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2012 AMC 10/12 B Discussion

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A discussion of problems from the AMC 10/12 B, which is administered February 22. We will cover the last 5 problems on each test, as well as requested earlier problems on the tests.

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Facilitator: Jeremy Copeland

copeland 2012-02-23 19:03:06
Welcome to the 2012 AMC 10B/12B Math Jam!
copeland 2012-02-23 19:03:12
I'm Jeremy Copeland, and I'll be leading our discussion tonight.
copeland 2012-02-23 19:03:17
Before we get started I would like to take a moment to explain our virtual classroom procedures to those who have not previously participated in a Math Jam or one of our online classes.
copeland 2012-02-23 19:03:23
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room.  These comments go to the instructors, who may choose to share your comments with the room.
copeland 2012-02-23 19:03:31
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.
copeland 2012-02-23 19:03:42
There are a lot of students here!  As I said, only (a fraction of the) well-written comments will be passed to the entire group.  Please do not take it personally if your comments do not get posted, and please do not complain about it.  I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!
AkshajK 2012-02-23 19:03:51
you type fast
copeland 2012-02-23 19:03:53
I have two keyboards.
copeland 2012-02-23 19:03:58
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the prerequisite material for every problem as we go.  Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask.  We usually do in our classes, but we have a large number of students tonight!  So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!
copeland 2012-02-23 19:04:29
We do have two assistants tonight who can help answer some of your questions: Samson Zhou (TowersFreak2006) and Joel Schneider (joelinia) .
joelinia 2012-02-23 19:04:39
Hello! :-D
towersfreak2006 2012-02-23 19:04:40
Hi!
basketballstar24 2012-02-23 19:04:52
Hi!
rlingineni 2012-02-23 19:04:52
Hi guys :)
duketip10 2012-02-23 19:04:52
HI!
tb1624 2012-02-23 19:04:52
HIII!
SHARKYBOY 2012-02-23 19:04:52
hi
copeland 2012-02-23 19:04:58
They can answer questions by whispering to you or by opening a window with you to chat 1-on-1.  However, due to the large size of the session tonight, they may not be able to get to you right away (or at all).  Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.
copeland 2012-02-23 19:05:08
Please also remember that the purpose of this Math Jam is to work through the solutions to AMC problems, and not to merely present the answers.  "Working through the solutions" includes discussing problem-solving tactics.  So please, when a question is posted, do not simply respond with the final answer.  That's not why we're here.  We're going to work through the problems step-by-step, and comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, won't be posted.
copeland 2012-02-23 19:05:43
We will work the last 5 problems from the AMC 10B, then 4 of the last 5 problems from the AMC 12B.  (There's an overlap: #25 on the 10B is also #22 on the 12B.)  After that, time permitting, I will take requests for some other problems for discussion.
copeland 2012-02-23 19:05:51
Let's get started!
teethpaste 2012-02-23 19:06:10
yay!
MathForFun 2012-02-23 19:06:10
YAY!!
rlingineni 2012-02-23 19:06:10
YAY!!!
copeland 2012-02-23 19:06:12
copeland 2012-02-23 19:06:22
You'll notice that I'll always "sticky" the current problem under discussion to the top of the classroom window.  You can resize the top region by dragging the horizontal gray bar that separates the top region from the main part of the classroom.
copeland 2012-02-23 19:06:31
Which number stands out to you and why?
Garchomp293 2012-02-23 19:07:10
2a, because it is different from the other a's.
christerson 2012-02-23 19:07:10
a + a = 2a
sammyMaX 2012-02-23 19:07:10
2a, because it is twice of many of the other sidelengths
martian179 2012-02-23 19:07:10
2a because three points must be collinear to work
copeland 2012-02-23 19:07:18
copeland 2012-02-23 19:07:21
(I'll label the points X, Y, Z, and W.)
copeland 2012-02-23 19:07:25
copeland 2012-02-23 19:07:40
Why is this interesting?
alex31415 2012-02-23 19:08:16
but the triangle is degenerate
AruKasera 2012-02-23 19:08:16
but it's degenarate!!
Seedleaf 2012-02-23 19:08:16
a + a = 2a, so they lie in a line.
mcdonalds106_7 2012-02-23 19:08:16
three points must be collinear
swimmer 2012-02-23 19:08:16
the "triangle" is degenerate
briantix 2012-02-23 19:08:16
they form a line
RotomPlasma 2012-02-23 19:08:16
Because then three of the points are collinear
negativebplusorminus 2012-02-23 19:08:16
A triangle like that is degenerate.
mathman500 2012-02-23 19:08:16
degenerate triangle
TeamJacob 2012-02-23 19:08:16
because that's just a line segment, degenerate triangle
copeland 2012-02-23 19:08:21
That triangle is degenerate.  Three of the points are collinear.
copeland 2012-02-23 19:08:24
copeland 2012-02-23 19:08:27
Now how do we find W?
somanywindows 2012-02-23 19:09:11
equilateral triangle
mcdonalds106_7 2012-02-23 19:09:11
it must be a away from 2 points
Macromiidae 2012-02-23 19:09:11
equilateral triangle
willwang123 2012-02-23 19:09:11
there's an equilateral triangle
martian179 2012-02-23 19:09:11
must be equidistant to two of other points
Flamel 2012-02-23 19:09:11
forms equilateral 'a' triangle
Iggy Iguana 2012-02-23 19:09:11
equalateral triangle
Jasonion 2012-02-23 19:09:11
make it equidistant from two of the points
HydraPyros 2012-02-23 19:09:11
Choose two adjacent points, that will be the base of an equilateral triangle of side length a. W is the vertex of this triangle.
swimmer 2012-02-23 19:09:11
it forms an equilateral triangle with two of the other points
copeland 2012-02-23 19:09:14
Since we still have 2 segments from W that are of length a, W is part of an equilateral triangle.  By symmetry we can assume X, Y, and W are the vertices of the equilateral triangle.
copeland 2012-02-23 19:09:17
copeland 2012-02-23 19:09:22
What is b/a?
mcqueen 2012-02-23 19:10:12
/whisper BigSams I can't see the problems. They're just blank message text. Any help?
Seedleaf 2012-02-23 19:10:12
sqrt3
teethpaste 2012-02-23 19:10:12
sqrt3
Lalagato 2012-02-23 19:10:12
rt(3)
christerson 2012-02-23 19:10:12
sqrt{3}
tan90 2012-02-23 19:10:12
sqrt3
sumkid25 2012-02-23 19:10:12
sqrt(3)
sammyMaX 2012-02-23 19:10:12
sqrt3
HydraPyros 2012-02-23 19:10:12
sqrt(3)
NeilOnnsu 2012-02-23 19:10:12
sqrt(3)
tc1729 2012-02-23 19:10:12
sqrt(3)
mathocean97 2012-02-23 19:10:12
\sqrt(3)
AruKasera 2012-02-23 19:10:12
sqrt3!
qaryaendpetyp 2012-02-23 19:10:12
.cpp 2012-02-23 19:10:12
sqrt(3)
azure55 2012-02-23 19:10:12
rt3
negativebplusorminus 2012-02-23 19:10:13
By the Law of Cosines, b^2=3a^2, so the answer is Sqrt[3]; A.
somanywindows 2012-02-23 19:10:13
hali323 2012-02-23 19:10:13
sqrt3
Xu12345 2012-02-23 19:10:13
sqrt 3
lucylai 2012-02-23 19:10:13
sqrt3
SDCZ6 2012-02-23 19:10:13
briantix 2012-02-23 19:10:13
Jasonion 2012-02-23 19:10:13
duketip10 2012-02-23 19:10:14
sqrt3
copeland 2012-02-23 19:10:18
copeland 2012-02-23 19:10:28
copeland 2012-02-23 19:10:32
The answer is (A).
copeland 2012-02-23 19:10:57
(I likely won't pass that many answers in the future.  There are a LOT of people here, though, so don't get sad if what you say doesn't get passed!)
copeland 2012-02-23 19:11:02
copeland 2012-02-23 19:11:51
More!
copeland 2012-02-23 19:11:51
copeland 2012-02-23 19:12:01
If you haven't seen this problem before, it's a little hard to know where to start.
copeland 2012-02-23 19:12:04
What should we try here?
.cpp 2012-02-23 19:12:38
Small cases.
christerson 2012-02-23 19:12:38
Small cases
martian179 2012-02-23 19:12:38
start with first number is 1
negativebplusorminus 2012-02-23 19:12:38
Smaller cases?
AkshajK 2012-02-23 19:12:38
making the problem easier
VIPMaster 2012-02-23 19:12:38
Smaller cases, in an attempt to find a recursion
ABCDE 2012-02-23 19:12:38
Smaller cases.
guilt 2012-02-23 19:12:38
Small cases.
copeland 2012-02-23 19:12:43
10 is a really big number.  Let's start with something smaller.  How many sequences like this are there using only the digits 1 and 2?
guilt 2012-02-23 19:13:14
2
Jasmine8925 2012-02-23 19:13:14
12, 21
swimmer 2012-02-23 19:13:14
2
VIPMaster 2012-02-23 19:13:14
2
wcao9311 2012-02-23 19:13:14
2
martian179 2012-02-23 19:13:14
2
mcdonalds106_7 2012-02-23 19:13:14
2.
numbertheory 2012-02-23 19:13:14
2
hi how are you doing toda 2012-02-23 19:13:14
2
.cpp 2012-02-23 19:13:14
2: (1,2) or (2,1)
mcdonalds106_7 2012-02-23 19:13:14
1,2 and 2,1
richard4912 2012-02-23 19:13:14
2
copeland 2012-02-23 19:13:16
Both sequences 12 and 21 satisfy the hypothesis.
copeland 2012-02-23 19:13:20
What about with the digits 1, 2, and 3?
guilt 2012-02-23 19:13:57
4
newchie123 2012-02-23 19:13:57
4
christerson 2012-02-23 19:13:57
4
Luckytoilet 2012-02-23 19:13:57
123, 321, 213, 231
mcdonalds106_7 2012-02-23 19:13:57
4- 1,2,3 2,1,3 2,3,1 3,2,1
teethpaste 2012-02-23 19:13:57
4
alex31415 2012-02-23 19:13:57
132, 312 are illegal=>4
RotomPlasma 2012-02-23 19:13:57
4
.cpp 2012-02-23 19:13:57
4: (1,2,3), (2,1,3), (2,3,1), or (3,2,1)
lucylai 2012-02-23 19:13:57
4
Iggy Iguana 2012-02-23 19:13:57
(1, 2, 3) (2, 3, 1) (2, 1, 3) (3, 2, 1)
AkshajK 2012-02-23 19:13:57
4
pjyoung 2012-02-23 19:13:57
123 132 213 231 312 321
willwang123 2012-02-23 19:13:59
123, 213, 231, 321
copeland 2012-02-23 19:14:00
copeland 2012-02-23 19:14:03
Working the other direction, how many sequences are there that use just the number 1?
pr0likethis 2012-02-23 19:14:25
1
briantix 2012-02-23 19:14:25
1
teethpaste 2012-02-23 19:14:25
1
MCL 2012-02-23 19:14:25
1.
joshxiong 2012-02-23 19:14:25
1
Jasonion 2012-02-23 19:14:25
1
alex31415 2012-02-23 19:14:25
1
sammyMaX 2012-02-23 19:14:25
1!
numbertheory 2012-02-23 19:14:25
1
coldsummer 2012-02-23 19:14:25
1
lucylai 2012-02-23 19:14:25
1
somanywindows 2012-02-23 19:14:25
1
copeland 2012-02-23 19:14:30
There is one such sequence, the sequence 1.
copeland 2012-02-23 19:14:33
What about with the digits 1, 2, 3, and 4?
ABCDE 2012-02-23 19:14:56
8
christerson 2012-02-23 19:14:56
8
va2010 2012-02-23 19:14:56
there are 8
super_pi314 2012-02-23 19:14:56
8
NeilOnnsu 2012-02-23 19:14:56
8
briantix 2012-02-23 19:14:56
8
hali323 2012-02-23 19:14:56
there are 8
MCL 2012-02-23 19:14:56
8.
joshxiong 2012-02-23 19:14:56
8
copeland 2012-02-23 19:15:00
copeland 2012-02-23 19:15:05
There are 8 of these.
copeland 2012-02-23 19:15:08
Now we can probably make a reasonable guess.  Our sequence starts 1, 2, 4, 8,. . . .
copeland 2012-02-23 19:15:15
What ought to be the 10th term in this sequence?
MathForFun 2012-02-23 19:15:44
16, 32, 64... 512
willwang123 2012-02-23 19:15:44
2^9 = 512
Iggy Iguana 2012-02-23 19:15:44
2^9=512
AkshajK 2012-02-23 19:15:44
(B) 512
Seedleaf 2012-02-23 19:15:44
2^9 = 512
martian179 2012-02-23 19:15:44
2^9=512
coldsummer 2012-02-23 19:15:44
512
Mattchu386 2012-02-23 19:15:44
512
sammyMaX 2012-02-23 19:15:44
512
Klu2014 2012-02-23 19:15:44
2^9
numbertheory 2012-02-23 19:15:44
512
copeland 2012-02-23 19:15:47
MCL 2012-02-23 19:16:06
How do we prove this?
mentalgenius 2012-02-23 19:16:06
I see a pattern (powers of 2), but how do we mathematically make sure that this is true?
swimmer 2012-02-23 19:16:06
how do you prove this though?
copeland 2012-02-23 19:16:15
Good!  Thinking like mathematicians now.
copeland 2012-02-23 19:16:30
How should we approach this?
alex31415 2012-02-23 19:17:08
prove the recursive pattern
Iggy Iguana 2012-02-23 19:17:08
recursion, induction
negativebplusorminus 2012-02-23 19:17:08
Recursion!
superpi83 2012-02-23 19:17:08
induction?
joshxiong 2012-02-23 19:17:08
Induction, maybe
newchie123 2012-02-23 19:17:08
INDUCTION
iwantcombo 2012-02-23 19:17:08
induction?
copeland 2012-02-23 19:17:13
Let's work recursively.
copeland 2012-02-23 19:17:14
Why are there always twice as many terms in an n-term sequence as there are in an (n-1)-term sequence?
copeland 2012-02-23 19:17:39
Let's look at our 8 again:
copeland 2012-02-23 19:17:43
copeland 2012-02-23 19:17:48
What do you notice about all these sequences?
.cpp 2012-02-23 19:18:23
End with 4 or 1.
super_pi314 2012-02-23 19:18:23
they end in 1 or 4
Iggy Iguana 2012-02-23 19:18:23
1 or 4 is at the end
MCL 2012-02-23 19:18:23
they end in 4 or 1
apple.singer 2012-02-23 19:18:23
all end in 1/4
myseokim 2012-02-23 19:18:23
ends in 1 or 4
fortenforge 2012-02-23 19:18:23
the last digit is either a 4 or a 1
VIPMaster 2012-02-23 19:18:23
they end with either 1 or 4
copeland 2012-02-23 19:18:27
We can start with any number we want.  Then each new number must be either (the lowest number that's greater than all the numbers we've placed) or (the highest number that's less than all the numbers we've placed).  (E.g. at some intermediate step we might have placed all of 3 through 8.  The next number must be 2 or 9.)  At the end this will leave us with either 1 or n.
copeland 2012-02-23 19:18:40
So. . . .
lucylai 2012-02-23 19:20:18
you can just attach a 4 to the end of a rearrangement of n-1
negativebplusorminus 2012-02-23 19:20:18
So if we delete that number, we either have the sequences for 9, or each one for 9, plus one.
copeland 2012-02-23 19:20:26
copeland 2012-02-23 19:20:40
In particular we can build a sequence from right-to-left by placing either 1 or n first, and then in every step picking either the highest or lowest term that we haven't placed yet.  There are 2^{n-1} ways to do this because there is no choice for the final digit placed.
copeland 2012-02-23 19:21:10
So we just proved that the answer is. . . .
alex31415 2012-02-23 19:21:35
Klu2014 2012-02-23 19:21:35
512
iwantcombo 2012-02-23 19:21:35
B
swimmer 2012-02-23 19:21:35
512
teethpaste 2012-02-23 19:21:35
512
ap1998 2012-02-23 19:21:35
512
mathgenius1982 2012-02-23 19:21:35
2^9=512
mcdonalds106_7 2012-02-23 19:21:35
512
christerson 2012-02-23 19:21:35
512
copeland 2012-02-23 19:21:38
(B) 512.
copeland 2012-02-23 19:21:54
Good job.  Let's see more!
copeland 2012-02-23 19:22:12
copeland 2012-02-23 19:22:21
Rule number 1 for a problem like this is. . . .
googol.plex 2012-02-23 19:22:51
Draw it!
alex31415 2012-02-23 19:22:51
Draw a diagram
mentalgenius 2012-02-23 19:22:51
Diagram please?
pr0likethis 2012-02-23 19:22:51
draw it out
superpi83 2012-02-23 19:22:51
draw it?
Turtle 2012-02-23 19:22:51
drawing
AmericanPi 2012-02-23 19:22:51
draw a diagram
va2010 2012-02-23 19:22:51
Draw a picture
sparkles257 2012-02-23 19:22:51
draw a diagram
Cortana 2012-02-23 19:22:51
Picture
Draco 2012-02-23 19:22:51
Draw a picture!
etothei 2012-02-23 19:22:51
draw a picture!
ken961996 2012-02-23 19:22:51
diagram
copeland 2012-02-23 19:22:55
Draw a diagram!
copeland 2012-02-23 19:22:59
Here's a cube:
copeland 2012-02-23 19:23:02
copeland 2012-02-23 19:23:06
Here are three vertices that fit the bill:
copeland 2012-02-23 19:23:09
copeland 2012-02-23 19:23:18
We're supposed to cut an rotate to get this picture:
copeland 2012-02-23 19:23:21
copeland 2012-02-23 19:23:35
We want to know the height, so we want to know the distance from the top vertex to the red plane.  How can we compute this height?
iwantcombo 2012-02-23 19:24:18
coordinate bash
mathdate 2012-02-23 19:24:18
the vertex to the centroid of the triangle
sparkles257 2012-02-23 19:24:18
find heigth of the tetrahedron and then subtract from psace daigonal
lucylai 2012-02-23 19:24:18
subtract height of tetrahedron from diagonal length
mentalgenius 2012-02-23 19:24:18
subtract the altitude of the tetrahedron from the space diag of the cube (sqrt(3))
mcdonalds106_7 2012-02-23 19:24:18
space diagonal of cube minus height of tetrahedron that was cut off
NeilOnnsu 2012-02-23 19:24:18
find interior diagnol - height of tetrahedron
iwantcombo 2012-02-23 19:24:18
or drop a perpendicular that hits the centroid of the triangle
va2010 2012-02-23 19:24:18
Take the diagnal and substract the pyramids height
c0mbusti0n1295 2012-02-23 19:24:18
height of the cube?
TeamJacob 2012-02-23 19:24:18
drawing a triangular pyramid using that base
iwantcombo 2012-02-23 19:24:18
then take the space diagonal and subtract
copeland 2012-02-23 19:24:23
That's a few ideas.
copeland 2012-02-23 19:24:28
We had a few ideas here, too.
copeland 2012-02-23 19:24:35
The most "geometric" approach is by computing the altitude of the tetrahedron we cut off and subtracting.
copeland 2012-02-23 19:24:44
How can we get at that altitude?
tan90 2012-02-23 19:25:18
Using volume of pyramid
Relativity1618 2012-02-23 19:25:18
finding volume
.cpp 2012-02-23 19:25:18
Find volume two ways.
Relativity1618 2012-02-23 19:25:18
finding volume
munygrubber 2012-02-23 19:25:18
find the volume of the pyramid in two ways
fortenforge 2012-02-23 19:25:18
Find the volume of the tetrahedron and divide by the base area
RotomPlasma 2012-02-23 19:25:18
It's a pyramid, with known base, and we know the volume
copeland 2012-02-23 19:25:22
We can compute the volume of the tetrahedron in 2 ways.  Let's go back to this picture:
copeland 2012-02-23 19:25:25
copeland 2012-02-23 19:25:29
Numerically what is the volume of the tetrahedron?
christerson 2012-02-23 19:26:01
1/6
mcdonalds106_7 2012-02-23 19:26:01
1/6
duketip10 2012-02-23 19:26:01
1/6
ss5188 2012-02-23 19:26:01
1/6
.cpp 2012-02-23 19:26:01
1/3 * 1/2 = 1/6
msinghal 2012-02-23 19:26:01
1/2*1*1/3=1/6
PChang 2012-02-23 19:26:01
1/6
copeland 2012-02-23 19:26:05
The tetrahedron has triangular base that is half a face of the cube and has height equal to 1.  Therefore it has volume 1/6.
copeland 2012-02-23 19:26:09
In terms of the altitude we care about, what is the volume?
alex31415 2012-02-23 19:27:03
1/3*b*h
Relativity1618 2012-02-23 19:27:03
H*B/3
Iggy Iguana 2012-02-23 19:27:03
sqrt3/2 * h / 3
tan90 2012-02-23 19:27:03
Bh/3
Relativity1618 2012-02-23 19:27:03
volume=height times base divided by 3
.cpp 2012-02-23 19:27:03
1/3 * sqrt{3}/2 * h
swimmer 2012-02-23 19:27:03
h * sqrt(3)/6
lucylai 2012-02-23 19:27:03
sqrt3/6*altitude
coldsummer 2012-02-23 19:27:03
1/2 basex height
copeland 2012-02-23 19:27:07
copeland 2012-02-23 19:27:15
What is the value of h?
briantix 2012-02-23 19:28:04
sqrt3 /3
RotomPlasma 2012-02-23 19:28:04
(sqrt 3)/3
christerson 2012-02-23 19:28:04
sqrt(3)/3
PChang 2012-02-23 19:28:04
sqrt(3)/3
ss5188 2012-02-23 19:28:04
rad3 /3
NeilOnnsu 2012-02-23 19:28:04
sqrt(3)/3
googol.plex 2012-02-23 19:28:04
\sqrt {3}/3
va2010 2012-02-23 19:28:04
sqrt(3)/3
Garchomp293 2012-02-23 19:28:04
1/sqrt3
ddot1 2012-02-23 19:28:04
sqrt{3}/3
mentalgenius 2012-02-23 19:28:04
1/sqrt(3)
sparkles257 2012-02-23 19:28:04
sqrt3/3
copeland 2012-02-23 19:28:08
copeland 2012-02-23 19:28:16
So what's the answer?
tan90 2012-02-23 19:28:48
(D)
Relativity1618 2012-02-23 19:28:48
D
msinghal 2012-02-23 19:28:48
D
negativebplusorminus 2012-02-23 19:28:48
So, since the space diagonal is Sqrt[3], the answer is D.
sparkles257 2012-02-23 19:28:48
subtract that from sqrt 3 to get 2sqrt3/3 D
Glaydus 2012-02-23 19:28:48
D
Iggy Iguana 2012-02-23 19:28:48
sqrt3 - sqrt3/3 = 2sqrt3/3
MNL9082 2012-02-23 19:28:48
2sqrt(3)/3
joshxiong 2012-02-23 19:28:48
D. 2sqrt3/3
Relativity1618 2012-02-23 19:28:48
D
iwantcombo 2012-02-23 19:28:48
D
copeland 2012-02-23 19:28:50
You guys just cannot be tricked!
copeland 2012-02-23 19:29:00
h is not what we want.  We want the height that's left once we cut that tetrahedron off.
copeland 2012-02-23 19:29:03
copeland 2012-02-23 19:29:08
The answer is (D).
mcdonalds106_7 2012-02-23 19:29:22
what were your other ideas?
copeland 2012-02-23 19:29:24
This is not how I solved the problem.
copeland 2012-02-23 19:29:36
I like coordinates with cubec.
copeland 2012-02-23 19:29:38
cubes.
copeland 2012-02-23 19:29:41
Cubes like coordinates.
copeland 2012-02-23 19:29:44
A more efficient solution goes like this.  We let the unit cube be all triples (a,b,c) where each of a, b, and c are either 0 or 1.  Then we place (0,0,0) at the top of this figure:
copeland 2012-02-23 19:29:48
copeland 2012-02-23 19:29:51
What are the three base vertices?
.cpp 2012-02-23 19:30:38
(1,1,0), (1,0,1), (0,1,1)
Bictor717 2012-02-23 19:30:38
two ones, one zero
nsun48 2012-02-23 19:30:38
110,101,001
superpi83 2012-02-23 19:30:38
(1,1,0) (1,0,1) (0,1,1)
tan90 2012-02-23 19:30:40
(1,1,0) (1,0,1) (0,1,1)
super_pi314 2012-02-23 19:30:40
(1, 0, 0); (0, 1, 0); (1,1,1)
swimmer 2012-02-23 19:30:40
(1,0,0), (0,1,0), (0,0,1)
Yongyi781 2012-02-23 19:30:40
(0, 1, 1) and permutations
azure55 2012-02-23 19:30:40
(1,0,0),(0,0,1),(0,1,0)
copeland 2012-02-23 19:30:43
The vertices of the base triangle are (1,1,0), (0,1,1), and (1,0,1).
copeland 2012-02-23 19:30:48
Where does the altitude meet the base triangle?
swimmer 2012-02-23 19:31:24
center
mentalgenius 2012-02-23 19:31:24
centroid
duketip10 2012-02-23 19:31:24
centroid
c0mbusti0n1295 2012-02-23 19:31:24
centroid?
superpi83 2012-02-23 19:31:24
at the centroid of the base triangle, which is clearly (2/3,2/3,2/3)
msinghal 2012-02-23 19:31:24
(2/3, 2/3, 2/3)
joshxiong 2012-02-23 19:31:24
At the centroid
AkshajK 2012-02-23 19:31:24
at 2/3, 2/3, 2/3
.cpp 2012-02-23 19:31:24
The very center (2/3, 2/3, 2/3).
copeland 2012-02-23 19:31:29
copeland 2012-02-23 19:31:32
What's the altitude then?
nsun48 2012-02-23 19:32:11
we use 3d pythag to get the answer
googol.plex 2012-02-23 19:32:11
2/3*sqrt 3
superpi83 2012-02-23 19:32:11
2sqrt(3)/3
duketip10 2012-02-23 19:32:11
the magnitude
swimmer 2012-02-23 19:32:11
2sqrt3/3
cerberus88 2012-02-23 19:32:11
2\sqrt{3}/3
somanywindows 2012-02-23 19:32:11
2sqrt3/3
Garchomp293 2012-02-23 19:32:11
2sqrt3/3
msinghal 2012-02-23 19:32:11
2/3*sqrt3
NeilOnnsu 2012-02-23 19:32:11
2/3*sqrt(3)
Yongyi781 2012-02-23 19:32:11
2sqrt(3)/3
lucylai 2012-02-23 19:32:11
sqrt(8/9)=2sqrt3/3
ddot1 2012-02-23 19:32:11
2sqrt{3}/3
copeland 2012-02-23 19:32:14
copeland 2012-02-23 19:32:18
That's what I did.
copeland 2012-02-23 19:32:24
We had someone else in the office go this way:
copeland 2012-02-23 19:32:28
Another approach is by using these same coordinates and noticing that the cut is on the plane x+y+z-2=0.  Then apply the formula for the distance from a point to a plane and you're done.
copeland 2012-02-23 19:32:37
copeland 2012-02-23 19:32:47
There's a prettier answer, though.
copeland 2012-02-23 19:33:03
I mean, why, after all, is this a rational multiple of the length of the diagonal?
copeland 2012-02-23 19:33:06
A fourth approach draws all the parallel planes x+y+z=n for integers n.  These planes intersect the cube at the top vertex, the bottom vertex, and pass through trios of points.  In particular one of these is the plane we are interested in.
copeland 2012-02-23 19:33:12
copeland 2012-02-23 19:33:22
c0mbusti0n1295 2012-02-23 19:33:52
wait, sqrt 3/2?
AkshajK 2012-02-23 19:33:52
you mean 2sqrt3/3
negativebplusorminus 2012-02-23 19:33:52
Not Sqrt[3]/2.
copeland 2012-02-23 19:33:57
Yeah, typing is hard.
copeland 2012-02-23 19:34:05
Yongyi781 2012-02-23 19:34:51
Why is it obvious that they trisect the space diagonal?
copeland 2012-02-23 19:34:53
There's another way to see this picture: line up a bunch of cubes in a row and cut them with parallel planes. That gives an explicit picture of this trisection by translation symmetry.
copeland 2012-02-23 19:35:07
The fact that it's a trisection, though, follows from the fact that the planes are evenly spaced.
copeland 2012-02-23 19:35:27
Next!
copeland 2012-02-23 19:35:33
copeland 2012-02-23 19:35:40
When I see a problem like this, I like to try to sketch an example.  We have two sets of objects and we have relationships between them.
copeland 2012-02-23 19:35:44
copeland 2012-02-23 19:35:54
Now we don't know which song is which and we don't know how many of the girls are fans of Happy Birthday (it could be 0, 1, or 2).
copeland 2012-02-23 19:36:03
How do we attack this?
joshxiong 2012-02-23 19:36:36
Casework.
alex31415 2012-02-23 19:36:36
casework
MNL9082 2012-02-23 19:36:36
Use casework
briantix 2012-02-23 19:36:36
casework!
Asmodeus1123 2012-02-23 19:36:36
casework?
va2010 2012-02-23 19:36:36
Casework
fishinsea 2012-02-23 19:36:36
casework?
Klu2014 2012-02-23 19:36:36
casework
pgmath 2012-02-23 19:36:36
Casework.
MathForFun 2012-02-23 19:36:36
Casework...
swimdad11 2012-02-23 19:36:36
we could do organized casework.
copeland 2012-02-23 19:36:39
We should use casework on the number of girls who liked the fourth song.
copeland 2012-02-23 19:36:43
I actually drew the less useful graph up there.  What should we draw instead?
superpi83 2012-02-23 19:37:41
graph of what they don't like
.cpp 2012-02-23 19:37:41
Connect dislikes.
lucylai 2012-02-23 19:37:41
like-dislike diagrams
copeland 2012-02-23 19:37:45
Let's pair the songs with the girls that dislike that song.  Our example above becomes:
copeland 2012-02-23 19:37:48
copeland 2012-02-23 19:38:03
Our cases are whether 1, 2, or 3 girls dislike the last song.
copeland 2012-02-23 19:38:09
Let's start with the case that all 3 girls dislike one of the songs.  How many ways are there to build a graph satisfying all the hypotheses where all three girls dislike one of the songs?  (and not necessarily such that Happy Birthday is the despicable song.)
flyrain 2012-02-23 19:39:04
24
[cube] 2012-02-23 19:39:04
4!
christerson 2012-02-23 19:39:04
24
.cpp 2012-02-23 19:39:04
4*4!
zheng 2012-02-23 19:39:04
24 since the songs can be ordered in any way
briantix 2012-02-23 19:39:04
24
joshxiong 2012-02-23 19:39:04
24.
moose97 2012-02-23 19:39:04
4! ways
AkshajK 2012-02-23 19:39:04
There are 24, 4 ways to pick what joe dislikes, 3 ways to think what bath dislikes, and 2 ways to blink what Emi dislikes.
Seedleaf 2012-02-23 19:39:04
24 ways
superpi83 2012-02-23 19:39:05
pick 1 song to be the despicable song (4 ways), then assign the remaining songs to the three girls (3!=6 ways); total 24 ways
somanywindows 2012-02-23 19:39:05
24
copeland 2012-02-23 19:39:09
There are 4 ways to pick the universally reviled song.  Then, since there still needs to be a unique song distasteful to each of the three girls, there are 3! ways to pair the three girls with the remaining three girls.  This gives a total of 24 graphs.
copeland 2012-02-23 19:39:13
How many graphs can we draw where 2 of the girls hate one of the songs?
briantix 2012-02-23 19:40:13
72
somanywindows 2012-02-23 19:40:13
72
Seedleaf 2012-02-23 19:40:13
72 ways
Luckytoilet 2012-02-23 19:40:13
24*3
negativebplusorminus 2012-02-23 19:40:13
72
superpi83 2012-02-23 19:40:13
4 ways to pick the hated song, 3 ways to pick the 2 haters of the hated song, then 6 ways to do the rest. So 72.
cerberus88 2012-02-23 19:40:13
48
tan90 2012-02-23 19:40:13
72
zheng 2012-02-23 19:40:13
3*24=72 since we must now orderthe girls also
copeland 2012-02-23 19:40:17
There are still 4 ways to pick doubly-hated song, but now there are 3 ways to pick the unique girl who likes that song.  After that we still have 3! ways to assign the remaining songs.  That gives 72 graphs with a doubly-hated song.
copeland 2012-02-23 19:40:25
How many graphs can we find where all of the songs are unpalatable to exactly one of the girls?
copeland 2012-02-23 19:40:32
Now an example of such a graph is
copeland 2012-02-23 19:40:39
lucylai 2012-02-23 19:41:37
3*12=36
Seedleaf 2012-02-23 19:41:37
36 ways
TeamJacob 2012-02-23 19:41:37
36
briantix 2012-02-23 19:41:37
36
Iggy Iguana 2012-02-23 19:41:37
3 * 4 * 3 = 36
AkshajK 2012-02-23 19:41:37
4 ways to pick what joe dislikes, 3 ways to think what bath dislikes, and 2 ways to blink what Emi dislikes. Now, 3 ways to pick the pair that likes the last song, BUT we have to divide by 2 because those same two also like another song, and we don't want to overcount. So 4*3*2*3/2 = 36
msinghal 2012-02-23 19:41:37
3*4C2*2*1=36
ss5188 2012-02-23 19:41:37
36
duketip10 2012-02-23 19:41:37
6*3*2
super_pi314 2012-02-23 19:41:37
36
.cpp 2012-02-23 19:41:37
36
christerson 2012-02-23 19:41:37
36
copeland 2012-02-23 19:41:47
copeland 2012-02-23 19:41:51
What's the final answer?
AkshajK 2012-02-23 19:42:32
132
somanywindows 2012-02-23 19:42:32
132
tan90 2012-02-23 19:42:32
132
alex31415 2012-02-23 19:42:32
132
Seedleaf 2012-02-23 19:42:32
132 --> B
willwang123 2012-02-23 19:42:32
(B)
thepisong 2012-02-23 19:42:32
132
sparkles257 2012-02-23 19:42:32
add em up B
va2010 2012-02-23 19:42:32
132 (B)
Polynomial 2012-02-23 19:42:32
132
pr0likethis 2012-02-23 19:42:32
132
joshxiong 2012-02-23 19:42:32
B. 132
RotomPlasma 2012-02-23 19:42:32
24+72+36=132
Relativity1618 2012-02-23 19:42:32
B.132
Polynomial 2012-02-23 19:42:32
B 132
Draco 2012-02-23 19:42:32
132
KeepingItReal 2012-02-23 19:42:32
6*36=132 so B
ss5188 2012-02-23 19:42:32
132
Jasonion 2012-02-23 19:42:32
132
copeland 2012-02-23 19:42:52
copeland 2012-02-23 19:42:54
The answer is (B).
munugala 2012-02-23 19:43:02
I like the songs you picked!!!
copeland 2012-02-23 19:43:04
Then you must not be one of the girls.
copeland 2012-02-23 19:43:35
Alright, the bug problem.
copeland 2012-02-23 19:43:45
Prepare yourselves for some mad graphing skillz.
copeland 2012-02-23 19:43:45
copeland 2012-02-23 19:43:52
mathdate 2012-02-23 19:44:07
this problem really bugged me
copeland 2012-02-23 19:44:09
:)
copeland 2012-02-23 19:44:20
This problem seems a little daunting.  Let's get our feet wet by solving an easier version.
copeland 2012-02-23 19:44:50
copeland 2012-02-23 19:44:54
copeland 2012-02-23 19:45:25
(I un-reversed the backwards arrows.)
copeland 2012-02-23 19:45:29
How can we solve this problem?
sammyMaX 2012-02-23 19:46:14
Well that was if you couldn't go backwards
sammyMaX 2012-02-23 19:46:14
Multiplication!
alex31415 2012-02-23 19:46:14
constructive counting
swimmer 2012-02-23 19:46:14
putting numbers at points
Garchomp293 2012-02-23 19:46:14
finding the amount of ways to get to each point, then adding them all up
cuttlefishhorn 2012-02-23 19:46:14
Go point by point! How many ways there are to get to each segment, and add them up to figure out the next segment
mentalgenius 2012-02-23 19:46:14
write down numbers at every corner
zheng 2012-02-23 19:46:14
Figuring out the number of paths to each point
Seedleaf 2012-02-23 19:46:14
there's 2 ways to go for the first intersection, then 3 ways, then 4 ways, then 5... so on.
copeland 2012-02-23 19:46:22
Now the little bug can only cross these vertical lines one time each:
copeland 2012-02-23 19:46:27
copeland 2012-02-23 19:46:29
How many ways can the bug cross the first line?
Macromiidae 2012-02-23 19:46:54
2
teethpaste 2012-02-23 19:46:54
2
c0mbusti0n1295 2012-02-23 19:46:54
dos
moose97 2012-02-23 19:46:54
2
timoteomo3 2012-02-23 19:46:54
2
flyrain 2012-02-23 19:46:54
2.
kenniky 2012-02-23 19:46:54
2
AkshajK 2012-02-23 19:46:54
2
va2010 2012-02-23 19:46:54
2
sammyMaX 2012-02-23 19:46:54
2
iamnew 2012-02-23 19:46:54
2
va2010 2012-02-23 19:46:54
2
zheng 2012-02-23 19:46:54
2
PChang 2012-02-23 19:46:54
2
pgmath 2012-02-23 19:46:54
2
177dum 2012-02-23 19:46:54
2
copeland 2012-02-23 19:46:58
2: either she goes through the top edge or the bottom edge.
copeland 2012-02-23 19:47:01
How many ways can the bug cross the first two lines?
Garchomp293 2012-02-23 19:47:30
6
Jzou 2012-02-23 19:47:30
6
willwang123 2012-02-23 19:47:30
2x3=6
PChang 2012-02-23 19:47:30
6
timoteomo3 2012-02-23 19:47:30
2*3=6!
pubado 2012-02-23 19:47:30
6
va2010 2012-02-23 19:47:30
2*3=6
ss5188 2012-02-23 19:47:30
6
trip 2012-02-23 19:47:30
6
martian179 2012-02-23 19:47:30
2 * 3 = 6
matthe44 2012-02-23 19:47:30
6
ptes77 2012-02-23 19:47:30
6
Mattchu386 2012-02-23 19:47:30
6
Draco 2012-02-23 19:47:30
6
copeland 2012-02-23 19:47:34
2*3.  For either choice crossing the first line, there are three ways to cross the second line.  Since the bug cannot pass along any edge more than once, there's exactly one path that uses one of the first edges and one of the second edges.
copeland 2012-02-23 19:47:36
How many ways can the bug cross the entire diagram?
iamnew 2012-02-23 19:48:23
2880
RotomPlasma 2012-02-23 19:48:23
2*3*4*5*4*3*2=2880 ways
PChang 2012-02-23 19:48:23
2*3*4*5*4*3*2
willwang123 2012-02-23 19:48:23
2 x 3 x 4 x 5 x 4 x 3 x 2 = <some big number>
InDe_eD 2012-02-23 19:48:23
2880
davidkim2106 2012-02-23 19:48:23
2880??
Seedleaf 2012-02-23 19:48:23
2*3*4*5*4*3*2 = 2880
Jzou 2012-02-23 19:48:23
2880
teethpaste 2012-02-23 19:48:23
2*3*4*5*4*3*2=2880
Macromiidae 2012-02-23 19:48:23
2*3*4*5*4*3*2=2880
christerson 2012-02-23 19:48:23
2880
MNL9082 2012-02-23 19:48:23
2*3*4*5*4*3*2
MathForFun 2012-02-23 19:48:23
2x3x4x5x4x3x2
eb8368 2012-02-23 19:48:23
2*3*4*5**4*3*2 = 2880
copeland 2012-02-23 19:48:32
Therefore there are 2*3*4*5*4*3*2 paths that the bug can take through this maze.
copeland 2012-02-23 19:48:45
Things get harder, though, when we look at the lattice with the reversed arrows.  Our toy problem was more approachable because the bug could only cross each line once, but, more importantly, what the bug could do to cross one blue line did not depend on what the bug did to cross any previous blue lines.
copeland 2012-02-23 19:48:50
Let's go back to the original problem.
copeland 2012-02-23 19:48:59
copeland 2012-02-23 19:49:04
copeland 2012-02-23 19:49:17
Now the bug can cross many of those lines more than once and sometimes the ways the bug can cross depends on what the bug did in the past.
copeland 2012-02-23 19:49:21
However it would still be nice if we could "factor" this problem.  Do you see anything that we can draw?
kenniky 2012-02-23 19:50:20
we can keep some of the lines from earlier
ksun48 2012-02-23 19:50:20
just the even vertical lines
Jasmine8925 2012-02-23 19:50:20
draw vertical lines in the places where there are no backwards arrows
briantix 2012-02-23 19:50:20
the lines from the previous problem escept the ones that cross a white arrow
Polynomial 2012-02-23 19:50:20
Draw lines in the columns where there are no backward arrows
christerson 2012-02-23 19:50:20
lines without backwards arrows?
copeland 2012-02-23 19:50:24
We can still draw a couple of the blue lines:
copeland 2012-02-23 19:50:26
copeland 2012-02-23 19:50:40
(There are a couple more we could draw, but, meh.)
copeland 2012-02-23 19:50:46
Once the bug passes either of these blue lines its options do not depend on what happened before it hit the blue line (except possibly where it crossed the line).
copeland 2012-02-23 19:50:56
Let's talk about the left region.  How are we going to count the ways that the bug can get from A to the left blue bar?
noedne 2012-02-23 19:51:50
casework
fishinsea 2012-02-23 19:51:50
if it uses reverse or not
gurev 2012-02-23 19:51:50
the ways not going backwards, then ways going backwards
Seedleaf 2012-02-23 19:51:50
with and without backwards arrow
RotomPlasma 2012-02-23 19:51:50
Only forwards and using the reverse to get to the other side
Jasmine8925 2012-02-23 19:51:50
two cases: bug goes backwards or doesn't go backwards
martian179 2012-02-23 19:51:50
casework (whether u use the backwards arrow)?
sammyMaX 2012-02-23 19:51:50
Casework: whether the bug goes backwards or not
copeland 2012-02-23 19:51:56
We should split it into cases.  Either the bug goes through the white arrow or it does not.
copeland 2012-02-23 19:52:00
How many paths can the bug take without going through the white arrow?
googol.plex 2012-02-23 19:52:31
16
Jzou 2012-02-23 19:52:31
16
qaryaendpetyp 2012-02-23 19:52:31
16
azure55 2012-02-23 19:52:34
2*2*4=16
qaryaendpetyp 2012-02-23 19:52:34
2*2*4=16
Iggy Iguana 2012-02-23 19:52:34
2*2*4=16
ksun48 2012-02-23 19:52:34
2*2*4=16
Seedleaf 2012-02-23 19:52:34
2*2*4 = 16
superpi83 2012-02-23 19:52:34
2*2*4=16
fishinsea 2012-02-23 19:52:34
2*2*4 = 16
AlcumusGuy 2012-02-23 19:52:34
16?
lucylai 2012-02-23 19:52:34
16
copeland 2012-02-23 19:52:36
There are 16 of these paths: 2 choices for the first segment, 2 for the second and 4 for the third.  (We might want to keep track, so notice that these choices are entirely symmetric.  Four of these paths go through each of the four segments on the first blue line.)
copeland 2012-02-23 19:52:45
Now how many paths cross the first white arrow on the way to the first blue line?
PChang 2012-02-23 19:53:29
2*2*4
AkshajK 2012-02-23 19:53:29
4
superpi83 2012-02-23 19:53:29
2*2=4
mcdonalds106_7 2012-02-23 19:53:29
4
Macromiidae 2012-02-23 19:53:29
4?
ksun48 2012-02-23 19:53:29
2*2=4
ifailedmathcounts 2012-02-23 19:53:29
4
MNL9082 2012-02-23 19:53:29
Four
InDe_eD 2012-02-23 19:53:29
4
teethpaste 2012-02-23 19:53:29
4
Seedleaf 2012-02-23 19:53:29
4 ways
copeland 2012-02-23 19:53:35
If a path crosses the top arrow first then it must start in this way:
copeland 2012-02-23 19:53:39
copeland 2012-02-23 19:53:43
There are only two choices from here to cross the middle line (one of the two bottom paths).
copeland 2012-02-23 19:53:47
By symmetry, there are 2 choices for a path that crosses the bottom segment first.
copeland 2012-02-23 19:53:50
There are a total of 4 paths.  Each of these four paths cross a different arrow.
copeland 2012-02-23 19:53:55
Now we take score.  There are exactly 4+1=5 ways to get from point A to each of those four arrows along the first blue line.
copeland 2012-02-23 19:54:08
copeland 2012-02-23 19:54:47
What does this picture tell us about the original problem?
Iggy Iguana 2012-02-23 19:55:28
so that means the answer is E because it is the only one that is divisible by 20
gurev 2012-02-23 19:55:28
Now you know that it has to be divisible byfive right. So your done
teethpaste 2012-02-23 19:55:28
it is a multiple of 5
kenniky 2012-02-23 19:55:28
It involves 5
Jasonion 2012-02-23 19:55:28
u don't have to do any more work, since E is the only multiple of 5?
AkshajK 2012-02-23 19:55:28
its a mulitple of 5
superpi83 2012-02-23 19:55:28
the answer is a multiple of 5. WOW. so E.
Iggy Iguana 2012-02-23 19:55:28
number of paths is divible by 20
somanywindows 2012-02-23 19:55:28
the answer has to be a multiple of 5
alex31415 2012-02-23 19:55:28
it must be a multiple of 5, so the answer must be 2400(E)?
fortenforge 2012-02-23 19:55:28
The answer must be divisible by 5
copeland 2012-02-23 19:55:37
BLAAH!! How lame. :(
copeland 2012-02-23 19:55:43
No matter what happens the rest of the way, the number of paths is guaranteed to be a multiple of 5.  The only solution given that is a multiple of 5 is the 2400, so that has to be the answer.
copeland 2012-02-23 19:55:57
Good going AMC dudes. . . .
copeland 2012-02-23 19:55:58
Oh well.  Let's see if this problem has anything more to teach us.
copeland 2012-02-23 19:56:07
Can we learn anything more from what we've done so far?
copeland 2012-02-23 19:57:10
My favorite thing in the mathiverse is symmetry.
copeland 2012-02-23 19:57:16
Do we have any symmetry here?
mentalgenius 2012-02-23 19:57:53
the last section of the graph is the same as the first one is
fractals 2012-02-23 19:57:53
Same for the other side, at the end
ifailedmathcounts 2012-02-23 19:57:53
it's the same from teh 2nd line to B, path is reversible
RotomPlasma 2012-02-23 19:57:53
Lololol. The entire field is symmetrical, so there are 5 solutions at the right, too.
azure55 2012-02-23 19:57:53
the rightmost part and the leftmost part we just calculated
sammyMaX 2012-02-23 19:57:53
Yes, the first section and the third section are the same
copeland 2012-02-23 19:58:00
There is symmetry!  We also know that there are 5 ways to get from any of the segments along the right blue line to the point B.
copeland 2012-02-23 19:58:04
copeland 2012-02-23 19:58:06
Now we just need to find all of the ways to get from any point on the first blue line to any point on the second blue line and multiply by 25.
copeland 2012-02-23 19:58:17
How many paths pass from one blue line to the other and DO NOT use the white arrow?
copeland 2012-02-23 19:58:41
(Considering ALL entry and exit points.)
fractals 2012-02-23 19:59:02
4 * 4 * 4 = 64
ptes77 2012-02-23 19:59:02
64
RotomPlasma 2012-02-23 19:59:02
4^3=64
mcdonalds106_7 2012-02-23 19:59:02
64
Goos 2012-02-23 19:59:02
4^3=64
lucylai 2012-02-23 19:59:02
64
fishinsea 2012-02-23 19:59:02
64
christerson 2012-02-23 19:59:02
64
Iggy Iguana 2012-02-23 19:59:02
4*4=16
NeilOnnsu 2012-02-23 19:59:02
64
copeland 2012-02-23 19:59:05
There are 4 choices for entry, 4 choices for exit and 4 choices for the middle path so there are 64 total choices in this case.
copeland 2012-02-23 19:59:08
How many paths are there from one blue line to the other that pass through the white arrow?
copeland 2012-02-23 20:00:05
This is more subtle: now the start and end paths are dependent.  The best way to count is by considering whether the white path is entered from the top or the bottom.  One case has paths that all use these edges:
copeland 2012-02-23 20:00:09
copeland 2012-02-23 20:00:18
How many paths are there like that?
copeland 2012-02-23 20:01:02
How many paths through the middle section are there that contain that subpath, I mean?
AkshajK 2012-02-23 20:01:51
16
ifailedmathcounts 2012-02-23 20:01:51
16
superpi83 2012-02-23 20:01:51
16
Iggy Iguana 2012-02-23 20:01:51
2*2*2*2=16
mcdonalds106_7 2012-02-23 20:01:51
16
noedne 2012-02-23 20:01:51
2*2*2*2=16
canadian 2012-02-23 20:01:51
2^4=16
copeland 2012-02-23 20:02:06
There are 2^4=16 such paths.  By symmetry there are 16 paths from blue to blue that go through the middle white arrow in the other direction as well.
copeland 2012-02-23 20:02:18
That gives a total of 96 paths through the middle section.  How many total paths can the bug take?
rlingineni 2012-02-23 20:02:54
2400
noedne 2012-02-23 20:02:54
2400E
mcdonalds106_7 2012-02-23 20:02:54
25*96=2400 or E
sparkles257 2012-02-23 20:02:54
5*5*96=2400
coldsummer 2012-02-23 20:02:54
2400
iamnew 2012-02-23 20:02:54
2400
Seedleaf 2012-02-23 20:02:54
96*25 = 2400 (E)
NeilOnnsu 2012-02-23 20:02:54
2400
negativebplusorminus 2012-02-23 20:02:54
2400
fishinsea 2012-02-23 20:02:54
96*25=2400
superpi83 2012-02-23 20:02:54
96*25=2400 E
matemyday1234567890 2012-02-23 20:02:54
2400
numberwiz 2012-02-23 20:02:54
2400
BOGTRO 2012-02-23 20:02:54
25*96=2400
noedne 2012-02-23 20:02:54
96*25=2400 so the answer is E
ringwraith10 2012-02-23 20:02:54
2400
va2010 2012-02-23 20:02:54
96*5*5=2400
martian179 2012-02-23 20:02:54
5 * 96 * 5 = 2400
ringwraith10 2012-02-23 20:02:54
25*96
timoteomo3 2012-02-23 20:02:54
2400! (E)
copeland 2012-02-23 20:02:58
copeland 2012-02-23 20:03:01
The answer is (E).
copeland 2012-02-23 20:03:04
Super.
sammy1615 2012-02-23 20:03:19
is that all for amc 10?
copeland 2012-02-23 20:03:32
Yeah, now we'll take a short break and then move on to the AMC12.
copeland 2012-02-23 20:03:49
It's 5:04 on my desk.  Meet me back on my desk at 5:09.
rlingineni 2012-02-23 20:04:15
That took 1 hr, on the AMC we're only given 75 min
copeland 2012-02-23 20:04:17
But we solved that last one about 15min ago.
BOGTRO 2012-02-23 20:05:03
we went through a lot more explanation than we would have on actual AMC yes?
copeland 2012-02-23 20:05:05
Yes. The AMC explanations consist of a few tiny bubbles.
Mrdavid445 2012-02-23 20:06:44
If we finish early, can you go over AMC 12 problems before 20?
copeland 2012-02-23 20:06:45
Yes, we may do that.
negativebplusorminus 2012-02-23 20:07:38
What do you think about number 12 of the AMC 12B, in terms of the interpretation? (there's a raging debate between D and E).
copeland 2012-02-23 20:07:42
The one with the sequences of ones and zeros?  I'm on the side of not having ones still makes you such a sequence.
copeland 2012-02-23 20:07:47
But I didn't think very hard about it.
Klu2014 2012-02-23 20:08:11
When will this jam end? At what time
zheng 2012-02-23 20:08:11
What time are we supposed to finish?
copeland 2012-02-23 20:08:13
Probably about an hour more?  Depends on how quickly we move.
copeland 2012-02-23 20:08:41
Breaktime's over!  TIME FOR MORE MATH!
copeland 2012-02-23 20:08:45
Sorry.  I get excited.
Klu2014 2012-02-23 20:08:59
YAY!
RotomPlasma 2012-02-23 20:08:59
YAY :D
copeland 2012-02-23 20:09:02
Looks like you two do, too.
copeland 2012-02-23 20:09:07
copeland 2012-02-23 20:09:14
copeland 2012-02-23 20:09:27
This was a funny point on the exam.  There's a chance that you came across this problem armed with WAY more tricks up your sleeve than is actually healthy, and spun your wheels with a lot of really sophisticated machines.
copeland 2012-02-23 20:09:37
Hopefully not, though.  Hopefully you remembered one of the most important rules: if you see a lot of equal angles you should look for. . . .
.cpp 2012-02-23 20:10:07
Similar triangles!
fishinsea 2012-02-23 20:10:07
similar triangles
negativebplusorminus 2012-02-23 20:10:07
Similar triangles.
iwantcombo 2012-02-23 20:10:07
similar triangles
mentalgenius 2012-02-23 20:10:07
similar + congruent triangles
tan90 2012-02-23 20:10:07
Similar triangles
azure55 2012-02-23 20:10:07
similar trangles
mdecross 2012-02-23 20:10:07
similar triangles
canadian 2012-02-23 20:10:07
Similar triangles
Iggy Iguana 2012-02-23 20:10:07
similar triangles
Mrdavid445 2012-02-23 20:10:07
similar stuff?
zheng 2012-02-23 20:10:07
similar triangles
coldsummer 2012-02-23 20:10:07
similarity
copeland 2012-02-23 20:10:10
Similar triangles!
copeland 2012-02-23 20:10:19
Super, good job.  We're almost about halfway home now.
copeland 2012-02-23 20:10:23
Do you see any similar triangles?
tan90 2012-02-23 20:11:28
ABX is congruent to YEZ
himym83 2012-02-23 20:11:28
abx, zey
briantix 2012-02-23 20:11:28
ABX~YEZ
Mrdavid445 2012-02-23 20:11:28
ABX and ZEY
christerson 2012-02-23 20:11:28
ZEY and XBA
TigerSneak1 2012-02-23 20:11:28
ABX and YEZ
negativebplusorminus 2012-02-23 20:11:28
ABX and ZEY, perhaps?
numberwiz 2012-02-23 20:11:28
abx and yez
copeland 2012-02-23 20:11:32
copeland 2012-02-23 20:11:46
Also, always put the vertices in your similar triangles in the same order.
copeland 2012-02-23 20:11:55
This will keep you from making a LOT of mistakes.
copeland 2012-02-23 20:12:07
copeland 2012-02-23 20:12:17
copeland 2012-02-23 20:12:38
At this point in solving, I decided to spin my paper so that that bit was at the bottom.
copeland 2012-02-23 20:12:43
copeland 2012-02-23 20:12:46
And the bits we actually know things about are:
copeland 2012-02-23 20:12:51
copeland 2012-02-23 20:12:56
That's a little cleaner.  Here's the actual data we know:
copeland 2012-02-23 20:13:00
copeland 2012-02-23 20:13:15
Phew OK.  We've written stuff down so we don't have to remember anymore.
copeland 2012-02-23 20:13:20
It's hard to think while remembering. . .
copeland 2012-02-23 20:13:24
That is nice, but we haven't really figured out how to use the information about the angles at E and F yet.
copeland 2012-02-23 20:13:40
How are these angles related?
Klu2014 2012-02-23 20:14:06
120 degrees
mentalgenius 2012-02-23 20:14:06
angle E = angle F = 120 degrees
numberwiz 2012-02-23 20:14:06
120
noedne 2012-02-23 20:14:06
congruent
negativebplusorminus 2012-02-23 20:14:06
They're equal.
Iggy Iguana 2012-02-23 20:14:06
they are equal, all 120 degrees
TeamJacob 2012-02-23 20:14:06
120degrees
RotomPlasma 2012-02-23 20:14:06
They're equal...?
AkshajK 2012-02-23 20:14:06
they are both 120
fractals 2012-02-23 20:14:06
They both equal 120 degrees
Seedleaf 2012-02-23 20:14:06
they are equal to 120
sammyMaX 2012-02-23 20:14:06
Congruent! (both 120)
copeland 2012-02-23 20:14:11
They are both equal to 120 degrees.  How do we use that?
alex31415 2012-02-23 20:14:58
law of cosines?
negativebplusorminus 2012-02-23 20:14:58
Law of Cosines?
numberwiz 2012-02-23 20:14:58
law of cos
briantix 2012-02-23 20:14:58
law of cosines?
tan90 2012-02-23 20:14:58
Law of cosines/sines?
zheng 2012-02-23 20:15:00
cosine law?
Seedleaf 2012-02-23 20:15:00
law of cosines to find other side (YZ or ZA)
mentalgenius 2012-02-23 20:15:00
law of cosines?
.cpp 2012-02-23 20:15:00
Law of cosines.
copeland 2012-02-23 20:15:17
We could try some Law of Sines or Law of Cosines shenanigans and that will work, eventually, but a lot of times the Law of Cosines for 120 degrees just boils down to. . . .
c0mbusti0n1295 2012-02-23 20:15:47
30 60 90 triangles?
coldsummer 2012-02-23 20:15:47
create 30-60-90 triangles
number.sense 2012-02-23 20:15:47
30-60-90 triangles
xxrxxhxx 2012-02-23 20:15:47
30-60-90 triangles
negativebplusorminus 2012-02-23 20:15:47
Pythagoras?
copeland 2012-02-23 20:15:59
Good.  Where are they?
tan90 2012-02-23 20:16:41
Draw altitudes.
RotomPlasma 2012-02-23 20:16:41
Draw altitudes
fractals 2012-02-23 20:16:41
Drop perpendiculars
c0mbusti0n1295 2012-02-23 20:16:41
lol i was thinking of drawing altitudes..
mcdonalds106_7 2012-02-23 20:16:41
draw the altitudes
KeepingItReal 2012-02-23 20:16:41
draw right angles
copeland 2012-02-23 20:16:42
Super, where?
copeland 2012-02-23 20:17:42
Notice that what we want to find is a 60 degree angle, remember?
fractals 2012-02-23 20:18:13
A and Y to EF
mdecross 2012-02-23 20:18:13
from A and Y
mentalgenius 2012-02-23 20:18:13
with hypotenuses YE and AF, with perpendiculars vertical
TeamJacob 2012-02-23 20:18:13
draw vertical lines from y and a
proglote 2012-02-23 20:18:13
Y to EF
silvernight 2012-02-23 20:18:13
extend EF and draw a perpedicular containing A to line EF
copeland 2012-02-23 20:18:17
If we drop perpendiculars from Y and from A to the line through E and F we get a couple of nice 30-60-90 triangles.
copeland 2012-02-23 20:18:20
(The picture will be a little easier if we ignore the top of the diagram entirely for a while.)
copeland 2012-02-23 20:18:24
copeland 2012-02-23 20:18:33
Now what distances do we know?
fractals 2012-02-23 20:19:12
LE = 20, YL = 20sqrt3
alex31415 2012-02-23 20:19:12
YL=20sqrt3, LE=20
PChang 2012-02-23 20:19:12
EL,LY
TigerSneak1 2012-02-23 20:19:12
LE and LY via 30-60-90
somanywindows 2012-02-23 20:19:12
LE = 20, YL = 20sqrt3
proglote 2012-02-23 20:19:12
YL and EL
Smart Pi 2012-02-23 20:19:12
YL and LE?
copeland 2012-02-23 20:19:20
Since YEL is a 30-60-90 triangle with hypotenuse 40 we know both of the legs:
copeland 2012-02-23 20:19:23
copeland 2012-02-23 20:20:21
What's left that we haven't incoroporated into the picture?
fractals 2012-02-23 20:20:35
angle YZA = 90
number.sense 2012-02-23 20:20:35
YZ = AZ and its a right angle
pjyoung 2012-02-23 20:20:35
YZ and AZ?
proglote 2012-02-23 20:20:35
the fact that YZ = AZ (sides of square)
mentalgenius 2012-02-23 20:20:35
YZAX is a square
.cpp 2012-02-23 20:20:35
YZ=AZ.
Yongyi781 2012-02-23 20:20:35
AZY is a right angle!
copeland 2012-02-23 20:20:45
We haven't used that AXYZ is a square.  What does that tell us?
copeland 2012-02-23 20:20:52
(It tells us 2 things.)
pgmath 2012-02-23 20:22:35
YZL is similar to ZAR and YZ = AZ
canadian 2012-02-23 20:22:35
Triangles YLZ and ZRA are similar
Iggy Iguana 2012-02-23 20:22:35
congruent triangles lyz, rza
zheng 2012-02-23 20:22:35
YLZ and ZRA are congruent since YZ=AZ and their angles are equal
Yongyi781 2012-02-23 20:22:35
YLZ is congruent to ZRA
mdecross 2012-02-23 20:22:35
also that YZ and AZ are equal in length
copeland 2012-02-23 20:22:42
Since AXYZ is a square, AZY is a right angle.  That forces YZL and AZR to be complementary angles.
copeland 2012-02-23 20:22:46
By AA similarity, YZL and ZAR are similar.  Furthermore, since YZ=ZA these are actually congruent!
copeland 2012-02-23 20:22:50
Therefore we can copy the distance YL=ZR.
copeland 2012-02-23 20:22:53
copeland 2012-02-23 20:23:31
Our diagram's looking good, but now we've run out of information.  This is to be expected: we should have to solve SOMETHING to figure out where A is.  In particular, we still need to find the scale of the triangle AFR.  So let's go ahead an bite the bullet and throw a variable in there.
copeland 2012-02-23 20:23:42
Here's a nice thing: waiting to introduce the variable allows us to pick one that will lead to the least headache when solving.
copeland 2012-02-23 20:24:55
Here's the philosophy.  We're trying to lock down this triangle on the right.  That's where the variable is going to be used.  What edge should we make the variable?
coldsummer 2012-02-23 20:25:22
FR
zheng 2012-02-23 20:25:22
FR
KeepingItReal 2012-02-23 20:25:22
FR
Yongyi781 2012-02-23 20:25:22
FR?
Iggy Iguana 2012-02-23 20:25:22
fr?
adamdai97 2012-02-23 20:25:22
FR
proglote 2012-02-23 20:25:22
FR
copeland 2012-02-23 20:25:26
copeland 2012-02-23 20:25:33
copeland 2012-02-23 20:25:47
copeland 2012-02-23 20:25:51
copeland 2012-02-23 20:26:01
Now we can find x!  I'll give you a moment.
number.sense 2012-02-23 20:27:18
21
Goos 2012-02-23 20:27:18
x=21
sammyMaX 2012-02-23 20:27:18
21
christerson 2012-02-23 20:27:18
21
fractals 2012-02-23 20:27:18
xsqrt3 + 20sqrt3 = 20 + 41(sqrt3 - 1) + x, so x(sqrt3 - 1) = (41 - 20)(sqrt3 - 1) = 21(sqrt3 - 1), so x = 21.
mentalgenius 2012-02-23 20:27:18
x = 21
Jzou 2012-02-23 20:27:18
x=21
Yongyi781 2012-02-23 20:27:18
x = 21!!!!!
copeland 2012-02-23 20:27:23
copeland 2012-02-23 20:27:39
copeland 2012-02-23 20:27:46
And what is the side-length of the square?
nikeballa96 2012-02-23 20:28:34
29sqrt3
noedne 2012-02-23 20:28:34
A
fractals 2012-02-23 20:28:34
29sqrt3
number.sense 2012-02-23 20:28:34
A
Seedleaf 2012-02-23 20:28:34
29sqrt3 -->A!
mentalgenius 2012-02-23 20:28:34
(A) 29 sqrt(3)
lucylai 2012-02-23 20:28:34
29sqrt3
noedne 2012-02-23 20:28:34
20-21-29 *sqrt 3
lucylai 2012-02-23 20:28:34
29sqrt3 (A)
christerson 2012-02-23 20:28:34
29 sqrt(3)
ksun48 2012-02-23 20:28:34
29sqrt{3}
.cpp 2012-02-23 20:28:34
29sqrt{3}
alex31415 2012-02-23 20:28:34
29sqrt2
KeepingItReal 2012-02-23 20:28:34
by pythag, length is 29sqrt(3) so answer is A
copeland 2012-02-23 20:29:11
The answer is (A).
copeland 2012-02-23 20:29:15
copeland 2012-02-23 20:29:21
copeland 2012-02-23 20:29:33
Problem 22 on the 12B is the same as 25 on the 10B so we already solved that one.
copeland 2012-02-23 20:29:41
copeland 2012-02-23 20:29:52
(They changed the answer choices from the original version on this problem.  These were the updated options.)
copeland 2012-02-23 20:30:05
I'll come clean with you on this one: the good way to do this problem is by writing all the answers down and calling yourself done.  Using math on this problem turns out not to be so helpful unless you use a LOT of math.
copeland 2012-02-23 20:30:18
Let's try that.  What z satisfy |z|=1 and are easy to work with?
c0mbusti0n1295 2012-02-23 20:30:43
1, -1
m1sterzer0 2012-02-23 20:30:43
-1
einstein3.14 2012-02-23 20:30:43
-1,1
somanywindows 2012-02-23 20:30:43
-1, 1
matemyday1234567890 2012-02-23 20:30:43
1
coldsummer 2012-02-23 20:30:43
1
KeepingItReal 2012-02-23 20:30:43
(1,0)
Animyth 2012-02-23 20:30:43
+or-1
knowmath 2012-02-23 20:30:43
-1,1
christerson 2012-02-23 20:30:43
1, -1
TigerSneak1 2012-02-23 20:30:43
1?
copeland 2012-02-23 20:30:47
First we have z=1.
copeland 2012-02-23 20:30:50
How many choices for a, b, c, and d give P(1)=0?
canadian 2012-02-23 20:31:26
non
fractals 2012-02-23 20:31:26
None
GoldenFrog1618 2012-02-23 20:31:26
it does not matter, it will give 0 to the sum
alex31415 2012-02-23 20:31:26
it doesnt matter
matemyday1234567890 2012-02-23 20:31:26
0
soccerfan 2012-02-23 20:31:26
0
alex31415 2012-02-23 20:31:26
it will always be 0
lucylai 2012-02-23 20:31:26
none
mentalgenius 2012-02-23 20:31:26
none
RotomPlasma 2012-02-23 20:31:26
0
mentalgenius 2012-02-23 20:31:26
ZERO
copeland 2012-02-23 20:31:31
copeland 2012-02-23 20:31:38
(Furthermore that doesn't contribute to the sum anyway.)
copeland 2012-02-23 20:31:43
The next simplest is z=-1.  What polynomials of the form above satisfy P(-1)=0?
knowmath 2012-02-23 20:32:20
a lot
iamnew 2012-02-23 20:32:20
-a+b-c+d=0
lucylai 2012-02-23 20:32:20
4-a+b-c+d=0
fractals 2012-02-23 20:32:20
4 - a + b - c + d = 0
alex31415 2012-02-23 20:32:20
b+d+4=a+c
.cpp 2012-02-23 20:32:20
a+c = 4+b+d
copeland 2012-02-23 20:32:34
Good, and how does the nondecreasing assumption narrow this?
copeland 2012-02-23 20:33:07
("Nonincreasing," I guess?  Their inequality is written d to a.)
copeland 2012-02-23 20:33:59
Here are some observations:
m1sterzer0 2012-02-23 20:34:01
4=a, b=c, d=0
KeepingItReal 2012-02-23 20:34:01
or 1,1,3,3
.cpp 2012-02-23 20:34:01
a= 4, b = 2, c=2, d=0; a=4, b=1, c=1, d=0
pr0likethis 2012-02-23 20:34:01
a must be four, because if a was smaller than b>=c>=d wouldnt be true
copeland 2012-02-23 20:34:19
Can we make that  formal?
.cpp 2012-02-23 20:34:56
4>=a; b>=c, so 4+b>=a+c.  Hence, 4=a, b=c, d=0.
copeland 2012-02-23 20:35:03
copeland 2012-02-23 20:35:17
KeepingItReal 2012-02-23 20:35:56
8+2b
Seedleaf 2012-02-23 20:35:56
8+2b
superwatchdog 2012-02-23 20:35:56
8+2b
iamnew 2012-02-23 20:35:56
8+2b
adamdai97 2012-02-23 20:35:56
8+2b
RotomPlasma 2012-02-23 20:35:56
8+2b
TigerSneak1 2012-02-23 20:35:56
8 + 2b
PChang 2012-02-23 20:35:56
8+2b
copeland 2012-02-23 20:35:58
And as b varies from 0 to 4, what is the total sum?
.cpp 2012-02-23 20:36:34
(8+2b) for b in range 0,4 = 40 + 20 = 60.
fractals 2012-02-23 20:36:34
40 + 2(1 + 2 + 3 + 4) = 60
Iggy Iguana 2012-02-23 20:36:34
60
pr0likethis 2012-02-23 20:36:34
8+10+12+14+16=60
number.sense 2012-02-23 20:36:34
60
Yongyi781 2012-02-23 20:36:34
60
azure55 2012-02-23 20:36:34
60
somanywindows 2012-02-23 20:36:34
60
iamnew 2012-02-23 20:36:34
60
pgmath 2012-02-23 20:36:34
60
super_pi314 2012-02-23 20:36:34
60
copeland 2012-02-23 20:36:38
copeland 2012-02-23 20:36:52
copeland 2012-02-23 20:36:57
What other z with |z|=1 might be good to try?
Yongyi781 2012-02-23 20:38:16
kth roots of unity for k=3, 4... and 5!!! (and probably 6, 7, 8, and 9). And also non-roots of unity
plokclop 2012-02-23 20:38:16
The cube roots of 1.
RotomPlasma 2012-02-23 20:38:16
+/- i?
number.sense 2012-02-23 20:38:16
cis45
azure55 2012-02-23 20:38:16
cos45+isin45
ksun48 2012-02-23 20:38:16
cube rts of unity
pr0likethis 2012-02-23 20:38:16
in other words, cis(30, 45, 60, etc)
copeland 2012-02-23 20:38:22
Let's look for other roots of unity.  Do you see any more that work?
fractals 2012-02-23 20:39:15
cis(2pi/5)
ksun48 2012-02-23 20:39:15
5th roots?
copeland 2012-02-23 20:39:26
What polynomial does a primitive fifth root solve?
m1sterzer0 2012-02-23 20:40:03
5th roots work for (z^5-1)/(z-1)
Yongyi781 2012-02-23 20:40:03
x^4 + x^3 + x^2 + x + 1 = 0
lucylai 2012-02-23 20:40:03
x^5-1=0
robinpark 2012-02-23 20:40:03
z^5 - 1 = 0
.cpp 2012-02-23 20:40:03
x^4+x^3+x^2+x+1 = 0.
fractals 2012-02-23 20:40:03
z^5 - 1 = 0, OR z^4 + z^3 + z^2 + z + 1 = 0 (it is primitive)
number.sense 2012-02-23 20:40:03
x^4 + x^3 + x^2 + x + 1
copeland 2012-02-23 20:40:11
copeland 2012-02-23 20:40:12
That goes on the list.
copeland 2012-02-23 20:40:20
Any other roots of unity in sight?
.cpp 2012-02-23 20:41:20
Cube roots work (e.g. 4x^2(x^2+x+1)).
dinoboy 2012-02-23 20:41:20
the third and fourth roots of unity generate meaningful polynomials
knowmath 2012-02-23 20:41:20
fourth root
copeland 2012-02-23 20:41:25
copeland 2012-02-23 20:41:36
These two add 12+20=32 to our sum.  That gives a running total of 92.
copeland 2012-02-23 20:41:42
No other roots of unity will work because the decreasing coefficients constraint forces our roots of unity equations to have consecutive powers of 4x^n.
copeland 2012-02-23 20:41:49
At this point I strongly suggest thinking a little more about how to make other polynomials, and in particular polynomials that will get us from a sum of 92 to one of the other values that are possible answers.  Then bubble in 92 and go on.
copeland 2012-02-23 20:42:11
However we can do a little better than that.  I'll run through a nice way to deal with this next, but a little more quickly than we've been going so far.
ahaanomegas 2012-02-23 20:42:51
(B)
coldsummer 2012-02-23 20:42:51
so the answer is B?
superwatchdog 2012-02-23 20:42:51
B
copeland 2012-02-23 20:42:54
We don't KNOW that yet.   At this point I hope you picked B and didn't spend any more time on this problem.  We haven't shown anything except that the answer is at least 92.
copeland 2012-02-23 20:43:11
Notice that all of the approaches we took work with roots of unity.
copeland 2012-02-23 20:43:13
copeland 2012-02-23 20:43:24
There's slightly better motivation for this.
copeland 2012-02-23 20:43:28
copeland 2012-02-23 20:43:37
A generating function approach allows us to combine all this data.
copeland 2012-02-23 20:43:45
copeland 2012-02-23 20:43:55
The coefficients of the bottom equation are always at least as large as the coefficients of the top equation, except for the leading 4z^5.
copeland 2012-02-23 20:44:29
copeland 2012-02-23 20:44:35
Now notice that if P(z)=0 then this new function is also 0.  However this new function also has a zero at z=1.  Since the coefficients of P are nonnegative, P(1) is at least 4, so 1 cannot be a root of P.
copeland 2012-02-23 20:44:49
Therefore we can instead find polynomials of the form (z-1)P(z) for which there is a root of modulus 1 other than z=1.
copeland 2012-02-23 20:44:53
What can we say about each of the coefficients of (z-1)P(z)?
fractals 2012-02-23 20:45:49
Negative except for 4z^5
.cpp 2012-02-23 20:45:49
All negative except for the 4x^5 term.
Iggy Iguana 2012-02-23 20:45:49
nonpositive
ksun48 2012-02-23 20:45:49
nonpositive, except for leading term
number.sense 2012-02-23 20:45:49
sign of constant term is reversed
mdecross 2012-02-23 20:45:49
They are <= 0, except the leading coefficient
mentalgenius 2012-02-23 20:45:49
it is negative
copeland 2012-02-23 20:45:53
copeland 2012-02-23 20:45:56
What else can we say?
copeland 2012-02-23 20:46:22
We have the numbers 4, (a-4), (b-a), (c-b), (d-c), and -d.
Goos 2012-02-23 20:47:25
sum of all coefficients is 0
Iggy Iguana 2012-02-23 20:47:25
sum is 0
sammyMaX 2012-02-23 20:47:25
They add to zero
.cpp 2012-02-23 20:47:25
Sum = 0.
mcdonalds106_7 2012-02-23 20:47:25
they sum to 0
coldsummer 2012-02-23 20:47:25
the sum is 0
superwatchdog 2012-02-23 20:47:25
the sum of the coefficients is zero
mentalgenius 2012-02-23 20:47:25
Adding them gives zero?
copeland 2012-02-23 20:47:30
copeland 2012-02-23 20:47:34
copeland 2012-02-23 20:47:37
The leading coefficient is 4 (which is big) and all the remaining coefficients add to -4.
copeland 2012-02-23 20:47:45
If z is of modulus 1 then all of the terms below the leading term have to work together to cancel with the leading 4 and make 0.
copeland 2012-02-23 20:47:50
copeland 2012-02-23 20:48:06
Now when comparing the sizes of elements, the best thing to try first is the triangle inequality.
copeland 2012-02-23 20:48:13
copeland 2012-02-23 20:48:39
That thing on the right is equal to 4, though, so we have equality.
copeland 2012-02-23 20:48:42
However when is the triangle inequality an actual equality?
Seedleaf 2012-02-23 20:49:37
when its in a straight line
fractals 2012-02-23 20:49:37
Everything is on a line through the origin
Yongyi781 2012-02-23 20:49:37
When the complex numbers are real multiples of each other
timoteomo3 2012-02-23 20:49:37
when its a line!
distortedwalrus 2012-02-23 20:49:37
when the points are collinear
number.sense 2012-02-23 20:49:37
when they all form a straight line
mentalgenius 2012-02-23 20:49:37
when there is a straight line
MathForFun 2012-02-23 20:49:37
When its a line... or degenerate
copeland 2012-02-23 20:49:42
copeland 2012-02-23 20:49:55
copeland 2012-02-23 20:50:15
Continuing down this road tells us that we have to have z be SOME root of unity or have all of the coefficients equal to zero.  However if all the coefficients are zero that contradicts that the sum of the coefficients is 4.  So the only possible values for z are roots of unity and the answer we gave (with a little more work to rule out polynomials like 4x^4+4) does count all of the possible sums.
Yongyi781 2012-02-23 20:51:03
4x^4 + 4 doesn't have nonincreasing coefficients
copeland 2012-02-23 20:51:15
Exactly.
copeland 2012-02-23 20:51:27
That's not precluded from the triangle inequality argument we gave, though.
copeland 2012-02-23 20:51:41
Polynomials like that will pop out and you have to rule them out by hand.
pr0likethis 2012-02-23 20:51:50
Is it worth noting that 1, -1, i, and -i are 4th roots of unity, so all of our solutions were roots of unity up to 5?
number.sense 2012-02-23 20:51:50
wait so the answer is B?
copeland 2012-02-23 20:51:56
Yes!
copeland 2012-02-23 20:52:02
Now.  I have no idea how you were expected to do any of that last bit on the exam.
copeland 2012-02-23 20:52:42
I haven't seen or found any argument that is even reasonably close to AMC12 level to get from what we did at the beginning to an honest proof that 92 is correct.
matemyday1234567890 2012-02-23 20:52:53
i would just guess b and move on instead of doing all the math after you said to bubble b
copeland 2012-02-23 20:52:56
Good!
copeland 2012-02-23 20:52:59
Always take my advice.
copeland 2012-02-23 20:53:02
Except when it's bad.
copeland 2012-02-23 20:53:06
Now it's time to bubble and move on, though.
copeland 2012-02-23 20:53:12
copeland 2012-02-23 20:53:27
Where should we start?
.cpp 2012-02-23 20:54:15
Try some small values first.
nikeballa96 2012-02-23 20:54:15
try some examples...
copeland 2012-02-23 20:54:18
It couldn't hurt to draw a table of the first few, just to see what's happening.  Here's my table.
copeland 2012-02-23 20:54:22
copeland 2012-02-23 20:54:43
If you're having trouble with a concept like "unbounded" when you hit a problem like this, if you have time play with it for a while and maybe the definition will clear itself up.
copeland 2012-02-23 20:54:50
That will happen for us now.
copeland 2012-02-23 20:54:54
copeland 2012-02-23 20:54:58
Hmm.  It looks like we are getting a lot of ones.  Let's make a wider table and go down to f_2.
copeland 2012-02-23 20:55:03
copeland 2012-02-23 20:55:24
Huh.  That's easy enough to extend since we already have f for these outputs.
copeland 2012-02-23 20:55:28
copeland 2012-02-23 20:55:30
We care about the vertical sequences.  Are any of these sequences unbounded?
.cpp 2012-02-23 20:55:52
16-27-16-27-... loops but is not unbounded.
ksun48 2012-02-23 20:55:52
nope!
canadian 2012-02-23 20:55:52
no
2718euler 2012-02-23 20:55:52
No
fractals 2012-02-23 20:55:52
None at all.
Iggy Iguana 2012-02-23 20:55:52
no
number.sense 2012-02-23 20:55:52
not yet...
Goos 2012-02-23 20:55:52
nope
copeland 2012-02-23 20:55:55
No.  All of them go to 1 except 16 which oscillates between 16 and 27.
copeland 2012-02-23 20:56:07
"Unbounded" means that the sequence gets bigger and bigger and bigger.
copeland 2012-02-23 20:56:22
16-27-16... never gets bigger than 27.
copeland 2012-02-23 20:56:28
Why do the others disappear?
matemyday1234567890 2012-02-23 20:57:12
it changes from one only when the number is a multiple of a square number
KeepingItReal 2012-02-23 20:57:12
no squares/cubes/etc
canadian 2012-02-23 20:57:12
All the e's eventually become 0
.cpp 2012-02-23 20:57:12
They slowly lose their exponential powers and become ordinary primes and then 1.
copeland 2012-02-23 20:57:17
copeland 2012-02-23 20:57:19
Let's get a couple more examples on the table before we can think about this.  What is the next sequence you're itching to try?
.cpp 2012-02-23 20:58:07
f(32)
2718euler 2012-02-23 20:58:07
n=32
iamnew 2012-02-23 20:58:07
32
copeland 2012-02-23 20:58:13
The next sequence I wanted to look at was the one that starts with N=32.  What does that sequence look like?
MathForFun 2012-02-23 20:59:46
81
pr0likethis 2012-02-23 20:59:46
3^4=81
lucylai 2012-02-23 20:59:46
32, 81, 64, 243, ...
.cpp 2012-02-23 20:59:46
32 => 3^4 => 4^3 = 2^6 => 3^5 => 4^4 = 2^8 ...
PChang 2012-02-23 20:59:46
3^4,4^3,3^5,... soon
fractals 2012-02-23 20:59:46
2^5  3^4  4^3=2^6  3^5  4^4=2^8, ..., and it seems unbounded
Goos 2012-02-23 20:59:46
81, 64, 3^5, 2^8,... unbounded!
number.sense 2012-02-23 20:59:46
32, 81, 64, 243...
copeland 2012-02-23 20:59:50
copeland 2012-02-23 20:59:55
It looks like these are unbounded.  Therefore 32 is our first case of an unbounded sequence, hopefully.  Maybe it'll shed some light to prove that.
copeland 2012-02-23 20:59:59
2718euler 2012-02-23 21:00:46
4^(n-2)
.cpp 2012-02-23 21:00:46
4^(n-2)
lucylai 2012-02-23 21:00:46
4^(n-2)
nackster12 2012-02-23 21:00:46
2^(2n-4)
PChang 2012-02-23 21:00:46
2^(n-2)sqrd
pgmath 2012-02-23 21:00:46
2^(2n-4)
infinity1 2012-02-23 21:00:46
2^2n-4
copeland 2012-02-23 21:00:50
copeland 2012-02-23 21:00:52
pr0likethis 2012-02-23 21:01:27
n>4
joshxiong 2012-02-23 21:01:27
n>4
infinity1 2012-02-23 21:01:27
n>4
nikeballa96 2012-02-23 21:01:27
n<2n-4 so n>4
pr0likethis 2012-02-23 21:01:27
so n can be anything from 5 to 8
number.sense 2012-02-23 21:01:27
when 2n>4
.cpp 2012-02-23 21:01:27
2n-4 > n => n>4,
Goos 2012-02-23 21:01:27
2n-4>n, so n>4
tan90 2012-02-23 21:01:27
n>=5
pgmath 2012-02-23 21:01:27
n>4
copeland 2012-02-23 21:01:31
If n>4 then 2n-4 > n.  Therefore this sequence is unbounded whenever we start with a power of 2 that is at least 2^5=32.
copeland 2012-02-23 21:01:34
Do we get any other unbounded sequences from knowing that 32 starts an unbounded sequence?
pr0likethis 2012-02-23 21:02:33
any multiple of 32!
ewcikewqikd 2012-02-23 21:02:33
all multiple of 32
PChang 2012-02-23 21:02:33
and multiples of those
joshxiong 2012-02-23 21:02:33
All other numbers in that sequnce and their multiples.
canadian 2012-02-23 21:02:33
Any multiple of 32 also starts an unbounded sequence.
copeland 2012-02-23 21:02:36
It turns out that if a|b then f(a)|f(b).
copeland 2012-02-23 21:02:40
copeland 2012-02-23 21:02:51
If all the exponents in a are at least as large as all the exponents in b then all the exponents in f(a) will be at least as large as the exponents in f(b).
copeland 2012-02-23 21:02:54
In particular, all the multiples of 32 will start unbounded sequences. How many multiples of 32 are there between 1 and 400?
super_pi314 2012-02-23 21:03:26
12
Yongyi781 2012-02-23 21:03:26
12
aerrowfinn72 2012-02-23 21:03:28
12
Iggy Iguana 2012-02-23 21:03:28
12
coldsummer 2012-02-23 21:03:28
12
pattycakechichi 2012-02-23 21:03:28
12
soccerfan 2012-02-23 21:03:28
12
matemyday1234567890 2012-02-23 21:03:28
12
RotomPlasma 2012-02-23 21:03:28
12
nackster12 2012-02-23 21:03:28
12
copeland 2012-02-23 21:03:30
copeland 2012-02-23 21:03:35
What else does the 32 sequence give us?
copeland 2012-02-23 21:03:48
Beyond multiples of 32 that is.
christerson 2012-02-23 21:04:30
81
Seedleaf 2012-02-23 21:04:30
all multiples of 3^4
ewcikewqikd 2012-02-23 21:04:30
3^4 = 81
.cpp 2012-02-23 21:04:30
3^4 and any multiple of it.
Iggy Iguana 2012-02-23 21:04:30
multiples of 81
aopsvd 2012-02-23 21:04:30
Multiples of 3^4.
munygrubber 2012-02-23 21:04:30
81 works too
copeland 2012-02-23 21:04:33
Since 3^4=81 is in this sequence, 81 and all of its multiples start unbounded sequences.
copeland 2012-02-23 21:04:37
There are 4 multiples of 81 below 400: 81, 162, 243, and 324.
copeland 2012-02-23 21:04:40
Therefore we're up to 16 good values for N.
copeland 2012-02-23 21:04:43
Powers of 2 and powers of 3 were nice.  Let's try powers of 5.
copeland 2012-02-23 21:04:47
We know that the N=5 sequence goes 5, 1, 1, 1,. . . .
copeland 2012-02-23 21:04:50
What about the N=25 sequence?
Iggy Iguana 2012-02-23 21:05:41
25, 6, 1
number.sense 2012-02-23 21:05:41
25, 6, 1,1,1,1
pgmath 2012-02-23 21:05:41
6, 1, 1, ...
va2010 2012-02-23 21:05:41
25,6,loses
Seedleaf 2012-02-23 21:05:41
25, 6, 1, 1...
mcdonalds106_7 2012-02-23 21:05:41
25,6,1,1,1,1,...
KeepingItReal 2012-02-23 21:05:41
5^2, 6,1,1,1...
Goos 2012-02-23 21:05:41
5^2, 6^1, 1,1,1,1
nackster12 2012-02-23 21:05:41
25,6,1
fractals 2012-02-23 21:05:41
5^2   6^1=2*3  1, so none
copeland 2012-02-23 21:05:45
copeland 2012-02-23 21:05:48
That's no good.  What about N=125?
Iggy Iguana 2012-02-23 21:06:51
125, 36, 12, 3, 1
va2010 2012-02-23 21:06:51
125, 36, 12, loses
mcdonalds106_7 2012-02-23 21:06:51
125,36,12,3,1,1,1,1,...
number.sense 2012-02-23 21:06:51
125, 36, 12, ... bounded
pgmath 2012-02-23 21:06:51
36, 12, 3, 1, 1, ...
nackster12 2012-02-23 21:06:51
125,36,3,1
super_pi314 2012-02-23 21:06:51
125,36,12,3,1,1,1...
soccerfan 2012-02-23 21:06:53
125, 36, 12, 3, 1...
copeland 2012-02-23 21:06:59
copeland 2012-02-23 21:07:02
OK, so powers of 5 are not helpful.  What about powers of 7?  We know 7 is no good.  What is the sequence for N=49?
Seedleaf 2012-02-23 21:07:57
49, 8, 9, 4, 1, ...
MathForFun 2012-02-23 21:07:57
8, 9, 4, 3
Iggy Iguana 2012-02-23 21:07:57
49, 8, 9, 4, 3, 1
bookie331 2012-02-23 21:07:57
49, 8, 9, 4, 3, 1, 1
copeland 2012-02-23 21:08:05
copeland 2012-02-23 21:08:08
However what is promising about 7?
Seedleaf 2012-02-23 21:08:39
it gives powers of 2. try 343?
Yongyi781 2012-02-23 21:08:39
7+1 = 8 which is 2^3
.cpp 2012-02-23 21:08:39
Powers of 2!
Goos 2012-02-23 21:08:39
it goes to 8 which is a power of 2!  343 works.
number.sense 2012-02-23 21:08:39
it gives us 8, which is 2^3 :D which gives us large powers of 2
nackster12 2012-02-23 21:08:39
you get powers of 2
copeland 2012-02-23 21:08:46
7 is promising because it gives a great big power of 2.
copeland 2012-02-23 21:08:48
copeland 2012-02-23 21:09:02
There is only one multiple of this number that is less than 400.
copeland 2012-02-23 21:09:06
Our count of Ns that work is now up to 17.  That's up there on the list, but it might be a distractor.  After all we've only thought about prime powers so far.  Now we should think about products of prime powers.
copeland 2012-02-23 21:09:26
number.sense 2012-02-23 21:09:41
1
pgmath 2012-02-23 21:09:41
1
Yongyi781 2012-02-23 21:09:41
1
fractals 2012-02-23 21:09:41
1
number.sense 2012-02-23 21:09:41
1
alex31415 2012-02-23 21:09:41
1?
lucylai 2012-02-23 21:09:41
1
copeland 2012-02-23 21:09:47
copeland 2012-02-23 21:09:52
So what should we really care about?
MathForFun 2012-02-23 21:10:30
The exponents
Seedleaf 2012-02-23 21:10:30
the bigger powers
pr0likethis 2012-02-23 21:10:30
squares of primes
pgmath 2012-02-23 21:10:30
Products of squares
aopsvd 2012-02-23 21:10:30
multiples of squares by factors of those squares.
ksun48 2012-02-23 21:10:30
at least squares
va2010 2012-02-23 21:10:30
products of squares
.cpp 2012-02-23 21:10:30
We need squares!
timoteomo3 2012-02-23 21:10:30
the largest exponent!
copeland 2012-02-23 21:10:40
The factors that are going to matter in the end are the primes with exponent greater than 1.  If N is not divisible by 3 then f_1(3N)=f_1(N), so the sequence for 3N is the same as the sequence for N. Therefore we should narrow our search to numbers that have exponents that are at least 2.
copeland 2012-02-23 21:10:49
va2010 2012-02-23 21:11:49
2
.cpp 2012-02-23 21:11:49
Can't be too big.
zinko1991 2012-02-23 21:11:49
2?
Yongyi781 2012-02-23 21:11:49
2
copeland 2012-02-23 21:11:54
copeland 2012-02-23 21:11:59
Let's start with 2 and 3.  We know that any multiple of 2^5 or 3^4 will give an unbounded sequence.  Therefore we ought to think about the number 2^43^3.  If this number's sequence is bounded then any factor of this number will have a bounded sequence as well.
copeland 2012-02-23 21:12:07
Iggy Iguana 2012-02-23 21:13:10
2^4*3^3
PChang 2012-02-23 21:13:10
same thing!!!
MathForFun 2012-02-23 21:13:10
432
KeepingItReal 2012-02-23 21:13:10
3^3 * 4^2
Seedleaf 2012-02-23 21:13:10
2^4*3^3
Yongyi781 2012-02-23 21:13:10
3^3 4^2 = 2^4 3^3
bookie331 2012-02-23 21:13:10
432
mz94 2012-02-23 21:13:10
itself; it cycles so its bounded
aopsvd 2012-02-23 21:13:10
3^3 2^4
Goos 2012-02-23 21:13:10
3^3*2^4
copeland 2012-02-23 21:13:16
copeland 2012-02-23 21:13:27
Some people are asking why we tested this one.
copeland 2012-02-23 21:13:51
Any number that's a factor of this one cannot possibly generate an unbounded sequence.
copeland 2012-02-23 21:14:03
We now that if a number generates an unbounded sequence then any multiple of it will as well.
copeland 2012-02-23 21:14:23
So by testing this guy we've simultaneously tested all other numbers of the form 3^a2^b.
copeland 2012-02-23 21:14:34
OK, fine.  The problem here, though, was that applying f_1 takes powers of 3 to powers of 2 and powers of 2 to powers of 3.  They never had a chance to combine and make a bigger number.
copeland 2012-02-23 21:14:43
Let's try 2 and 5 next.  Since 2^25^2=100, we can't use any more fives in our number.  Therefore we should think about numbers of the form 2^e5^2.
copeland 2012-02-23 21:14:48
number.sense 2012-02-23 21:15:33
3^e * 2
ksun48 2012-02-23 21:15:33
3^e*2
pgmath 2012-02-23 21:15:33
3^e * 2
.cpp 2012-02-23 21:15:33
3^(e)*2
pr0likethis 2012-02-23 21:15:33
6*3^e-1 = 2*3^e
nackster12 2012-02-23 21:15:33
2*3^e
copeland 2012-02-23 21:15:39
copeland 2012-02-23 21:15:48
We can apply f again to get
copeland 2012-02-23 21:15:48
copeland 2012-02-23 21:15:52
For what e is sequence unbounded?
pr0likethis 2012-02-23 21:16:32
so e=4 makes N=400, which works :D
ewcikewqikd 2012-02-23 21:16:32
4
alex31415 2012-02-23 21:16:32
4
MathForFun 2012-02-23 21:16:32
e>3/2
ksun48 2012-02-23 21:16:34
e >= 4
pgmath 2012-02-23 21:16:34
e = 4
fractals 2012-02-23 21:16:34
e >= 4
ksun48 2012-02-23 21:16:34
so 400
copeland 2012-02-23 21:16:43
copeland 2012-02-23 21:16:47
Check it out!  400=2^4*5^2.  It's the end of the range.  Therefore we've found another valid N.  That bumps the count up to 18.
copeland 2012-02-23 21:17:14
That's also annoying because there might be one more solution lurking around. . . .
copeland 2012-02-23 21:17:19
copeland 2012-02-23 21:17:30
copeland 2012-02-23 21:17:37
copeland 2012-02-23 21:17:54
copeland 2012-02-23 21:17:58
So what is the final answer?
ahaanomegas 2012-02-23 21:18:45
(D)
RotomPlasma 2012-02-23 21:18:45
so (D)!
Iggy Iguana 2012-02-23 21:18:45
D)18
MathForFun 2012-02-23 21:18:45
D
coldsummer 2012-02-23 21:18:45
D
pgmath 2012-02-23 21:18:45
(D) 18
Seedleaf 2012-02-23 21:18:45
D: 18
ksun48 2012-02-23 21:18:45
D18
super_pi314 2012-02-23 21:18:45
number.sense 2012-02-23 21:18:45
D
MathForFun 2012-02-23 21:18:45
D?
pr0likethis 2012-02-23 21:18:45
18 (D). 400 was very difficult to find though....
va2010 2012-02-23 21:18:45
18 /D
RunpengFAILS 2012-02-23 21:18:45
Eighteen
pattycakechichi 2012-02-23 21:18:45
18
alex31415 2012-02-23 21:18:45
18?
pattycakechichi 2012-02-23 21:18:45
D
fractals 2012-02-23 21:18:45
18, D
copeland 2012-02-23 21:18:47
We've found 18 total values of N that satisfy the problem so the answer is (D).
copeland 2012-02-23 21:19:07
Good work!
copeland 2012-02-23 21:19:09
We are awesome.
Iggy Iguana 2012-02-23 21:19:13
that took a long time
number.sense 2012-02-23 21:19:13
... too long...
copeland 2012-02-23 21:19:21
Yeah, but that problem was like the 25 on this test.
copeland 2012-02-23 21:19:28
And we'll call the 25 a 23. And we'll call 23 a 30.
matemyday1234567890 2012-02-23 21:19:39
how do you solve that quicker
copeland 2012-02-23 21:19:43
You can't, really.
copeland 2012-02-23 21:19:47
You can guess better than we did.
copeland 2012-02-23 21:19:58
Thinking about parity helps.
copeland 2012-02-23 21:20:43
Realizing that 7 and 3 are like anti-twos and that 5 is both a 2 and an anti-two put together.
copeland 2012-02-23 21:20:46
That kind of thing.
copeland 2012-02-23 21:20:50
But it only gets you so far.
copeland 2012-02-23 21:20:54
copeland 2012-02-23 21:21:08
And the problem came with a nice picture of S so that you don't actually have to figure out what all that notation means:
copeland 2012-02-23 21:21:17
copeland 2012-02-23 21:21:22
First off, what is the value of f(t) for this right triangle?
copeland 2012-02-23 21:21:31
fractals 2012-02-23 21:21:58
3/4
Yongyi781 2012-02-23 21:21:58
3/4
alex31415 2012-02-23 21:21:58
3/4
KeepingItReal 2012-02-23 21:21:58
3/4
RotomPlasma 2012-02-23 21:21:58
3/4
copeland 2012-02-23 21:22:04
copeland 2012-02-23 21:22:08
And the value of the tangent at angle B is 3/4.
copeland 2012-02-23 21:22:10
What is the value of f for this triangle?
copeland 2012-02-23 21:22:14
ksun48 2012-02-23 21:22:35
4/3
fractals 2012-02-23 21:22:35
4/3
KeepingItReal 2012-02-23 21:22:35
4/3
Iggy Iguana 2012-02-23 21:22:35
4/3
lucylai 2012-02-23 21:22:35
3/4
joshxiong 2012-02-23 21:22:35
4/3
noedne 2012-02-23 21:22:35
4/3
nackster12 2012-02-23 21:22:35
4/3
pgmath 2012-02-23 21:22:35
4/3
copeland 2012-02-23 21:22:41
Again we label:
copeland 2012-02-23 21:22:44
copeland 2012-02-23 21:22:47
And the tangent of B is 4/3.
copeland 2012-02-23 21:22:50
These are reflections of one another and the values of f are reciprocals.
copeland 2012-02-23 21:22:55
That's nice.  There's some symmetry in the problem, but there's not really any symmetry in the question because we're missing the point (0,0).
copeland 2012-02-23 21:23:00
Does this suggest any approach?
number.sense 2012-02-23 21:23:55
add the point 0,0
Yongyi781 2012-02-23 21:23:55
Pretend (0,0) is there and then account for the triangles that contain (0,0)
canadian 2012-02-23 21:23:55
exclusion
fractals 2012-02-23 21:23:55
Pretend (0,0) is in the set and the divide for the overcounting
Goos 2012-02-23 21:23:55
look at triangles whose counterparts (reflections) go through (0,0).
alex31415 2012-02-23 21:23:55
try including (0,0) and subtract the cases with (0,0)
copeland 2012-02-23 21:23:59
Let's solve a simpler problem again.  Throw the point (0,0) into the mix.
copeland 2012-02-23 21:24:02
copeland 2012-02-23 21:24:10
copeland 2012-02-23 21:24:13
How can we use the symmetry that we've found?
KeepingItReal 2012-02-23 21:24:52
each triangle has a conjugate, so tangents multiply to 1
fractals 2012-02-23 21:24:55
Pair up triangles by a reflection over x = 2
Yongyi781 2012-02-23 21:24:55
Can we pair up every triangle with its "conjugate"?
pgmath 2012-02-23 21:24:55
Every triangle has a reflection that it cancels out with.
ewcikewqikd 2012-02-23 21:24:55
1 because any right triangle reflect over the middle vertical line cancels itself
number.sense 2012-02-23 21:24:55
every triangle has its reflection over the line through the middle of the grid
newchie123 2012-02-23 21:24:55
since they cancel repricocals
copeland 2012-02-23 21:25:00
If we reflect our picture across the vertical midline, every right triangle is paired with another right triangle with a value of f that is inverse to the value of the original triangle.
copeland 2012-02-23 21:25:03
copeland 2012-02-23 21:25:05
Except not quite.  Which triangles have no partners?
nackster12 2012-02-23 21:25:59
the ones that are symmetrical about the center line
number.sense 2012-02-23 21:25:59
isosceles right triangles with right angle perfectly centered around our axis of symmetry
Yongyi781 2012-02-23 21:25:59
The ones with the midline as an axis of symmetry
Goos 2012-02-23 21:25:59
triangles symmetrical about the midline.  This means they have a right angle along the midline and two points on either side.
pgmath 2012-02-23 21:26:02
The isosceles ones with a vertex on the middle column
copeland 2012-02-23 21:26:06
There are some triangles that are fixed by this reflection:
copeland 2012-02-23 21:26:10
copeland 2012-02-23 21:26:12
However what do we know about f applied to these triangles?
Yongyi781 2012-02-23 21:26:33
But isosceles -> tan = 1 so that doesn't matter
ewcikewqikd 2012-02-23 21:26:33
equals 1
c0mbusti0n1295 2012-02-23 21:26:33
always 1
fractals 2012-02-23 21:26:33
But those are 45-45-90, which have tangent 1, so they don't matter
noedne 2012-02-23 21:26:33
it equals 1
joshxiong 2012-02-23 21:26:33
It is 1.
DF2222 2012-02-23 21:26:33
it's 1
Iggy Iguana 2012-02-23 21:26:33
its 1!
number.sense 2012-02-23 21:26:33
it already is 1
bookie331 2012-02-23 21:26:33
1
pr0likethis 2012-02-23 21:26:33
f is 1, so it doesnt need canceling
distortedwalrus 2012-02-23 21:26:33
it's 1
copeland 2012-02-23 21:26:36
Any triangle fixed by this reflection is isosceles and the value of f on an isosceles triangle must be 1.  (Alternatively, since these triangles are fixed by a reflection but f(t) is the reciprocal of the value of f on the reflection of t, we have to have f(t)=1 for these triangles.)
copeland 2012-02-23 21:26:40
Therefore what is the value of the big product we seek on this new S where we include the origin?
lucylai 2012-02-23 21:26:57
it's just 1
Seedleaf 2012-02-23 21:26:57
1
number.sense 2012-02-23 21:26:57
1
KeepingItReal 2012-02-23 21:26:57
1
noedne 2012-02-23 21:26:57
1
nackster12 2012-02-23 21:26:57
1
RunpengFAILS 2012-02-23 21:26:57
One.
mcdonalds106_7 2012-02-23 21:26:57
1
copeland 2012-02-23 21:27:03
The value is 1.  The symmetric triangles don't contribute to the product and the other triangles pair off into reciprocal factors.  Everything cancels leaving only the value 1.
copeland 2012-02-23 21:27:06
How can we use this to simplify the original problem?
nackster12 2012-02-23 21:27:34
find all the triangles that have a vertex at the origin
narto928 2012-02-23 21:27:34
just account for the ones that contain (0,0)
number.sense 2012-02-23 21:27:34
how many triangles now use the origin is the only question left to answer
pr0likethis 2012-02-23 21:27:34
our answer is the inverse of all f(t) with t including the origin
varunrocks 2012-02-23 21:27:34
look at all the triangles that pass through the origin and just take the reciprocal
copeland 2012-02-23 21:27:38
Since we know the product for all of the triangles is 1, and we want the product for all triangles that don't use the origin, we can compute the product for the triangles that do use the origin and take its reciprocal.
copeland 2012-02-23 21:27:43
(At this point when solving, I looked to see if any of the choices were reciprocals of one another to see if I had to pay attention to that.  PHEW!  None of them are.)
copeland 2012-02-23 21:27:53
Now we've reduced the problem to finding the value of the product for all triangles that use the origin (0,0).  There are still a TON of them.  Is there another reflection symmetry that might help?
copeland 2012-02-23 21:28:30
Notice here that we want a symmetry that permutes the set of triangles containing the origin as a vertex. Therefore we want a reflection that fixes the origin and mostly preserves the set of triangles.  We can't find a symmetry that does both, but preserving the origin is a LOT more important.
dinoboy 2012-02-23 21:28:54
x=y
KeepingItReal 2012-02-23 21:28:54
reflect the triangles about y=x?
dinoboy 2012-02-23 21:28:54
*x=y, except when some vertex on the top is used
canadian 2012-02-23 21:28:54
Along the diagonal
pr0likethis 2012-02-23 21:28:54
reflecting over the "line" y=x creates canceling triangles as well
Goos 2012-02-23 21:28:54
reflection about the diagonal (y=x, if it were the first quadrant.)
copeland 2012-02-23 21:28:58
Let's consider the reflection through this line:
copeland 2012-02-23 21:29:04
copeland 2012-02-23 21:29:05
What triangles are paired by this reflection?
fractals 2012-02-23 21:29:55
All but those with a vertex in the top row
mcdonalds106_7 2012-02-23 21:29:55
ones that don't have a vertex on y=5
Goos 2012-02-23 21:29:55
Anything not including the top five points.
copeland 2012-02-23 21:30:03
Any triangle that doesn't have a vertex along the top row will have a partner triangle under this reflection.
copeland 2012-02-23 21:30:12
So?
narto928 2012-02-23 21:31:02
we find the ones that use the top row
nackster12 2012-02-23 21:31:02
all triangles with a vertex at the origin and in the top row
number.sense 2012-02-23 21:31:02
we merely have to divide out those that use the top row?
KeepingItReal 2012-02-23 21:31:02
find ones with vertex on top row
lucylai 2012-02-23 21:31:02
their tangents multiply to 1
fractals 2012-02-23 21:31:02
Now we count those with one vertex on the origin and at least one in the top row.
Goos 2012-02-23 21:31:02
We know that 1 point is (0,0) and the other is on the top of the grid.  It's pretty simple from there.
copeland 2012-02-23 21:31:12
Those triangles also cancel in our product.
copeland 2012-02-23 21:31:16
That leaves only triangles that use the origin and use at least one of the topmost vertices.  Now we're in business.
copeland 2012-02-23 21:31:19
We seem to have 4 types of such triangles.  We have triangles with legs along the x- and y- axis:
copeland 2012-02-23 21:31:24
copeland 2012-02-23 21:31:31
We also have triangles with legs along the left side and along the top:
copeland 2012-02-23 21:31:34
copeland 2012-02-23 21:31:40
We also have triangles with legs parallel to the axes but right angle elsewhere in the grid:
copeland 2012-02-23 21:31:45
copeland 2012-02-23 21:31:50
Finally we have triangles that are just kinda wonky:
copeland 2012-02-23 21:31:56
copeland 2012-02-23 21:31:58
(Those purple guys make me nervous.)
copeland 2012-02-23 21:32:04
Is there any more symmetry that we can take advantage of?
KeepingItReal 2012-02-23 21:33:04
for the first three cases, yes. those *should* cancel i think?
pr0likethis 2012-02-23 21:33:04
each case that isnt purple has inverses
fractals 2012-02-23 21:33:04
Reds and blues are reflections of each other and therefore cancel out (to 1).
copeland 2012-02-23 21:33:06
Do those all just cancel and go away?
shengo8 2012-02-23 21:33:28
no
zinko1991 2012-02-23 21:33:28
no
copeland 2012-02-23 21:33:44
But some of them cancel, right?
sammyMaX 2012-02-23 21:34:02
No, two of them cancel out, leaving one behind along with the purple ones
pr0likethis 2012-02-23 21:34:02
oh, there are three of them. one pair cancels, but one remains
copeland 2012-02-23 21:34:06
Actually any 2 of the first three sets can be paired off by some reflection.  Let's get rid of the first two (because slopes in the third case look easy to compute).  The first two are interchanged by the reflection through the horizontal midline so they cancel.
copeland 2012-02-23 21:34:12
Now we are left computing the product for triangles of the form
copeland 2012-02-23 21:34:19
copeland 2012-02-23 21:34:21
and
copeland 2012-02-23 21:34:25
copeland 2012-02-23 21:34:28
What is f applied to the blue triangle?
copeland 2012-02-23 21:34:36
fractals 2012-02-23 21:35:04
3/5
KeepingItReal 2012-02-23 21:35:04
3/5
pr0likethis 2012-02-23 21:35:04
n/5 where n is 1, 2, 3, or 4
MathForFun 2012-02-23 21:35:04
3/5
Iggy Iguana 2012-02-23 21:35:04
3/5
dsj9999 2012-02-23 21:35:10
3/5
copeland 2012-02-23 21:35:11
We label the vertices counterclockwise from the right angle and then we compute the ratio of the opposite side to the adjacent side.  The adjacent side is 5 and the opposite side is 3.  The blue triangle has f(t)=3/5.
copeland 2012-02-23 21:35:17
copeland 2012-02-23 21:35:24
Super.  We've discovered one of the distractors!  However we need to deal with the wonky triangles, too.  The wonky triangles have right angle somewhere in the middle of the diagram.  What relationship is there between the legs of a wonky triangle that will let us locate the point with the right angle?
copeland 2012-02-23 21:35:45
sammyMaX 2012-02-23 21:36:18
The slopes of the legs are negative reciprocals
nackster12 2012-02-23 21:36:18
opposite reciprocoal slopes
c0mbusti0n1295 2012-02-23 21:36:18
slopes are negative reciprocals?
copeland 2012-02-23 21:36:26
The slopes of the legs are negative reciprocals!
copeland 2012-02-23 21:36:30
How many wonky triangles are there with slopes +1 and -1?
mcdonalds106_7 2012-02-23 21:37:26
2
fractals 2012-02-23 21:37:26
2
Iggy Iguana 2012-02-23 21:37:26
2.
KeepingItReal 2012-02-23 21:37:26
2
billybob42 2012-02-23 21:37:26
2
canadian 2012-02-23 21:37:26
2
copeland 2012-02-23 21:37:29
The pink triangle above is one of them.  Here's the other:
copeland 2012-02-23 21:37:34
copeland 2012-02-23 21:37:35
What is t for this second triangle?
copeland 2012-02-23 21:37:40
f
copeland 2012-02-23 21:37:46
What is f for this second triangle?
sammyMaX 2012-02-23 21:37:58
4
Iggy Iguana 2012-02-23 21:37:58
4.
KeepingItReal 2012-02-23 21:37:58
4
RunpengFAILS 2012-02-23 21:37:58
Four.
joshxiong 2012-02-23 21:37:58
4
number.sense 2012-02-23 21:37:58
4
alex31415 2012-02-23 21:37:58
4?
copeland 2012-02-23 21:38:02
copeland 2012-02-23 21:38:06
What is f for this triangle:
copeland 2012-02-23 21:38:10
alex31415 2012-02-23 21:38:44
3/2
joshxiong 2012-02-23 21:38:44
3/2
sammyMaX 2012-02-23 21:38:44
3/2
fractals 2012-02-23 21:38:44
3/2
bookfan 2012-02-23 21:38:44
3/2
dsj9999 2012-02-23 21:38:44
3/2
ABCDE 2012-02-23 21:38:44
3/2
RunpengFAILS 2012-02-23 21:38:44
Three halves
Yongyi781 2012-02-23 21:38:44
3/2
Seedleaf 2012-02-23 21:38:44
3/2
billybob42 2012-02-23 21:38:44
3/2
aopsvd 2012-02-23 21:38:44
3/2
MathForFun 2012-02-23 21:38:44
3/2
distortedwalrus 2012-02-23 21:38:44
3/2
KeepingItReal 2012-02-23 21:38:44
3/2
copeland 2012-02-23 21:38:49
copeland 2012-02-23 21:38:52
What are all the wonky triangles that we can make with slope 1/2 and 22 or with slope 2 and -1/2?
copeland 2012-02-23 21:39:11
What are all the wonky triangles that we can make with slope 1/2 and 2 or with slope 2 and -1/2?
copeland 2012-02-23 21:39:53
aargh!
copeland 2012-02-23 21:39:59
What are all the wonky triangles that we can make with slope 1/2 and -2 or with slope 2 and -1/2?
KeepingItReal 2012-02-23 21:40:04
2
Iggy Iguana 2012-02-23 21:40:04
just the sqrt5 by 2sqrt5 one
RotomPlasma 2012-02-23 21:40:04
2
KeepingItReal 2012-02-23 21:40:04
1 for each pair of slopes.
nackster12 2012-02-23 21:40:04
there are 2 of them
copeland 2012-02-23 21:40:08
Here are the only two:
copeland 2012-02-23 21:40:12
copeland 2012-02-23 21:40:18
Do we care?
Iggy Iguana 2012-02-23 21:40:41
but they multiply to 1
Seedleaf 2012-02-23 21:40:41
no, they cancel
vcez 2012-02-23 21:40:41
no they cancel
Yongyi781 2012-02-23 21:40:41
No! They're conjugates
fractals 2012-02-23 21:40:41
No, they multiply to 1
RotomPlasma 2012-02-23 21:40:41
No, symmetrical
c0mbusti0n1295 2012-02-23 21:40:41
they are reflections!
copeland 2012-02-23 21:40:45
No!  These two will cancel in the product.
copeland 2012-02-23 21:40:49
How about slopes of n and 1/n for n>2?
KeepingItReal 2012-02-23 21:42:10
oh the one with slope 1/4 and -4
nackster12 2012-02-23 21:42:10
theres one with slopes 1/4 and -4
canadian 2012-02-23 21:42:10
slope of 1/4 and -4
PChang 2012-02-23 21:42:10
4
Yongyi781 2012-02-23 21:42:10
Slopes 1/4 and -4*
fractals 2012-02-23 21:42:10
1/4 and -4 as slopes
copeland 2012-02-23 21:42:15
copeland 2012-02-23 21:42:17
What's the value of f for this guy?
ABCDE 2012-02-23 21:42:49
1
canadian 2012-02-23 21:42:49
1
Yongyi781 2012-02-23 21:42:49
1
sammyMaX 2012-02-23 21:42:49
1
PChang 2012-02-23 21:42:49
1
fishinsea 2012-02-23 21:42:49
1
joshxiong 2012-02-23 21:42:49
1.
nackster12 2012-02-23 21:42:49
1
Iggy Iguana 2012-02-23 21:42:49
1.
duketip10 2012-02-23 21:42:49
1
fractals 2012-02-23 21:42:49
1
aopsvd 2012-02-23 21:42:49
1, isosceles
bookfan 2012-02-23 21:42:49
1
copeland 2012-02-23 21:42:53
It's 1.  This doesn't contribute to the product either.
copeland 2012-02-23 21:42:56
So what's the answer?
aopsvd 2012-02-23 21:43:59
625/144
vcez 2012-02-23 21:43:59
B
Iggy Iguana 2012-02-23 21:43:59
(b)
tc1729 2012-02-23 21:43:59
B) 625/144
KeepingItReal 2012-02-23 21:43:59
625/144
fractals 2012-02-23 21:43:59
1/(144/625) = 625/144
fractals 2012-02-23 21:43:59
B
mdecross 2012-02-23 21:43:59
625/144, (B) (multiply the 6 into 24, take reciprocal)
alex31415 2012-02-23 21:43:59
625/144?
vcez 2012-02-23 21:43:59
$B$
numbertheorist17 2012-02-23 21:44:00
625/144
numbertheorist17 2012-02-23 21:44:00
(B) 625/144
copeland 2012-02-23 21:44:04
copeland 2012-02-23 21:44:07
We want the reciprocal of this, so the actual answer is 625/144 and that is (B).
copeland 2012-02-23 21:44:10
Super!
copeland 2012-02-23 21:44:21
Phew.
mcdonalds106_7 2012-02-23 21:45:09
what about slope 2/3?
copeland 2012-02-23 21:45:31
copeland 2012-02-23 21:45:41
Nice catch.
Iggy Iguana 2012-02-23 21:45:59
those don't contribute either
MathForFun 2012-02-23 21:45:59
1!
PChang 2012-02-23 21:45:59
1
freddylukai 2012-02-23 21:45:59
cancels with the other one
mcdonalds106_7 2012-02-23 21:45:59
its isosceles though
sammyMaX 2012-02-23 21:45:59
Tangent of that is one!
nackster12 2012-02-23 21:45:59
its still 1 though
sammyMaX 2012-02-23 21:45:59
(we were lucky)
fractals 2012-02-23 21:45:59
It is 1, though.
mdecross 2012-02-23 21:45:59
...but still 1.
copeland 2012-02-23 21:46:10
Good job! I totally missed that one.
copeland 2012-02-23 21:46:11
That triangle has f(t)=1 as well, so it doesn't contribute either.
copeland 2012-02-23 21:46:51
Alright I've kept you around for just about 3 hours.  Usually we pick off a problem or two from earlier in the exams, but usually we're done around an hour ago.
copeland 2012-02-23 21:47:17
So I think we'll shut this down and everyone can go to bed and sleep and be happy and dream of not math for a little while.
copeland 2012-02-23 21:47:34
Oh, incidentally, our own Richard Rusczyk has been making video solutions for many of this year's AMC 10A and 12A problems.  These will be up on our website in the next few days.  Click on the "Videos" tab of the website to see them when they're up (and of course we'll make an announcement on the Community when they are available).  They will probably be on YouTube too.
copeland 2012-02-23 21:47:42
If you qualify for the AIME, congratulations!  AoPS will be holding our annual Special AIME Problem Seminar on the weekend of March 3 and 4, from 3:30 to 6:30 (Eastern Time) both days, taught by Richard Rusczyk and Dave Patrick.  Each day consists of general approaches and important facts needed for problems within a given subject area, followed by a discussion of specific problems from past AIME competitions, or from other contests of a similar difficulty level. This course is largely a repeat of the weekend seminar we offered in 2010, and contains some material from the AIME Problem Series A course.  This course is appropriate for students who are hoping to pass the AIME and qualify for the USA(J)MO.  If a student already consistently scores above 10 on the AIME, this class is probably not necessary, and if a student is unlikely to answer more than 1 or 2 questions correctly, then that student should start with some of our Introduction series of classes.  The cost is $75.  You can enroll by clicking on the "School" tab of the website, and selecting "Special AIME Problem Seminar" from the list of courses.
copeland 2012-02-23 21:47:47
We will also have Math Jams for each of the two AIME contests in March -- check the website for specific dates and times.
copeland 2012-02-23 21:47:53
That's all for tonight -- thanks for coming!
c0mbusti0n1295 2012-02-23 21:48:32
good night Mr. Copeland!
ss5188 2012-02-23 21:48:32
How did you do on the AMC?
c0mbusti0n1295 2012-02-23 21:48:32
thanks for your help Samson!
ahaanomegas 2012-02-23 21:48:32
Thank you very much for this, Mr. Copeland, Duelist, Ms. Joelnia, and TowersFreak2006!
freddylukai 2012-02-23 21:48:32
Thank you!
blabla4198 2012-02-23 21:48:32
cya
fractals 2012-02-23 21:48:32
:bow:
ilovepink 2012-02-23 21:48:32
thank you!
cerberus88 2012-02-23 21:48:32
Thank you!!!! :bow:
lucylai 2012-02-23 21:48:32
:bow:
RotomPlasma 2012-02-23 21:48:32
Thanks!
Iggy Iguana 2012-02-23 21:48:32
thank you too!!!
timoteomo3 2012-02-23 21:48:32
thanks!
nackster12 2012-02-23 21:48:32
thanks!
cong989 2012-02-23 21:48:32
thanks!
Goos 2012-02-23 21:48:32
thanks!
quinntai1997 2012-02-23 21:48:32
Awesome, Thanks!
pattycakechichi 2012-02-23 21:48:32
Thank you!!!
copeland 2012-02-23 21:48:35
Thanks everyone, that was awesome!
sammyMaX 2012-02-23 21:49:25
There wasn't an AMC A Day math jam was there?
copeland 2012-02-23 21:49:28
Transcripts are here:
copeland 2012-02-23 21:49:29
http://www.artofproblemsolving.com/School/mathjams.php?past=1

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