Math Jams

copeland 2016-03-05 18:23:58
Hello and welcome to the Math Jam. We'll begin a minute or two after the hour.
The classroom is moderated. That means the things you type won't appear directly in the classroom. Don't worry, we do see you!

eisirrational 2016-03-05 18:51:09
Hello!

akaashp11 2016-03-05 18:51:09
Hello! How's your day been?

copeland 2016-03-05 18:51:12
Hi. Rocking.

copeland 2016-03-05 18:52:21
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.

epiclucario 2016-03-05 18:54:42
Will this math jam allow me to get a 16 on the AIME II?

copeland 2016-03-05 18:54:43
No guarantees.

ychen 2016-03-05 18:56:18
when will the results come out approximately?

copeland 2016-03-05 18:56:20
Speaking of "no guarantees. . . "

SHARKYBOY 2016-03-05 18:56:55
Hi AoPS!

SS00090 2016-03-05 18:57:16
How many problems are we gonna do?

copeland 2016-03-05 18:57:18
15!

SmallKid 2016-03-05 18:57:27
Hi SHARKYBOY!

chaoshadow37 2016-03-05 18:58:49
was there a problem with #9? if so how is it gonna affect the scoring?

copeland 2016-03-05 18:58:50
Yes, and I am officially refusing to predict that they won't do anything.

lidada 2016-03-05 18:59:10
15 factorial is a lot of problems...

copeland 2016-03-05 18:59:11
Yes, did you bring snacks?

v8599131275 2016-03-05 18:59:32
Wassup Mr. Copeland

copeland 2016-03-05 18:59:33
Yo.

ninjataco 2016-03-05 19:00:19
did you have a favorite problem?

copeland 2016-03-05 19:00:24
I loved all but #7 equally.

nukelauncher 2016-03-05 19:00:30
will we do the problems in order? this is my first math jam

copeland 2016-03-05 19:00:33
For sure.

popcorn1 2016-03-05 19:00:45
1307674368000 is a LOT of problems, what was the time limit?

SmallKid 2016-03-05 19:00:45
15! = $1307674368000$. Assuming we do one problem every 12 minutes (AIME speed), it will take us almost 30 million years.

copeland 2016-03-05 19:00:56
We need to be faster than that.

jfurf 2016-03-05 19:01:17
Lets do a problem every millisecond!

copeland 2016-03-05 19:01:23
That'll definitely speed things up.

SmallKid 2016-03-05 19:01:43
It'll still take us around 41 years.

copeland 2016-03-05 19:01:51
Getting better!

Locust 2016-03-05 19:01:57
Good, 30 million years, afterwards we can take the AIME again in the year 30002016

copeland 2016-03-05 19:02:15
Alright, ready to start?

brian6liu 2016-03-05 19:02:41
yeah

SimonSun 2016-03-05 19:02:41
yessir

walnutwaldo20 2016-03-05 19:02:41
Yup

sohappy 2016-03-05 19:02:41
yes

SHARKYBOY 2016-03-05 19:02:41
yup

Darth_Math 2016-03-05 19:02:41
yes

ychen 2016-03-05 19:02:41
YEAH!

blue8931 2016-03-05 19:02:41
yes

LoneConquerorer 2016-03-05 19:02:41
yeah!

math2fun 2016-03-05 19:02:41
oui

akaashp11 2016-03-05 19:02:41
Yessir

ompatel99 2016-03-05 19:02:41
Let's roll!

andsun19 2016-03-05 19:02:41
Yup

jfurf 2016-03-05 19:02:41
YEAH!!

hamup1 2016-03-05 19:02:41
Yes!

copeland 2016-03-05 19:02:57
Welcome to the 2016 AIME I Math Jam!

copeland 2016-03-05 19:02:57
I'm Jeremy Copeland, and I'll be leading our discussion tonight.

copeland 2016-03-05 19:03:03
I'm the school director here at AoPS.

copeland 2016-03-05 19:03:06
I like math.

andsun19 2016-03-05 19:03:17
I like math too!

sohappy 2016-03-05 19:03:17
I also like math

SmallKid 2016-03-05 19:03:17
Me too!

kikipet 2016-03-05 19:03:17
me too!

copeland 2016-03-05 19:03:19
Great!

copeland 2016-03-05 19:03:36
Um. . . bio. . .

copeland 2016-03-05 19:03:44
I have a Ph.D. from UChicago.

copeland 2016-03-05 19:03:48
and, uh. . .

copeland 2016-03-05 19:03:51
I taught at MIT for a while. . .

copeland 2016-03-05 19:04:02
and, uh, I work at AoPS.

copeland 2016-03-05 19:04:04
Great!

copeland 2016-03-05 19:04:06
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 2016-03-05 19:04:08
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 2016-03-05 19:04:15
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 2016-03-05 19:04:18
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 2016-03-05 19:04:32
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!

budu 2016-03-05 19:04:35
Jeremy joined AoPS in 2009. He earned his Ph.D. in mathematics from the University of Chicago in 2006 and was on the math faculty at MIT from 2006 to 2009. He is currently the Online School Director at AoPS. He once specialized in turning hard problems in geometry, algebra, and mathematical physics into easy problems in combinatorics and graph theory. Now he specializes in helping to redistribute mathematics from brilliant teachers to brilliant students. He gets his signs right 50 percent of the

copeland 2016-03-05 19:04:38
Ooh!

copeland 2016-03-05 19:04:41
Someone did his homework.

copeland 2016-03-05 19:04:51
We plan to have two teaching assistants with us tonight to help answer your questions.

copeland 2016-03-05 19:04:59
We sent one to the wrong room, unfortunately.

copeland 2016-03-05 19:05:06
The other is really good, though:

copeland 2016-03-05 19:05:09
Nicholas Yang (nackster): Nick is currently a sophomore at Princeton University. He has enjoyed competition math since elementary school, and has been particularly active in Florida's Mu Alpha Theta math organization. He has had success in math by qualifying for National MATHCOUNTS and the USA Math Olympiad. Nick loves to help others learn, whether it is in math or other subjects, and he's very excited to help out with AoPS classes!

nackster12 2016-03-05 19:05:15
Hi everyone!

popcorn1 2016-03-05 19:05:37
Time for # 1!

v8599131275 2016-03-05 19:05:37
Shout out to my man Mr. Yang

sohappy 2016-03-05 19:05:37
hi

SHARKYBOY 2016-03-05 19:05:37
Hi Nick!

Locust 2016-03-05 19:05:37
Hello!!!

kikipet 2016-03-05 19:05:37
Hi!

popcorn1 2016-03-05 19:05:37
Hello!

ThorJames 2016-03-05 19:05:37
Hi

ychen 2016-03-05 19:05:37
hi nackster12!

SmallKid 2016-03-05 19:05:37
Hi there!

SS00090 2016-03-05 19:05:37
hi

sxu 2016-03-05 19:05:37
hello

Aragorn66 2016-03-05 19:05:37
Hello

ryanyz10 2016-03-05 19:05:37
hi!

copeland 2016-03-05 19:05:43
He 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 2016-03-05 19:05:55
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 2016-03-05 19:05:58
Before we get started, I have a question for those of you who took the test: What was your favorite question on the test?

weilunsun28 2016-03-05 19:06:35
#1!

memc38123 2016-03-05 19:06:35
Number 1

SHARKYBOY 2016-03-05 19:06:35
Question #6

SmallKid 2016-03-05 19:06:35
15

ChickenOnRage 2016-03-05 19:06:35
8

budu 2016-03-05 19:06:35
number 10

popcorn1 2016-03-05 19:06:35
#2

SS00090 2016-03-05 19:06:35
#2 the easist

mewtwomew 2016-03-05 19:06:35
1

simon1221 2016-03-05 19:06:35
14

ThorJames 2016-03-05 19:06:35
3

exhonourated 2016-03-05 19:06:35
#7

ucap 2016-03-05 19:06:35
6

jwlw2014 2016-03-05 19:06:35
6

TheStrangeCharm 2016-03-05 19:06:35
3

jrl_ct 2016-03-05 19:06:35
5

kikipet 2016-03-05 19:06:35
not #7

math0127 2016-03-05 19:06:35
#14

exhonourated 2016-03-05 19:06:35
7 was littt

rcheng66 2016-03-05 19:06:35
NUMBER 1!!!!!!

copeland 2016-03-05 19:06:41
Really? 7?

copeland 2016-03-05 19:06:55
OK, let's rock.

copeland 2016-03-05 19:07:02
We're going to work through all 15 problems from the 2015 AIME I, in order.

copeland 2016-03-05 19:07:06
1. For $-1 < r < 1$, let $S(r)$ denote the sum of the geometric series \[12 + 12r + 12r^2 + 12r^3 + \cdots.\] Let $a$ between -1 and 1 satisfy $S(a)S(-a) = 2016$. Find $S(a) + S(-a)$.

copeland 2016-03-05 19:07:14
An infinite geometric series... What can we do with that?

calculus_riju 2016-03-05 19:07:31
I think u meant 2016?

copeland 2016-03-05 19:07:38
Let's get started! We're going to work through all 15 problems from the 2016 AIME I, in order.

copeland 2016-03-05 19:07:41
1. For $-1 < r < 1$, let $S(r)$ denote the sum of the geometric series \[12 + 12r + 12r^2 + 12r^3 + \cdots.\] Let $a$ between -1 and 1 satisfy $S(a)S(-a) = 2016$. Find $S(a) + S(-a)$.

copeland 2016-03-05 19:07:45
That's what I said. . .

copeland 2016-03-05 19:07:55
Now, here's an infinite geometric series... What can we do with that?

popcorn1 2016-03-05 19:08:11
a / 1-r

lidada 2016-03-05 19:08:11
12/(1-r)

SHARKYBOY 2016-03-05 19:08:11
the sum is 12/(1-r)

memc38123 2016-03-05 19:08:11
Use a/(1-r)

jfurf 2016-03-05 19:08:11
Write in closed form! $\frac{12}{1-r}$

MikoTennisPro 2016-03-05 19:08:11
use the formula a/(1-r) to express S(r)

blue8931 2016-03-05 19:08:11
write the sum as $\frac{12}{1-a}$

xwang1 2016-03-05 19:08:11
$\frac{a}{1-r}$

copeland 2016-03-05 19:08:14
We can write its sum in a closed form! \[12 + 12r + 12r^2 + \cdots = \frac{12}{1-r}.\]

copeland 2016-03-05 19:08:15
So what is $S(a) \cdot S(-a)$?

brian6liu 2016-03-05 19:08:49
144/(1-r^2)

SHARKYBOY 2016-03-05 19:08:49
144/(1-r^2)

hamup1 2016-03-05 19:08:49
$\dfrac{144}{1-a^2}.$

ChickenOnRage 2016-03-05 19:08:49
$144/(1 - r^2)$

math0127 2016-03-05 19:08:49
144/(1-r^2)

SimonSun 2016-03-05 19:08:49
$144/(1-r^2)$

ninjataco 2016-03-05 19:08:49
144/(1-r^2)

tennis1729 2016-03-05 19:08:49
144/(1-a^2)

copeland 2016-03-05 19:08:51
We have $S(a) = \dfrac{12}{1-a}$ and $S(-a) = \dfrac{12}{1+a}$. So, \[S(a) \cdot S(-a) = \frac{12}{1-a} \cdot \frac{12}{1+a} = \frac{144}{1 - a^2}.\]

copeland 2016-03-05 19:08:53
We know $S(a)S(-a) = 2016$. So we could use this to explicitly solve for $a$ right now if we wanted to. But, let's take a closer look at the quantity we're asked for first.

copeland 2016-03-05 19:08:54
What is $S(a) + S(-a)$?

ryanyz10 2016-03-05 19:09:43
24/(1-r^2)

ompatel99 2016-03-05 19:09:43
24/(1-a^2)

ninjataco 2016-03-05 19:09:43
24/(1-a^2)

hamup1 2016-03-05 19:09:43
$\dfrac{24}{1-a^2}$

andsun19 2016-03-05 19:09:43
24/(1-a)/(1+a)

thequantumguy 2016-03-05 19:09:43
$24/(1-a^2)$

ChickenOnRage 2016-03-05 19:09:43
$24/(1-a^2)$

alexli2014 2016-03-05 19:09:43
24/(1-a^2)

blue8931 2016-03-05 19:09:43
$\frac{24}{1-r^2}$

SimonSun 2016-03-05 19:09:43
$12/(1-a)+12/(1+a)$

mathmaniatoo 2016-03-05 19:09:43
24/1-a^2

copeland 2016-03-05 19:09:46
That's equal to \[S(a) + S(-a) = \frac{12}{1-a} + \frac{12}{1+a}.\]

copeland 2016-03-05 19:09:48
Putting everything over a common denominator, \[ S(a) + S(-a) = \frac{12(1+a)}{1-a^2} + \frac{12(1-a)}{1-a^2} = \frac{24}{1 - a^2}. \]

copeland 2016-03-05 19:10:05
That's way simpler than solving for $a.$ What's the sum?

copeland 2016-03-05 19:10:38
That is, what is $S(a)+S(-a)?$

blue8931 2016-03-05 19:10:43
oh... just divide 2016 by 6 to get the answer of $\boxed{336}$

sxu 2016-03-05 19:10:43
336

blue8931 2016-03-05 19:10:43
2016/6 = 336

calculus_riju 2016-03-05 19:10:43
336

tennis1729 2016-03-05 19:10:43
just divide 2016 by 6 to get 336

ChickenOnRage 2016-03-05 19:10:43
2016 / (144/24) = 2016/6 = 336

SimonSun 2016-03-05 19:10:43
2016/6=336

goodbear 2016-03-05 19:10:43
336

walnutwaldo20 2016-03-05 19:10:43
336

nosaj 2016-03-05 19:10:43
336!

copeland 2016-03-05 19:10:48
We see that $S(a) + S(-a)$ is just $\dfrac{24}{144}S(a) \cdot S(-a) = \dfrac16S(a) \cdot S(-a) = \dfrac16 \cdot 2016 =\boxed{336}.$

copeland 2016-03-05 19:10:54
OK, 1 down. 15 to go.

memc38123 2016-03-05 19:11:10
14 to go

ryanyz10 2016-03-05 19:11:10
14

brian6liu 2016-03-05 19:11:10
do you mean 14?

ThorJames 2016-03-05 19:11:10
14 to go

lidada 2016-03-05 19:11:10
14 to go

copeland 2016-03-05 19:11:19
OK, we'll only do 14 more, then.

copeland 2016-03-05 19:11:21
2. Two dice appear to be standard dice with their faces numbered from 1 to 6, but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$. The probability of rolling a 7 with this pair of dice is $\dfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

goodbear 2016-03-05 19:11:33
#15 is still left

copeland 2016-03-05 19:11:39
Thanks for the save.

copeland 2016-03-05 19:11:41
The probability of rolling a $k$ is directly proportional to $k$... Let's call the constant of proportionality $p$.

copeland 2016-03-05 19:11:43
Just to make sure we understand the question, what is the probability of rolling a 5?

brian6liu 2016-03-05 19:12:12
5p

calculus_riju 2016-03-05 19:12:12
5p

stan23456 2016-03-05 19:12:12
5p

blue8931 2016-03-05 19:12:12
5p

copeland 2016-03-05 19:12:15
The probability of rolling a 5 is $5p$. Similarly, the probability of rolling a 1 is $1p$, the probability of rolling a 2 is $2p$, etc.

copeland 2016-03-05 19:12:29
And what is $p?$

kikipet 2016-03-05 19:12:54
1/21

Locust 2016-03-05 19:12:54
1/21

ychen 2016-03-05 19:12:54
1/21

hamup1 2016-03-05 19:12:54
$\dfrac{1}{21}$

SimonSun 2016-03-05 19:12:54
1/21

simon1221 2016-03-05 19:12:54
1/21

idkanymath 2016-03-05 19:12:54
1/21

stan23456 2016-03-05 19:12:54
1/21

SmallKid 2016-03-05 19:12:54
Sum of all coefficients of p is 21, so 21p is 1. p has to be 1/21.

Aragorn66 2016-03-05 19:12:54
1/21, because 1+2+3+4+5+6=21

Prof.iHen 2016-03-05 19:12:54
1/21

TheStrangeCharm 2016-03-05 19:12:54
$\frac{1}{21}$

fishy15 2016-03-05 19:12:54
1/21

copeland 2016-03-05 19:12:58
If we roll a die, it's guaranteed to show one of the numbers 1,2,3,4,5,6, so \[ 1p + 2p + 3p + 4p + 5p + 6p = 1.\]

copeland 2016-03-05 19:13:00
Since $1 + 2 + 3 + 4 + 5+6 = 21$, we get $p = \dfrac 1{21}$.

copeland 2016-03-05 19:13:02
So the probability of rolling a 1 is $\dfrac{1}{21}$, the probability of rolling a 2 is $\dfrac{2}{21}$, etc.

copeland 2016-03-05 19:13:04
Now that we have the probability of rolling each number, let's think about the probability of rolling numbers that sum to 7.

copeland 2016-03-05 19:13:19
One of my favorite tools on the AIME. . .

fishy15 2016-03-05 19:13:46
chart

ThorJames 2016-03-05 19:13:46
chart.

TheStrangeCharm 2016-03-05 19:13:46
draw a chart

SK200 2016-03-05 19:13:46
table?

copeland 2016-03-05 19:13:49
A table! Let's start making a table to organize our counting.

copeland 2016-03-05 19:13:51
\[\begin{array}{c|c|c}
\text{First Die}& \text{Second Die} & \text{Probability} \\
\hline
1 && \\
2 && \\
3 && \\
4 && \\
5 && \\
6 && \\
\end{array}\]

copeland 2016-03-05 19:13:55
What can we roll on the second die to make a sum of 7 if we roll a 1 on the first die?

person754 2016-03-05 19:14:14
6

IQMathlete 2016-03-05 19:14:14
6\

Peggy 2016-03-05 19:14:14
6

jrl_ct 2016-03-05 19:14:14
6

mewtwomew 2016-03-05 19:14:14
6

alexli2014 2016-03-05 19:14:14
6

SmallKid 2016-03-05 19:14:14
6.

rcheng66 2016-03-05 19:14:14
6

akaashp11 2016-03-05 19:14:14
6

mewtwomew 2016-03-05 19:14:14
6.

copeland 2016-03-05 19:14:17
We need to roll a 6. What is the probability of rolling a 1 on the first die and 6 on the second die?

Math1331Math 2016-03-05 19:14:41
6/441

SimonSun 2016-03-05 19:14:41
6/441

MikoTennisPro 2016-03-05 19:14:41
6/441

goodbear 2016-03-05 19:14:41
6/21^2

ninjataco 2016-03-05 19:14:41
6/441

blue8931 2016-03-05 19:14:41
1/21 * 6/21 = 6/441

brian6liu 2016-03-05 19:14:41
6/21^2

jfurf 2016-03-05 19:14:41
a $6$! This occurs with probability $\frac{1}{21} \cdot \frac{6}{21} = \frac{6}{441} $.

goodbear 2016-03-05 19:14:41
6/441

copeland 2016-03-05 19:14:42
The probability of rolling a 1 is $\dfrac{1}{21}$ and the probability of rolling a 6 is $\dfrac{6}{21}$. So the probability is $\dfrac{6}{21^2}$.

copeland 2016-03-05 19:14:45
\[\begin{array}{c|c|c}
\text{First Die}& \text{Second Die} & \text{Probability} \\
\hline
1 & 6& \frac{6}{21^2} \\
2 && \\
3 && \\
4 && \\
5 && \\
6 && \\
\end{array}\]

copeland 2016-03-05 19:14:46
If we roll a 2 on the first die?

andsun19 2016-03-05 19:15:09
5

goodbear 2016-03-05 19:15:09
5

rcheng66 2016-03-05 19:15:09
5

ompatel99 2016-03-05 19:15:09
5 on second die

calculus_riju 2016-03-05 19:15:09
5

person754 2016-03-05 19:15:09
5

SimonSun 2016-03-05 19:15:09
5 on second die

sxu 2016-03-05 19:15:18
10/441

kikipet 2016-03-05 19:15:18
10/441

Aragorn66 2016-03-05 19:15:18
10/21^2

SimonSun 2016-03-05 19:15:18
10/441

goodbear 2016-03-05 19:15:18
10/441

jrl_ct 2016-03-05 19:15:18
10/441

blue8931 2016-03-05 19:15:18
you need a 5; this occurs with probability $\frac{10}{441}$

MikoTennisPro 2016-03-05 19:15:18
10/441

copeland 2016-03-05 19:15:20
Then we need a 5 on the second die. This happens with probability $\dfrac{2}{21} \cdot \dfrac{5}{21} = \dfrac{10}{21^2}.$

copeland 2016-03-05 19:15:20
\[\begin{array}{c|c|c}
\text{First Die}& \text{Second Die} & \text{Probability} \\
\hline
1 & 6& \frac{6}{21^2} \\
2 &5& \frac{10}{21^2} \\
3 && \\
4 && \\
5 && \\
6 && \\
\end{array}\]

copeland 2016-03-05 19:15:21
A 3?

Prof.iHen 2016-03-05 19:15:34
4

MikoTennisPro 2016-03-05 19:15:34
4 on second die

mewtwomew 2016-03-05 19:15:46
12/144

mewtwomew 2016-03-05 19:15:46
12/441

MikoTennisPro 2016-03-05 19:15:46
12/441

blue8931 2016-03-05 19:15:46
4; $\frac{12}{441}$

kikipet 2016-03-05 19:15:46
need a 4, P = 12/441

brian6liu 2016-03-05 19:15:46
you need a 4, so 12/21^2

fishy15 2016-03-05 19:15:46
4 so 12/441

idomath12345 2016-03-05 19:15:46
4 12/441

Math1331Math 2016-03-05 19:15:46
12/441

goodbear 2016-03-05 19:15:46
12/441

ompatel99 2016-03-05 19:15:50
You need a 4 on the second die with prob of 3/21*4/21=12/(21^2)

SmallKid 2016-03-05 19:15:50
We need a 4, 3/4 combo has 3/21 * 4/21 probability = 12/441 = 4/147

copeland 2016-03-05 19:15:52
Then we need a 4 on the second die. This happens with probability $\dfrac{3}{21} \cdot \dfrac{4}{21} = \dfrac{12}{21^2}.$

copeland 2016-03-05 19:15:53
\[\begin{array}{c|c|c}
\text{First Die}& \text{Second Die} & \text{Probability} \\
\hline
1 & 6& \frac{6}{21^2} \\
2 &5& \frac{10}{21^2} \\
3 &4& \frac{12}{21^2} \\
4 && \\
5 && \\
6 && \\
\end{array}\]

copeland 2016-03-05 19:15:58
Can we quickly fill up the table now?

akaashp11 2016-03-05 19:16:12
Symmetric after this point

ompatel99 2016-03-05 19:16:12
Now we just fill it in backwads

andsun19 2016-03-05 19:16:12
And... symmetry!

memc38123 2016-03-05 19:16:12
Now just use their reverses

ChickenOnRage 2016-03-05 19:16:12
yes, symmetry

brian6liu 2016-03-05 19:16:12
the second half is symmetric

jfurf 2016-03-05 19:16:12
Yes. It is symmetrical.

Math1331Math 2016-03-05 19:16:12
yes because of symmetry

mathmaniatoo 2016-03-05 19:16:12
yes, just double

copeland 2016-03-05 19:16:14
The next three rows are symmetric with the first three.

copeland 2016-03-05 19:16:15
\[\begin{array}{c|c|c}
\text{First Die}& \text{Second Die} & \text{Probability} \\
\hline
1 & 6& \frac{6}{21^2} \\
2 &5& \frac{10}{21^2} \\
3 &4& \frac{12}{21^2} \\
4 & 3& \frac{12}{21^2} \\
5 & 2&\frac{10}{21^2} \\
6 &1&\frac{6}{21^2} \\
\end{array}\]

copeland 2016-03-05 19:16:16
So what is our desired probability?

Math1331Math 2016-03-05 19:16:54
8/63

lidada 2016-03-05 19:16:54
8/63

memc38123 2016-03-05 19:16:54
8/63

idkanymath 2016-03-05 19:16:54
8/63

goseahawks 2016-03-05 19:16:54
8/63

calculus_riju 2016-03-05 19:16:54
$\frac{8}{63}$

Aragorn66 2016-03-05 19:16:54
=8/63

Locust 2016-03-05 19:16:54

weilunsun28 2016-03-05 19:16:54
8/63

copeland 2016-03-05 19:16:59
Since $56 = 8 \cdot 7$, this simplifies to $\dfrac{8}{7 \cdot 9} = \dfrac{8}{63}$. So what is our answer?

SmallKid 2016-03-05 19:17:14
But since we're adding m and n, our final answer is 71

andsun19 2016-03-05 19:17:14
71

weilunsun28 2016-03-05 19:17:14
71

ChickenOnRage 2016-03-05 19:17:14
071

fishy15 2016-03-05 19:17:14
71 yay

ninjataco 2016-03-05 19:17:14
071

sxu 2016-03-05 19:17:14
71

memc38123 2016-03-05 19:17:14
71

mewtwomew 2016-03-05 19:17:14
071

kikipet 2016-03-05 19:17:14
071

Math1331Math 2016-03-05 19:17:14
71

SimonSun 2016-03-05 19:17:14
071

idkanymath 2016-03-05 19:17:14
71

person754 2016-03-05 19:17:14
71

rocket13jg 2016-03-05 19:17:14
071

stan23456 2016-03-05 19:17:14
071

Aragorn66 2016-03-05 19:17:14
71

calculus_riju 2016-03-05 19:17:14
71

Peggy 2016-03-05 19:17:14
71

alexli2014 2016-03-05 19:17:14
071

nosaj 2016-03-05 19:17:14
071

Math1331Math 2016-03-05 19:17:14
8+63=71

thequantumguy 2016-03-05 19:17:14
71

ychen 2016-03-05 19:17:18
071, can't miss the zero guuys

copeland 2016-03-05 19:17:25
Mmm. Bubbles.

copeland 2016-03-05 19:17:30
3. A regular icosahedron is a 20-faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated.

copeland 2016-03-05 19:17:35

copeland 2016-03-05 19:17:46
(Deven stole the diagrams for this problem from MSTang.)

nosaj 2016-03-05 19:17:50
Is that MSTang's diagram?

copeland 2016-03-05 19:17:52
Yes it is, thank you MSTang!

copeland 2016-03-05 19:18:08
Any ideas about this problem?

andsun19 2016-03-05 19:18:48
Break it into steps!

ninjataco 2016-03-05 19:18:48
constructive counting

weilunsun28 2016-03-05 19:18:48
calculate each step then multiply together

blue8931 2016-03-05 19:18:48
constructive counting

thequantumguy 2016-03-05 19:18:48
we can count the ways to move for each plane

Aragorn66 2016-03-05 19:18:48
start from the top, how many ways can we moved down, etc

theeagle 2016-03-05 19:18:48
divide into layers

nosaj 2016-03-05 19:18:48
Let's think constructively about how to build a path to the bottom.

brian6liu 2016-03-05 19:18:48
constructive counting

copeland 2016-03-05 19:18:52
We can try constructive counting! We'll just start constructing the path step-by-step.

copeland 2016-03-05 19:19:10
Note: this is not "casework" here.

copeland 2016-03-05 19:19:12
Right now we're at the red vertex up-top. What can we do?

copeland 2016-03-05 19:19:13

memc38123 2016-03-05 19:19:44
5 ways downward

bestwillcui1 2016-03-05 19:19:44
five coihces

sxu 2016-03-05 19:19:44
5 ways to go down

Aragorn66 2016-03-05 19:19:44
we cna move down in 5 different ways

SK200 2016-03-05 19:19:44
go in 5 ways

MikoTennisPro 2016-03-05 19:19:44
choose any of the 5 paths to go down

fishy15 2016-03-05 19:19:44
go down 5 ways

goodbear 2016-03-05 19:19:44
first step has 5 choices

SimonSun 2016-03-05 19:19:44
we move to one of the 5 vertices

ChickenOnRage 2016-03-05 19:19:44
move down to 1 of 5 verticies

thequantumguy 2016-03-05 19:19:44
their are 5 ways to move to the next plane

ThorJames 2016-03-05 19:19:44
5 verticies

copeland 2016-03-05 19:19:46
We can move down to the pentagon (in blue below) in 5 different ways.

copeland 2016-03-05 19:19:47

copeland 2016-03-05 19:19:48
What can we do now that we're on the blue pentagon?

kikipet 2016-03-05 19:20:20
go to another vertex on the pentagon or go down

weilunsun28 2016-03-05 19:20:20
we can move around the pentagon or go down

SS00090 2016-03-05 19:20:20
Sideways or down

mewtwomew 2016-03-05 19:20:20
either go down or move horizontally

person754 2016-03-05 19:20:20
move horizontally or go down

walnutwaldo20 2016-03-05 19:20:56
stay at that vertex or got to any of the other 4

alexli2014 2016-03-05 19:20:56
stay or move right or left

SimonSun 2016-03-05 19:20:56
we can move to any of the 5 points with 9 ways

ninjataco 2016-03-05 19:20:56
stay in the same spot, or move horizontally

goodbear 2016-03-05 19:20:56
we can move to either one of the blue vertices in 9 ways

goodbear 2016-03-05 19:20:56
we can move to either one of the blue vertices in 2*5-1=9 ways

edwardneo 2016-03-05 19:20:56
there are 9 ways to move around the pentagon

calculus_riju 2016-03-05 19:20:56
we can stay there or move upto 4 vertices in clock or anticlock manner......9 ways

copeland 2016-03-05 19:21:17
If we go down, we're forced to the next pentagon. Before that we should wander around.

copeland 2016-03-05 19:21:30
We can move around the blue pentagon by either:
(a) Staying still
(b) Moving around counter-clockwise by up to 4 steps
(c) Moving around clockwise by up to 4 steps

copeland 2016-03-05 19:21:33
So we have 1+4+4=9 ways to move around the pentagon. Then how many ways are there to leave the blue pentagon?

ryanyz10 2016-03-05 19:21:56
2

blue8931 2016-03-05 19:21:56
2

math0127 2016-03-05 19:21:56
two ways

goodbear 2016-03-05 19:21:56
2

rcheng66 2016-03-05 19:21:56
2

hliu70 2016-03-05 19:21:56
2

person754 2016-03-05 19:21:56
2

hliu70 2016-03-05 19:21:56
2 for each way

Aragorn66 2016-03-05 19:21:56
2

jwlw2014 2016-03-05 19:21:56
2 from each vertex

copeland 2016-03-05 19:21:58
No matter where we land, there are two different ways. For example, if we are at the red vertex below, we can travel to either of the vertices in green along the blue edges:

copeland 2016-03-05 19:21:59

copeland 2016-03-05 19:22:01
Once we're at the lower pentagon, what can we do?

andsun19 2016-03-05 19:22:28
Same thing! 9 ways

sxu 2016-03-05 19:22:28
wander 9 ways

ompatel99 2016-03-05 19:22:28
Now another 9 ways because it's another pentagon

blue8931 2016-03-05 19:22:28
same as upper pentagon; 9 options

akaashp11 2016-03-05 19:22:28
9 ways horizontally

ryanyz10 2016-03-05 19:22:28
9 ways as well

Peggy 2016-03-05 19:22:28
move on the pentagon, 9 ways again

edwardneo 2016-03-05 19:22:28
there are 9 ways to move around the pentagon

rcheng66 2016-03-05 19:22:28
same as before wander around or go straight down in 9 ways

ninjataco 2016-03-05 19:22:28
9 ways to move around again

mathmaniatoo 2016-03-05 19:22:28
rotate around 9 different ways

copeland 2016-03-05 19:22:30
Again, we can move around the pentagon. We can either
(a) Stay still
(b) Move around counter-clockwise by up to 4 steps
(c) Move around clockwise by up to 4 steps

copeland 2016-03-05 19:22:31
So we have 1+4+4=9 ways to move around this pentagon as well. In how many ways can we leave this pentagon?

sxu 2016-03-05 19:22:59
1

SimonSun 2016-03-05 19:22:59
1

rcheng66 2016-03-05 19:22:59
1

andsun19 2016-03-05 19:22:59
1

MikoTennisPro 2016-03-05 19:22:59
1

ychen 2016-03-05 19:22:59
1

blue8931 2016-03-05 19:22:59
only 1

Peggy 2016-03-05 19:22:59
1

ompatel99 2016-03-05 19:22:59
Once we move down, we're done-zo with only 1 way

SK200 2016-03-05 19:22:59
1

fishy15 2016-03-05 19:22:59
1

alexli2014 2016-03-05 19:22:59
just 1

kikipet 2016-03-05 19:22:59
1 from each vertex

copeland 2016-03-05 19:23:02
There's only one vertex left for us to go to -- the bottom one -- so we have 1 way to do that.

copeland 2016-03-05 19:23:03
Let's consolidate all of our counting. How many ways are there to form such a path?

memc38123 2016-03-05 19:23:52
810

SimonSun 2016-03-05 19:23:52
$5*9*2*9=810$

rocket13jg 2016-03-05 19:23:52
810

MikoTennisPro 2016-03-05 19:23:52
5*9*2*9 = 810

blue8931 2016-03-05 19:23:52
5 * 9 * 2 * 9 * 1 = $\boxed{810}$

alexli2014 2016-03-05 19:23:52
5*9*2*9*1=810

Peggy 2016-03-05 19:23:52
5*9*2*9=810

jrl_ct 2016-03-05 19:23:52
5*9*9*2 = 810

SmallKid 2016-03-05 19:23:52
5*9*9*2=810

hliu70 2016-03-05 19:23:52
5*9*2*9=810

ompatel99 2016-03-05 19:23:52
5*9*2*9=810

Aragorn66 2016-03-05 19:23:52
810

copeland 2016-03-05 19:23:54
We have 5 options for leaving the top, then 9 options for the navigating the first pentagon, 2 options for traveling between the pentagons, 9 options for navigating the second pentagon, and one way to leave the second pentagon. For a total of \[5\cdot9\cdot2\cdot9 = \boxed{810}\text{ paths}.\]

copeland 2016-03-05 19:24:14
We finally managed to get our assistant into the right place.

copeland 2016-03-05 19:24:17
Dillon Liu (dtliu): Dillon graduated from Columbia University in 2013 with a BS in Applied Physics and is now working on a DPhil at Oxford in Theoretical Condensed Matter Physics on a Marshall Scholarship. At Columbia, Dillon was a teaching assistant for physics and mathematics courses, including quantum mechanics and introductory cryptography. He has also been a teaching assistant at the Summer Science Program. At Oxford, he teaches for the fourth-year theory option covering quantum field theory, statistical mechanics, and other advanced topics.

copeland 2016-03-05 19:24:24
Sorry, Dillon. Welcome!

dtliu 2016-03-05 19:24:25
Hey all, I'm happy to be here!

Locust 2016-03-05 19:24:37
Hi!!!!

Aragorn66 2016-03-05 19:24:37
dtliu hi long time no see

blue8931 2016-03-05 19:24:37
hello!

kikipet 2016-03-05 19:24:37
hi

edwardneo 2016-03-05 19:24:37
hi!

ychen 2016-03-05 19:24:41
hi mr dillon

Darth_Math 2016-03-05 19:24:41
Hi

sxu 2016-03-05 19:24:41
hi

copeland 2016-03-05 19:24:49
4. A right prism with height $h$ has bases that are regular hexagons with sides of length 12. A vertex $A$ of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain $A$ measures $60^\circ.$ Find $h^2.$

copeland 2016-03-05 19:24:53
We start with. . .

popcorn1 2016-03-05 19:25:28
picture!!!

Prof.iHen 2016-03-05 19:25:28
diagram

alexli2014 2016-03-05 19:25:28
diagram

calculus_riju 2016-03-05 19:25:28
diagram?!

LighteningAB 2016-03-05 19:25:28
A DIAGRAM

illogical_21 2016-03-05 19:25:28
diagram

brian6liu 2016-03-05 19:25:28
a diagram

simon1221 2016-03-05 19:25:28
diagram!

copeland 2016-03-05 19:25:30
. . . a nice diagram:

ompatel99 2016-03-05 19:25:33
Another stolen picture?

copeland 2016-03-05 19:25:38
I made this one.

copeland 2016-03-05 19:25:40

copeland 2016-03-05 19:25:54
You can click any picture to see how pro I am at Asymptote.

copeland 2016-03-05 19:26:01
You can also double-click it to make it stick around.

copeland 2016-03-05 19:26:03
And here is our plane:

copeland 2016-03-05 19:26:04

copeland 2016-03-05 19:26:23
Now what should we draw?

sxu 2016-03-05 19:26:51
the dihedral angle

copeland 2016-03-05 19:26:54
Sure, how?

ChickenOnRage 2016-03-05 19:27:31
an altitude from A

blue8931 2016-03-05 19:27:31
30-60-90 triangle

qwerty733 2016-03-05 19:27:31
The dihedral angle arms at the midpoint

memc38123 2016-03-05 19:27:31
A 30-60-90 triangle

hamup1 2016-03-05 19:27:31
foot from $A$ to that blue side

andsun19 2016-03-05 19:27:31
Draw the line connecting the midpoint and the vertex

andsun19 2016-03-05 19:27:31
Draw the line connecting the midpoint of the diagonal and A

TheStrangeCharm 2016-03-05 19:27:31
perpendicular from $A$ to the blue line in the bottom plane.

copeland 2016-03-05 19:27:34
We have a 30-60-90 triangle with right angle at $A,$ one vertex above $A$ and one vertex on the midpoint of the base edge:

copeland 2016-03-05 19:27:35

copeland 2016-03-05 19:27:38
Incidentally, to measure the dihedral angle between two planes, you first draw the intersection line ($CB$ in our diagram) and then you pick a point ($M$) on that line. From that point you draw lines in each of the two planes that are perpendicular to the original line. Here we drew $MA$ and $MV$, which are both perpendicular to $CB$. The angle between those lines is the dihedral angle between the planes - here it is $\angle VMA$.

copeland 2016-03-05 19:27:56
The height of $\triangle VAM$ is $h,$ which we're trying to find.

copeland 2016-03-05 19:27:57

copeland 2016-03-05 19:27:58
What else can we discover about $\triangle VAM?$

andsun19 2016-03-05 19:28:37
Find AM by looking at triangle ABC

SuperMaltese 2016-03-05 19:28:37
we can find AM

andsun19 2016-03-05 19:28:37
We can find AM

memc38123 2016-03-05 19:28:37
AM is part of another 30-60-90

Locust 2016-03-05 19:28:37
The length of AM

hliu70 2016-03-05 19:28:37
we can calculate AM

copeland 2016-03-05 19:28:40
The base is the height of an isosceles triangle with vertex angle $120^\circ$:

copeland 2016-03-05 19:28:40

copeland 2016-03-05 19:28:41
And the red length?

blue8931 2016-03-05 19:28:56
30-60-90 triangles and AM= 6

walnutwaldo20 2016-03-05 19:28:56
AM = 6

sophiazhi 2016-03-05 19:28:56
AM = AC/2 = 6

SHARKYBOY 2016-03-05 19:28:56
AM = 6

MikoTennisPro 2016-03-05 19:28:56
AM = 6

person754 2016-03-05 19:28:56
6

walnutwaldo20 2016-03-05 19:28:56
12/2 = 6

alexli2014 2016-03-05 19:28:56
6

ThorJames 2016-03-05 19:28:56
6

blizzard10 2016-03-05 19:28:56
6

copeland 2016-03-05 19:28:59
The red length is definitely 6.

copeland 2016-03-05 19:29:00
So what is $h^2?$

rcheng66 2016-03-05 19:29:33
108

alexli2014 2016-03-05 19:29:33
108

memc38123 2016-03-05 19:29:33
108

person754 2016-03-05 19:29:33
108

goodbear 2016-03-05 19:29:33
108

Peggy 2016-03-05 19:29:33
108

SuperMaltese 2016-03-05 19:29:33
(6sqrt3)^2 = 108

walnutwaldo20 2016-03-05 19:29:33
(6* sqrt(3))^2 = 36*3 = 108

Locust 2016-03-05 19:29:33

Niuniu01 2016-03-05 19:29:33
108

blue8931 2016-03-05 19:29:33
$(6\sqrt{3})^2=\boxed{108}$

ninjataco 2016-03-05 19:29:33
(6sqrt 3)^2 = 108

copeland 2016-03-05 19:29:37

copeland 2016-03-05 19:29:38
This is a 30-60-90 triangle so $h=6\sqrt{3}$ and $h^2=\boxed{108}.$

copeland 2016-03-05 19:29:54
Ready for more?

memc38123 2016-03-05 19:30:02
YEAH

blue8931 2016-03-05 19:30:02
yes

sxu 2016-03-05 19:30:02
yup

SuperMaltese 2016-03-05 19:30:02
YES

Aragorn66 2016-03-05 19:30:02
yes!

weilunsun28 2016-03-05 19:30:02
yesss

brian6liu 2016-03-05 19:30:02
yeah

Locust 2016-03-05 19:30:02
Yeah!!!

Niuniu01 2016-03-05 19:30:02
yeah!

jrl_ct 2016-03-05 19:30:02
Of course!

spicyray 2016-03-05 19:30:02
yeah

ninjataco 2016-03-05 19:30:02
yes

ThorJames 2016-03-05 19:30:02
yes

lleB_ocaT 2016-03-05 19:30:02
Ok

copeland 2016-03-05 19:30:05
Me too.

copeland 2016-03-05 19:30:06
5. Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n+1$ pages in $t+1$ minutes. Each day thereafter Ahn read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the 374 page book. It took her a total of 319 minutes to read the book. Find $n + t$.

copeland 2016-03-05 19:30:10
Let's start with small numbers and find a pattern.

copeland 2016-03-05 19:30:11
How many pages has Anh read in total after two days?

Niuniu01 2016-03-05 19:30:34
2n+1

mewtwomew 2016-03-05 19:30:34
2n+1

nosaj 2016-03-05 19:30:34
n + n + 1 = 2n+1

walnutwaldo20 2016-03-05 19:30:34
2n + 1

fishy15 2016-03-05 19:30:34
2n+1

SuperMaltese 2016-03-05 19:30:34
2n+1

ompatel99 2016-03-05 19:30:34
2n+1 pages

kikipet 2016-03-05 19:30:34
2n+1

Locust 2016-03-05 19:30:34
2n+1

rcheng66 2016-03-05 19:30:34
2n+1

person754 2016-03-05 19:30:34
2n+1

SimonSun 2016-03-05 19:30:34
2n+1

copeland 2016-03-05 19:30:37
After two days, Anh has read $n + (n+1)$, or $2n + 1$ pages.

copeland 2016-03-05 19:30:37
After three days?

SmallKid 2016-03-05 19:30:53
3n+3

person754 2016-03-05 19:30:53
3n+3

sxu 2016-03-05 19:30:53
3n+3

brian6liu 2016-03-05 19:30:53
3n+3

space_space 2016-03-05 19:30:53
3n+2

Locust 2016-03-05 19:30:53
3n+3

LighteningAB 2016-03-05 19:30:53
3n+3

ninjataco 2016-03-05 19:30:53
3n+3

fishy15 2016-03-05 19:30:53
3n+3

hliu70 2016-03-05 19:30:53
3n+3

copeland 2016-03-05 19:30:54
After three days, Anh has read $2n + 1 + (n+2) = 3n + (1 + 2)$ pages.

copeland 2016-03-05 19:30:55
Anyone seeing a pattern?

ompatel99 2016-03-05 19:31:41
Triangular numbers

memc38123 2016-03-05 19:31:41
Number of days * n + the number of days -1th triangular number

space_space 2016-03-05 19:31:41
1,3,6,10,15

SimonSun 2016-03-05 19:31:41
trianglular numbes

ompatel99 2016-03-05 19:31:41
kn+(k*(k-1))/2

goodbear 2016-03-05 19:31:41
kn+k(k-1)/2

Locust 2016-03-05 19:31:41
The amount of extra pages besides the n term are triangular numbers

blizzard10 2016-03-05 19:31:41
After m days, Anh has read $mn + \dfrac{(m-1)(m)}{2}$ pages.

kikipet 2016-03-05 19:31:41
after x days, total # of pages is xn+x(x-1)/2

ryanyz10 2016-03-05 19:31:41
kn + k(k-1)/2

brian6liu 2016-03-05 19:31:41
kn+k(k-1)/2, where k is the number of days

copeland 2016-03-05 19:31:43
After $k$ days, Anh has read $kn + T_{k-1}$ pages. Where $T_{k-1}$ is the $(k-1)^{\text{st}}$ triangle number.

copeland 2016-03-05 19:31:48
We know that $T_{k-1} = \dfrac{(k-1)k}{2}$. So, after $k$ days, Anh has read \[ kn + \frac{(k-1)k}{2} \] pages.

copeland 2016-03-05 19:31:50
How many minutes has Anh read for after $k$ days?

brian6liu 2016-03-05 19:32:25
kt+k(k-1)/2

SmallKid 2016-03-05 19:32:25
Same thing but replace n with t

LighteningAB 2016-03-05 19:32:25
same thing for n but for t

ninjataco 2016-03-05 19:32:25
kt + (k-1)k/2

goodbear 2016-03-05 19:32:25
kt+k(k-1)/2

nosaj 2016-03-05 19:32:25
*same thing except with $t$

andsun19 2016-03-05 19:32:25
kt+(k-1)k/2

kikipet 2016-03-05 19:32:25
$kt + \frac{(t - 1)t}{2}$

copeland 2016-03-05 19:32:27
Using the same argument, Anh has read for $kt + \dfrac{(k-1)k}{2}$ minutes.

copeland 2016-03-05 19:32:36
Now we get equations from the problem.

copeland 2016-03-05 19:32:37
If Anh read for $k$ days total, we have
\begin{align}
kn + \frac{(k-1)k}{2} &= 374 \\
kt + \frac{(k-1)k}{2} & = 319.
\end{align}

copeland 2016-03-05 19:32:40
What can we do with these two equations?

person754 2016-03-05 19:33:11
subtract

brian6liu 2016-03-05 19:33:11
subtract them

SimonSun 2016-03-05 19:33:11
subtract

jrl_ct 2016-03-05 19:33:11
subtraction!

goodbear 2016-03-05 19:33:11
subtract

sophiazhi 2016-03-05 19:33:11
subtract

jfurf 2016-03-05 19:33:11
Subtract them

weilunsun28 2016-03-05 19:33:11
subtract

space_space 2016-03-05 19:33:11
subtract them

Peggy 2016-03-05 19:33:11
subtract

ThorJames 2016-03-05 19:33:11
subtract?

copeland 2016-03-05 19:33:18
Since they both contain a $(k-1)k/2$ term, it makes sense to try subtracting them. That gives us \[k(n-t) = 55.\]

copeland 2016-03-05 19:33:19
Does that tell us anything interesting about $k$?

SmallKid 2016-03-05 19:33:49
k is a factor of 55

andsun19 2016-03-05 19:33:49
factor of 55

ninjataco 2016-03-05 19:33:49
k is a divisor of 55

blue8931 2016-03-05 19:33:49
divisor of 55

brian6liu 2016-03-05 19:33:49
k divides 55

sophiazhi 2016-03-05 19:33:49
factor of 55

kikipet 2016-03-05 19:33:49
k is a factor of 55

Peggy 2016-03-05 19:33:49
k is 1, 5, 11, 55

weilunsun28 2016-03-05 19:33:49
its a factor of 55

lidada 2016-03-05 19:33:49
k is a factor of 55

MikoTennisPro 2016-03-05 19:33:49
must be a factor of 55?

hliu70 2016-03-05 19:33:49
factor of 55

copeland 2016-03-05 19:33:53
This tells us that $k$ must be a divisor of 55. So Anh read for 1, 5, 11, or 55 days.

copeland 2016-03-05 19:33:54
Are there any of these we can eliminate immediately?

popcorn1 2016-03-05 19:34:32
1.

Peggy 2016-03-05 19:34:32
1

mewtwomew 2016-03-05 19:34:32
1

walnutwaldo20 2016-03-05 19:34:32
1

valmat01 2016-03-05 19:34:32
1

goseahawks 2016-03-05 19:34:32
1 day

math2fun 2016-03-05 19:34:32
1

space_space 2016-03-05 19:34:32
1

Prof.iHen 2016-03-05 19:34:32
1

qwerty733 2016-03-05 19:34:32
1

hliu70 2016-03-05 19:34:32
1 because the problem also says the next day, so this is impossible

blizzard10 2016-03-05 19:34:32
We know it took her more than one day. It says so in the problem.

illogical_21 2016-03-05 19:34:32
1

copeland 2016-03-05 19:34:36
The problem implicitly said that Anh read for a "second day", so we know $k\ne 1$.

copeland 2016-03-05 19:34:37
(I didn't like this step much.)

copeland 2016-03-05 19:34:39
Is there anything else we can do with these equations that might help?

ompatel99 2016-03-05 19:35:10
Add them

ninjataco 2016-03-05 19:35:10
add them

kikipet 2016-03-05 19:35:10
add them

sxu 2016-03-05 19:35:10
add

memc38123 2016-03-05 19:35:10
add them

Locust 2016-03-05 19:35:10
Add to get the n+t term

copeland 2016-03-05 19:35:13
We could also try adding them. That gives us \[k(n+t) + k(k-1) = 693.\]

copeland 2016-03-05 19:35:15
What's nice about that?

MikoTennisPro 2016-03-05 19:35:33
can factor out k again?

kikipet 2016-03-05 19:35:33
factor out k

ryanyz10 2016-03-05 19:35:33
factor k

brian6liu 2016-03-05 19:35:40
k divides 693 too

hamup1 2016-03-05 19:35:40
k divides 693

sxu 2016-03-05 19:35:40
k is a factor of 693

calculus_riju 2016-03-05 19:35:40
k is a factor of 693

Ericaops 2016-03-05 19:35:40
k is a factor of 693

copeland 2016-03-05 19:35:43
So?

mewtwomew 2016-03-05 19:36:09
11 is the only factor of 693 in common with 55

kikipet 2016-03-05 19:36:09
k is 11

SimonSun 2016-03-05 19:36:09
k=11

ompatel99 2016-03-05 19:36:09
k is 11

Locust 2016-03-05 19:36:09
k=11

space_space 2016-03-05 19:36:09
k is 11

math2fun 2016-03-05 19:36:09
k = 11

popcorn1 2016-03-05 19:36:09
not 11

ThorJames 2016-03-05 19:36:09
11

hamup1 2016-03-05 19:36:09
k must equal 11

blue8931 2016-03-05 19:36:09
k must be 11

TheStrangeCharm 2016-03-05 19:36:09
k = 11

Aragorn66 2016-03-05 19:36:09
k is a factor of both 693 and 55, so must be 11

copeland 2016-03-05 19:36:11
5 and 55 clearly don't divide 693. So we must have $k = 11$. What do we do now?

goodbear 2016-03-05 19:36:51
plug in

19bobhu 2016-03-05 19:36:51
sub it in

idomath12345 2016-03-05 19:36:51
sub in.

blizzard10 2016-03-05 19:36:51
Plug in to our expressions from earlier

space_space 2016-03-05 19:36:51
plug it into the equation

illogical_21 2016-03-05 19:36:51
plug it back in

copeland 2016-03-05 19:36:57
Plug in where?

ompatel99 2016-03-05 19:37:27
Plug it in and solve for n+t without solving for n and t

Locust 2016-03-05 19:37:27
Plug back into the determined equation to solve n+t

jfurf 2016-03-05 19:37:27
We can find $n+t$ quickly! Substitute in for $k$ in that equation.

Locust 2016-03-05 19:37:27
The equation with the term (n+t)

fishy15 2016-03-05 19:37:27
our added equation

idomath12345 2016-03-05 19:37:27
the sum eq.

sxu 2016-03-05 19:37:29
the second *new* equation

copeland 2016-03-05 19:37:32
We can just plug that in to this equation!

copeland 2016-03-05 19:37:35
We have \[11(n+t) + 11\cdot 10 = 693.\] So what is $n+t$?

jfurf 2016-03-05 19:38:11
$n+t=53$ Bubble in $\boxed{053}$ :)

calculus_riju 2016-03-05 19:38:11
53

kikipet 2016-03-05 19:38:11
053

ninjataco 2016-03-05 19:38:11
053

SimonSun 2016-03-05 19:38:11
53

goodbear 2016-03-05 19:38:11
53

ychen 2016-03-05 19:38:11
053

Prof.iHen 2016-03-05 19:38:11
53

math2fun 2016-03-05 19:38:11
53

fz2012 2016-03-05 19:38:11
53

SuperMaltese 2016-03-05 19:38:11
053

space_space 2016-03-05 19:38:11
53

tennis1729 2016-03-05 19:38:11
53

person754 2016-03-05 19:38:11
053

goodbear 2016-03-05 19:38:11
053

copeland 2016-03-05 19:38:22
We get $n+t = \boxed{53}$.

copeland 2016-03-05 19:38:25
Great!

copeland 2016-03-05 19:38:34
1/3 of the way done.

fishy15 2016-03-05 19:38:49
2/3 left

copeland 2016-03-05 19:38:51

copeland 2016-03-05 19:38:54
6. In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $\overline{AB}$ at $L$. The line through $C$ and $L$ intersects the cirumscribed circle of $\triangle ABC$ at the two points $C$ and $D$. If $LI = 2$ and $LD = 3$, then $IC = \frac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Quinn 2016-03-05 19:39:16
DIAGRAM!!!

blizzard10 2016-03-05 19:39:16
Diagram!

Aragorn66 2016-03-05 19:39:16
diagram

weilunsun28 2016-03-05 19:39:16
diagram

fishy15 2016-03-05 19:39:16
i think we need a diagram

space_space 2016-03-05 19:39:16
diagram plz

idomath12345 2016-03-05 19:39:16
DIAGRAM!

popcorn1 2016-03-05 19:39:16
Draw a picture

copeland 2016-03-05 19:39:17
Let's start with a diagram:

copeland 2016-03-05 19:39:18

andsun19 2016-03-05 19:39:26
I didn't get this one

nosaj 2016-03-05 19:39:26
the first really hard one

copeland 2016-03-05 19:39:53
Yeah, this problem was harder for me than it should have been. I think it's because it needed a more sophisticated theorem than 6 usually does.

copeland 2016-03-05 19:40:02
I've drawn $IC$ in red since that's what we're looking for.

copeland 2016-03-05 19:40:07
What else do we know about this diagram?

ychen 2016-03-05 19:40:37
IL = 2, LD = 3

nosaj 2016-03-05 19:40:37
IL = 2, LD = 3

mathmaniatoo 2016-03-05 19:40:37
LI=2, DL=3

person754 2016-03-05 19:40:37
li=2 ld-3

rcheng66 2016-03-05 19:40:37
LI and LD

goodbear 2016-03-05 19:40:37
2,3

sxu 2016-03-05 19:40:37
IL and LD

copeland 2016-03-05 19:40:39
Well, we know $IL = 2$ and $LD = 3$, we can add that in.

copeland 2016-03-05 19:40:40

copeland 2016-03-05 19:40:41
What else do we know?

hamup1 2016-03-05 19:41:13
D is the midpoint of arc BC b/c angle bisector

Locust 2016-03-05 19:41:13
Arcs AD and DB are congruent?

hamup1 2016-03-05 19:41:13
d is the midpoint of arc AB

ninjataco 2016-03-05 19:41:13
AD = BD

copeland 2016-03-05 19:41:15
We know that $D$ is the midpoint of arc $AB$. So $DA = DB$. I'll draw those in in blue so we remember they're the same.

copeland 2016-03-05 19:41:16

copeland 2016-03-05 19:41:19
Are there any other lines we might want to add to our diagram to help us figure out more?

copeland 2016-03-05 19:42:16
(That last bit is true since $\angle CAD=\angle BAD$ by the problem statement.)

ompatel99 2016-03-05 19:42:24
IA, IB

MikoTennisPro 2016-03-05 19:42:24
IA and IB?

r_b 2016-03-05 19:42:24
AI

hliu70 2016-03-05 19:42:24
IA and IB maybe?

nosaj 2016-03-05 19:42:24
AI, BI

Quinn 2016-03-05 19:42:24
AI and BI

copeland 2016-03-05 19:42:31
It might make sense to add the angle bisectors from $A$ or $B$, since we know those pass through $I$ as well. I'm going to add the bisector from $A$ into our diagram.

copeland 2016-03-05 19:42:34

copeland 2016-03-05 19:42:36
We have a lot of stuff in our diagram now. Anyone notice anything that looks interesting?

idomath12345 2016-03-05 19:43:07
Isosceles stuff?

MikoTennisPro 2016-03-05 19:43:07
AID is isoceles!

fishy15 2016-03-05 19:43:07
is AID isoceles?

nosaj 2016-03-05 19:43:07
ISOSCELES!

ompatel99 2016-03-05 19:43:07
IAD is a special triangle? isosceles?

copeland 2016-03-05 19:43:11
It looks a lot like $ADI$ is isosceles. Let's see if we can prove it. What would we want to show to prove that?

nosaj 2016-03-05 19:44:23
base angles equal

sxu 2016-03-05 19:44:23
congruent angles

ninjataco 2016-03-05 19:44:23
angle AID = angle DAI

Locust 2016-03-05 19:44:23
The bases angles are congruent

goodbear 2016-03-05 19:44:23
DAI=DIA

hamup1 2016-03-05 19:44:23
$\angle{AID}=\angle{IAD}$

calculus_riju 2016-03-05 19:44:23
the angles must be equal

Aragorn66 2016-03-05 19:44:23
angle IAD=angle AID

copeland 2016-03-05 19:44:33
We'll want to prove $\angle DAI = \angle DIA$ since we want to use angles to learn about lengths.

copeland 2016-03-05 19:44:37
How can we write $\angle DAI$ in a nicer way?

weilunsun28 2016-03-05 19:45:28
<DAB+<LAI

blue8931 2016-03-05 19:45:28
DAL + LAI

brian6liu 2016-03-05 19:45:28
<BAD+<BAI

goodbear 2016-03-05 19:45:28
BAI+BAD

spicyray 2016-03-05 19:45:28
BAD + LAI?

copeland 2016-03-05 19:45:30
We can split it up into $\angle DAB$ and $\angle BAI$. So $\angle DAI = \angle DAB + \angle BAI$.

copeland 2016-03-05 19:45:33
Can we express either of those angles more nicely?

ychen 2016-03-05 19:46:38
BAD = BCD by inscirbed arc

ompatel99 2016-03-05 19:46:38
DAB=DCB by subtended arc

copeland 2016-03-05 19:46:41
Yes, $\angle DAB = \angle DCB$, since they both open to arc $DB$.

copeland 2016-03-05 19:46:43
So $\angle DAI = \angle DCB + \angle BAI$.

copeland 2016-03-05 19:46:50
What's nice about both of those angles?

stan23456 2016-03-05 19:47:57
They are the result of bisected angles

19bobhu 2016-03-05 19:47:57
half of angles A and C

MikoTennisPro 2016-03-05 19:47:57
they both are created by angle bisectors

ychen 2016-03-05 19:47:57
they are both half of the angles on the original triangle

idomath12345 2016-03-05 19:47:57
they are 1/2 of other angles.

nosaj 2016-03-05 19:47:57
They're part of a bisected angle.

copeland 2016-03-05 19:48:00
They're both half of angles in our triangles because of the angle bisectors. $\angle DAI = \dfrac 12 \angle C + \dfrac 12 \angle A$.

copeland 2016-03-05 19:48:01
Can we show that this expression is also equal to $\angle AID$?

copeland 2016-03-05 19:48:15
Is there anything we know about $\angle AID$?

hamup1 2016-03-05 19:48:53
exterior angles

MikoTennisPro 2016-03-05 19:48:53
AID is exterior angle of CIA

sxu 2016-03-05 19:48:53
yah exterior angle

mewtwomew 2016-03-05 19:48:53
exterior angle of triangle cia

Ericaops 2016-03-05 19:48:53
exterior angle

copeland 2016-03-05 19:48:58
$\angle AID$ is external to triangle $AIC$. So $\angle AID = \angle ICA + \angle IAC$. Again, since $AI$ and $CI$ are angle bisectors, that is $\dfrac 12 C + \dfrac 12 A$.

copeland 2016-03-05 19:48:59
So $ADI$ is isosceles.

goodbear 2016-03-05 19:49:07
AID=180-(180-(A/2+C/2))=A/2+C/2

copeland 2016-03-05 19:49:11
By the way, the fact we just prove is really really well known. It's part of what's known colloquially as "Fact 5". So it really helps to know your common geometry configurations -- knowing that fact would have saved us quite a bit of time here.

copeland 2016-03-05 19:49:21
Let's simplify our diagram a bit now that we have that.

copeland 2016-03-05 19:49:22

copeland 2016-03-05 19:49:40
Cool. Now we have a lot more known lengths. We also have a lot of equal angles from the equal arcs. What can you find?

MikoTennisPro 2016-03-05 19:50:44
similar triangles!!!!

nosaj 2016-03-05 19:50:44
also, in cyclic quadrilateral, there are a lot of similar triangles!

eswa2000 2016-03-05 19:50:44
similar triangles

LighteningAB 2016-03-05 19:50:44
similar triangle lengths which match up

copeland 2016-03-05 19:51:07
Great! Do you see any equal angles near our known lengths?

brian6liu 2016-03-05 19:53:34
<ACI=<LAD

Aragorn66 2016-03-05 19:53:34
CBA, CDA

hliu70 2016-03-05 19:53:34
<CDA and <CBA

ThorJames 2016-03-05 19:53:34
<bcd = <bad

SimonSun 2016-03-05 19:53:34
ADI AND ABC

idomath12345 2016-03-05 19:53:34
ABD and ACD.

copeland 2016-03-05 19:53:50
So there are a lot and that is 6 different approaches to this argument.

copeland 2016-03-05 19:53:59
Here is one:

copeland 2016-03-05 19:54:01
We know that $\angle DBA = \angle DCB$ since arc $AD$ equals arc $BD$.

copeland 2016-03-05 19:54:02

copeland 2016-03-05 19:54:06
What does that tell us?

Locust 2016-03-05 19:55:44
Triangles LBD and BCD are similar

jfurf 2016-03-05 19:55:44
$\triangle{CDB}\sim\triangle{BDL}$

brian22 2016-03-05 19:55:44
$\triangle BLD \sim \triangle CBD$

sxu 2016-03-05 19:55:44
DCB similar to DBL

brian6liu 2016-03-05 19:55:44
BLD~CBD

SimonSun 2016-03-05 19:55:44
BDL similar CDB

copeland 2016-03-05 19:55:48
This tells us that $\triangle DBL \sim \triangle DCB$ since they share the angle at $D$.

copeland 2016-03-05 19:55:51
So?

nosaj 2016-03-05 19:57:04
Now we can get $CD$

Locust 2016-03-05 19:57:04

brian6liu 2016-03-05 19:57:04
CD=25/3

MikoTennisPro 2016-03-05 19:57:04
3/5 = 5/(5+x)

kikipet 2016-03-05 19:57:04
(IC + 5)/5 = 5/3

ompatel99 2016-03-05 19:57:04
DL/DB=3/5=DB/DC=5/(5+?)

goodbear 2016-03-05 19:57:04
CD=25/3

brian22 2016-03-05 19:57:04
$\frac{3}{5}=\frac{5}{5+CI}$

atmchallenge 2016-03-05 19:57:04
$\frac{BD}{DL} = \frac{CD}{BD} \implies 25 = 3 CD \implies CD=\frac{25}{3}$

blizzard10 2016-03-05 19:57:04
Therefore CD = $BD \cdot\frac{5}{3}$

memc38123 2016-03-05 19:57:04
5/(IC + 5) = 3/5

copeland 2016-03-05 19:57:07
That means that $\dfrac{DC}{DB} = \dfrac{DB}{DL}$. But we already know some of those...

copeland 2016-03-05 19:57:11
We have $\dfrac{5 + IC}{5} = \dfrac{5}{3}$.

copeland 2016-03-05 19:57:16
So what is $IC$?

blizzard10 2016-03-05 19:57:39
Which equals $\frac{10}{3}$.

brian6liu 2016-03-05 19:57:39
10/3

ompatel99 2016-03-05 19:57:39
10/3

SmallKid 2016-03-05 19:57:39
10/3?

tennis1729 2016-03-05 19:57:39
10/3

SimonSun 2016-03-05 19:57:39
10/3

jfurf 2016-03-05 19:57:39
$\frac{10}{3}$

idomath12345 2016-03-05 19:57:39
10/3

Ericaops 2016-03-05 19:57:39
10/3

ychen 2016-03-05 19:57:39
10/3

copeland 2016-03-05 19:57:41
We get $IC = \dfrac{10}{3}$.

copeland 2016-03-05 19:57:46
So what is our answer?

Aragorn66 2016-03-05 19:58:06
013

goodbear 2016-03-05 19:58:06
10/3,so 013

walnutwaldo20 2016-03-05 19:58:06
013

blue8931 2016-03-05 19:58:06
13

ychen 2016-03-05 19:58:06
013

popcorn1 2016-03-05 19:58:06
$013$

epiphany 2016-03-05 19:58:06
013

weilunsun28 2016-03-05 19:58:06
013

andsun19 2016-03-05 19:58:06
13

SK200 2016-03-05 19:58:06
013

copeland 2016-03-05 19:58:15
Our answer is $10+3=\boxed{13}.$

brian22 2016-03-05 19:58:59
Wait darn how did you add 10 and 3 so fast

copeland 2016-03-05 19:59:03
I converted to 1010+11 base 2 since the digits are smaller and easier to handle.

idomath12345 2016-03-05 19:59:20
Oh!!!!

walnutwaldo20 2016-03-05 19:59:20
Ah that makes sense

LighteningAB 2016-03-05 19:59:24
YAY ONTO SEVEN

popcorn1 2016-03-05 19:59:24
Time for #$7$

copeland 2016-03-05 19:59:27
7. For integers $a$ and $b$ consider the complex number \[ \frac{\sqrt{ab + 2016}}{ab + 100} - \left( \frac{\sqrt{|a + b|}}{ab + 100}\right) i. \] Find the number of ordered pairs of integers $(a,b)$ such that this complex number is a real number.

copeland 2016-03-05 19:59:39
How can we make this number real?

copeland 2016-03-05 19:59:59
Is it enough to have $\left( \dfrac{\sqrt{|a + b|}}{ab + 100}\right) i = 0$?

qwerty733 2016-03-05 20:00:45
imaginary part is 0, real part is real

blizzard10 2016-03-05 20:00:45
There are two options: the second part equals 0, or both are imaginary and they cancel out.

memc38123 2016-03-05 20:00:45
nooooo! You can have the imaginary parts cancel

brian6liu 2016-03-05 20:00:45
or the first term can equal the second

sxu 2016-03-05 20:00:45
no the first part must not be imaginary

Peggy 2016-03-05 20:00:45
no, because the thing inside the sq root has to be positive

walnutwaldo20 2016-03-05 20:00:45
No, left has to be real

qwerty733 2016-03-05 20:00:45
no "real" part must be real (i.e., sqrt geq 0)

copeland 2016-03-05 20:00:48
No, we also need $\sqrt{ab + 2016}$ to be real in that case.

copeland 2016-03-05 20:00:58
Apparently the real part is allowed to be non-real here. . .

copeland 2016-03-05 20:01:10
First things first.

copeland 2016-03-05 20:01:12
If $\left( \dfrac{\sqrt{|a + b|}}{ab + 100}\right) i = 0$, what can we say about $a$ and $b$?

blizzard10 2016-03-05 20:01:49
$a + b = 0$

sxu 2016-03-05 20:01:49
theyr opposites

ryanyz10 2016-03-05 20:01:49
a = -b

algebra_star1234 2016-03-05 20:01:49
a=-b

fishy15 2016-03-05 20:01:49
a=-b

memc38123 2016-03-05 20:01:49
a = -b

qwerty733 2016-03-05 20:01:49
a=-b

bestwillcui1 2016-03-05 20:01:49
they are negativs of each other

SuperMaltese 2016-03-05 20:01:49
a = -b

ChickenOnRage 2016-03-05 20:01:49
a = -b

copeland 2016-03-05 20:01:52
We must have $a = - b$. So our other radical becomes $\sqrt{2016 - a^2}$. When is that real?

memc38123 2016-03-05 20:02:19
when a^2 < 2016

sxu 2016-03-05 20:02:19
when a^2<2016

person754 2016-03-05 20:02:19
a^2<2016

epiphany 2016-03-05 20:02:19
when a^2 is less than 2016

ChickenOnRage 2016-03-05 20:02:19
when $a^2 <= 2016$

walnutwaldo20 2016-03-05 20:02:19
when a^2 < 2016

copeland 2016-03-05 20:02:21
That's real if $a^2 \le 2016$. How many $a$ are there like that?

SuperMaltese 2016-03-05 20:02:44
89

andsun19 2016-03-05 20:02:44
-44 to 44, so 89

fz2012 2016-03-05 20:02:44
89

nukelauncher 2016-03-05 20:02:44
89

ninjataco 2016-03-05 20:02:44
89

nosaj 2016-03-05 20:02:44
-44 to 44

Locust 2016-03-05 20:02:44
For integers, -44 through 44

nosaj 2016-03-05 20:02:44
89

copeland 2016-03-05 20:02:45
$a^2 \le 2016$ if $|a| \le 44$. So are there 89 solutions in this case?

MikoTennisPro 2016-03-05 20:03:10
-44 to 44 => 89 but can't be -10 or 10 b/c then denominator is 0

ninjataco 2016-03-05 20:03:10
89 but denominator needs to be nonzero so 87

MikoTennisPro 2016-03-05 20:03:10
87

ninjataco 2016-03-05 20:03:10
no, a is not +/- 10

andsun19 2016-03-05 20:03:10
nope., ab is not equal to -100

Locust 2016-03-05 20:03:10
no, a and b cannot be a combination of 10,-10

fz2012 2016-03-05 20:03:10
no, a=+-10 does not work

ompatel99 2016-03-05 20:03:10
But remember that ab cannot equal -100. So a cannot be 10 or -10. 87 solutions

azmath333 2016-03-05 20:03:10
ab cant be -100

copeland 2016-03-05 20:03:12
Not quite! Some of those solutions make our denominator $ab + 100 = 0$.

copeland 2016-03-05 20:03:13
We have to skip $a = -10$ and $a = 10$. For a total of 87 solutions here.

copeland 2016-03-05 20:03:17
Okay, so our answer is 87?

ChickenOnRage 2016-03-05 20:03:42
no

blizzard10 2016-03-05 20:03:42
No!

hliu70 2016-03-05 20:03:42
NOOOO

kikipet 2016-03-05 20:03:42
no

nukelauncher 2016-03-05 20:03:42
no one more case

ChickenOnRage 2016-03-05 20:03:42
no; the case when the entire expression evaluates to 0

bestwillcui1 2016-03-05 20:03:42
NO!

blizzard10 2016-03-05 20:03:42
There's a second part to consider. Both could be imaginary and cancel out.

mssmath 2016-03-05 20:03:44
Apparently stuff can cancel?

copeland 2016-03-05 20:03:47
No! We can also cancel the imaginary parts from both terms. That is, we can also have $\sqrt{ab+2016} = i\sqrt{|a+b|}$.

copeland 2016-03-05 20:03:51
Squaring both sides, we need to have $ab + 2016 = a + b$ (with $a + b < 0$) or $ab + 2016 = -a - b$ (with $a + b > 0$).

copeland 2016-03-05 20:03:53
What can we do with those?

atmchallenge 2016-03-05 20:04:22
SFFT

ompatel99 2016-03-05 20:04:22
SFFT

jfurf 2016-03-05 20:04:22
SFFT!!!

MikoTennisPro 2016-03-05 20:04:22
SFFT?

brian6liu 2016-03-05 20:04:22
sfft

Math1331Math 2016-03-05 20:04:22
SFFT

SimonSun 2016-03-05 20:04:22
SFFT

qwerty733 2016-03-05 20:04:22
SFFT

blue8931 2016-03-05 20:04:22
SFFT

andsun19 2016-03-05 20:04:22
sfft

nosaj 2016-03-05 20:04:22
I smell sfft

fz2012 2016-03-05 20:04:22
use SFFT and solve?

copeland 2016-03-05 20:04:27
SFFT! This is a classic set-up for Simon's favorite factoring trick: If you see terms like $xy, x,$ and $y$ in your equation, you probably want to think about using Simon's.

copeland 2016-03-05 20:04:31
Here's a link: http://www.artofproblemsolving.com/wiki/index.php/Simon's_Favorite_Factoring_Trick

copeland 2016-03-05 20:04:34
Let's deal with $ab + 2016 = a + b$ first. How does this factor?

brian6liu 2016-03-05 20:05:15
(a-1)(b-1)=-2015

algebra_star1234 2016-03-05 20:05:15
(a-1)(b-1) = -2015

andsun19 2016-03-05 20:05:15
(a-1)(b-1)=2015

blizzard10 2016-03-05 20:05:15
$(a-1)(b-1) + 2015 = 0$

mathguy5041 2016-03-05 20:05:15
$(a-1)(b-1)=5*13*31$

kikipet 2016-03-05 20:05:15
(a-1)(b-1)=-2015

memc38123 2016-03-05 20:05:15
(A-1)(b-1)=2015

brian22 2016-03-05 20:05:15
$(a-1)(b-1)=-2015$

SuperMaltese 2016-03-05 20:05:15
(a-1)(b-1) = -2015

copeland 2016-03-05 20:05:18
This factors as $(a-1)(b-1) = - 2015 = - 5 \cdot 13 \cdot 31$.

copeland 2016-03-05 20:05:19
How many such pairs are there?

tennis1729 2016-03-05 20:06:13
8

ninjataco 2016-03-05 20:06:13
8

andsun19 2016-03-05 20:06:13
8

idomath12345 2016-03-05 20:06:13
8.

person754 2016-03-05 20:06:13
8

ChickenOnRage 2016-03-05 20:06:13
8

memc38123 2016-03-05 20:06:13
8

idomath12345 2016-03-05 20:06:13
But 1/2 don't work.

qwerty733 2016-03-05 20:06:13
8

Locust 2016-03-05 20:06:13
8, but some of them don't satisfy a+b > 0

atmchallenge 2016-03-05 20:06:13
$8$ since $a+b<0$

copeland 2016-03-05 20:06:18
Since we must have $a + b < 0$, the signs are already determined (the negative must go on the bigger factor). So we simply need to assign 5, 13, and 31 to $a$ and $b$.

copeland 2016-03-05 20:06:20
There are 2 options for where we put the 5, 2 options for the 13, and again 2 options for the 31. That gives us $2\cdot 2 \cdot 2 = 8$ options.

copeland 2016-03-05 20:06:22
Okay, now let's deal with the other case: $ab + 2016 = -a - b$. How does this factor?

brian22 2016-03-05 20:06:53
$(a+1)(b+1)=-2015$

atmchallenge 2016-03-05 20:06:53
$(a+1)(b+1)=-2015$

brian6liu 2016-03-05 20:06:53
(a+1)(b+1)=-2015

memc38123 2016-03-05 20:06:53
(a+1)(b+1)=-2015

ninjataco 2016-03-05 20:06:53
(a+1)(b+1) = -2015

algebra_star1234 2016-03-05 20:06:53
(a+1)(b+1)=-2015

kikipet 2016-03-05 20:06:53
(a+1)(b+1)=-2015

copeland 2016-03-05 20:06:57
This factors as $(a + 1)(b+1) = -2015 = -5 \cdot 13 \cdot 31$.

copeland 2016-03-05 20:06:57
How many such pairs are there?

epiphany 2016-03-05 20:07:34
8

idomath12345 2016-03-05 20:07:34
8 again.

popcorn1 2016-03-05 20:07:34
8

nukelauncher 2016-03-05 20:07:34
8

brian22 2016-03-05 20:07:34
Same, $8$

fz2012 2016-03-05 20:07:34
8

atmchallenge 2016-03-05 20:07:34
$8$ again, since $a+b>0$

brian6liu 2016-03-05 20:07:34
8

jfurf 2016-03-05 20:07:34
$8$ again! Still have to watch out for the conditions, though :)

andsun19 2016-03-05 20:07:34
8

copeland 2016-03-05 20:07:37
Since we must have $a + b > 0$, the signs are already determined (the negative must go on the smaller factor). So we simply need to assign 5, 13, and 31 to $a$ and $b$.

copeland 2016-03-05 20:07:39
Just like before, there are 8 ways to do that.

copeland 2016-03-05 20:07:41
So, how many pairs are there in total?

fz2012 2016-03-05 20:08:16
so there are a total of 87+8+8=103 solutions

memc38123 2016-03-05 20:08:16
103

algebra_star1234 2016-03-05 20:08:16
103

mathguy5041 2016-03-05 20:08:16
87+8+8=$103$

nosaj 2016-03-05 20:08:16
103

andsun19 2016-03-05 20:08:16
16+87=103?

kikipet 2016-03-05 20:08:16
$\boxed{103}$

blue8931 2016-03-05 20:08:16
103

person754 2016-03-05 20:08:16
103

atmchallenge 2016-03-05 20:08:16
$87+8+8=\boxed{103}$.

walnutwaldo20 2016-03-05 20:08:16
16+87 = 103

copeland 2016-03-05 20:08:18
In total, we have $87 + 8 + 8 = \boxed{103}$ pairs.

copeland 2016-03-05 20:08:23
And as a public service announcement: Please do not learn anything about square roots from this problem. You should never, ever take the square root of a negative number and then just assume something about its sign: Any time you write $\sqrt{-x}=i\sqrt x$, you are very likely to make a huge mess. This is because you quickly get into situations like: \[\sqrt{x}=\sqrt{-(-x)}=i\sqrt{-x}=i^2\sqrt{x}=-\sqrt{x}.\]

copeland 2016-03-05 20:08:35
In fact, you are typically better off thinking about such an expression as having two values (like in the quadratic formula), and in that case, this problem deserves a footnote that we're looking for $a$ and $b$ where one of the values is zero.

fishy15 2016-03-05 20:09:08
the problems are getting longer

copeland 2016-03-05 20:09:10
I'm more worried about the solutions.

copeland 2016-03-05 20:09:17
8. For a permutation $p = (a_1, a_2, \ldots, a_9)$ of the digits $1,2,\ldots, 9$, let $s(p)$ denote the sum of the three 3-digit numbers $a_1a_2a_3, a_4a_5a_6,$ and $a_7a_8a_9$. Let $m$ be the minimum value of $s(p)$ subject to the condition that the units digit of $s(p)$ is 0. Let $n$ denote the number of permutations $p$ with $s(p) = m$. Find $|m - n|$.

copeland 2016-03-05 20:09:35
Case in point: this is a long problem.

copeland 2016-03-05 20:09:36
It's usually a good plan to deal with restrictive conditions first. What's a restrictive condition here?

idomath12345 2016-03-05 20:10:25
Go for the smallest.

blue8931 2016-03-05 20:10:25
units digit is 0

sxu 2016-03-05 20:10:25
units digit is0

janabel 2016-03-05 20:10:25
the units digit is 0

qwerty733 2016-03-05 20:10:25
units digit is 0

MikoTennisPro 2016-03-05 20:10:25
units digit is zero

algebra_star1234 2016-03-05 20:10:25
the units digit is 0

simon1221 2016-03-05 20:10:25
units digit = 0?

snowumbrella 2016-03-05 20:10:25
units digit of s(p) must be 0

illogical_21 2016-03-05 20:10:25
units digit is 0

copeland 2016-03-05 20:10:28
We probably want to deal with the fact that the units digit of $s(p)$ is 0 first. What does that tell us about $p$?

brian22 2016-03-05 20:11:17
$s(p) \pmod{10} \equiv 0$

hamup1 2016-03-05 20:11:17
$a_3+a_6+a_9=10$ or $a_3+a_6+a_9=20$

blue8931 2016-03-05 20:11:17
so you know that $a_3 + a_6 + a_9 = 10$ or $20$

brian22 2016-03-05 20:11:17
$a_3+a_6+a_9=10$ or $20$

memc38123 2016-03-05 20:11:17
a(3)+a(6)+a(9) = 10k

copeland 2016-03-05 20:11:20
Since \[a_1a_2a_3 + a_4a_5a_6 + a_7a_8a_9 = 100(a_1 + a_4 + a_7) + 10(a_2 + a_5 + a_8) + (a_3 + a_6 + a_9),\] this tells us that $a_3 + a_6 + a_9$ must be a multiple of 10.

copeland 2016-03-05 20:11:21
Can we get $a_3 + a_6 + a_9 = 10$?

blue8931 2016-03-05 20:12:39
$(7, 2, 1), (6, 3, 1), (5, 4, 1),$ or $(5, 3, 2)$

memc38123 2016-03-05 20:12:39
2+3+5

Aragorn66 2016-03-05 20:12:39
4,5,1

atmchallenge 2016-03-05 20:12:39
$1+2+7=1+3+6=1+4+5=2+3+5=10$

copeland 2016-03-05 20:12:42
Of course. For example, 2+3+5=10.

copeland 2016-03-05 20:12:46
Can we get $a_3 + a_6 + a_9 = 20$?

brian22 2016-03-05 20:13:39
Yes: $(3,8,9)$, $(4,7,9)$, $(5,6,9)$, $(5,7,8)$

epiphany 2016-03-05 20:13:39
yes, 9+8+3, 9+7+4, and 8+7+5

qwerty733 2016-03-05 20:13:39
Yes, e.g. 9+8+3

hamup1 2016-03-05 20:13:39
9+8+3

atmchallenge 2016-03-05 20:13:39
$9+8+3=9+7+4=9+6+5=8+7+5=20$

brian6liu 2016-03-05 20:13:39
389, 479, 569, 578

mssmath 2016-03-05 20:13:39
9+7+4

snowumbrella 2016-03-05 20:13:39
(9, 8, 3) (9, 7, 4) (9, 6, 5) (8, 7, 5)

copeland 2016-03-05 20:13:43
Yes. There are several ways.

copeland 2016-03-05 20:13:45
Up to permutations, the four ways to get 20 are for $(a_3,a_6,a_9)$ to be:
\begin{align*}
&(9,8,3) &(8,7,5) \\
&(9,7,4) \\
&(9,6,5)
\end{align*}

copeland 2016-03-05 20:13:47
Can we get $a_3 + a_6 + a_9 = 30$?

tennis1729 2016-03-05 20:14:03
no

Locust 2016-03-05 20:14:03
No

calculus_riju 2016-03-05 20:14:03
no

weilunsun28 2016-03-05 20:14:03
nope

jfurf 2016-03-05 20:14:03
No

kikipet 2016-03-05 20:14:03
no, it's too big

andsun19 2016-03-05 20:14:03
nope.

Ericaops 2016-03-05 20:14:03
no

thequantumguy 2016-03-05 20:14:03
no

GoldenPhi1618033 2016-03-05 20:14:03
No.

19bobhu 2016-03-05 20:14:03
no

SimonSun 2016-03-05 20:14:03
no

blizzard10 2016-03-05 20:14:03
No!

copeland 2016-03-05 20:14:05
No, that's too big. The biggest $a_3 + a_6 + a_9$ can be is 9+8+7 =24.

copeland 2016-03-05 20:14:07
Okay, what's the next restrictive condition we care about?

andsun19 2016-03-05 20:14:26
MINIMUM

snowumbrella 2016-03-05 20:14:26
m must be minimum

person754 2016-03-05 20:14:26
minimum value

weilunsun28 2016-03-05 20:14:26
minimize m

janabel 2016-03-05 20:14:26
minimum

blue8931 2016-03-05 20:14:26
we want the minimum value

copeland 2016-03-05 20:14:29
The next interesting restriction is that we need our permutation to achieve the minimum sum.

copeland 2016-03-05 20:14:33
How are we going to minimize \[a_1a_2a_3 + a_4a_5a_6 + a_7a_8a_9 = 100(a_1 + a_4 + a_7) + 10(a_2 + a_5 + a_8) + (a_3 + a_6 + a_9)?\]

mathguy5041 2016-03-05 20:15:08
Make hundreds digits $1, 2, 3$

qwerty733 2016-03-05 20:15:08
Minimize hundreds digits

andsun19 2016-03-05 20:15:08
well, a1 + a4 + a7, should be minimized

19bobhu 2016-03-05 20:15:08
make a1 a4 and a7 as small as possible?

jfurf 2016-03-05 20:15:08
Minimize $a_1+a_4+a_7$, first.

brian6liu 2016-03-05 20:15:08
minimize a_1+a_4+a_7

kikipet 2016-03-05 20:15:08
make $a_1, a_2, a_3$ be 1, 2, and 3

Locust 2016-03-05 20:15:08
first minimize $a_1,a_4,a_7$

fishy15 2016-03-05 20:15:08
minimize a_1, a_4, and a_7

nosaj 2016-03-05 20:15:08
hundreds should be 1,2,3

math1012 2016-03-05 20:15:08
make a_1, a_4, a_7 all 1,2,3

nukelauncher 2016-03-05 20:15:08
a(1) + a(4) +a(7) has to be 1, 2, and 3

copeland 2016-03-05 20:15:14
We'll need to make $a_1+a_4+a_7$ as small as possible. Up to permutations, we must have $(a_1,a_4,a_7) = (1,2,3)$.

copeland 2016-03-05 20:15:17
Note that replacing any of 1,2,3 with a number $k > 3$ will swap more 100's for fewer 10's or 1's, thus increasing the sum.

copeland 2016-03-05 20:16:03
Now if we use 1, 2, and 3 in the hundreds digit, we can't make 10 from the ones digit since we would need a or or a 2 for that. So we'll have to have $a_3+a_6+a_9 = 20$.

copeland 2016-03-05 20:16:05
So what will our minimum be?

weilunsun28 2016-03-05 20:16:39
810

MikoTennisPro 2016-03-05 20:16:39
810

jrl_ct 2016-03-05 20:16:39
810

qwerty733 2016-03-05 20:16:39
810

SimonSun 2016-03-05 20:16:39
810

nukelauncher 2016-03-05 20:16:39
810!

brian6liu 2016-03-05 20:16:39
810

algebra_star1234 2016-03-05 20:16:39
810

calculus_riju 2016-03-05 20:16:39
810

ninjataco 2016-03-05 20:16:39
810

copeland 2016-03-05 20:16:42
Since $a_1 + \cdots + a_9 = 45$ and $a_1 + a_4 + a_7 = 6$, $a_3 + a_6 + a_9 = 20$, we must have $a_2 + a_5 + a_8 = 19$. So our minimum sum is \[ 100 \cdot 6 + 10 \cdot 19 + 20 = 600 + 190 + 20 = 810.\]

copeland 2016-03-05 20:16:43
How many permutations achieve this sum?

Aragorn66 2016-03-05 20:17:31
a lot

copeland 2016-03-05 20:17:35
Indeed. Let's work it out.

copeland 2016-03-05 20:17:43
How many choices do we have for $(a_1,a_4,a_7)$?

fz2012 2016-03-05 20:18:08
6

MikoTennisPro 2016-03-05 20:18:08
3!

nukelauncher 2016-03-05 20:18:08
3!=6

andsun19 2016-03-05 20:18:08
3! = 6

brian6liu 2016-03-05 20:18:08
6

blue8931 2016-03-05 20:18:08
3! ways to order them

PurplePancakes 2016-03-05 20:18:08
6

simon1221 2016-03-05 20:18:08
3! = 6

copeland 2016-03-05 20:18:10
We have 3! choices for $(a_1, a_4, a_7)$ because it must be a permutation of $(1,2,3)$.

copeland 2016-03-05 20:18:11
How many choices do we have for $(a_3,a_6,a_9)$?

PurplePancakes 2016-03-05 20:18:52
6*3

hamup1 2016-03-05 20:18:52
three different choices, times 6 for permutations

nosaj 2016-03-05 20:18:52
3*6 = 18

MikoTennisPro 2016-03-05 20:18:52
6*3

person754 2016-03-05 20:18:52
18

calculus_riju 2016-03-05 20:18:52
3.3!

andsun19 2016-03-05 20:18:52
(9,7,4)(8,7,5)(5,6,9). So 3*3!

jwlw2014 2016-03-05 20:18:52
3*3!

copeland 2016-03-05 20:18:56
We can use permutations of any of $(9,7,4), (9,6,5), (8,7,5)$. For a total of 3*3!.

copeland 2016-03-05 20:18:57
How many choices for $(a_2,a_5,a_8)$?

mcmcphie 2016-03-05 20:19:32
6

MikoTennisPro 2016-03-05 20:19:32
3!, the remaining 3 #s

mathguy5041 2016-03-05 20:19:32
3!

nukelauncher 2016-03-05 20:19:32
3!=6

sxu 2016-03-05 20:19:32
3!'

idomath12345 2016-03-05 20:19:32
3!

copeland 2016-03-05 20:19:34
At this point, the three numbers available are determined. So we only get to choose their order. We can do that in 3! ways.

copeland 2016-03-05 20:19:35
So in total, we have 3*3!*3!*3! ways to achieve the minimum value.

copeland 2016-03-05 20:19:36
So what is our answer?

mewtwomew 2016-03-05 20:20:31
162

brian6liu 2016-03-05 20:20:31
162

idomath12345 2016-03-05 20:20:31
810-648

andsun19 2016-03-05 20:20:31
810-3*216=162

kikipet 2016-03-05 20:20:31
162

PurplePancakes 2016-03-05 20:20:31
810-3*6*6*6

blizzard10 2016-03-05 20:20:31
162

blue8931 2016-03-05 20:20:31
$810 - 648 = \boxed{162}$

algebra_star1234 2016-03-05 20:20:31
162

nukelauncher 2016-03-05 20:20:31
810-648=162!

jfurf 2016-03-05 20:20:31
$648$ ways so $810-648=\boxed{162}$

Aragorn66 2016-03-05 20:20:31
162

lleB_ocaT 2016-03-05 20:20:31
162

walnutwaldo20 2016-03-05 20:20:31
810-648= 162

lleB_ocaT 2016-03-05 20:20:31
162

copeland 2016-03-05 20:20:33
Multiplying this out, we have 3*3!*3!*3! = 648. So our answer is $810-648=\boxed{162}.$

copeland 2016-03-05 20:20:54
Okey-dokey!

copeland 2016-03-05 20:20:56
What now?

idomath12345 2016-03-05 20:21:13
9.

mathguy5041 2016-03-05 20:21:13
Onto 9?

hliu70 2016-03-05 20:21:13
#9!

Locust 2016-03-05 20:21:13
Problem 1+8

SuperMaltese 2016-03-05 20:21:13
number 8!!

brian22 2016-03-05 20:21:13
I think 9 comes after 8

copeland 2016-03-05 20:21:15
9. Triangle $ABC$ has $AB=40,$ $AC=31,$ and $\sin A=\dfrac15.$ This triangle is inscribed in rectangle $AQRS$ with $B$ on $\overline {QR}$ and $C$ on $\overline{RS}.$ Find the maximum possible area of $AQRS.$

copeland 2016-03-05 20:21:16
So what's the first step in a geometry problem?

Peggy 2016-03-05 20:21:36
diagram

SimonSun 2016-03-05 20:21:36
diagram

MikoTennisPro 2016-03-05 20:21:36
diagram

jrl_ct 2016-03-05 20:21:36
Diagram

nukelauncher 2016-03-05 20:21:36
diagram!!!

algebra_star1234 2016-03-05 20:21:36
DIAGRAM

blizzard10 2016-03-05 20:21:36
Diagram, of course!

blue8931 2016-03-05 20:21:36
draw a picture

tennis1729 2016-03-05 20:21:36
picture

ompatel99 2016-03-05 20:21:36
A diagram

hamup1 2016-03-05 20:21:36
diagram

Geode359 2016-03-05 20:21:36
draw a diagram

andsun19 2016-03-05 20:21:36
Draw a diagram.

copeland 2016-03-05 20:21:39
In most geometry problems, the first step is to draw a nice, accurate diagram.

copeland 2016-03-05 20:21:41
Here is a nice diagram:

copeland 2016-03-05 20:21:42

copeland 2016-03-05 20:21:45
. . . and here's a nice accurate diagram:

copeland 2016-03-05 20:21:46

copeland 2016-03-05 20:21:53
Unfortunately, this triangle is obtuse. It's impossible for $A$ and $B$ to lie on the same side of line $SR$.

copeland 2016-03-05 20:21:59
Hopefully, if you did draw an accurate diagram, it didn't throw you off your game too much.

blue8931 2016-03-05 20:22:04
ugh MAA...

andsun19 2016-03-05 20:22:10
So this problem is defective?

copeland 2016-03-05 20:22:14
Incidentally, the problem is perfectly legit and I believe it is what the AMC intended:

copeland 2016-03-05 20:22:18
9.b Triangle $ABC$ has $AB=40,$ $AC=31,$ and $\sin A=\dfrac15.$ Draw rectangle $AQRS$ with $B$ on line $QR$ and $C$ on line $RS.$ Find the maximum possible area of $AQRS.$

copeland 2016-03-05 20:22:29
Let's solve this problem instead.

copeland 2016-03-05 20:22:36
What should we focus on in the diagram?

brian6liu 2016-03-05 20:23:18
finding AQ*AS

nosaj 2016-03-05 20:23:18
lengths AQ, and AS, since their product gives the area

fishy15 2016-03-05 20:23:18
side lengtha

copeland 2016-03-05 20:23:20
That's a noble goal. What do we want to use for that?

andsun19 2016-03-05 20:23:55
trig

brian6liu 2016-03-05 20:23:55
trig

nosaj 2016-03-05 20:23:55
trig!

blue8931 2016-03-05 20:23:55
trig

PurplePancakes 2016-03-05 20:23:55
trig bash

copeland 2016-03-05 20:23:56
Trig needs angles, yo.

copeland 2016-03-05 20:24:04
What are the good angles?

hliu70 2016-03-05 20:24:51
sin A = 1/5 ?

ChickenOnRage 2016-03-05 20:24:51
A

hamup1 2016-03-05 20:24:51
$\angle{SAC},\angle{BAQ}$

brian6liu 2016-03-05 20:24:51
<SAC and <BAQ

PurplePancakes 2016-03-05 20:24:51
<CAB is basically given

mathguy5041 2016-03-05 20:24:51
Angle BAC=arcsin (0.2)

ninjataco 2016-03-05 20:24:51
the ones adjacent to angle A?

nosaj 2016-03-05 20:24:51
BAQ, CAS

copeland 2016-03-05 20:24:54
Let's think about the angles at $A$ since those determine the rest of the diagram. I've marked $\angle SAC$ and $\angle QAB$, as well as $\angle BAC$:

copeland 2016-03-05 20:24:56

copeland 2016-03-05 20:24:58
We know that $\sin z=\dfrac15$. We also know that $x+y+z=90^\circ.$

copeland 2016-03-05 20:25:17
To find the area we need the side length. What is $AS$ in terms of these angles?

Locust 2016-03-05 20:26:07
31 * cos x

snowumbrella 2016-03-05 20:26:07
31cosx

algebra_star1234 2016-03-05 20:26:07
AC cos x

SuperMaltese 2016-03-05 20:26:07
AS = ACcosx ?

brian6liu 2016-03-05 20:26:07
31cosx

memc38123 2016-03-05 20:26:07
cosx*AC

TheRealDeal 2016-03-05 20:26:07
cos(x)*31

fz2012 2016-03-05 20:26:07
31cosx

qwerty733 2016-03-05 20:26:07
31cosx

copeland 2016-03-05 20:26:09
$AS=AC\cos x=31\cos x$

copeland 2016-03-05 20:26:10
What is $AQ?$

snowumbrella 2016-03-05 20:26:52
40cosy

qwerty733 2016-03-05 20:26:52
40cosy

PurplePancakes 2016-03-05 20:26:52
40cos(y)

sxu 2016-03-05 20:26:52
40cosy

fz2012 2016-03-05 20:26:52
40cosy

algebra_star1234 2016-03-05 20:26:52
40cosy

andsun19 2016-03-05 20:26:52
40*cosy?

copeland 2016-03-05 20:26:57
$AQ=AB\cos y=40\cos y$.

copeland 2016-03-05 20:27:00
And we want the area which is
\[\text{Area}=AS\cdot AQ=(31\cos x)(40\cos y)=1240\cos x\cos y.\]

copeland 2016-03-05 20:27:06
That product is scary. There's no obvious way to minimize it right now. Can we write it any differently?

Ericaops 2016-03-05 20:27:46
product to sum

andsun19 2016-03-05 20:27:46
product of cosines

jfurf 2016-03-05 20:27:46
Yes! Use product to sum formula!

memc38123 2016-03-05 20:27:46
product to sum identity

mathguy5041 2016-03-05 20:27:46
Sum to product formulas!!!!

brian6liu 2016-03-05 20:27:46
product to sum

copeland 2016-03-05 20:27:49
We have the product-to-sum formula. What is $\cos x\cos y$ equal to?

PurplePancakes 2016-03-05 20:28:11
1/2(something I don't memorize because it's easy to derive)

copeland 2016-03-05 20:28:14
agreed.

andsun19 2016-03-05 20:29:13
use cosine sums to derive

weilunsun28 2016-03-05 20:29:13
lets derive it!

copeland 2016-03-05 20:29:15
That's better done alone. In a quiet, dark room. Thanks, though.

qwerty733 2016-03-05 20:29:28
1/2 (cos(x+y)+cos(x-y))

brian6liu 2016-03-05 20:29:28
(cos(x+y)+cos(x-y))/2

nosaj 2016-03-05 20:29:28
$\frac{1}{2}\left[\cos(x+y) + \cos(x-y)\right]$

Ericaops 2016-03-05 20:29:28
1/2(cos(x+y)+cos(x-y))

algebra_star1234 2016-03-05 20:29:28
1/2 (cos(x+y) + cos(x-y))

copeland 2016-03-05 20:29:31
\[\cos x\cos y=\frac12(\cos(x+y)+\cos(x-y)).\]

copeland 2016-03-05 20:29:33
In our case, \[\text{Area}=620(\cos (x+y)+\cos(x-y)).\]

copeland 2016-03-05 20:29:34
Do we know either of those?

PurplePancakes 2016-03-05 20:30:08
x+y=90-z

memc38123 2016-03-05 20:30:08
x+Y = 90-Z

qwerty733 2016-03-05 20:30:08
yeah cos(x+y)

fishy15 2016-03-05 20:30:08
x+y=90-z

brian6liu 2016-03-05 20:30:17
cos(x+y)=sin(A)=1/5

fz2012 2016-03-05 20:30:17
cos (x+y)=1/5

nukelauncher 2016-03-05 20:30:17
yes! cos(x+y) = sin z = 1/5!

bestwillcui1 2016-03-05 20:30:17
cos(x+y) = sin(z) = 1/5

PurplePancakes 2016-03-05 20:30:17
cos(x+y)=cos(90-z)=sin(z)

copeland 2016-03-05 20:30:20
We know $\cos(x+y)=\cos(90^\circ-z)=\sin(z)=\dfrac15,$ so\[\text{Area}=620\left(\frac15+\cos(x-y)\right).\]

copeland 2016-03-05 20:30:21
And how can we control $\cos(x-y)?$

fz2012 2016-03-05 20:31:31
Make x and y equal to maximize

brian6liu 2016-03-05 20:31:31
it's maximized when x=y and cos(x-y)=cos(0)=1

PurplePancakes 2016-03-05 20:31:31
maximized when x=y

algebra_star1234 2016-03-05 20:31:31
we can maximize by letting x=y

sxu 2016-03-05 20:31:31
make it 1

bestwillcui1 2016-03-05 20:31:31
x=y maximizes it

andsun19 2016-03-05 20:31:31
well, max = 1

nosaj 2016-03-05 20:31:31
it's maximized at cos(0) = 1

snowumbrella 2016-03-05 20:31:31
max possible is 1?

qwerty733 2016-03-05 20:31:31
Well is x=y then its 1

brian22 2016-03-05 20:31:31
Well it can be up to $1$

copeland 2016-03-05 20:31:34
If we make $x=y$ then this is equal to 1. That's definitely the maximum value. What is the final answer?

brian6liu 2016-03-05 20:32:23
744

andsun19 2016-03-05 20:32:23
744

mssmath 2016-03-05 20:32:23
744

nosaj 2016-03-05 20:32:23
744

blue8931 2016-03-05 20:32:23
744

blizzard10 2016-03-05 20:32:23
744

Axolotl 2016-03-05 20:32:23
744

MikoTennisPro 2016-03-05 20:32:23
744

copeland 2016-03-05 20:32:26
When $x=y,$\[\text{Area}=620\left(\frac15+1\right)=\frac{620\cdot6}5=\frac{3720}5=\boxed{744}.\]

LighteningAB 2016-03-05 20:34:19
NUMBER TENNNN

ThorJames 2016-03-05 20:34:19
10.

mathguy5041 2016-03-05 20:34:19
10

blizzard10 2016-03-05 20:34:19
Problem 10!

memc38123 2016-03-05 20:34:22
onward

copeland 2016-03-05 20:34:28
10. A strictly increasing sequence of positive integers $a_1, a_2, a_3, \ldots$ has the property that for every positive integer $k$, the subsequence $a_{2k-1}, a_{2k}, a_{2k+1}$ is geometric and the subsequence $a_{2k}, a_{2k+1},a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$. Find $a_1$.

copeland 2016-03-05 20:34:38
Hey, I've seen this problem before...

copeland 2016-03-05 20:34:40
2004II/9: A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all $n\ge1$, the terms $a_{2n-1}$, $a_{2n}$, $a_{2n+1}$ are in geometric progression, and the terms $a_{2n}$, $a_{2n+1}$, and $a_{2n+2}$ are in arithmetic progression. Let $a_n$ be the greatest term in this sequence that is less than 1000. Find $n+a_n$.

copeland 2016-03-05 20:34:47
Well, we have information about the end, so let's start there. What type of sequence is $a_{11}, a_{12}, a_{13}$?

ryanyz10 2016-03-05 20:35:12
geo

kikipet 2016-03-05 20:35:12
geometric

blue8931 2016-03-05 20:35:12
geometric

MikoTennisPro 2016-03-05 20:35:12
geometric

brian6liu 2016-03-05 20:35:12
geometric

atmchallenge 2016-03-05 20:35:12
Geometric

Axolotl 2016-03-05 20:35:12
geometric

PurplePancakes 2016-03-05 20:35:12
geometric

copeland 2016-03-05 20:35:14
That's a geometric sequence, so let's write $a_{12} = 2016r$ and $a_{11} = 2016 r^2$ for some $r = \dfrac mn$.

copeland 2016-03-05 20:35:19
Is there anything interesting we can say about $m$ or $n$?

copeland 2016-03-05 20:35:24
(Assume that $m$ and $n$ are relatively prime since this is likely to be about divisibility.)

mewtwomew 2016-03-05 20:36:04
n^2 is a factor of 2016

ryanyz10 2016-03-05 20:36:04
n is a factor of 2016 since all numbers are ints

Ericaops 2016-03-05 20:36:04
n^2 is a factor of 2016

qwerty733 2016-03-05 20:36:04
n^2 is a factor of 2016

brian6liu 2016-03-05 20:36:04
n^2 divides 2016

copeland 2016-03-05 20:36:09
Since $a_{11}$ is an integer, $n^2$ divides 2016. What is the prime factorization of 2016?

atmchallenge 2016-03-05 20:37:06
$2016=2^5\cdot 3^2 \cdot 7$

nukelauncher 2016-03-05 20:37:06
2^5*3^2*7

qwerty733 2016-03-05 20:37:06
2^5 * 3^2 *7

andsun19 2016-03-05 20:37:06
2^5*3^2*7

brian6liu 2016-03-05 20:37:06
2^5*3^2*7

sxu 2016-03-05 20:37:06
2^5 3^2 7 or something

Locust 2016-03-05 20:37:06

AlphaPi17 2016-03-05 20:37:06
2^5 * 3^2 * 7

fishy15 2016-03-05 20:37:06
2^5*3^2*7

copeland 2016-03-05 20:37:08
We have $2016 = 2^5 \cdot 3^2 \cdot 7$. So $n$ must be a divisor of $2^2 \cdot 3 =12$.

copeland 2016-03-05 20:37:18
And what else do we know about $m$ and $n$?

ryanyz10 2016-03-05 20:37:38
m < n

ChickenOnRage 2016-03-05 20:37:38
m/n < 1

Locust 2016-03-05 20:37:38
m<n

nosaj 2016-03-05 20:37:38
m<n

qwerty733 2016-03-05 20:37:38
m<n

copeland 2016-03-05 20:38:08
We need the sequence to decrease from right to left, so $r = \frac m{12}$ where $m$ is one of 1,2,3,...,11.

copeland 2016-03-05 20:38:11
Let's try our next three-term sequence: $a_{10}, a_{11}, a_{12}$ is arithmetic. What equation does that give us?

blue8931 2016-03-05 20:38:36
how do we know that n=12 for sure

copeland 2016-03-05 20:38:45
Sorry, we don't know $n=12$.

copeland 2016-03-05 20:38:48
We know $n$ divides 12.

copeland 2016-03-05 20:39:09
However, we know that the FRACTION $\frac mn$ is equal to something over 12.

copeland 2016-03-05 20:39:41
So if $m=2$ and $n=3$, we still have $\dfrac mn=\dfrac8{12}.$

copeland 2016-03-05 20:40:00
What equation do we get for $a_{10}$?

copeland 2016-03-05 20:41:01
We have $a_{12} = 2016r$ and $a_{11} = 2016 r^2.$

qwerty733 2016-03-05 20:42:06
a_10=2(2016(m/12)^2-2016(m/12)

atmchallenge 2016-03-05 20:42:06
$a_{10}=4032r^2-2016r$.

PurplePancakes 2016-03-05 20:42:06
2016r-2(2016r-2016r^2)=2*2016r^2-2016r

calculus_riju 2016-03-05 20:42:06
then $a_{10}= 4032r^2 - 2016r

ninjataco 2016-03-05 20:42:06
2*2016r^2 - 2016r

walnutwaldo20 2016-03-05 20:42:06
2016r(2r-1)

nukelauncher 2016-03-05 20:42:06
a(10)=4032r^2-2016r

copeland 2016-03-05 20:42:10
We get \[a_{10} = 2a_{11} - a_{12} = 2 \cdot 2016 r^2 - 2016r = 2016r(2r - 1).\]

copeland 2016-03-05 20:42:13
What does that tell us about $r$?

brian6liu 2016-03-05 20:42:46
r>1/2

mssmath 2016-03-05 20:42:46
r>1/2

hamup1 2016-03-05 20:42:46
greater than 1/2

calculus_riju 2016-03-05 20:42:46
r>.5

sophiazhi 2016-03-05 20:42:46
>1/2

copeland 2016-03-05 20:42:51
Since $a_{10}$ is a positive integer, this tells us that $r > \dfrac 12$. So $r = \dfrac m{12}$ where $m$ is one of 7,8,9,10,11.

copeland 2016-03-05 20:42:55
Next up, $a_9, a_{10}, a_{11}$ is geometric. What's its common ratio (again going downward, so the ratio of $a_{10}$ to $a_{11}$)?

atmchallenge 2016-03-05 20:44:32
$\frac{2r-1}{r}$.

sxu 2016-03-05 20:44:32
(2r-1)/r

xwang1 2016-03-05 20:44:32
$\frac{2r-1}{r}$

calculus_riju 2016-03-05 20:44:32
$\frac{2r-1}{r}$

blizzard10 2016-03-05 20:44:32
$\dfrac{2r-1}{r}$

nosaj 2016-03-05 20:44:34
(2r-1)/r

copeland 2016-03-05 20:44:36
Its common ratio is \[\frac{a_{10}}{a_{11}} = \frac{2016r(2r-1)}{2016r^2} = \frac{2r-1}{r}.\]

copeland 2016-03-05 20:44:38
In particular, $a_9 = 2016(2r - 1)^2$.

copeland 2016-03-05 20:44:44
Then $a_8, a_9, a_{10}$ is arithmetic. So we get
\begin{align*}
a_8 &= 2a_{9} - a_{10} \\
&= 2\cdot 2016 (2r-1)^2 - 2016 r(2r - 1) \\
&= 2016\cdot(2r-1) \cdot(4r - 2 - r) \\
&= 2016 \cdot (2r - 1) \cdot (3r - 2).
\end{align*}

copeland 2016-03-05 20:44:50
Does that tell us anything interesting?

brian6liu 2016-03-05 20:45:18
r>2/3

nosaj 2016-03-05 20:45:18
r>2/3

mssmath 2016-03-05 20:45:18
r>2/3

copeland 2016-03-05 20:45:22
Again, $a_8$ is a positive integer, so $r > \dfrac 23$.

copeland 2016-03-05 20:45:23
Now we have $m$ is one of 9, 10, 11.

deltaepsilon6 2016-03-05 20:45:33
pattern!

ryanyz10 2016-03-05 20:45:33
looks like a pattern kindof

SuperMaltese 2016-03-05 20:45:33
looks like a pattern

deltaepsilon6 2016-03-05 20:45:33
pattern recognition!

copeland 2016-03-05 20:45:39
It does look kind of like a pattern.

copeland 2016-03-05 20:45:43
Lt's try one mor.

copeland 2016-03-05 20:45:45
What is the common ratio of $a_7, a_8, a_9$? (Again, going downward.)

blizzard10 2016-03-05 20:47:24
$\dfrac{3r-2}{2r-1}$

SuperMaltese 2016-03-05 20:47:24
(3r-2)/(2r-1)

atmchallenge 2016-03-05 20:47:24
$\frac{3r-2}{2r-1}$.

memc38123 2016-03-05 20:47:24
(3r-2)/(2r-1)

nosaj 2016-03-05 20:47:24
(3r-2)/(2r-1)

copeland 2016-03-05 20:47:26
Its common ratio is \[ \frac{a_8}{a_9} = \frac{3r - 2}{2r - 1}. \]

copeland 2016-03-05 20:47:28
So $a_7 = 2016 (3r - 2)^2$.

copeland 2016-03-05 20:47:35
Anyone starting to notice a pattern?

blue8931 2016-03-05 20:48:09
yes

nukelauncher 2016-03-05 20:48:09
YES!

jfurf 2016-03-05 20:48:09
ME!

fishy15 2016-03-05 20:48:09
me!

Prof.iHen 2016-03-05 20:48:09
ja

sxu 2016-03-05 20:48:09
sort of

walnutwaldo20 2016-03-05 20:48:09
Me!

Aragorn66 2016-03-05 20:48:09
yup

copeland 2016-03-05 20:48:17
Let's compute a few more terms and see if we notice anything.

copeland 2016-03-05 20:48:21
\begin{align*}
a_6 &= 2a_7 - a_8 \\
&= 2\cdot 2016 \cdot (3r - 2)^2 - 2016 \cdot (2r - 1) \cdot (3r - 2) \\
& = 2016(3r - 2)(6r - 4 - 2r + 1) \\
&= 2016(3r - 2)(4r - 3)
\end{align*}

copeland 2016-03-05 20:48:27
The common ratio of $a_5, a_6, a_7$ is \[ \frac{a_6}{a_7} = \frac{4r -3}{3r - 2}.\]

copeland 2016-03-05 20:48:27
So $a_5 = 2016(4r - 3)^2$.

copeland 2016-03-05 20:48:31
What will $a_1$ be?

deltaepsilon6 2016-03-05 20:49:07
a1= 2016(6r-5)^2

deltaepsilon6 2016-03-05 20:49:07
a1=2016(6r-5)^2

nosaj 2016-03-05 20:49:07
2016(6r-5)^2

azmath333 2016-03-05 20:49:07
$2016(6r-5)^2$

blue8931 2016-03-05 20:49:07
$2016(6r-5)^2$

atmchallenge 2016-03-05 20:49:07
$a_1=2016(6r-5)^2

copeland 2016-03-05 20:49:09
We'll have $a_5 = 2016 (4r - 3)^2$, then $a_3 = 2016(5r - 4)^2$ and $a_1 = 2016(6r - 5)^2$.

copeland 2016-03-05 20:49:15
So what does this tell us about $r$?

brian6liu 2016-03-05 20:49:53
r>5/6

algebra_star1234 2016-03-05 20:49:53
r=11/12

walnutwaldo20 2016-03-05 20:49:53
r > 5/6

hamup1 2016-03-05 20:49:53
$r>\dfrac{5}{6}$

ryanyz10 2016-03-05 20:49:53
r > 5/6

sxu 2016-03-05 20:49:53
its 11/12

andsun19 2016-03-05 20:49:53
11/12

PurplePancakes 2016-03-05 20:49:53
r>5/6

hamup1 2016-03-05 20:49:53
or $r=\dfrac{11}{12}$!!

akaashp11 2016-03-05 20:49:53
11/12

Aragorn66 2016-03-05 20:49:53
r=11/12

copeland 2016-03-05 20:49:55
Since $a_1$ is a positive integer, we must have $r \ge \dfrac 56.$ That means $m$ can only be 11, and we have $r = \dfrac{11}{12}$.

copeland 2016-03-05 20:49:57
So what is $a_1$?

SuperMaltese 2016-03-05 20:50:31
504

tennis1729 2016-03-05 20:50:31
504

kikipet 2016-03-05 20:50:31
504

illogical_21 2016-03-05 20:50:31
504

nukelauncher 2016-03-05 20:50:31
504!

sxu 2016-03-05 20:50:31
$504$

bestwillcui1 2016-03-05 20:50:31
five hunnet 4

AlphaPi17 2016-03-05 20:50:31
504

brian6liu 2016-03-05 20:50:31
504

hamup1 2016-03-05 20:50:31
$\dfrac{2016}{4}=504$

copeland 2016-03-05 20:50:33
We must have $a_1 = 2016 \cdot \dfrac{36}{144} = \boxed{504}$.

copeland 2016-03-05 20:50:49
Incidentally, all the odd terms were squares of an arithmetic progression.

copeland 2016-03-05 20:51:05
The even terms were their geometric means (of course).

ChickenOnRage 2016-03-05 20:51:16
was this the best solution?

copeland 2016-03-05 20:51:33
Interestingly, this was probably the most "guessable" problem on the test. . .

copeland 2016-03-05 20:52:20
Now Problem 11:

copeland 2016-03-05 20:52:22
11. Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1) = (x+2)P(x)$ for every real $x$, and $(P(2))^2 = P(3)$. Then $P(\frac 72) = \frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

copeland 2016-03-05 20:52:36
What's a nice thing to do when you're given a functional equation?

themoocow 2016-03-05 20:52:52
plug values

sxu 2016-03-05 20:52:52
plug in values

mewtwomew 2016-03-05 20:52:52
plug and chug

brian6liu 2016-03-05 20:52:52
plug in values

blizzard10 2016-03-05 20:52:52
Plug in some numbers?

andsun19 2016-03-05 20:52:52
Plug in numbers that you have

copeland 2016-03-05 20:52:58
It's always a good idea to plug in nice values for the free variable. What's something good to plug in here?

atmchallenge 2016-03-05 20:53:12
$x=1$

AlphaPi17 2016-03-05 20:53:12
1

calculus_riju 2016-03-05 20:53:12
1

ChickenOnRage 2016-03-05 20:53:12
1

copeland 2016-03-05 20:53:15
Plugging in $x = 1$ gives us $0 = 3P(1)$. So?

atmchallenge 2016-03-05 20:53:32
$P(1)=0$.

themoocow 2016-03-05 20:53:32
1 is aroot of P

brian6liu 2016-03-05 20:53:32
P(1)=0

xwang1 2016-03-05 20:53:32
$P(1)=0$

PurplePancakes 2016-03-05 20:53:32
1 is a root

Locust 2016-03-05 20:53:32
P(1)=0

kikipet 2016-03-05 20:53:32
P(1) = 0

epgy8 2016-03-05 20:53:32
$P(1)=0$

copeland 2016-03-05 20:53:35
So we conclude that 1 is a root of $P(x)$. Is there anything else we might try plugging in?

PurplePancakes 2016-03-05 20:53:57
0

AlphaPi17 2016-03-05 20:53:57
0

epgy8 2016-03-05 20:53:57
0

hamup1 2016-03-05 20:53:57
or 0 while we're at it

Aragorn66 2016-03-05 20:53:57
0

copeland 2016-03-05 20:54:03
If we plug in $x = 0$, we get $-P(1) = 2P(0)$, or $0 = 2P(0)$. So 0 is also a root of $P(x)$.

copeland 2016-03-05 20:54:04
Anything else we might want to plug in?

Aragorn66 2016-03-05 20:54:26
-1

hamup1 2016-03-05 20:54:26
-1...

illogical_21 2016-03-05 20:54:26
-1

epgy8 2016-03-05 20:54:26
-1

AlphaPi17 2016-03-05 20:54:26
-1

copeland 2016-03-05 20:54:28
Now that we know 0 is a root, it makes sense to plug in $x = -1$. Then we get $-2P(0) = P(-1)$, or $0 = P(-1)$. So -1 is also a root of $P(x)$.

copeland 2016-03-05 20:54:49
Anything else?

calculus_riju 2016-03-05 20:55:38
-2

Aragorn66 2016-03-05 20:55:38
-2

algebra_star1234 2016-03-05 20:55:38
-2

ninjataco 2016-03-05 20:55:38
-2

PurplePancakes 2016-03-05 20:55:38
-2

atmchallenge 2016-03-05 20:55:38
$x=-2$

ChickenOnRage 2016-03-05 20:55:38
-2

azmath333 2016-03-05 20:55:38
-2

AlphaPi17 2016-03-05 20:55:38
-2

algebra_star1234 2016-03-05 20:55:38
-2?

walnutwaldo20 2016-03-05 20:55:38
-2

epgy8 2016-03-05 20:55:38
-2

19bobhu 2016-03-05 20:55:38
-2?

copeland 2016-03-05 20:55:40
If we substitute $-2$ we get $-3P(-1)=0P(-2).$

copeland 2016-03-05 20:55:44
What do we learn?

walnutwaldo20 2016-03-05 20:56:13
NOTHING!!!!!!!!!!

brian6liu 2016-03-05 20:56:13
nothing

epiphany 2016-03-05 20:56:13
0=0

sxu 2016-03-05 20:56:13
NOTHING!

hamup1 2016-03-05 20:56:13
nothing!

bestwillcui1 2016-03-05 20:56:13
nothing

copeland 2016-03-05 20:56:16
This tells us 0=0. Reassuring, guys.

copeland 2016-03-05 20:56:20
I don't see anything else that helps right away. But we do have $P(x) = (x-1)x(x+1)Q(x)$ for some polynomial $Q(x)$.

copeland 2016-03-05 20:56:27
What would be really nice?

xwang1 2016-03-05 20:56:58
P is a cubic

blue8931 2016-03-05 20:56:58
P(x) is cubic

xwang1 2016-03-05 20:56:58
If $Q(x)$ was a constant.

hamup1 2016-03-05 20:56:58
if $Q(x)$ was constant ;)

mewtwomew 2016-03-05 20:56:58
if it was a constant

blue8931 2016-03-05 20:56:58
Q(x) is constant

atmchallenge 2016-03-05 20:56:58
If $P(x)$ was cubic, so $Q(x)$ was a constant!

bestwillcui1 2016-03-05 20:56:58
if Q(x) is constant!!

deltaepsilon6 2016-03-05 20:56:58
if Q(x)=k

copeland 2016-03-05 20:56:59
Is it? That'd be cool. . .

copeland 2016-03-05 20:57:03
Since we're struggling to find more roots, it would be really nice if $Q(x)$ was just a constant. How can we figure out the degree of $P(x)$?

ninjataco 2016-03-05 20:57:42
finite differences

xwang1 2016-03-05 20:57:42
Finite differences!!!

nukelauncher 2016-03-05 20:57:42
use differences

SimonSun 2016-03-05 20:57:42
we have delta

andsun19 2016-03-05 20:57:42
differences... then differences of differences... etc etc until we get constants

deltaepsilon6 2016-03-05 20:57:42
list values of P(x), and find the difference of each term

SimonSun 2016-03-05 20:57:42
DELTA ALL THE WAY

copeland 2016-03-05 20:57:44
The given equation rearranges to look a lot like finite differences: \[x\left(P(x+1) - P(x) \right) = P(x+1) + 2P(x).\]

copeland 2016-03-05 20:57:52
What is interesting there?

copeland 2016-03-05 20:58:28
Hmm. . .

copeland 2016-03-05 20:58:41
The left side has the first finite difference.

copeland 2016-03-05 20:59:00
(The right doesn't.)

copeland 2016-03-05 20:59:14
What exactly is going to happen when we take the first finite difference?

copeland 2016-03-05 21:00:02
Say, for example, that $P(x)=5x^{13}+\cdots.$ Then what does $x(P(x+1)-P(x))$ look like?

brian6liu 2016-03-05 21:01:08
65x^13+...

ninjataco 2016-03-05 21:01:08
65x^13 + ...

copeland 2016-03-05 21:01:12
If $P(x)=5x^{13}+\cdots,$ then $$x(P(x+1)-P(x))=13\cdot 5x^{13}+\cdots$$

copeland 2016-03-05 21:01:25
Hmm.

copeland 2016-03-05 21:01:33
Maybe this finite differences thing isn't working so well.

copeland 2016-03-05 21:01:36
Let's come back to it in a bit.

copeland 2016-03-05 21:01:57
Our functional equation is\[(x-1)P(x+1) = (x+2)P(x).\]

copeland 2016-03-05 21:02:10
What does it "say"?

brian22 2016-03-05 21:03:10
copeland, I will NOT be putting a polynomial FE into words

copeland 2016-03-05 21:03:13

copeland 2016-03-05 21:03:15
Let me:

copeland 2016-03-05 21:03:16
The left side says, "I take $P$ and shift it to the right by one and then add a root at $1$."

copeland 2016-03-05 21:03:21
What does the right side say?

atmchallenge 2016-03-05 21:04:09
I take $P$ and add a root at $-2$.

TheRealDeal 2016-03-05 21:04:09
add a root at -2

walnutwaldo20 2016-03-05 21:04:09
add a root at -2

andsun19 2016-03-05 21:04:09
I take P, and add a root at -2?

ychen 2016-03-05 21:04:09
add a root at -2

nukelauncher 2016-03-05 21:04:09
add a root of -2

SuperMaltese 2016-03-05 21:04:09
i take P and add a root at -2

blizzard10 2016-03-05 21:04:09
I will take P and add a root at -2

kikipet 2016-03-05 21:04:09
I take P and add a root at -2

copeland 2016-03-05 21:04:11
The right side says, "I take $P$ and add a root at $-2$."

copeland 2016-03-05 21:04:46
So shifting right and adding a root at 1 is the same as doing nothing and adding a root at -2. What does that say about the roots of $P?$

brian6liu 2016-03-05 21:06:12
they're 0, 1, and -1

andsun19 2016-03-05 21:06:12
uh... 3 roots?

nosaj 2016-03-05 21:06:12
between 1 and -2

Aragorn66 2016-03-05 21:06:12
there are 3 of them

nukelauncher 2016-03-05 21:06:12
it has 3 root?

copeland 2016-03-05 21:06:14
The functional equation just says that the roots of $P$ are a sequence of integers starting at 1 and ending at -1.

copeland 2016-03-05 21:06:35
Here's a third approach:

calculus_riju 2016-03-05 21:06:42
substituting Q(x) in the mother equation we have Q(x+1)=Q(x). Thus its a constant function. So P(x) is cubic

copeland 2016-03-05 21:06:59
Is that right? What happens when we substitute our expression with $Q$ into the mother equation?

brian6liu 2016-03-05 21:08:45
x(x-1)(x+1)(x+2)Q(x+1)=x(x-1)(x+1)(x+2)Q(x)

PurplePancakes 2016-03-05 21:08:45
(x-1)(x)(x+1)(x+2)Q(x+1)=(x+2)(x-1)(x)(x+1)Q(x)

ninjataco 2016-03-05 21:08:45
Q(x+1) = Q(x) so Q is constant

copeland 2016-03-05 21:08:47
Indeed, Substituting $P(x) = (x-1)x(x+1)Q(x)$ into our original equation simplifies to $Q(x+1)=Q(x),$ so $Q$ is indeed constant (as it is a polynomial).

copeland 2016-03-05 21:08:56
So $P(x) = c(x-1)x(x+1)$. How can we find $c$?

SimonSun 2016-03-05 21:09:37
2/3

ninjataco 2016-03-05 21:09:37
solve for P(2)

epiphany 2016-03-05 21:09:37
we know that P(2)=4

walnutwaldo20 2016-03-05 21:09:37
Use P(2)^2 = P(3)

epiphany 2016-03-05 21:09:37
and P(3)=16

SimonSun 2016-03-05 21:09:37
plug in 2

AlcumusGuy 2016-03-05 21:09:37
(P(2))^2 = P(3)

PurplePancakes 2016-03-05 21:09:37
the other condition

ryanyz10 2016-03-05 21:09:37
use the other thing we're given

copeland 2016-03-05 21:09:40
We have another condition: $((P(2))^2 = P(3)$. Plugging into that we have \[ (6c)^2 = c\cdot 2 \cdot 3 \cdot 4 = 24c.\] So what is $c$?

bestwillcui1 2016-03-05 21:10:08
2/3

SimonSun 2016-03-05 21:10:08
2/3

ninjataco 2016-03-05 21:10:08
2/3

PurplePancakes 2016-03-05 21:10:08
2/3

brian6liu 2016-03-05 21:10:08
2/3

nosaj 2016-03-05 21:10:08
2/3

AlcumusGuy 2016-03-05 21:10:08
$c = \frac{2}{3}$

blue8931 2016-03-05 21:10:08
2/3

copeland 2016-03-05 21:10:10
We have $c = \dfrac 23$.

copeland 2016-03-05 21:10:48
So what is our answer?

blue8931 2016-03-05 21:11:23
just plug in 7/2 to get the answer of 105/4 ----> $\boxed{109}$

blizzard10 2016-03-05 21:11:23
Now plug in at $\frac{7}{2}$

illogical_21 2016-03-05 21:11:23
Plug in 7/2,

nosaj 2016-03-05 21:11:23
Finally, we have $P\left(\frac 72 \right) = \frac 23 \cdot \frac 52 \cdot \frac 72 \cdot \frac 92 = \frac{105}{4}$, so our answer is $\boxed{109}$.

mewtwomew 2016-03-05 21:11:23
109

andsun19 2016-03-05 21:11:23
109

algebra_star1234 2016-03-05 21:11:23
105/4 --> 109

SimonSun 2016-03-05 21:11:23
105/4=109

blizzard10 2016-03-05 21:11:23
109

copeland 2016-03-05 21:11:26
We can just substitute $\dfrac 72$ into $P(x) = \dfrac 23 (x-1)x(x+1)$ to get \[P\left(\frac 72\right) = \frac 23 \cdot \frac 52 \cdot \frac 72 \cdot \frac 92 = \frac{105}{4}.\] So our answer is $105+4=\boxed{109}.$

copeland 2016-03-05 21:11:40
Great.

copeland 2016-03-05 21:11:47
Also, to tie up that finite differences thing.

copeland 2016-03-05 21:12:21
When you take a polynomial of the form $P(x)=a_nx^n$ and you compute $\Delta P(x)=P(x+1)-P(x),$ you get a new polynomial.

copeland 2016-03-05 21:12:38
This new polynomial has degree $n-1$ and the leading term is $na_n$.

copeland 2016-03-05 21:12:57
We were here:\[x\left(P(x+1) - P(x) \right) = P(x+1) + 2P(x).\]

copeland 2016-03-05 21:13:11
The leading term on the left side gets a coefficient of 3.

copeland 2016-03-05 21:13:17
That means the degree of $P$ must be 3.

kikipet 2016-03-05 21:13:21
is finite differences like a calculus thing?

hamup1 2016-03-05 21:13:21
so basically the derivative

copeland 2016-03-05 21:13:27
More or less, yes.

copeland 2016-03-05 21:13:47
In fact, you can use calculus both to remember and to prove the leading term stuff I just mentioned.

copeland 2016-03-05 21:13:53
Where were we again?

blizzard10 2016-03-05 21:14:16
Problem 12

weilunsun28 2016-03-05 21:14:16
11+1=13...-1=12

Aragorn66 2016-03-05 21:14:16
problem 12

copeland 2016-03-05 21:14:22
12. Find the least positive integer $m$ such that $m^2-m+11$ is a product of at least four not necessarily distinct primes.

copeland 2016-03-05 21:14:26
OK, we're trying to minimize $m$ even though we're looking at some quadratic in $m$. Is that a problem?

mewtwomew 2016-03-05 21:15:16
no

illogical_21 2016-03-05 21:15:16
no

leeandrew1029gmail.com 2016-03-05 21:15:16
no

idomath12345 2016-03-05 21:15:16
No?

copeland 2016-03-05 21:15:22
Why not?

hliu70 2016-03-05 21:15:26
no, a lower m for the most part means a lower result.

copeland 2016-03-05 21:15:30
How can you tell?

kikipet 2016-03-05 21:16:03
Graph it!

ninjataco 2016-03-05 21:16:03
because the quadratic is increasing on the positive integers

illogical_21 2016-03-05 21:16:03
coefficient of m^2 is positive

idomath12345 2016-03-05 21:16:03
cuz strictly increasing

PurplePancakes 2016-03-05 21:16:03
m^2>m for sufficiently larg m (in this case 2)

qwerty733 2016-03-05 21:16:03
Because it's m(m-1)+11, and for positives, this increases

copeland 2016-03-05 21:16:06
No, the vertex of this parabola is at $\dfrac12$ so the function is strictly increasing on the positive integers. Minimizing $m$ is exactly the same as minimizing $m^2-m+11.$

copeland 2016-03-05 21:16:08
OK, so the smallest this can be is $2^4$. Is it possible for $m^2-m+11$ to be equal to $2^4$?

nosaj 2016-03-05 21:16:40
nope cuz it's odd

WL0410 2016-03-05 21:16:40
No, it's never divisible by 2

ninjataco 2016-03-05 21:16:40
no because it's 1 mod 2

hamup1 2016-03-05 21:16:40
no way, even + odd cannot equal even

walnutwaldo20 2016-03-05 21:16:40
no

copeland 2016-03-05 21:16:42
No! In fact, if $m$ is even then $m^2-m+11$ is a sum of two evens and an odd so it is odd. If $m$ is odd, $m^2-m+11$ is a sum of three odds so it is odd.

copeland 2016-03-05 21:16:43
Let's make that more formal since it sounds useful.

copeland 2016-03-05 21:17:02
Modulo 2, we have the polynomial congruence:\[m^2-m+11\equiv0\pmod{2}.\]This congruence has no solutions. That's the same as saying that we can't factor the polynomial (since it's quadratic).

copeland 2016-03-05 21:17:12
This gives us a model for looking for the primes in question. What do we check next?

blue8931 2016-03-05 21:17:41
3

AlcumusGuy 2016-03-05 21:17:41
Mod 3

19bobhu 2016-03-05 21:17:41
3^4

hamup1 2016-03-05 21:17:41
mod 3!

ninjataco 2016-03-05 21:17:41
mod 3

epiphany 2016-03-05 21:17:41
mod 3

copeland 2016-03-05 21:17:44
Let's look at \[m^2-m+11\equiv0\pmod3.\]That's the same as \[m^2+2m+2\equiv0\pmod3.\]

copeland 2016-03-05 21:17:47
Can you prove that doesn't factor?

algebra_star1234 2016-03-05 21:18:10
yes

jfurf 2016-03-05 21:18:10
Yep...

Aragorn66 2016-03-05 21:18:10
yes

PurplePancakes 2016-03-05 21:18:10
check the residues, and it doesn't work

copeland 2016-03-05 21:18:17
We can just substitute stuff, sure.

brian22 2016-03-05 21:18:24
$(m+1)^2+1$

WL0410 2016-03-05 21:18:24
$(m+1)^2\equiv -1 \mod 3$ has no solutions

copeland 2016-03-05 21:18:30
We can also complete the square. This is the same as \[(m+1)^2\equiv-1\pmod3.\]

copeland 2016-03-05 21:18:34
Are there any solutions?

kikipet 2016-03-05 21:18:54
no

ninjataco 2016-03-05 21:18:54
no, 2 is not a quadratic residue mod 3

blue8931 2016-03-05 21:18:54
not for mod 3

AlcumusGuy 2016-03-05 21:18:54
no since squares are never are only congruent to 0 or 1 mod 3

MikoTennisPro 2016-03-05 21:18:54
nope

copeland 2016-03-05 21:18:57
No! 0 and 1 are the only squares modulo 3 so that's out.

copeland 2016-03-05 21:18:58
Next?

mssmath 2016-03-05 21:19:22
5

hamup1 2016-03-05 21:19:22
mod 5

azmath333 2016-03-05 21:19:22
5

Locust 2016-03-05 21:19:22
5

epgy8 2016-03-05 21:19:22
Mod 5?

brian6liu 2016-03-05 21:19:22
mod 5

copeland 2016-03-05 21:19:24
Don't mind if I do!

copeland 2016-03-05 21:19:25
Next is 5:\[m^2-m+11\equiv0\pmod5\] is the same as \[m^2+4m+1\equiv0\pmod5.\]

copeland 2016-03-05 21:19:28
What can we do with that?

AlcumusGuy 2016-03-05 21:20:08
complete the square again

ninjataco 2016-03-05 21:20:08
add 3 to both sides to complete the square

rkm0959 2016-03-05 21:20:08
$(m+2)^2 \equiv 3 \pmod{5}$

brian6liu 2016-03-05 21:20:08
(m+2)^2=3 mod 5

hamup1 2016-03-05 21:20:08
add 3 and complete the square

WL0410 2016-03-05 21:20:08
$(m+2)^2\equiv 3 \mod 5$, no solutions

copeland 2016-03-05 21:20:10
To complete the square we write \[m^2+4m+1\equiv (m^2+4m+4)+2.\] Completing the square again gives\[(m+2)^2\equiv3\pmod5.\]

copeland 2016-03-05 21:20:11
Is 3 a square?

sxu 2016-03-05 21:20:26
NO

PurplePancakes 2016-03-05 21:20:26
no

Aragorn66 2016-03-05 21:20:26
no

TheRealDeal 2016-03-05 21:20:26
no

kikipet 2016-03-05 21:20:26
no

copeland 2016-03-05 21:20:30
No. The squares are 0, 1, and 4.

copeland 2016-03-05 21:20:30
On to 7!

rkm0959 2016-03-05 21:21:10
$m^2+6m+4 \equiv 0 \pmod{7}$ so $(m+3)^2 \equiv 5 \pmod{7}$, impossible as the squares are $0,1,2,4$

copeland 2016-03-05 21:21:12
We get \[m^2-m+11\equiv m^2+6m+4\equiv (m+3)^2-5\pmod7.\] This means we want to solve \[(m-3)^2\equiv 5\pmod7.\]

copeland 2016-03-05 21:21:55
The squares modulo 7 are {0,1,4,2}, so 7 does not divide our number either.

copeland 2016-03-05 21:21:56
On to 11. Do you see any solutions to $m^2-m+11\equiv0\pmod{11}?$

ninjataco 2016-03-05 21:22:13
m = 0 or 1 (mod 11)

PurplePancakes 2016-03-05 21:22:13
m is a multiple of 11

MikoTennisPro 2016-03-05 21:22:13
m=11

andsun19 2016-03-05 21:22:13
11, 12

blue8931 2016-03-05 21:22:13
yes! when m is a multiple of 11

andsun19 2016-03-05 21:22:13
anything divisible by 11 or one more than a multiple of 11

eswa2000 2016-03-05 21:22:13
m is divisible by 11

Ericaops 2016-03-05 21:22:13
11

copeland 2016-03-05 21:22:19
This simplifies to $m(m-1) \equiv 0 \pmod {11}$. This is clearly possible if $m$ or $m-1$ is divisible by 11, so $m^2 - m + 11$ can be divisible by 11. (And further, every prime dividing $m^2 - m + 11$ must be at least 11.)

copeland 2016-03-05 21:22:25
What should we check now?

ninjataco 2016-03-05 21:22:50
whether m^2 - m + 11 can be 11^4

lucylou 2016-03-05 21:22:50
11^4

blue8931 2016-03-05 21:22:50
if $11^4$ works

brian6liu 2016-03-05 21:22:50
if 11^4 works

AlcumusGuy 2016-03-05 21:22:50
$11^4$

copeland 2016-03-05 21:22:52
The smallest possible number is $11^4$. Now we should check whether we can solve $11^4=m^2-m+11$.

copeland 2016-03-05 21:22:54
This simplifies to $m(m-1)=11^4-11.$

copeland 2016-03-05 21:22:59
About how big is the $m$ that solves this equation?

nosaj 2016-03-05 21:23:22
121 ish?

sxu 2016-03-05 21:23:22
121

swagger21 2016-03-05 21:23:22
~121

andsun19 2016-03-05 21:23:22
11^2 or 121

Locust 2016-03-05 21:23:22
Somewhere around 11^2

Pot 2016-03-05 21:23:22
121

Aragorn66 2016-03-05 21:23:22
121

copeland 2016-03-05 21:23:24
The left side is a little less than $m^2$ and the right side is a little less than $121^2.$ Therefore $m\approx121$.

copeland 2016-03-05 21:23:25
Does $m=121$ solve the equation?

algebra_star1234 2016-03-05 21:23:39
no

Aragorn66 2016-03-05 21:23:39
no

ninjataco 2016-03-05 21:23:39
no

Pot 2016-03-05 21:23:39
no

epiphany 2016-03-05 21:23:39
no!

brian6liu 2016-03-05 21:23:39
no

nosaj 2016-03-05 21:23:39
nope

copeland 2016-03-05 21:23:41
No. $m(m-1)=121^2-121$ and $11^4-11=121^2-11$. This $m$ is too small.

copeland 2016-03-05 21:23:42
Does $m=122$ solve this equation?

Pot 2016-03-05 21:23:56
no

brian6liu 2016-03-05 21:23:56
no

ninjataco 2016-03-05 21:23:56
no

AlcumusGuy 2016-03-05 21:23:56
no

rkm0959 2016-03-05 21:23:56
Nope

leeandrew1029gmail.com 2016-03-05 21:23:56
no

copeland 2016-03-05 21:23:58
No. $m(m-1)=122\cdot121=121^2+121$ and this is too big. This shows that $11^4$ is not of this form.

copeland 2016-03-05 21:24:00
What now?

Pot 2016-03-05 21:24:26
11^3 * 13

mssmath 2016-03-05 21:24:26
11^3*13

brian6liu 2016-03-05 21:24:26
11^3*13

blue8931 2016-03-05 21:24:26
then we have no check if $11^3 * 13$ works

epiphany 2016-03-05 21:24:26
11^3*13

AlcumusGuy 2016-03-05 21:24:26
$11^3 \cdot 13$ is next smallest

azmath333 2016-03-05 21:24:26
11^3 * 13

Ericaops 2016-03-05 21:24:26
11^3*13

PurplePancakes 2016-03-05 21:24:26
11^3*13?

rkm0959 2016-03-05 21:24:26
So we move on to $13 \cdot 11^3$ since it's the next smallest one

blue8931 2016-03-05 21:24:26
have to check if $11^3 * 13$ works

copeland 2016-03-05 21:24:47
We could go the optimistic route. The next smallest value we could attain is $11^3\cdot13.$

copeland 2016-03-05 21:24:50
Let's try to solve $m^2-m+11=11^3\cdot13$.

copeland 2016-03-05 21:24:55
What is this equation approximately?

jfurf 2016-03-05 21:25:36
121 times 143

sxu 2016-03-05 21:25:36
m squared-m=11 cubed times 12

Aragorn66 2016-03-05 21:25:36
132 ish

calculus_riju 2016-03-05 21:25:36
11.(k^2-k+1) and check k to get three primes

brian6liu 2016-03-05 21:25:40
m=132

Ericaops 2016-03-05 21:25:40
132

epiphany 2016-03-05 21:25:40
132

copeland 2016-03-05 21:25:44
This is approximately $m^2\approx11^2(13\cdot11)\approx11^2\cdot12^2.$

copeland 2016-03-05 21:25:53
What happens if we plug $m=11\cdot12$ into our equation?

blue8931 2016-03-05 21:26:29
this actually works! m= $\boxed{132}$

brian6liu 2016-03-05 21:26:29
it works

SimonSun 2016-03-05 21:26:29
it works!!!!

AlcumusGuy 2016-03-05 21:26:29
132 works!

epiphany 2016-03-05 21:26:29
it works! we are done 132

blue8931 2016-03-05 21:26:29
it works!

Aragorn66 2016-03-05 21:26:29
it works

PurplePancakes 2016-03-05 21:26:29
it works

copeland 2016-03-05 21:26:33
OK, let's try it:\begin{align*}
m^2-m+11&=(11\cdot12)^2-11\cdot12+11\\
&=(11\cdot12)^2-11^2\\
&=(11\cdot12+11)(11\cdot12-11)\\
&=(11\cdot13)(11\cdot11)\\
&=11^3\cdot13.
\end{align*}

copeland 2016-03-05 21:26:41
And the answer?

illogical_21 2016-03-05 21:27:06
m=132

kikipet 2016-03-05 21:27:06
132

walnutwaldo20 2016-03-05 21:27:06
132!

ychen 2016-03-05 21:27:06
132

algebra_star1234 2016-03-05 21:27:06
132

andsun19 2016-03-05 21:27:06
Optimism FTW. 132

Ericaops 2016-03-05 21:27:06
132

SK200 2016-03-05 21:27:06
132

mssmath 2016-03-05 21:27:06
m=132

blue8931 2016-03-05 21:27:06
$\boxed{132}$

copeland 2016-03-05 21:27:08
The smallest value occurs at $m=11\cdot12=\boxed{132}.$

copeland 2016-03-05 21:27:11
Good.

copeland 2016-03-05 21:27:14
Ready for more?

calculus_riju 2016-03-05 21:27:36
#13

Transcranial 2016-03-05 21:27:36
Yea

brian6liu 2016-03-05 21:27:36
yeah

itised 2016-03-05 21:27:36
yep

Aragorn66 2016-03-05 21:27:36
yup

SimonSun 2016-03-05 21:27:36
yessir

algebra_star1234 2016-03-05 21:27:36
yes

andsun19 2016-03-05 21:27:36
yessir

ychen 2016-03-05 21:27:36
13 is my best question

illogical_21 2016-03-05 21:27:36
on to 13!

PurplePancakes 2016-03-05 21:27:36
13 was my favorite problem on the test... LET'S DO IT

copeland 2016-03-05 21:27:42
13. Freddy the frog is is jumping aroudn the coordinate plane searching for a river, which lies on the horizontal line $y=24.$ A fence is located at the horizontal line $y=0.$ On each jump Freddy randomly chooses a dirction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point where $y=0,$ with equal likelihoods he chooses one of three directions where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that would have him cross over the fence to where $y<0.$ Freddy starts his search at the point $(0,21)$ and will stop once he reaches a point on the river. Find the expected number of jumps it will take Freddy to reach the river.

copeland 2016-03-05 21:27:48
OK, this is a pretty standard set-up. What should we do?

idomath12345 2016-03-05 21:28:33
RECURSION.

ninjataco 2016-03-05 21:28:33
make one of those nifty state diagrams?

PurplePancakes 2016-03-05 21:28:33
I solved it with recursion on the actual test

AlcumusGuy 2016-03-05 21:28:33
Let $E(n)$ be the expected number of steps starting from y = n

brian6liu 2016-03-05 21:28:33
let E_i be the expected steps it takes freddy when he starts out i steps away from the river

copeland 2016-03-05 21:28:36
We want the expected number of jumps Freddy will take to get to the river starting at $(0,21)$.

copeland 2016-03-05 21:28:37
This value depends on the expected number of steps Freddy takes from $(1,21)$, $(-1,21)$, $(0,22)$ and $(0,20)$.

copeland 2016-03-05 21:28:39
And what simplification is immediate?

PurplePancakes 2016-03-05 21:29:23
all expected values for (n,21) are the same

copeland 2016-03-05 21:29:26
The expected number of steps Freddy takes from $(a,n)$ is independent of $a$, since the river and the fence are both horizontal.

copeland 2016-03-05 21:29:29
Specifically, the expected time to reach the river is the same for any point on the line $y=n$. Let's let $e_n$ be the expected number of jumps to the river from the line $y=n$.

copeland 2016-03-05 21:29:31
Which value do we know immediately?

idomath12345 2016-03-05 21:30:04
e24 is 0

idomath12345 2016-03-05 21:30:04
e24

Pot 2016-03-05 21:30:04
when n=24, en = 0

ninjataco 2016-03-05 21:30:04
e_24 = 0

sxu 2016-03-05 21:30:04
n=24

andsun19 2016-03-05 21:30:04
24

deltaepsilon6 2016-03-05 21:30:04
n=24

copeland 2016-03-05 21:30:07
Of course $e_{24}=0$ since if Freddy is at the river, he doesn't need to move at all to get to the river.

copeland 2016-03-05 21:30:09
What can we say about $e_{23}?$

AlcumusGuy 2016-03-05 21:31:12
e(23) = 1/2 e(23) + 1/4 e(24) + 1/4 e(22) + 1

idomath12345 2016-03-05 21:31:12
1/2e23+1/4e22+1/4e24+1

ninjataco 2016-03-05 21:31:12
e_23 = 1/4 e_22 + 1/2 e_23 + 1

copeland 2016-03-05 21:31:15
When Freddy is on the line $y=23,$ he will hop one of each of the four directions with equal probability. The time it takes him to get to the river after the hop is the average of the times it takes from the four possible places he lands. That means the expected time from his current position is one more than the average of the expected times from the new positions.

copeland 2016-03-05 21:31:24
He hops to $y=24$ with probability $\dfrac14$, he hops to $y=22$ with probability $\dfrac14$ and he hops horizontally to $y=23$ with probability $\dfrac24$.

copeland 2016-03-05 21:31:27
Therefore

copeland 2016-03-05 21:31:30
\[e_{23}=1+\frac14(e_{24}+e_{22}+2e_{23}).\]

copeland 2016-03-05 21:31:34
What about $e_{22}$?

PurplePancakes 2016-03-05 21:32:17
basically the same thing

sxu 2016-03-05 21:32:17
same, excep with the #s switched

MikoTennisPro 2016-03-05 21:32:17
same thing except with 23,21,22

brian6liu 2016-03-05 21:32:17
same thing with all the indices shifted

andsun19 2016-03-05 21:32:17
e22 can be same done the same way

AlcumusGuy 2016-03-05 21:32:17
same structure: e22 = 1 + 1/4(e23 + e21 + 2e22)

hamup1 2016-03-05 21:32:17
same thing

copeland 2016-03-05 21:32:21
\[e_{22}=1+\frac14(e_{23}+e_{21}+2e_{22}).\]

copeland 2016-03-05 21:32:23
Also:\[e_{21}=1+\frac14(e_{22}+e_{20}+2e_{21}).\]

copeland 2016-03-05 21:32:28
This keeps going until. . .

sxu 2016-03-05 21:32:50
y=0!

epiphany 2016-03-05 21:32:50
e0

mewtwomew 2016-03-05 21:32:50
e0

idomath12345 2016-03-05 21:32:50
Keep going until e0?

Pot 2016-03-05 21:32:50
e_0

hamup1 2016-03-05 21:32:50
$e_{0}$

copeland 2016-03-05 21:32:52
At $y=1$ we still have\[e_{1}=1+\frac14(e_{2}+e_{0}+2e_{1}).\]

copeland 2016-03-05 21:33:00
And what about at $y=0?$

sxu 2016-03-05 21:33:51
1/3 this tiem

walnutwaldo20 2016-03-05 21:33:51
e0 = 1 + 1/3*(e1+2*e0)

AlcumusGuy 2016-03-05 21:33:51
e0 = 2/3 e0 + 1/3 e1 + 1

ychen 2016-03-05 21:33:51
e0 = 1 + 1/3(2e0 + e1)

deltaepsilon6 2016-03-05 21:33:51
e0=1+1/3(e1+2e0)

hamup1 2016-03-05 21:33:51
$e_0 = 1+ \frac{2}{3}e_0+\frac{1}{3}e_1$

copeland 2016-03-05 21:33:54
At $y=0$, he can't hop downward. Therefore he hops left or right with probability $\dfrac13$ each and he hops up with probability $\dfrac13$. This gives\[e_0=1+\dfrac13(e_1+2e_0).\]

copeland 2016-03-05 21:33:55
Let me write all of our equations and clear the denominators:

copeland 2016-03-05 21:33:56
\begin{align*}
e_{24}&=0\\
4e_{23}&=4+e_{24}+e_{22}+2e_{23}\\
4e_{22}&=4+e_{23}+e_{21}+2e_{22}\\
&\vdots\\
4e_{2}&=4+e_{3}+e_{1}+2e_{2}\\
4e_{1}&=4+e_{2}+e_{0}+2e_{1}\\
3e_0&=3+e_1+2e_0
\end{align*}

copeland 2016-03-05 21:33:59
What should we do with those?

Ericaops 2016-03-05 21:34:26
add

fz2012 2016-03-05 21:34:26
Add them up

Transcranial 2016-03-05 21:34:26
Add them

MikoTennisPro 2016-03-05 21:34:26
add?

illogical_21 2016-03-05 21:34:26
add!

mssmath 2016-03-05 21:34:26
add?

deltaepsilon6 2016-03-05 21:34:26
add

copeland 2016-03-05 21:34:28
Since there's a lot of symmetry and the coefficients on the variables on both sides almost add to 4, something cool ought to happen if we add all of the equations.

copeland 2016-03-05 21:34:32
However, everything doesn't quite line up correctly and I'm a little worried. Is there something we can do to make things look a little cleaner?

blue8931 2016-03-05 21:35:44
shift the e0 equation over a bit

azmath333 2016-03-05 21:35:44
add e_0 to both sides of the last equation

brian22 2016-03-05 21:35:44
Add some e0 to the bottom-most equation

copeland 2016-03-05 21:35:47
We could add $e_0=e_0$ to the bottom equation to get another 4 on the left side and make the right side a little neater.

copeland 2016-03-05 21:35:53
And what about the top equation?

sxu 2016-03-05 21:36:26
x4

walnutwaldo20 2016-03-05 21:36:26
multiply by 4

kikipet 2016-03-05 21:36:26
*4

azmath333 2016-03-05 21:36:26
multiply by 4

SuperMaltese 2016-03-05 21:36:26
multiply by 4

leeandrew1029gmail.com 2016-03-05 21:36:26
multiply by 4

copeland 2016-03-05 21:36:29
We could multiply the top equation by 4 to get another 4 on the left side.

copeland 2016-03-05 21:36:30
\begin{align*}
4e_{24}&=0\\
4e_{23}&=4+e_{24}+e_{22}+2e_{23}\\
4e_{22}&=4+e_{23}+e_{21}+2e_{22}\\
&\vdots\\
4e_{2}&=4+e_{3}+e_{1}+2e_{2}\\
4e_{1}&=4+e_{2}+e_{0}+2e_{1}\\
4e_0&=3+e_1+e_0+2e_0
\end{align*}

copeland 2016-03-05 21:36:34
Adding the left sides gives\[4(e_0+e_1+\cdots+e_n).\] This looks like an important quantity, so let's call that $4S$. What is the sum on the right side?

fz2012 2016-03-05 21:38:45
4(e_0+e_1+...+e_22)+3e_23+95

TheRealDeal 2016-03-05 21:38:45
95+4(e0+e1+...)

deltaepsilon6 2016-03-05 21:38:45
95+4S-e23

hamup1 2016-03-05 21:38:45
$4\cdot 23 + 3 + 2S -(e_{23}+e_{24}) + 2S.$

copeland 2016-03-05 21:38:48
Summing everything gives
\[\begin{array}{rcccccccc}
4e_{24}&=&0&&&&&&\\
4e_{23}&=&4&+&e_{24}&+&e_{22}&+&2e_{23}\\
4e_{22}&=&4&+&e_{23}&+&e_{21}&+&2e_{22}\\
\vdots&&\vdots&&\vdots&&\vdots&&\vdots\\
4e_{2}&=&4&+&e_{3}&+&e_{1}&+&2e_{2}\\
4e_{1}&=&4&+&e_{2}&+&e_{0}&+&2e_{1}\\
4e_0&=&3&+&e_1&+&e_0&+&2e_0\\
\hline
4S&=&95&+&(S-e_{0})&+&(S-e_{24}-e_{23}+e_0)&+&2(S-e_{24}).
\end{array}\]

swagger21 2016-03-05 21:38:55
95 + 4(e0 + e1 + .... + e23) - e23

copeland 2016-03-05 21:38:57
There's $4S$ on each side, so those cancel. We also have $e_{24}=0$, so this simplifies to $0=95-e_{23}$ so \[e_{23}=95.\]

copeland 2016-03-05 21:39:02
What is $e_{22}?$

brian6liu 2016-03-05 21:39:41
186

deltaepsilon6 2016-03-05 21:39:41
186

MikoTennisPro 2016-03-05 21:39:41
186

swagger21 2016-03-05 21:39:41
186

calculus_riju 2016-03-05 21:39:41
186

fz2012 2016-03-05 21:39:41
186

copeland 2016-03-05 21:39:43
The equation\[4e_{23}=4+e_{24}+e_{22}+2e_{23}\] becomes \[4\cdot95=4+0+e_{22}+2\cdot95.\]This makes \[e_{22}=2\cdot95-4=186.\]

copeland 2016-03-05 21:39:44
What is $e_{21}?$

PurplePancakes 2016-03-05 21:40:19
273

brian6liu 2016-03-05 21:40:19
273, so our answer is 273

swagger21 2016-03-05 21:40:19
273

fz2012 2016-03-05 21:40:19
273

calculus_riju 2016-03-05 21:40:19
273

Pot 2016-03-05 21:40:19
\boxed{\text{273}}

blue8931 2016-03-05 21:40:19
273

nosaj 2016-03-05 21:40:19
273

deltaepsilon6 2016-03-05 21:40:19
273

copeland 2016-03-05 21:40:21
The equation\[4e_{22}=4+e_{23}+e_{21}+2e_{22}\] becomes \[4\cdot186=4+95+e_{22}+2\cdot186.\]This makes \[e_{21}=2\cdot186-4-95=\boxed{273}.\]

copeland 2016-03-05 21:40:24
Rad.

copeland 2016-03-05 21:40:37
Two more. Dos moro.

copeland 2016-03-05 21:40:42
14. Centered at each lattice point in the coordinate plane are a circle radius $\dfrac1{10}$ and a square with sides of length $\dfrac15$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001,429)$ intersects $m$ of the squaers and $n$ of the circles. Find $m+n.$

copeland 2016-03-05 21:40:50
Here we have a line working its way up and to the right and we want to know how often it hits a bunch of circles and squares centered at lattice points.

copeland 2016-03-05 21:40:54
OK, everybody knows about 1001. What is special about 1001?

Ericaops 2016-03-05 21:41:11
7*11*13

sxu 2016-03-05 21:41:11
7,11,13

MikoTennisPro 2016-03-05 21:41:11
11 * 7 * 13

TheRealDeal 2016-03-05 21:41:11
factors as 7*11*13

hamup1 2016-03-05 21:41:11
$1001 = 7\cdot 11\cdot 13$

ompatel99 2016-03-05 21:41:14
7*11*13?

ChickenOnRage 2016-03-05 21:41:14
7 * 11 * 13

copeland 2016-03-05 21:41:18
This number comes up on all sorts of math contests because it factors nicely as $1001=7\cdot11\cdot13.$ Is that useful here?

blue8931 2016-03-05 21:41:48
7 * 11 *13, which cancels with 429 to become 3/7

hamup1 2016-03-05 21:41:48
$429 = 11\cdot 13 \cdot 3\implies \gcd(1001,429) = 11\cdot 13$

illogical_21 2016-03-05 21:41:48
yes common factors with 429

deltaepsilon6 2016-03-05 21:41:48
yes because 429=11*13*3

Pot 2016-03-05 21:41:48
useful since 429 also has 11*13

ChickenOnRage 2016-03-05 21:41:48
429 = 3 * 11 * 13

copeland 2016-03-05 21:41:50
I see that since $4+9-2=11,$ we have 11 as a factor of 429! Dividing gives $429=11\cdot39.$

copeland 2016-03-05 21:41:54
Ooh, also we have a 13 in there so $429=11\cdot13\cdot3.$ Both of those big numbers are divisible by $11\cdot13=143$.

copeland 2016-03-05 21:41:55
Our top-right vertex is $(143\cdot7,143\cdot3).$

copeland 2016-03-05 21:41:57
What's that tell us about our line?

Axolotl 2016-03-05 21:42:19
slope=7/3

kikipet 2016-03-05 21:42:19
slope of 3/7

deltaepsilon6 2016-03-05 21:42:19
slope of 3/7

algebra_star1234 2016-03-05 21:42:19
slope is 3/7

copeland 2016-03-05 21:42:22
First off, it has slope $\dfrac37$.

hamup1 2016-03-05 21:42:47
therefore just consider a 7 vy 3 box

illogical_21 2016-03-05 21:42:47
repeats in 3 by 7 rectangles

idomath12345 2016-03-05 21:42:47
We only need to go til 7,3

blue8931 2016-03-05 21:42:47
so just examine one rectangle and find a pattern

TheRealDeal 2016-03-05 21:42:47
3*7 box 143 times

copeland 2016-03-05 21:42:55
The line goes through $(7,3),$ and in fact it goes through every lattice point of the form $(7a,3b)$.

copeland 2016-03-05 21:42:56

copeland 2016-03-05 21:42:59
This rectangle repeats:

copeland 2016-03-05 21:43:00

copeland 2016-03-05 21:43:01
So what?

Ericaops 2016-03-05 21:43:36
just check one of the rectangles

Peggy 2016-03-05 21:43:36
just focus on one rectangle

TheRealDeal 2016-03-05 21:43:36
do a 7*3 box and then multiply by 143

brian6liu 2016-03-05 21:43:36
draw in squares and circles

andsun19 2016-03-05 21:43:36
think about one box!

hamup1 2016-03-05 21:43:36
rejoice because i brought graph paper

blue8931 2016-03-05 21:43:36
focus on one rectangle and then mutiply

deltaepsilon6 2016-03-05 21:43:36
consider one rectangle

Pot 2016-03-05 21:43:36
draw a very accurate diagram on graph paper XD

AlcumusGuy 2016-03-05 21:43:36
just focus on (0, 0) to (7, 3) and the multiply by 143

PurplePancakes 2016-03-05 21:43:36
count the ones that apply in one rectangle and multiply it by 143

copeland 2016-03-05 21:43:39
So all we need to do is solve the problem on the rectangle with corners $(0,0)$ and $(7,3)$ and multiply by the number of rectangles!

copeland 2016-03-05 21:43:40

copeland 2016-03-05 21:43:55
OK, so let's figure out if any of the points on the line $y=\dfrac37x$ falls into a box or circle around any of these points.

copeland 2016-03-05 21:44:08
Are any of them obvious?

TheRealDeal 2016-03-05 21:44:35
(0,0) has square and circle of course

andsun19 2016-03-05 21:44:35
well, the one at (0,0)

Pot 2016-03-05 21:44:35
only falls on circles at (0, 0) and (7,3)

Ericaops 2016-03-05 21:44:35
endpoints

kikipet 2016-03-05 21:44:35
the 2 corners

TheRealDeal 2016-03-05 21:44:35
(0,0) and (7,3)

brian6liu 2016-03-05 21:44:35
the 2 corners

walnutwaldo20 2016-03-05 21:44:35
(0,0) and (7,3)

blizzard10 2016-03-05 21:44:35
(0,0), (7,3)

copeland 2016-03-05 21:44:39
The line definitely falls in the both the square and circle centered at $(0,0)$ and $(7,3)$ since it passes through these two points.

copeland 2016-03-05 21:44:46
How should we organize checking all these points?

andsun19 2016-03-05 21:45:17
Think about x=something lines and when they interesect

SuperMaltese 2016-03-05 21:45:17
pair by column?

copeland 2016-03-05 21:45:20
That works.

copeland 2016-03-05 21:45:24
Any better ideas?

TheRealDeal 2016-03-05 21:45:58
(2,1) (3,1) (4,2) and (5,2) are the only point that need to be checked

AlcumusGuy 2016-03-05 21:45:58
look at y = blah

copeland 2016-03-05 21:46:01
There are fewer rows than columns. . .

copeland 2016-03-05 21:46:02
Why don't we use casework on the $y$-coordinate of the points.

copeland 2016-03-05 21:46:03
Start with $y=0$:

copeland 2016-03-05 21:46:06
We know we're in the circle and square at $(0,0).$ Can we pass through the square at $(1,0)$?

illogical_21 2016-03-05 21:46:30
no

kikipet 2016-03-05 21:46:30
no

fishy15 2016-03-05 21:46:30
no . . .

brian6liu 2016-03-05 21:46:30
no

nosaj 2016-03-05 21:46:30
nope

copeland 2016-03-05 21:46:33
No! The slope of the line is $\dfrac37$. However, look at two neighboring squares:

copeland 2016-03-05 21:46:36

copeland 2016-03-05 21:46:36
What's the maximum slope of a line containing a point in the left square and a point in the right square?

brian6liu 2016-03-05 21:47:09
1/4

Pot 2016-03-05 21:47:09
0.2/0.8 = 1/4

deltaepsilon6 2016-03-05 21:47:09
1/4

kikipet 2016-03-05 21:47:09
1/4?

copeland 2016-03-05 21:47:12
The maximum slope is between these two points (I've put the centers at $(0,0)$ and $(1,0)$ in this example):

copeland 2016-03-05 21:47:13

copeland 2016-03-05 21:47:15
This maximum slope is $\dfrac{0.2}{0.8}=\dfrac14$. This is less than $\dfrac37$ so two horizontally neighboring squares can't both contain a point on the line.

copeland 2016-03-05 21:47:19
So, we can hit at most one square per row.

copeland 2016-03-05 21:47:22
Since the circles are inscribed in the squares, we can't hit a circle without hitting its square.

copeland 2016-03-05 21:47:23
Thus we hit one square at height $y=0$ and we hit one circle at height $y=0$.

copeland 2016-03-05 21:47:25
Next for $y=1$:

copeland 2016-03-05 21:47:30
As before, we can either hit the square containing $(2,1)$ or the square containing $(3,1)$. Which should we check first?

nosaj 2016-03-05 21:47:56
(2,1)

brian6liu 2016-03-05 21:47:56
(2, 1)

blue8931 2016-03-05 21:47:56
(2,1)

illogical_21 2016-03-05 21:47:56
(2,1)

Ericaops 2016-03-05 21:47:56
(2,1)

TheRealDeal 2016-03-05 21:47:56
2,1

kikipet 2016-03-05 21:47:56
(2,1)

copeland 2016-03-05 21:47:59
Well, it looks like the line gets closer to $(2,1)$ than $(3,1)$ in our picture. If you want to be more algebraic, then for $y=1$ the line contains the point $\left(\dfrac73,1\right),$ and $\dfrac73$ is indeed closer to 2 than to 3.

copeland 2016-03-05 21:48:04
What is the bottom-right point of the square in question?

Pot 2016-03-05 21:49:03
2.1, 0.9

swagger21 2016-03-05 21:49:03
21/10, 9/10

brian6liu 2016-03-05 21:49:03
(2.1, 0.9)

copeland 2016-03-05 21:49:05
The bottom-right point is $(2.1,0.9).$ What do you notice about this point?

leeandrew1029gmail.com 2016-03-05 21:49:44
also on the line

sxu 2016-03-05 21:49:44
it works!

brian6liu 2016-03-05 21:49:44
it lies on the line

Pot 2016-03-05 21:49:44
wow it has slope 3/7 if you extend through origin!1!!!

andsun19 2016-03-05 21:49:44
it exactly matches?

SimonSun 2016-03-05 21:49:44
it matches

SuperMaltese 2016-03-05 21:49:44
7/3 ratio!

Ericaops 2016-03-05 21:49:44
it is on the line

copeland 2016-03-05 21:49:46
This point is on the line! $\dfrac{0.9}{2.1}=\dfrac37.$ Therefore we hit this square.

copeland 2016-03-05 21:49:48
Do we hit the circle inside this square?

kikipet 2016-03-05 21:50:03
no

Pot 2016-03-05 21:50:03
nope

memc38123 2016-03-05 21:50:03
no

eswa2000 2016-03-05 21:50:03
no

sxu 2016-03-05 21:50:03
NOPE

PurplePancakes 2016-03-05 21:50:03
no

leeandrew1029gmail.com 2016-03-05 21:50:03
no

brian6liu 2016-03-05 21:50:03
no, it barely touches the square

copeland 2016-03-05 21:50:05
No. ONLY the corner of the square is on the line. Since the circle we've defined is the incircle of the square, it does not contain the corner.

copeland 2016-03-05 21:50:09
Therefore we hit one square at height $y=1$ and zero circles at height $y=1$.

copeland 2016-03-05 21:50:10
Next for $y=2$:

copeland 2016-03-05 21:50:11
How do we handle the $y=2$ case?

copeland 2016-03-05 21:51:07
We could do the same thing, but stare at the diagram a moment. Stare deep into its soul. . .

brian6liu 2016-03-05 21:51:35
it's symmetric to y=1, just turn the rectangle

PurplePancakes 2016-03-05 21:51:35
It's symmetric

simon1221 2016-03-05 21:51:35
symmetry!

hamup1 2016-03-05 21:51:35
symmetric

copeland 2016-03-05 21:51:36
The diagram is rotationally symmetric. So?

Transcranial 2016-03-05 21:52:05
Its same

AlcumusGuy 2016-03-05 21:52:05
it will pass through corner of a square

kikipet 2016-03-05 21:52:05
it touches the top left corner of (5,2)'s square

hamup1 2016-03-05 21:52:05
exact same amount as y=1

sxu 2016-03-05 21:52:05
just the square

simon1221 2016-03-05 21:52:05
it intersects the square at (5,2) but not the circle

leeandrew1029gmail.com 2016-03-05 21:52:05
square on 5,2 touches the line

Ericaops 2016-03-05 21:52:05
(5,2) works for the square but not the circle

andsun19 2016-03-05 21:52:05
so 1

brian6liu 2016-03-05 21:52:05
(2, 1) corresponds to (5, 2), so (5, 2) works, and the corner is barely on the libne

copeland 2016-03-05 21:52:08
So we also hit one square at height $y=2$ and zero circles at height $y=2$.

copeland 2016-03-05 21:52:09
Finally we hit one circle at $y=3$ and one square at $y=3$.

copeland 2016-03-05 21:52:10
In all, we hit 4 squares and 2 circles in this rectangle.

copeland 2016-03-05 21:52:16
Now we multiply! Since there are 143 of these rectangles, we hit $4\cdot143$ squares and $2\cdot143$ circles.

brian6liu 2016-03-05 21:52:43
no

sxu 2016-03-05 21:52:43
Wait, won't you overcount?

PurplePancakes 2016-03-05 21:52:43
WAIT we overcounted

blue8931 2016-03-05 21:52:43
no no no

copeland 2016-03-05 21:52:48
No?

kikipet 2016-03-05 21:52:54
Doesn't that overcount?

hamup1 2016-03-05 21:52:54
corners!

copeland 2016-03-05 21:52:55
Oh, I counted the corners a lot of times. How many intersections are there actually?

brian6liu 2016-03-05 21:53:55
430 squares and 144 circles

PurplePancakes 2016-03-05 21:53:55
4*143+2

hamup1 2016-03-05 21:53:55
3*143 + 1*143 + 1 + 1

Stephenpiano 2016-03-05 21:53:55
We could not take the upper square and circle and then multiply by 143 and finally add 1 circle and 1 square for an easier counting method.

simon1221 2016-03-05 21:53:55
3*143 squares, 1*143 circles, +2 for the ones at the end

kikipet 2016-03-05 21:53:55
143* (3 squares, 1 circle) + 1 each

Pot 2016-03-05 21:53:55
3 squares and 1 circle, so 4*143 then add origin = 574

copeland 2016-03-05 21:53:58
We hit the interior squares $2\cdot143$ times and we hit the interior circles 0 times.

copeland 2016-03-05 21:53:58
We hit the corner squares 144 times and we hit the corner circles 144 times.

copeland 2016-03-05 21:53:59
There are a total of \[2\cdot143+2\cdot144=2\cdot287=\boxed{574}\] intersections.

copeland 2016-03-05 21:54:03
OK, now what?

kikipet 2016-03-05 21:54:46
Problem 15!

Pot 2016-03-05 21:54:46
on to number 15!

copeland 2016-03-05 21:54:51
Yup.

copeland 2016-03-05 21:54:53
15. Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y.$ Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B,$ respectively, with line $AB$ closer to point $X$ than to $Y.$ Circle $\omega$ passes through $A$ and$B$ intersecting $\omega_1$ again at $D\neq A$ and intersecting $\omega_2$ again at $C\neq B.$ The three points $C,$ $Y,$ and $D$ are collinear, $XC=67,$ $XY=47,$ and $XD=37.$ Find $AB.$

copeland 2016-03-05 21:55:00
Wut?

copeland 2016-03-05 21:55:02
So. . .

brian22 2016-03-05 21:55:08
Well, Problem 15 is trivial, so we're done guys!

copeland 2016-03-05 21:55:12
Almost.

copeland 2016-03-05 21:56:53
And our diagram:

copeland 2016-03-05 21:56:55

copeland 2016-03-05 21:56:58
Obviously this diagram isn't so great: $XY$ is supposd to be bigger than $XD$.

copeland 2016-03-05 21:57:03
Let's start over. The real diagram looks a bit more like this:

copeland 2016-03-05 21:57:05

copeland 2016-03-05 21:57:20
The lengths we know are the three edges emanating from $X.$ I've left those off because we're clearly going to need to know a lot more before we can us them. Don't forget, though, that those are the edges we know about. We're trying to find $AB.$

copeland 2016-03-05 21:57:25
First off, what do we get from those three circles?

copeland 2016-03-05 21:58:49
What is the line $XY?$

copeland 2016-03-05 21:59:07
Oh, oops, there's also a typo in the problem. We want:

copeland 2016-03-05 21:59:14
15. Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y.$ Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B,$ respectively, with line $AB$ closer to point $X$ than to $Y.$ Circle $\omega$ passes through $A$ and$B$ intersecting $\omega_1$ again at $D\neq A$ and intersecting $\omega_2$ again at $C\neq B.$ The three points $C,$ $Y,$ and $D$ are collinear, $XC=67,$ $XY=47,$ and $XD=37.$ Find $AB^2.$

copeland 2016-03-05 21:59:39
Note that we want to find $AB^2,$ not $AB$.

hamup1 2016-03-05 22:00:16
by radical axis lemma we have that da, xy, and bc intersect at a point p

PurplePancakes 2016-03-05 22:00:16
radial axis

brian22 2016-03-05 22:00:16
Radical axis

sxu 2016-03-05 22:00:16
radical axis, wait, wait

AlcumusGuy 2016-03-05 22:00:16
radical axis

ninjataco 2016-03-05 22:00:16
XY is the radical axis of w1 and w2

copeland 2016-03-05 22:00:21
The line $XY$ is the radical axis of $\omega_1$ and $\omega_2$. This is the set of points for which the power with respect to $\omega_1$ is equal to the power with respect to $\omega_2$.

copeland 2016-03-05 22:00:26
Likewise, $DA$ is the radical axis of $\omega_1$ and $\omega$ and $BC$ is the radical axis of $\omega_2$ and $\omega$.

copeland 2016-03-05 22:00:29
What do we know about those?

Ericaops 2016-03-05 22:01:15
concurrent

brian6liu 2016-03-05 22:01:15
they're concurrent

hamup1 2016-03-05 22:01:15
intersect at a point! call it $P$ :)

AlcumusGuy 2016-03-05 22:01:15
concurrent at radical center

sxu 2016-03-05 22:01:15
concurrent?

copeland 2016-03-05 22:01:17
They're concurrent. Specifically, call $Z$ the intersection of $AD$ and $BC$ (clearly they intersect). Since $Z$ lies on both lines, $\text{Pow}_{\omega_1}(Z)=\text{Pow}_{\omega}(Z)$ and $\text{Pow}_{\omega_2}(Z)=\text{Pow}_{\omega}(Z).$ Putting those together gives $\text{Pow}_{\omega_1}(Z)=\text{Pow}_{\omega_2}(Z),$ so $Z$ also lies on $XY$.

copeland 2016-03-05 22:01:23
But it's obviously $Z$ and not $P.$

copeland 2016-03-05 22:01:25

copeland 2016-03-05 22:01:27
What else do you see here?

copeland 2016-03-05 22:02:12
Here, check this out:

copeland 2016-03-05 22:02:14

deltaepsilon6 2016-03-05 22:02:47
cyclic quadrilateral!

hamup1 2016-03-05 22:02:47
cyclic quads!

PurplePancakes 2016-03-05 22:02:47
cyclic quads

deltaepsilon6 2016-03-05 22:02:47
2 cyclic quadrilaterals

copeland 2016-03-05 22:02:49
Two cyclic quadrilaterals!

hamup1 2016-03-05 22:03:00
xazb might be cyclic

copeland 2016-03-05 22:03:05
Huh, might be.

SohamSchwarz119 2016-03-05 22:03:27
Is AXBZ cyclic?

copeland 2016-03-05 22:03:29
Is it?

swagger21 2016-03-05 22:04:06
miquel!

fishy15 2016-03-05 22:04:06
lets check

PurplePancakes 2016-03-05 22:04:06
YES IT IS angle chase for a bit

Ericaops 2016-03-05 22:04:06
yes

hamup1 2016-03-05 22:04:06
yes!

calculus_riju 2016-03-05 22:04:06
we can find out by angle chasing

copeland 2016-03-05 22:04:18
$X$ is actually the Miquel point of $CDZ$ with respect to $A,$ $B,$ and $Y.$

copeland 2016-03-05 22:04:21
Specifically, $AXYD$ and $BXYC$ are cyclic. Let me prove really quickly that $AXBZ$ is also cyclic for those of you who haven't seen it.

copeland 2016-03-05 22:04:27
Since the other two quadrilaterals are cyclic, we get
\begin{align*}
\angle D+\angle YXA=180^\circ\\
\angle C+\angle YXB=180^\circ\\
\end{align*}

copeland 2016-03-05 22:04:32
We would like to compute\[\angle Z+\angle AXB=\theta\] and show that $\theta=180^\circ$.

copeland 2016-03-05 22:04:37
If we add the three equations we get \[(\angle D+\angle C+\angle Z)+(\angle YXA+\angle YXB+\angle AXB)=180^\circ+180^\circ+\theta.\]

copeland 2016-03-05 22:04:40
This simplifies to \[180^\circ+360^\circ=180^\circ+180^\circ+\theta,\] so $\theta=180^\circ,$ as promised.

hamup1 2016-03-05 22:04:56
$180=\angle{DYX} + \angle{XYC} = 180-\angle{DAX} + 180-\angle{CXB} = \angle{XAZ} + \angle{XBZ} = 180.$

deltaepsilon6 2016-03-05 22:04:56
xazb is cyclic

copeland 2016-03-05 22:05:02

deltaepsilon6 2016-03-05 22:05:05
how does that help

copeland 2016-03-05 22:05:23
No idea, but the principle of "More circles makes you look smart" definitely applies here.

copeland 2016-03-05 22:05:28
Now, we have a lot of angles and circles. We should probably be looking for similar inscribed angles that will help. Now is the time to remember that we know $XD,$ $XY,$ and $XC.$

copeland 2016-03-05 22:05:37
What do you got?

copeland 2016-03-05 22:07:13
Not a lot yet, huh? I'm sure it's time to look for similar triangles now.

copeland 2016-03-05 22:07:28
Where can we get some equal angles that might help? Which trianglees are similar?

copeland 2016-03-05 22:09:08
This is definitely the hard part.

copeland 2016-03-05 22:09:40
We have lengths $DX,$ $YX,$ and $CX.$ Where can we find similar triangles that us some of those lengths?

copeland 2016-03-05 22:12:37
A lot of people are excited about $\triangle CDX.$ This is not a bad triangle, except that it's hard to move either that angle at $D$ or that angle at $C$ around nicely.

copeland 2016-03-05 22:13:05
However, we do have a lot of things we can do with $\angle ADX.$

calculus_riju 2016-03-05 22:13:45
can u join XA and XB?

calculus_riju 2016-03-05 22:13:45
DXZ and CXZ

deltaepsilon6 2016-03-05 22:13:45
XDA and DZX

copeland 2016-03-05 22:14:18
Here are a couple of good ideas.

copeland 2016-03-05 22:14:42
Considering $\omega_1,$ we have equal inscribed angles $\angle XDA=\angle XAB.$ Similarly, $\omega_2$ tells us $\angle XCB=\angle XBA.$

copeland 2016-03-05 22:14:46

copeland 2016-03-05 22:15:14
It would be nice if we had all three of these triangles similar, $\triangle DXZ\sim\triangle AXB\sim\triangle ZXC,$ and we're close with a handful of matching angles.

copeland 2016-03-05 22:15:16
Let's try to show the rest. Can you argue that $\angle BZX=\alpha?$

PurplePancakes 2016-03-05 22:17:52
yes, same arc

hamup1 2016-03-05 22:17:52
$\angle{BAX}$ and $\angle{BZX}$ both intercept arc $BX$, thus equal to $\alpha$

deltaepsilon6 2016-03-05 22:17:52
properties of cyclic quadrilaterals

danzhi 2016-03-05 22:17:52
[asy]\angle BZX = \angle XAB = \alpha[\asy]

deltaepsilon6 2016-03-05 22:17:52
if XAB=alpha, then cyclic quadrilaterals state that BZX is also alpha

copeland 2016-03-05 22:17:56
These angles are also inside the lower circle: $\angle XAB$ and $\angle XZB$ subtend the same arc, so $\angle XZB=\alpha$ as well. Symmetrically, $\angle XZA=\beta.$

copeland 2016-03-05 22:17:58

copeland 2016-03-05 22:18:05
Now we have similar triangles! How can we use that?

ninjataco 2016-03-05 22:18:42
length ratios?

fishy15 2016-03-05 22:18:42
side lengths

copeland 2016-03-05 22:19:03
Which ratios? What length do we get instantly, knowing $DX,$ $YX,$ and $XC?$

deltaepsilon6 2016-03-05 22:19:58
XD/XZ=XZ/XC

PurplePancakes 2016-03-05 22:19:58
ZX is the geometric mean of DX and CX

hamup1 2016-03-05 22:19:58
XZ?

copeland 2016-03-05 22:20:01
We know $DX$ and $CX.$ Since $\triangle DXZ\sim\triangle ZXC,$ we get\[\frac{DX}{XZ}=\frac{ZX}{XC}.\]

copeland 2016-03-05 22:20:03
This tells us

copeland 2016-03-05 22:20:04
\[ZX=\sqrt{DX\cdot XC}=\sqrt{37\cdot67}.\]

copeland 2016-03-05 22:20:11
Can you see any more triangles that are similar to these three?

copeland 2016-03-05 22:20:24

deltaepsilon6 2016-03-05 22:20:33
XAB

copeland 2016-03-05 22:20:38
Yup, that's on our list.

copeland 2016-03-05 22:21:55
Do you see $\alpha$ anywhere else in the diagram?

brian22 2016-03-05 22:22:30
AYZ

hamup1 2016-03-05 22:22:36
YAZ?

hamup1 2016-03-05 22:22:36
YAZ and YBZ

brian22 2016-03-05 22:22:36
AYZ

andsun19 2016-03-05 22:22:36
yes, ayz

copeland 2016-03-05 22:22:50
Great (some of those are angles and some are triangles, note).

copeland 2016-03-05 22:23:06
I see another triangle! In $\omega_1,$ the angles $\angle ADX$ and $\angle AYX$ subtend the same arc. Therefore $\angle AYX=\alpha.$ Symmetrically, $\angle BYX=\beta.$

copeland 2016-03-05 22:23:23
Now, we could try to construct more ratios, but it looks a lot like we'd end up reproving Power of a Point six different ways if we did. I'd rather not.

copeland 2016-03-05 22:23:28
Let me drop those new angles on the diagram, ignoring $C$ and $D$ for now since we just used them.

copeland 2016-03-05 22:23:29

copeland 2016-03-05 22:23:35
What do you see?

Ericaops 2016-03-05 22:24:17
parallelogram

danzhi 2016-03-05 22:24:17
parallelgram AZBY

calculus_riju 2016-03-05 22:24:17
parallelogram

brian22 2016-03-05 22:24:17
Parallelogram!

swagger21 2016-03-05 22:24:17
parallelogram?

copeland 2016-03-05 22:24:19
I see a parallelogram!

copeland 2016-03-05 22:24:19

copeland 2016-03-05 22:24:21
Now we know $ZX$ and we know $XY,$ so we seem to know quite a bit about the segment $ZY.$

copeland 2016-03-05 22:24:31
What else should we draw?

hamup1 2016-03-05 22:25:36
midpoint of $AB$!!

copeland 2016-03-05 22:25:46
Let's label the intersection of $AB$ and $YZ$ as $M.$ (It's the midpoint of both $AB$ and $YZ$, of course. A couple of you noticed that it is the midpoint of $AB$ a long time ago since it's on the radical axis of $\omega_1$ and $\omega_2.$)

copeland 2016-03-05 22:25:48

copeland 2016-03-05 22:26:04
Now, what do we have?

deltaepsilon6 2016-03-05 22:26:39
XM

danzhi 2016-03-05 22:26:39
XM

fishy15 2016-03-05 22:26:39
similar triangles?

Ericaops 2016-03-05 22:26:39
similar triangles

copeland 2016-03-05 22:26:41
Similar triangles WITH MEDIANS. That's a little more information.

copeland 2016-03-05 22:26:49
Now we have medians of similar triangles so we get a new equation.

copeland 2016-03-05 22:27:26
Specifcially, $\triangle AYZ$ with median $AM$ is similar to $\triangle XAB$ with median $XM.$

copeland 2016-03-05 22:27:31
What can we do?

sxu 2016-03-05 22:28:29
ratios to find lengths

hamup1 2016-03-05 22:29:43
$\dfrac{YZ}{AX} = \dfrac{AB}{XM}$?

danzhi 2016-03-05 22:29:43
AB : YZ = XM : AM

copeland 2016-03-05 22:29:47
Here's the one I see:\[\frac{AM}{XM}=\frac{YZ}{AB}.\]

copeland 2016-03-05 22:30:41
Let's think about all these ideas.

hamup1 2016-03-05 22:31:02
wait this is good

copeland 2016-03-05 22:31:06
Yeah, this is good.

copeland 2016-03-05 22:31:22
Why did I pick the ratios I picked?

copeland 2016-03-05 22:31:33
I wanted to get $AB$ (or $AM$) from both triangles.

copeland 2016-03-05 22:31:49
It's the base in one and the median in the other, so we had to take that ratio.

copeland 2016-03-05 22:31:53
Can we write $YZ$ in terms of the known values $XY$ and $XZ?$

hamup1 2016-03-05 22:32:32
XY + XZ

sxu 2016-03-05 22:32:32
XY+XZ

brian6liu 2016-03-05 22:32:32
XY+XZ

MathLearner01 2016-03-05 22:32:32
YZ=XY+XZ

copeland 2016-03-05 22:32:34
Simple, it's the sum: $YZ=YX+XZ$.

copeland 2016-03-05 22:32:35
Can we write $XM$ in terms of the known values $XY$ and $XZ?$

deltaepsilon6 2016-03-05 22:33:45
XM=.5*(XZ-XY)

MathLearner01 2016-03-05 22:33:45
(XZ-XY)/2

copeland 2016-03-05 22:33:47
Since $YM=ZM,$ we have $XM+YX=XZ-XM$ so $XM=(XZ-YX)/2$.

copeland 2016-03-05 22:33:50
Now we can rewrite that equation, \[\frac{AM}{XM}=\frac{YZ}{AB}.\] It becomes, \[\frac{AM}{(XZ-YX)/2}=\frac{YX+XZ}{AB}.\]

copeland 2016-03-05 22:33:57
Cross-multiplying gives \[2AM\cdot AB=(XZ-YX)(XZ+YX)=XZ^2-YX^2.\]

copeland 2016-03-05 22:33:58
Now what?

deltaepsilon6 2016-03-05 22:35:10
let AM=.5AB and substitute

hamup1 2016-03-05 22:35:10
$AB^2 = XZ^2-YX^2$!

brian6liu 2016-03-05 22:35:10
our answer is just XZ^2-YX^2=37*67-47^2=270

calculus_riju 2016-03-05 22:35:10
2AM=AB^2

calculus_riju 2016-03-05 22:35:10
we know RHS

calculus_riju 2016-03-05 22:35:10
jst compute

calculus_riju 2016-03-05 22:35:10
tada!

nosaj 2016-03-05 22:35:10
plug in stuff!

copeland 2016-03-05 22:35:12
Well, $2AM=AB,$ so the left side is what we want, and we know the right side, too:

copeland 2016-03-05 22:35:14
\[AB^2=\left(\sqrt{37\cdot67}\right)^2-\left(47\right)^2.\]

copeland 2016-03-05 22:35:18
What does that simplify to?

brian6liu 2016-03-05 22:35:44
270

brian22 2016-03-05 22:35:44
270 I suppose

swagger21 2016-03-05 22:35:44
270

fclvbfm934 2016-03-05 22:35:44
270

fz2012 2016-03-05 22:35:49
270

copeland 2016-03-05 22:35:51
\begin{align*}
AB^2&=37\cdot67-47^2\\
&=(47-10)(47+20)-47^2\\
&=47\cdot10-200\\
&=\boxed{270}.
\end{align*}

copeland 2016-03-05 22:35:55
Great!

copeland 2016-03-05 22:35:59
That's all. That's it.

copeland 2016-03-05 22:36:05
Time to go have dinner?

blizzard10 2016-03-05 22:36:45
That was long.

copeland 2016-03-05 22:36:51
Possibly a Math Jam record, yes.

copeland 2016-03-05 22:37:00
Thanks, Problem 15 geo writers.

copeland 2016-03-05 22:37:07
Don't forget to come back for the AIME II Math Jam, tentatively planned for March 18.

calculus_riju 2016-03-05 22:37:12
see u on the nect jam i.e. AIME II

copeland 2016-03-05 22:37:17
See you around.

violu 2016-03-05 22:38:00
Is the math jam finished?

copeland 2016-03-05 22:38:06
Yes, all done. Nothing else.

copeland 2016-03-05 22:38:10
Go away!

2016 AIME I Math Jam

Facilitator: Jeremy Copeland