| Transcript
for the Math
Jam "AoPS Classes Math Jam"
on Mar 3. |
| Math Jam hosted by rrusczyk
(Richard Rusczyk ). |
rrusczyk19:34:04
Hello, and welcome to an Art of Problem Solving Math Jam. Today we'll be discussing the Intermediate Counting & Probability course that starts next week.
rrusczyk19:34:14
We will go through a thorough example problem and then discuss how the course works.
rrusczyk19:34:20
My name is Richard Rusczyk. I founded Art of Problem Solving and have written several Art of Problem Solving textbooks.
rrusczyk19:34:26
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.
rrusczyk19:34:35
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.
rrusczyk19:35:13
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.
rrusczyk19:35:23
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.
rrusczyk19:35:39
You do not need to learn LaTeX to use our classes or our classroom!
rrusczyk19:35:48
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
rrusczyk19:35:51
rrusczyk19:35:56
This message will look like this when posted in the classroom:
rrusczyk19:36:00
rrusczyk19:36:09
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 the course, and then go through an example problem. Then, I will hold a question-and-answer session about the class.
rrusczyk19:36:40
Before we get going with the counting, one more time:
rrusczyk19:36:52
If you are trying to attend the AIME Problem Series, the Calculus class, or the Introduction to Geometry course, you are in the wrong room.
rrusczyk19:37:11
Intermediate Counting & Probability
rrusczyk19:37:18
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.
rrusczyk19:37:48
I'm going to dive right into the sample problem, and then we'll discuss the class. This will probably be a relatively short Math Jam -- about 30 minutes or so.
rrusczyk19:37:56
Here is the sample problem:
rrusczyk19:38:04
rrusczyk19:38:22
You'll note that I have "stuck" the problem to the top of the room.
rrusczyk19:38:41
You can increase the size of the window that the problem is in by sliding the bar between that window and the main classroom.
rrusczyk19:38:58
(You can also change the font size in the room with the buttons above the input box is you like.)
rrusczyk19:39:05
And now, back to our story.
rrusczyk19:39:12
We're going to do this problem a couple different ways. Are there any suggestions about how to approach it?
rrusczyk19:39:44
Please don't just shout out numerical answers -- provide justification!
nikeballa9619:40:08
categorize the ways to get 4 straight
queenez0819:40:08
start with counting with either rows, columns, or diagonals?
rrusczyk19:40:17
We can try cases. What cases should we consider?
rrusczyk19:40:49
(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.)
ecnerwal19:41:38
rows, columns, diagonals
nsreenivas19:41:38
Width, Length, Height, diagonal
nikeballa9619:41:38
surface horizontal, surface vertical, surface diagonals, internal horizontal, interntal vertical, internal diagonal, assuming that you wont rotate the cube.
queenez0819:41:38
first on one outer surface and then count the internal ones
rrusczyk19:42:34
We'd like to refine these cases a bit. Can we define them in ways that we are sure that we count everything once and only once?
gh62519:43:03
left-to-right, top-to-bottom, front-to-back, 2D diagonal, and 3D diagonal
rrusczyk19:43:42
That's a nice organization. I'll group the first three into one case.
rrusczyk19:43:48
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.
rrusczyk19:43:52
rrusczyk19:44:00
rrusczyk19:44:05
swsim0019:44:19
when you say "how many possible winning sets," do you mean that for winning sets of both X and O?
rrusczyk19:44:43
Great question -- I'm looking for just the number of 4-in-a-rows we can make. Above, we just saw 3.
rrusczyk19:44:48
So, we start with our first case:
rrusczyk19:44:53
rrusczyk19:45:09
How many "not diagonal 4-in-a-rows" are there? (And why?)
gh62519:46:37
4*4 left to right, 4*4 top to bottom, and 4*4 front to back, for a total of 4*4+4*4+4*4=48 total
alligator11219:46:37
there are 4*4 front to back, 4*4 left to right, and 4*4 up to down, so a total of 48
rrusczyk19:46:57
Nice. Notice how they did this -- they broke this case into three sub-cases. Starting kind of like this:
nikeballa9619:47:02
there are 16 verticals, 4*4 *they can start on 16 spots, all on the top)
rrusczyk19:47:11
And then they noticed there are three dimensions:
nsreenivas19:47:13
48 since there are 16 heights widhts and lengths
nikeballa9619:47:18
rrusczyk19:47:25
rrusczyk19:47:33
In the direction shown, there are 4 on each level. Therefore, there are 4^2 = 16 with that direction since there are 4 levels.
rrusczyk19:47:37
There are two other directions we could have chosen, examples of which are here in red and blue:
rrusczyk19:47:42
rrusczyk19:47:46
There are 4^2 in each of these directions, as well. So our total for the 'not diagonal' case is 3(4^2) = 48.
rrusczyk19:47:51
How about the '2D Diagonal' case?
rrusczyk19:47:56
rrusczyk19:49:20
A number of you are off by a factor of 2. Be careful!!!
gh62519:49:35
There are 8 for each of the three directions, for a total of 8*3=24 2D diagonals.
alligator11219:49:36
there are 2 for each"plane" and there are 12 such "planes" so 2*12 = 24
alexhhmun19:49:36
There are 4 bottom right to top left, and 4 bottom left to top right. We can shift our perspective 2 more times without double counting. 24.
queenez0819:49:46
oh 2*4 front to back, 2*4 left to right, 2*4 top to bottom
rrusczyk19:49:59
We can basically look at these face diagonals as diagonals of squares.
rrusczyk19:50:03
We note that in any given square, there are two of these 2D diagonals:
rrusczyk19:50:09
ecnerwal19:50:12
there are 2 diagonals in each "plane", and there are 4 lenghth+ 4 width+ 4 height= 12 "planes", 12*2=24
nikeballa9619:50:17
rrusczyk19:50:25
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.
rrusczyk19:50:39
Almost there.
rrusczyk19:50:43
What about the 3D case:
rrusczyk19:50:48
rrusczyk19:50:53
How many such diagonals are there?
gh62519:51:19
There are 4 3D diagonals.
rorys19:51:19
4
pieater19:51:19
4
Ispxye19:51:19
4?
venkataraman19:51:19
4
prakhar2319:51:19
There are 4 diagonals
rrusczyk19:51:32
There are 4 interior diagonals of the cube (one for each vertex on top), so here we have 4.
rrusczyk19:51:39
So what is the total number of winning sets?
venkataraman19:52:07
48+4+24=76
swsim0019:52:07
76 winning sets
alligator11219:52:07
4+24+48 = 76!
gh62519:52:07
The total is 48+24+4=76
prakhar2319:52:07
24+48+4=76
rrusczyk19:52:11
rrusczyk19:52:19
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.
rrusczyk19:52:32
Does anyone *see* that slick way to do it?
saszs19:53:09
are there casework problems on the states for mathcounts
rrusczyk19:53:21
Casework problems are often the hardest MATHCOUNTS problems.
skaterpigsusa19:53:24
find a pattern for increasing sizes of cubes?
rrusczyk19:54:05
This is a very powerful problem-solving strategy for counting problems -- you'll see many examples in the course. For this problem, it will prove pretty tough (but you can try that on your own).
rrusczyk19:54:17
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 count.
rrusczyk19:54:28
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.
rrusczyk19:54:43
What might we relate our 4x4x4 cube to that might help us perform our count?
limac19:55:21
a 2x2x2 cube?
queenez0819:55:27
2x2x2 cube?
rrusczyk19:55:53
Interesting idea -- but are there 4-in-a-rows in the 4x4x4 that are basically impossible to relate to a 2x2x2 cube?
rrusczyk19:55:57
Here's the 4x4x4 again:
rrusczyk19:56:06
rrusczyk19:57:21
We had the idea of relating our 4x4x4 cube to a 2x2x2 cube.
rrusczyk19:57:56
But, if we take a 2x2x2 cube, are there 4-in-a-rows in our 4x4x4 cube that are impossible to relate to some 2x2x2 cube inside the 4x4x4 cube?
alligator11219:58:16
yes
nikeballa9619:58:16
yes, there are .
rrusczyk19:58:33
No matter what 2x2x2 cube we take, some of the 4-in-a-rows will miss the 2x2x2 cube entirely. So, it will be hard to relate those to something about our 2x2x2 cube.
rorys19:58:37
a 3x3x3 cube
nsreenivas19:58:37
What about a 3x3x3
rrusczyk19:58:45
Will we have the same problem with a 3x3x3 cube?
nikeballa9619:58:56
same problem...?
rfms9619:58:56
yES
alligator11219:58:56
yes
rrusczyk19:59:20
Yep, we have the same problem -- we can't relate all of our 4x4x4's to a 3x3x3 cube.
rrusczyk19:59:28
Hmmm.... Any other ideas?
azhang19:59:46
would it be easier if we did an 8*8*8?
prakhar2319:59:46
What about a 5x5x5
rrusczyk19:59:53
Now we're thinking outside the box!
rrusczyk20:00:24
Suppose you take a 4-in-a-row. How might you relate it to a *larger* box?
rrusczyk20:00:33
rrusczyk20:00:45
There's a 4-in-a-row. How might we relate this to a *larger* box?
rrusczyk20:01:46
What would we do with this 4-in-a-row to relate it to a bigger cube than the 4x4x4 it sits in?
IsTvOn20:02:12
we enlarge it
rrusczyk20:02:24
How do we "enlarge" a 4-in-a-row?
nikeballa9620:02:50
turn it into an 8 in a row? hm.
rorys20:02:50
make it 5 in a row
rrusczyk20:02:58
We can simply lengthen it.
nsreenivas20:03:00
make a 5-in-a-row in a 5x5x5
rrusczyk20:03:20
A 5-in-a-row in a 5x5x5, or a 6 in a row in a 6x6x6, or . . .
rrusczyk20:03:27
We just extend the 4-in-a-row.
rrusczyk20:03:30
limac20:03:34
but wouldn't that make the problem more complicated?
rrusczyk20:03:37
Let's see!
rrusczyk20:03:53
Maybe we can relate this to something that is easy to count.
rrusczyk20:04:27
The problem with using 5x5x5 is that a 5x5x5 cube is not symmetric with respect to the 4x4x4 cube. What larger cube would be symmetric with respect to the 6x6x6 cube?
rrusczyk20:04:42
Oops! I meant 4x4x4 cube :)
nikeballa9620:04:53
8x8x8?
azhang20:04:53
8*8*8!!! woot
leicao69620:04:53
8x8x8
venkataraman20:04:53
8*8*8 cube
rrusczyk20:04:59
Can we get by with something smaller?
nikeballa9620:05:12
6x6x6?
venkataraman20:05:12
6*6*6
limac20:05:15
a 6x6x6.
queenez0820:05:16
6x6x6?
tekjam20:05:37
6x6x6
saszs20:05:37
yep 6x6x6
rrusczyk20:05:38
And where might we find a convenient 4x4x4 cube from a 6x6x6 cube?
swsim0020:05:54
?in the center
nikeballa9620:05:54
dead center, with 1 inch on each side.
rfms9620:05:58
Right in the center of it
gh62520:05:58
The center!
ecnerwal20:06:03
in the center
rrusczyk20:06:04
Ispxye20:06:10
convient? what do you mean by that?
rrusczyk20:06:15
Let's see what I mean!
rrusczyk20:06:27
Here's a link you can open in your own browser:
rrusczyk20:06:41
Our 4x4x4 cube is embedded in a 6x6x6 cube. How can we relate our 4-in-a-rows to the 6x6x6 cube?
saszs20:06:44
the black borders represent the 4x4x4 right
rrusczyk20:06:48
Correct.
yjlim5520:07:19
so how does this relate with our problem
rrusczyk20:07:25
Great question. How?
rrusczyk20:07:47
Imagine you have a single 4-in-a-row inside the 4x4x4 cube. What would you do with it to relate it to the larger 6x6x6 cube?
rorys20:08:26
extend it
ecnerwal20:08:26
extend it both ways
skaterpigsusa20:08:26
extend it by 2?
gh62520:08:26
It can be extended to make a 6-in-a-row in the 6x6x6 cube.
rrusczyk20:08:32
Any 4 in the row when extended hits our big cube in two places:
rrusczyk20:08:37
Duelist20:08:54
All the 6 in a rows in the 6x6x6 that are not contained in an outer face correspond 1-1 with the 4 in a rows in the inner 4x4x4
nsreenivas20:08:54
the 6ina row will draw a 4 in a row as well
rrusczyk20:09:12
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.
rrusczyk20:09:26
How does this help?
nsreenivas20:09:55
the number of points divided by 2
queenez0820:10:03
count how many places these extended lines hit?
nsreenivas20:10:03
the number of points on the surface divided by 2
rrusczyk20:10:27
Interesting idea. Is it true? Is it true that we can just count the number of points on the surface of the cube and divide by 2?
rrusczyk20:10:56
rrusczyk20:11:17
Notice that each 6-in-a-row hits 2 outside dots on the 6x6x6 cube.
tekjam20:11:51
how does this info help us with the problem
rrusczyk20:12:05
Good question -- how does this observation help?
rfms9620:12:07
So, no other 4 in a row could be hitting the same dot as another
rrusczyk20:12:16
Bold claim!!!! Is it true?
hjoon012520:12:42
yes
nikeballa9620:12:42
yes.
saszs20:12:42
yes
HiDN42820:12:42
yes?
alexhhmun20:12:46
How would we prove so if needed?
rrusczyk20:13:03
Good question -- how do we know that no two 4-in-a-rows can hit the same outside dot?
rrusczyk20:13:07
rrusczyk20:14:57
Hint: instead of starting with a 4-in-a-row, start with one of the points on the surface of the 6x6x6. We started with the 4-in-a-row to see that we could extend to the surface of the 6x6x6. Now, we must go the other way -- start with the 6x6x6 and show that there's only one corresponding 4-in-a-row for each point on the outside of the cube.
leicao69620:15:03
look at a corner of the large cube
rrusczyk20:15:22
OK, suppose we took the corner of the 6x6x6. Is there only one 4-in-a-row that can be extended to it?
venkataraman20:15:55
yes
nsreenivas20:15:55
yes
rfms9620:15:55
Yes and that is a 3D diagnol
rrusczyk20:16:20
Only the corresponding 3D diagonal can hit a given corner of the 6x6x6 cube.
rrusczyk20:16:30
Is this true for any point on the outside of the 6x6x6 cube?
rorys20:16:51
yes
rfms9620:16:51
Yes
HiDN42820:16:51
yea
tekjam20:16:51
yes
venkataraman20:16:51
Yes
skaterpigsusa20:16:51
yes
rrusczyk20:17:04
If we 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:
rrusczyk20:17:10
benzi45520:17:13
yes -- if it is along an edge of the 6x6x6, then it must go diagonally through the 4x4x4; if it is not, then it must go straight through perpendicular to a face
rrusczyk20:17:19
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.
rrusczyk20:18:10
That's the previous diagram -- start from each colored point, and try to make a 6-in-a-row that contains a 4-in-a-row inside the little cube.
rrusczyk20:18:29
You'll see that there's only 1 way to do it for each, as benzi describes nicely.
rrusczyk20:18:34
So? How does this give us a way to count our 4-in-a-rows?
venkataraman20:19:51
Number of points on outside divided by two for overcounting
rfms9620:19:51
We divide the total number of points on the big cube by 2, because each 4 in a row has 2 points when it is extended
benzi45520:19:51
each point on the surface of the 6x6x6 gives one winning set, and each winning set is given by two points on the surface of the 6x6x6
rrusczyk20:19:56
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.
nsreenivas20:20:08
take the number of outside points and divide by 2
rrusczyk20:20:12
Therefore, to count our 4-in-a-rows, we just count the points on the outside of the cube and divide by two. How many points are on the outside of the cube?
nsreenivas20:20:36
152
venkataraman20:20:36
152
benzi45520:20:36
6^3 - 4^3 = 152
theone14285720:20:45
152
rrusczyk20:20:51
Which gives us how many 4-in-a-rows?
nsreenivas20:21:00
76
venkataraman20:21:00
76
yjlim5520:21:00
76
nsreenivas20:21:00
76
theone14285720:21:00
76
IsTvOn20:21:00
76
rrusczyk20:21:06
rrusczyk20:21:28
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.)
nsreenivas20:21:30
can we use this for all cubes?
rrusczyk20:21:37
GREAT QUESTION!
rrusczyk20:21:39
Can we ?
theone14285720:22:06
yea
hjoon012520:22:06
yes
rorys20:22:06
yes
yjlim5520:22:06
yeah i think
HoratioK20:22:06
yes?
benzi45520:22:06
sure, why not, as long as you have to get n in a row for an n-cube
rrusczyk20:22:25
Exactly -- we can extend this to any size cube -- the reasoning is the same.
rrusczyk20:22:28
But wait, there's more.
nsreenivas20:22:31
the number of X-in-a-rows is (X+2)^2-X^2 all over 2
rrusczyk20:22:40
There's your formula.
rrusczyk20:22:45
Take a look at our two answers to the original problem:
rrusczyk20:22:49
rrusczyk20:22:54
Why are these the same?
rrusczyk20:23:01
(Algebraically speaking)
alexhhmun20:23:14
Do you mean (X+2)^3-X^3?
rrusczyk20:23:19
Yes, nsreen does.
Duelist20:23:54
Factor x^3-y^3, and then they're the same
rrusczyk20:23:59
That works, too:)
rrusczyk20:24:05
I was thinking of something a little different.
rrusczyk20:24:09
That these are the same is a result of the Binomial Theorem.
rrusczyk20:24:17
rrusczyk20:24:25
We'll use the Binomial Theorem and many other tools to develop all sorts of neat relationships when we talk about combinatorial identities.
benzi45520:24:38
if I come out of this class able to do that, that would be great
rrusczyk20:24:50
We spend a fair amount of time discussing how to find "magical" solutions like this one.
HiDN42820:24:58
so, algerbraically, these are really the same thing
rrusczyk20:25:00
Exactly.
rrusczyk20:25:04
Here are a couple more problems that can be solved with clever correspondences:
rrusczyk20:25:11
(We won't solve these now.)
rrusczyk20:25:12
Define the alternating sum of a set as follows:
rrusczyk20:25:15
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.
rrusczyk20:25:19
Find the sum of the alternating sums of all of the subsets of {1,2,3,4,5,6,7,8,9,10}.
rrusczyk20:25:23
And here's another:
rrusczyk20:25:28
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?
HiDN42820:25:47
we arent doing this now, are we
rrusczyk20:25:50
No.
rrusczyk20:25:54
These problems are roughly middle-level difficulty for the course. The example we did was slightly easier than the average problem in the course.
rrusczyk20:25:59
You can find more questions like those we cover in the course by checking out the Post Test for the course here:
rrusczyk20:26:10
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.)
rrusczyk20:26:18
The course will meet for 18 weeks on Mondays, starting March 9, at 7:30 PM Eastern / 4:30 PM Pacific. The last class date is July 13. Each class is 90 minutes.
rrusczyk20:26:23
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.
rrusczyk20:26:36
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:
rrusczyk20:26:58
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.
rrusczyk20:27:18
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.
venkataraman20:27:28
Where would this problem be, AIME, Mathcounts, AMC10, AMC12, or USAMO?
rrusczyk20:27:35
I believe the original question was from the AIME.
rrusczyk20:27:44
Before we get started taking questions, I'd like to remind everyone that we have a Special AIME Problem Seminar this weekend, March 7-8, 3:30-6:30 PM ET both days. This is a concentrated preparation course for the upcoming American Invitational Math Exam.
saszs20:27:46
will the class help for arml also
rrusczyk20:27:51
Absolutely.
yjlim5520:27:53
is AIME harder than Mathcounts
rrusczyk20:27:57
Much harder.
rrusczyk20:28:07
I will now take questions about the class. We are finished doing math tonight.
rfms9620:28:11
Are there grades?
rrusczyk20:28:29
You can choose to receive a grade if you need one for school, but most students take the class ungraded.
Nerd_of_the_Ages20:28:37
When will the boards be up?
rrusczyk20:28:45
The class message board will be activated later this week.
benzi45520:28:49
hmm... for that AIME problem seminar... who is it targeted at?
rrusczyk20:29:10
It is targeted at students who qualified for the AIME this year, and are very confident they can tackle at least a few problems.
prakhar2320:29:18
How much does it cost
rrusczyk20:29:22
Course fees are here:
nsreenivas20:29:28
Really quick, can you explain how to do the last question on the AMC 10 2009
rrusczyk20:29:37
You can find that on the Math Jam Transcript for the AMC:
HiDN42820:29:49
what art of problem solving books do you think will great for mathcounts states and nats
rrusczyk20:30:01
The Intro level books, and Volume 1 of the classic series.
benzi45520:30:05
I usually get about 10 solutions to AIME problems and then a few mistakes in the execution
rrusczyk20:30:08
Read this:
rrusczyk20:30:28
That's how I aced the AIME. I talk more about this in the Special AIME Seminar.
andrewnycpops20:30:49
you got a 15 on AIME?
rrusczyk20:30:58
Yes; I had the only perfect score on the AIME in 1989.
andrewnycpops20:31:14
what is like the score on AMC needed for AIME problem series?
rrusczyk20:31:26
A passing score for the AIME
alex96053120:31:29
were you a 12th grader then?
rrusczyk20:31:30
Yes.
theone14285720:31:57
I make some mistakes,can I stop it
rrusczyk20:32:01
See that link above.
rfms9620:32:05
What would you do to improve your math counts countdown round?/
rrusczyk20:32:11
Play For the Win on our site!
rfms9620:32:15
How did you prepare for this stuff?
rrusczyk20:32:23
I did many, many problems.
nsreenivas20:33:11
Do you teach any one-on-one classes?
rrusczyk20:33:21
No; we do not currently do any tutoring.
tekjam20:33:24
what types of problems did you do
rrusczyk20:33:29
Problems from old contests.
nsreenivas20:33:49
If I get only 5 questions on the AIME as a Freshman, where would that put me?
rrusczyk20:34:08
I think both the Intermediate Counting and the Special AIME class would be a fit for you (and you're doing quite well).
andrewnycpops20:34:10
i have trouble finding old AMC's, how can i find them?
rrusczyk20:34:24
We have books of them available in the bookstore, and many AMC problems are on the website.
prakhar2320:34:27
where did you get old contests?
rrusczyk20:34:37
My teacher. Now, you can find many on the AoPS website.
HiDN42820:34:38
how can i best prepare for my mathcounts? other than old tests.
rrusczyk20:34:49
Art of Problem Solving's Introduction series of books.
lifeisacircle20:34:52
About what score should we be getting on practice AIMEs for the AIME seminar?
rrusczyk20:35:11
At least 3. Not more than 12 (if you're hitting 12 or so, you should be doing USAMOs).
HiDN42820:35:19
do you get paid for this class?
rrusczyk20:35:23
I run the company :)
leicao69620:35:39
whats play for the win
rrusczyk20:35:51
Click For the Win on the sidebar of the site and find out :)
andrewnycpops20:36:00
what would be a passing score for MC Nats if you are from NY?
rrusczyk20:36:03
No idea!
rrusczyk20:36:09
Changes a lot from year to year.
pieater20:36:12
Will FTW help me for AIME, or more for mathcounts?
rrusczyk20:36:17
Only MATHCOUNTS.
alexhhmun20:36:20
What is ARML?
rrusczyk20:36:24
www.arml.com
nsreenivas20:36:31
What score do I need on the AIME to get into IMO
rrusczyk20:36:41
The AIME isn't used to pick the IMO team.
nsreenivas20:36:45
To get into USAMO what score do u need on AIME?
rrusczyk20:36:54
Varies a lot from year to year. Impossible to say.
HiDN42820:37:20
when do you guys think you will add extra things to the Alcumus?
rrusczyk20:37:26
Hopefully later this year.
saszs20:37:28
I got only 2 questions right on a practice AIME and am only in 6th grade, How can I improve for 7th grade.
rrusczyk20:38:07
Get one year older :) (Seriously, you're doing fine.) Master the material in the Intro series, and then move on to the Intermediate series of texts, which is where you'll learn the fundamentals for the harder AIME problems.)
leicao69620:38:10
what sources should i use to prepare im currently a 7 grader i want 2 prepare for things like aime and so on amc etc
rrusczyk20:38:25
Those mentioned above -- work through the Intro series of books, and then move on to the Intro Series.
rfms9620:38:38
Whats the alcumus?
rrusczyk20:38:46
Click Alcumus on our sidebar and find out!
HiDN42820:38:49
i was wondering what your professions are, since you did so well in math in high school?
rrusczyk20:39:18
I was a bond trader at DE Shaw for 4 years. I founded Art of Problem Solving in 2003. I co-wrote the original AoPS books in my last year of college and the year after.
benzi45520:39:24
hmm... here's a question: do you think that there is any part of a standard American high school curriculum that holds a candle over the current AoPS offerings?
rrusczyk20:39:36
Not for high-performing students.
rfms9620:39:51
How long did you practice daily
rrusczyk20:40:03
It varied a lot -- I tended to practice in spurts.
hjoon012520:40:22
spurts?
rrusczyk20:40:39
Lots of practice during the summer when I was bored. And in history class when I was bored :)
alexhhmun20:41:06
What would you recommend for someone just starting proofs? Which books/courses?
rrusczyk20:41:18
We're going to put together a course on that, hopefully for next spring.
rrusczyk20:41:33
Do the USAMTS, and read the article on how to write in our Resources section.
leicao69620:41:35
i have many problems in trig during comps what should i do for that?
rrusczyk20:41:51
Post them in the Intermediate forum and read the solutions closely.
saszs20:41:54
when does aops do logorithm
rrusczyk20:42:14
Introductory material is in Intro Algebra, more advanced material in Interm Algebra.
benzi45520:42:16
are college courses as good as aops?
rrusczyk20:42:46
Depends on the college. At top tier colleges, the classes are designed for top tier students, like AoPS classes are, so they're at the right level for students like you.
zamboni22320:42:49
Is there any books or resources for geometry past the introduction level?
rrusczyk20:43:20
Not yet from AoPS -- it's on my todo list. We'll have a book from the UK IMO team training on the AoPS site in a week or two.
saszs20:43:31
where's the messageboard
rrusczyk20:43:36
Click Community.
tekjam20:43:38
do you have a calculus book:P
rrusczyk20:43:49
We will by the end of the year.
benzi45520:44:43
Where do you see AoPS going in the future? How far do you plan to expand?
rrusczyk20:45:00
We'll keep developing both upward and downward in age range. No firm plans yet.
rrusczyk20:45:21
We also expect to expand Alcumus, and we might reach into computer science.
carberry20:45:25
how many books do you have? 9not counting solution manuals)
rrusczyk20:45:45
Currently, we have 8 math texts. We'll add precalc and calc this year, and prealgebra next year.
andrewnycpops20:45:47
what is alcumus?
rrusczyk20:45:53
Click Alcumus on the site and see.
HiDN42820:45:55
if i make it to MC nationals, will colleges see this when i'm older?
rrusczyk20:46:03
If you list it on your application, yes.
HiDN42820:46:05
I am in 8th grade and i took a practice AIME and got a solid 8, if i keep practicing, do you think i might get a perfect score on my senior or junior year?
rrusczyk20:46:25
Sure -- you have to learn how to get rid of careless mistakes, which comes with practice.
harukikara20:46:29
what should an average ninth grader get on the AIME
rrusczyk20:46:42
Average for the United States would be . . . 0
carberry20:46:44
which books do you recommend for somone studying for mathcounts?
rrusczyk20:46:51
Our Introduction Series of texts.
carberry20:46:53
are you guys going to continue working closely with math competitions?
rrusczyk20:46:55
Yes.
rrusczyk20:47:12
All right -- that's it for the Math Jam. If you have any more questions about our courses, please write classes@artofproblemsolving.com.
benzi45520:47:36
Wow... computer science, didn't expect that one. Anyhow, I have a question: how useful is olympiad math for other math? I mean, olympiad-level geometry seems kind of useless for current areas of study.
rrusczyk20:48:47
Hard to say -- I would say it mainly helps you learn how to think deeply about complex problems, which is useful for anything. I'd say that *nothing* in my collegiate education trained me for running a company as well as solving Olympiad level geometry problems. In these problems, there are many, many ways to start, and many dead ends. You have to find the most productive line. This is exactly what I do every day.
rrusczyk20:49:38
That's it for class tonight; please email me if you have more questions about the classes. (classes@artofproblemsolving.com).
