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2013 AIME I Math Jam

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AoPS instructors discuss all 15 problems of the 2013 AIME I.

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Facilitator: Dave Patrick

DPatrick 2013-03-16 18:57:32
Before we get started (in about 2 minutes), I have a question: for those of you who took the test, what was your favorite question on the test.
DPatrick 2013-03-16 18:57:35
?
DPatrick 2013-03-16 18:59:12
#11 seems to be winning the vote. I liked it too. We'll get to it...
anthonyjang 2013-03-16 19:00:02
When are the official answers coming out?
DPatrick 2013-03-16 19:00:11
I believe on Monday. But you'll see all 15 answers tonight.
DPatrick 2013-03-16 19:00:18
Let's get started!
DPatrick 2013-03-16 19:00:23
Welcome to the 2013 AIME I Math Jam!
DPatrick 2013-03-16 19:00:30
I'm Dave Patrick, and I'll be leading our discussion tonight.
DPatrick 2013-03-16 19:00:38
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 2013-03-16 19:00:48
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 2013-03-16 19:01:03
This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick 2013-03-16 19:01:27
There are a lot of students here, much more than our regular online classes! 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. This Math Jam is much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!
DPatrick 2013-03-16 19:01:43
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!
DPatrick 2013-03-16 19:01:55
We do have two teaching assistants with us tonight to help answer your questions: Luis Ares (Duelist) and Lizard (Elizabeth) Reiland (Entropy).
DPatrick 2013-03-16 19:02:02
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 2013-03-16 19:02:31
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 2013-03-16 19:02:48
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.
DPatrick 2013-03-16 19:03:07
Let's get started! We're going to work through all 15 problems, in order.
DPatrick 2013-03-16 19:03:14
DPatrick 2013-03-16 19:03:19
(You'll notice that the current problem will always be posted to the top of the window. You can resize the top part of the window by dragging on the horizontal bar separating it from the main part of the window.)
kiwiluver75 2013-03-16 19:03:51
d=rt
cxiong 2013-03-16 19:03:51
Use a system of equations, D = RT
btilm305 2013-03-16 19:03:51
Distance = rate * time
DPatrick 2013-03-16 19:04:03
Of course, you know that distance = rate * time. We'll obviously want to use that!
teranz0 2013-03-16 19:04:19
Express the rate at which he swims as x, runs as 5x, and bikes as 10x
pieslinger 2013-03-16 19:04:19
set a variable for swiming speed
kdokmeci 2013-03-16 19:04:19
We make our variables
akalykid012 2013-03-16 19:04:19
Let x equal speed when swimming
ClassicCalculator 2013-03-16 19:04:19
Express the swimming rate (d/t) as "x".
DPatrick 2013-03-16 19:04:36
I like setting the problem up in terms of the swimming rate, because the other two rates are then multiples of it.
DPatrick 2013-03-16 19:04:42
So let's set $r$ to be the rate (in miles per hour) that Tom swims.
DPatrick 2013-03-16 19:04:56
He runs 5 times faster than he swims, so he runs at a rate of $5r$.
DPatrick 2013-03-16 19:05:02
And he bikes twice as fast as he runs, so he bikes at a rate of $10r$.
DPatrick 2013-03-16 19:05:09
So how long does Tom take to complete the triathlon?
billgates42 2013-03-16 19:05:30
4 adn 1/4 hr
Wickedestjr 2013-03-16 19:05:30
4 and a quarter hours
bugegg3 2013-03-16 19:05:30
4.25 hours
phoenix827 2013-03-16 19:05:30
17/4 hr
jeff10 2013-03-16 19:05:30
17/4 hours: I think writing this as an improper fraction helps later on
DPatrick 2013-03-16 19:05:45
Right, on the one hand the problem tells us: 4.25 or 17/4 hours.
DPatrick 2013-03-16 19:05:57
But on the other hand, we can compute his time in terms of r.
akalykid012 2013-03-16 19:06:31
1/2r + 8/5r + 30/10r hours
Annabeth 2013-03-16 19:06:31
0.5/r+30/10r+8/5r
btilm305 2013-03-16 19:06:31
1/2r, 3/r, and 8/5r respectively
joshxiong 2013-03-16 19:06:31
1/2r+8/5r+30/10r
ucantbeatmario 2013-03-16 19:06:31
t=.5/r+8/5r+30/10r
DPatrick 2013-03-16 19:06:36
The swimming takes $\frac{0.5}{r}$ hours.
DPatrick 2013-03-16 19:06:45
The cycling takes $\frac{30}{10r} = \frac{3}{r}$ hours.
DPatrick 2013-03-16 19:06:51
The running takes $\frac{8}{5r} = \frac{1.6}{r}$ hours.
DPatrick 2013-03-16 19:06:59
So the whole thing takes $\frac{0.5}{r} + \frac{3}{r} + \frac{1.6}{r} = \frac{5.1}{r}$ hours.
DPatrick 2013-03-16 19:07:17
(All of those used "time = distance / rate".)
kdokmeci 2013-03-16 19:07:30
5.1/r=17/4
ninjataco 2013-03-16 19:07:30
then 5.1/r = 17/4
DPatrick 2013-03-16 19:07:46
Exactly: the two times are equal, so $\frac{5.1}{r} = \frac{17}{4}$.
mg13 2013-03-16 19:08:26
r=1.2 miles per hours
poponono 2013-03-16 19:08:26
r=1.2
csmath 2013-03-16 19:08:26
r=1.2
amoghgaitonde 2013-03-16 19:08:26
r = 1.2
25120012jsy21 2013-03-16 19:08:26
r=1.2
willabc 2013-03-16 19:08:26
r=1.2
DPatrick 2013-03-16 19:08:40
Right: we solve this to get $r = \frac65$ (or $r=1.2$) (miles per hour).
DPatrick 2013-03-16 19:08:51
And what's the finish?
mathwizard888 2013-03-16 19:09:20
Tom spends 3/1.2 hrs=150 minutes biking
csmath 2013-03-16 19:09:20
Cycling takes 2.5 hours which is 150 minutes
sahilp 2013-03-16 19:09:20
3/1.2=2.5 hr= 150 mins
avi1234 2013-03-16 19:09:20
30/1.2*60
sunny2000 2013-03-16 19:09:20
We plug this into the other equation
ProbaBillity 2013-03-16 19:09:20
The answer is 3/r = 2.5 hours, or 150 minutes.
DPatrick 2013-03-16 19:09:28
Right. The bicycling time is $\frac{3}{r} = \frac{3}{\frac65} = \frac52$ hours, which is $\boxed{150}$ minutes.
DPatrick 2013-03-16 19:09:54
DPatrick 2013-03-16 19:10:05
What do we know about $n$?
thkim1011 2013-03-16 19:10:22
Since 0 can't be the first digit, we must have 5 as the first and last digits.
minimario 2013-03-16 19:10:22
The first and last digits must be 5
swe1 2013-03-16 19:10:22
First and last digits have to be 5
54math 2013-03-16 19:10:22
first and last digits of n have to be 5
sparkles257 2013-03-16 19:10:22
has to begin and end with 5
lcamhie142857 2013-03-16 19:10:22
first and last digits are 5
anthonyjang 2013-03-16 19:10:22
5_ _ _ 5 is the form
mathcool2009 2013-03-16 19:10:22
n = 5_ _ _5
DPatrick 2013-03-16 19:10:27
Condition (a) says that $n$ ends in 0 or 5.
DPatrick 2013-03-16 19:10:33
But condition (b) says that $n$ begins with the same digit that it ends with, and a number can't begin with 0.
DPatrick 2013-03-16 19:10:43
So $n$ starts and ends with 5.
DPatrick 2013-03-16 19:10:54
Or in other words, $n = 5xyz5$ for digits $x$, $y$, and $z$.
DPatrick 2013-03-16 19:10:59
How do we count these that satisfy condition (c)?
cxiong 2013-03-16 19:11:26
abcde, b+c+d=5,10,15,20,25
ProbaBillity 2013-03-16 19:11:26
bobthesmartypants 2013-03-16 19:11:26
casework
simon1221 2013-03-16 19:11:26
x+y+z is divisible by 5
andrewlin 2013-03-16 19:11:26
x+y+z=5,10,15,20 or 25 because 27 is the largest sum
wjli 2013-03-16 19:11:26
the sum of xyz must be a multiple of 5
az_phx_brandon_jiang 2013-03-16 19:11:26
x+y+z=5, 10, 15, 20, 25
DPatrick 2013-03-16 19:11:32
We could do cases: $x+y+z=0$, $x+y+z=5$, $x+y+z=10$, and so on, but is there a more clever way?
lucylai 2013-03-16 19:12:02
x and y can be anything, and z just has to be congruent to -x-y (mod 5)
lcamhie142857 2013-03-16 19:12:02
Use mod 5 on the 4th digit
ucantbeatmario 2013-03-16 19:12:02
yes. Fore very choice of x and y, there two choices for z
ABCDE 2013-03-16 19:12:02
2 possibilities for z for every pair of x and y
anwang16 2013-03-16 19:12:02
its 1/5 of the total ways to have x,y,z
billgates42 2013-03-16 19:12:02
there are 2 possible values of c for every a and b
footballer 2013-03-16 19:12:02
for each pair of digits, there is 2 possible third digits for the sum to be a multiple of 5
DPatrick 2013-03-16 19:12:07
Right!
DPatrick 2013-03-16 19:12:21
Given any choices for $x$ and $y$, how many $z$'s are there?
apple.singer 2013-03-16 19:12:33
2
kiwiluver75 2013-03-16 19:12:33
2
akalykid012 2013-03-16 19:12:33
2
bugegg3 2013-03-16 19:12:33
2
noobynoob 2013-03-16 19:12:33
$2$
Wickedestjr 2013-03-16 19:12:33
TWO!
DPatrick 2013-03-16 19:12:48
Indeed, there are exactly 2.
DPatrick 2013-03-16 19:12:55
For example, if $x+y=2$, then we need $z$ to be 3 or 8.
DPatrick 2013-03-16 19:13:13
No matter what x and y are, there will be two z's that make x+y+z a multiple of 5.
DPatrick 2013-03-16 19:13:30
(In the language of modular arithmetic, we need $z \equiv -(x+y) \pmod{5}$, and there are always two digits that satisfy this.)
number.sense 2013-03-16 19:13:57
so we have 10*10*2 = 200 choices
andrewlin 2013-03-16 19:13:57
so there are 10 choices for x, 10 for y and 2 for z. 10*10*2=200
billgates42 2013-03-16 19:13:57
There are 10*10 pairs for a and b because a can be 0.
csmath 2013-03-16 19:13:57
There are 10 ways to pick each of x and y so 100 total *2=200 (final answer)
DPatrick 2013-03-16 19:14:12
Right: for every pair of digits $(x,y)$, there are two such $z$.
DPatrick 2013-03-16 19:14:24
And since there are $10 \cdot 10 = 100$ pairs $(x,y)$, there are $2 \cdot 100 = \boxed{200}$ numbers satisfying all the conditions.
DPatrick 2013-03-16 19:15:01
Of course, counting all the x+y+z = 0, x+y+z = 5, and so on by brute-force works too.
mathnerd101 2013-03-16 19:15:12
2 down 13 more to go
DPatrick 2013-03-16 19:15:16
ninjataco 2013-03-16 19:15:34
draw a picture!
katelynnsmith 2013-03-16 19:15:34
draw a diagram first
DPatrick 2013-03-16 19:15:38
Sure, let's sketch a quick picture:
DPatrick 2013-03-16 19:15:47
DPatrick 2013-03-16 19:16:07
(You might have had F closer to B than to C -- it doesn't matter)
ninjataco 2013-03-16 19:16:27
set AE equal to "x"
jeff10 2013-03-16 19:16:27
mg13 2013-03-16 19:16:27
label sides and assign variables
mathman500 2013-03-16 19:16:27
WLOG let AB=1
DPatrick 2013-03-16 19:16:44
That's how I did it too. You could also use two variables but I like setting things to equal 1 if I can.
DPatrick 2013-03-16 19:16:54
So let's set $AB = 1$ and $AE = x$.
DPatrick 2013-03-16 19:17:05
This gives $EB = 1-x$, $BF = x$, and $FC = 1-x$.
DPatrick 2013-03-16 19:17:18
n1000 2013-03-16 19:17:52
cxiong 2013-03-16 19:17:52
x² + (1-x)² = 9/10
TMTOLBTWNTOF 2013-03-16 19:17:52
x^2 + 1-x ^2 = 9/10
Mrdavid445 2013-03-16 19:17:52
(1-x)^2+x^2=(9/10)
DPatrick 2013-03-16 19:18:23
Right, the sum of the areas of the two smaller squares is $x^2 + (1-x)^2$.
DPatrick 2013-03-16 19:18:27
And we know this equals 9/10.
DPatrick 2013-03-16 19:18:36
So we have
$$
2x^2 - 2x + 1 = \frac{9}{10}.
$$
DPatrick 2013-03-16 19:18:54
Let's not solve for $x$ just yet, because that's not what we're asked for. What are we asked for?
cxiong 2013-03-16 19:19:17
find x/1-x + 1-x/x
lucylai 2013-03-16 19:19:17
(1-x)/x+x/(1-x)
avi1234 2013-03-16 19:19:17
x/(1-x)+(1-x)/x?
mathcool2009 2013-03-16 19:19:17
x/1-x + 1-x/x
JRY 2013-03-16 19:19:17
x/(1-x)+(1-x)/x
VietaFan 2013-03-16 19:19:17
x/(1-x)+(1-x)/x
DPatrick 2013-03-16 19:19:28
Yes: we want $\dfrac{AE}{EB} + \dfrac{EB}{AE} = \dfrac{x}{1-x} + \dfrac{1-x}{x}$.
DPatrick 2013-03-16 19:19:40
This simplifies to
$$
\frac{x^2 + (1-x)^2}{x(1-x)} = \frac{2x^2 - 2x + 1}{x - x^2}.
$$
DPatrick 2013-03-16 19:19:48
Aha, I recognize the numerator!
kdokmeci 2013-03-16 19:20:16
9/10
promathist99 2013-03-16 19:20:16
It's the same!
SuperSnivy 2013-03-16 19:20:16
9/10 for numerator
csmath 2013-03-16 19:20:16
That = 9/10
billgates42 2013-03-16 19:20:16
9/10
DPatrick 2013-03-16 19:20:22
The numerator is just $\frac{9}{10}$.
DPatrick 2013-03-16 19:20:27
What about the denominator?
akalykid012 2013-03-16 19:20:57
denominator = 1/20
mapletree14 2013-03-16 19:20:57
And subtract one and divide by 2 to get the denominator.
nikoma 2013-03-16 19:20:57
Isolate 2x^2 - 2x
ucantbeatmario 2013-03-16 19:20:57
x-x^2=(1-9/10)/2=1/20
DPatrick 2013-03-16 19:21:04
Rearranging our area equation a little bit gives us
$$
2x^2 - 2x = -\frac{1}{10},
$$
so $x - x^2 = \frac{1}{20}$.
lcamhie142857 2013-03-16 19:21:35
dividing gives 18
mathwizard888 2013-03-16 19:21:35
(9/10)/(1/20)=018 answer
Nitzuga 2013-03-16 19:21:35
So our answer is $018$
DPatrick 2013-03-16 19:21:40
Therefore, the quantity we want is
$$
\frac{\frac{9}{10}}{\frac{1}{20}} = \frac{180}{10} = \boxed{018}.
$$
DPatrick 2013-03-16 19:22:32
You can also get a perhaps more elegant solution by using two variables: letting AE = x and EB = y. It requires you to be slightly more clever. You can post it on the forum if you like.
DPatrick 2013-03-16 19:22:49
DPatrick 2013-03-16 19:22:58
DPatrick 2013-03-16 19:23:09
What exactly does the rotation condition mean?
steve123456 2013-03-16 19:23:40
all the corresponding positions have to be the same color
TMTOLBTWNTOF 2013-03-16 19:23:40
All the L's on the outside need the same arrangement
andrewlin 2013-03-16 19:23:40
it means that all of the L's have the same colouring
pedronr 2013-03-16 19:23:40
Each of the sets of three has to have the same pattern.
mathman98 2013-03-16 19:23:40
It's the same on all four "sides"
n1000 2013-03-16 19:23:40
that the four little Ls have the same coloring
DPatrick 2013-03-16 19:23:47
Right. In other words, all of the lettered squares in the picture below must be the same color (that is, all the X's must be the same, all the Y's must be the same, and all the Z's must be the same):
DPatrick 2013-03-16 19:23:52
DPatrick 2013-03-16 19:24:05
And how is that possible with the given colors (8 red and 5 blue)?
swe1 2013-03-16 19:24:25
Middle has to be blue
mathnerd101 2013-03-16 19:24:25
blue has to be in the middle
timoteomo3 2013-03-16 19:24:25
center must be blue
thkim1011 2013-03-16 19:24:25
Middle must be blue
mathman27 2013-03-16 19:24:25
middle is blue
bugegg3 2013-03-16 19:24:25
the center must be blue
ClassicCalculator 2013-03-16 19:24:42
The center square must be blue, and each surrounding group must have 2 reds and 1 blue.
sahilp 2013-03-16 19:24:42
x y or z is blue
countingkg 2013-03-16 19:24:42
so either X = blue, Y = blue, or Z=blue
hi how are you doing toda 2013-03-16 19:24:42
so each coloring of L can only be 2 red and 1 blue
DPatrick 2013-03-16 19:24:51
Right. We need to pick one of X, Y, or Z to be 4 of the five blues (along with the center square). The other two letters will all be red.
DPatrick 2013-03-16 19:25:10
So how many colorings work?
ninjataco 2013-03-16 19:25:25
3 ways to do this
distortedwalrus 2013-03-16 19:25:25
3
btilm305 2013-03-16 19:25:25
3
Math_Kirby 2013-03-16 19:25:25
3
sparkles257 2013-03-16 19:25:25
3
anwang16 2013-03-16 19:25:25
3
number.sense 2013-03-16 19:25:25
3 total
Wickedestjr 2013-03-16 19:25:25
Three
anthonyjang 2013-03-16 19:25:25
3
lucylai 2013-03-16 19:25:25
3
willabc 2013-03-16 19:25:25
3
DPatrick 2013-03-16 19:25:42
Right: just 3. Either X+center blue, or Y+center blue, or Z+center blue.
kouhong 2013-03-16 19:25:49
now we find how many possibilites there are total
DPatrick 2013-03-16 19:25:58
Right...and how many ways to color the 13 squares overall?
ABCDE 2013-03-16 19:26:18
13 choose 5 total ways
mg13 2013-03-16 19:26:18
13C5
SuperSnivy 2013-03-16 19:26:18
54math 2013-03-16 19:26:18
13C5
VietaFan 2013-03-16 19:26:18
$\binom{13}{5}$
ryanyoo 2013-03-16 19:26:18
13c5
koel17 2013-03-16 19:26:18
13 choose 8
noobynoob 2013-03-16 19:26:18
$\frac{13!}{8!5!}$
moana314 2013-03-16 19:26:18
13 choose 5
DPatrick 2013-03-16 19:26:22
There are 13 squares, and we must pick any 5 of them to be blue.
DPatrick 2013-03-16 19:26:37
So there are
$$
\binom{13}{5} = \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 13 \cdot 11 \cdot 9 = 1287
$$
total colorings.
DPatrick 2013-03-16 19:26:55
(I wrote it out because that's what you had to do on the contest too -- no calcuators!)
DPatrick 2013-03-16 19:27:01
...or calcuLators even!
ucantbeatmario 2013-03-16 19:27:21
and 3/1287=1/429, so n=429
gaberen 2013-03-16 19:27:21
3/1287 = 1/429 so n = 429
SurrealisticStranger 2013-03-16 19:27:21
Divide by 3, we get 429
kdokmeci 2013-03-16 19:27:21
Our probability is 3/1287=1/429 so the answer is 429
Annabeth 2013-03-16 19:27:21
3/1287=1/429, so n=429
DPatrick 2013-03-16 19:27:26
Hence, the probability of a successful coloring is $\frac{3}{1287} = \frac{1}{429}$, and the answer is $\boxed{429}$.
DPatrick 2013-03-16 19:27:51
DPatrick 2013-03-16 19:28:15
Since there's no easy formula for the roots of a cubic, what should we do?
cxiong 2013-03-16 19:28:46
(x + 1)³ = x³ + 3x² + 3x + 1
Wickedestjr 2013-03-16 19:28:46
Notice the last three terms of the equation
mathman500 2013-03-16 19:28:46
The polynomial almost looks like -(x+1)^3
timoteomo3 2013-03-16 19:28:46
-3x^2-3x-1 looks just like (x+1)^3
DPatrick 2013-03-16 19:29:08
Aha, most of the coefficients look nice. In particular the 3,3,1 in the final three terms are suggestive of a perfect cube.
DPatrick 2013-03-16 19:29:17
But the 8 up front is annoying.
mathman98 2013-03-16 19:29:49
Write 8x^3 as 9x^3-x^3
az_phx_brandon_jiang 2013-03-16 19:29:49
9x^3-x^3
photondragon314 2013-03-16 19:29:49
rewrite as: 9x^3 - x^3 - 3x^2 - 3x - 1 = 0
DPatrick 2013-03-16 19:30:08
Let's force the perfect cube in there!
DPatrick 2013-03-16 19:30:15
Let's write it as
$$
9x^3 {\color{red}{- x^3 - 3x^2 - 3x - 1}} = 0.
$$
What good did that do?
Hydroxide 2013-03-16 19:30:36
it's -(x+1)^3
sparkles257 2013-03-16 19:30:36
-(x+1)^3
pedronr 2013-03-16 19:30:36
So 9x^3-(x+1)^3=0.
Aplus95 2013-03-16 19:30:36
9x^3 = (x+1)^3
mathwizard888 2013-03-16 19:30:36
9x^3=(x+1)^3
DPatrick 2013-03-16 19:30:41
Now it's really nice:
$$
9x^3 = x^3 + 3x^2 + 3x + 1 = (x+1)^3.
$$
mathnerd101 2013-03-16 19:31:06
cube root both sides
billgates42 2013-03-16 19:31:06
cubed root time!
mjoshi 2013-03-16 19:31:06
take the cube root
n1000 2013-03-16 19:31:06
take the cube root of both sides
DPatrick 2013-03-16 19:31:19
So taking the cube root of both sides, we get $\sqrt[3]{9}\cdot x = x+1$.
Wickedestjr 2013-03-16 19:31:51
Solve for x!
trophies 2013-03-16 19:31:51
now solve for x!
mssmath 2013-03-16 19:31:51
x=1/(9^(1/3)-1)
ninjataco 2013-03-16 19:31:51
subtract x from each side and factor it out
number.sense 2013-03-16 19:31:51
solve for x: 1/(cbrt9-1)
DPatrick 2013-03-16 19:32:00
Solving for $x$ gives $x = \dfrac{1}{\sqrt[3]{9} - 1}$.
DPatrick 2013-03-16 19:32:08
So far, so good, but it's not in the form that the answer requires. What do we do?
anwang16 2013-03-16 19:32:42
use difference of cubes backwards
sindennisz 2013-03-16 19:32:42
Complete the cube on the bottom using difference of cubes.
tiger21 2013-03-16 19:32:42
multiply to get a difference of cubes in the denominator
DPatrick 2013-03-16 19:32:53
Maybe if we let $y = \sqrt[3]{9}$ it's easier to see: $x = \dfrac{1}{y-1}$. We need something with just $y^3$ and 1 in the denominator.
DPatrick 2013-03-16 19:33:11
Since we know $y^3,$ we should multiply numerator and denominator by $y^2 + y + 1$, noting that
$$
(y-1)(y^2+y+1) = y^3 - 1.
$$
DPatrick 2013-03-16 19:33:21
(That's a special case of the difference-of-cubes factorization.)
Math_Kirby 2013-03-16 19:33:45
multiply top and bottom with (sqrt[3](81)+sqrt[3]{9}+1)
awesomeusername 2013-03-16 19:34:00
multiply numerator and denominator by y^2 + y + 1
mathawesomeness777 2013-03-16 19:34:00
multiply by y^2 + y + 1
DPatrick 2013-03-16 19:34:05
So now we have
$$
x = \frac{y^2 + y + 1}{y^3 - 1} = \frac{\sqrt[3]{81} + \sqrt[3]{9} + 1}{8}.
$$
zhuangzhuang 2013-03-16 19:34:29
81+9+8=98=ans
ucantbeatmario 2013-03-16 19:34:29
so answer =81+9+8=98
amoghgaitonde 2013-03-16 19:34:29
a = 81, b = 9, c = 8 so a + b + c = 98
Annabeth 2013-03-16 19:34:29
81+9+8=98
coldsummer 2013-03-16 19:34:29
so the answer is 81 + 9 + 8 = 098
chiku 2013-03-16 19:34:29
81+9+8=98
DPatrick 2013-03-16 19:34:36
This is the form we need, and the answer is $81 + 9 + 8 = \boxed{098}$.
DPatrick 2013-03-16 19:35:00
DPatrick 2013-03-16 19:35:29
There are many many ways to solve this problem -- the main hitch is to keep track of your assumptions carefully.
TMTOLBTWNTOF 2013-03-16 19:35:48
does order within the box matter?
DPatrick 2013-03-16 19:35:56
That's one assumption we have to decide upon.
DPatrick 2013-03-16 19:36:08
I found it easier to count if we assume that it DOES matter.
DPatrick 2013-03-16 19:36:16
(You can do it the other way too.)
timoteomo3 2013-03-16 19:36:26
indistinguishable math textbooks! much more convenient
DPatrick 2013-03-16 19:36:52
Yes, I also found it easiest to assume the books themselves are indistinguishable.
DPatrick 2013-03-16 19:37:08
Let's think about it as arranging the 12 books in order -- 3 M's (for math) and 9 O's (for other) -- from which the first 3 books will go in the first box, and then the next 4 books will go in the second box, and the last 5 books into the third box.
DPatrick 2013-03-16 19:37:41
First, how many ways to arrange the books in any order?
gaberen 2013-03-16 19:38:09
12!/9!3!
54math 2013-03-16 19:38:09
12C3
QuantumandMath 2013-03-16 19:38:09
(12 3)
Annabeth 2013-03-16 19:38:09
12!/9!*3!
cerberus88 2013-03-16 19:38:09
12!/3!9!
bugegg3 2013-03-16 19:38:09
12 choose 3
Flamewire 2013-03-16 19:38:09
12!/9!3! since they're indistinguishable
billgates42 2013-03-16 19:38:09
There are 12 choose 3
TheUnChosenOne 2013-03-16 19:38:09
12C3
DPatrick 2013-03-16 19:38:22
Right. I'm just thinking of arranging 3 M's and 9 O's into a 12-letter "word".
DPatrick 2013-03-16 19:38:32
We only need to choose the 3 slots (among the 12) for the math books.
DPatrick 2013-03-16 19:38:43
So there are $\dbinom{12}{3} = \dfrac{12 \cdot 11 \cdot 10}{6} = 220$ arrangements.
DPatrick 2013-03-16 19:39:03
And now we count "successes": where the M's all end up in the same box.
QuantumandMath 2013-03-16 19:39:23
ror left box, 1 way for math
kdokmeci 2013-03-16 19:39:23
Casework into wwhich box
d95776 2013-03-16 19:39:23
Count the number of cases.
mathnerd101 2013-03-16 19:39:23
the first box with 3 only has one way
DPatrick 2013-03-16 19:39:30
Right, there are 3 cases, depending on which box they end up in.
DPatrick 2013-03-16 19:39:44
The first case of the first box has just 1 way: MMMOOOOOOOOO (where O is an "other" non-math book).
DPatrick 2013-03-16 19:39:55
How about the second box?
jeff10 2013-03-16 19:40:14
the number of ways to put the 3 math books into the "4 box" is 4
cxiong 2013-03-16 19:40:14
second has 4
mikhailgromov 2013-03-16 19:40:14
second 4 ways
csmath 2013-03-16 19:40:14
There are 4 ways for the second box.
JRY 2013-03-16 19:40:14
4C3=4
DVA6102 2013-03-16 19:40:14
4 C 3 = 4
cjquines0 2013-03-16 19:40:14
4C3
VietaFan 2013-03-16 19:40:14
4 choose 3 or 4 ways
DPatrick 2013-03-16 19:40:20
We must have OOOXXXXOOOOO where XXXX is the middle box. 3 of the X's have to be M's.
DPatrick 2013-03-16 19:40:30
So there are $\binom43 = 4$ ways (choose 3 of the 4 X's to be M's, or alternatively just choose one of them to be an O).
DPatrick 2013-03-16 19:40:48
How about the case for the last box?
tiger21 2013-03-16 19:41:10
5C3 or 10 ways
phoenix827 2013-03-16 19:41:10
5C3
SuperSnivy 2013-03-16 19:41:10
10 ways for last box
sahilp 2013-03-16 19:41:10
5C3=10
flyrain 2013-03-16 19:41:10
5C3=10
davidkim2106 2013-03-16 19:41:10
5C3
ninjataco 2013-03-16 19:41:10
5C3=10 ways
AlcumusGuy 2013-03-16 19:41:10
5C3 = 10 ways
DPatrick 2013-03-16 19:41:22
Now we must have OOOOOOOXXXXX where XXXXX is the last box. 3 of the X's have to be M's.
DPatrick 2013-03-16 19:41:27
So there are $\binom53 = 10$ ways (choose 3 of the 5 X's to be M's).
Annabeth 2013-03-16 19:41:49
In all, 15 ways
kdokmeci 2013-03-16 19:41:49
The total number of ways is 1+4+10=15 ways.
pedronr 2013-03-16 19:41:53
So (1+4+10)/220 is the answer
mathwizard888 2013-03-16 19:41:58
probability is 15/220=3/44, so 047 answer
DPatrick 2013-03-16 19:42:02
All together, there are $1+4+10 = 15$ successful orderings of the books.
promathist99 2013-03-16 19:42:52
Answer: 047
distortedwalrus 2013-03-16 19:42:52
15/220 = 3/44 --> 47
n1000 2013-03-16 19:42:52
Sun7777 2013-03-16 19:42:52
47
kouhong 2013-03-16 19:42:52
so3+44=47 (answer)
DPatrick 2013-03-16 19:43:00
Thus the probability is $\frac{15}{220} = \frac{3}{44}$ in lowest terms, so the answer is $3+44 = \boxed{047}$.
DPatrick 2013-03-16 19:43:52
Mattchu386 2013-03-16 19:44:07
Diagram!
mathman500 2013-03-16 19:44:07
Draw a diagram
andrewlin 2013-03-16 19:44:07
should we attempt a diagram?
billgates42 2013-03-16 19:44:07
Draw a picture!!!
DPatrick 2013-03-16 19:44:14
I would certainly attempt to sketch a picture.
DPatrick 2013-03-16 19:44:26
Fortunately, I have a whizzy drawing program. With my whizzy drawing program, I can make this:
DPatrick 2013-03-16 19:44:31
btilm305 2013-03-16 19:44:37
height = h
DPatrick 2013-03-16 19:44:57
Indeed, I've called the height h (what they called m/n in the problem).
DPatrick 2013-03-16 19:45:12
OK. That's not very illuminating. Point A is the corner where the three faces mentioned in the problem meet, and the dashed red triangle is the triangle with area 30. But that red triangle is still just plain hard to think about.
DPatrick 2013-03-16 19:45:18
Any ideas?
Knightone 2013-03-16 19:45:53
dilate the midpoints to the corners and use area(FCH)=120
DPatrick 2013-03-16 19:46:25
I like this idea. If we blow up the triangle to FHC, we get a triangle whose side lengths are two as big, so whose area is 2^2 = 4 times as big:
DPatrick 2013-03-16 19:46:31
DPatrick 2013-03-16 19:46:56
I find that triangle CFH is way easier to think about than PQR, because we already have a mess of right triangles. (Note that we could have reached an essentially equivalent position by considering the box with A, P, R, and Q as vertices. But blowing things up is more fun.)
DPatrick 2013-03-16 19:47:14
So now the area of the blue triangle is 4*30 = 120, and we still want to find h.
Wickedestjr 2013-03-16 19:47:31
Then note that FH = 20, thus the altitude to FH = 12.
csmath 2013-03-16 19:47:31
FH=20
Polynomial 2013-03-16 19:47:31
FH is 20
bugegg3 2013-03-16 19:47:31
12^2+16^2=20^2
Wickedestjr 2013-03-16 19:47:31
The altitude from C to FH must be equal to 12.
DPatrick 2013-03-16 19:47:35
Aha!
DPatrick 2013-03-16 19:47:43
We know that FE = 12 and EH = 16.
DPatrick 2013-03-16 19:48:07
So FEH is a 12-16-20 right triangle! And FH = 20. (You could use the Pythagorean Theorem too if you didn't see it right away.)
DPatrick 2013-03-16 19:48:13
We don't need PQR anymore, so I'll wipe that out to make our diagram cleaner:
DPatrick 2013-03-16 19:48:22
Polynomial 2013-03-16 19:48:57
And the altitude from C to FH is 12. Use variables and solve for h
kdokmeci 2013-03-16 19:49:00
Connect C to FH
billgates42 2013-03-16 19:49:00
FIND THE ALTITUDE from C
DPatrick 2013-03-16 19:49:27
Right, as mentioned a bit earlier, the area of the blue triangle is 120, and if we think of FH as the base, with length 20, then the height from C to FH must be 12:
DPatrick 2013-03-16 19:49:36
DPatrick 2013-03-16 19:49:45
All right, how are we going to get h?
QuantumandMath 2013-03-16 19:50:11
CGS
forthegreatergood 2013-03-16 19:50:11
make a triangle
cxiong 2013-03-16 19:50:11
pythagorean theorem from SG, SC, GC
DPatrick 2013-03-16 19:50:22
We go back to that classic strategy of "build right triangles and hope good things happen". We draw GS to make right triangle CGS. Once we find GS, we're home free.
DPatrick 2013-03-16 19:50:27
akalykid012 2013-03-16 19:51:06
GS * 20 = 12 * 16
chiku 2013-03-16 19:51:19
SG=9.6
tiger21 2013-03-16 19:51:19
GS=12*16/20
DPatrick 2013-03-16 19:51:28
Aha! Look at the area of the top face EFGH.
DPatrick 2013-03-16 19:51:57
On the one hand, it's 12*16. (Sorry, EH = 16 got lost from the diagram somewhere.)
DPatrick 2013-03-16 19:52:31
On the other hand, FH is a diagonal and GS is an altitude to FH, so the area of 2 * (1/2 * FH * SG) = 20 * SG.
DPatrick 2013-03-16 19:52:58
So we have $12 \cdot 16 = 20 \cdot SG$, which makes $SG = \frac{48}{5}$.
DPatrick 2013-03-16 19:53:10
And to finish?
cerberus88 2013-03-16 19:53:34
We get that h=36/5, so the answer is 041
akalykid012 2013-03-16 19:53:34
h= 36/5
ssilwa 2013-03-16 19:53:34
now use pythagorean theorem to find GC or h
mathwizard888 2013-03-16 19:53:34
CG=36/5, so 041 answer
cerberus88 2013-03-16 19:53:34
h^2=12^2-9.6^2=7.2^2, so h=36/5->041
anwang16 2013-03-16 19:53:34
so h=36/5
cxiong 2013-03-16 19:53:34
(48/5)² + h² = 12²
SuperSnivy 2013-03-16 19:53:34
h=36/5, so ans is 041
cjquines0 2013-03-16 19:53:34
sqrt(12^2 - (48/5)^2) = h
DPatrick 2013-03-16 19:54:17
Right. We know CS = 12 and GS = 48/5 and CGS is a right triangle, so we can use the Pythagorean Theorem again to find CG = h, or again notice that CGS is similar to a 3-4-5 triangle with sides (36/5)-(48/5)-(60/5).
DPatrick 2013-03-16 19:54:37
In any event, h = 36/5, so our answer is $36 + 5 = \boxed{041}$.
DPatrick 2013-03-16 19:55:01
mathawesomeness777 2013-03-16 19:55:20
nasty expression
DPatrick 2013-03-16 19:55:37
Often the nastiest-looking AIME questions are really not that hard once you get past the notation.
DPatrick 2013-03-16 19:55:42
What are the conditions for this function to be defined?
lcamhie142857 2013-03-16 19:55:56
domain of arcsin is [-1,1]
AlcumusGuy 2013-03-16 19:55:56
start with the big picture or arcsin, whose domain is [-1, 1]
Aplus95 2013-03-16 19:55:56
domain of arcsin is [-1,1]
lucylai 2013-03-16 19:55:56
the domain of arcsin(x) is [-1, 1]
Wickedestjr 2013-03-16 19:55:56
-1 </= arcsin </= 1
steve123456 2013-03-16 19:55:56
log(m)nx must be between -1 and 1
VietaFan 2013-03-16 19:55:56
arcsin's domain is [-1, 1]
billgates42 2013-03-16 19:55:56
The domain of arcsin=[-1,1]
mathman500 2013-03-16 19:55:56
Domain of arcsin(x) is [-1,1]
DPatrick 2013-03-16 19:56:22
Right. Since sines are always between -1 and 1, the arcsin function must take an input between -1 and 1 (since sine's output is between -1 and 1).
DPatrick 2013-03-16 19:56:29
So we must have $-1 \le \log_m(nx) \le 1$.
DPatrick 2013-03-16 19:56:36
What does that mean?
distortedwalrus 2013-03-16 19:57:08
raise everything to the mth power
Hydroxide 2013-03-16 19:57:08
1/m <= nx <= m
swe1 2013-03-16 19:57:08
Raise both sides with base m
QuantumandMath 2013-03-16 19:57:08
1/m<=nx<=m
JRY 2013-03-16 19:57:08
1/m <= nx <=m
bromine 2013-03-16 19:57:08
1/m <= nx <= m
DPatrick 2013-03-16 19:57:24
Right, we get rid of the log by raising m to all the terms in our inequality.
DPatrick 2013-03-16 19:57:30
This means that $\frac{1}{m} \le nx \le m$.
lucylai 2013-03-16 19:57:47
divide everything by n
billgates42 2013-03-16 19:57:47
ISOLATE X!!!!!!
QuantumandMath 2013-03-16 19:57:47
1/nm<=x<=m/n
DPatrick 2013-03-16 19:58:04
This gives us $\frac{1}{mn} \le x \le \frac{m}{n}$.
DPatrick 2013-03-16 19:58:12
So what is the length of the interval?
DVA6102 2013-03-16 19:58:34
1/2013
promathist99 2013-03-16 19:58:34
1/2013
pedronr 2013-03-16 19:58:34
1/2013
bugegg3 2013-03-16 19:58:34
1/2013
sparkles257 2013-03-16 19:58:34
1/2013
d95776 2013-03-16 19:58:34
1/2013
mathematician153 2013-03-16 19:58:34
m/n-1/mn
VietaFan 2013-03-16 19:58:34
That means that m/n-1/mn = 1/2013
joshxiong 2013-03-16 19:58:34
m/n-1/mn
DPatrick 2013-03-16 19:58:39
We must have $\frac{1}{mn} \le x \le \frac{m}{n}$, so the interval has length
$$
\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}.
$$
DPatrick 2013-03-16 19:58:51
Now what?
AlcumusGuy 2013-03-16 19:59:19
Clear the denominators and solve for n
kdokmeci 2013-03-16 19:59:19
Solve for n or m
DPatrick 2013-03-16 19:59:30
Well, we want to examine m+n, and that's hard to do at present.
DPatrick 2013-03-16 19:59:40
It'll probably help to solve for one of m or n in terms of the other.
DPatrick 2013-03-16 19:59:53
And n looks easier to solve for because it only appears in denominators.
DPatrick 2013-03-16 20:00:10
Let's isolate the two variables by multiplying by $n$ and simplifying the left side:
$$
\frac{m^2-1}{m} = \frac{n}{2013}.
$$
DPatrick 2013-03-16 20:00:33
And now we can solve for $n$ in terms of $m$:
$$
n = \frac{2013(m^2-1)}{m}.
$$
DPatrick 2013-03-16 20:00:40
What does this tell us?
ssilwa 2013-03-16 20:01:01
so m|2013
Knightone 2013-03-16 20:01:01
m divides 2013
ABCDE 2013-03-16 20:01:01
m is a factor of 2013
number.sense 2013-03-16 20:01:01
m divides 2013
calculatorwiz 2013-03-16 20:01:01
as m increases so does n
cerberus88 2013-03-16 20:01:01
m divides 2013
tiger21 2013-03-16 20:01:01
m is a factor of 2013
mathwizard888 2013-03-16 20:01:01
m is a factor of 2013
DPatrick 2013-03-16 20:01:11
Right -- it tells us two important things.
DPatrick 2013-03-16 20:01:31
Remember that $m$ and $n$ must be positive integers, and $m>1$. So $m$ must be a factor of 2013 other than 1.
DPatrick 2013-03-16 20:01:49
Of course, you've all memorized that $2013 = 3 \cdot 11 \cdot 61$ (or you can easily work it out again).
DPatrick 2013-03-16 20:02:15
But also, as calculatorwiz pointed out, as m increases, so does n.
DPatrick 2013-03-16 20:02:35
So making m as small as possible will also make n as small as possible.
steve123456 2013-03-16 20:02:51
so to minimize m+n, m=3
tc1729 2013-03-16 20:02:51
so minimizing m minimizes n, so take m=3
n1000 2013-03-16 20:02:51
so we want the smallest value of m, namely three
cxiong 2013-03-16 20:02:51
m = 3
csmath 2013-03-16 20:02:51
m=3?
DPatrick 2013-03-16 20:03:04
...and the smallest m we can take that's >1 and a factor of 2013 is m=3/
sahilp 2013-03-16 20:03:16
so m=3 and n=671*8
DPatrick 2013-03-16 20:03:23
Indeed, for $m=3$ we get
$$
n = \frac{2013(3^2-1)}{3} = 671 \cdot 8 = 5368.
$$
SuperSnivy 2013-03-16 20:03:48
m = 3, so n = 5368 ==> ans 371
number.sense 2013-03-16 20:03:48
m+n = 5371, so our answer is 371
AlcumusGuy 2013-03-16 20:03:48
so m + n = 5371, so answer is 371
lucylai 2013-03-16 20:03:48
371
davidkim2106 2013-03-16 20:03:48
5368+3=5371
SuperSnivy 2013-03-16 20:03:48
3+368 = 371
billgates42 2013-03-16 20:03:48
368+3=371!!!
DPatrick 2013-03-16 20:03:54
So $m+n = 5371$, and the answer is $\boxed{371}$.
DPatrick 2013-03-16 20:04:22
Let's do one more (#9), and then I'm going to take a little break before going on to the double-digit problems.
DPatrick 2013-03-16 20:04:29
DPatrick 2013-03-16 20:04:44
Here's the picture they gave:
DPatrick 2013-03-16 20:04:49
DPatrick 2013-03-16 20:04:57
Let's blow up that second picture a bit, and label the points, angles, and lengths that we know:
DPatrick 2013-03-16 20:05:02
DPatrick 2013-03-16 20:05:12
Anything else that we know that we can label?
teranz0 2013-03-16 20:05:37
Call XB 12 -x, XA : x
ssilwa 2013-03-16 20:05:37
let BX = x and XA = 12-x same for YC and AY
clear 2013-03-16 20:05:37
label Bx as x, and AX as 12-x, and same for CY and AY
VietaFan 2013-03-16 20:05:37
AB+BX = 12
DPatrick 2013-03-16 20:06:01
Right: we know that the folded sides AB = 12 and AC = 12 before the folding, so we know AX + XB = 12 and AY + YC = 12 after the folding.
DPatrick 2013-03-16 20:06:07
So let's call $AX = x$ and $AY = y$, and label the other lengths accordingly:
DPatrick 2013-03-16 20:06:12
DPatrick 2013-03-16 20:06:23
And let's keep in mind what we're trying to find: we want the length $XY$.
DPatrick 2013-03-16 20:06:27
I think we've collected the data -- how do we proceed?
noobynoob 2013-03-16 20:06:42
LoC
d95776 2013-03-16 20:06:42
Use the law of cosines
mathcool2009 2013-03-16 20:06:42
LAW OF COSINES
joshxiong 2013-03-16 20:06:42
We can use the Law of Cosines
mprashker 2013-03-16 20:06:42
law of cosines!!!
trumpetjean 2013-03-16 20:06:42
law of cosines
noobynoob 2013-03-16 20:06:42
LoC
thkim1011 2013-03-16 20:06:42
laws of cosine?
cire_il 2013-03-16 20:06:42
LOC to find x and y
sunny2000 2013-03-16 20:06:42
laws of cosine!
DPatrick 2013-03-16 20:07:01
I can't think of a more elegant way to proceed than to just bash with the Law of Cosines.
DPatrick 2013-03-16 20:07:36
We have two triangles ($ABX$ and $ACY$) in which we have all three sides and a 60-degree angle. That means we can use the Law of Cosines in each triangle to solve for $x$ and $y$. (60-degree angles are especially nice with Law of Cosines, of course, since $\cos(60^\circ) = \frac12$.)
DPatrick 2013-03-16 20:07:51
What does triangle $ABX$ give us?
steve123456 2013-03-16 20:08:46
x=39/5
kdokmeci 2013-03-16 20:08:46
x=39/5
msinghal 2013-03-16 20:08:46
x=39/5
ws5188 2013-03-16 20:08:46
x^2=81+(12-x)^2-9(12-x)
DPatrick 2013-03-16 20:08:56
Plugging in the data gives us
$$
x^2 = 9^2 + (12-x)^2 - 2(9)(12-x)(\cos 60^\circ).
$$
DPatrick 2013-03-16 20:09:10
Conveniently, $\cos 60^\circ = \frac12$, so this is just
$$
x^2 = 81 + 144 - 24x + x^2 - 108 + 9x.
$$
DPatrick 2013-03-16 20:09:29
The $x^2$'s cancel, and we have $0 = 117 - 15x$, which solves to give $x = \frac{39}{5}$.
DPatrick 2013-03-16 20:09:48
(I'm not belaboring the routine algebra -- you all can manage that I'm sure!)
brandbest1 2013-03-16 20:10:00
same thing for y
cxiong 2013-03-16 20:10:00
Find y, use cosine law again
DPatrick 2013-03-16 20:10:13
In triangle $ACY$, the Law of Cosines gives us
$$
y^2 = 3^2 + (12-y)^2 - 2(3)(12-y)(\cos 60^\circ).
$$
lcamhie142857 2013-03-16 20:10:22
y=39/7
lucylai 2013-03-16 20:10:22
y=39/7
distortedwalrus 2013-03-16 20:10:22
similarly we can get y=39/7 using the other triangle
DPatrick 2013-03-16 20:10:26
This simplifies to $0 = 117 - 21y$, so $y = \frac{39}{7}$.
DPatrick 2013-03-16 20:10:34
And how do we finish?
davidkim2106 2013-03-16 20:10:48
law of cosines again
ryanyoo 2013-03-16 20:10:48
one more time
onta 2013-03-16 20:10:48
Law of Cosines AGAIN!
pedronr 2013-03-16 20:10:48
Use LOC again for side XY.
Wickedestjr 2013-03-16 20:10:48
Law of cosines again!
DPatrick 2013-03-16 20:10:51
We again use Law of Cosines, on triangle $XAY$:
$$
(XY)^2 = x^2 + y^2 - 2xy(\cos60^\circ) = x^2 + y^2 - xy.
$$
DPatrick 2013-03-16 20:11:04
This gives
$$
(XY)^2 = \left(\frac{39}{5}\right)^2 + \left(\frac{39}{7}\right)^2 - \left(\frac{39}{5}\right)\left(\frac{39}{7}\right).
$$
DPatrick 2013-03-16 20:11:19
How do we simplify this most efficiently?
pickten 2013-03-16 20:11:35
factor out 39^2
calculatorwiz 2013-03-16 20:11:35
factor out 39^2
scgorantla 2013-03-16 20:11:35
factor out 39^2
ws5188 2013-03-16 20:11:35
factor out 39^2
kdokmeci 2013-03-16 20:11:35
Remove the 39's
DPatrick 2013-03-16 20:11:39
We can pull out the $39^2$ term:
$$
(XY)^2 = (39)^2 \left(\frac{1}{5^2} + \frac{1}{7^2} - \frac{1}{5 \cdot 7}\right).
$$
number.sense 2013-03-16 20:11:48
factor out the 39^2 and common denominator of 25*49
DPatrick 2013-03-16 20:12:02
We see that the common denominator of the expression in the parentheses is $5^2 \cdot 7^2$, so we have:
$$
(XY)^2 = (39)^2 \left(\frac{7^2 + 5^2 - 5 \cdot 7}{5^2 \cdot 7^2}\right) = (39)^2 \left(\frac{39}{5^2 \cdot 7^2}\right).
$$
cerberus88 2013-03-16 20:12:15
so XY=39sqrt(39)/35
DPatrick 2013-03-16 20:12:23
Taking the square root gives us
$$
XY = \frac{39\sqrt{39}}{35},
$$
which is in the required format, so the answer is $39 + 39 + 35 = \boxed{113}$.
DPatrick 2013-03-16 20:12:50
I'm going to take a quick 2-minute break to stretch my legs...we'll resume at :15 past!
DPatrick 2013-03-16 20:15:06
A couple of quick reminders before we continue...
DPatrick 2013-03-16 20:15:46
First, if you're new to AoPS, please note that this Math Jam is not like our usual online classes! Our regular classes have many fewer students and we're able to answer all your questions during regular classtime.
DPatrick 2013-03-16 20:16:24
Second, there will be a complete transcript of this Math Jam posted on the website once we're finished, so if you came late or just want to go back and review, you'll be able to see everything we did.
DPatrick 2013-03-16 20:16:38
And now, on to the double-digit problems!
DPatrick 2013-03-16 20:16:43
DPatrick 2013-03-16 20:17:12
What do we know about $P(x)$?
onta 2013-03-16 20:17:27
vieta?
billgates42 2013-03-16 20:17:27
Use VIETA!!!
QuantumandMath 2013-03-16 20:17:27
a is sum of zero's
TMTOLBTWNTOF 2013-03-16 20:17:32
r - si is therefore also a root
tc1729 2013-03-16 20:17:32
r-si is also a root of P(x)
kdokmeci 2013-03-16 20:17:32
If r+si is a root, so is r-si
ssilwa 2013-03-16 20:17:32
it also has root r-si
calculatorwiz 2013-03-16 20:17:32
it has roots r+si and r-si
distortedwalrus 2013-03-16 20:17:38
it has three roots
anthonyjang 2013-03-16 20:17:38
it has 2 imaginary roots and one real root
lazorpenguin27143 2013-03-16 20:17:38
r-si is also a zero since the coefficients are rational
Flamewire 2013-03-16 20:17:38
Two of its zeros are r+si and r-si
pickten 2013-03-16 20:17:38
p_A,B = a
AlcumusGuy 2013-03-16 20:17:41
sum of roots = a and product = 65
DVA6102 2013-03-16 20:17:41
65 is product of zeroes
DPatrick 2013-03-16 20:17:51
Right, all good observations! Let me organize them,
DPatrick 2013-03-16 20:18:00
The sum of its roots is $a$.
DPatrick 2013-03-16 20:18:04
The product of its roots is 65.
DPatrick 2013-03-16 20:18:15
(These are both consequences of Vieta's Formulas for polynomials.)
DPatrick 2013-03-16 20:18:41
We also know that if $r+si$ is a root, then so is $r-si$, and the third root is real.
DPatrick 2013-03-16 20:18:57
So what does this tell us about $r$ and $s$?
cerberus88 2013-03-16 20:19:28
we know that r^2+s^2 divides 65
OCed 2013-03-16 20:19:28
r^2+s^2|65
calculatorwiz 2013-03-16 20:19:28
r^2 + s^2 is the product of those too roots
DVA6102 2013-03-16 20:19:28
(r^2+s^2)k=65, where k is the third real root
lucylai 2013-03-16 20:19:28
the third root is 65/(r^2+s^2)
DPatrick 2013-03-16 20:19:51
Aha!
DPatrick 2013-03-16 20:19:58
Let's call the third (real) root $t$.
DPatrick 2013-03-16 20:20:08
Then we have $(r+si)(r-si)t = (r^2+s^2)t = 65$.
DPatrick 2013-03-16 20:20:20
So $r^2+s^2$ is a factor of 65.
DPatrick 2013-03-16 20:20:29
What are the possibilites?
minimario 2013-03-16 20:20:34
what is t isn't an integer?
DPatrick 2013-03-16 20:21:07
That's a good question. But we certainly know that t is rational, so one way we can tell is by the Rational Root Theorem that it must be an integer (because the coefficient of x^3 is 1).
DPatrick 2013-03-16 20:21:15
We can also tell by looking at the sum of the roots.
DPatrick 2013-03-16 20:21:37
DPatrick 2013-03-16 20:22:02
In fact, let's file that fact up top for future use...
DPatrick 2013-03-16 20:22:25
But back to the product...what are the possibilities for $r$ and $s$ given that $r^2 + s^2$ is a factor of 65?
number.sense 2013-03-16 20:22:45
r^2+s^2 = 5, 13, 65 (cannot be 1 because r,s >0)
kdokmeci 2013-03-16 20:22:45
1,5,13,65
davidkim2106 2013-03-16 20:22:45
13,5 1,65
DPatrick 2013-03-16 20:23:03
Right: the only positive integer factors of 65 are 1, 5, 13, and 65. And we can't have $r^2 + s^2 = 1$ because they both have to be nonzero.
DPatrick 2013-03-16 20:23:18
So we only need to look for nonzero perfect squares that sum to 5, 13, or 65.
DPatrick 2013-03-16 20:23:25
And there aren't very many...
Flamewire 2013-03-16 20:23:47
5 = 1^2 + 2^2
JRY 2013-03-16 20:23:47
5:1+4, 13:4+9
ssilwa 2013-03-16 20:23:47
5 = 1+4
Flamewire 2013-03-16 20:23:47
13 = 2^2 + 3^2
QuantumandMath 2013-03-16 20:23:47
5=1^2+2^2
mathcool2009 2013-03-16 20:24:05
r,s = 8,1 7,4 1,2 2,3
JRY 2013-03-16 20:24:05
65:64+1
Flamewire 2013-03-16 20:24:05
65 = 7^2 + 4^2
billgates42 2013-03-16 20:24:05
DONT FORGET that r can be negative!!!
Knightone 2013-03-16 20:24:05
(r,s)=plus/minus (1,8),(8,1) (2,3),(3,2), (1,2),(2,1) (4,7),(7,4)
DPatrick 2013-03-16 20:24:13
Indeed, the only possibilities are:
\begin{align*}
1^2 + 2^2 &= 5, \\
2^2 + 3^2 &= 13, \\
1^2 + 8^2 &= 65, \\
4^2 + 7^2 &= 65.
\end{align*}
DPatrick 2013-03-16 20:24:24
Of course, as you mention, r and/or s can be negative too!
DPatrick 2013-03-16 20:24:53
Perhaps its better to think in terms of the polynomials rather than choices for r and s. How many different cubic polynomials does each of these give us?
d95776 2013-03-16 20:25:19
4
sjag 2013-03-16 20:25:19
4
davidkim2106 2013-03-16 20:25:19
4?
jasonmathcounts 2013-03-16 20:25:19
4
DPatrick 2013-03-16 20:25:32
Right. Each solution above gives 4 cubics, for the four different choices of r. (So there are 16 cubics total.)
DPatrick 2013-03-16 20:25:46
For example, the equation $1^2 + 2^2 = 5$ gives a different cubic for $r = 1$, $r=-1$, $r=2$, and $r=-2$.
DPatrick 2013-03-16 20:26:11
Or to expand on this a bit further, each of the four triples below are a possible set of roots of the polynomial:
\begin{align*}
(+1+2i,~~ +1-2i,~~& 13), \\
(-1+2i,~~ -1-2i,~~& 13), \\
(+2+1i,~~ +2-1i,~~& 13), \\
(-2+1i,~~ -2-1i,~~& 13).
\end{align*}
DPatrick 2013-03-16 20:26:27
What do we get when we sum all these roots?
DPatrick 2013-03-16 20:26:38
(Just for the four cubics above that I've listed.)
calculatorwiz 2013-03-16 20:26:56
52
sahilp 2013-03-16 20:26:56
52
AlcumusGuy 2013-03-16 20:26:56
imaginary parts simplify to 0!
cerberus88 2013-03-16 20:26:56
52
lucylai 2013-03-16 20:26:56
52
csmath 2013-03-16 20:26:56
13*4=52
tiger21 2013-03-16 20:26:56
52
VietaFan 2013-03-16 20:26:56
52 = 13*4
mikhailgromov 2013-03-16 20:26:56
52
sunny2000 2013-03-16 20:26:56
52
jaymo 2013-03-16 20:26:56
52
flyrain 2013-03-16 20:26:56
52
DPatrick 2013-03-16 20:27:14
Right! Notice that not only do all the imaginary parts cancel out, but all the real parts of the complex roots cancel out too!
DPatrick 2013-03-16 20:27:26
Except for the 13's, everything cancels. $1+2i$ can be paired with $-1-2i$, $-1+2i$ can be paired with $1-2i$, etc. So we're just left with $4 \cdot 13 = 52$.
DPatrick 2013-03-16 20:27:59
So the four cubics coming from $1^2 + 2^2 = 5$ give roots that sum to $4 \cdot 13 = 52$.
DPatrick 2013-03-16 20:28:12
How about the four cubics coming from $2^2 + 3^2 = 13$? What do their roots sum to?
distortedwalrus 2013-03-16 20:28:32
20
number.sense 2013-03-16 20:28:32
20
joshxiong 2013-03-16 20:28:32
4*5=20
mathwizard888 2013-03-16 20:28:32
4*5=20
Sanqiang3 2013-03-16 20:28:32
20?
Superwiz 2013-03-16 20:28:32
20
DPatrick 2013-03-16 20:29:02
Right! Each of those four cubics will have two complex roots and a real root of 5 (so that 13*5 = 65 gives the correct product). But when we sum them, all the complex roots will cancel out, and we'll just have 4*5 = 20.
DPatrick 2013-03-16 20:29:20
And how about the third and fourth equations above?
SuperSnivy 2013-03-16 20:29:35
4 each
empoleon 2013-03-16 20:29:35
4 and 4 each
akalykid012 2013-03-16 20:29:35
4, 4
mathcool2009 2013-03-16 20:29:35
4 and 4
kdokmeci 2013-03-16 20:29:35
2*(4*1)=8
anthonyjang 2013-03-16 20:29:35
8, total
DPatrick 2013-03-16 20:30:01
Right: each has four cubics with two complex roots and a real root of 1. The complex roots cancel when we sum them, so we're left with 4*1 = 4 for each set of four cubics.
cerberus88 2013-03-16 20:30:15
the answer is 080
number.sense 2013-03-16 20:30:15
so our answer is 52 + 20 + 8 = 80
distortedwalrus 2013-03-16 20:30:15
for a total of 80 = 52+20+4+4
VietaFan 2013-03-16 20:30:15
The sum is 20+52+8 = 080.
DPatrick 2013-03-16 20:30:25
Adding them all up, we see that all the roots over all the cubics sum to $52 + 20 + 4 + 4 = \boxed{080}$.
DPatrick 2013-03-16 20:30:48
DPatrick 2013-03-16 20:31:06
First let's look at (a). What does that tell us?
mathworld1 2013-03-16 20:31:35
N divisible by 14,15,16
andrewlin 2013-03-16 20:31:35
N is div by 16,15 and 14
wjq8g6 2013-03-16 20:31:35
N = 0 mod 16, 15 and 14
gengkev 2013-03-16 20:31:35
N is a multiple of lcm(16,15,14)
pedronr 2013-03-16 20:31:35
N is a multiple of LCM(14,15,16)
ksun48 2013-03-16 20:31:35
14|N, 15|N, 16|N
TheUnChosenOne 2013-03-16 20:31:35
multiple of 16 15 and 14
ABCDE 2013-03-16 20:31:35
N is a multiple of 2^4*3*5*7
Hydroxide 2013-03-16 20:31:35
lcm(14, 15, 16) | N
DPatrick 2013-03-16 20:31:41
Right, $N$ must be a multiple of the least common multiple of 16, 15, and 14.
DPatrick 2013-03-16 20:31:47
So $N = 2^4 \cdot 3 \cdot 5 \cdot 7 \cdot k$ for some positive integer $k$.
DPatrick 2013-03-16 20:32:11
(Let's leave the primes as-is instead of multiplying them out right now -- we may not have to.)
DPatrick 2013-03-16 20:32:17
Now on to part (b).
DPatrick 2013-03-16 20:32:21
What numbers less than 14 are not already divisors of $N$?
cire_il 2013-03-16 20:32:48
9 11 13
mathematician153 2013-03-16 20:32:48
9,11,13
billgates42 2013-03-16 20:32:48
13 11 9
willabc 2013-03-16 20:32:48
9,11,13
Flamewire 2013-03-16 20:32:48
9, 11, 13
Knightone 2013-03-16 20:32:48
9,11,13
sparkles257 2013-03-16 20:32:48
9 and 11 and 13
lastenemytobeconquered 2013-03-16 20:32:48
9 11 and 13
lcamhie142857 2013-03-16 20:32:48
9, 11, 13
ws5188 2013-03-16 20:32:48
9, 11, 13
DPatrick 2013-03-16 20:33:04
1,2,3,4,5,6,7,8,10,12 are already divisors of $2^4 \cdot 3 \cdot 5 \cdot 7$, so the blocks will get distributed evenly to those numbers of students.
DPatrick 2013-03-16 20:33:12
This just leaves 9,11,13. So we must have $x=9$, $y=11$, and $z=13$.
DPatrick 2013-03-16 20:33:27
In particular, we now know that $N$ is 3 more than a multiple of 9, and 3 more than a multiple of 11, and 3 more than a multiple of 13.
mathworld1 2013-03-16 20:33:43
So N leaves a remainder of 3 when divided by 9,11,13
fmasroor 2013-03-16 20:33:43
and thus 3+(9*11*13)k
DPatrick 2013-03-16 20:33:52
Right...but I've already used "k".
DPatrick 2013-03-16 20:34:03
We can write this as $N = 9 \cdot 11 \cdot 13 \cdot m + 3$, for some positive integer $m$.
DPatrick 2013-03-16 20:34:13
So now we have two equations for $N$:
\begin{align*}
N &= 2^4 \cdot 3 \cdot 5 \cdot 7 \cdot k, \\
N &= 9 \cdot 11 \cdot 13 \cdot m + 3,
\end{align*}
where $m$ and $k$ are some positive integers.
DPatrick 2013-03-16 20:34:23
How do we find the minimal such $N$?
billgates42 2013-03-16 20:34:41
factor out a 3!!!
andrewlin 2013-03-16 20:34:41
well we can divide by 3 on both sides
DPatrick 2013-03-16 20:34:47
Getting rid of the factor of 3 might help: let $N' = N/3$. (Both equations are multiples of 3.)
DPatrick 2013-03-16 20:34:54
Now we have
$$
N' = 2^4 \cdot 5 \cdot 7 \cdot k = 3 \cdot 11 \cdot 13 \cdot m + 1.
$$
mprashker 2013-03-16 20:35:08
euclidean algorithm
noobynoob 2013-03-16 20:35:08
Euclidean algorithm?
DPatrick 2013-03-16 20:35:40
One very good idea at this point is to use the Euclidean Algorithm to solve this equation. You can use this algorithm to solve ax + by = 1 where a and b are relatively prime.
DPatrick 2013-03-16 20:35:49
In fact that's how I solved the problem the first time.
DPatrick 2013-03-16 20:36:14
But I think I'll solve it using modular arithmetic right now, because it's a little less messy.
DPatrick 2013-03-16 20:36:40
We're trying to solve for k, so we can look at mod 3, mod 11, and mod 13 separately.
DPatrick 2013-03-16 20:36:55
What do we get if we look at our equation mod 3?
ssilwa 2013-03-16 20:37:32
k == 2 mod 3
SuperSnivy 2013-03-16 20:37:32
2k = 1 mod 3
number.sense 2013-03-16 20:37:32
we get that the lhs = 1 mod 3
mathwizard888 2013-03-16 20:37:32
2^4*5*7*k==1 (mod 3)
kdokmeci 2013-03-16 20:37:32
2k=1mod3
nikoma 2013-03-16 20:37:32
2^4*5*7*k = 1 (mod 3)
cerberus88 2013-03-16 20:37:32
DPatrick 2013-03-16 20:37:39
We get
$$
2^4 \cdot 5 \cdot 7 \cdot k \equiv 1 \cdot 2 \cdot 1 \cdot k \equiv 2k \equiv 1 \pmod{3}.
$$
So $k \equiv 2 \pmod{3}$.
DPatrick 2013-03-16 20:38:03
How about mod 11?
cerberus88 2013-03-16 20:38:36
SuperSnivy 2013-03-16 20:38:36
10k = 1 mod 11
kdokmeci 2013-03-16 20:38:36
10k=1mod11, or -1k=1mod11
n1000 2013-03-16 20:38:36
10k=1(mod11)
DPatrick 2013-03-16 20:38:43
We get
$$
5 \cdot 5 \cdot 7 \cdot k \equiv 3 \cdot 7 \cdot k \equiv 10k \equiv 1 \pmod{11},
$$
DPatrick 2013-03-16 20:39:16
And 10k is the same as -k mod 11, so we have that k = -1 mod 11.
DPatrick 2013-03-16 20:39:22
Before we go on to mod 13, what do you notice about the mod 3 and mod 11 results?
mathwizard888 2013-03-16 20:39:48
both -1
fmasroor 2013-03-16 20:39:48
both =-1 mod M
mathworld1 2013-03-16 20:39:48
k = -1 of both mods
anwang16 2013-03-16 20:39:48
both -1
mathworld1 2013-03-16 20:39:48
so k = -1 mod 33
DPatrick 2013-03-16 20:39:52
We note that $k \equiv -1 \pmod{3}$ and $k \equiv -1 \pmod{11}$, so $k \equiv -1 \pmod{33}$. That is, $k$ is one less than a multiple of 33.
DPatrick 2013-03-16 20:40:22
How about mod 13?
goran 2013-03-16 20:40:40
k=1 mod 13
Annabeth 2013-03-16 20:40:40
k=1 (mod 13)
cerberus88 2013-03-16 20:40:40
cire_il 2013-03-16 20:40:40
k=1 mod 13
flyrain 2013-03-16 20:40:40
k=1 mod 13
DPatrick 2013-03-16 20:40:45
We get
$$
3 \cdot 5 \cdot 7 \cdot k \equiv 2 \cdot 7 \cdot k \equiv k \equiv 1 \pmod{13}.
$$
So $k \equiv 1 \pmod{13}$.
DPatrick 2013-03-16 20:40:56
So we've concluded that k is 1 more than a multiple of 13 but 1 less than a multiple of 33.
DPatrick 2013-03-16 20:41:03
You could guess-and-check from here, but what's a systematic way to solve it?
anthonyjang 2013-03-16 20:41:27
CRT?
mathworld1 2013-03-16 20:41:27
chinese remainder thm!
nikoma 2013-03-16 20:41:27
Chinese remainder theorem
mprashker 2013-03-16 20:41:27
chinese remainder theorem
Hydroxide 2013-03-16 20:41:27
CRT
pickten 2013-03-16 20:41:27
CRT
DPatrick 2013-03-16 20:41:31
What does that mean?
DPatrick 2013-03-16 20:41:55
The Chinese Remainder Theorem says that there is a solution, but how do we find it?
nsun48 2013-03-16 20:42:17
substitute 1+13a in for k, and use another one of the equations
DPatrick 2013-03-16 20:42:36
That's a good idea. I like working with the bigger number though (but it doesn't matter much).
DPatrick 2013-03-16 20:42:46
In other words, we can let $k = 33c - 1$ for some positive integer $c$. Then we have
$$
33c - 1 \equiv 1 \pmod{13},
$$
which reduces to $7c \equiv 2 \pmod{13}$.
Hydroxide 2013-03-16 20:43:20
so c=4 (mod 13)
anwang16 2013-03-16 20:43:20
c==4
ws5188 2013-03-16 20:43:20
c==4(mod 13)
DPatrick 2013-03-16 20:43:24
Now multiplying by 2 gives $14c \equiv c \equiv 4 \pmod{13}$.
DPatrick 2013-03-16 20:43:38
Now we just unravel everything we've done...
DPatrick 2013-03-16 20:43:48
We have $c = 4$ and $k = 4(33)-1 = 131$ is the smallest positive solution.
kdokmeci 2013-03-16 20:44:06
131 is prime
DPatrick 2013-03-16 20:44:12
And indeed 131 is prime.
DPatrick 2013-03-16 20:44:16
Thus $N = 2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 131$.
lucylai 2013-03-16 20:44:39
2+3+5+7+131=148, so 148 is the answer
mathwizard888 2013-03-16 20:44:39
c=4 gives k=131, and 131 is prime, so 2+3+5+7+131=148 answer
kdokmeci 2013-03-16 20:44:39
2+3+5+7+131=148
cerberus88 2013-03-16 20:44:39
2+3+5+7+131=148'
viker 2013-03-16 20:44:39
$ 2+3+5+7+131 =\boxed{148} $
akalykid012 2013-03-16 20:44:39
2+3+5+7+131= 148
wjq8g6 2013-03-16 20:44:39
131 + 2 + 3 + 5 + 7 = 148
DPatrick 2013-03-16 20:44:50
And the sum of the distinct primes of $N$ is $2+3+5+7+131 = \boxed{148}$.
jasonmathcounts 2013-03-16 20:45:16
I thought this problem was kind of unfair because part B in the question doesn't specifically rule out x=1 . If there was one student present then the blocks could technically be distributed equally (the one student gets total number - 3 blocks). If you do that you'll get 17 (like me and a lot of other people did).
DPatrick 2013-03-16 20:45:27
I do not agree with this interpretation, but if you feel strongly about it, you should write to the AMC.
DPatrick 2013-03-16 20:45:42
DPatrick 2013-03-16 20:46:02
Before I try to draw the picture, which is the point on side RP: is it E or F?
SuperSnivy 2013-03-16 20:46:28
F
AayushGupta 2013-03-16 20:46:28
F
distortedwalrus 2013-03-16 20:46:28
F
sindennisz 2013-03-16 20:46:28
F
jeff10 2013-03-16 20:46:28
F
DPatrick 2013-03-16 20:46:32
Well, to be honest, I couldn't tell until I drew both pictures and looked at the angles. If you try to put E on side RP, you get too much angle at point E (you can draw it and see for yourself). It is point F that is on RP.
DPatrick 2013-03-16 20:46:53
So the picture looks something like this:
DPatrick 2013-03-16 20:46:58
DPatrick 2013-03-16 20:47:08
All the side lengths of the hexagon are 1. Any other lengths we know?
lucylai 2013-03-16 20:47:32
CB=CQ=BQ=1
superpi83 2013-03-16 20:47:32
CBQ is equilateral with side length 1
billgates42 2013-03-16 20:47:32
BQ=1
cerberus88 2013-03-16 20:47:32
CQ and BQ are also 1
Hydroxide 2013-03-16 20:47:32
BQ=CQ=1
AceOfDiamonds 2013-03-16 20:47:32
BQ, CQ
Polynomial 2013-03-16 20:47:32
BQ and CQ
OCed 2013-03-16 20:47:32
BC=CQ=QB=1
DPatrick 2013-03-16 20:47:37
CBQ is equilateral, so $CQ = BQ = 1$ too.
DPatrick 2013-03-16 20:48:24
Any other length in this picture that we can quickly figure out?
DPatrick 2013-03-16 20:48:54
We can do a lot with the Law of Sines, but there's a simpler (in my opinion) non-trig solution.
davidkim2106 2013-03-16 20:49:03
RD=RQ-2
QuantumandMath 2013-03-16 20:49:08
rd?
DPatrick 2013-03-16 20:49:23
We actually also know $RD$ too! We just need to draw in $DF:$
DPatrick 2013-03-16 20:49:30
Flamewire 2013-03-16 20:49:51
Then we get a 45-45-90 triangle
Piya31415 2013-03-16 20:49:51
45 45 90 triangle!
mathworld1 2013-03-16 20:49:51
45-45-90
DPatrick 2013-03-16 20:50:02
Right! DEF is 120-30-30. So angle RDF is 60+30 = 90 degrees, a right angle!
cerberus88 2013-03-16 20:50:18
so RD=FD=sqrt(3)
DPatrick 2013-03-16 20:50:24
So RDF is an isosceles right triangle, and hence $RD = DF = \sqrt3$.
DPatrick 2013-03-16 20:50:36
We're actually almost done: how do we finish?
kdokmeci 2013-03-16 20:50:52
Make ALTITUDE to RQ
DPatrick 2013-03-16 20:51:14
Yes! We want the area, and side $RQ = 2 + \sqrt3$, so if we can compute the length of the altitude from $P$ to $RQ$, we'll be done.
noobynoob 2013-03-16 20:51:26
30-60-90 and 45-45-90
mdlu 2013-03-16 20:51:26
altitude forms a 30-60-90 and a 45-45-90
DPatrick 2013-03-16 20:51:36
DPatrick 2013-03-16 20:51:50
Oh, now we're in great shape! PXR is 45-45-90 and PXQ is 30-60-90!
DPatrick 2013-03-16 20:52:16
Let's call $PX = h$. What do we know about $RX$ and $XQ$?
csmath 2013-03-16 20:52:39
They sum to 2 +sqrt 3
cjquines0 2013-03-16 20:52:39
RX + XQ = 2 + sqrt(3)
VietaFan 2013-03-16 20:52:39
RX+XQ = 2+sqrt3
mondi 2013-03-16 20:52:42
RX=h
anthonyjang 2013-03-16 20:52:46
RX=h, XQ=h/\sqrt{3}
DPatrick 2013-03-16 20:52:53
Right, and that's all the data we need!
DPatrick 2013-03-16 20:52:58
$PXR$ is an isosceles right triangle angle, so $RX = PX = h$.
DPatrick 2013-03-16 20:53:17
...(I meant to say "right triangle *again*" above)...
DPatrick 2013-03-16 20:53:27
And...$PXQ$ is 30-60-90! So $XQ = \frac{h}{\sqrt3}$.
DPatrick 2013-03-16 20:53:37
Thus, since $RX + XQ = RQ$, we have
$$
h + \frac{h}{\sqrt3} = 2 + \sqrt3.
$$
VietaFan 2013-03-16 20:53:43
just algebra now
Flamewire 2013-03-16 20:53:52
h
viker 2013-03-16 20:53:52
solve for h
DPatrick 2013-03-16 20:54:00
Indeed, this solves to gives us
$$
h = \frac{2+\sqrt3}{1+\frac{1}{\sqrt3}}.
$$
DPatrick 2013-03-16 20:54:14
We can multiply numerator and denominator by $\sqrt3$ to get the slightly nicer
$$
h = \frac{3 + 2\sqrt3}{1 + \sqrt3}.
$$
VietaFan 2013-03-16 20:54:35
rationalize denominator
Flamewire 2013-03-16 20:54:35
multiply top and bottom by 1 - sqrt3
n1000 2013-03-16 20:54:35
multiply top and bottom by 1-sqrt(3)
DPatrick 2013-03-16 20:54:55
We could do that now, but let's write the expression for the area of the triangle first. (Doesn't really matter.)
DPatrick 2013-03-16 20:55:05
Our desired area is
$$
\frac12h(RQ) = \frac{(3+2\sqrt3)(2+\sqrt3)}{2(1+\sqrt3)}.
$$
DPatrick 2013-03-16 20:55:27
I'll brush through the algebra: this cleans up a bit to
$$
\frac{12 + 7\sqrt3}{2 + 2\sqrt3},
$$
and we can rationalize the denominator by multiplying by $2\sqrt3 - 2$:
$$
\frac{(12+7\sqrt3)(2\sqrt3-2)}{12-4} = \frac{18 + 10\sqrt3}{8} = \frac{9 + 5\sqrt3}{4}.
$$
cerberus88 2013-03-16 20:55:55
so the answer is 021
mathworld1 2013-03-16 20:55:55
a+b+c+d=021
kdokmeci 2013-03-16 20:55:55
Now our answer is 9+5+3+4=21
flyrain 2013-03-16 20:55:55
021
mssmath 2013-03-16 20:55:55
or 021
DPatrick 2013-03-16 20:56:00
This is in the proper format, so the answer is $9+5+3+4 = \boxed{021}$.
DPatrick 2013-03-16 20:56:27
DPatrick 2013-03-16 20:56:40
Yikes. So many subscripts.
DPatrick 2013-03-16 20:56:51
Let's try to draw a picture. Note that the angle at B is obtuse. I'll also use colors to label the angles that are equal (via the similar triangles):
DPatrick 2013-03-16 20:57:01
DPatrick 2013-03-16 20:57:20
I shaded $B_0C_1B_1$, because that's what we want to try to keep track of the area of.
DPatrick 2013-03-16 20:57:31
There are of course many many more triangles, but let's start with this picture.
DPatrick 2013-03-16 20:57:48
What length can we try to compute?
thecmd999 2013-03-16 20:58:23
C_0C_1
mdlu 2013-03-16 20:58:23
c1-b0 or c1-c0
DPatrick 2013-03-16 20:59:17
$C_0C_1$ is common to the two similar triangles $B_0C_1C_0$ and $AB_0C_0$, so let's set $x = C_0C_1$ and try to compute it.
DPatrick 2013-03-16 20:59:23
mathwizard888 2013-03-16 20:59:47
x/17=17/25
superpi83 2013-03-16 20:59:47
17/25=x/17 --> x=289/25
DPatrick 2013-03-16 20:59:56
We get that the ratio of the sides opposite red angles equals the ratio of sides opposite blue angles, or
$$
\frac{x}{17} = \frac{17}{25}.
$$
DPatrick 2013-03-16 21:00:17
So $x = \frac{289}{25}$.
DPatrick 2013-03-16 21:00:30
Well, that's a bit of data, but how does it help?
goran 2013-03-16 21:00:45
We can find AC_{1}
mondi 2013-03-16 21:00:45
so now we know AC1
cerberus88 2013-03-16 21:00:59
then we can find AC_1 and then use similar triangles to find B_1C_1
viker 2013-03-16 21:01:08
we can find $ AC_! $
jigglypuff 2013-03-16 21:01:08
find a c1
DPatrick 2013-03-16 21:01:29
These are probably all good ideas, but before we get bogged in too many computations, what's the general strategy here?
DPatrick 2013-03-16 21:01:42
Maybe we should draw some more shaded triangles to get an idea of the big picture:
DPatrick 2013-03-16 21:01:49
DPatrick 2013-03-16 21:01:55
Notice anything helpful?
flyrain 2013-03-16 21:02:27
all the shadeds are similar
Math99 2013-03-16 21:02:27
similar triangles
Math_Kirby 2013-03-16 21:02:27
infinite geometric series and sim. triangles
VietaFan 2013-03-16 21:02:27
infinite geometric series of triangles
fz0718 2013-03-16 21:02:27
they're summing to a geometric series
calculatorwiz 2013-03-16 21:02:27
they're all similar to each other
lucylai 2013-03-16 21:02:27
the areas form geometric sequence
DPatrick 2013-03-16 21:02:50
There's a lot of similarity, and it suggested that the areas of the shaded triangles will be a geometric series.
DPatrick 2013-03-16 21:03:00
But there's another clever idea that I like...
superpi83 2013-03-16 21:03:09
The fraction of grey in AB1C1 is, by similarity, the same as the fraction of grey in AB0C0, and therefore also equal to the fraction of grey in B0B1C1C0. So, we can find the total grey area by finding the fraction of grey in B0B1C1C0 and multiplying by the area of triangle AB0C0.
DPatrick 2013-03-16 21:03:40
That's a great idea (and I should also credit by AoPS colleague Palmer Mebane for suggesting this idea to me too).
DPatrick 2013-03-16 21:03:51
Let me try to explain what this means in the picture.
DPatrick 2013-03-16 21:04:10
Let's break up the picture into quadrilaterals:
DPatrick 2013-03-16 21:04:15
DPatrick 2013-03-16 21:04:42
Notice that the ratio of (shaded area) : (total area) within each quadrilateral is the same, by the similarity.
DPatrick 2013-03-16 21:05:16
So if we can compute that ratio just for the first quadrilateral, that's all we need! We then just take that ratio times the total are of $AB_0C_0$ to get our answer.
DPatrick 2013-03-16 21:05:43
So we don't need to bother with a geometric series: we just do it once for the first quadrilateral.
DPatrick 2013-03-16 21:05:55
In other words, we want the ratio of shaded area to the total area in the brown-bordered quadrilateral below:
DPatrick 2013-03-16 21:05:59
DPatrick 2013-03-16 21:06:11
Multiplying this ratio by the total area of $AB_0C_0$ will give us the total shaded area of all the little triangles, and hence our answer.
DPatrick 2013-03-16 21:06:36
(This might get a little clearer as we go. Or you may need to come back and think about it more later.)
DPatrick 2013-03-16 21:06:43
So how do we compute this ratio?
billgates42 2013-03-16 21:07:03
Ratio of C1B1 to 17
QuantumandMath 2013-03-16 21:07:03
find C1B1
pickten 2013-03-16 21:07:10
fractions of the entire area?
DPatrick 2013-03-16 21:07:21
I like this last idea. Note that our original triangle has three pieces:
DPatrick 2013-03-16 21:07:26
DPatrick 2013-03-16 21:07:56
We want the ratio of (gray) / (gray + purple). Rather than compute the actual areas, we can find out the fraction that each piece makes up of the whole.
DPatrick 2013-03-16 21:08:04
What fraction of the total triangle's area is the purple piece?
cire_il 2013-03-16 21:08:35
289/625
mathcool2009 2013-03-16 21:08:35
289/625
cerberus88 2013-03-16 21:08:35
289/25
wjq8g6 2013-03-16 21:08:35
17/25 ^2
dandedude 2013-03-16 21:08:35
289/625?
superpi83 2013-03-16 21:08:35
(17/25)^2
distortedwalrus 2013-03-16 21:08:35
289/625
DPatrick 2013-03-16 21:09:01
This is not so bad, because the purple triangle is similar to the entire large triangle. The ratio of similarity is $\frac{B_0C_0}{AC_0} = \frac{17}{25}$, so it is $\left(\frac{17}{25}\right)^2$ of the total triangle's area.
DPatrick 2013-03-16 21:09:23
The gray one is hard to get at directly, but the yellow one is also similar to the whole.
DPatrick 2013-03-16 21:09:27
What fraction of the total triangle's area is the yellow piece?
wjq8g6 2013-03-16 21:09:57
(25 - 289/25)/ 25 squared
DPatrick 2013-03-16 21:10:06
Unfortunately, yeah.
DPatrick 2013-03-16 21:10:11
The ratio of similarity is a little more complicated. It's:
$$
\frac{AC_1}{AC_0} = \frac{25-\frac{289}{25}}{25} = \frac{625 - 289}{625} = \frac{336}{625}.
$$
So it is $\left(\frac{336}{625}\right)^2$ of the total area.
dandedude 2013-03-16 21:10:24
ouch
fmasroor 2013-03-16 21:10:24
eww
DPatrick 2013-03-16 21:10:28
This is getting complicated. What can we do to make it simpler?
lcamhie142857 2013-03-16 21:10:47
don't calculate anything out until the end
DPatrick 2013-03-16 21:11:04
And what's a really good way to avoid the temptation of calculating anything?
jameswangisb 2013-03-16 21:11:09
give it a varible?
zqjx 2013-03-16 21:11:16
use variables
gengkev 2013-03-16 21:11:16
assign vars!
AayushGupta 2013-03-16 21:11:16
variables!
DPatrick 2013-03-16 21:11:19
The quantity 17/25 comes up a lot so let's give it a name. Let's call it r.
DPatrick 2013-03-16 21:11:33
So now the purple piece is $r^2$ of the total area, and the yellow piece is what in terms of $r$?
VietaFan 2013-03-16 21:12:10
$\frac{25-17r}{25}$
DPatrick 2013-03-16 21:12:26
That's the ratio of the lengths, so we'll have to square that to get the fraction of the area.
superpi83 2013-03-16 21:12:37
(1-r^2)^2
DPatrick 2013-03-16 21:12:57
Indeed, let's write it out:
$$
\frac{AC_1}{AC_0} = \frac{25 - 17r}{25} = 1 - \frac{17}{25}r = 1-r^2,
$$
so the yellow piece's area is $(1-r^2)^2$ of the total area.
DPatrick 2013-03-16 21:13:17
And thus the gray area is what's left, which is $1 - r^2 - (1-r^2)^2$.
DPatrick 2013-03-16 21:13:39
So the ratio that we wanted is:
$$
\frac{\text{gray}}{\text{gray} + \text{purple}} = \frac{1-r^2-(1-r^2)^2}{1-(1-r^2)^2}.
$$
DPatrick 2013-03-16 21:14:13
So we just compute this ratio, and multiply it by the area of the original triangle, and we've got our total area that we want!
DPatrick 2013-03-16 21:14:33
By the way, what is the area of the original triangle?
DVA6102 2013-03-16 21:15:06
By Heron, the area of the original is 90.
davidkim2106 2013-03-16 21:15:06
herons
mondi 2013-03-16 21:15:06
herons formula
cerberus88 2013-03-16 21:15:06
90
lucylai 2013-03-16 21:15:06
90 by Heron's formula
bestwillcui1 2013-03-16 21:15:06
use herons
DPatrick 2013-03-16 21:15:11
One option is to use Heron's Formula, noting that the semiperimeter is (12+17+25)/2 = 27:
$$
[AB_0C_0] = \sqrt{27(27-12)(27-17)(27-25)}.
$$
DPatrick 2013-03-16 21:15:20
This simplifies to
$$
\sqrt{27 \cdot 15 \cdot 10 \cdot 2} = \sqrt{2^2 \cdot 3^4 \cdot 5^2} = 2 \cdot 9 \cdot 5 = 90.
$$
DPatrick 2013-03-16 21:15:41
Or, you might notice that the triangle sits as the difference of two nice right triangles:
ryanyoo 2013-03-16 21:15:46
or you could think of it as a 15-20-25 triangle minus a 8-15-17 one
mdlu 2013-03-16 21:15:46
we can see that fitting in an 8-15-17 on the right hand side works
DPatrick 2013-03-16 21:15:51
DPatrick 2013-03-16 21:16:12
So $AB_0C_0$ is the difference between a 15-20-15 right triangle and a 8-15-17 right triangle, and its area is thus the difference of areas: 150 - 60 = 90.
davidkim2106 2013-03-16 21:16:29
multiply the ratio with 90
DPatrick 2013-03-16 21:16:47
Thus the area of all the shaded triangles is $$
\frac{90(1-r^2-(1-r^2)^2)}{1-(1-r^2)^2},
$$
where $r = \frac{17}{25}$.
pedronr 2013-03-16 21:16:58
then do all of the calculation
jigglypuff 2013-03-16 21:17:09
simplify
DPatrick 2013-03-16 21:17:21
It simplifies quite a bit:
$$
\frac{90(1-r^2-(1-2r^2+r^4))}{1-(1-2r^2+r^4)} = \frac{90(r^2-r^4)}{2r^2-r^4} = \frac{90(1-r^2)}{2-r^2}.
$$
DPatrick 2013-03-16 21:17:36
(Using r instead of 17/25 was a very good idea!)
DPatrick 2013-03-16 21:17:49
But now we don't have much choice but to plug r back in.
DPatrick 2013-03-16 21:17:59
We plug $r = \frac{17}{25}$ in, and multiply numerator and denominator by $25^2$ to clear fractions:
$$
\frac{90(25^2 - 17^2)}{2 \cdot 25^2 - 17^2}.
$$
DPatrick 2013-03-16 21:18:13
This is $\dfrac{90 \cdot 336}{961}$.
viker 2013-03-16 21:18:40
superpi83 2013-03-16 21:18:40
therefore 961
kdokmeci 2013-03-16 21:18:40
Answer;961
mathmaster2012 2013-03-16 21:18:40
so 961
fmasroor 2013-03-16 21:18:40
961
kdokmeci 2013-03-16 21:18:40
961
DPatrick 2013-03-16 21:18:45
Since 961 doesn't have any factors in common with the numerator, and we only want the denominator as our final answer, we're done. The final answer is $\boxed{961}$.
distortedwalrus 2013-03-16 21:19:12
so it's 31^2...coincidence that it's a perfect square?
DPatrick 2013-03-16 21:19:19
I think so, but I'm not sure...
DPatrick 2013-03-16 21:19:37
I thought #13 was the hardest and the least elegant problem on the contest. #14 and #15 are much prettier!
DPatrick 2013-03-16 21:19:46
Sun7777 2013-03-16 21:20:05
that looks complex
DPatrick 2013-03-16 21:20:10
...in more than one way.
DPatrick 2013-03-16 21:20:21
There are a lot of clues here for what to try...
DPatrick 2013-03-16 21:20:25
...trig functions...
DPatrick 2013-03-16 21:20:35
...alternating signs...in particular in cycles of 4...
DPatrick 2013-03-16 21:20:40
...powers...
mathmaster2012 2013-03-16 21:21:06
thecmd999 2013-03-16 21:21:06
complex number BASH
kdokmeci 2013-03-16 21:21:06
complex numbers!
lucylai 2013-03-16 21:21:06
complex numbers
vlchen888 2013-03-16 21:21:06
i
lcamhie142857 2013-03-16 21:21:06
Demoivre's?
lucylai 2013-03-16 21:21:06
complex numbers with demoivre's
danielguo94 2013-03-16 21:21:06
de moivre's
DPatrick 2013-03-16 21:21:16
Right. We want to convert the trig functions to something that works nicely with powers.
DPatrick 2013-03-16 21:21:31
This (plus the other clues) strongly suggest trying to use complex exponentials (or deMoivre's Theorem if you prefer) in some way.
DPatrick 2013-03-16 21:21:49
In particular, we might want to try looking at some combination of $P \pm iQ$ or $Q \pm iP$ and seeing which one works.
DPatrick 2013-03-16 21:22:11
You have to experiment a little bit to see exactly what combination works.
QuantumandMath 2013-03-16 21:22:22
Q+I*p
DPatrick 2013-03-16 21:22:31
A little experimentation will lead you to look at $Q + iP$:
DPatrick 2013-03-16 21:22:38
\begin{align*}
Q + iP &= 1 \\
&+ \frac12(-\sin\theta + i\cos\theta) \\
&+ \frac14(-\cos 2\theta - i\sin 2\theta) \\
&+ \frac18(\sin 3\theta -i\cos 3\theta) \\
&+ \frac{1}{16}(\cos 4\theta + i\sin 4\theta) \\
&+ \frac{1}{32}(-\sin 5\theta + i\cos 5\theta) \\
&+ \cdots.
\end{align*}
DPatrick 2013-03-16 21:22:46
Notice the nice pattern?
QuantumandMath 2013-03-16 21:23:18
geometric powers
viker 2013-03-16 21:23:18
kdokmeci 2013-03-16 21:23:18
Math99 2013-03-16 21:23:21
geometric serieis?
DPatrick 2013-03-16 21:23:29
Conveniently, all of the terms are of the form
$$
\frac{1}{2^k} i^k (\cos k\theta + i\sin k\theta)
$$
where $k$ is a nonnegative integer.
DPatrick 2013-03-16 21:23:53
Look at the coefficients of the cosine term: 1, i, -1, -i, 1, i, ...
DPatrick 2013-03-16 21:23:59
They're the powers of i.
DPatrick 2013-03-16 21:24:12
And the sine coefficients: -1, -i, 1, i, -1, ...
DPatrick 2013-03-16 21:24:28
Also the powers of i, just shifted ahead by an extra multiple of i.
DPatrick 2013-03-16 21:24:41
So that's pretty cool!
DPatrick 2013-03-16 21:24:56
And we also know that $\cos k\theta + i\sin k\theta = e^{ik\theta} = (e^{i\theta})^k$.
DPatrick 2013-03-16 21:26:41
Or in the language of deMoivre's Theorem,
$$
\cos k\theta + i \sin k\theta = \left(\cos\theta + i\sin\theta\right)^k.
$$
DPatrick 2013-03-16 21:27:03
(that was exciting!)
superpi83 2013-03-16 21:27:17
geometric series with first term 1 and common ratio e^(itheta)*i/2
DPatrick 2013-03-16 21:27:36
Right! Our sum $Q+iP$ is now an infinite geometric series!
DPatrick 2013-03-16 21:27:41
$$
Q + iP = \sum_{k=0}^\infty \left(\frac{ie^{i\theta}}{2}\right)^k.
$$
DPatrick 2013-03-16 21:27:57
What is its sum?
QuantumandMath 2013-03-16 21:28:16
1/(1-x)
Flamewire 2013-03-16 21:28:37
1/(1-ie^itheta/2)
cerberus88 2013-03-16 21:28:37
superpi83 2013-03-16 21:28:37
1/(1-(ie^(itheta))/2)
pickten 2013-03-16 21:28:37
1/(1-ie^{itheta}/2)
DPatrick 2013-03-16 21:28:47
Right. The sum $1 + r + r^2 + r^3 + \cdots = \frac{1}{1-r}$, provided $|r| < 1$. But $\left|\frac{ie^{i\theta}}{2}\right| = \frac12$, so we're safe.
DPatrick 2013-03-16 21:28:55
Thus we have
$$
Q + iP = \frac{1}{1-\frac{ie^{i\theta}}{2}} = \frac{2}{2-ie^{i\theta}}.
$$
DPatrick 2013-03-16 21:29:04
Now what?
AceOfDiamonds 2013-03-16 21:29:27
find P/Q
DVA6102 2013-03-16 21:29:27
We haven't used P/Q=2root2/7 yet.
DPatrick 2013-03-16 21:29:50
Correct, we still have that data to use. What does $\frac{P}{Q} = \frac{2\sqrt2}{7}$ mean in terms of the complex number $Q + iP$?
mapletree14 2013-03-16 21:30:17
argument?
thecmd999 2013-03-16 21:30:17
it's the tangent of the angle formed by the complex number with the real axis in the complex plane
centralbs 2013-03-16 21:30:17
the tangent of the angle is 2sqrt2/7
mathmaster2012 2013-03-16 21:30:17
argument is arctan(2sqrt2/7)
DPatrick 2013-03-16 21:30:25
Not exactly, but you all have the right idea...
DPatrick 2013-03-16 21:30:56
Well, let's see..
DPatrick 2013-03-16 21:31:10
It means that $7P = 2\sqrt2Q$. In other words, 7 times the imaginary part of $Q+iP$ equals $2\sqrt2$ times the real part of $Q+iP$.
DPatrick 2013-03-16 21:31:32
And this means that $Q + iP$ is a real multiple of $7 + i2\sqrt2$. (The key is that it's a real multiple.)
DPatrick 2013-03-16 21:32:01
The "imaginary"/"real" ratio is the constant $2\sqrt2 / 7$.
DPatrick 2013-03-16 21:32:31
In other words, now we know that
$$
Q + iP = \frac{2}{2 - ie^{i\theta}} = c(7 + i2\sqrt2)
$$
for some real constant $c$.
DPatrick 2013-03-16 21:32:51
How does this help?
zqjx 2013-03-16 21:33:30
we can separate the real and imaginary parts?
pickten 2013-03-16 21:33:30
expand using cis?
DPatrick 2013-03-16 21:33:37
It's annoying to have the polar form on the left side but the rectangular form on the right side, so let's go back to cosine and sine:
DPatrick 2013-03-16 21:33:44
$$
Q + iP = \frac{2}{(2 + \sin\theta) - i\cos\theta)} = c(7 + i2\sqrt2)
$$
for some real constant $c$.
nsun48 2013-03-16 21:34:01
cross multiply
nsun48 2013-03-16 21:34:11
and both sides should be real
DPatrick 2013-03-16 21:34:20
Right. This means that the imaginary part of
$$
\left((2 + \sin\theta) - i\cos\theta\right)\left(7 + i2\sqrt2\right)
$$
must be 0, because this product must be real (it must equal 2/c).
DPatrick 2013-03-16 21:34:34
But what is that imaginary part?
QuantumandMath 2013-03-16 21:35:01
-7cos+2sqrt2(2+sin)
DPatrick 2013-03-16 21:35:10
Multiply it out: it's $2\sqrt2(2 + \sin\theta) - 7\cos\theta$.
DPatrick 2013-03-16 21:35:23
Since this must be 0, we now have the equation
$$
4\sqrt2 + 2\sqrt2\sin\theta = 7\cos\theta.
$$
DPatrick 2013-03-16 21:35:33
Now what?
lucylai 2013-03-16 21:35:49
square both sides
thecmd999 2013-03-16 21:35:49
square and solve for sine with the pythagorean identity
DPatrick 2013-03-16 21:35:54
I'd square both sides: both to get rid of the square roots, and to hopefully be able to apply $\sin^2\theta + \cos^2\theta = 1$.
DPatrick 2013-03-16 21:36:02
We get:
$$
32 + 32\sin\theta + 8\sin^2\theta = 49\cos^2\theta.
$$
gengkev 2013-03-16 21:36:15
cos^2=1-sin^2?
DPatrick 2013-03-16 21:36:22
Aha, now that's
$$
32 + 32\sin\theta + 8\sin^2\theta = 49 - 49\sin^2\theta.
$$
DVA6102 2013-03-16 21:36:35
solve for sin(theta)
cerberus88 2013-03-16 21:36:35
quadratic so solve fro sin(theta)
AayushGupta 2013-03-16 21:36:38
assign sin a variable and solve
DPatrick 2013-03-16 21:36:44
Since $\sin\theta$ is what we want to solve for, let's call it $x$, and what we have is just a quadratic in $x$:
$$
32 + 32x + 8x^2 = 49 - 49x^2.
$$
DPatrick 2013-03-16 21:36:58
This simplifies to $57x^2 + 32x - 17 = 0$.
DVA6102 2013-03-16 21:37:12
factor!
SkinnySanta 2013-03-16 21:37:12
rational root theorem
DPatrick 2013-03-16 21:37:16
Since we're told it has rational roots, it's probably not too difficult to factor.
DPatrick 2013-03-16 21:37:29
And since 17 is prime, there aren't too many possibilities.
mathmaster2012 2013-03-16 21:37:40
(19x+17)(3x-1)=0
DPatrick 2013-03-16 21:37:42
Indeed, we get $(19x + 17)(3x - 1) = 0$.
cire_il 2013-03-16 21:37:49
-17/19 happens to work
DPatrick 2013-03-16 21:37:56
To finish, note that we want the negative root, so we have $x = -\frac{17}{19}$. Our final answer is $17 + 19 = \boxed{036}$.
DPatrick 2013-03-16 21:38:22
The finish line is in sight!
DPatrick 2013-03-16 21:38:27
DPatrick 2013-03-16 21:38:47
We normally try to write arithmetic sequences in terms of their common differences; how can we do that here?
DPatrick 2013-03-16 21:39:23
This is where a clever choice can simplify things later quite a bit.
lucylai 2013-03-16 21:39:31
let x and y be common differences for (A, B, C) and (b, a, c), respectively
DPatrick 2013-03-16 21:39:44
That's a good idea, but to keep better track, I'm going to call them D and d.
SuperSnivy 2013-03-16 21:40:15
(B-D , B , B+D) and (b-d , b , b+d)
DPatrick 2013-03-16 21:40:32
That's what I normally like to do, and indeed on my first attempt I did it that way. It was needlessly messy.
DPatrick 2013-03-16 21:40:44
Then my AoPS colleague Jeremy Copeland gave be a better idea.
DPatrick 2013-03-16 21:41:05
Notice that in part (d) of the problem statement, the $C$'s line up ($C$ with $c$), whereas the $A$'s and $B$'s are mismatched ($b$ is in $A$'s position and $a$ is in $B$'s position).
DPatrick 2013-03-16 21:41:22
So this suggests making the third term the "base" term, and writing the other two terms using that base.
DPatrick 2013-03-16 21:41:35
In other words, we'll write
$$
(A,B,C) = (C-2D,C-D,C)
$$
where $D>0$ is the common difference, and
$$
(b,a,c) = (c-2d,c-d,c)
$$
where $d>0$ is the common difference.
DPatrick 2013-03-16 21:42:03
Now what does condition (c) say?
distortedwalrus 2013-03-16 21:42:27
A == a (mod p) and similarly for the others
kdokmeci 2013-03-16 21:42:36
A-a, B-b, C-c have a factor of p
DPatrick 2013-03-16 21:42:47
Right. But let's write them all out in terms of C,c,D,d.
DPatrick 2013-03-16 21:42:58
It says that
\begin{align*}
A-a &= (C-2D) - (c-d) = (C-c) - (2D-d), \\
B-b &= (C-D) - (c-2d) = (C-c) - (D-2d), \\
C-c
\end{align*}
are all multiples of $p$.
superpi83 2013-03-16 21:43:18
so 2D-d and D-2d are both multiples of p
mathworld1 2013-03-16 21:43:24
So p divides 2D-d and D-2d!
kdokmeci 2013-03-16 21:43:26
2D-d, D-2d are multiples of p
DPatrick 2013-03-16 21:43:42
Aha! It tells us that $2D-d$ and $D-2d$ are also multiples of $p$. (These could possibly be negative or 0, depending on what $D$ and $d$ are.)
mathmaster2012 2013-03-16 21:43:58
3d is a multiple of p
superpi83 2013-03-16 21:43:58
This means p divides 3d
DPatrick 2013-03-16 21:44:14
Right! We also then get $p\mid (2D-d)-2(D-2d)=3d.$
DPatrick 2013-03-16 21:44:29
So $3d$ is a multiple of $p$. What does that mean?
countingkg 2013-03-16 21:45:12
because d < p, p = 3
DPatrick 2013-03-16 21:45:25
Right!
DPatrick 2013-03-16 21:45:32
We know that $p$ is prime, and $0 < d < p$, and $3d$ is a multiple of $p$.
DPatrick 2013-03-16 21:46:10
The only possibility is that $p = 3$.
mathwizard888 2013-03-16 21:46:27
(b,a,c)=(0,1,2)
DPatrick 2013-03-16 21:46:47
Right: we have to have $d=1$, and the little sequence $(b,a,c) = (0,1,2)$ is the only one that works.
DPatrick 2013-03-16 21:47:08
So now let's go back and use this data in the problem statement...
DPatrick 2013-03-16 21:47:22
We have an arithmetic sequence $(A,B,C)$ with $0 \le A < B < C \le 99$, such that $A-1$, $B$, and $C-2$ are all multiples of 3.
fmasroor 2013-03-16 21:47:50
write in terms of B, R
DPatrick 2013-03-16 21:48:09
I agree: now $B$ is the nice term (it's just a multiple of 3), so I'd write the sequence in terms of $B$.
DPatrick 2013-03-16 21:48:15
That is, we want to count the ordered pairs $(B,X)$ of positive integers such that:
(i) $0 \le B-X < B < B+X \le 99$, and
(ii) $B$ is a multiple of 3, and
(iii) $X \equiv 1 \pmod 3$
DPatrick 2013-03-16 21:48:38
(I'm calling the common difference $X$ now, for no good reason -- I probably should have just stuck with $D$.)
DPatrick 2013-03-16 21:48:56
How do we count these?
VietaFan 2013-03-16 21:49:21
Count the B first.
mathworld1 2013-03-16 21:49:28
For each B, find the number of possible X
DPatrick 2013-03-16 21:49:32
The choices for B are easy: 3, 6, 9, 12, ..., 96=3*32. There are 32 of them.
DPatrick 2013-03-16 21:49:42
How many X's for each B?
pedronr 2013-03-16 21:49:54
The number of possible values of X per every B is like a pyramid centering at closest to 50
DPatrick 2013-03-16 21:50:04
Right. Note that if $B < 50$, then the $0 \le B-X$ is going to be the limiting condition on $X$; but if $B > 50$, then the $B+X \le 99$ is the limiting condition on $X$, and indeed the two cases are symmetric.
superpi83 2013-03-16 21:50:11
3 has 1, 6 has 2, 9 has 3...
DPatrick 2013-03-16 21:50:32
Exactly. If $B = 3k$ (where $1 \le k \le 16$), then $X$ can be $1, 4, 7, \ldots$, up to $3k-2$. So there are $k$ choices for $X$.
DPatrick 2013-03-16 21:50:48
Hence, there are $1+2+3+\cdots+16 = \frac{16 \cdot 17}{2} = 136$ pairs $(B,X)$ with $B < 50$.
mathworld1 2013-03-16 21:51:06
Symmetric for B>50
DPatrick 2013-03-16 21:51:22
Right. If $B = 99 - 3k$ (where $1 \le k \le 16$), then again $X$ can be $1, 4, 7, \ldots$, again up to $3k-2$. So again there are $k$ choices for $X$.
DPatrick 2013-03-16 21:51:29
So that gives another 136 pairs $(B,X)$ with $B>50$.
countingkg 2013-03-16 21:51:35
272!!!!!!!!!!!!!!!
cerberus88 2013-03-16 21:51:35
so the answer is 272
kdokmeci 2013-03-16 21:51:35
272 total
DPatrick 2013-03-16 21:51:38
Hence our answer is $2(136) = \boxed{272}$.
DPatrick 2013-03-16 21:52:00
And that's it for the AIME I! We just got in under 3 hours. Thanks for hanging out with my on a Saturday night.
DPatrick 2013-03-16 21:52:04
*me*!
DPatrick 2013-03-16 21:52:09
And please join us again for the AIME II Math Jam on Friday, April 5, when we'll do all this again for the AIME II.
anwang16 2013-03-16 21:52:26
wait are we allowed to take both tests
DPatrick 2013-03-16 21:52:37
No, you're not. But we're certainly allowed to discuss them both!
zero.destroyer 2013-03-16 21:52:59
what would you predict the usamo cutoff is?
DPatrick 2013-03-16 21:53:03
I wouldn't. I never do.
distortedwalrus 2013-03-16 21:53:19
is there a reason why we can take both AMCs but not both AIMEs?
DPatrick 2013-03-16 21:53:26
I have no idea. You'd have to ask the AMC that.
apple.singer 2013-03-16 21:53:36
What was your favorite problem?
DPatrick 2013-03-16 21:53:49
I liked #11 and #14.
DPatrick 2013-03-16 21:53:55
#10 was kind of nice too.
DPatrick 2013-03-16 21:54:01
I hated #13, to be honest.
Klu2014 2013-03-16 21:54:37
Is this a new rule? That you can only take one AIME?
DPatrick 2013-03-16 21:54:40
That's always been the rule.
tRIG 2013-03-16 21:56:01
would you say this years Aime was easier than previous years?
Wickedestjr 2013-03-16 21:56:01
It seemed like this was a particularly easy AIME in comparison to previous years.
mapletree14 2013-03-16 21:56:01
Was this an easy AIME?
DPatrick 2013-03-16 21:56:35
It's hard for me to say. There didn't seem to be a "killer" question in my opinion.
DPatrick 2013-03-16 21:57:10
The geometry seemed a bit easier than usual (to me anyway), except perhaps for #13.
DPatrick 2013-03-16 21:57:21
...which was more messy than really hard.
forthegreatergood 2013-03-16 21:57:33
when will transcripts be up
DPatrick 2013-03-16 21:57:45
About 5 minutes or so after I close the classroom.
minimario 2013-03-16 21:57:50
Will Richard be making videos?
DPatrick 2013-03-16 21:57:53
I'm not sure.
csmath 2013-03-16 21:58:31
When does the classroom close?
DPatrick 2013-03-16 21:58:43
When I decide to close it, which is going to be in just a few minutes.
mapletree14 2013-03-16 21:58:48
What did you score on the AIME?
DPatrick 2013-03-16 21:59:10
I honestly do not remember. (And perhaps that's a commentary on how important your AIME scores is in the long term.) Not a 15.
scgorantla 2013-03-16 21:59:28
Are AIME 1 and 2 on the same difficulty, or how is the difficulty of each exam defined?
DPatrick 2013-03-16 21:59:34
Two contests are written, and then they flip a coin,
DPatrick 2013-03-16 21:59:36
Really.
mathmaster2012 2013-03-16 21:59:56
How do you know they flip a coin?
DPatrick 2013-03-16 22:00:03
Because I've been told so by the Director of the AMC>
DPatrick 2013-03-16 22:01:15
OK, I'm going to start the process for closing the classroom and doing the transcript, in part because I'm hungry for dinner (it's 7 PM here), so very shortly you'll all be ejected from the room.
DPatrick 2013-03-16 22:01:25
Good night and enjoy St. Patrick's Day tomorrow!
DPatrick 2013-03-16 22:02:44
Bye!

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