## 2011 AIME I Math Jam

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AoPS instructors discuss all 15 problems of the 2011 AIME I.

#### Facilitator: Richard Rusczyk

rrusczyk 2011-03-19 19:02:53
Welcome to the 2011 AIME I Math Jam!
rrusczyk 2011-03-19 19:02:58
I'm Richard Rusczyk, founder of AoPS, and I'll be leading our discussion tonight.
rrusczyk 2011-03-19 19:03:01
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.
rrusczyk 2011-03-19 19:03:07
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.
rrusczyk 2011-03-19 19:03:15
This helps keep the class organized and on track.  This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
rrusczyk 2011-03-19 19:03:26
There are a lot of students here!  As I said, only (a very 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!!
rrusczyk 2011-03-19 19:03:46
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the prerequisite material for every problem as we go.  Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask.  We usually do in our classes, but we have a large number of students tonight!  So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!  This Math Jam will be much more of a zoo than our usual classes.
rrusczyk 2011-03-19 19:04:03
We have an assistant to help us out tonight -- Miles Dillon Edwards.  Miles is an alumnus of Canada/USA Mathcamp, the Research Science Institute, HCSSiM, and MOP.  He is a top-25 scorer on the Putnam exam, and he was also a four-time USAMO qualifier, a two-time ARML high scorer, and a participant at Mathcounts nationals (representing Georgia).  He discovered a new proof for Heron's formula in 2007.  (You can find it in our Precalculus text!)  He is currently studying math and cello performance at Indiana University.  His username is Boy Soprano II.
Boy Soprano II 2011-03-19 19:04:13
Hello!
rrusczyk 2011-03-19 19:04:35
He can answer questions by whispering to you or by opening a window with you to chat 1-on-1.
rrusczyk 2011-03-19 19:04:45
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.  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 posted.
apple.singer 2011-03-19 19:05:19
will we go through multiple solutions to some problems?
rrusczyk 2011-03-19 19:05:28
Yes, but not many.  We do have 15 problems to cover!
rrusczyk 2011-03-19 19:05:35
Before we get started, I have a question: for those of you who took the test, what was your favorite question on the test?
SuperNerd123 2011-03-19 19:06:05
#4
hsm174 2011-03-19 19:06:05
10
calvinhobbesliker 2011-03-19 19:06:05
Number 9
cwein3 2011-03-19 19:06:05
number 4
policecap 2011-03-19 19:06:05
#15
1
knowmath 2011-03-19 19:06:05
#8
KingSmasher3 2011-03-19 19:06:05
8
hchen023 2011-03-19 19:06:05
Question 15
hrithikguy 2011-03-19 19:06:05
#4 because it was so epic
inquisitivity 2011-03-19 19:06:05
#7
tan90 2011-03-19 19:06:05
Number 6
leekspeak 2011-03-19 19:06:05
5
policecap 2011-03-19 19:06:05
#12
JSGandora 2011-03-19 19:06:05
number 7!
pi091997 2011-03-19 19:06:05
1.  the easiest one :D
#1-it was the only one I could easily do.
dlennon 2011-03-19 19:06:05
I liked problem 11.
mathelete 2011-03-19 19:06:05
#11
willwang123 2011-03-19 19:06:05
Number 1
apple.singer 2011-03-19 19:06:05
#5
hrithikguy 2011-03-19 19:06:05
#4 because it was so epically hard!
BOGTRO 2011-03-19 19:06:05
7 i guess
wtRiViaL 2011-03-19 19:06:05
not number 3
.cpp 2011-03-19 19:06:05
Number 4.
crobogc 2011-03-19 19:06:05
#3, because of the graph
rrusczyk 2011-03-19 19:06:12
rrusczyk 2011-03-19 19:06:26
Here's a question I'm guessing won't have many different answers:
rrusczyk 2011-03-19 19:06:28
OK, what was your least favorite problem?
leekspeak 2011-03-19 19:06:43
3
mhy123 2011-03-19 19:06:43
#4
inquisitivity 2011-03-19 19:06:43
#3
zazz4 2011-03-19 19:06:43
3
prophet886 2011-03-19 19:06:43
the same ones :)
bojobo 2011-03-19 19:06:43
number 4
minirafa 2011-03-19 19:06:43
13
SuperNerd123 2011-03-19 19:06:43
#1
agejiageji 2011-03-19 19:06:43
number 3
hrithikguy 2011-03-19 19:06:43
#4!! I hated that!!
girishvar12 2011-03-19 19:06:43
#3
jeff10 2011-03-19 19:06:43
4
wtRiViaL 2011-03-19 19:06:43
number 3 haha
apple.singer 2011-03-19 19:06:43
#4
cwein3 2011-03-19 19:06:43
3
mathelete 2011-03-19 19:06:43
#4
sincostanseccsccot 2011-03-19 19:06:43
#1 I got it wrong
KingSmasher3 2011-03-19 19:06:43
definately 4
willwang123 2011-03-19 19:06:43
All the other ones
minirafa 2011-03-19 19:06:43
13
JoeJack 2011-03-19 19:06:43
14
ahaanomegas 2011-03-19 19:06:43
None, all were pretty interesting
DavidTong 2011-03-19 19:06:43
3
bojobo 2011-03-19 19:06:43
definitely
omega1 2011-03-19 19:06:43
number 7
hsm174 2011-03-19 19:06:43
num 5
vcez 2011-03-19 19:06:43
3/4, too hard
theone142857 2011-03-19 19:06:43
12
policecap 2011-03-19 19:06:43
7
calvinhobbesliker 2011-03-19 19:06:43
4 and 3
policecap 2011-03-19 19:06:43
#7
rrusczyk 2011-03-19 19:06:56
I was guessing I'd see a lot of 3&4 on that one.
rrusczyk 2011-03-19 19:07:01
Let's get to the problems.
rrusczyk 2011-03-19 19:07:06
At a couple points during the Math Jam, I will get thirsty or hungry, or my fingers will get tired, and I'll take a 1-2 minute break.  Other than that, we will simply plow through all 15 problems.
rrusczyk 2011-03-19 19:07:17
rrusczyk 2011-03-19 19:07:28
What do we find first?
Ttocs45 2011-03-19 19:07:54
k
Kinata12 2011-03-19 19:07:54
k?
.cpp 2011-03-19 19:07:54
Solve for k.
willwang123 2011-03-19 19:07:54
k?
vcez 2011-03-19 19:07:54
k
rrusczyk 2011-03-19 19:08:02
And how do we find k quickly
rrusczyk 2011-03-19 19:08:03
?
connaissance 2011-03-19 19:08:48
look at the acid before and after
KingSmasher3 2011-03-19 19:08:49
set up an equation relating k with the final ratio of 50%
.cpp 2011-03-19 19:08:49
There must be a total of 5 liters of acid.
JSGandora 2011-03-19 19:08:49
we see that the total amount of acid will be 5 liters so we just add up the acid from each jar to get 5 and then solve for k
superpi83 2011-03-19 19:08:49
jar A has 1.8L of acid, jar B has 2.4L acid. we need a total of 5L acid so jar C must have 0.8L acid
rrusczyk 2011-03-19 19:08:53
We can find k pretty quickly.  We know that we have 10 liters total in the end, and that 5 liters of this total is acid.
rrusczyk 2011-03-19 19:09:08
sincostanseccsccot 2011-03-19 19:09:16
k=80
rrusczyk 2011-03-19 19:09:18
Now, we must find the number of liters of jar C we add to jar A.  Suppose we let this amount be t.  Then what?
rrusczyk 2011-03-19 19:10:36
There are a lot of ways to finish from here!
BOGTRO 2011-03-19 19:10:41
1.8+4/5t=1/2(4+t)
superpi83 2011-03-19 19:10:41
1.8+0.8t=0.5(4+t)
.cpp 2011-03-19 19:10:41
Jar A must have 50% acid. .8t + .45(4) = 1/2 (t+4).
Luminescence 2011-03-19 19:10:41
1.8 + .8t = .5(4 + t)
BOGTRO 2011-03-19 19:10:47
1.8+.8t=1/2(4+t)=2+.5t --> .3t=.2 --> t=2/3
rrusczyk 2011-03-19 19:10:52
rrusczyk 2011-03-19 19:11:04
And what do we find?
tiger21 2011-03-19 19:11:20
t=2/3
hrithikguy 2011-03-19 19:11:20
t = 2/3
tiger21 2011-03-19 19:11:20
t=2/3
JSGandora 2011-03-19 19:11:20
t=2/3
KingSmasher3 2011-03-19 19:11:20
t=2/3
apple.singer 2011-03-19 19:11:26
t=2/3
centralbs 2011-03-19 19:11:26
rrusczyk 2011-03-19 19:11:29
Rearranging this equation gives 0.6t = 0.4, so t = 2/3.
prezcoin 2011-03-19 19:11:43
k+m+n= 80+2+3= 85
j_f_c_w 2011-03-19 19:11:43
m+n+k=85
superpi83 2011-03-19 19:11:43
vcez 2011-03-19 19:11:43
.cpp 2011-03-19 19:11:43
Thus, k+m+n = 80+2+3 = 085.
rrusczyk 2011-03-19 19:11:47
rrusczyk 2011-03-19 19:11:53
1 down.  14 to go.
rrusczyk 2011-03-19 19:12:05
mhy123 2011-03-19 19:12:17
diagram
calvinhobbesliker 2011-03-19 19:12:17
Draw a picture
rrusczyk 2011-03-19 19:12:20
rrusczyk 2011-03-19 19:12:24
rrusczyk 2011-03-19 19:12:36
Any bright ideas?
rrusczyk 2011-03-19 19:12:51
What should we build here?
willwang123 2011-03-19 19:13:05
Draw triangles?
agejiageji 2011-03-19 19:13:05
a right triangle
jellymoop 2011-03-19 19:13:05
right triangle?
mhy123 2011-03-19 19:13:05
right triangles
rrusczyk 2011-03-19 19:13:10
We have a rectangle.  We want a length.  We should build a right triangle.  We should build a smart right triangle.
Kinata12 2011-03-19 19:13:56
Move ED upwards so that F is on E
Luminescence 2011-03-19 19:13:56
extend BE and FD
dlennon 2011-03-19 19:13:56
Extend BE and DF.
Ttocs45 2011-03-19 19:13:56
Extend BE so it meets CD
rrusczyk 2011-03-19 19:14:01
We extend BE to hit the extension of CD.  This may look like magic, but it builds a right triangle in which we know two sides, and the third has the length we want as part of it:
rrusczyk 2011-03-19 19:14:05
rrusczyk 2011-03-19 19:14:12
And just like that, we're done.  Why?
policecap 2011-03-19 19:14:50
GD=EF
zazz4 2011-03-19 19:14:50
hrithikguy 2011-03-19 19:14:50
GD = EF
prezcoin 2011-03-19 19:14:50
EF=GD
wtRiViaL 2011-03-19 19:14:50
now BG=17, BC=10, and you can find GC. THen you subtract 12 to get GD=EF
SuperNerd123 2011-03-19 19:14:50
Because GD=EF
girishvar12 2011-03-19 19:14:50
pythogeoren theorem
Luminescence 2011-03-19 19:14:50
pythagorean theorem!
BOGTRO 2011-03-19 19:14:50
right triangle, 10^2+(x+12)^2=(9+8)^2
Kinata12 2011-03-19 19:14:50
10^2+(12+x)^2=17^2. Easy to solve
tekgeek 2011-03-19 19:14:50
EG=FD=8
xD13G0x 2011-03-19 19:14:57
we can calculate GC, we know CD so we know GD=EF
rrusczyk 2011-03-19 19:15:01
EFDG is a parallelogram, so EG = FD = 8 and GD = EF.  Now we just have to find GD.
stickfigure 2011-03-19 19:15:17
since GD = EF, we can do sqrt(17^2-10^2) to find GC and subtract away 12 to get the answer
.cpp 2011-03-19 19:15:17
GC = sqrt(289-100) = sqrt(189) by Pythagorean theorem, so the answer is 3sqrt21 - 12.
vcez 2011-03-19 19:15:22
3sqrt21-10
policecap 2011-03-19 19:15:26
sqrt{189}-12
rrusczyk 2011-03-19 19:15:32
apple.singer 2011-03-19 19:15:37
and 3+21+12=36
ahaanomegas 2011-03-19 19:15:37
Answer = 3 + 21 + 12 = 3 + 33 = 036
rrusczyk 2011-03-19 19:15:54
2 down.  The next one is a bit of a monster.
rrusczyk 2011-03-19 19:16:05
rrusczyk 2011-03-19 19:16:17
This problem gave a lot of people problems on the test, so we're going to go through 3 solutions here with pretty different fundamental approaches.  One is going to be grind-it-out-analytic geometry solution that requires knowing a somewhat obscure formula.  One is going to be an elegant linear algebra approach that requires knowing some pretty advanced (and cool) math.  Both of these require knowledge that we think is too obscure for a #3 on the AIME (but the first is lamentably the official solution).  So, we're going to also find a solution using very basic geometry.
rrusczyk 2011-03-19 19:16:40
All three solutions start with the same insight.  What are we really looking for when we say we want the coordinates of P with respect to the new axes.
leekspeak 2011-03-19 19:17:05
distance to each line
tan90 2011-03-19 19:17:05
The distance from P to the lines.
policecap 2011-03-19 19:17:05
the lengtsh of the perpendiculars to the axes
Luminescence 2011-03-19 19:17:05
the distance from P to the two lines
calvinhobbesliker 2011-03-19 19:17:05
We want the distance from P to each of the 2 lines
wtRiViaL 2011-03-19 19:17:05
distance from point P to lines L and M
rrusczyk 2011-03-19 19:17:10
The coordinates of a point in any rectangular coordinate system are simply the signed distances to the axes of the coordinate system (the "signed" means points on one side of an axis have a negative coordinate and those on the other side have a positive coordinate.)
rrusczyk 2011-03-19 19:17:19
To find the coordinates of (-14,27) with respect to our new axes, we just figure out how far (-14,27) is from each axis.  Again, what do we have to be careful about?
leekspeak 2011-03-19 19:17:28
signs
AceOfDiamonds 2011-03-19 19:17:28
signs
SuperNerd123 2011-03-19 19:17:28
signs
mhy123 2011-03-19 19:17:30
signs?
numberwiz 2011-03-19 19:17:37
sign
PiquantPeppers 2011-03-19 19:17:37
signs
ahaanomegas 2011-03-19 19:17:37
Signs*
ahaanomegas 2011-03-19 19:17:39
Sings
rrusczyk 2011-03-19 19:17:58
It's OK to sing.  Well, maybe OK for Boy Soprano to sing.  You don't want me to sing.
rrusczyk 2011-03-19 19:18:06
rrusczyk 2011-03-19 19:18:13
All right, we have a plan.  Let's execute it carefully and make sure we don't make any mistakes.
rrusczyk 2011-03-19 19:18:16
First, we'll do the grimy analytic geometry solution.
calvinhobbesliker 2011-03-19 19:18:21
Can we use distance from point to line formula?
jellymoop 2011-03-19 19:18:27
distance from a point to a line
rrusczyk 2011-03-19 19:18:40
First line L.  How do we figure out how far (-14,27) is from L?
tan90 2011-03-19 19:19:04
Find the equation of L first.
hrithikguy 2011-03-19 19:19:04
find the equation of L
Jasmine8925 2011-03-19 19:19:04
first let's find line L
rrusczyk 2011-03-19 19:19:11
We first figure out an equation for L in the original coordinate system.  What is it?
leekspeak 2011-03-19 19:19:40
L is y=(5/12)x-11
jsani0102 2011-03-19 19:19:40
L=5/12x-11
hrithikguy 2011-03-19 19:19:40
y = 5/12 x  - 11
apple.singer 2011-03-19 19:19:40
y=5/12x-11
policecap 2011-03-19 19:19:40
y=5x/12-11
hchen023 2011-03-19 19:19:40
y + 1 = (5/12)(x - 24)
rrusczyk 2011-03-19 19:19:44
rrusczyk 2011-03-19 19:19:47
And how do figure out where (-14, 27) is with respect to this line?
anonymous0 2011-03-19 19:20:02
you use magic
rrusczyk 2011-03-19 19:20:06
Pretty close.
hrithikguy 2011-03-19 19:20:09
iti s above line L
rrusczyk 2011-03-19 19:20:19
So, we know that the coordinate is positive.
.cpp 2011-03-19 19:20:21
Use the |Ax+By+C|/sqrt(A^2+B^2) formula.
jellymoop 2011-03-19 19:20:25
the brutal formula!
rrusczyk 2011-03-19 19:20:40
First, we note that this point is above the line.  The y-coordinate of the point on the line with x = -14 is the solution to 5(-14) - 12y -132 =0.  The solution to this equation is clearly way less than 27.  So, (-14,27) is "above" this line in our new coordinate system.
rrusczyk 2011-03-19 19:20:43
rrusczyk 2011-03-19 19:20:51
This is the grimy formula people are talking about.
rrusczyk 2011-03-19 19:21:12
This is why I was not very happy to see this at #3, and that this was the official solution.
SuperNerd123 2011-03-19 19:21:19
How is that derived?
rrusczyk 2011-03-19 19:21:33
It's proved in a couple of our books :)  A nice way is with vectors.
GoldenFrog1618 2011-03-19 19:21:35
when did the AMC assume everyone had knowledge of that?
mentalgenius 2011-03-19 19:21:49
which books?
rrusczyk 2011-03-19 19:21:57
Precalculus, and possibly Intro Geometry.
rrusczyk 2011-03-19 19:22:09
Fair question, GoldenFrog.  Now, let's finish the problem with this formula.
rrusczyk 2011-03-19 19:22:16
What does this give us for the distance?
tiger21 2011-03-19 19:23:14
526/13
dlennon 2011-03-19 19:23:33
526/13
526/13
ahaanomegas 2011-03-19 19:23:33
526/13
rrusczyk 2011-03-19 19:23:43
rrusczyk 2011-03-19 19:23:51
Since the point is 526/13 above our new x-axis (line L), its y-coordinate is 526/13.
rrusczyk 2011-03-19 19:23:54
OK, on to line M.  What is the equation of line M?
jeff10 2011-03-19 19:24:25
y=-12x/5+18
apple.singer 2011-03-19 19:24:25
y=-12/5x+18
hrithikguy 2011-03-19 19:24:25
y = -12/5 x + 18
tekgeek 2011-03-19 19:24:25
hrithikguy 2011-03-19 19:24:25
y = -12/5 x + 18
rrusczyk 2011-03-19 19:24:28
Since line M is perpendicular to L, it has slope -1/(5/12) = -12/5.
rrusczyk 2011-03-19 19:24:32
rrusczyk 2011-03-19 19:24:44
What do we have to be careful about here?
zazz4 2011-03-19 19:24:55
signs
hrithikguy 2011-03-19 19:24:55
the signs!!
sindennisz 2011-03-19 19:24:55
SIgns
anonymous0 2011-03-19 19:24:55
signs
calvinhobbesliker 2011-03-19 19:24:55
Signs
vcez 2011-03-19 19:24:55
signs'
mentalgenius 2011-03-19 19:24:58
signs
JSGandora 2011-03-19 19:24:58
signs
prezcoin 2011-03-19 19:24:58
the sign is negative
policecap 2011-03-19 19:24:58
sings
jzz 2011-03-19 19:24:58
signs?
rrusczyk 2011-03-19 19:25:01
Here, we have to be a little careful when figuring out where (-14,27) is with respect to this line.  The point on line M with x = -14 in our original coordinate system has a y-coordinate that is the solution to 12(-14) + 5y - 90 = 0.  The solution to this equation is greater than 50, so M passes above (14,-27) is our original coordinate system.  That means that when we make M our y-axis in the new coordinate system, (-14,27) is to the left of our y-axis.  In other words, its x-coordinate in the new system is negative.
rrusczyk 2011-03-19 19:25:32
How far is the point from M?
KingSmasher3 2011-03-19 19:25:51
tiger21 2011-03-19 19:25:51
123/13
tekgeek 2011-03-19 19:25:51
superpi83 2011-03-19 19:25:51
123/13
tan90 2011-03-19 19:25:58
123/13
dlennon 2011-03-19 19:25:58
123/13
vcez 2011-03-19 19:25:58
123/13
superpi83 2011-03-19 19:26:00
well signs doesnt really matter much in this problem because if you add both instead of subtracting one from the other you get a non-integral answer...
rrusczyk 2011-03-19 19:26:04
rrusczyk 2011-03-19 19:26:09
Since the point is to the left of our new y-axis (line m), the new x-coordinate of the point is -123/13.  (This would have been a particularly evil problem if taking the positive value here had made the sum of the new coordinates an integer.)
rrusczyk 2011-03-19 19:26:16
theone142857 2011-03-19 19:26:36
31
vcez 2011-03-19 19:26:36
31
policecap 2011-03-19 19:26:36
31
KingSmasher3 2011-03-19 19:26:36
031
JSGandora 2011-03-19 19:26:36
031
.cpp 2011-03-19 19:26:36
Adding (watching signs), we get (526-123)/13 = 031.
gh625 2011-03-19 19:26:36
031
mentalgenius 2011-03-19 19:26:36
31
tiger21 2011-03-19 19:26:36
31
policecap 2011-03-19 19:26:36
mhy123 2011-03-19 19:26:40
31
PiquantPeppers 2011-03-19 19:26:40
31
omega1 2011-03-19 19:26:40
31
tekgeek 2011-03-19 19:26:40
vcez 2011-03-19 19:26:40
031
rrusczyk 2011-03-19 19:26:43
rrusczyk 2011-03-19 19:26:47
All right, that wasn't much fun.  Let's try to do something a little more elegant.
rrusczyk 2011-03-19 19:26:52
We want the distance from P to the line through A.
rrusczyk 2011-03-19 19:26:58
rrusczyk 2011-03-19 19:27:01
What else do we want to add to this diagram?
pi091997 2011-03-19 19:27:15
perp
hrithikguy 2011-03-19 19:27:15
a perp from P to line A
vcez 2011-03-19 19:27:15
a perpendicular line
numberwiz 2011-03-19 19:27:15
perp. form p to a
perpendicular line from P to the line through A
GoldenFrog1618 2011-03-19 19:27:15
foot of P to A
AlphaMath1 2011-03-19 19:27:15
perpendicular from P to A
rrusczyk 2011-03-19 19:27:21
We add in a perpendicular from P to the line, and we go ahead and draw PA.  Since we know P and A, we can figure out stuff about PA if we need to.
rrusczyk 2011-03-19 19:27:25
rrusczyk 2011-03-19 19:27:32
Now, for those of you who have some advanced experience with vectors, this sort of thing should look familiar.  Vector XP is the projection of vector AP onto the vector normal to the blue line.  (If that sentence was gibberish to you, it's OK to check out for a few minutes while we finish this solution.  In the meantime, you can think about where you might learn such cool stuff: http://www.artofproblemsolving.com/Store/viewitem.php?item=precalc)
rrusczyk 2011-03-19 19:28:06
What is a vector normal to the blue line?  (That is, tell me the components of a vector that is normal to the blue line.)
willwang123 2011-03-19 19:28:10
bulutcocuk 2011-03-19 19:28:10
parenthesis in the url :P
rrusczyk 2011-03-19 19:28:27
Thank you for helping me advertise :)
Jasmine8925 2011-03-19 19:28:29
normal to means perpendicular to?
rrusczyk 2011-03-19 19:28:32
Yes.
JSGandora 2011-03-19 19:28:54
PX
centralbs 2011-03-19 19:28:54
Vector XP
rrusczyk 2011-03-19 19:28:59
I'm looking for numbers here.
rrusczyk 2011-03-19 19:29:11
Is (2 1) perpendicular to the blue line?
anonymous0 2011-03-19 19:29:22
no
gh625 2011-03-19 19:29:22
No
rrusczyk 2011-03-19 19:29:27
Right.  What is?
policecap 2011-03-19 19:29:39
no, <-5,12> is
Potential5 2011-03-19 19:29:39
(5, -12)
carmelninja 2011-03-19 19:29:39
<5, -12>
gh625 2011-03-19 19:29:39
(-5,12)
policecap 2011-03-19 19:29:41
(-5,12)
rrusczyk 2011-03-19 19:29:45
rrusczyk 2011-03-19 19:29:56
And what is vector AP?
rrusczyk 2011-03-19 19:30:01
(The vector from A to P?)
GoldenFrog1618 2011-03-19 19:30:41
(-38,28)
gh625 2011-03-19 19:30:41
(-38,28)
rrusczyk 2011-03-19 19:30:45
rrusczyk 2011-03-19 19:30:54
And now what do we want to do?
AceOfDiamonds 2011-03-19 19:31:18
fancy vector thing you were talking about earlier
SuperNerd123 2011-03-19 19:31:18
Use the projection formula (Express the cosine as the dot product)
rrusczyk 2011-03-19 19:31:28
Exactly, here's where we use a projection.
rrusczyk 2011-03-19 19:31:32
rrusczyk 2011-03-19 19:31:57
This is the super fancy thing that you have to know some pretty advanced linear algebra (vector math) to know.  (It's in the Precalculus book, of course!)
ahaanomegas 2011-03-19 19:32:03
When you say || n ||, do you mean |n|?
rrusczyk 2011-03-19 19:32:07
Yes.
rrusczyk 2011-03-19 19:32:14
Now, we crank through the computation.
rrusczyk 2011-03-19 19:32:18
PiquantPeppers 2011-03-19 19:32:32
beast
rrusczyk 2011-03-19 19:32:38
And that's all there is to it with this approach!
GoldenFrog1618 2011-03-19 19:32:43
this looks the same as above
hrithikguy 2011-03-19 19:32:43
this was basically the same formula, right?
rrusczyk 2011-03-19 19:33:01
Exactly!  The distance from a point to a line formula can be proved with this same machinery.
rrusczyk 2011-03-19 19:33:15
With this approach, though, we didn't have to mess around with equations of lines and all that.
rrusczyk 2011-03-19 19:33:24
Computationally, this way is a good deal faster.
rrusczyk 2011-03-19 19:33:32
But it requires knowing some very fancy stuff.
rrusczyk 2011-03-19 19:33:38
Now, here's a good question:
mentalgenius 2011-03-19 19:33:41
Is there an easier way to do this problem?
rrusczyk 2011-03-19 19:34:01
We have seen two solutions that require knowing either advanced math or obscure formulas.
policecap 2011-03-19 19:34:29
is there a solution with rotation?
rrusczyk 2011-03-19 19:34:31
Some of you have mentioned using rotation of axes.  That works, but that's pretty advanced, too (and you can do it pretty quickly with a rotation matrix!)
.cpp 2011-03-19 19:34:33
Now, let's just use normal geometry.
rrusczyk 2011-03-19 19:34:42
OK, let's try that.
rrusczyk 2011-03-19 19:34:46
Let's see a solution that doesn't require any fancy machinery: just good old middle school geometry.
rrusczyk 2011-03-19 19:34:52
We start with the target.  We want the distance from P to the blue line, and we know where P and A are:
rrusczyk 2011-03-19 19:34:56
rrusczyk 2011-03-19 19:35:17
Notice that I've removed the axes.  We're not going to write any equations of lines here.  I want to find a purely geometric solution.  What sort of distances do we know besides the distance from P to A?
rrusczyk 2011-03-19 19:35:41
No distance formula.  No nothing.
rrusczyk 2011-03-19 19:35:45
What do we know?
rrusczyk 2011-03-19 19:35:58
What do the coordinates tell us about P relative to A?
connaissance 2011-03-19 19:36:03
we know the differences between x-coordinates
rrusczyk 2011-03-19 19:36:12
We know the horizontal distance and vertical distance from P to A.  That is, we can draw a rectangle with P and A at opposite vertices such that the sides are parallel to the old axes.  From the coordinates we are given for P and A, we know that the rectangle's dimensions are 38 and 28.
rrusczyk 2011-03-19 19:36:23
ahaanomegas 2011-03-19 19:36:44
How did we get 28 and 38, sorry!
rrusczyk 2011-03-19 19:37:01
38 is the difference in x-coordinates, 28 the difference in y coordinates
hrithikguy 2011-03-19 19:37:10
i spy similar triangles!
Jasmine8925 2011-03-19 19:37:11
i see similar triangles
rrusczyk 2011-03-19 19:37:26
We draw in this rectangle, and we extend the sides of the rectangle to hit the blue lines because we like right triangles.
rrusczyk 2011-03-19 19:37:40
We particularly like these because they are all sorts of similar.
rrusczyk 2011-03-19 19:37:50
What more do we know about all these similar right triangles?
AlphaMath1 2011-03-19 19:38:10
5-12-13 right triangles
they're similar to 5-12-13 triangles
Kinata12 2011-03-19 19:38:13
The ratios of their sides.
Ttocs45 2011-03-19 19:38:13
Their leg ratio is 5 to 12 since the slope of the line is 5/12
rrusczyk 2011-03-19 19:38:17
These are 5-12-13 right triangles, from the given slope of the blue line.  In fact, we have a bunch of similar 5-12-13 right triangles.
rrusczyk 2011-03-19 19:38:27
We want PX.  What length will allow us to get PX fastest?
GoldenFrog1618 2011-03-19 19:39:01
PK
dlennon 2011-03-19 19:39:01
PK
policecap 2011-03-19 19:39:10
PK
ahaanomegas 2011-03-19 19:39:11
PK
rrusczyk 2011-03-19 19:39:24
Since PKX is a 5-12-13 right triangle, we have PX = (12/13)(PK).  And how do we get PK?
theone142857 2011-03-19 19:39:52
28+NK
mhy123 2011-03-19 19:39:52
PN + NK
rrusczyk 2011-03-19 19:39:57
And how do we get NK?
policecap 2011-03-19 19:40:05
28+38*5/12
.cpp 2011-03-19 19:40:05
NK = 5/12 (PM), so PK= 5/12(38) + 28.
AlphaMath1 2011-03-19 19:40:05
Find NK using the same 5-12-13 ratio in Triangle NAK
theone142857 2011-03-19 19:40:05
28+38*5/12
rrusczyk 2011-03-19 19:40:11
We find NK from 5-12-13 right triangle ANK.  We have AN = PM = 38, so NK = (5/12)(38) = (5/6)(19).
rrusczyk 2011-03-19 19:40:19
And now we compute.
rrusczyk 2011-03-19 19:40:23
dlennon 2011-03-19 19:40:34
Same number
rrusczyk 2011-03-19 19:40:41
Phew!  We didn't make a mistake.
rrusczyk 2011-03-19 19:40:48
Tada!  And we can do the same thing for the other coordinate.  Because we still have 12 more problems to go, I'll leave that for you to work through for practice!
ahaanomegas 2011-03-19 19:40:54
That's nice!
bojobo 2011-03-19 19:40:54
the best solution
rrusczyk 2011-03-19 19:40:57
:)
flamingmath 2011-03-19 19:40:59
Could we have solved this with graph paper and a ruler?
rrusczyk 2011-03-19 19:41:03
YES!
rrusczyk 2011-03-19 19:41:08
But it is very tricky how.
rrusczyk 2011-03-19 19:41:17
Your ruler is not going to tell you 526/13.
rrusczyk 2011-03-19 19:41:32
But there is a way to get a line with length 31 in the problem that you could measure!!
GoldenFrog1618 2011-03-19 19:41:41
who doesn't have a 1/13th inch ruler?
rrusczyk 2011-03-19 19:41:44
:)
numberwiz 2011-03-19 19:41:46
Where?
rrusczyk 2011-03-19 19:42:06
I'll tell you how, but you will have to figure out why it works.
rrusczyk 2011-03-19 19:42:12
rrusczyk 2011-03-19 19:42:24
The blue lines are the new axes.
rrusczyk 2011-03-19 19:42:44
Draw a line through P that makes a 45 degree angle with both of the positive parts of the blue axes.
rrusczyk 2011-03-19 19:43:00
Let T be the point where that line hits one of the axes.
rrusczyk 2011-03-19 19:43:13
Think about the distance between T and the new origin!
JSGandora 2011-03-19 19:43:30
clever
rrusczyk 2011-03-19 19:43:51
If you find a quick solution with that approach, let me know.  I so wanted that to lead to a nice solution!
rrusczyk 2011-03-19 19:44:00
rrusczyk 2011-03-19 19:44:11
AlphaMath1 2011-03-19 19:44:14
it's 31 because subtracting TY is the same thing as subtracting PY
numberwiz 2011-03-19 19:44:15
Have you found that sol?
rrusczyk 2011-03-19 19:44:20
Not yet :(
mhy123 2011-03-19 19:44:29
oh oh #4
rrusczyk 2011-03-19 19:44:32
rrusczyk 2011-03-19 19:44:39
rrusczyk 2011-03-19 19:44:52
(If you drew your diagram with a straightedge and compass, then you would have a huge clue right away -- MN just plain looks parallel to AB.)
rrusczyk 2011-03-19 19:45:27
Even if you didn't, you have a couple right triangles sitting there, and equal angles from the angle bisectors.  What do these just beg you to do?
calvinhobbesliker 2011-03-19 19:45:44
Extend the perpendiculars
vcez 2011-03-19 19:45:44
extend the perpendiculars
rrusczyk 2011-03-19 19:45:48
They just plain want you to continue CM and CN, since you know you'll get nice pretty congruent right triangles.
.cpp 2011-03-19 19:45:59
Extend CM and CN to intersect line AB.
rrusczyk 2011-03-19 19:46:02
mentalgenius 2011-03-19 19:46:24
congruent
MathTwo 2011-03-19 19:46:24
angle bisectors/ right angles=> isosceles triangles
aerrowfinn72 2011-03-19 19:46:24
2 isosceles triangles
rrusczyk 2011-03-19 19:46:42
As advertised, we have congruent right triangles.  For example, triangles CMB and PMB are congruent by ASA.  What notable bit of information does this give us?
aerrowfinn72 2011-03-19 19:47:26
AC = AQ and BC = BP
BOGTRO 2011-03-19 19:47:27
PB=BC
gh625 2011-03-19 19:47:27
CB=PB
tan90 2011-03-19 19:47:27
PB = CB
rrusczyk 2011-03-19 19:47:48
Please use capital letters when referring to points.  I don't read the posts with lowercase point labels.
rrusczyk 2011-03-19 19:47:51
This tells us that BP = CB = 120.  So?
alligator112 2011-03-19 19:48:08
AP = 5
Jasmine8925 2011-03-19 19:48:08
we can find AP
centralbs 2011-03-19 19:48:08
AP = 5
rrusczyk 2011-03-19 19:48:15
So, AP = AB - BP = 5.
calvinhobbesliker 2011-03-19 19:48:19
Find AQ similarly, then  PQ can be found
mathepic 2011-03-19 19:48:32
We do the same thing to find BQ, then subtract both
rrusczyk 2011-03-19 19:48:38
What do we get for PQ?
theone142857 2011-03-19 19:49:06
PQ=112
IsTvOn 2011-03-19 19:49:06
112
Alex Song 2011-03-19 19:49:06
112
calvinhobbesliker 2011-03-19 19:49:06
112
tan90 2011-03-19 19:49:06
112
.cpp 2011-03-19 19:49:06
PQ = AB-AP-BQ = 112.
centralbs 2011-03-19 19:49:06
112
Kinata12 2011-03-19 19:49:06
112
rrusczyk 2011-03-19 19:49:20
Similar to how we found AP, we have AQ = AC = 117, so BQ = AB - AQ = 8.
rrusczyk 2011-03-19 19:49:36
So, PQ = AB - AP - BQ = 112.
rrusczyk 2011-03-19 19:49:38
Now what?
ParallelProcess 2011-03-19 19:50:15
M is the midpoint of CP, and similarly for the opposite side, so MN = 1/2*PQ
theone142857 2011-03-19 19:50:15
MN=1/2PQ
v_Enhance 2011-03-19 19:50:15
MN = 1/2 PQ
calvinhobbesliker 2011-03-19 19:50:15
MN=PQ/2 by similar triangles
Kinata12 2011-03-19 19:50:15
Use CMN and CPQ similar triangles!
Alex Song 2011-03-19 19:50:15
MN is a midline of triangle CPQ; MN = PQ/2
ahaanomegas 2011-03-19 19:50:15
rrusczyk 2011-03-19 19:50:33
GoldenFrog1618 2011-03-19 19:50:40
I solved this problem initially with the "measure the line" method
rrusczyk 2011-03-19 19:51:01
Funny.  Don't underestimate that : I got into MOP my sophomore year largely because of it :)
ahaanomegas 2011-03-19 19:51:11
What is the "measure the line" method?
tiger21 2011-03-19 19:51:25
you use a ruler
calvinhobbesliker 2011-03-19 19:51:25
Draw the triangle precisely and measure
rrusczyk 2011-03-19 19:51:26
Draw to scale.  Measure with ruler.
centralbs 2011-03-19 19:51:28
How do you know its the midline?
rrusczyk 2011-03-19 19:51:51
Because CM = MP (triangles CMB and CMP are congruent) and CN = NQ.
ahaanomegas 2011-03-19 19:51:54
Congruence
pi091997 2011-03-19 19:51:54
congruent triangles
GoldenFrog1618 2011-03-19 19:51:57
M and N are midpoints
ParallelProcess 2011-03-19 19:51:59
on the AIME, if something "looks" parallel like in this diagram, is it usually actually parallel?
rrusczyk 2011-03-19 19:52:11
Dangerous -- it tells you that you should try to prove that it is parallel.
rrusczyk 2011-03-19 19:52:32
rrusczyk 2011-03-19 19:52:38
Where do we start?
take every number mod 3
bulutcocuk 2011-03-19 19:52:51
mod 3
zero.destroyer 2011-03-19 19:52:51
use mods
policecap 2011-03-19 19:52:51
mod 3
superpi83 2011-03-19 19:52:51
reduce in mod 3
theone142857 2011-03-19 19:52:51
mod 3
hrithikguy 2011-03-19 19:52:51
rename the digits as 1,2,0,1,2,0,1,2,0
carmelninja 2011-03-19 19:52:54
look at it in terms of modulo 3
rrusczyk 2011-03-19 19:52:57
We start by thinking about the numbers mod 3 (that is, their remainders when divided by 3).  When determining if the sums of the numbers are multiples of 3, their remainders when divided by 3 are all that matters.
rrusczyk 2011-03-19 19:53:01
So, now we have 3 0's, 3 1's, and 3 2's to place around the nonagons.  Suppose we start with a 0.  Then, what are our options for the next number going clockwise?
hrithikguy 2011-03-19 19:53:16
1 or 2
tekgeek 2011-03-19 19:53:16
1 or 2
pi.guy3.14 2011-03-19 19:53:16
1,2 or 2,1
we can have 1 or 2
EggyLv.999 2011-03-19 19:53:16
2 or 1
zero.destroyer 2011-03-19 19:53:16
1, or 2
Spring 2011-03-19 19:53:16
1 or 2
pi091997 2011-03-19 19:53:16
1 or 2
superpi83 2011-03-19 19:53:16
1 or 2
hrithikguy 2011-03-19 19:53:16
1 or 2 (NOT zero)!
1 or 2
zazz4 2011-03-19 19:53:16
2 or 1
rrusczyk 2011-03-19 19:53:19
We can't have a 0 next.  If we place a 0 next, then the following number must also be a 0 to make the sum of the three numbers divisible by 3.  But so must the next number, and then next number, and so on.
rrusczyk 2011-03-19 19:53:22
What if we place a 1 next?
KeepingItReal 2011-03-19 19:53:34
2 must be next
jellymoop 2011-03-19 19:53:34
2
arigao2007 2011-03-19 19:53:34
then 2 next
EggyLv.999 2011-03-19 19:53:34
2 next
pi091997 2011-03-19 19:53:34
then 2
mthcz11 2011-03-19 19:53:34
wowzo
Kinata12 2011-03-19 19:53:34
The next is 2
centralbs 2011-03-19 19:53:34
it has to be a 2
then the next number is 2
superpi83 2011-03-19 19:53:34
then it goes 012012012
theone142857 2011-03-19 19:53:34
2 next
.cpp 2011-03-19 19:53:34
Then a 2 goes next.
alligator112 2011-03-19 19:53:34
then a 2 next!
jeff10 2011-03-19 19:53:34
0 1 2 0 1 2 0 1 2
zero.destroyer 2011-03-19 19:53:34
it's 2
fortenforge 2011-03-19 19:53:34
Then the number after that is a 2
AlphaMath1 2011-03-19 19:53:34
next has to be 2
alanzee 2011-03-19 19:53:34
next 2, then 0, then 1, 2, ...
rrusczyk 2011-03-19 19:53:38
If we start 0, 1, then the next number must be 2 to make the sum a multiple of 3.  Then we have 0, 1, 2.
rrusczyk 2011-03-19 19:53:41
So, the next number must be a 0, and the next a 1, the next a 2, and so on, so we have the pattern
0, 1, 2, 0, 1, 2, 0, 1, 2.
rrusczyk 2011-03-19 19:53:45
How many ways can we stick the numbers around the circle in this pattern?
rrusczyk 2011-03-19 19:54:06
(I'm talking about the original 9 numbers)
.cpp 2011-03-19 19:54:29
72.
KeepingItReal 2011-03-19 19:54:29
72
bulutcocuk 2011-03-19 19:54:29
72
superpi83 2011-03-19 19:54:29
fix 9 to some position so there are 2!*3!*3!=72 ways
mathepic 2011-03-19 19:54:29
3!3!2!
kangchangood 2011-03-19 19:54:29
3x3x2x2x2?
tekgeek 2011-03-19 19:54:29
We can fix the zero to avoid rotation so there are 3!*3!*2=72 ways
pi.guy3.14 2011-03-19 19:54:32
72
jeff10 2011-03-19 19:54:32
72
rrusczyk 2011-03-19 19:54:35
In order to not worry about rotations, we just imagine that we place the 3 (which has remainder 0 when divided by 3) first.  From there, we have 3*2*1 = 6 ways to place the 1,4, and 7 in the "remainder 1" slots, and 3*2*1 = 6 ways to place the 2,5, and 8 in the "remainder 2" slots.  Finally, we have 2 ways to place the 6 and the 9 in the other two "remainder 0" slots.  From 6*6*2 = 72 total ways to place the numbers.
rrusczyk 2011-03-19 19:54:43
Are we finished?
centralbs 2011-03-19 19:55:07
no
pi091997 2011-03-19 19:55:07
no
delta1 2011-03-19 19:55:07
the other way
theone142857 2011-03-19 19:55:07
*2
KeepingItReal 2011-03-19 19:55:07
no
policecap 2011-03-19 19:55:07
no *2 again because you can flip it
pi.guy3.14 2011-03-19 19:55:07
no, there are 2 patterns, so 2x72=144
jellymoop 2011-03-19 19:55:07
no, 021021
EggyLv.999 2011-03-19 19:55:07
nope
chessmaster7 2011-03-19 19:55:07
the order can be reversed
delta1 2011-03-19 19:55:07
no
BOGTRO 2011-03-19 19:55:07
could also be 021
superpi83 2011-03-19 19:55:07
no we still need the 021021021 case
kangchangood 2011-03-19 19:55:07
0,2,1 case
tan90 2011-03-19 19:55:07
It could go 0 2 1 0 2 1
mthcz11 2011-03-19 19:55:07
not yet, still need to multipy by 2 to get 144
rrusczyk 2011-03-19 19:55:18
No.  We have considered starting 0,0 going clockwise and 0,1 going clockwise.  We still have to deal with 0,2.
j_f_c_w 2011-03-19 19:55:23
No there are 72 ways of arranging the numbers but 2 patters
mathepic 2011-03-19 19:55:26
no, double it
filetmignon821 2011-03-19 19:55:28
we have to multiply by 2
rrusczyk 2011-03-19 19:55:31
We get the same thing as 0,1, but the pattern now going clockwise is
0,2,1,0,2,1,0,2,1
rrusczyk 2011-03-19 19:55:36
rrusczyk 2011-03-19 19:55:45
Way easier than the previous two!
Jasmine8925 2011-03-19 19:55:54
that was a cool problem
mthcz11 2011-03-19 19:55:54
yay!
icecreamcakepie 2011-03-19 19:55:54
nice
anonymous0 2011-03-19 19:55:54
brilliant!
AlphaMath1 2011-03-19 19:55:54
I agree :)
leekspeak 2011-03-19 19:55:54
YES!
mthcz11 2011-03-19 19:55:54
yes it was :)
rrusczyk 2011-03-19 19:55:58
rrusczyk 2011-03-19 19:56:12
leekspeak 2011-03-19 19:56:17
put the quadratic into vertex form
Jasmine8925 2011-03-19 19:56:18
let's write vertex form
theone142857 2011-03-19 19:56:20
vertex
tekgeek 2011-03-19 19:56:29
carmelninja 2011-03-19 19:56:29
Jasmine8925 2011-03-19 19:56:30
y = a(x-1/4)^2 - 9/8
rrusczyk 2011-03-19 19:56:34
rrusczyk 2011-03-19 19:56:39
We could bash this out and compare it to y = ax^2 + bx + c, but is there a more clever way to proceed?
Z91 2011-03-19 19:56:56
a+b+c=f(1)
centralbs 2011-03-19 19:56:56
"Keep your eye on the ball" - we want a + b + c, which is the same as p(1)
hrithikguy 2011-03-19 19:56:56
f(1) = a+b+c
zero.destroyer 2011-03-19 19:56:56
f(1)=a+b+c
rrusczyk 2011-03-19 19:57:03
The one key piece of information we haven't used yet is that a+b+c is an integer.
delta1 2011-03-19 19:57:09
a+b+c is when x=1
rrusczyk 2011-03-19 19:57:18
To use that, we simply let x = 1 in y =ax^2 + bx + c!  Doing so gives us y = a + b + c.  So, y is an integer when x = 1.  How does this help?
BOGTRO 2011-03-19 19:57:56
9/16a-9/8 is integral
KeepingItReal 2011-03-19 19:57:57
Plug 1 into vertex form.
Jasmine8925 2011-03-19 19:57:57
now plug 1 into the vertex form
hrithikguy 2011-03-19 19:57:57
plug this in the vertext form
Kinata12 2011-03-19 19:57:57
Plug x=1 into vertex form
kangchangood 2011-03-19 19:57:57
a(3/4)^2 - 9/8 is integer.
alanzee 2011-03-19 19:57:57
a*(3/4)^2-9/8 is an integer
.cpp 2011-03-19 19:57:57
Thus a(9/16) - 9/8 is an integer.
Jasmine8925 2011-03-19 19:57:57
a(3/4)^2-9/8
rrusczyk 2011-03-19 19:58:05
We stick x = 1 into our earlier equation with the vertex:
rrusczyk 2011-03-19 19:58:09
rrusczyk 2011-03-19 19:58:19
calvinhobbesliker 2011-03-19 19:59:04
set it equal to -1
Spring 2011-03-19 19:59:04
y has to be at least -1 for a to be positive
etude 2011-03-19 19:59:04
a is positive, so y > -9/8 (parabola opens upwards). Thus y = -1.
prezcoin 2011-03-19 19:59:04
we get a= (16y-18)/9 is positive, so y= -1 to minimize a
rrusczyk 2011-03-19 19:59:15
And what do we get?
alanzee 2011-03-19 19:59:27
take the fractional part of 9/8, so (9a-2)/16 is an integer => a=2/9
9a-18 must be divisible by 16, so 9a = 2 mod 16, a = 2/9 minimum
theone142857 2011-03-19 19:59:27
2/9
centralbs 2011-03-19 19:59:27
2/9 so 11
mthcz11 2011-03-19 19:59:27
2/9! :)
hrithikguy 2011-03-19 19:59:27
a =2/9
skyhog 2011-03-19 19:59:27
a=2/9
BOGTRO 2011-03-19 19:59:27
a=2/9
carmelninja 2011-03-19 19:59:27
a = 2/9
vcez 2011-03-19 19:59:27
2/9
rrusczyk 2011-03-19 19:59:32
theone142857 2011-03-19 19:59:35
so 011
vcez 2011-03-19 19:59:35
ahaanomegas 2011-03-19 19:59:37
2 + 9 = 011 = Answer
rrusczyk 2011-03-19 19:59:42
rrusczyk 2011-03-19 19:59:56
As an extra challenge, see if you can find a geometric inspiration for this solution.
rrusczyk 2011-03-19 20:00:08
(On your own.  We have 9 more problems to do!
rrusczyk 2011-03-19 20:00:18
rrusczyk 2011-03-19 20:00:26
This is a pretty confusing problem at first glance.  What's one way we can play with the problem a bit to get used to it?
leekspeak 2011-03-19 20:00:36
Blindly guess, then run home and make a sacrifice to Jupiter
rrusczyk 2011-03-19 20:00:40
That might work.
policecap 2011-03-19 20:00:49
try out small m
.cpp 2011-03-19 20:00:49
Try small values!
Spring 2011-03-19 20:00:49
Try some small integers for m
Kinata12 2011-03-19 20:00:53
Try various values of m
rrusczyk 2011-03-19 20:00:58
We can pick a specific m and mess around with it. We'll pick a small m, since big numbers are scary.  Where should we start?
EggyLv.999 2011-03-19 20:01:07
2
carmelninja 2011-03-19 20:01:07
m=2
hrithikguy 2011-03-19 20:01:07
2
calvinhobbesliker 2011-03-19 20:01:07
2
hrithikguy 2011-03-19 20:01:07
m = 2
delta1 2011-03-19 20:01:07
2
rrusczyk 2011-03-19 20:01:14
rrusczyk 2011-03-19 20:01:21
rrusczyk 2011-03-19 20:01:28
What might we wonder first about the terms on the right?
Spring 2011-03-19 20:02:29
how many 2^0=1 terms
rrusczyk 2011-03-19 20:02:47
We might wonder if we can have any of the x_i on the right be 0's.  That is, can any of the terms on the right be simply 1?
rrusczyk 2011-03-19 20:02:50
delta1 2011-03-19 20:03:10
two 1s
skyhog 2011-03-19 20:03:10
No.  There have to be an even number of 1's
AlphaMath1 2011-03-19 20:03:17
we must have another 1 because 2^{x_0} is even
rrusczyk 2011-03-19 20:03:19
rrusczyk 2011-03-19 20:03:23
Now what do we need?
theone142857 2011-03-19 20:03:44
2
delta1 2011-03-19 20:03:44
a 2
.cpp 2011-03-19 20:03:44
2^1 seems like a good idea.
bojobo 2011-03-19 20:03:48
and repeat
carmelninja 2011-03-19 20:03:48
we need one 2^1 so the LHS will be divisible by 4
EggyLv.999 2011-03-19 20:03:48
some 2^1
rrusczyk 2011-03-19 20:03:51
Now we need another 2.  This could from two more 1's or from a 2^1 term, but we definitely need another 2 somewhere.  Otherwise, the right-hand side will not be divisible by 4.
rrusczyk 2011-03-19 20:03:56
rrusczyk 2011-03-19 20:04:00
Now what do we need?
delta1 2011-03-19 20:04:12
and a 4
pi.guy3.14 2011-03-19 20:04:12
4
hrithikguy 2011-03-19 20:04:12
a 4
EggyLv.999 2011-03-19 20:04:12
4
SuperNerd123 2011-03-19 20:04:12
4
skyhog 2011-03-19 20:04:12
Another 4
policecap 2011-03-19 20:04:12
4, 8 and so on
dlennon 2011-03-19 20:04:12
A 4.
delta1 2011-03-19 20:04:12
and so on
rrusczyk 2011-03-19 20:04:16
Now, if we don't have another 4 from somewhere, the right side can't be a multiple of 8.  We can get a 4 in a lot of ways (1+1+2), (2+2), etc., but it's easiest if we just have a 2^2 term.
rrusczyk 2011-03-19 20:04:23
icecreamcakepie 2011-03-19 20:05:25
then a 16,32,64,128, and so on
rrusczyk 2011-03-19 20:05:37
PiquantPeppers 2011-03-19 20:05:37
we repeat this process
rrusczyk 2011-03-19 20:05:41
hrithikguy 2011-03-19 20:06:08
Boy SOPRANO!! Can you start a chat with me? I am TOTALLY lost!!
centralbs 2011-03-19 20:06:08
We can figure that out from the geometric sequence formula, you get 2^{n+1} -1, and then we added the extra 1
delta1 2011-03-19 20:06:08
yes
numberwiz 2011-03-19 20:06:08
yes!
.cpp 2011-03-19 20:06:08
Yes, absolutely!
superpi83 2011-03-19 20:06:08
yes.
JSGandora 2011-03-19 20:06:08
yes, yes there are
Kinata12 2011-03-19 20:06:08
Yes, both sides are 2^2010
yes, there is a set
rrusczyk 2011-03-19 20:06:13
rrusczyk 2011-03-19 20:06:36
Now, this may not be the only way to make it work for 2.  We were just trying to see if we could find a way to make it work for 2.
theone142857 2011-03-19 20:06:38
m=3!!!!!!!!!!!!!!!!!!
rrusczyk 2011-03-19 20:06:41
Maybe 2 is special.  Let's see if we can do the same thing with 3.
rrusczyk 2011-03-19 20:06:48
rrusczyk 2011-03-19 20:06:54
We'll start out with some 1's, since that worked so well when m=2.  How many do we need?
centrino 2011-03-19 20:07:20
3
theone142857 2011-03-19 20:07:20
3
Jasmine8925 2011-03-19 20:07:20
3 1s
JSGandora 2011-03-19 20:07:20
we need three ones
Kinata12 2011-03-19 20:07:23
3
skyhog 2011-03-19 20:07:23
3 of them (they're already there...)
superpi83 2011-03-19 20:07:23
3
rrusczyk 2011-03-19 20:07:27
If we have any 1's, we need 3, otherwise we won't have a multiple of 3 on the right-hand side.  (We could have more, but we'll start with three 1's, just like we started with two 1's for m=2.)
rrusczyk 2011-03-19 20:07:30
rrusczyk 2011-03-19 20:07:42
This looks familiar.
rrusczyk 2011-03-19 20:07:47
What are we going to need on the right-hand side?
pi.guy3.14 2011-03-19 20:07:59
wups, we need 2 3's
Jasmine8925 2011-03-19 20:07:59
two 3s
dlennon 2011-03-19 20:07:59
We then need 2 more 3s
tan90 2011-03-19 20:07:59
Two more 3's
rrusczyk 2011-03-19 20:08:03
We're going to need some more 3's to go with our first 3, otherwise the right side won't be a multiple of 9.  Specifically, we'll need a pair of them.  As with our m=2 case, there are a lot of ways to get our needed extra terms.  We're lazy, so we'll take the easiest way, by including a couple 3^1 terms on the right side.
rrusczyk 2011-03-19 20:08:06
.cpp 2011-03-19 20:08:27
Two more 3 terms, then two more 9 terms, ...
policecap 2011-03-19 20:08:27
2 9's, and so on
delta1 2011-03-19 20:08:27
two 9's
Kinata12 2011-03-19 20:08:27
Just add 2 at a time of powers of 3
dlennon 2011-03-19 20:08:27
And 2 more 9s
tiger21 2011-03-19 20:08:27
and then 2 9's
jeff10 2011-03-19 20:08:27
2 more 3s
KeepingItReal 2011-03-19 20:08:27
Now we need 2 9's.
rrusczyk 2011-03-19 20:08:30
And now we need a couple more 3^2 terms, and then we'll need a couple more 3^3 terms, and so on.
rrusczyk 2011-03-19 20:08:34
Will this work?  Does it generate a set of x_i that satisfy the original equation when m = 3?
delta1 2011-03-19 20:08:53
yes
swimfreak07059 2011-03-19 20:08:54
yes
.cpp 2011-03-19 20:08:54
Yes.
EggyLv.999 2011-03-19 20:08:54
yes
Kinata12 2011-03-19 20:08:54
Yes, 2011=3+2+2+2...
dlennon 2011-03-19 20:08:55
Yes it works.
2011 = 3 + 2(1004) yes
rrusczyk 2011-03-19 20:09:02
rrusczyk 2011-03-19 20:09:07
We have another winner.
rrusczyk 2011-03-19 20:09:12
We should stop now and think about why this process is finding sets of x_i that work for a given m.  Then, maybe we can figure out which m will work.  (Of course, this process may not give us the only set of x_i that work, but, first things first -- let's at least see if we can figure out the m that work with this process.)
rrusczyk 2011-03-19 20:09:16
What exactly is our process here for a given value of m?
rrusczyk 2011-03-19 20:09:20
A) We start with m terms of the form m^0.  These are our 1's, and they add up to m.
rrusczyk 2011-03-19 20:09:35
B) Then, we need m-1 terms of the form m^1 to combine with this initial m to give us an m^2.
rrusczyk 2011-03-19 20:09:39
C) Then, we need m-1 terms of the form m^2 to combine with this initial m^2 to give us an m^3.
rrusczyk 2011-03-19 20:09:56
D) And so on.
rrusczyk 2011-03-19 20:10:01
We want this process to produce 2011 terms total, and it did for m=2 and m=3.  Will it always produce 2011 terms?
rrusczyk 2011-03-19 20:10:11
(That is, can we always hit 2011 exactly)
delta1 2011-03-19 20:10:18
no
Jasmine8925 2011-03-19 20:10:18
no
centrino 2011-03-19 20:10:18
no
.cpp 2011-03-19 20:10:18
No.
gh625 2011-03-19 20:10:18
no
theone142857 2011-03-19 20:10:18
no
leekspeak 2011-03-19 20:10:20
no
rrusczyk 2011-03-19 20:10:23
No, it won't.  Can we write an expression for how many terms it will produce?
numberwiz 2011-03-19 20:10:52
in general, if a can be an int., x + a(x-1) = 2011
Kinata12 2011-03-19 20:10:52
m+j(m-1)
policecap 2011-03-19 20:10:52
m+(m-1)*k for some k
Spring 2011-03-19 20:10:57
m+n(m-1)
carmelninja 2011-03-19 20:10:57
m + n(m-1)
rrusczyk 2011-03-19 20:11:02
Yes.  We have m terms initially, and then a bunch of groups of (m-1) terms.  So, for some value of k, there are
m+k(m-1)
terms.
rrusczyk 2011-03-19 20:11:10
Therefore, we need there to be some value of k for which
m + k(m-1) = 2011
rrusczyk 2011-03-19 20:11:17
What can we do with this?
Jasmine8925 2011-03-19 20:11:47
so 2011 = 1 (mod m-1)
leekspeak 2011-03-19 20:11:47
factor!
.cpp 2011-03-19 20:11:47
(k+1)(m-1) = 2010.
billybob42 2011-03-19 20:11:47
find all m such that m|2010
prezcoin 2011-03-19 20:11:47
m= (m-1) +1, so 2010= (k+1)(m-1)
pi091997 2011-03-19 20:11:51
(k+1)(m-1)=2010
jeff10 2011-03-19 20:11:51
FACTOR!
rrusczyk 2011-03-19 20:11:54
Subtracting 1 from both sides allows us to factor:
(m-1) + k(m-1) = 2010,
so
(m-1)(k+1) = 2010
rrusczyk 2011-03-19 20:12:08
And what does this tell us?
bojobo 2011-03-19 20:12:33
all of the factors of 2010 work
ahaanomegas 2011-03-19 20:12:33
m - 1 and k + 1 are both factors of 2010
superpi83 2011-03-19 20:12:33
every m corresponds to a factor of 2010
JoeJack 2011-03-19 20:12:33
we're looking for the number of factors of 2010
JSGandora 2011-03-19 20:12:33
This tell us there are as many ms as there are as many factors of 2010
girishvar12 2011-03-19 20:12:33
they have to be factors of 2010
leekspeak 2011-03-19 20:12:39
the possible values of m are the number of factors of 2010
.cpp 2011-03-19 20:12:39
m-1 is a factor of 2010, of which there are 16. So m has 16 values here.
pi091997 2011-03-19 20:12:39
m-1 divides 2010
rrusczyk 2011-03-19 20:12:46
So, if m-1 divides 2010, then our process will give us a set of x_i for that value of m.  Therefore, all of the positive divisors (including 1 and 2010) give us solutions.
Jasmine8925 2011-03-19 20:13:13
2010 = 2*3*5*67
delta1 2011-03-19 20:13:13
16 possible values of m-1, so 16 possible values of m
tan90 2011-03-19 20:13:13
2010 = 2 * 3 * 5 * 67
policecap 2011-03-19 20:13:13
so theres 2^4=16
superpi83 2011-03-19 20:13:13
2010=2*3*5*67 so there are 2*2*2*2=16 factors of 2010
and 2010 has 16 factors
girishvar12 2011-03-19 20:13:13
so 16 numbers work, 016
KeepingItReal 2011-03-19 20:13:13
There are 16 positive divisors
policecap 2011-03-19 20:13:13
2*3*5*67=2010, so 2*2*2*2 divisors
NoWayHaze 2011-03-19 20:13:25
how doesn't  2011^2 = 2011^1 + 2011^1 + 2011^1 ..... 2011^1  work?
rrusczyk 2011-03-19 20:13:36
Yep.  m=2011 means 2011 - 1 divides 2010.
.cpp 2011-03-19 20:13:39
We're using m-1, not m
rrusczyk 2011-03-19 20:13:42
Exactly.
Jasmine8925 2011-03-19 20:13:47
is this the only way to get solutions?
rrusczyk 2011-03-19 20:13:53
Excellent question.
SuperNerd123 2011-03-19 20:14:07
Wait, don't we have to disclude 1?
carmelninja 2011-03-19 20:14:36
2-1 = 1
skyhog 2011-03-19 20:14:36
m-1=1 which is just the m=2 case we already showed works
superpi83 2011-03-19 20:14:39
1 was never a valid m. 1 was a valid m-1, which corresponded to 2
rrusczyk 2011-03-19 20:14:41
No: m=2 when m-1 is 1.
rrusczyk 2011-03-19 20:14:57
I want to get back to Jasmine's question.
rrusczyk 2011-03-19 20:15:06
We're not finished.  There might be some other process that gives us more values of m that work.  That said, you'd be forgiven for deciding the work we did so far was about the difficulty level of a #7 on the AIME, assuming the 16 we have found are the only ones that work, and moving on.  But let's go ahead and see if we can figure out what's going on here in the whole problem, and prove that these are the only 16 that work.
rrusczyk 2011-03-19 20:15:24
rrusczyk 2011-03-19 20:15:30
In our work so far, we discovered that we have a solution when m-1 divides 2010.  This m-1 seems important.  Can we jam m-1 terms into this equation?  If so, how?
rrusczyk 2011-03-19 20:16:10
I want to start from the original equation and get some m-1 terms in there.
Jasmine8925 2011-03-19 20:16:38
(m-1)^x_o
rrusczyk 2011-03-19 20:16:47
We'd like there to be factors of m-1, yes
Kinata12 2011-03-19 20:16:49
If you subtract 1 from m^x_k, you can factor out m-1
rrusczyk 2011-03-19 20:16:53
And that will make them!
rrusczyk 2011-03-19 20:16:58
rrusczyk 2011-03-19 20:17:32
And what do we do about the left side?
.cpp 2011-03-19 20:18:05
[m^(x_0)-1]-2010.
delta1 2011-03-19 20:18:05
m^(x_0)-1-2010
carmelninja 2011-03-19 20:18:05
(m^x_0 - 1) - 2010
Kinata12 2011-03-19 20:18:05
Subtract 1 from m^x_0 and add it to -2011
rrusczyk 2011-03-19 20:18:10
We'd like an m-1 over there, and we want a 2010 somewhere, so we break 2011 into 1 + 2010:
rrusczyk 2011-03-19 20:18:14
rrusczyk 2011-03-19 20:18:29
rrusczyk 2011-03-19 20:18:36
rrusczyk 2011-03-19 20:18:48
leekspeak 2011-03-19 20:19:03
woah thats brilliant
.cpp 2011-03-19 20:19:03
Thus, (m-1) is a factor of 2010, so our solutions are the only ones.
rrusczyk 2011-03-19 20:19:13
If this were the USAMO, that's what you'd have to do.
rrusczyk 2011-03-19 20:19:26
We will do one more problem, and then take a short break
Is that a USAMO level problem?
Jasmine8925 2011-03-19 20:19:48
could that be an easy USAMO problem?
rrusczyk 2011-03-19 20:19:54
Maybe early USAJMO
rrusczyk 2011-03-19 20:20:00
rrusczyk 2011-03-19 20:20:12
Jasmine8925 2011-03-19 20:20:18
are we given a diagram?
rrusczyk 2011-03-19 20:20:23
Yes, it was on the test.
policecap 2011-03-19 20:20:38
the heights from A to UV, B to YZ, and C to WX are all h
Kinata12 2011-03-19 20:20:55
Each small triangle has the same height
rrusczyk 2011-03-19 20:20:58
Exactly.  That's what we mean by the table coming out to be level.
rrusczyk 2011-03-19 20:21:10
What's our first observation we can make that will simplify our work for finding the maximal height.
Jasmine8925 2011-03-19 20:21:22
what is a "right angle fold"?
rrusczyk 2011-03-19 20:21:46
Folded so that the plane of the fold is perpendicular to the plane of the original table
rrusczyk 2011-03-19 20:22:11
Now, what are we going to do about finding the largest possible height?
ParallelProcess 2011-03-19 20:22:14
the max height is when you have a collision of smaller triangles
rrusczyk 2011-03-19 20:22:18
Exactly.
tan90 2011-03-19 20:22:20
Two of the sub-triangles (like AVU) must touch at one point to maximize height.
rrusczyk 2011-03-19 20:22:24
If h is maximized then two of the folds must pass through the same point on the perimeter of the triangle.  Otherwise, we can increase h a little bit by moving all three folds closer to the center of ABC.  This makes life a lot easier.  Here's our new diagram:
rrusczyk 2011-03-19 20:22:29
Maxima 2011-03-19 20:22:33
What do they mean by maximum possible height?
rrusczyk 2011-03-19 20:22:44
We want to be able to fold as far in as possible.
rrusczyk 2011-03-19 20:23:00
I've just included the two folds that hit at a point.  We don't know for sure which two folds it will be that meet at a common point, but we'll go ahead and work out the case above and leave the side lengths as variables (a=BC, b=AC, c=AB).  Hopefully our work will reveal which folds intersect.
skyhog 2011-03-19 20:23:09
How do we know which of the two triangles "collide" ?
rrusczyk 2011-03-19 20:23:19
We don't know for sure; we'll come back to that.
rrusczyk 2011-03-19 20:23:46
For now, we'll just work with this in terms of side lengths a,b,c, and then hope our result will tell us which folds collide.
rrusczyk 2011-03-19 20:23:49
What should we do with this?
rrusczyk 2011-03-19 20:24:02
What should we add to this diagram?
skyhog 2011-03-19 20:24:12
altitudes
kangchangood 2011-03-19 20:24:12
h things?
carmelninja 2011-03-19 20:24:12
the altitudes
rrusczyk 2011-03-19 20:24:16
We include the heights from C and B to their respective folds.  These heights have length h_A, the desired height of the table.
rrusczyk 2011-03-19 20:24:19
rrusczyk 2011-03-19 20:24:29
Where do we go from here?  We'd like to use the fact that these heights are the same; how can we relate these two small triangles to each other?
AlphaMath1 2011-03-19 20:24:44
similar triangles
leekspeak 2011-03-19 20:24:44
they are similar
billybob42 2011-03-19 20:24:44
similarity
Cortana 2011-03-19 20:24:44
similar triangles
carmelninja 2011-03-19 20:24:44
similarity
apple.singer 2011-03-19 20:24:44
similar?
dlennon 2011-03-19 20:24:44
They are similar
hrithikguy 2011-03-19 20:24:44
same area?
rrusczyk 2011-03-19 20:24:52
There are a lot of ways to go from here.  We can note that these two little triangles are similar and go from there.  A somewhat slicker solution goes through the larger triangle.  How can we relate these little triangles to the larger triangle?
leekspeak 2011-03-19 20:25:18
similar
centralbs 2011-03-19 20:25:18
similar
Titandrake 2011-03-19 20:25:18
Each little triangle similar to larger triangle. Find heights of large triangle
KeepingItReal 2011-03-19 20:25:18
similar
jeff10 2011-03-19 20:25:18
They are both similar to the large triangle.
apple.singer 2011-03-19 20:25:18
similar again?
rrusczyk 2011-03-19 20:25:23
Each little triangle is similar to the larger triangle.  So, what does that tell us about CX?
.cpp 2011-03-19 20:26:17
CX/CB = h_a/h_(C to AB).
rrusczyk 2011-03-19 20:26:24
rrusczyk 2011-03-19 20:26:44
And XB?
.cpp 2011-03-19 20:27:25
XB/BC = h_A/h_b.
KeepingItReal 2011-03-19 20:27:25
XB/BC = h_A/ h _ (B to CA)
rrusczyk 2011-03-19 20:27:30
policecap 2011-03-19 20:28:12
.cpp 2011-03-19 20:28:12
CX+XB = CB.
fortenforge 2011-03-19 20:28:12
rrusczyk 2011-03-19 20:28:16
We simply add CX and XB.  We have CX + XB = BC, so
rrusczyk 2011-03-19 20:28:21
rrusczyk 2011-03-19 20:28:25
And how will we deal with h_b and h_c?
Cortana 2011-03-19 20:28:53
find the area
etude 2011-03-19 20:28:53
We find them using our area formulae.
s.homberg 2011-03-19 20:28:53
find them with Heron's and 1/2 bh
AlphaMath1 2011-03-19 20:28:57
we can calculate that using herons then multiply by 2 then divide by the base
rrusczyk 2011-03-19 20:29:18
Do we actually have to find h_b and h_c?
iamthygod!!!!!! 2011-03-19 20:29:31
no
ChipDale 2011-03-19 20:29:31
no
danielguo94 2011-03-19 20:29:31
no
rrusczyk 2011-03-19 20:29:34
Why not?
rrusczyk 2011-03-19 20:29:45
How can we just skip around that?
policecap 2011-03-19 20:30:00
area/side = height, use that for all h's
rrusczyk 2011-03-19 20:30:14
Exactly.  Let K be the area, and what do we get for h_A?
.cpp 2011-03-19 20:31:19
2K/(b+c).
rrusczyk 2011-03-19 20:31:25
rrusczyk 2011-03-19 20:31:39
rrusczyk 2011-03-19 20:32:22
These are the three heights we get if we have folds meet on each pair of sides.  h_A is what we get when the folds meet opposite A, h_B is what we get when the folds meet opposite B, etc.
rrusczyk 2011-03-19 20:32:26
Now, which one is the winner?
rrusczyk 2011-03-19 20:32:35
Do we want the largest or the smallest?
GoldenFrog1618 2011-03-19 20:32:52
pick the minimal h_A
carmelninja 2011-03-19 20:32:53
the smallest
calvinhobbesliker 2011-03-19 20:32:53
The smallest one
iamthygod!!!!!! 2011-03-19 20:32:53
smallest
mathcountsloser 2011-03-19 20:32:53
Smallest.
etude 2011-03-19 20:32:53
smallest.
zero.destroyer 2011-03-19 20:32:53
smallest
Kinata12 2011-03-19 20:32:58
The smallest, it doesn't have any overlaps
JSGandora 2011-03-19 20:32:58
We want the smallest.
rrusczyk 2011-03-19 20:33:03
It looks like we maximize h by choosing b and c to be the smallest lengths of the triangle.  But we have to be careful.  We actually want the smallest of these three heights.  If the height of the table is larger than h_A, for example, it will cause the folds corresponding to h_A to overlap, since h_A is the greatest the height can be to prevent those folds from overlapping.  So, the height of the table can be no greater than the smallest of these three expressions.
rrusczyk 2011-03-19 20:33:23
The minimum of these three expressions occurs when we have the longest sides in the denominator.
rrusczyk 2011-03-19 20:33:41
So, how do we finish?
GoldenFrog1618 2011-03-19 20:34:23
2K/57
Spring 2011-03-19 20:34:23
2K/57, use Heron's to find K
Cortana 2011-03-19 20:34:23
2K/57
.cpp 2011-03-19 20:34:23
2K/(27+30) gives 2K/57 is the maximum height.
rrusczyk 2011-03-19 20:34:30
rrusczyk 2011-03-19 20:34:55
All right, we'll take a short break to rest my fingers :)
mathcountsloser 2011-03-19 20:36:07
That was hard. Do you think elementary geometry covers this?
ahaanomegas 2011-03-19 20:36:07
Was there an easier way to do that one?
rrusczyk 2011-03-19 20:36:18
I don't think you'll see this in a typical geometry class!
rrusczyk 2011-03-19 20:36:27
I don't think there's really an easier way to do this.
mathepic 2011-03-19 20:37:03
my school doesn't discuss how to do anything past 5 on an AMC...
rrusczyk 2011-03-19 20:37:11
That's why we started AoPS :)
Jasmine8925 2011-03-19 20:37:54
haha my school doesn't discuss anything that would ever be on any math competition
mathcountsloser 2011-03-19 20:37:55
My school doesn't discuss anything.
rrusczyk 2011-03-19 20:38:08
All right, back to work.  Er, fun.
rrusczyk 2011-03-19 20:38:12
rrusczyk 2011-03-19 20:38:23
Where do we start
BOGTRO 2011-03-19 20:38:43
(24sinx)^3/2=24cosx
ahaanomegas 2011-03-19 20:38:43
fortenforge 2011-03-19 20:38:43
get rid of the log
superpi83 2011-03-19 20:38:43
convert to exponential form first
Titandrake 2011-03-19 20:38:43
expand the log
vcez 2011-03-19 20:38:43
GeorgiaTechMan 2011-03-19 20:38:43
(24sinx)^3/2=24cosx
tan90 2011-03-19 20:38:43
Exponential form
delta1 2011-03-19 20:38:43
get rid of the logs
Potential5 2011-03-19 20:38:46
unlog it
rrusczyk 2011-03-19 20:39:00
unlog.  Wonderful.  My vocabulary has just expanded by one word.
rrusczyk 2011-03-19 20:39:04
We start by getting rid of the logarithm, and writing the equation in exponential form:
rrusczyk 2011-03-19 20:39:09
eb8368 2011-03-19 20:39:28
Square both sides
tan90 2011-03-19 20:39:28
Now square both sides.
GoldenFrog1618 2011-03-19 20:39:31
square
VIPMaster 2011-03-19 20:39:31
square it!!!
hrithikguy 2011-03-19 20:39:31
square both sides
vcez 2011-03-19 20:39:31
square both sides
pi091997 2011-03-19 20:39:31
square both sides?
fortenforge 2011-03-19 20:39:31
I squared both sides here
hsm174 2011-03-19 20:39:38
move the 24 to the lhs
calvinhobbesliker 2011-03-19 20:39:38
Square both sides and divide by 24^2
rrusczyk 2011-03-19 20:39:45
mathcountsloser 2011-03-19 20:39:52
this seems to be getting easier by the moment
rrusczyk 2011-03-19 20:39:56
That's the goal.
policecap 2011-03-19 20:40:01
cos^2 = 1-sin^2
GeorgiaTechMan 2011-03-19 20:40:01
cosx=1-sin^2x
Cortana 2011-03-19 20:40:03
replace cos^2(x) with 1-sin^2(x)
esque 2011-03-19 20:40:07
substitue 1-sin^x for cos^x
alanzee 2011-03-19 20:40:07
cos^2 x=1-sin^2 x
rrusczyk 2011-03-19 20:40:12
etude 2011-03-19 20:40:23
cos^2 x = 1 - sin^2 x. Substitute y = sin x and solve for y.
ahaanomegas 2011-03-19 20:40:24
fortenforge 2011-03-19 20:40:28
Now write sin x = y,
rrusczyk 2011-03-19 20:40:32
ahaanomegas 2011-03-19 20:40:34
pinkmuskrat 2011-03-19 20:40:45
for the cubic in sin x, 1/3 is a root
kangchangood 2011-03-19 20:40:45
then we test.. the value 1/3
And can we factor this?
mathepic 2011-03-19 20:40:45
IIRC y = 3
theone142857 2011-03-19 20:40:45
y=1/3 by rational root:whistle:
pinkmuskrat 2011-03-19 20:40:45
1/3 is a root
Z91 2011-03-19 20:40:45
y=1/3
SuperNerd123 2011-03-19 20:40:52
y=1/3 Rational Root therom
rational root theorem, y = 1/3 happens to work'
rrusczyk 2011-03-19 20:40:55
We don't know for sure yet that (sin x) must be rational, but we check for that just in case.  From the Rational Root Theorem and the fact that y is positive (since x is between 0 and pi/2), we know that we only have to test fractions of the form 1/n, where n divides 24, for y.  We quickly find that y=1/3 works, and we have the factorization
rrusczyk 2011-03-19 20:40:58
(3y-1)(8y^2+3y +1) = 0
rrusczyk 2011-03-19 20:41:03
The quadratic has no positive roots, so we know that sin x = 1/3.
rrusczyk 2011-03-19 20:41:06
Now, we just have to find 24cot^2x.  How do we finish?
Cortana 2011-03-19 20:41:50
find cos^2(x)
Cortana 2011-03-19 20:41:50
cot^2(X)=cos^2(x)/sin^2(x)
tan90 2011-03-19 20:41:53
Find cos^2 (x)
rrusczyk 2011-03-19 20:42:01
OK, finish it off.
rrusczyk 2011-03-19 20:42:35
There are a ton of ways to finish from here:
eb8368 2011-03-19 20:42:38
24cot^2x = 24(1-sin^2x)/(sin^2x)
Spring 2011-03-19 20:42:38
24cot^2x=24(cos2^x/sin^2x)=24^2sinx=24^2/3=192
calvinhobbesliker 2011-03-19 20:42:38
cos^2=8/9; sin^2=1/9 Answer is thus 24*8
Z91 2011-03-19 20:42:38
Since 24sin^3(x)=cos^2(x), cot(x)=24sin(x) and 24cot(x)=576sin(x)
policecap 2011-03-19 20:42:48
alanzee 2011-03-19 20:42:58
24(1/3)sin^2 x=cos^2 x =>24cos^2 x/sin^2x=192
ahaanomegas 2011-03-19 20:42:58
etude 2011-03-19 20:43:01
cot^2 x = (8/3)/(1/3) = 8. 8*24 = 192 answer.
rrusczyk 2011-03-19 20:43:06
billybob42 2011-03-19 20:43:12
i got scared by the cubic. is there an easier way to solve this?
rrusczyk 2011-03-19 20:43:43
The first time I did it, I converted everything to cotangent, and you get a cubic with an integer root, which is less scary (and was simpler to see the solution.)
rrusczyk 2011-03-19 20:43:58
But, this problem was particularly guessable: we might have guessed that 24 sin x and 24 cos x need be rational powers of either 2 or 3, and then just tried values of sin x with 3 or 8 as the denominator.  3 seems like a better first guess, and trying 1/3 gets you to the answer right away!
rrusczyk 2011-03-19 20:44:31
In almost every AIME, there is a problem from 7 to 12 that has lots of notation (logs, trig, floor function) that turns out to be way easier than it looks.
GoldenFrog1618 2011-03-19 20:44:47
how do you convert sin^3x/cos^2 x to cot?
rrusczyk 2011-03-19 20:45:12
I started with the cot^2 + 1 = csc^2 identity.
rrusczyk 2011-03-19 20:46:03
rrusczyk 2011-03-19 20:46:33
Rather than going after obtuse, what might we go after?
rdj5933mile5 2011-03-19 20:47:01
Consider probability of finding acute triangle or right?
pinkmuskrat 2011-03-19 20:47:01
right and acute
pi.guy3.14 2011-03-19 20:47:01
right and acute?
Spring 2011-03-19 20:47:01
right or acute
Right or acute?
rrusczyk 2011-03-19 20:47:08
When I did this problem, my first thought was to do the complement: find n such that the probability that a random triangle is acute or right is 32/125.  (Because 32/125 is a smaller number, so perhaps easier.)
skyhog 2011-03-19 20:47:17
Break into cases:  n is odd or n is even
calvinhobbesliker 2011-03-19 20:47:17
Casework with n even and odd?
rrusczyk 2011-03-19 20:47:22
We can only get a right angle if n is even (so that one of the sides of the triangle is a diameter).
rrusczyk 2011-03-19 20:47:25
This suggests we may have to look at the odd and even cases separately.
rrusczyk 2011-03-19 20:47:33
So let's do the odd case first.  We let n = 2k+1 for some positive k.
rrusczyk 2011-03-19 20:47:57
What might we do as a first step to simplify thinking about the problem?
calvinhobbesliker 2011-03-19 20:48:28
Fix a vertex
soulspeedy 2011-03-19 20:48:28
fix a point on the figure
rrusczyk 2011-03-19 20:48:31
We can arbitrarily choose the first vertex to be any vertex of the n-gon (by the rotational symmetry).   Suppose for the sake of ease of discussion that it's the top vertex:
rrusczyk 2011-03-19 20:48:35
rrusczyk 2011-03-19 20:48:41
How many possible ways to complete the triangle?
rrusczyk 2011-03-19 20:49:01
Remember, we are letting n = 2k+1 here.
Spring 2011-03-19 20:49:29
GeorgiaTechMan 2011-03-19 20:49:29
or (2k)C2
etude 2011-03-19 20:49:29
2k C 2
tiger21 2011-03-19 20:49:29
pick 2 from 2k
AlphaMath1 2011-03-19 20:49:29
(2k 2)
pi091997 2011-03-19 20:49:29
C(2k,2)
centralbs 2011-03-19 20:49:29
(2k C 2)
rrusczyk 2011-03-19 20:49:40
We have to choose two of the remaining 2k vertices, so it's C(2k,2).
rrusczyk 2011-03-19 20:49:44
How many ways to make an acute triangle?
kangchangood 2011-03-19 20:49:53
well... triangles that include that point.
rrusczyk 2011-03-19 20:50:06
Right, triangles that include that point.
rrusczyk 2011-03-19 20:50:33
Remember that we are aiming at acute triangles here.
rrusczyk 2011-03-19 20:50:46
So, what do we know about the two other points?
Kinata12 2011-03-19 20:51:00
The 2 points must be on opposite sides of the diameter
rrusczyk 2011-03-19 20:51:05
We need one vertex from the left side and one from the right side.  (For example, if both are from the left side, then the middle vertex there will give an obtuse triangle.)
rrusczyk 2011-03-19 20:51:15
Also, we need that the two new vertices are separated by no more than half the triangle.
rrusczyk 2011-03-19 20:51:26
So, in particular, if we pick the top vertex on the left, we must pick the bottom vertex on the right:
rrusczyk 2011-03-19 20:51:30
rrusczyk 2011-03-19 20:51:44
What if we pick the second highest vertex on the left?
VIPMaster 2011-03-19 20:52:31
the bottom vertex on the right or the one right above it
rdj5933mile5 2011-03-19 20:52:31
2 choices
rrusczyk 2011-03-19 20:52:35
If we pick the 2nd highest vertex on the left, we must pick one of the bottom two vertices on the right.
rrusczyk 2011-03-19 20:52:38
rrusczyk 2011-03-19 20:52:41
rrusczyk 2011-03-19 20:52:46
And so on...
rrusczyk 2011-03-19 20:52:51
So how many acute triangles are there?
ahaanomegas 2011-03-19 20:53:20
But, in your diagram, you just arbitrarily picked the number of sides in the polygon, is that correct?
rrusczyk 2011-03-19 20:53:27
Right, I want an answer in terms of k
skyhog 2011-03-19 20:53:31
k(k+1)/2
etude 2011-03-19 20:53:32
k(k+1)/2?
rrusczyk 2011-03-19 20:53:37
As we pick a vertex from top to bottom on the left side, we have 1, 2, ..., k choices for a vertex on the right side.
rrusczyk 2011-03-19 20:53:40
So there are 1+2+...+k acute triangles.
rrusczyk 2011-03-19 20:53:46
rrusczyk 2011-03-19 20:53:59
How does this simplify?
pinkmuskrat 2011-03-19 20:54:38
k(k+1)/2k(2k-1)
eb8368 2011-03-19 20:54:39
(k+1)/2(k-1)
skyhog 2011-03-19 20:54:39
(k+1)/[2(2k-1)]
rdj5933mile5 2011-03-19 20:54:39
(k+1)/(4k-2)
Spring 2011-03-19 20:54:42
(k+1)/(2(k-1))
rrusczyk 2011-03-19 20:54:45
The numerator is equal to (1/2)k(k+1).
rrusczyk 2011-03-19 20:54:48
The denominator is equal to (1/2)(2k)(2k-1).
rrusczyk 2011-03-19 20:54:52
policecap 2011-03-19 20:54:56
=32/125
rrusczyk 2011-03-19 20:54:59
We want this to be equal to 32/125.
rrusczyk 2011-03-19 20:55:02
infinity1 2011-03-19 20:55:31
k=63
jeff10 2011-03-19 20:55:31
k=63
DystopianStriker 2011-03-19 20:55:31
k=63
calvinhobbesliker 2011-03-19 20:55:35
k=63?
super_pi314 2011-03-19 20:55:35
63
ahaanomegas 2011-03-19 20:55:40
VIPMaster 2011-03-19 20:55:44
125k + 125 = 128k - 64 => k = 63
rrusczyk 2011-03-19 20:55:47
AlphaMath1 2011-03-19 20:56:16
BUT we need n
tan90 2011-03-19 20:56:16
so n = 127
skyhog 2011-03-19 20:56:16
so n = 127
BarbieRocks 2011-03-19 20:56:20
so 2k+1=127
ahaanomegas 2011-03-19 20:56:20
etude 2011-03-19 20:56:20
Thus n = 127
rrusczyk 2011-03-19 20:56:23
Recalling that n = 2k+1, we see that n = 2(63)+1 = 127 is a possible value.
kangchangood 2011-03-19 20:56:25
there come even cases
rrusczyk 2011-03-19 20:56:28
Now the even case.  Let n = 2k.
rrusczyk 2011-03-19 20:56:32
Again, we can assume the first vertex picked is the "top" vertex.  We can pick two other vertices in C(2k-1,2) ways.
rrusczyk 2011-03-19 20:56:38
Jasmine8925 2011-03-19 20:57:17
if we pick the bottom vertex, any other vertex we pick will make a right triangle
rrusczyk 2011-03-19 20:57:21
If we have the opposite vertex, then we can pick any of the other 2k-2 vertices and get a right triangle.
rrusczyk 2011-03-19 20:57:29
Otherwise, we have one from the left side and one from the right side (just like in the odd case).
rrusczyk 2011-03-19 20:57:51
OK, so what is the number of acute or right triangles we get in this case?
etude 2011-03-19 20:59:25
2k-2 + k(k-1)/2?
.cpp 2011-03-19 20:59:25
(k-1)k/2 + 2k-2.
rrusczyk 2011-03-19 20:59:39
And also as in the odd case, if we take the top vertex on the left, we must take the bottom vertex on the right; if we take the next highest vertex on the left, we must take one of the 2 bottom vertices on the right, and so on.
rrusczyk 2011-03-19 20:59:57
So there are 1 + 2 + ... + (k-1) ways to take a vertex from either side and get an acute or right triangle.  We put these together with the 2k-2 that we already found.
rrusczyk 2011-03-19 21:00:03
rrusczyk 2011-03-19 21:00:11
And how does this sample?
rrusczyk 2011-03-19 21:00:18
simplify?
rdj5933mile5 2011-03-19 21:01:27
(k+4)/(4k-2)
DystopianStriker 2011-03-19 21:01:27
(k+4)/2(2k-1)?
policecap 2011-03-19 21:01:27
rrusczyk 2011-03-19 21:01:30
rrusczyk 2011-03-19 21:01:34
rrusczyk 2011-03-19 21:01:37
policecap 2011-03-19 21:02:24
k=188
DystopianStriker 2011-03-19 21:02:24
188
jeff10 2011-03-19 21:02:24
k=188
super_pi314 2011-03-19 21:02:24
188
ahaanomegas 2011-03-19 21:02:30
VIPMaster 2011-03-19 21:02:30
125k + 500 = 128k - 64 => k = 188
rrusczyk 2011-03-19 21:02:32
policecap 2011-03-19 21:02:52
376 is a solution
ahaanomegas 2011-03-19 21:02:52
kangchangood 2011-03-19 21:02:52
then n
prezcoin 2011-03-19 21:02:52
n= 376
icecreamcakepie 2011-03-19 21:02:52
n=2K
VIPMaster 2011-03-19 21:02:58
n = 188*2 = 376
.cpp 2011-03-19 21:02:58
2k = 376, so our answer is 376+127 = 503.
rrusczyk 2011-03-19 21:03:00
This yields n = 2k = 376 as our other possible value.
rrusczyk 2011-03-19 21:03:06
RobRoobiks 2011-03-19 21:03:14
is there a way to count obtuse triangles?
rrusczyk 2011-03-19 21:03:23
Sure!  See if you can find one on your own.
rrusczyk 2011-03-19 21:03:40
(There are probably discussions about it on the message board.  It will look a lot like this one.
rrusczyk 2011-03-19 21:03:49
All right.  We've rounded the turn!
rrusczyk 2011-03-19 21:03:51
5 more to go.
rrusczyk 2011-03-19 21:03:56
rrusczyk 2011-03-19 21:04:26
The "divided by 1000" appearing twice suggests what?
jeff10 2011-03-19 21:04:37
mod1000 basically
Lemon123 2011-03-19 21:04:37
mod 1000
pi.guy3.14 2011-03-19 21:04:37
mod 1000
Jasmine8925 2011-03-19 21:04:37
mod 1000
policecap 2011-03-19 21:04:37
mod 1000
prezcoin 2011-03-19 21:04:37
mod 1000
rrusczyk 2011-03-19 21:04:39
We probably want to be working mod 1000.
rrusczyk 2011-03-19 21:04:44
So we want to determine all the different residues of 2^n mod 1000 and then add them up mod 1000.
rrusczyk 2011-03-19 21:05:12
.cpp 2011-03-19 21:05:25
All remainders are multiples of 8, except for 1, 2, and 4.
prezcoin 2011-03-19 21:05:35
after 1,2,4, the rest of the numbers are divisible by 8, as is 1000
carmelninja 2011-03-19 21:05:40
we actually only really need to consider it mod125
ahaanomegas 2011-03-19 21:05:45
All remainders are multiples of 8, except for 1, 2, and 4
rdj5933mile5 2011-03-19 21:05:47
work mod 125?
rrusczyk 2011-03-19 21:05:53
Often with mods, we break up the modulus into its prime power components.
rrusczyk 2011-03-19 21:05:56
So we can look separately as residues mod 8 and mod 125.
rrusczyk 2011-03-19 21:05:59
Mod 8 is easy: the powers start 1,2,4, and then all the rest of the powers are multiples of 8.
rrusczyk 2011-03-19 21:06:08
So our sum looks like 1 + 2 + 4 + 8(1+2+4+...), and we need to analyze the sum inside the parentheses mod 125.
ahaanomegas 2011-03-19 21:06:11
Why 125?
soulspeedy 2011-03-19 21:06:13
note that 1000=2^3*5^3
rrusczyk 2011-03-19 21:06:26
But what do we know about that sum in parentheses there?
Jasmine8925 2011-03-19 21:06:58
geometric series
VIPMaster 2011-03-19 21:06:58
its a geometric series!
calvinhobbesliker 2011-03-19 21:06:58
It's 2^n-1
eb8368 2011-03-19 21:06:58
It cycles after 2^100
RelaxationUtopia 2011-03-19 21:06:58
Geometric sequence?
rrusczyk 2011-03-19 21:07:06
It is indeed a geometric series.
VIPMaster 2011-03-19 21:07:27
i like eb8368's observation better
rrusczyk 2011-03-19 21:07:44
We care about the sum mod 125 (we'll multiply the result by 8 after that to get the sum mod 1000).
rrusczyk 2011-03-19 21:07:52
Now, eb's observation is interesting.
centralbs 2011-03-19 21:07:54
How do you know it cycles after 2^100?
rrusczyk 2011-03-19 21:08:03
We could use the Euler-Fermat Theorem, which tells us that because 2 and 125 are relatively prime, we have 2^100 = 1 (mod 125). But we don't even need such a fancy theorem.  We know for sure that the powers of 2 mod 125 will eventually repeat, since there are infinitely many powers of 2.  So, we'll get 2^a = 2^b (mod 125) with a>b at some point.  Then, because 2 and 125 are relatively prime, we have 2^(a-b) = 1 (mod 125).
rrusczyk 2011-03-19 21:08:37
We don't necessarily need to know 2^100 = 1; it's enough to know that 2^k = 1 (mod 125) for some k.  Why is this useful?  That is, why is it important that it will repeat mod 125?
jeff10 2011-03-19 21:09:35
2^100 is not 1.
rrusczyk 2011-03-19 21:09:41
That's mod 125...
eb8368 2011-03-19 21:09:49
2^k = 2^(k+1) - 1
etude 2011-03-19 21:09:49
Therefore, that series is 2^k - 1 which is 0 mod 125. Thus the sum is 7 mod 125.
.cpp 2011-03-19 21:09:49
Sum in parenthesis is 2^k - 1 = 0 mod 125.
rrusczyk 2011-03-19 21:09:58
Exactly.
rrusczyk 2011-03-19 21:10:02
Because when we hit 2^k, our sum will repeat mod 125, and thus repeat mod 1000.
rrusczyk 2011-03-19 21:10:07
rrusczyk 2011-03-19 21:10:20
All we are using here is the fact that this sum will repeat at some point mod 125.
rrusczyk 2011-03-19 21:10:28
We don't even care where!
rrusczyk 2011-03-19 21:10:37
As noted before, the expression is a geometric series.
rrusczyk 2011-03-19 21:10:40
It's equal to 2^k - 1.
rrusczyk 2011-03-19 21:10:43
Titandrake 2011-03-19 21:10:45
that is pretty tricky
mathcountsloser 2011-03-19 21:10:52
nice
centralbs 2011-03-19 21:10:53
ooo wow thats soo clever
rrusczyk 2011-03-19 21:11:06
Yes, very tricky (DPatrick came up with this; I can't claim credit...)
Titandrake 2011-03-19 21:11:19
so then 2^k -1 = 0 mod 125, so 8 times that is 0 mod 1000
rrusczyk 2011-03-19 21:11:21
2^k = 1 (mod 125), so 2^k - 1 is a multiple of 125.
rrusczyk 2011-03-19 21:11:30
Thus 8(2^k - 1) is a multiple of 1000, and hence goes away mod 1000.
VIPMaster 2011-03-19 21:11:35
then we get 1 + 2 + 4 + 0 = 7
hchen023 2011-03-19 21:11:35
so S = 1 + 2 + 4 + 0 = 007?
jeff10 2011-03-19 21:11:40
that means that S is congruent to 7 (mod 1000)
soulspeedy 2011-03-19 21:11:40
DystopianStriker 2011-03-19 21:11:53
1+2+4=7
ahaanomegas 2011-03-19 21:11:53
007
tan90 2011-03-19 21:11:53
James Bond... 007
girishvar12 2011-03-19 21:11:53
$007$
mathcountsloser 2011-03-19 21:11:53
007 = james bond
eb8368 2011-03-19 21:11:53
S= 007
rrusczyk 2011-03-19 21:11:56
So we're left with just S = 1 + 2 + 4 (mod 1000) = 7 (mod 1000).
rrusczyk 2011-03-19 21:12:00
rrusczyk 2011-03-19 21:12:13
rrusczyk 2011-03-19 21:12:33
So, what do we have to do here?
.cpp 2011-03-19 21:12:52
Probability = desired/total possibilities.
ahaanomegas 2011-03-19 21:13:02
P = desired/total
jeff10 2011-03-19 21:13:17
P=desired/total possible
rrusczyk 2011-03-19 21:13:19
We treat this as a usual "desired/possible" counting probability problem.
ksun48 2011-03-19 21:13:27
casework?
AlphaMath1 2011-03-19 21:13:27
Start caseworking?
calvinhobbesliker 2011-03-19 21:13:27
Split into cases
rrusczyk 2011-03-19 21:13:46
And we'll have to break into careful casework.
rrusczyk 2011-03-19 21:13:50
rrusczyk 2011-03-19 21:13:54
Let's list the cases.  If every man must be next to another man, what types of configurations can we have?
policecap 2011-03-19 21:15:05
2/2/2, 2/4, 3/3, 4/2, 6
carmelninja 2011-03-19 21:15:05
2-2-2,   2-4,   3-3,   4-2,   6
rrusczyk 2011-03-19 21:15:07
Notice that there are two with a 4 and a 2!
rrusczyk 2011-03-19 21:15:15
rrusczyk 2011-03-19 21:15:24
Each asterisk represents some unknown number of women.
rrusczyk 2011-03-19 21:15:29
Now we need a variable.  Let w be the number of women.
rrusczyk 2011-03-19 21:15:41
OK, how do we could the first case?
willwang123 2011-03-19 21:15:53
3rd case?
Kinata12 2011-03-19 21:15:53
You drew #3 wrong
rrusczyk 2011-03-19 21:16:06
The invisible man is in that group.  Look closely.
rrusczyk 2011-03-19 21:16:17
How do we count the first case?
mthcz11 2011-03-19 21:16:21
omg! i c him!!!
rrusczyk 2011-03-19 21:16:25
That's better
rrusczyk 2011-03-19 21:16:39
How do we could the first case
ksun48 2011-03-19 21:16:47
w+1 because there does not need to be a man in each *
.cpp 2011-03-19 21:16:47
(w+1 choose 1) possibilities.
rrusczyk 2011-03-19 21:16:55
There are w+1 slots for the men to take (including the ends), so there are w+1 such arrangements.
rrusczyk 2011-03-19 21:17:12
(We first line the women up from 1 to w)
rrusczyk 2011-03-19 21:17:19
rrusczyk 2011-03-19 21:17:22
How many ways can we place  *MMM*MMM*?
centralbs 2011-03-19 21:17:35
Are we counting the men as indistinguishable?
rrusczyk 2011-03-19 21:17:37
Yes.
rrusczyk 2011-03-19 21:17:44
Way easier that way!
Jasmine8925 2011-03-19 21:17:50
what's w?
rrusczyk 2011-03-19 21:17:54
Number of women
rrusczyk 2011-03-19 21:18:30
We line them up and then put the men in the line afterwards.  How many ways can we tackle the *MMM*MMM* case?
etude 2011-03-19 21:18:53
w+1 C 2
Kinata12 2011-03-19 21:18:53
w(w+1)/2
rrusczyk 2011-03-19 21:18:57
We need to pick two of these w+1 slots in the line of w women (including the ends!).
rrusczyk 2011-03-19 21:19:09
rrusczyk 2011-03-19 21:19:15
rrusczyk 2011-03-19 21:19:24
pi.guy3.14 2011-03-19 21:19:49
the next two are the same
policecap 2011-03-19 21:19:49
same thing
ParallelProcess 2011-03-19 21:19:49
and then that applies for the next two as well
GoldenFrog1618 2011-03-19 21:19:49
same
VIPMaster 2011-03-19 21:19:49
same as the last one
Kinata12 2011-03-19 21:19:49
The same, w+1C2
Lemon123 2011-03-19 21:19:49
same
tan90 2011-03-19 21:19:56
Same as case 2
eb8368 2011-03-19 21:19:56
w+1 C 2
rrusczyk 2011-03-19 21:19:58
We know that n_24 and n_42 must be equal to n_33 since it's the same computation:
rrusczyk 2011-03-19 21:20:02
rrusczyk 2011-03-19 21:20:11
And the last case?
Jasmine8925 2011-03-19 21:20:22
the last case is w+1 C 3
policecap 2011-03-19 21:20:23
n+1 C 3
etude 2011-03-19 21:20:23
The last one is w+1 C 3.
VIPMaster 2011-03-19 21:20:23
w + 1 choose 3
dlennon 2011-03-19 21:20:26
C(w+1,3)
Kinata12 2011-03-19 21:20:26
w+1C3
rdj5933mile5 2011-03-19 21:20:26
w+1c3
rrusczyk 2011-03-19 21:20:28
We need to place three groups of men.  There are C(w+1,3) ways to place these groups.
rrusczyk 2011-03-19 21:20:31
rrusczyk 2011-03-19 21:20:41
Now what is p?
VIPMaster 2011-03-19 21:20:55
the sum of all of these
kangchangood 2011-03-19 21:20:55
sum?
vcez 2011-03-19 21:20:55
probability is desired over total
aerrowfinn72 2011-03-19 21:20:57
rrusczyk 2011-03-19 21:21:06
We add all of these, and that gives the denominator.
rrusczyk 2011-03-19 21:21:10
What is the numerator?
vcez 2011-03-19 21:21:35
cases 1,3,4
v_Enhance 2011-03-19 21:21:35
F = n6 + n24 + n42
.cpp 2011-03-19 21:21:35
n_6+n_24+n_42
vcez 2011-03-19 21:21:35
first, third, and fourth cases in the list
oops 1,3,4
carmelninja 2011-03-19 21:21:35
n_6 + n_24 + n_42
Kinata12 2011-03-19 21:21:38
The sum of cases 1, 3, and 4
rrusczyk 2011-03-19 21:21:40
We want to divide the cases where four men stand together by the total number of cases:
rrusczyk 2011-03-19 21:21:46
ahaanomegas 2011-03-19 21:21:59
Huge w bashing, here!
BarbieRocks 2011-03-19 21:21:59
cancel w+1
.cpp 2011-03-19 21:21:59
Factor out w+1.
rrusczyk 2011-03-19 21:22:01
I'll spare you the algebra.
rrusczyk 2011-03-19 21:22:05
Jasmine8925 2011-03-19 21:22:09
p is less than or equal to 1/100
rrusczyk 2011-03-19 21:22:16
More bashing.
rrusczyk 2011-03-19 21:22:17
rrusczyk 2011-03-19 21:22:49
What do we want to find here?
eb8368 2011-03-19 21:23:08
The least possible value of w
tan90 2011-03-19 21:23:08
The minimum value for w
calvinhobbesliker 2011-03-19 21:23:08
Lower bound
The least number of w.
Jasmine8925 2011-03-19 21:23:11
minimize w
centrino 2011-03-19 21:23:11
smallest value of w
policecap 2011-03-19 21:23:11
w minimum
rrusczyk 2011-03-19 21:23:21
And how can we do that?
rrusczyk 2011-03-19 21:24:14
There are lots of ways to do this
GoldenFrog1618 2011-03-19 21:24:28
594<=w(w-592)
.cpp 2011-03-19 21:24:28
Complete the square for (w-296)^2 - (296^2 + 594).
BarbieRocks 2011-03-19 21:24:28
w(w-592)>594 so it makes sense to trie numbers near 592
edisonchew240 2011-03-19 21:24:28
Use completing the square
vcez 2011-03-19 21:24:28
find the roots
rrusczyk 2011-03-19 21:24:50
Here's one way to do this thinking about roots:
rrusczyk 2011-03-19 21:24:51
rrusczyk 2011-03-19 21:25:03
Since the numbers -594 and 592 are large and pretty close in magnitude, it looks like the negative root must be really close to -1 and the positive root has to be a little more than 592.
rrusczyk 2011-03-19 21:25:15
What do we find as our minimal w?
bridgemaster 2011-03-19 21:25:29
594
eb8368 2011-03-19 21:25:29
w=594 !
prezcoin 2011-03-19 21:25:29
w= 594
DystopianStriker 2011-03-19 21:25:31
594
theone142857 2011-03-19 21:25:31
594
.cpp 2011-03-19 21:25:33
rrusczyk 2011-03-19 21:25:36
Let's see where the root is in relation to 593.
rrusczyk 2011-03-19 21:25:39
rrusczyk 2011-03-19 21:25:45
So f is still negative at 593.
rrusczyk 2011-03-19 21:25:48
rrusczyk 2011-03-19 21:25:51
BarbieRocks 2011-03-19 21:26:05
why on earth would 6 men and 594 women stand in a line
rrusczyk 2011-03-19 21:26:19
Justin Bieber concert
rrusczyk 2011-03-19 21:26:35
OK, on we rollll
.cpp 2011-03-19 21:26:48
6 = 0.01 (594+6), oddly.
ksun48 2011-03-19 21:26:48
also there are 600 total people. Is that significant
rrusczyk 2011-03-19 21:26:59
Interesting; may want to check that out for some other numbers...
rrusczyk 2011-03-19 21:27:04
rrusczyk 2011-03-19 21:27:12
At first glance this looks icky.  At second glance too.
GoldenFrog1618 2011-03-19 21:27:40
use coordinates
Jasmine8925 2011-03-19 21:28:00
coordinates will still be icky
rrusczyk 2011-03-19 21:28:20
How can we make coordinates less icky ?
tan90 2011-03-19 21:28:47
If A is at the origin?
kangchangood 2011-03-19 21:28:47
put origin at proper location..
jeff10 2011-03-19 21:28:50
change x, y, and z axis to the side lengths
rrusczyk 2011-03-19 21:29:03
There are lots of ways to solve this problem; some are messier than others.
rrusczyk 2011-03-19 21:29:21
We have three important lines that are mutually perpendicular -- each perpendicular to the other two.
rrusczyk 2011-03-19 21:29:24
We're in space.
aerrowfinn72 2011-03-19 21:29:36
those are x,y,z axes
tan90 2011-03-19 21:29:39
Make them the axes
rrusczyk 2011-03-19 21:29:44
So, we make these three lines (the edges of the cube) the axes.
rrusczyk 2011-03-19 21:29:51
So I would let A = (0,0,0) and let our three points P,Q,R be (10,0,0), (0,10,0), and (0,0,10).  Suppose these points have distance 10, 11, 12 from some plane.
rrusczyk 2011-03-19 21:29:55
What's the generic equation for a plane in 3-space?
.cpp 2011-03-19 21:30:27
Ax+By+Cz=D.
karatemagic7 2011-03-19 21:30:28
ax +by+cz=d
NoWayHaze 2011-03-19 21:30:28
ax+by+cz+d=0
centrino 2011-03-19 21:30:28
ax+by+cz=d
rrusczyk 2011-03-19 21:30:35
ax + by + cz+ d = 0.
rrusczyk 2011-03-19 21:30:47
And what's the distance from a point (p,q,r) to that plane?
AlphaMath1 2011-03-19 21:30:56
Don't tell me we have to use that formula again!
rrusczyk 2011-03-19 21:31:04
Sadly, this is the cleanest solution.
BarbieRocks 2011-03-19 21:31:48
|pa+qb+rc+d|/sqrt(a^2+b^2+c^2)
karatemagic7 2011-03-19 21:31:48
(ap+bq+cr+d)/(a^2+b^2+c^2)^.5
gh625 2011-03-19 21:31:48
|ax+by+cz+d|/sqrt{a^2+b^2+c^2}
js2082 2011-03-19 21:31:48
|ax + by + cz + d|/(a*a+b*b+c*c)**0.5
rrusczyk 2011-03-19 21:32:03
We can think of this as "plugging" the point into the equation of the plane, and then dividing by a normalizing factor:
.cpp 2011-03-19 21:32:05
|Ax+By+Cz+D|/sqrt(A^2+B^2+C^2)
rrusczyk 2011-03-19 21:32:09
rrusczyk 2011-03-19 21:32:18
Note this is a signed distance: it's positive for all points on one side of the plane, and negative for all points on the other side.
rrusczyk 2011-03-19 21:32:21
(We used the similar formula for the distance from a point to a line in Problem 3.)
JSGandora 2011-03-19 21:32:51
there's an absolute value for the numerator right?
.cpp 2011-03-19 21:33:13
It's a signed distance here.
JSGandora 2011-03-19 21:33:13
oh, sorry.  it's a signed distance.
rrusczyk 2011-03-19 21:33:14
We're taking signed distance here --> "above the plane" is positive and "below the plane" is negative.
rrusczyk 2011-03-19 21:33:23
Now, let's look at our plane equation:
rrusczyk 2011-03-19 21:33:29
ax + by + cz + d = 0.
rrusczyk 2011-03-19 21:33:46
Can we make any assumptions here that will make that distance formula less nasty
AlphaMath1 2011-03-19 21:34:02
a=1
rrusczyk 2011-03-19 21:34:14
We can scale the constants a,b,c,d by anything.
rrusczyk 2011-03-19 21:34:34
What would be a really nice scaling to choose, to make the distance formula not nasty?
karatemagic7 2011-03-19 21:34:40
We can assume that (a^2+b^2+c^2)=1
rrusczyk 2011-03-19 21:34:48
We can multiply the equation of the plane by any nonzero constant we want!
rrusczyk 2011-03-19 21:35:06
So, we can scale it such that a^2 + b^2 + c^2 = 1!
Jasmine8925 2011-03-19 21:35:21
how do we know it's possible to scale it like that?
gh625 2011-03-19 21:35:40
Multiply both sides by the same number.
rrusczyk 2011-03-19 21:35:47
Because we can multiply the equation of a plane by any constant we want to and it won't change the plane.
rrusczyk 2011-03-19 21:35:51
x+y+z = 9
rrusczyk 2011-03-19 21:35:55
is the same plane as
rrusczyk 2011-03-19 21:36:00
3x+3y+3z = 27
rrusczyk 2011-03-19 21:36:02
And so on:
GoldenFrog1618 2011-03-19 21:36:05
let a^2+b^2+c^2=e^2, divide each number by e
rrusczyk 2011-03-19 21:36:15
That sure simplifies our distance formula.
rrusczyk 2011-03-19 21:36:42
If we have a^2 + b^2 +c^2 =1 , we have ap+bq+cr + d as our distance formula.
rrusczyk 2011-03-19 21:36:57
So, what equations do we get now?
gh625 2011-03-19 21:37:59
10a+d=10, 10b+d=11, 10c+d=12
.cpp 2011-03-19 21:37:59
10 = a(10)+d, 11 = b(10)+d, 12 = c(10)+d.
tan90 2011-03-19 21:37:59
10a + d = 10, 10b + d = 11, 10c + d = 12
etude 2011-03-19 21:37:59
10a+d = 10, 10b+d = 11, 10c+d = 12.
rrusczyk 2011-03-19 21:38:43
We just cram the other vertices into our new distance formula.
rrusczyk 2011-03-19 21:38:49
They would have looked like this:
etude 2011-03-19 21:38:51
We have (10a + d)/sqrt(a^2+b^2+c^2) = 10, (10b + d)/sqrt(a^2+b^2+c^2) = 11, and (10c + d)/sqrt(a^2+b^2+c^2) = 12.
rrusczyk 2011-03-19 21:39:01
But we cleverly let a^2 + b^2 + c^2 = 1
rrusczyk 2011-03-19 21:39:16
Jasmine8925 2011-03-19 21:39:25
we're trying to find d
.cpp 2011-03-19 21:39:25
Now, we just need to find d!
BarbieRocks 2011-03-19 21:39:25
we need to find d
rrusczyk 2011-03-19 21:39:29
We want the distance from A to the plane.
rrusczyk 2011-03-19 21:39:32
But this is just d, because A=(0,0,0) and a^2+b^2+c^2 = 1.
rrusczyk 2011-03-19 21:39:42
So we have to solve the above system for d.
rrusczyk 2011-03-19 21:39:54
What's the quickest way to do this?
prezcoin 2011-03-19 21:40:27
(10-d)/10=a, (11-d)/10=b, (12-d)/10=c
BarbieRocks 2011-03-19 21:40:27
(10-d)^2+(11-d)^2+(12-d)^2=100
aerrowfinn72 2011-03-19 21:40:27
write everyhthing in terms of d
kangchangood 2011-03-19 21:40:27
100 (a^2 + b^2 + c^2) = 100
rrusczyk 2011-03-19 21:40:46
We solve for a,b,c in terms of d in the linear equation, and plug them into the quadratic a^2+b^2+c^2 = 1.
rrusczyk 2011-03-19 21:40:49
Life is a lot simpler if we get 10a, 10b, 10c into the quadratic first. This allows us to substitute for 10a, 10b, and 10c rather than a,b,c.   Since a^2+b^2+c^2 = 1, we have
rrusczyk 2011-03-19 21:40:55
rrusczyk 2011-03-19 21:40:56
rrusczyk 2011-03-19 21:41:04
The rest is just bashing.
rrusczyk 2011-03-19 21:41:13
We're closing on 3 hours now, so I will bash for you.
rrusczyk 2011-03-19 21:41:18
rrusczyk 2011-03-19 21:41:22
rrusczyk 2011-03-19 21:41:31
rrusczyk 2011-03-19 21:41:37
Which answer to we want, + or -?
VIPMaster 2011-03-19 21:41:56
-
tan90 2011-03-19 21:41:56
-
ahaanomegas 2011-03-19 21:41:56
-
policecap 2011-03-19 21:41:56
-
SDLucifer 2011-03-19 21:41:56
-
carmelninja 2011-03-19 21:41:56
the minus
GoldenFrog1618 2011-03-19 21:41:56
- because the answer says so
gh625 2011-03-19 21:41:56
- because the answer has a - in it
tan90 2011-03-19 21:41:56
rrusczyk 2011-03-19 21:42:09
Because the test said so, cute.
rrusczyk 2011-03-19 21:42:14
Here's a better reason:
js2082 2011-03-19 21:42:18
as it shd be less than 10
carmelninja 2011-03-19 21:42:18
positive value is way to big anyway
rrusczyk 2011-03-19 21:42:22
Given the form of the answer required in the problem, we clearly had better have -.  Indeed, A has to be closer to the plane than P,Q,R, so we need an answer between 0 and 10.
rrusczyk 2011-03-19 21:42:26
rrusczyk 2011-03-19 21:42:48
You can think later about what the other sign gives you (what significance does that other root have?)
rrusczyk 2011-03-19 21:42:53
OK, 2 more problems.
rrusczyk 2011-03-19 21:42:58
rrusczyk 2011-03-19 21:43:10
alanzee 2011-03-19 21:43:30
square!
carmelninja 2011-03-19 21:43:30
a crooked square
VIPMaster 2011-03-19 21:43:30
a square!!!
calvinhobbesliker 2011-03-19 21:43:30
Square
we will have a square that's lopsided
rrusczyk 2011-03-19 21:43:33
From the symmetry of the problem, the quadrilateral formed by the B_i is a square.
rrusczyk 2011-03-19 21:43:37
rrusczyk 2011-03-19 21:43:44
OK, now what?  Can we at least simplify what we're looking for?
Jasmine8925 2011-03-19 21:44:09
double angle cosine formula
rrusczyk 2011-03-19 21:44:14
Cosine of double an angle is kind of a pain.  We can use the double-angle identity to give:
rrusczyk 2011-03-19 21:44:20
rrusczyk 2011-03-19 21:44:24
That's a nicer target.  That's still a not-very-fun angle to work with, but is part of that angle easy?
etude 2011-03-19 21:44:54
Draw M3M1. Then A3M3M1 is just 45 degrees.
rrusczyk 2011-03-19 21:45:01
rrusczyk 2011-03-19 21:45:04
rrusczyk 2011-03-19 21:45:18
tan90 2011-03-19 21:45:42
right triangle trig
etude 2011-03-19 21:45:44
Find two sides of the right triangle.
rrusczyk 2011-03-19 21:45:47
We want sine and/or cosine of this angle, so let's label some sides.
let M3B1 = x.  Then B1M1 = 1 + x
rrusczyk 2011-03-19 21:46:20
I'm going to let the sides of the square be 2.
rrusczyk 2011-03-19 21:46:21
Why?
.cpp 2011-03-19 21:46:37
To avoid fractions.
etude 2011-03-19 21:46:44
That eliminates fractions.
ParallelProcess 2011-03-19 21:46:44
midpoint divides into 1 and 1
ahaanomegas 2011-03-19 21:46:44
Because then we won't have fractions
rrusczyk 2011-03-19 21:46:46
We can choose the scale of the octagon, so we let the side length of the octagon be 2 (if we choose 1, we get fractions, ick).  We also know that B_1M_3 = B_7M_1, so we let those lengths be x:
rrusczyk 2011-03-19 21:46:50
rrusczyk 2011-03-19 21:46:57
Now it is awfully tempting to pull out our hammer and bash, bash, bash.
rrusczyk 2011-03-19 21:47:04
jeff10 2011-03-19 21:47:14
i think we should go for cosine
rrusczyk 2011-03-19 21:48:05
rrusczyk 2011-03-19 21:48:13
And we broke that up:
rrusczyk 2011-03-19 21:48:31
jeff10 2011-03-19 21:48:33
which is 45 + unknown angle
rrusczyk 2011-03-19 21:48:40
And what do we do with this?
kangchangood 2011-03-19 21:48:51
identity?..
DavidTong 2011-03-19 21:48:51
sum formula
ahaanomegas 2011-03-19 21:48:54
Sum of angle formula
etude 2011-03-19 21:48:54
rrusczyk 2011-03-19 21:48:58
rrusczyk 2011-03-19 21:49:03
Keep on rolling...
kangchangood 2011-03-19 21:49:13
pull out 45 thing.
rrusczyk 2011-03-19 21:49:17
Keep going...
carmelninja 2011-03-19 21:49:21
and the x terms should cancel out
rrusczyk 2011-03-19 21:49:25
Do they?
GoldenFrog1618 2011-03-19 21:50:30
yes!
VIPMaster 2011-03-19 21:50:30
Yes, they do - we're left with 2/hypotenuse
etude 2011-03-19 21:50:30
sqrt(2)/2 [x/c + 2-x/c] = sqrt(2)/2 *2/c, where c is M3M1.
rrusczyk 2011-03-19 21:50:46
rrusczyk 2011-03-19 21:50:51
Booyah!  Touchdown!  We make miracles happen!
jeff10 2011-03-19 21:51:06
negative number?
rrusczyk 2011-03-19 21:51:13
Yes, the angle was obtuse
rrusczyk 2011-03-19 21:51:31
Now, how do we finish?
carmelninja 2011-03-19 21:51:54
now find double angle
sindennisz 2011-03-19 21:51:54
square it?
dlennon 2011-03-19 21:51:54
Now find cos2<A3M3B1
calvinhobbesliker 2011-03-19 21:51:54
Double angle formula
VIPMaster 2011-03-19 21:51:56
plug everything back into our original cosine double angle formula equation
rrusczyk 2011-03-19 21:52:11
rrusczyk 2011-03-19 21:52:23
eb8368 2011-03-19 21:52:37
5-Root 32 5+32 = 37
ahaanomegas 2011-03-19 21:52:37
037 is what you would bubble in
GoldenFrog1618 2011-03-19 21:52:45
last one!
rrusczyk 2011-03-19 21:52:47
rrusczyk 2011-03-19 21:52:56
There are a ton of ways to do this one...
eb8368 2011-03-19 21:52:58
Vieta's + Bash !
mhy123 2011-03-19 21:53:00
Vieta's bash
rrusczyk 2011-03-19 21:53:07
rrusczyk 2011-03-19 21:53:14
Before we start substituting and cranking through algebra, does a quick glance at these equations tell you anything about a,b,c?
BarbieRocks 2011-03-19 21:53:38
one even, 2 odd
carmelninja 2011-03-19 21:53:38
two are odd, one is even
rrusczyk 2011-03-19 21:53:44
The first equation tells us that either all three of a,b,c are even, or one is even and the other two odd.  The second equation tells us that we cannot have all three even.  So, we know that one is even and the other two are odd.
rrusczyk 2011-03-19 21:53:48
We might substitute c = 2k at this point, but let's hold off on that, so as not to ruin our nice symmetry yet.
AlphaMath1 2011-03-19 21:53:55
a=-b-c
a+b=-c
BarbieRocks 2011-03-19 21:53:55
plug in c=-a-b and get a^2+ab+b^2=2011
rrusczyk 2011-03-19 21:54:06
About the only thing left to do is substitute, so we let a=-b-c in the second equation.
rrusczyk 2011-03-19 21:54:12
rrusczyk 2011-03-19 21:54:16
rrusczyk 2011-03-19 21:54:32
(I'm going quickly through this because I think a lot of you probably got that far that fast on the test.)
rrusczyk 2011-03-19 21:54:49
Now, you might just start guessing here, and you might get there pretty quick.
rrusczyk 2011-03-19 21:54:57
I'm going to talk a bit about how to narrow the search.
rrusczyk 2011-03-19 21:55:00
What kind of equation is each of these?
anonymous0 2011-03-19 21:55:12
VIPMaster 2011-03-19 21:55:12
prezcoin 2011-03-19 21:55:12
rrusczyk 2011-03-19 21:55:17
rrusczyk 2011-03-19 21:55:23
What can we do with quadratics?
gh625 2011-03-19 21:55:34
bulutcocuk 2011-03-19 21:55:34
tan90 2011-03-19 21:55:34
VIPMaster 2011-03-19 21:55:36
rrusczyk 2011-03-19 21:56:01
This can be a particularly useful tool in quadratic Diophantine equations (equations we must solve in integers)
bojobo 2011-03-19 21:56:13
we express one value in terms of the other
rrusczyk 2011-03-19 21:56:17
karatemagic7 2011-03-19 21:57:06
4*2011-3c^2 is a square.
.cpp 2011-03-19 21:57:07
(-c+-sqrt(c^2+8044-c^2))/2 = root.
calvinhobbesliker 2011-03-19 21:57:07
-c+-sqrt(c^2-4(c^2-2011))/2
gh625 2011-03-19 21:57:07
b=(-c+/-sqrt{-3c^2+8044})/2
rrusczyk 2011-03-19 21:57:09
JSGandora 2011-03-19 21:57:14
rrusczyk 2011-03-19 21:57:17
We could start cranking through trial-and-error to find a value of c that makes the discriminant a perfect square, but how can we make life a little easier?
GoldenFrog1618 2011-03-19 21:57:38
assume c is even
calvinhobbesliker 2011-03-19 21:57:38
C is even
VIPMaster 2011-03-19 21:57:38
c is even
karatemagic7 2011-03-19 21:57:39
Assume c is the even one
rrusczyk 2011-03-19 21:57:42
We remember that one of a,b,c is even, and that if we choose c to be the even one, that will simplify the radical.
rrusczyk 2011-03-19 21:57:48
rrusczyk 2011-03-19 21:58:04
Now life is a lot easier.  Can we make life even easier before we start guessing k's?
rrusczyk 2011-03-19 21:58:41
I'm trying to narrow how much I have to guess.  What mod is good for squares?
panjia123 2011-03-19 21:59:01
4
etude 2011-03-19 21:59:01
mod 4
BarbieRocks 2011-03-19 21:59:01
4
calvinhobbesliker 2011-03-19 21:59:01
4
vcez 2011-03-19 21:59:01
4
Jasmine8925 2011-03-19 21:59:01
4?
superrook 2011-03-19 21:59:01
4
NoWayHaze 2011-03-19 21:59:01
4
munygrubber 2011-03-19 21:59:01
4
JSGandora 2011-03-19 21:59:01
4
vcez 2011-03-19 21:59:01
mod 4
Kinata12 2011-03-19 21:59:01
mod 4
rrusczyk 2011-03-19 21:59:03
Squares must be 0 or 1 mod 4.
rrusczyk 2011-03-19 21:59:15
What do we know immediately about k, then?
karatemagic7 2011-03-19 21:59:39
k has to be odd in order for the discrimant to be 0 or 1  (mod 4)
fractals 2011-03-19 21:59:40
must be odd
VIPMaster 2011-03-19 21:59:40
its odd!
rrusczyk 2011-03-19 21:59:43
We notice that because 2011 is 3 mod 4, and no square is congruent to 3 mod 4, we cannot have k be even.
rrusczyk 2011-03-19 21:59:49
So, now we really have narrowed our search.  We want 2011-3k^2 to be a perfect square, and we know that k is odd.  Can we narrow it even more at a glance?
.cpp 2011-03-19 22:00:10
k has to be pretty small.
rrusczyk 2011-03-19 22:00:11
(I'm really just teaching you tricks of the trade here; we've already narrowed the search a lot.)
panjia123 2011-03-19 22:00:16
lower than 26
rrusczyk 2011-03-19 22:00:36
So, we don't have many to test.  We can just start going up through the odds.
rrusczyk 2011-03-19 22:00:44
We also know that k cannot end in 1 or 9, since either of these would have 2011-3k^2 ending in 8, and no square ends in 8.
rrusczyk 2011-03-19 22:00:47
All right, now we only have a small handful of options to check: odd numbers ending in 3,5, or 7, and we know k can't be 25 or larger, since 2011-3k^2 would then be negative.
eb8368 2011-03-19 22:00:59
k = 5 works
SDLucifer 2011-03-19 22:00:59
k=5
.cpp 2011-03-19 22:01:03
k = 5 works.
jeff10 2011-03-19 22:01:03
k=5
rrusczyk 2011-03-19 22:01:05
We start from the beginning, and see that 2011 - 3(5^2) = 1936 = 44^2.  Success!
rrusczyk 2011-03-19 22:01:16
And how do we finish?
fractals 2011-03-19 22:02:09
kangchangood 2011-03-19 22:02:09
from c= 10 just plug that in the other equations and get value of a and b
eb8368 2011-03-19 22:02:12
Find c , and plug it back into the original equation
calvinhobbesliker 2011-03-19 22:02:12
c is thus 10
gh625 2011-03-19 22:02:14
c=10, then solve for a and b
kangchangood 2011-03-19 22:02:18
plug and get the values
AlphaMath1 2011-03-19 22:02:20
c=2k=10
karatemagic7 2011-03-19 22:02:20
c=10, so b=39, so a=-49, so the sum of the abs. is 98.
.cpp 2011-03-19 22:02:22
Then, |a| + |b| + |c| = 39+49+10 = 098.
etude 2011-03-19 22:02:24
c = 10, b = 39, a = -49 by plug-and-chug. Thus 098.
rrusczyk 2011-03-19 22:02:31
rrusczyk 2011-03-19 22:02:36
policecap 2011-03-19 22:02:49
98
eb8368 2011-03-19 22:02:49
098
ahaanomegas 2011-03-19 22:02:49
danielguo94 2011-03-19 22:02:49
$098$
DystopianStriker 2011-03-19 22:02:49
ans is 098
rrusczyk 2011-03-19 22:03:03
calculatorwiz 2011-03-19 22:03:05
and that's how u ace an aime in 3  hours
rrusczyk 2011-03-19 22:03:12
Yay, we made it at the wire!!!
danielguo94 2011-03-19 22:03:16
WOOT
ahaanomegas 2011-03-19 22:03:27
WOOT!
Lemon123 2011-03-19 22:03:27
break included
eb8368 2011-03-19 22:03:27
YAY!
GoldenFrog1618 2011-03-19 22:03:27
thanks!
edisonchew240 2011-03-19 22:03:27
Woot
mathelete 2011-03-19 22:03:38
wow...that really was perfect timing
GoldenFrog1618 2011-03-19 22:03:39
any other solutions to 15?
calculusinmath 2011-03-19 22:03:39
Thank you
jeff10 2011-03-19 22:03:39
thank you Mr. Rusczyk
wow
JSGandora 2011-03-19 22:03:39
Thank you so much! :)
KeepingItReal 2011-03-19 22:03:39
Dang, that was amazing rrusczyk!
bojobo 2011-03-19 22:03:39
thank you
.cpp 2011-03-19 22:03:39
Thank you very much, Mr. Rusczyk!
DystopianStriker 2011-03-19 22:03:39
ty very much
AlphaMath1 2011-03-19 22:03:39
Are all the solutions on here similar to the official ones?
sincostanseccsccot 2011-03-19 22:03:39
thx
Lemon123 2011-03-19 22:03:39
thanks!
rrusczyk 2011-03-19 22:03:58
Best of luck to all of you who made the USA(J)MO, and good luck to those of you who are taking the AIME II
jsperling 2011-03-19 22:04:07
W00t
bojobo 2011-03-19 22:04:07
hope the next is easier
kelwenp123 2011-03-19 22:04:07
thanks!
jsperling 2011-03-19 22:04:07
Thank you Richard!
hrithikguy 2011-03-19 22:04:07
hi
rasgorial 2011-03-19 22:04:07
THANK YOU!
hrithikguy 2011-03-19 22:04:07
HI
sindennisz 2011-03-19 22:04:07
Thank you Mr. Rusczyk
hrithikguy 2011-03-19 22:04:07
THANKS
carmelninja 2011-03-19 22:04:07
thanks
jellymoop 2011-03-19 22:04:07
That was heartbreakingly beautiful....
dlennon 2011-03-19 22:04:07
Thanks!:bow:
panjia123 2011-03-19 22:04:07
thanks
eb8368 2011-03-19 22:04:07
Thanks!
centralbs 2011-03-19 22:04:07
Thanks!!!
hrithikguy 2011-03-19 22:04:07
Thank yoU!!
pi.guy3.14 2011-03-19 22:04:07
thank you!!!
gh625 2011-03-19 22:04:07
thank you
SDLucifer 2011-03-19 22:04:07
thank you!
chessmaster7 2011-03-19 22:04:07
thank you
apple.singer 2011-03-19 22:04:07
nice job guys
rdj5933mile5 2011-03-19 22:04:07
Thank you
anonymous0 2011-03-19 22:04:07
thanks you very much, sir!
rasgorial 2011-03-19 22:04:07
thank you thank you thank you
girishvar12 2011-03-19 22:04:07
thanks
sincostanseccsccot 2011-03-19 22:04:07
bye
edisonchew240 2011-03-19 22:04:07
Thanks man!
THANK YOU :bow:
policecap 2011-03-19 22:04:07
ttocs
apple.singer 2011-03-19 22:04:07
and thanks Mr. Ruscyzk
BarbieRocks 2011-03-19 22:04:12
im scared these problems end ened up asier than i thought--its just the order--now idk what the index is
rrusczyk 2011-03-19 22:04:24
Nah -- they only look easy because there are 200 of us working on them!
VIPMaster 2011-03-19 22:04:35
Thank you so much for your time Mr. Rusczyk. As always, ArtOfProblemSolving is my first source for everything math-related!
hrithikguy 2011-03-19 22:04:36
hi
Jasmine8925 2011-03-19 22:04:36
thank you!
inquisitivity 2011-03-19 22:04:36
Thank you!
july5117 2011-03-19 22:04:36
wow
ahaanomegas 2011-03-19 22:04:36
Thank you for the class, Mr. Rusczyk!
calculusinmath 2011-03-19 22:05:14
rrusczyk 2011-03-19 22:05:21
KeepingItReal 2011-03-19 22:05:24
This is my first time at one of these classes, does AOPS have many of these sessions?
rrusczyk 2011-03-19 22:05:33
We do all the AMC and AIME tests like this.
GoldenFrog1618 2011-03-19 22:05:35
Wow! you were able to attract 200 people to do math on a Saturday for 3 hours.
rrusczyk 2011-03-19 22:06:03
Yep.  If I could, I think I would invest in the futures of these 200 people over any other group of 200 people you might get online at one time.
AlphaMath1 2011-03-19 22:06:15
Are there any plans for USA(J)MO math jams?
rrusczyk 2011-03-19 22:06:39
We probably won't do that, in part because these things take forever to put together, and in part because the AMC publishes solutions the day after the test.
iamthygod!!!!!! 2011-03-19 22:06:42
My math teacher already ordered me copies of all your books a long time ago! :bow: They are all amazing!
rrusczyk 2011-03-19 22:06:48
You have a good teacher :)
rrusczyk 2011-03-19 22:07:26
All right, it's been three hours!  Thanks for coming.  The transcript will be up shortly, and the room will close in a minute.  It's time for dinner out here in California.  Math is hungry work!
rrusczyk 2011-03-19 22:08:35
There's the schedule for future Math Jams.