| Transcript
for the Math
Jam "AoPS Classes Math Jam"
on Oct 10. |
| Math Jam hosted by Valentin Vornicu
(Valentin Vornicu ). |
DPatrick19:29:21
Hello, and welcome to another Art of Problem Solving Fall 2007 Classes Math Jam!
DPatrick19:29:26
Tonight we will be discussing the AMC 10 and AMC 12 Problem Series classes.
DPatrick19:29:35
My name is Dave Patrick, and I am an instructor at the AoPS Online School.
DPatrick19:29:43
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.
DPatrick19:29:55
The classroom is moderated: students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. This helps keep the class organized and on track.
DPatrick19:30:10
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.
DPatrick19:30:17
Also, only moderators can enter into private chats with other people in the classroom.
DPatrick19:30:32
Note that it is not possible for the instructor to personally respond to every comment that you submit -- please do not take it personally if your comment is not posted or responded to! I will try to respond to all questions to the extent that I can.
DPatrick19:30:56
As I mentioned, today we'll be discussing our two AMC Problem Series classes that we'll be running this fall and winter.
DPatrick19:31:07
I'll start with the AMC 10 class.
DPatrick19:31:16
The AMC 10 class starts on October 24, and meets every Wednesday from 7:30-9:00 PM Eastern. The class meets for 12 weeks and ends on February 6. (There are no classes on November 21 nor December 19-January 2.) The course is designed to cover a large portion of the curriculum tested on the AMC 10 exam.
DPatrick19:31:31
This class is a Problem Series class, meaning that the major focus of the class will be working through various AMC problems.
DPatrick19:31:41
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. (As a result, these classes are somewhat less expensive that our regular subject classes.)
DPatrick19:31:59
The following problems are excerpts of a couple of the areas of problem solving covered in the AMC 10 Problem Series.
DPatrick19:32:12
DPatrick19:32:28
How can we use the given information to find the number of sides of the convex polygon?
Altheman19:32:58
We should consider the sum of the angles...
DPatrick19:33:16
Right. How can we compute the sum of the angles of the polygon?
Quickster9419:33:45
180(n-2)
3.14159man19:33:59
A convex polygon's interior measure= 180(n-2)
DPatrick19:34:00
Yes: conveniently, we have a formula for the sum of the interior angles of a convex n-sided polygon: 180(n - 2).
DPatrick19:34:13
How do we relate this to the arithmetic progression data given in the problem?
DPatrick19:34:48
In other words: once we calculate the sum of the arithmetic progression we can set it equal to 180(n - 2). But how do we calculate the sum?
3.14159man19:34:58
well the largest measure is 160, each successive is 5 smaller..
DPatrick19:35:15
OK, so what then is the measure of the smallest angle?
3.14159man19:35:51
160- 5(n-1)
DPatrick19:36:05
Right. A couple people said 160-5n, and it's easy to make that mistake.
DPatrick19:36:18
To get down to the smallest angle, we need to subtract 5 from 160 n-1 times.
DPatrick19:36:24
So the smallest angle is 160-5(n-1).
DPatrick19:36:38
So we have the sequence 160, 155, ..., 160-5(n-1). What is its sum?
edwarddinh19:37:15
((320-5(n-1))n/2
masonliang19:37:30
the largest angle is 160 and the smallest is 160-5(n-1) so the sum is {160+160-5(n-1)}n/2
DPatrick19:37:50
Good. An easy way to sum an arithmetic series is to take the average term -- in this case, it's (160+160-5(n-1))/2 -- and multiply by the number of terms -- in this case, n.
DPatrick19:38:09
So, simplifying a bit, we get (320-5(n-1))n/2.
DPatrick19:38:30
So what do we do now?
edwarddinh19:39:01
set that equal to 180(n-2)
masonliang19:39:07
set that equal to (n-2)180
DPatrick19:39:22
Exactly. We set that sum equal to 180(n-2), which is also the sum of the angles (by our formula).
DPatrick19:39:39
I'll save us a little time and tell you that when we simplify, we get:
DPatrick19:39:44
masonliang19:39:59
A:9
cascadee19:39:59
n=9
DPatrick19:40:08
Right.
We can solve the quadratic in a number of ways including factoring it into
(n - 9)(n + 16) = 0, so n = 9 or -16.
DPatrick19:40:15
We know that a polygon cannot have a negative number of sides and so we have our answer, n = 9. (A).
DPatrick19:40:28
This problem shows us that we can often draw upon information not directly mentioned in the problem to help us create an equation to solve. Of course, we also needed to sum an arithmetic series to give us the second side of this equation.
DPatrick19:40:50
Here's another sample problem from the course:
DPatrick19:40:58
DPatrick19:41:16
How can we count the three-digit numbers with digits in increasing or decreasing order?
3.14159man19:41:52
anything with two of the same number in any two spots doesn't work..
DPatrick19:42:10
That's a useful observation: all such numbers must have 3 different digits.
DPatrick19:42:22
How can we use that to help us?
cognos59919:42:43
case by case analysis
DPatrick19:42:50
That's a good idea: what are the different cases?
Lawrence Wu19:43:41
increasing and decreasing order
DPatrick19:43:57
Good. We can count the numbers in increasing order, then count the numbers in decreasing order.
DPatrick19:44:10
Let's go back to our observation that each number has 3 different digits.
DPatrick19:44:29
Suppose we have three different digits such as 2, 4, and 7. We can arrange them in increasing order in exactly one way (247) and in decreasing order in exactly one way (742).
DPatrick19:44:49
So does this mean that the answer is the 2*(the number of ways to choose 3 different digits)?
cascadee19:45:15
no
cascadee19:45:15
because 210=/=012
edwarddinh19:45:15
no because if theres a zero than theres only the decreasing choice
DPatrick19:45:31
Right -- we've got a problem if 0 is one of our digits, since a 3-digit number can't start with 0.
DPatrick19:45:43
So indeed we have to do the two cases separately.
DPatrick19:45:59
Let's begin with three digit numbers with increasing digits. How many are there going to be?
DPatrick19:46:23
As we mentioned, we can only choose from digits 1 through 9 (no 0).
cognos59919:46:48
9 choose 3?
edwarddinh19:46:48
84
Quickster9419:46:48
84 of them
DPatrick19:46:56
DPatrick19:47:08
(Don't worry if you don't know this notation -- it will be covered in the class.)
DPatrick19:47:21
Now, how many three digit numbers will have all their digits in decreasing order?
Altheman19:47:36
i) increasing: 9 digits, choose 3
ii) decreasing: 10 digits, choose 3
Lawrence Wu19:47:37
10 choose 3
DPatrick19:47:51
Right -- now we're allowed to choose 0, so we have 10 digits to choose from.
DPatrick19:47:58
DPatrick19:48:14
What is our final answer?
Quickster9419:48:26
so u add them up and 204
edwarddinh19:48:26
204
360_Fan19:48:27
204
DPatrick19:48:31
We now add the total numbers from each case:
84 + 120 = 204, so our answer is (C).
DPatrick19:48:40
The key to this problem was recognizing that the case for increasing digits is different from the case for decreasing digits. When solving combinatorial problems it is necessary to consider exactly how we are choosing members of a group. When the methods are different we must set up cases and calculate them separately.
DPatrick19:49:14
These questions are roughly medium-level difficulty of the problems that will be considered in the class.
DPatrick19:49:24
The class will be taught by Naoki Sato. He won first place in the 1993 Canadian Mathematical Olympiad, and represented Canada at the 1992 and 1993 International Mathematical Olympiads, winning a bronze and silver medal, respectively. He has also served as deputy leader for the Canadian IMO team in 1997, 2002, and 2006.
DPatrick19:50:07
Now I'll discuss the AMC 12 class. (I'll take questions at the end, since the questions for both classes are likely to be similar.)
DPatrick19:50:22
The AMC 12 class starts on October 22, and meets every Monday from 7:30-9:00 PM Eastern. The class meets for 12 weeks and ends on February 4. (There are no classes on November 19 nor December 17-31.) The course is designed to cover a large portion of the curriculum tested on the AMC 12 exam.
DPatrick19:50:39
Just like the AMC 10 class, the AMC 12 class is a Problem Series class, 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. (As a result, these classes are somewhat less expensive that our regular subject classes.)
DPatrick19:50:57
The following is an excerpt of one of the areas of problem solving covered in the AMC 12 Problem Series.
DPatrick19:51:06
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
panjia12319:51:37
graph it
DPatrick19:52:00
How do we graph the region that it closer to one point (0,0) than another (3,1)?
DPatrick19:52:05
...is closer...
yanglunj19:52:40
Draw perpendicular bisector of line connecting the points
Lawrence Wu19:52:54
a line perpendicular to the line formed by those two points that intersects it at the midpoint?
DPatrick19:53:27
Good -- the perpendicular bisector of the segment between (0,0) and (3,1) is the set of points an equal distance from the two points. So the region we want will be on the side this bisector that's closer to (0,0).
DPatrick19:53:35
Here's a picture:
DPatrick19:53:41
DPatrick19:54:07
Our rectangular region is shown in red. The slanted line is the perpendicular bisector of the segment from (0,0) to (3,10.
DPatrick19:54:08
(3,1).
DPatrick19:54:18
So what do we need to do now?
360_Fan19:54:36
See the ratio it divides the rectangle into
DPatrick19:55:03
Right. We need the ratio of the area of the grey shaded area to the area of the rectangle. That will be the answer.
Quickster9419:55:14
so now we just find the area of that trapazoid, 1 1/2, and divide by the area of the rectangle, 2, and get 3/4?
360_Fan19:55:14
Which is 3/4
Lawrence Wu19:55:14
3/4
DPatrick19:55:23
How did you get that the area of the grey region is 3/2?
yanglunj19:55:58
Second square is divided in half because it passes througgh midpoint
masonliang19:55:58
it's height is 1 and it's side lengths average to 1/2
Lawrence Wu19:55:58
b/c the mid point of that line earlier was (1.5,0.5)
360_Fan19:55:58
The line divides the right unit square in half
DPatrick19:56:33
Right. We could compute the equation of the bisector, or we could observe that since the bisector passes through (1.5,0.5), which is the center of the square, then it divide the square in half.
DPatrick19:56:43
To sum up: The total rectangular region has area 2.
The area of the region close to the origin than (3, 1) within that rectangle is 3/2.
Our probability is (3/2)/2 = 3/4. Our answer is (C).
DPatrick19:57:02
As we just saw, probability problems are sometimes geometry problems and this problem gives us a direct view into using coordinates to help us with such problems.
DPatrick19:57:14
Let's do one more that's similar:
DPatrick19:57:20
DPatrick19:57:46
How can we begin determining the area of R?
Altheman19:58:27
Well the function notation is not that easy to deal with so we should look at the inequalities in terms of x and y...
cascadee19:58:27
Try to visualize/draw it?
DPatrick19:58:42
Yes -- we'll want to try to graph it, but it's hard to work with in this form.
DPatrick19:58:52
So let's write the inequalities in terms of x and y.
DPatrick19:58:56
DPatrick19:59:07
What can we do with these inequalities?
panjia12319:59:35
They're inequalities with ellipses sort of...
cascadee19:59:35
Turn them into equations of ellipses?
Altheman19:59:35
The first one looks like a completing the square problem...
DPatrick19:59:53
These inequalities represent graphical regions and our job is to determine the intersection of those regions. We can best work with the inequalities by organizing terms.
DPatrick20:00:09
In the first inequality we can complete the square to get it into something a little more recognizable, like a circle.
DPatrick20:00:22
DPatrick20:00:42
(Again, I'm omitting the algebra today to save time -- we'd do it more step-by-step in the class.)
DPatrick20:00:48
What is this the graph of?
edwarddinh20:01:04
a circle
360_Fan20:01:04
a circle
panjia12320:01:04
a circle with center (-3,-3) and radius 4.
ChaosTheory20:01:17
Circle centered at (-3, -3) with radius 4.
masonliang20:01:17
a circle with it's center at (-3,-3) and radius 4
panjia12320:01:17
that is, the interior of it
DPatrick20:01:32
Exactly. This region is the interior of a circle centered at (-3,-3) and with radius 4.
DPatrick20:01:43
Let's file that away for now, and look at the other inequality.
DPatrick20:01:52
DPatrick20:01:59
What can we do with this?
360_Fan20:02:24
This is a hyperbola, right?
whylime20:02:24
maybe another circle
Goldentide20:02:24
complete the square
phs194520:02:25
1's cancel each other out
DPatrick20:02:39
Lots of suggestions...let's cancel the 1's first since that's quick and easy.
DPatrick20:02:48
panjia12320:02:53
factor
DPatrick20:02:58
How can we factor it?
Goldentide20:03:32
x(x+6)-y(y+6)
360_Fan20:03:32
Factor out x's and y's to get x(x + 6) - y(y + 6)
DPatrick20:03:42
That's a possibility, but it's still hard to work with in that form.
phs194520:03:49
x^2-y^2 is (x+y)(x-y)
masonliang20:03:49
(x-y)(6+x+y)
edwarddinh20:03:49
(x+y+6)(x-y)
DPatrick20:03:56
Bingo.
DPatrick20:04:03
DPatrick20:04:10
What's the graph of this?
DPatrick20:04:54
Well, what's the graph of x-y?
Goldentide20:05:11
line
panjia12320:05:11
y=x
edwarddinh20:05:11
a line
DPatrick20:05:21
And what's the graph of the other term?
edwarddinh20:05:31
another line
Goldentide20:05:31
a line
cascadee20:05:31
another line
DPatrick20:05:34
Right.
DPatrick20:05:47
So the graph of their product is two lines.
panjia12320:05:51
that means one of the quantities is negative and one is positive
x-y<=0 and x+y+6>=0
or
x-y>=0 and x+y+6<=0
DPatrick20:05:57
Exactly.
DPatrick20:06:18
So to put it all together, the intersection of the two inequalities looks like this:
DPatrick20:06:25
Altheman20:06:31
We are happy to notice that both of these lines pass through (-3,-3), which is the center of our circle.
DPatrick20:06:37
DPatrick20:06:41
Indeed, that's very helpful!
DPatrick20:06:54
What is the shaded area?
360_Fan20:07:09
half of the circle
DPatrick20:07:28
Right. The lines are perpendicular (one has slope 1, the other has slope -1), and meet at the center of the circle.
DPatrick20:07:37
So each quadrant is 1/4 of the circle, and the shaded area is 1/2 of the circle.
edwarddinh20:07:50
8pi
yanglunj20:07:50
8pi
yanglunj20:07:51
8 * 3.1415926535897...
DPatrick20:08:01
Yes, the area we want is 8*pi.
edwarddinh20:08:15
so the answer is E?
cascadee20:08:18
so the answer is 25?
DPatrick20:08:24
The area is 8pi and using 3.14 as an approximation we get 8(3.14) = 25.12, so the answer is (E).
DPatrick20:08:33
Most students who can score well on the AMC 12 are familiar with relating equation and graphs, but fewer are familiar with relating inequalities and graphs. Growing confident with separating regions defined by an inequality helps in evaluating this kind of problem confidently. Once we deduced the locations and shaped of the regions, the rest was easy.
DPatrick20:08:45
The first problem we did was medium difficulty.
DPatrick20:08:51
The second problem was very hard. :)
DPatrick20:09:04
The class will be taught by Valentin Vornicu. Valentin was a Gold Medal and Special Award winner at the Junior Balkan Math Olympiad in 1998 and a Silver Medal winner at the Balkan Math Olympiad in 2001. He also took 2nd Prize at the IMC in 2003 and 2004. He won first or second prize in the Romanian National Olympiad 1997-2002, and was a member of the Romanian International Mathematical Olympiad (IMO) team in 2001 and 2002, winning a bronze medal in 2002. Valentin has coached numerous successful Olympiad students in Romania and abroad and authored the widely used Romanian text "The Math Olympiad, from Challenge to Experience". He is also one of the instructors of the Romanian IMO Team since 2003, and proposed a number of problems at the different math Olympiads.
DPatrick20:09:24
So that's a sampler of both the AMC 10 and AMC 12 classes.
DPatrick20:09:29
Are there any questions about the classes?
ChaosTheory20:09:57
If you can solve some of the harder AMC 10/12 problems, will you be OK for some of the AIME as well?
DPatrick20:10:18
Yeah -- the easiest AIME problems are roughly equal in difficulty to the hardest AMC 10/12 problems.
The Visionary20:10:42
Are AMC 12 MCQ-type?
DPatrick20:10:45
Multiple choice? yes.
Quickster9420:10:54
how much harder is AIME than AMC-12?
DPatrick20:11:03
Much harder.
Quickster9420:11:26
wat exactly are the scoring rules this year, cuz they changed again didn't they?
DPatrick20:11:45
Check with the AMC. I don't know for sure. I'm here mainly to talk about our classes, not the contests themselves.
360_Fan20:11:51
Calculators aren't allowed now, right?
DPatrick20:11:55
That is correct.
unimpossible20:11:58
will we need to know the square root algorithm because calculators are no longer allowed?
DPatrick20:12:09
No. No problem will require that much precision with square roots.
phs194520:12:14
what is the difference between AMC10 and AMC12?
DPatrick20:12:28
AMC 12 has trig, advanced algebra, and formal geometry.
AMC 10 only has basic algebra, basic geometry, and intro-level number theory and combinatorics (counting & probability).
The two tests typically have about half of their questions in common.
ChaosTheory20:12:46
Will using AoPS Volume 2 prepare you adequately for AMC 10 and 12?
DPatrick20:13:14
Volumes 1 and 2 cover just about anything you'd find on the AMC 10 or 12, and a lot of AIME-level material as well.
masonliang20:13:29
how much is usually needed on the AIME to pass on to the USAMO?
DPatrick20:13:41
It varies by year. Check the AMC for historical data. www.unl.edu/amc
Lawrence Wu20:13:47
what grade level are they? can 7th graders take it?
DPatrick20:14:16
The AMC 10 is "designed" for grades 9 and 10, the AMC 12 for high school. But anybody can take them (except, of course, 11th and 12th graders can't take the AMC 10).
phs194520:14:24
can you only take one?
DPatrick20:14:35
The tests are offered on two dates. You can take one test on each date.
yanglunj20:14:37
So will we need to memorize trig tables now?
DPatrick20:14:56
No. For the AMC 10, there's no trig at all. For the AMC 12, you should know 30-60-90 and 45-45-90 triangles.
Quickster9420:15:09
wat do u need to score to get into the AIME, is it based on score or percentile?
DPatrick20:15:25
Both. AMC 10 takers need 120 or to be in the top 1%, whichever is larger.
DPatrick20:15:33
AMC 12 takers need 100 (out of 150) or to be in the top 5%, whichever is larger.
DPatrick20:15:42
Check with the AMC for past years' statistics.
cascadee20:15:48
When would you suggest to read "Art and Craft of Problem Solving"?
DPatrick20:15:58
When you're starting to look at the USAMO as a legitimate possibility.
Lawrence Wu20:16:04
i saw something called AHSME in the AoPS books what is it?
DPatrick20:16:14
That's the name of the AMC prior to 2000: the American High School Math Exam.
3.14159man20:16:20
What's the highest score for AMC's 10 and 12?
DPatrick20:16:21
150
The Visionary20:16:27
What are the best books to prepare for the AIME then?
DPatrick20:16:32
AoPS books, of course! :)
DPatrick20:16:40
See our website for specific recommendations.
ChaosTheory20:16:45
Is it good chance that if we do well in these AMC 10/12 courses, then we can pass on to AIME?
DPatrick20:17:19
Yes...most students who take these course are near the AIME-qualification level, and one of the goals of these courses is to get you over that hump and onto the AIME>
ChaosTheory20:17:41
Do you suggest buying the past problem books and "First Steps for Math Olympians" as a supplement to AoPS Volume I and II if you're doing AMC10/12?
DPatrick20:18:07
That's up to you -- there's a lot of overlap. Ask people on the message board who might have used them what they think. It's hard for me to give an unbiased answer.
phs194520:18:20
will there be homework from the class?if so, how many per week?
DPatrick20:18:28
Roughly 10-20 problems per week will be posted on the message board.
DPatrick20:18:47
You don't need to do them all (or any of them in fact), and you don't turn in your solutions.
Lawrence Wu20:18:54
how much trig is there do we need to know inverse trigonometric functions?
DPatrick20:19:21
Just the basics -- you don't need to memorize a trig table, but you should know, for example, when cosine is positive and when it's negative (again, for the AMC 12 only; there's no trig on the 10).
Altheman20:19:26
What changes do you expect in the types of problems considering the banning the use of calculators on the AMC 10/12?
DPatrick20:19:33
I expect hardly any changes whatsoever.
Quickster9420:19:42
will the homework be similar to the mathcounts series?
DPatrick20:19:52
Yes -- if you've taken that class, then the way the course will run is very similar.
phs194520:20:02
do you think taking both Intermediate Counting and Possibilities and AMC 12 class will be hard?
DPatrick20:20:15
Not necessarily -- the AMC 12 class is not nearly as time-consuming as our subject classes.
whylime20:20:19
Where can we find the message board?
DPatrick20:20:46
You'll get instructions when you enroll in the class. (If you're already enrolled, visit the "My Classes" page by clicking on "My Classes" on the left side of the web page, and look for further information there.)
phs194520:20:53
do you turn in the problems?
wgpark092420:20:53
Are you going to grade the problems?
DPatrick20:20:54
No and no.
yanglunj20:21:09
What level problem is the USAMTS compared to AMC 12, AIME, or USAMO
DPatrick20:21:33
It's not really comparable to AMC or AIME since the problems are proof-style. But USAMTS is much easier than USAMO.
ChaosTheory20:21:42
Do the AMC 10/12 classes have roughly the same level of geometry as offered in your Introduction to Geometry class?
DPatrick20:21:48
Yes, that's about right.
Quickster9420:21:59
which do u think is better, taking on of each, or two of the same contest?
DPatrick20:22:39
Check the message boards -- there is a lot of debate about this. I don't really have an strong opinion. If you don't know trig, then take the 10 twice. If you do, I'd say take the 10 first, and if you qualify for the AIME, then take the 12 for practice.
ChaosTheory20:23:22
So if we take the problem class and use the Introduction to Geometry textbook and AoPS Volume I and II books, will we be set with regard to geometry for the AMC 10/12?
DPatrick20:24:01
I'm not comfortable with the word "set", but you'll have the required knowledge of geometry facts, and you should have a lot of problem-solving practice too.
Quickster9420:24:05
is there still time to register for test B after getting A scores?
DPatrick20:24:18
I have no information about the AMC's registration procedures -- check with them directly.
ChaosTheory20:24:25
What is the rough individual subject breakdown for the AMC 10/12? How much of it is counting, how much is number theory, etc.?
DPatrick20:24:33
The week-by-week syllabi are posted on our website.
DPatrick20:24:55
Click "Online classes", then "Class List/Enroll", then click on the class you want.
The Visionary20:25:05
Is the book art and craft of problem solving at the level of AMC 12, AIME, USAMO or IMO?
DPatrick20:25:11
Advanced AIME through IMO.
3.14159man20:25:43
Do you have to enroll through your school or can you enroll by yourself?
(to take the tests)
DPatrick20:25:49
Through your school.
DPatrick20:26:15
Again, I don't have a lot of knowledge about the AMC's procedures -- they are a separate organization. Check with them directly with registration questions.
ChaosTheory20:26:22
If you are homeschooled, how can you take the AMC 10/12?
DPatrick20:26:40
The AMC has procedures for homeschool administration of the contests. Check with them.
DPatrick20:26:48
www.unl.edu/amc
Quickster9420:27:00
so r the fees paid by u or the school?
DPatrick20:27:11
Almost always the school. Check with your school.
3.14159man20:27:21
so would taking the amc12 course help you prepare for the amc10 or not?
DPatrick20:27:44
Yes. The two contests have a lot of questions in common each year.
DPatrick20:28:00
But you'd also do stuff in the AMC 12 class -- like trig problems -- that would not show up at all on the AMC 10.
The Visionary20:28:09
Are USAMO of the same level to that IMO?
DPatrick20:28:12
Roughly, yes.
ChaosTheory20:28:26
Around when is the AIME? Can't find it on the website.
DPatrick20:28:29
March.
phs194520:28:39
if i'm taking it for the first time and i'm in Math Analysis a class right now.. should i take both AMC 10 and 12?
DPatrick20:28:55
I don't know what that class is. Ask on the message board -- there are a variety of opinions on this subject.
phs194520:29:49
will the class go over the homework problems?
DPatrick20:29:57
No -- they'll be discussed solely on the class's message board.
ChaosTheory20:30:06
How much extra preparation is needed for the AIME after you've done AMC 10/12?
DPatrick20:30:26
None is "needed" -- it depends how much you want to do.
DPatrick20:30:55
We will offer a special weekend class in late February/early March (between the AMC and the AIME) for students who want to do some last-minute AIME preparation.
3.14159man20:31:04
is there a textbook associated with teh amc10 class? like the number theory and geometry classes
have a book that we can read on the side..
DPatrick20:31:27
No, there's no book. Many students and top math teams use our Introduction series of texts and our Art of Problem Solving Volumes 1 to prepare for the AMC 10.
phs194520:32:00
Is volume 2 for AMC 12 and higher?
DPatrick20:32:04
Generally, yes.
Quickster9420:32:37
so if u master everything in book 1 u should be set for amc-10?
DPatrick20:32:55
Depends what you mean by "master", but Volume 1 has most (if not all) of the material you'd find on the AMC 10.
DPatrick20:33:32
OK, that's all the time we have tonight. You can always post your questions on the forum if you have further questions about the AMC classes or the contests themselves.
DPatrick20:33:35
Thanks for coming!
