| Transcript
for the Math
Jam "AoPS Classes Math Jam"
on Aug 27. |
| Math Jam hosted by rrusczyk
(Richard Rusczyk ). |
rrusczyk19:32:35
Hello, and welcome to an Art of Problem Solving Math Jam. Today we'll be discussing the MATHCOUNTS/AMC 8 Basics, Advanced MATHCOUNTS/AMC 8, AMC 10, and AMC 12 Problem Series. (These are four different courses.)
rrusczyk19:33:11
My name is Richard Rusczyk. I founded Art of Problem Solving and have written several Art of Problem Solving textbooks.
rrusczyk19:33:30
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.
rrusczyk19:33:36
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.
rrusczyk19:33:56
In general in our classes, we have assistant instructors in all of our classes, and all math questions get answered by the primary instructor of the assistants. Tonight, since there are so many of you, we might not be able to answer *every single question*, but we get them all in the classes.
rrusczyk19:34:09
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.
rrusczyk19:34:38
Not now :)
rrusczyk19:34:42
We'll do some math first.
rrusczyk19:35:11
But before that, I'll introduce my assistant, Justin Yan, whose username is justinian.
rrusczyk19:35:28
If you ask questions while we are doing math, sometimes he will answer you.
rrusczyk19:36:03
In this Math Jam, I will briefly describe the course, then go through an example problem or two. Then, I will hold a question-and-answer session about the class.
rrusczyk19:36:11
Before we get started, I'd like to note that the mathematics we will discuss today cover a *very* wide range of difficulty. Moreover, I know that many of you are here just to check out the classroom before your classes start this summer.
ak4719:36:27
how many problems will there be???
rrusczyk19:36:34
7 or 8 total.
rrusczyk19:36:40
A couple of you have asked about sound. There is no sound in the classroom. This webpage explains why:
rrusczyk19:36:48
You can read that later; you don't have to check it out now. There will be a full transcript of this class available on the Math Jam pages about an hour after we finish.
rrusczyk19:36:58
Please understand that if you are enrolled in one of our introductory classes, or haven't much experience yet with advanced problem solving, then much of the material we cover when discussing the AMC 10 and AMC 12 classes will be well beyond you right now. We won't be able to teach you all the math you need to understand this material in one night! So, don't be frustrated if you don't understand the problems we discuss for those classes -- your time will come!
rrusczyk19:37:17
MATHCOUNTS
rrusczyk19:37:29
This fall, we are offering two different MATHCOUNTS/AMC 8 classes: MATHCOUNTS/AMC 8 Basics and Advanced MATHCOUNTS/AMC 8.
rrusczyk19:37:36
The MATHCOUNTS/AMC 8 Basics course will meet for 12 weeks on Mondays, starting September 14, at 7:30 PM Eastern / 4:30 PM Pacific. Each class is 90 minutes, and each is 7:30 - 9 PM ET (4:30 - 6 PM PT). This class is for students just getting started with the type of problem solving required for success in MATHCOUNTS and the AMC 8.
rrusczyk19:38:03
Our Advanced MATHCOUNTS/AMC 8 is for more experienced students, such as those who are training for State MATHCOUNTS, with hopes of qualifying for National MATHCOUNTS. The Advanced MATHCOUNTS/AMC 8 course will meet for 12 weeks on Fridays, starting September 4, at 7:30 PM Eastern / 4:30 PM Pacific. Each class is 90 minutes, and each is 7:30 - 9 PM ET (4:30 - 6 PM PT).
rrusczyk19:38:15
Each course will be offered one more time this fall, the basics class starting in October, and the Advanced class starting in November. Visit this page for details: http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php
rrusczyk19:38:41
Both classes include some material from our previous MATHCOUNTS Problem Series class. The Advanced MATHCOUNTS/AMC 8 class will be a little more challenging than the old class, and the MATHCOUNTS/AMC 8 Basics course will be a little easier.
rrusczyk19:39:04
I will be teaching the MATHCOUNTS/AMC 8 Basics. (As background, I went to National MATHCOUNTS a long, long time ago, and have taught a large number of students how to do the same. I also started AoPS and wrote a number of the AoPS texts.)
rrusczyk19:39:20
The Advanced MATHCOUNTS/AMC 8 class will be team-taught by David Patrick and Mike Greenberg.
rrusczyk19:39:27
David Patrick was a 2-time Math Olympiad Summer Program invitee and a winner of the USA Math Olympiad back in high school. He has a Ph.D. in Mathematics from MIT and is the author of two of Art of Problem Solving's textbooks: Introduction to Counting & Probability and Intermediate Counting & Probability.
Mike Greenberg participated in several math competitions in middle and high school, attending National MATHCOUNTS and the Math Olympiad Summer Program. He earned a BA in mathematics from New York University and an MS in mathematics from Brown University. In his free time, he takes flying trapeze lessons.
rrusczyk19:39:47
You can learn more about MATHCOUNTS here:
rrusczyk19:39:51
www.mathcounts.org
all4math19:39:56
Did you make the AoPS website?
rrusczyk19:40:08
Originally, yes. We have much better developers building it now :)
rrusczyk19:40:34
While there is overlap in topics between the two classes, there will be almost no overlap in problems. Topics covered include methods of counting, probability, algebraic techniques, geometry, word problems, number theory, and more.
rrusczyk19:40:45
This class is a Problem Series class, meaning that the major focus of the class will be working through various contest problems. Although there will be weekly problem sets for each class posted on the message board, students do not submit their homeworks to be graded, and there is no personalized instructor feedback on the solutions. (However, the instructors will be monitoring the message board to answer questions and comment occasionally on the students' discussions there.)
rrusczyk19:41:08
I will take questions about the courses after we do some math.
rrusczyk19:41:13
I'll let you know when to start asking.
rrusczyk19:41:21
Each course will also include a whole class that is a single giant Countdown Round contest!
rrusczyk19:41:27
We'll now look at a few sample problems from the MATHCOUNTS/AMC 8 courses. Hopefully these problems will give you an idea of which course is right for you.
rrusczyk19:41:36
We'll start with a couple problems that are examples of easier problems. These are problems that would be included in the Basics class, but would only be used as a warm-up (or not at all) for the Advanced MATHCOUNTS/AMC 8 course.
math1896619:41:59
wait, so I was just kind of confused.......you're going to ask us the questions and then we're going to answer them, right?
rrusczyk19:42:04
Exactly.
cheeseyicecream19:42:06
Bring it on!
rrusczyk19:42:12
Let's go!
rrusczyk19:42:15
EpicFailure19:42:55
38
MathTwo19:42:55
38 digits
all4math19:42:55
38
rrusczyk19:42:59
How did you do that?
rrusczyk19:43:04
Where do we start?
AwesomeToad19:43:15
Look for factors of 10.
ChickenMitsupishi19:43:15
factor out 10's
rrusczyk19:43:21
How do we find factors of 10?
cheeseyicecream19:43:45
4^20=2^40 and 2*5=10, so we could pair them up
EpicFailure19:43:45
find factors of 5 and 2
number.sense19:43:45
make all the 4's into 2
mz9419:43:45
we write 4^20 as 2^40. now since every group of 5 and 2 produce one digit (a 0) there are 36 0s. then there is a 2^4 term left over which is 16 which adds 2 digits. therefore there are 38 factors
AwesomeToad19:43:45
2x5=10
meowmix19:43:45
and a two and a five is 10
RubiksPro19:43:45
4 = 2 * 2, and 2 * 5 = 10
ak4719:43:59
how will that help??
rrusczyk19:44:18
Indeed - how does that help? What do we get after we write the 4^(20) as 2 raised to some power?
shrig9419:45:12
Therefore we have 2^4*10^36
jsrunnels19:45:21
mz94's explanation makes alot of sense to me. Thanks!
rrusczyk19:45:28
turak19:45:35
we get 2^40*5^36, which is 10^36*2^4, which is 16 followed by 36 zeros => 38 digits
calculatorwiz19:45:48
that makes a lot of sense
rrusczyk19:45:58
Yes it does :) Let's try another problem.
rrusczyk19:46:17
rrusczyk19:46:24
The exclamation point is a 'factorial'. 4! = 4 x 3 x 2 x 1, and factorials for all other positive integers are defined similarly - we multiply all the numbers from the given positive integer down to 1.
copeland19:46:28
This was probably a countdown round problem, so there wouldn't have been a calculator available.
rrusczyk19:46:35
Do we have to multiply all that out to evaluate the given expression?
Krrish19:47:02
No!
stickfigure19:47:02
factor out the 5!
number.sense19:47:02
no! factor 5! from the top and bottom
limac19:47:02
no, we can use factorization
brightmz19:47:02
no, we can factor out 5! on the bottom
faush10119:47:02
smplfy 5! from everything first
RubiksPro19:47:02
factor out 5! from all terms on the denominator
rrusczyk19:47:06
We don't have to multiply. We can factor instead.
rrusczyk19:47:09
How can we factor the denominator?
AwesomeToad19:47:58
limac19:47:58
each of the terms have 5!, thus we have 5!(6+1+1)
shadowmath19:47:58
5!(6+1+1)
RubiksPro19:47:58
5!(6 + 1 + 1)
brightmz19:47:58
5!(6+1+1)=5!(8)
Ruboks19:47:58
5!(6+1+1)=5!(8)
red.hot.spikerXOXO19:48:03
5!(6+1+1)=5!*8
rrusczyk19:48:17
Excellent; that's how we factor the denominator.
rrusczyk19:48:23
So, how do we finish the problem?
junichi94@hotmail.com19:48:29
then we can cross out the 5! from the numerator and the denominator
Masashi19:48:29
since 5! is = 5*4*3*2*1 we could simplify this by dividing the top and the bottom with 5!
rrusczyk19:48:33
And what do we get?
number.sense19:48:55
720/8=90
victorzhou19:48:55
so its 6!/8 which is 90
shadowmath19:48:55
So then we can cancel out the 5!s to get 6!/8=90
twin7719:48:55
720/8=90
math1896619:48:55
90
turak19:48:55
so thats 6!/8
archemides779019:48:55
720/8 which is 90
thyang19:48:55
cancel the 5! and there's a 6!/ 8
rrusczyk19:49:00
rrusczyk19:49:16
Also, clever factoring is extremely useful in math problems, and often is helpful in MATHCOUNTS.
rrusczyk19:49:28
The next problem is an example of a problem that would be on the harder end of the Basics class and the easier end of the Advanced MATHCOUNTS/AMC 8 class.
rrusczyk19:49:36
rrusczyk19:49:50
Wow!!!! What will we do with all those numbers?
rrusczyk19:49:54
Some of them are HUGE
shrig9419:50:00
Forget about 5! and beyond
rrusczyk19:50:04
Why can we do that?
mz9419:50:20
hint: every factorial from 5! up ends in 0
ChickenMitsupishi19:50:20
most of those end in 0
Ruboks19:50:20
5! and after, the units digit is 0
shadowmath19:50:20
Starting from 5! the units digits are 0
turak19:50:20
well all from 5! and beyond end in 0
AwesomeToad19:50:20
melodicangel19:50:20
anything after 5! would have 0 as units digit
rrusczyk19:50:24
We don't get intimidated by the 99!. We only want the last digit. Most of these terms have last digit of 0.
rrusczyk19:50:30
Only the first 4 terms have a nonzero units digit.
rrusczyk19:50:49
So, what's the last digit of that expression? What last digits are left?
AlphaBetaTheta19:50:57
but be careful
rrusczyk19:51:10
Yes, we have to be careful. Why?
all4math19:51:13
So can't we just calculate the first four easily?
rrusczyk19:51:33
Yes, we can, but what do we have to be careful about?
lightbluemathangel19:51:39
negatives
winner200919:51:39
there are negatives
rrusczyk19:51:51
Exactly. We have to be careful about that. And how do we watch out for that?
cheeseyicecream19:52:15
we have 1-2+6-24=-19+(some huge number)=a units digit of 1
MuffinMan0919:52:16
1-2+6-24=-19; but because it's negative the unit's digit is one.
number.sense19:52:16
adding 120
mz9419:52:16
lets add 5!=120 to it
rrusczyk19:52:29
Here are the first four terms:
faush10119:52:31
rrusczyk19:52:36
1 - 2 + 6 - 24 = -19.
rrusczyk19:53:01
But the sum we are given obviously isn't negative, since we add 99! as the last step.
Cliu030119:53:10
so then add -19 to 120 and you get 101...so since every last digit after 5! is 0...the units digit is 1
AwesomeToad19:53:25
rrusczyk19:53:29
The answer isn't 9 because our sum is clearly positive. All those terms with a zero units digit clearly have a positive sum, so our expression equals -19 + (something big that ends in zero). Thus we have a final digit of 1, not 9.
lzdg888819:53:33
So if the last factorial is being subtracted then it's negative?
rrusczyk19:53:36
Exactly!
rrusczyk19:54:05
Before we go on, our regular classes are a lot smaller than this, and a good bit less hectic :)
ak4719:54:10
so the answer is one??
rrusczyk19:54:12
Yes
rrusczyk19:54:29
The last two problems are problems that would only appear in the Advanced class.
rrusczyk19:54:41
ak4719:54:59
what is intergal mean??
smartalec1719:55:01
what is an integral
rrusczyk19:55:04
Integers.
AlphaBetaTheta19:55:16
prime factorize 792
RubiksPro19:55:16
find the prime factorization of 792
MathTwo19:55:16
factor 792 first
shadowmath19:55:16
Prime factorization of 792
rrusczyk19:55:28
What is the prime factorization?
cheeseyicecream19:55:42
792=2^3*3^2*11
mathboy10119:55:42
to find the total number of divisors, we prime factorize it. we get 2^3*3^2*11
melodicangel19:55:42
2^3(3^2)(11)
shadowmath19:55:42
MuffinMan0919:55:42
2^3*3^2*11
rrusczyk19:55:49
rrusczyk19:56:04
What makes this problem a little harder than the usual divisor counting problem?
Ruboks19:56:23
even factors
Cliu030119:56:23
you count all the even numbers!
AlphaBetaTheta19:56:23
even divisors
potatopi19:56:23
the divisor's have to be even
shadowmath19:56:23
it wants only even
rrusczyk19:56:39
Right--- how would we deal with the problem if we wanted to count *all* divisors?
number.sense19:57:03
number of factors is (3+1)(2+1)(1+1)
pianist-goomba19:57:03
add one to each power and multiply them
shrig9419:57:03
add 1 to each exponent and multiply
rrusczyk19:57:16
We explain why this works in detail in the Introduction to Number Theory course.
rrusczyk19:57:24
Here's what it looks like with 792.
AwesomeToad19:57:27
rrusczyk19:57:31
Very good.
rrusczyk19:57:37
limac19:58:17
But a has to be at least 1.
melodicangel19:58:17
a can''t be 0
rrusczyk19:58:47
pianist-goomba19:59:04
there are 3 choices for a, 3 for b, and 2 for c
yingggao19:59:05
so just dont add the 1?
ak4719:59:05
only 3
AwesomeToad19:59:05
So we have 3 choices for the a.
RubiksPro19:59:05
3 possibilities
rrusczyk19:59:17
suryabrata19:59:41
18 even divisors
sunilkjain19:59:41
18 is the answer
hyperddude19:59:41
There are 18 even divisors.
poke9619:59:41
18 divisors
meowmix19:59:41
3.3.2
EpicFailure19:59:41
18
AwesomeToad19:59:47
rrusczyk19:59:52
rrusczyk20:00:10
If you found this problem impossibly hard, then the Basics class is probably better for you.
rrusczyk20:00:22
Here's another clever solution:
number.sense20:00:25
I have another way: You can find find the all the factors of 792/2, and multiply each one by two to get even factors of 792. Therefore the number of factors of 396 is the same as the number of even factors of 792. 396 = 2^2 * 3^2 * 11, so the number of factors is 18.
rrusczyk20:00:28
Very nice :)
rrusczyk20:00:45
Also, some of you counted the number of odd divisors and subtracted that from 24.
AlphaBetaTheta20:00:48
u could use complementary counting
AwesomeToad20:00:48
Complementary counting? Can't we use that?
rrusczyk20:01:04
And we call the method I just mentioned, "Complementary counting".
Cliu030120:01:07
can you apply your clever solution to other problems like this one?
rrusczyk20:01:13
Absolutely.
calculatorwiz20:01:21
why would you divide792 by 2
cheeseyicecream20:01:42
because it removes one two as a factor
meowmix20:01:48
you are multiplying 396 by two
turak20:01:51
because if you multiply those factors by 2 you know they're even
rrusczyk20:01:53
To make an even divisor of 792, you just take a divisor of 792/2 and multiply it by 2!
rrusczyk20:02:12
One more problem for MATHCOUNTS, and then I will take questions about the class.
rrusczyk20:02:18
rrusczyk20:03:08
I don't have enough time to teach base numbers tonight! So, this problem will only be for those who know base numbers already.
rrusczyk20:03:17
Some of you are suggesting converting to base 10 first.
meowmix20:03:29
don't
cheeseyicecream20:03:29
turning this into base ten is too tedious...
rrusczyk20:03:30
That would take a loooooooong time.
rrusczyk20:03:35
(No calculators.)
mz9420:03:42
since 3^2=9, we can take each group of 2 and write it in base 9. (ex 02 is still 02 in base 9). then we are done
meowmix20:03:42
3 is square root of 9
victorzhou20:03:42
base 3 convert to base 3^2
shadowmath20:03:59
AlphaBetaTheta20:04:03
3 and 9 have a relationship, you could use that relationship
melodicangel20:04:15
2(3^6)+2(3^5)+3^2+2
rrusczyk20:04:18
shadowmath20:04:51
Carry
cheeseyicecream20:04:52
just multiply them by 3
shrig9420:04:52
Just add them to the power ahead
rrusczyk20:05:20
What do you mean? How would you deal with 2*3^5?
Ruboks20:06:01
6x3^4
snail220:06:01
2x3^5 = 6 x 3^4 = 6 x 9^2
limac20:06:01
we can factor a 3^4 out of 2*3^5.
number.sense20:06:01
6*9^2
turak20:06:01
its 6*9^2
Ruboks20:06:01
6x3^4=6x9^2
rrusczyk20:06:05
rrusczyk20:06:12
So, what is the answer to the original problem?
siva20:06:26
2612 is the answer
Ruboks20:06:26
2612
MuffinMan0920:06:26
2612_9
poke9620:06:26
2612
caroltao20:06:26
2612
rrusczyk20:06:31
rrusczyk20:06:57
If the last 2 problems we did seem impossible to you, then the MATHCOUNTS/AMC 8 Basics class is probably the right class for you. If the first 2 problems seem way, way easy to you, and you understood the solutions to the last 2-3 problems, then the Advanced MATHCOUNTS/AMC 8 class is the right class for you.
rrusczyk20:07:17
Are there any questions about either of the two MATHCOUNTS courses?
jsrunnels20:08:04
Is the pace of these examples reflective of the pace in the actual MATHCOUNTS class? It seems that everyone here already knows how to do all of these problems...
rrusczyk20:08:07
We're going faster tonight, partly because there are a lot of students here for the AMC class who know this material already, and partly because there are so many of you.
mathboy10120:08:14
will we have to know bases for the advanced class, or will we learn them?
rrusczyk20:08:38
We won't spend a ton of time on bases for the advanced class, so you don't *need* them. You can learn about them in the Introduction to Number Theory class.
twin7720:08:45
what will we be covering in the Advanced class?
rrusczyk20:08:54
Harder problems from MATHCOUNTS and the AMC 8
cheeseyicecream20:09:00
What does the "homework" consist of?
rrusczyk20:09:16
After each class, we post around a dozen problems on the message board for students to discuss.
mkbobba20:09:38
Do you have any classes for elmentary grade students?
rrusczyk20:09:40
Not yet, though some elementary-age students are ready for our Algebra 1 class.
smartalec1720:09:52
is there going to be a pre-test
rrusczyk20:09:54
No pretest for the Problem Series.
eginsber20:10:25
So if we sign up for the Basica class will most people know all the answers (like in this jams), or is it much more like a regular classroom of smart people that are learning things.
rrusczyk20:10:27
The Basics class will not be as blindingly fast as this might have seemed for you. There will also be more time for you to ask questions and soak up the material.
all4math20:10:46
What are the grade levels that usually go to them
rrusczyk20:10:48
These classes are typically grades 6-8 (the MATHCOUNTS/AMC 8 classes)
EpicFailure20:10:59
the last question is about what level in the advandced mathcounts
rrusczyk20:11:01
Averagish.
meowmix20:11:12
If you got to the Countdown round at State last year, is advanced the course for you?
rrusczyk20:11:13
Depends on the state, but almost certainly.
always.recycle.things20:11:45
what if we signed up for the advanced one and think it might be too hard
rrusczyk20:11:47
You can drop a course before the third class for a full refund, or switch to the other class before the third class.
dinapro20:12:28
will geometry be covered in the courses?
rrusczyk20:12:30
Yes, there will be some geometry.
win20:12:32
im in 6th grade so i should take the basic class right?
rrusczyk20:12:34
Probably
potatopi20:12:51
Are calculators allowed on the AMC or MATHCOUNTS?
rrusczyk20:12:53
The rules change from year to year. I'm not sure what they are in 2010.
number.sense20:12:58
Are there any prerequisites?
rrusczyk20:13:00
No
firepower42120:13:18
How many problem will be covered in one class
calculatorwiz20:13:18
how many problems are there in a problem set
rrusczyk20:13:19
There are 10-15 problems per class. Sometimes more.
sammybh20:14:04
i understand the solutions, but only after you explain them. which class id right for me?
rrusczyk20:14:06
Possibly -- depends on how you like to learn. If you like the challenge and ask questions when you don't understand, you may be fine.
chipmunk8820:14:24
When you get problems to work out by yourself after the courses, is there anyplace you can get their solutions?
rrusczyk20:14:26
On the message board, yes.
AwesomeToad20:14:30
I took a practice nationals before the Advanced and after week 10. I got 24 before and 37(!) after.
rrusczyk20:14:32
:)
Cliu030120:14:55
what if you didn't know the first two but understand the rest of them?
rrusczyk20:14:57
Tough to say -- you can try one and switch if it doesn't fit.
suryabrata20:15:19
How much time should we devote to this each week?
rrusczyk20:15:21
3-4 hours for the Problem Series.
firepower42120:15:36
In regular class, how many people will be there ?
rrusczyk20:15:37
50-70
siva20:15:51
is it true that mathcount nationals is as hard as amc 10
rrusczyk20:15:53
Varies a lot from year to year.
jsrunnels20:16:23
Is there prep material for each of the classes or do you come into a session "cold?"
rrusczyk20:16:24
No prep material in the Problem Series; only in the subject classes (Intro classes, Algebra 1, etc.)
bubmaster20:16:46
which one is harder: the algebra 2 class or the advanced mathcounts/amc 8 class?
rrusczyk20:16:48
Tough to say -- they're very different. They're probably comparable.
always.recycle.things20:16:55
what is covered in the basics
rrusczyk20:16:56
Easier MATHCOUNTS problems.
junichi94@hotmail.com20:17:05
the last two problems could actually be amc problems
rrusczyk20:17:07
Correct.
ak4720:17:19
if we have the AoPS book, is that as good as the class
rrusczyk20:17:20
Tough to ask a book questions :)
Cliu030120:17:26
Are there going to be count down tips from the Advanced Class?
rrusczyk20:17:27
Yes
eginsber20:17:37
Will any math problems similar to these come up in the basic course?
rrusczyk20:17:38
Sure.
he534220:17:58
is the advanced class based on certain subjects?
rrusczyk20:18:00
It covers a wide range of topics.
mkbobba20:18:13
In general which grade students take the MATHCOUNTS/AMC 8 Basics class?
rrusczyk20:18:15
Grade 6-8
meowmix20:18:38
what formulas should we already know?
rrusczyk20:18:40
Don't worry about formulas!!! Worry about understanding math.
lstar20:19:04
i took mathcounts basics, i wus just wondering if there was a AMC=>8 basics?
rrusczyk20:19:06
The MATHCOUNTS class is MATHCOUNTS/AMC 8. It covers both.
maybach20:19:15
Willl we get challenge sets?
rrusczyk20:19:16
Not in the Problem Series class.
yingggao20:19:23
should the questions for the advanced class be states or national level?
rrusczyk20:19:24
Both
chesskid20:19:31
will you post homework tonight?
rrusczyk20:19:33
No
LittleMaster20:19:56
Will the class be posted if you cant make it?
rrusczyk20:19:58
Yes, there is a full transcript that you can review later if you miss class (or if you don't!)
thunderbirdz20:20:11
are the questions available for everyone or only the people signed up for the class
rrusczyk20:20:12
Only people in class.
poke9620:20:21
What is the difference between AMC8 and AMC 10
rrusczyk20:20:22
AMC 10 is harder.
Ruboks20:20:28
are there going to be tests?
rrusczyk20:20:29
No
Krrish20:20:57
im finishing my algebra 1 which would ou recommend?
rrusczyk20:20:58
Depends on your problem solving experience! Take a look at the problems we did tonight and the comments I made after them.
Cliu030120:21:17
what will happen if i can't attend a class?
rrusczyk20:21:18
Catch up with the transcript I mentioned, and you can use the message board to ask any questions you have.
all4math20:21:24
Do you have to be in a cirtain grade to be in the classes
rrusczyk20:21:25
No
ak4720:21:34
if we have the AoPs book is that good for mathcounts
rrusczyk20:21:35
Yes :)
always.recycle.things20:22:02
what if its too hard for you
rrusczyk20:22:04
You can drop any course and get a full refund if you do so before the third class.
cheeseyicecream20:22:49
Is this class enough to get to me nats in a *relatively* big state?
rrusczyk20:22:51
Depends on how much you are working. The class alone probably isn't enough; you should be doing other things too, like doing practice tests, studying AoPS texts, etc.
dinapro20:22:56
is geometry covered in the classes?
rrusczyk20:22:58
Yes
lzdg888820:23:17
If I want to take the AMC Test this November, should I take the Basic or Advanced class?
rrusczyk20:23:19
Depends -- if you're a beginner, the Basics; if you're advanced, the advanced.
rrusczyk20:24:06
There is a list of general subjects that will be covered, but that's it. In our subject classes, there is specific direction and reading to do to prepare for class. So, if that's the structure you seek, you should take the subject classes.
jsrunnels20:24:09
Is there a way to familiarize yourself with the class topic BEFORE each lesson, so as to be more prepared to undestand/ask questions?
rrusczyk20:24:19
(Answered before question, sorry)
rrusczyk20:24:49
Right now, I have over 200 questions in my queue. I obviously won't get to them all tonight. I will answer questions for another 5 minutes, and then do the AMC classes and take more questions.
maybach20:25:08
Which is better for State MC Preparation: Basics or Advanced?
rrusczyk20:25:09
Depends on the state. In your state, advanced, for sure.
abhilash20:25:23
do they cost money?
rrusczyk20:25:28
Yes; details are here:
shrig9420:25:56
Do you restrain students from answering to fast to allow the rest of us to think about the problem?
rrusczyk20:25:58
Yes; the classes are at a more sane pace.
yingggao20:26:11
so even the advanced class should be a bit less hectic?
rrusczyk20:26:13
Yes.
sammybh20:26:24
is it ok that i did not know many of the questions in this session
rrusczyk20:26:26
Certainly. I am sure you are not alone.
eginsber20:27:09
In that class (Algebra 1) do all the students know everything like in this Jams, or is it much more teaching topics, like you might find in a normal classroom?
rrusczyk20:27:11
The Algebra 1 is more like a "normal" class in that the fundamentals that are needed for problems are taught as part of the class.
jsrunnels20:27:29
How do you pronounce your name? :)
rrusczyk20:27:31
RUH-sick
smartalec1720:27:53
what would be the appropriate score on the state for the advanced class?
rrusczyk20:27:54
Mid-high 20s in 2008, if you've done some work since then.
faush10120:28:20
If you just did AMC 8 Basics, is AMC 8 Advanced a good place to be?
rrusczyk20:28:21
If you didn't find the class terribly challenging, yes. You might also consider one of the Intro classes.
Cliu030120:28:56
If i didn't make states last year but am at state level...should i take this course?
rrusczyk20:28:58
Possibly; use the problems we did earlier as a guide. If you did a lot of work since then, you may be ready for Advanced (or if you are in a killer chapter)
AwesomeToad20:29:31
If I finish Counting and Probability Introduction, will I be in good shape for Mathcounts national level?
rrusczyk20:29:33
If you really understand it, that will help a great deal.
cheeseyicecream20:30:25
My parents are were very reluctant to enroll me in one of these classes...anything to convince them that this is better than the traditional tutor?
rrusczyk20:30:26
Show them the teacher's credentials :) And also, I'll note that I believe the top 6 students or so at Nationals in 2009 were all our students. (And so were several of the rest of the top 20)
all4math20:30:52
Can the classrooms be filled up?
rrusczyk20:30:53
Yes. It is possible the classes would fill.
maybach20:31:00
Is AoPS Vol. 1 similar to the stuff covered in the advanced class?
rrusczyk20:31:01
Yes
dinapro20:31:15
in what grade are you eligible to take amc10?
rrusczyk20:31:16
Any up to 10th
always.recycle.things20:31:27
how long is each class
rrusczyk20:31:28
90 minutes.
abhilash20:31:31
are we going to do more problems?
rrusczyk20:31:48
Yes. I am now going to talk about the AMC classes, and we will do more problems.
rrusczyk20:31:53
I will then take more questions.
rrusczyk20:31:57
Sorry I didn't get to them all.
rrusczyk20:32:03
There are now 263 in my queue.
rrusczyk20:32:21
If I didn't get to yours, you can email me at classes@artofproblemsolving.com, and I'll answer.
rrusczyk20:32:33
AMC 10 and AMC 12 Problem Series
rrusczyk20:32:38
The AMC 10 Problem Series starts on September 10, and meets every Thursday from 7:30-9:00 PM Eastern. The class meets for 12 weeks and ends on December 3. The course is designed to cover a large portion of the curriculum tested on the AMC 10 exam.
rrusczyk20:32:48
This class is taught by Paul Ryu. He was a third-place winner at National MATHCOUNTS, a two-time perfect scorer on the American Mathematics Contest (AMC) 12, a participant at the Research Science Institute, and a prize winner at the Intel International Science and Engineering Fair. He has also helped students achieve success at Intel Science Talent Search and Siemens Westinghouse Science and Technology Competition.
rrusczyk20:32:59
The AMC 12 class starts on September 9, and meets every Wednesday from 9:00-10:30 PM Eastern (note the later start time) . The class meets for 12 weeks and ends on December 2. The course is designed to cover a large portion of the curriculum tested on the AMC 12 exam.
rrusczyk20:33:20
The class will be taught by Valentin Vornicu. Valentin was a 2-time participant at the International Math Olympiad, and won prizes at numerous national math competitions in Romania. Valentin has a Master's degree in Mathematics from the University of Bucharest and founded the MathLinks website in 2002 (which merged with Art of Problem Solving in 2004).
rrusczyk20:33:29
These classes are Problem Series courses, meaning that the major focus of the class will be working through various AMC problems. Although there will be weekly problem sets for each class, students do not submit their homeworks to be graded, and there is no personalized instructor feedback.
Cliu030120:33:53
2 time PERFECT SCORER ON AMC 12?! geez
Cliu030120:33:54
:O :O :O!!!! these ppl are too smart
rrusczyk20:33:57
They're OK :)
rrusczyk20:34:01
Both classes will be offered again starting in October.
rrusczyk20:34:07
Tonight, I will cover material from both the AMC 10 and AMC 12 Problem Series classes before taking questions about the classes.
rrusczyk20:34:13
The following are excerpts of a couple of the areas of problem solving covered in the AMC 10 Problem Series.
rrusczyk20:34:27
ARITHMETIC SEQUENCES
rrusczyk20:34:34
We have all probably seen many arithmetic sequences, but I would like you to pay close attention to the ways in which we manipulate facts according to our understanding of arithmetic sequences in the following problems. In particular, arithmetic sequences involve common differences that are constant. Constants are our friends and we should remember how useful they can be.
rrusczyk20:34:52
rrusczyk20:35:14
Where might we start? What are the angles?
hyperddude20:35:27
One of them is 160 degrees...
rrusczyk20:35:33
Indeed. And the others?
potatopi20:35:45
they are 160, 155, 150, etc
junichi94@hotmail.com20:35:46
160,155,150...
smartalec1720:35:46
they are 5 less
ThreeRivers20:35:46
5 less
maybach20:35:46
155, 150, 145, 140..................
rrusczyk20:35:52
OK...
archemides779020:35:55
160-5 until you have n sides
rrusczyk20:36:07
Until we have n sides... So, what will the last one be?
EpicFailure20:36:34
160-5(n-1)
cheeseyicecream20:36:34
160-(n-1)5
number.sense20:36:34
160-5(n-1)
brightmz20:36:34
160-5(n-1)
shadowmath20:36:34
160-5(n-1)
rrusczyk20:36:37
rrusczyk20:36:42
Remember that the last value should be 160 - 5(n-1), not 160 - 5n because we started with 160 in the list.
rrusczyk20:36:53
So, there are our angles. Now what?
AceOfDiamonds20:37:19
add them up
junichi94@hotmail.com20:37:19
add them up
rrusczyk20:37:32
rrusczyk20:37:42
Now what?
ChickenMitsupishi20:37:57
sum of the angles = 160 - 5*n(n-1)/2
AceOfDiamonds20:37:57
160n-5(n)(n-1)(1/2)
number.sense20:37:57
This is just 160n-5(n)(n-1)/2
rrusczyk20:38:11
rrusczyk20:38:25
(It's late, and we have a ways to go, so I gave you a little of the algebra there ;) )
rrusczyk20:38:34
What will we do with this?
Wenghim20:38:48
Find n so that the sum of the sequence is 180*(n-2)
shadowmath20:38:48
They should add up to (n-2)180
turak20:38:48
total degrees in n-gon is 180(n-2)
number.sense20:38:48
Now, another way to express the sum of the angles in terms of n is 180(n-2)
AceOfDiamonds20:38:48
=180(n-2)
rrusczyk20:38:54
Conveniently, we also have a formula for the sum of the interior angles of a convex n-sided polygon: 180(n - 2).
shadowmath20:39:14
So we have an equation!
potatopi20:39:14
solve the equation now
rrusczyk20:39:15
hyperddude20:39:37
And NOW we simplify!
lightbluemathangel20:39:37
multiply both sides bvy 2 to get rid of fractions
abhilash20:39:37
miltiply by 2 to get rid of fraction
rrusczyk20:39:46
What else can we do to simplify?
hi how are you doing toda20:40:09
divide 5
archemides779020:40:10
divide by 5
number.sense20:40:10
divide by 5 too
brightmz20:40:10
multiply all sides by 2/5
carlito30020:40:10
divide by 5 on both sides
meowmix20:40:10
divide by 5
hyperddude20:40:10
Multiply each side by 2, then divide both sides by 5.
rrusczyk20:40:19
potatopi20:40:23
if we muliply things out, we get a quadratic equation
rrusczyk20:40:31
What quadratic do we get?
maybach20:41:04
65n-n^2 = 72n - 144
shadowmath20:41:04
n^2+7n-144=0
potatopi20:41:04
65n-n^2 = 72n - 144
junichi94@hotmail.com20:41:04
n^2+7n-144=0
rrusczyk20:41:07
rrusczyk20:41:10
What are the solutions to this quadratic?
brightmz20:41:35
-16 and 9
AceOfDiamonds20:41:35
16,9
meowmix20:41:35
-16, 9
Wenghim20:41:35
n=-16 or 9, but n>0, so n=9
hyperddude20:41:39
n= -16 or 9
aunch20:41:44
-16,9
rrusczyk20:41:45
We can solve the quadratic in a number of ways including factoring it into
(n - 9)(n + 16) = 0, so n = 9 or -16.
number.sense20:41:49
0=(n+16)(n-9), so since n cant be negative, n=9
mathq20:41:49
therefore n=9 since it has to be positive
cheeseyicecream20:41:58
the answer is 9 since it must be positive
lightbluemathangel20:41:58
n has to equal 9
rrusczyk20:41:59
We know that a polygon cannot have a negative number of sides and so we have our answer, n = 9. (A).
rrusczyk20:42:04
This problem is a good example of how important it is to be able to turn words in a problem into an equation. A great many problems are solved this way.
rrusczyk20:42:21
Now, we'll switch over to another topic from the AMC 10 class.
rrusczyk20:42:24
Many problems involve combinations. One simple example is the problem of finding the number of triangles that can be formed from 6 points in space, no 3 of which are collinear.
rrusczyk20:42:40
We can solve this problem by considering the arbitrary points A, B, C, D, E, and F. Since no three of the points are collinear, we know that any set of three points forms a (nondegenerate) triangle. We might start by counting the number of three letter strings we can form by choosing one of each letter:
rrusczyk20:42:44
ABC
ACB
BAC
BCA
and so on.
rrusczyk20:42:59
How many of these strings are there?
number.sense20:43:24
120
archemides779020:43:24
120
brightmz20:43:24
6*5*4=120
AceOfDiamonds20:43:24
6*5*4=120
he534220:43:31
120
rrusczyk20:43:34
We can choose 6 possible letters to be the first letter, 5 to be the second, and 4 to be the third, so there are 6*5*4 = 120 such strings.
rrusczyk20:43:41
That gives us a count of the 3-letter strings. But to count the triangles, we only want three letter groups in which the order does not matter. Triangle ABC is the same as ACB is the same as CAB, etc.
rrusczyk20:44:06
How do we correct our count for this?
number.sense20:44:27
but we are way overcounting! the order of the vertices doesn't matter
turak20:44:27
divide by 3!
he534220:44:27
divide by 3!
AIME15USAMO20:44:27
divide bye 3! or 6
suryabrata20:44:27
divide by 6
lightbluemathangel20:44:27
divide by 3*2*1
rrusczyk20:44:32
There are 3! possible strings with A, B, C, so we count triangle ABC 6 times in our count of 120 ordered strings above. Similarly, we count every group of three letters exactly 6 times, so to count the total number of different groups, we divide our count of 120 ordered strings by 6 to get 120/6 = 20.
rrusczyk20:44:42
This is what we mean by 'combinations': the number of ways in which we can choose r objects from a group of n objects if we don't care about order.
rrusczyk20:44:48
rrusczyk20:44:54
We have already seen how to compute a combination. We first do a count of selecting r items from the n if order matters, and then we divide by r!.
rrusczyk20:45:00
rrusczyk20:45:12
If you are unfamiliar with combinations, you can learn all about them with our Introduction to Counting & Probability book.
rrusczyk20:45:28
Wenghim20:46:14
28
vliu20:46:14
28
number.sense20:46:14
D, 8C2=28
vliu20:46:14
D or 28
RubiksPro20:46:14
there are 8 vertices and 2 points needed for a line segment, so use C(8,2)
junichi94@hotmail.com20:46:14
8C2
EpicFailure20:46:14
28
meowmix20:46:14
8 C 2
rrusczyk20:46:20
rrusczyk20:46:26
Understanding combinations can directly lead us to quick solutions to some AMC problems!
rrusczyk20:46:38
(That's really all there is to this problem!!)
rrusczyk20:46:47
Those two problems were on the easy end of the AMC 10 class.
rrusczyk20:46:52
The following is an excerpt of one of the areas of problem solving covered in the AMC 12 Problem Series.
rrusczyk20:46:59
A point P is randomly selected from the rectangular region with vertices (0, 0), (2, 0), (2, 1), (0, 1). What is the probability that P is closer to the origin than it is to the point (3, 1)?
(A) 1/2
(B) 2/3
(C) 3/4
(D) 4/5
(E) 1
rrusczyk20:47:18
So, do we count the number of possible outcomes?
abhilash20:47:28
no
brightmz20:47:28
no
rrusczyk20:47:31
Why not?
he534220:47:43
there's infinite
cheeseyicecream20:47:43
there are infinite
meowmix20:47:43
there is infinite amount
rrusczyk20:47:52
There are infinitely many ways we can pick a point!
rrusczyk20:47:57
So, how do we do this?
mathq20:48:05
no it's probability using the area
jeffreyyan820:48:05
geometric
shadowmath20:48:05
Geometric probability
maybach20:48:05
Do we do area of possible?
suryabrata20:48:05
we measure the area
rrusczyk20:48:16
Geometry gives us tools to measure infinite sets of points.
rrusczyk20:48:21
Here, we'll use area.
rrusczyk20:48:28
The possible region is obvious.
rrusczyk20:48:35
We have to find the desired region.
rrusczyk20:48:41
How do we know when a point is closer to one location or another in a plane?
vliu20:48:52
draw a line connecting the points 3,1 and 0,0
rrusczyk20:49:01
OK, and what do we do with that line segment?
Wenghim20:49:11
find the midpoint
RubiksPro20:49:12
find the midpoint
hyperddude20:49:12
split it in 2
rrusczyk20:49:16
And then what?
mathq20:49:26
draw line between two points, and then draw the perpendiicular line at the midpoint
number.sense20:49:26
the perpendicular bisector of the segment between the two points divides the region
RubiksPro20:49:26
then draw a perpendicular line through the midpoint
aunch20:49:26
and draw a perpendicular line
dkboy20:49:26
perpendicular bisector?
rrusczyk20:49:31
The set of points (in a plane) equidistant to two given points is the perpendicular bisector of the line segment with the two points as endpoints.
rrusczyk20:49:51
This bisector will divide the possible region into points closer to the origin and points closer to (3,1).
rrusczyk20:50:03
So, can we find this line?
mathq20:50:25
find the equation
rrusczyk20:50:30
How will we do that?
rrusczyk20:50:34
What do we know about the line?
Wenghim20:50:42
point and slope
rrusczyk20:50:50
We need a point on the line and the line's slope.
rrusczyk20:50:58
What is a point on the line?
rrusczyk20:51:05
(We're looking at the perpendicular bisector here)
turak20:51:27
passes through (1.5,.5)
maybach20:51:28
(3/2, 1/2)
number.sense20:51:28
3/2, 1/2
Wenghim20:51:28
1.5,.5
AIME15USAMO20:51:28
(1.5,.5)
rrusczyk20:51:32
The midpoint of the segment between (0, 0) and (3, 1) is (1.5, .5).
rrusczyk20:51:43
OK, what is the slope of the perpendicular bisector, and why?
carlito30020:52:37
negative reciprocal of 1/3 = -3
chipmunk8820:52:37
-3, it is perpendicular to the line segment.
RubiksPro20:52:37
this gives a slope of -3
maybach20:52:37
-3, because the slope for the bisector is -1/m and -1/(1/3) is -3
AIME15USAMO20:52:37
opposite of the reciprocal of the slope of the line connexting (0,0) and (3,1)
rrusczyk20:52:44
The equation of the line through (0, 0) and (3, 1) is y = x/3.
rrusczyk20:52:47
This line has slope 1/3
rrusczyk20:52:52
The slope of a line perpendicular to that line is -1/(1/3) = -3. (The product of the slopes of two perpendicular lines in a plane is -1 if neither line is vertical.)
rrusczyk20:53:06
So, what is the point-slope form of the perpendicular bisector?
Wenghim20:53:51
y-.5=-3(x-1.5)
RubiksPro20:53:51
y - 0.5 = -3(x - 1.5)
mathq20:53:51
y-1/2=-3(x-3/2)
rrusczyk20:53:54
We want the line through (1.5,0.5) with slope -3. We can use point-slope form:
y-0.5 = -3(x-1.5).
rrusczyk20:54:13
Rearranging this gives y = -3x + 5.
rrusczyk20:54:17
The equation of the points equidistant between (0, 0) and (3, 1) is y = -3x + 5.
Now, what do we do with all this information?
number.sense20:54:42
graph !
RubiksPro20:54:48
draw the line passing throught the rectangle and find the areas of the newly formed polygons
aunch20:54:49
draw the line and see how much area falls on the side of the box that you are looking for
turak20:54:51
find area of region on either side
rrusczyk20:54:56
It's picture time!
rrusczyk20:55:00
rrusczyk20:55:16
The red rectangle is the possible region.
twin7720:55:24
i still think that it is 3/4
Wenghim20:55:24
3/4
aunch20:55:24
3/4
turak20:55:24
3/4
hi how are you doing toda20:55:24
yes it is 3/4
rrusczyk20:55:38
How can you see so quickly? Don't we have to find the area of that trapezoid?
Wenghim20:55:59
the line divides the square on the right exactly in half, so we never needed the equation, just the point
meowmix20:55:59
no, just use congruency
mathq20:55:59
flip it
archemides779020:55:59
no the line shows half the area
number.sense20:55:59
no, the line cuts through the center of the square
RubiksPro20:55:59
the line divides the second square in half
rrusczyk20:56:02
Exactly!!!!
rrusczyk20:56:18
The line goes through the center of the little square, so it cuts that square in half.
rrusczyk20:56:33
So, the white area inside the red rectangle is 1/2.
rrusczyk20:56:40
The grey area then is 3/2.
cheeseyicecream20:56:42
but how do we know? we probably won't draw a nice picture in a competition..
rrusczyk20:56:58
Great question; how do we know that the line cuts the square in half?
twin7720:57:09
the trapezoids in the second square are congruent
rrusczyk20:57:21
The trapezoids are congruent.
rrusczyk20:57:35
We can see that by rotating everything 180 degrees around the center.
rrusczyk20:58:05
Then, the grey trapezoid rotates onto the white one (because the bisector rotates to itself, and so does the square).
hi how are you doing toda20:58:12
yeah, well rotate the paper
rrusczyk20:58:32
(That's the way you should think about this, and you can do that without even having to draw the diagram)
twin7720:58:40
it is easy. first you extend the rectangle so that its verticies are (0,0), (3,0), (3,1), (0,1). Then you draw a line perpendicular to the rectangle at (1.5,0). The area of that half is 1.5 sq. units. 1.5/2=3/4.
rrusczyk20:58:44
That's clever :)
meowmix20:58:48
but then you would only know they were congruent if you drew a nice picture
rrusczyk20:59:21
Knowing that the line goes through the center is enough to deduce this, but I agree that it isn't obvious!
AIME15USAMO20:59:23
How do we know the perpendicular bisector intersects the center of the second square?
rrusczyk20:59:39
When we found the midpoint of (0,0) and (3,1), we see that it is (1.5,0.5).
Wenghim20:59:45
It passes through (1.5,.5)
hi how are you doing toda20:59:45
center = 1.5,.5
rrusczyk20:59:59
The .5s tell us that something is up!
maybach21:00:01
Could that also be a problem on the last page of a state MC test?
rrusczyk21:00:03
Sure.
rrusczyk21:00:10
The AMC 10 problems we discussed tonight are a bit on the easy end of the AMC 10 course, and the AMC 12 problem was also on the easier end of the AMC 12 class.
potatopi21:00:28
was that a level AMC 12 problem?
rrusczyk21:00:36
That was an easier-level AMC 12 problem.
rrusczyk21:00:53
That's the last problem for tonight. I will stay and answer questions about the classes.
RubiksPro21:01:05
will the AMC 12 class cover mostly problems harder than this?
rrusczyk21:01:12
Yes, considerably in most cases.
AwesomeToad21:01:26
If I finish Introduction to Counting/Probability (including challenge problems), how will I be in shape for AMC10/12?
rrusczyk21:01:28
You should be fine on the counting problems.
potatopi21:01:36
so would it appear on an AMC 10 test?
rrusczyk21:01:37
Sure
junichi94@hotmail.com21:02:21
Can you explain the new AMC policy?
rrusczyk21:02:23
No; check the AMC website and ask on the AMC forum if you have questions. I don't know the ins and outs well enough to be able to explain every detail.
potatopi21:02:29
is a calculator required for the AMC 10 class?
rrusczyk21:02:42
No - it is discouraged since there is no calculator allowed on the AMC 10
EpicFailure21:03:16
i got a 117 on both the amc 10's so it this course good for me because i was thinking of doing the aime problem series
rrusczyk21:03:21
If you have worked on math since the AMC 10, my guess is that you are at the AMC 12 or AIME Problem Series level now. It's been a long time since the AMC 10.
AceOfDiamonds21:03:36
which class/book is suitable for improving AMC 12 score from 130 to 150?
rrusczyk21:03:38
The Intermediate classes and books, and the AIME Problem Series.
junichi94@hotmail.com21:03:56
Will the materials in the AIME class and the AMC12 class overlap?
rrusczyk21:03:58
The subjects will overlap some, but not the problems.
Wenghim21:04:09
Is WOOT good preparation for the AMC 12?
rrusczyk21:04:10
No, WOOT is well beyond AMC 12
EpicFailure21:04:22
Also i have a book "First step for Math Olympians" for amc 10 and 12 so if i finish that which course should i take
rrusczyk21:04:24
Move on to the Intermediate level of books and classes.
ChessGeek1221:05:14
What should you know for the AMC 10 class?
rrusczyk21:05:16
Be at the point where you think you can get at least 80 on the AMC 10, but below 130. If you're at 144 already, go on to harder stuff.
AIME15USAMO21:05:26
What level would the problems be in the AMC 12 course? What number range on AMC 12's?
rrusczyk21:05:27
Almost entirely 15-25
firepower42121:05:36
What do you use to make the geometry graphs?
rrusczyk21:05:44
Asymptote, usually. You can read about it on the AoPSWiki
potatopi21:05:49
does the homework work the same way in these classes as in the AMC 8 classe?
rrusczyk21:05:51
Yes
thunderbirdz21:06:04
how much longer is this clas
rrusczyk21:06:05
There is no more math, only question and answer.
siviba21:06:17
are the real classes going to go this fast?
rrusczyk21:06:18
No, somewhat slower.
chrome21:06:35
which class should i take if im not good at the problem we just did's topic?
rrusczyk21:06:37
Introduction to Counting & Probability would help.
junichi94@hotmail.com21:07:02
Are there any tests?
rrusczyk21:07:04
No tests
maxotter21:07:26
What books would be helpful (Aopswise) for the Basics class? I have intro to number theory and counting & probability.
rrusczyk21:07:28
Those and Intro Algebra. Maybe Intro Geometry when you're ready for a little harder stuff.
AwesomeToad21:07:41
Also, if I take 3 tests and get an average of 134 on AMC 10, do I need that? (Booh hoo got a 121.5 on 2001 AMC 10)
rrusczyk21:07:43
Move on to harder stuff. AMC 12 and AIME.
abhilash21:07:59
are calculators allowed on mathcounts tests?
rrusczyk21:08:00
Some parts; I'm not sure which.
hi how are you doing toda21:08:21
are calculaters allowed for the advanced?
rrusczyk21:08:23
We discourage calculator use with our classes. Do the math!
junichi94@hotmail.com21:08:33
Can we send emails to the instructor if we didn't understand something in class?
rrusczyk21:08:35
Use the class message board
potatopi21:08:38
will classes ever run late?
rrusczyk21:08:47
Occasionally, but usually only 5 minutes or so.
turak21:08:56
what class is best for amc series level geometry?
rrusczyk21:08:57
Intro Geometry.
vliu21:09:00
Will the AMC12 include any problems with logrithims and trigonometry
rrusczyk21:09:02
Yes
AIME15USAMO21:09:04
Are the intermediate books recommended before doing volume 2?
rrusczyk21:09:06
Yes
chrome21:09:12
should i take the intro to geometry class if i didnt really understand the problem we just did?
rrusczyk21:09:19
It would probably help.
mathq21:09:33
how much harder is amc 12 compared to amc 10?
rrusczyk21:09:35
Considerably harder, more advanced topics.
RubiksPro21:09:37
is it common for students to take the AIME problem series A and B at the same time?
rrusczyk21:09:47
This fall is the first time we have offered them at the same time.
abhilash21:09:55
is amc12 harder than MC?
rrusczyk21:09:57
Way harder
hi how are you doing toda21:10:01
is for the win good for mathcounts
hi how are you doing toda21:10:01
What is "for the win" anyway?
rrusczyk21:10:08
Yes, it helps with countdown round.
AwesomeToad21:10:20
How hard do you think the AMC 10 from 2001 was? And if I finish the Intermediate Series/Vol. 2, am I in good shape for AIME?
rrusczyk21:10:34
No idea how hard that test was. Yes, those books will help a lot with AIME.
AIME15USAMO21:10:43
Would it be recommended to do the intro series and/or the intermediate series before taking the AIME problem solving class?
rrusczyk21:11:05
Don't have to finish them all before that class, but you should have a lot of the materials down.
RubiksPro21:11:30
what is recommended work to do outside of class along with AMC 12 for improving on this?
rrusczyk21:11:31
Subject books in week areas and practice tests.
cheeseyicecream21:11:44
what subjects do you think are most emphasized in the amc's?
rrusczyk21:11:46
They balance pretty well usual.
maybach21:11:55
After Advanced Mathcounts/AMC8 which would help more for MC: Introduction to C&P or Intro to Geometry?
rrusczyk21:11:57
C&P
Wenghim21:12:30
What is the average AMC 12 score for those who have taken a class compared to those who haven't?
rrusczyk21:12:31
No idea -- we can't collect that data. The average for those who took it would be way, way higher due to selection bias.
potatopi21:12:33
would the AoPS volume 2 book be good prep for the AMC 10 class?
rrusczyk21:12:38
No, Volume 1 would.
faush10121:12:57
in class, what formulas are worth memorizing?
rrusczyk21:13:39
That's a question to ask on the class message board. It varies from class to class, and while the list is pretty short (understanding is more important than memorizing), we don't have time tonight to talk about what's worth memorizing or not.
MathTwo21:13:57
Will there be homework for the last class in the course?
rrusczyk21:14:10
In the AMC classes, yes, there will be message board problems after the last class.
firepower42121:14:15
We already did all the chapter and state mathcounts problems, which class should I take and will it help me?
rrusczyk21:14:26
If you understand them, then the Advanced MATHCOUNTS class.
RubiksPro21:14:56
are there topics that will be taught/explained, or only problems covering certain topics?
rrusczyk21:14:57
For the problem series, it will be almost totally problems, not concepts. The subject classes cover the concepts.
maybach21:15:11
Would Algebra II help with that last problem? I understood it, but would probably choke under the pressure in a contest.
rrusczyk21:15:16
Experience helps with choking. You know the math you need for that problem now.
Wenghim21:15:19
How much worse can I expect to score on the AMC 12 than the AMC 10?
rrusczyk21:15:28
Varies wildly. 20-30 points?
faush10121:15:31
when is the real AMC 8 test, usually?
rrusczyk21:15:33
November
siviba21:15:37
is AMC 10 harder the\an mathcounts
rrusczyk21:15:39
Yes
vliu21:15:48
I noticed that in MC they tend to repeat some questions. Why?
rrusczyk21:16:04
They have different writers over time. They can't all know what questions have been asked.
MathTwo21:16:09
I saw several AMC10 problems put on AMC12s, and vice versa. How is AMC12 yhat much harder?
rrusczyk21:16:21
Because the problems on AMC 12 but not AMC 10 are much harder.
potatopi21:16:25
if i have considerable trouble with the AoPS Volume 2, should i take AMC 10 class?
rrusczyk21:16:45
That would probably be a good place to start.
faush10121:16:47
Is AMC 12 harder than WOOT? What is WOOT for?
rrusczyk21:16:58
WOOT is for the AIME and national olympiads. It is way harder than AMC 12
AIME15USAMO21:17:04
What books or classes are helpful beyond the intermediate series and volume 2?
rrusczyk21:17:12
Art and Craft of Problem Solving by Paul Zeitz.
rrusczyk21:17:20
Problem Solving Strategies by Arthur Engel.
rrusczyk21:17:30
After those, any of the Titu Andreescu books.
rrusczyk21:17:39
We also have some good British books on our site.
AwesomeToad21:17:41
is there a lot of C&P and Geometry on AIME?
rrusczyk21:17:43
yes
Cliu030121:17:45
is the chat transcript available after the end of this Math Jam? I missed part of it...
rrusczyk21:17:52
yes; in the Math Jams section.
potatopi21:18:03
will we ever get whole practice tests in the class?
rrusczyk21:18:10
No; there are many on the site.
dinapro21:18:18
are 8th grade alg. 1 students eligible to participate in amc8?
rrusczyk21:18:21
Absolutely.
aunch21:18:33
What does WOOT stand for
aunch21:18:33
What grade is WOOT for
Wenghim21:18:38
What percent of AMC 10 and AMC 12 problems are the same?
rrusczyk21:18:43
Around 25 or 30?
Cliu030121:18:47
What concepts should I know for AMC 10?
rrusczyk21:19:05
Take a look at the tables of contents for our Introduction level books. Those are the concepts.
EpicFailure21:19:11
where are the practice tests
rrusczyk21:19:19
Look in the AMC forum and on the wiki.
AwesomeToad21:19:38
Sorry for asking so many questions, but how much AIME-level counting/probability is covered in Introduction to C&P. I have a lot less time than before.
rrusczyk21:19:40
Some, but most of the AIME counting is in Intermediate Counting.
mathq21:19:45
in amc 12, which typically is harder ? test a or b
rrusczyk21:19:48
Random
Cliu030121:19:50
How difficult are the problems on AMC 10 compared to Math Counts? Chapter, School, State, Nationals...?
rrusczyk21:20:08
Roughly Nationals maybe? Varies a lot from year to year.
dinapro21:20:10
where can i get questions from previous amc tests?
rrusczyk21:20:22
Books of them are available on our site. There are some on the forum, too.
AIME15USAMO21:21:02
AoPS books consist of the introduction series, and the intermediate series. To me, introduction means "beggining", and intermediate means "middle", are you considering a third level of subject books? Something like "the advanced series"?
rrusczyk21:21:08
Maybe someday. Not soon!
vliu21:21:21
Why test a/b every year?
rrusczyk21:21:32
Lots of schools have a week off in February.
jeffreyyan821:21:36
what is the difference between problem set a and problem set b
jeffreyyan821:21:36
the class there are 2 classes A and B
rrusczyk21:22:04
Ah, you mean the AIME Problem Series -- the only difference is that there are different problems in the two. They are the same difficulty and roughly same subjects.
vliu21:22:06
do you have to take both a and b?
rrusczyk21:22:16
No, but you can if you want. They take your higher score.
AwesomeToad21:22:23
If I study the later chapters (e.g.17) of intro to geometry what shape am I in for AIME-level Euclidean Geometry?
rrusczyk21:22:44
If you master the Intro Geometry, you should be able to do at least half the non-trig geometry on the AIME.
AIME15USAMO21:22:48
Would taking A and B help you or would it be repetetive? (AIME course)
rrusczyk21:23:01
It would help, but the Intermediate level classes would probably help more.
Cliu030121:23:45
Can some AMC 10 problems be solved using trigonometry? (if so, yay for me)
rrusczyk21:23:46
Yes, but I think they try to avoid having problems that are made easier with trig (that is, the trig might be the hard way)
AIME1521:23:48
What book(s) would you recommend (besides Geometry Revisited) for Geometry study past Intro. Geometry?
rrusczyk21:23:59
There's a British book on our site that's decent.
rrusczyk21:24:07
Challenging Problems in Geometry also has some nice problems.
rrusczyk21:24:23
That's it for the Math Jam tonight. If you have any more questions about the classes, you can write me at classes@artofproblemsolving.com
