Math Jams

DPatrick 2012-02-08 19:10:04
Welcome to the 2012 AMC 10A/12A Math Jam!

DPatrick 2012-02-08 19:10:13
I'm Dave Patrick, and I'll be leading our discussion tonight.

DPatrick 2012-02-08 19:10:20
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.

DPatrick 2012-02-08 19:10:29
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.

DPatrick 2012-02-08 19:10:43
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.

DPatrick 2012-02-08 19:10:53
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!!

DPatrick 2012-02-08 19:11:15
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the 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!

DPatrick 2012-02-08 19:11:30
We do have two assistants tonight who can help answer some of your questions: Ariana Levin (ArianaL) and Luyi Zhang (redcomet46).

DPatrick 2012-02-08 19:11:43
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.

DPatrick 2012-02-08 19:12:12
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, will likely be ignored.

DPatrick 2012-02-08 19:12:31
We will work the last 5 problems from the AMC 10A, then the last 4 problems from the AMC 12A. (There's an overlap: #24 on the 10A is also #21 on the 12A.) After that, time permitting, I will take requests for some other problems for discussion.

DPatrick 2012-02-08 19:12:44
Let's get started!

DPatrick 2012-02-08 19:13:00
We'll start with #21 on the AMC 10A:

DPatrick 2012-02-08 19:13:04

DPatrick 2012-02-08 19:13:13
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.

hutyputy66 2012-02-08 19:13:35
Make a diagram!

Polynomial 2012-02-08 19:13:35
First Draw a Diagram

pieofdoom51413 2012-02-08 19:13:35
Diagram

Binomial-theorem 2012-02-08 19:13:35
To solve this one, I drew a picture.

DPatrick 2012-02-08 19:13:45
You might have done it by computing the coordinates of all the points and applying some formula. That works but is unnecessarily complicated.

DPatrick 2012-02-08 19:13:49
We can start by trying to sketch a picture.

DPatrick 2012-02-08 19:13:54

DPatrick 2012-02-08 19:14:02
(Presumably you drew a similar picture. Yours might look different than mine if you oriented the axes differently.)

andrewjjiang97 2012-02-08 19:14:19
its a rectangle

ForeverAce 2012-02-08 19:14:19
Oh look! It's a RECTANGLE!

negativebplusorminus 2012-02-08 19:14:19
It's a rectangle!

Orange-2728 2012-02-08 19:14:19
that looks suspiciously like a rectangle to me

mathawesome 2012-02-08 19:14:19
Blue Rectangle!

DPatrick 2012-02-08 19:14:25
It sure looks like a rectangle.

DPatrick 2012-02-08 19:14:32
Is it really a rectangle? Do we know for sure?

Klu2014 2012-02-08 19:14:45
Do you have to prove that it is a rectangle?

DPatrick 2012-02-08 19:14:59
That's the nice(?) thing about the AMCs: you don't necessarily have to *prove* anything.

DPatrick 2012-02-08 19:15:07
You could guess it's a rectangle and go on from there.

DPatrick 2012-02-08 19:15:22
In fact, let's do that and see what we get, and we'll come back to the question of whether it's a rectangle or not.

DPatrick 2012-02-08 19:15:35
Are any of the side lengths of EFGH easy to compute?

mathlover3737 2012-02-08 19:15:58
Find the length of GF!

eb8368 2012-02-08 19:15:58
GF

anastasia84 2012-02-08 19:15:58
GF

Xu12345 2012-02-08 19:15:58
GF

DPatrick 2012-02-08 19:16:08
GF is pretty easy to compute. How?

iwantcombo 2012-02-08 19:16:37
they are just halves of the original side lengths

mentalgenius 2012-02-08 19:16:38
yeah, GF = CB/2, and we find CB with pythag

bookie331 2012-02-08 19:16:38
GF is half of CB

DPatrick 2012-02-08 19:16:46
Notice that AFG is similar to ABC and half the size.

DPatrick 2012-02-08 19:16:54

Iggy Iguana 2012-02-08 19:17:26
same thing with EF: 2 * EF = AD

Watermelon876 2012-02-08 19:17:27
Similarly FE is half of AD

pieofdoom51413 2012-02-08 19:17:27
we can compute GH as well

DPatrick 2012-02-08 19:17:58
Also CHG is similar to CDA and half the size. (Or BEF is similar to BAD and half the size.) Either way GH = FE = (1/2)(AD) = 3/2.

DPatrick 2012-02-08 19:18:11

DPatrick 2012-02-08 19:18:23
So let's go back to the earlier question: do we know that it's a rectangle? Or is it not?

RelaxationUtopia 2012-02-08 19:19:05
Prove that the sides are perpendicular?

joshy1227 2012-02-08 19:19:05
HG and EF are equal and parallel so yes

theone142857 2012-02-08 19:19:05
use the fact that FG and HE are both 1/2 and parallel to BC

DPatrick 2012-02-08 19:19:23
Right, we just saw that EF and GH are parallel and the same length (they're both parallel to AD and half its length).

DPatrick 2012-02-08 19:19:31
So it's at least a parallelogram.

PestofWest 2012-02-08 19:20:06
Now we have to prove that the sides are perpendicular...

jeff10 2012-02-08 19:20:07
prove one angle to be a right angle

2718euler 2012-02-08 19:20:07
GF lies in the XZ plane and so it is perpendicular to both GH and FE which lie in the YZ and XZ planes respectively.

DPatrick 2012-02-08 19:20:17
Furthermore, FG is on the xy-plane, and EF is parallel to the z-axis, so they're perpendicular.

DPatrick 2012-02-08 19:20:26
Thus EFGH is a rectangle as we suspected, and our area computation was correct. The answer is (C).

DPatrick 2012-02-08 19:20:58
Let's move on to #22:

DPatrick 2012-02-08 19:21:03

DPatrick 2012-02-08 19:21:29
We could just start plugging in formulas, but let's think for a moment first.

Hydroxide 2012-02-08 19:21:38
the sum of the first m positive odd integers is m^2

NextEinstein 2012-02-08 19:21:38
sum of first m odd integers is m^2

andrewjjiang97 2012-02-08 19:21:38
1+3+5+...+2m-1=m^2

ss5188 2012-02-08 19:21:38
sum of m integers is m squared

DPatrick 2012-02-08 19:21:48
Good start: the sum of the first m positive odd integers is just m^2.

MCL 2012-02-08 19:22:12
sum of first n even integers is n(n+1)

scgorantla 2012-02-08 19:22:13
sum of first n positive even integers n(n+1)

henrypickle 2012-02-08 19:22:13
sum of first n even integers is n(n+1)

Relativity1618 2012-02-08 19:22:13
sum of first n even integers is n(n+1)

DPatrick 2012-02-08 19:22:30
Right: another way to say "the sum of the first n positive even integers" is "twice the sum of the first n integers".

DPatrick 2012-02-08 19:22:36

andrewjjiang97 2012-02-08 19:22:58
m^2=212+n(n+1)

King6997 2012-02-08 19:22:59
m^2=212 + n(n+1)

DPatrick 2012-02-08 19:23:08

DPatrick 2012-02-08 19:23:18
What do we do now?

geoistrivial 2012-02-08 19:23:40
we form a quadratic with respect to n

willwang123 2012-02-08 19:23:40
shove everything to one side and quadratic bash?

fermat007 2012-02-08 19:23:40
quadratic in terms of n

deepankar 2012-02-08 19:23:40
make a quadratic with n as the variable

DPatrick 2012-02-08 19:23:47
I like this idea. Since we're trying to solve for n, we could write it as a quadratic in n:

DPatrick 2012-02-08 19:23:52

flyrain 2012-02-08 19:24:13
use quadratic formula

electron 2012-02-08 19:24:13
and apply the quadratic formula?

Hydroxide 2012-02-08 19:24:15
The discriminant must be a perfect square.

DPatrick 2012-02-08 19:24:20

DPatrick 2012-02-08 19:24:43

Iggy Iguana 2012-02-08 19:25:04
we can say that 4m^2-847=a^2

theone142857 2012-02-08 19:25:04
set it to be b^2

DPatrick 2012-02-08 19:25:11
I like giving it a name too.

DPatrick 2012-02-08 19:25:15

King6997 2012-02-08 19:25:30
difference of squares

2718euler 2012-02-08 19:25:31
Difference of Squares

Relativity1618 2012-02-08 19:25:31
difference of squares

DPatrick 2012-02-08 19:25:36
You said the magic words! :)

DPatrick 2012-02-08 19:25:42

DPatrick 2012-02-08 19:25:57
(a **difference** of squares would be even better than what I wrote)

fermat007 2012-02-08 19:26:12
847 = 7*11^2

va2010 2012-02-08 19:26:13
Factor 847

LightningStreak 2012-02-08 19:26:13
Find factors of 847!

DPatrick 2012-02-08 19:26:23
Again, both m and a are integers! So we can look at factors of 847.

DPatrick 2012-02-08 19:26:30
847 = 7 * 11 * 11.

So we have 847 = 1*847 = 7*121 = 77*11. (There are negative factorizations too, but they don't give us any different values of m^2.)

mistere 2012-02-08 19:27:02
Find the average of the factors

Iggy Iguana 2012-02-08 19:27:14
to find m, sum the factor pairs and divide by 4

DPatrick 2012-02-08 19:27:23
We could, but we really care about "a".

DPatrick 2012-02-08 19:27:41
We notice that 2a is the difference of the factors.

DPatrick 2012-02-08 19:27:56

yankeefan6795 2012-02-08 19:28:36
255

Iggy Iguana 2012-02-08 19:28:36
211+28+16=255

BookwormB 2012-02-08 19:28:36
then just add the possible values of n, and you're done :)

ktbroborg 2012-02-08 19:28:36
211+28+16=255

DPatrick 2012-02-08 19:28:40
So our answer is 211 + 28 + 16 = 255, answer (A).

DPatrick 2012-02-08 19:29:09
A quick reminder since lots of people have been asking:

DPatrick 2012-02-08 19:29:39
A complete transcript of the session will be up on the website after we're done. So if you have to leave, or parts are going too fast and you want to review later, look on the website in a couple of hours.

DPatrick 2012-02-08 19:29:55
Let's move on to #23 on the 10A, which is also #19 on the 12A:

DPatrick 2012-02-08 19:30:01

DPatrick 2012-02-08 19:30:17
We need to first take a moment to make sure we understand the problem.

DPatrick 2012-02-08 19:30:29
There are 6 people (A,B,C,D,E,F so I don't have to write the names all the time). Each is friends with the same number of people.

joshy1227 2012-02-08 19:30:38
well they all dont have 5 internet friends, because then they all would be friends with eachother

Iggy Iguana 2012-02-08 19:30:41
everyone can have 1, 2, 3, or 4 friends

DPatrick 2012-02-08 19:30:56
Indeed, let's call the number of people that each is friends with n. We clearly have 0 <= n <= 5.

DPatrick 2012-02-08 19:31:02
But the "Some, but not all" clause rules out n=0 and n=5. So we only have n=1, n=2, n=3, or n=4.

willwang123 2012-02-08 19:31:11
casework!

Klu2014 2012-02-08 19:31:11
Use casework

Relativity1618 2012-02-08 19:31:11
casework

DPatrick 2012-02-08 19:31:23
This looks like a job for casework. But do we have to work through all four cases?

Iggy Iguana 2012-02-08 19:31:36
we only care about the 1 and 2 cases, since they are symetric to 4 and 3

gurev 2012-02-08 19:31:36
1 and 2 is symmetrical to 3 and 4

DPatrick 2012-02-08 19:31:44
How so?

noedne 2012-02-08 19:32:03
n=1 and n=4, n=2 and n=3 are complementary

RelaxationUtopia 2012-02-08 19:32:04
Chosing 2 be friends is same as 2 not vice versa

2718euler 2012-02-08 19:32:04
The number of ways for everyone to have 2 friends is the same as the number of ways for everyone to have 3 friends. Here's why: If everyone has 3 friends then they each have 2 friends missing.

DPatrick 2012-02-08 19:32:26
Right! For example, suppose we have a way for n=1.

DPatrick 2012-02-08 19:32:31
We just flip all the "friend"/"not friend" relationships! That is, if A and B are friends in the n=1 configuration, then they're not friends in the n=4 configuration, and vice versa.

DPatrick 2012-02-08 19:32:43
We get a new arrangement in which everyone has 4 friends.

Iggy Iguana 2012-02-08 19:32:51
so now we do casework on n=1, 2

noedne 2012-02-08 19:32:51
so there are only two cases!

DPatrick 2012-02-08 19:32:58
Right, we get that (# of arrangements with n=1) = (# of arrangements with n=4), and also (# of arrangements with n=2) = (# of arrangements with n=3).

DPatrick 2012-02-08 19:33:04
This means we only need count the n=1 and n=2 arrangements, and then double the total to get our answer.

DPatrick 2012-02-08 19:33:20
Let's start with the first case -- how many ways are possible with n=1 (that is, each person has a single friend)?

theartof 2012-02-08 19:33:40
Start with 1

andrewjjiang97 2012-02-08 19:33:40
first choose A's friend

DPatrick 2012-02-08 19:33:54
Right, I like to count this "constructively" by actually building a valid arrangment.

DPatrick 2012-02-08 19:34:10
Start with A, and pick a friend for him. That's 5 choices. Those two people now have their total "friend allotment" for an n=1 arrangment, so we set them aside,

Iggy Iguana 2012-02-08 19:34:46
now there are 3 choices for someone else friend

noedne 2012-02-08 19:34:46
3 remaining ways

alex31415 2012-02-08 19:34:46
3 more choices for another person

joshy1227 2012-02-08 19:34:46
oh so for the next person theres only 3 choices

DPatrick 2012-02-08 19:35:02
Right: take a remaining person (say the first alphabetically-remaining person left), and pick a friend for him/her. That's 3 choices.

DPatrick 2012-02-08 19:35:10
Then the two who are left must be friends.

mdecross 2012-02-08 19:35:23
5 x 3 x 1 = 15 for n = 1 as a result

batter8642 2012-02-08 19:35:25
15 in total

mintchip 2012-02-08 19:35:25
5*3=15

flyrain 2012-02-08 19:35:25
5*3*1=15

DPatrick 2012-02-08 19:35:30
So there are 5*3 = 15 arrangements where each has exactly 1 friend. That's the n=1 case.

DPatrick 2012-02-08 19:35:38
Next, let's count n=2 arrangements.

AlcumusGuy 2012-02-08 19:35:57
Same approach

mathnerd101 2012-02-08 19:35:58
pick a's 2 friends?

DPatrick 2012-02-08 19:36:05
Sure, we'll try the same approach. First, start at A and pick two friends. In how many ways can we do that?

anaverageaopser 2012-02-08 19:36:19
There are C(5,2) ways to choose friends for A.

Hydroxide 2012-02-08 19:36:19
C(5,2) ways to choose friends for A

math-rules 2012-02-08 19:36:19
5c2=10 ways for a

superbob 2012-02-08 19:36:19
for A, you have 5C2 = 10 combos of friends

DPatrick 2012-02-08 19:36:44
We have C(5,2) = 10 choices. Suppose they're B and C. (The same argument that follows works no matter which two they are.)

DPatrick 2012-02-08 19:36:49
Now what?

Iggy Iguana 2012-02-08 19:37:15
subcase 1: b and c are friends

pinkmuskrat 2012-02-08 19:37:15
two cases: b and c friends or not

jeff10 2012-02-08 19:37:15
B and C need to fill their other friend

andrewjjiang97 2012-02-08 19:37:18
B and C could be friends or they could be friends with other people

DPatrick 2012-02-08 19:37:26
Exactly. We have two subcases, depending on whether B and C are themselves friends with each other.

DPatrick 2012-02-08 19:37:46
If B and C are themselves friends, then ABC are all friends with each other, and DEF must also all be friends with each other. So that's 1 way to finish.

DPatrick 2012-02-08 19:37:57
Otherwise, B and C are friends with different people. Can they be friends with the same person?

Iggy Iguana 2012-02-08 19:38:22
no because then you have 2 people left with 2 friendship pairs

superpi83 2012-02-08 19:38:25
no. Then the last two people could only have one friend each.

DPatrick 2012-02-08 19:38:32
No. If B and C were both friends with D, then A,B,C,D would all have 2 friends, and there wouldn't be enough left for E and F.

DPatrick 2012-02-08 19:38:42
So B must be friends with one of the three remaining people, and C with a different remaining person.

sparkles257 2012-02-08 19:39:05
there are 6

PestofWest 2012-02-08 19:39:05
so there are 6 ways

King6997 2012-02-08 19:39:05
3*2=6

mentalgenius 2012-02-08 19:39:05
that is 3x2 = 6

DPatrick 2012-02-08 19:39:14
Right, that's 3*2 = 6 choices.

DPatrick 2012-02-08 19:39:29
Once we make those choices, there's only one way to finish: the person left over who still has no friends must be friends with the two most recent choices.

BarbieRocks 2012-02-08 19:39:41
isn't it easier to think of it as a hexagon case and a two triangles case?

jeff10 2012-02-08 19:39:42
It could be two triangle groups of friends or one huge hexagon.

DPatrick 2012-02-08 19:39:58
Right: if you drew a graph of the friendships, you'd either get two triangles (in the first subcase) or a hexagon (in the second subcase).

DPatrick 2012-02-08 19:40:17
But let's finish up: how many total arrangements in our n=2 case?

Iggy Iguana 2012-02-08 19:40:39
so 10(1+6)=70

noedne 2012-02-08 19:40:39
10*(1+6)

superpi83 2012-02-08 19:40:39
10(6+1)=70 arrangements.

ashgabat 2012-02-08 19:40:39
10*(1+6)=10*7=70

DPatrick 2012-02-08 19:40:53
Right, we have C(5,2) = 10 ways to pick the first two friends, then 1+6 = 7 ways to finish after we pick friends for the first person.

DPatrick 2012-02-08 19:41:00
Hence, there are 10*7 = 70 possible arrangements for n=2.

DPatrick 2012-02-08 19:41:07
And to finish the problem?

jeff10 2012-02-08 19:41:26
15+70=85, and 85*2=170.

Relativity1618 2012-02-08 19:41:27
so 2(70+15)=(B) 170

ashgabat 2012-02-08 19:41:27
2*(70+15)=170

eshoyfer 2012-02-08 19:41:27
70+15 = 85, then double that.

DPatrick 2012-02-08 19:41:30
We had a count of 15 for n=1 and 70 for n=2.

DPatrick 2012-02-08 19:41:37
Thus our final answer is 2(15+70) = 2(85) = 170. Answer (B).

DPatrick 2012-02-08 19:42:05
Don't forget to double at the end to capture the n=3 and n=4 cases that we skipped? (Of course, 85 isn't an answer choice.)

DPatrick 2012-02-08 19:42:30
Let's continue with #24 on the AMC 10A, which is also #21 on the 12A:

DPatrick 2012-02-08 19:42:38

DPatrick 2012-02-08 19:42:52
Yikes. These look ugly. What can we do?

DPatrick 2012-02-08 19:43:18
Many many of you said "add the equations together", which is the right thing to do.

DPatrick 2012-02-08 19:43:38
If you don't see this right away, you still have to try something. Don't just stare blankly at a problem -- try something!

DPatrick 2012-02-08 19:43:41
Ideally we'd like to have something we can factor on the left side.

DPatrick 2012-02-08 19:43:56
Adding them together is at least something worth trying if you don't see right away that it will help.

DPatrick 2012-02-08 19:44:07

noedne 2012-02-08 19:44:32
(a-b)^2+(b-c)^2+(c-a)^2=14

xjava 2012-02-08 19:44:33
(a-b)^2+(b-c)^2+(c-a)^2=14

lucylai 2012-02-08 19:44:33
you get (a-b)^2+(b-c)^2+(a-c)^2=14

DPatrick 2012-02-08 19:44:38
Happy day, this factors!

DPatrick 2012-02-08 19:44:44

DPatrick 2012-02-08 19:44:56
Note how we wrote this so that the term inside each set of parentheses is a nonnegative integer.

DPatrick 2012-02-08 19:45:06
How can we write 14 as the sum of three nonnegative perfect squares?

lucylai 2012-02-08 19:45:29
the pairs have to be 1, 9, and 4

Watermelon876 2012-02-08 19:45:30
14=1+9+4

number.sense 2012-02-08 19:45:30
they must be 1, 4, 9'

yankeefan6795 2012-02-08 19:45:30
3^2+2^2+1^2=14

DPatrick 2012-02-08 19:45:36
There's only one way: 1 + 4 + 9 = 14.

DPatrick 2012-02-08 19:45:45
What can we conclude?

jeff10 2012-02-08 19:46:10
a-c has to be 3 because it is 3^2+2^2+1^2 and 3 is the biggest of 1, 2, and 3.

Hydroxide 2012-02-08 19:46:10
a-b, a-c, and b-c have to 1,2, or 3

AceOfDiamonds 2012-02-08 19:46:10
a-c = 3

mcdonalds106_7 2012-02-08 19:46:10
a-c is the biggest, so (a-c)^2=9!

DPatrick 2012-02-08 19:46:25
We must have a-c = 3 since that's the biggest difference, and one each of a-b and b-c must be 1 or 2. We can't tell which is which though.

Iggy Iguana 2012-02-08 19:46:39
casework

xjava 2012-02-08 19:46:40
try both cases

DPatrick 2012-02-08 19:46:46
Yep, we're back to casework.

DPatrick 2012-02-08 19:46:55

number.sense 2012-02-08 19:47:19
plug into the top equation

yankeefan6795 2012-02-08 19:47:19
plug into first original equation

scgorantla 2012-02-08 19:47:19
we plug both cases in to the original equation

mcdonalds106_7 2012-02-08 19:47:19
plug it into the first equation!

DPatrick 2012-02-08 19:47:24
We plug these into one of the equations and solve for a. My choice would be to use the first equation, since it's a lot simpler.

DPatrick 2012-02-08 19:47:37
Now it's just algebra crunching.

DPatrick 2012-02-08 19:47:46

DPatrick 2012-02-08 19:48:01

PestofWest 2012-02-08 19:48:17
2021/7

Relativity1618 2012-02-08 19:48:17
2021 is not divisible by 7

DPatrick 2012-02-08 19:48:24
This gives 7a = 2021, but then a = 2021/7, which is not an integer. So no solution in this case.

DPatrick 2012-02-08 19:48:32
The other case had better work!

DPatrick 2012-02-08 19:48:37

DPatrick 2012-02-08 19:48:50

DPatrick 2012-02-08 19:49:11
This gives 8a = 2024, so a = 253. Answer (E).

DPatrick 2012-02-08 19:49:39
The "clever" step was adding the equations and recognizing the factorization.

DPatrick 2012-02-08 19:50:00
Recognizing when something factors is not always easy, but is a skill that gets better with practice and experience.

DPatrick 2012-02-08 19:50:29
Let finish up the end of the AMC 10A with #25:

DPatrick 2012-02-08 19:50:35

DPatrick 2012-02-08 19:50:55
In general, how do we calculate this sort of probability, when there are continuously-many infinite possibilities?

lucylai 2012-02-08 19:51:17
geometric

Lalagato 2012-02-08 19:51:17
geometry!

yankeefan6795 2012-02-08 19:51:17
geometrically

alberthyang 2012-02-08 19:51:17
geometric probability?

talkinaway 2012-02-08 19:51:17
Areas and/or volumes

DPatrick 2012-02-08 19:51:30

DPatrick 2012-02-08 19:51:51
The denominator is easy: what is the "possible" region?

scgorantla 2012-02-08 19:52:08
n^3

number.sense 2012-02-08 19:52:08
n^3

Lalagato 2012-02-08 19:52:08
n^3

Binomial-theorem 2012-02-08 19:52:08
n^3

BarbieRocks 2012-02-08 19:52:08
n^3

hyperbolictangent 2012-02-08 19:52:08
A cube of side n

scgorantla 2012-02-08 19:52:08
cube with side length of n

DPatrick 2012-02-08 19:52:15

DPatrick 2012-02-08 19:52:19

DPatrick 2012-02-08 19:52:36
The successful region is trickier. (Obviously, or this wouldn't be a #25!)

DPatrick 2012-02-08 19:52:45
What can we do to simplify its computation?

Binomial-theorem 2012-02-08 19:53:16
WLOG let x>y>z

noedne 2012-02-08 19:53:16
0<=x<=y<=z<=n

BarbieRocks 2012-02-08 19:53:16
assume x>=y>=z and multiply by 6 later

GoldenFrog1618 2012-02-08 19:53:16
assume x>y>z, and then multiply by 6

DPatrick 2012-02-08 19:53:29
Right. The "successful" region is actually made up of six identical pieces, depending on the order of x, y, and z.

DPatrick 2012-02-08 19:53:35

DPatrick 2012-02-08 19:53:40
Otherwise these regions have identical shapes, so if we find the volume of one of them, we can multiply by 6 to get our numerator.

DPatrick 2012-02-08 19:53:53
So let's focus on x < y < z.

DPatrick 2012-02-08 19:53:58
What is the condition for "success"?

lucylai 2012-02-08 19:54:35
z-y>1, y-x>1

memphis_tiger 2012-02-08 19:54:35
z>y+1, y>x+1

Binomial-theorem 2012-02-08 19:54:35
z-y>1 and y-x>1

DPatrick 2012-02-08 19:54:44

DPatrick 2012-02-08 19:54:55
You could actually determine this region directly and compute its volume, but what's a slick way to compute the volume of this region?

pl210741 2012-02-08 19:55:44
Its a pyramid

Iggy Iguana 2012-02-08 19:55:44
transformation

DPatrick 2012-02-08 19:55:58
Indeed, it is a pyramid, and you could directly determine its base and its height.

DPatrick 2012-02-08 19:56:06
But there is a clever transformation that makes it easy to compute.

DPatrick 2012-02-08 19:56:17

ClassClown 2012-02-08 19:56:32
You just move the base from the original one

DPatrick 2012-02-08 19:56:51
That's exactly what this is doing: it's just sliding the region 1 unit in the y-direction and 2 units in the z-direction.

DPatrick 2012-02-08 19:57:12
But now it's clear that what we have is just the part of the cube of side length (n-2) with x' <= y' <= z'.

DPatrick 2012-02-08 19:57:37
This is 1/6 of the cube of side length (n-2).

AkshajK 2012-02-08 19:57:59
SO WHEN YOU MULTIPLY BY SIX YOU GET (n-2)^3 / n^3

DPatrick 2012-02-08 19:58:24
Right. There are 6 pieces (for the 6 different orderings of x, y, and z), and each piece is 1/6 of the cube of side length (n-2).

DPatrick 2012-02-08 19:58:35

DPatrick 2012-02-08 19:58:49

jeff10 2012-02-08 19:58:57
Just keep testing until you get that fraction is greater than 1/2.

yankeefan6795 2012-02-08 19:59:05
which needs to be greater than 1/2

Relativity1618 2012-02-08 19:59:08
that must exceed 1/2

DPatrick 2012-02-08 19:59:17

alex31415 2012-02-08 19:59:36
n=10 will lead to 0.512, which is greater than 1/2. So 10(D) is the answer.

Relativity1618 2012-02-08 19:59:36
so n=10

pieofdoom51413 2012-02-08 19:59:36
n = 10

kaiz99 2012-02-08 19:59:36
n=10

number.sense 2012-02-08 19:59:36
9 is not big enough

DPatrick 2012-02-08 19:59:41
Note that (0.8)^3 = 0.512 > 0.5, so n=10 definitely works.

DPatrick 2012-02-08 19:59:52
We check n=9: (7/9)^3 = 343/729 < 1/2. So n=9 doesn't work.

DPatrick 2012-02-08 19:59:59
Hence the smallest n that works is n=10. Answer (D).

DPatrick 2012-02-08 20:00:31
So that wraps up the last five problems on the AMC 10A.

DPatrick 2012-02-08 20:00:38
Now let's go over to the 12A!

DPatrick 2012-02-08 20:00:59
We already did #21: it was #24 on the 10A as well (the (a-b)^2+... problem).

DPatrick 2012-02-08 20:01:05
So we'll pick up with #22:

DPatrick 2012-02-08 20:01:12

DPatrick 2012-02-08 20:01:32
What the heck is going on here?

AlcumusGuy 2012-02-08 20:01:40
What does the big U mean?

alex31415 2012-02-08 20:01:40
What does that big "U" mean?

DPatrick 2012-02-08 20:01:50
It's "union". It means take all the planes together.

Watermelon876 2012-02-08 20:02:00
Apparently a cube is being cut up. the u is the sum of all the planes by the sum axiom

mdecross 2012-02-08 20:02:02
Draw a pretty picture

DPatrick 2012-02-08 20:02:27
What's going on in this problem is buried under a lot of notation.

2718euler 2012-02-08 20:02:36
The planes must intersect the cube ONLY at the segments joining the midpoints of pairs of edges, or else the intersection would include some other points.

DPatrick 2012-02-08 20:02:47
Right. We're chopping a cube with a bunch of planes. The places where the planes cut into the cube have to be these special segments.

DPatrick 2012-02-08 20:02:55

DPatrick 2012-02-08 20:03:01
The blue lines are the segments that we have to hit.

DPatrick 2012-02-08 20:03:15
Notice there are 6 segments on each face, for a total of 6*6 = 36 segments that we have to hit.

DPatrick 2012-02-08 20:03:28
So how do we attack this problem?

2718euler 2012-02-08 20:03:45
What are the different kinds of planes?

number.sense 2012-02-08 20:03:46
what kinds of planes can we have is the question

DPatrick 2012-02-08 20:04:04
Indeed. Let's start with trying to find the maximum, because that's doesn't require quite as much cleverness -- we just have to determine how many possible planes there are. (And, determining all the possible planes will probably help us with the minimum argument too.)

DPatrick 2012-02-08 20:04:43
There are probably lots of ways to go about this, but the key is to be methodical and organized, and have a plan. Don't start aimlessly counting planes without a plan. (You can't spell "plane" with "plan", after all.)

fmasroor 2012-02-08 20:04:55
without

DPatrick 2012-02-08 20:05:03
Yeah, I ruined my own joke. :(

2718euler 2012-02-08 20:05:22
What kinds of segments do the planes go through?

DPatrick 2012-02-08 20:05:34
That's a good idea. What I did was focus on a single point:

DPatrick 2012-02-08 20:05:38

DPatrick 2012-02-08 20:05:44
(I circled it above)

DPatrick 2012-02-08 20:05:51
How many planes (satisfying the conditions of the problem) pass through this point?

ziv 2012-02-08 20:06:25
9: pick one line on top face, one on front face

DPatrick 2012-02-08 20:06:29
Right!

DPatrick 2012-02-08 20:06:40
Each valid plane through the circled point must contain one of the 3 segments passing through the circled point and lying on the top face.

DPatrick 2012-02-08 20:06:49
Each valid plane must also contain one of the 3 segments passing through the circled point and lying on the front face.

DPatrick 2012-02-08 20:07:00
So there are 3*3 = 9 possible planes through the circled point.

DPatrick 2012-02-08 20:07:32
And we know that there are 12 points like the circled one (one per edge of the cube), so does that mean there are 9*12 = 108 possible planes?

hyperbolictangent 2012-02-08 20:07:49
Overcounting

number.sense 2012-02-08 20:07:49
NO! We're vastly overcounting

fro116 2012-02-08 20:07:49
no, we are overcounting

2718euler 2012-02-08 20:07:49
NO! We've overcounted

DPatrick 2012-02-08 20:07:58
Indeed, since all those planes pass through multiple points, so each plane is counted multiple times in our dumb count of 108.

DPatrick 2012-02-08 20:08:19
How much each plane is overcounted depends on the type of plane, meaning it depending on the choices of segments that we make.

number.sense 2012-02-08 20:08:41
we can't evenly divide that number either, because some planes pass thorugh 4 or 3

DPatrick 2012-02-08 20:08:49
Right...and maybe some other number too...

DPatrick 2012-02-08 20:09:07
For example, we might choose a plane using two of the shorter "diagonal" segments on the same side, like this:

DPatrick 2012-02-08 20:09:11

DPatrick 2012-02-08 20:09:33
How many our of special points does that plane pass through?

number.sense 2012-02-08 20:09:47
3

mathnerd101 2012-02-08 20:09:47
3

Flamel 2012-02-08 20:09:47
3

VIPMaster 2012-02-08 20:09:47
3

Asmodeus1123 2012-02-08 20:09:47
3

yankeefan6795 2012-02-08 20:09:47
3

mathlover3737 2012-02-08 20:09:47
3

sparkles257 2012-02-08 20:09:47
3

DPatrick 2012-02-08 20:10:00
Right, 3. (Basically the three that we can see in the above picture.)

DPatrick 2012-02-08 20:10:14
And how many of the 9 planes through the circled point are of this type?

mathnerd101 2012-02-08 20:10:36
2

yankeefan6795 2012-02-08 20:10:36
2

Flamel 2012-02-08 20:10:36
2

Iggy Iguana 2012-02-08 20:10:36
2.

mathlover3737 2012-02-08 20:10:36
2

Shoelace Thm. 2012-02-08 20:10:36
2

DPatrick 2012-02-08 20:10:49
There are 2 of this type of plane at the circled point, one on each side.

DPatrick 2012-02-08 20:11:16
So each circled point has 2 planes of this type, and there are 12 of these points (one on each edge), so that's a preliminary overcount of 2*12 = 24.

mathlover3737 2012-02-08 20:11:40
24/3=8 of these

DPatrick 2012-02-08 20:11:57
But we just said that each of these planes passes through 3 points, so it gets counted 3 times.

DPatrick 2012-02-08 20:12:14
So there are a total of 24/3 = 8 planes of this type (that intersect 3 of our special points).

DPatrick 2012-02-08 20:12:46
Another type of plane through the circled points uses a shorter diagonal segment, but on opposite sides of the circled point:

DPatrick 2012-02-08 20:12:50

DPatrick 2012-02-08 20:13:05
How many of our special points does this intersect?

number.sense 2012-02-08 20:13:28
6 points

leekspeak 2012-02-08 20:13:28
6

mathnerd101 2012-02-08 20:13:28
6

mdecross 2012-02-08 20:13:28
6

ChipDale 2012-02-08 20:13:28
6

x31415926535x 2012-02-08 20:13:28
6

DPatrick 2012-02-08 20:13:37
The plane intersects the cube in a hexagon. (There are two more intersection points that we can't see in the back of the cube, and there's one side of the hexagon on each of the six faces of the cube.)

DPatrick 2012-02-08 20:14:00
And again, there are 2 of these through each circled point, one in each direction.

DPatrick 2012-02-08 20:14:08
So in total, how many planes of this type are there?

Iggy Iguana 2012-02-08 20:14:33
so we have 24/6=4

math-rules 2012-02-08 20:14:33
12*2/6=4

number.sense 2012-02-08 20:14:33
24/6 = 4

proglote 2012-02-08 20:14:33
24/6 = 4 planes

DPatrick 2012-02-08 20:14:45
Right: we have a preliminary overcount of 2*12 = 24 planes, but each passes through 6 points, so there are 24/6 = 4 of them.

DPatrick 2012-02-08 20:15:01
That leaves 9-2-2=5 choices for a pair of segments at the circled points; each of these choices contains one or two of the longer segments (parallel to the edges of the cube).

DPatrick 2012-02-08 20:15:06

DPatrick 2012-02-08 20:15:10

DPatrick 2012-02-08 20:15:30
I give you a sec to digest these new pics...how many special points do these red planes intersect the cube at?

Iggy Iguana 2012-02-08 20:16:01
each one passes through 4 poitns

number.sense 2012-02-08 20:16:02
4 special points

yankeefan6795 2012-02-08 20:16:02
4 each

Rocksolid 2012-02-08 20:16:02
4 both

DPatrick 2012-02-08 20:16:15
They come in two types, but either type intersects 4 points.

number.sense 2012-02-08 20:16:27
5*12/4 = 15 total planes

Iggy Iguana 2012-02-08 20:16:27
so 60/4=15

DPatrick 2012-02-08 20:16:43
Right. There are 5 planes at each of the 12 points, and each passes through 4 points. So there are 5*12/4 = 15 of them.

supermathman 2012-02-08 20:16:58
so max is gone be 27

number.sense 2012-02-08 20:16:58
27 seems to be the maximum then

DPatrick 2012-02-08 20:17:09
We've methodically counted all the planes. We add the cases: 8 + 4 + 15 = 27 planes.

DPatrick 2012-02-08 20:17:16
These are all distinct and are all the legal planes, so 27 is the maximum number of planes.

DPatrick 2012-02-08 20:17:58
That was just methodical casework, and while it may have taken some geometric insight, it didn't require any special degree of "cleverness", just an organized plan.

DPatrick 2012-02-08 20:18:03
Now for the more subtle question: what is the minimum number that we need?

DPatrick 2012-02-08 20:18:19
There are a total of 36 segments that we need to hit (6 on each side).

DPatrick 2012-02-08 20:18:24
How can we hit all 36 segments most efficiently?

mathlover3737 2012-02-08 20:18:30
use the 6-point planes

number.sense 2012-02-08 20:18:36
now for the minimum; it seems like the hexagons can get a lot of bang for their buck :)

Iggy Iguana 2012-02-08 20:18:36
using hexagons

DPatrick 2012-02-08 20:19:02
Indeed, using the 4 hexagon planes will give us 6 segments each, so we'll get a total of 24 segments -- all the "diagonal" segments in fact.

DPatrick 2012-02-08 20:19:10
Here's the hexagon pic again:

DPatrick 2012-02-08 20:19:15

DPatrick 2012-02-08 20:19:37
So we're left with just the segments parallel to the edges of the cube.

mdecross 2012-02-08 20:19:57
For the min, you can quickly eliminate the + on every face with only 3 planes used

314159Nerd 2012-02-08 20:19:57
3 more planes for vertical and horizontal

yankeefan6795 2012-02-08 20:19:57
3 more!

superpi83 2012-02-08 20:19:57
3 planes to cover the parallel segments.

Iggy Iguana 2012-02-08 20:19:57
we can use 3 planes parallel to faces

DPatrick 2012-02-08 20:20:06
Right, we need the 3 planes that look like this:

DPatrick 2012-02-08 20:20:10

DPatrick 2012-02-08 20:20:20
Using the three planes that are parallel to the sides of the cube will give all 12 longer segments.

number.sense 2012-02-08 20:20:31
so 7 is the minimum

Iggy Iguana 2012-02-08 20:20:31
so that's 7 in total

DPatrick 2012-02-08 20:20:35
So 7 planes is the minimum.

number.sense 2012-02-08 20:20:45
and 27-7 = (C) is our answer

yankeefan6795 2012-02-08 20:20:45
27-7=20

Cortana 2012-02-08 20:20:45
27-7=20

Iggy Iguana 2012-02-08 20:20:45
the answer is 27-7=20 (C)

DPatrick 2012-02-08 20:20:48
Thus the answer is 27 - 7 = 20. Answer (C).

DPatrick 2012-02-08 20:21:26
I liked this problem -- it was a nice blend of geometric intuition and careful counting with overcounting.

DPatrick 2012-02-08 20:21:54
(I didn't particularly like the way the question was phrased though -- I think it could have probably been phrased more simply.)

DPatrick 2012-02-08 20:22:01
Anyway, on to #23:

DPatrick 2012-02-08 20:22:07

Asmodeus1123 2012-02-08 20:22:48
picture?

Iggy Iguana 2012-02-08 20:22:48
draw the square

DPatrick 2012-02-08 20:22:52
We can sketch a picture:

DPatrick 2012-02-08 20:22:55

DPatrick 2012-02-08 20:23:10
S is the square in bold.

DPatrick 2012-02-08 20:23:20
Now what?

supermathman 2012-02-08 20:23:45
consider the unit square U with vertices 0,0 1,0 1,1 0,1

number.sense 2012-02-08 20:23:57
how can we cover 2

Shoelace Thm. 2012-02-08 20:23:57
FInd conditions on how it can contain two lattice points

Iggy Iguana 2012-02-08 20:23:57
it can only contain 2 adjacent lattice points at the most

DPatrick 2012-02-08 20:24:09
There are a few simplifying observations we can make.

DPatrick 2012-02-08 20:24:22
The 2012 seems irrelevant, and in fact it is. By symmetry, we can assume 0 <= x <= 1 and 0 <= y <= 1, or indeed we can just look at any lattice square.

DPatrick 2012-02-08 20:25:07
You might notice that the square S has area 1 and that its sides have slope 3/4 and -4/3.

DPatrick 2012-02-08 20:25:14
Let's just tuck that factoid away for later.

DPatrick 2012-02-08 20:25:42
You might also notice that if translation of our square covers two lattice points, they have to be adjacent. (Any two non-adjacent points are too far apart.)

DPatrick 2012-02-08 20:26:21
So let's look at the unit square with lower-left corner (0,0), and find the translations of S that cover both the points (0,0) and (1,0).
(By symmetry, there will be 3 equal regions covering the other three adjacent pairs of corners.)

DPatrick 2012-02-08 20:26:58
Here's the question that breaks open the problem: where must the center of S be in order for S to contain the point (0,0)?

Cortana 2012-02-08 20:27:42
Inside the square?

DPatrick 2012-02-08 20:28:07
Right. To say it in the technical language of the problem, v must be in S for T(v) to contain (0,0).

DPatrick 2012-02-08 20:28:36
To say it more intuitively, we can slide S so that it's centered at any point that's inside the original S.

DPatrick 2012-02-08 20:28:51
This will still keep the origin (0,0) inside of the slid square.

DPatrick 2012-02-08 20:29:20
In other words, the center of S (after we move it) must be in the red square below. Note the red square is the same as the original square!

DPatrick 2012-02-08 20:29:25

DPatrick 2012-02-08 20:29:47
Similarly, the blue square above is the locations at which the center of the tilted square captures the point (1,0).

DPatrick 2012-02-08 20:30:23
So the purple area (where the tilted squares overlap) is the region where we can center the tilted square and get both (0,0) and (1,0).

DPatrick 2012-02-08 20:30:36
(I admit this is a subtle argument. You may have to think about it more later on your own.)

Iggy Iguana 2012-02-08 20:30:46
find area of overlap

brokenfixer 2012-02-08 20:30:50
two purple rectangles total

number.sense 2012-02-08 20:31:03
2 times that purple area / white area is what we want

DPatrick 2012-02-08 20:31:28
Exactly. We only want to count what works that inside the unit square, so we only want half of that purple region to capture the bottom two points.

DPatrick 2012-02-08 20:31:48
But there are four such regions (one on each side of the square, capturing the two points on that side). Thus, 4 times half of the purple area is our answer. (That is, twice its area is our answer.)

saprmarks 2012-02-08 20:31:57
is this where we use the slopes?

DPatrick 2012-02-08 20:32:16
Indeed, this is where we use the slopes! There's a really easy way to find the area of the purple rectangle if we are clever.

DPatrick 2012-02-08 20:32:42
We can get all coordinate-y and compute the actual points and bash from there, but there's a vastly simpler way.

DPatrick 2012-02-08 20:33:28
Let me shade a particularly interesting part of the picture:

Yongyi781 2012-02-08 20:33:31
0.6-0.8-1.0 "hole" at bottom

DPatrick 2012-02-08 20:33:34
Exactly!

DPatrick 2012-02-08 20:33:38

DPatrick 2012-02-08 20:34:12
Because we know the slopes, we know that the grey triangle is similar to a 3-4-5 right triangle!

DPatrick 2012-02-08 20:34:23
So since the hypotenuse has length 1, its red side has length 0.6 and its blue side has length 0.8.

Iggy Iguana 2012-02-08 20:34:33
so the sides of rectangle are .2 and .4

DPatrick 2012-02-08 20:34:45
Yes! The purple rectangle now has sides of length 0.4 (the continuation of the red square) and 0.2 (the continuation of the blue square).

DPatrick 2012-02-08 20:34:55
So its area is (0.4)(0.2) = 0.08.

Draco 2012-02-08 20:35:06
Therefore .2*.4*2=.16.

Iggy Iguana 2012-02-08 20:35:06
so 2*0.2*0.4=0.16 (C)

number.sense 2012-02-08 20:35:06
so 0.08 is the area of one rectangle, giving 0.16 (C)

DPatrick 2012-02-08 20:35:10
Hence our final answer is 2*0.08 = 0.16. Answer (C).

DPatrick 2012-02-08 20:35:31
This was pretty hard I thought.

DPatrick 2012-02-08 20:35:44
But I liked its pretty solution at the end.

DPatrick 2012-02-08 20:36:12
My editorial comment is that I thought 22-25 were harder than usual, but I thought that 15-20 (or so) were easier than usual, so it kind of balances out.

DPatrick 2012-02-08 20:36:28
Let's press on with #24:

DPatrick 2012-02-08 20:36:34

DPatrick 2012-02-08 20:36:45
Double yikes!

DPatrick 2012-02-08 20:37:03
Computing these numbers seems really really unpleasant.

DPatrick 2012-02-08 20:37:27
In fact, so unpleasant that it's probably unreasonable to expect that computing these numbers is part of the solution.

DPatrick 2012-02-08 20:37:50
So instead of computing, let's try to make some qualitative observations first. (Meaning, what can we observe without bogging down into computations?)

DPatrick 2012-02-08 20:38:07
Let's start simple: what do we know about the a's?

rdj5933mile5 2012-02-08 20:38:24
always less than 1

brokenfixer 2012-02-08 20:38:24
positive, less than 1

DPatrick 2012-02-08 20:38:34
First, we can observe that all the a's are positive numbers between 0 and 1. (This is easy to prove by induction: at each step we're taking a number between 0 and 1 and raising it to a power between 0 and 1.)

DPatrick 2012-02-08 20:39:11
What do we know about a_2? I don't want to compute it exactly of course, but in a vague sense what does it look like?

brokenfixer 2012-02-08 20:39:40
(1/5)^(1/5)

hyperbolictangent 2012-02-08 20:39:42
almost 1

Iggy Iguana 2012-02-08 20:39:44
.2^.2 is about .7 or .8?

DPatrick 2012-02-08 20:39:49
It's about 0.2^0.2. This is roughly the fifth root of 0.2. We expect this number to be close to 1, definitely well over 1/2.

DPatrick 2012-02-08 20:40:05
So what does this tell us about a_3?

number.sense 2012-02-08 20:40:38
a_3 is closer to 0.2

mentalgenius 2012-02-08 20:40:38
closer to 0.2

DPatrick 2012-02-08 20:40:45
a_3 is basically 0.2 raised to a number close to 1. Therefore it's still pretty close to 0.2.

DPatrick 2012-02-08 20:41:07
Here's a key question, since we're concerned about ordering the numbers (to create the b's later): is a_3 smaller or bigger than a_1, and why?

strategist 2012-02-08 20:41:46
bigger because a_3 is bigger and a_2 is pretty much 1

superpi83 2012-02-08 20:41:46
bigger because its exponent is less than 1.

brokenfixer 2012-02-08 20:41:46
a_3 has bigger base and smaller power = bigger

DPatrick 2012-02-08 20:42:18
Exactly, and that last comment by brokenfixer is the key: a_3 has a bigger base than a_1 and is raised to a smaller power than a_1 (think of a_1 as raised to the 1 power).

DPatrick 2012-02-08 20:42:49
Both of those facts mean that a_3 is larger than a_1. (Since these numbers are between 0 and 1, the smaller the power, the larger the number.)

DPatrick 2012-02-08 20:43:01
What does this tell us about a_4? Can we use a similar argument?

DPatrick 2012-02-08 20:43:34
Is a_4 bigger than a_2 or smaller than a_2?

superpi83 2012-02-08 20:44:13
less than a_2 because it has a smaller base and a bigger power.

mathlover3737 2012-02-08 20:44:13
it's a larger exponent, so it's smaller

ClassClown 2012-02-08 20:44:13
small base, big power smaller overall

DPatrick 2012-02-08 20:44:47
Right: a_4 is smaller than a_2 because it is being raised to a bigger power (namely, a_3) than a_2 was raised to (namely, a_1).

DPatrick 2012-02-08 20:45:02
What might we conjecture at this point?

saprmarks 2012-02-08 20:45:13
so are the a_(odds) getting larger as the a_(evens) getting smaller?

ClassClown 2012-02-08 20:45:17
it alternates smaller, larger, smaller, larger

DPatrick 2012-02-08 20:45:27
Exactly.

DPatrick 2012-02-08 20:45:37
The odd-position a's are increasing and the even-position a's are decreasing, and all the evens are bigger than all the odds.

DPatrick 2012-02-08 20:45:55

DPatrick 2012-02-08 20:46:19
Can we prove this?

DPatrick 2012-02-08 20:46:35
Well, we can. But it's gory. I'm going to leave it as a homework exercise. :)

DPatrick 2012-02-08 20:47:03
When actually taking the AMC's, I'd probably just convince myself that it's true but not rigorously prove it.

DPatrick 2012-02-08 20:47:41
So now we can go and try to finish the problem.

DPatrick 2012-02-08 20:48:15
Let's use what we've learned, and list the original sequence and the new ordered sequence:

DPatrick 2012-02-08 20:48:19

DPatrick 2012-02-08 20:48:43
Read the question carefully: the b's are in decreasing order. So a_2 (the biggest) comes first and a_1 (the smallest) comes last.

DPatrick 2012-02-08 20:48:52
What goes in the question marks in the middle?

strategist 2012-02-08 20:49:11
1005 1006 1007 lol

rdj5933mile5 2012-02-08 20:49:11
a_1005,a_1006,a_1007

osmosis92 2012-02-08 20:49:11
1005,1006,1007

DPatrick 2012-02-08 20:49:16
There are 1005 even-indexed terms, so the first "?" is 1005:

DPatrick 2012-02-08 20:49:19

strategist 2012-02-08 20:49:28
there will only be an intersection when b_k is coming down

DPatrick 2012-02-08 20:49:46
Aha, we can see that none of the terms match up in the first 1005, so the other match(es) are in the later terms.

Iggy Iguana 2012-02-08 20:50:01
and there will only be one such intersection

delta1 2012-02-08 20:50:03
only one match

DPatrick 2012-02-08 20:50:07
Right.

DPatrick 2012-02-08 20:50:39
We could use a little test-taking strategy and see that the only plausible answer choice is the choice in the "middle" of the range from 1005 to 2011, so it must be (C) 1341.

DPatrick 2012-02-08 20:50:55
But it's easy to verify it (and it's a good check that we didn't make a mistake).

DPatrick 2012-02-08 20:51:15

DPatrick 2012-02-08 20:51:29

DPatrick 2012-02-08 20:51:42
And it is (C) just like we suspected.

DPatrick 2012-02-08 20:52:08
Lastly, AMC 12A #25.

DPatrick 2012-02-08 20:52:13

DPatrick 2012-02-08 20:52:34
How can we approach this problem?

dlennon 2012-02-08 20:52:58
Graph f(x)?

DPatrick 2012-02-08 20:53:15
That's a good idea. We want to try to understand what this function f(x) is doing, and graphing it is one way to do that.

DPatrick 2012-02-08 20:53:32

DPatrick 2012-02-08 20:54:01

narto928 2012-02-08 20:54:14
it's a V shape

brokenfixer 2012-02-08 20:54:14
V shape

mentalgenius 2012-02-08 20:54:14
it is a V-shape

number.sense 2012-02-08 20:54:23
its an absolute value function that hits the x axis at 1/2

ClassClown 2012-02-08 20:54:23
It goes down until 1/2, then back p

DPatrick 2012-02-08 20:54:33

DPatrick 2012-02-08 20:54:46

Iggy Iguana 2012-02-08 20:54:58
it is repeating every increment of 1

DPatrick 2012-02-08 20:55:04
Right, so the graph looks like this:

DPatrick 2012-02-08 20:55:08

DPatrick 2012-02-08 20:55:26

DPatrick 2012-02-08 20:55:41
We also notice that the range of f is just [0,1]. It's 0 at the points in H and it's 1 at the integers.

brokenfixer 2012-02-08 20:55:50
what about xf(x)

number.sense 2012-02-08 20:55:50
now whats x times f(x)

DPatrick 2012-02-08 20:55:59
Yep, that's next. What does the graph of xf(x) look like?

number.sense 2012-02-08 20:56:24
its scaled

strategist 2012-02-08 20:56:25
like a staircase with valleys in it

brokenfixer 2012-02-08 20:56:25
looks like x at integers, 0 at half ints

narto928 2012-02-08 20:56:25
vertically stretched*

DPatrick 2012-02-08 20:56:55
Right. We're multiplying by x, so the "peaks", instead of being at 1, are now at x (for any positive integer x). It's very hard to graph it exactly but we can sketch it roughly:

DPatrick 2012-02-08 20:57:02

DPatrick 2012-02-08 20:57:32
I drew them dashed because they're definitely not straight lines (in fact the pieces are parabolas), but they rise up from the x-axis at a point in H, to the next peak at an integer, and then back down to the x-axis at the next point in H. The only points we really know for sure are the points on the x-axis at the half-integers (where x is in H and y=0), and the points (x,x) where x is an integer.

ClassClown 2012-02-08 20:57:52
how come the first one isn't a triangle

DPatrick 2012-02-08 20:58:24
At x=0 we're multiply by 0, so we get the point (0,0) on the graph of xf(x). Then we get another zero at x=1/2 since f(1/2) = 0, and from there it proceeds "normally" (if you can call this normal).

DPatrick 2012-02-08 20:58:54
I don't know how high the "peak" is (compared with the other peaks, which we do know accurately), so I drew it curved instead.

DPatrick 2012-02-08 20:59:16
Now what happens when we apply x again? What's the graph of f(xf(x)) look like?

number.sense 2012-02-08 20:59:31
you mean apply f again?

DPatrick 2012-02-08 20:59:37
Yeah, sorry, we apply **f** again.

DPatrick 2012-02-08 21:00:05
Again, remember that the integers evaluate to 1 when we apply f. So the point x=0 and all the points in H, since they are zero in xf(x), will evaluate to 1 in f(xf(x)).

number.sense 2012-02-08 21:00:15
wherever the function was a half integer, it goes down to 0, wheever it is 0, it goes up to 1

DPatrick 2012-02-08 21:00:38
Right, but moreover, wherever the function xf(x) was any integer (not just 0), the function f(xf(x)) will be 1.

DPatrick 2012-02-08 21:00:45
It might be clearer if I put some red and blue bands in my previous graph:

DPatrick 2012-02-08 21:00:48

DPatrick 2012-02-08 21:01:11
The points on the red bands (with y-values in the integers) will go to 1 when I apply f again.

brokenfixer 2012-02-08 21:01:14
red 1 blue 0

DPatrick 2012-02-08 21:01:17
Yep.

DPatrick 2012-02-08 21:01:29
The points on the blue bands (with y-values in H, the half-integers) will go to 0 when I apply f again.

DPatrick 2012-02-08 21:01:49
So we just count where we cross the red and blue bands.

DPatrick 2012-02-08 21:01:57
Again, the red points go to 1, and the blue points go to 0.

DPatrick 2012-02-08 21:02:07
We get one cycle from 1 to 0 to 1 in the interval [0,1]
We get three cycles from 1 to 0 to 1 to 0 to 1 to 0 to 1 in the interval [1,2]
We get five cycles from 1 to 0 to 1 to 0 to 1 to 0 to 1 to 0 to 1 to 0 to 1 in the interval [2,3]
And so on.

DPatrick 2012-02-08 21:02:23
...and it looks something like this (I'm going to widen it and only show up to x=3 to hopefully make it look a little clearer):

DPatrick 2012-02-08 21:02:26

DPatrick 2012-02-08 21:02:37
Again, they're not straight lines, and they're not this "regular".

DPatrick 2012-02-08 21:03:04
I'm only trying to capture the broad behavior: where it's 0 and where it's 1.

DPatrick 2012-02-08 21:03:20
This is only a qualitative picture of the function f(xf(x)), and that's good enough for the next step.

brokenfixer 2012-02-08 21:03:30
compare to y = x/n?

number.sense 2012-02-08 21:03:33
now we want this to intersect the line y = x/n 2012 times

DPatrick 2012-02-08 21:03:47
Exactly -- we now figure out when this equals x/n for some n.

DPatrick 2012-02-08 21:04:06
Since n is pretty big, x/n is a line through (0,0) with pretty low slope. I'll add it to the previous picture:

DPatrick 2012-02-08 21:04:10

DPatrick 2012-02-08 21:04:19
What do you notice?

KeepingItReal 2012-02-08 21:04:50
intersects 2 times per triangle

Iggy Iguana 2012-02-08 21:04:50
intersections are 2, 6, 10, 14, ...

brokenfixer 2012-02-08 21:04:50
2,6,10,14,...

314159Nerd 2012-02-08 21:04:50
It intersects 2 * x^2 times by the time it reaches an x value

DPatrick 2012-02-08 21:04:58
There are 2 solutions on the interval [0,1].
There are 6 solutions on the interval [1,2].
There are 10 solutions on the interval [2,3].

DPatrick 2012-02-08 21:05:10

DPatrick 2012-02-08 21:05:38
...since we have the sum 2(1+3+5+...+(2k-1)), which is 2k^2.

DPatrick 2012-02-08 21:06:10
Of course, we also know that when nx > 1, there can't be any more solutions.

narto928 2012-02-08 21:06:20
Set 2k^2 = 2012 and solve, round up for "at least"

alex31415 2012-02-08 21:06:20
find k such that 2k^2>=2012?

DPatrick 2012-02-08 21:06:41

DPatrick 2012-02-08 21:06:48
Those are weird symbols.... :(

DPatrick 2012-02-08 21:07:07
Just to be sure, we should check around our answer.

DPatrick 2012-02-08 21:07:12
If we use n=31, we'll only get 2(31)^2 = 1922 solutions, not enough.

DPatrick 2012-02-08 21:07:19
If we use n=32, we'll get 2(32)^2 = 2048 solutions, more than enough!

number.sense 2012-02-08 21:07:39
which gives us 32

alex31415 2012-02-08 21:07:39
the answer is 32?

DPatrick 2012-02-08 21:07:43
So n=32 is the smallest n giving us at least 2012 solutions. Answer (C).

Iggy Iguana 2012-02-08 21:07:47
there are a lot of (c) answers

DPatrick 2012-02-08 21:07:57
Yes, the answer to each of 22 through 25 on the AMC 12A was (C).

DPatrick 2012-02-08 21:08:29
Notice the strategy -- we didn't exactly determine what f(xf(x)) looks liked, and we certainly didn't compute it. We just got an accurate-enough qualitative picture to allow us to count the roots.

DPatrick 2012-02-08 21:08:58
We've run for 2 hours, but I will let you request **one** additional problem (say which test it is from), and I'll pick the one that gets the most vote and/or I like the most. :)

DPatrick 2012-02-08 21:10:00
Hang on, I'm tabulating the votes...

DPatrick 2012-02-08 21:10:40
Sadly, no one picked my favorite (which was #16 on the 12A), and the overwhelming winner was the "Paula the Painter" problem, so here goes:

DPatrick 2012-02-08 21:10:53

DPatrick 2012-02-08 21:11:09
oops, wrong problem!

DPatrick 2012-02-08 21:11:20

ss5188 2012-02-08 21:11:54
set up systems of equations

DPatrick 2012-02-08 21:12:13
The trick with this sort of "work" problem is setting up the right equations. From there the algebra is usually pretty easy.

Draco 2012-02-08 21:12:17
We can have the rates of the helpers be 1 variable.

DPatrick 2012-02-08 21:12:25
That simplifies things a little bit.

soccerfan 2012-02-08 21:12:29
3 equations, 3 variables

DPatrick 2012-02-08 21:12:36
Say that Paula paints p% of the house in 1 minute, and the helpers (together) paint h% of the house in 1 minute. Also say the lunch break is m minutes.

DPatrick 2012-02-08 21:12:51
We might as well use percentages for p and h since the given data in the problem is also in percentages.

ahaanomegas 2012-02-08 21:13:05
(p+h)(480-m) = 50

DPatrick 2012-02-08 21:13:14

DPatrick 2012-02-08 21:13:33

DPatrick 2012-02-08 21:13:43

DPatrick 2012-02-08 21:13:53
Those are the three equations, and what's left is to solve for m.

strategist 2012-02-08 21:14:31
after a bit of tedious algebra, it is trivialized

DPatrick 2012-02-08 21:14:55
That's kind of my opinion. I didn't actually work out the algebra from this point.

DPatrick 2012-02-08 21:15:09
I think that most of the difficulty is the setup.

DPatrick 2012-02-08 21:15:20

DPatrick 2012-02-08 21:15:55
We're now a bit over 2 hours, so I think I'll stop here with some closing remarks.

DPatrick 2012-02-08 21:16:13
First, I have an exciting (I hope) announcement!

DPatrick 2012-02-08 21:16:21
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.

DPatrick 2012-02-08 21:17:01
Second, as you probably know, the AMC 10B/12B is on Wednesday, February 22 (exactly two weeks from today). As we did today, we'll have our Math Jam the day after, Feb 23, at 7 PM Eastern.

DPatrick 2012-02-08 21:17:18
I believe that my colleague Jeremy Copeland will be leading the discussion on the 23rd.

DPatrick 2012-02-08 21:17:47
Next, a shameless plug for AoPS. :)

DPatrick 2012-02-08 21:17:51
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 myself. 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.l

DPatrick 2012-02-08 21:18:26
We will also have Math Jams for each of the two AIME contests in March -- check the website for specific dates and times.

DPatrick 2012-02-08 21:19:01
I will stick around for a little while to answer questions, but please: I will not speculate on the cutoff score for AIME qualification, so please don't ask me to.

DPatrick 2012-02-08 21:19:26
And a reminder: the transcript of our entire session will be posted on the website as soon as I wrap up and finish answering questions.

DPatrick 2012-02-08 21:19:35
That's all for tonight -- thanks for coming!

hrithikguy 2012-02-08 21:19:45
will this year's AIME seminar be different from last year's?

DPatrick 2012-02-08 21:20:11
Yes, it is completely different material than 2011 -- it cycles every 2 years (so if you took it in 2010 it's largely a repeat).

Klu2014 2012-02-08 21:20:42
So they are based on the same topics?

DPatrick 2012-02-08 21:21:05
It's the same four general topics (algebra, counting, geometry, and number theory), but the problems used to illustrate the material and techniques are all different in 2012 than from 2011.

dsj9999 2012-02-08 21:21:15
Do you know when we'll learn the cutoff though?

vliu 2012-02-08 21:21:15
when should the official scores for our AMC's come out?

DPatrick 2012-02-08 21:21:21
I don't know.

KeepingItReal 2012-02-08 21:21:27
Do you think this year's contest was harder than normal, or was it trivial as strategist said?

BOGTRO 2012-02-08 21:21:27
What do you think the relative difficulty of this year was?

strategist 2012-02-08 21:21:27
how hard was this compared to last years amc?

DPatrick 2012-02-08 21:21:51
This is a topic I'd prefer not to comment on, because "harder" for some people is "easier" for others. Also I haven't read all the problems. :)

Klu2014 2012-02-08 21:22:00
Do you write the AMC tests?

DPatrick 2012-02-08 21:22:19
No. I also didn't see the test until yesterday, just like you.

Flamewire 2012-02-08 21:22:28
What's the best way to prepare for the AMC12, by doing old tests? What types of topics should we learn?

DPatrick 2012-02-08 21:22:47
Taking old tests is a great way to practice, both for the material and for "test-taking" skills like pacing yourself, checking your work, and so on.

Burtleberry 2012-02-08 21:23:10
Do you think the AOPS Volume II is good preparation for AIME? Any other reading recommendations?

hrithikguy 2012-02-08 21:23:10
how do you get faster at doing computation on AMC problems?

distortedwalrus 2012-02-08 21:23:10
what should we do when we run out of old tests to take?

DPatrick 2012-02-08 21:23:40
You should ask these questions on the AMC Forum on our website. Other students will have better recommendations than I.

KeepingItReal 2012-02-08 21:24:12
If the AIME qualifying index for 12A is lower than normal, will the USAMO qualifying index be lower than normal?

DPatrick 2012-02-08 21:24:42
Administrative questions like this should be directed to the AMC themselves, or ask on the forum. I try not to get bogged down in AMC minutia.

breakeroice 2012-02-08 21:24:58
If I am a sophomore in high school, should I take the AMC10B or 12B if given the option to take either?

DPatrick 2012-02-08 21:25:33
You should search for this on the forum (or re-ask it): lots of people have lots of different opinions. My only short answer is "it depends" (on what your realistic goals are).

monarchb 2012-02-08 21:25:54
Do you get the official answers from MAA for the AMC10 or do you work them out?

DPatrick 2012-02-08 21:26:23
We got the official answers, the same as your school did. But I didn't look at them until after I had solved the problem, as a check of my work and to see if they had done anything more clever than I did.

DPatrick 2012-02-08 21:26:59
Fair warning: I am going to stop at :30 past, and you will be forcefully evicted from the classroom so I can generate the transcript. (Sorry.)

math-rules 2012-02-08 21:27:37
is this Q and A part recorded on the transcript of this math jam?

DPatrick 2012-02-08 21:27:39
It will be.

KeepingItReal 2012-02-08 21:28:08
wait, my school has the official answers?

mentalgenius 2012-02-08 21:28:09
how can I get a hold of the solutions for this year's AMC 10 A? (They usually hand out the solutions at the actual contest, but they didn't at school this time for some reason)

DPatrick 2012-02-08 21:28:22
Every school that administered the contest has a copy of the solutions.

DPatrick 2012-02-08 21:28:44
They cannot be posted online, as they are copyrighted by the AMC.

ahaanomegas 2012-02-08 21:28:55
It's called the "Solutions Pamphlet", in blue.

Bictor717 2012-02-08 21:29:01
The rules say that solutions are not to be given out until the answer sheets are mailed.

DPatrick 2012-02-08 21:29:10
Yes -- your school should have mailed them today.

breakeroice 2012-02-08 21:29:14
Is the reason why the Math Jam is today because you aren't allowed to talk about the test until the day after?

DPatrick 2012-02-08 21:29:16
Yes.

distortedwalrus 2012-02-08 21:29:24
what did you get on the AMC when you took it?

DPatrick 2012-02-08 21:29:30
My senior year, 150. :)

DPatrick 2012-02-08 21:29:41
That seems like a great place to stop. Good night!

2012 AMC 10/12 A Discussion

Facilitator: Dave Patrick