnam is a college undergraduate-level math contest). Rather than do the same basic exercises over and over, we will stretch our understanding with difficult, challenging problems.
rrusczyk20:58:32
On the other hand, the course is not "rigorous" in the sense that a college-level Real Analysis course would be. Although we will try to gain a thorough conceptual understanding of many calculus concepts, we will not commit to rigorously proving every result.
mathq20:58:50
are we learning like olympiad calculus?
rrusczyk20:59:13
There isn't really "olympiad" calculus, but there is a college competition called the Putnam that we will draw from.
$LaTeX$.20:59:15
olympiad is rigorous, isn't it?
rrusczyk20:59:25
Yes, but calculus is not required to solve olympiad problems.
rrusczyk20:59:28
A course like the one that I've described requires a highly self-motivated student in order to succeed. Students will have to take it upon themselves to do a lot of work on their own between class sessions in order to master the material.
rrusczyk20:59:37
The course will meet once per week for 25 weeks, on Mondays from 7:30-9:30 PM Eastern / 4:30-6:30 PM Pacific, starting on October 12. (We won't meet on Nov 23, Dec 21, or Dec 28; the last day of class is April 19.)
rrusczyk20:59:46
We are in the late stages of writing textbook for this class. It should be available about midway through the class in December or January. You will be sent the textbook free of charge, when it is available, if you are enrolled in the class.
rrusczyk20:59:57
There will 2 different types of "homework" for this class.
rrusczyk21:00:02
1. Each class will have a weekly problem set that will be posted on the class's message board. Many of the problems on this set will be "routine" problems to reinforce basic concepts. As mentioned above, we will not be doing very much of this in class, so it is important to solve these basic review problems in order to master the material. Other problems on the weekly sets will be more difficult and will challenge students' understanding of the material. Students should work on these problems on their own, but do not turn them in for evaluation
rrusczyk21:00:17
2. There will be 6 longer Challenge Problem sets (one set every 4 weeks). These will be almost entirely composed of more difficult problems (as opposed to "routine" exercises). Students should write up complete solutions to these problems and submit them for instructor feedback.
rrusczyk21:00:28
Calculus has significantly more outside-of-class work than other AoPS classes. Students should be prepared for this.
GoldenFrog161821:00:41
How many people are currently enrolled?
rrusczyk21:00:51
We have 24 enrolled now, and expect to have 30-40.
rrusczyk21:00:55
Also, and we cannot stress this enough, students who do not have a solid algebra background will not be successful in this course. It is vital that students have mastered the basic high-school math curriculum -- algebra, geometry, trigonometry -- before attempting a calculus course.
rrusczyk21:01:02
To this end, there is a diagnostic test on our website:
rrusczyk21:01:14
Students should be able to solve most (if not all) of the problems on the diagnostic, with little difficulty, before considering enrolling in the calculus class. We will be doing only a very brief review of precalculus concepts, and students who do not have a solid precalculus background will quickly be behind.
rrusczyk21:01:25
Taking calculus as the "next" course after an algebra or trigonometry course is not the right decision for many students. See our articles "The Calculus Trap"and "Why Discrete Math Is Important" in the Resources section of our website for more discussion of this.
rrusczyk21:01:36
We recommend that students have a graphing calculator to use as a tool in their study of calculus, and to have it with them at each class session. However, this is not required, and we will not be emphasizing calculator techniques. Note that the AP Calculus exam (which Art of Problem Solving does not offer nor endorse) requires a graphing calculator; a list of AP-approved calculators is on the College Board website.
rrusczyk21:01:51
We don't have any sample problems for the calculus class tonight, since most of the material in the course requires fundamental definitions and concepts that are taught at the start of the course, but which very students have been exposed to prior to a study of calculus.
rrusczyk21:02:00
Are there any questions about the course?
mathq21:02:36
how much is this course?
rrusczyk21:02:38
The enrollment fee is $545, and this covers the full year enrollment and the textbook.
aunch21:02:58
what is the textbook?
rrusczyk21:03:54
The Art of Problem Solving Calculus textbook, which is not yet available. It will be finished during the course, and sent to all enrolled students.
GoldenFrog161821:04:20
how many handouts does the class have?
rrusczyk21:04:23
There will be 1-2 per month until the Calculus book is available, at which point all the reading material will be in the text.
Mycroft21:04:48
Will this course start at the very beginning of Calculus - explaining what derivatives are, etc. - or is some amount of basic knolwedge of the subject expected?
rrusczyk21:04:50
Yes, it will start from the beginning; no knowledge of calculus is expected.
mathq21:04:52
is antiderivative unit included?
rrusczyk21:04:54
yes
GoldenFrog161821:05:30
How much of a loss would it be to not use a graphing calculator?
rrusczyk21:05:32
Not much; it won't be necessary for the class. You can find things online to graph functions, like WolframAlpha or geogebra.
$LaTeX$.21:05:46
is this inclusive of multivariable calc.?
rrusczyk21:05:48
No; this is not a multivariable calculus class.
Mycroft21:06:07
So what will we be reading before the
rrusczyk21:06:26
There will be some handouts before the text is available that will have some of the material from the book. For the most part, you'll be learning the fundamentals in class, and from the class transcripts/forum.
Mycroft21:06:59
Will there be anything other than the forums as out -of-class support type things (to look things up in, etc.) until the book comes out?
rrusczyk21:07:17
No, but you can use the internet to look up most of the concepts we'll be discussing if you need another source. We also have a very active forum full of people who like to help people learn math.
mathq21:07:35
will it be hard(throughout)
rrusczyk21:07:38
It will be much more challenging than your typical school class.
GoldenFrog161821:08:04
How much do you recommend MIT OpenCourseWare's Single Variable Calculus resources?
rrusczyk21:08:08
We haven't investigated it in detail, so we can't really say.
mathq21:08:22
I liked it though
rrusczyk21:08:30
There's an endorsement :)
rrusczyk21:09:34
Thanks for coming to the Math Jam. If you think of more questions later, write us at classes@artofproblemsolving.com
