Transcript for the Math Jam "AoPS AMC 12 Problem Series" on Oct 17.

Math Jam hosted by nsato (Naoki Sato ).

nsato (19:30:56)
Welcome to the AMC 12 Problem Series math jam!

nsato (19:31:04)
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.

nsato (19:31:11)
The classroom is moderated meaning that 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.

nsato (19:31:25)
Also, only moderators can enter into private chats with other people in the classroom. You can copy the contents of a private chat by clicking on the notebook icon on the private chat window and copying and pasting the contents into a file on your own computer.

nsato (19:31:39)
Finally, the virtual classroom is TeX enabled. TeX allows users to make nice equations and other math expressions. If you would like to learn how to write in TeX/LaTeX, click on the tab on the left side panel of our site and there is a tutorial and reference guide there.

nsato (19:31:46)
Using TeX in the virtual classroom is slightly different than using it on the message board or in a TeX editor. If anything you type up in a post that uses TeX then you must use a semicolon (;) to begin your post. For example, if you type

nsato (19:32:01)


nsato (19:32:10)
This message will look like this when posted in the classroom:

nsato (19:32:13)


nsato (19:32:20)
Just remember, if your post uses TeX, use the semicolon (;) to begin your post!

nsato (19:32:55)
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.

nsato (19:33:02)
The following is an excerpt of one of the areas of problem solving covered in the AMC 12 Problem Series.

nsato (19:33:15)
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

nsato (19:33:43)
Sorry, that should say ""closer"", not close.

nsato (19:34:03)
How do we know when a point is closer to one location or another in a plane?

galbatorix (19:34:10)
distance formula

bpms (19:34:14)
The distance formula

nsato (19:35:14)
Is there a more geometric description? For example, what is the set of points equidistant from A and B?

bpms (19:35:39)
The perpendicular bisector

gtf (19:35:53)
The perpendicular bisector of line segment AB

nsato (19:36:10)
Correct.

nsato (19:36:17)
The set of points (in a plane) equidistant to two given points is the perpendicular bisector of the two points. 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?

bpms (19:36:49)
So we just find the equation of the perpendicular bisector, and see how much of the area of the triangle is above it divided by 2 since that is the area of the rectangle

gtf (19:37:12)
This segment will divide the original rectangle into two separate regions.

nsato (19:37:26)
Good.

nsato (19:37:32)
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?

nsato (19:38:00)
How can we find the equation of the perpendicular bisector? What do we need?

tzmonster (19:38:12)
slope

bpms (19:38:15)
It has the opposite reciprocal slope of the segment from (0,0) to (3,1) and hits the midpoint (1.5,.5)

gtf (19:38:18)
Point (midpoint) and slope

nsato (19:38:55)
We need two things: A point on the line, and the slope.

nsato (19:39:07)
We know that the midpoint lies on the line.

nsato (19:39:12)
The midpoint of the segment between (0, 0) and (3, 1) is (1.5, .5).

nsato (19:39:34)
What is the slope between points (0,0) and (3,1)?

galbatorix (19:39:44)
1/3

LarseD (19:39:51)
1/3

nsato (19:40:13)
So what is the slope of the perpendicular bisector?

bpms (19:40:17)
-3

tzmonster (19:40:21)
-3

LarseD (19:40:23)
-3

galbatorix (19:40:25)
-3?

nsato (19:40:39)
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.)

nsato (19:40:56)
So what is the equation of the perpendicular bisector?

LarseD (19:41:34)
(y-.5)=-3(x-1.5)

bpms (19:42:15)
y=-3x+5

nsato (19:42:23)
Right, and this simplifies to y = -3x + 5.

nsato (19:42:48)
We must now find the area of the rectangular region. This will be the denominator in our geometric probability problem.

nsato (19:42:55)
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.

nsato (19:43:04)
What are these two quantities?

galbatorix (19:43:16)
The Area of the Rectangle is 2

nsato (19:43:40)
We can easily see that the rectangular region was height 1 and width 2, so it's area is 2.

nsato (19:43:49)
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:

nsato (19:43:58)


nsato (19:43:59)
http://www.artofproblemsolving.com/Classes/AMC/Images/2002amc12B18region.gif

nsato (19:44:13)
What are the dimensions of the trapezoid that we need?

nsato (19:45:24)
We need the height of the trapezoid, and the lengths of the two bases.

nsato (19:45:27)
What is the height of the trapezoid?

galbatorix (19:45:43)
1

nsato (19:46:02)
We need the height which is 1.

nsato (19:46:07)
We also need the bases. What are they?

nsato (19:46:24)
We need to find where y = 0 and y = 1 for the line y = -3x + 5.

nsato (19:46:34)
If y = 0, what is x?

galbatorix (19:46:45)
5/3

tzmonster (19:46:47)
5/3

LarseD (19:46:57)
1 2/3

nsato (19:47:04)
Right!

nsato (19:47:08)
If y = 1, what is x?

LarseD (19:47:15)
4/3

tzmonster (19:47:22)
4/3

galbatorix (19:47:25)
4/3

LarseD (19:45:55)
base 1=1 1/3 base 2=1 2/3 height=1

nsato (19:47:48)
We can see that 4/3 and 5/3 are the lengths of the bases.

nsato (19:47:59)
What is the area of the trapezoid?

bpms_2 (19:48:15)
1.5

galbatorix (19:48:25)
3/2

LarseD (19:48:28)
3/2

nsato (19:48:39)
The area of the trapezoid is 1 times the average of the bases which is 3/2.

nsato (19:48:52)
The total rectangular region has area 2.

nsato (19:48:56)
The area of the region close to the origin than (3, 1) within that rectangle is 3/2.

nsato (19:48:58)
What is our final answer?

bpms_2 (19:49:05)


Jay Clark (19:34:03)
C

nsato (19:49:46)
Our probability is (3/2)/2 = 3/4. Our answer is (C).

nsato (19:50:03)
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.

bpms_2 (19:50:21)
Could this problem have been solved with the shoestring as a shortcut?

nsato (19:51:33)
You could have used the shoestring formula, but it would have been longer. The formula for the area of a trapezoid is actually easier.

LarseD (19:51:51)
what is the shoestring formula?

Jay Clark (19:52:20)
what is the shoestring formula?

nsato (19:53:53)
It's a formula for the area of a triangle in terms of its coordinates.

nsato (19:54:15)
It's at the bottom of this link, for example:

nsato (19:54:20)
http://mathforum.org/library/drmath/view/55063.html

frostmourne (19:54:19)
is it the one with the matrix?

nsato (19:54:27)
Yes, more or less.

nsato (19:56:11)


nsato (19:56:41)
I'll give you a minute to read the problem.

nsato (19:57:15)
How can we begin determining the area of R?

nsato (19:57:24)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-101090981.gif

bpms (19:57:20)
We are looking for negative values of the f(x)

nsato (19:58:08)
That's true, but not exslusively. We need to involve y as well.

nsato (19:58:18)
...exclusively...

nsato (19:58:23)


bpms (19:58:41)
Add them

nsato (19:59:35)
But then we lose information about y.

nsato (19:59:50)
What does the graph of the first inequality represent?

bpms (19:59:23)
Complete the square

nsato (20:00:25)
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.

nsato (20:00:47)
Something interesting happens with the second inequality - does anybody see what?

bpms (20:00:45)


nsato (20:01:36)


nsato (20:02:01)
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?

nsato (20:02:20)
First, what points satisfy the first inequality?

bpms (20:02:37)
A circle with center (-3,-3) and radius 4

mgao (20:03:05)
a radius of 4 around the point (-3, -3).,

nsato (20:03:18)
The first inequality defines the interior of a circle with radius 4 centered at (-3, -3).

nsato (20:03:24)
The second inequality says that the product of two quantities is no greater than 0, so we can rewrite that inequality as

nsato (20:03:30)


nsato (20:03:36)
What regions does this separation of the second inequality carve out?

nsato (20:04:13)
The second inequality defines to opposite ""quadrants"" carved out by two perpendicular lines that intersect at . What does the region R look like?

nsato (20:04:19)
The intersection of the two regions carved out by the inequalities looks like this:

nsato (20:04:25)


nsato (20:04:26)
http://www.artofproblemsolving.com/Classes/AMC/Images/2002amc12B25region.gif

nsato (20:04:34)
Now what?

nsato (20:04:51)
We need to find the area of the shaded region. What is that?

mgao (20:05:04)
we divide the whole circle's area by 2

bpms (20:04:52)
So the area is half the circles, so the area is 8pi which is 25 to the nearest int

nsato (20:05:26)
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).

Jay Clark (20:05:30)
8 pie

nsato (20:05:59)
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.

nsato (20:06:30)
So that was a taste of the problems we'll be covering in the AMC 12 Problem Series.

nsato (20:06:33)
Any questions?

nsato (20:07:08)
A very quiet crowd...

bpms (20:07:12)
I suck at trig really bad. I don't mind the law of cosines/sines but the complicated manipulations drive me nuts. Is this class good for me?

nsato (20:08:49)
There is one class that will cover trig. That may not be a lot, but it will probably help.

tzmonster (20:07:13)
Will there be any extra time needed outside of the Monday classes?

nsato (20:10:03)
You don't have to spend any time outside the classes, but the students who do spend time thinking about the problems and working the message board problems are the ones that get the most out of the classes.

LarseD (20:07:18)
and the AMC 12 problem series is a class?

nsato (20:10:11)
Yes.

SmitSchu (20:07:49)
so this was problem 25 in 2002?

nsato (20:10:32)
Yes.

Jay Clark (20:10:33)
Can we do more problems?

nsato (20:10:59)
I'm afraid that's all the problems we'll be doing for tonight.

bpms (20:10:34)
If I am planning on taking the AIME problem series, would this class be good, seeing as it doesn't matter how I do on the AMC as long as I qualify (I am in <10th grade)? Also, will we normally spend this long on problems in the class?

nsato (20:13:01)
I would say that if you are confident in qualifying for the AIME, then that's what you should be focusing on, and you don't need this class. If you are not sure, and think you need to brush up on AMC 12-type problems, then I would recommend you take it.

nsato (20:13:34)
And this is typical of the pace in the classes.

LarseD (20:11:00)
are there problems on the message boards for people who aren't in specific classes?

nsato (20:14:13)
Sure, there's the High School and Olympiad Section, and they are very active.

bpms (20:14:27)
What books do you recommend for AIME/hard AMC level trig? Where can I get them?

nsato (20:16:40)
My first recommendation is AoPS Vol. 2. Others would be ""First Steps for Math Olympians"", ACoPS, Engel. The first two are available from our online bookstore.

tzmonster (20:14:41)
What is the minimum level of math we should have taken in order to fully understand this class?

nsato (20:20:03)
You should have taken geometry, and algebra 2. The best indicator is to look at an old AMC 12. If you can get at least 85, then you will be able to understand the course.

bpms (20:17:28)
Um, I meant books that are pure trig and start at that level.

nsato (20:21:06)
The only book I can think of that only does trig is ""103 Trigonometry Problems"", by Andreescu and Feng.

nsato (20:21:14)
http://www.amazon.com/103-Trigonometry-Problems-Training-Team/dp/0817643346

bpms (20:19:39)
Will the problems we use be old AMC problems, or problems that the teacher makes up? Also, will there be any mocks?

nsato (20:22:05)
The course will use old AMC 12 problems. No mocks, but there will be lots of message board problems each week.

LarseD (20:21:19)
are there old AMC 12 tests on this site?

nsato (20:22:19)
No, but I bet you can find some on the AMC site.

SmitSchu (20:22:18)
how high does this course go - about what score would be expected from someone who would not benefit from taking it?

nsato (20:25:19)
It covers most of the topics that appear on the AMC 12, and draws on old AMC 12 contests.

nsato (20:26:13)
Hard to give an exact figure for the second part - if you can get, say, 135 consistently, like I said above, don't worry about AMC 12 problems anymore and move onto AIME level problems.

tzmonster (20:22:23)
Where can we find old AMC tests to practice on?

nsato (20:26:27)
First place I would look is the AMC website.

nsato (20:26:33)
http://www.unl.edu/amc/

nsato (20:26:44)
Any other questions?

nsato (20:27:34)
In that case, that concludes tonight's math jam. Thanks for coming.