Transcript for the Math Jam "AoPS Classes Math Jam" on Jun 4.

Math Jam hosted by nsato (Naoki Sato ).

nsato (19:32:56)
Hello, and welcome to the AoPS Summer 2007 Classes Math Jam!

nsato (19:33:07)
Tonight, we will be discussing the AMC 12 Problem Series, the AIME A Problem Series, and the Intermediate Trigonometry/Complex Numbers Subject Class.

nsato (19:33:12)
We'll go through some sample problems from each class. If you have any questions, we'll take them after all three classes have been discussed.

nsato (19:33:20)
Valentin Vornicu will begin by discussing the AMC 12 Problem Series.

Valentin Vornicu (19:33:30)
Greetings all!

Valentin Vornicu (19:33:42)
[b]AMC-12[/b]

Valentin Vornicu (19:33:47)
The AMC 12 Problem Series is a 12-week course designed to cover a large portion of the curriculum tested on the AMC 12 exam.

Valentin Vornicu (19:34:01)
The following is an excerpt of one of the areas of problem solving covered in the AMC 12 Problem Series.

Valentin Vornicu (19:34:22)
[b]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 close 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[/b]

Valentin Vornicu (19:34:40)
How do we know when a point is closer to one location or another in a plane?

harbinger (19:34:54)
distance formula?

Valentin Vornicu (19:35:23)
We could indeed compare the two distances. Any other ways though?

rnwang2 (19:35:01)
perpendicular b i s e c t o r

Valentin Vornicu (19:35:35)
The set of points (in a plane) equidistant to two given points is a straight line. The regions on either side of that line are the points close to one or the other of the given points. How does this fact relate to this problem?

avcamethyst (19:35:51)
you can take the area on one side of the line.

avcamethyst (19:35:56)
and set it over the total area.

Valentin Vornicu (19:36:26)
Our goal is to find the line that is the perpendicular bisector of the segment between (0, 0) and (3, 1). We must then find the proportion of the region inside the given rectangle that is also in the region of points closer to (0, 0). How can accomplish this task?

avcamethyst (19:37:19)
find the slope of the line between (0,0) and (3,1) and find the midpoint of that line too.

Valentin Vornicu (19:37:36)
The midpoint of the segment between (0, 0) and (3, 1) is (1.5, .5).
The equation of the line through (0, 0) and (3, 1) is y = x/3.

Valentin Vornicu (19:37:48)
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.)

Valentin Vornicu (19:38:01)
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?

1/(ln x) (19:38:14)
Find areas

pythag011 (19:38:43)
area of portion of rectangle that is on one side of perpendicular bisecter divided by area of rectangle

Valentin Vornicu (19:39:07)
We must now find the area of the rectangular region. This will be the denominator in our geometric probability problem.

Valentin Vornicu (19:39:12)
We must also find the area of the region within the rectangle that is on the origin's side of the line y = -3x + 5. This will be the numerator of our probability.
What are these two quantities?

avcamethyst (19:38:40)
could we draw a diagram with the rectangle and the line?

pythag011 (19:39:14)
2

Valentin Vornicu (19:39:47)
We can easily see that the rectangular region was height 1 and width 2, so it's area is 2.
The area of the region cut by the line is a trapezoid that we must find the area of. It might help if we make a picture:

Valentin Vornicu (19:39:55)


Valentin Vornicu (19:39:58)
http://www.artofproblemsolving.com/Classes/AMC/Images/2002amc12B18region.gif

Valentin Vornicu (19:40:09)
What are the dimensions of the trapezoid that we need?

bizzielilbee101 (19:40:27)
the two bases and the heights

Valentin Vornicu (19:40:40)
We need the height which is 1.
We also need the bases. What are they?

avcamethyst (19:40:23)
5/3, 4/3 and 1

Valentin Vornicu (19:41:06)
We need to find where y = 0 and y = 1 for the line y = -3x + 5.
This happens at (5/3, 0) and (4/3, 1).

Valentin Vornicu (19:41:11)
We can see that 4/3 and 5/3 are the lengths of the bases.
What is the area of the trapezoid?

avcamethyst (19:41:29)
1.5

bluesky (19:41:39)
3/2

pythag011 (19:41:43)
3/2

Valentin Vornicu (19:41:53)
The area of the trapezoid is 1 times the average of the bases which is 3/2.
What is our final answer?

pythag011 (19:42:00)
3/4

avcamethyst (19:42:07)
3/4?

aufha (19:42:15)
3/4

bizzielilbee101 (19:42:15)
3/4

Valentin Vornicu (19:42:25)
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).

Valentin Vornicu (19:42:34)
We have learned that probability problems are sometimes geometry problems and this problem gives us a direct view into using coordinates to help us with such problems.

Valentin Vornicu (19:43:31)


Valentin Vornicu (19:43:47)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-193682548.gif

Valentin Vornicu (19:43:54)
How can we begin determining the area of R?

Valentin Vornicu (19:44:38)
It seems reasonable to begin by analyzing the given inequalities. We can replace the functional parts of the inequalities with the definition of the functions:

Valentin Vornicu (19:44:45)


Valentin Vornicu (19:44:52)
What can we do with these inequalities?

Valentin Vornicu (19:45:42)
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. In the first inequality we can complete the square to best evaluate the circle the graph produces. What happens with the second inequality?

Valentin Vornicu (19:46:52)


Valentin Vornicu (19:47:02)
The left hand side of the second inequality turned out to factor nicely. This makes sense as the corresponding terms in f(x) - f(y) match up in the form a(x^n - y^n) where we could always factor out a (x - y). Now, what is the intersection of the regions defined by these inequalities?

Valentin Vornicu (19:48:13)
The first inequality defines the interior of a circle with radius 4 centered at (-3, -3). The second inequality says that the product of two quantities is no greater than 0, so we can rewrite that inequality as

Valentin Vornicu (19:48:18)


Valentin Vornicu (19:48:44)
The second inequality defines to opposite ""quadrants"" carved out by two perpendicular lines that intersect at .

Valentin Vornicu (19:48:50)
The intersection of the two regions carved out by the inequalities looks like this:

Valentin Vornicu (19:48:55)


Valentin Vornicu (19:48:58)
http://www.artofproblemsolving.com/Classes/AMC/Images/2002amc12B25region.gif

Valentin Vornicu (19:49:28)
Now what?

avcamethyst (19:49:45)
area of shaded region

Valentin Vornicu (19:50:10)
How do we calculate it?

bizzielilbee101 (19:50:22)
find the area of the circle and divde by 2?

Valentin Vornicu (19:50:31)
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).

Valentin Vornicu (19:50:50)
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.

Valentin Vornicu (19:51:04)
Are there any questions about this class?

bizzielilbee101 (19:51:33)
what regular math classes should we take to help us prepare for this class?

rrusczyk (19:53:03)
The AMC 12 class assumes students have taken a geometry class (or know all the geometry in a regular geometry class), have a good algebra background up through factoring polynomials, and have a basic background in counting and number theory.

aufha (19:52:19)
How do you know if you're ready to take the class?

rrusczyk (19:54:01)
If you are consistently scoring over, say, 80 on AMC12s for practice or for real, than you're ready. If you're consistently getting 130+ on AMC12s, then you should stop worrying about the AMC12 and move on to the AIME.

rrusczyk (19:54:09)
Any more questions?

1/(ln x) (19:54:17)
What about low 100s?

rrusczyk (19:54:29)
This would be a good class for you.

mihail911 (19:54:13)
what AoPS books are expected to go along with the class?

rrusczyk (19:54:56)
Studying AoPS Volume 1 would be good. If you've already covered most of it, then working through parts of Volume 2 this summer would be good.

avcamethyst (19:54:14)
will we be using AMC12 style questions or real former AMC12 questions?

rrusczyk (19:55:25)
Both. (Mainly old AMC12 and AHSME questions, but we do take problems from other sources.)

aufha (19:55:51)
How is the class different from the AMC 10 class?

rrusczyk (19:56:42)
First, it's a little harder. Slightly harder geometry and algebra problems, etc. Second, there are some subjects like trigonometry that are covered in the AMC 12 class but not in the AMC 10 class (because they are on the AMC 12 but not the AMC 10).

mihail911 (19:57:03)
wait, there are trigonometry concepts on the AMC 12?

rrusczyk (19:57:15)
Yes.

IntellectMage (19:57:09)
If I just missed the AMC 10 Cutoff by one question should I take this course or the AMC 10 course?

rrusczyk (19:57:36)
Your call, really. Both would probably be helpful.

bluesky (19:57:22)
What do the AoPS Volume 1 & 2 cover in terms of contests?

rrusczyk (19:58:27)
Vol 1: State + National MATHCOUNTS, AMC (10 and 12), some beginning concepts for the AIME. Vol 2: Advanced AMC problems, AIME, very beginning Olympiad

aufha (19:58:19)
I've never heard of a cutoff to take the AMC 10, do you mean AIME?

rrusczyk (19:58:38)
Means the cutoff for passing the AMC 10.

mihail911 (19:58:01)
would taking this class help you score better on the amc 10?

rrusczyk (19:59:03)
Yes. But it would also include some content that won't help on the AMC 10 (trig, for example), but will help on AMC 12 and AIME.

avcamethyst (19:58:06)
if I qualified on the AMC12 by 5 points but missed the AMC10 cutoff by 1.5, should I take the 12 or 10 if I can only take one?

rrusczyk (19:59:40)
I'd probably opt for the one that best fits your schedule. If you're particularly weak in geometry, lean towards the AMC 10 class.

Alex Whatley (19:59:19)
How long should you study on this course to get ready for the AMC 12?

rrusczyk (20:00:11)
We recommend spending 3-4 hours a week on the course, and spending time studying books and past AMC test.

mihail911 (20:00:13)
is there any particular area that the amc 12 class focuses on (e.g. counting & prob, number theory)

rrusczyk (20:00:58)
It covers nearly all the main subject areas that are covered on the AMC 12.

og4life42592 (20:00:45)
where can we find past AMCs

rrusczyk (20:01:38)
Click AoPSWiki on the sidebar, or Contests on the Forum and look around. You can also buy them from our site, or from the AMC. There are a couple free tests on the AMC page as well (www.unl.edu/amc)

rrusczyk (20:02:12)
Now, I'm going to turn the room over to Naoki Sato, 2-time IMO medalist, who is the instructor of the AIME Problem Series. You can ask more questions when he is finished.

nsato (20:02:27)
The AIME A Problem Series will be taught by me, Naoki Sato. The course begins Tuesday, June 12, and classes are Tuesday nights from 7:30 to 9 PM ET (4:30 to 6 PM PT).

nsato (20:02:44)
In the AIME A Problem Series, we will be looking at problems that almost all come from the AIME contest.

nsato (20:02:46)
The following problems are taken directly from the AIME Problem Series.

nsato (20:02:58)


nsato (20:03:03)
http://www.artofproblemsolving.com/Classes/AIME/Images/1983aime6k14.gif

nsato (20:03:09)


pythag011 (20:03:19)
6=7-1, 8=7+1

nsato (20:03:56)


nsato (20:04:08)


nsato (20:04:23)


aufha (20:04:21)
terms cancel

nsato (20:04:34)
When we add these two quantities, which part of the sum is not divisible by 7^2?

harbinger (20:05:20)
any part with an odd exponent?

pythag011 (20:05:07)
2*(83 82)*7

1/(ln x) (20:05:23)
2 times (83 choose 82) times 7

nsato (20:05:37)


nsato (20:05:47)
What is our final answer?

pythag011 (20:06:12)
35

bizzielilbee101 (20:06:32)
35

avcamethyst (20:06:32)
35?

nsato (20:06:42)
Our final answer is the remainder when 1162 is divided by 49 which is 35.

nsato (20:06:53)


nsato (20:07:06)
http://www.artofproblemsolving.com/Classes/AIME/Images/1990aime9NoHH.gif

nsato (20:07:10)
There are 2^10 = 1024 possible outcomes for the series of ten tosses. How can we count the number of tosses that include no consecutive heads?

aufha (20:07:23)
try smaller example

nsato (20:07:39)
We could begin by examining smaller sequences of tosses.

nsato (20:07:44)
One toss outcomes without HH:

nsato (20:07:46)
H
T

nsato (20:07:51)
Two toss outcomes without HH:

nsato (20:08:04)
HT
TH
TT

nsato (20:08:10)
In what ways could we toss a coin three times without ending up with HH in the sequence?

nsato (20:09:01)
Three toss outcomes without HH:

pythag011 (20:08:46)
HTH, HTT, THT, TTT, TTH

bluesky (20:08:49)
TTT
TTH
THT
HTT
HTH

pyrite417 (20:09:05)
htt hth tht tth ttt

nsato (20:09:18)
HTH
HTT
THT
TTH
TTT

nsato (20:09:22)
In what ways could we toss a coin four times without ending up with HH in the sequence?

avcamethyst (20:09:59)
TTTT, HTTT, THTT, TTHT, TTTH, HTHT, THTH, HTTH

nsato (20:10:38)
Four toss outcomes without HH:

nsato (20:10:42)
HTHT
HTTH
HTTT
THTH
THTT
TTHT
TTTH
TTTT

nsato (20:10:51)
Do we notice anything from examining these smaller cases?

nsato (20:11:11)
How are we constructing each sequence as we add an additional flip?

nsato (20:11:30)
Each flip of n coins either ends in an H or a T. How can we construct new sequences from these possibilities?

nsato (20:11:59)
First we should note that any sequence of n + 1 flips without HH occurring is a sequence of n flips without HH occurring plus one more flip (such that the new sequence does not end in HH).

nsato (20:12:08)
If we add one flip to a sequence that ends in H, the new flip must be a T or the sequence will end in HH.

nsato (20:12:14)
If we add one flip to a sequence that ends in T, the new flip can be either a T or an H and there is no new occurrence of HH.

nsato (20:12:18)
What does this tells us about the total number of sequences of length n + 1 without an occurrence of HH?

nsato (20:12:44)


nsato (20:13:00)
What have we learned?

aufha (20:10:58)
fibonacci

avcamethyst (20:11:04)
it's going 2,3,5,8, which is +1, +2, +3

pythag011 (20:11:22)
2+1=3, 3+2=5, 5+3=8

bluesky (20:11:29)
the sum of the previous two possibilities

nsato (20:13:44)


nsato (20:14:37)


nsato (20:14:44)
How can we use these facts to help us solve the problem?

nsato (20:15:32)


nsato (20:17:02)
We can now say that i/j = 144/1024 = 9/64, so i + j = 73.

nsato (20:17:19)
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.

nsato (20:17:38)
Are there any questions about the AIME A Problem Series?

aufha (20:18:06)
Would a person who didn't even qualify for the AIME be able to keep up with the class?

rrusczyk (20:19:20)
We would recommend such a student take the AMC classes and do some study over the summer, then take the AIME B Problem Series this fall/winter.

avcamethyst (20:18:16)
were these problems examples of harder or easier ones in the course?

nsato (20:19:55)
These problems are a bit on the easier side; most are more challening than this.

mihail911 (20:18:23)
what former knowledge is needed to do well/understand this class?

rrusczyk (20:20:26)
Probably the best indicator is past performance in contests. If you are able to regularly pass the AMC, then this would probably be a good course for you. If you don't have a good background in basic geometry, then this wouldn't be a good class for you.

bluesky (20:18:33)
What books are useful for AIME preparation?

rrusczyk (20:21:08)
Art of Problem Solving Volume 2. Art and Craft of Problem Solving by Paul Zeitz. We will also have Intermediate level texts coming out later this year that will be good AIME and early Olympiad preparation.

Jennie93 (20:21:04)
were these problems examples of what's in the AMC10, AMC12, or the AIME?

rrusczyk (20:21:18)
Yes.

mihail911 (20:19:26)
up to what math level does the AIME span (e.g. algebra 2, ectr)?

rrusczyk (20:21:46)
Up through trigonometry/pre-calculus (though only a little of the test requires trig, typically)

Alex Whatley (20:18:41)
How long should we take this course to get ready for the AIME A?

rrusczyk (20:22:05)
Expect to spend 3.5-4.5 hours per week on the course.

Jennie93 (20:21:51)
Some of each?

rrusczyk (20:22:22)
AMC 10 for the AMC 10 class, AMC 12 for the AMC 12 class, AIME for the AIME class

mihail911 (20:22:02)
what books will you be releasing, intermediate algebra, intermediate geometry?

rrusczyk (20:23:31)
Counting & Probability and Algebra (hopefully) this year. Others will come next year and the year thereafter.

Alex Whatley (20:22:17)
which course should we take if we failed the AIME tests by .5 points and we failed the AMC tests by 4 points?

rrusczyk (20:24:24)
Not sure what you mean, but if you're at the borderline of passing the AMC, you'll get more out of the AMC class (10 or 12) than the AIME class

aufha (20:24:02)
How does algebra 2 compare to intermediate algebra?

rrusczyk (20:24:46)
Intermediate Algebra is a good deal more in-depth and more challenging (and more interesting) than the typical Algebra 2 class.

bluesky (20:24:11)
is the only way to purchase the AoPS books through the AoPS site online?

rrusczyk (20:25:04)
The only easy way, I think. You might be able to get them on Amazon sometimes.

bizzielilbee101 (20:24:58)
what if you had geometry but still feel weak in it (even though you passed the course)

rrusczyk (20:25:18)
Study AoPS Volume 1 closely.

avcamethyst (20:25:16)
if we only got 4 right on the AIME this year, should we take this course or focus on the AMC?

rrusczyk (20:25:45)
Take the AIME class, Intermediate classes, and consider WOOT in the fall if you do a lot of preparation during the summer.

rrusczyk (20:25:59)
Speaking of Intermediate classes, I'll talk about the trig class now, then take more questions.

rrusczyk (20:26:09)
Intermediate Trigonometry/Complex Numbers

rrusczyk (20:26:14)
The Intermediate Trigonometry/Complex Numbers course is for students who have mastered basic algebra and geometry, including: the Pythagorean Theorem, solving basic equations and systems of equations, graphing simple functions, the geometry of triangles, circles, and arcs, and the basic arithmetic of complex numbers. We do not assume any past experience working with trigonometry, but it would be helpful if you have some.

rrusczyk (20:26:27)
Here's an example of a trigonometry problem that we'll be doing in the course. The solution of this problem may use a couple of trig identities that you're not familiar with. Don't worry -- we'll be covering them in the class!

rrusczyk (20:26:44)


rrusczyk (20:26:49)
No calculators!!!

rrusczyk (20:27:05)
What should we do? Should we try to compute sin(20) and all the others?

harbinger (20:27:19)
no.

rrusczyk (20:27:31)
No; there is no way we can compute sin(20) by hand. So what should we do instead?

pythag011 (20:27:15)
sin 20= 2sin10cos10

rrusczyk (20:28:00)
Let's put everything in terms of 10s rather than 10s and 20s. (In general, we always want to reduce the number of different angles we're dealing with if possible.)

rrusczyk (20:28:06)
We know that sin(20)=2sin(10)cos(10). What does our expression then become?

harbinger (20:28:54)
2sin(10)cos(10)(tan(10)+cot(10))

rrusczyk (20:29:03)


rrusczyk (20:29:07)
The ugliest thing left is the tan(10)+cot(10). What can we do about that?

majesticman (20:28:12)
convert all tan's and cot to sin and cos

avcamethyst (20:29:14)
tan=sin/cos

avcamethyst (20:29:20)
cot=cos/sin

rrusczyk (20:29:25)
We write it terms of sines and cosines (sines and cosines, sines and cosines; not always the way to the solution, but it's usually the best way to bet).

rrusczyk (20:29:31)
We get:

rrusczyk (20:29:34)


rrusczyk (20:29:41)
Now what?

pythag011 (20:29:58)
sin^2+cos^2=1

aufha (20:30:00)
disribute?

avcamethyst (20:30:14)
2(sin squared 10 + cos squared 10)

bluesky (20:30:17)
add the two fractions by making a common base

rrusczyk (20:30:29)
We get a common denominator, and we have:

rrusczyk (20:30:34)


rrusczyk (20:30:39)
So what is the answer to the problem?

kirkhuang (20:30:25)
Distribute 2sin(10)cos(10) into the expression.

avcamethyst (20:30:25)
which equals 2.

bluesky (20:30:30)
when multiplied, the base is cancelled out

pythag011 (20:30:42)
2

kirkhuang (20:30:47)
2

rrusczyk (20:30:54)


rrusczyk (20:31:04)
Ta-da!

rrusczyk (20:31:10)
This problem is on the easier end of the spectrum of problems we'll be doing. Here are a few examples of more difficult trigonometry problems that we'll be covering in the class:

rrusczyk (20:31:17)
(We won't be solving these right now!)

rrusczyk (20:31:21)


rrusczyk (20:31:29)


rrusczyk (20:31:48)


rrusczyk (20:32:07)
I repeat, those are among the harder problems we'll be doing!

rrusczyk (20:32:16)
They won't all look like these!

rrusczyk (20:32:23)
Trigonometry and complex numbers are not only very interesting topics in their own right, they also are extremely useful problem solving tools, and can be used to solve many problems which, on the surface, don't appear to involve trigonometry or complex numbers at all.

rrusczyk (20:32:28)
The following two problems are examples of such problems. The first one is solved using trigonometry, the second one is solved using complex numbers. They are very, very hard to solve without these tools. We won't work through the solutions here, but we'll cover them in the course:

rrusczyk (20:32:39)


rrusczyk (20:32:43)


rrusczyk (20:33:21)
Again, these are on the harder end of the problems we'll be doing. We'll be taking gradual steps starting from the basics, through moderately hard problems, then on to these really tough problems in the course.

rrusczyk (20:33:27)
In the course, we will cover the definitions and properties of the trigonometric functions, various trigonometric identities, the Law of Sines and Law of Cosines, the definition and properties of complex numbers and the complex plane, the exponential form of complex numbers, De Moivre's Theorem, the geometry of complex numbers, and roots of unity.

rrusczyk (20:33:35)
Any questions?

mihail911 (20:33:49)
are these topics covered in AoPS volume 2?

rrusczyk (20:34:25)
Some are. I will suggest readings in AoPS Volume 2 to help students prepare for classes. The Interm class will go a good bit beyond AoPS 2, however.

avcamethyst (20:34:24)
does this focus more on trig in competition-math or trig in general?

rrusczyk (20:35:09)
We'll cover both. We'll go through the basics that you have in a trig class pretty quickly, then move on to tougher stuff.

bluesky (20:34:48)
are there any books that go beyond AoPS volume 2?

rrusczyk (20:35:31)
Paul Zeitz's Art and Craft of Problem Solving is what we recommend when you finish AoPS Vol 2

aufha (20:34:39)
Which contests will the course be most useful for?

rrusczyk (20:36:02)
Hard AMC problems, AIME problems, even a few olympiad problems. HMMT, too, and ARML.

harbinger (20:34:53)
What would be a good judgment of whether or not I need this class?

rrusczyk (20:36:23)
There is an 'Are You Ready?' test on the Enrollment page. This is a good guide.

IntellectMage (20:34:59)
When will the next Intermediate Trig & Complex course be held?

rrusczyk (20:36:37)
We may offer it next spring.

Alex Whatley (20:35:52)
When will the course end?

rrusczyk (20:36:51)
August 30

bizzielilbee101 (20:34:59)
is there an AoPS 3

rrusczyk (20:37:06)
Nope. Some people call Zeitz's book AoPS 3

mihail911 (20:36:15)
what contests/competitions is the art and craft of problem solving most appropriate for?

rrusczyk (20:37:16)
AIME + Olympiad

rrusczyk (20:37:45)
Any more questions?

bluesky (20:37:48)
What level is HMMT at? AIME or Olympiad?

rrusczyk (20:38:01)
AMC/AIME, mostly

aufha (20:37:57)
What do you recommend for help with memorizing identities?

rrusczyk (20:38:32)
Take this class, and through it learn how to derive all these identities very quickly whenever you need them.

mihail911 (20:38:12)
what is HMMT?

rrusczyk (20:38:44)
Harvard - MIT Math Tournament

rrusczyk (20:38:59)
Google for it - they have some excellent tests from past years available

rrusczyk (20:39:29)
Any more questions?

Alex Whatley (20:39:44)
Will taking this course help you improve your grades on the AIME?

rrusczyk (20:40:15)
It should help with trigonometric and complex numbers problem on the AIME. We will do a few AIME problems in the course, too.

rrusczyk (20:41:20)
That's it for the Math Jam. Thanks for attending! If you have any more questions about the classes, please write us at classes@artofproblemsolving.com.

rrusczyk (20:41:33)
We look forward to working with you!