Math Jams

copeland 2015-03-27 18:39:40
This classroom is moderated. That means you won't be able to see the things you type. We do see those things, however, so don't worry. The Math Jam will begin at around 7pm ET, 4pm PT.

Guendabiaani 2015-03-27 18:52:22
Hello

copeland 2015-03-27 18:52:24
Hi!

1023ong 2015-03-27 18:52:59
I'm so pumped for this!

copeland 2015-03-27 18:53:00
Me too! I just fell out of my chair.

nosaj 2015-03-27 18:53:28
How long does it take for AoPS to prepare the Math Jam?

copeland 2015-03-27 18:53:29
About 20 hours of work, maybe?

copeland 2015-03-27 18:54:02
I lied, actually. I don't have a chair. I had one once but I don't know where it went off to.

nosaj 2015-03-27 18:54:24
Well, thanks for all the hard work!

beanielove2 2015-03-27 18:54:24
Wow! Thanks for doing this!

copeland 2015-03-27 18:54:25
Of course, but don't get the wrong impression. I'm not big on "working hard."

vinayak-kumar 2015-03-27 18:54:50
which did you think was harder, aime i or aime ii?

copeland 2015-03-27 18:54:51
I don't want to speculate on things like that. This one was harder for me since I was in charge of it, though.

nosaj 2015-03-27 18:55:41
Why does Richard never lead Math Jams anymore?

copeland 2015-03-27 18:55:42
He does the classes ones. It all depends on this and that. This year he didn't do any because of the site launch.

Dayranger 2015-03-27 18:56:35
Yay! I love math.

copeland 2015-03-27 18:56:36
Gosh, this is the right place for you.

hlasker1 2015-03-27 18:56:59
Richard did do the AIME I Math Jam

copeland 2015-03-27 18:57:01
Dave did that one.

unskillfulnoob 2015-03-27 18:57:21
Who exactly is copeland?

copeland 2015-03-27 18:57:23
That's me! I'm the School Director here at AoPS.

copeland 2015-03-27 18:57:33
My name's Jeremy Copeland.

unskillfulnoob 2015-03-27 18:59:08
like your qualifications, and do you do or come up with the imo problems or something?

copeland 2015-03-27 18:59:10
Oh.

copeland 2015-03-27 18:59:30
I was an undergrad at Reed college, Ph.D. from University of Chicago. I was an instructor at MIT for 3 years before joining AoPS.

copeland 2015-03-27 18:59:58
I don't write IMO problems. I've written several USAMTS problems if you find that interesting.

ImpossibleCube 2015-03-27 19:00:07
Is that a unicorn in your avatar? http://www.artofproblemsolving.com/community/user/66362

copeland 2015-03-27 19:00:10
Yes.

copeland 2015-03-27 19:00:14
He loves math, too.

unskillfulnoob 2015-03-27 19:01:30
lastly, do you work in a room at home by yourself or is it like you in a building and there are other aops people with you near you right now?

copeland 2015-03-27 19:01:32
I'm standing among 15-20 other AoPSers at HQ right now.

Tommy2000 2015-03-27 19:02:02
standing?

copeland 2015-03-27 19:02:03
(See above)

Rmehtany 2015-03-27 19:02:30
where is HQ

copeland 2015-03-27 19:02:31
Sunny San Diego.

1023ong 2015-03-27 19:02:59
Greetings other AoPSers!

henryweng 2015-03-27 19:02:59
is it start?

copeland 2015-03-27 19:03:05
Hey, how about we start?

C-bass 2015-03-27 19:03:22
Hello!

shiningsunnyday 2015-03-27 19:03:22
Good morning from China guys

henryweng 2015-03-27 19:03:22
good!

C-bass 2015-03-27 19:03:22
YEEEEEEEEEEEEEEEEEEEEEES

jamesxie 2015-03-27 19:03:22
yay

xayy 2015-03-27 19:03:22
yay

MathStudent2002 2015-03-27 19:03:22
Yeah!

RedDevil 2015-03-27 19:03:22
perfect!

AlcumusGuy 2015-03-27 19:03:22
great idea!

Dayranger 2015-03-27 19:03:22
great idea

bhildak 2015-03-27 19:03:22
yeah

Tommy2000 2015-03-27 19:03:22
Sure

copeland 2015-03-27 19:03:25
Alright.

copeland 2015-03-27 19:03:31
The classroom is moderated, meaning that you can type into the classroom, but these comments will not go directly into the room.

copeland 2015-03-27 19:03:33
Please do not ask about administrative aspects of the contests, and please do not ask me to speculate about the results. I have no idea what the index will be for the USAJMO or the USAMO.

copeland 2015-03-27 19:03:37
Welcome to the 2014 AIME II Math Jam!

copeland 2015-03-27 19:03:38
I'm Jeremy Copeland and I'll be leading tonight's discussion.

copeland 2015-03-27 19:03:41
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 2015-03-27 19:03:47
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 2015-03-27 19:03:52
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 2015-03-27 19:04:12
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 responses 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 2015-03-27 19:04:17
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 always to try do so in our regular online 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 2015-03-27 19:04:28
We do have two teaching assistants with us tonight to help answer your questions: Luis (Duelist) and Alex (BOGTRO).

copeland 2015-03-27 19:04:34
Luis Ares took many math contests in high school, and still enjoys solving math problems as well as playing ultimate frisbee in his spare time.

BOGTRO 2015-03-27 19:04:49
Hi!

copeland 2015-03-27 19:04:50
BOGTRO is Alexander Katz, better known in some parts as "Alex the Kat", a three-time olympiad qualifier who will attend MIT next year. When not doing math, Alex busies himself with chess, programming, and creating his own attempts at competition writing.

Duelist 2015-03-27 19:05:12
Hi!

FractalMathHistory 2015-03-27 19:05:20
Ares the god of war!

copeland 2015-03-27 19:05:21
Katz, the god of, um,. . .

copeland 2015-03-27 19:05:29
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 2015-03-27 19:05:32
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 2015-03-27 19:05:43
Let's get started! We're going to work through all 15 problems from the 2014 AIME II, in order.

ThePiPie 2015-03-27 19:06:01
2015...?

copeland 2015-03-27 19:06:04
Hm.

copeland 2015-03-27 19:06:10
Which one do you want? 2014 or 2015?

swirlykick 2015-03-27 19:06:33
2015

Rmehtany 2015-03-27 19:06:33
2015

acegikmoqsuwy2000 2015-03-27 19:06:33
2016.

derpyuniverse 2015-03-27 19:06:33
2015 lol

urmilla 2015-03-27 19:06:33
2015

Tommy2000 2015-03-27 19:06:33
2016

awesome123 2015-03-27 19:06:33
isn't it the 2015 AIME II

Jyzhang12 2015-03-27 19:06:33
2015!

caressezhu 2015-03-27 19:06:33
2015

jamesxie 2015-03-27 19:06:33
2015!

amburger66 2015-03-27 19:06:33
2015 would be nice

epiclucario 2015-03-27 19:06:33
2016 please

Abeymom 2015-03-27 19:06:33
2015

mathperson9 2015-03-27 19:06:33
2015

copeland 2015-03-27 19:06:39
Should have voted for '16.

copeland 2015-03-27 19:06:42
Let's get started! We're going to work through all 15 problems from the 2015 AIME II, in order.

copeland 2015-03-27 19:06:44
1. Let $N$ be the least positive integer that is both 22 percent less than one integer and 16 percent greater than another integer. Find the remainder when $N$ is divided by 1000.

copeland 2015-03-27 19:06:52
How do we write "$N$ is 22 percent less than one integer"?

Gina 2015-03-27 19:07:15
N = .78x

ompatel99 2015-03-27 19:07:15
N=0.78*m

acegikmoqsuwy2000 2015-03-27 19:07:15
N=0.78x

Aang 2015-03-27 19:07:15
N=0.78x

shiningsunnyday 2015-03-27 19:07:15
0.78X

epiclucario 2015-03-27 19:07:15
.78x=N

beanielove2 2015-03-27 19:07:15
$N=0.78m$, where $m$ is an integer

copeland 2015-03-27 19:07:18
There is some integer $a$ such that $N=0.78a$.

copeland 2015-03-27 19:07:21
How do we write "$N$ is 16 percent more than another integer"?

AlisonH 2015-03-27 19:07:41
N=1.16y

ompatel99 2015-03-27 19:07:41
N=1.16k

vinayak-kumar 2015-03-27 19:07:41
N=1.16x$

WhaleVomit 2015-03-27 19:07:41
$N=1.16b$

nosaj 2015-03-27 19:07:41
$N=1.16b$

gxah 2015-03-27 19:07:41
N=1.16b

Rmink41 2015-03-27 19:07:41
N=1.16y

cxiong 2015-03-27 19:07:41
N=1.16y

copeland 2015-03-27 19:07:43
There is some integer $b$ such that $N=1.16b$.

copeland 2015-03-27 19:07:47
Therefore \[0.78a=1.16b\] or $78a=116b$. Those are both even so we really have

copeland 2015-03-27 19:07:49
$39a=58b$.

copeland 2015-03-27 19:07:53
What does that tell us?

Darn 2015-03-27 19:08:28
We must have that $29$ and $39$ both divide $N$.

Jyzhang12 2015-03-27 19:08:28
Multiple of 58 and 39

cxiong 2015-03-27 19:08:28
b is a multiple of 39, a is a multiple of 58

amburger66 2015-03-27 19:08:28
a is a multiple of 58, while b is a multiple of 39

jam10307 2015-03-27 19:08:28
divisible by 39 and 58

copeland 2015-03-27 19:08:31
Since 39 and 58 are coprime, $a$ is a multiple of 58 and $b$ is a multiple of 39. We can write $a=58c$ and $b=39c$.

copeland 2015-03-27 19:08:41
What is $N$ in terms of $c$?

copeland 2015-03-27 19:09:54
Be careful that you're actually using an expression for $N$ here.

acegikmoqsuwy2000 2015-03-27 19:10:01
N=0.78*58c=1.16*39c

ompatel99 2015-03-27 19:10:01
N=29*39c/25

henryweng 2015-03-27 19:10:01
N=58*0.78c

swirlykick 2015-03-27 19:10:01
.78*58c

firemike 2015-03-27 19:10:01
N = .78 * 58c

Dayranger 2015-03-27 19:10:01
0.78x58c

copeland 2015-03-27 19:10:04
We have \[N=0.78a=\frac{78}{100}\cdot58c=\frac{39\cdot29}{25}\cdot c.\]

copeland 2015-03-27 19:10:06
What is the smallest possible value for $N$?

xayy 2015-03-27 19:10:39
39*29

AlisonH 2015-03-27 19:10:39
1131

C-bass 2015-03-27 19:10:39
39 x 29

ZZmath9 2015-03-27 19:10:39
39*29=1131, when c is 25

swirlykick 2015-03-27 19:10:39
39*29=1131 so 131

xayy 2015-03-27 19:10:39
39*29 or 1131

ompatel99 2015-03-27 19:10:39
39*29=1131

WhaleVomit 2015-03-27 19:10:39
39*29

AlcumusGuy 2015-03-27 19:10:41
39*29 = 1131

copeland 2015-03-27 19:10:43
The smallest integer multiple of $\dfrac{39\cdot29}{25}$ is \[39\cdot29=(40-1)(30-1)=1200-40-30+1=1131.\]

copeland 2015-03-27 19:10:44
This multiple is achieved at $c=25$ so $a=58c$ and $b=39c$ are both integers.

copeland 2015-03-27 19:10:47
The final answer is $\boxed{131}$.

copeland 2015-03-27 19:11:01
Kind of a tough problem to start with. Let's make up for that. . .

copeland 2015-03-27 19:11:02
2. In a new school 40 percent of the students are freshmen, 30 percent are sophomores, 20 percent are juniors, and 10 percent are seniors. All freshmen are required to take Latin, and 80 percent of the sophomores, 50 percent of the juniors, and 20 percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

copeland 2015-03-27 19:11:07
Not much to this problem. What do we do?

maverick8 2015-03-27 19:11:51
WLOG, let there be 100 students

cxiong 2015-03-27 19:11:51
assume the school has 1000 people

Jyzhang12 2015-03-27 19:11:51
Assume total students is 100

WhaleVomit 2015-03-27 19:11:51
assume there are 100 people in the school

vinayak-kumar 2015-03-27 19:11:51
assume there are 100 students

gxah 2015-03-27 19:11:51
let there be 100 students

copeland 2015-03-27 19:11:52
We could do that. It's a good approach.

amburger66 2015-03-27 19:12:04
find out the total number of Latin students, and how many of them are sophomores

24iam24 2015-03-27 19:12:04
find percent of all people who take latin

shiningsunnyday 2015-03-27 19:12:04
Count what we want/Total

swirlykick 2015-03-27 19:12:04
Find out how many total people take latin, and how many are sophmores

copeland 2015-03-27 19:12:15
And my favorite technique:

burunduchok 2015-03-27 19:12:17
make a chart

copeland 2015-03-27 19:12:19
We divide the number of Latin sophomores by the total number of Latin students.

copeland 2015-03-27 19:12:22
I love tables. Let's make a table of the percents in each category.

copeland 2015-03-27 19:12:24

copeland 2015-03-27 19:12:25
(Note that all of these values are in the unit "percent of students".)

copeland 2015-03-27 19:12:28
What's the answer?

Poseidon2001 2015-03-27 19:13:13
25

Tommy2000 2015-03-27 19:13:13
6/19

henryweng 2015-03-27 19:13:13
=24/76=6/19

maverick8 2015-03-27 19:13:13
25

Rocksolid 2015-03-27 19:13:13
025

cxiong 2015-03-27 19:13:13
24/76 = 6/19

caressezhu 2015-03-27 19:13:13
25

nosaj 2015-03-27 19:13:13
24/76=6/19

xayy 2015-03-27 19:13:13
24/76=6/19 or 25

ShadowQueenPeach 2015-03-27 19:13:13
6/19

copeland 2015-03-27 19:13:22
The probability we are looking for is
\[
\dfrac{\text{Latin sophomores}}{\text{total Latin students}}=
\dfrac{24}{40+24+10+2}=\dfrac{24}{76}=\dfrac{6}{19}.\] The answer is $6+19=\boxed{25}$.

copeland 2015-03-27 19:13:39
Don't worry about AIMEifying at the end. That step is (usually) pretty simple.

copeland 2015-03-27 19:13:50
6/19 is a much clearer answer for this. I know you can mostly add.

copeland 2015-03-27 19:13:55
3. Let $m$ be the least positive integer divisible by 17 whose digits sum to 17. Find $m$.

copeland 2015-03-27 19:14:08
Any suggestions?

maverick8 2015-03-27 19:14:44
m is a 3 digit number because 89 and 98 don't work

math-rules 2015-03-27 19:14:44
since 89 and 98 don't work, let m = 100a+10b+c

C-bass 2015-03-27 19:14:44
Find numbers whose digits sum to 17, since this is the AIME and the only possible numbers are either 2- or 3- digit #s

nosaj 2015-03-27 19:14:44
You know that $m$ can't be four digits because this is the AIME. :P

mathperson9 2015-03-27 19:14:44
If m's digits are xyz, then x+y+z=17 and 100x+10y+z mod 17 is 0

shiningsunnyday 2015-03-27 19:14:44
Put in the form 100a+10b+c

copeland 2015-03-27 19:14:47
The only possible 2-digit numbers whose digits sum to 17 are 89 and 98, but neither of those are multiples of 17.

copeland 2015-03-27 19:14:48
So $m$ is going to be a 3-digit number. (Or more, but $m$ is also an AIME answer!)

copeland 2015-03-27 19:14:50
We'll probably want to name the digits. So let's say $m$ is the number $abc$, where $a$ is the hundreds digit, $b$ is the tens digit, and $c$ is the units digit.

copeland 2015-03-27 19:15:00
Now we can write $m = 100a + 10b + c$.

copeland 2015-03-27 19:15:10
But what else do we know about $m$?

owm 2015-03-27 19:15:53
a+b+c=17

Rmehtany 2015-03-27 19:15:53
a+b+c =17

flyrain 2015-03-27 19:15:53
a+b+c=17

C-bass 2015-03-27 19:15:53
a + b + c = 17

dhruv 2015-03-27 19:15:53
a+b+c=17

mssmath 2015-03-27 19:15:53
a+b+c=17

copeland 2015-03-27 19:16:06
The digits of $m$ sum to 17, so $a+b+c = 17$.

xayy 2015-03-27 19:16:35
divisible by 17

Dayranger 2015-03-27 19:16:35
m = 17n

Poseidon2001 2015-03-27 19:16:35
divisible by 17

copeland 2015-03-27 19:16:37
$m$ is a multiple of 17, so $100a + 10b + c = 17n$ for some positive integer $n$.

copeland 2015-03-27 19:16:45
So we have the system of equations
\begin{align*}
100a + 10b + c &= 17n, \\
a+b+c &= 17.
\end{align*}

copeland 2015-03-27 19:16:46
Now what?

Rmehtany 2015-03-27 19:17:12
subtract them

ompatel99 2015-03-27 19:17:12
Get rid of c

Tommy2000 2015-03-27 19:17:12
hmm, subtract

math-rules 2015-03-27 19:17:12
subtract second from first

mathperson9 2015-03-27 19:17:12
SUbtract

mssmath 2015-03-27 19:17:12
Subtract!

Aang 2015-03-27 19:17:12
Eliminate

owm 2015-03-27 19:17:12
subtract?

sicilianfan 2015-03-27 19:17:12
subtract them

copeland 2015-03-27 19:17:16
We can subtract the equations and get rid of $c$. That leaves us with $99a + 9b = 17(n-1)$.

copeland 2015-03-27 19:17:16
How does that help?

Tommy2000 2015-03-27 19:17:54
n is 1 mod 9

Rmehtany 2015-03-27 19:17:54
n-1 divisble by 9

Darn 2015-03-27 19:17:54
We know that n-1 is divisible by 9

mxgo 2015-03-27 19:17:54
We know n-1 is divisble by 9

jam10307 2015-03-27 19:17:54
n is 10+9x

copeland 2015-03-27 19:17:57
Factoring the left side gives $9(11a+b) = 17(n-1)$.

copeland 2015-03-27 19:18:00
This means that $n-1$ must be a multiple of 9, or equivalently that $n$ is 1 plus a multiple of 9. Therefore $17n$ is $17$ plus a multiple of $17\cdot9=153$.

copeland 2015-03-27 19:18:07
Know the answer yet?

dhruv 2015-03-27 19:18:45
476

maverick8 2015-03-27 19:18:45
476

caressezhu 2015-03-27 19:18:45
476=17*28

burunduchok 2015-03-27 19:18:45
476

gxah 2015-03-27 19:18:45
476

24iam24 2015-03-27 19:18:45
476

ompatel99 2015-03-27 19:18:45
476

maverick8 2015-03-27 19:18:45
$ 476 $

ZZmath9 2015-03-27 19:18:45
the answer is 17 + 3 * 153 = 17 + 459 = 476.

C-bass 2015-03-27 19:18:45
476

copeland 2015-03-27 19:18:48
It's easy to finish just by checking by hand, by starting at $10\cdot17 = 170$ and adding $9\cdot17 = 153$ until we get one that has a digit sum of 17.

copeland 2015-03-27 19:18:50
$10\cdot17 = 170$, nope

copeland 2015-03-27 19:18:54
$19\cdot17 = 170+153 = 323$, nope

copeland 2015-03-27 19:18:57
$28\cdot 17 = 323+153 = 476$, success!

copeland 2015-03-27 19:18:58
The answer is $\boxed{476}$.

copeland 2015-03-27 19:19:09
A fifth of the way done! (By some metric.)

Darn 2015-03-27 19:19:25
you could've taken a modular approach

copeland 2015-03-27 19:19:26
True.

copeland 2015-03-27 19:19:51
17 and 9 are both interesting in some way. However that doesn't turn out to help as much as all that for this particular problem.

copeland 2015-03-27 19:19:59
4. In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$, and the altitude to those bases has length $\log 16$. the perimeter of the trapezoid can be written in the form $\log 2^p3^q$, where $p$ and $q$ are positive integers. Find $p + q$.

copeland 2015-03-27 19:20:15
(The base of the log probably doesn't matter here since they didn't tell us. In fact it won't.)

copeland 2015-03-27 19:20:18
I sense that this is one of those problems that is a lot easier than it looks as long as you know your logarithm facts.

copeland 2015-03-27 19:20:20
What do we do first?

maverick8 2015-03-27 19:20:33
Draw a diagram, of course

Darn 2015-03-27 19:20:33
Draw a diagram

shiningsunnyday 2015-03-27 19:20:33
Diagram!

simon1221 2015-03-27 19:20:33
diagram!!

tdeng 2015-03-27 19:20:33
diagram

math-rules 2015-03-27 19:20:33
draw a diagram

Dracae8 2015-03-27 19:20:33
diagram

cxiong 2015-03-27 19:20:33
draw the diagram

Rocksolid 2015-03-27 19:20:33
picture

copeland 2015-03-27 19:20:37
Draw a diagram, of course. This is obviously not to scale but the labels are going to fit.

copeland 2015-03-27 19:20:38

copeland 2015-03-27 19:20:40
Now what?

Jyzhang12 2015-03-27 19:21:13
Draw a picture and add 2 heights

24iam24 2015-03-27 19:21:13
drop altitudes

simon1221 2015-03-27 19:21:13
make 2 triangles

Jyzhang12 2015-03-27 19:21:13
Draw 2 heights

AlcumusGuy 2015-03-27 19:21:13
draw altitudes

ompatel99 2015-03-27 19:21:13
Drop altitudes from the endpoints of the top base to the bottom base

Darn 2015-03-27 19:21:13
Draw the altitude to make two right triangles

hlasker1 2015-03-27 19:21:13
make a right triangle

sicilianfan 2015-03-27 19:21:13
form a right triangle by dropping altitude from one of the top vertices to the base

MathStudent2002 2015-03-27 19:21:13
Two right triangles and a rectangle

rt03 2015-03-27 19:21:13
Make two right triangles on the ends

copeland 2015-03-27 19:21:15
Let's drop some altitudes from the vertices.

copeland 2015-03-27 19:21:16

copeland 2015-03-27 19:21:20
And the base tells us that $a+\log3+a=\log192$. What is $a$?

Darn 2015-03-27 19:21:48
So $a=\log 8$

cxiong 2015-03-27 19:21:48
log8

maverick8 2015-03-27 19:21:48
log 8

dantx5 2015-03-27 19:21:48
log 8

countingarithmetic 2015-03-27 19:21:48
log 8

NextEinstein 2015-03-27 19:21:48
log 8

azmath333 2015-03-27 19:21:48
log 8

simon1221 2015-03-27 19:21:48
log 8?

BPM14 2015-03-27 19:21:48
log8

Abeymom 2015-03-27 19:21:48
log 8

sicilianfan 2015-03-27 19:21:48
log 8

copeland 2015-03-27 19:21:50
$a=\frac12(\log192-\log3)=\frac12\log64=\log8$.

copeland 2015-03-27 19:21:52

copeland 2015-03-27 19:21:53
Very not to scale.

copeland 2015-03-27 19:21:56
Now what?

Darn 2015-03-27 19:22:39
We can split the logs and we find that it's a 3-4-5 right triangle, so $h= 5\log 2 = \log 32$.

math-rules 2015-03-27 19:22:39
we have 3-4-5 triangle (scaled by log 2)

Dracae8 2015-03-27 19:22:39
3-4-5 triangle!

Tommy2000 2015-03-27 19:22:39
3-4-5 triangle so h=log32

simon1221 2015-03-27 19:22:39
the triangles are 3 4 5 * log2

24iam24 2015-03-27 19:22:39
log8=3log2 log16=4log2 so 3-4-5 right triangle and h=5log2

vinayak-kumar 2015-03-27 19:22:39
the sides are 3-4-5 triangles

ompatel99 2015-03-27 19:22:39
log8=3log2. log16=4log2. h=5log2

azmath333 2015-03-27 19:22:39
use 3-4-5 right triangles

copeland 2015-03-27 19:22:42
We could use the Pythagorean Theorem right away, but we can also simplify those lengths:

copeland 2015-03-27 19:22:44

copeland 2015-03-27 19:22:47
That's a 3-4-5 triangle:

copeland 2015-03-27 19:22:48

copeland 2015-03-27 19:22:55
What is the perimeter?

tdeng 2015-03-27 19:23:40
16log2+2log3

Rmink41 2015-03-27 19:23:40
16log2+2log3

nosaj 2015-03-27 19:23:40
Now just add: $16 \log 2 + 2\log 3$.

firemike 2015-03-27 19:23:40
16log2 + 2log3

mathperson9 2015-03-27 19:23:40
16log2+2log3

Dayranger 2015-03-27 19:23:40
16 log 2 + 2 log 3

temp8909 2015-03-27 19:23:40
16log2+2log3

ShadowQueenPeach 2015-03-27 19:23:49
log (2^16 * 3^2)

flyrain 2015-03-27 19:23:49
log 3+log 192+10 log 2= log 3^2*2^16 so 018

amburger66 2015-03-27 19:23:49
log (192*3*32*32) = log (2^16*3^2)

copeland 2015-03-27 19:23:55
The perimeter is
\[5\log2+\log3+5\log2+3\log2+\log3+3\log2=16\log2+2\log3=\log2^{16}3^2.\]

copeland 2015-03-27 19:23:56
And the final answer?

AlisonH 2015-03-27 19:24:21
18

Poseidon2001 2015-03-27 19:24:21
18

xayy 2015-03-27 19:24:21
018

maverick8 2015-03-27 19:24:21
eighteen

Rmehtany 2015-03-27 19:24:21
18

slenderestman 2015-03-27 19:24:21
18

koteswari 2015-03-27 19:24:21
18

mxgo 2015-03-27 19:24:21
18

mathwizard888 2015-03-27 19:24:21
18

linmichael 2015-03-27 19:24:21
18

mathperson9 2015-03-27 19:24:21
18

xiaodongxi 2015-03-27 19:24:21
18

copeland 2015-03-27 19:24:23
The final answer is $16+2=\boxed{18}$.

copeland 2015-03-27 19:24:24
By the way, to scale that diagram looks like this:

copeland 2015-03-27 19:24:25

1023ong 2015-03-27 19:24:46
that one doesn't look as pretty

copeland 2015-03-27 19:24:52
Thanks. I like mine better, too.

copeland 2015-03-27 19:24:56
This one is easier to solve.

copeland 2015-03-27 19:24:58
5. Two unit squares are selected at random without replacement from an $n\times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected squares are horizontally or vertically adjacent is less than $\frac 1{2015}$.

copeland 2015-03-27 19:25:03
In probability problems I usually like to count the denominator first.

copeland 2015-03-27 19:25:10
How many ways are there to select 2 squares?

IsabeltheCat 2015-03-27 19:25:55
n^2 choose 2

tdeng 2015-03-27 19:25:55
$\dbinom{n^2}{2}$

vinayak-kumar 2015-03-27 19:25:55
$\binom{n^2}2$

Eugenis 2015-03-27 19:25:55
$\binom{n^2}{2}$

Aang 2015-03-27 19:25:55
N^2C2

math-rules 2015-03-27 19:25:55

Poseidon2001 2015-03-27 19:25:55
n^2 choose 2

lcsmart 2015-03-27 19:25:55
$N^2*(N^2-1)/2$

owm 2015-03-27 19:25:55
n^2 chose 2

danusv 2015-03-27 19:25:55
n^2 Choose 2

burunduchok 2015-03-27 19:25:55
n^2 choose 2

Eugenis 2015-03-27 19:25:55
$\binom{n^2}{2}$

copeland 2015-03-27 19:25:57
There are $n^2$ squares, so there are $\dbinom{n^2}{2} = \dfrac{n^2(n^2-1)}{2}$ ways to choose 2 of them (without caring about the order).

copeland 2015-03-27 19:25:59
How many pairs of squares are horizontally or vertically adjacent?

copeland 2015-03-27 19:26:03
A simpler question: how many pairs within a single row are horizontally adjacent?

cxiong 2015-03-27 19:26:25
n-1

SockFoot 2015-03-27 19:26:25
$n-1$

slenderestman 2015-03-27 19:26:25
n-1

nosaj 2015-03-27 19:26:25
n-1

NextEinstein 2015-03-27 19:26:25
n-1

maverick8 2015-03-27 19:26:25
n-1

vinayak-kumar 2015-03-27 19:26:25
$n-1$

copeland 2015-03-27 19:26:27
There are $n-1$ such pairs: the first two, the next two, and so on, up to squares $(n-1)$ and $n$ at the end of the row.

copeland 2015-03-27 19:26:29
(Or, you can think about selecting one of the $n-1$ interior vertical edges inside the row, and then taking the squares on either side.)

copeland 2015-03-27 19:26:33
So each row has $n-1$ pairs.

copeland 2015-03-27 19:26:39
And, there are $n$ rows. So there are $n(n-1)$ pairs of horizontally-adjacent squares.

copeland 2015-03-27 19:26:40
And, the same logic works for columns too, right?

C-bass 2015-03-27 19:26:49
same with each column

SockFoot 2015-03-27 19:26:49
you are correct sir

Abeymom 2015-03-27 19:26:49
yes

copeland 2015-03-27 19:26:51
So there are also $n(n-1)$ pairs of vertically-adjacent squares.

copeland 2015-03-27 19:26:53
That makes $2n(n-1)$ total pairs of adjacent square.

copeland 2015-03-27 19:26:56
Thus, our probability is $\dfrac{2n(n-1)}{\frac{n^2(n^2-1)}{2}} = \dfrac{4n(n-1)}{n^2(n^2-1)}$.

copeland 2015-03-27 19:27:04
Can this be simplified further?

24iam24 2015-03-27 19:27:47
4/n(n+1)

Rmehtany 2015-03-27 19:27:47
4/n(n+1)

IsabeltheCat 2015-03-27 19:27:47
yes; 4/n(n+1)

SockFoot 2015-03-27 19:27:47
$\dfrac{4}{n(n+1)}$

AlcumusGuy 2015-03-27 19:27:47
$\frac{4}{n(n+1)}$

dhruv 2015-03-27 19:27:47
4/n(n+1)

gxah 2015-03-27 19:27:47
$\frac{4}{n^{2}+n}$

copeland 2015-03-27 19:27:51
A factor of $n$ cancels, and $(n-1)$ cancels with $(n^2-1)$ leaving just $(n+1)$.

copeland 2015-03-27 19:27:53
So our probability is $\dfrac{4}{n(n+1)}$.

copeland 2015-03-27 19:27:57
How do we finish from here?

ompatel99 2015-03-27 19:28:31
It's <1/2015

cxiong 2015-03-27 19:28:31
set that to less than 1/2015

countingarithmetic 2015-03-27 19:28:31
less than 2015

ImpossibleTriangle 2015-03-27 19:28:31
4/n(n+1) < 1/2015

shiningsunnyday 2015-03-27 19:28:31
<1/2015, then finish off the equation

copeland 2015-03-27 19:28:33
We're looking for the least positive $n$ such that $\dfrac{4}{n(n+1)} < \dfrac{1}{2015}$.

C-bass 2015-03-27 19:28:43
n(n+1) > 8060

TheEconomist 2015-03-27 19:28:43
n(n+1) > 8060

copeland 2015-03-27 19:28:45
Since $n$ is positive, this simplifies to $n(n+1) > 8060$.

copeland 2015-03-27 19:29:05
Since $90^2 = 8100$, that's about where we need to look.

swirlykick 2015-03-27 19:29:14
090

BPM14 2015-03-27 19:29:14
n = 90

henryweng 2015-03-27 19:29:14
n=90

caressezhu 2015-03-27 19:29:14
90

copeland 2015-03-27 19:29:17
Indeed,
\begin{align*}
89(90) = 8010 &< 8060, \\
90(91) = 8190 &> 8060.
\end{align*}

copeland 2015-03-27 19:29:18
So $n = \boxed{090}$ is our answer.

copeland 2015-03-27 19:29:26
Alright, a third of the way done!

copeland 2015-03-27 19:29:37
How are you guys doing? Having a good time?

cxiong 2015-03-27 19:29:49
yep

Eugenis 2015-03-27 19:29:49
Yes sir

gxah 2015-03-27 19:29:49
yup

Poseidon2001 2015-03-27 19:29:49
yes

Dracae8 2015-03-27 19:29:49
yes

cxiong 2015-03-27 19:29:49

bluephoenix 2015-03-27 19:29:49
YAH

dhruv 2015-03-27 19:29:49
yessir

mikromgrom 2015-03-27 19:29:49
Yes

Dayranger 2015-03-27 19:29:49
yes!!!

MathStudent2002 2015-03-27 19:29:49
Yeah sure

firemike 2015-03-27 19:29:49
yse!

hlasker1 2015-03-27 19:29:49
yeah

ArgusKelvoy 2015-03-27 19:29:49
Sure.

ompatel99 2015-03-27 19:29:49
It's straightforward so far

rt03 2015-03-27 19:29:49
Yes!

copeland 2015-03-27 19:29:51
Me too.

copeland 2015-03-27 19:30:02
I think the last 2/3 of the exam are going to be a little more work.

copeland 2015-03-27 19:30:05
6. Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form
$$P(x) = 2x^3 - 2ax^2 + (a^2-81)x - c$$
for some positive integers $a$ and $c$. Can you tell me the values of $a$ and $c$?"

copeland 2015-03-27 19:30:06
After some calculations, Jon says, "There is more than one such polynomial."

copeland 2015-03-27 19:30:07
Steve says, "You're right. Here is the value of $a$." He writes down a positive integer and asks, "Can you tell me the value of $c$?"

copeland 2015-03-27 19:30:08
Jon says, "There are still two possible values of $c$."

copeland 2015-03-27 19:30:09
Find the sum of the two possible values of $c$.

copeland 2015-03-27 19:30:16
What a problem. . .

copeland 2015-03-27 19:30:30
What's going to be our key tool here?

ompatel99 2015-03-27 19:30:46
Vieta's equations

maverick8 2015-03-27 19:30:46
Use Vieta's Formulas

SockFoot 2015-03-27 19:30:46
probably going to use vietas

Eugenis 2015-03-27 19:30:46
Vieta's

cxiong 2015-03-27 19:30:46
vieta's!

SockFoot 2015-03-27 19:30:46
V I E T A S

bluephoenix 2015-03-27 19:30:46
vietas

ompatel99 2015-03-27 19:30:46
Vieta

mohanxue612 2015-03-27 19:30:46
vieta

tdeng 2015-03-27 19:30:46
vietas?

24iam24 2015-03-27 19:30:46
vieta

vinayak-kumar 2015-03-27 19:30:46
Vieta!

gxah 2015-03-27 19:30:46
vieta's

ImpossibleTriangle 2015-03-27 19:30:46
vieta

High 2015-03-27 19:30:46
vieta's

amburger66 2015-03-27 19:30:46
vieta's?

Borealis 2015-03-27 19:30:46
Vieta's

copeland 2015-03-27 19:30:50
Ding ding.

copeland 2015-03-27 19:30:51
Vieta's formulas! We'll want to try to relate the roots of the cubic (which we know to be integers!) to the unknown coefficients.

copeland 2015-03-27 19:30:55
Let's call the roots $p$, $q$, and $r$. What equations can we write?

copeland 2015-03-27 19:31:01
Pick one at random to write.

24iam24 2015-03-27 19:31:29
p+q+r=a

Eugenis 2015-03-27 19:31:29
$p+q+r=a$

vinayak-kumar 2015-03-27 19:31:29
p+q+r=a

swirlykick 2015-03-27 19:31:29
p+q+r=a

acegikmoqsuwy2000 2015-03-27 19:31:29
p+q+r=a

C-bass 2015-03-27 19:31:29
p + q + r = a

SockFoot 2015-03-27 19:31:29
$p+q+r=a$

High 2015-03-27 19:31:29
p+q+r=a

acegikmoqsuwy2000 2015-03-27 19:31:41
pq+qr+pr=(a^2-81)/2

swirlykick 2015-03-27 19:31:41
pr+pq+qr=(a^2 -81)/2

24iam24 2015-03-27 19:31:41
pq+qr+pr=a^2-81

maverick8 2015-03-27 19:31:41
pq+qr+pr=(a^2-81)/2

math-rules 2015-03-27 19:31:59
pqr=c/2

dhruv 2015-03-27 19:31:59
pqr=c/2

ShadowQueenPeach 2015-03-27 19:31:59
pqr = c/2

vinayak-kumar 2015-03-27 19:31:59
pqr=c/2

copeland 2015-03-27 19:32:02
Vieta's formulas give us:
\begin{align*}
p+q+r &= a, \\
pq + pr + qr &= \dfrac{a^2-81}{2}, \\
pqr &= \dfrac{c}{2}.
\end{align*}

copeland 2015-03-27 19:32:05
(Notice how they carefully arranged the minus signs in the original cubic so that we don't have any negative terms above!)

copeland 2015-03-27 19:32:10
Any suggestions for how to work with these?

Eugenis 2015-03-27 19:32:55
Square the first equation?

High 2015-03-27 19:32:55
square p+q+r

shiningsunnyday 2015-03-27 19:32:55
Square the first?

NextEinstein 2015-03-27 19:32:55
square first one

slenderestman 2015-03-27 19:32:55
square the frst equation

Borealis 2015-03-27 19:32:55
Square the first!

24iam24 2015-03-27 19:32:55
square first

copeland 2015-03-27 19:32:58
That last one doesn't look too helpful yet, but perhaps the first two can be combined in a fruitful way to eliminate $a$ somehow.

copeland 2015-03-27 19:33:00
How about we square the first equation? That will give us lots of terms that appear in the second equation, and both right hand sides will be in terms of $a^2$.

copeland 2015-03-27 19:33:04
Specifically,\[a^2=(p+q+r)^2 = p^2 + q^2 + r^2 + 2pq + 2pr + 2qr.\]

maverick8 2015-03-27 19:33:49
Substitute the first equation into the second

vinayak-kumar 2015-03-27 19:33:49
substitute pq+pr+qr

ShadowQueenPeach 2015-03-27 19:33:49
multiply the next equation by 2

dragon_summation 2015-03-27 19:33:49
Multiply second equation by 2

countingarithmetic 2015-03-27 19:33:49
$p^2 + q^2 + r^2 = 81$

ompatel99 2015-03-27 19:33:56
p^2+q^2+r^2=81

Rmehtany 2015-03-27 19:33:56
p^2+q^2+r^2 = 81

copeland 2015-03-27 19:33:58
Ah -- we see $2(pq+pr+qr)$ in our second equation. We can substitute $a^2 - 81$ in for it.

copeland 2015-03-27 19:33:59
Now we have
\[
p^2 + q^2 + r^2 + a^2 - 81 = a^2.
\]

copeland 2015-03-27 19:34:01
That's great! The $a^2$'s cancel and we're left with just
\[
p^2 + q^2 + r^2 = 81.
\]

copeland 2015-03-27 19:34:08
Now what?

room456 2015-03-27 19:34:30
bash out triples of p q and r

High 2015-03-27 19:34:30
find integer values that work

maverick8 2015-03-27 19:34:30
Guess and check

cxiong 2015-03-27 19:34:30
just try to find squares that add up to 81

ompatel99 2015-03-27 19:34:30
not that many cases

dantx5 2015-03-27 19:34:30
bash

Tommy2000 2015-03-27 19:34:30
Find the integer solutions

Eugenis 2015-03-27 19:34:30
List out the triples that when squared are $81$

tdeng 2015-03-27 19:34:30
find 3 squares whose sum is 81

copeland 2015-03-27 19:34:32
Oh yeah.

copeland 2015-03-27 19:34:33
This equation probably does not have a lot of positive integer solutions. Perhaps we can just list them all.

copeland 2015-03-27 19:34:36
How can we organize listing all the solutions in a sensible way?

NextEinstein 2015-03-27 19:35:11
largest square

vinayak-kumar 2015-03-27 19:35:11
WLOG assume $p\ge q\ge r$

mikromgrom 2015-03-27 19:35:11
casework on 1 variable

andrewlin 2015-03-27 19:35:11
Assume p<=q<=r

nosaj 2015-03-27 19:35:11
by the largest value

copeland 2015-03-27 19:35:14
Let's assume that $p \ge q \ge r$. (It doesn't matter what order we write the roots in, since all our equations are symmetric.)

copeland 2015-03-27 19:35:24
What do we know about $p$?

slenderestman 2015-03-27 19:36:03
less than 9

tdeng 2015-03-27 19:36:03
<9

C-bass 2015-03-27 19:36:03
p>5

vinayak-kumar 2015-03-27 19:36:03
p can range from 6 to 8

cxiong 2015-03-27 19:36:03
0 < p < 9

slenderestman 2015-03-27 19:36:03
less than 9 and greater than 5

ShadowQueenPeach 2015-03-27 19:36:03
it is less than 9

xayy 2015-03-27 19:36:03
less than 9

High 2015-03-27 19:36:03
p < 9

dantx5 2015-03-27 19:36:03
p>=6

rt03 2015-03-27 19:36:03
it is < than 9

copeland 2015-03-27 19:36:06
$p=9$ is too big, and $p=5$ is too small (since $3 \cdot 5^2$ isn't big enough). So $p$ must be 6, 7, or 8.

copeland 2015-03-27 19:36:07
If $p=8$, are there any solutions?

SockFoot 2015-03-27 19:36:26
$(4,1)$

Dayranger 2015-03-27 19:36:26
8, 4, 1

maverick8 2015-03-27 19:36:26
(1,4,8)

koteswari 2015-03-27 19:36:26
Yes, 64,16,1

mathwizard888 2015-03-27 19:36:26
4,1

slenderestman 2015-03-27 19:36:26
8, 4, 1

Abeymom 2015-03-27 19:36:26
q=4 r=1

Darn 2015-03-27 19:36:26
$841$

copeland 2015-03-27 19:36:28
We must have $q^2 + r^2 = 81 - 64 = 17$. The only solution is $(p,q,r) = (8,4,1)$.

copeland 2015-03-27 19:36:29
How about $p=7$?

henryweng 2015-03-27 19:37:05
(4,4,7)

ImpossibleTriangle 2015-03-27 19:37:05
7,4,4

Rocksolid 2015-03-27 19:37:05
7,4,4

azmath333 2015-03-27 19:37:05
(7,4,4)

bluephoenix 2015-03-27 19:37:05
wait, 7, 4, 4

amburger66 2015-03-27 19:37:05
7,4,4

copeland 2015-03-27 19:37:07
We must have $q^2 + r^2 = 81 - 49 = 32$. The only solution is $(p,q,r) = (7,4,4)$.

copeland 2015-03-27 19:37:08
How about $p=6$?

swirlykick 2015-03-27 19:37:26
6,6,3

ShadowQueenPeach 2015-03-27 19:37:26
6,6,3

24iam24 2015-03-27 19:37:26
6, 6, 3

sjag 2015-03-27 19:37:26
(6,6,3)

maverick8 2015-03-27 19:37:26
6,6,3

ThePiPie 2015-03-27 19:37:26
6, 6, 3

copeland 2015-03-27 19:37:28
We must have $q^2 + r^2 = 81 - 36 = 45$. The only solution is $(p,q,r) = (6,6,3)$.

copeland 2015-03-27 19:37:31
So these three are our only solutions. Here's a chart:
\[\begin{array}{c|c|c}
p&q&r \\ \hline
8 & 4 & 1 \\ \hline
7 & 4 & 4 \\ \hline
6 & 6 & 3
\end{array}\]

copeland 2015-03-27 19:37:32
How do we finish?

dhruv 2015-03-27 19:38:24
we find the values of c that correspond to the roots

shiningsunnyday 2015-03-27 19:38:24
Find values of C

24iam24 2015-03-27 19:38:24
so thats a and we can find c

andrewlin 2015-03-27 19:38:24
calculate c

Darn 2015-03-27 19:38:24
$pqr=\frac{c}{2}$

Dayranger 2015-03-27 19:38:24
find all possible values for c

copeland 2015-03-27 19:38:29
We can compute $a$ and $c$ for these. Recall that $a$ is their sum, and $c$ is twice their product:

copeland 2015-03-27 19:38:30
\[\begin{array}{c|c|c|c|c}
p&q&r&a&c \\ \hline
8 & 4 & 1 & 13 & 64 \\ \hline
7 & 4 & 4 & 15 & 224 \\ \hline
6 & 6 & 3 & 15 & 216
\end{array}\]

copeland 2015-03-27 19:38:31
Knowing the value of $a$ didn't help Jon, so what can we conclude?

MathStudent2002 2015-03-27 19:38:53
Then $a=15$

C-bass 2015-03-27 19:38:53
a is 15

ompatel99 2015-03-27 19:38:53
a must be 15

xayy 2015-03-27 19:38:53
a=15

Abeymom 2015-03-27 19:38:53
a=15

shiningsunnyday 2015-03-27 19:38:53
It must be 15

gxah 2015-03-27 19:39:05
224+216=440

AlisonH 2015-03-27 19:39:05
c=216 and 224

mathmaster2012 2015-03-27 19:39:05
a = 15, answer is 224+216=440

firemike 2015-03-27 19:39:05
224;216

caressezhu 2015-03-27 19:39:05
440

firemike 2015-03-27 19:39:05
224 + 216

danusv 2015-03-27 19:39:05
224+216 = 440

copeland 2015-03-27 19:39:08
$a$ must be 15 -- there are two possible solutions.

copeland 2015-03-27 19:39:10
We're asked for the sum of the two $c$'s for these solutions, so our answer is $224 + 216 = \boxed{440}$.

Darn 2015-03-27 19:39:31
2/5 of the way there!

copeland 2015-03-27 19:39:33
Gogo fractions!

copeland 2015-03-27 19:39:36
7. Triangle $ABC$ has side lengths $AB=12$, $BC=25$, and $CA=17$. Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$, vertex $Q$ on $\overline{AC}$, and vertices $R$ and $S$ on $\overline{BC}$. In terms of the side length $PQ=w$, the area of $PQRS$ can be expressed as the quadratic polynomial \[\operatorname{Area}(PQRS)=\alpha w-\beta w^2.\]Then the coefficient $\beta=\dfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

copeland 2015-03-27 19:39:46
When I first saw this problem it took me a second to decide whether I thought the area formula was a restriction on the problem or a consequence of the rest of the statement. That is, I took a moment to decide whether that fact was needed to solve the problem or if it was just required to AIME-ify the answer.

copeland 2015-03-27 19:39:59
It is a consequence of the rest of the setup, but I will point out along the way the place where you can use that statement to shortcut a bit of the solution. That is, if you assume that the area has to have that form, you can shorten some of the computations.

copeland 2015-03-27 19:40:07
What do we do first?

maverick8 2015-03-27 19:40:20
DiAgRaM

Dracae8 2015-03-27 19:40:20
picture!

swirlykick 2015-03-27 19:40:20
Draw a diagram

tdeng 2015-03-27 19:40:20
diagram

Dayranger 2015-03-27 19:40:20
diagram

hlasker1 2015-03-27 19:40:20
diagram!

SockFoot 2015-03-27 19:40:20
please give diagram

bluephoenix 2015-03-27 19:40:20
draw a diagram!

slenderestman 2015-03-27 19:40:20
draw diagram

dhruv 2015-03-27 19:40:20
draw a diagram

maverick8 2015-03-27 19:40:20
diagram

ompatel99 2015-03-27 19:40:20
Picture

_--__--_ 2015-03-27 19:40:20
Draw a diagram?

copeland 2015-03-27 19:40:23
Draw a diagram!

copeland 2015-03-27 19:40:24

copeland 2015-03-27 19:40:25

copeland 2015-03-27 19:40:27

copeland 2015-03-27 19:40:30
Now what do we do?

copeland 2015-03-27 19:40:54
The area is $wh$ so we need to find $h$.

copeland 2015-03-27 19:40:55
How can we compute $h$?

ShadowQueenPeach 2015-03-27 19:41:37
use similar triangles

Tommy2000 2015-03-27 19:41:37
similar triangles

mssmath 2015-03-27 19:41:37
Similiarity

slenderestman 2015-03-27 19:41:37
similar triangles?

copeland 2015-03-27 19:41:40
There are similar triangles here.

copeland 2015-03-27 19:41:42

copeland 2015-03-27 19:41:43
$\triangle ABC\sim\triangle PBQ$. If the height of $\triangle ABC$ is $H$, then what is $h$?

mathwrath 2015-03-27 19:42:44
H * (25-w)/25

dhruv 2015-03-27 19:42:44
(25-w)/25 * H

High 2015-03-27 19:42:44
H - Hw/25

SockFoot 2015-03-27 19:42:44
$H-H\cdot\dfrac{w}{25}$

swirlykick 2015-03-27 19:42:44
(H-h)/H = W/25

vinayak-kumar 2015-03-27 19:42:44
$H-wH/25$

copeland 2015-03-27 19:42:50
\[\dfrac {H-h}H=\dfrac w{25}\]or\[H-h=\dfrac{Hw}{25}\] Therefore $h=H-\dfrac{H}{25}\cdot w$.

copeland 2015-03-27 19:42:55
The area is \[w\cdot h=wH-\frac H{25}w^2.\] The value we are looking for is $\dfrac H{25}$.

copeland 2015-03-27 19:43:02
How can we find the height of $\triangle ABC$?

mohanxue612 2015-03-27 19:43:20
finding the area of abc

vinayak-kumar 2015-03-27 19:43:20
Find the area of the triangle

simon1221 2015-03-27 19:43:20
area two ways

copeland 2015-03-27 19:43:26
Good. Tell me more.

awesome123 2015-03-27 19:43:43
use herons formula

24iam24 2015-03-27 19:43:43
we can find H with heron's

maverick8 2015-03-27 19:43:43
Heron

awesome123 2015-03-27 19:43:43
herons formula

swirlykick 2015-03-27 19:43:43
Herons formula

Borealis 2015-03-27 19:43:43
Let's use Heron's formula and then find the altitude as area = 1/2 * b * h

Dracae8 2015-03-27 19:43:43
herons --> bh/2

dhruv 2015-03-27 19:43:43
heron's formula!

cxiong 2015-03-27 19:43:43
heron's formula

1023ong 2015-03-27 19:43:43
herons was beautiful

copeland 2015-03-27 19:43:47
There are two reasonable approaches and one silly one.

copeland 2015-03-27 19:43:49
The first reasonable approach is to drop an altitude and set up a large number of equations. That's what I did originally because I'm lazy.

copeland 2015-03-27 19:43:52
The second reasonable approach is to hope that this is a nice triangle and throw Heron's formula at it to try to find the area.

copeland 2015-03-27 19:43:57
The silly one is Stewart's Theorem. If you know a lot about Stewart's Theorem then you know it reduces to Heron where the cevian is an altitude. If you don't know that much about Stewart's Theorem then you are a lot like me.

copeland 2015-03-27 19:44:07
I think the most enlightening approach is Heron, especially given how many other problems today that we have solved/will solve by setting up massive systems of quadratic equations.

copeland 2015-03-27 19:44:11
What is the area of this triangle by Heron?

cxiong 2015-03-27 19:44:53
90

vinayak-kumar 2015-03-27 19:44:53
90

caressezhu 2015-03-27 19:44:53
90

maverick8 2015-03-27 19:44:53
90

andrewlin 2015-03-27 19:44:53
90

swirlykick 2015-03-27 19:44:53
90

awesome123 2015-03-27 19:44:53
90

nosaj 2015-03-27 19:44:53
90

Rmehtany 2015-03-27 19:44:53
90

copeland 2015-03-27 19:44:56
The perimeter is $12+17+25=54$ so the semiperimeter is $s=27$.

copeland 2015-03-27 19:45:01
The area solves
\begin{align*}
A^2
&=s(s-12)(s-17)(s-25)\\
&=27\cdot15\cdot10\cdot2\\
&=(3^3)(3\cdot5)(2\cdot5)\cdot2\\
&=(3^2\cdot5\cdot2)^2\\
&=90^2.
\end{align*}
The area is 90.

copeland 2015-03-27 19:45:11
What is the height?

maverick8 2015-03-27 19:45:58
36/5

24iam24 2015-03-27 19:45:58
so H=180/25=36/5

gxah 2015-03-27 19:45:58
36/5

simon1221 2015-03-27 19:45:58
180/25 or 36/5

dhruv 2015-03-27 19:45:58
36/5

mathwizard888 2015-03-27 19:45:58
36/5

countingarithmetic 2015-03-27 19:45:58
36/5

hlasker1 2015-03-27 19:45:58
36/5

swe1 2015-03-27 19:45:58
36/5

copeland 2015-03-27 19:46:03
The height solves $90=\frac{25H}2$ so the height is $H=\dfrac{180}{25}=\dfrac{36}5$

copeland 2015-03-27 19:46:07
What's the answer to the problem?

Darn 2015-03-27 19:46:45
36+125=161

shiningsunnyday 2015-03-27 19:46:45
36+125=161

dhruv 2015-03-27 19:46:45
H/25=36/125 =>161

High 2015-03-27 19:46:45
36/5*1/25=36/125 --> 161

azmath333 2015-03-27 19:46:45
161

ryanyz10 2015-03-27 19:46:45
161

owm 2015-03-27 19:46:45
36+125

mathwrath 2015-03-27 19:46:45
36/125=>161

24iam24 2015-03-27 19:46:45
36/125 so 161

copeland 2015-03-27 19:46:49
We want to compute $\dfrac{H}{25}=\dfrac{36}{125}$. The answer is $36+125=\boxed{161}$.

copeland 2015-03-27 19:46:56
Incidentally, notice if we assume the formula from the problem right at the beginning then we know that $h=\dfrac{\operatorname{Area}(PQRS)}{w}=\alpha-\beta w$. How could we finish from there?

vinayak-kumar 2015-03-27 19:48:13
let w=25

mssmath 2015-03-27 19:48:13
We know a from w=0, h=max cases

copeland 2015-03-27 19:48:16
If we know the value of $h$ at two different widths we can get two equations in $\alpha$ and $\beta$ that we can solve for $\beta$.

copeland 2015-03-27 19:48:18
If $w=25$ then $h=0$. That gives\[0=\alpha+25\beta.\]

copeland 2015-03-27 19:48:18
What other point would you choose?

cxiong 2015-03-27 19:48:57
w = 12.5

vinayak-kumar 2015-03-27 19:48:57
w=25/2? seems the easiest

24iam24 2015-03-27 19:48:57
w=0

1023ong 2015-03-27 19:48:57
midsegment?

swe1 2015-03-27 19:48:57
$w = \frac{25}{2}$

ompatel99 2015-03-27 19:48:57
w=0

acegikmoqsuwy2000 2015-03-27 19:48:57
at the top of the triangle when w=0

copeland 2015-03-27 19:49:10
You can choose $w=\dfrac{25}2$. That makes the $h$ half the height of the triangle so we're reduced to finding the height of the triangle.

copeland 2015-03-27 19:49:11
You could also choose $w=0$. Then the rectangle has area 0 and the height is equal to the height of the triangle. Again you need the height of the triangle but the computation is easier.

copeland 2015-03-27 19:49:15
However note that we divided by $w$ to find $h=\dfrac{\operatorname{Area}(PQRS)}{w}$ so you really need to mumble something about continuity before this is kosher.

copeland 2015-03-27 19:49:24
Also, this is an awesome triangle:

copeland 2015-03-27 19:49:25

copeland 2015-03-27 19:49:38
Also, yes, I swapped two of the vertices in my diagram. My bad.

copeland 2015-03-27 19:49:57

copeland 2015-03-27 19:50:11
The guy on the left is a 3-4-5 triangle and the one on the right is everybody's favorite 36-77-85 Pythagorean triple.

Tommy2000 2015-03-27 19:50:26
Yes, I knew that one!

copeland 2015-03-27 19:50:39
Raise your hand if you knew that one!

copeland 2015-03-27 19:50:49
Liars. All.

copeland 2015-03-27 19:50:56
8. Let $a$ and $b$ be positive integers satisfying $\dfrac{ab+1}{a+b}<\dfrac32$. The maximum possible value of $\dfrac{a^3b^3+1}{a^3+b^3}$ is $\dfrac pq$, where $p$ and $q$ are realtively prime positive integers. Find $p+q$.

copeland 2015-03-27 19:51:10
Notice anything about the expression $\dfrac{ab+1}{a+b}$? How strict is it?

1915933 2015-03-27 19:51:41
"Relatively" is misspelled.

copeland 2015-03-27 19:51:56
realtively. That's what I meant.

Poseidon2001 2015-03-27 19:52:20
what do you mean by strict?

copeland 2015-03-27 19:52:23
Good question.

copeland 2015-03-27 19:53:03
What about the inequality $\dfrac{x^6+3x^2}{x^2}<1$? Does that have a "lot" of solutions?

ryanyz10 2015-03-27 19:53:44
no.. very quickyl is greater than 1

Rmehtany 2015-03-27 19:53:44
no

henryweng 2015-03-27 19:53:44
no solution

dhruv 2015-03-27 19:53:44
no

copeland 2015-03-27 19:53:48
Bad example.

copeland 2015-03-27 19:54:01
$\dfrac{x^6+3x^2}{x^2+1}<1$?

tdeng 2015-03-27 19:54:36
how do you define a "lot"

copeland 2015-03-27 19:54:42
The area of the region, say.

vinayak-kumar 2015-03-27 19:54:57
it doesnt have a lot either

copeland 2015-03-27 19:55:00
Why not?

mssmath 2015-03-27 19:55:17
x going to infinity is bad

copeland 2015-03-27 19:55:19
It's degree 6 over degree 2. Eventually the numerator wins out and we fail.

copeland 2015-03-27 19:55:27
Back to our problem:

copeland 2015-03-27 19:55:35
$\dfrac{ab+1}{a+b}$ How strict is it?

SockFoot 2015-03-27 19:55:54
well $ab$ is gonna be a lot bigger than $a+b$

SockFoot 2015-03-27 19:55:54
most of the time

ompatel99 2015-03-27 19:55:57
Very strict. Product triumphs over sum later

copeland 2015-03-27 19:56:00
Right.

copeland 2015-03-27 19:56:10
This problem is trying to be tricky, but we're on to it.

copeland 2015-03-27 19:56:26
It's quadratic over linear so it should usually be big for positive integers. There aren't going to be very many solutions to this equation. That's nice.

copeland 2015-03-27 19:56:31
(I also notice that this looks like the hyperbolic cotangent sum formula, but that helps less than zero.)

copeland 2015-03-27 19:56:35
What should we do?

nosaj 2015-03-27 19:57:06
should we cross multiply?

cpma213 2015-03-27 19:57:06
Multiply a+b over?

vinayak-kumar 2015-03-27 19:57:06
cross multiply?

C-bass 2015-03-27 19:57:06
multiply both sides by a + b and 2

copeland 2015-03-27 19:57:08
Let's multiply by $2(a+b)$ and see where that gets us.

copeland 2015-03-27 19:57:10
Since $a$ and $b$ are positive, $\dfrac{ab+1}{a+b}<\dfrac32$ if and only if \[2ab+2<3a+3b.\]Bringing everything to one side gives \[2ab-3a-3b+2<0.\]

copeland 2015-03-27 19:57:10
Now what?

cpma213 2015-03-27 19:57:28
SFFT?

acegikmoqsuwy2000 2015-03-27 19:57:28
2ab+2<3a+3b then SFFT/

Tommy2000 2015-03-27 19:57:28
SFFT

bluephoenix 2015-03-27 19:57:28
SFFT

ryanyz10 2015-03-27 19:57:28
SFFT

gxah 2015-03-27 19:57:28
SFFT

acegikmoqsuwy2000 2015-03-27 19:57:28
simons favorite factoring trick?

_--__--_ 2015-03-27 19:57:28
SFFT?

copeland 2015-03-27 19:57:31
Simon's Favorite Factoring Trick. Of course we need that somewhere on the AIME.

copeland 2015-03-27 19:57:32
You can read about SFFT on our Wiki later here:
http://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick

copeland 2015-03-27 19:57:36
SFFT is hard for this expression. What do we do to make it easier?

NextEinstein 2015-03-27 19:58:01
multiply by 2 and SFFT

cpma213 2015-03-27 19:58:01
Multiply by 2 again for SFFT

vinayak-kumar 2015-03-27 19:58:01
multiply by 2 hehe

nosaj 2015-03-27 19:58:01
mulitply by 2

Aviously 2015-03-27 19:58:01
Multiply ny 2

copeland 2015-03-27 19:58:04
The middle terms are symmetric so we would benefit from the leading term being a square. What product of binomials do we want to pull out of \[4ab-6a-6b+4<0?\]

BPM14 2015-03-27 19:58:57
(2a-3)(2b-3)<5

Darn 2015-03-27 19:58:57
$(2a-3)(2b-3) < 5$

_--__--_ 2015-03-27 19:58:57
(2a-3)(2b-3)<5

ryanyz10 2015-03-27 19:58:57
(2a - 3) (2b - 3)

Aviously 2015-03-27 19:58:57
(2a+3)(2b+3)

nosaj 2015-03-27 19:58:57
Now it is (2a-3)(2b-3)-5<0

vinayak-kumar 2015-03-27 19:58:57
(2a-3)(2b-3)

AlisonH 2015-03-27 19:58:57
(2a-3)(2b-3)

copeland 2015-03-27 19:59:01
We can rewrite this quadratic as \[(2a-3)(2b-3)<5.\] $a$ and $b$ are both positive. By symmetry we can assume that $a\leq b$.

copeland 2015-03-27 19:59:02
If $a=1$ what are the solutions?

cpma213 2015-03-27 19:59:44
b is anything

vinayak-kumar 2015-03-27 19:59:44
any b

SockFoot 2015-03-27 19:59:44
$b=$anything

henryweng 2015-03-27 19:59:44
b can be anything

tdeng 2015-03-27 19:59:44
b can be anything

Rmehtany 2015-03-27 19:59:44
b>-1

copeland 2015-03-27 19:59:47
If $a=1$ then we want to solve $-(2b-3)<5$ or $2b>-8$. That's every $b$.

copeland 2015-03-27 19:59:53
If $a=2$ what are the solutions?

cpma213 2015-03-27 20:00:29
b=1,2,3

Dayranger 2015-03-27 20:00:29
b < 4

amburger66 2015-03-27 20:00:29
b<4

Poseidon2001 2015-03-27 20:00:29
b<4

chessderek 2015-03-27 20:00:29
anything less than 4

vinayak-kumar 2015-03-27 20:00:29
2 and 3

Darn 2015-03-27 20:00:29
wait i mean b=2,3

mathperson9 2015-03-27 20:00:29
b less than 4

countingarithmetic 2015-03-27 20:00:29
b = 2, 3

copeland 2015-03-27 20:00:31
If $a=2$ then we want to solve $(2b-3)<5$ or $b<4$. That's $(a,b)=(2,2),(2,3)$.

copeland 2015-03-27 20:00:32
If $a=3$ what are the solutions?

swirlykick 2015-03-27 20:01:14
no solutions

awesomemathlete 2015-03-27 20:01:14
none

_--__--_ 2015-03-27 20:01:14
b<7/3, so no solutions.

dantx5 2015-03-27 20:01:14
none

henryweng 2015-03-27 20:01:14
but a<=b so no solution

Rmink41 2015-03-27 20:01:14
None

24iam24 2015-03-27 20:01:14
no solution

copeland 2015-03-27 20:01:16
If $a=3$ then we want to solve $3(2b-3)<5$ or $6b-9<5$. That simplifies to \[b<\dfrac{7}3\] so there aren't any new solutions with $b\geq 3$.

copeland 2015-03-27 20:01:20
We're done. If $a,b>3$ then $(2a-3)(2b-3)>9$ and we fail.

copeland 2015-03-27 20:01:22
The only solutions to the first inequality with $a\leq b$ are
$\bullet$ $(1,b)$ for all $b$
$\bullet$ $(2,2)$, $(2,3)$

copeland 2015-03-27 20:01:27
What is the maximum possible value of $\dfrac{1^3+b^3}{1^3b^3+1}$?

ShadowQueenPeach 2015-03-27 20:02:00
1

tdeng 2015-03-27 20:02:00
1

swirlykick 2015-03-27 20:02:00
1

Darn 2015-03-27 20:02:00
1 lol

joey8189681 2015-03-27 20:02:00
1

simon1221 2015-03-27 20:02:00
1

nosaj 2015-03-27 20:02:00
1

rt03 2015-03-27 20:02:00
1

Aang 2015-03-27 20:02:00
1

vinayak-kumar 2015-03-27 20:02:00
it will always be 1 haha

dhruv 2015-03-27 20:02:00
1

czhu000 2015-03-27 20:02:00
1

hlasker1 2015-03-27 20:02:00
1

mathwizard888 2015-03-27 20:02:00
1

jam10307 2015-03-27 20:02:00
1

copeland 2015-03-27 20:02:02
This is always equal to 1 so the maximum is 1.

copeland 2015-03-27 20:02:05
What is the maximum possible value if $a=2$?

Eugenis 2015-03-27 20:03:00
Hmm, do we really have to chug through all these cases?

24iam24 2015-03-27 20:03:00
217/35=31/5

swirlykick 2015-03-27 20:03:00
oops, i mean 217/35=31/5

henryweng 2015-03-27 20:03:00
max=31/5

chessderek 2015-03-27 20:03:00
31/5

simon1221 2015-03-27 20:03:00
31/5?

cpma213 2015-03-27 20:03:00
31/5

High 2015-03-27 20:03:00
b= 3 and (a^3*b^3+1)(a^3+b^3) = 217/35 = 31/5

Eugenis 2015-03-27 20:03:00
$\frac{31}{5}$

ompatel99 2015-03-27 20:03:00
31/5

joey8189681 2015-03-27 20:03:00
217/35

vinayak-kumar 2015-03-27 20:03:00
(216+1)/(8+27)=31/5

bluephoenix 2015-03-27 20:03:00
217/35

tdeng 2015-03-27 20:03:00
217/35

copeland 2015-03-27 20:03:04
For $b=2$ we have \[\dfrac{2^32^3+1}{2^3+2^3}=\dfrac{8\cdot8+1}{8+8}=\dfrac{65}{16}\]
and $b=3$ gives \[\dfrac{2^33^3+1}{2^3+3^3}=\dfrac{217}{35}.\] The first one is around 4 and the second one is around 6 so the second is larger.

copeland 2015-03-27 20:03:08
The maximum value is \[\dfrac{217}{35}=\dfrac{31}{5}\] so the answer is $31+5=\boxed{36}$.

bluephoenix 2015-03-27 20:03:48
Yay, 8/15 of the way there!!

copeland 2015-03-27 20:03:51
Alright.

copeland 2015-03-27 20:03:58
Let's keep rocking.

copeland 2015-03-27 20:04:02
9. A cylindrical barrel with radius 4 feet and height 10 feet is full of water. A solid cube with side length 8 feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$.

copeland 2015-03-27 20:04:05

copeland 2015-03-27 20:04:08
Thanks to AoPS community member chezbgone2 for the great picture that he or she posted in the community and that we stol...er, I mean, appropriated (under the "publicly perform" subclause of article 7 of the AoPS Terms of Service of course) for this Math Jam.

copeland 2015-03-27 20:04:19
Everybody say, "Way to go, chez!"

High 2015-03-27 20:04:49
way to go chez

RegretDragunity 2015-03-27 20:04:49
Way to go, chez!!

henryweng 2015-03-27 20:04:49
way to go

ThePiPie 2015-03-27 20:04:49
way to go, chez!

bluephoenix 2015-03-27 20:04:49
Way to go, chez!!!!!!!!!!!!!!!!!!1

_--__--_ 2015-03-27 20:04:49
Way to go, chez!

flyrain 2015-03-27 20:04:49
Way to go, chez!

ssk9208 2015-03-27 20:04:49
Way to go, chez!

Tommy2000 2015-03-27 20:04:49
GO CHEZ!!!!!!!

cobbler 2015-03-27 20:04:49
Way to go, cheese!

tdeng 2015-03-27 20:04:49
way to go chez!

azmath333 2015-03-27 20:04:49
Way to go, chez!

chessderek 2015-03-27 20:04:49
Way to go, chezbgone2!

C-bass 2015-03-27 20:04:49
way to go, chez

SockFoot 2015-03-27 20:04:49
WAY TO G0O, CHEZ!!!!!!!!!

IsaacZ123 2015-03-27 20:04:49
CHEZ I LOVE YOU

cobbler 2015-03-27 20:04:49
chez, sorry

vinayak-kumar 2015-03-27 20:04:49
Way to go, chez!

yunchao 2015-03-27 20:04:49
Way to go, chez!

rt03 2015-03-27 20:04:49
Way to go, chez!

24iam24 2015-03-27 20:04:49
Go chez!

pops2724 2015-03-27 20:04:49
way to go, chez1

owm 2015-03-27 20:04:54
Way to go Chez!

xayy 2015-03-27 20:04:54
way to go chez

caressezhu 2015-03-27 20:04:54
Way to go, chez

copeland 2015-03-27 20:04:58
The volume we're looking for is the volume of the portion of the cube that lies below the rim of the cylinder.

copeland 2015-03-27 20:05:08
What's the cross-section of the cube look like in the plane of the cylinder's top?

math_cool 2015-03-27 20:05:42
Equilateral triangle

hnkevin42 2015-03-27 20:05:42
an equilateral triangle?

Borealis 2015-03-27 20:05:42
An equilateral triangle!

Abeymom 2015-03-27 20:05:42
triangle equilateral

shiningsunnyday 2015-03-27 20:05:42
Equilateral triangle

kaserav 2015-03-27 20:05:42
An equilateral triangle.

swe1 2015-03-27 20:05:42
An equilateral triangle

math-rules 2015-03-27 20:05:42
equilateral triangle

copeland 2015-03-27 20:05:45
By the symmetry of the cube, the three edges that are adjacent to the sunken vertex hit the cylinder at points that are equally-spaced around the rim.

vinayak-kumar 2015-03-27 20:05:55
An equilateral triangle with its circumcircle of radius 4

copeland 2015-03-27 20:05:57
So the cross-section is an equilateral triangle inscribed in the rim, which is a circle of radius 4:

copeland 2015-03-27 20:06:01

copeland 2015-03-27 20:06:05
What's the side length of that triangle?

phultrix 2015-03-27 20:06:35
4sqrt(3)

cpma213 2015-03-27 20:06:35
4sqrt(3)

Darn 2015-03-27 20:06:35
4sqrt3

acegikmoqsuwy2000 2015-03-27 20:06:35
4sqrt3

czhu000 2015-03-27 20:06:35
$4\sqrt{3}$

lovejj 2015-03-27 20:06:35
4sqrt3

_--__--_ 2015-03-27 20:06:35
4sqrt(3)

Poseidon2001 2015-03-27 20:06:35
4sqrt3

Aviously 2015-03-27 20:06:35
$4\sqrt{3}$

lcsmart 2015-03-27 20:06:37
4sqrt3

copeland 2015-03-27 20:06:39
If you don't see it right away, extend an altitude from the center of circle to one of the sides:

copeland 2015-03-27 20:06:41

copeland 2015-03-27 20:06:43
The small triangle is 30-60-90, so half of the big triangle's side length is $2\sqrt3$, and thus the big triangle has side length $4\sqrt3$.

copeland 2015-03-27 20:06:59
(Sanity check: It's definitely between 4 and 8.)

copeland 2015-03-27 20:07:01
Great -- now what?

SockFoot 2015-03-27 20:07:50
find height of the tetrahedron

High 2015-03-27 20:07:50
Find the height of the pyramid

uberminecraft722 2015-03-27 20:07:50
Construct a pyramid using the triangle

acegikmoqsuwy2000 2015-03-27 20:07:50
now we find the volume of the tetrahedron inside the cylinder by finding the height

raptorw 2015-03-27 20:07:50
find the height of the tetrahedron

Rmehtany 2015-03-27 20:07:50
find the height of the pyramid

shiningsunnyday 2015-03-27 20:07:50
Volume of terahedron

Rmehtany 2015-03-27 20:07:50
tetrahedron

copeland 2015-03-27 20:07:59
Imaging slicing the piece of the cube that's below the rim of the cylinder.

copeland 2015-03-27 20:08:01
If we turn it upside-down and set it on a table, it's a pyramid whose base is an equilateral triangle with side length $4\sqrt3$.

copeland 2015-03-27 20:08:02
What do the other three faces look like?

nosaj 2015-03-27 20:08:30
right iscoscles

IsaacZ123 2015-03-27 20:08:30
45-45-90 triangles

High 2015-03-27 20:08:30
isosceles right triangles?

swe1 2015-03-27 20:08:30
Isosceles right triangles

raptorw 2015-03-27 20:08:30
45-45-90 triangles

maverick8 2015-03-27 20:08:30
45-45-90 right triangles

SockFoot 2015-03-27 20:08:30
45 45 90 right triangles

copeland 2015-03-27 20:08:33
They're all isosceles right triangles. (They're triangular corners of faces of the original cube.)

copeland 2015-03-27 20:08:34
So what are the edge lengths?

maverick8 2015-03-27 20:09:06
2 sqrt 6

Rmehtany 2015-03-27 20:09:06
2rad6

uberminecraft722 2015-03-27 20:09:06
2sqrt(6)

Abeymom 2015-03-27 20:09:06
2 sqrt 6

Aviously 2015-03-27 20:09:06
$2\sqrt{6}$

malarm 2015-03-27 20:09:06
2rt6

Darn 2015-03-27 20:09:06
So it's $2\sqrt6$

copeland 2015-03-27 20:09:08
The legs of an isosceles right triangle are each $\dfrac{1}{\sqrt2}$ times the hypotenuse. Thus, they're $\dfrac{1}{\sqrt2} \cdot 4\sqrt{3} = 2\sqrt6$.

copeland 2015-03-27 20:09:12
How do we finish?

Abeymom 2015-03-27 20:10:03
look for height using pyth

uberminecraft722 2015-03-27 20:10:03
find the height, then the volume

maverick8 2015-03-27 20:10:03
Base*height/3

rt03 2015-03-27 20:10:03
find the height

copeland 2015-03-27 20:10:09
Base and height and, yeah.

copeland 2015-03-27 20:10:11
What's the base?

copeland 2015-03-27 20:10:30
I mean, qualitatively, which base?

vinayak-kumar 2015-03-27 20:10:55
the equilateral triangle

Rmehtany 2015-03-27 20:10:55
equilateral

azmath333 2015-03-27 20:10:55
Equilateral triangle

uberminecraft722 2015-03-27 20:10:55
The equilateral triangle base

IsaacZ123 2015-03-27 20:10:55
the 45-45-90 base

Darn 2015-03-27 20:10:55
oh it's the 45-45-90

maverick8 2015-03-27 20:10:55
A 45-45-90

swe1 2015-03-27 20:10:55
One of the isosceles right triangles, it's easier

copeland 2015-03-27 20:11:06
Either works, actually. Let's use the 45-45-90 base.

copeland 2015-03-27 20:11:14
If we set our pyramid flat onto one of the bases that was originally a side of the cube, we have a right triangular pyramid: a pyramid in which three edges are mutually perpendicular.

copeland 2015-03-27 20:11:16

copeland 2015-03-27 20:11:20
What's the volume of this?

simon1221 2015-03-27 20:12:07
1/3 base height

Darn 2015-03-27 20:12:07
Equal to $(2\sqrt{6})^3/6$

Rmehtany 2015-03-27 20:12:07
8rad6

rt03 2015-03-27 20:12:07
8sqrt6

tdeng 2015-03-27 20:12:07
8sqrt6?

_--__--_ 2015-03-27 20:12:07
8sqrt(6)?

vinayak-kumar 2015-03-27 20:12:07
2rt(6)^3/6=8rt(6)

Tommy2000 2015-03-27 20:12:07
$(2\sqrt6)^3/6=8sqrt6$

nosaj 2015-03-27 20:12:07
1/3 bh = 1/3*1/2*(2sqrt6)^3

swe1 2015-03-27 20:12:07
$\frac{(2\sqrt{6})^3}{6} = 8\sqrt{6}$

copeland 2015-03-27 20:12:10
It's just $\frac16$ times the product of the lengths of the three mutually perpendicular sides.

copeland 2015-03-27 20:12:11
These sides all have length $2\sqrt6$, so our volume is
\[
\frac16 \cdot (2\sqrt6)^3 = 8\sqrt6.
\]

copeland 2015-03-27 20:12:16
Our final answer is the volume squared, so it's $(8\sqrt6)^2 = 64 \cdot 6 = \boxed{384}$.

Darn 2015-03-27 20:12:35
3/5 done!!

chessderek 2015-03-27 20:12:35
3/5 done

bluephoenix 2015-03-27 20:12:35
oh boy, 3/5 of the way

Dayranger 2015-03-27 20:12:35
3/5 way through!

copeland 2015-03-27 20:12:42
You guys are killing the fractions part today, too.

copeland 2015-03-27 20:12:53
10. Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1,2,\ldots, n$ quasi-increasing if $a_k \le a_{k+1} + 2$ for each $1 \le k \le n - 1$. For example, $54321$ and $14253$ are quasi-increasing permutations of the integers $1,2,3,4,5,$ but $45123$ is not. Find the number of quasi-increasing permutations of the integers $1,2, \ldots, 7$.

copeland 2015-03-27 20:12:58
How might we get started?

swirlykick 2015-03-27 20:13:15
small cases

ompatel99 2015-03-27 20:13:15
Small cases?

ShadowQueenPeach 2015-03-27 20:13:15
try smaller cases

BPM14 2015-03-27 20:13:15
Make small cases with 1, then 2 numbers, etc.

SockFoot 2015-03-27 20:13:15
test small cases

copeland 2015-03-27 20:13:18
Trying a few smaller cases sounds like a good idea. How many quasi-increasing permutations are there of length 1?

Aviously 2015-03-27 20:13:35
I

AlisonH 2015-03-27 20:13:35
1

tdeng 2015-03-27 20:13:35
1

czhu000 2015-03-27 20:13:35
1

Abeymom 2015-03-27 20:13:35
1

gxah 2015-03-27 20:13:35
1

Superwiz 2015-03-27 20:13:36
1

copeland 2015-03-27 20:13:38
Every permutation of length 1 is quasi-increasing. In fact, for $n \le 3$ we see every permutation of length $n$ is quasi-increasing.

copeland 2015-03-27 20:13:42
Things really start to get interesting at $n = 4$. How many quasi-increasing permutations of length 4 are there?

copeland 2015-03-27 20:13:52
That's a bit more difficult, but it still seems like most sequences are quasi-increasing. How can a permutation of $1,2,3,4$ not be quasi-increasing?

IsaacZ123 2015-03-27 20:14:49
if 1 is next to 4

czhu000 2015-03-27 20:14:49
If 4 is followed by 1

ShadowQueenPeach 2015-03-27 20:14:49
4 is behind 1

maverick8 2015-03-27 20:14:49
A jump from 4 to 1

RegretDragunity 2015-03-27 20:14:49
if 4 is followed by 1

simon1221 2015-03-27 20:14:49
4123 or anything with 41

High 2015-03-27 20:14:49
1 comes after 4

Rocksolid 2015-03-27 20:14:49
If it contains the pair 41

henryweng 2015-03-27 20:14:49
4 before 1

mathwrath 2015-03-27 20:14:49
If the 4 is right before the 1

copeland 2015-03-27 20:15:00
The only way a permutation of $1,2,3,4$ can fail to be quasi-increasing is if the 4 immediately precedes the 1.

copeland 2015-03-27 20:15:01
Does this give any insight on how we might count the number of quasi-increasing permutations of length 4?

Aviously 2015-03-27 20:15:27
$24-6=18$ for length of 4.

Tommy2000 2015-03-27 20:15:27
total - failures

swe1 2015-03-27 20:15:27
Complementary counting

uberminecraft722 2015-03-27 20:15:27
count the wrong answers and subtract

shiningsunnyday 2015-03-27 20:15:27
Complement

_--__--_ 2015-03-27 20:15:27
Casework, 24-6 = 18

High 2015-03-27 20:15:27
4! - 3! = 18

copeland 2015-03-27 20:15:32
Well, we could just take the total number of permutations (there are 24) and subtract the number with 1 immediately preceding 4 (there are 6). This gives us $24 - 6 = 18$ permutations of length 4.

copeland 2015-03-27 20:15:41
However, is there any way we could count this more constructively given the information we know?

copeland 2015-03-27 20:15:59
I get the feeling that recursion will be more valuable than complementary counting when $n$ becomes large.

copeland 2015-03-27 20:16:04
How might we get from a permutation of length 3 to a permutation of length 4?

swirlykick 2015-03-27 20:16:24
we start with a permutation of 3 that works, then we add 4 at the end, or before 2,3

vinayak-kumar 2015-03-27 20:16:24
find a way to insert the largest numer

ShadowQueenPeach 2015-03-27 20:16:24
insert a 4

temp8909 2015-03-27 20:16:24
add 4

SockFoot 2015-03-27 20:16:24
add a $4$ somewhere that isn't bad

cxiong 2015-03-27 20:16:24
add a 4 somewhere

Tommy2000 2015-03-27 20:16:24
We have locations to place the 4 in

lcsmart 2015-03-27 20:16:24
implanting a term $4$

copeland 2015-03-27 20:16:30
If we take any permutation of length 3, we can create a quasi-increasing permutation of length 4 by inserting the 4 in any position except immediately before the 1.

copeland 2015-03-27 20:16:32
For example, if we had the permutation 123, we could stick the 4 in any blank except the first: __1__2__3__.

copeland 2015-03-27 20:16:38
This gives us 3 quasi-increasing permutations of length 4 for each permutation of length 3.

copeland 2015-03-27 20:16:40
Can we generalize this to permutations of length $n$?

swirlykick 2015-03-27 20:18:06
P(n)=3P(n-1)

swe1 2015-03-27 20:18:06
Add next number to end, or before $n-1$ or $n-2$

High 2015-03-27 20:18:06
permutations of n = 3*permutations of n - 1

mathwrath 2015-03-27 20:18:06
Yes: Before n-2, n-1, or at the end

Abeymom 2015-03-27 20:18:06
multiply by three evertytime

24iam24 2015-03-27 20:18:06
is it always 3f(n-1)?

czhu000 2015-03-27 20:18:06
It's $3 \times$ (the number of ways for $n-1$)

DivideBy0 2015-03-27 20:18:06
there are always 3 places to insert n

chessderek 2015-03-27 20:18:06
It seems like a geometric sequence with a ratio of 3.

Tommy2000 2015-03-27 20:18:06
$S_n=3S_{n-1}$

copeland 2015-03-27 20:18:11
Yes! In fact, it generalizes immediately. If we take any quasi-increasing permutation of length $n-1$, we can place $n$ before $n-1$ or $n-2$, or at the end of the permutation to produce a quasi-increasing permutation of length $n$.

copeland 2015-03-27 20:18:13
So we get 3 quasi-increasing permutations of length $n$ for each quasi-increasing permutation of length $n-1$.

copeland 2015-03-27 20:18:17
Moreover, if we erase $n$ in any quasi-increasing permutation of length $n$, we get a quasi-increasing permutation of length $n - 1$. So this gives us all of the quasi-increasing permutations of length $n$.

copeland 2015-03-27 20:18:29
That is, if we let $S_n$ be the number of quasi-increasing permutations of length $n$, we have $S_n = 3\cdot S_{n-1}$.

copeland 2015-03-27 20:18:33
(Note that we need $n\ge 3$ here -- our argument clearly breaks down for $n = 1$ or $2$.)

copeland 2015-03-27 20:18:34
So how do we get $S_{7}$?

Darn 2015-03-27 20:19:34
Keep multiplying by $3$!

BPM14 2015-03-27 20:19:34
468 = 3^6*2

24iam24 2015-03-27 20:19:34
18x3x3x3

_--__--_ 2015-03-27 20:19:34
S7 = S4 * 3 * 3 * 3 = 18 * 3 * 3 * 3 = 486

Tommy2000 2015-03-27 20:19:34
so our answer 6*3^4=486

chessderek 2015-03-27 20:19:34
we can multiply 18 by 3^3 to get 486

DivideBy0 2015-03-27 20:19:34

Abeymom 2015-03-27 20:19:34
6*3^4

maverick8 2015-03-27 20:19:34
18*27

Dracae8 2015-03-27 20:19:34
2*3^5

High 2015-03-27 20:19:34
S_4 = 18, s_5 = 54, S_6= 162, S_7=486

copeland 2015-03-27 20:19:36
We already know $S_4 = 18$, so we just start multiplying by 3 until we get to $S_{7} = 18\cdot 3^{3} = \boxed{486}$.

24iam24 2015-03-27 20:20:06
2/3 of the way there!

chessderek 2015-03-27 20:20:06
2/3!

IsaacZ123 2015-03-27 20:20:28
This problem looked tougher than it actually was

copeland 2015-03-27 20:20:41
There were a lot of problems on this that were either easier or harder than expected.

copeland 2015-03-27 20:20:51
You never now until you solve them I guess.

copeland 2015-03-27 20:20:54
11. The circumcircle of acute $\triangle ABC$ has center $O$. The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ respectively. Also $AB=5$, $BC=4$, $BQ=4.5$, and $BP=\dfrac mn$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

copeland 2015-03-27 20:20:59
Let's start with a diagram, of course.

copeland 2015-03-27 20:21:05

copeland 2015-03-27 20:21:10
The diagram uses that $\triangle ABC$ is acute: If $P$ is not between $A$ and $B$ then $A$, $B$, and $C$ are all "above" the diameter meaning the triangle is obtuse.

copeland 2015-03-27 20:21:21
(Those lengths on $BC$ and $BQ$ are a little confusing, sorry.)

copeland 2015-03-27 20:21:25
OK, so we have a circle. We have lines intersecting at points and we have distances. What tool pops into mind now?

_--__--_ 2015-03-27 20:21:37
You said "now", isn't it "know"?

tdeng 2015-03-27 20:21:37
we never now?

SockFoot 2015-03-27 20:21:37
yeah man you never now

copeland 2015-03-27 20:21:40
Never now.

copeland 2015-03-27 20:21:42
Always later.

bluephoenix 2015-03-27 20:21:50
Power of a Point

Rmehtany 2015-03-27 20:21:50
power of point

_--__--_ 2015-03-27 20:21:50
Power of a Point

tdeng 2015-03-27 20:21:50
power of a point

cxiong 2015-03-27 20:21:50
power of a point?

Darn 2015-03-27 20:21:50
Power of a Point

HANRUI11122001 2015-03-27 20:21:50
POWER OF A POINT

maverick8 2015-03-27 20:21:50
Pop

IsaacZ123 2015-03-27 20:21:50
power of a opint

copeland 2015-03-27 20:21:54
Power of a point! What point?

va2010 2015-03-27 20:22:35
point Q?

maverick8 2015-03-27 20:22:35
Q

nosaj 2015-03-27 20:22:35
Q

bluephoenix 2015-03-27 20:22:35
Point Q

vinayak-kumar 2015-03-27 20:22:35
Q

azmath333 2015-03-27 20:22:35
Q?

copeland 2015-03-27 20:22:40
Q is good. What else is good?

Rmehtany 2015-03-27 20:23:01
P for point

nosaj 2015-03-27 20:23:01
P

henryweng 2015-03-27 20:23:01
P

Rmehtany 2015-03-27 20:23:01
P

va2010 2015-03-27 20:23:01
P

copeland 2015-03-27 20:23:04
Both $P$ and $Q$ lie on two lines so let's use POP on $P$ and $Q$. Let me label some more edges.

copeland 2015-03-27 20:23:06
Let's actually get the interesting distances in there, $BC$, $CQ$, $PA$, and $PB$:

copeland 2015-03-27 20:23:08

copeland 2015-03-27 20:23:13
What other length do we need?

Rmehtany 2015-03-27 20:24:10
OP

amburger66 2015-03-27 20:24:10
radius?

zvxcqu 2015-03-27 20:24:10
OB

Dayranger 2015-03-27 20:24:10
OB

acegikmoqsuwy2000 2015-03-27 20:24:10
PQ

DivideBy0 2015-03-27 20:24:10
extension of PQ till it hits circle on the other side of O

copeland 2015-03-27 20:24:13
In order to use POP we're also going to need the radius.

copeland 2015-03-27 20:24:16

copeland 2015-03-27 20:25:53
With POP it's nice to have the distance from the center because difference of squares factorizations pop out. We'll see.

copeland 2015-03-27 20:26:14
So let's also name $PO=u$ and $OQ=v$. What equations do we have?

Tommy2000 2015-03-27 20:26:34
x+y=5

copeland 2015-03-27 20:26:37
Clearly we have \[x+y=5.\]

IsaacZ123 2015-03-27 20:26:58
.5*4.5=(v-r)(v+r)

mathwrath 2015-03-27 20:26:58
v^2-r^2=9/4

copeland 2015-03-27 20:27:02
POP on $Q$ gives $(v+r)(v-r)=4.5\cdot0.5$ or \[v^2-r^2=\frac94.\]

zvxcqu 2015-03-27 20:27:16
u^2+r^2=y^2

RegretDragunity 2015-03-27 20:27:16
u^2+r^2 = y^2

math-rules 2015-03-27 20:27:31
r^2 + v^2 = (4.5)^2

copeland 2015-03-27 20:27:34
The Pythagorean Theorem on $\triangle POB$ gives \[u^2+r^2=y^2.\]
The Pythagorean Theorem on $\triangle QOB$ gives \[v^2+r^2=\left(\frac92\right)^2=\frac{81}4.\]

copeland 2015-03-27 20:28:05
We need one more. . .

copeland 2015-03-27 20:28:21
What about POP on $P$?

zvxcqu 2015-03-27 20:28:42
xy = r^2-u^2

IsaacZ123 2015-03-27 20:28:42
xy=(r-u)(r+u)

AlisonH 2015-03-27 20:28:42
xy=(r-u)(u+v)

copeland 2015-03-27 20:28:48
POP on $P$ gives $(r-u)(r+u)=xy$ or \[r^2-u^2=xy.\]

copeland 2015-03-27 20:28:50
So we get

copeland 2015-03-27 20:28:51
\begin{eqnarray}
x+y&=5&\qquad(1)\\
r^2-u^2&=xy&\qquad(2)\\
v^2-r^2&=\frac94&\qquad(3)\\
u^2+r^2&=y^2&\qquad(4)\\
v^2+r^2&=\frac{81}4&\qquad(5)
\end{eqnarray}

cobbler 2015-03-27 20:29:15
What tool POPS into mind. I see what you did there ;-]

copeland 2015-03-27 20:29:16
Thank you. Finally. . .

copeland 2015-03-27 20:29:22
We want to find $BP=y$.

copeland 2015-03-27 20:29:23
Which pair of equations gets us closer to $y$?

dhruv 2015-03-27 20:29:51
2 and 4

tdeng 2015-03-27 20:29:51
2,4

NextEinstein 2015-03-27 20:29:51
2, 4

acegikmoqsuwy2000 2015-03-27 20:29:51
(4) and (2)

copeland 2015-03-27 20:29:53
Adding (2) and (4) gives
\[2r^2=xy+y^2.\]

copeland 2015-03-27 20:29:54
Now what?

nosaj 2015-03-27 20:30:31
Factor RHS

ShadowQueenPeach 2015-03-27 20:30:31
factor out y and use equation 1

DivideBy0 2015-03-27 20:30:31
factor out y and plug in x+y

gxah 2015-03-27 20:30:31
substitute x+y=5

copeland 2015-03-27 20:30:35
We can substitute $x+y=5$ to get
\[2r^2=xy+y^2=y(x+y)=5y.\]

copeland 2015-03-27 20:30:37
OK, now all we need is $r^2$. How do we get that?

tdeng 2015-03-27 20:31:00
subtract 3 and 5

IsaacZ123 2015-03-27 20:31:00
3 and 5?

24iam24 2015-03-27 20:31:00
subtract 3 and 5

Tommy2000 2015-03-27 20:31:00
subtract 5 and 3

Darn 2015-03-27 20:31:00
equations (3) and (5)

amburger66 2015-03-27 20:31:00
3 and 5

copeland 2015-03-27 20:31:02
Subtracting (3) from (5) gives
\[
2r^2=\frac{81}4-\frac94=\frac{72}4=18.
\]

copeland 2015-03-27 20:31:03
And what is $y$?

acegikmoqsuwy2000 2015-03-27 20:31:52
18/5

_--__--_ 2015-03-27 20:31:52
y = 18/5

IsaacZ123 2015-03-27 20:31:52
18/5

SockFoot 2015-03-27 20:31:52
$\dfrac{18}{5}$

DivideBy0 2015-03-27 20:31:52

High 2015-03-27 20:31:52
18/5

owm 2015-03-27 20:31:52
18/5

malarm 2015-03-27 20:31:52
18/5

Mlux 2015-03-27 20:31:52
2*18*18/5

Rocksolid 2015-03-27 20:31:52
18/5

copeland 2015-03-27 20:31:55
\[y=\frac{2r^2}5=\frac{18}5.\]

copeland 2015-03-27 20:31:56
The answer is $18+5=\boxed{23}$.

copeland 2015-03-27 20:32:01
There's a second solution to this problem using similar triangles. To get there you begin with the other possible first instinct on the problem: circumcircles want radii:

copeland 2015-03-27 20:32:07

copeland 2015-03-27 20:32:14
Now you have 3 isosceles triangles whose angles are related somehow. Time for an angle chase!

copeland 2015-03-27 20:32:20
I'll leave that for you to think about on your own.

ompatel99 2015-03-27 20:32:34
11/15 done. It doesnt simplify

copeland 2015-03-27 20:32:37
Tough.

copeland 2015-03-27 20:32:46
Hopefully the rest of them will.

copeland 2015-03-27 20:32:48
12. There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an $A$ or a $B$. Find the number of such strings that do not have more than 3 adjacent letters that are identical.

24iam24 2015-03-27 20:33:21
this was an amc 12 vid

High 2015-03-27 20:33:21
AMC 12A #22

vinayak-kumar 2015-03-27 20:33:21
Next up... featuring the easy version of AMC 12 problem 22

Tommy2000 2015-03-27 20:33:21
AMC 12 #22

nosaj 2015-03-27 20:33:21
The repeat question.

thequantumguy 2015-03-27 20:33:21
just like AMC 12 2015 number 22

swirlykick 2015-03-27 20:33:21
AMC12 A?

bluephoenix 2015-03-27 20:33:21
Isn't this like one of the AMC 12 this year?

danusv 2015-03-27 20:33:21
Misplaced problem LOL

math_cool 2015-03-27 20:33:21
AMC 12 lol

mssmath 2015-03-27 20:33:21
MAA copies itself yeah

copeland 2015-03-27 20:33:24
Freebie for you 12A folks. A significantly harder version of this problem appeared as 2015 AMC12A Problem 22. We will speed through this one.

copeland 2015-03-27 20:33:34
Let's try some small cases. Let's set $S(n)$ to be the number of valid strings of length $n$.

copeland 2015-03-27 20:33:36
What is $S(1)$?

vinayak-kumar 2015-03-27 20:34:05
2

High 2015-03-27 20:34:05
2.

mathwizard888 2015-03-27 20:34:05
2

Rmehtany 2015-03-27 20:34:05
2

nosaj 2015-03-27 20:34:05
2

ShadowQueenPeach 2015-03-27 20:34:05
2

copeland 2015-03-27 20:34:07
$S(1)=2.$ We can either write $A$ or $B.$

copeland 2015-03-27 20:34:08
What is $S(2)?$

swirlykick 2015-03-27 20:34:30
4

24iam24 2015-03-27 20:34:30
4

lcsmart 2015-03-27 20:34:30
4

RegretDragunity 2015-03-27 20:34:30
4

xayy 2015-03-27 20:34:30
4

rt03 2015-03-27 20:34:30
4

meeptopia 2015-03-27 20:34:30
4

Aviously 2015-03-27 20:34:30
4

simon1221 2015-03-27 20:34:30
4

ImpossibleTriangle 2015-03-27 20:34:30
4

Dayranger 2015-03-27 20:34:30
4

copeland 2015-03-27 20:34:34
$S(2)=4.$ We can have $\{AA,AB,BA,BB\}.$

copeland 2015-03-27 20:34:35
What is $S(3)?$

swirlykick 2015-03-27 20:34:49
8

Aviously 2015-03-27 20:34:49
8

AlisonH 2015-03-27 20:34:49
8

ucap 2015-03-27 20:34:49
8

rt03 2015-03-27 20:34:49
8

asp7479 2015-03-27 20:34:49
8

copeland 2015-03-27 20:34:52
$S(3)=8.$ We can have $\{AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB\}.$

copeland 2015-03-27 20:34:53
OK, so $S(n)=2^n.$ Clearly.

24iam24 2015-03-27 20:35:17
no

danusv 2015-03-27 20:35:17
NOPE!

Rmehtany 2015-03-27 20:35:17
no

cpma213 2015-03-27 20:35:17
Nope...

24iam24 2015-03-27 20:35:17
never

RegretDragunity 2015-03-27 20:35:17
Nope, not just yet

lcsmart 2015-03-27 20:35:19
very funny

copeland 2015-03-27 20:35:20
(I recycled that joke, too.)

copeland 2015-03-27 20:35:26
No. . . that's wrong. You can usually put $A$ at the end, unless you already have 3 $A$s to begin with.

copeland 2015-03-27 20:35:30
There's an even deeper problem with naive recursion in that we can't just chop off the last character and talk in terms of $S(n-1)$. Is there something we can do?

nosaj 2015-03-27 20:36:22
Use recursion based on how many consecutive letters there are at the end.

simon1221 2015-03-27 20:36:22
blocks of length 1 2 and 3?

DivideBy0 2015-03-27 20:36:22
chop off the last set of equal characters

acegikmoqsuwy2000 2015-03-27 20:36:22
use S(n-1), S(n-2), and S(n-3)

mssmath 2015-03-27 20:36:22
Chop of last set of same integers

copeland 2015-03-27 20:36:25
We could instead chop off the last block of equal characters. What do we get?

Rmehtany 2015-03-27 20:37:12
S(n-1)+S(n-2)+S(n-3)

vinayak-kumar 2015-03-27 20:37:12
you will get a sequence of either n-1, n-2, n-3

cpma213 2015-03-27 20:37:12
S(n-1), S(n-2), S(n-3), they sum to S(n)

copeland 2015-03-27 20:37:15
We can get a valid string by taking any string of length $n$ and deleting as many identical characters as possible from the end of the string (1, 2, or 3).

copeland 2015-03-27 20:37:16
Therefore the strings of length $n$ ends in one of three ways:
* math=inline]$\cdots BA$[/math A single character of some parity following a string of length $n-1$ that ends in the opposite parity
* math=inline]$\cdots BAA$[/math Two characters of some parity following a string of length $n-2$ that ends in the opposite parity
* math=inline]$\cdots BAAA$[/math Three characters of some parity following a string of length $n-3$ that ends in the opposite parity

copeland 2015-03-27 20:37:24
What happened!

copeland 2015-03-27 20:37:36
Therefore the strings of length $n$ ends in one of three ways:
* [ $\cdots BA$ ] A single character of some parity following a string of length $n-1$ that ends in the opposite parity
* [ $\cdots BAA$ ] Two characters of some parity following a string of length $n-2$ that ends in the opposite parity
* [ $\cdots BAAA$ ] Three characters of some parity following a string of length $n-3$ that ends in the opposite parity

copeland 2015-03-27 20:38:01
Furthermore note that this gives a bijection between the set of strings of length $n$ and the union of the sets of strings of length $n-1,$ $n-2,$ and $n-3$ given by deleting as many identical characters as possible from the end of the string.

copeland 2015-03-27 20:38:05
This sets up a bijection. Every string of length $n$ gives a string of length $n-1,$ $n-2,$ or $n-3$ by deleting the final characters. We get all such strings uniquely in this way. So the set counted by $S(n)$ is in correspondence with the union of the 3 sets counted by $S(n-1),$ $S(n-2),$ and $S(n-3).$

copeland 2015-03-27 20:38:19
Deleting as many identical characters as possible from the end of each of our strings of length $n$ gives a bijection between the set of strings of length $n$ and the union of the sets of strings of length $n-1,$ $n-2,$ and $n-3.$

copeland 2015-03-27 20:38:20
What recursive formula does that give us?

ImpossibleTriangle 2015-03-27 20:39:18
s(n)=s(n-1)+s(n-2)+s(n-3)

acegikmoqsuwy2000 2015-03-27 20:39:18
S(n)=S(n-1)+S(n-2)+S(n-3)

High 2015-03-27 20:39:18
S(n)=S(n-1)+S(n-2)+S(n-3)

cpma213 2015-03-27 20:39:18
S(n) = S(n-1)+S(n-2)+S(n-3)

countingarithmetic 2015-03-27 20:39:18
S(n) = S(n-1)+S(n-2)+S(n-3)

vinayak-kumar 2015-03-27 20:39:18
s(n)=s(n-1)+s(n-2)+s(n-3)

ImpossibleTriangle 2015-03-27 20:39:18
s(n)=s(n-1)+s(n-2)+s(n-3

24iam24 2015-03-27 20:39:18
add those 3

dhruv 2015-03-27 20:39:18
S(n)=S(n-1)+S(n-2)+S(n-3)

Dayranger 2015-03-27 20:39:18
S(n) = S(n - 1) + S(n - 2) + S(n - 3)

swirlykick 2015-03-27 20:39:18
S(N)=S(N-1)+S(N-2)+S(N-3)

owm 2015-03-27 20:39:18
S(n)=S(n-1)+S(n-2)+S(n-3)

copeland 2015-03-27 20:39:23
This tells us that

copeland 2015-03-27 20:39:24
\[S(n)=S(n-1)+S(n-2)+S(n-3).\]

copeland 2015-03-27 20:39:28
Now we just start listing. Fortunately we computed the first three already, and then each number is just the sum of the previous three:

copeland 2015-03-27 20:39:30
\[\begin{array}{c|cccccccccc}
n &1&2&3& 4&5 &6&7&8&9&10\\
\hline
S(n)&2&4&8&14&26&48&88&162&298&\boxed{548}\\
\end{array}\]

copeland 2015-03-27 20:39:44
Alrighty.

ompatel99 2015-03-27 20:39:55
12/15 Done. 4/5 or 80%!

copeland 2015-03-27 20:40:56
13. Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum_{k=1}^n \sin(k)$, where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$.

copeland 2015-03-27 20:41:02
What methods do we have in general for evaluating long series (not just trig series)? What sorts of tools have you used in problems before?

tdeng 2015-03-27 20:41:45
telescoping

DivideBy0 2015-03-27 20:41:45
telescoping

andrewlin 2015-03-27 20:41:45
telescoping?

High 2015-03-27 20:41:45
Telescoping series

math-rules 2015-03-27 20:41:45
cancellation/telescoping

ImpossibleTriangle 2015-03-27 20:41:45
cancell

nosaj 2015-03-27 20:41:45
telescoping

mathwizard888 2015-03-27 20:41:45
telescopign

macandcheese 2015-03-27 20:41:45
telescoping

thequantumguy 2015-03-27 20:41:48
from your precalculus book

copeland 2015-03-27 20:41:55
(Yeah, this problem is in the precalc book. Too bad that's not an approved tool.)

copeland 2015-03-27 20:42:11
Unfortunately, this series doesn't look geometric or arithmetic, so the simplest tools don't do much. However, telescoping is an interesting idea.

copeland 2015-03-27 20:42:16
I might think of telescoping when I see big summations of trig functions, because many trig identities turn things into differences.

copeland 2015-03-27 20:42:28
Are there any trig identities we might use to turn each of our terms into $\text{something} - \text{something else}$?

nosaj 2015-03-27 20:43:16
What is up with problem 13 on both AIMEs being from the precalc book?

copeland 2015-03-27 20:43:18
I don't know. You can find the precalc book in the online bookstore here after the Math Jam:
http://www.artofproblemsolving.com/store/item/precalculus

ryanyz10 2015-03-27 20:43:54
pythag identities?

copeland 2015-03-27 20:44:00
Which one?

nosaj 2015-03-27 20:44:19
sum of angles forumla?

copeland 2015-03-27 20:44:31
Sum of angles is good but it gives a quadratic stuff.

copeland 2015-03-27 20:44:48
We want to multiply by something, say, in order to get degree 1 things.

Dayranger 2015-03-27 20:44:56
sin 2k = 2 sin k cos k

copeland 2015-03-27 20:44:59
Yeah, like that one!

copeland 2015-03-27 20:45:06
Except there's no sum.

copeland 2015-03-27 20:45:19
But I multiply a sine by a cosine and I get a single sine out.

tdeng 2015-03-27 20:45:40
product to sum?

copeland 2015-03-27 20:45:43
Yeah, that's great.

copeland 2015-03-27 20:45:51
The product to sum formula for sine has this form: \[\sin(x)\sin(y) = \frac12 \left(\cos(x-y) - \cos(x+y)\right).\]

copeland 2015-03-27 20:45:55
How can we make our sum look more like this?

copeland 2015-03-27 20:47:05
I have a sum of sines.

copeland 2015-03-27 20:47:16
I somehow want to turn each sine into a pair of things.

copeland 2015-03-27 20:47:23
(In order to telescope)

q12 2015-03-27 20:47:36
multiply everything by sin(1)

copeland 2015-03-27 20:47:43
We sure want to multiply by sin(something).

copeland 2015-03-27 20:47:54
Let's just leave it as a variable. Maybe that will help us figure out what the something is.

copeland 2015-03-27 20:48:00
For example, we can multiply by 1 in the form of $\dfrac{\sin(y)}{\sin(y)}$ for some $y$. Then our sum becomes \[ a_n = \frac{1}{\sin(y)}\sum_{k=1}^n \sin(k)\sin(y).\]

copeland 2015-03-27 20:48:11
Are there any choices for $y$ that would make this series telescope?

acegikmoqsuwy2000 2015-03-27 20:49:04
multiply by sin(pi/2)

math-rules 2015-03-27 20:49:04
1/2

thequantumguy 2015-03-27 20:49:04
1/2

ucap 2015-03-27 20:49:04
1/2

copeland 2015-03-27 20:49:07
This sum evaluates cosine at $k-y,k+y$ then $k+1-y,k+1+y$ then $k+2-y,k+2+y$ then $k+3-y,k+3+y$, etc.

copeland 2015-03-27 20:49:19
In order for it to telescope we ought to try to arrange that neighboring terms are equal, so $k+y=k+1-y$ and $k+1+y=k+2-y$, etc.

copeland 2015-03-27 20:49:28
These all simplify to $y=\dfrac12$.

copeland 2015-03-27 20:49:42
When we plut in $y=\dfrac12$, all the intermediate terms cancel and we're left with: \[ a_n = \frac{1}{2\sin(\frac12)}\left(\cos\left(\frac12\right) - \cos\left(n + \frac 12\right)\right). \]

SockFoot 2015-03-27 20:50:00
great

copeland 2015-03-27 20:50:01
Indeed.

copeland 2015-03-27 20:50:14
Now we see that $a_n < 0$ is equivalent to $\cos\left(\frac12\right) <\cos\left(n + \frac12\right)$.

copeland 2015-03-27 20:50:24
Let's draw the region on the unit circle where $\cos(\theta)$ is greater than $\cos\left(\frac 12 \right)$ to get an idea of what's going on.

copeland 2015-03-27 20:50:25

copeland 2015-03-27 20:50:30
What do we know about $n$ if $n + \frac 12$ is in this region?

copeland 2015-03-27 20:51:47
What is this region anyway?

thequantumguy 2015-03-27 20:52:19
$n \in [2k\pi - 1, 2k\pi]$

simon1221 2015-03-27 20:52:19
a triangle!

ompatel99 2015-03-27 20:52:19
A triangle

nosaj 2015-03-27 20:52:19
a isosceles triangle!

copeland 2015-03-27 20:52:33
I drew a triangle, yes.

copeland 2015-03-27 20:52:52
But the region we are talking about is the angle.

copeland 2015-03-27 20:53:00
What angles are in that region?

cpma213 2015-03-27 20:53:41
A sector of the unit circle capped at 1/2 radians above and below

swirlykick 2015-03-27 20:53:41
1 radian?

copeland 2015-03-27 20:53:58
Notice that we're not really using $\pi$ at all. 1 radian is a truly weird number.

copeland 2015-03-27 20:54:11
This is the angle from -1/2 to 1/2.

copeland 2015-03-27 20:54:26
.3 is in there, -.22 is in there, etc.

copeland 2015-03-27 20:54:36
What other numbers fall in there?

Dayranger 2015-03-27 20:54:54
-.4

Dayranger 2015-03-27 20:54:54
.4

Dayranger 2015-03-27 20:54:54
0

math_cool 2015-03-27 20:54:54
0

Tommy2000 2015-03-27 20:54:54
0

gxah 2015-03-27 20:54:54
(-1/2, 1/2)

ShadowQueenPeach 2015-03-27 20:54:54
anything from -1/2 to 1/2

copeland 2015-03-27 20:54:57
OK, great.

copeland 2015-03-27 20:55:05
Are those the only numbers that fall in there?

copeland 2015-03-27 20:55:13
What about 6?

copeland 2015-03-27 20:55:19
Is 6 in there?

saagar 2015-03-27 20:55:54
?YES

andrewlin 2015-03-27 20:55:54
yep!

Rocksolid 2015-03-27 20:55:54
yes

DivideBy0 2015-03-27 20:55:54
well, mod 2pi

ryanyz10 2015-03-27 20:55:54
oh wait yes

uberminecraft722 2015-03-27 20:55:54
yes

DivideBy0 2015-03-27 20:55:54
yeah; look at everything mod 2pi

ompatel99 2015-03-27 20:55:54
Yes because it will come around twice then be in that region

tdeng 2015-03-27 20:55:54
yes

High 2015-03-27 20:55:54
yes

Rmink41 2015-03-27 20:55:54
YEAH

SmartYuRuo 2015-03-27 20:55:54
yes

xayy 2015-03-27 20:55:54
yes?

copeland 2015-03-27 20:55:56
6 is between $2\pi-1/2$ and $2\pi+1/2$.

copeland 2015-03-27 20:55:59
It is in there.

copeland 2015-03-27 20:56:06
What are all the real numbers that land in there?

Mlux 2015-03-27 20:57:46
you mean intergral

nosaj 2015-03-27 20:57:46
must be between 2pi-1/2 and 2pi+1/2

ompatel99 2015-03-27 20:57:46
2pi*n-0.5 to 2pi*n+0.5 for integer n

DivideBy0 2015-03-27 20:57:46
2pik - 1/2 to 2pik +1/2 for all integers k

cpma213 2015-03-27 20:57:46
In General, between 2\pi x-1/2 and 2\pi x+1/2

24iam24 2015-03-27 20:57:46
from 2kpi-1/2 to 2kpi+1/2 where k is any integer

ImpossibleTriangle 2015-03-27 20:57:46
(2kpi -1/2, 2kpi + 1/2)

uberminecraft722 2015-03-27 20:57:46
[2(pi)k-(1/2),2(pi)k+(1/2)]

copeland 2015-03-27 20:58:30
Right, exactly. This is the union of all regions of the form $(2\pi-1/2,2\pi+1/2)$.

copeland 2015-03-27 20:58:31
What do we know about $n$ if $n + \frac 12$ is in this region?

thequantumguy 2015-03-27 20:58:49
n is in [2kpi - 1, 2kpi]

Rocksolid 2015-03-27 20:58:49
n is between 2pi and 2pi -1

ryanyz10 2015-03-27 20:58:49
in (2pi - 1, 2pi)?

copeland 2015-03-27 20:59:17
Right. And I have a really special name for "the integer that is just less than a number."

Rmehtany 2015-03-27 20:59:35
floor

swirlykick 2015-03-27 20:59:35
Floor?

dhruv 2015-03-27 20:59:35
floor

vinayak-kumar 2015-03-27 20:59:35
floor function

thequantumguy 2015-03-27 20:59:35
floor

Rocksolid 2015-03-27 20:59:35
floor

copeland 2015-03-27 20:59:40
We see $n+ \frac 12$ must be between $2\pi m - \frac 12$ and $2\pi m + \frac 12$ for some positive integer $m$.

copeland 2015-03-27 20:59:42
Isolating $n$ that says $2\pi m - 1 < n < 2\pi m$ for some $m$, or $n = \lfloor 2\pi m \rfloor$.

ShadowQueenPeach 2015-03-27 20:59:50
the copeland?

copeland 2015-03-27 20:59:55
This is Palmer's vote.

copeland 2015-03-27 21:00:09
What is the index of the 11th term?

copeland 2015-03-27 21:00:31
gosh.

copeland 2015-03-27 21:00:37
What is the index of the 100th term?

thequantumguy 2015-03-27 21:01:11
which is 628

swirlykick 2015-03-27 21:01:11
628

ompatel99 2015-03-27 21:01:11
floor(200pi)=628

saagar 2015-03-27 21:01:11
628?

thequantumguy 2015-03-27 21:01:11
floor of 200 pi

uberminecraft722 2015-03-27 21:01:11
628

thequantumguy 2015-03-27 21:01:11
which is 628

raptorw 2015-03-27 21:01:11
628

andrewlin 2015-03-27 21:01:11
628

czhu000 2015-03-27 21:01:11
628

copeland 2015-03-27 21:01:22
So the index of the 100th term less than 0 is just $\lfloor 2\pi \cdot 100 \rfloor$ or $\boxed{628}$.

copeland 2015-03-27 21:01:40
Incidentally, my original solution to this problem used complex numbers to write the series as the imaginary part of a geometric series. It looked a bit like this:
\begin{align*}
a_n
&=\sum_{k=1}^n\sin(k)\\
&=\operatorname{Im}\left(\sum_{k=1}^ne^{ik}\right)\\
&=\operatorname{Im}\left(\frac{e^{i(n+1)}-e^i}{e^{i}-1}\right)\\
&=\operatorname{Im}\left(\frac{e^{i(n+1/2)}-e^{i/2}}{e^{i/2}-e^{-i/2}}\right)\\
&=\operatorname{Im}\left(\frac{e^{i(n+1/2)}-e^{i/2}}{2i\sin(1/2)}\right)\\
&=\frac{-1}{2\sin(1/2)}\operatorname{Re}\left(e^{i(n+1/2)}-e^{i/2}\right)\\
&=\frac{\cos(n+\frac12)-\cos(\frac12)}{2\sin(\frac12)}\\
\end{align*}

acegikmoqsuwy2000 2015-03-27 21:02:09
that's what I was going to do too

tdeng 2015-03-27 21:02:09
i think the way we did was easier

copeland 2015-03-27 21:02:21
To each his own.

copeland 2015-03-27 21:02:32
14. Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$. Evaluate \[2x^3+(xy)^3+2y^3.\]

copeland 2015-03-27 21:02:42
What basic fact about polynomials pops into your head immediately when you see this problem?

ompatel99 2015-03-27 21:03:03
Symmetry?

copeland 2015-03-27 21:03:04
What does symmetry tell us?

nosaj 2015-03-27 21:03:26
roots are interchangable?

DivideBy0 2015-03-27 21:03:26
replace x and y and get the same result

copeland 2015-03-27 21:03:39
That is the definition. What do we know about symmetric polynomials?

vinayak-kumar 2015-03-27 21:03:57
sum and product

copeland 2015-03-27 21:04:07
We KNOW that any symmetric polynomial in $x$ and $y$ can be expressed in terms of $s=x+y$ and $p=xy$.

copeland 2015-03-27 21:04:12
Rewriting all three expressions in terms of $s$ and $p$ is a fabulous first step and will take you to the answer with some work.

copeland 2015-03-27 21:04:15
This is Problem 14, though, so maybe there's something useful we can do first.

copeland 2015-03-27 21:04:26
Can we combine the two given equations in a nice way? They have the same degree after all.

Tommy2000 2015-03-27 21:05:37
divide?

24iam24 2015-03-27 21:05:37
divide them?

Rmink41 2015-03-27 21:05:37
Divide

DivideBy0 2015-03-27 21:05:37
divide?

copeland 2015-03-27 21:05:39
Dividing is neat. A lot of stuff will cancel. Any other things we could do?

MathStudent2002 2015-03-27 21:06:27
$x^3y^3(x+y)^3=810\cdot 3+945$

nosaj 2015-03-27 21:06:27
Multiply the first equation by 3, then add.

mathwrath 2015-03-27 21:06:27
Add 3 times the first to the second and factor

countingarithmetic 2015-03-27 21:06:27
945+3*810 = x^3*y^3*(x+y)^3

copeland 2015-03-27 21:06:34
If we triple the first and add it to the second we get something nice:\[3(x^4y^5+y^4x^5)+(x^3y^6+y^3x^6)=3(810)+945.\]
My reflex here is to factor the integers a little on the right and expand the left side:\[x^3y^6+3x^4y^5+3y^4x^5+y^3x^6=(3\cdot6+7)\cdot135.\]

copeland 2015-03-27 21:06:44
How does the left side simplify?

mathwrath 2015-03-27 21:07:27
x^3y^3(x+y)^3

acegikmoqsuwy2000 2015-03-27 21:07:27
$[xy(x+y)]^3$

DivideBy0 2015-03-27 21:07:27

nosaj 2015-03-27 21:07:27
$x^3y^3(x+y)^3$

copeland 2015-03-27 21:07:28
Yeah, I guess I passed that already.

copeland 2015-03-27 21:07:29
The left side simplifies to \[x^3y^3(y^3+3xy^2+3x^2y+x^3)=x^3y^3(x+y)^3\]

copeland 2015-03-27 21:07:31
Putting it all together gives \[p^3s^3=3^3\cdot5^3.\]

copeland 2015-03-27 21:07:32
And?

countingarithmetic 2015-03-27 21:08:05
15= x*y*(x+y)

andrewlin 2015-03-27 21:08:05
ps=15

24iam24 2015-03-27 21:08:05
ps=15

Dayranger 2015-03-27 21:08:05
ps = 15

copeland 2015-03-27 21:08:09
Since $x$ and $y$ are real numbers, the cube roots are well-defined.

copeland 2015-03-27 21:08:10
We get $ps=15$.

copeland 2015-03-27 21:08:17
Now what should we do?

DivideBy0 2015-03-27 21:09:34
factor the first equation and plug in p and s

andrewlin 2015-03-27 21:09:34
we have p^4*s=810 so we can solve for p and s

copeland 2015-03-27 21:09:38
Let's go back to the first equation,\[x^4y^5+x^5y^4=810.\]

copeland 2015-03-27 21:09:43
This equation is divisible by $ps$. Specifically we have \[p^4s=810.\] So. . .

ompatel99 2015-03-27 21:10:25
p^3=54

BPM14 2015-03-27 21:10:25
p^3 = 54

gxah 2015-03-27 21:10:25
p^3 = 54

henryweng 2015-03-27 21:10:25
p3=54

acegikmoqsuwy2000 2015-03-27 21:10:25
p^3 = 54

copeland 2015-03-27 21:10:28
Dividing by $ps=15$ gives us $p^3=54$.

copeland 2015-03-27 21:10:32
Now what?

nosaj 2015-03-27 21:11:04
so that means we can find $p$ and $s$!

math_cool 2015-03-27 21:11:04
Solve for s

thequantumguy 2015-03-27 21:11:04
solve for p to find s

guilt 2015-03-27 21:11:04
We can solve for p and then subsequently q

Ramanan369 2015-03-27 21:11:04
solve for s

copeland 2015-03-27 21:11:09
Sure that also tells us $p$ and $s$ but hang on for a second with that - we might not need them.

copeland 2015-03-27 21:11:27
What can we do with our newly discovered $p^3$?

24iam24 2015-03-27 21:11:50
we just got the middle part of what we wanted

mssmath 2015-03-27 21:11:50
(xy)^3 is done

Dayranger 2015-03-27 21:11:50
(xy)^3 = 54

RJ810 2015-03-27 21:11:50
put it into the final question

saagar 2015-03-27 21:11:54
(xy)^3=54

copeland 2015-03-27 21:12:07
That's great. We have the middle part of what we want now.

copeland 2015-03-27 21:12:09
What are we missing?

saagar 2015-03-27 21:12:31
2(x^3+y^3)

ryanyz10 2015-03-27 21:12:31
2(x^3 + y^3)

RJ810 2015-03-27 21:12:31
x^3+y^3

math_cool 2015-03-27 21:12:31
2(x^3 + y^3)

High 2015-03-27 21:12:31
2(x^3+y^3)

copeland 2015-03-27 21:12:38
Now we just need to find $2(x^3+y^3)$. How can we find that?

andrewlin 2015-03-27 21:13:06
we can use p^3 in the second equation to get x^3+y^3=35/2

swirlykick 2015-03-27 21:13:06
second equation

czhu000 2015-03-27 21:13:06
Factor $p^3$ out of the second equation to get the remaining part

guilt 2015-03-27 21:13:06
divide the 2nd equation by p^3

czhu000 2015-03-27 21:13:06
Factor $p^3$ out of the second equation and multiply by 2

Rmink41 2015-03-27 21:13:06
Using eq 2

czhu000 2015-03-27 21:13:06
Divide $p^3$ out of the second equation and multiply by 2

copeland 2015-03-27 21:13:13
The second equation is divisible by $p^3$. It is \[p^3(x^3+y^3)=945.\]

copeland 2015-03-27 21:13:16
Now we know \[x^3+y^3=\frac{945}{54}=\frac{9\cdot105}{9\cdot6}=\frac{105}6=\frac{35}2.\]

copeland 2015-03-27 21:13:23
What do we do now?

nosaj 2015-03-27 21:13:57
add

Rmink41 2015-03-27 21:13:57
Multiply by 2 and add the middle

dhruv 2015-03-27 21:13:57
35+54=89

bluephoenix 2015-03-27 21:13:57
multiply by 2 to get 35

24iam24 2015-03-27 21:13:57
35+54=89

caressezhu 2015-03-27 21:13:57
89

RJ810 2015-03-27 21:13:57
mult by 2 and add by 54

math_cool 2015-03-27 21:13:57
35 + 54 = 089

Dayranger 2015-03-27 21:13:57
35 + 54 = 89

xayy 2015-03-27 21:13:57
multiply it by 2

BPM14 2015-03-27 21:13:57
35+54 = 89

RJ810 2015-03-27 21:13:57
answer is 89

Mlux 2015-03-27 21:13:57
35 + 54

copeland 2015-03-27 21:13:59
OH, hey, we're done!

copeland 2015-03-27 21:14:01
We want \[2x^3+(xy)^3+2y^3=2(x^3+y^2)+p^3=35+54=\boxed{89}.\]

copeland 2015-03-27 21:14:04
Awesome.

copeland 2015-03-27 21:14:10
Alright. Ready to go home?

Mlux 2015-03-27 21:14:24
14/15

tdeng 2015-03-27 21:14:24
14/15 of the way home

acegikmoqsuwy2000 2015-03-27 21:14:24
Nope 1 more!! only 14/15 done

firemike 2015-03-27 21:14:26
no!

copeland 2015-03-27 21:14:34
OK, let's do 15 then.

copeland 2015-03-27 21:14:38
15. Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii 1 and 4, respectively, and are externally tangent at point $A$. Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$. Points $B$ and $C$ lie on the same side of $\ell$, and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

copeland 2015-03-27 21:14:40

copeland 2015-03-27 21:14:58
Cool diagram!

copeland 2015-03-27 21:14:59
I like this diagram since it has a lot of different pieces that you can focus on and start playing with. Let's see if we can get to the right start for this problem.

copeland 2015-03-27 21:15:05
The line $\ell$ is the variable here. If you ignore it, all that is left in the configuration are two circles and a line, all of which are mutually tangent. If you've seen a lot of Olympiad-level geometry, what might you be inclined to do with such a diagram?

copeland 2015-03-27 21:16:02
Look again at $\ell$. As you vary it, it creates a lot of paired objects. It creates points $D$ and $E$, it creates lines $DA$ and $AE$, and it creates arcs $DA$ and $AE$.

copeland 2015-03-27 21:16:21
Notice anything interesting along those lines?

saagar 2015-03-27 21:17:03
AE=4DA

acegikmoqsuwy2000 2015-03-27 21:17:03
measure of the minor arcs DA and AE are the same?

elf123 2015-03-27 21:17:03
same arc length

copeland 2015-03-27 21:17:07
Hey, that's crazy.

copeland 2015-03-27 21:17:08
You might just know that those two arcs are the same angle measure. That comes from comparing those angles to the angles with the tangent line:

copeland 2015-03-27 21:17:13

copeland 2015-03-27 21:17:20
The arcs have equal inscribed angles so the arcs are equal angles. (The ratio of the length is then $1:4$.)

copeland 2015-03-27 21:17:22
How else could we compare those two arcs?

copeland 2015-03-27 21:18:57
So here's some more intuition: I have all kinds of paired objects all over the place.

copeland 2015-03-27 21:19:03
Paired points, paired arcs, etc.

copeland 2015-03-27 21:19:24
Is there some nice transformation that let's us discuss them together?

saagar 2015-03-27 21:20:00
dilation?

24iam24 2015-03-27 21:20:00
dilation?

thequantumguy 2015-03-27 21:20:00
rotatiom

tdeng 2015-03-27 21:20:00
rotation

math-rules 2015-03-27 21:20:00
dilation?

copeland 2015-03-27 21:20:05
Neither one is quite enough.

Rmehtany 2015-03-27 21:20:37
together

SockFoot 2015-03-27 21:20:37
homothety

math-rules 2015-03-27 21:20:37
homothety?

chezbgone 2015-03-27 21:20:37
spiral similarity?

bluephoenix 2015-03-27 21:20:37
homotheicy?

andrewlin 2015-03-27 21:20:37
BOTH?

Rmehtany 2015-03-27 21:20:37
dilate and rotate

copeland 2015-03-27 21:20:46
We could either perform a homothety through $A$ or a rotation and dilation about $A$. (In fact, those are the same operation.)

copeland 2015-03-27 21:20:48
If homothety is new to you, it means we take every point, stretch it by some factor, and in this case also reflect it through $A$. Just keep saying 'rotation and stretch'.

copeland 2015-03-27 21:21:29
In general, the nice rule of thumb is that if you have tangent circles and lines and whatnot, try to find a transformation that interchanges some of the elements.

copeland 2015-03-27 21:21:40
So really, this diagram kinda wants us to perform one of these operations because even $\ell$ finds them to be interesting - rotation by $\pi$ fixes $\ell$ as well.

copeland 2015-03-27 21:21:42
Which is better? We have options:
$\bullet$ Homothety by $-4$ taking the little circle to the big circle
$\bullet$ Homothety by $-\dfrac14$ taking the big circle to the little circle

copeland 2015-03-27 21:21:50
Which of these two operations is best?

Rmehtany 2015-03-27 21:22:05
little to big

High 2015-03-27 21:22:05
little to big

SKundu13 2015-03-27 21:22:05
Homothety by -4

firemike 2015-03-27 21:22:05
small to big

24iam24 2015-03-27 21:22:05
first no fractions

saagar 2015-03-27 21:22:05
make the smaller one bigger

ryanyz10 2015-03-27 21:22:05
first one... less messy

copeland 2015-03-27 21:22:08
We ought to look at the other elements of the diagram to answer this question.

copeland 2015-03-27 21:22:10
Both take one of the circles to the other. (Little to big or big to little.)
Both turn the remaining circle into nonsense. (Big to really big or little to really little.)
Both take $\ell$ and give back $\ell$.
Both take one of the arcs to the other arc.

copeland 2015-03-27 21:22:23
Both operations are kind to the tangent line. The big homothety does this:

copeland 2015-03-27 21:22:30

copeland 2015-03-27 21:22:36

copeland 2015-03-27 21:22:39
The little homothety does this:

copeland 2015-03-27 21:22:41

copeland 2015-03-27 21:22:45
Those give us effectively the same diagram, but the first is clearly nicer since it is easier to read.

copeland 2015-03-27 21:22:45
Why does this transformation jive well with the tangent line?

chezbgone 2015-03-27 21:23:54
B'C is a diameter

xayy 2015-03-27 21:23:54
parallel?

nosaj 2015-03-27 21:23:54
red line is parallel to tangent line?

saagar 2015-03-27 21:23:54
tangents are perpendicular and so the two lines are parallel

copeland 2015-03-27 21:24:03
Like we said before, this is also a rotation by $\pi$ and then a rescaling.

copeland 2015-03-27 21:24:05
A rotation by $\pi$ takes lines to parallel lines. The top and bottom are parallel lines. That means $B'$ and $C$ are antipodes. $BC$ is is a diameter!

copeland 2015-03-27 21:24:10
Now we have a cyclic quadrilateral in which one diagonal is a diameter. That is really awesome. Let me clean up the diagram for you:

copeland 2015-03-27 21:24:15

copeland 2015-03-27 21:24:19
What do we know about the red region?

mathwrath 2015-03-27 21:24:48
16 times the area of the gray region

24iam24 2015-03-27 21:24:48
it is 16 times the gray

High 2015-03-27 21:24:48
area of little triangle*16

countingarithmetic 2015-03-27 21:24:48
16 times the size of the gray region

acegikmoqsuwy2000 2015-03-27 21:24:48
its 16 times the grey region

saagar 2015-03-27 21:24:48
16 times the gray triangle?

copeland 2015-03-27 21:24:51
Since the original grey regions were the same size and we dilated by a factor of 4, the red region is 16 times the size of the grey region.

copeland 2015-03-27 21:24:55
We also ought to know exactly what the dimensions of $\triangle AB'C$ are. Remember that $A,$ $B,$ and $C,$ are fixed. We were a little bit too quick when we abandoned the original figure. Let's circle back.

copeland 2015-03-27 21:25:03
What should we draw on the original figure to figure out where $A$ is?

copeland 2015-03-27 21:25:48

Tommy2000 2015-03-27 21:26:13
radii

saagar 2015-03-27 21:26:13
draw a radius?

copeland 2015-03-27 21:26:15
Let's connect the centers with radii:

copeland 2015-03-27 21:26:16

copeland 2015-03-27 21:26:25
What kind of triangle is that?

Dayranger 2015-03-27 21:26:41
haha, CIRCLE back

copeland 2015-03-27 21:26:43
Thanks.

rcbg0713 2015-03-27 21:27:08
right

bluephoenix 2015-03-27 21:27:08
right triangle

saagar 2015-03-27 21:27:08
im not sure, but it looks like the RIGHT triangle

copeland 2015-03-27 21:27:11
per force.

copeland 2015-03-27 21:27:19
I just dropped a perpendicular.

copeland 2015-03-27 21:27:29
But read the problem. Think about the radii.

High 2015-03-27 21:27:51
3,4,5

swe1 2015-03-27 21:27:51
3-4-5

acegikmoqsuwy2000 2015-03-27 21:27:51
3,4,5 triangle

amburger66 2015-03-27 21:27:51
3,4,5

acegikmoqsuwy2000 2015-03-27 21:27:51
its a 3,4,5 right triangle

ryanyz10 2015-03-27 21:27:51
is it not a 3-4-5?

High 2015-03-27 21:27:51
3-4-5 right triangle

danzhi 2015-03-27 21:27:51
3,4,5

DivideBy0 2015-03-27 21:27:51
3-4-5

bengals 2015-03-27 21:27:51
3-4-5?

math_cool 2015-03-27 21:27:51
3-4-5

copeland 2015-03-27 21:27:53
That is a 3-4-5 triangle.

copeland 2015-03-27 21:27:57

copeland 2015-03-27 21:28:14
What about this little triangle?

copeland 2015-03-27 21:28:44
(The a-b-4 one.)

caressezhu 2015-03-27 21:29:11
3-4-5

amburger66 2015-03-27 21:29:11
also a 3,4,5

saagar 2015-03-27 21:29:11
its proportions are 4/5 of the 3-4-5

ryanyz10 2015-03-27 21:29:11
also a 3-4-5?

copeland 2015-03-27 21:29:25
It is also 3-4-5. What is $a$?

ryanyz10 2015-03-27 21:29:39
16/5

acegikmoqsuwy2000 2015-03-27 21:29:39
16/5

derpyuniverse 2015-03-27 21:29:39
16/5

24iam24 2015-03-27 21:29:39
16/5

meeptopia 2015-03-27 21:29:39
16/5

malarm 2015-03-27 21:29:39
16/5

copeland 2015-03-27 21:29:41
$a$ is the "4 side" of a 3-4-5 triangle. The hypotenuse has length 4 so the side length is $\dfrac45\cdot4=\dfrac{16}5$.

copeland 2015-03-27 21:29:42
Likewise $b$ is the "3 side" so $b=\dfrac45\cdot3=\dfrac{12}5$.

copeland 2015-03-27 21:29:48
What is $c$?

24iam24 2015-03-27 21:30:54
16sqrt5/5

acegikmoqsuwy2000 2015-03-27 21:30:54
$\frac{16sqrt 5}{5}$

ryanyz10 2015-03-27 21:30:54
16sqrt(5)/5

amburger66 2015-03-27 21:30:54
(16sqrt5)5

copeland 2015-03-27 21:30:58
We have right triangle $APB'$ with base $a=\dfrac{16}5$ and height $b+r=\dfrac{12}5+4=\dfrac{32}5$. That is a $1-2-\sqrt5$ triangle so \[c=\dfrac{16}5\sqrt5=\dfrac{16}{\sqrt5}.\]

copeland 2015-03-27 21:31:04
What is $d$?

ryanyz10 2015-03-27 21:32:24
8/sqrt(5)

mathwrath 2015-03-27 21:32:24
8sqrt5/5

24iam24 2015-03-27 21:32:24
8sqrt5/5

Tommy2000 2015-03-27 21:32:24
8sqrt5/5

acegikmoqsuwy2000 2015-03-27 21:32:24
$\frac{8\sqrt 5} {5}$

amburger66 2015-03-27 21:32:24
8sqrt5/5

Rmink41 2015-03-27 21:32:24
I mean 8sqrt5

tdeng 2015-03-27 21:32:24
8/sqrt5

AlisonH 2015-03-27 21:32:24
8\sqrt{5}/5

copeland 2015-03-27 21:32:27
We could similarly use Pythagoras, but notice that $\triangle AB'P\sim \triangle CAP$ so \[\dfrac{AC}{CP}=\dfrac{B'A}{AP}=\sqrt5.\] Therefore
\[d=(r-b)\sqrt5=\left(4-\frac{12}5\right)\sqrt5=\frac{8\sqrt5}5=\frac8{\sqrt5}.\]

copeland 2015-03-27 21:32:31
Back to our diagram:

copeland 2015-03-27 21:32:34

copeland 2015-03-27 21:32:39
Let's give all these regions names. I'm also going to label the other 2 edge lengths.

copeland 2015-03-27 21:32:40

acegikmoqsuwy2000 2015-03-27 21:32:58
yay colors!

copeland 2015-03-27 21:33:12
You'd think I could make less 80s colors but rgb space is weird space.

copeland 2015-03-27 21:33:13
Now what does our ratios statement tell us?

tdeng 2015-03-27 21:33:35
what if someone is color blind

copeland 2015-03-27 21:33:36
If you're colorblind, this diagram probably looks better.

elf123 2015-03-27 21:34:45
1:16 blue to purple and green to light blue

ryanyz10 2015-03-27 21:34:45
w + x = 16(z + y)

math_cool 2015-03-27 21:34:45
AEB' = 16 AEC

copeland 2015-03-27 21:34:48
We know that $\dfrac{W+X}{Z+Y}=16$.

copeland 2015-03-27 21:34:49
But look at $Z$ and $W$. What do we know about how their areas are related?

High 2015-03-27 21:36:12
ratio of areas is BZ/ZC

amburger66 2015-03-27 21:36:12
it's equal to CZ:ZB'

xayy 2015-03-27 21:36:12
the heights' ratio

Tommy2000 2015-03-27 21:36:12
ratio of the bases

copeland 2015-03-27 21:36:17
Since these two triangles have the same altitude to the diameter, their ratio is equal to the ratio of the lengths of the sections that $AE$ dissects $B'C$ into.

copeland 2015-03-27 21:36:29
(That point is not $Z$. I meant for $Z$ to be the area.)

copeland 2015-03-27 21:36:31
What else?

copeland 2015-03-27 21:36:55
What can we say about $X$ and $Y$?

Tommy2000 2015-03-27 21:37:20
Same for X and Y

ryanyz10 2015-03-27 21:37:20
same ratio!!!

meeptopia 2015-03-27 21:37:20
same thing?

amburger66 2015-03-27 21:37:20
same ratio

24iam24 2015-03-27 21:37:20
has the same ratio

copeland 2015-03-27 21:37:22
The same argument works or $X$ and $Y$. Their areas are also in the same ratio as the ratio of the lengths of the two sections of $B'C$.

copeland 2015-03-27 21:37:23
Specifically, we know that $\dfrac WZ=\dfrac XY$ is some constant $k$.

copeland 2015-03-27 21:37:25
And what can we conclude?

math_cool 2015-03-27 21:38:24
Similar triangles: W = 16 Y and X = 16 Z

bengals 2015-03-27 21:38:24
w/z = 16

DivideBy0 2015-03-27 21:38:24
k = 16

amburger66 2015-03-27 21:38:24
that ratio is 16?

Tommy2000 2015-03-27 21:38:24
k=16

DivideBy0 2015-03-27 21:38:24
W/Z = 16

mathwrath 2015-03-27 21:38:24
k=16

caressezhu 2015-03-27 21:38:37
W+X=16Z+16Y

copeland 2015-03-27 21:38:38
We can conclude that \[16=\frac{W+X}{Z+Y}=\frac{kZ+kY}{Z+Y}=k.\] Therefore $k=16$ and the ratios $\dfrac WZ=\dfrac XY=16$.

copeland 2015-03-27 21:38:44
Cool.

copeland 2015-03-27 21:38:45
How can we finish from here?

copeland 2015-03-27 21:40:11
For example, we could go analytic geometry all the way. We know $A$ and we know the point where $B'C$ intersects $AE$ is $\dfrac{16}{17}$ of the way down. Therefore we could find $E$ by intersecting the circle with the line. That might be the easiest way but let's do something a little different.

copeland 2015-03-27 21:40:22
We already have some lengths in this diagram. We also have area ratios. Let's just try to force out the rest of those areas.

copeland 2015-03-27 21:40:28
Remember that we have a cyclic quadrilateral. We haven't really used that very much yet. What is the nice property of cyclic quadrilaterals (beyond that they live in circles)?

DivideBy0 2015-03-27 21:41:36
ptolomy's

thequantumguy 2015-03-27 21:41:36
ptomely

BPM14 2015-03-27 21:41:36
ABD = ACD

tdeng 2015-03-27 21:41:36
ptolemys theorem?

High 2015-03-27 21:41:36
Ptolemy's theorem

acegikmoqsuwy2000 2015-03-27 21:41:36
ptolemy's theorem?

24iam24 2015-03-27 21:41:36
supplementary angles opposite angles

lazboy 2015-03-27 21:41:36
Ptolemy?

copeland 2015-03-27 21:41:41
Here are two facts. Let's think about the angles statement.

copeland 2015-03-27 21:41:42
The opposite angles are supplementary.

copeland 2015-03-27 21:41:44
We used that $A$ and $E$ are right angles. We haven't used that $C$ and $B'$ are supplementary. Does that tell us anything about the area?

copeland 2015-03-27 21:42:53
Remember we can compute the area from angles. . .

24iam24 2015-03-27 21:43:29
area of triangle equals absin(x)/2

Tommy2000 2015-03-27 21:43:29
sines are equal so use absinc/2

vinayak-kumar 2015-03-27 21:43:29
i meant area of triangle with sines

copeland 2015-03-27 21:43:31
\begin{align*}
[\triangle AB'E]=\frac12AB'\cdot B'E\sin(B')\\
[\triangle AEC]=\frac12AC\cdot CE\sin(C)
\end{align*}

copeland 2015-03-27 21:43:32
However. . .

24iam24 2015-03-27 21:43:48
sin(x)=sin(180-x)

countingarithmetic 2015-03-27 21:43:48
same sine

High 2015-03-27 21:43:48
the sines are equal

copeland 2015-03-27 21:43:51
Supplementary angles have equal sines. Therefore
\begin{align*}
\frac{[\triangle AB'E]}{AB'\cdot B'E}=\frac12\sin(B')\\
\frac{[\triangle ACE]}{AC\cdot CE}=\frac12\sin(C)
\end{align*}
so
\[
\frac{[\triangle AB'E]}{AB'\cdot B'E}=\frac{[\triangle ACE]}{AC\cdot CE}
\]

copeland 2015-03-27 21:43:59
What else can we use?

Tommy2000 2015-03-27 21:44:35
The ratio of the areas is 16

24iam24 2015-03-27 21:44:35
also ratio of areas is 16

countingarithmetic 2015-03-27 21:44:35
AB' = 2AC

rlz 2015-03-27 21:44:35
We know the ratio of the areas

copeland 2015-03-27 21:44:38
We know the ratios of those areas:\[16=
\frac{[\triangle AB'E]}{[\triangle ACE]}
=\frac{AB'\cdot B'E}{AC\cdot CE}
=\frac{\frac{16}{\sqrt5}\cdot x}{\frac{8}{\sqrt5}\cdot y}
=\frac{2x}{y}.\]

copeland 2015-03-27 21:44:44
That means $x=8y$.

copeland 2015-03-27 21:44:46
Where do we get another relation between $x$ and $y$?

countingarithmetic 2015-03-27 21:45:42
$x^2 + y^2 = 64$

24iam24 2015-03-27 21:45:42
x^2+y^2=64

Tommy2000 2015-03-27 21:45:42
pythagorean theorem!

Dayranger 2015-03-27 21:45:42
x^2 + y^2 = 64

acegikmoqsuwy2000 2015-03-27 21:45:42
x^2+y^2 = 64

Tommy2000 2015-03-27 21:45:42
x^2+y^2=64

copeland 2015-03-27 21:45:45
The Pythagorean Theorem on $\triangle B'EC$ gives us $x^2+y^2=8^2$

copeland 2015-03-27 21:45:47
Solving that tells us \[64=x^2+y^2=\left(8y\right)^2+y^2=65y^2.\]

copeland 2015-03-27 21:45:51
That gives us $y$.

copeland 2015-03-27 21:45:53
Finally we want to compute the area $Z+Y$. That's hard. What can we compute?

24iam24 2015-03-27 21:46:30
X+Y

copeland 2015-03-27 21:46:36
And what's so great about X+Y?

BPM14 2015-03-27 21:47:06
they have a ratio

elf123 2015-03-27 21:47:06
it equals 17y

copeland 2015-03-27 21:47:09
Right.

copeland 2015-03-27 21:47:18
$X+Y$ lets us compute
\[17(Z+Y)=(Z+Y)+(X+W)=(Z+W)+(X+Y).\]

copeland 2015-03-27 21:47:23
What is $Z+W$?

24iam24 2015-03-27 21:48:03
64/5

Tommy2000 2015-03-27 21:48:03
64/5

acegikmoqsuwy2000 2015-03-27 21:48:03
64/5

copeland 2015-03-27 21:48:06
That's a right triangle. Its area is $\dfrac12\cdot\dfrac{16}{\sqrt5}\cdot\dfrac{8}{\sqrt5}=\dfrac{64}5$.

copeland 2015-03-27 21:48:07
What is $X+Y$?

copeland 2015-03-27 21:48:47
(Remember that $x=8y$ and $64=65y^2$.)

tdeng 2015-03-27 21:49:36
256/65

Tommy2000 2015-03-27 21:49:36
256/65

mathwizard888 2015-03-27 21:49:36
256/65

copeland 2015-03-27 21:49:40
That's also a right triangle. Its area is
\begin{align*}
\frac12xy
&=\frac12\cdot8y^2\\
&=4\cdot\frac{64}{65}\\
&=\frac{256}{65}.
\end{align*}

copeland 2015-03-27 21:49:47
So what is the answer?

caressezhu 2015-03-27 21:50:46
129

Tommy2000 2015-03-27 21:50:46
64/65

derpyuniverse 2015-03-27 21:50:46
64/65

24iam24 2015-03-27 21:50:46
add them and /17

copeland 2015-03-27 21:50:50
The answer is
\[
\frac1{17}\left(\frac{64}{5}+\frac{256}{65}\right)
=\frac{64}{17}\left(\frac{13}{65}+\frac{4}{65}\right)
=\frac{64\cdot17}{17\cdot65}
=\frac{64}{65}.
\]

copeland 2015-03-27 21:50:51
The final answer is $64+65=\boxed{129}$.

copeland 2015-03-27 21:51:12
Alright, next problem. . .

nosaj 2015-03-27 21:51:14
we are 1/1 of the way done!

copeland 2015-03-27 21:51:20
Oh wait. It's time to go home now.

copeland 2015-03-27 21:51:40
Thanks for coming, I had a great time.

_--__--_ 2015-03-27 21:51:57
we are already at home (most of us)

countingarithmetic 2015-03-27 21:51:57
Thank you!

lazboy 2015-03-27 21:51:57
Thank you, Mr. Copeland!

xayy 2015-03-27 21:51:57
bye

Dayranger 2015-03-27 21:51:57
ME TOO!!!

BPM14 2015-03-27 21:52:23
Can we use you on our next AIME?

copeland 2015-03-27 21:52:25
Maybe not the best idea. I did not include (most of) the arithmetic errors I made in today's class.

DivideBy0 2015-03-27 21:52:35
thanks!!

czhu000 2015-03-27 21:52:35
Thanks, bye!

1915933 2015-03-27 21:52:35
Thank you! Have a good evening!

math_cool 2015-03-27 21:52:35
thanks!

ryanyz10 2015-03-27 21:52:35
thanks!

caressezhu 2015-03-27 21:52:35
Thank you, I enjoy it!

malarm 2015-03-27 21:52:35
thank YOU

BPM14 2015-03-27 21:52:35
Thanks!

Guendabiaani 2015-03-27 21:52:35
Thanks!

Dayranger 2015-03-27 21:52:35
Thank you, Mr. Copeland

bengals 2015-03-27 21:52:35
thank you

acegikmoqsuwy2000 2015-03-27 21:52:39
AoPS is my home.

2015 AIME II Math Jam

Facilitator: Jeremy Copeland