## 2012 AIME II Math Jam

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AoPS instructors discuss all 15 problems of the 2012 AIME 2.

#### Facilitator: Jeremy Copeland

copeland 2012-03-30 19:01:15
Welcome to the 2012 AIME II Math Jam!
copeland 2012-03-30 19:01:20
I'm Jeremy Copeland, and I'll be leading our discussion tonight.
copeland 2012-03-30 19:01:26
Before we get started I would like to take a moment to explain our virtual classroom to those who have not previously participated in a Math Jam or one of our online classes.
copeland 2012-03-30 19:01:35
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 2012-03-30 19:01:41
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 2012-03-30 19:01:51
There are a lot of students here! As I said, only a relatively small 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!!
copeland 2012-03-30 19:02:04
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the necessary 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 2012-03-30 19:02:21
We do have two teaching assistants with us tonight to help answer your questions: Carl Lian (CatalystOfNostalgia) and Kevin Hu (Aryth).
copeland 2012-03-30 19:02:28
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 2012-03-30 19:02:46
Please also remember that the purpose of this Math Jam is to work through the solutions to AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. Also on occasion we may stop to prove things that you wouldn't necessary need to prove while doing the contest. 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 acknowledged.
copeland 2012-03-30 19:03:06
At a couple points during the Math Jam, I might get thirsty or hungry, or my fingers will get tired, and I'll take a 1-2 minute break. Other than that, we will simply plow through all 15 problems.
Navigator 2012-03-30 19:03:18
oh good! copeland!
copeland 2012-03-30 19:03:20
I think the same thing every morning when I wake up!
DaChickenInc 2012-03-30 19:03:26
copeland 2012-03-30 19:03:28
I only rarely think that.
copeland 2012-03-30 19:03:36
Before we get started, I have a question: for those of you who took the test, what was your favorite question on the test.
copeland 2012-03-30 19:03:47
That's not a question.
copeland 2012-03-30 19:03:49
Before we get started, I have a question: for those of you who took the test, what was your favorite question on the test?
mathswimmer 2012-03-30 19:04:28
6
ABCDE 2012-03-30 19:04:28
14
alex31415 2012-03-30 19:04:28
#10! It was SOOOOOOOO easy!!!
EuclidGenius 2012-03-30 19:04:28
#16
MNL9082 2012-03-30 19:04:28
number 5: it was way too easy and made me feel smart :)
nackster12 2012-03-30 19:04:28
13
kmc927 2012-03-30 19:04:28
#1
ABCDE 2012-03-30 19:04:28
14
awesomemathlete 2012-03-30 19:04:28
10
googol.plex 2012-03-30 19:04:28
I liked number 3
deepankar 2012-03-30 19:04:28
#9
zinko1991 2012-03-30 19:04:28
#2
ptea 2012-03-30 19:04:28
14. So simple, yet difficult.
Voytek 2012-03-30 19:04:28
2 - Geometric series!
awesomemathlete 2012-03-30 19:04:28
number 10
lucylai 2012-03-30 19:04:28
7
Imsacred 2012-03-30 19:04:28
number 14
AndroidFusion 2012-03-30 19:04:28
13 because it was easy once you drew the diagram
centralbs 2012-03-30 19:04:28
12
lucylai 2012-03-30 19:04:28
7!!
kmc927 2012-03-30 19:04:28
and #5
awesomemathlete 2012-03-30 19:04:28
numer 10, it was easy, but elegant
natec 2012-03-30 19:04:28
The one with the money exchange between students
numbertheorist17 2012-03-30 19:04:28
14
shreyash 2012-03-30 19:04:28
12
va2010 2012-03-30 19:04:28
#8, YAY
ahaanomegas 2012-03-30 19:04:28
Hardest: #15
copeland 2012-03-30 19:04:42
And many of you voted for 14:
numbertheorist17 2012-03-30 19:04:43
14
numbertheorist17 2012-03-30 19:04:43
14
numbertheorist17 2012-03-30 19:04:43
14
numbertheorist17 2012-03-30 19:04:43
14
numbertheorist17 2012-03-30 19:04:43
14
numbertheorist17 2012-03-30 19:04:43
14
numbertheorist17 2012-03-30 19:04:43
14
numbertheorist17 2012-03-30 19:04:43
14
numbertheorist17 2012-03-30 19:04:43
14
numbertheorist17 2012-03-30 19:04:43
14
copeland 2012-03-30 19:04:55
Early and often, right?
MNL9082 2012-03-30 19:05:01
seems fishy too me
copeland 2012-03-30 19:05:03
Let's get started.
copeland 2012-03-30 19:05:07
1. Find the number of ordered pairs of positive integer solutions $(m,n)$ to the equation
$20m + 12n = 2012.$
copeland 2012-03-30 19:05:16
(Note that I'll always put the current problem statement up at the top of the window. You can resize the region at the top of the window by dragging the gray horizontal bar.)
copeland 2012-03-30 19:05:22
See anything right off the bat?
lazorpenguin27143 2012-03-30 19:05:41
divide the equation by 4
theone142857 2012-03-30 19:05:41
divide by 4
pleekey 2012-03-30 19:05:41
div by 4 first
Hydroxide 2012-03-30 19:05:41
divide both sides by 4
Flamewire 2012-03-30 19:05:41
Divide by 4
divide by 4
himym83 2012-03-30 19:05:41
divide by 4
pleekey 2012-03-30 19:05:41
everything divisible by 4
joshxiong 2012-03-30 19:05:41
Divide both sides by 4.
robinpark 2012-03-30 19:05:41
All coefficients are divisible by 4.
ahaanomegas 2012-03-30 19:05:41
We can divide each side of the equation by 4 to get: 5m + 3n = 503.
copeland 2012-03-30 19:05:46
All of the numbers are divisible by 4, so the solutions to this equation are exactly the solutions to $5m+3n=503.$
copeland 2012-03-30 19:05:49
What are the positive integer solutions to this equation?
AndroidFusion 2012-03-30 19:06:42
shreyash 2012-03-30 19:06:42
100,1
googol.plex 2012-03-30 19:06:42
the most obvious one is m=100, n=1
vic317yeh 2012-03-30 19:06:42
100,1
PERFECTION 2012-03-30 19:06:42
first, if we maximize the value of m, we have (100,1)
ptea 2012-03-30 19:06:42
well (100, 1) is kind of a freebie.
copeland 2012-03-30 19:06:47
We can start with the easiest solution, $5\cdot100+3\cdot1=503.$ This has to be the solution with smallest value for $n$.
lucylai 2012-03-30 19:07:13
(100, 1), (97, 6), ... (1, 166)
alex31415 2012-03-30 19:07:13
(100,1), (97,6),...,(1,166)
centralbs 2012-03-30 19:07:13
n=1, 6, ... 166
Hydroxide 2012-03-30 19:07:13
m=100, 97, ..., 1 and the corresponding n
ahaanomegas 2012-03-30 19:07:13
(m, n) = (100, 1) is a solution and we see that all positive n with units digit of 1 and 6 work. To keep m positive, n goes up to 166. So, the n that work are 1, 6, 11, 16, ..., 166.
shreyash 2012-03-30 19:07:13
100 - 3k, 1 + 5k for k from 0 to 33
va2010 2012-03-30 19:07:13
100, 1 97, 6 I really see a pattern all
Hydroxide 2012-03-30 19:07:13
keep going down by 3
Flamewire 2012-03-30 19:07:13
Decrement m by 3
pleekey 2012-03-30 19:07:13
every time m goes down three, n goes up five
copeland 2012-03-30 19:07:17
From here, every time we increase $n$ we increase $5m+3n$ by a multiple of 3. Therefore to get to another solution we need to decrease $5m$ by a multiple of 3 as well.
copeland 2012-03-30 19:07:20
Therefore we must always decrease $m$ by some multiple of 3 to get from $(100,1)$ to another solution. Furthermore, subtracting 3 from $m$ and adding 5 to $n$ will always give us another solution. All solutions are below:
copeland 2012-03-30 19:07:30
copeland 2012-03-30 19:07:31
How many pairs are on that list?
ahaanomegas 2012-03-30 19:07:57
mjseaman1 2012-03-30 19:07:57
(1, 100) , (6, 97) ..... (165, 1) are the solutions. There are 34 in all.
alex31415 2012-03-30 19:07:57
34
MNL9082 2012-03-30 19:07:57
34
claudiafeng 2012-03-30 19:07:57
34
ABCDE 2012-03-30 19:07:57
034
Seedleaf 2012-03-30 19:07:57
34 pairs! so 034
silvernight 2012-03-30 19:07:57
34
Flamewire 2012-03-30 19:07:57
99/3 = 33 + 1 = 34
shreyash 2012-03-30 19:07:57
34!
ashgabat 2012-03-30 19:07:57
34 pairs.
kangchangood 2012-03-30 19:07:57
34
sonchecky 2012-03-30 19:07:57
34
copeland 2012-03-30 19:08:00
We can count the number of copies of $m$: The list $1,4,7,10,\ldots,97,100$ has the same number of elements as the list we get by adding 2. Therefore we can count the number of elements on the list $3,6,9,12,\ldots,99,102$.
copeland 2012-03-30 19:08:06
If we divide all the terms on this list by 3 we get the list $1,2,3,4,\ldots34$. There are $\boxed{034}$ solutions to our equation.
copeland 2012-03-30 19:08:32
Alrighty! Now we're rolling.
copeland 2012-03-30 19:08:33
More math!
copeland 2012-03-30 19:09:01
2. Two geometric sequences $a_1$, $a_2$, $a_3, \dots$ and $b_1$, $b_2$, $b_3, \dots$ have the same common ratio, with $a_1 = 27$, $b_1 = 99$, and $a_{15} = b_{11}$. Find $a_9$.
copeland 2012-03-30 19:09:17
Formula time. What's the formula for the terms of a geometric sequence?
lazorpenguin27143 2012-03-30 19:09:58
by definition of geometric sequences write each term as a, ar, ar^2, etc
shreyash 2012-03-30 19:09:58
a * r^(n-1)
Voytek 2012-03-30 19:09:58
an=a1*r^n-1
tc1729 2012-03-30 19:09:58
a(1-r^n)/(1-r)
zinko1991 2012-03-30 19:09:58
a, ar, ar^2, ar^3, ...
deepankar 2012-03-30 19:09:58
Seedleaf 2012-03-30 19:09:58
a, ar, ar^2, ar^3....... when a is starting term and r is the ratio
pl210741 2012-03-30 19:09:58
$a_n=(First term)(Common ration)^{n-1}$
copeland 2012-03-30 19:10:09
We have two suggestions. I don't like using the -1, personally.
copeland 2012-03-30 19:10:19
(And I'm in charge, so. . . .)
copeland 2012-03-30 19:10:21
copeland 2012-03-30 19:10:27
How can we use the constraints $a_1=27$ and $b_1=99$?
Seedleaf 2012-03-30 19:12:14
27 and 99 can be the starting terms
kangchangood 2012-03-30 19:12:14
simply plug in 1 for n
sonchecky 2012-03-30 19:12:14
$a/b=a_1/b_1=27/99=3/11$
lily6481 2012-03-30 19:12:14
plug in values
Voytek 2012-03-30 19:12:14
27=ar and 99=br
shreyash 2012-03-30 19:12:14
a1 = a * r
shreyash 2012-03-30 19:12:14
b1 = b * r
theone142857 2012-03-30 19:12:14
get a and b in terms of r
AbsoluteFriend 2012-03-30 19:12:14
I think 27/r=a and 99/r=b
copeland 2012-03-30 19:12:24
My reflex is to divide these we get $\frac{27}{99}=\frac{ar^1}{br^1}=\frac ab.$
copeland 2012-03-30 19:12:27
So we have:
copeland 2012-03-30 19:12:30
$\frac3{11}=\frac ab$.
copeland 2012-03-30 19:12:37
Now how do we use $a_{15}=b_{11}$?
ahaanomegas 2012-03-30 19:13:25
So: r^4 = 11/3.
nmandi7 2012-03-30 19:13:25
r^4 = 99/27
nmandi7 2012-03-30 19:13:25
a(r^4) = b
googol.plex 2012-03-30 19:13:25
3/11*r^4=1
lucylai 2012-03-30 19:13:25
r^4=11/3
Imsacred 2012-03-30 19:13:25
since a15 = b11, a5 = b1
substitute n=15 into the equation for 'a' and n=11 into the equation for 'b'
ahaanomegas 2012-03-30 19:13:25
Set the formulas for a_15 and b_11 equal using the expressions in terms of r. Then, we solve for r^4.
va2010 2012-03-30 19:13:31
a(r^15)=b(r^11)
copeland 2012-03-30 19:13:32
We divide them again. If two numbers are equal, then their ratio is 1, so $1=\frac{a_{15}}{b_{11}}=\frac{ar^{15}}{br^{11}}=\frac abr^4.$
copeland 2012-03-30 19:13:36
$r^4=\frac ba=\frac{11}{3}.$
copeland 2012-03-30 19:13:38
And what is $a_9$?
alex31415 2012-03-30 19:14:40
363
kmc927 2012-03-30 19:14:40
363
cerberus88 2012-03-30 19:14:40
363
TigerSneak1 2012-03-30 19:14:40
33
joshxiong 2012-03-30 19:14:40
27*(11/3)*(11/3)=363
silvernight 2012-03-30 19:14:40
363
numbertheorist17 2012-03-30 19:14:40
27*121/9=363
TigerSneak1 2012-03-30 19:14:40
363
cerberus88 2012-03-30 19:14:40
(r^4)^2*a=(11/3)^2*27=363
copeland 2012-03-30 19:14:45
We know that $a_1=27$, so $a_9=ar^9=(ar)(r^8)=a_1(r^4)^2=27\left(\frac{11}{3}\right)^2=363.$
copeland 2012-03-30 19:14:52
$\boxed{363}$ is the answer.
copeland 2012-03-30 19:15:12
3. At a certain university, the division of mathematical sciences consists of the departments of mathematics, statistics, and computer science. There are two male and two female professors in each department. A committee of six professors is to contain three men and three women and must also contain two professors from each of the three departments. Find the number of possible committees that can be formed subject to these requirements.
copeland 2012-03-30 19:15:26
OK, counting problem. We're trying to build a committee, so this looks like a nice place for a constructive approach.
pl210741 2012-03-30 19:15:46
Casework!
Hydroxide 2012-03-30 19:15:46
ughh casework
ABCDE 2012-03-30 19:15:46
Casework.
MNL9082 2012-03-30 19:15:46
casework time
ytao 2012-03-30 19:15:46
Casework!
lucylai 2012-03-30 19:15:46
or casework!!!
deepankar 2012-03-30 19:15:46
casework
copeland 2012-03-30 19:15:48
Let's think just about the genders of the committee members. The committee could, for example, look like MW MW MW or it could be MM MW WW or many other things.
copeland 2012-03-30 19:15:54
The committee we want to construct has 3 men and 3 women. How do we count the ways to pick 3 of each from the 3 departments where 2 professors come from each department?
mathswimmer 2012-03-30 19:17:05
from each department, you either take two of the same gender, or one of each
joshxiong 2012-03-30 19:17:05
First case: MW from all 3 departments.
googol.plex 2012-03-30 19:17:05
There are 2 cases, one where all of the departments are male and the other are female and one where 1 has 2 male, 1 has 1 of each, and 1 has 2 females
lazorpenguin27143 2012-03-30 19:17:05
One man and one woman from each department is on the committee and One department has two men, another has two women, and the last has one of each.
ABCDE 2012-03-30 19:17:05
MM FF MF and MF MF MF
ahaanomegas 2012-03-30 19:17:05
Case 1: One male and one female from each subject.
19oshawott98 2012-03-30 19:17:05
mm,ff,mf or mfmfmf
copeland 2012-03-30 19:17:09
We have cases. Either all of the women come from the different departments, or 2 women come from a single department.
copeland 2012-03-30 19:17:12
How many ways can we pick one woman and one man from each department?
copeland 2012-03-30 19:17:27
(total.)
centralbs 2012-03-30 19:17:52
4^3=64
lazorpenguin27143 2012-03-30 19:17:52
4^3 = 64
mjseaman1 2012-03-30 19:17:52
2^6=64 ways.
EuclidGenius 2012-03-30 19:17:52
4^3
nmandi7 2012-03-30 19:17:52
4^3=64
va2010 2012-03-30 19:17:52
2*2*2*2*2*2=64
joshxiong 2012-03-30 19:17:52
2^6=64.
EuclidGenius 2012-03-30 19:17:52
$4^3=64$ total ways
sinnoel 2012-03-30 19:17:52
64
nackster12 2012-03-30 19:17:52
2^6=64
zinko1991 2012-03-30 19:17:52
2^6=64
copeland 2012-03-30 19:17:57
Each department has 2 choices for each of the male and female representatives it contributes. So each department has 4 possibilities. Since there are 3 departments, there are $4^3=64$ ways to pick one man and one woman from each department.
copeland 2012-03-30 19:18:04
How many total committees have 2 women from the same department?
lazorpenguin27143 2012-03-30 19:18:58
6*4 = 24
kmc927 2012-03-30 19:18:58
6*4=24
ABCDE 2012-03-30 19:18:58
6*2*2=24
va2010 2012-03-30 19:18:58
3*2*4=24
nmandi7 2012-03-30 19:18:58
24
3 ways to pick department with all women, 2 with all men, and 2*2 ways to pick the last two. 3*2*2*2=24
nackster12 2012-03-30 19:18:58
6*(1*4*1)=24
awesomemathlete 2012-03-30 19:18:58
24
joshxiong 2012-03-30 19:18:58
3*2*2*2=24
copeland 2012-03-30 19:19:03
The only way to distribute the committee members with 2 women from the same department is by having one department contribute 2 women, one department contribute 1 woman and one department contribute 0 women.
copeland 2012-03-30 19:19:13
There are 3 choices for which department contributes the 2 women and there are 2 choices for which other department contributes the other woman. After these choices are made, the 2 women from the same department and the 2 men from the other department are determined. This leaves only 4 ways to choose the man and woman from the department that contributes one of each. There are $3\cdot2\cdot4=24$ total committees that can be formed in this way.
copeland 2012-03-30 19:19:17
And the total?
sonchecky 2012-03-30 19:19:53
88
alex31415 2012-03-30 19:19:53
64+24=88
Flamewire 2012-03-30 19:19:53
88
Seedleaf 2012-03-30 19:19:53
64+24 = 88
Imsacred 2012-03-30 19:19:53
24 + 64 = 088
kmc927 2012-03-30 19:19:53
88
64+24= $\boxed{088}$
CantonMathGuy 2012-03-30 19:19:53
88
PERFECTION 2012-03-30 19:19:53
64+24=88 total possible comittees
jiyer99 2012-03-30 19:19:53
88
Mary_Posa 2012-03-30 19:19:53
88
ABCDE 2012-03-30 19:19:53
64+24=088
Hydroxide 2012-03-30 19:19:53
24+64=88
sarvottam 2012-03-30 19:19:53
88
copeland 2012-03-30 19:19:57
There are $64+24=\boxed{088}$ total ways to form such a committee.
copeland 2012-03-30 19:20:05
3 down. . .
ABCDE 2012-03-30 19:20:36
12 to go!
MNL9082 2012-03-30 19:20:36
12 to go
EuclidGenius 2012-03-30 19:20:36
12 left
awesomemathlete 2012-03-30 19:20:36
12 to go
theone142857 2012-03-30 19:20:36
12 to go
copeland 2012-03-30 19:20:38
Yep. You don't actually get any points for that, though.
copeland 2012-03-30 19:20:39
4. Ana, Bob, and Cao bike at constant rates of 8.6 meters per second, 6.2 meters per second, and 5 meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point $D$ on the south edge of the field. Cao arrives at point $D$ at the same time that Ana and Bob arrive at $D$ for the first time. The ratio of the field's length to the field's width to the distance from the point $D$ to the southeast corner of the field can be represented as $p:q:r$, where $p$, $q$, and $r$ are positive integers with $p$ and $q$ relatively prime. Find $p + q + r$.
copeland 2012-03-30 19:20:52
Yuck, yuck, yuck.
copeland 2012-03-30 19:20:56
That is a LOT of words.
copeland 2012-03-30 19:21:05
What should we do?
WAGE 2012-03-30 19:21:33
Diagram.
deepankar 2012-03-30 19:21:33
draw a picture
ahaanomegas 2012-03-30 19:21:33
Draw a diagram!
joshxiong 2012-03-30 19:21:33
Draw a diagram.
Imsacred 2012-03-30 19:21:33
diagram!
shreyash 2012-03-30 19:21:33
make a diagram
Mattchu386 2012-03-30 19:21:33
Draw a diagram!
TigerSneak1 2012-03-30 19:21:33
Draw a pretty picture
lucylai 2012-03-30 19:21:33
diagram
lily6481 2012-03-30 19:21:33
draw a picture
lucylai 2012-03-30 19:21:33
picture
ptea 2012-03-30 19:21:33
draw a picture!
mathnerd101 2012-03-30 19:21:33
draw a picture
copeland 2012-03-30 19:21:37
Draw a diagram!
AbsoluteFriend 2012-03-30 19:21:44
EuclidGenius 2012-03-30 19:21:44
copeland 2012-03-30 19:21:47
Also, don't forget to read the problem carefully. Since we want the the longer side to run due west, we should draw our diagram to reflect that.
copeland 2012-03-30 19:21:55
copeland 2012-03-30 19:22:05
The each start biking from the top right corner in the indicated directions. They meet at the point $D$.
copeland 2012-03-30 19:22:16
We should also label the distances we want. I have no idea what the words "length" and "width" mean, but we want their sum anyway, so it doesn't matter which one is which. I labeled my diagram by letting the length be the up-and-down direction:
copeland 2012-03-30 19:22:35
copeland 2012-03-30 19:22:46
In terms of $p$, $q$, and $r$, how far did Bob bike?
lucylai 2012-03-30 19:23:15
p+r
nackster12 2012-03-30 19:23:15
p+r
Flamewire 2012-03-30 19:23:15
p + r
centralbs 2012-03-30 19:23:15
p+r
zinko1991 2012-03-30 19:23:15
p+r
proglote 2012-03-30 19:23:15
p + r
King6997 2012-03-30 19:23:15
r+p
ytao 2012-03-30 19:23:15
p+r
claudiafeng 2012-03-30 19:23:15
p+r
lily6481 2012-03-30 19:23:15
p+r
copeland 2012-03-30 19:23:16
Bob biked down the right side and over to $D$. His path has length $p+r$.
copeland 2012-03-30 19:23:21
How far did Ana bike in terms of $p$, $q$, and $r$?
sonchecky 2012-03-30 19:24:13
$2q+p-r$
lucylai 2012-03-30 19:24:13
p+2q-r
19oshawott98 2012-03-30 19:24:13
q+p+q-r
AbsoluteFriend 2012-03-30 19:24:13
r+p
googol.plex 2012-03-30 19:24:13
p+2q-r
p+q+(q-r)=2q+p-r
Flamewire 2012-03-30 19:24:13
2q + p -r
ABCDE 2012-03-30 19:24:13
p+2q-r
copeland 2012-03-30 19:24:15
Ana biked across the top ($q$), then down the side ($p$), and then across the segment of length $q-r$ from the southwest corner to $D$. So Ana biked a distance of $q+p+(q-r)=p+2q-r.$
copeland 2012-03-30 19:24:18
As a check we note that Ana and Bob combined biked $(p+r)+(p+2q-r)=2p+2q$, which is the entire perimeter of the field.
copeland 2012-03-30 19:24:23
How far did Cao bike?
claudiafeng 2012-03-30 19:24:53
$\sqrt{p^2+r^2}$
ahaanomegas 2012-03-30 19:24:53
alex31415 2012-03-30 19:24:53
sqrt(p^2+r^2)
himym83 2012-03-30 19:24:53
sqrt(p^2+r^2)
deepankar 2012-03-30 19:24:53
Seedleaf 2012-03-30 19:24:53
sqrt(p^2+r^2)
joshxiong 2012-03-30 19:24:53
$\sqrt{p^2+r^2}$
PERFECTION 2012-03-30 19:24:53
sqrt(r^2+p^2)
sinnoel 2012-03-30 19:24:53
sqrt(p^2+r^2)
sonchecky 2012-03-30 19:24:57
$\sqrt{q^2+r^2} by the Pythagorean Theorem admin25 2012-03-30 19:24:57$ \sqrt{p^2+r^2} $, by pythagorean theorem (blech, radicals) copeland 2012-03-30 19:25:01 By the Pythagorean Theorem, Cao biked a total distance of$\sqrt{p^2+r^2}$. copeland 2012-03-30 19:25:05 So the ratio of the distances that they traveled is $a:b:c=p+2q-r:p+r:\sqrt{p^2+q^2}.$ copeland 2012-03-30 19:25:08 Do we know anything else about this ratio? copeland 2012-03-30 19:25:59 Typo up there: copeland 2012-03-30 19:26:03 So the ratio of the distances that they traveled is $a:b:c=p+2q-r:p+r:\sqrt{p^2+r^2}.$ ABCDE 2012-03-30 19:26:08 8.6:6.2:5 TigerSneak1 2012-03-30 19:26:08 It's also 8.6 to 6.2 to 5 googol.plex 2012-03-30 19:26:08 it should be 8.6:6.2:5 mathswimmer 2012-03-30 19:26:08 same as velocity centralbs 2012-03-30 19:26:08 since d=rt and the time is equal, d/r is constant MNL9082 2012-03-30 19:26:08 it equals the ratio of their speeds brokenfixer 2012-03-30 19:26:08 8.6:6.2:5 CantonMathGuy 2012-03-30 19:26:08 use speeds copeland 2012-03-30 19:26:13 It's given in the problem! We know the ratios of their speeds, and if they travel for the same amount of time, that will also be the ratio of the distances that they travel. $a:b:c=8.6:6.2:5.$ copeland 2012-03-30 19:26:17 I'm compelled to clean that up by multiplying by 5. copeland 2012-03-30 19:26:22 $a:b:c=43:31:25.$ copeland 2012-03-30 19:26:33 Now what? sparkle123 2012-03-30 19:27:12 Set r=1. AbsoluteFriend 2012-03-30 19:27:17 Equate them. Flamewire 2012-03-30 19:27:17 Set up a system of equations googol.plex 2012-03-30 19:27:17 system of equations shreyash 2012-03-30 19:27:17 make equations ptea 2012-03-30 19:27:17 solve that mess kmc927 2012-03-30 19:27:17 solve for p q & r nackster12 2012-03-30 19:27:17 set the ratios equal to each other and solve (ew) flyrain 2012-03-30 19:27:17 use these in p, q, and r, so we can system of equations copeland 2012-03-30 19:27:22 copeland 2012-03-30 19:27:24 Hmm. This makes things harder for us than they need to be. What should we do instead? alex31415 2012-03-30 19:27:51 Can we assume the width is 1 and multiply by an appropriate scale factor at the end? ahaanomegas 2012-03-30 19:27:51 On the actual test, I called the time t and, not too long after, I assumed WLOG that t = 10. shreyash 2012-03-30 19:27:51 set k = 1 Flamewire 2012-03-30 19:27:51 WLOG assume k = 1 nackster12 2012-03-30 19:27:51 let k=1 robinpark 2012-03-30 19:27:51 let k = 1 nmandi7 2012-03-30 19:27:51 k=1 copeland 2012-03-30 19:27:55 We don't really need to worry about units at this point. It's actually fair just to throw the$kaway. After all we want a ratio for our final answer and not actual values. copeland 2012-03-30 19:28:00 \begin{align*} p+2q-r&=43,\\ p+r&=31,\\ \sqrt{p^2+r^2}&=25. \end{align*} copeland 2012-03-30 19:28:06 Three equations in three unknowns so we should be able to solve. Where should we start? ytao 2012-03-30 19:28:47 Square the latter. Flamewire 2012-03-30 19:28:47 Square the third equation CantonMathGuy 2012-03-30 19:28:47 square the third equation lazorpenguin27143 2012-03-30 19:28:47 square the last equation zinko1991 2012-03-30 19:28:47 3rd equation Hydroxide 2012-03-30 19:28:47 square last equation Flamewire 2012-03-30 19:28:47 Square the third equation, set p^2 + r^2 = (p+r)^2 - 2pr = 25 sinnoel 2012-03-30 19:28:47 2nd and 3rd equation MNL9082 2012-03-30 19:28:47 use the bottom two equations copeland 2012-03-30 19:28:52 Since the last two equations only deal withp$and$r, we should start with those. What's the first step in dealing with \begin{align*} p+r&=31,\\ \sqrt{p^2+r^2}&=25? \end{align*} copeland 2012-03-30 19:28:56 Let's square the bottom equation:$p^2+r^2=625.$ copeland 2012-03-30 19:29:01 What else? ABCDE 2012-03-30 19:29:43 square the first one and subtrack brokenfixer 2012-03-30 19:29:43 square the top equation lazorpenguin27143 2012-03-30 19:29:43 square p+r = 31 centralbs 2012-03-30 19:29:43 square the top equation and subtract DF2222 2012-03-30 19:29:43 square the top equation robinpark 2012-03-30 19:29:43 Squarep+r$then subtract$p^2 + r^2Seedleaf 2012-03-30 19:29:43 square the top one to create square values lily6481 2012-03-30 19:29:43 square p+r pl210741 2012-03-30 19:29:45 square the first equation copeland 2012-03-30 19:29:48 This is quadratic, so it might help to make the other equation quadratic, too: copeland 2012-03-30 19:29:53 \begin{align*} p^2+2pr+r^2&=961,\\ p^2+r^2&=625. \end{align*} copeland 2012-03-30 19:29:57 So we could subtract those to get2pr$, and that might be helpful, but what else could we get that would be more helpful? brokenfixer 2012-03-30 19:30:42 (p-r)^2 AbsoluteFriend 2012-03-30 19:30:42 A difference of two squares. admin25 2012-03-30 19:30:42$ p-r $, so we can substitute that into the first equation! copeland 2012-03-30 19:30:44 Good! copeland 2012-03-30 19:30:46 Wait, how? brokenfixer 2012-03-30 19:31:54 625 - (961-625) proglote 2012-03-30 19:31:54 get 2pr first sonchecky 2012-03-30 19:31:54 Subtract twice the second equation minus the first. admin25 2012-03-30 19:31:54 subtract$ 4pr $from$ (p+r)^2 $to get$ (p-r)^2 $nackster12 2012-03-30 19:31:54 find 2pr then subtract 2pr from the bottom equation Imsacred 2012-03-30 19:31:54 find 2pr, then subtract 4pr from the first equation? alex31415 2012-03-30 19:31:54 (p-r)^2=289 copeland 2012-03-30 19:31:58 We notice that $2(p^2+r^2)-(p^2+2pr+r^2)=p^2-2pr+r^2=(p-r)^2.$ That should be useful. copeland 2012-03-30 19:32:05 Therefore$(p-r)^2=2(p^2+r^2)-(p^2+2pr+r^2)=2\cdot625-961=289=17^2.$ copeland 2012-03-30 19:32:12 So what is$p-r$? wmcho1007 2012-03-30 19:33:08 p-r=17 flyrain 2012-03-30 19:33:08 17 trident 2012-03-30 19:33:08 17 pl210741 2012-03-30 19:33:08 17 Hydroxide 2012-03-30 19:33:08 17 kmc927 2012-03-30 19:33:08 17 sonchecky 2012-03-30 19:33:08 17 lazorpenguin27143 2012-03-30 19:33:08 17 sinnoel 2012-03-30 19:33:08 17 matholympiad25 2012-03-30 19:33:08 17 ashgabat 2012-03-30 19:33:08 17. googol.plex 2012-03-30 19:33:08 17, but we know p+r is 31 so we have r! alex31415 2012-03-30 19:33:08 p-r=17, p+r=31, solve! FunnyBounny 2012-03-30 19:33:08 17 ytao 2012-03-30 19:33:08$\sqrt{17^2}=17$TigerSneak1 2012-03-30 19:33:08 17 EuclidGenius 2012-03-30 19:33:08 17 sarvottam 2012-03-30 19:33:08 17 bunniesrcute 2012-03-30 19:33:08 17 Leafpelt 2012-03-30 19:33:08 17 shreyash 2012-03-30 19:33:08 17 EuclidGenius 2012-03-30 19:33:08 17 zinko1991 2012-03-30 19:33:08 17 ahaanomegas 2012-03-30 19:33:08 17 iamthygod!!!!!! 2012-03-30 19:33:08 the square root of 17^2, which is 17 copeland 2012-03-30 19:33:10 No! copeland 2012-03-30 19:33:15 Hrm. copeland 2012-03-30 19:33:36 This is why I missed this problem, too. :) nackster12 2012-03-30 19:33:53 17 or -17 DF2222 2012-03-30 19:33:53 plus or minus 17 MNL9082 2012-03-30 19:33:53 plus or minus 17? nackster12 2012-03-30 19:33:53 could be negative matholympiad25 2012-03-30 19:33:53 COULD IT BE -17? kmc927 2012-03-30 19:33:53 or -17? dantx5 2012-03-30 19:33:53 +/- 17? wmcho1007 2012-03-30 19:33:53 or maybe -17? waveahu 2012-03-30 19:33:53 plus or minus 17? shreyash 2012-03-30 19:33:53 it can also be -17 matholympiad25 2012-03-30 19:33:53 Couldn't it also be -17? mathnerd101 2012-03-30 19:33:53 it could be -17 too copeland 2012-03-30 19:34:00 Careful! This is the trickiest part of this problem. We know that$p-r=\pm17$, but we don't yet know the sign. copeland 2012-03-30 19:34:04 What are$p$and$r$? alex31415 2012-03-30 19:34:56 24 and 7, respectively matholympiad25 2012-03-30 19:34:56 if p-r = -17, p+r = 31, p = 7, r = 24 TigerSneak1 2012-03-30 19:34:56 one is 24 and the other 7, but we don't know which. wmcho1007 2012-03-30 19:34:56 7 and 24? MNL9082 2012-03-30 19:34:56 24 and 7 googol.plex 2012-03-30 19:34:56 well r is either 24 or 7 nmandi7 2012-03-30 19:34:57 24 and 7 copeland 2012-03-30 19:35:01 If the sum of two numbers is 31 and the difference is 17, then the two numbers must be 24 and 7. We just don't know the order yet. copeland 2012-03-30 19:35:08 If$p=7$and$r=24$then what is$q$? lazorpenguin27143 2012-03-30 19:35:38 30 lucylai 2012-03-30 19:35:38 30 Hydroxide 2012-03-30 19:35:38 30 ytao 2012-03-30 19:35:38 30 Seedleaf 2012-03-30 19:35:38 q = 30 Imsacred 2012-03-30 19:35:38 30 zinko1991 2012-03-30 19:35:38 30 kmc927 2012-03-30 19:35:38 30 admin25 2012-03-30 19:35:38$ 60/3=30 $copeland 2012-03-30 19:35:42 If$p=7$and$r=24$then $q=\frac{43+r-p}2=\frac{43+17}2=30.$ copeland 2012-03-30 19:35:51 If$p=24$and$r=7$then what is$q$? Fire-Sword 2012-03-30 19:36:41 13 ptes77 2012-03-30 19:36:41 13 Seedleaf 2012-03-30 19:36:41 q = 13 lucylai 2012-03-30 19:36:41 13 joshxiong 2012-03-30 19:36:41 13 mathswimmer 2012-03-30 19:36:41 13 iamthygod!!!!!! 2012-03-30 19:36:41 13 copeland 2012-03-30 19:36:47 If$p=24$and$r=7$then $q=\frac{43+r-p}2=\frac{43-17}2=13.$ copeland 2012-03-30 19:36:50 Hmm. Are there two solutions? nackster12 2012-03-30 19:37:29 13 but q cant be less than p according to the problem Flamewire 2012-03-30 19:37:29 Which can't work because q > p GeorgiaTechMan 2012-03-30 19:37:29 but p>q Mary_Posa 2012-03-30 19:37:29 no, 13 is less than 24 himym83 2012-03-30 19:37:29 no, we are told that q>p sonchecky 2012-03-30 19:37:29 No - q must be greater than p because the field's longer side runs west. shreyash 2012-03-30 19:37:29 q has to greater than p therefore q is 30, p = 7, r = 24 sarvottam 2012-03-30 19:37:29 no m1sterzer0 2012-03-30 19:37:29 long side of the rectangles runs east to west copeland 2012-03-30 19:37:33 No! We also assumed that$p<q$. Since 24 is greater than 13, we cannot have$p=24$and$q=13$. Therefore$p:q:r=7:30:24$. copeland 2012-03-30 19:37:36 What is the answer? EuclidGenius 2012-03-30 19:38:07 061 flyrain 2012-03-30 19:38:07 61 proglote 2012-03-30 19:38:07 061 claudiafeng 2012-03-30 19:38:07 61 ABCDE 2012-03-30 19:38:07 7+30+24=061 trident 2012-03-30 19:38:07 7+30+24 = 061 ytao 2012-03-30 19:38:07 30+24+7=061 CantonMathGuy 2012-03-30 19:38:07 61 hatchguy 2012-03-30 19:38:07 61 AbsoluteFriend 2012-03-30 19:38:07 61 ptes77 2012-03-30 19:38:07 61 copeland 2012-03-30 19:38:11 Since these integers are relatively prime, the answer is the sum:$p+q+r=7+30+24=\boxed{061}$. copeland 2012-03-30 19:38:26 5. In the accompanying figure, the outer square$S$has side length 40. A second square$S'$of side length 15 is constructed inside$S$with the same center as$S$and with sides parallel to those of$S$. From each midpoint of a side of$S$, segments are drawn to the two closest vertices of$S'$. The result is a four-pointed starlike figure inscribed in$S$. The star figure is cut out and then folded to form a pyramid with base$S'$. Find the volume of this pyramid. copeland 2012-03-30 19:38:34 copeland 2012-03-30 19:38:41 Hope you guys brought your scissors. copeland 2012-03-30 19:39:03 I like building these things out of cardboard and filling them with pudding. mathnerd101 2012-03-30 19:39:23 do we have to cut? copeland 2012-03-30 19:39:24 Yes. Safety scissors for you, though! copeland 2012-03-30 19:39:28 What's the easiest formula for the volume of a pyramid? Flamewire 2012-03-30 19:39:55 Bh/3 alex31415 2012-03-30 19:39:55 1/3*b*h googol.plex 2012-03-30 19:39:55 b*h/3 matholympiad25 2012-03-30 19:39:55 1/3 * base area * height Voytek 2012-03-30 19:39:55 1/3 * base * height lily6481 2012-03-30 19:39:55 1/3bh Past 2012-03-30 19:39:55 bh/3 coldsummer 2012-03-30 19:39:55 1/3bh TeamJacob 2012-03-30 19:39:55 bh/3 copeland 2012-03-30 19:40:00 The volume is$\frac{Bh}3$, where$B$is the area of the base, and$h$is the height. copeland 2012-03-30 19:40:06 Do we know either of these values? copeland 2012-03-30 19:40:24 (Wait, I'm being told that pudding is not allowed on the AIME.) wmcho1007 2012-03-30 19:40:41 we have the base, which is 15*15=225 brokenfixer 2012-03-30 19:40:41 225 admin25 2012-03-30 19:40:41 Area of base=$ 15^2=225 $centralbs 2012-03-30 19:40:41 B=225 deepankar 2012-03-30 19:40:41 we know side of square is 15, so B = 15^2 = 225 MNL9082 2012-03-30 19:40:41 the base, 225 TigerSneak1 2012-03-30 19:40:41 we know the base, its 225 ahaanomegas 2012-03-30 19:40:46 We know B, since the base is a square with side length 15. copeland 2012-03-30 19:40:49 The base is a square of side length 15, so$B=15^2.$copeland 2012-03-30 19:40:53 All we need is the height! copeland 2012-03-30 19:40:57 Time to visualize: copeland 2012-03-30 19:41:02 copeland 2012-03-30 19:41:05 We know the length of the base. What else can we compute easily? GeorgiaTechMan 2012-03-30 19:41:42 lateral height trident 2012-03-30 19:41:42 find the lateral height of one of the triangles centralbs 2012-03-30 19:41:42 The slant height heliootrope 2012-03-30 19:41:42 slant height Potla 2012-03-30 19:41:42 Height of the side face Imsacred 2012-03-30 19:41:42 slant height kmc927 2012-03-30 19:41:42 height of triangle nmandi7 2012-03-30 19:41:42 the slant height lazorpenguin27143 2012-03-30 19:41:42 the altitude of the triangle faces Voytek 2012-03-30 19:41:42 The distance from one edge to the top copeland 2012-03-30 19:41:46 We can compute the distance from the vertex to the midpoint of a side of the base: copeland 2012-03-30 19:41:51 copeland 2012-03-30 19:41:53 copeland 2012-03-30 19:41:54 What is that distance? joshxiong 2012-03-30 19:42:33 Lateral Height=25/2 Hydroxide 2012-03-30 19:42:33 25/2 lucylai 2012-03-30 19:42:33 25/2 jasonmathcounts 2012-03-30 19:42:33 12.5 wmcho1007 2012-03-30 19:42:33 (40-15)/2=25/2 ashgabat 2012-03-30 19:42:33 12.5 Seedleaf 2012-03-30 19:42:33 (40 - 15)/2 = 25/2 n2010Math 2012-03-30 19:42:33 25/2 DF2222 2012-03-30 19:42:33 12.5 copeland 2012-03-30 19:42:38 If that distance is$x$, then$x+15+x=40$, so$x=\frac{25}2$. copeland 2012-03-30 19:42:41 What other distance can we compute easily? himym83 2012-03-30 19:43:35 distance from edge to center of square lazorpenguin27143 2012-03-30 19:43:35 the distance from the center of the circle to the side danielguo94 2012-03-30 19:43:35 center to edge of square Past 2012-03-30 19:43:35 distance from a side to the center deepankar 2012-03-30 19:43:35 apothem of base joshxiong 2012-03-30 19:43:35 distance from the center of the square to a midpoint mathnerd101 2012-03-30 19:43:35 half the side length of the base copeland 2012-03-30 19:43:38 We can compute the distance from the center of the base to the midpoint of any base edge: copeland 2012-03-30 19:43:42 copeland 2012-03-30 19:43:46 copeland 2012-03-30 19:43:47 How long is that segment? EuclidGenius 2012-03-30 19:44:15 It is 15/2 iamthygod!!!!!! 2012-03-30 19:44:15 7.5 shreyash 2012-03-30 19:44:15 7.5 ptes77 2012-03-30 19:44:15 15/2 Fire-Sword 2012-03-30 19:44:15 15/2 thunderPA 2012-03-30 19:44:15 7.5 ABCDE 2012-03-30 19:44:15 15/2 sarvottam 2012-03-30 19:44:15 15/2 sonchecky 2012-03-30 19:44:15$15/2 = 7.5$claudiafeng 2012-03-30 19:44:15 7.5 m1sterzer0 2012-03-30 19:44:15 15/2 admin25 2012-03-30 19:44:15$ \frac{15}{2} $, half the side length of the square wlpj11 2012-03-30 19:44:15 15/2=7.5 copeland 2012-03-30 19:44:19 This is half the square, so has length$\frac{15}2$. copeland 2012-03-30 19:44:23 What is the height of the pyramid? copeland 2012-03-30 19:44:27 mathswimmer 2012-03-30 19:45:08 15/2 3-4-5 triangle! Larklight 2012-03-30 19:45:08 25/2^2-15/2^2=20/2=10 deepankar 2012-03-30 19:45:08 20/2 = 10 (it's a 3-4-5 right triangle) m1sterzer0 2012-03-30 19:45:08 20/2 (scaled 3-4-5 triangle) mathswimmer 2012-03-30 19:45:08 it's a 3-4-5 triangle, so 20/5=10 flyrain 2012-03-30 19:45:08 20/2=10 newchie123 2012-03-30 19:45:08 10 Flamewire 2012-03-30 19:45:08 sqrt(625/4 - 225/4) = sqrt(400/4) = 10 ytao 2012-03-30 19:45:08 By the Pythagorean Theorem, 20/2=10 Mattchu386 2012-03-30 19:45:08 it's a 3-4-5 right triangle, so 20/2=10. coldsummer 2012-03-30 19:45:08 we can now use 7.5=a and 12.5=c and use the pythagorean theorem, which shows b=10 hrq 2012-03-30 19:45:08 10 theone142857 2012-03-30 19:45:08 10 copeland 2012-03-30 19:45:11 We have a right triangle here. it has hypotenuse$\frac{25}2$and the green base is$\frac{15}2$. We notice that these are the dimensions of a$3-4-5$triangle, with edges$\frac{15}2-\frac{20}2-\frac{25}2$. Therefore the height is$\frac{20}2=10$. copeland 2012-03-30 19:45:19 What is the volume? ahaanomegas 2012-03-30 19:45:44 brokenfixer 2012-03-30 19:45:44 (1/3) * 225 * 10 = 750 ! alex31415 2012-03-30 19:45:44 TeamJacob 2012-03-30 19:45:44 750 claudiafeng 2012-03-30 19:45:44 750 CantonMathGuy 2012-03-30 19:45:44 750 Imsacred 2012-03-30 19:45:44 1/3 * 10 * 15 * 15 = 750 Tigeraops 2012-03-30 19:45:44 750 nackster12 2012-03-30 19:45:44 225*10/3=750 matholympiad25 2012-03-30 19:45:44 1/3 * 10 * 15^2 = 750 TeamJacob 2012-03-30 19:45:44 225(10)/3 = 750 copeland 2012-03-30 19:45:49 The volume is$\frac{15^2\cdot10}3=\boxed{750}.$Past 2012-03-30 19:46:13 can I have your pudding recipe? copeland 2012-03-30 19:46:14 No. My mom would KILL me. Family secrets. You understand, I'm sure. copeland 2012-03-30 19:46:22 6. Let$z = a + bi$be the complex number with$|z| = 5$and$b > 0$such that the distance between$(1 + 2i) z^3$and$z^5$is maximized, and let$z^4 = c + di$. Find$c + d$. copeland 2012-03-30 19:46:36 Let's start with the distance statement. What is the distance from point$a$to point$b$in the complex plane? proglote 2012-03-30 19:47:21 |a-b| mathswimmer 2012-03-30 19:47:21 |a-b| admin25 2012-03-30 19:47:21 |a-b| sonchecky 2012-03-30 19:47:21 |a-b| nackster12 2012-03-30 19:47:21 |a-b| Imsacred 2012-03-30 19:47:21 |b-a| Potla 2012-03-30 19:47:21 |a-b| copeland 2012-03-30 19:47:29 The distance between two complex numbers is the modulus of their difference$|a-b|$. copeland 2012-03-30 19:47:42 The distance between$(1+2i)z^3$and$z^5$is $|(1+2i)z^3-z^5|.$ copeland 2012-03-30 19:47:45 Do you see an easy way to simplify this? mathswimmer 2012-03-30 19:48:14 z^3 pops otu lucylai 2012-03-30 19:48:14 factor out z^3 lazorpenguin27143 2012-03-30 19:48:14 factor out |z^3| Hydroxide 2012-03-30 19:48:14 factor out the z^3 admin25 2012-03-30 19:48:14 We can factor out a$ z^3 $kmc927 2012-03-30 19:48:14 factor a z^3 centralbs 2012-03-30 19:48:14 Factor out a abs (z^3) copeland 2012-03-30 19:48:20 copeland 2012-03-30 19:48:25 Can we simplify this quantity further? shreyash 2012-03-30 19:49:06 Factor out the magnitude of z^3 which is 125 brokenfixer 2012-03-30 19:49:06 |z^3| is known matholympiad25 2012-03-30 19:49:06 Yes |z^3|=125 shreyash 2012-03-30 19:49:06 Magnitude of z^3 is 5^3 = 125 lucylai 2012-03-30 19:49:06 |z^3|=|z|^3=125 nmandi7 2012-03-30 19:49:06 125 TeamJacob 2012-03-30 19:49:06 magnitude of z^3 is 5^3 Hydroxide 2012-03-30 19:49:06 |z^3|=|z|^3=5^3=125 nmandi7 2012-03-30 19:49:08 mag of z^3 = 125 copeland 2012-03-30 19:49:14 We know that$|z|=5$and since$|z^3|=|z|^3=125$, we can say $|(1+2i)z^3-z^5|=125|(1+2i)-z^2|.$ copeland 2012-03-30 19:49:17 Now we want to maximize this value. Specifically we want to find the point$z^2$such that$|z^2|=25$and$z^2$is as far away as possible from the point$1+2i$. copeland 2012-03-30 19:49:26 What is the set$|u|=25$, where$u$is a complex number? himym83 2012-03-30 19:49:55 the circle of radius 25 danielguo94 2012-03-30 19:49:55 circle w/ center at origin and radius 25 sonchecky 2012-03-30 19:49:55 A circle with radius 25 centered around the point 0+0i lucylai 2012-03-30 19:49:55 circle of radius 25 centered on origin nackster12 2012-03-30 19:49:55 the circle with radius 25 centered at the origin heliootrope 2012-03-30 19:49:55 A circle centered at the origin with radius 25 copeland 2012-03-30 19:49:59 This set is a circle of radius 25 centered at the origin. copeland 2012-03-30 19:50:03 copeland 2012-03-30 19:50:05 What point on the circle is farthest from the given point? brokenfixer 2012-03-30 19:50:55 make sure that z^2 points opposite to (1+2i) ... z^2 = - C * (1+2i) where C real >0 theone142857 2012-03-30 19:50:55 line through 1+2i and origin hits circle lz^2l=25 in third quadrant. That's z ptes77 2012-03-30 19:50:55 the diameter adamdai97 2012-03-30 19:50:55 oppostite on the diameter admin25 2012-03-30 19:50:55 If you draw a diameter through the point, it's the point where the diameter intersects the circle claudiafeng 2012-03-30 19:50:55 The point farther from 1+2i on the diameter that contains 1+2i. ptes77 2012-03-30 19:50:55 the point on the opposite side in quadrant III AbsoluteFriend 2012-03-30 19:50:55 A point on the circumference in the lower left. trident 2012-03-30 19:50:55 point where the diameter through 1 + 2i hits the circle copeland 2012-03-30 19:50:59 The distance is maximized at the antipodal point: copeland 2012-03-30 19:51:04 copeland 2012-03-30 19:51:12 Incidentally, the synthetic way to show that this point is farthest is by drawing (in blue below) the circle with center$1+2i$that contains the antipodal point. By symmetry the two circles must be tangent, so every other point on the black circle is closer than this point of tangency. copeland 2012-03-30 19:51:17 copeland 2012-03-30 19:51:32 Now how does this help us write$z^2$? brokenfixer 2012-03-30 19:52:39 -(1+2i)*25 / |1+2i| googol.plex 2012-03-30 19:52:39 we know the slope of the line and the length of the line mathswimmer 2012-03-30 19:52:39 x^2+(2x)^2=25^2 nackster12 2012-03-30 19:52:39 we can find that point by coordinate bashing? copeland 2012-03-30 19:52:46 That's a lot of fancy ideas. copeland 2012-03-30 19:53:14 What do we KNOW about this point? How is it related to 1+2i? matholympiad25 2012-03-30 19:53:57 it is a real number * (1+2i) heliootrope 2012-03-30 19:53:57 a and b have the same ratio? m1sterzer0 2012-03-30 19:53:57 on the line through 0 and 1+2i admin25 2012-03-30 19:53:57 the ratio of the imaginary part to the real part is the same, since it's on the same line through the origin nackster12 2012-03-30 19:53:57 the real part is half of the imaginary part TeamJacob 2012-03-30 19:53:57 the coefficients are proportional shreyash 2012-03-30 19:53:57 -1 - 2i joshxiong 2012-03-30 19:53:57 b=2a copeland 2012-03-30 19:54:03 We know that$z^2$has to be a positive multiple of$-1-2i$. We also know that$|z^2|=|z|^2=25.$What is$z^2$? GeorgiaTechMan 2012-03-30 19:55:00 5sqrt5(-1-2i) iamthygod!!!!!! 2012-03-30 19:55:00 -25/(sqrt5) iamthygod!!!!!! 2012-03-30 19:55:00 (-25/root5)*(1+2i) matholympiad25 2012-03-30 19:55:00 -sqrt(125) - sqrt(500) * i lazorpenguin27143 2012-03-30 19:55:00$ z^{2}=-5\sqrt{5}-10\sqrt{5}i $sonchecky 2012-03-30 19:55:00 (-1-2i)(25/sqrt(5)) = -5sqrt(5) - 10sqrt(5)i brokenfixer 2012-03-30 19:55:00 -5sqrt(5) -10sqrt(5) i m1sterzer0 2012-03-30 19:55:07 copeland 2012-03-30 19:55:08 Since$|-1-2i|=\sqrt{1^2+2^2}=\sqrt5$, We need$z^2=(5\sqrt5)(-1-2i)$. As a check, this does indeed give us $|z^2|=|(5\sqrt5)(-1-2i)|=5\sqrt5|-1-2i|=25.$ copeland 2012-03-30 19:55:12 Now what is$z^4$? lazorpenguin27143 2012-03-30 19:56:08$ z^{4}=-375+500i $Duncanyang 2012-03-30 19:56:08 125(-3+4i) brokenfixer 2012-03-30 19:56:08 125(-3+4i) alex31415 2012-03-30 19:56:08 -375+500i? Duncanyang 2012-03-30 19:56:08 -375+500i copeland 2012-03-30 19:57:02 Squaring$z^2gives us \begin{align*} z^4&=(z^2)^2\\ &=(5\sqrt5)^2(-1-2i)^2\\ &=125(1+2i)^2\\ &=125(1+4i-4)\\ &=-375+500i. \end{align*} copeland 2012-03-30 19:57:07 What does that give us for the solution? matholympiad25 2012-03-30 19:57:30 -375+500i, c+d = 125 kmc927 2012-03-30 19:57:30 125 himym83 2012-03-30 19:57:30 500-375=125 lucylai 2012-03-30 19:57:30 125 19oshawott98 2012-03-30 19:57:30 125 Voytek 2012-03-30 19:57:30 125 centralbs 2012-03-30 19:57:30 125 ptes77 2012-03-30 19:57:30 125 ahaanomegas 2012-03-30 19:57:30 500 - 375 = 125. n2010Math 2012-03-30 19:57:30 125 jasonmathcounts 2012-03-30 19:57:30 125 ytao 2012-03-30 19:57:30 500+(-375)=125 lucylai 2012-03-30 19:57:30 125 ABCDE 2012-03-30 19:57:30 125 claudiafeng 2012-03-30 19:57:30 125 CantonMathGuy 2012-03-30 19:57:30 125 Hydroxide 2012-03-30 19:57:30 -375+500=125 copeland 2012-03-30 19:57:32500+(-375)=\boxed{125}$. nackster12 2012-03-30 19:57:49 why were we given b>0? did we use that at all? copeland 2012-03-30 19:57:52 Good question. copeland 2012-03-30 19:58:03 We never needed to find$z$to solve this problem. copeland 2012-03-30 19:58:33 In fact there are TWO possibilities for$z$that solve this problem and that restriction narrows it down to a unique value. copeland 2012-03-30 19:58:56 7. Let$S$be the increasing sequence of positive integers whose binary representation has exactly 8 ones. Let$N$be the 1000th number in$S$. Find the remainder when$N$is divided by 1000. copeland 2012-03-30 19:59:11 What information might be useful to know? lucylai 2012-03-30 20:00:08 how many zeros N has CantonMathGuy 2012-03-30 20:00:08 the number on numbers with a number of zeros ytao 2012-03-30 20:00:08 How many 0s copeland 2012-03-30 20:00:13 We might want to know how many digits are in the number we want. copeland 2012-03-30 20:00:17 How can we determine this? copeland 2012-03-30 20:00:36 How many$k$-digit binary numbers are in$S$? admin25 2012-03-30 20:01:34$ \binom{k}{8}-\binom{k-1}{8} $nackster12 2012-03-30 20:01:34 k-1 choose 7 alex31415 2012-03-30 20:01:34 (k-1)C7? copeland 2012-03-30 20:01:39 The first digit of any number has to be a 1, so the remaining$k-1$digits consist of 7 ones and$k-8$zeroes. copeland 2012-03-30 20:01:46 So there are$\dbinom{k-1}{7}such numbers. copeland 2012-03-30 20:01:51 That is, there are: \begin{align*} \tbinom77 & \text{8-digit numbers}\\ \tbinom87 & \text{9-digit numbers}\\ \tbinom97 & \text{10-digit numbers}\\ \end{align*} etc. copeland 2012-03-30 20:01:58 How do we use this information? I don't want to compute all those numbers if I don't have to. copeland 2012-03-30 20:02:40 If we sum them, what do we get? himym83 2012-03-30 20:02:46 hockey stick identity alex31415 2012-03-30 20:02:46 Hockey Stick Identity flyrain 2012-03-30 20:02:46 hockey stick identity joshxiong 2012-03-30 20:02:46 Hockey stick identity. admin25 2012-03-30 20:02:46 Hockey stick identity! lucylai 2012-03-30 20:02:46 hockey stick rule ahaanomegas 2012-03-30 20:02:46 Is this the Hockey Stick Identity? copeland 2012-03-30 20:02:49 We can use the Hockey Stick Identity! copeland 2012-03-30 20:02:59 $\binom77 + \binom87 + \cdots + \binom{n}{7} = \binom{n+1}{8}.$ copeland 2012-03-30 20:03:04 What does this mean for us? alex31415 2012-03-30 20:03:47 You could put beginning 0's, so 10101 would be 010101, or 0010101, etc. copeland 2012-03-30 20:03:51 It means that there are\binom{k}{8}$numbers in$S$with$k$digits or less. copeland 2012-03-30 20:03:56 We actually could have seen this directly: think of arranging 8 ones and$k-8$zeroes, where we allow leading zeroes. (This illustrates a nice rule of thumb: whenever you find yourself using the Hockey Stick Identity, there's a really good chance you're doing things the hard way.) himym83 2012-03-30 20:04:20 (n+1)C8 > 1000 Hydroxide 2012-03-30 20:04:20 find smallest such that n+1C 8>1000 mathswimmer 2012-03-30 20:04:20 first n such that it is greater than 1000, that is the number of digits flyrain 2012-03-30 20:04:20 solve n when n+1 choose 8 is 1000 ptes77 2012-03-30 20:04:20 we find the least n that it will be over 1000 wmcho1007 2012-03-30 20:04:20 we find the largest n+1 such that c(n+1,8) is less than 1000 claudiafeng 2012-03-30 20:04:20 We need to find when (n+1)C8 is greater/equal to 1000? copeland 2012-03-30 20:04:24 Now we can test some values of$k.$copeland 2012-03-30 20:04:27 A little experimentation gives us$\binom{12}{8} = 495$and$\binom{13}{8} = 1287.$ copeland 2012-03-30 20:04:33 So what can we conclude? awesomemathlete 2012-03-30 20:05:06 13 digits nmandi7 2012-03-30 20:05:06 13 digit TigerSneak1 2012-03-30 20:05:06 the number has 13 digits Larklight 2012-03-30 20:05:06 13 digits ABCDE 2012-03-30 20:05:06 N is a 13 digit binary number nmandi7 2012-03-30 20:05:06 k is 13 digit # proglote 2012-03-30 20:05:06 the 1000th number has 13 digits trident 2012-03-30 20:05:06 the 1000th number has 13 digits copeland 2012-03-30 20:05:10 The number we want (the 1000th number in$S$) is a 13-digit number. copeland 2012-03-30 20:05:13 Is there something stronger we can say? googol.plex 2012-03-30 20:05:54 so it's the 505th number with 13 digits. shreyash 2012-03-30 20:05:54 It is the 505th number with 13 digits brokenfixer 2012-03-30 20:05:54 it is the 505th? lucylai 2012-03-30 20:05:54 it's the 505th 13 digit number with 8 ones Imsacred 2012-03-30 20:05:54 it is the 505th 13 digit number nmandi7 2012-03-30 20:05:54 505th copeland 2012-03-30 20:05:58 The number we want is the 1000-495 = 505th 13-digit number in$S.$copeland 2012-03-30 20:06:02 This number is the digit 1 placed in front of the 505th largest number with 7 ones and 5 zeroes. copeland 2012-03-30 20:06:07 Now the problem becomes a matter of careful counting and bookkeeping. copeland 2012-03-30 20:06:09 What next? alex31415 2012-03-30 20:06:55 Apply that to every digit? mathswimmer 2012-03-30 20:06:55 if the 12th digit is 0, see if filling in the remaining 11 is enough to get up to 505. if not, then the 12th digit is 1. continue in this manner proglote 2012-03-30 20:06:55 consider the second digit nmandi7 2012-03-30 20:06:55 split it into lower combinations wmcho1007 2012-03-30 20:06:57 determine if the second digit is a 1 or a 0 copeland 2012-03-30 20:07:07 We can determine if the 2nd digit (from the left) is a 0 or 1. copeland 2012-03-30 20:07:13 The 13-digit numbers in$S$consist of all the 13-digit numbers in S starting with 10, followed by all the 13-digit numbers in$S$starting with 11. copeland 2012-03-30 20:07:16 How many numbers in the former group: that start with 10? mathswimmer 2012-03-30 20:08:26 11C7 Seedleaf 2012-03-30 20:08:26 330 admin25 2012-03-30 20:08:26$ \binom{11}{7} $wmcho1007 2012-03-30 20:08:26 C(11,7)=330 nackster12 2012-03-30 20:08:26 11 choose 7=330 kmc927 2012-03-30 20:08:26 330 heliootrope 2012-03-30 20:08:26 11 choose 7 iamthygod!!!!!! 2012-03-30 20:08:26 330 copeland 2012-03-30 20:08:32 Thus there are$\dbinom{11}{4} = 330$such numbers. copeland 2012-03-30 20:08:35 So since we wanted the 505th 13-digit number, we first have to exhaust the 330 13-digit numbers starting with 10. copeland 2012-03-30 20:08:41 Then we want the 505-330 = 175th 13-digit number starting with 11. copeland 2012-03-30 20:08:46 Now what? nackster12 2012-03-30 20:09:20 find the third digit himym83 2012-03-30 20:09:20 do the same thing for the next digit lucylai 2012-03-30 20:09:20 find numbers starting with 110 coldsummer 2012-03-30 20:09:20 110 or 111 deepankar 2012-03-30 20:09:20 break into start with 110 or 111 eccfcco15 2012-03-30 20:09:20 repeat the above process to get the next digit mathnerd101 2012-03-30 20:09:20 find the 3rd digit? copeland 2012-03-30 20:09:27 Ugh, really? copeland 2012-03-30 20:09:28 OK. copeland 2012-03-30 20:09:32 Again, we split our group into 13-digit numbers starting with 110 and 13-digit numbers starting with 111. copeland 2012-03-30 20:09:35 How many of the former: 13-digit numbers starting with 110? admin25 2012-03-30 20:10:16 There are$ \binom{10}{6} $numbers that begin with 110 joshxiong 2012-03-30 20:10:16 210. claudiafeng 2012-03-30 20:10:16 10C6 wmcho1007 2012-03-30 20:10:16 C(10,6)=210 googol.plex 2012-03-30 20:10:16 10C4 lucylai 2012-03-30 20:10:16 C(10, 4)=210 iamthygod!!!!!! 2012-03-30 20:10:16 210? copeland 2012-03-30 20:10:20 There are 6 ones and 4 zeroes to place, so there are$\dbinom{10}{4} = 210$of them. copeland 2012-03-30 20:10:24 Since we only want the 175th 13-digit number starting with 11, we know it will also be the 175th 13-digit number among the 210 starting with 110. fredgauss 2012-03-30 20:10:41 is this the easiest way to do this? copeland 2012-03-30 20:10:42 Trust me. If I knew a better answer we would NOT be doing this. copeland 2012-03-30 20:10:46 So now we want the 175th 13-digit number in$S$starting with 110. copeland 2012-03-30 20:10:55 Now what? coldsummer 2012-03-30 20:11:19 1100 or 1101? ptes77 2012-03-30 20:11:19 do the same to the 4th digit shreyash 2012-03-30 20:11:19 Find the 4th digit wmcho1007 2012-03-30 20:11:19 the fourth digit...... eccfcco15 2012-03-30 20:11:19 continue claudiafeng 2012-03-30 20:11:19 Break it.. further.. nmandi7 2012-03-30 20:11:19 do the next digit ... ugh sonchecky 2012-03-30 20:11:19 consider all 13-digit numbers starting with 1100 copeland 2012-03-30 20:11:23 Same game: consider numbers starting with 1100 and numbers starting with 1101. How many of the former? -Transcend- 2012-03-30 20:12:14 9C3 claudiafeng 2012-03-30 20:12:14 9C5 joshxiong 2012-03-30 20:12:14 84. kmc927 2012-03-30 20:12:14 84 coldsummer 2012-03-30 20:12:14 9C6 Seedleaf 2012-03-30 20:12:14 9C6 = 84 lucylai 2012-03-30 20:12:14 C(9, 3)=84 deepankar 2012-03-30 20:12:14 nackster12 2012-03-30 20:12:14 9C6=84 copeland 2012-03-30 20:12:18 We have 6 ones and 3 zeroes remaining, so there are$\binom93 = 84$of them. copeland 2012-03-30 20:12:21 That means we want the 175 - 84 = 91st 13-digit number in$S$starting with 1101. copeland 2012-03-30 20:12:30 And again: split into numbers beginning with 11010 and 11011. How many of the former? sonchecky 2012-03-30 20:12:53 8 choose 3 = 56 Duncanyang 2012-03-30 20:12:53 8 choose 5 claudiafeng 2012-03-30 20:12:53 8C5=56 Relativity1618 2012-03-30 20:12:53 8C3 kmc927 2012-03-30 20:12:53 56 coldsummer 2012-03-30 20:12:53 therefore we do it again.......8C5, or 56 copeland 2012-03-30 20:12:55 We have 5 ones and 3 zeroes remaining, so there are$\binom83 = 56$of them. copeland 2012-03-30 20:12:59 That means we want the 91 - 56 = 35th 13-digit number in$S$starting with 11011. copeland 2012-03-30 20:13:05 And again(!): split into starting with 110110 and 110111. How many of the former? joshxiong 2012-03-30 20:13:42 35. flyrain 2012-03-30 20:13:42 110110 7C3 shreyash 2012-03-30 20:13:42 7C3 mathnerd101 2012-03-30 20:13:42 7C3 Mary_Posa 2012-03-30 20:13:42 35 sinnoel 2012-03-30 20:13:42 35 sonchecky 2012-03-30 20:13:42 7 choose 3 = 35 nackster12 2012-03-30 20:13:42 7C3=35 Seedleaf 2012-03-30 20:13:42 7C4 = 35 jinguluninja 2012-03-30 20:13:42 35 copeland 2012-03-30 20:13:46 We have 4 ones and 3 zeroes remaining, so there are$\binom73 = 35$of them. sinnoel 2012-03-30 20:14:08 yyayyy! brokenfixer 2012-03-30 20:14:08 hooray copeland 2012-03-30 20:14:12 Aha! We want the 35th such number (beginning with 110110), and there are exactly 35 of them. copeland 2012-03-30 20:14:17 So we want the biggest one: the one with all the remaining 1's pushed all the way to the left. copeland 2012-03-30 20:14:20 That means that the number we're looking for is 1101101111000. copeland 2012-03-30 20:14:25 Now how do we finish? Relativity1618 2012-03-30 20:15:20 =7032 sonchecky 2012-03-30 20:15:20 Convert to base 10 Imsacred 2012-03-30 20:15:20 base 10! GeorgiaTechMan 2012-03-30 20:15:20 and that is 7032 base ten so our answer is 032! joshxiong 2012-03-30 20:15:20 Convert to base 10. ytao 2012-03-30 20:15:20 Convert to base 10. sinnoel 2012-03-30 20:15:20 convert to base 10 and add? -Transcend- 2012-03-30 20:15:20 Convert to base 10? deepankar 2012-03-30 20:15:20 Find base 10 value and divide by 1000 alex31415 2012-03-30 20:15:20 2^3+2^4+2^5+2^6+2^8+2^9+2^11+2^12=7032, so the answer is 032 thunderPA 2012-03-30 20:15:20 convert it into base ten and divide by 1000 HeatDynasty 2012-03-30 20:15:20 convert it to base 10 number 19oshawott98 2012-03-30 20:15:26 unencrypt copeland 2012-03-30 20:15:33 We need the last 3 digits of the base-10 representation of the binary number $1101101111000.$ copeland 2012-03-30 20:15:37 This is the last 3 digits of$8 + 16 + 32 + 64 + 256 + 512 + 2048 + 4096.$ copeland 2012-03-30 20:15:40 Since we can ignore thousands digits, this is just the last three digits of$8 + 16 + 32 + 64 + 256 + 512 + 48 + 96.$ copeland 2012-03-30 20:15:43 These numbers sum to 1032, so our answer is$\boxed{032}$. copeland 2012-03-30 20:16:01 I notice that nobody said 7 when I asked what your favorite problems were. . . lucylai 2012-03-30 20:16:28 i did copeland 2012-03-30 20:16:37 Oh, there was one vote. copeland 2012-03-30 20:16:44 Yay. Love for #7. copeland 2012-03-30 20:16:45 8. The complex numbers$z$and$wsatisfy the system \begin{align*} z + \frac{20i}{w} &= 5 + i \\ w + \frac{12i}{z} &= -4 + 10i. \end{align*} Find the smallest possible value of|zw|^2$. mathnerd101 2012-03-30 20:17:11 ugh complex numbers copeland 2012-03-30 20:17:12 "Woohoo complex numbers!" you mean? copeland 2012-03-30 20:17:13 How can we get something manageable from these equations? Relativity1618 2012-03-30 20:17:33 multiply both equations claudiafeng 2012-03-30 20:17:33 Multiply the equations theone142857 2012-03-30 20:17:33 mutiply them together proglote 2012-03-30 20:17:33 multiply the two equations himym83 2012-03-30 20:17:33 multiply the equations together jinguluninja 2012-03-30 20:17:33 multiply them sonchecky 2012-03-30 20:17:33 Multiply them together ABCDE 2012-03-30 20:17:33 multiply them zinko1991 2012-03-30 20:17:33 multiply them? CantonMathGuy 2012-03-30 20:17:33 multiply! GeorgiaTechMan 2012-03-30 20:17:33 multiply! esque 2012-03-30 20:17:39 multiply them together~~~~ copeland 2012-03-30 20:17:41 Notice that if we take the product on the left side the cross-terms are going to be constants:$\left(z+\frac{20i}w\right)\left(w+\frac{12i}z\right)=zw+20i\frac ww+12i\frac zz+\frac{20i\cdot12i}{wz}.$ copeland 2012-03-30 20:17:49 Therefore if we multiply these equations, left-by-left and right-by-right, we get $zw+32i-\frac{240}{wz}=(5+i)(-4+10i)=-20-4i+50i-10=-30+46i.$ copeland 2012-03-30 20:17:58 We notice that$zw$cannot be zero or the original system is not well-defined. copeland 2012-03-30 20:18:06 Multiplying this equation by$zw$and simplifying gives us$(zw)^2+(30-14i)(zw)-240=0.$ copeland 2012-03-30 20:18:14 This polynomial has two roots. How do we find them? Relativity1618 2012-03-30 20:18:41 quadratic formula ABCDE 2012-03-30 20:18:41 quadratic formula lucylai 2012-03-30 20:18:41 quadratic formula Xcellence 2012-03-30 20:18:41 vietas formulas Duncanyang 2012-03-30 20:18:41 quadratic formulae!!! coldsummer 2012-03-30 20:18:41 quadratic formula? Hydroxide 2012-03-30 20:18:41 quadratoc formula shreyash 2012-03-30 20:18:41 Quadratic equation admin25 2012-03-30 20:18:41 Vietas? nmandi7 2012-03-30 20:18:41 quadratic formula googol.plex 2012-03-30 20:18:41 we use quadratic formula adamdai97 2012-03-30 20:18:41 quadratic formula? Seedleaf 2012-03-30 20:18:41 quadratic formula? copeland 2012-03-30 20:18:45 The roots are actually guessable if you're really lucky (noticing that the product of the roots is real and negative is a big hint). The quadratic formula is probably the fastest way to go, though. copeland 2012-03-30 20:18:53 By the quadratic formula we must have $zw=\frac{-30+14i\pm\sqrt{(30-14i)^2+4\cdot240}}2,$which simplifies to $zw=-15+7i\pm\sqrt{416-210i}.$ copeland 2012-03-30 20:19:02 Yuck. We need a square root,$a+bi=\sqrt{416-210i}$. We can find this square root by squaring,$(a+bi)^2=416-210i. copeland 2012-03-30 20:19:13 Expanding this and equating the real and imaginary parts tells us we want to solve \begin{align*} a^2-b^2&=416\\ 2ab&=-210. \end{align*} copeland 2012-03-30 20:19:26 Do you see the solutions? copeland 2012-03-30 20:20:15 Check your answers! You don't want to be losing points on sign errors! himym83 2012-03-30 20:20:46 21 and -5 esque 2012-03-30 20:20:46 a=-21, b=5 or a=21, b=-5 ABCDE 2012-03-30 20:20:46 a=21 and b=-5 claudiafeng 2012-03-30 20:20:46(\pm 21, \mp 5)$theone142857 2012-03-30 20:20:46 21 and -5 admin25 2012-03-30 20:20:46 (a,b)=(21,-5) or (-21,5) vjnmath 2012-03-30 20:20:46 21 and -5 ABCDE 2012-03-30 20:20:46 a=21 and b=-5 -Transcend- 2012-03-30 20:20:46 (a, b) = (21, -5) karatemagic7 2012-03-30 20:20:46 a=21, b=-5 or a=-21, b=5 nackster12 2012-03-30 20:20:46 (21, -5) or (-21,5) alex31415 2012-03-30 20:20:46 -21,5? ThEeLiTePaRk 2012-03-30 20:20:46 21 and -5 :D copeland 2012-03-30 20:20:52 Since$a^2>416$, we know that$a$must be a factor of 105 and that$|a|$must also be at least 20. That narrows things down a lot. Indeed we have to have$a=\pm21$or$a=\pm35$or$a=\pm105$. copeland 2012-03-30 20:21:02 Checking we find that$21^2-416=441-416=25=5^2$, so indeed$a=\pm21$and$b=\pm5$. copeland 2012-03-30 20:21:10 Since$ab$is negative, the square roots of$416-210i$are$\pm(21-5i).$copeland 2012-03-30 20:21:16 Therefore$zw=-15+7i\pm(21-5i)$, so$zw=6+2i$or$zw=-36+12i$. copeland 2012-03-30 20:21:20 What is the minimum value for$|zw|^2$? copeland 2012-03-30 20:22:07 I'll give you a hint: This is an AIME problem. The answer is probably at least real. sinnoel 2012-03-30 20:22:19 040! wmcho1007 2012-03-30 20:22:19 therefore its 40? GeorgiaTechMan 2012-03-30 20:22:19 6^2+2^2=040 karatemagic7 2012-03-30 20:22:19 6^2+2^2=40 Hydroxide 2012-03-30 20:22:19 40 brokenfixer 2012-03-30 20:22:19 040 admin25 2012-03-30 20:22:19$ 6^2+2^2=\boxed{040} $Seedleaf 2012-03-30 20:22:19 40 sonchecky 2012-03-30 20:22:19$6^2 + 2^2 = 36+4 = 040$proglote 2012-03-30 20:22:19 040 MNL9082 2012-03-30 20:22:19 40 copeland 2012-03-30 20:22:23 The minimum value for$|zw|^2$is$6^2+2^2=\boxed{40}$. copeland 2012-03-30 20:22:28 Notice that we didn't actually solve our original system, but only solved an equation that is implied by our original system. In particular we don't know that there actually is a pair$(z,w)$such that this system is solved. However we can use$zw=6+2i$to eliminate$w$from the first equation and solve for$z$. This shows that we can solve$zw=6+2i$simultaneously with the first equation. copeland 2012-03-30 20:22:48 If we can solve the first equation and we can solve the product of the equations, then we can solve the second equation as well (since the right side of both equations is never 0). copeland 2012-03-30 20:23:26 9. Let$x$and$y$be real numbers such that$\dfrac{\sin x}{\sin y} = 3$and$\dfrac{\cos x}{\cos y} = \frac{1}{2}$. The value of$\dfrac{\sin 2x}{\sin 2y} + \dfrac{\cos 2x}{\cos 2y}$can be expressed in the form$\dfrac{p}{q}$, where$p$and$q$are relatively prime positive integers. Find$p + q$. copeland 2012-03-30 20:23:35 Perhaps we can solve this without actually having to compute any of the sines and cosines directly? copeland 2012-03-30 20:23:39 Can we use trig identities to get at any part of the answer? shreyash 2012-03-30 20:24:10 First find sin2x/sin2y esque 2012-03-30 20:24:10 the first part is immediately 3/2 Imsacred 2012-03-30 20:24:10 sin2x/sin2y sonchecky 2012-03-30 20:24:10 sin 2x = 2 sin x cos x nmandi7 2012-03-30 20:24:10 sin2x = 2sinxcosx GeorgiaTechMan 2012-03-30 20:24:10 sin(2x)/sin(2y) is easy deepankar 2012-03-30 20:24:10 sin(2x)/sin(2y) = 3/2 morpheus44 2012-03-30 20:24:10 sin2x/sin2y=sinxcosx/sinycosy=3/2 eccfcco15 2012-03-30 20:24:10 yes, we can get sin(2x)/sin(2y) ABCDE 2012-03-30 20:24:10 multiply them to get sin(2x)/sin(2y) copeland 2012-03-30 20:24:14 The first fraction in the answer is accessible if we use the sine double-angle formula. copeland 2012-03-30 20:24:19 Specifically, we have$\frac{\sin 2x}{\sin 2y} = \frac{2 \sin x \cos x}{2 \sin y \cos y}.$ copeland 2012-03-30 20:24:24 So this term is just$\frac{\sin x}{\sin y} \cdot \frac{\cos x}{\cos y} = 3 \cdot \frac12 = \frac32.$ copeland 2012-03-30 20:24:27 Hopefully the other term will be just as easy! copeland 2012-03-30 20:24:34 . . . . copeland 2012-03-30 20:24:43 Well, maybe not. The cosine double-angle formula doesn't work as nicely for us. copeland 2012-03-30 20:24:49 So I think maybe we sure try to solve for the sines and cosines directly. copeland 2012-03-30 20:24:52 Let's set$c = \cos y$and$s = \sin y$. copeland 2012-03-30 20:24:57 (I chose these to keep the denominators as simple as possible and to force potentially more complicated stuff in the numerators.) copeland 2012-03-30 20:25:04 Then what are the trig functions for$x?$mathswimmer 2012-03-30 20:26:14 sin(x)=3y, cos(x)=c/2 admin25 2012-03-30 20:26:14$ \sin x=3s $,$ \cos x=\frac{c}{2} $nackster12 2012-03-30 20:26:14 sin(x)=3s and cos(x)=c/2 ABCDE 2012-03-30 20:26:14 sinx=3s and 2cosx=c mathswimmer 2012-03-30 20:26:14 sin(x)=3s, cos(x)=c/2 lucylai 2012-03-30 20:26:14 3c, s/2 nmandi7 2012-03-30 20:26:14 sinx=3s nmandi7 2012-03-30 20:26:14 cosx=c/2 joshxiong 2012-03-30 20:26:14 cos(x)=c/2, sin(x)=3s copeland 2012-03-30 20:26:18$\sin x = 3s$and$\cos x = \frac12c$. copeland 2012-03-30 20:26:22 But what else do we know? Relativity1618 2012-03-30 20:27:22 set up systems of equations using pythagorean identity theone142857 2012-03-30 20:27:22 sum of squares=1 brokenfixer 2012-03-30 20:27:22 s^2+c^2=1 ABCDE 2012-03-30 20:27:22 s^2+c^2=1 lucylai 2012-03-30 20:27:22 s^2+c^2=1 admin25 2012-03-30 20:27:22$ s^2+c^2=1 $, and also$ (3s)^2+(\frac{c}{2})^2=1 $deepankar 2012-03-30 20:27:22 sin^2(x) + cos^2(x) = 1 GeorgiaTechMan 2012-03-30 20:27:22 sin(a)^2+cos(a)^2=1 for all a sonchecky 2012-03-30 20:27:22$s^2 + c^2 = 1$and$(3s)^2 + (c/2)^2 = 1$nmandi7 2012-03-30 20:27:22 9s^2 + c^2/4 = 1 morpheus44 2012-03-30 20:27:22 9s^2+1/4c^2=1 mathswimmer 2012-03-30 20:27:22 s^2+c^2=1, (3s)^2+(c/2)^2=1 joshxiong 2012-03-30 20:27:22 (c/2)^2+9s^2=1 and c^2+s^2=1 lucylai 2012-03-30 20:27:22 9s^2+c^2/4=s^2+c^2=1 copeland 2012-03-30 20:27:24 Yep. You lose your licenses for doing trig if you don't try that one first. copeland 2012-03-30 20:27:27$\sin^2 + \cos^2 = 1!copeland 2012-03-30 20:27:30 So we have the system of equations \begin{align*} s^2 + c^2 &= 1, \\ (3s)^2 + \left(\frac12c\right)^2 &= 1. \end{align*} copeland 2012-03-30 20:27:35 We can easily solve this system! copeland 2012-03-30 20:27:41 For example, multiply the second equation by 4, then subtract the first from the second, leaving35s^2 = 3.$copeland 2012-03-30 20:27:50 This gives the solution$s^2 = \frac{3}{35},\, c^2 = \frac{32}{35}.$ copeland 2012-03-30 20:28:03 I'd leave the solution in terms of the squares, because we know that the cosine double-angle formula involves squares. copeland 2012-03-30 20:28:12 What is the cosine double angle formula? ahaanomegas 2012-03-30 20:28:24 There are a few. :P copeland 2012-03-30 20:28:26 Goo! copeland 2012-03-30 20:28:28 er, copeland 2012-03-30 20:28:29 Good! copeland 2012-03-30 20:28:35 Which one is best? alex31415 2012-03-30 20:29:08 use cos2x=2cos^2x-1 himym83 2012-03-30 20:29:08 cos 2x = 1-2sin^2x ahaanomegas 2012-03-30 20:29:08 Xcellence 2012-03-30 20:29:08 cos(2u)=1-2sin^2u theone142857 2012-03-30 20:29:08 2cos^2-1 shreyash 2012-03-30 20:29:08 2cos(x)^2 - 1 Relativity1618 2012-03-30 20:29:08 1-2sin^2x theone142857 2012-03-30 20:29:08 2cos^2-1 and 1-2sin^2 copeland 2012-03-30 20:29:20 These are both useful here. copeland 2012-03-30 20:29:25 Indeed,$\frac{\cos 2x}{\cos 2y} = \frac{2 \cos^2 x - 1}{2 \cos^2 y - 1} = \frac{2(c/2)^2 - 1}{2c^2-1}.$ copeland 2012-03-30 20:29:35 I won't torture you with arithmetic. copeland 2012-03-30 20:29:36 We plug in our known$c^2 = \frac{32}{35}$and we get$\frac{\frac{16}{35} - 1}{\frac{64}{35} - 1}.$ copeland 2012-03-30 20:29:42 This simplifies to$\frac{16 - 35}{64 - 35} = -\frac{19}{29}.$ copeland 2012-03-30 20:29:51 And the solution? matholympiad25 2012-03-30 20:31:20 3/2-19/29=49/58 => 107 Relativity1618 2012-03-30 20:31:20 49+58=107 brokenfixer 2012-03-30 20:31:20 need 3/2 - 19/29 joshxiong 2012-03-30 20:31:20 3/2-19/29=49/58, so p+q=107 admin25 2012-03-30 20:31:20$ \frac{3}{2}-\frac{19}{29}=\frac{49}{58} $ABCDE 2012-03-30 20:31:20 3/2-19/29=49/58 so 107 brokenfixer 2012-03-30 20:31:20 (87 - 38) / 58 theone142857 2012-03-30 20:31:20 49/58 googol.plex 2012-03-30 20:31:20 107 deepankar 2012-03-30 20:31:26 107 Xcellence 2012-03-30 20:31:26 107 copeland 2012-03-30 20:31:36 Our final quantity is$\frac32 - \frac{19}{29} = \frac{87 - 38}{58} = \frac{49}{58},$these are coprime, so the final answer is$49 + 58 = \boxed{107}$. copeland 2012-03-30 20:32:16 10. Find the number of positive integers$n$less than 1000 for which there exists a positive real number$x$such that$n = x \lfloor x \rfloor$. Note:$\lfloor x \rfloor$is the greatest integer less than or equal to$x$. copeland 2012-03-30 20:32:22 What's the first trick to try when dealing with floor? tc1729 2012-03-30 20:33:07 express it with x and fractional part? Hydroxide 2012-03-30 20:33:07 write x=floor(x)+a Voytek 2012-03-30 20:33:07 round down to integers adamdai97 2012-03-30 20:33:07 write it as x-n copeland 2012-03-30 20:33:13 We should write$x=a+r$, where$a$is an integer and$0\leq r< 1$. The number$a$is the integer part of$x$and$r$is the fractional part. copeland 2012-03-30 20:33:18 Now what is$n$in terms of$a$and$r$? himym83 2012-03-30 20:34:02 n = a(a+r) lucylai 2012-03-30 20:34:02 a^2+ar brokenfixer 2012-03-30 20:34:02 a*(a+r) sonchecky 2012-03-30 20:34:02$a(a+r)$nmandi7 2012-03-30 20:34:02 a(a+r) 19oshawott98 2012-03-30 20:34:02 a^2+ar nackster12 2012-03-30 20:34:02 a^2+a*r claudiafeng 2012-03-30 20:34:02 a^2+ar Larklight 2012-03-30 20:34:02 a^2+ar edisonchew240 2012-03-30 20:34:02 a^{2}+ar nmandi7 2012-03-30 20:34:02 a^2+ar alex31415 2012-03-30 20:34:02 (a+r)a copeland 2012-03-30 20:34:07$n=x\lfloor x\rfloor=(a+r)\lfloor a+r\rfloor=(a+r)a.\]
copeland 2012-03-30 20:34:10
It will help us to have a name for this function, so let $f(x)=x\lfloor x\rfloor$. As above we have
copeland 2012-03-30 20:34:13
$f(a+r)=a(a+r)$ when $a$ is an integer and $0\leq r<1$.
copeland 2012-03-30 20:34:18
I don't quite see what's going on here yet. What should we do if we don't see what's going on?
theone142857 2012-03-30 20:34:55
small cases
wmcho1007 2012-03-30 20:34:55
put in some numbers as a test
centralbs 2012-03-30 20:34:55
plug in some values into f(a+r)
Relativity1618 2012-03-30 20:34:55
plug in values
-Transcend- 2012-03-30 20:34:55
Plug in some numbers?
Seedleaf 2012-03-30 20:34:55
Test a few examples.
copeland 2012-03-30 20:34:59
Let's try the smallest examples.
copeland 2012-03-30 20:35:03
Now we are assuming that $x$ is positive, so the smallest that $a$ can be is zero. What $n$ can we get when $a=0$?
shreyash 2012-03-30 20:35:43
0
nackster12 2012-03-30 20:35:43
0
coldsummer 2012-03-30 20:35:43
0
19oshawott98 2012-03-30 20:35:43
0
nmandi7 2012-03-30 20:35:43
0
wmcho1007 2012-03-30 20:35:43
then its 0
n2010Math 2012-03-30 20:35:43
0
copeland 2012-03-30 20:35:47
When $a=0$, we always get $0(0+x)=0$. This is not a positive integer so we can ignore this case.
copeland 2012-03-30 20:35:50
What do we get if $a=1$?
copeland 2012-03-30 20:36:28
What do we get for $f$, that is?
Duncanyang 2012-03-30 20:36:31
f(1+r)=1+r
shreyash 2012-03-30 20:36:31
1+r
Relativity1618 2012-03-30 20:36:31
1+r
centralbs 2012-03-30 20:36:31
1+r
copeland 2012-03-30 20:36:46
When $a=1$, we get $f(1+r)=1(1+r)=1+r$. Since $0\leq r<1$, we get $1\leq f(a+r)<2$, so the possible outputs of $f$ are exactly the numbers in the interval $[1,2)$.
copeland 2012-03-30 20:36:54
How many integers are in this interval?
zinko1991 2012-03-30 20:37:20
n=1
ytao 2012-03-30 20:37:20
1
flyrain 2012-03-30 20:37:20
1
himym83 2012-03-30 20:37:20
1
Imsacred 2012-03-30 20:37:20
1; 1
Just one.
Xcellence 2012-03-30 20:37:20
one
joshxiong 2012-03-30 20:37:20
1, that is 1.
sinnoel 2012-03-30 20:37:20
`1
deepankar 2012-03-30 20:37:20
1
copeland 2012-03-30 20:37:27
Just $n=1$.
copeland 2012-03-30 20:37:30
What do we get when $a=2$?
theone142857 2012-03-30 20:37:53
4+2r
Relativity1618 2012-03-30 20:37:53
4+2r
ptes77 2012-03-30 20:37:53
4+2r
Hydroxide 2012-03-30 20:37:53
2(2+r)=4+2r
Duncanyang 2012-03-30 20:37:53
2(2+r)
nackster12 2012-03-30 20:37:53
4+2r
alex31415 2012-03-30 20:37:53
2(2+r)=4+2r
copeland 2012-03-30 20:38:07
When $a=2$, we get $f(2+r)=2(2+r)=2+2r$. Since $0\leq 2r<2$, we get $4\leq f(a+r)<6$, so the possible outputs of $f$ are exactly the numbers in the interval $[4,6)$.
copeland 2012-03-30 20:38:11
How many integers are in this interval?
googol.plex 2012-03-30 20:38:42
2
mathswimmer 2012-03-30 20:38:42
2
sonchecky 2012-03-30 20:38:42
2 (namely 4 and 5)
CantonMathGuy 2012-03-30 20:38:42
2
mathnerd101 2012-03-30 20:38:42
2
joshxiong 2012-03-30 20:38:42
4,5, so 2.
Imsacred 2012-03-30 20:38:42
2; 4 and 5
copeland 2012-03-30 20:38:47
Two: 4 and 5.
copeland 2012-03-30 20:38:50
What do we get when $a=3$?
theone142857 2012-03-30 20:39:20
9+3r
-Transcend- 2012-03-30 20:39:20
9+3r
Duncanyang 2012-03-30 20:39:20
3(3+r)
shreyash 2012-03-30 20:39:20
Relativity1618 2012-03-30 20:39:20
9+3r
flyrain 2012-03-30 20:39:20
f(3+r)= 9+3r
Hydroxide 2012-03-30 20:39:20
3(3+r)=9+3r
nackster12 2012-03-30 20:39:20
9+3r
copeland 2012-03-30 20:39:24
When $a=3$, we get $f(3+r)=3(3+r)=3+3r$. Since $0\leq 3r<3$, we get $9\leq f(a+r)<12$, so the possible outputs of $f$ are exactly the numbers in the interval $[9,12)$.
copeland 2012-03-30 20:39:27
How many integers are in this interval?
lucylai 2012-03-30 20:40:02
3
n2010Math 2012-03-30 20:40:02
[9,12), so 3 - 9,10,11
sonchecky 2012-03-30 20:40:02
3 (9, 10, and 11)
nackster12 2012-03-30 20:40:02
3
shreyash 2012-03-30 20:40:02
3-9,10,11
-Transcend- 2012-03-30 20:40:02
3
nmandi7 2012-03-30 20:40:02
wmcho1007 2012-03-30 20:40:02
9,10,11, so 3
Duncanyang 2012-03-30 20:40:02
3
copeland 2012-03-30 20:40:06
Three: 9, 10, and 11.
copeland 2012-03-30 20:40:09
And for an arbitrary $a$, what do we get?
sinnoel 2012-03-30 20:40:38
a
Duncanyang 2012-03-30 20:40:38
a(a+r)
nmandi7 2012-03-30 20:40:38
googol.plex 2012-03-30 20:40:38
it should be a
Xcellence 2012-03-30 20:40:38
a integers
Duncanyang 2012-03-30 20:40:38
and a integers possible
there are a numbers that work
sonchecky 2012-03-30 20:40:38
a integers (a^2, a^2 + 1, ... a(a-1) - 1)
awesomemathlete 2012-03-30 20:40:38
a
copeland 2012-03-30 20:40:57
For arbitrary (integers) $a$ we get we get $f(a+r)=a(a+r)=a+ar$. Since $0\leq ar<a$, we get $a^2\leq f(a+r)<a^2+a$, so the possible outputs of $f$ are exactly the numbers in the interval $[a^2,a^2+a)$.
copeland 2012-03-30 20:41:01
The integers in this interval are $\{a^2,a^2+1,\ldots,a^2+a-1\}$. There are $a$ integers in this interval.
copeland 2012-03-30 20:41:10
Can we just add these up and be done?
copeland 2012-03-30 20:41:49
Two things:
alex31415 2012-03-30 20:41:52
floor(sqrt1000)=31
sonchecky 2012-03-30 20:41:52
We need the upper limit for a
theone142857 2012-03-30 20:41:52
less than 1000
nmandi7 2012-03-30 20:41:52
must be <1000
nackster12 2012-03-30 20:41:52
we need to check where we add up to
Imsacred 2012-03-30 20:41:52
well we have to check to make sure all of the numbers are under 1000
copeland 2012-03-30 20:41:56
We'll get to that.
centralbs 2012-03-30 20:42:01
We must ensure that there is no overlap in the intervals
heliootrope 2012-03-30 20:42:01
check for double-counted integers?
copeland 2012-03-30 20:42:10
This is scarier for me, though.
copeland 2012-03-30 20:42:14
We need to check that all of the intervals are disjoint.
copeland 2012-03-30 20:42:16
Are they?
brokenfixer 2012-03-30 20:42:54
a^2 + a - 1 < (a+1)^2 yep
himym83 2012-03-30 20:42:54
yes: a^2+a < a^2+2a+1
sonchecky 2012-03-30 20:42:54
Yes because $(a+1)^2 = a^2 + 2a + 1 > a^2 + a$
theone142857 2012-03-30 20:42:54
But that's obvious due to the fact that a^2-a+1<(a+1)^2
nackster12 2012-03-30 20:42:54
yes since (a)(a+1) is less than (a+1)(a+1)
Yes: $a^2+a-1<(a+1)^2$
copeland 2012-03-30 20:43:00
$a(a+1)$ is the rightmost endpoint of the $a^{th}$ interval and $(a+1)(a+1)$ is the leftmost endpoint of the next interval, so these intervals are disjoint.
copeland 2012-03-30 20:43:05
The first interval is $[1,2)$. What is the last interval?
alex31415 2012-03-30 20:43:53
[961,992)
himym83 2012-03-30 20:43:53
[961,992)
Imsacred 2012-03-30 20:43:53
[961,992)
shreyash 2012-03-30 20:43:53
[961,992)
sonchecky 2012-03-30 20:43:53
[961, 992)
19oshawott98 2012-03-30 20:43:53
961-991
brokenfixer 2012-03-30 20:43:53
31^2 to 31^2+31-1
copeland 2012-03-30 20:44:01
Since $31^2=961$ and $32^2=1024$, the last interval we need to deal with is $[31^2,31\cdot32)$.
copeland 2012-03-30 20:44:14
Here was something that should have made us nervous:
googol.plex 2012-03-30 20:44:18
There might be a partial thing at the end
copeland 2012-03-30 20:44:36
Fortuitously, all 31 of the numbers are less than 1000 so the counting is much easier.
copeland 2012-03-30 20:44:40
Relativity1618 2012-03-30 20:45:18
so 31*32/2=496
Relativity1618 2012-03-30 20:45:18
496
alex31415 2012-03-30 20:45:18
496
GeorgiaTechMan 2012-03-30 20:45:18
496
Duncanyang 2012-03-30 20:45:18
496
shreyash 2012-03-30 20:45:18
496
theone142857 2012-03-30 20:45:18
496
proglote 2012-03-30 20:45:18
31*32/2 = 496
himym83 2012-03-30 20:45:18
1+2+3+...+31 = (31)(32)/2 = 496
-Transcend- 2012-03-30 20:45:18
496
eccfcco15 2012-03-30 20:45:18
496
wmcho1007 2012-03-30 20:45:18
1+2+3+4+....+30+31=496
vjnmath 2012-03-30 20:45:18
31*16 = 496
numbertheorist17 2012-03-30 20:45:18
496
GeorgiaTechMan 2012-03-30 20:45:21
yay perfect numbers
Relativity1618 2012-03-30 20:45:21
3rd perfect number
copeland 2012-03-30 20:45:26
Since the intervals are disjoint we need to add the number of elements in all of them:$1+2+3+\cdots+31=\frac{31\cdot32}2=\boxed{496}.$
copeland 2012-03-30 20:45:34
Phew. 10 down.
copeland 2012-03-30 20:45:37
Man we're good.
danielguo94 2012-03-30 20:46:02
Yay.
-Transcend- 2012-03-30 20:46:02
yay!
Duncanyang 2012-03-30 20:46:02
yay.
copeland 2012-03-30 20:46:06
11. Let $f_1(x) = \frac{2}{3} - \frac{3}{3x + 1}$, and for $n \ge 1$, define $f_n(x) = f_1 (f_{n - 1}(x))$. The value of $x$ that satisfies $f_{1001}(x) = x - 3$ can be expressed in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
copeland 2012-03-30 20:46:15
Does anybody know the name of this type of function?
brokenfixer 2012-03-30 20:46:47
moebius transformations
copeland 2012-03-30 20:46:52
This type of of function is a called a Mobius transformation.
copeland 2012-03-30 20:46:56
(There should be dots: Möbius, but they're annoying to type.)
copeland 2012-03-30 20:47:00
A Mobius transformation has the form $g(x)=\frac{ax+b}{cx+d}$. Ours is in disguise though:
copeland 2012-03-30 20:47:06
$f_1(x)=\frac23-\frac{3}{3x+1}=\frac{6x+2}{9x+3}-\frac{9}{9x+3}=\frac{6x-7}{9x+3}.$
copeland 2012-03-30 20:47:12
(Henceforth I'm going to use $f$ to denote $f_1$, since that's easier.)
copeland 2012-03-30 20:47:16
When you see a Mobius transformation on a math contest, what's the first thing you should assume?
danielguo94 2012-03-30 20:48:07
cyclicity
himym83 2012-03-30 20:48:07
function is cyclic
ABCDE 2012-03-30 20:48:07
it repeats?
Imsacred 2012-03-30 20:48:07
it cycles?
sonchecky 2012-03-30 20:48:07
$f_n(x) = x$ for some n
CantonMathGuy 2012-03-30 20:48:07
it repeats
wmcho1007 2012-03-30 20:48:07
it's going to repeat after some time
lucylai 2012-03-30 20:48:07
it's cyclic
copeland 2012-03-30 20:48:11
It's probably periodic (meaning applying it the right number of times gives back the identity). Most math contest Mobius transformations are periodic. There's a really good hint to this in the problem, too: we want to iterate the function 1001 times.
copeland 2012-03-30 20:48:21
Don't get the wrong idea here. Mobius transformations are very interesting and useful in mathematics. If you pull one out of the blue, it almost certainly will not be periodic. If you meet one at your next swim meet, it's probably not periodic. If you run across one on a math contest, however, it is likely that it will turn out to be periodic.
copeland 2012-03-30 20:48:32
So our goal now should be to find the period.
copeland 2012-03-30 20:48:45
We could iterate our function, but that sounds tedious. is there a nice number we should start with instead?
centralbs 2012-03-30 20:49:24
0
theone142857 2012-03-30 20:49:24
x=0
Relativity1618 2012-03-30 20:49:24
0
zinko1991 2012-03-30 20:49:24
0
theone142857 2012-03-30 20:49:24
x=7/6
brokenfixer 2012-03-30 20:49:24
x = 0
copeland 2012-03-30 20:49:28
There are actually two nice choices:
copeland 2012-03-30 20:49:32
We can apply the function to $x=0$ to get $f(0)=\frac{-7}3.$
copeland 2012-03-30 20:49:36
We could also apply the function to $x=\frac76$ to get $\frac{6\cdot\frac76-7}{\text{stuff}}=\frac{7-7}{\text{stuff}}=0.$
copeland 2012-03-30 20:49:59
Yeah, I'm down with stuff, too. Stuff rules.
zinko1991 2012-03-30 20:50:04
i like stuff
copeland 2012-03-30 20:50:10
You know what else rules? Things rule.
copeland 2012-03-30 20:50:13
What is $f\left(-\tfrac73\right)$?
lucylai 2012-03-30 20:51:17
7/6 wow
alex31415 2012-03-30 20:51:17
7/6
7/6
ABCDE 2012-03-30 20:51:17
7/6
Woah, $\frac{7}{6}$
nackster12 2012-03-30 20:51:17
7/6
googol.plex 2012-03-30 20:51:17
7/6
ptes77 2012-03-30 20:51:17
7/6
Duncanyang 2012-03-30 20:51:17
2/3+1/2=7/6
Relativity1618 2012-03-30 20:51:17
$\frac{7}{6}$
Yongyi781 2012-03-30 20:51:17
7/6
copeland 2012-03-30 20:51:19
$f\left(-\tfrac73\right)=\frac{6\left(-\tfrac73\right)-7}{9\left(-\tfrac73\right)+3}=\frac{6(-7)-7\cdot3}{9(-7)+3\cdot3}=\frac{9(-7)}{9(-6)}=\frac76$
copeland 2012-03-30 20:51:32
And what is $f\left(\tfrac76\right)$?
Relativity1618 2012-03-30 20:51:56
and f(7/6)=0
Relativity1618 2012-03-30 20:51:56
so the period is 3
numbertheorist17 2012-03-30 20:51:56
0
numbertheorist17 2012-03-30 20:51:56
0
nmandi7 2012-03-30 20:51:56
0
nackster12 2012-03-30 20:51:56
0
himym83 2012-03-30 20:51:56
zero
theone142857 2012-03-30 20:51:56
0
ABCDE 2012-03-30 20:51:56
0!
Xcellence 2012-03-30 20:51:56
0
0
flyrain 2012-03-30 20:51:56
0!
kmc927 2012-03-30 20:51:56
0
shreyash 2012-03-30 20:51:56
0
loserboy 2012-03-30 20:51:56
0
19oshawott98 2012-03-30 20:51:56
0
$0$! It's repeating! :D
copeland 2012-03-30 20:52:00
We already know that this gives 0.
copeland 2012-03-30 20:52:04
This does not prove that $f$ has period 3. This proves only that if $f$ is periodic at all, then the period is a multiple of 3. That's OK. For now let's work on the assumption that $f$ does have period 3. Therefore $f(f(f(x)))=x$ for every $x$.
copeland 2012-03-30 20:52:28
(Incidentally, I conjectured the period was 6 when I first looked at this problem.)
copeland 2012-03-30 20:52:36
Periodicity means that $f_{n+3}=f_n$. In particular we are conjecturing that $f_{1001}=f_2$.
copeland 2012-03-30 20:52:48
So next we ought to try to solve the equation $f_2(x)=x-3$.
copeland 2012-03-30 20:52:53
Darn! This means we have to apply $f$ twice, right?
brokenfixer 2012-03-30 20:53:43
take f(x-3)
Duncanyang 2012-03-30 20:53:43
iterate this.
copeland 2012-03-30 20:53:46
If $f$ is periodic, then we can apply $f$ to this equation to get $f_3(x)=f(x-3)$ so $f(x-3)=x$. That should be easier to solve.
copeland 2012-03-30 20:53:58
copeland 2012-03-30 20:54:01
We want this to be equal to $x$, so we want to solve $x=\frac{6x-25}{9x-24}.$
copeland 2012-03-30 20:54:08
Cross-multiplying gives $9x^2-24x=6x-25$or$9x^2-30x+25=0.$
copeland 2012-03-30 20:54:13
What are the solutions to this quadratic?
alex31415 2012-03-30 20:54:42
x=5/3 is the only solution
Duncanyang 2012-03-30 20:54:42
x=5/3 only
success 2012-03-30 20:54:42
5/3
sonchecky 2012-03-30 20:54:42
(3x-5)^2 = 0, so x = 5/3
ytao 2012-03-30 20:54:42
5/3 with multiplicity 2.
kmc927 2012-03-30 20:54:42
5/3
herro66 2012-03-30 20:54:42
5/3
lucylai 2012-03-30 20:54:42
double root, 5/3
nackster12 2012-03-30 20:54:42
5/3
19oshawott98 2012-03-30 20:54:42
5/3
It's a perfect square! So $x=\frac{5}{3}$
theGCF 2012-03-30 20:54:42
5/3
herro66 2012-03-30 20:54:42
5/3
joshxiong 2012-03-30 20:54:42
(3x-5)^2=0, so 5/3
coldsummer 2012-03-30 20:54:42
it has solution 5/3
copeland 2012-03-30 20:54:46
The quadratic is a square! $9x^2-30x+25=(3x-5)^2.$ The only solution is $\frac53$.
copeland 2012-03-30 20:54:49
Remember, though, that we assumed $f$ was periodic. We actually don't need anything that strong to show that $f_{1001}(\frac53)=\frac53-3.$ We can look just as what successively applications of $f$ do to the value $\frac53$.
copeland 2012-03-30 20:55:03
Notice $f_1\left(\frac53\right)=\frac23-\frac3{3\cdot\frac53+1}=\frac23-\frac12=\frac16$.
copeland 2012-03-30 20:55:22
Then $f_2\left(\frac53\right)=f\left(\frac16\right)=\frac23-\frac3{3\cdot\frac16+1}\frac23-2=-\frac43$.
copeland 2012-03-30 20:55:32
Finally we note that $-\frac43=\frac53-3$ and $f\left(\frac53-3\right)=\frac53$, so these three values do repeat, and $f_{1001}\left(\frac53\right)=-\frac43=\frac53-3$.
copeland 2012-03-30 20:55:42
So. . . .
shreyash 2012-03-30 20:56:37
theone142857 2012-03-30 20:56:37
008
alex31415 2012-03-30 20:56:37
zinko1991 2012-03-30 20:56:37
-Transcend- 2012-03-30 20:56:37
008
GeorgiaTechMan 2012-03-30 20:56:37
mathswimmer 2012-03-30 20:56:37
008
googol.plex 2012-03-30 20:56:37
the answer is $\boxed {8}$
copeland 2012-03-30 20:56:40
The answer to the problem is $5+3=\boxed{008}$.
copeland 2012-03-30 20:57:19
12. For a positive integer $p$, define the positive integer $n$ to be $p$-safe if $n$ differs in absolute value by more than 2 from all multiples of $p$. For example, the set of 10-safe numbers is $\{3, 4, 5, 6, 7, 13, 14, 15, 16, 17, 23, \dots\}$. Find the number of positive integers less than or equal to 10,000 which are simultaneously 7-safe, 11-safe, and 13-safe.
copeland 2012-03-30 20:57:28
How can we tell if a number is 7-safe?
Relativity1618 2012-03-30 20:58:03
3 or 4 mod 7
GeorgiaTechMan 2012-03-30 20:58:03
congruent to 3 or 4 mod 7
alex31415 2012-03-30 20:58:03
if it is 3 or 4 mod 7
numberdance 2012-03-30 20:58:03
3 or 4 mod 7
googol.plex 2012-03-30 20:58:03
It's 3 or 4 mod 7
awesomemathlete 2012-03-30 20:58:03
3 or 4 mod 7
nackster12 2012-03-30 20:58:03
the number is congruent to 3 or 4 mod 7
ABCDE 2012-03-30 20:58:03
n==3 or 4 (mod 7)
Imsacred 2012-03-30 20:58:03
n is 7-safe if n-3 or n-4 is a multiple of 7
copeland 2012-03-30 20:58:06
The 7-safe numbers are the ones that are congruent to 3 or 4 mod 7.
copeland 2012-03-30 20:58:09
How can we tell if a number is 11-safe?
alex31415 2012-03-30 20:58:44
3-8 mod 11
himym83 2012-03-30 20:58:44
3,4,5,6,7,8 mod 7
hodgeheg 2012-03-30 20:58:44
if it is congruent to 3, 4, 5, 6, 7, 8, mod 11
ahaanomegas 2012-03-30 20:58:44
It is 3, 4, 5, 6, 7, or 8 (mod 11).
googol.plex 2012-03-30 20:58:44
3, 4, 5, 6, 7, 8 mod 11
-Transcend- 2012-03-30 20:58:44
3, 4, 5, 6, 7, 8 mod 11
19oshawott98 2012-03-30 20:58:44
3,4,5,6,7,8mod11
copeland 2012-03-30 20:58:48
The 11-safe numbers are the ones that are congruent to 3, 4, 5, 6, 7, or 8 mod 11.
copeland 2012-03-30 20:58:51
How can we tell if a number is 13-safe?
awesomemathlete 2012-03-30 20:59:40
3-10 mod 13
ptes77 2012-03-30 20:59:40
3-10 mod 13
sonchecky 2012-03-30 20:59:40
congruent to 3-10 mod 13
Mary_Posa 2012-03-30 20:59:40
3-10 mod 13
-Transcend- 2012-03-30 20:59:40
3 to 10 mod 13
brokenfixer 2012-03-30 20:59:40
same: congruent to 3,4,5,6,7,8,9,10 mod 13
3,4,5,6,7,8,9,10 mod 13
ahaanomegas 2012-03-30 20:59:40
It is 3, 4, 5, 6, 7, 8, 9, or 10 (mod 13).
numberdance 2012-03-30 20:59:40
Congruent to 3, 4, 5, 6, 7, 8, 9, or 10 mod 13
Xcellence 2012-03-30 20:59:40
must be congruent to 3,4,5,6,7,8,9,10 mod 13
wmcho1007 2012-03-30 20:59:40
3,4,5,6,7,8,9,10 mod 13
copeland 2012-03-30 20:59:45
The 13-safe numbers are the ones that are congruent to 3, 4, 5, 6, 7, 8, 9, and 10 mod 13.
copeland 2012-03-30 20:59:52
What theorem do we want to apply here?
Relativity1618 2012-03-30 21:00:35
chinese remainder theorem
GeorgiaTechMan 2012-03-30 21:00:35
Chinese remainder theorem kills this problem
19oshawott98 2012-03-30 21:00:35
we can use the CRT?
Relativity1618 2012-03-30 21:00:35
chinese remainder theorem
brokenfixer 2012-03-30 21:00:35
chinese remainder
mcdonalds106_7 2012-03-30 21:00:35
CRT
himym83 2012-03-30 21:00:35
chinese remainder theorem?
alex31415 2012-03-30 21:00:35
CHinese remainder theorem?
19oshawott98 2012-03-30 21:00:35
Chinese remainder theorem
nackster12 2012-03-30 21:00:35
chinese remainder theorem
MNL9082 2012-03-30 21:00:35
chinese remainder theorem
theone142857 2012-03-30 21:00:35
Chinese remainder thereom
danielguo94 2012-03-30 21:00:35
chinese remainder
copeland 2012-03-30 21:00:45
The Chinese Remainder Theorem tells us that for every $x$, $y$, and $z$, there is exactly one residue class modulo $7\cdot11\cdot13=1001$ such that the numbers $N$ in this residue class simultaneously solve
\begin{align*}
N\equiv x\pmod 7\\
N\equiv y\pmod 11\\
N\equiv z\pmod 13
\end{align*}
copeland 2012-03-30 21:00:51
What does this tell us about the number of integers from 1-1001that are simultaneously 7-safe, 11-safe, and 13-safe?
alex31415 2012-03-30 21:02:40
There are 96 of them?
danielguo94 2012-03-30 21:02:40
2*6*8 numbers
mcdonalds106_7 2012-03-30 21:02:40
there are 2*6*8 of them
sonchecky 2012-03-30 21:02:40
There are 2 * 6 * 8 = 96 between 1 and 1001.
There are 2*6*8=96
numbertheorist17 2012-03-30 21:02:40
96
nackster12 2012-03-30 21:02:40
there are 2*6*8=96 of them
numberdance 2012-03-30 21:02:40
There are 2 (for 7)*6 (for 11)*8 (for 13) from 1-1001, so 96.
awesomemathlete 2012-03-30 21:02:40
there are 2*6*8
Relativity1618 2012-03-30 21:02:40
they satisfy one of 96 linear congruences
CantonMathGuy 2012-03-30 21:02:40
2*6*8
brokenfixer 2012-03-30 21:02:40
there are 2 * 6 * 8 = 96 distinct possibilities from 1 ... 1001
copeland 2012-03-30 21:02:42
There are 2 choices (either 3 or 4) for the number's residue class mod 7. There are 6 choices (3-8) for the number's residue class mod 11. There are 8 choices (3-10) for the number's residue class mod 13. For each of these $2\cdot6\cdot8=96$ choices, there is exactly one number from 1 through 1001 that is 7-, 11-, and 13-safe (we'll call these numbers safe.
copeland 2012-03-30 21:02:46
What else?
alex31415 2012-03-30 21:04:10
there are 960 of them from 1-10010
brokenfixer 2012-03-30 21:04:10
they repeat every 1001
nackster12 2012-03-30 21:04:10
from 1-10010 there are 96*10=960 safe numbers
PERFECTION 2012-03-30 21:04:10
from 1 to 10010, there are 960 numbers that match the conditions
This repeats for the next multiples of 1001, so there are 96*10=960 through 10010
mcdonalds106_7 2012-03-30 21:04:10
so there are 960 from 1 to 10010... now cut out the end cases
copeland 2012-03-30 21:04:14
The same holds for 1002-2002. There are 96 safe integers in this interval as well.
copeland 2012-03-30 21:04:17
There are 96 safe integers in each of the intervals 2003-3003, 3004-4004, all the way up to 8009-9009, and 9010-10010.
copeland 2012-03-30 21:04:21
The first 9 intervals give us $9\cdot96$ total safe integers. However the last interval is only a partial interval. How can we count the number of safe integers in 9010-10000?
googol.plex 2012-03-30 21:05:27
subtract the ones from 10000-10010
GeorgiaTechMan 2012-03-30 21:05:27
just count the numbers from 10001-10010 that are p-safe and take 96- that number
CantonMathGuy 2012-03-30 21:05:27
subtract ones from 10000-10010
brokenfixer 2012-03-30 21:05:27
find the safe integers in 10,001 to 10,010.
sonchecky 2012-03-30 21:05:27
Go through 9010 - 10010 and then subtract out the extra ones above 10,000
ABCDE 2012-03-30 21:05:27
subtract the number of safe integers in 10000-10010
nackster12 2012-03-30 21:05:27
960 minus the number of safe numbers in [10001, 10010]
copeland 2012-03-30 21:05:31
We know that there are 96 safe integers in 9010-10010. All we need to do is subtract the safe integers from 10001 to 10010.
copeland 2012-03-30 21:05:34
Is 10010 safe?
himym83 2012-03-30 21:06:11
no, multiple of 1001
wmcho1007 2012-03-30 21:06:11
uh. no.
coldsummer 2012-03-30 21:06:11
NO WAY!!!!!!!!!!!!!!!!11
-Transcend- 2012-03-30 21:06:11
no
lucylai 2012-03-30 21:06:11
nope
numberdance 2012-03-30 21:06:11
No -- it's divisible by all of 7, 11, and 13.
nackster12 2012-03-30 21:06:11
no its divisible by 7, 11, and 13
awesomemathlete 2012-03-30 21:06:11
of course not
copeland 2012-03-30 21:06:12
No! This is a multiple of all 3.
copeland 2012-03-30 21:06:15
Is 10009 safe?
noo
sinnoel 2012-03-30 21:07:06
No
heliootrope 2012-03-30 21:07:06
No, it's too close to 10010
mathnerd101 2012-03-30 21:07:06
no it is one less than a multiple of 7, 11, 13
no, too close to 10010
Relativity1618 2012-03-30 21:07:06
within 1 of a multiple of all 3
alex31415 2012-03-30 21:07:06
No, it is dangerous to 7
nackster12 2012-03-30 21:07:06
no, it's congruent to 6 modulo 7
copeland 2012-03-30 21:07:13
No. This is within 1 of each of them.
copeland 2012-03-30 21:07:16
Now is a good time for a table:
copeland 2012-03-30 21:07:20
copeland 2012-03-30 21:07:22
Which of these numbers are safe?
Relativity1618 2012-03-30 21:08:10
10006 and 10007
himym83 2012-03-30 21:08:10
10006, 10007
mathnerd101 2012-03-30 21:08:10
10007 and 10008
vjnmath 2012-03-30 21:08:10
1006,1007
linpaws 2012-03-30 21:08:10
10006 and 10007
10006 and 10007
coldsummer 2012-03-30 21:08:10
10007, 10006,
zinko1991 2012-03-30 21:08:10
10007, 10006
MNL9082 2012-03-30 21:08:10
10006 10007
copeland 2012-03-30 21:08:15
copeland 2012-03-30 21:08:17
Only 2 of the numbers are safe. How many total safe numbers are there from 1 to 10000?
sonchecky 2012-03-30 21:09:18
960 - 2 = 958
Tuxianeer 2012-03-30 21:09:18
958
19oshawott98 2012-03-30 21:09:18
958!!
Relativity1618 2012-03-30 21:09:18
960-2=958
96*10-2=958
ahaanomegas 2012-03-30 21:09:18
958
loserboy 2012-03-30 21:09:18
958
hodgeheg 2012-03-30 21:09:18
There are 960-2=958 safe numbers. HOORAY... I understood the question
bojobo 2012-03-30 21:09:18
958
copeland 2012-03-30 21:09:25
There are $9\cdot96+96-2=\boxed{958}$ total safe numbers through 10000.
ptes77 2012-03-30 21:09:49
3 to go.
copeland 2012-03-30 21:09:52
Phew.
copeland 2012-03-30 21:10:13
13. Equilateral $\triangle ABC$ has side length $\sqrt{111}$. There are four distinct triangles $AD_1 E_1$, $AD_1 E_2$, $AD_2 E_3$, and $AD_2 E_4$, each congruent to $\triangle ABC$, with $BD_1 = BD_2 = \sqrt{11}$. Find $\displaystyle \sum_{k = 1}^4 (CE_k)^2$.
copeland 2012-03-30 21:10:30
Where do we start?
GoldenFrog1618 2012-03-30 21:11:08
Since Phi_6(x) is the largest quadratic cyclotomic polynomial, can mobious transformations only have a period of 1 or 3?
edisonchew240 2012-03-30 21:11:08
Diagram
ABCDE 2012-03-30 21:11:08
draw a diagram but look at the triangles separately
Mary_Posa 2012-03-30 21:11:08
draw a diagram!
coldsummer 2012-03-30 21:11:08
diagram?
brokenfixer 2012-03-30 21:11:08
draw a picture
Tuxianeer 2012-03-30 21:11:08
diagram
numbertheorist17 2012-03-30 21:11:08
diagram
copeland 2012-03-30 21:11:12
We should draw a picture.
copeland 2012-03-30 21:11:17
copeland 2012-03-30 21:11:19
That's our first triangle. Where are the points $D_1$ and $D_2$?
mathswimmer 2012-03-30 21:12:38
they all share point A and have same sidelengths-> circle and rotation
herro66 2012-03-30 21:12:38
equidistant from B
ABCDE 2012-03-30 21:12:38
draw circle with center c and radius sqrt(111) and circle with center b with radius sqrt(11) and D_1 and D_2 are the intersections
coldsummer 2012-03-30 21:12:38
on a circle with radius 11^1/2 with center B?
brokenfixer 2012-03-30 21:12:38
draw circle around A of radius sqrt(111)
sonchecky 2012-03-30 21:12:38
$sqrt(11)$ units away from B on separate arcs around C and A
copeland 2012-03-30 21:12:41
These two points are on the circle of radius $\sqrt{11}$ centered at $B$, and they are also a distance of $\sqrt{111}$ from $A$. Therefore these two points are the intersection points of the circles of radius $\sqrt{11}$ and $\sqrt{111}$ centered at $B$ and $A$ respectively.
copeland 2012-03-30 21:12:46
copeland 2012-03-30 21:12:48
Now we just draw equilateral triangles.
copeland 2012-03-30 21:12:53
copeland 2012-03-30 21:12:59
Wow. Inscrutable. Let's focus on just one of those triangles:
copeland 2012-03-30 21:13:05
copeland 2012-03-30 21:13:08
How can we compute the distance $CE_1$?
GeorgiaTechMan 2012-03-30 21:13:54
that's easy. sqrt11!
edisonchew240 2012-03-30 21:13:54
CE_{1}=BD_{1}
brokenfixer 2012-03-30 21:13:54
BD1
Tuxianeer 2012-03-30 21:13:54
it is sqrt(11)
ABCDE 2012-03-30 21:13:54
it's symetrical so it's equal to BD_1 so sqrt(11)
GeorgiaTechMan 2012-03-30 21:13:54
ACE1 similar to ABD1
heliootrope 2012-03-30 21:13:54
it's rotated the same distance away from C as D_1 is from B
nackster12 2012-03-30 21:13:54
its just sqrt(11) since ABD_1 is congruent to ACE_1
sonchecky 2012-03-30 21:13:54
$CE_1 = BD_1 = \sqrt(11)$
herro66 2012-03-30 21:13:54
its sqrt{11}
copeland 2012-03-30 21:13:57
This one's easy! $CE_1$ is equal to $BD_1$, since these two triangles are related by a rotation by some angle. The triangles $BAD_1$ and $CAE_1$ are SAS congruent, so
copeland 2012-03-30 21:14:05
$CE_1^2=BD_1^2=11$.
copeland 2012-03-30 21:14:10
Let's actually give this angle a name, since it looks like it's going to be important. Let $\theta=\angle CAE_1$.
copeland 2012-03-30 21:14:15
Good. Maybe the others will be this easy, too.
copeland 2012-03-30 21:14:20
copeland 2012-03-30 21:14:25
Or maybe not. . .
copeland 2012-03-30 21:14:29
What can we use to compute this distance?
himym83 2012-03-30 21:14:55
law of cosines
theone142857 2012-03-30 21:14:55
law of cos
Relativity1618 2012-03-30 21:14:55
Law of Cosines
Duncanyang 2012-03-30 21:14:55
law of cosines
ahaanomegas 2012-03-30 21:14:55
Law of Cosines, maybe?
eccfcco15 2012-03-30 21:14:55
law of cosines
alex31415 2012-03-30 21:14:55
law of cosines
copeland 2012-03-30 21:15:02
Since we have a lot of distances we know and angles we either know or can compute, it looks like the Law of Cosines could be very effective here.
copeland 2012-03-30 21:15:24
We want $CE_2$ so we ought to think about $\triangle ACE_2$.
copeland 2012-03-30 21:15:43
The Law of Cosines tells us that $CE_2^2=AE_2^2+AC^2-2AE_2\cdot AC\cos\angle CAE_2.$
copeland 2012-03-30 21:15:47
All of those other lengths are known. In fact they're all $\sqrt{111}$. Therefore we have$CE_2^2=111+111+2\cdot\sqrt{111}\sqrt{111}\cos\angle CAE_2$ so $CE_2^2=222(1-\cos\angle CAE_2).$
copeland 2012-03-30 21:15:57
Another angle! Can we write $\angle CAE_2$ in terms of $\theta$?
himym83 2012-03-30 21:17:13
120-theta
ABCDE 2012-03-30 21:17:13
120-theta
proglote 2012-03-30 21:17:13
yes, 120 - theta
nackster12 2012-03-30 21:17:13
120 degrees minus theta
sonchecky 2012-03-30 21:17:13
$120 - \theta$
brokenfixer 2012-03-30 21:17:13
120 - theta i think
-Transcend- 2012-03-30 21:17:13
120-theta?
copeland 2012-03-30 21:17:21
Since $\theta=\angle BAD_1$, we know $\angle BAE_2=\frac\pi3-\theta$. Therefore $\angle CAE_2=\angle CAB+\angle BAE_2=\frac\pi3+\frac\pi3-\theta.$
copeland 2012-03-30 21:17:23
$CE_2^2=222(1-\cos(\tfrac{2\pi}3-\theta)).$
copeland 2012-03-30 21:17:35
I don't want to deal with that cosine yet. A lot of times in problems like these terms like that will cancel and we don't want to waste time. (Plus we can compare our expressions later and maybe find a mistake!)
copeland 2012-03-30 21:17:39
2 down. . . .
copeland 2012-03-30 21:17:49
copeland 2012-03-30 21:18:05
Again we want the Law of Cosines. What is $\angle CAE_3$?
sonchecky 2012-03-30 21:20:06
$\theta$
alex31415 2012-03-30 21:20:06
Is it the same as theta?
it's equal to $\angle BAD_2$
numbertheorist17 2012-03-30 21:20:06
proglote 2012-03-30 21:20:06
isn't CE_3 = BD_2 = sqrt(11)?
numbertheorist17 2012-03-30 21:20:06
copeland 2012-03-30 21:20:09
Notice that $AD_1BD_2$ is a kite.
copeland 2012-03-30 21:20:13
copeland 2012-03-30 21:20:25
Therefore $\angle CAD_2=\angle CAD_1=\theta$. By symmetry $\angle CAE_3=\angle BAD_2=\theta$.
copeland 2012-03-30 21:20:35
The LOC was a red herring.
copeland 2012-03-30 21:20:36
We again have congruence, $\triangle CAE_3\cong \triangle BAD_2$, so
copeland 2012-03-30 21:20:39
$CE_3^2=BD_2^2=11$.
copeland 2012-03-30 21:21:09
One more;
copeland 2012-03-30 21:21:12
copeland 2012-03-30 21:21:14
Finally, $CE_4$. What is $\angle CAE_4$?
himym83 2012-03-30 21:22:11
120+theta
heliootrope 2012-03-30 21:22:11
120 + theta
ABCDE 2012-03-30 21:22:11
120+theta
proglote 2012-03-30 21:22:11
120 + theta
-Transcend- 2012-03-30 21:22:11
120 + theta
sonchecky 2012-03-30 21:22:11
$120 + \theta$
120+theta
Tuxianeer 2012-03-30 21:22:11
120+theta
nackster12 2012-03-30 21:22:11
120+theta
copeland 2012-03-30 21:22:14
$\angle CAE_4=\angle CAB+\angle BAD_2+\angle D_2AE_4=\frac{2\pi}3+\theta.$ Therefore
copeland 2012-03-30 21:22:19
$CE_4^2=222(1-\cos(\tfrac{2\pi}3+\theta)).$
copeland 2012-03-30 21:22:24
When we add these four values we get
copeland 2012-03-30 21:23:27
\begin{align*}CE_1^2+CE_2^2&+CE_3^2+CE_4^2=11+11\\&+222(1-\cos(\tfrac{2\pi}3-\theta))+222(1-\cos(\tfrac{2\pi}3+\theta)).\end{align*}
copeland 2012-03-30 21:23:31
$466-222(\cos(\tfrac{2\pi}3-\theta)+\cos(\tfrac{2\pi}3+\theta)).$
Relativity1618 2012-03-30 21:23:52
now we use the sum to product formula
theone142857 2012-03-30 21:23:52
2cos(2pi/3)cos(theta)
brokenfixer 2012-03-30 21:23:52
cos(a+b) + cos(a-b)
Relativity1618 2012-03-30 21:23:52
sum to product formula
copeland 2012-03-30 21:23:54
And. . .
mathswimmer 2012-03-30 21:26:02
-cos(theta)
GeorgiaTechMan 2012-03-30 21:26:02
sorry, -cos(theta)
-cos theta
copeland 2012-03-30 21:26:07
We could use $\cos(x+y)+\cos(x-y)=2\cos x\cos y$, to write$\cos(\tfrac{2\pi}3-\theta)+\cos(\tfrac{2\pi}3+\theta)=2\cos\tfrac{2\pi}3\cos\theta=-\cos\theta.$
copeland 2012-03-30 21:26:10
(Incidentally, this makes sense geometrically since the points $e^{i\theta}$, $e^{i(\theta+\frac{2\pi}3)}$ and $e^{i(\theta-\frac{2\pi}3)}$ are the vertices of an equilateral triangle with center of mass 0, so the real parts of these numbers must add to 0.)
copeland 2012-03-30 21:26:21
$466+222\cos\theta.$
copeland 2012-03-30 21:26:30
Hm. Of course we were going to have to figure something out about $\theta$. I guess now's the time. How can we find $\cos\theta$?
GeorgiaTechMan 2012-03-30 21:26:57
and we can use law of cosines on the sqrt111 sqrt111 sqrt 11 triangle to find out what cos theta is!
ABCDE 2012-03-30 21:26:57
law of cosines
LOC on $BAD_1$
himym83 2012-03-30 21:26:57
use law of cosines on sqrt(11), sqrt(111), sqrt(111) triangle
proglote 2012-03-30 21:26:57
law of cosines in triangle BAD_1
copeland 2012-03-30 21:27:01
We use the Law of Cosines on $\triangle ABD_1$.
copeland 2012-03-30 21:27:06
copeland 2012-03-30 21:27:10
$\cos\theta=\frac{\sqrt{111}^2+\sqrt{111}^2-\sqrt{11}^2}{2\sqrt{111}\cdot\sqrt{111}}=\frac{111+111-11}{222}=\frac{211}{222}$.
copeland 2012-03-30 21:27:15
So what is the sum?
numbertheorist17 2012-03-30 21:28:01
Ans. 677
alex31415 2012-03-30 21:28:01
Duncanyang 2012-03-30 21:28:01
677
nackster12 2012-03-30 21:28:01
466+211=677
numbertheorist17 2012-03-30 21:28:01
proglote 2012-03-30 21:28:01
677
sonchecky 2012-03-30 21:28:01
$466-222(-211/222) = 466+211=677$
edisonchew240 2012-03-30 21:28:01
Ans=466+211=677
numbertheorist17 2012-03-30 21:28:01
466+211=677
math-rules 2012-03-30 21:28:01
677
-Transcend- 2012-03-30 21:28:01
677
copeland 2012-03-30 21:28:04
The sum is $466+222\cdot{211}{222}=466+211=\boxed{677}$.
copeland 2012-03-30 21:28:13
Phew, that was 13. We have 2 to go.
copeland 2012-03-30 21:28:18
Still trucking?
brokenfixer 2012-03-30 21:28:52
in need of pudding
mathnerd101 2012-03-30 21:28:52
man im hungry for pudding :P
copeland 2012-03-30 21:28:54
I THINK IT AND YOU SAY IT! AMAZING!
copeland 2012-03-30 21:29:08
14. In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements to be different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when $N$ is divided by 1000.
GeorgiaTechMan 2012-03-30 21:29:23
whats up with pudding?
copeland 2012-03-30 21:29:25
Pudding rules, yo.
copeland 2012-03-30 21:29:37
Hmm. Again it's not clear what to do here, so where should we start?
mathnerd101 2012-03-30 21:30:21
picture?
Imsacred 2012-03-30 21:30:21
picture? at least thats what i did
draw a diagram?
ahaanomegas 2012-03-30 21:30:21
Read the problem at least 3 times. Process the words.
Titandrake 2012-03-30 21:30:21
draw some graphs?
copeland 2012-03-30 21:30:26
Yeah, let's draw some stuff.
brokenfixer 2012-03-30 21:30:31
what's with the "consider handshakings different" clause?
copeland 2012-03-30 21:30:47
Yeah, right? I read that 4 times before I decided it didn't say anything.
copeland 2012-03-30 21:31:13
When I started drawing pictures, I decided that 9 was WAY too big and started with smaller numbers instead.
copeland 2012-03-30 21:31:28
So let's begin with a smaller example. Let $n$ be the number of people.
copeland 2012-03-30 21:31:32
Let's start with $n=3$. How many ways are there for 3 people to shake hands where each person shakes hands with two others?
Tuxianeer 2012-03-30 21:32:08
1
heliootrope 2012-03-30 21:32:08
1
lucylai 2012-03-30 21:32:08
1
MathTwo 2012-03-30 21:32:08
1 way
proglote 2012-03-30 21:32:08
only one, each person shakes the hands of the other two
mcdonalds106_7 2012-03-30 21:32:08
1 way
copeland 2012-03-30 21:32:11
There's only one way to do this.
copeland 2012-03-30 21:32:16
copeland 2012-03-30 21:32:17
If $n=4$ how many ways are there to shake hands?
GeorgiaTechMan 2012-03-30 21:32:46
3
lucylai 2012-03-30 21:32:46
3
Duncanyang 2012-03-30 21:32:46
3
Tuxianeer 2012-03-30 21:32:46
3
brokenfixer 2012-03-30 21:32:46
two bowties and a loop?
nackster12 2012-03-30 21:32:46
3
3, right?
copeland 2012-03-30 21:32:51
Each person shakes hands with 2 people so he does not shake hands with one other person. There are 3 ways to match 4 people into pairs of uncordial pairs.
copeland 2012-03-30 21:32:57
copeland 2012-03-30 21:33:01
What do you notice?
it's a closed loop?
Imsacred 2012-03-30 21:34:01
or you can always represent this problem with a closed loop or a system of closed loops
copeland 2012-03-30 21:34:06
Since every person is connected to two other people, all of our graphs are "loops."
copeland 2012-03-30 21:34:10
By this we mean we start with some Person A. Person A shakes hand with Person B. Person B shakes hands with one other person, Person C. Person C shakes with one other person, Person D. Since there are only finitely many people, eventually we must come around to the other person who shook hands with Person A.
copeland 2012-03-30 21:34:21
When $n=9$, will all of the graphs be loops?
GeorgiaTechMan 2012-03-30 21:34:53
so we can either have 1 loop, 2 loops, or 3 loops
lucylai 2012-03-30 21:34:53
no there can be separate loops
nackster12 2012-03-30 21:34:53
you could have multiple loops
proglote 2012-03-30 21:34:53
there might be disjoint loops
copeland 2012-03-30 21:34:56
No, it's possible that the graph breaks into multiple loops:
copeland 2012-03-30 21:35:01
copeland 2012-03-30 21:35:03
However all 9 of the people must be in some loop or other.
copeland 2012-03-30 21:35:07
The graph could be a single loop of all 9 people.
copeland 2012-03-30 21:35:12
If the graph breaks into two loops, what are the possibilites for the number of people in each loop?
GeorgiaTechMan 2012-03-30 21:35:45
3 and 6, or 4 and 5
math-rules 2012-03-30 21:35:45
3-6, 4-5
ABCDE 2012-03-30 21:35:45
4&5 or 3&6
heliootrope 2012-03-30 21:35:45
3 and 6 or 4 and 5 in some order
proglote 2012-03-30 21:35:45
(3,6) and (4,5)
Imsacred 2012-03-30 21:35:45
3 and 6 or 4 and 5
dantx5 2012-03-30 21:35:45
3 and 6, 4 and 5
copeland 2012-03-30 21:35:49
Since every loop needs at least 3 people, two loops can be created with sizes 3 and 6 or 4 and 5.
copeland 2012-03-30 21:35:53
brokenfixer 2012-03-30 21:36:13
3 3 3
proglote 2012-03-30 21:36:13
only (3, 3, 3)
ABCDE 2012-03-30 21:36:13
3, 3, 3
himym83 2012-03-30 21:36:13
3, 3, and 3
alex31415 2012-03-30 21:36:13
3, 3, and 3
3,3,3
success 2012-03-30 21:36:13
3,3,3
copeland 2012-03-30 21:36:18
With three loops, all of the loops must be of size 3.
copeland 2012-03-30 21:36:21
Therefore we're down to our casework. We can either have a single loop of size 9, we can have two loops of sizes {6,3} or {5,4} or we could have three loops of sizes {3,3,3}.
copeland 2012-03-30 21:36:27
OK, good. Now we are set up to start counting.
copeland 2012-03-30 21:36:32
How many possible ways are there for 9 people to shake hands in a single loop?
GeorgiaTechMan 2012-03-30 21:37:48
8 factorial divided by 2
GeorgiaTechMan 2012-03-30 21:37:48
or 20160
ABCDE 2012-03-30 21:37:48
8!/2=20160
lucylai 2012-03-30 21:37:48
20,160
dantx5 2012-03-30 21:37:48
8!/2 ?
lucylai 2012-03-30 21:37:48
8!/2=20160
sonchecky 2012-03-30 21:37:48
8! / 2 = 20160
copeland 2012-03-30 21:37:52
We can construct a loop of 9 people in the same way as we would place 9 people around a round table. There are $8!$ ways to create such an arrangement, and we count these by finding where Person 1 is sitting and placing the remaining 8 people relative to him in any permutation. However there's a difference here: our loops don't have orientation, so reversing the order of the table gives the same loop.
copeland 2012-03-30 21:37:58
There are $\frac{8!}2$ ways to create a single loop of people shaking hands.
copeland 2012-03-30 21:38:05
We can compute this completely, or notice that $8\cdot7\cdot6\cdot5\cdot4\cdot3=(7\cdot3)(6\cdot8)(5\cdot4)=21\cdot48\cdot20=21\cdot960.$
copeland 2012-03-30 21:38:12
We only care about this value mod 1000, so$21\cdot960\equiv21\cdot(-40)\equiv-840\equiv160\pmod{1000}.$
copeland 2012-03-30 21:38:23
How many ways are there to create a pair of loops if one has size 3 and the other has size 6?
copeland 2012-03-30 21:39:42
Hmm. Maybe that's too many steps at once.
copeland 2012-03-30 21:39:48
First how many ways can we form the group of 3?
-Transcend- 2012-03-30 21:40:23
9 choose 3
brokenfixer 2012-03-30 21:40:23
9 choose 3
himym83 2012-03-30 21:40:23
9C3
Duncanyang 2012-03-30 21:40:23
9 choose 3
dantx5 2012-03-30 21:40:23
9 C 3
ABCDE 2012-03-30 21:40:23
9C3=84
math-rules 2012-03-30 21:40:23
9c3=84
proglote 2012-03-30 21:40:23
9C3 = 84
Tuxianeer 2012-03-30 21:40:23
9C3=84
wmcho1007 2012-03-30 21:40:26
9*8*7/6=84
copeland 2012-03-30 21:40:34
First we pick which people will be in the loop of size 3. There are $\binom93$ ways to do this.
copeland 2012-03-30 21:40:46
Once we make that choice, we need to arrange the remaining people into a loop of 6.
copeland 2012-03-30 21:40:51
How many ways are there to do that?
brokenfixer 2012-03-30 21:41:20
2!/2 * 5!/2 which is 5*4*3 = 60 ways
Tuxianeer 2012-03-30 21:41:20
1 way to loop group of 3, and 5!/2=60 WAYS FOR THE GROUP OF 6
lucylai 2012-03-30 21:41:20
5!/2=60
himym83 2012-03-30 21:41:20
5!/2
GeorgiaTechMan 2012-03-30 21:41:20
5!/2=60
nackster12 2012-03-30 21:41:20
5!/2=60
proglote 2012-03-30 21:41:20
5!/2 = 60
copeland 2012-03-30 21:41:35
Once the trio has been chosen, there are $\frac{5!}2$ ways to organize the subgroup of 6 people and only $\frac{2!}2=1$ way to arrange the subgroup of size 3.
copeland 2012-03-30 21:41:40
There are $\binom93\cdot\frac{5!}2\cdot1=\frac{9!5!}{2\cdot3!6!}$ ways to organize 9 people into a loop of size 6 and a loop of size 3.
copeland 2012-03-30 21:41:45
Again for some arithmetic. This one is easy enough to compute directly:$\frac{(9\cdot8\cdot7)(5\cdot4)}2=72\cdot7\cdot10=5040.$So we get another $5040\equiv40\pmod{1000}.$
copeland 2012-03-30 21:41:51
Likewise there are $\binom94\cdot\frac{4!}2\cdot\frac{3!}2=\frac{9!4!3!}{2\cdot2\cdot5!\cdot4!}$ ways to organize 9 people into a loop of size 5 and a loop of size 4.
copeland 2012-03-30 21:42:05
$\frac{9!4!3!}{2\cdot2\cdot5!\cdot4!}=\frac{9\cdot8\cdot7\cdot6\cdot3}2=(9\cdot6)(7\cdot3)\cdot\frac82=54\cdot21\cdot4=1134\cdot4.$
copeland 2012-03-30 21:42:10
Mod 1000 we get $1134\cdot4\equiv134\cdot4=536\pmod{1000}.$
copeland 2012-03-30 21:42:40
What about three loops of size 3? How many ways can we organize 9 people into such groups?
himym83 2012-03-30 21:44:13
9C3 * 6C3 * 3C3/3!
nackster12 2012-03-30 21:44:13
9C3*6C3/3!
ABCDE 2012-03-30 21:44:13
9C3*6C3/6
-Transcend- 2012-03-30 21:44:13
9C3 * 6C3 / 3! ?
wmcho1007 2012-03-30 21:44:21
(9C3*6C3*3C3)/6
brokenfixer 2012-03-30 21:44:21
9C3 * 6C3 / 3!
copeland 2012-03-30 21:44:22
We can paint the 9 people in three colors in $\binom 96\cdot\binom63$ ways. however the colors are irrelevant so we need to divide by $3!$ to handle overcounting.
copeland 2012-03-30 21:45:12
There is a total of $\frac{9\cdot8\cdot7\cdot6\cdot5\cdot4}{3!\cdot3!\cdot3!}=\frac{8\cdot7\cdot5\cdot4}{2\cdot2}=280$ ways for the 9 people to be partitioned into groups of 3.
copeland 2012-03-30 21:45:26
Our grand total is $160+40+536+280\equiv016\pmod{1000}.$
copeland 2012-03-30 21:45:32
The answer is $\boxed{016}$.
ahaanomegas 2012-03-30 21:45:38
copeland 2012-03-30 21:45:40
Whee!
copeland 2012-03-30 21:45:52
Hey, guess what?
Fire-Sword 2012-03-30 21:46:12
time for the last one
19oshawott98 2012-03-30 21:46:12
LAAST PROBLEM!!!!
-Transcend- 2012-03-30 21:46:12
Last one!
numbertheorist17 2012-03-30 21:46:12
pudding is good
alex31415 2012-03-30 21:46:12
Last Problem!
brokenfixer 2012-03-30 21:46:12
pudding!
nackster12 2012-03-30 21:46:12
last question!
ahaanomegas 2012-03-30 21:46:12
Last one, best one.
mcdonalds106_7 2012-03-30 21:46:12
Last problem~
-Transcend- 2012-03-30 21:46:12
You like pudding?
Duncanyang 2012-03-30 21:46:12
one more problem to go!!!!
MathTwo 2012-03-30 21:46:12
we have one problem left!
copeland 2012-03-30 21:46:18
Yeah, last problem.
copeland 2012-03-30 21:46:29
copeland 2012-03-30 21:46:47
copeland 2012-03-30 21:46:53
copeland 2012-03-30 21:47:00
That's clearly not enough. Do we know anything interesting about $E$ or $F?$
numbertheorist17 2012-03-30 21:48:27
equilateral triangle
EB=EC
proglote 2012-03-30 21:48:27
E is in the perpendicular bisector of BC
GeorgiaTechMan 2012-03-30 21:48:27
we can also do angle bisector and stewarts and then figure out EBC is equilateral.
sonchecky 2012-03-30 21:48:27
BEC is an equilateral triangle.
copeland 2012-03-30 21:48:31
We know that $E$ is the midpoint of arc $BEC,$ since $AE$ bisects $\angle CAB.$
copeland 2012-03-30 21:48:45
Let's keep that in mind.
copeland 2012-03-30 21:48:55
What other point wants to be labeled?
Duncanyang 2012-03-30 21:50:22
they interstctions.
dantx5 2012-03-30 21:50:22
bottom of small circle?
math-rules 2012-03-30 21:50:22
intersection of AF and smaller circle
copeland 2012-03-30 21:50:31
OK, that's two suggestions of points that are already on the diagram.
copeland 2012-03-30 21:50:50
I like the point on CB because it has more to do with the things we know (ABC).
copeland 2012-03-30 21:50:59
We will think about the other point where the little circle hits $BC.$ Let's call that $M.$
copeland 2012-03-30 21:51:04
copeland 2012-03-30 21:51:07
Let's think about the little circle for a second.
copeland 2012-03-30 21:51:11
copeland 2012-03-30 21:51:19
Does the little circle tell us anything interesting about $M?$
Duncanyang 2012-03-30 21:52:00
EM perpendicular to CB
brokenfixer 2012-03-30 21:52:00
EMD is right
himym83 2012-03-30 21:52:00
angle EMD is 90
Fire-Sword 2012-03-30 21:52:00
ummm its perpendicular to CB?
$\angle EDM=90^\circ$
EDM is right
brokenfixer 2012-03-30 21:52:00
EM is perpendicular to BC because EMD is right angle
heliootrope 2012-03-30 21:52:02
EMD is right
copeland 2012-03-30 21:52:05
We know that $\angle EMD$ is a right angle because $ED$ is a diameter of the small circle. So?
numbertheorist17 2012-03-30 21:52:46
M is midpoint?
MathTwo 2012-03-30 21:52:46
M is the midpoint of BC
sonchecky 2012-03-30 21:52:46
M is the midpoint of BC
proglote 2012-03-30 21:52:46
so M is the midpoint of BC
so EM is perpendicular to BC!
brokenfixer 2012-03-30 21:52:46
EM is perpendicular bisector to BC
copeland 2012-03-30 21:52:52
Since $E$ is the midpoint of arc $BEC$ and $EM$ is perpendicular to $BC,$ we know that $EM$ bisects chord $BC.$ That is, $M$ is the midpoint of $BC.$
copeland 2012-03-30 21:52:56
copeland 2012-03-30 21:53:01
We've stumbled on a couple interesting facts here: that $\angle EMD$ is a right angle, and that $M$ is the midpoint of $BC.$ When we stumble on interesting facts, we should check out where they lead. Where might we go from here?
GeorgiaTechMan 2012-03-30 21:54:24
DFE is right?
The center of the large circle lies on EM?
dantx5 2012-03-30 21:54:24
draw diameter of big circle
copeland 2012-03-30 21:54:29
There are a couple directions we could go. First, seeing the right angle at $\angle EMD,$ we realize that there's a right angle at $\angle DFE,$ again because $ED$ is a diameter (or because FEMD is cyclic).
copeland 2012-03-30 21:54:40
Since $M$ is the midpoint of $BC$ and $E$ is the midpoint of arc $BEC,$ extending $EM$ to hit the large circle again gives us a diameter of the large circle:
copeland 2012-03-30 21:54:52
copeland 2012-03-30 21:55:00
What does this diagram make us wonder?
proglote 2012-03-30 21:55:56
so F, D, S are collinear
sonchecky 2012-03-30 21:55:56
if F, D, and S are collinear
eccfcco15 2012-03-30 21:56:01
are F, D, and S colinear
brokenfixer 2012-03-30 21:56:01
does FD continue through S
copeland 2012-03-30 21:56:07
It sure looks like the extension of $FD$ hits $S.$ Does it?
proglote 2012-03-30 21:57:05
yes, since EF is perpendicular to FD, it should hit the circle at the antipode of E, which is just S
himym83 2012-03-30 21:57:05
yes: EFS is a right angle since ES is the diameter
ES is a diameter, EFD is a right angle, and F is on the circle, so yes!
sonchecky 2012-03-30 21:57:05
Yes because EFD is right so it must subtend a semicircle and the endpoints must be a diameter of the large circle
brokenfixer 2012-03-30 21:57:05
90 degree angle at F must hit both ends of the diameter ES
dantx5 2012-03-30 21:57:05
yes, because FED is a right angle, and ES is a diameter.
copeland 2012-03-30 21:57:09
Because $ES$ is a diameter of the large circle, $\angle EFS$ is a right angle. But we already know that $\angle EFD$ is right, so $D$ is on $FS:$
copeland 2012-03-30 21:57:15
copeland 2012-03-30 21:57:24
We've found another surprise in that $EM$ and $FD$ meet on the circle at point $S,$ so we should investigate that point and how it relates to the rest of the diagram. See anything interesting?
copeland 2012-03-30 21:59:32
Several of you have found interesting similar triangles. These can be helpful for various tactics, but most of the suggestions give us relationships between unknown lengths.
copeland 2012-03-30 21:59:37
Can we find any more right angles in here?
himym83 2012-03-30 22:00:36
EAS
proglote 2012-03-30 22:00:36
EAS is right since F, E, A, S are concyclic
sonchecky 2012-03-30 22:00:36
angle EAS
edisonchew240 2012-03-30 22:00:36
<EAS
math-rules 2012-03-30 22:00:36
EAS is right
copeland 2012-03-30 22:00:40
Right angles lead to cyclic quadrilaterals and diameters lead to right angles. $\angle DMS = 90^\circ$, and $\angle SAE = 90^\circ$ because $ES$ is a diameter. $ADMS$ is cyclic!
copeland 2012-03-30 22:00:47
copeland 2012-03-30 22:00:54
Of course, we still haven't gotten around to $AF.$ We could chase and discover more stuff about this diagram (there's a lot of stuff lurking about), but it's not clear we're even getting closer to $AF.$ We have found lots of good stuff. At some point, we have to think about what we want, $AF,$ and focus our efforts. Let's try that now. We'd like to relate $AF$ to parts of the diagram that we know more about. What might we do to help us with that?
copeland 2012-03-30 22:01:53
Triangles $AFD$ and $SED$ are similar, so we can relate $AF$ to $AD,$ $DF,$ and the sides of $SED.$ With a whole lot of effort, we can work out some of those lengths, but we sure would like to find something slicker to do. We'd prefer to relate AF to sides whose lengths we already know. What side lengths do we already know?
ABCDE 2012-03-30 22:02:52
AB BC and CA
proglote 2012-03-30 22:02:52
the side lengths of ABC
alex31415 2012-03-30 22:02:52
AC=3, AB=5, and BC=7
Tuxianeer 2012-03-30 22:02:52
AB=5,BC=7,AC=3
copeland 2012-03-30 22:02:57
We know the side lengths of $\triangle ABC,$ we know $BM$ and $CM,$ and we can use the Angle Bisector Theorem to get at $DC$ and $DB.$
copeland 2012-03-30 22:03:01
We want to relate $AF$ to parts of $ABC.$ Let's connect $F$ to the other two vertices:
copeland 2012-03-30 22:03:06
copeland 2012-03-30 22:03:12
Too much. We want to focus on $AF$ and relating it to parts of $\triangle ABC.$ Let's finish out a couple of these triangles that use lengths we know we can get to.
copeland 2012-03-30 22:03:18
copeland 2012-03-30 22:03:31
Now that's a little less terrifying.
copeland 2012-03-30 22:03:40
We should be looking at $\triangle AFC$ and $\triangle AFB,$ hoping to relate them to other triangles in the diagram. I'd start with $\triangle AFC,$ since it spans less of the diagram and seems easier to focus on. What are we trying to find?
copeland 2012-03-30 22:03:57
Xcellence 2012-03-30 22:04:17
AF
brokenfixer 2012-03-30 22:04:17
eventually we want AF
ahaanomegas 2012-03-30 22:04:17
AF
Duncanyang 2012-03-30 22:04:17
AF.
eccfcco15 2012-03-30 22:04:17
AF
copeland 2012-03-30 22:04:23
Great.
copeland 2012-03-30 22:04:29
What will help us find that?
Fire-Sword 2012-03-30 22:05:28
arent triangles ABC and CFA similar?
brokenfixer 2012-03-30 22:05:28
a similar triangle and AC works
copeland 2012-03-30 22:05:33
Perhaps $\triangle AFC$ and $\triangle CBA$ are similar. There are a few clues that this isn't the case. Most notably, that would lead us to an answer that doesn't fit the format of the problem. A less "cheating" way to check real quick that $AFC$ and $CBA$ aren't similar is to note that if $AFC$ and $CBA$ are similar, then they're congruent. A quick measurement with a compass or ruler would show that $AB$ and $FC$ aren't equal.
copeland 2012-03-30 22:06:06
However $\angle AFC = \angle ABC,$ since they're inscribed in the same arc. That's a really good start.
copeland 2012-03-30 22:06:15
We do have $\angle AFC = \angle ABC,$ so we have something to look more closely at. Perhaps $\triangle AFC$ is similar to a triangle that includes $\angle ABC.$ There aren't many candidates! There are no notable points on $AB$ besides $A$ and $B,$ so we look at triangles with $AB$ as a side and the third vertex on $BC.$ The only candidates after dismissing $\triangle ABC$ are $\triangle ADB$ and $\triangle AMB.$
copeland 2012-03-30 22:06:23
copeland 2012-03-30 22:06:39
Which one looks like the better candidate?
MathTwo 2012-03-30 22:07:20
AMB
brokenfixer 2012-03-30 22:07:20
AFC -> ABM
Fire-Sword 2012-03-30 22:07:20
AMB
numbertheorist17 2012-03-30 22:07:20
AMB
ABCDE 2012-03-30 22:07:20
AMB
Duncanyang 2012-03-30 22:07:20
AMB
copeland 2012-03-30 22:07:24
$\angle AMB$ looks much more likely to be similar to $\angle AFC.$ (This is one example of why it's good to draw very precise diagrams!)
copeland 2012-03-30 22:07:28
So, we'll go after $\angle AMB.$ We'd like to show that $\angle AMB = \angle ACF$ or $\angle MAB = \angle FAC.$ Which looks easier to chase with, $\angle AMB$ or $\angle MAB?$
Duncanyang 2012-03-30 22:08:09
AMB
proglote 2012-03-30 22:08:09
AMB
numbertheorist17 2012-03-30 22:08:09
AMB
dantx5 2012-03-30 22:08:09
AMB
Xcellence 2012-03-30 22:08:09
AMB
math-rules 2012-03-30 22:08:11
AMB
copeland 2012-03-30 22:08:15
It looks easier to relate $\angle AMB$ to other parts of the diagram. It's hard to relate $\angle MAB$ to anything outside of $\triangle MAB$ (though it can be done-- think about extending $AM$ later).
copeland 2012-03-30 22:08:24
What can we do with $\angle AMB?$
copeland 2012-03-30 22:09:16
I bet I put that little circle in there for something. . . .
proglote 2012-03-30 22:10:04
it is 180 - CMA
brokenfixer 2012-03-30 22:10:04
look at supplemental AMD
copeland 2012-03-30 22:10:14
We have $\angle AMB = 180 - \angle DMA.$ What next?
brokenfixer 2012-03-30 22:11:28
matches ASD
ABCDE 2012-03-30 22:11:28
angle DMA is equal to angle DSA
Yongyi781 2012-03-30 22:11:28
AMB - 180 - AMD = 180 - ASD
sonchecky 2012-03-30 22:11:28
DMA = DSA
copeland 2012-03-30 22:11:33
Well, we have that surprising little circle there, and we haven't used it yet... Now looks like a good time.
copeland 2012-03-30 22:11:39
copeland 2012-03-30 22:11:46
We have $\angle AMB = 180 - \angle DMA = 180 - \angle DSA.$ What else can we do with $\angle DSA$?
Potla 2012-03-30 22:12:51
AMB=180-DMA=180-DSA=FCA
Potla 2012-03-30 22:12:51
DMA=DSA=180-FCA
$\angle DSA$ is the supplement to $\angle ACF$, since $ACFS$ is cyclic
copeland 2012-03-30 22:12:59
copeland 2012-03-30 22:13:03
From cyclic quad $FCAS,$ we have $\amgle AMB=180-\angle DSA = 180-\angle FSA = \angle FCA.$ SUCCESS!
copeland 2012-03-30 22:13:09
We now have $\angle AMB=\angle FCA$ and $\angle AFC = \angle ABM,$ so $\triangle AFC \sim\triangel ABM.$
copeland 2012-03-30 22:13:12
gosh
copeland 2012-03-30 22:13:15
Let me try that again.
copeland 2012-03-30 22:13:23
From cyclic quad $FCAS,$ we have $\angle AMB=180-\angle DSA = 180-\angle FSA = \angle FCA.$ SUCCESS!
copeland 2012-03-30 22:13:32
We now have $\angle AMB=\angle FCA$ and $\angle AFC = \angle ABM,$ so $\triangle AFC \sim\triangle ABM.$
copeland 2012-03-30 22:13:39
What expression does that give us for $AF?$
Duncanyang 2012-03-30 22:14:49
AB/AF=AM/AC
brokenfixer 2012-03-30 22:14:49
AC / AF = AM / AB
sonchecky 2012-03-30 22:14:49
AF / AC = AB / AM, so AF = 2(AB)(AC)/(BC)
copeland 2012-03-30 22:14:53
We have $AF/AC = AB/AM,$ so $AF = AC\cdot AB/AM.$ Now we just have to find $AM.$
copeland 2012-03-30 22:15:01
copeland 2012-03-30 22:15:07
Now we know $AC=3$, $AB=5$, and $CB=7$. There's a fancy theorem that tells us $AM$. The space in my brain is way too precious for me to place something as obscure as that in it, though.
copeland 2012-03-30 22:15:19
copeland 2012-03-30 22:15:27
Furthermore, there's a symmetry argument that will tell us $M$ quickly what should we draw?
ABCDE 2012-03-30 22:16:12
parallelogram?
Fire-Sword 2012-03-30 22:16:12
another triangle with identical lines on the other side sharing side CB?
brokenfixer 2012-03-30 22:16:12
A' reflecting A over ES
heliootrope 2012-03-30 22:16:12
a parallelogram!
copeland 2012-03-30 22:16:16
$M$ is the midpoint. We can rotate the entire triangle by an angle of $180^\circ$ around $M$:
copeland 2012-03-30 22:16:22
copeland 2012-03-30 22:16:34
$AM$ is half the diagonal of this parallelogram. The sum of the squares of the diagonals is the sum of the squares of the (four) side lengths. (Try proving that on your own! There are several pretty quick proofs.) What is $AM^2$?
copeland 2012-03-30 22:17:18
AC=3, AB=5, BC=7.
brokenfixer 2012-03-30 22:18:31
3^2+5^2+3^2+5^2 = 68 = 49+ 4*AM^2
Duncanyang 2012-03-30 22:18:31
2(34)=49^2+(2AM)^2
nackster12 2012-03-30 22:18:31
19/2
sonchecky 2012-03-30 22:18:31
$AM^2 = 1/4 (AA')^2 = 1/4 (3^2 + 3^2 + 5^2 + 5^2 - 7^2) = 19/4$
brokenfixer 2012-03-30 22:18:31
19/4
Potla 2012-03-30 22:18:37
19/4
copeland 2012-03-30 22:19:24
$AM^2=\frac{(AA')^2}4=\frac{2AC^2+2AB^2-CB^2}4=\frac{18+50-49}{4}=\frac{19}4$.
copeland 2012-03-30 22:19:27
And what is $AF^2?$
Potla 2012-03-30 22:20:12
900/19
numbertheorist17 2012-03-30 22:20:12
900/19
sonchecky 2012-03-30 22:20:12
$AF^2 = (AB)^2 (AC)^2 / (AM)^2 = \frac{3^2 (5^2)}{19/4} = \frac{900}{19}$
brokenfixer 2012-03-30 22:20:12
3^2 5^2 * (4/19)
heliootrope 2012-03-30 22:20:12
900/19
numbertheorist17 2012-03-30 22:20:12
900/19, So 919
copeland 2012-03-30 22:20:38
$AF^2 = \frac{AC^2\cdot AB^2}{AM^2}=\frac{3^2\cdot5^2}{\frac{19}4}=\frac{3^2\cdot10^2}{19}=\frac{900}{19}.$
copeland 2012-03-30 22:20:43
These integers are relatively prime, so the answer is $\boxed{919}$.
copeland 2012-03-30 22:20:48
Woohoo!
copeland 2012-03-30 22:20:49
We win.
ahaanomegas 2012-03-30 22:21:08
That was spectular!
brokenfixer 2012-03-30 22:21:08
hooray
alex31415 2012-03-30 22:21:08
Wow! That Math Jam was probably the longest one in history!
copeland 2012-03-30 22:21:11
Yeah, that's what I was thinking.
copeland 2012-03-30 22:21:19
That's what we get for having me do the AIME. :)
ahaanomegas 2012-03-30 22:21:25
That was spectacular!
19oshawott98 2012-03-30 22:21:25
WE win PUDDING!!!!
ahaanomegas 2012-03-30 22:21:25
What a problem!
-Transcend- 2012-03-30 22:21:25
wow, that was an intense geometry problem
Fire-Sword 2012-03-30 22:21:25
pudding time!
Potla 2012-03-30 22:21:25
Great
copeland 2012-03-30 22:21:42
OK, now it's time for me to sleep.
copeland 2012-03-30 22:21:55
Thanks, everyone, for sticking around. It's been rad.
sonchecky 2012-03-30 22:22:00
This was even longer than the actual AIME, but thank you for your time doing this!
math-rules 2012-03-30 22:22:00
ok. goodnight
alex31415 2012-03-30 22:22:00
Good night!
Potla 2012-03-30 22:22:10
Good night
ahaanomegas 2012-03-30 22:22:10
Good night!
proglote 2012-03-30 22:22:10
bye :)
math-rules 2012-03-30 22:23:04
Good night
eccfcco15 2012-03-30 22:23:04
Thanks, and good night
ahaanomegas 2012-03-30 22:23:04
Thanks a lot, Mr. Copeland, and, of course, the TAs such as Aryth and Mr. Lian!
copeland 2012-03-30 22:23:58
Alright, I'm going to shut down the room now. Thanks everyone!
copeland 2012-03-30 22:25:24
Once I figure out how.
copeland 2012-03-30 22:26:50
OK, I can't purge the room.
copeland 2012-03-30 22:26:56
But you should go now.
copeland 2012-03-30 22:26:59
Really.
copeland 2012-03-30 22:27:02
The show's over.
copeland 2012-03-30 22:27:03
Thanks a lot.