| Transcript
for the Math
Jam "AoPS Classes Math Jam"
on May 27. |
| Math Jam hosted by rrusczyk
(Richard Rusczyk ). |
rrusczyk19:26:18
Hello, and welcome to an Art of Problem Solving Math Jam. Today we'll be discussing the AMC 10 Problem Series, the AMC 12 Problem Series, and the AIME Problem Series A. We will go through a couple example problems from each class, and discuss both what these classes cover and how they work.
rrusczyk19:26:33
My name is Richard Rusczyk. I founded Art of Problem Solving and have written several Art of Problem Solving textbooks.
rrusczyk19:26:45
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.
rrusczyk19:26:53
The classroom is moderated: students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read. Also, only moderators can enter into private chats with other people in the classroom.
rrusczyk19:27:19
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.
eshen19:27:33
Who are the moderators?
rrusczyk19:28:00
The instructor and the assistant instructors. Tonight, I am the primary instructor, and nsato may be assisting me as needed. (I'll be introducing him later.)
rrusczyk19:28:22
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. You do not need to use LaTeX if you don't want to.
rrusczyk19:28:30
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
rrusczyk19:28:37
rrusczyk19:28:42
This message will look like this when posted in the classroom:
rrusczyk19:28:47
rrusczyk19:29:03
Just remember, if your post uses LaTeX, use the semicolon (;) to begin your post!
rrusczyk19:29:18
In this Math Jam, I will briefly describe a course, then go through a few example problems. Then, I will hold a question-and-answer session about that class.
Ruby509999119:29:21
is the order going to be AMC10 AMC12 and then AIME?
rrusczyk19:29:24
Yes.
ChocolateChipCookie19:29:47
what about amc8
rrusczyk19:29:48
You can take the MATHCOUNTS Problem Series for that.
Geliva19:29:59
What if we want to combine text and LaTex? Can we do that in the same post?
rrusczyk19:30:28
Sure. If I write,
;This is $2+2=4$
I get:
rrusczyk19:30:34
rrusczyk19:31:12
I'll take questions about LaTeX later. For now, I'd like to focus on the math classes.
rrusczyk19:31:19
You do not need LaTeX for the class.
rrusczyk19:31:46
AMC 10
rrusczyk19:31:54
The AMC 10 class starts on June 5, and meets every Thursday from 7:30-9:00 PM Eastern. The class meets for 12 weeks and ends on August 21. The course is designed to cover a large portion of the curriculum tested on the AMC 10 exam.
rrusczyk19:32:31
This 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.
rrusczyk19:32:40
This class is taught by Sean Markan, an MIT graduate who was invited to the Math Olympiad Program in 2001 and was a member of the US Physics Team in 2000 and 2002.
ChocolateChipCookie19:33:01
if i am traveling, is it possible for me to do the class
rrusczyk19:33:03
Yes! Full transcripts are posted for each class, so you can review *everything* that happens in class if you miss it.
isabella229619:33:49
What kind of things will be covered?
rrusczyk19:34:19
We will cover all the topics that commonly occur on the AMC 10.
rrusczyk19:34:21
Here is a list:
mathgenius19:35:13
If I am in Algebra II and in 8th grade, should I take the class on the AMC 10 or 12?
rrusczyk19:35:19
I recommend the AMC 10 class.
tsai_daniel19:35:35
How are AMC 10, AMC 12, and AIME problems proposed?
rrusczyk19:35:38
The AMC has committees of mathematicians and teachers who produce them.
andrewnycpops19:35:58
If you got a 96 on the 10 last year which course should i take
rrusczyk19:35:59
AMC 10
Jessie Shin19:36:31
are there seperate classes for each of the AMCs?
rrusczyk19:36:41
Separate classes for the AMC 10 and for the AMC 12.
BoxinaJack19:36:45
If we're in a different country, can we access the classroom?
rrusczyk19:36:49
Yes.
mahi9819:37:05
If you are reviewing past classes,can we still ask questions?
rrusczyk19:37:06
Yes. There is a class message board for questions you have at any time.
vvipul19:37:42
so these classes are sort of like a "bridge" between the introduction and intermediate classes?
rrusczyk19:37:43
You can think of them that way, or as something relatively separate. You don't have to take these between the Intro and Intermediate classes.
ChocolateChipCookie19:38:24
how would we take the test if our school doesnt have the test what do i do
rrusczyk19:38:38
Contact the AMC - they will help you. See their website: www.unl.edu/amc
isabella229619:38:42
Are there any tests or exams?
rrusczyk19:38:49
Not in the Problem Series classes.
Lulze19:38:52
Is there a deadline / late fee for signing up past a certain date?
rrusczyk19:39:03
The deadline is roughly 10 days after the first class. There is no late fee.
Geliva19:39:22
Is it useful to take both the AMC 10 class and the AMC 12? I know the tests are very similar, but I would like as much practice as possible.
rrusczyk19:39:38
Yes, particularly if you are taking both.
vvipul19:39:48
Is it true that it's easier to make AIME on the 12 than the 10? I've heard that in several places.
rrusczyk19:39:57
Tough to say. We recommend taking both if you can.
hello12319:40:02
are we going to do some sample problems?
rrusczyk19:40:10
Let's do some now, and then I will answer more of your questions.
rrusczyk19:40:16
The following are excerpts of a couple of the areas of problem solving covered in the AMC 10 Problem Series.
rrusczyk19:40:21
ARITHMETIC SEQUENCES
rrusczyk19:40:35
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.
rrusczyk19:40:50
We'll now discuss a problem:
rrusczyk19:40:51
rrusczyk19:41:29
Where do we start?
nikki19:42:07
work backwards from 160
iloveham4719:42:07
from 160deg and sub 5 going down
rrusczyk19:42:18
OK, the angles are 160, 155, 150, . . .
rrusczyk19:42:25
What is the smallest angle in terms of n?
BoxinaJack19:43:14
160 - 5(n-1)
abshin382219:43:14
160-5(n-1)
abshin382219:43:14
160-5(n-1)
bisbmard919:43:14
160-5(n-1)
ppzuan19:43:14
160-5*(n-1)
rrusczyk19:43:19
rrusczyk19:43:41
Remember that the last value should be 160 - 5(n-1), not 160 - 5n because we started with 160 in the list.
rrusczyk19:43:59
OK, now what?
Lolligal10119:44:10
well, the sum of the interior angles is 180(n-2)
nikki19:44:10
there are a total of (n-2)180 degrees in the polygon
Geliva19:44:10
It is also useful to determine the total sum of the interior angles. This will be the sum of our sequence. In this case, 180(n-2)
rrusczyk19:44:26
Aha! And what will we do with this?
Ruby509999119:44:39
form an equation with n
nikki19:44:39
write an equation
istos19:44:39
set up an equation with the number of sides and interior angle sum equation
ppzuan19:44:39
sum of arithmetic series = sum of interior angles
isabella229619:44:41
write equation
rrusczyk19:45:14
Great. The sum of our angles must equal 180(n-2). What is the sum of that arithmetic sequence we found for the angles?
Jakey19:46:20
(325-5n)n/2
abshin382219:46:20
(320-5(n-1))n/2
hello12319:46:20
(325n-5n^2)/2
rrusczyk19:46:35
rrusczyk19:47:03
How can we simplify that sum on the right?
hello12319:47:44
use the formula for sum of consecutive numbers
vvipul19:47:44
n(n-1)/2
tsai_daniel19:47:44
1 + 2 + ... + (n - 1) = 1/2(n - 1)n.
rrusczyk19:47:54
rrusczyk19:48:06
And what do we do with this final expression?
metafor19:48:35
= 180(n-2)
bl.html19:48:35
set it equal to 180(n-2)
Jakey19:48:35
equate it to (n-2)180
isabella229619:48:35
set equal to 180(n-2)
rrusczyk19:48:38
rrusczyk19:48:49
And what do we find for n?
ABNoRMaLY_NoT_QuieT19:49:11
simplify?
isabella229619:49:11
simplify both sides
rrusczyk19:49:19
What do we get when we simplify?
hello12319:49:31
we get the quadratic equation N^2 +7n-144+0
rrusczyk19:49:36
rrusczyk19:49:48
And what do we find for n?
hello12319:49:58
n=9
BoxinaJack19:49:58
eshen19:49:58
9
Lolligal10119:49:58
n=9
BoxinaJack19:50:01
(n+16)(n-9)=0 so n=-16 or n=9, but n=-16 won't work for sides
rrusczyk19:50:04
We can solve the quadratic in a number of ways including factoring it into
(n - 9)(n + 16) = 0, so n = 9 or -16.
hello12319:50:06
we reject -16
rrusczyk19:50:14
We know that a polygon cannot have a negative number of sides and so we have our answer, n = 9. (A).
rrusczyk19:50:47
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.
jhangil19:50:54
Is this an actual problem on the AMC?
rrusczyk19:50:57
Yes.
rrusczyk19:51:02
COMBINATIONS
rrusczyk19:51:06
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.
rrusczyk19:51:29
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.
Jakey19:51:48
6C3
isabella229619:51:48
20 triangles
bl.html19:51:48
6 C 3 ??
Lolligal10119:51:48
20
rrusczyk19:51:54
Correct, and we'll see why.
rrusczyk19:52:01
We can now consider each possible grouping:
ABC, ABD, ABE, ABF, ACD, etc.
rrusczyk19:52:08
These three-letter strings are really part of any 6 letter string of all six letters except that we chop off the last three letters. This gives us a way to count.
rrusczyk19:52:54
How many ways can we form a 3-letter sequence by taking 3 of these 6 letters?
Lolligal10119:53:27
120
istos19:53:27
120
ChocolateChipCookie19:53:27
would u use permutations
termantiger19:53:27
oops. 120?
ChocolateChipCookie19:53:30
does order matter
rrusczyk19:53:40
First, we'll look at what happens when order matters.
rrusczyk19:53:43
Then, we want:
Jakey19:53:44
6x5x4?
Geliva19:53:47
Six possibilities for the first letter * five for the second * four for the third
rrusczyk19:54:13
Indeed, there are 6 possible letters to be the first letter, 5 to be the second, 4 to be the third, so the total number of three letter strings is 6*5*4 = 120.
rrusczyk19:54:26
But when we are forming triangles, order doesn't matter!
rrusczyk19:54:31
ABC is the same as ACB is the same as CAB, etc
rrusczyk19:54:39
How do we deal with this?
MathAndKnowledge19:54:51
So we divide by 3!
bisbmard919:54:51
divide by 3!=6 to get 20
Jakey19:54:51
so we divide 3!
BoxinaJack19:54:51
and because of 3 letters, there are 3x2x1 ways to order each one or 6, so divide by 6
nikki19:55:02
divide by 3!, which is the number of ways to order 3 letters
rrusczyk19:55:26
Each group of 3 letters can be ordered 3! = 6 ways, so each group of 3 is counted 6 times in our count of 120 strings.
rrusczyk19:55:52
Therefore, to count the number of triangles, we divide 120 by 6 so that we count each group of three letters once: 120/6 = 20.
rrusczyk19:55:57
rrusczyk19:56:28
If you are not familiar with combinations, or that symbol in the middle of that equation, we recommend looking into our Introduction to Counting & Probability class and book.
rrusczyk19:56:37
Combinations lies at the heart of a great many AMC problems and so we will now examine a number of different kinds of problems that require us to use combinations.
rrusczyk19:56:48
hello12319:57:25
do we include sides of the cube?
rrusczyk19:57:27
Yep.
MathAndKnowledge19:57:35
8 choose 2...?
BoxinaJack19:57:35
Well, it'd be 8x7 to start off, then divide by 2
FibonacciFan19:57:35
8 C 2
ppzuan19:57:35
8C2
eshen19:57:35
8*7/2
abshin382219:57:35
(D) 28
istos19:57:35
28, D
rrusczyk19:57:42
rrusczyk19:57:55
Understanding combinations can directly lead us to quick solutions to some AMC problems.
rrusczyk19:58:16
Here's a somewhat tougher problem:
rrusczyk19:58:19
rrusczyk19:58:39
Where should we start?
vvipul19:59:13
use constructive counting?
rrusczyk19:59:24
Interesting idea: how would you make a number with digits in increasing order?
ppzuan19:59:35
any group of 3 distinct digits can be arranged into increasing order
tsai_daniel19:59:35
Choose three digits from the total ten and consider restrictions in counting.
FibonacciFan19:59:35
choose 3 digits and then realize there is one way to put them in increasing order and one way to put them in decreasing order
hello12319:59:40
pick 3 digits put them in order
rrusczyk20:00:10
How does this help?
Lulze20:00:48
But groups which include 0 can't have 0 as the first digit
muskrat194520:01:04
but we must watch for 0
muskrat194520:01:06
we must watch out for 0 in the hundreds
Ruby509999120:01:10
find all the ways to choose 3 digits, except 0 for first
BoxinaJack20:01:10
Groups with zero can only have 0 as the last digit
rrusczyk20:01:11
Ooooh. Always have to be careful to look out for special cases!!!
rrusczyk20:01:17
So, what do we do about that?
andrewnycpops20:01:19
so you take them out
rrusczyk20:01:35
All right, how many numbers are there with digits in increasing order?
MathAndKnowledge20:02:44
9 choose 3 (excluding 0)
Lolligal10120:02:44
84
abshin382220:02:44
84
rrusczyk20:02:49
rrusczyk20:03:14
So, is the final answer 84?
GautamR20:03:35
no
istos20:03:35
no.
iloveham47_220:03:35
No
vvipul20:03:35
no. count decreasing
metafor20:03:39
no, we still have those in decreasing order
Lolligal10120:03:39
no we have to remember decreasing order
rrusczyk20:03:42
Now, how many three digit numbers will have all their digits in decreasing order?
rrusczyk20:04:01
Are there just 84 again?
Ruby509999120:04:13
for this, 0 can be used
FibonacciFan20:04:13
no
zephyredx20:04:13
same as increasing, but can have zeroes
rrusczyk20:04:23
Ah, yeah, we have those 0's to include.
rrusczyk20:04:29
So, how many such numbers are there?
MathAndKnowledge20:04:41
10 choose 3
FibonacciFan20:04:41
10 C 3
Lulze20:04:41
10 c 3 = 120, because now we can use 0, because when arranged in increasing, it won't be the hundreds digit
abshin382220:04:45
10C3=120
rrusczyk20:04:49
GautamR20:05:34
120+84=204
Lolligal10120:05:34
the answer is C (204)
termantiger20:05:34
120 + 84 = 204
KLA228620:05:34
120+84
Wun Dum Guy20:05:36
i say it's 204
qwertythecucumber20:05:37
120+84=204
rrusczyk20:05:41
We now add the total numbers from each case:
84 + 120 = 204, so our answer is (C).
rrusczyk20:05:45
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.
rrusczyk20:06:10
That's it for sample problems for the AMC 10 class (we'll do some for the AMC 12 and AIME class after questions about the classes).
rrusczyk20:06:22
Does anyone have any more questions about the AMC 10 class in general?
Wun Dum Guy20:06:55
when do u take the amc 10
andrewnycpops20:07:11
february
tenniskidperson320:07:11
sometime in february
rrusczyk20:07:13
It's in February. See the AMC website for exact dates (www.unl.edu/amc)
metafor20:07:19
rrusczyk, what is the code for the combinations (n-take-k)?k
rrusczyk20:07:30
In LaTeX, I use \binom{n}{k}.
mathemonster20:07:39
I can make AIME but am looking to boost my score by learning to solve those last 22-25. Is this class right for me?
rrusczyk20:08:04
This class does spend a lot of time on the harder problems of the AMC 10, but I'll also note that AMC 12 and AIME training will help you with those tough questions.
Aneadi20:08:12
how do you find out your score in AMC 10
rrusczyk20:08:16
They contact your school.
KLA228620:08:19
if i got a 65 on AMC 12, what class should i take
rrusczyk20:08:43
I would recommend some of the Intro subject classes to start with -- probably the Counting class.
abshin382220:08:47
I heard that there are AMC10A and AMC10B. Is there any difference?
rrusczyk20:08:55
The AMC tries to make them of the same difficulty.
GautamR20:08:58
what are the scores out of (the max points on theAMC)?
rrusczyk20:09:00
150
Lulze20:09:07
If you don't qualify for AIME, will they still contact your school with your score?
rrusczyk20:09:08
yes
Wun Dum Guy20:09:16
what do you take after the amc 12
rrusczyk20:09:28
If you pass, you get to take the AIME. Passing is usually around 100 out of 150 (25 questions)
abshin382220:09:40
Can we take both AMC10A and AMC10B at the same time?
rrusczyk20:09:42
They are on different days, so you can take both.
istos20:09:54
what is the score you have to get to qualify for the AIME?
rrusczyk20:09:56
120 on the AMC 10 (sometimes they lower it if the test turns out to be hard).
mathemonster20:10:03
Is doing AIME and AMC 12 excessive?
rrusczyk20:10:06
No.
termantiger20:10:37
Can you take both the AMC 10A and the AMC 10B (and, in extension, AMC 12A/12B)?
rrusczyk20:10:39
You can take either the 10A or the 12A, and then on a different day take either the 10B or the 12B.
MathAndKnowledge20:10:42
If you get a 100 on AMC 12 A/B or in the first 5%, then you move on to AIME.
rrusczyk20:10:44
Correct.
jin Li20:10:46
are all the questions worth the same points?
rrusczyk20:11:03
Yes - see the website for details. It changes from year to year.
hello12320:11:14
what is the typical USAMO qualifier?
rrusczyk20:11:15
There is no typical - it changes a lot from year to year.
suncrestmath20:11:18
which test is harder A or B
rrusczyk20:11:19
Same.
KLA228620:11:22
what class should i take if i got a 100 on the AMC 10?
rrusczyk20:11:24
AMC 10
abshin382220:11:29
Is it more common to take both AMC10 and AMC12?
rrusczyk20:11:35
If you can, we recommend taking both.
mathemonster20:11:37
All 3 would be expensive i think. how much does each class cost.
rrusczyk20:11:43
Each class is $150.
tsai_daniel20:11:45
Could not one also qualify for the AIME via the USAMTS?
rrusczyk20:11:51
Yes; see www.usamts.org for details.
iloveham47_220:12:03
Would the Intro books help more than the intermediates?
rrusczyk20:12:05
For the AMC 10 or AMC 12, yes.
jouyang20:12:12
Are calculators allowed?
rrusczyk20:12:13
No.
bisbmard9_220:12:15
Is there a percent cutoff for the AMC10?
rrusczyk20:12:20
1% (or 120)
ChocolateChipCookie20:12:22
is there a certain age limit
rrusczyk20:12:38
You cannot be beyond 10th grade to take the AMC 10. You can be any age younger.
MathAndKnowledge20:12:47
what class should i take if i got 117 on AMC 10 and 5 on AIME
rrusczyk20:13:20
Tough call - I generally recommend going for subject classes if you have time. You might be well off with the AMC 12 class now.
mathemonster20:13:26
do we need recursion and ball and urns for the 10?
rrusczyk20:13:47
Not sure what you mean by ball and urns, but probably not recursion (though it might show up - it definitely shows on the 12)
ChocolateChipCookie20:14:14
is it 150 per one hour session
rrusczyk20:14:22
No - it is 150 for the full 12 weeks.
suncrestmath20:14:27
Do you know what score qualified for the AIME this year, my teacher thought i qualified but she wasnt sure
rrusczyk20:14:37
It is posted in the AMC Forum on our website. Check that out.
tsai_daniel20:14:56
I find that I am able to solve the problems on the AIME, but only very slowly. How should one train for being faster?
rrusczyk20:15:18
More practice problems. Don't worry too much about speed. Solving the problems is way more important in the long run.
abshin382220:15:21
How is AIME different from AMC 12 or AMC 10
rrusczyk20:15:24
Harder.
mathemonster20:15:40
I got a 112.5 on the 10a without studying. I plan to do aops v1 and v2 before the 10, as well as the online classes. I am goal oriented, as are many people on the forums. What is a realistic goal?
rrusczyk20:16:16
Definitely realistic to aim for passing. If you are very serious about studying problem solving, set your high goal to be qualifying for the USAMO.
rrusczyk20:16:30
That shouldn't be an expected goal, but it's not a totally unrealistic one.
mathgenius20:16:33
What class should i take if I haven't taken any of the AMCs, and all of this is completely new to me?
rrusczyk20:17:15
There are some AMC tests in the AoPSWiki - take a couple of those and see how you fare. Then email me the results and I'll advise you further.
lightning20:17:18
What class should I take if I got a 114 on the AMC 10, but a 6 on the AIME?
rrusczyk20:17:44
Focus more on AIME training in general. Maybe the AMC 12 class, but I wouldn't study a whole lot more on the AMC 10.
Wun Dum Guy20:17:48
when does the class start?
rrusczyk20:17:51
Next week.
jin Li20:17:52
do you have to write proofs for any of the tests?
rrusczyk20:17:57
Only the USAMO.
KLA228620:18:04
how is amc 10 different form amc 12
rrusczyk20:18:14
AMC 12 is harder and covers a few more advanced topics.
andrewnycpops20:18:17
what is v1 and v2
rrusczyk20:18:23
Art of Problem Solving books.
abshin382220:18:25
If I want to buy a book for studying AMC12 or AIME, which book would you recommend?
rrusczyk20:18:50
AMC 10/12: Art of Problem Solving Volume 1. AIME: Volume 2, then Paul Zeitz's Art and Craft of Problem Solving.
Wun Dum Guy20:18:52
Is the ARML related to the AMCs.
rrusczyk20:18:54
No.
jin Li20:18:55
do any of the tests use statistics?
rrusczyk20:19:02
Nothing beyond mean, median, mode.
plumbst20:19:11
will there be a USAMO class in the future
rrusczyk20:19:15
WOOT is our USAMO class.
jhangil20:19:17
Is there a difference between AIME problem series A and B?
rrusczyk20:19:25
Different problems, same general difficulty.
mathemonster20:19:27
I can do a 3-5 on the AIME, would u advise me to focus more on the 12 as well
rrusczyk20:19:45
Yes, though if you're particularly low on the AMC 10 due to geometry, you should spend some extra time on that.
Wun Dum Guy20:19:47
What is the maximum score you can get on the AMC 10?
rrusczyk20:19:50
150
termantiger20:19:53
Will graphing be used in the AMC 12 now that calculators are banned?
rrusczyk20:19:55
Not sure.
qwertythecucumber20:19:56
which book is recommended for AMC in general?
rrusczyk20:20:03
Art of Problem Solving Volume 1.
termantiger20:20:05
Will topics such as derivatives and integrals be used in the AMC 12 or 10?
rrusczyk20:20:06
No
lshare20:20:14
What would you recommend if I haven't done any of these classes but pretty much get what's being talked about?
rrusczyk20:20:27
Email me your math background (classes@artofproblemsolving.com), and I can advise you.
bl.html20:20:31
will AMC 10 or 12 test arithmetic mean or geometric mean?
rrusczyk20:20:33
Yes.
Jakey20:20:35
if i got 5/8 in ARML, is it ok for AIME?
rrusczyk20:20:40
Probably.
andrewnycpops20:20:44
is there and AMC 8 course
rrusczyk20:20:54
The MATHCOUNTS Problem Series helps with that.
Ruby509999120:21:16
what class would you recommend us to take after the AIME preparation classes?
rrusczyk20:21:17
WOOT or any of the Intermediate classes
Jessie Shin20:21:20
will anything about precal be on amc 10?
rrusczyk20:21:22
Probably not.
mathemonster20:21:27
Just to clarify geometric mean is the sqrt of the product of two numbers, right?
rrusczyk20:21:39
That's the GM of two numbers - look in the AoPSWiki for a more thorough definition.
termantiger20:21:41
Would you recommend I study AoPS book 1 or book 2 while I am taking a class? I have both
rrusczyk20:21:48
for AMC: 1, for AIME: 2
istos20:21:50
MATHCOUNTS would be around AMC 8 level?
rrusczyk20:21:52
Yes.
ChocolateChipCookie20:21:56
i am probably taking the amc 8 would you recommend taking any of these classes
rrusczyk20:22:02
The MATHCOUNTS Problem Series would help.
Wun Dum Guy20:22:05
will there be trig on amc 10?
rrusczyk20:22:07
No
andrewnycpops20:22:20
on the 12?
rrusczyk20:22:23
Yes.
JesusFreak20:22:25
Will there be an AMC 10 class during the fall, or is it just a summer class?
rrusczyk20:22:29
There will be a fall class.
termantiger20:22:32
Now that we know geometric mean and arithmatic mean will be used in the AMCs, will harmonic mean be used as well?
rrusczyk20:22:43
Possibly, but they'll define it in the test if they use it.
Jakey20:22:46
can we start with the AMC 12 problems? :)) LOL
rrusczyk20:22:50
Good idea :)
rrusczyk20:22:53
AMC-12
rrusczyk20:22:59
The AMC 12 class starts on June 2, and meets every Monday from 7:30-9:00 PM Eastern. The class meets for 12 weeks and ends on August 18. The course is designed to cover a large portion of the curriculum tested on the AMC 12 exam.
rrusczyk20:23:07
This 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.
rrusczyk20:23:18
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.
rrusczyk20:23:53
We'll do a couple problems:
rrusczyk20:24:03
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
rrusczyk20:24:33
Uh-oh. What do we do?
qwertythecucumber20:24:35
graph it
rrusczyk20:24:40
What do we graph?
Ruby509999120:25:04
the region enclosed by the 4 points
rrusczyk20:25:10
What is this region
rrusczyk20:25:11
?
rrusczyk20:25:56
Don't just send me an answer to the problem - I'm expecting explanations
qwertythecucumber20:25:58
a rectangle
metafor20:25:59
A rectangle?
rrusczyk20:26:08
The region we draw from is a rectangle.
rrusczyk20:26:18
What else must we determine?
Ruby509999120:26:51
the boundary for which the point will be closer to origin or (3,1)
Geliva20:26:51
What amount of that rectangle is closer to the origin than 3,1
rrusczyk20:27:10
Exactly - we must find what points in this rectangle are closer to the origin than to (3,1).
rrusczyk20:27:15
How will we do that?
Wun Dum Guy20:27:37
we can draw a boundary line!
ppzuan20:27:42
find the line that is the equidistant from both
istos20:27:50
boundary line.
rrusczyk20:27:53
And what line is this?
plumbst20:28:39
perpendicular bisector
zephyredx20:28:41
perpendicular to the line connecting them and passing through the midpoint
rrusczyk20:29:10
The set of points (in a plane) equidistant to two given points is a the perpendicular bisector of the line segment with the two points as endpoints.
rrusczyk20:29:40
What is the perpendicular bisector of the segment connecting (0,0) and (3,1)?
bisbmard9_220:30:06
y=-3x+5, perpendicular bisector
abshin382220:30:06
y=-3x+5
rrusczyk20:30:13
How do we find this?
termantiger20:30:50
the line connecting them is y=x/3, so the line containing (1.5, 0.5) and perpendicular to y=x/3 is y=-3x+5
bisbmard9_220:30:50
The midpoint is (1.5,0.5)
rrusczyk20:30:59
The midpoint of the segment between (0, 0) and (3, 1) is (1.5, .5).
rrusczyk20:31:03
The equation of the line through (0, 0) and (3, 1) is y = x/3.
bisbmard9_220:31:10
The slope of the line connecting (0,0) and (3,1) is 1/3, so the slope of the perpendicular is -3
rrusczyk20:31:15
The slope of a line perpendicular to that line is -1/(1/3) = -3. (Students should know that the product of the slopes of two perpendicular lines in a plane is -1.)
rrusczyk20:31:52
So, we want the line through (1.5,0.5) with slope -3.
qwertythecucumber20:32:21
point-slope form
rrusczyk20:32:22
This is y - 0.5 = -3(x - 1.5), which simplifies to y = -3x + 5.
rrusczyk20:32:32
So, what do we do with this line?
plumbst20:33:17
find where it intersects with rectangle
Ruby509999120:33:17
find where it crosses the rectangle
rrusczyk20:33:23
We can graph it:
rrusczyk20:33:31
rrusczyk20:33:47
Now, what?
mathemonster20:34:04
find where it intersects the rectangle?
Wun Dum Guy20:34:04
we can find the intersection
rrusczyk20:34:18
We could do this, but someone has a very clever observation:
plumbst20:34:24
the square on the right is divided in half
rrusczyk20:34:39
(abshin pointed this out, too)
rrusczyk20:35:02
The line goes through (1.5,0.5), so it bisects the square on the right!
rrusczyk20:35:32
(Because the line goes through the center of the square - it cuts the square into two congruent pieces).
rrusczyk20:35:35
How does this help?
Wun Dum Guy20:35:55
the area of the trapezoid is then 3/2
mathemonster20:35:55
So we can easily calculate the area of the shaded region to be 1 x 1 / 2 + 1 = 1.5
termantiger20:35:55
1.5 units of the rectangle are closer to the origin, so that the probability is 1.5/2 which is that portion divided by the total area, which turns out to be...3/4!
Jakey20:35:55
1/2 divided by 2 = 1/4 the area of the unshaded trapezoid. which means that the probability that a point is closer to (0,0) is 1-.25 = 3/4
FibonacciFan20:35:58
it tells us the answer is 3/4
rrusczyk20:36:03
rrusczyk20:36:36
Exactly! Half of one square is unshaded, so the unshaded region has area 0.5. That means the shaded region has area 1.5, so the probability is 1.5/2 = 3/4.
Wun Dum Guy20:37:20
great problem
jin Li20:37:20
magic!
rrusczyk20:37:26
Let's try one more.
rrusczyk20:37:34
rrusczyk20:38:29
Yikes - that looks complicated. Here's a tip: often the most complicated-looking problems on the AMC are way easier than they look. Don't panic!
rrusczyk20:38:37
So, after we don't panic, what do we do?
plumbst20:38:42
plug in and simplify first
FibonacciFan20:39:15
plug in what?
rrusczyk20:39:21
Good question - what do we do?
Lulze20:39:24
Can we graph f(x) + f(y) and f(x) - f(y)?
rrusczyk20:39:39
Good observation - and how will we figure out what those graphs are?
termantiger20:39:46
plug in x and y into f(x) and make it f(x) and f(y)
rrusczyk20:40:08
Let's do that. We just replace f(x) and f(y) with our function definition.
bl.html20:40:10
x^2 + 6x + 1 + y^2 + 6y + 1 <= 0
rrusczyk20:40:19
There's one inequality. What's the other one?
ChocolateChipCookie20:40:47
the negative one
termantiger20:40:47
so it is x^2+6x-y^2-6y<=0
bisbmard9_220:40:50
x^2+6x+1-y^2-6y-1<=0
rrusczyk20:41:00
All right, we now have two inequalities:
rrusczyk20:41:04
rrusczyk20:41:09
What do we do with the first one?
aggarwal20:41:21
first is cricle
istos20:41:21
simplify
zephyredx20:41:21
complete the square
rrusczyk20:41:27
What does that give us?
hello12320:41:53
we get (x+3)^2+(y+3)^2 less than equal to 16
Jakey20:41:56
its a circle with center (-3,-3) with radius 4, The area is the inside of the circle
rrusczyk20:41:57
Completing the square gives:
termantiger20:42:05
(x+3)^2 + (y+3)^2 <=16
rrusczyk20:42:06
rrusczyk20:42:12
That's our first inequality.
rrusczyk20:42:24
As Jakey noted, this defines the interior of a circle with radius 4 centered at (-3, -3).
rrusczyk20:42:32
What about the second inequality, which we can simplify to:
rrusczyk20:42:39
rrusczyk20:42:51
rrusczyk20:43:05
What might we do with that left side?
iloveham47_220:43:35
group them
hello12320:43:36
factor as differenc of squares
rrusczyk20:43:45
rrusczyk20:43:48
Then what?
mround20:44:21
(x-y)(x+y+6)<=0
Ruby509999120:44:21
(x - y)(x+y+6)
Bigmo561120:44:24
(x + y + 6)(x-y) <= 0
rrusczyk20:44:29
rrusczyk20:44:37
We also could have gotten to this with:
termantiger20:44:42
becomes (x+3)^2 - (y+3)^2 <= 0? because the -9 and +9 cancel out?
Jakey20:44:42
(x+3)^2 - (y+3)^2 <=0
rrusczyk20:44:53
And what is the graph of this inequality?
mround20:44:56
two stright lines crossing each other
rrusczyk20:45:11
Where do these lines cross?
aggarwal20:45:46
how do you know that
rrusczyk20:45:47
Look at the boundary - it's where the left side is 0. That means one of the factors is 0. Setting each factor equal to 0 gives a line.
termantiger20:45:51
at (-3,-3)
zephyredx20:45:51
(-3,-3)
rrusczyk20:45:58
Aha! What's so nice about that?
Jakey20:46:00
at the center of the circle
rrusczyk20:46:03
Indeed!
rrusczyk20:46:31
So, when we graph, what do we get as the region of points that satisfy the two inequalities?
Lulze20:46:38
Will multiplying two lines always make a graph which is the two lines crossing?
rrusczyk20:46:58
Notice that the right side of the inequality is 0 -- this is why we get two lines as the boundary of the inequality.
mathemonster20:47:44
these will be sectors of the circle!
rrusczyk20:47:55
What size will these sectors be (what portion of the circle?)
Ruby509999120:48:07
R = 1/2 area of the circle??
FibonacciFan20:48:07
half circle actually
termantiger20:48:13
1/4?
termantiger20:48:13
each
rrusczyk20:48:21
rrusczyk20:48:37
The two lines are perpendicular, so the desired regions are quarter circles.
rrusczyk20:48:55
So, what is the area of the shaded region?
Bigmo561120:49:22
sorry, 8pi
abshin382220:49:22
8pi?
bl.html20:49:22
8pi
plumbst20:49:22
8pi
rrusczyk20:49:38
Now we simply calculate the area of the circle and take half. The area is 8pi and using 3.14 as an approximation we get 8(3.14) = 25.12, so the answer is (E).
rrusczyk20:49:47
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.
mathemonster20:49:53
how do we know that the shaded area of the lines is above and below the intersection?
rrusczyk20:50:07
Good question. How can we figure that out quickly?
rrusczyk20:50:22
FibonacciFan20:50:27
test a point
mo357820:50:27
cant we plug in a point?
qwertythecucumber20:50:27
test points
rrusczyk20:50:42
That's what I'd do to be sure. But you can also break into cases.
rrusczyk20:51:02
One is x - y >= 0, x+y+6 >= 0; graph these and find the intersection, and then do where both less than 0.
rrusczyk20:51:18
Any questions about the AMC 12 class?
mathemonster20:52:07
what topics does it cover? and is Valentin also teaching the 10 class?
rrusczyk20:52:14
Sean Markan is teaching the 10 class.
rrusczyk20:52:35
Topics are here:
Ruby509999120:52:41
were those problems moderate level AMC 12?
rrusczyk20:52:59
First was probably in the 10-15 range, the second in the 16-20 range.
termantiger20:53:02
It will cover most of the topics in AoPS 1 right?
rrusczyk20:53:10
Yes, and some topics in AoPS Volume 2
Jakey20:53:12
if you are not in the USA, but your country hosts AMC12, what are the requirements to take the AMC?
rrusczyk20:53:21
You haven't graduated high school
abshin382220:53:23
Is calculus enough for both AMC10, AMC12 and AMIE?
rrusczyk20:53:32
Calculus is irrelevant for these tests.
rrusczyk20:53:42
(And is, in fact, far easier than most of these tests).
termantiger20:53:45
Is it possible for us to do one of the harder problems on the AMC 12? because there are definitely problems on the AMC 12 that are harder (experience)!
rrusczyk20:54:04
There will be many harder questions in the class. And we will do one harder questions when discussing the AIME class.
ChocolateChipCookie20:54:15
is it mostly geometry
rrusczyk20:54:23
No. Maybe 1/4-1/3
ChocolateChipCookie20:54:27
what grade would you reccomend for starting either the 10 or 12
rrusczyk20:54:41
Depends a lot on the student. Email me your math background and I'll make recommendations.
F-86K Sabre Dog20:54:47
What topics does AMC 12 cover
rrusczyk20:54:51
See this link:
mo357820:55:02
how much number theory and counting/probability is on the amc12 and aime?
rrusczyk20:55:10
Together, they are around 30-40% of test.
rrusczyk20:55:19
It is on the AMC 12.
rrusczyk20:55:23
Right now!
rrusczyk20:55:31
AIME
rrusczyk20:55:37
The AIME Problem Series A class starts on June 3, and meets every Tuesday from 7:30-9:00 PM Eastern. The class meets for 12 weeks and ends on August 19. The course is designed to cover a large portion of the curriculum tested on the AIME exam.
rrusczyk20:55:43
The AIME Problem Series 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.
rrusczyk20:55:48
Almost all of the problems in the AIME Problem Series A, come from past AIME contests.
metafor20:55:57
Did you take any of these exams, Rusczyk?
rrusczyk20:56:05
Yes, I had the only perfect score on the 1989 AIME.
rrusczyk20:56:30
We'll look at a couple problems.
rrusczyk20:56:48
Some of these will involve tools that are pretty advanced - you can ask on the message board about these tools if you've never seen them.
rrusczyk20:57:03
We don't have time tonight to go into some of these tools in detail for less experienced students.
rrusczyk20:57:09
Here's the first problem:
rrusczyk20:57:12
rrusczyk20:57:27
mround20:58:04
(7-1)^n + (7+1)^n
zephyredx20:58:04
(7-1)^n + (7+1)^n
trigfan20:58:04
6=7-1; 8=7+1
rrusczyk20:58:10
mround20:58:18
use bionomial theorem to expand them
rrusczyk20:58:22
rrusczyk20:58:28
rrusczyk20:58:31
Now what?
nikki20:59:05
sum them
Lulze20:59:05
most of these are divisible by 7^2
hello12320:59:05
anything with 7 squared or higher can be removed
rrusczyk20:59:10
When we add these two quantities, which part of the sum is not divisible by 7^2?
ChocolateChipCookie20:59:40
why do we remove anything with 7 squared or higher
rrusczyk20:59:56
We want the remainder when dividing by 7^2. All those terms contribute 0 to the remainder.
mathemonster21:00:01
the 7^1's and the 7^0's
nikki21:00:01
ones with 7^1 and 7^0
mo357821:00:01
the last 2 terms
rrusczyk21:00:12
And what are we left with?
Jakey21:01:04
2(83)(7)
FibonacciFan21:01:04
2*83*7
nikki21:01:04
2(83C82*7^1)
Lulze21:01:14
2 * (83 c 82) * 7 = 2 * (83 c 1) * 7 = 2 * 83 * 7
rrusczyk21:01:15
rrusczyk21:01:21
What is our final answer?
zephyredx21:02:20
35?
plumbst21:02:20
35
trigfan21:02:20
35
aggarwal21:02:20
35
mo357821:02:20
35
rrusczyk21:02:23
Our final answer is the remainder when 1162 is divided by 49, which is 35.
rrusczyk21:02:44
Let's try a tougher one:
rrusczyk21:02:47
rrusczyk21:03:11
What's a good strategy in general on tougher counting problems?
nikki21:03:28
there are 2^10 possible series of outcomes
rrusczyk21:03:40
Indeed, but counting the desired ones is a little tough!
metafor21:03:44
Simplify them to easier counting problemS?
zephyredx21:03:44
start with simple cases?
Lulze21:03:44
Evaluate the first few terms (tossed 1 times, 2 times, 3 times), and look for a pattern of some sorts
mround21:03:44
start with small cases
rrusczyk21:04:02
Play with simple cases. The best problem solvers are playful, and are good at finding patterns:
rrusczyk21:04:05
We could begin by examining smaller sequences of tosses.
rrusczyk21:04:08
One toss outcomes without HH:
H
T
rrusczyk21:04:10
Boring.
rrusczyk21:04:14
Let's try 2:
rrusczyk21:04:19
HT
TH
TT
rrusczyk21:04:25
In what ways could we toss a coin three times without ending up with HH in the sequence?
Lulze21:05:06
HTH
THT
TTH
HTT
TTT
plumbst21:05:06
TTT, TTH, HTH, HTT, THT
termantiger21:05:08
5 ways
rrusczyk21:05:10
Three toss outcomes without HH:
HTH
HTT
THT
TTH
TTT
rrusczyk21:05:16
In what ways could we toss a coin four times without ending up with HH in the sequence?
zephyredx21:06:00
TTTT, TTTH, TTHT, THTT, HTTT, HTHT, THTH, HTTH
termantiger21:06:00
8 ways
rrusczyk21:06:25
Four toss outcomes without HH:
HTHT
HTTH
HTTT
THTH
THTT
TTHT
TTTH
TTTT
rrusczyk21:06:35
Do we notice anything interesting from examining these smaller cases?
termantiger21:06:43
Fibonnaci!
mathemonster21:06:43
fibbonaci's
mo357821:06:43
like fibonaci sequence?
hello12321:06:43
the numbers form the fibbonci sequence
lightning21:06:43
Is it a fibbonacci sequence
rrusczyk21:06:59
Aha! But we should prove it! What strategy are we likely to use?
hello12321:07:14
we use recursion
Bigmo561121:07:14
recursion
rrusczyk21:07:27
The Fibonacci numbers stem from a recursion.
rrusczyk21:07:32
Let's try to use a recursion.
rrusczyk21:07:57
How can we relate the number of n-flip sequences with no HH to (n-1)-flip and (n-2)-flip sequences with no HH?
termantiger21:08:30
u(n) = u(n-1) + u(n-2)
Bigmo561121:08:30
f(n-1) + f(n-2) = f(n)
rrusczyk21:08:36
But why? Have to prove it!
hello12321:10:30
in n tosses if the 1st is heads the next must be tails and the rest can be tossed in the number of ways to toss n-2 coins if the 1st is tails the rest can be tossed in any of the ways we toss n-1 coins
rrusczyk21:10:47
Let f(k) be the number of sequences of k flips with no HH.
rrusczyk21:10:52
We consider two possibilities
rrusczyk21:11:03
If the first of our n flips is H, then the next must be T, and then there are f(k-2) ways to flip the rest of the k-2 coins and get no HH.
rrusczyk21:11:52
If the first of our flips is T, then the remaining k-1 can be any of the f(k-1) sequences of k-1 flips with no HH.
hello12321:12:07
letting f(n) = #ways to toss n coins such that there are no consecutive heads thus f(n) =f(n-1) +f(n-2)
rrusczyk21:12:36
Adding these two cases, we have a total of
rrusczyk21:12:49
f(k-1) + f(k-2) ways to flip k coins and get no HH!
rrusczyk21:12:59
So, we have f(k) = f(k-1) + f(k-2)
Bigmo561121:13:12
Don't forget to mention that the two cases are disjoint so we can add them.
rrusczyk21:13:27
Indeed - the two cases don't overlap, and they cover all possibilities.
aggarwal21:13:56
why do they not overlap
rrusczyk21:14:05
Case 1: Start with H. Case 2: Start with T.
hello12321:14:10
f(1)=2 f(2) =3 thus we are ready to calculate f(10)
rrusczyk21:14:20
We already saw that f(1) = 2, f(2) = 3, so continuing with Fibonacci, what do we find for f(10)?
lightning21:15:05
144
aggarwal21:15:05
144
FibonacciFan21:15:09
144
rrusczyk21:15:18
We just keep cranking out the Fibonacci numbers:
rrusczyk21:15:19
2, 3, 5, 8, 13, 21, 34, 55, 89, 144
rrusczyk21:15:36
We start with 2,3 since f(1) = 2 and f(2) = 3.
rrusczyk21:15:45
144 is the tenth number we his. So, what is our final answer?
FibonacciFan21:16:15
144/1024=9/64 so the answer is 73
abshin382221:16:15
144/1024=9/64
d199421:16:15
73
hello12321:16:31
73
Jakey21:16:31
73
rrusczyk21:16:33
We can now say that i/j = 144/1024 = 9/64, so i + j = 73.
rrusczyk21:16:44
This is one of those problems that shows how important it is to test small cases and then to observe the way that we generate those cases. Making the connection between the way we generated each new sequence and the recursion that we derived gave us a direct calculation to our answer.
Lulze21:16:46
Will we typically have to find relations like this on most AIME problems?
rrusczyk21:17:05
Yes, as AIME problems get harder, you'll have to get more clever about finding surprising connections.
rrusczyk21:17:14
Are there any questions about the AIME Problem Series?
istos21:17:39
it seemed really difficult! what are some methods that you recommend on connecting these relations?
rrusczyk21:17:59
Lots of practice -- our Intermediate books are packed with suggestions and strategies for finding these.
termantiger21:18:09
Will most AIME problems contain intricate details in the question that require deep reading to find?
rrusczyk21:18:19
You definitely have to pay attention to the details in the questions.
Jakey21:18:23
this method is also taught in the First steps for math olympians by douglas!!!
rrusczyk21:18:31
That is a nice book for preparing for some of these tests.
Lulze21:18:58
I've noticed that AIME problems seem to be what you would call "computationally difficult", or in other words, involve lots of non-calculator number crunching. Does the class discuss any techniques to make this number crunching easier?
rrusczyk21:19:02
I suggest three techniques:
rrusczyk21:19:23
1) Do problems 2 ways, which helps you cut down on errors (particuarly in counting problems)
rrusczyk21:19:43
2) If very computationally intensive, step back and look for a simpler approach to the problem.
rrusczyk21:19:47
But the most important is
rrusczyk21:19:52
3) Read this article:
rrusczyk21:20:27
This describes how I cut down on careless errors and learned how to carefully manage these computations.
abshin382221:20:29
Is this book good for AIME too?
rrusczyk21:20:34
It's more for the AMCs
rrusczyk21:20:39
Any more questions?
Jakey21:20:42
by the way, rusczyk, what's your email ad?
rrusczyk21:20:47
rusczyk@artofproblemsolving.com
abshin382221:21:49
Is AMC10 harder than MATHCOUNTS Nationals?
rrusczyk21:22:05
It's different. There's less time pressure, but the hardest problems are harder.
Ruby509999121:22:22
anything you recommend we should definitely know for the classes for AIME?
rrusczyk21:22:41
You should know to speak up and ask questions when you're stuck!
Jakey21:22:44
what book can you refer which is very friendly and very easy to understand to prepare for AIME's, USAMOs and even IMO's?
rrusczyk21:22:50
No book hits that whole range.
rrusczyk21:22:56
AMC: AoPS Volume 1.
rrusczyk21:23:13
AIME: AoPS Volume 2. Art and Craft of Problem Solving by Paul Zeitz.
rrusczyk21:23:21
USAMO: The Zeitz book.
rrusczyk21:23:44
If you have plenty of time for more thorough preparation, our subject books are even better.
rrusczyk21:24:05
Intro for AMC, Intermediate for advanced AMC, AIME, and very beginning Olympiad.
Jakey21:24:07
oh...but in the book by Zeitz doesnt show solutions to its exercises
rrusczyk21:24:11
Ask on the message board.
aggarwal21:24:14
how should we prepare for proofs
rrusczyk21:24:17
By writing proofs.
rrusczyk21:24:19
Do the USAMTS.
rrusczyk21:24:24
Do the Mandelbrot Team Round.
rrusczyk21:24:29
Do the ARML Power Round.
iloveham47_221:24:32
What is the normal score for Aime + AMC to qualify for USAMO?
rrusczyk21:24:42
Varies a great deal from year to year.
Jakey21:25:11
is ARML great for preparing for AIME's and beginners at USAMO?
rrusczyk21:25:26
Yes. And it's just plain fun (more details at www.arml.com for those who don't know about it)
ChocolateChipCookie21:25:42
have you been to the usamo and what is it like
rrusczyk21:25:51
Yes, I was a winner in 1989. It's long.
rrusczyk21:26:02
9 hours over 2 days, only 6 questions. And they're hard!
