Math Jams

DPatrick 2013-04-05 19:00:10
Welcome to the 2013 AIME II Math Jam!

DPatrick 2013-04-05 19:00:15
I'm Dave Patrick, and I'll be leading our discussion tonight.

DPatrick 2013-04-05 19:00:21
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.

DPatrick 2013-04-05 19:00:28
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.

DPatrick 2013-04-05 19:00:38
This helps keep the session organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.

DPatrick 2013-04-05 19:01:00
There are a lot of students here already, and there will likely be more people arriving as we go. As I said, only a relatively small fraction of the well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!

DPatrick 2013-04-05 19:01:18
Also, we won't be going through all 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.

DPatrick 2013-04-05 19:01:28
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.

DPatrick 2013-04-05 19:01:38
We do have two teaching assistants with us tonight to help answer your questions: Catherine Sheard (greekpanda) and Elena Sizikova (Anna Smith).

DPatrick 2013-04-05 19:01:53
They can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.

DPatrick 2013-04-05 19:02:10
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.

DPatrick 2013-04-05 19:02:30
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 most likely won't be acknowledged.

DPatrick 2013-04-05 19:02:44
Let's get started! We're going to work through all 15 problems, in order.

DPatrick 2013-04-05 19:02:55

DPatrick 2013-04-05 19:03:06
Notice that the current problem under discussion will always be placed at the top of the window. You can resize that top region by dragging the horizontal bar separating the top pane from the main pane.

mathmaster2012 2013-04-05 19:03:47
Fractional parts of the day

sunny2000 2013-04-05 19:03:47
make proportions and use the information that they give you already! (metric, etc)

DPatrick 2013-04-05 19:03:59
Right, that's a productive way to think about it.

DPatrick 2013-04-05 19:04:11
In terms of a fraction of a day, what does the time A:BC mean?

sibirica 2013-04-05 19:04:41
(100*A+10*B+10*C)/1000ths of a day

genesis2 2013-04-05 19:04:41
ABC/1000

sarvottam 2013-04-05 19:04:41
ABC/1000?

Tommy2000 2013-04-05 19:04:46
100A+10B+c/1000

DPatrick 2013-04-05 19:04:50
Yes: $\dfrac{\text{ABC}}{1000}$ is the fraction of the day that has gone by. (Here, we're thinking of ABC as a 3-digit number, which conveniently happens to be what we want our final answer to be.)

DPatrick 2013-04-05 19:05:10
So the question becomes: at 6:36 AM, what fraction of the day has passed?

mathmaster2012 2013-04-05 19:05:42
11/40

bobthesmartypants 2013-04-05 19:05:42
(6*60+36)/1440

countyguy 2013-04-05 19:05:42
11/40

JWK750 2013-04-05 19:05:42
11/40

stellalau 2013-04-05 19:05:42
ABC/1000 = (60*6 + 36) / (1440)

kdokmeci 2013-04-05 19:05:42
1/4+(3/5)/24=11/40

DPatrick 2013-04-05 19:05:46
Right.

DPatrick 2013-04-05 19:05:51
The 6 hours have consumed $\frac{6}{24} = \frac14$ of the day.

DPatrick 2013-04-05 19:06:01
The 36 minutes have consumed $\frac{36}{60} = \frac35$ of the next hour...

DPatrick 2013-04-05 19:06:05
...but that hour itself is $\frac{1}{24}$ of the day.

DPatrick 2013-04-05 19:06:14
So the 36 minutes consume another $\frac35 \cdot \frac{1}{24} = \frac{1}{40}$ of the day.

DPatrick 2013-04-05 19:06:30
Thus, at 6:36 AM, $\frac{1}{4} + \frac{1}{40} = \frac{11}{40}$ of the day is gone.

DPatrick 2013-04-05 19:06:46
But we need this as thousandths of the day in order to find the metric time.

mathmaster2012 2013-04-05 19:07:01
1000*(11/40)=275

lazorpenguin27143 2013-04-05 19:07:01
multiply by 1000 to get answer

kdokmeci 2013-04-05 19:07:01
11/40=0.275, so answer is 275

sunny2000 2013-04-05 19:07:01
proportions!

ZZmath9 2013-04-05 19:07:01
$\frac{275}{1000}$

countyguy 2013-04-05 19:07:01
11/40=275/1000

DPatrick 2013-04-05 19:07:11
Yep: \[
\frac{11}{40} = \frac{11 \cdot 25}{40 \cdot 25} = \frac{275}{1000},
\]
so $\dfrac{275}{1000}$ of the day is gone at 6:36 AM.

DPatrick 2013-04-05 19:07:20
So we set the alarm for 2:75 metric time. Answer $\boxed{275}$.

DPatrick 2013-04-05 19:07:34

sunny2000 2013-04-05 19:08:10
first simplify by getting the logs out

genesis2 2013-04-05 19:08:10
Try to turn these into exponents.

AayushGupta 2013-04-05 19:08:10
remove the log 2

DPatrick 2013-04-05 19:08:23
Right, we want to get rid of the logs and write this in a way that we can better understand.

DPatrick 2013-04-05 19:08:42
Let's work from the outside in.

distortedwalrus 2013-04-05 19:08:49
this means that the part inside the argument of the first log expression must be equal to 1.

bobthesmartypants 2013-04-05 19:08:49
If $\log_2{x}=0$, then $x$ must equal $1$

DPatrick 2013-04-05 19:09:09
Good start. If $\log_2(x) = 0$, then $x = 1$, since $2^0 = 1$.

DPatrick 2013-04-05 19:09:26
So now we have
$$
\log_{2^a}(\log_{2^b}(2^{1000})) = 1.
$$

DPatrick 2013-04-05 19:09:57
Now what?

DPatrick 2013-04-05 19:10:07
Next: if $\log_{2^a}(y) = 1$, then what's $y$?

countyguy 2013-04-05 19:10:21
2^a

sunny2000 2013-04-05 19:10:21
2^a

lazorpenguin27143 2013-04-05 19:10:21
2^a

Superwiz 2013-04-05 19:10:21
2^a

tc1729 2013-04-05 19:10:21
2^a

DPatrick 2013-04-05 19:10:30
Yes: $y = 2^a$, since $(2^a)^1 = 2^a$.

DPatrick 2013-04-05 19:10:37
So now we have
\[
\log_{2^b}(2^{1000}) = 2^a.
\]

DPatrick 2013-04-05 19:10:55
But what is the log that remains?

kdokmeci 2013-04-05 19:11:15
Simplify log: It equals 1000/b

wpk 2013-04-05 19:11:15
2^1000 = (2^b)^(1000/b)

matholympiad25 2013-04-05 19:11:15

turkeybob777 2013-04-05 19:11:15
1000/b

ZZmath9 2013-04-05 19:11:15
$\frac{1000}{b}$

mathwrath 2013-04-05 19:11:15
1000/b

DPatrick 2013-04-05 19:11:23

sibirica 2013-04-05 19:11:35
simplify as 1000/b=2^a

DPatrick 2013-04-05 19:11:46
Thus our equation is simply $\dfrac{1000}{b} = 2^a$.

DPatrick 2013-04-05 19:11:55
What now?

MathPerson060375 2013-04-05 19:12:03
So 1000 = b*2^a

AlcumusGuy 2013-04-05 19:12:03
multiply both sides by b

DPatrick 2013-04-05 19:12:08
We need to solve $1000 = 2^a \cdot b$ where $a$ and $b$ are positive integers.

bobthesmartypants 2013-04-05 19:12:29
now we factor 1000=2^3 x 5^3

wpk 2013-04-05 19:12:29
1000 = 2^3 * 5^3

TheStrangeCharm 2013-04-05 19:12:29
1000 = 2^3*5^3

DPatrick 2013-04-05 19:12:39
Aha, perhaps writing it as $2^3 \cdot 5^3 = 2^a \cdot b$ will help.

vyzw 2013-04-05 19:13:00
a = 1,2,3

AlcumusGuy 2013-04-05 19:13:00
so a = 1, 2, or 3

DPatrick 2013-04-05 19:13:23
Right. We see that $a$ must be 1, 2, or 3. (Note that $a$ must be positive, so $a \not= 0$.)

Tommy2000 2013-04-05 19:13:47
a=1,2,3 and b= 500, 250,125 respectively

mathmaster2012 2013-04-05 19:13:47
(1,500), (2,250), and (3,125) are our solution sets

DPatrick 2013-04-05 19:13:55
Right: this gives the solutions
\[
(a,b) = \{ (1,500), (2,250), (3,125) \}.
\]

ZZmath9 2013-04-05 19:14:12
$1 + 500 + 2 + 250 + 3 + 125 = \boxed{881}$ is the answer

matholympiad25 2013-04-05 19:14:12
1+2+3+500+250+125=881

ninjataco 2013-04-05 19:14:12
so we have 881

genesis2 2013-04-05 19:14:12

DPatrick 2013-04-05 19:14:16
The sum of all of these is
\[
1+500+2+250+3+125 = \boxed{881}.
\]

DPatrick 2013-04-05 19:14:48
By the way, I should mention that there will be a full transcript of this entire session posted on the web page once we're done.

DPatrick 2013-04-05 19:15:01
So you don't have to take notes, and you'll be able to go back and review anything you might have missed.

DPatrick 2013-04-05 19:15:09

kdokmeci 2013-04-05 19:15:32
Find T

mathwrath 2013-04-05 19:15:32
Find the time it takes to burn down!

DPatrick 2013-04-05 19:15:39
Seems like a good place to start. What is $T$?

ninjataco 2013-04-05 19:16:00
T = 10(1+2+3+...+118+119)

JWK750 2013-04-05 19:16:00
T=10(1+2..+119)

piguy314 2013-04-05 19:16:00
(10+20+30+...+1190)

DPatrick 2013-04-05 19:16:06
We know that
\begin{align*}
T &= 10 + 20 + 30 + \cdots + 1190 \\
&= 10(1+2+3+\cdots+119).
\end{align*}

DPatrick 2013-04-05 19:16:13
What is that sum?

ninjataco 2013-04-05 19:16:32
T=10* (119*120)/2

sibirica 2013-04-05 19:16:32
119*120*10/2

fmasroor 2013-04-05 19:16:32
5(119)(120)

MathPerson060375 2013-04-05 19:16:32
Which is (119*120/2)*10

viva 2013-04-05 19:16:32
119*120*5

Ronnicus 2013-04-05 19:16:32
10*(119*120/2)

countyguy 2013-04-05 19:16:40
triangular numbers

DPatrick 2013-04-05 19:16:46
It's $T = 10 \cdot\frac{(119)(120)}{2} = 5(119)(120)$.

DPatrick 2013-04-05 19:17:12
(Let's not multiply it out -- it doesn't look like it'll help us at the moment.)

sunny2000 2013-04-05 19:17:17
divide that time by half

DPatrick 2013-04-05 19:17:24
So $T/2 = 5(119)(60)$, and we need to find out how much has burned (and hence how much is left) after this many seconds.

DPatrick 2013-04-05 19:17:32
What now?

kdokmeci 2013-04-05 19:17:48
T/2 is also a triangular number.

ninjataco 2013-04-05 19:17:48
another triangular number because it is 10(1+2+...+x)

DPatrick 2013-04-05 19:18:17
Right. If $x$ is an integer, then it takes $10(1+2+\cdots+x) = 5x(x+1)$ seconds to burn $x$ centimeters.

DPatrick 2013-04-05 19:18:43
(If $x$ is not an integer we might have a problem. But let's assume it is for the moment and hope we get lucky.)

DPatrick 2013-04-05 19:18:55
So if we get lucky, we might be able to solve
$$
5x(x+1) = 5(119)(60).
$$

ninjataco 2013-04-05 19:19:09
x(x+1) = 119 * 60

sunny2000 2013-04-05 19:19:09
divide by 5 on both sides

DPatrick 2013-04-05 19:19:13
This is $x(x+1) = 119(60)$.

DPatrick 2013-04-05 19:19:21
How do we solve this (without a calculator)?

matholympiad25 2013-04-05 19:19:41
estimate the square root

piguy314 2013-04-05 19:19:51
119=17*7, then use 17 for some trial and error to get 84*85

fmasroor 2013-04-05 19:19:51
factor 119 as 7*17, maybe something comes up

DPatrick 2013-04-05 19:19:56
That's basically what I did.

DPatrick 2013-04-05 19:20:13
The left side is almost a perfect square...so we can guess what perfect square is the right side close to.

DPatrick 2013-04-05 19:20:46
$119(60) = 7140$ (you don't even need to compute it exactly, just estimate that it's about $120(60) = 7200$, and that's between $80^2 = 6400$ and $90^2 = 8100$. Probably close to halfway.

DPatrick 2013-04-05 19:21:20
And we'll need a factor of 5, so $x=84$ and $x=85$ are the only reasonable things to try.

ZZmath9 2013-04-05 19:21:29
The last digit is $0$, so it's either $84 \times 85$ or $85 \times 86$. Checking, we find it's $84 \times 85$

kdokmeci 2013-04-05 19:21:29
84*85

matholympiad25 2013-04-05 19:21:29

DPatrick 2013-04-05 19:21:38
And indeed, $x=84$ works, since $84 \cdot 85 = 7140$.

DPatrick 2013-04-05 19:21:44
We got lucky. After exactly 84 centimeters have burned, we've used up $\frac{T}{2}$ seconds.

epicpwn314 2013-04-05 19:21:58
yeah so x=84 and 119-84=35 so the answer is 350

mathwrath 2013-04-05 19:21:58
119-84=35

wpk 2013-04-05 19:22:01
leaving 35

DPatrick 2013-04-05 19:22:07
Be careful at the end! The final answer is 10 times how much is left. There are $119 - 84 = 35$ centimeters remaining, so the final answer is $\boxed{350}$.

DPatrick 2013-04-05 19:22:37

DPatrick 2013-04-05 19:22:55
If you brought graph paper and a ruler and protractor to the contest (which I hope you did!), you can draw a pretty accurate picture:

DPatrick 2013-04-05 19:23:02

DPatrick 2013-04-05 19:23:13
(There's already a quick sanity check for our answer for $P$: it looks like it's close to $(\frac52,\frac32)$.)

DPatrick 2013-04-05 19:23:20
What will make this problem a little easier to solve?

stellalau 2013-04-05 19:23:57
Easy to shift A to the origin to work with it. So A' = (0,0), B' = (1,2root3)

TheStrangeCharm 2013-04-05 19:23:57
shifting the triangle over 1 so A is the origin

DPatrick 2013-04-05 19:24:07
I like the idea of moving everything in the diagram 1 unit to the left, so that A becomes the origin. But if we do this: DO NOT FORGET TO MOVE IT BACK BEFORE COMPUTING THE FINAL ANSWER!!

DPatrick 2013-04-05 19:24:16

DPatrick 2013-04-05 19:24:26
I've given them new names. $A' = (0,0)$ and $B' = (1,2\sqrt3)$ now.

DPatrick 2013-04-05 19:24:37
How do we find $C'$?

tc1729 2013-04-05 19:25:14
toss onto the complex plane

matholympiad25 2013-04-05 19:25:14
Complex numbers!

TheStrangeCharm 2013-04-05 19:25:14
complex number!

sunny2000 2013-04-05 19:25:14
rotation of 60 degrees=polar coordinates

tc1729 2013-04-05 19:25:14
C' is a -60 degree rotation of B', which is easy to deal with on the complex plane

DPatrick 2013-04-05 19:25:24
There are many ways to proceed (in fact the AMC published 3 different "official" solutions for this problem). For example, you could compute the midpoint of $A'B'$ (call it $M'$), and then compute the line through $M$ perpendicular to $A'B'$, and then go down this line the correct distance (specifically, $\sqrt3(A'B')$) to find $C'$.

DPatrick 2013-04-05 19:25:42
But I like the method of thinking of these as points in the complex plane. (Or in terms of polar coordinates, which is essentially the same thing.)

DPatrick 2013-04-05 19:25:56
What point is $B'$ if we think of it as a complex number?

bobthesmartypants 2013-04-05 19:26:20
1+2isqrt3

ZZmath9 2013-04-05 19:26:20
$1 + 2i\sqrt{3}$

matholympiad25 2013-04-05 19:26:20

kdokmeci 2013-04-05 19:26:20
1+2i root 3

DPatrick 2013-04-05 19:26:24
We have $B' = 1 + 2\sqrt{3}i$. (The $x$-coordinate becomes the real part, and the $y$-coordinate becomes the imaginary part.)

DPatrick 2013-04-05 19:26:39
And what do we do to B' to get C'?

matholympiad25 2013-04-05 19:27:04
Multiply by $\cos(-60^{\circ}) + i\sin(-60^{\circ})$

fmasroor 2013-04-05 19:27:04
rotate by -60 degrees

twin77 2013-04-05 19:27:04
rotate -60 degrees

TheStrangeCharm 2013-04-05 19:27:04
multiply by cis(-pi/3)

DPatrick 2013-04-05 19:27:18
We rotate $B'$ by 60 degrees clockwise. (Which is easy to do now that A' is the origin!)

DPatrick 2013-04-05 19:27:41
That's the same as multiplying $B'$ by the complex number $e^{-\frac{\pi}{3}i}$. (Or $\cis(-60^\circ)$ if you prefer.)

DPatrick 2013-04-05 19:28:07
Either way, \[
e^{-\frac{\pi}{3}i} = \cos\left(-\frac\pi3\right) + i\sin\left(-\frac\pi3\right).
\]

DPatrick 2013-04-05 19:28:26
What is this complex number?

twin77 2013-04-05 19:29:03
1/2 - sqrt3/2 i

vyzw 2013-04-05 19:29:06
1/2-isqrt3/2

theGoodGuy 2013-04-05 19:29:13
1/2 - isqrt3/2

matholympiad25 2013-04-05 19:29:13

DPatrick 2013-04-05 19:29:33
This is $\frac12 - \frac{\sqrt3}{2}i$. (The real part is $\cos(-60^\circ)$, and the imaginary part is $\sin(-60^\circ)$.)

DPatrick 2013-04-05 19:29:45
So we multiply $B'$ by this to get:
\[
C' = (1 + 2\sqrt3i)\left(\frac12 - \frac{\sqrt3}{2}i\right).
\]

DPatrick 2013-04-05 19:29:51
What does this simplify to?

theGoodGuy 2013-04-05 19:30:37
7/2+isqrt3/2

MathPerson060375 2013-04-05 19:30:43
7/2 + (sqrt(3)/2)i

DPatrick 2013-04-05 19:31:17
Multiplying it out gives \[ C' = \frac12 - \frac{\sqrt3}{2}i + \frac{2\sqrt3}{2}i - \frac{2\sqrt3\sqrt3}{2}i^2.\]

DPatrick 2013-04-05 19:31:30
This simplifies to \[
C' = \frac72 + \frac{\sqrt3}{2}i.
\]

DPatrick 2013-04-05 19:31:44
(Note that the last term in the expansion is $+3$.)

DPatrick 2013-04-05 19:32:11
Also this is a good point to check that this $C'$ is consistent with our original picture, which it is!

kdokmeci 2013-04-05 19:32:22
C'=(7/2, root 3 /2)

DPatrick 2013-04-05 19:32:30
Right. Which is about where it looks to be.

DPatrick 2013-04-05 19:32:35
Now, what is $P'$?

matholympiad25 2013-04-05 19:32:54

vyzw 2013-04-05 19:32:54
average the 3 points

TheStrangeCharm 2013-04-05 19:32:54
centroid or just average of all the corrdinates

DPatrick 2013-04-05 19:33:00
Right. It's the average of $A'$, $B'$, and $C'$.

DPatrick 2013-04-05 19:33:20
That is,
\[
P' = \left(\frac13\left(0 + 1 + \frac72\right),\frac13\left(0+2\sqrt3+\frac{\sqrt3}{2}\right)\right).
\]

DPatrick 2013-04-05 19:33:46
This simplifies to $P' = \left(\frac32,\frac{5\sqrt3}{6}\right)$.

vyzw 2013-04-05 19:33:57
(3/2,5sqrt3/6) but then shift it back

twin77 2013-04-05 19:34:06
shift it 1 to the right

MathPerson060375 2013-04-05 19:34:06
And move back to P (not prime)

DPatrick 2013-04-05 19:34:07
BUT DON'T FORGET TO SHIFT IT BACK!

DPatrick 2013-04-05 19:34:12
So we have $P = \left(\frac52,\frac{5\sqrt3}{6}\right)$.

DPatrick 2013-04-05 19:34:23
(And looking at our sanity check from the picture, we indeed see that this is close to $(\frac52,\frac32)$.)

DPatrick 2013-04-05 19:34:41
(From the original picture, I mean!)

ninjataco 2013-04-05 19:34:51
then we have 25root3/12, and then the answer is 040

ZZmath9 2013-04-05 19:34:51
Product: $ \frac{25\sqrt{3}}{12} $, answer: $25 + 3 + 12 = \boxed{40}$

kdokmeci 2013-04-05 19:34:51
x*y=25root3 /12; 25+3+12=040

DPatrick 2013-04-05 19:34:58
And to finish: the product of the coordinates of $P$ is $\dfrac{25\sqrt{3}}{12}$, so our final answer is $25 + 3 + 12 = \boxed{040}$.

DPatrick 2013-04-05 19:35:23
On to another back-to-back problem involving an equilateral triangle...

DPatrick 2013-04-05 19:35:28

DPatrick 2013-04-05 19:35:40
Let's draw a picture.

DPatrick 2013-04-05 19:35:49
But there's a clever tactic we can use...

mathwrath 2013-04-05 19:35:56
I plugged in values.

DPatrick 2013-04-05 19:36:05
Right. Since we have flexibility with the lengths, let's pick some convenient ones. I made the trisected lengths 2 and the big triangle's side length 6, so as to avoid fractions:

DPatrick 2013-04-05 19:36:12

DPatrick 2013-04-05 19:36:27
How can we get data for $\sin(DAE)$?

epicpwn314 2013-04-05 19:36:46
draw the altitude

sunny2000 2013-04-05 19:36:46
draw altitude

Lemon123 2013-04-05 19:36:46
Draw Altitude/Median

tenniskidperson3 2013-04-05 19:36:46
Area

ssilwa 2013-04-05 19:36:46
Use area

Tommy2000 2013-04-05 19:36:46
make a right triangle?

bobthesmartypants 2013-04-05 19:36:46
First draw height from A to BC

DPatrick 2013-04-05 19:37:05
First let's think about the area. What do we know about this picture?

kdokmeci 2013-04-05 19:37:28
Area is trisected

ZZmath9 2013-04-05 19:37:28
All three triangles have same area

sparkles257 2013-04-05 19:37:28
three triangles have equal areas

DPatrick 2013-04-05 19:37:33
Right. Each small triangle is one-third the area of the big triangle.

DPatrick 2013-04-05 19:37:51
What's the area of the big triangle (using the lengths I've chosen)?

countyguy 2013-04-05 19:38:12
$9sqrt3$

theGoodGuy 2013-04-05 19:38:12
9sqrt3

AayushGupta 2013-04-05 19:38:12
9sqrt3 is the area of the big one

fmasroor 2013-04-05 19:38:12
9sqrt3

az_phx_brandon_jiang 2013-04-05 19:38:12
9sqrt3

NumberGiant 2013-04-05 19:38:12
9sqrt3

DPatrick 2013-04-05 19:38:16
\[
[ABC] = \frac{6^2 \cdot \sqrt{3}}{4} = 9\sqrt3.
\]

DPatrick 2013-04-05 19:38:24
So each small triangle has area $3\sqrt3$.

DPatrick 2013-04-05 19:38:37
How does that help?

vyzw 2013-04-05 19:38:53
so the height is 3sqrt3.

bobthesmartypants 2013-04-05 19:38:53
so the height is 3sqrt3/2*2=3sqrt3

yrushi 2013-04-05 19:38:53
so altitude is $3\sqrt{3}$

ZZmath9 2013-04-05 19:38:53
We now can find the height

DPatrick 2013-04-05 19:39:15
Indeed, the height of any of them is $3\sqrt3$.

DPatrick 2013-04-05 19:39:18

kdokmeci 2013-04-05 19:39:49
So we can find AD

Lemon123 2013-04-05 19:39:49
pythogras Theorem for AE or AD

noobynoob 2013-04-05 19:39:49
Area = $(AD)(DE)\sin DAE

DPatrick 2013-04-05 19:39:55
Aha, all the pieces are falling into place!

yrushi 2013-04-05 19:40:14
find $AD=2\sqrt{7}$

vyzw 2013-04-05 19:40:14
AD=2sqrt7

DPatrick 2013-04-05 19:40:16
We have
$$
(AD)^2 = 1^2 + (3\sqrt3)^2 = 28,
$$
so $AD = \sqrt{28} = 2\sqrt7$.

DPatrick 2013-04-05 19:40:43
But now the thing we want -- $\sin(DAE)$ -- is part of the area formula for the middle triangle.

matholympiad25 2013-04-05 19:40:54

az_phx_brandon_jiang 2013-04-05 19:40:54
[ADE]=3sqrt3=1/2 sin(DAE)*AD*DE

mxie 2013-04-05 19:41:02
now use the area formula, Area = AD * AD * sin DAE * 1/2

DPatrick 2013-04-05 19:41:11
We use the area formula for triangle $ADE$:
\[
3\sqrt3 = [ADE] = \frac12(AD)^2\sin(DAE).
\]

ninjataco 2013-04-05 19:41:22
and we can substitute in the values we know

DPatrick 2013-04-05 19:41:31
We know $(AD)^2 = 28$, so
\[
3\sqrt3 = 14\sin(DAE),
\]
hence $\sin(DAE) = \dfrac{3\sqrt3}{14}$.

ninjataco 2013-04-05 19:41:43
and the answer is 020

matholympiad25 2013-04-05 19:41:49
3+3+14=020

twin77 2013-04-05 19:41:49
the answer is 020

mathmaster2012 2013-04-05 19:41:49
3+3+14=20, which is our answer!

DPatrick 2013-04-05 19:41:53
Our final answer is $3+3+14 = \boxed{020}$.

sunny2000 2013-04-05 19:41:56
wait, couldn't you just have drawn altitude, use pythagorean, and use double angle sine formula?

DPatrick 2013-04-05 19:42:02
Certainly!

DPatrick 2013-04-05 19:42:19
But I trust myself to remember the area formula more than the double-angle formula.

DPatrick 2013-04-05 19:42:38

DPatrick 2013-04-05 19:42:57
Just to agree on some terminology, let's call a "block" a set of 1000 consecutive digits beginning with $1000N$ for some $N$. These are all the numbers that share the same digits except for the final three.

DPatrick 2013-04-05 19:43:05
How high do we have to go before the perfect squares get at least 1000 apart?

mathwrath 2013-04-05 19:43:21
500^2 and 501^2 are the first two more than 1000 apart.

yrushi 2013-04-05 19:43:21
we have $(n+1)^2-n^2=2n+1$

ssilwa 2013-04-05 19:43:21
500

davidkim2106 2013-04-05 19:43:21
500?

epicpwn314 2013-04-05 19:43:21
500

JWK750 2013-04-05 19:43:21
N=500

genesis2 2013-04-05 19:43:21
250000

DPatrick 2013-04-05 19:43:33
Right. Notice that $(k+1)^2 - k^2 = 2k+1$.

DPatrick 2013-04-05 19:43:40
So we need $k \ge 500$ for $k^2$ and $(k+1)^2$ to be at least 1000 apart.

DPatrick 2013-04-05 19:43:52
So 500 is the smallest we need to even consider looking at.

DPatrick 2013-04-05 19:43:58
Does $k=500$ work? That is, does the block after $k^2$ avoid perfect squares?

kdokmeci 2013-04-05 19:44:20
500^2=250000, 501^2=2510001

math-rules 2013-04-05 19:44:20
no

AayushGupta 2013-04-05 19:44:20
no because it includes 500^2

mathmaster2012 2013-04-05 19:44:26
No, 501^2=500^2+1001 so no avoidance

DPatrick 2013-04-05 19:44:36
Right. $500^2 = 250000$ doesn't work.

DPatrick 2013-04-05 19:44:56
And $501^2 = 251001$, so we don't avoid a perfect square in the $251xxx$ block.

mathwrath 2013-04-05 19:45:12
(500+x)^2=500^2+2*500*x+x^2=500^2+100x+x^2

bobthesmartypants 2013-04-05 19:45:12
can we represent $k$ as $x+500$, such that $x\ge 0$

twin77 2013-04-05 19:45:12
use (500+n)^2 to test

DPatrick 2013-04-05 19:45:24
Good idea. $(500+a)^2 = 250000 + 1000a + a^2$, so choosing $k = 500+a$ will give us a perfect square in the block beginning with $1000 \cdot (250+a)$:

DPatrick 2013-04-05 19:45:29
\begin{align*}
501^2 &= 251001, \\
502^2 &= 252004, \\
503^2 &= 253009, \\
504^2 &= 254016, \\
\text{etc.}
\end{align*}

swimmerstar 2013-04-05 19:45:45
wait the last 3 digits are perfect squares

ZZmath9 2013-04-05 19:45:45
There's a pattern

DPatrick 2013-04-05 19:45:50
Aha.

DPatrick 2013-04-05 19:46:10
The last three digits of $(500+a)^2$ are just $a^2$.

DPatrick 2013-04-05 19:46:25
So if $a^2 \ge 1000$, then we might be able to skip a block.

Tommy2000 2013-04-05 19:46:46
it will stop when it gets to 961 or 31^2

Superwiz 2013-04-05 19:46:46
a^2>1000[

mathworld1 2013-04-05 19:46:46
we need a >= 32

vyzw 2013-04-05 19:46:46
so then since 31^2<1000 we check 32

stellalau 2013-04-05 19:46:46
so then we choose a were a^2 > 1000, smallest of this would be 32^2

DPatrick 2013-04-05 19:46:54
Notice that $31^2 = 961$ and $32^2 = 1024$.

DPatrick 2013-04-05 19:47:01
$531^2 = (500+31)^2 = 250000 + 31000 + 961 = 281961$

DPatrick 2013-04-05 19:47:06
$532^2 = (500+32)^2 = 250000 + 32000 + 1024 = 283024$

DPatrick 2013-04-05 19:47:16
We win!

mathwrath 2013-04-05 19:47:38
282000 has none.

ninjataco 2013-04-05 19:47:38
it skipped 282000

bobthesmartypants 2013-04-05 19:47:38
It skips 282000, so $\boxed{282}$ is the answer

NumberGiant 2013-04-05 19:47:38

DPatrick 2013-04-05 19:47:46
The $282xxx$ block doesn't have any perfect squares!

DPatrick 2013-04-05 19:47:52
So our answer is $\boxed{282}$.

DPatrick 2013-04-05 19:48:14

ninjataco 2013-04-05 19:48:39
assign variables for the original number of clerks and the number that are reassigned

matholympiad25 2013-04-05 19:48:39
Let original number of clerks = c, number removed every hour = r.

DPatrick 2013-04-05 19:48:46
Good idea. Suppose we start with $c$ clerks, and reassign $r$ of them every hour.

DPatrick 2013-04-05 19:48:59
How many files get sorted in the first hour?

MathPerson060375 2013-04-05 19:49:14
30c

19bobhu 2013-04-05 19:49:14
30c

az_phx_brandon_jiang 2013-04-05 19:49:14
30c

countyguy 2013-04-05 19:49:14
30c

dan2012 2013-04-05 19:49:14
30c

DPatrick 2013-04-05 19:49:18
In the first hour, they sort $30c$ files.

DPatrick 2013-04-05 19:49:23
How about the second hour?

number.sense 2013-04-05 19:49:34
30c-30r

JWK750 2013-04-05 19:49:34
30(c-r)

wpk 2013-04-05 19:49:34
30(c-r)

yrushi 2013-04-05 19:49:34
30(c-r)

mxie 2013-04-05 19:49:34
30(c-r)

fmasroor 2013-04-05 19:49:34
30(c-r)

DPatrick 2013-04-05 19:49:38
In the second hour, there are only $c-r$ clerks remaining, so they sort another $30(c-r)$ files.

DPatrick 2013-04-05 19:49:43
How about the third hour?

sparkles257 2013-04-05 19:50:00
30(c-2r)

genesis2 2013-04-05 19:50:00
30(c-2r)

ssilwa 2013-04-05 19:50:00
30c-60r

davidkim2106 2013-04-05 19:50:00

distortedwalrus 2013-04-05 19:50:00
30(c-2r)

atmath2011 2013-04-05 19:50:00
30(c-2r)

DPatrick 2013-04-05 19:50:04
In the third hour, there are only $c-2r$ clerks remaining, so they sort another $30(c-2r)$ files.

DPatrick 2013-04-05 19:50:11
And how about the final 10 minutes?

AayushGupta 2013-04-05 19:50:30
5(c-3r)

matholympiad25 2013-04-05 19:50:30
10/60*30(c-3r)=5(c-3r)

kdokmeci 2013-04-05 19:50:30
5(c-3r)

mathmaster2012 2013-04-05 19:50:30
5(c-3r)

math-rules 2013-04-05 19:50:30
5(c-3r)

ilikepie333 2013-04-05 19:50:30
(30(c-3r))/6

DPatrick 2013-04-05 19:50:35
In the final 10 minutes (which is $\frac16$ of an hour), there are only $c-3r$ clerks remaining, so they sort the final $5(c-3r)$ files.

fmasroor 2013-04-05 19:50:44
add them up, we should get 1775

DPatrick 2013-04-05 19:50:50
Hence, the total number of files sorted is
\[
30c + 30(c-r) + 30(c-2r) + 5(c-3r) = 1775.
\]'

DPatrick 2013-04-05 19:51:02
This simplifies to $95c - 105r = 1775$.

mathmaster2012 2013-04-05 19:51:11
Divide by 5

twin77 2013-04-05 19:51:11
divide by 5

mathworld1 2013-04-05 19:51:11
19c-21r = 355

DPatrick 2013-04-05 19:51:14
We can divide by 5 to get $19c - 21r = 355$.

DPatrick 2013-04-05 19:51:20
How do we solve this?

matholympiad25 2013-04-05 19:51:33
c-3r>0, since there were still some clerks in the last 10 minutes

DPatrick 2013-04-05 19:51:47
That's an important condition to note. We must have c and r positive integers, and c-3r > 0.

yrushi 2013-04-05 19:52:20
just add 21 to 355 until we get a multiple of 19

ninjataco 2013-04-05 19:52:20
c= (21r+355)/19, create a table?

DPatrick 2013-04-05 19:52:29
You could certainly guess-and-check to find the answer from here.

number.sense 2013-04-05 19:52:41
consider modulo 19

ssilwa 2013-04-05 19:52:41
take mod 19

bobthesmartypants 2013-04-05 19:52:41
use modular arithmetic

DPatrick 2013-04-05 19:52:52
A more systematic way to solve is to look at the equation either mod 19 or mod 21.

DPatrick 2013-04-05 19:53:20
Funnily enough, you all suggested mod 19. I actually used mod 21 at the time. But let's do mod 19 since you all suggested it.

DPatrick 2013-04-05 19:54:42

DPatrick 2013-04-05 19:54:51
And what is 355 mod 19?

stellalau 2013-04-05 19:55:12
13

Tommy2000 2013-04-05 19:55:12
13

mathwrath 2013-04-05 19:55:12
also -6

hjia1 2013-04-05 19:55:12
13

Debdut 2013-04-05 19:55:12
13

Superwiz 2013-04-05 19:55:12
-6 (mod 19)

MathPerson060375 2013-04-05 19:55:12
Is 13 mod 19

DPatrick 2013-04-05 19:56:03
You could divide it out to compute that 355 = 13 (mod 19).

DPatrick 2013-04-05 19:56:44
But 13 is the same as -6 (mod 19), so our equation is just $-2r \equiv -6 \pmod{19}$.

math-rules 2013-04-05 19:56:58
r=3

vyzw 2013-04-05 19:56:58
r=3 is possible from -6

dilei 2013-04-05 19:56:58
r=3

19bobhu 2013-04-05 19:56:58
r=3

DPatrick 2013-04-05 19:57:11
This gives r=3 as the smallest possible solution. Does it work?

countyguy 2013-04-05 19:57:29
plug r=3 into the original equation

ninjataco 2013-04-05 19:57:29
yes and then c = 22

wpk 2013-04-05 19:57:29
c = 22

DPatrick 2013-04-05 19:57:35
It does!

DPatrick 2013-04-05 19:58:01

DPatrick 2013-04-05 19:58:15
So we get a solution with 22 clerks, with 3 being removed every hour.

noobynoob 2013-04-05 19:58:26
what if r were bigger?

DPatrick 2013-04-05 19:58:38
That's a good question to ask. You can observe that this is the only solution, since to get any other potential solution we'd have to increase the number of clerks by 21 and the number removed by 19. But 43 clerks, removing 22 per hour, would leave a negative number of clerks after 2 hours! And if we went up to 64 clerks, they'd finish all the sorting in the first hour!

DPatrick 2013-04-05 19:59:04
So (c,r) = (22,3) is in fact the only solution that works.

DPatrick 2013-04-05 19:59:11
So we have 22 clerks in the first hour, with 3 being removed every hour. What's the final answer?

yrushi 2013-04-05 19:59:20
now we need $30c+15(c-r)$

twin77 2013-04-05 19:59:36
so the answer is 30(22) + 15(19) = 945

bobthesmartypants 2013-04-05 19:59:36
now to find the 1.5 hour we do 30(22)+15(19)

dan2012 2013-04-05 19:59:40
the answer is 945 then

DPatrick 2013-04-05 19:59:43
In 1.5 hours, they sort $30c + 15(c-r) = 45c - 15r = 15(3c-r)$ files. Hence our answer is
\[
15(3(22)-3) = 15(63) = \boxed{945}.
\]

DPatrick 2013-04-05 20:00:06

DPatrick 2013-04-05 20:00:17
Let's sketch a picture:

DPatrick 2013-04-05 20:00:22

DPatrick 2013-04-05 20:00:26
Let's also call the unknown radius $r$.

DPatrick 2013-04-05 20:00:28
Now what?

ssilwa 2013-04-05 20:01:11
draw some lines

MSTang 2013-04-05 20:01:11
draw radii to the vertices

kdokmeci 2013-04-05 20:01:11
Draw radii

AayushGupta 2013-04-05 20:01:11
connect the center to each of the vertices

DPatrick 2013-04-05 20:01:33
There are a lot of complicated things you could do, or one really simple thing you could do.

DPatrick 2013-04-05 20:01:56
In fact, we only need to draw two radii:

DPatrick 2013-04-05 20:02:02

DPatrick 2013-04-05 20:02:11
What do we know about this triangle?

matholympiad25 2013-04-05 20:02:28
it is isosceles

sparkles257 2013-04-05 20:02:28
isosclees

sunny2000 2013-04-05 20:02:28
it's isoceles

vyzw 2013-04-05 20:02:28
isosceles

fmasroor 2013-04-05 20:02:28
isosceles

stellalau 2013-04-05 20:02:28
isosceles

DPatrick 2013-04-05 20:02:48
Sure: both blue radii have length $r$. But surely we know a little more than that...

genesis2 2013-04-05 20:03:04
the triangle seems to cry out to draw an altitude

DPatrick 2013-04-05 20:03:22

DPatrick 2013-04-05 20:03:39
So what?

Apollo13 2013-04-05 20:04:11
use pythagarean theroem

Apollo13 2013-04-05 20:04:11
use Pythagorean theorem

zhuangzhuang 2013-04-05 20:04:11
Pythagorean theo

Tommy2000 2013-04-05 20:04:11
call the height h and solve for it using the pythagorean theorm

MathPerson060375 2013-04-05 20:04:11
pythag!

math-rules 2013-04-05 20:04:11
pythag for the altitude on both right triangles and equate

ico_the_psycho 2013-04-05 20:04:11
Equate the two expressions for the altitude, obtained by Pythagoras' Thm?

DPatrick 2013-04-05 20:04:23
If we call the height $h$, then by the Pythagorean Theorem (twice):
\[
h^2 = r^2 - 10^2 = 22^2 - (r-10)^2.
\]

ilikepie333 2013-04-05 20:04:49
simplify for r

DPatrick 2013-04-05 20:04:54
This gives
$$
r^2 - 100 = 484 - r^2 + 20r - 100,
$$
or $2r^2 - 20r - 484 = 0$.

HYP135peppers 2013-04-05 20:05:12
divide by 2

bobthesmartypants 2013-04-05 20:05:12
divide by 2

DPatrick 2013-04-05 20:05:16
Dividing by 2 gives $r^2 - 10r - 242 = 0$.

DPatrick 2013-04-05 20:05:53
(Note: several people are asking where the 10 and r-10 came from. The "10" is half of the "20" from the top side.)

mathwrath 2013-04-05 20:06:05
quadratic formula

19bobhu 2013-04-05 20:06:05
Quadratic

DPatrick 2013-04-05 20:06:10
Now the quadratic formula gives
\[
r = \frac{10 \pm \sqrt{100 + 4(242)}}{2} = 5 \pm \sqrt{25 + 242} = 5 \pm \sqrt{267}.
\]

tapir1729 2013-04-05 20:06:22

MSTang 2013-04-05 20:06:22
disregard negative solution

DPatrick 2013-04-05 20:06:27
We clearly want the positive value of $r$, so $r = 5 + \sqrt{267}$, and our final answer is $5 + 267 = \boxed{272}$.

DPatrick 2013-04-05 20:07:00

DPatrick 2013-04-05 20:07:15
I found that a little picture helped me visualize the problem. Here's the sample that they gave:

DPatrick 2013-04-05 20:07:20

DPatrick 2013-04-05 20:07:24
And here's the blank board:

DPatrick 2013-04-05 20:07:29

DPatrick 2013-04-05 20:07:36
How do we count the tilings?

DPatrick 2013-04-05 20:07:56
Let's suppose we tried to build one of these tilings. How could we do so?

HYP135peppers 2013-04-05 20:08:30
fill in with different colors and size

AayushGupta 2013-04-05 20:08:36
assign a first color, then a second, and so on, adding tile breaks as well

DPatrick 2013-04-05 20:08:57
Let's just move left-to-right along the board.

DPatrick 2013-04-05 20:09:06
How many choices for the first square?

yrushi 2013-04-05 20:09:21
3

dan2012 2013-04-05 20:09:21
3

kdokmeci 2013-04-05 20:09:21
3

ssilwa 2013-04-05 20:09:21
3

stellalau 2013-04-05 20:09:21
3

19JasonH 2013-04-05 20:09:21
3

DPatrick 2013-04-05 20:09:28
We need to pick a color for the first square. That's 3 choices.

DPatrick 2013-04-05 20:09:37
Then, for the next square, what are the choices?

MathPerson060375 2013-04-05 20:09:57
4 choices

zhuangzhuang 2013-04-05 20:09:57
same or new tile

AayushGupta 2013-04-05 20:09:57
4

MSTang 2013-04-05 20:10:04
extend tile from first square, or build another separate tile in 3 ways

DPatrick 2013-04-05 20:10:16
Right: We could continue the previous tile, or we could start a new tile in any of the 3 colors. So there are 4 choices for the next square.

DPatrick 2013-04-05 20:10:46
The same is true for every subsequent square: we could continue the previous tile, or we could start a new tile in any of the 3 colors. So there are 4 choices for each subsequent square.

DPatrick 2013-04-05 20:10:58
For example, the sample tiling above is from the choices:
red, green, continue, green, blue, continue, green

AayushGupta 2013-04-05 20:11:13
so 3 x 4^6

viva 2013-04-05 20:11:13
3*(4^6)

19bobhu 2013-04-05 20:11:13
so it's 3*4^6

dan2012 2013-04-05 20:11:13
3x4^6

DPatrick 2013-04-05 20:11:40
Right, there are 3 choices for the first square, then 4 choices for each of the 6 subsequent square, so that's $3 \cdot 4^6 = 3 \cdot 4096 = 12288$ tilings total.

number.sense 2013-04-05 20:11:50
have to correct for not using all 3 colors

TheStrangeCharm 2013-04-05 20:11:50
but we need to use each color exactly once?

math-rules 2013-04-05 20:11:50
but you need to make sure all colors are used

sume 2013-04-05 20:11:50
now count tilings that dont have three colors

DPatrick 2013-04-05 20:12:16
Indeed, it's not quite so simple. We've counted some tilings that don't use all three colors, as required.

DPatrick 2013-04-05 20:12:37
We need to ensure that we use each color at least once.

DPatrick 2013-04-05 20:12:54
So we must subtract off the colorings that don't use some color.

DPatrick 2013-04-05 20:12:58
How many tilings don't use red?

kdokmeci 2013-04-05 20:13:24
2x3^6

atmath2011 2013-04-05 20:13:24
2*3^6

shreyash 2013-04-05 20:13:24
2*3^6

Tommy2000 2013-04-05 20:13:24
2*3^6

twin77 2013-04-05 20:13:24
2*3^6

DPatrick 2013-04-05 20:13:36
Right, we can use the same logic: now we have 2 choices (blue or green) for the first square, and 3 choices (continue, blue, or green) for each of the 6 subsequent squares.

DPatrick 2013-04-05 20:13:44
So there are $2 \cdot 3^6 = 2 \cdot 729 = 1458$ tilings that don't use red.

Tommy2000 2013-04-05 20:13:50
same for blue and green

matholympiad25 2013-04-05 20:13:53
similar for blue, green

JoshH 2013-04-05 20:13:53
Same with other two colors

DPatrick 2013-04-05 20:13:57
Similarly, there are 1458 that don't use blue, and 1458 that don't use green.

ilikepie333 2013-04-05 20:14:04
4374 tilings that are not good

DPatrick 2013-04-05 20:14:08
So our tiling count is now $12288 - 3(1458) = 12288 - 4374 = 7914$.

DPatrick 2013-04-05 20:14:14
Are we done?

matholympiad25 2013-04-05 20:14:40
now we count back those that have only one color.

twin77 2013-04-05 20:14:40
but we need to add the ones with only 1 color

bobthesmartypants 2013-04-05 20:14:40
no, undercounted

DPatrick 2013-04-05 20:14:49
We're not done! We've doubly-subtracted the tilings that only use one color.

DPatrick 2013-04-05 20:14:53
(For example, a tiling that's all blue will have been subtracted twice, once in the "no red" group and once in the "no green" group.)

DPatrick 2013-04-05 20:15:05
So we need to add the one-color tilings back in.

DPatrick 2013-04-05 20:15:11
How many are there?

MathPerson060375 2013-04-05 20:15:38
So add in 3*2^6

MSTang 2013-04-05 20:15:38
There are 3*2^6 of them

HYP135peppers 2013-04-05 20:15:38
2^6 for each color

willabc 2013-04-05 20:15:38
3*2^6

countyguy 2013-04-05 20:15:38
3*2^6=192

coldsummer 2013-04-05 20:15:38
for red its 2^6, multiply by 3 to get 192 not counted cases

DPatrick 2013-04-05 20:15:48
Right. There are 3 choices for which color to use, and then $2^6 = 64$ choices for whether to start a new tile or not at each subsequent square.

DPatrick 2013-04-05 20:15:57
So there are $3 \cdot 64 = 192$ one-color tilings.

DPatrick 2013-04-05 20:16:03
We add these back in to get $7914 + 192 = 8106$ legal tilings.

DPatrick 2013-04-05 20:16:08
Are we done?

sume 2013-04-05 20:16:26
yes

mathwrath 2013-04-05 20:16:26
Yes

TheStrangeCharm 2013-04-05 20:16:26
yes

MathPerson060375 2013-04-05 20:16:26
I hope

biddyxm 2013-04-05 20:16:26
Yes

ilikepie333 2013-04-05 20:16:26
106=final answer

Tommy2000 2013-04-05 20:16:26
8106mod 100= 106

twin77 2013-04-05 20:16:26
yes, so the answer is 106

DPatrick 2013-04-05 20:16:38
Now we're done! There are 8106 tilings, so mod 1000 our answer is $\boxed{106}$.

DPatrick 2013-04-05 20:17:13
We've finish all the single-digit problems (1-9). I'm going to take a quick break to rest my typing hands, and we'll resume at :22 past the hours to look at problems 10-15.

DPatrick 2013-04-05 20:21:39
:30 second warning

DPatrick 2013-04-05 20:22:16

DPatrick 2013-04-05 20:22:36
Obviously we want a picture!

DPatrick 2013-04-05 20:22:41

DPatrick 2013-04-05 20:22:44
We want to maximize the area of the red triangle.

DPatrick 2013-04-05 20:22:51
Any ideas?

DPatrick 2013-04-05 20:23:58
To me, the given lengths suggest to try to set up areas of triangles with bases OB or AB or AO. We want to be able to isolate [BKL] but let's see what we can find out.

math-rules 2013-04-05 20:24:05
[BKL]=[ABL]-[ABK]

DPatrick 2013-04-05 20:24:31
Indeed, so let's look at ABL and ABK. And let's get O in there too. So let me add OK and OL to the picture.

DPatrick 2013-04-05 20:24:37

DPatrick 2013-04-05 20:25:10

DPatrick 2013-04-05 20:25:29
What triangle(s) can we relate [ABK] to easily?

DPatrick 2013-04-05 20:25:42
(By the way, for those who don't know, [ABK] is a shorthand for the area of ABK.)

MSTang 2013-04-05 20:26:07
OBK?

ilikepie333 2013-04-05 20:26:07
OBK?

sparkles257 2013-04-05 20:26:07
ako

matholympiad25 2013-04-05 20:26:07
AKO

HYP135peppers 2013-04-05 20:26:07
[OBK]

atmath2011 2013-04-05 20:26:07
obk

DPatrick 2013-04-05 20:26:34
Either OBK or OAK share the same height from K down to the line A--B--O, and we also know how the lengths of the bases relate.

DPatrick 2013-04-05 20:27:00
It turns out that using OAK will be better in the end (though admittedly it's not so easy maybe to see that now).

DPatrick 2013-04-05 20:27:09
What's the ratio [KAB]/[KAO]?

ninjataco 2013-04-05 20:27:20
their areas are in the same proportion as their bases

They 2013-04-05 20:27:33
4/(4+rt13)

kdokmeci 2013-04-05 20:27:33
4/(4+root 13)

mathwrath 2013-04-05 20:27:33
4/(4+sqrt13)

ssilwa 2013-04-05 20:27:33
4/(4+sqrt13)

MSTang 2013-04-05 20:27:33

DPatrick 2013-04-05 20:27:51
$KAB$ and $KAO$ have the same height from $K$, so the ratio of their areas is the ratio of their bases:
\[
\frac{[KAB]}{[KAO]} = \frac{4}{4+\sqrt{13}}.
\]

DPatrick 2013-04-05 20:28:15
So far, so good.

DPatrick 2013-04-05 20:28:57
We earlier observed that [BKL] is the difference between [KAB] and [LAB], so maybe we should try to do the same thing with [LAB]?

matholympiad25 2013-04-05 20:29:20
yes, indeed, the areas are in same ratio

bobthesmartypants 2013-04-05 20:29:23
[LAB] with [LOB]

DPatrick 2013-04-05 20:29:37
Right, we can do the same thing with point $L$:
\[
\frac{[LAB]}{[LAO]} = \frac{4}{4+\sqrt{13}}.
\]

DPatrick 2013-04-05 20:30:03
And keep in mind that what's nice is that the red triangle is the difference between LAB and KAB, the two numerators of our ratios of areas.

DPatrick 2013-04-05 20:30:13
So we have two ratios that are the same. What does that mean?

DPatrick 2013-04-05 20:30:34
...and we want [LAB] - [KAB]....

bobthesmartypants 2013-04-05 20:30:47
subtract them

kdokmeci 2013-04-05 20:30:53
subtract ratios?

DPatrick 2013-04-05 20:31:02
Right! If two ratios are the same, we're allowed to subtract them!

DPatrick 2013-04-05 20:31:08
That is, if $\frac{a}{b} = \frac{c}{d} = k$, then $\frac{a-c}{b-d} = k$ too. (This is worth learning, and proving, if you haven't seen it before!)

DPatrick 2013-04-05 20:31:25
So let's use that with our area ratios!

DPatrick 2013-04-05 20:31:32
\[
\frac{[LAB]-[KAB]}{[LAO]-[KAO]} = \frac{[LKB]}{[LKO]} = \frac{4}{4+\sqrt{13}}.
\]

DPatrick 2013-04-05 20:31:45
What have we accomplished by doing this?

zhuangzhuang 2013-04-05 20:32:03
maximise area of LKO

number.sense 2013-04-05 20:32:03
so you want to maximize LKO's area.... which is much easier to think about

DPatrick 2013-04-05 20:32:28
Exactly. Our goal is to maximize [LKB]. But now we can maximize [LKO] instead and use it to compute [LKB].

DPatrick 2013-04-05 20:32:41
How do we maximize the area of LKO?

mathwrath 2013-04-05 20:33:12
best with 90 degree arc KL

noobynoob 2013-04-05 20:33:12
measure of $\angle LOK = 90$

They 2013-04-05 20:33:12
Remember it's isosceles

matholympiad25 2013-04-05 20:33:12
LKO area is easy to maximize, it is just 1/2*OK*OL*sin LKO, so max sin LKO is just 1, so it is just 13/2

bobthesmartypants 2013-04-05 20:33:12
LKO is maximized in area when LOK=90 degrees

zhuangzhuang 2013-04-05 20:33:12
make it right

shreyash 2013-04-05 20:33:19
OK and OL are perpendicular

DPatrick 2013-04-05 20:33:28
Exactly. $LKO$ is isosceles with sides $LO = KO = \sqrt{13}$.

DPatrick 2013-04-05 20:33:44
So its area is maximized when $\angle LOK$ is a right angle. Is that in fact possible to do in our construction?

ws0414 2013-04-05 20:34:08
yes?

ssilwa 2013-04-05 20:34:08
yes

sunny2000 2013-04-05 20:34:08
yes

MSTang 2013-04-05 20:34:08
yes

They 2013-04-05 20:34:08
Why not?

ico_the_psycho 2013-04-05 20:34:26
Yes, because there is enough wiggle room for the angle to vary from 0 to 180 degrees.

DPatrick 2013-04-05 20:34:28
Sure -- we just draw the ray from A to cut off a 90-degree arc of the circle.

DPatrick 2013-04-05 20:34:51
(The picture I drew looks a little too big an arc is cut off, but I was close!)

ninjataco 2013-04-05 20:35:02
then the area is 13/2

fmasroor 2013-04-05 20:35:02
[LOK]=13/2

Tommy2000 2013-04-05 20:35:02
lko's area is 13\2

kdokmeci 2013-04-05 20:35:02
Maximum area of LOK is 13/2

DPatrick 2013-04-05 20:35:16
So we get a maximum area of $[LKO] = \frac12(\sqrt{13})^2 = \frac{13}{2}$.

ninjataco 2013-04-05 20:35:24
so, knowing that [LKO] is 13/2, we can substitute and solve for [LKB]

ssilwa 2013-04-05 20:35:24
now just cross multiply ansd solve for LKB

DPatrick 2013-04-05 20:35:37
Right, we go back to our ratio to get:
\[
[LKB] = \left(\frac{4}{4+\sqrt{13}}\right)[LKO] = \frac{26}{4+\sqrt{13}}.
\]

DPatrick 2013-04-05 20:35:52
Almost there, one small step to finish...

Debdut 2013-04-05 20:36:07
Rationalize

countyguy 2013-04-05 20:36:07
conjugates to rationalize

distortedwalrus 2013-04-05 20:36:07
woohoo rationalize!

Hamliet 2013-04-05 20:36:07
CONJUGATE

muhaboug 2013-04-05 20:36:07
no sqrts in the denoms

AayushGupta 2013-04-05 20:36:07
rationalize the denominator

DPatrick 2013-04-05 20:36:16
We've still got to rationalize the denominator:
\[
[LKB] = \frac{26}{4+\sqrt{13}} \cdot \frac{4-\sqrt{13}}{4-\sqrt{13}} = \frac{104 - 26\sqrt{13}}{3}.
\]

kdokmeci 2013-04-05 20:36:30
104+26+13+3=146

DPatrick 2013-04-05 20:36:33
Now it's in the correct format, so our answer is $104+26+13+3 = \boxed{146}$.

DPatrick 2013-04-05 20:36:55
This may have been my favorite problem, and I rarely say that about geometry problems!

DPatrick 2013-04-05 20:37:09

DPatrick 2013-04-05 20:37:26
As with the tiling problem, let's think about how we would construct such a function.

DPatrick 2013-04-05 20:37:36
What do we have to decide?

ico_the_psycho 2013-04-05 20:38:01
What the constant is going to be.

MSTang 2013-04-05 20:38:01
which constant f(f(x)) should be

DPatrick 2013-04-05 20:38:06
We first need to pick the constant. Let's call it $c$, so that $f(f(x)) = c$ for all $x \in A$.

DPatrick 2013-04-05 20:38:14
There are obviously 7 choices for $c$.

DPatrick 2013-04-05 20:38:28
Now what?

matholympiad25 2013-04-05 20:39:00
now we pick the numbers n such that f(n)=c

ico_the_psycho 2013-04-05 20:39:00
Now we decide which subset of A collapses to c under the action of f.

DPatrick 2013-04-05 20:39:41
Right. Next we think about the rest of the numbers in A. They either map to c directly in one step (that is, $f(x) = c$), or in two steps (so that $f(x) \not=c$ but $f(f(x)) = c$).

matholympiad25 2013-04-05 20:39:47
c needs to map to c

DPatrick 2013-04-05 20:39:55
This may seem obvious, but how do we know that?

DPatrick 2013-04-05 20:40:03
That is, how do we know that $f(c) = c$?

DPatrick 2013-04-05 20:40:40
Why can't we have $f(c) = d$ for some $d \not= c$ and then $f(d) = c$?

q12 2013-04-05 20:41:13
then f(f(d))=d

They 2013-04-05 20:41:13
Then f(d) will be c and the second round will make it d.

DPatrick 2013-04-05 20:42:12
Right, if $d \not= c$, then we get $f(f(d)) = f(c) = d$, which isn't allowed.

DPatrick 2013-04-05 20:42:28
So we must have $f(c) = c$.

DPatrick 2013-04-05 20:42:52
But for each of the other 6 elements, they might map to $c$ in one step or two steps.

matholympiad25 2013-04-05 20:43:00
now we pick the other ones that map to c

DPatrick 2013-04-05 20:43:23
Right. And we have to keep in mind that the ones that don't map directly to $c$ then have to map to one of the elements that does map to $c$.

DPatrick 2013-04-05 20:43:38
This is getting a little confusing in words, so I decided to make a little chart to keep track of everything.

DPatrick 2013-04-05 20:43:43
\[
\begin{array}{c||c|c|c}
\text{# of add'l elements} & \text{# of choices for} & \text{# of choices for} & \\
\text{mapping to $c$} & \text{elements mapping to $c$} & \text{other elements} & \text{total} \\ \hline
1 \\
2 \\
3 \\
4 \\
5 \\
6
\end{array}
\]

MSTang 2013-04-05 20:43:59
at least one must map to c

mathwrath 2013-04-05 20:43:59
At least one maps directly to c.

DPatrick 2013-04-05 20:44:20
Indeed, we can't have all other 6 elements not map to $c$ in one step, because they'd never map to $c$ in two steps either.

DPatrick 2013-04-05 20:44:29
So I don't have a "0" row in my left column.

DPatrick 2013-04-05 20:44:45
For each row, in how many ways can we choose the elements that map to $c$?

DPatrick 2013-04-05 20:45:01
If we need $k$ elements, how many ways are there to pick them?

DPatrick 2013-04-05 20:45:15
(Note that "add'l" is an abbreviation for "additional" in the chart.)

math-rules 2013-04-05 20:45:26
6ck, where k is the number in the left column

zhuangzhuang 2013-04-05 20:45:26
if the number in the row is x, then we can do 6CX

distortedwalrus 2013-04-05 20:45:26
6Ck

TheStrangeCharm 2013-04-05 20:45:26
6 choose k

zhuangzhuang 2013-04-05 20:45:26
6Ck

PlatinumFalcon 2013-04-05 20:45:26
6Ck

DPatrick 2013-04-05 20:45:42
If we need $k$ elements from the other 6 elements to map to $c$ (besides $c$ itself), then there are $\binom{6}{k}$ choices.

DPatrick 2013-04-05 20:45:47
So let's fill in the second column of the table:

DPatrick 2013-04-05 20:45:53
\[
\begin{array}{c||c|c|r}
\text{# of add'l elements} & \text{# of choices for} & \text{# of choices for} & \\
\text{mapping to $c$} & \text{elements mapping to $c$} & \text{other elements} & \text{total} \\ \hline
1 & 6 \\
2 & 15 \\
3 & 20 \\
4 & 15 \\
5 & 6 \\
6 & 1
\end{array}
\]

DPatrick 2013-04-05 20:46:07
How about the third column?

DPatrick 2013-04-05 20:46:15
Each of the other items must map to one of the other elements mapping to $c$.

ico_the_psycho 2013-04-05 20:46:38
Then the remaining $6-k$ elements must map to these $k$ elements. This can be done in $k^{6-k}$ ways.

DPatrick 2013-04-05 20:46:47
Right. Let's see how this works.

DPatrick 2013-04-05 20:46:58
The first row has 5 other elements, but they each only have 1 choice: they must map to the element (other than $c$) that maps to $c$. So that's $1^5$ choices.

DPatrick 2013-04-05 20:47:24
The second row has 4 other elements, and each has 2 choices of where to map to: one of the 2 elements (other than $c$) that maps to $c$. So that's $2^4$ choices.

DPatrick 2013-04-05 20:47:33
And so on.

DPatrick 2013-04-05 20:47:37
We continue like this and fill in the third column:

DPatrick 2013-04-05 20:47:41
\[
\begin{array}{c||c|c|r}
\text{# of add'l elements} & \text{# of choices for} & \text{# of choices for} & \\
\text{mapping to $c$} & \text{elements mapping to $c$} & \text{other elements} & \text{total} \\ \hline
1 & 6 & 1^5 \\
2 & 15 & 2^4 \\
3 & 20 & 3^3 \\
4 & 15 & 4^2 \\
5 & 6 & 5^1 \\
6 & 1 & 6^0
\end{array}
\]

DPatrick 2013-04-05 20:48:03
What's the fourth column? What is the total possibilities for each row?

Tommy2000 2013-04-05 20:48:15
multipy

number.sense 2013-04-05 20:48:15
multiple pairwise the second and third columns to get the total column

sparkles257 2013-04-05 20:48:15
multiply them

Tommy2000 2013-04-05 20:48:19
multiply column2 *column 3

DPatrick 2013-04-05 20:48:58
Right. To construct a legal function we must make the choices indicated in both the second *and* third columns, so we multiply to get the total choices for that row:

DPatrick 2013-04-05 20:49:03
\[
\begin{array}{c||c|c|l}
\text{# of add'l elements} & \text{# of choices for} & \text{# of choices for} & \\
\text{mapping to $c$} & \text{elements mapping to $c$} & \text{other elements} & \text{total} \\ \hline
1 & 6 & 1^5 & 6 \cdot 1 = 6 \\
2 & 15 & 2^4 & 15 \cdot 16 = 240 \\
3 & 20 & 3^3 & 20 \cdot 27 = 540 \\
4 & 15 & 4^2 & 15 \cdot 16 = 240 \\
5 & 6 & 5^1 & 6 \cdot 5 = 30 \\
6 & 1 & 6^0 & 1 \cdot 1 = 1
\end{array}
\]

MSTang 2013-04-05 20:49:22
add totals

distortedwalrus 2013-04-05 20:49:22
add em up

DPatrick 2013-04-05 20:49:26
So that gives a total of
\[
6 + 240 + 540 + 240 + 30 + 1 = 1057
\]
ways to finish constructing the function, once we've chosen $c$.

zhuangzhuang 2013-04-05 20:49:40
*7.

mathwrath 2013-04-05 20:49:40
7 possibilities for c...

DPatrick 2013-04-05 20:49:53
And there were 7 choices for $c$ at the beginning. Thus, there are $7 \cdot 1057 = 7399$ such functions.

DPatrick 2013-04-05 20:50:00
Our answer is $\boxed{399}$.

DPatrick 2013-04-05 20:50:29

DPatrick 2013-04-05 20:50:45
Broadly speaking, what do we know about the roots of a cubic?

countyguy 2013-04-05 20:51:12
three of them

kdokmeci 2013-04-05 20:51:12
Either all real, or 2 complex one real

TheStrangeCharm 2013-04-05 20:51:12
either all 3 are real, or one is real and the others are complex conjugates

zhuangzhuang 2013-04-05 20:51:12
they have at least one real root

fmasroor 2013-04-05 20:51:12
all real or 2 imaginaries 1 real

matholympiad25 2013-04-05 20:51:12
complex conjugates if the coefficients are real

DPatrick 2013-04-05 20:51:26
Right. I suppose I should have asked "a cubic with real coefficients" to be more clear, sorry.

DPatrick 2013-04-05 20:51:30
We either have 3 real roots, or 1 real root and a conjugate pair of complex roots.

kdokmeci 2013-04-05 20:51:37
Casework?

DPatrick 2013-04-05 20:51:44
Yeah: we'll count cubics in each case, and add them.

DPatrick 2013-04-05 20:52:03
(Note that since the cubics in $S$ are monic, they are uniquely determined by their roots.)

DPatrick 2013-04-05 20:52:08
How many possibilities are there for three real roots?

DPatrick 2013-04-05 20:52:19
Note that the roots have to be in the set $\{-20,-13,13,20\}$, and we can take any three elements of this set (with repetition) and get a different cubic.

DPatrick 2013-04-05 20:52:33
How many are there?

DPatrick 2013-04-05 20:52:54
I mean, how many sets of 3 roots from the set (with possible repetition)?

giftedbee 2013-04-05 20:53:18
20

googol.plex 2013-04-05 20:53:18
we have 20, 4 from n^3, 12 from n^2*m, and 4 from taking 3 different ones

countyguy 2013-04-05 20:53:18
20

DPatrick 2013-04-05 20:53:45
Right. You could do something semi-complicated (like a "stars and bars" or "balls and urns" argument) to count them, but I found it easy to just think about the cases.

DPatrick 2013-04-05 20:53:57
There are $\binom43 = 4$ with three distinct roots.

DPatrick 2013-04-05 20:54:09
There are 4*3=12 with a double root (4 choices) and a third different root (3 choices).

DPatrick 2013-04-05 20:54:15
And there are 4 with a triple root.

DPatrick 2013-04-05 20:54:19
So there are 4+12+4 = 20 with three real roots.

DPatrick 2013-04-05 20:54:39
Now on to the other case: a complex (non-real) conjugate pair and a real root.

DPatrick 2013-04-05 20:54:57
There are obviously 4 choices for the real root.

DPatrick 2013-04-05 20:55:02
How about the conjugate pair?

DPatrick 2013-04-05 20:55:12
What do we know about it?

MSTang 2013-04-05 20:55:26
have the same magnitude

countyguy 2013-04-05 20:55:26
absolute value of 20 or 13

kdokmeci 2013-04-05 20:55:29
Their magnitude is 20 or 13

DPatrick 2013-04-05 20:55:34
Right. If our conjugate pair is $p \pm qi$, then we must have $p^2 + q^2 = 13^2$ or $p^2 + q^2 = 20^2$.

DPatrick 2013-04-05 20:55:46
What are the conditions on $p$ and/or $q$?

DPatrick 2013-04-05 20:56:22
A lot of you are saying they have to be integers...why?

DPatrick 2013-04-05 20:56:47
In fact there's no reason either of them has to be an integer.

zhuangzhuang 2013-04-05 20:57:07
2p is an integer(by vieta) and q is not 0

Sayan 2013-04-05 20:57:17
2p is an integer

DPatrick 2013-04-05 20:57:41
We do know that the sum of the three roots is an integer: it's $-a$ from the original cubic, by Vieta's Formulas.

DPatrick 2013-04-05 20:58:07
But if $p \pm qi$ and $r$ are our roots (where $r$ is some real number), what is their sum?

mathmaster2012 2013-04-05 20:58:28
2p+r!

matholympiad25 2013-04-05 20:58:28
2p+r

ssilwa 2013-04-05 20:58:28
2p+r

shreyash 2013-04-05 20:58:28
2p + r

JWK750 2013-04-05 20:58:28
2p+r

DPatrick 2013-04-05 20:58:44
Their sum is \[ (p + qi) + (p - qi) + r = 2p + r.\]

DPatrick 2013-04-05 20:59:04
We already know $r$ is an integer, so $2p$ must be an integer too. (We sometimes say that $p$ is a half-integer.)

DPatrick 2013-04-05 20:59:34
There actually is no condition at all on $q$, except that it's non-zero (so that the roots are non-real).

DPatrick 2013-04-05 20:59:54
So if $p^2 + q^2 = 13^2$, how many choices are there for the complex roots?

MSTang 2013-04-05 21:00:34
from -12.5 to 12.5 on p, so 51

aux770 2013-04-05 21:00:34
51

DPatrick 2013-04-05 21:00:54
We must have $|p| < 13$. (We can't have $|p| = 13$ because that would make $q = 0$.)

DPatrick 2013-04-05 21:01:00
So $p$ must be in the set
\[
\left\{ -\frac{25}{2},-\frac{24}{2},\ldots,0,\ldots,\frac{24}{2},\frac{25}{2}\right\}.
\]

DPatrick 2013-04-05 21:01:14
So there are 51 choices for $p$, and once we choose $p$, the imaginary part $q$ is uniquely determined by $p^2 + q^2 = 13^2$.

DPatrick 2013-04-05 21:01:25
How about if $p^2 + q^2 = 20^2$? How many choices?

mathwrath 2013-04-05 21:01:43
79

tapir1729 2013-04-05 21:01:43
79

Tommy2000 2013-04-05 21:01:43
79

AayushGupta 2013-04-05 21:01:43
79

DPatrick 2013-04-05 21:01:47
Now $p$ must be in the set
\[
\left\{ -\frac{39}{2},-\frac{38}{2},\ldots,0,\ldots,\frac{38}{2},\frac{39}{2}\right\}.
\]

DPatrick 2013-04-05 21:01:57
So there are 79 choices for $p$, and once we choose $p$, the imaginary part $q$ is uniquely determined.

DPatrick 2013-04-05 21:02:22
Hence combined there are $51 + 79 = 130$ possible non-real conjugate pairs.

DPatrick 2013-04-05 21:02:32
So how many cubics with non-real roots?

kdokmeci 2013-04-05 21:02:54
130x4=520

piguy314 2013-04-05 21:02:54
4*130 = 520

DPatrick 2013-04-05 21:03:04
Right, don't forget each of these has 4 choices for the third (real) root!

DPatrick 2013-04-05 21:03:28
So there are $4 \cdot 130 = 520$ cubics in the case of "1 real, 2 non-real conjugate roots".

zhuangzhuang 2013-04-05 21:03:40
520+20--=540 as our answer

mathmaster2012 2013-04-05 21:03:40
520+20=540, which is our answer!

shreyash 2013-04-05 21:03:40
520 + 20 = 540

googol.plex 2013-04-05 21:03:40
so adding the cases we have 540

DPatrick 2013-04-05 21:03:47
And to finish: adding the 20 cubics with real roots, we get an answer of $520 + 20 = \boxed{540}$.

DPatrick 2013-04-05 21:04:12
I actually liked that problem a lot too. #10 and #12 were my co-favorites.

MSTang 2013-04-05 21:04:29
which was your least favorite?

DPatrick 2013-04-05 21:04:36
#15. And we'll soon see why.

DPatrick 2013-04-05 21:04:43

DPatrick 2013-04-05 21:04:58
Here's a picture. I've used $x$ and $y$ for unknown lengths, incorporating the given data.

DPatrick 2013-04-05 21:05:03

DPatrick 2013-04-05 21:05:33
There are some really clever ways to solve this by extending lots of lines, or by using some high-powered machinery, but I just did a lot of Law of Cosines bashing.

DPatrick 2013-04-05 21:05:51
I've got lots of lengths and lots of triangles that share either common angles or supplementary angles.

matholympiad25 2013-04-05 21:06:15
Stewart's Theorem on ACD and CEB

DPatrick 2013-04-05 21:06:38
That's essentially what I did, except without actually using Stewart's Theorem, for the very good reason that I can never remember it.

DPatrick 2013-04-05 21:06:49
And you don't actually need it if you know the Law of Cosines.

RelaxationUtopia 2013-04-05 21:06:53
Law of Cos on supplementary angles is essentially Stewart's Theorem in disguise xD

kdokmeci 2013-04-05 21:06:58
cos x =-cos (180-x)

DPatrick 2013-04-05 21:07:02
Indeed it is.

DPatrick 2013-04-05 21:07:08
One place to start is the supplementary angles at $E$: $AEC$ and $DEC$. Note that $\cos(AEC) = -\cos(DEC)$.

DPatrick 2013-04-05 21:07:20
So we can set up two Law of Cosines expressions, using triangles $AEC$ and $DEC$ and the common cosine.

DPatrick 2013-04-05 21:07:25
\begin{align*}
(4x)^2 &= y^2 + 7 - 2y\sqrt{7}\cos(AEC) \\
(3x)^2 &= y^2 + 7 - 2y\sqrt{7}\cos(DEC)
\end{align*}

DPatrick 2013-04-05 21:07:59
We substitute our observation about the supplementary angles:

DPatrick 2013-04-05 21:08:03
\begin{align*}
(4x)^2 &= y^2 + 7 + 2y\sqrt{7}\cos(DEC) \\
(3x)^2 &= y^2 + 7 - 2y\sqrt{7}\cos(DEC)
\end{align*}

DPatrick 2013-04-05 21:08:18
And it should be "obvious" what to do next...

fmasroor 2013-04-05 21:08:35
add?

googol.plex 2013-04-05 21:08:35
if we add the equations they go away

AayushGupta 2013-04-05 21:08:35
add both

Sayan 2013-04-05 21:08:35
add them

genesis2 2013-04-05 21:08:35
addd them together

DPatrick 2013-04-05 21:08:54
If we add the equations, all the icky parts go away!

DPatrick 2013-04-05 21:08:59
\[
25x^2 = 2y^2 + 14
\]

Tommy2000 2013-04-05 21:09:11
solve for x^2 in terms of y^2

DPatrick 2013-04-05 21:09:19
Or, to match my notes, solve for y^2 in terms of x^2.

DPatrick 2013-04-05 21:09:35
It doesn't really matter but my notes for the rest of the problem will be hopeless otherwise.

DPatrick 2013-04-05 21:09:45
We get $y^2 = \frac{25}{2}x^2 - 7$.

DPatrick 2013-04-05 21:10:00
Now what?

kdokmeci 2013-04-05 21:10:13
Law of cosine on new triangle

ssilwa 2013-04-05 21:10:17
again on CEB

DPatrick 2013-04-05 21:10:41
I'd like to find triangles with a common angle, to use the same sort of trick. Since I don't know AB (yet?) I can't use the other two supplementary angles at E.

AayushGupta 2013-04-05 21:10:49
do the law of cosines on DEB and DEC, and add again

DPatrick 2013-04-05 21:11:18
That's an idea too. And there are lots of ways that work. But (again, to match my notes) I chose ECD and ECB.

DPatrick 2013-04-05 21:11:26
Let's use the Law of Cosines on the two triangles $ECD$ and $ECB$ with the common angle at $C$.

DPatrick 2013-04-05 21:11:42
\begin{align*}
y^2 &= 7 + (3x)^2 - 6x\sqrt{7}\cos(ECD) \\
3^2 &= 7 + (4x)^2 - 8x\sqrt{7}\cos(ECB)
\end{align*}

herro66 2013-04-05 21:12:06
make the cos parts equal

DPatrick 2013-04-05 21:12:10
Now we'll use our info about $y^2$ from above, and also the fact the $ECD$ and $ECB$ are the same angle.

DPatrick 2013-04-05 21:12:14
\begin{align*}
\frac{25}{2}x^2 - 7 &= 7 + 9x^2 - 6x\sqrt{7}\cos(ECD) \\
9 &= 7 + 16x^2 - 8x\sqrt{7}\cos(ECD)
\end{align*}

19bobhu 2013-04-05 21:12:30
add?

kdokmeci 2013-04-05 21:12:30
add as planned

hatchguy 2013-04-05 21:12:30
multiply first by 4, second by two 3

kdokmeci 2013-04-05 21:12:34
Multiply top by 4, bottom by 3

DPatrick 2013-04-05 21:12:47
It's not quite as straightforward as before, but we multiply the first by 4, the second by -3, and then add:
\begin{align*}
50x^2 - 28 &= 28 + 36x^2 - 24x\sqrt{7}\cos(ECD) \\
-27 &= -21 - 48x^2 + 24x^2\sqrt{7}\cos(ECD) \\ \hline
50x^2 - 55 &= 7 - 12x^2
\end{align*}

DPatrick 2013-04-05 21:13:07
Again, the icky goes away.

zhuangzhuang 2013-04-05 21:13:21
so x=1!

mathwrath 2013-04-05 21:13:21
x=1

kdokmeci 2013-04-05 21:13:21
x=1! 1! NOT SOME CRAZY VALUE!

AayushGupta 2013-04-05 21:13:21
x^2 = 1

MSTang 2013-04-05 21:13:21
x=1

DPatrick 2013-04-05 21:13:26
Well, whad'ya know?

DPatrick 2013-04-05 21:13:30
$62x^2 = 62$, so $x = 1$.

matholympiad25 2013-04-05 21:13:40
x=1, y=sqrt(11/2)

DPatrick 2013-04-05 21:13:44
This makes $y^2 = \frac{25}{2} - 7 = \frac{11}{2}$, so $y = \sqrt{\frac{11}{2}}$.

DPatrick 2013-04-05 21:13:51
Let's update the picture with the now-known lengths:

DPatrick 2013-04-05 21:13:58

DPatrick 2013-04-05 21:14:30
Now we can continue to do ugly things from here to finish, or if we're sufficiently observant we might notice something interesting.

zhuangzhuang 2013-04-05 21:14:53
so the triangle CEB is right!

az_phx_brandon_jiang 2013-04-05 21:14:53
So CEB is right?

matholympiad25 2013-04-05 21:14:53
CEB is right!!!!!!!!!!!!!!!!!!1

hatchguy 2013-04-05 21:14:53
CEB has a right angle, by pythagoras

mathwrath 2013-04-05 21:14:53
BEC=90 degrees

DPatrick 2013-04-05 21:15:07
$CEB$ is a $3$-$\sqrt7$-$4$ right triangle!

DPatrick 2013-04-05 21:15:32
So how do we finish from here, now that we've noticed this really really useful fact?

mathwrath 2013-04-05 21:15:50
CEB=half of ACB

DPatrick 2013-04-05 21:16:19
Exactly. Note that CED and CEA have the same area (same bases AE = ED, and same altitude from C to AED).

DPatrick 2013-04-05 21:16:28
Also BED and BEA have the same area for the same reason.

DPatrick 2013-04-05 21:16:42
So CEB is exactly half the entire triangle.

DPatrick 2013-04-05 21:17:00
And since CEB is a right triangle, its area is pretty easy to compute.

mxie 2013-04-05 21:17:11
so [ACB] = 2 * (1/2) * sqrt7 * 3

cheese1000 2013-04-05 21:17:11
So 3sqrt7 is the answer?

googol.plex 2013-04-05 21:17:11
so the answer is 3sqrt 7

DPatrick 2013-04-05 21:17:21
Hence our area of $ABC$ is
\[
[ABC] = 2[CEB] = 2\left(\frac12 \cdot 3 \cdot \sqrt7\right) = 3\sqrt{7}.
\]

DPatrick 2013-04-05 21:17:30
The final answer is $3 + 7 = \boxed{010}$.

noobynoob 2013-04-05 21:17:55
easily guessable

ssilwa 2013-04-05 21:17:55
its funny how we had teh 3 and the sqrt7 all along

DPatrick 2013-04-05 21:18:15
Unfortunately, yes. I would have preferred something more interesting to happen along the way.

DPatrick 2013-04-05 21:18:23

DPatrick 2013-04-05 21:18:43
Alright, I read it and I don't get it. Anybody have any good ideas?

matholympiad25 2013-04-05 21:19:04
try out a few n starting from 20

kdokmeci 2013-04-05 21:19:04
Pattern?

twin77 2013-04-05 21:19:04
try some numbers

trophies 2013-04-05 21:19:08
try a few cases.

DPatrick 2013-04-05 21:19:30
I don't have any clever ideas, so let's just pick a few small n's and compute the f(n,k)'s and see if we learn anything.

DPatrick 2013-04-05 21:19:43
I'm going to save us some time...I did a bunch of these ahead of time:

DPatrick 2013-04-05 21:19:47
\[\begin{array}{c|cc cc cc cc cc}
\substack{n \rightarrow \\ k \downarrow} &~20~&~21~&~22~&~23~&~24~&~25~&~26~&~27\\\hline\hline
1&0&0&0&0&0&0&0&0\\
2&0&1&0&1&0&1&0&1\\
3&2&0&1&2&0&1&2&0\\
4&0&1&2&3&0&1&2&3\\
5&0&1&2&3&4&0&1&2\\
6&2&3&4&5&0&1&2&3\\
7&6&0&1&2&3&4&5&6\\
8&4&5&6&7&0&1&2&3\\
9&2&3&4&5&6&7&8&0\\
10&0&1&2&3&4&5&6&7\\
11& & &0&1&2&3&4&5\\
12& & & & &0&1&2&3\\
13& & & & &&&0&1\\
\end{array}\]

DPatrick 2013-04-05 21:20:06
Notice that the rows just cycle through the residues mod $k$, and each new row starts with a 0 (because $k$ divides $n=2k$ evenly).

DPatrick 2013-04-05 21:20:35
So the rows are easy to understand, but F(n) is (by definition) the largest value in n's column.

matholympiad25 2013-04-05 21:20:51
wow it happens when k=7,8,8,8,9,9,9,10,10,10,...

DPatrick 2013-04-05 21:20:58
Yeah, let's mark them in the table:

DPatrick 2013-04-05 21:21:02
\[\begin{array}{c|cc cc cc cc cc}
\substack{n \rightarrow \\ k \downarrow}&~20~&~21~&~22~&~23~&~24~&~25~&~26~&~27\\
\hline\hline
1&0&0&0&0&0&0&0&0\\
2&0&1&0&1&0&1&0&1\\
3&2&0&1&2&0&1&2&0\\
4&0&1&2&3&0&1&2&3\\
5&0&1&2&3&4&0&1&2\\
6&2&3&4&5&0&1&2&3\\
7&\boxed6&0&1&2&3&4&5&6\\
8&4&\boxed5&\boxed6&\boxed7&0&1&2&3\\
9&2&3&4&5&\boxed6&\boxed7&\boxed8&0\\
10&0&1&2&3&4&5&6&\boxed7\\
11& & &0&1&2&3&4&5\\
12& & & & &0&1&2&3\\
13& & & & &&&0&1\\
\end{array}\]

DPatrick 2013-04-05 21:21:19
It looks kinda like a staircase with slope -1/3. If we're not confident yet, we can collect a little more data:

DPatrick 2013-04-05 21:21:29
\[\begin{array}{c|cc cc cc cc cc}
\substack{n \rightarrow \\ k \downarrow}&~20~&~21~&~22~&~23~&~24~&~25~&~26~&~27~&~28~&~29~&~30\\
\hline\hline
1&0&0&0&0&0&0&0&0&0&0&0\\
2&0&1&0&1&0&1&0&1&0&1&0\\
3&2&0&1&2&0&1&2&0&1&2&0\\
4&0&1&2&3&0&1&2&3&0&1&2\\
5&0&1&2&3&4&0&1&2&3&4&0\\
6&2&3&4&5&0&1&2&3&4&5&0\\
7&\boxed6&0&1&2&3&4&5&6&0&1&2\\
8&4&\boxed5&\boxed6&\boxed7&0&1&2&3&4&5&6\\
9&2&3&4&5&\boxed6&\boxed7&\boxed8&0&1&2&3\\
10&0&1&2&3&4&5&6&\boxed7&\boxed8&\boxed9&0\\
11& & &0&1&2&3&4&5&6&7&\boxed8\\
12& & & & &0&1&2&3&4&5&6\\
13& & & & &&&0&1&2&3&4\\
14& & & & &&&&&0&1&2\\
15& & & & &&&&&&&0\\
\end{array}\]

DPatrick 2013-04-05 21:21:39
OK, I'm convinced.

DPatrick 2013-04-05 21:22:18
On the actual contest, if you were confident that the pattern continues, you could assume such and proceed.

DPatrick 2013-04-05 21:22:35
But let's at least try to quickly convince ourselves why this pattern holds.

DPatrick 2013-04-05 21:22:54
What do you notice as you read up each column from the bottom?

thehanbro 2013-04-05 21:23:26
+2

ssilwa 2013-04-05 21:23:26
0246 or 1357

jmvc003 2013-04-05 21:23:26
it's increasing by 2

19JasonH 2013-04-05 21:23:26
0 or 1

DPatrick 2013-04-05 21:23:50
Right. Every column starts at the bottom with either 0 or 1 (if n is even or odd, respectively), then the numbers increase by 2 for a while.

DPatrick 2013-04-05 21:24:27
It's hopefully pretty clear why we get the 0's and 1's at the bottom: we know that $n / (n/2)$ is an integer if $n$ is even, so it has remainder 0.

DPatrick 2013-04-05 21:24:46
If $n$ is odd, then $n = 2\left(\frac{n-1}{2}\right) + 1$, so the remainder is 1.

DPatrick 2013-04-05 21:25:05
Why does it go up by 2 every time as we climb from the bottom towards the maximum?

DPatrick 2013-04-05 21:26:14
For example, 27/12 has a remainder of 3. Why does this mean that 27/11 has a remainder of 5, and why does this generalize?

DPatrick 2013-04-05 21:27:26
At times in number theory, it pays to appeal to algebra. Suppose $n = 2k + r$ for some "large" $k$ close to $\lfloor \frac{n}{2} \rfloor$ (that is, some $k$ close to the bottom of a column).

DPatrick 2013-04-05 21:27:42
What equation do we write when we try to divide $n$ by $k-1$?

giftedbee 2013-04-05 21:28:09
2(k-1)+r+2

kdokmeci 2013-04-05 21:28:26
n=2(k-1)+r+2

DPatrick 2013-04-05 21:28:33
We get $n = 2(k-1) + (r+2)$, assuming that $k$ is still "large enough".

DPatrick 2013-04-05 21:28:49
So by decreasing the divisor from $k$ to $k-1$, we've increased the remainder from $r$ to $r+2$.

DPatrick 2013-04-05 21:29:04
So that's why the remainders in the columns are increasing by 2's near the bottom.

DPatrick 2013-04-05 21:29:11
But at some point this fails...when?

centralbs 2013-04-05 21:29:22
below n/3

AayushGupta 2013-04-05 21:29:22
when it goes over

DPatrick 2013-04-05 21:29:34
Right, this only works as long as the quotient part of n/k is 2.

DPatrick 2013-04-05 21:29:47
When the quotient increases to 3, then the remainder drops back to 0, 1, or 2 again.

kdokmeci 2013-04-05 21:29:53
And thats why slope is -1/3!

DPatrick 2013-04-05 21:29:55
Indeed.

DPatrick 2013-04-05 21:30:10
Here's the chart again, with the "groups of threes" highlighted a little more:

DPatrick 2013-04-05 21:30:13
\[\begin{array}{c|ccc|ccc|ccc|ccc}
\substack{n \rightarrow \\ k \downarrow}&~20~&~21~&~22~&~23~&~24~&~25~&~26~&~27~&~28~&~29~&~30~&~31\\
\hline\hline
1&0&0&0&0&0&0&0&0&0&0&0&0\\
2&0&1&0&1&0&1&0&1&0&1&0&1\\
3&2&0&1&2&0&1&2&0&1&2&0&1\\
4&0&1&2&3&0&1&2&3&0&1&2&3\\
5&0&1&2&3&4&0&1&2&3&4&0&1\\
6&2&3&4&5&0&1&2&3&4&5&0&1\\
7&\boxed6&0&1&2&3&4&5&6&0&1&2&3\\
8&4&\boxed5&\boxed6&\boxed7&0&1&2&3&4&5&6&7\\
9&2&3&4&5&\boxed6&\boxed7&\boxed8&0&1&2&3&4\\
10&0&1&2&3&4&5&6&\boxed7&\boxed8&\boxed9&0&1\\
11& & &0&1&2&3&4&5&6&7&\boxed8&\boxed{9}\\
12& & & & &0&1&2&3&4&5&6&7\\
13& & & & &&&0&1&2&3&4&5\\
14& & & & &&&&&0&1&2&3\\
15& & & & &&&&&&&0&1\\
\end{array}\]

DPatrick 2013-04-05 21:30:47
Notice that the maximum value for $n$'s column always is as $\lfloor \frac{n}{3} \rfloor + 1$, right before the quotient of $n/k$ jumps from 2 to 3.

DPatrick 2013-04-05 21:31:06
("as" should have been "at" in the previous sentence!)

DPatrick 2013-04-05 21:31:49
OK, so I'm more convinced now! Again, using test-taking strategy, your better option during the contest probably is to assume the pattern and try to answer the problem.

DPatrick 2013-04-05 21:31:56
Now that we've established the pattern, how do we compute the sum of the boxed numbers over $20 \le n \le 100$?

kdokmeci 2013-04-05 21:32:26
6+5+6+(7+6+7)+8+7+8+(9+8+9(+...

DPatrick 2013-04-05 21:33:04
Right, I've grouped the columns into groups of three, with each group centered at some multiple of 3. (For example, the first group is centered at $n = 21 = 3 \cdot 7$.)

DPatrick 2013-04-05 21:33:18
For the group centered at $n = 3m$, what is the sum of the numbers?

Sayan 2013-04-05 21:33:53
3m-4

twin77 2013-04-05 21:33:53
3m - 4

DPatrick 2013-04-05 21:34:05
Right: their sum is
$$
F(3m-1) + F(3m) + F(3m+1) = (m-1) + (m-2) + (m-1) = 3m - 4.
$$

DPatrick 2013-04-05 21:34:15
So each group sums to $3m-4$.

DPatrick 2013-04-05 21:34:20
(As a quick check: when $m=7$ we get $3(7) - 4 = 17$, which is $6+5+6$; when $m=8$ we get $3(8) - 4 = 20$, which is $7+6+7$, and so on.)

DPatrick 2013-04-05 21:34:29
So the triples sum to
$$
17 + 20 + 23 + \cdots
$$
Where does it stop?

twin77 2013-04-05 21:35:08
when it is centered at n=99

matholympiad25 2013-04-05 21:35:08
stops at 3(33)-4=95

DPatrick 2013-04-05 21:35:27
Right. The last triple is centered at $n=99$.

DPatrick 2013-04-05 21:35:33
This last triple is for $m=33$, where we get
$$
F(98) + F(99) + F(100) = 3(33)-4 = 95.
$$

DPatrick 2013-04-05 21:35:42
So our answer is just the arithmetic series
$$
17 + 20 + 23 + \cdots + 95.
$$

DPatrick 2013-04-05 21:35:49
What is this sum?

kdokmeci 2013-04-05 21:36:18
1512

zhuangzhuang 2013-04-05 21:36:18
56*27=1512, ans=512

matholympiad25 2013-04-05 21:36:18
1512, so answer is 512

ZZmath9 2013-04-05 21:36:18
1512

distortedwalrus 2013-04-05 21:36:18
56*# of terms...

DPatrick 2013-04-05 21:36:23
We run from $m=7$ to $m=33$, so there are $33-7+1 = 27$ terms.

DPatrick 2013-04-05 21:36:27
And the average term is $\frac{17 + 95}{2} = 56$.

DPatrick 2013-04-05 21:36:34
So our sum is $(27)(56) = 1512$. Our final answer is $\boxed{512}$.

DPatrick 2013-04-05 21:36:53
And finally...

DPatrick 2013-04-05 21:36:58

AayushGupta 2013-04-05 21:37:15
the problem of doom!

MSTang 2013-04-05 21:37:17
I now see why you didn't like this problem.

DPatrick 2013-04-05 21:37:39
Well, I could have liked this problem, but we'll see as we go why I ultimately didn't like it.

DPatrick 2013-04-05 21:37:56
First of all, let's assign variables to the right sides of the equations, so I don't have to keep writing them out:

DPatrick 2013-04-05 21:38:02
\begin{align*}
\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= x, \\
\cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= y, \\
\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B &= z.
\end{align*}

DPatrick 2013-04-05 21:38:13
We're given $x$ and $y$, and we want to find $z$.

kdokmeci 2013-04-05 21:38:19
Symmetry!

DPatrick 2013-04-05 21:38:26
There is a lot of symmetry here.

trophies 2013-04-05 21:38:38
the first thought that comes into my head right now (I didnt take the AIME II) is subtracting the equations

DPatrick 2013-04-05 21:39:05
That's often a very good idea with a symmetric-looking system of equations: either add or subtract them together and see if something nicer comes out.

borntobeweild 2013-04-05 21:39:15
maybe add equations and use triangle identities

DPatrick 2013-04-05 21:39:22
Let's add the first two and see what happens.

DPatrick 2013-04-05 21:39:31
\[
\cos^2 A + 2\cos^2 B + \cos^2 C + 2\sin B(\sin A \cos C + \sin C \cos A) = x+y
\]

DPatrick 2013-04-05 21:39:45
What's the term in the parentheses?

kdokmeci 2013-04-05 21:39:59
Sin addtion formula!

kdokmeci 2013-04-05 21:39:59
sin(A+C)

RelaxationUtopia 2013-04-05 21:39:59
sin (A+C)

Debdut 2013-04-05 21:39:59
sin(A+c)

number.sense 2013-04-05 21:39:59
sin (A+C)

teethpaste 2013-04-05 21:39:59
the stuff in parentheses is sin(A+C)!

DPatrick 2013-04-05 21:40:08
It's $\sin(A+C)$! That's good, but what's even better?

Sayan 2013-04-05 21:40:24
sin(A+C)=sinB

matholympiad25 2013-04-05 21:40:24
sin A cos C + sin C cos A = sin(A+C)=sin (180-B)=sin B

zhuangzhuang 2013-04-05 21:40:24
its also sin B!

AayushGupta 2013-04-05 21:40:24
sin A+C = sin 180-B

teethpaste 2013-04-05 21:40:24
A+C=180-B!

DPatrick 2013-04-05 21:40:32
That's even better: it's $\sin B$! Note that
\[
\sin B = \sin(180^\circ - B) = \sin(A+C).
\]

DPatrick 2013-04-05 21:40:44
So now our equation is
\[
\cos^2 A + 2\cos^2 B + \cos^2 C + 2\sin B(\sin B) = x+y
\]

DPatrick 2013-04-05 21:40:57
And it keeps getting better and better!

AayushGupta 2013-04-05 21:41:12
sin^2 B + cos^2 B = 1

Tuxianeer 2013-04-05 21:41:12
sin^2+cos^2=1

zhuangzhuang 2013-04-05 21:41:12
get rid of sin^2 +cos^2 b

kdokmeci 2013-04-05 21:41:12
sin^2+cos^2=1

cheese1000 2013-04-05 21:41:12
cos^2+sin^2 equals 1

DPatrick 2013-04-05 21:41:16
Well, that's convenient! $2\cos^2 B + 2 \sin^2 B$ is just 2.

DPatrick 2013-04-05 21:41:22
So our equation is just
\[
\cos^2 A + \cos^2 C + 2 = x + y.
\]

DPatrick 2013-04-05 21:41:38
That is way way way better than anything we started with.

number.sense 2013-04-05 21:41:45
no reason these two equations were special

centralbs 2013-04-05 21:41:49
we can get 3 equations like this by symmetry

DPatrick 2013-04-05 21:42:00
Right, we can do the same thing by adding the other two pairs of equations.

DPatrick 2013-04-05 21:42:09
We get a new system of 3 equations:
\begin{align*}
\cos^2 A + \cos^2 C + 2 = x+y, \\
\cos^2 A + \cos^2 B + 2 = y+z, \\
\cos^2 B + \cos^2 C + 2 = x+z.
\end{align*}

DPatrick 2013-04-05 21:42:38
Life seems good so far. Now what?

kdokmeci 2013-04-05 21:42:53
Add them up again?

19bobhu 2013-04-05 21:42:53
subtract, or add

borntobeweild 2013-04-05 21:42:53
We can add them all and subtract each one

Sayan 2013-04-05 21:42:53
subtract first 2 and add it with the third

number.sense 2013-04-05 21:42:53
what if we added them all up?

DPatrick 2013-04-05 21:43:05
Now there are lots of ways to isolate the individual cosine terms.

DPatrick 2013-04-05 21:43:12
For example, add the first two and subtract the third.

DPatrick 2013-04-05 21:43:34
The B and C parts cancel, and we're left with
\[
2\cos^2 A + 2 = (x+y)+(y+z)-(x+z) = 2y.
\]

kdokmeci 2013-04-05 21:44:05
cos^2 A+1=y

mathwrath 2013-04-05 21:44:05
cos^2A+1=y

fprosk 2013-04-05 21:44:05
cos^2A=y-1

DPatrick 2013-04-05 21:44:06
Sweet! So $\cos^2 A + 1 = y$, and hence $\cos^2 A = y-1$.

DPatrick 2013-04-05 21:44:13
Similarly $\cos^2 B = z-1$ and $\cos^2 C = x-1$.

DPatrick 2013-04-05 21:44:41
Remember, we know x = 15/8 and y = 14/9, and we want to find z.

kdokmeci 2013-04-05 21:44:48
Plug in x, y

DPatrick 2013-04-05 21:44:54
So we know that $\cos^2 A = \frac59$ and $\cos^2 C = \frac78$, and we want to find $z = \cos^2 B + 1$.

AayushGupta 2013-04-05 21:45:07
B = 180-C-A, so we can solve

DPatrick 2013-04-05 21:45:16
Right, recall that $\cos B = \cos(180^\circ - (A+C)) = -\cos(A+C)$.

DPatrick 2013-04-05 21:45:34
So we can use the cosine angle-sum formula:
\[
\cos B = -\left(\cos A \cos C - \sin A \sin C \right)
\]

sparkles257 2013-04-05 21:45:54
substitute in values of cosA and c

DPatrick 2013-04-05 21:46:07
We plug in numbers, using the fact that $\sin^2 A = 1 - \cos^2 A = \frac49$ and $\sin^2 C = 1 - \cos^2 C = \frac18$.

DPatrick 2013-04-05 21:46:40
So the sines and cosines are the square roots of all these numbers, and recall that we were given that the angles are acute, so we use the positive square root for the cosine terms.

DPatrick 2013-04-05 21:46:55
This gives us
\[
\cos B = -\left(\sqrt{\frac59}\cdot\sqrt{\frac78} - \sqrt{\frac49}\cdot\sqrt{\frac18}\right)
\]

DPatrick 2013-04-05 21:47:36
A little algebra that I triple-checked (believe me) shows that this simplifies to $$\cos B = \dfrac{2 - \sqrt{35}}{\sqrt{72}}.$$

matholympiad25 2013-04-05 21:47:51
WHAT!? THE COSINE IS NEGATIVE!!!!!!!!!!!!!????????????

henrikjb 2013-04-05 21:47:51
How can that be negative?

wfu 2013-04-05 21:47:51
But why is cos B negative...

DPatrick 2013-04-05 21:48:00
Uh-oh. Now we see the flaw.

DPatrick 2013-04-05 21:48:13
This cosine is negative. That means that angle $B$ is obtuse, contrary to what the problem statement says.

trophies 2013-04-05 21:48:27
What did we do wrong?

DPatrick 2013-04-05 21:48:40
We didn't do anything wrong. There's no way to get around this. The problem statement is incorrect -- there is no solution with $A$, $B$, $C$ all acute. This also brings into question our assumption a moment ago to use the positive square roots for $\cos A$ and $\cos C$, but I'll address that at the end.

Sayan 2013-04-05 21:49:04
So that's why it was your least favourite!

DPatrick 2013-04-05 21:49:20
Yes, I prefer my AIME problems to have solutions. This one doesn't as written.

DPatrick 2013-04-05 21:49:27
For now, let's ignore (or pretend we didn't notice) that $B$ is obtuse, and finish the problem.

distortedwalrus 2013-04-05 21:49:58
now just square this and add 1 and you're done.

DPatrick 2013-04-05 21:50:08
Right: to finish, we compute $z = \cos^2 B + 1$:
\[
z = \left( \dfrac{2 - \sqrt{35}}{\sqrt{72}} \right)^2 + 1.
\]

DPatrick 2013-04-05 21:50:25
This simplifies to
\[
z = \frac{4 - 4\sqrt{35} + 35 + 72}{72} = \frac{111 - 4\sqrt{35}}{72}.
\]

DPatrick 2013-04-05 21:50:37
This is in the correct format, so our final "answer" is $111 + 4 + 35 + 72 = \boxed{222}$.

19bobhu 2013-04-05 21:51:01
is that actually the answer in the answer sheet

DPatrick 2013-04-05 21:51:04
It is.

DPatrick 2013-04-05 21:51:09
Unfortunately, as I mentioned, there is no solution with A,B,C all acute, contrary to the problem statement.

DPatrick 2013-04-05 21:51:21
Equally unfortunately, there are actually two solutions to the system of equations.

DPatrick 2013-04-05 21:51:33
As you may recall, we assumed that $A$ and $C$ were acute when we did our $\cos(A+C)$ calculation. If you instead assume that $A$ is obtuse, you get a different triangle and a different answer for $z$, though not one that can be put in the required format. (It doesn't work with $C$ obtuse.)

MSTang 2013-04-05 21:51:46
could they have removed the word acute and fixed the problem?

DPatrick 2013-04-05 21:52:10
Well...there still would have been two solutions. They could have worded it to force one or the other I suppose.

number.sense 2013-04-05 21:52:26
what's the amc doing with this problem?

19oshawott98 2013-04-05 21:52:26
what are they going to do about the problem for AIME test-takers?

DPatrick 2013-04-05 21:52:37
I can report that the AMC is aware of the error in this problem, because many people (including myself) have told them. I do not know what (if anything) they intend to do.

DPatrick 2013-04-05 21:53:11
Anyway...that's it for the AIME II!

DPatrick 2013-04-05 21:53:15
Good luck on the USA(J)MO if you qualify for it!

Superwiz 2013-04-05 21:53:59
What do you think the cutoff score for USA(J)MO will be?

DPatrick 2013-04-05 21:54:10
No idea, and I prefer not to speculate.

henrikjb 2013-04-05 21:54:29
When do we find out the cutoff?

DPatrick 2013-04-05 21:54:41
Probably next week. Maybe the week after, I'm not sure.

DPatrick 2013-04-05 21:55:08
They tend to score the AIME II fairly quickly, because the USA(J)MO is only 4 weeks away.

atmath2011 2013-04-05 21:55:36
anybody feel the probs on the AIME II (even though i didn't take it) were easier than the AIME I?

number.sense 2013-04-05 21:55:36
What do you think the difficulty of AIME 2 was relative to 1

Knightone 2013-04-05 21:55:36
Which AIME did you find harder?

DPatrick 2013-04-05 21:55:54
I thought they were about the same.

DPatrick 2013-04-05 21:56:20
The AIME I had an unusually-difficult early problem (#5 if I remember, with a cubic polynomial) that the AIME II didn't really have.

centralbs 2013-04-05 21:56:38
Which problem did you find the most difficult?

DPatrick 2013-04-05 21:56:47
Besides the one that was actually impossible?

DPatrick 2013-04-05 21:57:06
#13 was ugly to do correctly, but as we mentioned perhaps easy to cut corners and/or guess.

DPatrick 2013-04-05 21:57:30
#11 was hard because it was a little technical with the language.

DPatrick 2013-04-05 21:57:51
But I've seen so many counting problems in my lifetime that I can do them almost in my sleep.

DPatrick 2013-04-05 21:57:53

DPatrick 2013-04-05 21:58:47
#10 is still probably my favorite. It had a nice elegant solution.

DPatrick 2013-04-05 21:59:21
OK, I'm going to close up the classroom now so that I can generate and post the transcript of the session.

DPatrick 2013-04-05 21:59:32
Have a nice weekend!

2013 AIME II Math Jam

Facilitator: Dave Patrick