| Transcript
for the Math
Jam "AoPS Classes Math Jam"
on Oct 14. |
| Math Jam hosted by rrusczyk
(Richard Rusczyk ). |
rrusczyk19:29:21
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 Intermediate Number Theory seminar. We will go through a couple example problems from each class, and discuss both what these classes cover and how they work.
rrusczyk19:29:31
My name is Richard Rusczyk. I founded Art of Problem Solving and have written several Art of Problem Solving textbooks.
rrusczyk19:29:36
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.
rrusczyk19:29:59
The classroom is moderated: students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. So, when you send a message, it will not appear immediately, and may not appear at all. This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read. Also, only moderators can enter into private chats with other people in the classroom.
rrusczyk19:30:14
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. I will let you know when to start asking questions about the classes.
rrusczyk19:30:35
The virtual classroom is LaTeX enabled. LaTeX allows users to make nice equations and other math expressions. If you would like to learn how to write in LaTeX, click on the tab on the left side panel of our site and there is a tutorial and reference guide there.
rrusczyk19:30:42
You do not need to learn LaTeX to use our classes or our classroom!
rrusczyk19:30:50
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
rrusczyk19:30:57
rrusczyk19:31:01
This message will look like this when posted in the classroom:
rrusczyk19:31:04
rrusczyk19:31:13
Just remember, if your post uses LaTeX, use the semicolon (;) to begin your post!
rrusczyk19:31:23
One last thing: we recommend not to use a wireless connection while in the classroom. These have a tendency to cause disconnections. Please use a wired connection if possible.
rrusczyk19:31:32
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.
mewto5555519:31:43
i cant see anything to the right of \pm1 in your one post
rrusczyk19:31:54
You can widen the classroom.
rrusczyk19:32:15
Click on the bar between where the class text is and where the usernames are. You can also enlarge the whole window with the classroom.
rrusczyk19:32:19
We'll be discussing the AMC 10 Problem Series, then the AMC 12 Problem Series, and then the Intermediate Number Theory class.
rrusczyk19:32:24
For those of you who are here because you are enrolled in the MATHCOUNTS class or the Introduction to Algebra class, and are just testing out the classroom, nearly all of the math we will be doing tonight will be well beyond your level right now. Don't panic -- the MATHCOUNTS class and the Algebra class start at a much earlier point mathematically, and you'll be up to the point of doing the kind of problems we do tonight in due time!
rrusczyk19:32:45
AMC 10
rrusczyk19:32:50
The AMC 10 class starts on October 20, and meets every Monday from 7:30-9:00 PM Eastern. The class meets for 12 weeks and ends on January 26. The course is designed to cover a large portion of the curriculum tested on the AMC 10 exam.
rrusczyk19:32:59
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.
rrusczyk19:33:12
This class is taught by David Patrick, who was a 2-time Math Olympiad Summer Program invitee and a winner of the USA Math Olympiad back in high school. He has a Ph.D. in Mathematics from MIT and is the author of two of Art of Problem Solving's textbooks: Introduction to Counting & Probability and Intermediate Counting & Probability.
rrusczyk19:33:28
The following are excerpts of a couple of the areas of problem solving covered in the AMC 10 Problem Series.
rrusczyk19:33:30
ARITHMETIC SEQUENCES
rrusczyk19:33:38
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.
rrusczyk19:33:56
rrusczyk19:34:14
Where should we start?
mewto5555519:34:43
set up an equation
rrusczyk19:34:54
We have a word problem, so we'd like to convert it to math.
james4l19:35:00
sum of angles = (n-2)180
herefishyfishy119:35:00
Use the formula for the sum of the measures of the angles of an n-gon
rrusczyk19:35:14
Conveniently, we have a formula for the sum of the interior angles of a convex n-sided polygon: 180(n - 2).
rrusczyk19:35:25
How else can we express the angles of the polygon?
lowtopology19:35:32
We know that each angle is of the form 160-5n degrees
rrusczyk19:35:51
We know that the largest angle is 160, the next largest is 155, and so on . . . down to what?
rrusczyk19:36:18
We don't necessarily go all the way down to 5!
mewto5555519:36:23
down to 160-5(n-1)
xsk1319:36:23
160 - 5(n-1)
rrusczyk19:36:53
rrusczyk19:36:56
Remember that the last value should be 160 - 5(n-1), not 160 - 5n because we started with 160 in the list.
rrusczyk19:37:04
So, what now?
james4l19:37:33
add the terms up
mewto5555519:37:33
Well we can sum that because it is an arithmetic sequence
xsk1319:37:39
Set the sum of the angles equal to (n-2)180
rrusczyk19:37:53
We add these angles and set the result equal to 180(n-2).
rrusczyk19:38:02
rrusczyk19:38:10
How can we simplify this sum?
modx0719:38:58
160n-5(n(n-1)/2))
james4l19:38:58
(1+2+...+(n-1))=(n-1)n/2
lowtopology19:39:11
1+2+...+(n-1) = (n-1)(n)/2
rrusczyk19:39:14
rrusczyk19:39:41
rrusczyk19:39:45
What now?
techsam2k819:40:31
solve for n
lowtopology19:40:31
solve for n
Jesse19:40:31
solve for n
rrusczyk19:40:34
We'll start with:
futbill202519:40:35
multiply by 2/5
rrusczyk19:40:49
With a little more algebra we get to:
techsam2k819:40:50
its becomes 0=n^2 +7n-144
rrusczyk19:40:52
FantasyLover19:41:34
(n+16)(n-9)=0 thus n=9
lowtopology19:41:34
(n+16)(n-9) = 0
mewto5555519:41:34
(n+16)(n-9)=0, so n=9
rrusczyk19:41:42
We can solve the quadratic in a number of ways including factoring it into
(n - 9)(n + 16) = 0, so n = 9 or -16.
lowtopology19:41:47
n=9
gu_co19:41:47
9
ericmwalsh19:41:47
rrusczyk19:41:51
We know that a polygon cannot have a negative number of sides and so we have our answer, n = 9. (A).
rrusczyk19:41:55
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.
rrusczyk19:42:05
We'll now move from algebra to counting.
rrusczyk19:42:13
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.
rrusczyk19:42:16
We can solve this problem by considering the arbitrary points A, B, C, D, E, and F. Since no three of the points are collinear, we know that any set of three points forms a (nondegenerate) triangle. We might start by counting the number of three letter strings we can form by choosing one of each letter:
rrusczyk19:42:22
ABC
ACB
BAC
BCA
and so on.
rrusczyk19:42:32
How many of these strings are there?
rrusczyk19:42:57
(Here, we are not counting the triangles yet, just the number of 3-letter strings we can form with these 6 letters.)
james4l19:43:23
6*5*4=120
FantasyLover19:43:23
6*5*4=120
styrofoam199419:43:23
6*5*4
mewto5555519:43:25
rrusczyk19:43:31
We can choose 6 possible letters to be the first letter, 5 to be the second, and 4 to be the third, so there are 6*5*4 = 120 such strings.
rrusczyk19:43:54
That gives us a count of the 3-letter strings. But to count the triangles, we only want three letter groups in which the order does not matter. Triangle ABC is the same as ACB is the same as CAB, etc.
rrusczyk19:43:59
How do we correct our count for this?
feuxfollets19:44:18
zepiphanus19:44:18
divide by 6
FantasyLover19:44:18
divide by 6
hyperfusion19:44:18
120/6 = 20
Yimocool19:44:18
120/6=20
modx0719:44:18
divide by 3!
xsk1319:44:25
we divide by the number of times we can get repeats - (3!)
rrusczyk19:44:27
There are 3! possible strings with A, B, C, so we count triangle ABC 6 times in our count of 120 ordered strings above. Similarly, we count every group of three letters exactly 6 times, so to count the total number of different groups, we divide our count of 120 ordered strings by 6 to get 120/6 = 20.
rrusczyk19:44:42
This is what we mean by 'combinations': the number of ways in which we can choose r objects from a group of n objects if we don't care about order.
rrusczyk19:44:53
MathMasta_219:45:13
n choose r?
rrusczyk19:45:27
rrusczyk19:45:30
We have already seen how to compute a combination. We first do a count of selecting r items from the n if order matters, and then we divide by r!.
rrusczyk19:45:36
rrusczyk19:45:58
If you are unfamiliar with combinations, you can learn all about them with our Introduction to Counting & Probability book.
rrusczyk19:46:18
Let's try a few problems with combinations.
rrusczyk19:46:19
xsk1319:47:21
there are 8 vertices of a cube, and a line segment is made from 2 vertices, so 8C2
FantasyLover19:47:21
james4l19:47:21
8C2=28, so D
mewto5555519:47:21
rrusczyk19:47:34
rrusczyk19:47:40
Understanding combinations can directly lead us to quick solutions to some AMC problems!
rrusczyk19:47:54
Next problem:
rrusczyk19:47:55
rrusczyk19:48:15
If it's not obvious how to do a counting problem, a good way to start is to play with it.
rrusczyk19:48:31
What might we do here?
james4l19:48:55
when they mean "increasing order" is it strictly increasing order, (e.g. does 224 count as increasing order?)
rrusczyk19:49:02
No -> strictly increasing.
rrusczyk19:49:16
One way I like to play with counting problems is to try to construct one of the items we're trying to count.
rrusczyk19:49:32
How can we construct a number that has its digits in increasing order?
rivmath19:50:40
would 137 work or would it hav to be strictly 123, 234, etc..?
rrusczyk19:50:45
137 works.
pianogirl19:50:54
take a number that is the 1st digit, then add one, that is the second, add one, you have the 3rd: 123
rrusczyk19:51:09
That's one way to make one. Are there other ways to come up with numbers with increasing digits?
MathMasta_219:51:15
137 works because the number after is more than the one before
rrusczyk19:51:17
True.
silvercharcoal519:51:27
Maybe let the number be ABC, where A<B<C?
rrusczyk19:51:36
OK, suppose I give you A, B, and C. Then what?
herefishyfishy119:51:58
Put them in order
rrusczyk19:52:26
Yep, if I give you an A, B, and C, then you put them in order to make an increasing number. Does this help you count how many increasing numbers there are?
zepiphanus19:53:07
there is only one permutation strictly increasing for any set of A, B, C
zepiphanus19:53:11
for every A, B, C chosen without replacement, there is only one permutation of A, B, C that is strictly increasing
rrusczyk19:53:42
Aha! How does this observation help us count the number of increasing numbers?
rrusczyk19:54:09
Zepip had noted that whatever three digits I give you, there is only one way to put the three of them in increasing order.
rrusczyk19:54:15
How does this help?
zepiphanus19:54:52
which is just the same thing as 9C3
MathMasta_219:54:52
we find the combinations of all the A,B,C can be
vahalla19:54:52
We use a combination. There are 9*8*7/3*2*1 possibilities to pick three different numbers.
rrusczyk19:55:01
Why is it C(9,3) and not C(10,3)?
FantasyLover19:55:21
because first digit can't be 0
futbill202519:55:21
you can't use 0
MathMasta_219:55:21
because 0 cannot be the first digit
styrofoam199419:55:21
because a cannot be 0
herefishyfishy119:55:24
We cannot choose 0 for the first digit
modx0719:55:24
0 doesnt work as the first digit
lowtopology19:55:24
because hundreds digit cannot be 0
rrusczyk19:55:36
rrusczyk19:56:24
(A lot of you were attacking the increasing numbers using casework based on the first digit -- if you keep going with that reasoning, you'll prove a nice little combinatorial identity. Ask about it on the message board later on if you don't figure it out on your own.)
rrusczyk19:56:27
Now, how many three digit numbers will have all their digits in decreasing order?
late20s19:56:57
but for the decreasing digits we can include 0
FantasyLover19:56:57
modx0719:56:57
so do the same exact thing for decreasing...except that 0 CAN be the last digit
herefishyfishy119:56:57
late20s19:56:57
10*9*8/3*2*1
rrusczyk19:57:03
Yimocool19:57:27
204
lowtopology19:57:27
204
infinitypi19:57:27
204
mewto5555519:57:27
120+84=204
infinitypi19:57:27
204
futbill202519:57:27
120 + 84 = 204
james4l19:57:27
10c3+9c3=204, so C
alligator11219:57:27
120 + 84 = 204
rrusczyk19:57:32
We now add the total numbers from each case:
84 + 120 = 204, so our answer is (C).
rrusczyk19:57:43
That does it for the problems from the AMC 10 course.
rrusczyk19:57:48
Are there any questions about the course?
futbill202519:58:10
can you use a calculator
rrusczyk19:58:27
During the class, sure, but I recommend that you don't, because they don't allow calculators on the AMC anymore.
modx0719:58:30
What is the difference in difficulty between the AMC 10 & 12 courses?
rrusczyk19:58:39
Roughly the same as the difficulty level between the tests.
pianogirl19:58:54
are the problems from the beginning, middle or end of the test?
rrusczyk19:59:17
In the course, we have few problems from the start of the test. We focus much more on problems among the last 10 problems of the test.
wz111819:59:19
does amc 10 mean you take it in tenth grade?
rrusczyk19:59:34
It means that you cannot take it after 10th grade. You can take it before 10th grade.
modx0719:59:37
What book do we use?
rrusczyk20:00:00
There's no book required for the course; we recommend our Art of Problem Solving Volume 1 for AMC 10 preparation.
silvercharcoal520:00:05
What benefits will the AMC 10 have in your high school career?
rrusczyk20:00:33
That link describes what problem solving is, and why it is important.
rrusczyk20:00:49
I will note that my preparation for the AMC and similar contests made college *very, very* easy.
wz111820:00:59
can you take it during tenth grade
rrusczyk20:01:03
Yes.
gu_co20:01:05
Could you recommend a strong Combinatorics book?
mewto5555520:01:40
here comes the AOPS book sales pitch
rrusczyk20:01:42
:)
rrusczyk20:01:43
Beginning level: our Introduction to Counting & Probability. More advanced (AIME, beginning USAMO): our Intermediate Counting & Probability.
silvercharcoal520:02:05
Will there be a final exam?
rrusczyk20:02:14
There are no exams in any of the classes we are discussing tonight.
futbill202520:02:20
what score should you get on the AMC 10 to try and score well on the AMC 12
rrusczyk20:02:38
This is a tough call -- I usually recommend most students take both after grade 8.
Yimocool20:02:48
I am taking this course to prepare for the AMC 10. Is this a good idea or should i switch to the AMC 10 course?
rrusczyk20:02:57
The problems we just discussed were from the AMC 10 class.
rrusczyk20:03:32
The AMC 12 class will help with the AMC 10 as well, but there are 2-3 days of advanced material in the AMC 12 class that are not relevant to the AMC 10 (trig, polynomials, etc).
futbill202520:03:34
is there a course to prepare you for the AIME's
rrusczyk20:03:41
Yes; it starts in December.
silvercharcoal520:03:43
Without us taking tests and submitting homework, how do you track our progress?
rrusczyk20:04:31
These Problem Series classes are not graded like your typical class -- they're for extra practice for the contests. Our subject classes have homeworks you turn in and get feedback on.
rrusczyk20:04:45
We do monitor student work on the course message board, and comment on student solutions there.
Math46_220:04:59
Is it crazy for a 6th grader to take the AMC 10?
rrusczyk20:05:08
No - it is not that uncommon at all.
rivmath20:05:10
If I am only able to choose one class to participate in, which one would be most important for the AMC 10?
rrusczyk20:05:14
The AMC 10 Problem Series.
rrusczyk20:05:30
Are there any more questions about the AMC 10 Problem Series?
futbill202520:05:44
will we be solving som AMC 12 problems tonight
rrusczyk20:05:48
Yes. Let's talk about that now.
rrusczyk20:05:56
AMC-12
rrusczyk20:06:00
The AMC 12 class starts on October 24, and meets every Friday from 7:30-9:00 PM Eastern. The class meets for 12 weeks and ends on January 30. The course is designed to cover a large portion of the curriculum tested on the AMC 12 exam.
rrusczyk20:06:03
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.
rrusczyk20:06:16
The class will be taught by David Patrick and Naoki Sato.
rrusczyk20:06:19
David Patrick was a 2-time Math Olympiad Summer Program invitee and a winner of the USA Math Olympiad back in high school. He has a Ph.D. in Mathematics from MIT and is the author of two of Art of Problem Solving's textbooks: Introduction to Counting & Probability and Intermediate Counting & Probability.
rrusczyk20:06:23
Naoki Sato 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.
mewto5555520:06:40
he is a beast
rrusczyk20:06:43
Indeed :)
wz111820:06:47
who is teaching the amc10 class?
rrusczyk20:06:50
David Patrick.
rrusczyk20:07:09
The following is an excerpt of one of the areas of problem solving covered in the AMC 12 Problem Series.
rrusczyk20:07:21
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
rrusczyk20:08:27
Hurm. Where do we start? How can we tell if a point is closer to the origin than it is to (3,1)?
futbill202520:08:51
draw the halfway line between (3,1) and the origin
herefishyfishy120:08:51
Find the perpendicular bisector of (0,0)-(3,1)
james4l20:08:51
perpendicular bisector
rrusczyk20:09:01
The set of points (in a plane) equidistant to two given points is the perpendicular bisector of the line segment with the two points as endpoints.
rrusczyk20:09:37
So, the perpendicular bisector of the segment that connects (0,0) and (3,1) splits the plane into points closer to the origin and points closer to (3,1).
rrusczyk20:10:17
So, we should find this perpendicular bisector. How?
mewto5555520:10:43
well we draw the line that seperates the region closer to 3,1 than 0,0. This line goes through (3/2,1/2) the midpoint, and has a slope perpendicular to the line connecting 3,1 and 0,0. THis line has slope -1/3
futbill202520:10:43
find the midpoint
ericmwalsh_620:10:43
midpoint
styrofoam199420:10:43
reciprocal of slope of OP
xsk1320:11:00
the slope will be the negative reciprocal of 1/3, which is -3
FantasyLover20:11:00
the slope of the line connecting the origin and (3,1) is 1/3, so the perpendicular bisector's slope is -3
rrusczyk20:11:07
The midpoint of the segment between (0, 0) and (3, 1) is (1.5, .5).
rrusczyk20:11:11
The equation of the line through (0, 0) and (3, 1) is y = x/3.
rrusczyk20:11:27
The slope of a line perpendicular to that line is -1/(1/3) = -3. (The product of the slopes of two perpendicular lines in a plane is -1 if neither line is vertical.)
rrusczyk20:11:45
Now what?
rrusczyk20:12:44
Yikes, I see a lot of wrong answers.
rrusczyk20:12:45
Be careful.
FantasyLover20:13:16
put (3/2,1/2) in y=-3x+a
FantasyLover20:13:16
therefore a=5, so the equation of the line is y=-3x+5
mewto5555520:13:16
y-.5=-3(x-1.5)-> y=-3x+5
rrusczyk20:13:21
We want the line through (1.5,0.5) with slope -3. We can use point-slope form:
y-0.5 = -3(x-1.5).
rrusczyk20:13:32
Rearranging this gives y = -3x + 5.
rrusczyk20:13:44
Now what do we do with this?
xsk1320:14:07
Find the areas of the two regions the line splits the rectangle into.
MathMasta_220:14:07
you draw the line to separate the rectangle in 2.
ericmwalsh_620:14:08
find the area between this shape and the rectangle and the other area and compare
rrusczyk20:14:21
Exactly. We plot our rectangle, and draw our line.
rrusczyk20:14:33
rrusczyk20:15:01
The red rectangle is the region we must choose P from in the problem.
rrusczyk20:15:13
The grey area is the portion of the rectangle closer to the origin.
rrusczyk20:15:49
How can we find the area of that region?
mewto5555520:16:06
this intersects the rectangle at point -3x+5=1. -3x=-4, x=4/3
mewto5555520:16:06
it also intersects at y=0, x=5/3 so we have a trapezoid
techsam2k820:16:06
area of trapezoid is (4/3 + 5/3)/2
mewto5555520:16:06
obviously, the bases are 4/3 and 5/3 and the height is 1. Using the formula for the area of a trapezoid we have 1x(4/3+5/3)/2=9/6 we divide this by 2 because the rectangle is area 2. thus 3/4
vahalla20:16:13
Use the formula for area of a trapezoid.
rrusczyk20:16:18
rrusczyk20:16:27
We can indeed find the area of the trapezoid.
rrusczyk20:16:34
We need the height, which is 1.
rrusczyk20:16:38
We also need the bases.
rrusczyk20:16:59
We can determine those by finding where the lines y=0 and y=1 intersect y = -3x+5 (the slanted line).
rrusczyk20:17:03
This happens at (5/3, 0) and (4/3, 1).
We can see that 4/3 and 5/3 are the lengths of the bases.
rrusczyk20:17:13
So, what is the area of the trapezoid?
herefishyfishy120:17:47
3/2
futbill202520:17:47
3/2
hyperspace122120:17:47
3/2
ericmwalsh_620:17:47
rrusczyk20:17:56
The area of the trapezoid is 1 (the height) times the average of the bases which is 3/2.
rrusczyk20:18:08
What is our final answer for the probability?
herefishyfishy120:18:55
MathMasta_220:18:55
3/4
hyperfusion20:18:55
3/4
FantasyLover20:18:55
the answer is (3/2)/2, which is 3/4.
infinitypi20:18:55
3/4
N134920:18:55
3/4
hyperspace122120:18:55
3/4
rrusczyk20:19:00
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).
rrusczyk20:19:06
rrusczyk20:19:08
Does anyone see a faster way to find the area of the trapezoid?
simo1420:20:03
the line cuts the second square in half
infinitypi20:20:03
we know that the area of the trapezoid is half of the square so it's half of 1 which is 1/2
modx0720:20:03
well, isnt the part taken out 1/2?
rrusczyk20:20:07
The line passes through (1.5,0.5), which is the center of the square that point is in. Therefore, the line splits that square in half. So, the trapezoid is one full square plus one half square, and we see that we didn't even have to find the equation of the perpendicular bisector!
futbill202520:20:09
how do you find out the length of hte bases. i'm confused
rrusczyk20:20:20
The slanted line is the graph of y = -3x + 5.
rrusczyk20:20:34
We found where that intersects y = 0 and y= 1 (these two lines contain the bases of the trapezoid.
futbill202520:20:53
I get it. you plug in 1 an 0 for y
rrusczyk20:20:55
Exactly.
skier20:21:09
yes, but that formula only works if the diagram is accurate, they don't always make accurate diagrams on the AMC tests.
rrusczyk20:21:44
You aren't given a diagram here, but you do need to at least know something about the slope of the perpendicular bisector to be confident you are correct.
rrusczyk20:22:08
Let's try another problem
rrusczyk20:22:16
rrusczyk20:22:34
That link is just a copy of the problem.
rrusczyk20:22:38
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!
rrusczyk20:22:50
Where do we start?
rrusczyk20:23:32
I see that some of you are suggesting adding the inequalities.
rrusczyk20:23:35
Does that really help?
lowtopology20:24:25
f(x)< or =0
zepiphanus20:24:25
it shows that f(x)<=0
rrusczyk20:24:27
That just gives you f(x) <= 0.
rrusczyk20:24:31
Not so clear what to do with that.
techsam2k820:24:34
the left is a circle and the right is a line
rrusczyk20:24:40
How did you figure that out?
rrusczyk20:25:14
MathMasta_220:25:35
if u plug in the quadratics it makes forms of a line and circle
rrusczyk20:25:54
Exactly. All we do is go ahead and write the two inequalties using our definition of f.
rrusczyk20:25:59
We get:
rrusczyk20:26:02
rrusczyk20:26:19
All we did here was replace f(x) and f(y) using the definition of the function.
rrusczyk20:26:37
What happens with the equation on the left?
herefishyfishy120:26:51
Complete the square
hyperfusion20:26:51
complete the square
lowtopology20:26:51
complete the square
ericmwalsh_620:26:51
complete the square .
rrusczyk20:26:54
And what do we get?
mewto5555520:27:26
(x+3)^2+(y+3)^2=16
styrofoam199420:27:26
(x+3)^2+(y+3)^2=16
lowtopology20:27:45
(x+3)^2 + (y+3)^2 =16
rrusczyk20:27:53
Be careful; we're not doing an equation!
hyperfusion20:28:22
<=
mewto5555520:28:29
oh gah <=
lowtopology20:28:31
<=16...oops again
futbill202520:28:34
(x+3)^2 + (y+3) ^ 2 <= 16
rrusczyk20:28:50
rrusczyk20:29:24
What is the graph of this region?
MathMasta_220:29:40
this is a form of a circle so we can graph it to find its location on the graph
ericmwalsh_620:29:40
circle
MathMasta_220:29:40
a circle
herefishyfishy120:29:40
A circle
simo1420:29:40
center (-3,-3) and radius 4
herefishyfishy120:29:44
(x,y) must be inside or on the circle, because the distance from (-3,-3) is less than 4.
james4l20:29:48
A circle bounded at radius 4, center (-3,-3)
ycz900020:29:48
A circlular region of radius 4, centered at (-3,-3)
rrusczyk20:30:09
The graph of this inequality is a circular disk with center (-3,-3) and radius 4.
rrusczyk20:30:24
rrusczyk20:31:29
There are a lot of ways to deal with this one.
rrusczyk20:31:50
We can complete the square again, and we'll get:
ericmwalsh_620:31:56
(x+3)^2 <= (y+3)^2
lowtopology20:31:56
(x+3)^2 - (y+3)^2 <= 0
mewto5555520:31:56
(x+3)^2<=(y+3)^2
rrusczyk20:32:37
What can we do with lowtopology's inequality?
xsk1320:32:59
difference of squares
rrusczyk20:33:35
Note: we don't want to take the square root! We can't assume from a^2 <= b^2 that a <= b, since a or b might be negative!
rrusczyk20:33:49
We use difference of squares:
simo1420:33:51
(x+3+y+3)(x+3-y-3)<=0
rrusczyk20:33:56
Simplifying this gives:
mewto5555520:34:00
(x+3+y+3)(x-y)=(x+y+6)(x-y)<=0
gu_co20:34:00
(x-y)(x+y+6)<=0
lowtopology20:34:00
(x+y+6)(x-y) <= 0
rrusczyk20:34:09
rrusczyk20:34:29
What is the graph of this inequality?
gu_co20:35:01
two lines
herefishyfishy120:35:01
a pair of lines
MathMasta_220:35:15
2 lines
infinitypi20:35:33
two liens
rrusczyk20:35:43
The graph of where the left side equals 0 is a pair of lines.
rrusczyk20:35:48
What do we know about those lines?
MathMasta_220:36:46
they are perpendicular
mewto5555520:36:46
PERPENDICULAR!
rrusczyk20:36:48
What else?
alligator11220:36:53
intersect at (-3,-3)
rrusczyk20:36:57
And what's nice about that?
hyperfusion20:37:10
that's where the circle's centered
futbill202520:37:10
center of the circle
mewto5555520:37:10
center of our circle
alligator11220:37:11
center of circle
gu_co20:37:11
the center of the circle
rrusczyk20:37:16
And that makes us happy!
techsam2k820:37:49
ah, so we take area of a quarter of circle?
rrusczyk20:37:56
Is it just a quarter?
ericmwalsh_620:38:07
half.
late20s20:38:07
half
rrusczyk20:38:12
alligator11220:38:15
half of area of circle
infinitypi20:38:15
it's a half
rrusczyk20:38:49
rrusczyk20:38:53
We want half the circle.
alligator11220:39:01
16pi/2=8pi
styrofoam199420:39:01
futbill202520:39:01
8pi
mewto5555520:39:10
thuse the area is 16pi/2=8pi=25.2 so E. Hm I guess it isn't C again xD
kuznetsivan_220:39:20
8pi so about 25
rrusczyk20:39:24
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).
rrusczyk20:39:31
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.
rrusczyk20:39:33
Are there any questions?
AN LE20:39:36
I have one question, I am in 10th grade and i want to score well on the AMC 10, what do you suggest?
rrusczyk20:39:52
I suggest spending a fair amount of time with Art of Problem Solving Volume 1, along with:
MathMasta_220:39:53
study past tests
rrusczyk20:40:10
Take past tests under test conditions.
simo1420:40:30
should we master aops v2 for the amc 12?
rrusczyk20:40:44
For the hardest problems in AMC 12; Volume 1 will help a great deal for that, too.
modx0720:40:50
past test conditions (like calculator)?
rrusczyk20:41:08
No - don't use a calculator. Even the tests that allowed calculators didn't require them.
futbill202520:41:12
do we do proofs in class
rrusczyk20:41:21
Not in the AMC classes -- no proofs on the AMC's
infinitypi20:41:23
what if you're a sixth grader and want to score well on AMC 10
rrusczyk20:41:24
Same!
hyperspace122120:41:29
What does volume 2 primarily get you ready for?
rrusczyk20:41:36
Hard AMC problems and the AIME
lowtopology20:41:39
What kinds of subjects show up most on AMC 12?
rrusczyk20:41:56
Algebra, Geometry, Counting, Number Theory. A wide range of each.
poodleuw20:41:58
I have to leave the class room, is there a way to save the class.
rrusczyk20:42:12
There will be a transcript tomorrow in the Math Jams section of the community.
simo1420:42:13
how much can we expect to improve by the end of this class?
rrusczyk20:42:42
Depends on how hard you work. If you work a great deal in the class, you should pass the AMC 12 handily (assuming you come in at least knowing something!)
ericmwalsh_620:42:47
why is it that (x-y) (x+y+6) <= 0 makes half though? i thought it was supposed to be below y>=x
rrusczyk20:43:12
Consider x - y > 0, x + y+ 6 < 0, graph that, and then consider x - y < 0, x+ y+6>0, and graph that.
futbill202520:43:16
should we take the AMC 10 or 12 if we have taken precal
rrusczyk20:43:19
I recommend both.
AN LE20:43:21
how about AMC 10?
rrusczyk20:43:30
Same as the 12, but no trig, logs, or polynomials.
hyperspace122120:43:35
How do we get LaTeX on a Mac? I read the directions, but I'm confused!
rrusczyk20:43:37
Ask on the message board.
lowtopology20:43:40
What kind of score could I expect on AIME if I used this class & study hard, using just knowledgy from this class and Volume 2?
rrusczyk20:44:30
Tough to say -- if you're aiming at the AIME, I'd recommend the AIME class and Intermediate classes. If you work hard and get through Vol 2 and one of the Interm books (and take some practice tests), you should have a fighting chance.
monilshah20:44:34
do the regular classes have audio, or do you just type it in
rrusczyk20:44:36
No audio.
rrusczyk20:44:44
That explains why.
modx0720:45:06
What is the difference between AMC 10/12 and AIME?
rrusczyk20:45:12
AIME is harder. A lot harder.
skier20:45:14
how much do the volumes cost?
rrusczyk20:45:26
You can look up the books you're interested in here:
wz111820:45:50
is there algebra 2 covered in anc10?
rrusczyk20:45:53
Some, but not all.
modx0720:45:58
But whatabout material wise? Does AIME use calculus or something?
rrusczyk20:46:04
No calculus, some more advanced topics.
AN LE20:46:07
so the transcript for this session will be on the Math Jams part of the forum tomorrow?
rrusczyk20:46:15
Yes, on the Math Jams area of the community.
myamc1020:46:19
Should i buy a book to prepare for AMC 10?
rrusczyk20:46:25
If you do, we recommend our Volume 1
rrusczyk20:46:46
Very hard to say. There's a lot of volatility in scores from year to year.
simo1420:46:48
If we score 105-110 on 1980 AHSME's, how comparable is that to current ones?
gu_co20:46:50
is calculus used at all at usa olympiads( I mean directly said in the problem)
rrusczyk20:47:00
It is not required for the USAMO
wz111820:47:01
are there stats on AIME
rrusczyk20:47:05
On the AMC website.
AN LE20:47:07
i have First Steps for Math Olympians, is this a good supplement to go along with art of problem solving v1?
rrusczyk20:47:12
That's a nice book, too.
futbill202520:47:27
what topics are on AMC12 that arent on AMC10
rrusczyk20:47:35
Logs, trig, polynomials.
rrusczyk20:47:42
probably some others.
wz111820:47:44
i meant statistics
rrusczyk20:47:51
Nothing more than mean, median, mode, range.
rrusczyk20:47:56
Let's do so number theory!
rrusczyk20:48:06
Intermediate Number Theory
rrusczyk20:48:10
The Intermediate Number Theory Seminar is an 8 week course for students who have both a strong foundation in basic number theory and solid algebraic skills. The course starts October 21 and meets on Tuesdays, 7:30 PM - 9 PM ET, until December 16. It is recommended that students take the Introductory Number Theory course and Intermediate Algebra course before enrolling, or be confident with the material covered in those classes.
rrusczyk20:48:15
The course is taught by Valentin Vornicu. Valentin joined AoPS in 2004. He founded the MathLinks site for Olympiad students in 2002. 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 was a member of the Romanian International Mathematical Olympiad (IMO) team in 2001 and 2002, winning a bronze medal in 2002.
rrusczyk20:48:37
Topics covered in the Intermediate Number Theory Seminar include algebraic methods of problem solving in number theory, base number problems involving algebra, counting, and qualitative problem solving techniques, divisibility and divisor problems involving algebra, Diophantine equations, modular arithmetic with an emphasis on algebraic applications, perfect squares, an introduction to Pell's equations, Fermat's Little Theorem, Euler's Phi Function, and Euler's Theorem.
rrusczyk20:48:40
As with all AoPS classes, the Intermediate Number Theory Seminar will have its own private message board in the AoPS Forum. The Intermediate Number Theory class will include more message board problems per week of class than other AoPS courses to be sure that students get ample practice with the concepts discussed in each lesson.
rrusczyk20:49:08
We will now work a few problems from different days in the class. We will start with an easier class problem and move into harder ones. Unfortunately, we cannot sample some of the most important topics covered in class because it would take too much time developing the background to discuss them here.
rrusczyk20:49:49
rrusczyk20:50:22
Where do we start?
james4l20:50:42
Factor f(x) = (x+2)(x+1)</span>
FantasyLover20:50:42
factor out x^2+3x+2
rrusczyk20:50:49
rrusczyk20:50:56
How does that help?
gu_co20:51:15
yeah, divisible by two
mewto5555520:51:15
we know either one will be a multiple of 2
rrusczyk20:51:22
We can now see that f(s) will always be even because either x + 1 or x + 2 must be even. This examination of the evenness or oddness of the integers is called a "parity" argument.
rrusczyk20:51:25
Our goal now is to find out exactly when f(s) will be a multiple of 3.
rrusczyk20:51:29
When will f(s) be a multiple of 3?
techsam2k820:51:57
when s is not a multiple of 3 itself
herefishyfishy120:51:57
When s is not equal to 0 mod 3
james4l20:51:57
if s is not a multiple of 3
rrusczyk20:52:03
f(s) = (s+1)(s+2) will be a multiple of 3 when either s + 1 or s + 2 is a multiple of 3. This is the same thing as saying that f(s) will be a multiple of 3 when s is NOT a multiple of 3.
rrusczyk20:52:08
So, what is the answer to the problem?
rrusczyk20:52:12
Here it is again:
rrusczyk20:52:15
mewto5555520:53:01
which is 2/3 of 1-24 or 16 and also when n=25 so 17 members
alligator11220:53:01
17
ericmwalsh_620:53:01
17
alligator11220:53:01
26-9=17
james4l20:53:01
26-9=17
rrusczyk20:53:15
We now count the members of S that aren't multiples of 3 and we count a total of 17 of them. (There are 26 numbers in the set, 9 are multiples of 3.)
rrusczyk20:53:18
The key to this problem was factoring the function. Once we did that, the actual divisibility arguments were relatively simple.
rrusczyk20:53:26
This problem is on the easy end of the problems we cover in the course.
rrusczyk20:53:31
Let's look at another problem.
rrusczyk20:53:39
rrusczyk20:53:58
Where do we start?
mewto5555520:54:34
start listing integers and pray for a pattern?
rrusczyk20:54:46
:) Not a bad place to start if you're totally lost.
rrusczyk20:54:52
But we might start with a general strategy like this:
infinitypi20:54:55
write an equation
rrusczyk20:55:13
We have a bunch of words. And we'd sure like to have mathy stuff to work with.
rrusczyk20:55:21
In other words, we'd like an equation.
rrusczyk20:55:32
rrusczyk20:55:48
All we're doing here is writing down n in terms of some variables that will allow us to make an equation for n.
rrusczyk20:55:56
What equation can we write using this definition?
rrusczyk20:56:46
All I'm looking for here is an equation to write down: n = <something>, and we hope to be able to use that in some way.
herefishyfishy120:57:02
ericmwalsh_620:57:02
2^0 * a0 + 2^1 * a1 + .... 2^ k * ak = n
rrusczyk20:57:06
rrusczyk20:57:32
Exactly -- we just write n in terms of powers of 2, since we have its binary representation.
rrusczyk20:57:37
Now, can we write another equation?
rrusczyk20:58:03
(Notice our general strategy here: all we are doing is trying to make equations out of the given information, in the hopes that we can do something with them.)
MathMasta_220:58:05
the equation for base 3
mewto5555520:58:05
i guess the same thing buit with 2n in base 3?
rrusczyk20:58:11
And what equation do we get?
xsk1320:58:24
rrusczyk20:58:33
And we turn this into an equation just like before:
FantasyLover20:58:36
ericmwalsh_620:58:36
3^0 *a 0 + 3^1 * a1 + .... 3^k * ak =2n
rrusczyk20:58:38
rrusczyk20:58:44
Now we are getting somewhere!
rrusczyk20:58:46
Er, where?
rrusczyk20:58:55
What can we do with these two equations?
herefishyfishy120:59:37
Multiply both sides of the first equation by 2
leafiness020:59:37
put them both in terms of 2n, substitute, and solve
leafiness020:59:37
multiply the first equation by 2, then substitute it into the second equation
rrusczyk20:59:42
We can now apply what I like to call the Fundamental Principle of Algebraic Problem Solving -- find two ways to express the same thing and set them equal!
rrusczyk20:59:48
rrusczyk20:59:56
rrusczyk21:00:09
How does this help?
james4l21:00:53
2^(k+1)-3^k < 0, bad
FantasyLover21:00:53
everything after a_2 is 0
rrusczyk21:00:59
Interesting observation!
rrusczyk21:01:51
The coefficients of everything from a_3 on are big (and growing) negative numbers. So, we can't have that (remember a_0, a_1, and a_2 can be no more than 1).
rrusczyk21:02:07
So, what does that tell us about a_3, a_4, and so on?
FantasyLover21:02:24
they are all 0
infinitypi21:02:24
they are all 0
leafiness021:02:24
they are all zero
rrusczyk21:02:26
We know that a_3, a_4, and so on must all be 0.
james4l21:02:47
so it is just a_0+a_1-a_2=0
james4l21:02:57
Or, a_0+a_1=a_2
rrusczyk21:03:00
rrusczyk21:03:04
Oops
rrusczyk21:03:19
rrusczyk21:03:23
And we are almost home.
rrusczyk21:03:34
What's our answer?
rrusczyk21:03:39
Here is the question again:
rrusczyk21:03:43
james4l21:04:18
000, 101, 110... 3 answers
herefishyfishy121:04:18
3
MathMasta_221:04:18
3
rrusczyk21:04:20
We can have a_0 = a_1 = a_2 = 0, or a_2 can be 1 and either of a_0 or a_1 can be 1 (but not both). So, we get n = 0 , 5, or 6.
rrusczyk21:04:31
(I saw some 2's; don't forget zero!!)
rrusczyk21:04:43
The key to this problem was using a sequence of variables to represent the digits of n so that we could use these variables to compare the two different ways we had to express the value of n. Once we did that, we could fall back on our algebraic skills using equations.
rrusczyk21:05:22
One more question, and then we're finished.
rrusczyk21:05:35
rrusczyk21:06:02
Yikes. Where do we start with this?
ericmwalsh_621:06:24
define relatively prime :P
rrusczyk21:06:30
No divisors in common besides 1.
vahalla21:06:39
Prime factorization of 1000.
rrusczyk21:06:44
OK:
rrusczyk21:06:54
rrusczyk21:07:06
What do we know about divisors of this?
futbill202521:08:36
are in the form 2^a * 5^b
rrusczyk21:08:58
Exactly -- all divisors are of this form. And what do we know about the exponents of 2 and 5 in such a divisor?
herefishyfishy121:09:40
They are not greater than 3
ericmwalsh_621:09:40
cant exceed 3
alligator11221:09:40
a<=3, b<=3
FantasyLover21:09:43
a<=3 and b<=3
futbill202521:09:45
either 0, 1, 2, 3
gu_co21:09:45
0, 1, 2, or 3
rrusczyk21:09:49
rrusczyk21:10:09
How does this help us describe all the values of a/b? How can we describe all the possible values of a/b?
herefishyfishy121:11:03
Since they are relatively prime, one must be a power of 2 and the other a power of 5
rrusczyk21:11:11
OK - tell me about the resulting a/b's
FantasyLover21:11:26
leafiness021:11:26
a/b can either be 2^m/5^n or 5^n/2^m
rrusczyk21:11:34
How might we put these two possibilities together?
james4l21:11:39
basically the sum of all numbers 2^a*5^b where 3 >= a,b >= -3
rrusczyk21:11:43
Exactly.
rrusczyk21:11:49
rrusczyk21:11:55
The new range of the exponents takes into account the fact that powers of 2 and 5 can now be in the denominator.
rrusczyk21:12:05
How can we easily sum these?
james4l21:13:34
This is (2^3+2^2+2^1+2^0+2^-1+2^-2+2^-3)(5^3+5^2+5^1+5^0+5^-1+5^-2+5^-3)
The first part is slightly less than 16 (15 7/8)
The second part is (125+25+5+1+0.2+0.04+0.008) = (156.248) [about 625/4]
127/8*(625/4)= 2480.~
rrusczyk21:13:59
rrusczyk21:14:21
Take some time to consider the expansion of that product to see how we get the desired sum.
rrusczyk21:14:28
As james41 noted, it's just computation now:
rrusczyk21:14:32
rrusczyk21:14:36
rrusczyk21:14:54
That clever algebraic manipulation is explored more in the number theory class.
rrusczyk21:15:31
That's it for the math problems tonight.
rrusczyk21:15:56
A significant portion of the number theory class focuses on putting our algebra skills together with number theory to solve problems.
futbill202521:16:09
do you reccomend taking both the amc 10 and 12
rrusczyk21:16:17
Yes, I recommend taking both of those tests.
silvercharcoal521:16:23
where can we get help with Latex
rrusczyk21:16:38
There is a tutorial on our site. When you get stuck, use the LaTeX forum on our site.
MathMasta_221:16:42
10a and 12b or the other way around
rrusczyk21:16:45
Doesn't matter.
futbill202521:16:46
so, you reccomend taking both the classes at the same time
rrusczyk21:16:49
Only if you have time.
xsk1321:16:56
when you recommend taking both of the tests, do you mean taking the AMC10A and the AMC12B, or vice versa?
rrusczyk21:17:10
Either is fine. I'd probably take the 12 first, but it really doesn't matter.
gu_co21:17:12
If you would express in points from 1 to 10, how much would this problem be given?
rrusczyk21:17:22
I would say the problem we did is a mid-level AIME problem.
futbill202521:17:25
are the amc 10 and amc 12 on the same date
rrusczyk21:17:29
They are both offered twice.
skier21:17:50
on the same day?
rrusczyk21:18:21
Both are offered on the same day at the same, on two different days (check the AMC website for details).
MathMasta_221:18:25
week apart?
rrusczyk21:18:32
A couple weeks apart.
Math46_221:18:37
If you got slightly lost with those examply's then is the Interm number theory class to hard or just right?
rrusczyk21:19:13
If you were lost with the first example, then the Interm class will be too hard. If you were a little lost with the last two, but think you could read the transcript for a while and get it, then you should be fine.
gu_co21:19:17
the Number theory, in how many problems from 100 would we need to use it? at olympiads
rrusczyk21:19:42
I'm not sure I understand this question, but you can typically count on around 2 problems a year on the USAMO with a significant Number Theory component.
gu_co21:20:35
usaually how many problems are at USAmo?
rrusczyk21:20:37
6
rrusczyk21:20:56
Usually 2 will have a significant number theory component. They are all proofs.
futbill202521:20:58
around what time of year are the AMC's
rrusczyk21:21:01
February.
wz111821:21:03
how many problem are on the amc 10 and the amc 12>
rrusczyk21:21:07
25
