2009 Alternate AIME Math Jam
Go back to the Math Jam ArchiveAoPS instructors will lead a discussion on all 15 problems from the 2009 Alternate AIME.
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Facilitator: Richard Rusczyk
rrusczyk
2009-04-03 18:59:49
Hello and welcome to the 2009 AIME II Math Jam!
Hello and welcome to the 2009 AIME II Math Jam!
rrusczyk
2009-04-03 18:59:56
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
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
2009-04-03 19:00:04
The classroom is moderated, meaning that students can type into the classroom, but only the moderator can choose a comment to drop into the classroom. 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.
The classroom is moderated, meaning that students can type into the classroom, but only the moderator can choose a comment to drop into the classroom. 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
2009-04-03 19:00:17
There are a lot of students here! As I said, only 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!
There are a lot of students here! As I said, only 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
2009-04-03 19:00:32
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the prerequisite material for every problem as we go (and certainly not to 200+ students -- this Math Jam won't be as orderly or as thorough as our typical online classes.)! If you do not know a particular concept necessary for a problem, I probably will not be able to explain it to you tonight.
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 (and certainly not to 200+ students -- this Math Jam won't be as orderly or as thorough as our typical online classes.)! If you do not know a particular concept necessary for a problem, I probably will not be able to explain it to you tonight.
rrusczyk
2009-04-03 19:01:08
(I don't think we'll get 200+ students tonight like we had for the first AMC Math Jam. Phew.)
(I don't think we'll get 200+ students tonight like we had for the first AMC Math Jam. Phew.)
rrusczyk
2009-04-03 19:01:18
Please also remember that the purpose of this Math Jam is to work through the solutions to the AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. So please, when a problem is posted, do not simply respond with the answer. That's not why we're here. We're going to work through the problems step-by-step, and people who post comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, are going to be ignored.
Please also remember that the purpose of this Math Jam is to work through the solutions to the AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. So please, when a problem is posted, do not simply respond with the answer. That's not why we're here. We're going to work through the problems step-by-step, and people who post comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, are going to be ignored.
rrusczyk
2009-04-03 19:01:59
I don't claim that the solutions that we'll present today are the "definitive" or "best" solutions to each problem. Many of the problems have more than one plausible approach. We don't have time to consider every possible approach to every problem, but you can discuss other approaches on the Forum or post them in the AoPSWiki.
I don't claim that the solutions that we'll present today are the "definitive" or "best" solutions to each problem. Many of the problems have more than one plausible approach. We don't have time to consider every possible approach to every problem, but you can discuss other approaches on the Forum or post them in the AoPSWiki.
rrusczyk
2009-04-03 19:02:12
Our agenda is to work through all 15 problems on the 2009 AIME II, in order.
Our agenda is to work through all 15 problems on the 2009 AIME II, in order.
rrusczyk
2009-04-03 19:02:30
Let's get to the problems!
Let's get to the problems!
rrusczyk
2009-04-03 19:02:32
rrusczyk
2009-04-03 19:03:07
Here are a couple ways to tackle the problem:
Here are a couple ways to tackle the problem:
Recursive Acronym
2009-04-03 19:03:12
Since you need the white, red, and blue paint levels to be equal *before* painting the white, red, and blue stripes, you need to use up enough paint in the pink stripe to bring all three levels to 130.
Since you need the white, red, and blue paint levels to be equal *before* painting the white, red, and blue stripes, you need to use up enough paint in the pink stripe to bring all three levels to 130.
GoldenFrog1618
2009-04-03 19:03:12
let x be the amount of paint for each stripe
let x be the amount of paint for each stripe
Hullohihell0
2009-04-03 19:03:12
set up an equation
set up an equation
rrusczyk
2009-04-03 19:03:33
We could set up an equation and do some algebra. Or, we can reason our way to the answer pretty quickly.
We could set up an equation and do some algebra. Or, we can reason our way to the answer pretty quickly.
rrusczyk
2009-04-03 19:03:43
Let's look at Recursive's solution first:
Let's look at Recursive's solution first:
Recursive Acronym
2009-04-03 19:03:45
So you use 34 ounces of red and 58 ounces of white on the pink.
So you use 34 ounces of red and 58 ounces of white on the pink.
rrusczyk
2009-04-03 19:04:13
We use 164 - 130 = 34 more ounces of red and 188 - 130 = 58 more ounces of white paint in making the stripes, since in the end we must have the same amount of all three colors.
We use 164 - 130 = 34 more ounces of red and 188 - 130 = 58 more ounces of white paint in making the stripes, since in the end we must have the same amount of all three colors.
Recursive Acronym
2009-04-03 19:04:22
Which means 92 ounces a stripe.
Which means 92 ounces a stripe.
GoldenFrog1618
2009-04-03 19:04:22
92 ounces per stripe
92 ounces per stripe
rrusczyk
2009-04-03 19:04:34
So there's a total of 34 + 58 = 92 ounces in the pink stripe
So there's a total of 34 + 58 = 92 ounces in the pink stripe
tinytim
2009-04-03 19:04:38
Thus, you use 92 ounces of blue because the stripes are the same size
Thus, you use 92 ounces of blue because the stripes are the same size
Recursive Acronym
2009-04-03 19:05:00
So then you use 92 on the pink stripe, and are left with 130 of each, which means that after using 92 on Red, White, and Blue, you have 38 left on each stripe.
So then you use 92 on the pink stripe, and are left with 130 of each, which means that after using 92 on Red, White, and Blue, you have 38 left on each stripe.
rrusczyk
2009-04-03 19:05:03
Therefore, there are 92 ounces of paint in the blue stripe, leaving 130 - 92 = 38 ounces of blue paint.
Therefore, there are 92 ounces of paint in the blue stripe, leaving 130 - 92 = 38 ounces of blue paint.
GoldenFrog1618
2009-04-03 19:05:22
130*3-92*3=114
130*3-92*3=114
mathspee
2009-04-03 19:05:22
so there are 114 ounces left
so there are 114 ounces left
geoffreywong
2009-04-03 19:05:22
so subtract 130-92 and multiply by three
so subtract 130-92 and multiply by three
Recursive Acronym
2009-04-03 19:05:22
38 ounces left of each color times 3 colors equals 114 ounces left.
38 ounces left of each color times 3 colors equals 114 ounces left.
tinytim
2009-04-03 19:05:22
Which gives an answer of 38*3=114 ounces
Which gives an answer of 38*3=114 ounces
rrusczyk
2009-04-03 19:05:29
This means we have 3(38) = 114 ounces of paint left over.
This means we have 3(38) = 114 ounces of paint left over.
rrusczyk
2009-04-03 19:05:37
We also could have tackled this with algebra.
We also could have tackled this with algebra.
rrusczyk
2009-04-03 19:05:58
We let the amount of paint in each stripe be t.
We let the amount of paint in each stripe be t.
rrusczyk
2009-04-03 19:06:12
Then what?
Then what?
GoldenFrog1618
2009-04-03 19:06:52
(188-t)+(164-t)-t=2(130-t)
(188-t)+(164-t)-t=2(130-t)
tinytim
2009-04-03 19:07:04
2(180-t)=(164-t)+(188-t)-t
2(180-t)=(164-t)+(188-t)-t
rrusczyk
2009-04-03 19:07:08
Where does this come from?
Where does this come from?
tinytim
2009-04-03 19:07:18
130 I mean
130 I mean
tinytim
2009-04-03 19:07:38
We know that the same amount of each paint are left
We know that the same amount of each paint are left
rrusczyk
2009-04-03 19:07:42
Then, we have used t ounces of blue, so there is 130-t blue left.
Then, we have used t ounces of blue, so there is 130-t blue left.
chessmaster7
2009-04-03 19:07:45
the stripes are all the same size, so two stripes should equal two stripes
the stripes are all the same size, so two stripes should equal two stripes
rrusczyk
2009-04-03 19:07:49
We must also have 130-t red and 130-t white, for a total of 260-2t.
We must also have 130-t red and 130-t white, for a total of 260-2t.
GoldenFrog1618
2009-04-03 19:07:51
t ounces is removed from each stripe and another t ounces is removed from the red and white
t ounces is removed from each stripe and another t ounces is removed from the red and white
MathWhiz2009
2009-04-03 19:08:04
2(130-t)=188+164-t-t-t
2(130-t)=188+164-t-t-t
rrusczyk
2009-04-03 19:08:06
We started with 164 + 188 = 352 ounces of red and white combined, and used 3t total on the red, white, and pink stripes, so we have 352 - 3t left.
We started with 164 + 188 = 352 ounces of red and white combined, and used 3t total on the red, white, and pink stripes, so we have 352 - 3t left.
rrusczyk
2009-04-03 19:08:10
Therefore, we must have 352 - 3t = 260 - 2t.
Therefore, we must have 352 - 3t = 260 - 2t.
GoldenFrog1618
2009-04-03 19:08:28
t=92
t=92
tinytim
2009-04-03 19:08:28
so t=92 as we earlier found
so t=92 as we earlier found
mathq
2009-04-03 19:08:37
t=352-260=92
t=352-260=92
saszs
2009-04-03 19:08:40
t=92
t=92
chessmaster7
2009-04-03 19:08:49
this provides t=92, and 130*3-92*3=114
this provides t=92, and 130*3-92*3=114
MathWhiz2009
2009-04-03 19:08:49
then solve the equation to find t, then subtract 4t from 130+164+188
then solve the equation to find t, then subtract 4t from 130+164+188
GoldenFrog1618
2009-04-03 19:08:49
188+130+164-92*4=114
188+130+164-92*4=114
rrusczyk
2009-04-03 19:08:50
From here, we find t=92, just like before, and we have a total of 3(130-t) = 114 ounces left.
From here, we find t=92, just like before, and we have a total of 3(130-t) = 114 ounces left.
rrusczyk
2009-04-03 19:09:01
On to #2!
On to #2!
rrusczyk
2009-04-03 19:09:02
rrusczyk
2009-04-03 19:09:11
There sure doesn't look like any easy way to combine the three numbers we have to add, so let's see if we can compute one individually.
There sure doesn't look like any easy way to combine the three numbers we have to add, so let's see if we can compute one individually.
rrusczyk
2009-04-03 19:09:18
Recursive Acronym
2009-04-03 19:10:02
Rewrite x^(y^2) as (x^y)^y
Rewrite x^(y^2) as (x^y)^y
dgreenb801
2009-04-03 19:10:02
a^{x^2}=(a^x)^x
a^{x^2}=(a^x)^x
GoldenFrog1618
2009-04-03 19:10:02
Hyperboloid
2009-04-03 19:10:02
rrusczyk
2009-04-03 19:10:17
rrusczyk
2009-04-03 19:10:19
Then what?
Then what?
spelljack825
2009-04-03 19:10:47
substitute
substitute
mathq
2009-04-03 19:10:48
then it becomes 27^log7(base 3)
then it becomes 27^log7(base 3)
chessmaster7
2009-04-03 19:10:48
then substitute in 27 for 27^(log_3(7))
then substitute in 27 for 27^(log_3(7))
rrusczyk
2009-04-03 19:10:53
rrusczyk
2009-04-03 19:11:06
All we did here was substitute the given information.
All we did here was substitute the given information.
rrusczyk
2009-04-03 19:11:12
Can we simplify this further?
Can we simplify this further?
GoldenFrog1618
2009-04-03 19:11:34
27^(log_3 7)=7^3=343
27^(log_3 7)=7^3=343
spelljack825
2009-04-03 19:11:34
27 = 3^3
27 = 3^3
m1sterzer0
2009-04-03 19:11:34
27=3^3
27=3^3
tinytim
2009-04-03 19:11:34
3 to "something" equals 7. Thus, 27 to that "something equals 7^3=343.
3 to "something" equals 7. Thus, 27 to that "something equals 7^3=343.
maxschindler
2009-04-03 19:11:34
3^3log_3{7}
3^3log_3{7}
rrusczyk
2009-04-03 19:11:52
rrusczyk
2009-04-03 19:12:01
tinytim
2009-04-03 19:12:18
11^2
11^2
Hyperboloid
2009-04-03 19:12:18
121
121
GoldenFrog1618
2009-04-03 19:12:18
121
121
rrusczyk
2009-04-03 19:12:21
rrusczyk
2009-04-03 19:12:39
This is basically the same thing we did for the first one.
This is basically the same thing we did for the first one.
rrusczyk
2009-04-03 19:12:43
sj0201
2009-04-03 19:12:54
then 5
then 5
tinytim
2009-04-03 19:12:54
25^(1/2)=5
25^(1/2)=5
kohjhsd
2009-04-03 19:12:54
25^.5
25^.5
nechesskid13
2009-04-03 19:12:54
5
5
GoldenFrog1618
2009-04-03 19:12:54
e714212835
2009-04-03 19:12:54
25^1/2, or 5
25^1/2, or 5
Hyperboloid
2009-04-03 19:13:00
rrusczyk
2009-04-03 19:13:01
rrusczyk
2009-04-03 19:13:10
So what's our answer?
So what's our answer?
tinytim
2009-04-03 19:13:24
Adding these gives an answer of 469.
Adding these gives an answer of 469.
geoffreywong
2009-04-03 19:13:24
469
469
sj0201
2009-04-03 19:13:24
343+121+5 = 469
343+121+5 = 469
GoldenFrog1618
2009-04-03 19:13:24
469
469
Hyperboloid
2009-04-03 19:13:24
So the answer is 469.
So the answer is 469.
sherlock
2009-04-03 19:13:24
469
469
rrusczyk
2009-04-03 19:13:29
Adding our results gives 343 + 121 + 5 = 469.
Adding our results gives 343 + 121 + 5 = 469.
rrusczyk
2009-04-03 19:13:38
Two down. 13 to go.
Two down. 13 to go.
rrusczyk
2009-04-03 19:13:43
mathq
2009-04-03 19:13:55
make a diagram
make a diagram
sj0201
2009-04-03 19:13:57
diagram!!
diagram!!
Recursive Acronym
2009-04-03 19:13:59
Draw a picture!
Draw a picture!
rrusczyk
2009-04-03 19:14:04
rrusczyk
2009-04-03 19:14:17
What are we likely to use to solve this problem?
What are we likely to use to solve this problem?
dgreenb801
2009-04-03 19:14:24
similar triangles
similar triangles
tinytim
2009-04-03 19:14:24
Bunch of similar tirangles
Bunch of similar tirangles
Recursive Acronym
2009-04-03 19:14:24
So many similar right triangles...
So many similar right triangles...
rrusczyk
2009-04-03 19:14:30
We have lots of right triangles, and we have similar triangles all over the place.
We have lots of right triangles, and we have similar triangles all over the place.
rrusczyk
2009-04-03 19:14:47
We let AE = x so we can start building equations. (Notice that this is easier than letting AD = x, since we avoid fractions.)
We let AE = x so we can start building equations. (Notice that this is easier than letting AD = x, since we avoid fractions.)
rrusczyk
2009-04-03 19:14:55
What next?
What next?
all4math
2009-04-03 19:15:26
we use ratios
we use ratios
rrusczyk
2009-04-03 19:15:29
Which ones?
Which ones?
dgreenb801
2009-04-03 19:15:59
ABE and CBA are similar
ABE and CBA are similar
dgreenb801
2009-04-03 19:15:59
AE/AB=AB/AC
AE/AB=AB/AC
MathWhiz2009
2009-04-03 19:15:59
ABE~ABC
ABE~ABC
m1sterzer0
2009-04-03 19:15:59
AB/AE=BC/AC
AB/AE=BC/AC
motoe123
2009-04-03 19:15:59
AE:AB=AB:BC
AE:AB=AB:BC
rrusczyk
2009-04-03 19:16:11
Here's a reason we focus on these:
Here's a reason we focus on these:
Hyperboloid
2009-04-03 19:16:13
One involving AB, since we know its length.
One involving AB, since we know its length.
MathWhiz2009
2009-04-03 19:16:22
actually ABE~BCA
actually ABE~BCA
rrusczyk
2009-04-03 19:16:29
Since AE = x, we have BC = 2x.
Since AE = x, we have BC = 2x.
rrusczyk
2009-04-03 19:16:42
So what equation do we get?
So what equation do we get?
MathWhiz2009
2009-04-03 19:16:58
x/100=100/(2x)
x/100=100/(2x)
GoldenFrog1618
2009-04-03 19:16:58
2x^2=BA^2
2x^2=BA^2
sj0201
2009-04-03 19:16:58
2x/100 = 100/x
2x/100 = 100/x
mathq
2009-04-03 19:16:58
x:100=100:2x
x:100=100:2x
chessmaster7
2009-04-03 19:16:58
2x/100=100/x
2x/100=100/x
rrusczyk
2009-04-03 19:17:01
From ABC ~ EAB, we have AE/AB = AB/BC, so x/100 = 100/(2x).
From ABC ~ EAB, we have AE/AB = AB/BC, so x/100 = 100/(2x).
rrusczyk
2009-04-03 19:17:27
So?
So?
tinytim
2009-04-03 19:17:53
x=50sqrt{2}
x=50sqrt{2}
mismatchtea
2009-04-03 19:17:54
2x^2 = 10000
2x^2 = 10000
sj0201
2009-04-03 19:17:54
x^2 = 5000
x^2 = 5000
GoldenFrog1618
2009-04-03 19:17:54
x=50*sqrt(2)
x=50*sqrt(2)
sa314
2009-04-03 19:17:54
x^2 = 5000
x^2 = 5000
dgreenb801
2009-04-03 19:17:54
x=100/sqrt{2}=50sqrt{2}
x=100/sqrt{2}=50sqrt{2}
mathspee
2009-04-03 19:17:54
10000=2x^2
10000=2x^2
rrusczyk
2009-04-03 19:17:57
rrusczyk
2009-04-03 19:18:03
Keep going . . .
Keep going . . .
rrusczyk
2009-04-03 19:18:09
What is AD?
What is AD?
Recursive Acronym
2009-04-03 19:18:24
Double.
Double.
tinytim
2009-04-03 19:18:24
so 2x is 100sqrt{2}
so 2x is 100sqrt{2}
e^i_penguin
2009-04-03 19:18:24
100sqrt2
100sqrt2
mathspee
2009-04-03 19:18:24
2x=100 root2
2x=100 root2
GoldenFrog1618
2009-04-03 19:18:24
AD=100*sqrt(2)
AD=100*sqrt(2)
rrusczyk
2009-04-03 19:18:26
rrusczyk
2009-04-03 19:18:29
And what is our answer?
And what is our answer?
Yongyi781
2009-04-03 19:18:43
2x = 100sqrt(2) = 141
2x = 100sqrt(2) = 141
sj0201
2009-04-03 19:18:43
141
141
GoldenFrog1618
2009-04-03 19:18:43
141
141
chundai
2009-04-03 19:18:43
141.
141.
mathspee
2009-04-03 19:18:43
141
141
tinytim
2009-04-03 19:18:43
Hullohihell0
2009-04-03 19:18:43
141
141
rrusczyk
2009-04-03 19:18:46
rrusczyk
2009-04-03 19:18:56
Three down.
Three down.
rrusczyk
2009-04-03 19:18:59
rrusczyk
2009-04-03 19:19:34
Note: If you are here for the Intro Geometry class, please leave and then refresh the Classroom page. This is the AIME Math Jam.
Note: If you are here for the Intro Geometry class, please leave and then refresh the Classroom page. This is the AIME Math Jam.
rrusczyk
2009-04-03 19:19:40
We could just set this up like a typical arithmetic series problem. We haven + (n-2) + (n-4) + . . . + (n-2(m-1)) = 2009,where there are m children in the contest.
We could just set this up like a typical arithmetic series problem. We haven + (n-2) + (n-4) + . . . + (n-2(m-1)) = 2009,where there are m children in the contest.
rrusczyk
2009-04-03 19:19:44
Ick.
Ick.
rrusczyk
2009-04-03 19:19:50
Anyone have a nicer way to approach this?
Anyone have a nicer way to approach this?
rrusczyk
2009-04-03 19:20:29
What is typically an easier way to think about the sum of an arithmetic series?
What is typically an easier way to think about the sum of an arithmetic series?
Recursive Acronym
2009-04-03 19:21:01
First term plus last term divided by two times number of terms.
First term plus last term divided by two times number of terms.
chundai
2009-04-03 19:21:01
the average of the first and last terms multiplied by the number of terms?
the average of the first and last terms multiplied by the number of terms?
rrusczyk
2009-04-03 19:21:39
A nicer way of thinking about the sum of an arithmetic series is the average of the terms times the number of terms. Letting m be the number of students and t be the average, we now have simply mt = 2009.
A nicer way of thinking about the sum of an arithmetic series is the average of the terms times the number of terms. Letting m be the number of students and t be the average, we now have simply mt = 2009.
rrusczyk
2009-04-03 19:21:46
It appears that m and t must divide 2009, but first we must make sure both are integers. Are they?
It appears that m and t must divide 2009, but first we must make sure both are integers. Are they?
rrusczyk
2009-04-03 19:22:38
We have to make sure the *average*, t, is an integer.
We have to make sure the *average*, t, is an integer.
Hyperboloid
2009-04-03 19:23:04
Yes
Yes
GoldenFrog1618
2009-04-03 19:23:04
n and k must have the same parity
n and k must have the same parity
nechesskid13
2009-04-03 19:23:04
The numbers all have the same parity, so the average is an integer.
The numbers all have the same parity, so the average is an integer.
rrusczyk
2009-04-03 19:23:07
Yes -- the average of the terms equals the average of the first and last terms. These two terms have the same parity (both odd or both even), so their average is an integer.
Yes -- the average of the terms equals the average of the first and last terms. These two terms have the same parity (both odd or both even), so their average is an integer.
mismatchtea
2009-04-03 19:23:12
the prime factors of 2009 are 41 and 7 and 7
the prime factors of 2009 are 41 and 7 and 7
Hyperboloid
2009-04-03 19:23:12
rrusczyk
2009-04-03 19:23:18
rrusczyk
2009-04-03 19:23:22
We don't have many options for (m,t) to check. Do we have to try every factor of 2009 for m?
We don't have many options for (m,t) to check. Do we have to try every factor of 2009 for m?
rrusczyk
2009-04-03 19:24:06
Are there any we know we'll be able to exclude?
Are there any we know we'll be able to exclude?
mismatchtea
2009-04-03 19:24:09
only 3*2/2 possibilities
only 3*2/2 possibilities
rrusczyk
2009-04-03 19:24:20
Why do we only have to test half the possibilities for m?
Why do we only have to test half the possibilities for m?
rrusczyk
2009-04-03 19:24:46
m and t are not the same -- they mean different things.
m and t are not the same -- they mean different things.
BenM
2009-04-03 19:25:41
t must be greater than m
t must be greater than m
Hyperboloid
2009-04-03 19:25:41
m<t
m<t
m1sterzer0
2009-04-03 19:25:41
we can't have anyone having to eat 'negative grapes' to make the average too low
we can't have anyone having to eat 'negative grapes' to make the average too low
tinytim
2009-04-03 19:25:41
The average must be greater than the number of people.
The average must be greater than the number of people.
rrusczyk
2009-04-03 19:25:47
Students can't eat a negative number of grapes, so we only need to check those cases for which the last-place student eats a positive number of grapes.
Students can't eat a negative number of grapes, so we only need to check those cases for which the last-place student eats a positive number of grapes.
tinytim
2009-04-03 19:26:09
Thus, we only need to try 1,7,and 41 for m.
Thus, we only need to try 1,7,and 41 for m.
rrusczyk
2009-04-03 19:26:14
Those cases are (m,t) = (1,2009); (7, 287); (41,49). If (m,t) = (49, 41), then we have 49 kids eating an average of 41 grapes, so the last place kid eats 41 - 2(24) = -7 grapes. Ooops! Similarly, if we have even more kids and a lower average of grapes, we'll have a kid eating a negative number of grapes, which would not be pretty.
Those cases are (m,t) = (1,2009); (7, 287); (41,49). If (m,t) = (49, 41), then we have 49 kids eating an average of 41 grapes, so the last place kid eats 41 - 2(24) = -7 grapes. Ooops! Similarly, if we have even more kids and a lower average of grapes, we'll have a kid eating a negative number of grapes, which would not be pretty.
rrusczyk
2009-04-03 19:26:34
What do we get for n in each of these cases?
What do we get for n in each of these cases?
tinytim
2009-04-03 19:27:31
rrusczyk
2009-04-03 19:27:36
So, all that's left is to find the smallest n. If (m,t) = (1,2009), we have n = 2009.
So, all that's left is to find the smallest n. If (m,t) = (1,2009), we have n = 2009.
rrusczyk
2009-04-03 19:27:43
If (m,t) = (7,287), then 7 kids average 287 grapes, so n = 287 + 3*2 = 293.
If (m,t) = (7,287), then 7 kids average 287 grapes, so n = 287 + 3*2 = 293.
rrusczyk
2009-04-03 19:27:48
If (m,t) = (41,49), then 41 kids average 49 grapes, and n = 49 + 2*20 = 89.
If (m,t) = (41,49), then 41 kids average 49 grapes, and n = 49 + 2*20 = 89.
rrusczyk
2009-04-03 19:27:50
So. . . .
So. . . .
sucker0923
2009-04-03 19:28:03
smallest
smallest
Hullohihell0
2009-04-03 19:28:03
89 is the answer
89 is the answer
sucker0923
2009-04-03 19:28:03
=89
=89
lukezli
2009-04-03 19:28:03
89 is the answer.
89 is the answer.
rrusczyk
2009-04-03 19:28:14
The smallest of these is obviously 89, so that's our answer.
The smallest of these is obviously 89, so that's our answer.
rrusczyk
2009-04-03 19:28:38
Four down in a half hour. Still on track to get that perfect 15!
Four down in a half hour. Still on track to get that perfect 15!
rrusczyk
2009-04-03 19:28:43
rrusczyk
2009-04-03 19:28:57
rrusczyk
2009-04-03 19:29:12
Where do we start?
Where do we start?
Hullohihell0
2009-04-03 19:29:24
where's the triangle?
where's the triangle?
rrusczyk
2009-04-03 19:29:42
The triangle is formed by connecting the outer points of tangency.
The triangle is formed by connecting the outer points of tangency.
dgreenb801
2009-04-03 19:30:07
Canter of big circle, center of smaller circle, point of tangency collinear
Canter of big circle, center of smaller circle, point of tangency collinear
Recursive Acronym
2009-04-03 19:30:07
Draw a lot of lines.
Draw a lot of lines.
rrusczyk
2009-04-03 19:30:16
Let's play connect the dots.
Let's play connect the dots.
m1sterzer0
2009-04-03 19:30:18
usually i start by drawing a lot of lines
usually i start by drawing a lot of lines
rrusczyk
2009-04-03 19:30:30
What lines are the first ones to start with here?
What lines are the first ones to start with here?
tinytim
2009-04-03 19:30:56
AED(or AEC) is a good triangle to try and use.
AED(or AEC) is a good triangle to try and use.
chessmaster7
2009-04-03 19:30:57
AC, CD, AD, BA, BC, and BD
AC, CD, AD, BA, BC, and BD
GoldenFrog1618
2009-04-03 19:30:57
CE, DE, BE
CE, DE, BE
tinytim
2009-04-03 19:31:17
Connect centers of circles
Connect centers of circles
rrusczyk
2009-04-03 19:31:21
rrusczyk
2009-04-03 19:31:29
We definitely want to start by connecting up the centers.
We definitely want to start by connecting up the centers.
rrusczyk
2009-04-03 19:31:39
What's going to be our general strategy here?
What's going to be our general strategy here?
chessmaster7
2009-04-03 19:32:25
now we can write in lengths using the given radiuses, and maybe try pythagoras or trig
now we can write in lengths using the given radiuses, and maybe try pythagoras or trig
tinytim
2009-04-03 19:32:25
Find expressions for lengths in terms of r, the radius of E.
Find expressions for lengths in terms of r, the radius of E.
rrusczyk
2009-04-03 19:32:41
Our goal is to build an equation for r, the radius of E.
Our goal is to build an equation for r, the radius of E.
rrusczyk
2009-04-03 19:33:04
So, let's find some lengths.
So, let's find some lengths.
rrusczyk
2009-04-03 19:33:10
What lengths can we find?
What lengths can we find?
geoffreywong
2009-04-03 19:33:43
AC
AC
mismatchtea
2009-04-03 19:33:43
AC
AC
rrusczyk
2009-04-03 19:33:46
What is AC?
What is AC?
tinytim
2009-04-03 19:34:17
10-2=8
10-2=8
geoffreywong
2009-04-03 19:34:17
8
8
mismatchtea
2009-04-03 19:34:17
10-2=8
10-2=8
chessmaster7
2009-04-03 19:34:17
8
8
daydreamer1116
2009-04-03 19:34:22
10-2?
10-2?
rrusczyk
2009-04-03 19:34:48
We have AC = 10-2 = 8, since the radius of C is 2, and if we continue AC, it will hit the point of tangency of C and A.
We have AC = 10-2 = 8, since the radius of C is 2, and if we continue AC, it will hit the point of tangency of C and A.
daydreamer1116
2009-04-03 19:34:54
and AC=AD
and AC=AD
rrusczyk
2009-04-03 19:34:58
True.
True.
rrusczyk
2009-04-03 19:35:20
So, we have AC = 8. What other sides can we find (either find the length, or find them in terms of r).
So, we have AC = 8. What other sides can we find (either find the length, or find them in terms of r).
mismatchtea
2009-04-03 19:35:26
AB = 10-3=7
AB = 10-3=7
daydreamer1116
2009-04-03 19:35:26
and doesn't AB=10-3=7?
and doesn't AB=10-3=7?
rrusczyk
2009-04-03 19:35:37
Exactly. For the same reason, we have AB = 10 - 3 = 7.
Exactly. For the same reason, we have AB = 10 - 3 = 7.
GoldenFrog1618
2009-04-03 19:35:41
E is on line BA
E is on line BA
rrusczyk
2009-04-03 19:35:51
True -- by symmetry, E is on AB.
True -- by symmetry, E is on AB.
MathWhiz2009
2009-04-03 19:35:54
but how does this help us
but how does this help us
rrusczyk
2009-04-03 19:35:59
How does this help?
How does this help?
chessmaster7
2009-04-03 19:36:10
EC = r+2
EC = r+2
rrusczyk
2009-04-03 19:36:21
EC = r + 2, since it is the sum of the radii of E and C.
EC = r + 2, since it is the sum of the radii of E and C.
rrusczyk
2009-04-03 19:36:37
(Please use capital letters for points - I don't read the ones that have lowercase letters for points.)
(Please use capital letters for points - I don't read the ones that have lowercase letters for points.)
rrusczyk
2009-04-03 19:36:49
Are there any other lengths we can find?
Are there any other lengths we can find?
tinytim
2009-04-03 19:37:04
m1sterzer0
2009-04-03 19:37:04
EA = EB-AB = (r+3)-7=r-4
EA = EB-AB = (r+3)-7=r-4
GoldenFrog1618
2009-04-03 19:37:04
AE=r-4
AE=r-4
rrusczyk
2009-04-03 19:37:41
We have AE = r-4, since the distance from A to the point of tangency of A and B is 7-3 = 4.
We have AE = r-4, since the distance from A to the point of tangency of A and B is 7-3 = 4.
rrusczyk
2009-04-03 19:37:45
Now what?
Now what?
tinytim
2009-04-03 19:38:16
Now we can use Law of Cosines
Now we can use Law of Cosines
GoldenFrog1618
2009-04-03 19:38:17
EAD=60 degrees
EAD=60 degrees
dgreenb801
2009-04-03 19:38:17
EAD=60?
EAD=60?
tinytim
2009-04-03 19:38:20
We have enough information to use Law of Cosines.
We have enough information to use Law of Cosines.
m1sterzer0
2009-04-03 19:38:20
we know angle EAD is 60 degrees
we know angle EAD is 60 degrees
rrusczyk
2009-04-03 19:38:22
We have three sides of ACE: AC = 8, AE = r-4, CE = 2+r. . .
We have three sides of ACE: AC = 8, AE = r-4, CE = 2+r. . .
rrusczyk
2009-04-03 19:38:27
If we're stuck here, we ask ourselves what we haven't used yet. We haven't used the information about the equilateral triangle. The outer points of tangency on circle A are vertices of an equilateral triangle. The rays from A through B, C, and D will pass through these three points, so <BAC = <CAD = <DAB = 120. By symmetry, AE bisects <CAD, so <CAE = 60. Now, we're set.
If we're stuck here, we ask ourselves what we haven't used yet. We haven't used the information about the equilateral triangle. The outer points of tangency on circle A are vertices of an equilateral triangle. The rays from A through B, C, and D will pass through these three points, so <BAC = <CAD = <DAB = 120. By symmetry, AE bisects <CAD, so <CAE = 60. Now, we're set.
rrusczyk
2009-04-03 19:38:38
In triangle ACE, we have AC = 8, CE = 2 + r, AE = r - 4, and <CAE = 60.
In triangle ACE, we have AC = 8, CE = 2 + r, AE = r - 4, and <CAE = 60.
rrusczyk
2009-04-03 19:39:22
Now we break out the law of cosines.
Now we break out the law of cosines.
chessmaster7
2009-04-03 19:39:33
(r+2)^2=8^2+(r-4)^2-2(8)(r-4)*1/2
(r+2)^2=8^2+(r-4)^2-2(8)(r-4)*1/2
tinytim
2009-04-03 19:39:33
rrusczyk
2009-04-03 19:39:38
rrusczyk
2009-04-03 19:39:46
And what do we find?
And what do we find?
chessmaster7
2009-04-03 19:40:38
we get 20r=108
we get 20r=108
m1sterzer0
2009-04-03 19:40:38
r=108/20=27/5
r=108/20=27/5
charles.du
2009-04-03 19:40:38
27/5 so answer is 32
27/5 so answer is 32
rrusczyk
2009-04-03 19:40:43
tinytim
2009-04-03 19:41:14
or we could have Pythagoras bashed...
or we could have Pythagoras bashed...
rrusczyk
2009-04-03 19:41:25
You can indeed build some right triangles and bash away.
You can indeed build some right triangles and bash away.
charles.du
2009-04-03 19:41:30
yeah, extend BA to meet CD
yeah, extend BA to meet CD
rrusczyk
2009-04-03 19:41:39
Like that. You can work out that solution on your own.
Like that. You can work out that solution on your own.
rrusczyk
2009-04-03 19:41:45
For now, we're on to #6!!
For now, we're on to #6!!
HiDN428
2009-04-03 19:41:55
we would get 32 using the pythag as well
we would get 32 using the pythag as well
rrusczyk
2009-04-03 19:41:59
I sure hope so :)
I sure hope so :)
rrusczyk
2009-04-03 19:42:00
rrusczyk
2009-04-03 19:42:10
"At least" in a counting problem. What should that make us think?
"At least" in a counting problem. What should that make us think?
jeffreyyan8
2009-04-03 19:42:21
well right off the bat, i used PIE-principle of inclusion exclusion
well right off the bat, i used PIE-principle of inclusion exclusion
rrusczyk
2009-04-03 19:42:32
That's one thing I sometimes think, but usually I think this first:
That's one thing I sometimes think, but usually I think this first:
BenM
2009-04-03 19:42:42
find the opposite
find the opposite
spelljack825
2009-04-03 19:42:42
complement counting
complement counting
chundai
2009-04-03 19:42:42
complimentary counting!
complimentary counting!
tinytim
2009-04-03 19:42:42
Count the total number of 5-element subsets and then use complementary counting.
Count the total number of 5-element subsets and then use complementary counting.
GoldenFrog1618
2009-04-03 19:42:42
complimentary counting
complimentary counting
maxschindler
2009-04-03 19:42:42
complementary counting
complementary counting
rrusczyk
2009-04-03 19:42:49
Any time we see "at least" in a counting problem, we should consider counting what we don't want instead of directly counting what we want. (We at AoPS sometimes call this "complementary counting".)
Any time we see "at least" in a counting problem, we should consider counting what we don't want instead of directly counting what we want. (We at AoPS sometimes call this "complementary counting".)
rrusczyk
2009-04-03 19:43:04
Complimentary counting is nice, but it's complementary counting that we want to do here.
Complimentary counting is nice, but it's complementary counting that we want to do here.
GoldenFrog1618
2009-04-03 19:43:17
without restriction there are C(14,5)
without restriction there are C(14,5)
tinytim
2009-04-03 19:43:17
The total number of sets is easy. Simply 14C5=2002.
The total number of sets is easy. Simply 14C5=2002.
rrusczyk
2009-04-03 19:43:23
rrusczyk
2009-04-03 19:43:39
Next, we have to count the number of five-element subsets that don't satisfy the problem. In words, what are these?
Next, we have to count the number of five-element subsets that don't satisfy the problem. In words, what are these?
tinytim
2009-04-03 19:44:11
Now we think of how to "construct" a set such no two members are consecutive.
Now we think of how to "construct" a set such no two members are consecutive.
e714212835
2009-04-03 19:44:11
no two numbers are consecutive
no two numbers are consecutive
GoldenFrog1618
2009-04-03 19:44:11
no two numbers are consecutive
no two numbers are consecutive
Hullohihell0
2009-04-03 19:44:11
ones where no two numbers are consecutive
ones where no two numbers are consecutive
happysunnyshine
2009-04-03 19:44:11
no consecutive numbers
no consecutive numbers
rrusczyk
2009-04-03 19:44:15
These are the five-element subsets such that no two numbers are consecutive.
These are the five-element subsets such that no two numbers are consecutive.
rrusczyk
2009-04-03 19:44:21
How many of these are there?
How many of these are there?
Hyperboloid
2009-04-03 19:45:05
10C5 = 252
10C5 = 252
BenM
2009-04-03 19:45:05
10 chose 5
10 chose 5
lameneses
2009-04-03 19:45:05
rrusczyk
2009-04-03 19:45:15
qwang
2009-04-03 19:45:32
If we don't want any 5 numbers to be consecutive, consider 5 placemarkers that we place numbers around. for every arrangement of numbers around the placeholders, there's one set.
If we don't want any 5 numbers to be consecutive, consider 5 placemarkers that we place numbers around. for every arrangement of numbers around the placeholders, there's one set.
dgreenb801
2009-04-03 19:45:32
We draw a bijection: for each possibility, we can take out a number in between each of the selected numbers
We draw a bijection: for each possibility, we can take out a number in between each of the selected numbers
BenM
2009-04-03 19:45:32
because you insert 4 spaces between the numbers and 14-4=10
because you insert 4 spaces between the numbers and 14-4=10
rrusczyk
2009-04-03 19:45:44
I'm going to do a quick counting lesson to go through what these answers mean.
I'm going to do a quick counting lesson to go through what these answers mean.
rrusczyk
2009-04-03 19:45:49
There are a number of ways to see this.
There are a number of ways to see this.
rrusczyk
2009-04-03 19:45:59
We can view the first n integers as n chairs in a row. So, we have 14 chairs in a row, and we must pick 5 chairs such that no two chosen chairs are adjacent. Here's an example:
We can view the first n integers as n chairs in a row. So, we have 14 chairs in a row, and we must pick 5 chairs such that no two chosen chairs are adjacent. Here's an example:
rrusczyk
2009-04-03 19:46:03
OXXOXXXOXOXXOX
OXXOXXXOXOXXOX
rrusczyk
2009-04-03 19:46:07
The O's are the chosen chairs.
The O's are the chosen chairs.
tinytim
2009-04-03 19:46:11
Since no two are consective we can "take out" one number between each of the 5 numbers.
Since no two are consective we can "take out" one number between each of the 5 numbers.
rrusczyk
2009-04-03 19:46:15
If we take away one un-chosen chair between each successive pair of O's, we have
If we take away one un-chosen chair between each successive pair of O's, we have
rrusczyk
2009-04-03 19:46:19
OXOXXOOXOX
OXOXXOOXOX
rrusczyk
2009-04-03 19:46:33
We have 10 chairs now, and 5 of them are chosen. What happens when we reverse this process, and jam an extra X between each successive pair of O's?
We have 10 chairs now, and 5 of them are chosen. What happens when we reverse this process, and jam an extra X between each successive pair of O's?
spelljack825
2009-04-03 19:47:12
you get the original
you get the original
tinytim
2009-04-03 19:47:13
We get a set that has no two consecutive O's
We get a set that has no two consecutive O's
rrusczyk
2009-04-03 19:47:15
We get back the original arrangement of 14 chairs:
We get back the original arrangement of 14 chairs:
rrusczyk
2009-04-03 19:47:20
OXXOXXXOXOXXOX
OXXOXXXOXOXXOX
rrusczyk
2009-04-03 19:47:25
How does this help us count the number of ways we can choose 5 of the 14 chairs such that no two are adjacent?
How does this help us count the number of ways we can choose 5 of the 14 chairs such that no two are adjacent?
tinytim
2009-04-03 19:47:49
So this corresponds o arranging 5 x's and 5 o's which is 10C5.
So this corresponds o arranging 5 x's and 5 o's which is 10C5.
spelljack825
2009-04-03 19:47:49
the number of combinations is equivalent
the number of combinations is equivalent
rrusczyk
2009-04-03 19:48:02
For any such arrangement, we can take out 4 X's, one from between each successive pair of O's (which we can always do because no two chairs are adjacent!). We then get a row of 5 X's and 5 O's. We can reverse the process. For any row of 5 X's and 5 O's, we can add an X between each successive pair of O's and produce a row with 9 X's and 5 O's such that no two O's are adjacent (since we added an X between each successive pair).
For any such arrangement, we can take out 4 X's, one from between each successive pair of O's (which we can always do because no two chairs are adjacent!). We then get a row of 5 X's and 5 O's. We can reverse the process. For any row of 5 X's and 5 O's, we can add an X between each successive pair of O's and produce a row with 9 X's and 5 O's such that no two O's are adjacent (since we added an X between each successive pair).
rrusczyk
2009-04-03 19:48:16
Since each 14-letter string of 9 X's and 5 O's and no O's adjacent can be turned into a single 10-letter string of 5 X's and 5 O's, and vice-versa, we say that we have a 1-to-1 correspondence between the two sets of strings.
Since each 14-letter string of 9 X's and 5 O's and no O's adjacent can be turned into a single 10-letter string of 5 X's and 5 O's, and vice-versa, we say that we have a 1-to-1 correspondence between the two sets of strings.
chessmaster7
2009-04-03 19:48:23
so there are 10C5 possbilities on ways to be able to removes X's without removing O's, or in other words, having chairs that are not adjacent
so there are 10C5 possbilities on ways to be able to removes X's without removing O's, or in other words, having chairs that are not adjacent
rrusczyk
2009-04-03 19:48:30
Therefore, the number of 5-element subsets of the first 14 integers with no two consecutive integers equals the number of 5-element subsets of the first 10 integers.
Therefore, the number of 5-element subsets of the first 14 integers with no two consecutive integers equals the number of 5-element subsets of the first 10 integers.
rrusczyk
2009-04-03 19:48:42
So, how do we finish?
So, how do we finish?
Hyperboloid
2009-04-03 19:48:49
charles.du
2009-04-03 19:48:50
you just do 2002-252=1750
you just do 2002-252=1750
tinytim
2009-04-03 19:48:50
And the answer is 14C5-10C5=2002-252=1750=750 (mod 1000)
And the answer is 14C5-10C5=2002-252=1750=750 (mod 1000)
BenM
2009-04-03 19:48:50
so the answer is 14C5-10C5
so the answer is 14C5-10C5
Hyperboloid
2009-04-03 19:48:54
Subtract the complement form the total.
Subtract the complement form the total.
rrusczyk
2009-04-03 19:48:59
rrusczyk
2009-04-03 19:49:22
And that takes care of #6!!
And that takes care of #6!!
rrusczyk
2009-04-03 19:49:32
Speaking of !!, it's time for #7!!
Speaking of !!, it's time for #7!!
rrusczyk
2009-04-03 19:49:36
rrusczyk
2009-04-03 19:49:54
Icky notation. Whenever I see that on problem 6-10 on the AIME, my first thought is, "I'll bet this problem is way easier than it looks."
Icky notation. Whenever I see that on problem 6-10 on the AIME, my first thought is, "I'll bet this problem is way easier than it looks."
rrusczyk
2009-04-03 19:50:01
Let's see if I'm right. Where should we start?
Let's see if I'm right. Where should we start?
all4math
2009-04-03 19:50:09
what does that where e thing mean
what does that where e thing mean
rrusczyk
2009-04-03 19:50:13
It's a summation.
It's a summation.
rrusczyk
2009-04-03 19:50:33
It means you put in i=1, then i=2, then i=3, all the way up to 2009, and add up all the results you get.
It means you put in i=1, then i=2, then i=3, all the way up to 2009, and add up all the results you get.
rrusczyk
2009-04-03 19:50:50
What did you do when you first saw the problem?
What did you do when you first saw the problem?
rrusczyk
2009-04-03 19:51:01
Did you immediately start writing general formulas?
Did you immediately start writing general formulas?
Hullohihell0
2009-04-03 19:51:04
panic
panic
rrusczyk
2009-04-03 19:51:24
Gotta get over that! I used to, too, but then I realized the nasty notation problems are often easiest!
Gotta get over that! I used to, too, but then I realized the nasty notation problems are often easiest!
chessmaster7
2009-04-03 19:51:28
calculated the first few n's
calculated the first few n's
federernadal
2009-04-03 19:51:29
look for a pattern
look for a pattern
m1sterzer0
2009-04-03 19:51:29
started trying the first few terms to see if I saw a pattern
started trying the first few terms to see if I saw a pattern
BenM
2009-04-03 19:51:29
write a few terms
write a few terms
happysunnyshine
2009-04-03 19:51:29
try a few
try a few
rrusczyk
2009-04-03 19:51:33
This is what I usually do.
This is what I usually do.
rrusczyk
2009-04-03 19:51:41
Let's look at the first few terms, if only to get used to the notation. What is the first term?
Let's look at the first few terms, if only to get used to the notation. What is the first term?
tinytim
2009-04-03 19:51:54
1/2
1/2
nechesskid13
2009-04-03 19:51:54
1/2
1/2
rrusczyk
2009-04-03 19:51:58
rrusczyk
2009-04-03 19:52:00
Not too interesting. Next term?
Not too interesting. Next term?
BenM
2009-04-03 19:52:13
3/8
3/8
tinytim
2009-04-03 19:52:13
3/8
3/8
mathq
2009-04-03 19:52:13
3/8
3/8
jweisblat14
2009-04-03 19:52:13
3/8
3/8
rrusczyk
2009-04-03 19:52:16
rrusczyk
2009-04-03 19:52:17
Next?
Next?
rrusczyk
2009-04-03 19:52:43
Simplify your result!
Simplify your result!
GoldenFrog1618
2009-04-03 19:52:47
5/16
5/16
mathq
2009-04-03 19:52:47
5/16
5/16
jweisblat14
2009-04-03 19:52:47
5/16
5/16
nechesskid13
2009-04-03 19:52:47
15/48 = 5/16
15/48 = 5/16
rrusczyk
2009-04-03 19:52:50
rrusczyk
2009-04-03 19:53:06
Next?
Next?
geoffreywong
2009-04-03 19:53:45
35/128
35/128
jweisblat14
2009-04-03 19:53:45
35/128
35/128
e714212835
2009-04-03 19:53:45
35/128
35/128
late20s
2009-04-03 19:53:45
35/128
35/128
rrusczyk
2009-04-03 19:53:47
rrusczyk
2009-04-03 19:53:52
Are we ready to make any conjectures yet?
Are we ready to make any conjectures yet?
charles.du
2009-04-03 19:54:05
all the numerators are odd, and all the denominators are even
all the numerators are odd, and all the denominators are even
mismatchtea
2009-04-03 19:54:06
so the top appears to be consecutive odds, while the bottom is 2 raised to the nth power
so the top appears to be consecutive odds, while the bottom is 2 raised to the nth power
GoldenFrog1618
2009-04-03 19:54:06
denominator is always a power of 2
denominator is always a power of 2
GoldenFrog1618
2009-04-03 19:54:16
do you mean i=4 not 7?
do you mean i=4 not 7?
rrusczyk
2009-04-03 19:54:25
Yes, I meant i=4 for that last case.
Yes, I meant i=4 for that last case.
rrusczyk
2009-04-03 19:54:43
And i=2, i=3 for the previous 2.
And i=2, i=3 for the previous 2.
tinytim
2009-04-03 19:54:48
if the denominator is always a power of 2, then b=1
if the denominator is always a power of 2, then b=1
rrusczyk
2009-04-03 19:54:51
It sure looks like the denominator will always be a power of 2. Will the next one be a power of 2?
It sure looks like the denominator will always be a power of 2. Will the next one be a power of 2?
rrusczyk
2009-04-03 19:55:18
How can we easily see what the next one is?
How can we easily see what the next one is?
jweisblat14
2009-04-03 19:55:20
10 has a five
10 has a five
rrusczyk
2009-04-03 19:55:28
Yes it does, and what happens to that five?
Yes it does, and what happens to that five?
spelljack825
2009-04-03 19:55:33
yes, 5|10
yes, 5|10
charles.du
2009-04-03 19:55:33
but it cancels
but it cancels
e714212835
2009-04-03 19:55:33
we multiply the denomonator by 10, and the 5 gets cancelled
we multiply the denomonator by 10, and the 5 gets cancelled
rrusczyk
2009-04-03 19:55:37
Yes. The next term is 9/10 the previous term. The 5 in the 10 cancels with the 5 in the numerator.
Yes. The next term is 9/10 the previous term. The 5 in the 10 cancels with the 5 in the numerator.
rrusczyk
2009-04-03 19:55:42
Will this continue forever?
Will this continue forever?
GoldenFrog1618
2009-04-03 19:55:54
yes
yes
geoffreywong
2009-04-03 19:55:54
yes
yes
jweisblat14
2009-04-03 19:55:54
yes
yes
nechesskid13
2009-04-03 19:55:54
The numerator will simplify all odd factors in the denominator.
The numerator will simplify all odd factors in the denominator.
tinytim
2009-04-03 19:55:54
Yes.
Yes.
qwang
2009-04-03 19:56:18
the only way the denominator isn't a power of two is if it has 2*odd somewhere in there. however, the 'odd' will always cancel with one on top.
the only way the denominator isn't a power of two is if it has 2*odd somewhere in there. however, the 'odd' will always cancel with one on top.
rrusczyk
2009-04-03 19:56:28
A true observation and .. . .
A true observation and .. . .
tinytim
2009-04-03 19:56:30
Because this is the AIME we have no need to prove it.:-D
Because this is the AIME we have no need to prove it.:-D
rrusczyk
2009-04-03 19:56:42
If this were the USAMO, qwang's statement would need to be proved.
If this were the USAMO, qwang's statement would need to be proved.
rrusczyk
2009-04-03 19:57:01
It's not enough to say that for every 2n in the denom, there's an n in the numerator. What happens if n is even?
It's not enough to say that for every 2n in the denom, there's an n in the numerator. What happens if n is even?
rrusczyk
2009-04-03 19:57:10
We can get a sense for why it is true like this:
We can get a sense for why it is true like this:
rrusczyk
2009-04-03 19:57:16
rrusczyk
2009-04-03 19:57:20
For each odd integer k greater than 1, we have:
For each odd integer k greater than 1, we have:
rrusczyk
2009-04-03 19:57:23
k appears in the numerator before 2k appears in the denominator (since the denominator of the term with k in the numerator is just k+1).
k appears in the numerator before 2k appears in the denominator (since the denominator of the term with k in the numerator is just k+1).
rrusczyk
2009-04-03 19:57:27
3k appears in the numerator before 4k appears in the denominator.
3k appears in the numerator before 4k appears in the denominator.
rrusczyk
2009-04-03 19:57:30
5k appears in the numerator before 6k appears in the denominator
5k appears in the numerator before 6k appears in the denominator
rrusczyk
2009-04-03 19:57:34
And so on.
And so on.
rrusczyk
2009-04-03 19:57:38
This isn't perfectly rigorous, since more factors of k will appear in the coefficients of k in our argument. But then we can apply the same argument to those coefficients to take care of those factors, and so on.
This isn't perfectly rigorous, since more factors of k will appear in the coefficients of k in our argument. But then we can apply the same argument to those coefficients to take care of those factors, and so on.
rrusczyk
2009-04-03 19:58:01
I mention this argument because it is *exactly* the kind of convincing argument some of you will write on the USAMO, thinking you have a proof.
I mention this argument because it is *exactly* the kind of convincing argument some of you will write on the USAMO, thinking you have a proof.
rrusczyk
2009-04-03 19:58:10
And then you'll get a 0/7 and wonder what happened.
And then you'll get a 0/7 and wonder what happened.
rrusczyk
2009-04-03 19:58:47
Everything I wrote above is true, but the "and so on" part is not rigorous. We would have to prove that the "and so on" does continue forever.
Everything I wrote above is true, but the "and so on" part is not rigorous. We would have to prove that the "and so on" does continue forever.
charles.du
2009-04-03 19:58:49
wait so what gets you partial credit on USAMO
wait so what gets you partial credit on USAMO
rrusczyk
2009-04-03 19:59:03
It's very hard to get partial credit on the USAMO -- ask about it on the message board.
It's very hard to get partial credit on the USAMO -- ask about it on the message board.
rrusczyk
2009-04-03 19:59:14
Does anyone have a more rigorous explanation for why the odds always cancel out? (At this point, the observation above is enough to finish the problem, but let's really nail this one down by proving we're right for sure.)
Does anyone have a more rigorous explanation for why the odds always cancel out? (At this point, the observation above is enough to finish the problem, but let's really nail this one down by proving we're right for sure.)
GoldenFrog1618
2009-04-03 20:01:22
2^i*i!
2^i*i!
tinytim
2009-04-03 20:01:23
Start by noting (2i)!!=2^i(i!)... Is that still a typo...?
Start by noting (2i)!!=2^i(i!)... Is that still a typo...?
rrusczyk
2009-04-03 20:01:47
We have (2i)!! = 2^i * i!
We have (2i)!! = 2^i * i!
rrusczyk
2009-04-03 20:02:05
Can we relate (2i)!! and (2i-1)! to something that we are more comfortable with?
Can we relate (2i)!! and (2i-1)! to something that we are more comfortable with?
rrusczyk
2009-04-03 20:02:25
Sorry, that should be (2i)!! and (2i-1)!!
Sorry, that should be (2i)!! and (2i-1)!!
spelljack825
2009-04-03 20:02:27
factorials!
factorials!
rrusczyk
2009-04-03 20:02:37
And how can we relate these to factorials?
And how can we relate these to factorials?
spelljack825
2009-04-03 20:03:10
(2i)!! * (2i-1)!! = 2i!
(2i)!! * (2i-1)!! = 2i!
rrusczyk
2009-04-03 20:03:29
rrusczyk
2009-04-03 20:03:55
Where have you seen (2m)! and m! together before?
Where have you seen (2m)! and m! together before?
tinytim
2009-04-03 20:04:20
2mCm!
2mCm!
chessmaster7
2009-04-03 20:04:20
combinations?
combinations?
rrusczyk
2009-04-03 20:04:24
rrusczyk
2009-04-03 20:04:45
rrusczyk
2009-04-03 20:04:47
How else can we write that?
How else can we write that?
rrusczyk
2009-04-03 20:04:56
(Using our !! notation.)
(Using our !! notation.)
rrusczyk
2009-04-03 20:05:37
(Also, please write (2m)! rather than 2m!. There's a big difference!)
(Also, please write (2m)! rather than 2m!. There's a big difference!)
chundai
2009-04-03 20:05:39
why is (2i)!!*(2i-1)!!=2i!
why is (2i)!!*(2i-1)!!=2i!
rrusczyk
2009-04-03 20:06:07
The (2i)!! has all the evens and the (2i-1)!! has all the odds. Together, they have all the numbers from 1 to 2i. That's (2i)!
The (2i)!! has all the evens and the (2i-1)!! has all the odds. Together, they have all the numbers from 1 to 2i. That's (2i)!
rrusczyk
2009-04-03 20:06:31
Just above, we saw a way to write (2m)! in terms of !! -- what was that?
Just above, we saw a way to write (2m)! in terms of !! -- what was that?
happysunnyshine
2009-04-03 20:07:00
2m!!*(2m-1)!!
2m!!*(2m-1)!!
rrusczyk
2009-04-03 20:07:11
rrusczyk
2009-04-03 20:07:26
OK. Keep going. We saw a way to bring m! into the mix, too. What was that?
OK. Keep going. We saw a way to bring m! into the mix, too. What was that?
e714212835
2009-04-03 20:08:01
(2^m)(m!) = (2m)!!
(2^m)(m!) = (2m)!!
rrusczyk
2009-04-03 20:08:03
We had (2m)!! = 2^m * m!.
We had (2m)!! = 2^m * m!.
rrusczyk
2009-04-03 20:08:07
So...
So...
GoldenFrog1618
2009-04-03 20:08:08
2^m(2m-1)!!/m!
2^m(2m-1)!!/m!
rrusczyk
2009-04-03 20:08:13
rrusczyk
2009-04-03 20:08:17
How does this help us?
How does this help us?
rrusczyk
2009-04-03 20:09:06
What do we know is true about this number?
What do we know is true about this number?
late20s
2009-04-03 20:09:08
a integer
a integer
rrusczyk
2009-04-03 20:09:24
This number is an integer because all C(n,r) are integers. Why does that help?
This number is an integer because all C(n,r) are integers. Why does that help?
happysunnyshine
2009-04-03 20:09:26
why is (2^m)(m!) = (2m)!!?
why is (2^m)(m!) = (2m)!!?
rrusczyk
2009-04-03 20:10:00
(2m)!! = 2m * (2m-2) * . . . * 1 = 2^m[(m)*(m-1) *(m-2)* . . . *1]
(2m)!! = 2m * (2m-2) * . . . * 1 = 2^m[(m)*(m-1) *(m-2)* . . . *1]
rrusczyk
2009-04-03 20:10:41
How does the fact that this number is an integer help us? What does it tell us about (2m-1)!! and m!?
How does the fact that this number is an integer help us? What does it tell us about (2m-1)!! and m!?
geoffreywong
2009-04-03 20:11:24
they cancel out all odd numbers for m!
they cancel out all odd numbers for m!
charles.du
2009-04-03 20:11:44
we know m!|2^m(2m-1)!!
we know m!|2^m(2m-1)!!
rrusczyk
2009-04-03 20:11:55
Exactly. Because 2^m(2m-1)!! / m! is an integer, all the odd factors of m! cancel with (2m-1)!!.
Exactly. Because 2^m(2m-1)!! / m! is an integer, all the odd factors of m! cancel with (2m-1)!!.
rrusczyk
2009-04-03 20:12:16
Going back to (2m-1)!! / (2m)!!, what do we now know?
Going back to (2m-1)!! / (2m)!!, what do we now know?
tinytim
2009-04-03 20:13:04
Which leaves us with just powers of 2
Which leaves us with just powers of 2
tinytim
2009-04-03 20:13:04
that just equals (random stuff)/(2^i)
that just equals (random stuff)/(2^i)
GoldenFrog1618
2009-04-03 20:13:04
the odd factors cancel
the odd factors cancel
rrusczyk
2009-04-03 20:13:16
rrusczyk
2009-04-03 20:13:23
We now know that all the denominators are powers of 2, so all we have to do is find the highest one.
We now know that all the denominators are powers of 2, so all we have to do is find the highest one.
rrusczyk
2009-04-03 20:13:48
That's the 7/7 proof. Didn't need it for the AIME, but now you should remember that "and so on" won't get you far on the USAMO next month.
That's the 7/7 proof. Didn't need it for the AIME, but now you should remember that "and so on" won't get you far on the USAMO next month.
rrusczyk
2009-04-03 20:14:01
rrusczyk
2009-04-03 20:14:05
This is a problem we know how to handle. How do we do it?
This is a problem we know how to handle. How do we do it?
geoffreywong
2009-04-03 20:14:39
pull out the 2^i first
pull out the 2^i first
rrusczyk
2009-04-03 20:14:47
rrusczyk
2009-04-03 20:14:50
And how do we do that?
And how do we do that?
GoldenFrog1618
2009-04-03 20:15:21
[2009/2]+[2009/4]+[2009/8]+...
[2009/2]+[2009/4]+[2009/8]+...
kevmo314
2009-04-03 20:15:21
Count those divisible by 2, then 4, then 8, and so on, and add them up.
Count those divisible by 2, then 4, then 8, and so on, and add them up.
Hullohihell0
2009-04-03 20:15:21
count 2s then 4s then 8s and so on
count 2s then 4s then 8s and so on
spelljack825
2009-04-03 20:15:27
[2009/2] + [2009/2^2] +...
[2009/2] + [2009/2^2] +...
chessmaster7
2009-04-03 20:15:30
count factors of largest factor of 2 smaller than 2009, then 2nd largest factor, 3rd largest, and so on
count factors of largest factor of 2 smaller than 2009, then 2nd largest factor, 3rd largest, and so on
rrusczyk
2009-04-03 20:15:33
rrusczyk
2009-04-03 20:15:55
So. . .
So. . .
tinytim
2009-04-03 20:16:11
add 2009 to that
add 2009 to that
rrusczyk
2009-04-03 20:16:20
tinytim
2009-04-03 20:16:53
Which gives 2^{4010} in the denominator. So a/10=401
Which gives 2^{4010} in the denominator. So a/10=401
chessmaster7
2009-04-03 20:16:53
so a=4010 from the original question
so a=4010 from the original question
jweisblat14
2009-04-03 20:16:53
4001*1/10=401
4001*1/10=401
rrusczyk
2009-04-03 20:16:56
charles.du
2009-04-03 20:16:59
I just went ahead and found a for this problem to be 401 and realized that b had to be one, or ab/10 would be over 1000 :D
I just went ahead and found a for this problem to be 401 and realized that b had to be one, or ab/10 would be over 1000 :D
rrusczyk
2009-04-03 20:17:08
Funny -- that's a bit of a bug in the problem :)
Funny -- that's a bit of a bug in the problem :)
rrusczyk
2009-04-03 20:17:34
(This would have been a better USAMTS problem than an AIME problem -- you can get the answer without doing the most interesting part of the problem.)
(This would have been a better USAMTS problem than an AIME problem -- you can get the answer without doing the most interesting part of the problem.)
rrusczyk
2009-04-03 20:17:42
But now we're ready for number 8!
But now we're ready for number 8!
rrusczyk
2009-04-03 20:17:48
rrusczyk
2009-04-03 20:17:54
One approach is to set up a big infinite series indexed by the number of rolls it takes each person to roll a 6.
One approach is to set up a big infinite series indexed by the number of rolls it takes each person to roll a 6.
rrusczyk
2009-04-03 20:18:00
That seems ugly to me.
That seems ugly to me.
rrusczyk
2009-04-03 20:18:11
What's another way to approach the problem?
What's another way to approach the problem?
rrusczyk
2009-04-03 20:18:48
I see some pretty complicated suggestions!
I see some pretty complicated suggestions!
nechesskid13
2009-04-03 20:19:02
Geometric series
Geometric series
rrusczyk
2009-04-03 20:19:18
That will work, but let's see if we can find a simpler approach.
That will work, but let's see if we can find a simpler approach.
rrusczyk
2009-04-03 20:19:22
What can happen on the first roll?
What can happen on the first roll?
m1sterzer0
2009-04-03 20:19:29
recognize that if neither person rolls a 6 on the first roll, you are back to where you started
recognize that if neither person rolls a 6 on the first roll, you are back to where you started
rrusczyk
2009-04-03 20:19:35
Very interesting observation!
Very interesting observation!
rrusczyk
2009-04-03 20:19:38
Let's investigate.
Let's investigate.
rrusczyk
2009-04-03 20:19:54
So, what are the things that could happen on the first roll?
So, what are the things that could happen on the first roll?
charles.du
2009-04-03 20:19:58
6 or not 6 for either person
6 or not 6 for either person
tinytim
2009-04-03 20:20:05
6 or no 6
6 or no 6
Hullohihell0
2009-04-03 20:20:05
6 or not a 6
6 or not a 6
rrusczyk
2009-04-03 20:20:14
If they both get a 6 right way, they win.
If they both get a 6 right way, they win.
jweisblat14
2009-04-03 20:20:19
1/36 for both 6
1/36 for both 6
rrusczyk
2009-04-03 20:20:24
This occurs with probability 1/36.
This occurs with probability 1/36.
rrusczyk
2009-04-03 20:20:33
What about neither get 6?
What about neither get 6?
tinytim
2009-04-03 20:20:46
25/36
25/36
nechesskid13
2009-04-03 20:20:46
25/36
25/36
late20s
2009-04-03 20:20:46
25/36 for reatarting the game
25/36 for reatarting the game
Hullohihell0
2009-04-03 20:20:46
25/36
25/36
charles.du
2009-04-03 20:20:46
25/36
25/36
rrusczyk
2009-04-03 20:20:48
If they both don't get a 6, they'll have to roll again.
If they both don't get a 6, they'll have to roll again.
rrusczyk
2009-04-03 20:20:52
This occurs with probability 25/36.
This occurs with probability 25/36.
geoffreywong
2009-04-03 20:20:56
goes back to original setup
goes back to original setup
rrusczyk
2009-04-03 20:21:02
And then we're back where we started.
And then we're back where we started.
rrusczyk
2009-04-03 20:21:06
So the rest of the time, with probability 10/36, exactly one of them will roll a 6.
So the rest of the time, with probability 10/36, exactly one of them will roll a 6.
rrusczyk
2009-04-03 20:21:09
Then what happens?
Then what happens?
nechesskid13
2009-04-03 20:21:24
The other has to roll a six
The other has to roll a six
Hullohihell0
2009-04-03 20:21:24
the other has to roll
the other has to roll
deltaren
2009-04-03 20:21:24
the other has to roll a 6
the other has to roll a 6
rrusczyk
2009-04-03 20:21:27
They win if the other person rolls a 6 on their next roll.
They win if the other person rolls a 6 on their next roll.
chessmaster7
2009-04-03 20:21:33
then 10/36*1/6=10/216=5/108 for then to win
then 10/36*1/6=10/216=5/108 for then to win
rrusczyk
2009-04-03 20:21:54
This occurs with probability 1/6, so the probability is (10/36)(1/6) that they'll win this way.
This occurs with probability 1/6, so the probability is (10/36)(1/6) that they'll win this way.
rrusczyk
2009-04-03 20:22:09
How can we use all these observations *to finish the problem*.
How can we use all these observations *to finish the problem*.
tinytim
2009-04-03 20:22:37
Let x be the answer. x=1/36+10/36*1/6+25/36x
Let x be the answer. x=1/36+10/36*1/6+25/36x
sa314
2009-04-03 20:22:37
write an equation
write an equation
GoldenFrog1618
2009-04-03 20:22:37
5/108+1/6+25/36p=p
5/108+1/6+25/36p=p
rrusczyk
2009-04-03 20:22:43
Exactly.
Exactly.
rrusczyk
2009-04-03 20:22:49
We let p be the probability they win.
We let p be the probability they win.
rrusczyk
2009-04-03 20:23:21
We have p = (win on first roll) + (win on one rolling a 6 first and the other rolling a 6 second) + (win on neither rolling a 6 first).
We have p = (win on first roll) + (win on one rolling a 6 first and the other rolling a 6 second) + (win on neither rolling a 6 first).
rrusczyk
2009-04-03 20:23:25
In other words:
In other words:
rrusczyk
2009-04-03 20:23:32
rrusczyk
2009-04-03 20:23:40
The first term is "win on first roll"
The first term is "win on first roll"
rrusczyk
2009-04-03 20:23:58
The second term is when one rolls a 6 on the first turn and the other rolls a 6 on the next turn.
The second term is when one rolls a 6 on the first turn and the other rolls a 6 on the next turn.
rrusczyk
2009-04-03 20:24:10
Otherwise, with probability 25/36, they get to try again and win with probability p --- that's the third term.
Otherwise, with probability 25/36, they get to try again and win with probability p --- that's the third term.
rrusczyk
2009-04-03 20:24:39
We add up the probability of winning in each case, and we get the overall probability.
We add up the probability of winning in each case, and we get the overall probability.
rrusczyk
2009-04-03 20:24:46
So, we have that equation!
So, we have that equation!
rrusczyk
2009-04-03 20:24:52
Now we just have to solve for p.
Now we just have to solve for p.
chessmaster7
2009-04-03 20:25:25
we get 11p/36=8/108=4/54=2/27, so 72=297p and p=24/99=8/33
we get 11p/36=8/108=4/54=2/27, so 72=297p and p=24/99=8/33
tinytim
2009-04-03 20:25:25
11/36p=8/108 gives p=8/33
11/36p=8/108 gives p=8/33
rrusczyk
2009-04-03 20:25:32
Now it's just an algebra problem.
Now it's just an algebra problem.
rrusczyk
2009-04-03 20:25:38
Multiply through by 36 and collect terms:
Multiply through by 36 and collect terms:
rrusczyk
2009-04-03 20:25:43
rrusczyk
2009-04-03 20:25:46
rrusczyk
2009-04-03 20:25:48
Thus the answer is 8 + 33 = 041.
Thus the answer is 8 + 33 = 041.
mismatchtea
2009-04-03 20:25:51
wow, nice....just like solving an infinitely continuing fraction
wow, nice....just like solving an infinitely continuing fraction
rrusczyk
2009-04-03 20:26:16
We go through problems like these in our Intermediate Counting book.
We go through problems like these in our Intermediate Counting book.
m1sterzer0
2009-04-03 20:26:21
not bad --- beats pounding through the geometric series
not bad --- beats pounding through the geometric series
ThinkFlow
2009-04-03 20:26:21
So... when the problem can return to a previous state, we can usually do something like this?
So... when the problem can return to a previous state, we can usually do something like this?
rrusczyk
2009-04-03 20:26:39
Exactly -- we call the strategy we use for problems like this "state diagrams".
Exactly -- we call the strategy we use for problems like this "state diagrams".
asdfjklasdfjkl
2009-04-03 20:26:40
im buying that!
im buying that!
rrusczyk
2009-04-03 20:26:43
:)
:)
rrusczyk
2009-04-03 20:26:50
And now for number 9
And now for number 9
rrusczyk
2009-04-03 20:26:58
spelljack825
2009-04-03 20:27:25
bijection!
bijection!
dgreenb801
2009-04-03 20:27:25
relate the two
relate the two
rrusczyk
2009-04-03 20:27:34
How can we relate the two?
How can we relate the two?
Hyperboloid
2009-04-03 20:27:40
What is a bijection?
What is a bijection?
tinytim
2009-04-03 20:27:40
1-1 correspondence
1-1 correspondence
rrusczyk
2009-04-03 20:27:55
A bijection is a 1-1 correspondence, like we did in an earlier problem.
A bijection is a 1-1 correspondence, like we did in an earlier problem.
asdfjklasdfjkl
2009-04-03 20:27:57
notice that when y=3 or more its the same problem
notice that when y=3 or more its the same problem
Wikipedian1337
2009-04-03 20:27:58
In the second equation, if y > 3, there it's the same as the first equation.
In the second equation, if y > 3, there it's the same as the first equation.
ThinkFlow
2009-04-03 20:28:33
why?
why?
rrusczyk
2009-04-03 20:28:38
Good question -- why?
Good question -- why?
dgreenb801
2009-04-03 20:29:10
For each solution x,y,z in the second equation, there exists a solution x,y+3,z in the first
For each solution x,y,z in the second equation, there exists a solution x,y+3,z in the first
charles.du
2009-04-03 20:29:10
because if (x,y,z) solves the second one, (x,y+3,z) solves the first one
because if (x,y,z) solves the second one, (x,y+3,z) solves the first one
chessmaster7
2009-04-03 20:29:12
3*3 (which is y) yields 9, which can be added to 2000 to get 2009. but since y is not fixed, the equations become the same again
3*3 (which is y) yields 9, which can be added to 2000 to get 2009. but since y is not fixed, the equations become the same again
rrusczyk
2009-04-03 20:29:35
We can pair up solutions:(x,y+3,z) to the first <---> (x,y,z) to the second for y > 0.
We can pair up solutions:(x,y+3,z) to the first <---> (x,y,z) to the second for y > 0.
tinytim
2009-04-03 20:29:52
For EVERY soltuion to the second equation, we can add 3 to the y value to get a soltuion to the first eqaution. Thus we count the number of soltuions to the first equation that we CANT get back to the first equation by subtracting 3 from the y value.
For EVERY soltuion to the second equation, we can add 3 to the y value to get a soltuion to the first eqaution. Thus we count the number of soltuions to the first equation that we CANT get back to the first equation by subtracting 3 from the y value.
rrusczyk
2009-04-03 20:29:54
So we have a 1-1 correspondence of solutions:(x,y',z) to the first with y' > 3 <----> (x,y,z) to the secondwhere y'= y+3.
So we have a 1-1 correspondence of solutions:(x,y',z) to the first with y' > 3 <----> (x,y,z) to the secondwhere y'= y+3.
rrusczyk
2009-04-03 20:30:00
This pairs every solution of the second equation to a solution of the first equation with y>3, and vice versa.
This pairs every solution of the second equation to a solution of the first equation with y>3, and vice versa.
rrusczyk
2009-04-03 20:30:09
So what are the "left over" solutions to the first equation?
So what are the "left over" solutions to the first equation?
mathq
2009-04-03 20:30:22
so we solve for y=1,2,3
so we solve for y=1,2,3
ThinkFlow
2009-04-03 20:30:22
y<=3
y<=3
BenM
2009-04-03 20:30:22
y=1,2,3
y=1,2,3
rrusczyk
2009-04-03 20:30:36
We have the solutions where y is 1, 2 or 3.
We have the solutions where y is 1, 2 or 3.
rrusczyk
2009-04-03 20:30:57
(We have to include the y=3 case because we are interested in *positive* integer solutions, so we can't have y=0 in the second equation.)
(We have to include the y=3 case because we are interested in *positive* integer solutions, so we can't have y=0 in the second equation.)
charles.du
2009-04-03 20:31:03
and y cant be 2 because then the whole LHS would be even, and 2009 is odd
and y cant be 2 because then the whole LHS would be even, and 2009 is odd
funtwo
2009-04-03 20:31:03
y cant be 2 because of parity
y cant be 2 because of parity
rrusczyk
2009-04-03 20:31:07
Noting that y must be odd, we see we must have y=1 or y=3.
Noting that y must be odd, we see we must have y=1 or y=3.
rrusczyk
2009-04-03 20:31:15
So, now we have a much simpler problem.
So, now we have a much simpler problem.
tinytim
2009-04-03 20:31:26
If y=1 then we must find the number of solutions to 2(x+2z)=2009-3 or x+2z=1003
If y=1 then we must find the number of solutions to 2(x+2z)=2009-3 or x+2z=1003
rrusczyk
2009-04-03 20:31:49
rrusczyk
2009-04-03 20:32:22
How do we count the solutions to 4x + 2z = 2006? (We can divide by 2 to get 2x + z = 1003.)
How do we count the solutions to 4x + 2z = 2006? (We can divide by 2 to get 2x + z = 1003.)
BenM
2009-04-03 20:32:55
x can range from 1 to 501 and there is a unique z for each x
x can range from 1 to 501 and there is a unique z for each x
mathq
2009-04-03 20:32:55
total 501 solutions
total 501 solutions
FantasyLover
2009-04-03 20:32:55
z=1, 3, ..., 1001
z=1, 3, ..., 1001
charles.du
2009-04-03 20:33:03
everything from (501,1) to (1,1001) works, so we have 501 solutions
everything from (501,1) to (1,1001) works, so we have 501 solutions
rrusczyk
2009-04-03 20:33:05
We have z = 1003-2x, so x can be anything from 1 to 501.
We have z = 1003-2x, so x can be anything from 1 to 501.
rrusczyk
2009-04-03 20:33:19
And for 4x + 2z = 2000 (otherwise known as 2x + z = 1000)?
And for 4x + 2z = 2000 (otherwise known as 2x + z = 1000)?
tinytim
2009-04-03 20:33:51
Similarily for 2x+z=1000, x can be anything from 1 to 499.
Similarily for 2x+z=1000, x can be anything from 1 to 499.
funtwo
2009-04-03 20:33:51
x=1-499 because z must be positive
x=1-499 because z must be positive
BenM
2009-04-03 20:33:51
x ranges from 1 to 499
x ranges from 1 to 499
e714212835
2009-04-03 20:33:51
z=1000 - 2x, so x can be anything from 1 to 499
z=1000 - 2x, so x can be anything from 1 to 499
mathq
2009-04-03 20:33:51
(1,998) to (499,2), 499 solutions
(1,998) to (499,2), 499 solutions
rrusczyk
2009-04-03 20:33:53
We have z = 1000 - 2x, so x can be anything from 1 to 499.
We have z = 1000 - 2x, so x can be anything from 1 to 499.
rrusczyk
2009-04-03 20:34:05
So /. . . .
So /. . . .
jweisblat14
2009-04-03 20:34:09
1000
1000
mathq
2009-04-03 20:34:09
501+499=1000
501+499=1000
charles.du
2009-04-03 20:34:11
our answer is 499+501/1000 = OMG 000
our answer is 499+501/1000 = OMG 000
mathq
2009-04-03 20:34:11
sothe answer is 0
sothe answer is 0
rrusczyk
2009-04-03 20:34:14
So we have a total of 501 + 499 = 1000 solutions.
So we have a total of 501 + 499 = 1000 solutions.
tinytim
2009-04-03 20:34:17
Adding the number of solutions to the two equations gives 1000. THis means that m-n=1000 and the answer is 000!:-o
Adding the number of solutions to the two equations gives 1000. THis means that m-n=1000 and the answer is 000!:-o
BenM
2009-04-03 20:34:27
000 mod 1000
000 mod 1000
jweisblat14
2009-04-03 20:34:27
answer = 000
answer = 000
spelljack825
2009-04-03 20:34:27
499 + 501 = 0 (mod 1000) :-c
499 + 501 = 0 (mod 1000) :-c
rrusczyk
2009-04-03 20:34:28
The answer is thus 000.
The answer is thus 000.
mathq
2009-04-03 20:34:43
rare to get 0 for the answer...
rare to get 0 for the answer...
rrusczyk
2009-04-03 20:34:58
It happens more than average. It has happened twice.
It happens more than average. It has happened twice.
Smartguy
2009-04-03 20:35:20
wasn't there one in 2000?
wasn't there one in 2000?
tinytim
2009-04-03 20:35:20
Once on the 2000 AIME i believe
Once on the 2000 AIME i believe
rrusczyk
2009-04-03 20:35:24
One more problem, and then I'm going to take a short break.
One more problem, and then I'm going to take a short break.
rrusczyk
2009-04-03 20:35:29
rrusczyk
2009-04-03 20:35:38
http://www.artofproblemsolving.com/Forum/latexrender/pictures/c/4/2/c42a1c3234061db99aa6fa3e1cc04816e8033176.gif
http://www.artofproblemsolving.com/Forum/latexrender/pictures/c/4/2/c42a1c3234061db99aa6fa3e1cc04816e8033176.gif
rrusczyk
2009-04-03 20:35:52
Not exactly an appealing problem with all those words.
Not exactly an appealing problem with all those words.
geoffreywong
2009-04-03 20:35:55
diagram?
diagram?
e^i_penguin
2009-04-03 20:35:55
draw a picture
draw a picture
Hullohihell0
2009-04-03 20:35:55
draw a diagram
draw a diagram
rrusczyk
2009-04-03 20:36:01
A lot of words for a pretty simple diagram. Of course, I drew it wrong the first time I saw the problem.
A lot of words for a pretty simple diagram. Of course, I drew it wrong the first time I saw the problem.
rrusczyk
2009-04-03 20:36:04
In the interests of not being here all night, here's the correct diagram:
In the interests of not being here all night, here's the correct diagram:
rrusczyk
2009-04-03 20:36:09
mathq
2009-04-03 20:36:12
ABC makes an right triangle
ABC makes an right triangle
rrusczyk
2009-04-03 20:36:24
We could do this with analytic geometry and/or bash it out with trig, but let's see if we can find a nicer solution without those crutches.
We could do this with analytic geometry and/or bash it out with trig, but let's see if we can find a nicer solution without those crutches.
rrusczyk
2009-04-03 20:36:32
We know that AB = 5, BC = 12, AC = 13, and the angles are marked as in the diagram. What else can we determine?
We know that AB = 5, BC = 12, AC = 13, and the angles are marked as in the diagram. What else can we determine?
tinytim
2009-04-03 20:37:21
Extend DC so it meets line AB at sy E.
Extend DC so it meets line AB at sy E.
charles.du
2009-04-03 20:37:21
extend CD to AB to get a nice isosceles triangle
extend CD to AB to get a nice isosceles triangle
wangsacl
2009-04-03 20:37:21
I prefer to intersect the extension of both AB and CD
I prefer to intersect the extension of both AB and CD
rrusczyk
2009-04-03 20:38:33
Interesting observation -- write me a full solution with that and I'll post it at the end :)
Interesting observation -- write me a full solution with that and I'll post it at the end :)
rrusczyk
2009-04-03 20:38:43
I see many of you suggesting this:
I see many of you suggesting this:
GoldenFrog1618
2009-04-03 20:38:48
angle bisector
angle bisector
Hullohihell0
2009-04-03 20:38:48
well you could use angle bisector theorem possibly
well you could use angle bisector theorem possibly
Wikipedian1337
2009-04-03 20:38:48
angle bisector theorem
angle bisector theorem
rrusczyk
2009-04-03 20:38:58
We have an angle bisector of an angle in triangle ABC.
We have an angle bisector of an angle in triangle ABC.
rrusczyk
2009-04-03 20:39:02
rrusczyk
2009-04-03 20:39:10
Applying the Angle Bisector Theorem to angle bisector AF in triangle ABC, we have BF/CF = AB/AC. So, what are BF and CF?
Applying the Angle Bisector Theorem to angle bisector AF in triangle ABC, we have BF/CF = AB/AC. So, what are BF and CF?
charles.du
2009-04-03 20:39:50
10/3 and 26/3
10/3 and 26/3
chessmaster7
2009-04-03 20:39:56
10/3 = BF and 26/3 = CF
10/3 = BF and 26/3 = CF
rrusczyk
2009-04-03 20:40:04
Since AB is 5/18 of AB+AC, we know that BF is 5/18 of BF + FC, which is 12. Therefore, BF = 12*(5/18) = 10/3, and FC = 12 - BF= 26/3.
Since AB is 5/18 of AB+AC, we know that BF is 5/18 of BF + FC, which is 12. Therefore, BF = 12*(5/18) = 10/3, and FC = 12 - BF= 26/3.
rrusczyk
2009-04-03 20:40:18
Now what might we be on the lookout for?
Now what might we be on the lookout for?
ThinkFlow
2009-04-03 20:40:33
similar triangles
similar triangles
tinytim
2009-04-03 20:40:33
Similar Tiangles
Similar Tiangles
chessmaster7
2009-04-03 20:40:38
similar triangles?
similar triangles?
rrusczyk
2009-04-03 20:40:40
We have a right angle. We have equal angles. We should look for similar triangles. Do we have any?
We have a right angle. We have equal angles. We should look for similar triangles. Do we have any?
chessmaster7
2009-04-03 20:41:07
not yet...
not yet...
rrusczyk
2009-04-03 20:41:11
That's the spirit.
That's the spirit.
rrusczyk
2009-04-03 20:41:15
not *yet*
not *yet*
billy314
2009-04-03 20:41:20
we could draw some auxillary lines
we could draw some auxillary lines
happysunnyshine
2009-04-03 20:41:21
No, but we can create
No, but we can create
ThinkFlow
2009-04-03 20:41:26
Draw some
Draw some
rrusczyk
2009-04-03 20:41:30
How can we make some easily?
How can we make some easily?
rrusczyk
2009-04-03 20:42:06
(I think there are a lot of ways to finish from here.)
(I think there are a lot of ways to finish from here.)
billy314
2009-04-03 20:42:09
Draw the segment from D parallel to segment BA
Draw the segment from D parallel to segment BA
ThinkFlow
2009-04-03 20:42:10
Draw D perpendicular to BC
Draw D perpendicular to BC
rrusczyk
2009-04-03 20:42:13
We draw the altitude from D to BC.
We draw the altitude from D to BC.
rrusczyk
2009-04-03 20:42:17
rrusczyk
2009-04-03 20:42:26
Now what?
Now what?
tinytim
2009-04-03 20:42:45
ABF is similar to DHF
ABF is similar to DHF
chessmaster7
2009-04-03 20:42:45
find DF and FA using similar triangles and we have answer!
find DF and FA using similar triangles and we have answer!
funtwo
2009-04-03 20:42:45
BAF~HDF
BAF~HDF
ThinkFlow
2009-04-03 20:42:50
We find the ratio of similarity, and we can find via pythag AF, so we find our answer
We find the ratio of similarity, and we can find via pythag AF, so we find our answer
rrusczyk
2009-04-03 20:42:57
We now have a problem we know how to handle.
We now have a problem we know how to handle.
rrusczyk
2009-04-03 20:43:02
Let's knock off the details.
Let's knock off the details.
rrusczyk
2009-04-03 20:43:05
We have DHF ~ ABF.
We have DHF ~ ABF.
rrusczyk
2009-04-03 20:43:07
What else?
What else?
rrusczyk
2009-04-03 20:43:15
(Any other similarities?)
(Any other similarities?)
chessmaster7
2009-04-03 20:43:28
CHD and CBA
CHD and CBA
ThinkFlow
2009-04-03 20:43:28
DHC similar to ABC
DHC similar to ABC
funtwo
2009-04-03 20:43:28
CDH and CAB
CDH and CAB
rrusczyk
2009-04-03 20:43:30
We also have DHC ~ ABC.
We also have DHC ~ ABC.
rrusczyk
2009-04-03 20:43:50
OK, let's get to work with these. How? What's our first step in setting up equations?
OK, let's get to work with these. How? What's our first step in setting up equations?
chessmaster7
2009-04-03 20:44:19
we need to find DH?
we need to find DH?
rrusczyk
2009-04-03 20:44:26
Sure. Let's call that x.
Sure. Let's call that x.
rrusczyk
2009-04-03 20:44:38
What else do we have?
What else do we have?
mathq
2009-04-03 20:45:28
CH:CB=x:AB which equals CH:12=x:5
CH:CB=x:AB which equals CH:12=x:5
ThinkFlow
2009-04-03 20:45:28
FH = 2/3x
FH = 2/3x
GoldenFrog1618
2009-04-03 20:45:28
x/5=CH/12
x/5=CH/12
rrusczyk
2009-04-03 20:45:35
From DHF ~ ABF, we have FH/DH = BF/AB = 2/3, so FH = 2x/3.
From DHF ~ ABF, we have FH/DH = BF/AB = 2/3, so FH = 2x/3.
rrusczyk
2009-04-03 20:45:41
From DHC ~ ABC, we have CH/DH = BC/AB = 12/5, so CH = 12x/5.
From DHC ~ ABC, we have CH/DH = BC/AB = 12/5, so CH = 12x/5.
rrusczyk
2009-04-03 20:46:07
And what do we do with this?
And what do we do with this?
ThinkFlow
2009-04-03 20:46:15
FH + HC = 26/3
FH + HC = 26/3
tinytim
2009-04-03 20:46:49
so x=65/23
so x=65/23
rrusczyk
2009-04-03 20:46:57
rrusczyk
2009-04-03 20:47:03
Now, we're set. How do we find AD?
Now, we're set. How do we find AD?
mismatchtea
2009-04-03 20:47:43
pythagorean's theorem?
pythagorean's theorem?
ThinkFlow
2009-04-03 20:47:43
pythag
pythag
GoldenFrog1618
2009-04-03 20:47:43
sqrt((AB+HD)^2+(BH)^2)
sqrt((AB+HD)^2+(BH)^2)
rrusczyk
2009-04-03 20:47:52
rrusczyk
2009-04-03 20:47:57
rrusczyk
2009-04-03 20:48:03
We could pound this out with the Pythagorean Theorem now. Anyone see an easier way to do it?
We could pound this out with the Pythagorean Theorem now. Anyone see an easier way to do it?
tinytim
2009-04-03 20:48:25
continue using similar traingles
continue using similar traingles
chessmaster7
2009-04-03 20:48:25
similar triangles again? :)
similar triangles again? :)
rrusczyk
2009-04-03 20:48:35
Yep. Similar triangles to the rescue again. DXA ~ FBA.
Yep. Similar triangles to the rescue again. DXA ~ FBA.
rrusczyk
2009-04-03 20:48:45
How does that get us to the answer relatively quickly?
How does that get us to the answer relatively quickly?
ThinkFlow
2009-04-03 20:49:12
pythag on ABF
pythag on ABF
e714212835
2009-04-03 20:49:19
BA and BF are nice numbers, so we can easily find AF, and use ratios
BA and BF are nice numbers, so we can easily find AF, and use ratios
rrusczyk
2009-04-03 20:49:26
rrusczyk
2009-04-03 20:49:46
Now it's just arithmetic.
Now it's just arithmetic.
rrusczyk
2009-04-03 20:50:25
We could do it this way:
We could do it this way:
chessmaster7
2009-04-03 20:50:27
do pythagoras to find AF then multiply by AX or AB
do pythagoras to find AF then multiply by AX or AB
ThinkFlow
2009-04-03 20:50:27
AF = 5\sqrt{13}/3
AF = 5\sqrt{13}/3
rrusczyk
2009-04-03 20:50:31
Or this way:
Or this way:
rrusczyk
2009-04-03 20:50:35
rrusczyk
2009-04-03 20:50:59
Here are solutions with the "extend a side" approach:
Here are solutions with the "extend a side" approach:
wangsacl
2009-04-03 20:51:03
Extend AB and CD, so both of them intersect at E. We get AE=10, CE=13, AC=13. By applying the angle bisector theorem: DE=130/23, DC=169/23. Then, we could compute AD using this formula: AD^2=AC x AE - CD x ED. Then, we get: (let's skip the rigorous part): AD=60\sqrt{13}/23. Then, we get p+q+r=096
Extend AB and CD, so both of them intersect at E. We get AE=10, CE=13, AC=13. By applying the angle bisector theorem: DE=130/23, DC=169/23. Then, we could compute AD using this formula: AD^2=AC x AE - CD x ED. Then, we get: (let's skip the rigorous part): AD=60\sqrt{13}/23. Then, we get p+q+r=096
wangsacl
2009-04-03 20:51:03
you could derive that from stewart's theorem
you could derive that from stewart's theorem
dgreenb801
2009-04-03 20:51:03
Full solution with extension: Let AB and CD intersect at A'. We have, A'B=5 and A'C=13, by the angle bisector theorem we have CD=169/23 and DA'=130/23. Now Stewart's on triangle ABC finds AD (a little messy but not too bad) is 60sqrt{13}/23.
Full solution with extension: Let AB and CD intersect at A'. We have, A'B=5 and A'C=13, by the angle bisector theorem we have CD=169/23 and DA'=130/23. Now Stewart's on triangle ABC finds AD (a little messy but not too bad) is 60sqrt{13}/23.
rrusczyk
2009-04-03 20:51:25
(You can see why I didn't continue with that approach -- looking for a solution without the heavy machinery of Stewart.)
(You can see why I didn't continue with that approach -- looking for a solution without the heavy machinery of Stewart.)
Wikipedian1337
2009-04-03 20:51:28
I did the problem using coordinates and found it much more straightforward than pure geometry.
I did the problem using coordinates and found it much more straightforward than pure geometry.
rrusczyk
2009-04-03 20:51:40
There's another way to approach the problem.
There's another way to approach the problem.
chessmaster7
2009-04-03 20:51:42
theres another (more complicated) way to extend CD to meet AB, then inscribe a circle and find the radius to be 10/3 (which is BF) and find AD from finding AF and FD from there.
theres another (more complicated) way to extend CD to meet AB, then inscribe a circle and find the radius to be 10/3 (which is BF) and find AD from finding AF and FD from there.
rrusczyk
2009-04-03 20:51:45
And another.
And another.
ThinkFlow
2009-04-03 20:51:47
Should I know stewart's theorem?
Should I know stewart's theorem?
rrusczyk
2009-04-03 20:52:02
It's really not that important. But I would try to prove it for fun.
It's really not that important. But I would try to prove it for fun.
spelljack825
2009-04-03 20:52:04
trig bashing is quick and straightforward
trig bashing is quick and straightforward
m1sterzer0
2009-04-03 20:52:04
the trig solution seems pretty elegant as well (perhaps even moreso than the geometry approach)
the trig solution seems pretty elegant as well (perhaps even moreso than the geometry approach)
rrusczyk
2009-04-03 20:52:12
Beauty is in the eye of the beholder :)
Beauty is in the eye of the beholder :)
MathAndKnowledge
2009-04-03 20:52:26
just the derivation, because you can easily derive it when you need a cevian length
just the derivation, because you can easily derive it when you need a cevian length
rrusczyk
2009-04-03 20:52:32
True: (about Stewart)
True: (about Stewart)
rrusczyk
2009-04-03 20:52:36
We'll take a 5 minute break and then knock off the last 5 problems.
We'll take a 5 minute break and then knock off the last 5 problems.
Hullohihell0
2009-04-03 20:56:53
how much longer is this math jam going to last?
how much longer is this math jam going to last?
rrusczyk
2009-04-03 20:57:01
Maybe 75 minutes.
Maybe 75 minutes.
ThinkFlow
2009-04-03 20:57:05
Just looked at http://en.wikipedia.org/wiki/Stewart's_theorem. Stewart's theorem seems pretty useful to me...
Just looked at http://en.wikipedia.org/wiki/Stewart's_theorem. Stewart's theorem seems pretty useful to me...
rrusczyk
2009-04-03 20:57:22
So - so. I think that any math contest problem that requires Stewart is just a bad problem.
So - so. I think that any math contest problem that requires Stewart is just a bad problem.
rrusczyk
2009-04-03 20:58:01
On to problem 11!
On to problem 11!
rrusczyk
2009-04-03 20:58:06
rrusczyk
2009-04-03 20:58:18
Where do we start with this?
Where do we start with this?
charles.du
2009-04-03 20:58:56
|log(m/k)| < log n
|log(m/k)| < log n
geoffreywong
2009-04-03 20:58:56
log(m/k)
log(m/k)
dgreenb801
2009-04-03 20:58:56
log m-log k=log(m/k)
log m-log k=log(m/k)
rrusczyk
2009-04-03 20:59:13
GoldenFrog1618
2009-04-03 20:59:17
turn the abs value into 2 inequalities
turn the abs value into 2 inequalities
m1sterzer0
2009-04-03 20:59:17
rewrite the equation for positive and negative (log(m) - log(k))
rewrite the equation for positive and negative (log(m) - log(k))
charles.du
2009-04-03 20:59:30
and then get rid of the absolute value
and then get rid of the absolute value
rrusczyk
2009-04-03 20:59:41
We have two inequalities to worry about to get rid of the absolute value.
We have two inequalities to worry about to get rid of the absolute value.
rrusczyk
2009-04-03 20:59:47
What are the cases we have to worry about?
What are the cases we have to worry about?
late20s
2009-04-03 20:59:51
get rid of the log as well?
get rid of the log as well?
rrusczyk
2009-04-03 20:59:56
We'd love to do that :)
We'd love to do that :)
rrusczyk
2009-04-03 21:00:14
Let's get rid of the absolute value first.
Let's get rid of the absolute value first.
rrusczyk
2009-04-03 21:00:15
How?
How?
m1sterzer0
2009-04-03 21:00:23
m > k and m < k
m > k and m < k
rrusczyk
2009-04-03 21:00:37
If m>=k, then what do we have?
If m>=k, then what do we have?
tinytim
2009-04-03 21:01:11
m/k<n
m/k<n
happysunnyshine
2009-04-03 21:01:11
m/k< n
m/k< n
rrusczyk
2009-04-03 21:01:31
happysunnyshine
2009-04-03 21:02:05
k/m<n
k/m<n
funtwo
2009-04-03 21:02:05
k/m<n, k<mn
k/m<n, k<mn
tinytim
2009-04-03 21:02:12
If k>m, we have m/k>1/n (from log(1/n)=-logn)
If k>m, we have m/k>1/n (from log(1/n)=-logn)
Wikipedian1337
2009-04-03 21:02:12
m/k > 1/n
m/k > 1/n
rrusczyk
2009-04-03 21:02:15
rrusczyk
2009-04-03 21:02:41
So, we must have both m/k < n and k/m < n. Now what?
So, we must have both m/k < n and k/m < n. Now what?
funtwo
2009-04-03 21:03:25
dangit i already forgot how i got m/n<k<mn
dangit i already forgot how i got m/n<k<mn
funtwo
2009-04-03 21:03:25
oh i remember it now. rearrange the first inequality to get m/n<k
oh i remember it now. rearrange the first inequality to get m/n<k
tinytim
2009-04-03 21:03:25
so we have m/n<k<mn
so we have m/n<k<mn
wangsacl
2009-04-03 21:03:25
restrict the value of k
restrict the value of k
rrusczyk
2009-04-03 21:03:40
From m/k < n, we have k > m/n, and from k/m < n, we have k < nm, so the values of k that satisfy the inequality range from m/n to mn.
From m/k < n, we have k > m/n, and from k/m < n, we have k < nm, so the values of k that satisfy the inequality range from m/n to mn.
rrusczyk
2009-04-03 21:03:42
Now what?
Now what?
mathq
2009-04-03 21:03:47
use the fact that there are only 50 solutions?
use the fact that there are only 50 solutions?
rrusczyk
2009-04-03 21:03:50
We know that there must be exactly 50 integers k between m/n and mn (exclusive, meaning we can't include m/n or mn). How does that help?
We know that there must be exactly 50 integers k between m/n and mn (exclusive, meaning we can't include m/n or mn). How does that help?
Wikipedian1337
2009-04-03 21:04:55
mn = floor(m/n) + 51
mn = floor(m/n) + 51
wangsacl
2009-04-03 21:04:55
floor(m/n)+1<=k<=mn-1, so there are mn-floor(m/n)-1 possible values of k
floor(m/n)+1<=k<=mn-1, so there are mn-floor(m/n)-1 possible values of k
rrusczyk
2009-04-03 21:05:10
We can say essentially this without the fancy notation, like this:
We can say essentially this without the fancy notation, like this:
rrusczyk
2009-04-03 21:18:19
rrusczyk
2009-04-03 21:05:39
(That's often how I avoid floor notation -- just make an inequality instead.)
(That's often how I avoid floor notation -- just make an inequality instead.)
rrusczyk
2009-04-03 21:05:58
All that says is "there are 50 integers from m/n to mn, exclusive".
All that says is "there are 50 integers from m/n to mn, exclusive".
rrusczyk
2009-04-03 21:06:04
There's nothing fancy there.
There's nothing fancy there.
Wikipedian1337
2009-04-03 21:06:13
That's why I was stuck on mn = floor(m/n) + 51 during the test and couldn't go any further!
That's why I was stuck on mn = floor(m/n) + 51 during the test and couldn't go any further!
rrusczyk
2009-04-03 21:06:33
Yeah, notation can really trap you!
Yeah, notation can really trap you!
rrusczyk
2009-04-03 21:06:47
Going to inequalities is a good way to get away from the floor notation.
Going to inequalities is a good way to get away from the floor notation.
rrusczyk
2009-04-03 21:06:48
Now what?
Now what?
tinytim
2009-04-03 21:06:52
We only need to try n=1,2,...,7 because m>=n
We only need to try n=1,2,...,7 because m>=n
rrusczyk
2009-04-03 21:07:02
This immediately places some restrictions on n. Since m >= n, if n > 7, then m(n-1/n) is more than 52. So, we can only have n = 2,3,4,5,6, or 7. (Obviously, n=1 won't work!)
This immediately places some restrictions on n. Since m >= n, if n > 7, then m(n-1/n) is more than 52. So, we can only have n = 2,3,4,5,6, or 7. (Obviously, n=1 won't work!)
tinytim
2009-04-03 21:07:05
n=1 obviously doesnt work
n=1 obviously doesnt work
rrusczyk
2009-04-03 21:07:09
That doesn't give us too much to test!
That doesn't give us too much to test!
rrusczyk
2009-04-03 21:07:13
What happens for n=7?
What happens for n=7?
rrusczyk
2009-04-03 21:07:56
If n=7, we have 52 > m(7-1/7) >= 50. Any luck?
If n=7, we have 52 > m(7-1/7) >= 50. Any luck?
tinytim
2009-04-03 21:08:00
no solution
no solution
Wikipedian1337
2009-04-03 21:08:00
nope
nope
funtwo
2009-04-03 21:08:03
nothing
nothing
rrusczyk
2009-04-03 21:08:06
No solutions here; if m=7, then m(7 - 1/7) is 48, and if m = 8, then the product is 48/7 more, which gives us a number that is greater than 52.
No solutions here; if m=7, then m(7 - 1/7) is 48, and if m = 8, then the product is 48/7 more, which gives us a number that is greater than 52.
rrusczyk
2009-04-03 21:08:12
And we just keep cranking down.
And we just keep cranking down.
rrusczyk
2009-04-03 21:08:15
If n=6, we have 52 > m(6 - 1/6) >= 50. Any luck?
If n=6, we have 52 > m(6 - 1/6) >= 50. Any luck?
tinytim
2009-04-03 21:08:26
nope
nope
mathq
2009-04-03 21:08:26
no solutions?nope
no solutions?nope
funtwo
2009-04-03 21:08:26
no
no
rrusczyk
2009-04-03 21:08:28
Again, no. Here, we see that m(6-1/6) < 50 if m = 8 and m(6-1/6) >= 52 if m=9.
Again, no. Here, we see that m(6-1/6) < 50 if m = 8 and m(6-1/6) >= 52 if m=9.
rrusczyk
2009-04-03 21:08:34
If n=5, we have 52 > m(5 - 1/5) >= 50. Any luck?
If n=5, we have 52 > m(5 - 1/5) >= 50. Any luck?
mathq
2009-04-03 21:08:56
no again
no again
chessmaster7
2009-04-03 21:08:56
no
no
rrusczyk
2009-04-03 21:09:01
Nope. m = 10 gives us 48 and m = 11 gives us a number greater than 52. No luck.
Nope. m = 10 gives us 48 and m = 11 gives us a number greater than 52. No luck.
rrusczyk
2009-04-03 21:09:05
If n=4, we have 52 > m(4 - 1/4) >= 50. Any luck?
If n=4, we have 52 > m(4 - 1/4) >= 50. Any luck?
tinytim
2009-04-03 21:09:11
no
no
mathq
2009-04-03 21:09:12
no...
no...
rrusczyk
2009-04-03 21:09:13
Nope. m = 13 gives us 48 3/4 and m = 14 gives us a number greater than 52. No luck.
Nope. m = 13 gives us 48 3/4 and m = 14 gives us a number greater than 52. No luck.
spelljack825
2009-04-03 21:09:16
is there no better way than trying all the remaining cases?
is there no better way than trying all the remaining cases?
rrusczyk
2009-04-03 21:09:41
I don't know -- I don't think there's a way that's easier to see than bashing this out :)
I don't know -- I don't think there's a way that's easier to see than bashing this out :)
chessmaster7
2009-04-03 21:09:43
well trying the cases isn't necessarily what you would call "time consuming"
well trying the cases isn't necessarily what you would call "time consuming"
rrusczyk
2009-04-03 21:09:46
Exactly :)
Exactly :)
rrusczyk
2009-04-03 21:09:54
If n=4, we have 52 > m(4 - 1/4) >= 50. Any luck?
If n=4, we have 52 > m(4 - 1/4) >= 50. Any luck?
charles.du
2009-04-03 21:10:19
we just did that
we just did that
rrusczyk
2009-04-03 21:10:21
lol
lol
rrusczyk
2009-04-03 21:10:27
If n=3, we have 52 > m(3 - 1/3) >= 50. Any luck?
If n=3, we have 52 > m(3 - 1/3) >= 50. Any luck?
funtwo
2009-04-03 21:10:38
yes! 3*19=57, 19/3>6
yes! 3*19=57, 19/3>6
tinytim
2009-04-03 21:10:39
Yay,m=19
Yay,m=19
chessmaster7
2009-04-03 21:10:39
m can be 19
m can be 19
GoldenFrog1618
2009-04-03 21:10:40
19
19
rrusczyk
2009-04-03 21:10:42
Yes! m = 19 gives us m(3 - 1/3) = 50 2/3. We test that (m,n) = (19,3) satisfies the problem.
Yes! m = 19 gives us m(3 - 1/3) = 50 2/3. We test that (m,n) = (19,3) satisfies the problem.
rrusczyk
2009-04-03 21:10:47
late20s
2009-04-03 21:12:08
7 - 56
7 - 56
charles.du
2009-04-03 21:12:08
7 through 56
7 through 56
andrewda
2009-04-03 21:12:08
7to56
7to56
spelljack825
2009-04-03 21:12:08
56>=k>=7
56>=k>=7
rrusczyk
2009-04-03 21:12:10
Any k that is more than a third of 19 and less than 3 times 19 satisfies the inequality. That gives us 7, 8, 9, . . . , 56. Indeed, there are 50 such k.
Any k that is more than a third of 19 and less than 3 times 19 satisfies the inequality. That gives us 7, 8, 9, . . . , 56. Indeed, there are 50 such k.
rrusczyk
2009-04-03 21:12:15
And finally, n = 2, which gives us 52 > m(2 - 1/2) >= 50. Any luck?
And finally, n = 2, which gives us 52 > m(2 - 1/2) >= 50. Any luck?
tinytim
2009-04-03 21:12:47
yay again, m=34
yay again, m=34
wangsacl
2009-04-03 21:12:47
34
34
rrusczyk
2009-04-03 21:12:50
Yes. Here, m=34 is the lucky number, since 34(2-1/2) = 51.
Yes. Here, m=34 is the lucky number, since 34(2-1/2) = 51.
rrusczyk
2009-04-03 21:12:54
rrusczyk
2009-04-03 21:12:59
And our answer is ?
And our answer is ?
tinytim
2009-04-03 21:13:39
So the sum of all possible mn is 19*3+34*2=125.
So the sum of all possible mn is 19*3+34*2=125.
GoldenFrog1618
2009-04-03 21:13:39
68+57=125
68+57=125
charles.du
2009-04-03 21:13:39
68+57=125
68+57=125
wangsacl
2009-04-03 21:13:39
68+57=125?
68+57=125?
rrusczyk
2009-04-03 21:13:41
The answer to the problem then is 3(19) + 2(34) = 125.
The answer to the problem then is 3(19) + 2(34) = 125.
ThinkFlow
2009-04-03 21:14:44
So we didn't like absolute values or logs so we found a way to get rid of them?
So we didn't like absolute values or logs so we found a way to get rid of them?
rrusczyk
2009-04-03 21:14:47
Exactly.
Exactly.
ThinkFlow
2009-04-03 21:18:55
next problem?
next problem?
rrusczyk
2009-04-03 21:18:58
Good idea.
Good idea.
rrusczyk
2009-04-03 21:19:04
rrusczyk
2009-04-03 21:19:19
Any ideas where to start?
Any ideas where to start?
rrusczyk
2009-04-03 21:20:03
One way to start with this would be to try a smaller set of integers, like 1 through 5 or 1 through 7. Unfortunately, no patterns jump out when we try this.
One way to start with this would be to try a smaller set of integers, like 1 through 5 or 1 through 7. Unfortunately, no patterns jump out when we try this.
rrusczyk
2009-04-03 21:20:07
We might also try constructing pairs in some smart way, but that runs into trouble pretty quickly. There are all sorts of strategies to take in making the pairs, and it's not clear which is best or why.
We might also try constructing pairs in some smart way, but that runs into trouble pretty quickly. There are all sorts of strategies to take in making the pairs, and it's not clear which is best or why.
tinytim
2009-04-03 21:20:13
Try to find the upper bound of k?
Try to find the upper bound of k?
rrusczyk
2009-04-03 21:20:21
We have a maximization problem -- we want to find the maximum value of k. Finding a maximum means that we need to show that k <= (some number), and that it is possible for k to equal that number. In other words, we need to build an inequality that involves k.
We have a maximization problem -- we want to find the maximum value of k. Finding a maximum means that we need to show that k <= (some number), and that it is possible for k to equal that number. In other words, we need to build an inequality that involves k.
rrusczyk
2009-04-03 21:20:40
How might we go after finding an inequality that involves k?
How might we go after finding an inequality that involves k?
spelljack825
2009-04-03 21:20:47
sum the sums?
sum the sums?
funtwo
2009-04-03 21:20:47
consider the total sum of all the pairs
consider the total sum of all the pairs
rrusczyk
2009-04-03 21:21:01
Can we bound this sum in any useful way?
Can we bound this sum in any useful way?
math154
2009-04-03 21:22:39
For k sets, the minimum sum of the numbers in all is 1+2+...+2k, and the maximum sum is 2009+2008+...+(2010-k), and clearly the maximum sum has to be greater than the minimum for k to work.
For k sets, the minimum sum of the numbers in all is 1+2+...+2k, and the maximum sum is 2009+2008+...+(2010-k), and clearly the maximum sum has to be greater than the minimum for k to work.
rrusczyk
2009-04-03 21:22:55
We can bound the sum of all the numbers in the pairs. If we add all 2k numbers in the k pairs, the smallest the sum can possibly be is 1+2+3+ . . . + 2k, which equals k(2k+1). Letting the sum of the 2k numbers in the pairs be s, we have s >= k(2k+1).
We can bound the sum of all the numbers in the pairs. If we add all 2k numbers in the k pairs, the smallest the sum can possibly be is 1+2+3+ . . . + 2k, which equals k(2k+1). Letting the sum of the 2k numbers in the pairs be s, we have s >= k(2k+1).
rrusczyk
2009-04-03 21:23:07
Where does the upper bound math154 identified come from?
Where does the upper bound math154 identified come from?
funtwo
2009-04-03 21:23:35
all sums must be distinct and <=2009
all sums must be distinct and <=2009
happysunnyshine
2009-04-03 21:23:37
all sums are distinct and less than or equal to 2009
all sums are distinct and less than or equal to 2009
rrusczyk
2009-04-03 21:23:41
Since each pair of numbers has sum no greater than 2009, and the sums are all distinct, we have s <= 2009 + 2008 + 2007 + . . . + (2010 - k).
Since each pair of numbers has sum no greater than 2009, and the sums are all distinct, we have s <= 2009 + 2008 + 2007 + . . . + (2010 - k).
rrusczyk
2009-04-03 21:23:49
Can we write that more simply?
Can we write that more simply?
GoldenFrog1618
2009-04-03 21:24:22
2010k-k(k+1)/2
2010k-k(k+1)/2
rrusczyk
2009-04-03 21:24:25
We can write this more simply as s <= 2010k - (1+2+3+. . . +k) = 2010k - k(k+1)/2.
We can write this more simply as s <= 2010k - (1+2+3+. . . +k) = 2010k - k(k+1)/2.
rrusczyk
2009-04-03 21:24:31
And now?
And now?
funtwo
2009-04-03 21:24:47
minimum must be less than maximum
minimum must be less than maximum
ThinkFlow
2009-04-03 21:24:47
write our inequality
write our inequality
tinytim
2009-04-03 21:24:47
Combine and solve the inequality
Combine and solve the inequality
rrusczyk
2009-04-03 21:24:53
rrusczyk
2009-04-03 21:25:00
And what does this tell us?
And what does this tell us?
tinytim
2009-04-03 21:25:48
Solving for k gives k<=803
Solving for k gives k<=803
funtwo
2009-04-03 21:25:48
a regular quadratic inequality, k>0<804
a regular quadratic inequality, k>0<804
rrusczyk
2009-04-03 21:25:59
Nothing fancy here. Just a bit of algebra.
Nothing fancy here. Just a bit of algebra.
rrusczyk
2009-04-03 21:26:04
rrusczyk
2009-04-03 21:26:15
rrusczyk
2009-04-03 21:26:20
Are we finished?
Are we finished?
rrusczyk
2009-04-03 21:26:44
We have a guess. But on the USAMO, we wouldn't nearly be done.
We have a guess. But on the USAMO, we wouldn't nearly be done.
rrusczyk
2009-04-03 21:26:50
We have to:
We have to:
tinytim
2009-04-03 21:26:56
Now we must find 803 pairs that satisfy the conditions...
Now we must find 803 pairs that satisfy the conditions...
ThinkFlow
2009-04-03 21:26:56
no; that's a bound
no; that's a bound
charles.du
2009-04-03 21:26:56
need to find a way to use 803 pairs
need to find a way to use 803 pairs
#H34N1
2009-04-03 21:26:56
Prove that value is achievable?
Prove that value is achievable?
tinytim
2009-04-03 21:26:56
we cant just assume that 803 works.
we cant just assume that 803 works.
m1sterzer0
2009-04-03 21:26:56
need to show that 803 can be achieved
need to show that 803 can be achieved
rrusczyk
2009-04-03 21:27:02
Not quite finished. We have to show that it is possible to come up with 803 pairs that have different sums. All we have shown is that it is not possible to have more than 803 pairs. (So, at the very least, we have a decent *guess* at the answer.)
Not quite finished. We have to show that it is possible to come up with 803 pairs that have different sums. All we have shown is that it is not possible to have more than 803 pairs. (So, at the very least, we have a decent *guess* at the answer.)
rrusczyk
2009-04-03 21:27:06
Where might we start with building the pairs?
Where might we start with building the pairs?
rrusczyk
2009-04-03 21:27:16
What might your first step be in trying to build 803 pairs?
What might your first step be in trying to build 803 pairs?
rrusczyk
2009-04-03 21:28:09
If you were to write out 803 blank pairs, what numbers might you write as the first in each?
If you were to write out 803 blank pairs, what numbers might you write as the first in each?
happysunnyshine
2009-04-03 21:28:55
1,2,3..803
1,2,3..803
GoldenFrog1618
2009-04-03 21:28:55
1-803
1-803
billy314
2009-04-03 21:28:55
1,2,3,...,803
1,2,3,...,803
rrusczyk
2009-04-03 21:29:03
We can start by trying to build 803 pairs in which a number from 1 to 803 is one of the numbers in each pair. So, the pairs are:(1, ?)(2, ?)(3, ?). . .(803, ?)
We can start by trying to build 803 pairs in which a number from 1 to 803 is one of the numbers in each pair. So, the pairs are:(1, ?)(2, ?)(3, ?). . .(803, ?)
rrusczyk
2009-04-03 21:29:21
What are we pretty sure at least one of these sums will equal?
What are we pretty sure at least one of these sums will equal?
billy314
2009-04-03 21:29:43
2009
2009
chessmaster7
2009-04-03 21:29:43
2009?
2009?
Bl00d
2009-04-03 21:29:43
2009
2009
GoldenFrog1618
2009-04-03 21:29:43
2009
2009
tinytim
2009-04-03 21:29:43
2009?
2009?
rrusczyk
2009-04-03 21:30:08
OK, we start with the extremes. Are we going to make an extreme low sum with 803?
OK, we start with the extremes. Are we going to make an extreme low sum with 803?
Bl00d
2009-04-03 21:30:18
no
no
spelljack825
2009-04-03 21:30:21
no
no
rrusczyk
2009-04-03 21:30:25
Certainly not.
Certainly not.
Bl00d
2009-04-03 21:30:27
so that would probably = 2009
so that would probably = 2009
rrusczyk
2009-04-03 21:30:36
Let's start with the extremes. We won't make a little sum using the 803, since we've already used all the numbers from 1 to 803. So, we'll make a big sum there:(1, ?)(2, ?)(3, ?). . .(803, 1206)
Let's start with the extremes. We won't make a little sum using the 803, since we've already used all the numbers from 1 to 803. So, we'll make a big sum there:(1, ?)(2, ?)(3, ?). . .(803, 1206)
cool4ever
2009-04-03 21:31:22
figure out the sums
figure out the sums
rrusczyk
2009-04-03 21:31:30
What do we think the other sums will be?
What do we think the other sums will be?
ThinkFlow
2009-04-03 21:31:45
sums: 2008, 2007, 2006, etc.?
sums: 2008, 2007, 2006, etc.?
GoldenFrog1618
2009-04-03 21:31:52
2008,2007,2006...
2008,2007,2006...
rrusczyk
2009-04-03 21:31:55
All the way down to what?
All the way down to what?
chessmaster7
2009-04-03 21:32:10
1207
1207
charles.du
2009-04-03 21:32:10
1207
1207
rrusczyk
2009-04-03 21:32:12
No sum can be greater than 2009, so we try to build pairs that will give us the 803 greatest possible sums. That is, the sums 2009, 2008, 2007, . . ., 1207.
No sum can be greater than 2009, so we try to build pairs that will give us the 803 greatest possible sums. That is, the sums 2009, 2008, 2007, . . ., 1207.
rrusczyk
2009-04-03 21:32:27
So, we made the biggest sum with 803.
So, we made the biggest sum with 803.
rrusczyk
2009-04-03 21:32:32
Can we make the smallest with 1?
Can we make the smallest with 1?
ThinkFlow
2009-04-03 21:32:40
no
no
spelljack825
2009-04-03 21:32:40
no
no
charles.du
2009-04-03 21:32:40
no
no
rrusczyk
2009-04-03 21:32:48
Nope; we already used 1206.
Nope; we already used 1206.
rrusczyk
2009-04-03 21:32:56
So, what will we do to make 1207?
So, what will we do to make 1207?
chessmaster7
2009-04-03 21:33:07
so we go to 2 with 1205
so we go to 2 with 1205
tinytim
2009-04-03 21:33:07
2,1205
2,1205
e714212835
2009-04-03 21:33:12
2 and 1205?
2 and 1205?
rrusczyk
2009-04-03 21:33:13
We can't make 1207 with 1, since we've already used 1206. So, we use 2:(1, ?)(2, 1205)(3, ?). . .(803, 1206)
We can't make 1207 with 1, since we've already used 1206. So, we use 2:(1, ?)(2, 1205)(3, ?). . .(803, 1206)
rrusczyk
2009-04-03 21:33:16
What next?
What next?
tinytim
2009-04-03 21:34:02
try to make the second largest sum? (2008)
try to make the second largest sum? (2008)
rrusczyk
2009-04-03 21:34:12
How are we going to make 2008? With 802?
How are we going to make 2008? With 802?
tinytim
2009-04-03 21:34:31
we cant to 802,1206 so we must to 801,1207
we cant to 802,1206 so we must to 801,1207
ThinkFlow
2009-04-03 21:34:31
uses 1206, so no
uses 1206, so no
GoldenFrog1618
2009-04-03 21:34:31
no, 801
no, 801
funtwo
2009-04-03 21:34:31
we cant. 801
we cant. 801
rrusczyk
2009-04-03 21:34:34
We take another step, building 2008. We can't do so with 802, since 1206 is taken, so:(1, ?)(2, 1205)(3, ?). . .(801, 1207)(802, ?)(803, 1206)
We take another step, building 2008. We can't do so with 802, since 1206 is taken, so:(1, ?)(2, 1205)(3, ?). . .(801, 1207)(802, ?)(803, 1206)
rrusczyk
2009-04-03 21:34:42
What's next?
What's next?
m1sterzer0
2009-04-03 21:34:52
back to the small end
back to the small end
tinytim
2009-04-03 21:34:52
then do the second smallest sum
then do the second smallest sum
happysunnyshine
2009-04-03 21:34:56
1208
1208
e714212835
2009-04-03 21:34:56
1208?
1208?
chessmaster7
2009-04-03 21:34:56
now can we do the second smallest sum
now can we do the second smallest sum
rrusczyk
2009-04-03 21:35:00
Which is?
Which is?
tinytim
2009-04-03 21:35:09
3,1205 doesnt work so 4,1204
3,1205 doesnt work so 4,1204
chessmaster7
2009-04-03 21:35:10
4 and 1204 = 1208
4 and 1204 = 1208
m1sterzer0
2009-04-03 21:35:10
4 + 1204
4 + 1204
e714212835
2009-04-03 21:35:10
4, 1204
4, 1204
rrusczyk
2009-04-03 21:35:11
Next, we make 1208. We can't use 3, but we can use 4:(1, ?)(2, 1205)(3, ?)(4, 1204). . .(801, 1207)(802, ?)(803, 1206)
Next, we make 1208. We can't use 3, but we can use 4:(1, ?)(2, 1205)(3, ?)(4, 1204). . .(801, 1207)(802, ?)(803, 1206)
rrusczyk
2009-04-03 21:35:20
Aha!
Aha!
tinytim
2009-04-03 21:35:24
a pattern!
a pattern!
chessmaster7
2009-04-03 21:35:43
since we're going up along even numbers and down along odds, all the sums will work like teeth of a comb fitting together
since we're going up along even numbers and down along odds, all the sums will work like teeth of a comb fitting together
rrusczyk
2009-04-03 21:35:47
And now the pattern emerges!(2,1205)(4,1204)(6,1203)(8,1202). . .(802, 805)(803,1206)(801,1207)(799, 1208). . .(1, 1607)
And now the pattern emerges!(2,1205)(4,1204)(6,1203)(8,1202). . .(802, 805)(803,1206)(801,1207)(799, 1208). . .(1, 1607)
rrusczyk
2009-04-03 21:35:59
There is joy in Mudville.
There is joy in Mudville.
rrusczyk
2009-04-03 21:36:00
We thereby form the sums 1207, 1208, . . . 1607,1608,1609, . . ., 2009, as desired.
We thereby form the sums 1207, 1208, . . . 1607,1608,1609, . . ., 2009, as desired.
rrusczyk
2009-04-03 21:36:09
The maximum value of k is therefore 803.
The maximum value of k is therefore 803.
rrusczyk
2009-04-03 21:36:23
Now we're finished. (There are other ways to form 803 sums -- see if you can find some.)
Now we're finished. (There are other ways to form 803 sums -- see if you can find some.)
rrusczyk
2009-04-03 21:36:44
But for now, we're on to #13
But for now, we're on to #13
rrusczyk
2009-04-03 21:36:48
rrusczyk
2009-04-03 21:37:05
rrusczyk
2009-04-03 21:37:20
We could go after this with all sorts of trig. But . . .
We could go after this with all sorts of trig. But . . .
funtwo
2009-04-03 21:37:23
this immediately reminds of roots of unity
this immediately reminds of roots of unity
rrusczyk
2009-04-03 21:37:42
Regular polygons. Already in a circle! This really makes us think of complex numbers.
Regular polygons. Already in a circle! This really makes us think of complex numbers.
rrusczyk
2009-04-03 21:38:00
How can we relate the problem to complex numbers?
How can we relate the problem to complex numbers?
dragon96
2009-04-03 21:38:24
the real imaginary plane
the real imaginary plane
rrusczyk
2009-04-03 21:38:41
And how can we view these on the complex plane?
And how can we view these on the complex plane?
funtwo
2009-04-03 21:38:54
C1 is the first 14th root of unity, C2 is the second 14th root of unity...
C1 is the first 14th root of unity, C2 is the second 14th root of unity...
tinytim
2009-04-03 21:38:54
Let the Center of the semicircle be the origin
Let the Center of the semicircle be the origin
charles.du
2009-04-03 21:38:54
let 0+0i be the midpoint of AB
let 0+0i be the midpoint of AB
rrusczyk
2009-04-03 21:38:58
rrusczyk
2009-04-03 21:39:14
rrusczyk
2009-04-03 21:39:19
ThinkFlow
2009-04-03 21:39:58
But how do we find the lengths of the chords?
But how do we find the lengths of the chords?
rrusczyk
2009-04-03 21:40:16
Good question. Do we want to stick with the diagram we have up there?
Good question. Do we want to stick with the diagram we have up there?
hello123
2009-04-03 21:40:29
1-w^i
1-w^i
rrusczyk
2009-04-03 21:40:42
Will this work for the chords at B?
Will this work for the chords at B?
all4math
2009-04-03 21:41:01
no
no
GoldenFrog1618
2009-04-03 21:41:04
they are the same for B
they are the same for B
rrusczyk
2009-04-03 21:41:12
The chords have the same lenth, yes.
The chords have the same lenth, yes.
m1sterzer0
2009-04-03 21:41:32
we can replace the chords from B with chords drawn to the lower half of the circle
we can replace the chords from B with chords drawn to the lower half of the circle
rrusczyk
2009-04-03 21:41:43
Geometrically speaking, we can use this fact to make the diagram prettier:
Geometrically speaking, we can use this fact to make the diagram prettier:
rrusczyk
2009-04-03 21:41:48
rrusczyk
2009-04-03 21:42:39
We want the product of all of these lengths except for the middle one (the diameter), which is not included in the problem.
We want the product of all of these lengths except for the middle one (the diameter), which is not included in the problem.
rrusczyk
2009-04-03 21:42:51
Now, how can we write that product of lengths in terms of complex numbers?
Now, how can we write that product of lengths in terms of complex numbers?
funtwo
2009-04-03 21:44:12
the distance between two points is the magnitude of their difference
the distance between two points is the magnitude of their difference
benzi455
2009-04-03 21:44:12
abs of product of (2-2w^n) as n ranges from 1 to 13 but does not equal 7
abs of product of (2-2w^n) as n ranges from 1 to 13 but does not equal 7
rrusczyk
2009-04-03 21:44:15
The distance between w and z in the complex plane is |w-z|. So, we can write our desired product as:
The distance between w and z in the complex plane is |w-z|. So, we can write our desired product as:
rrusczyk
2009-04-03 21:44:19
rrusczyk
2009-04-03 21:44:31
all4math
2009-04-03 21:44:33
It looks cleaner
It looks cleaner
rrusczyk
2009-04-03 21:44:39
It looks a lot clearner!
It looks a lot clearner!
rrusczyk
2009-04-03 21:44:45
Cleaner, too.
Cleaner, too.
rrusczyk
2009-04-03 21:44:51
How can we compute this product?
How can we compute this product?
xelalex1231030
2009-04-03 21:44:57
but it helps to add it in
but it helps to add it in
rrusczyk
2009-04-03 21:45:26
Then what will the product look like -- that is, what product that we know how to handle will it look like?
Then what will the product look like -- that is, what product that we know how to handle will it look like?
xelalex1231030
2009-04-03 21:45:56
put that factor back in, and note that it's 1 + z + z^2 + ... + z^13 evaluated at 1
put that factor back in, and note that it's 1 + z + z^2 + ... + z^13 evaluated at 1
hello123
2009-04-03 21:45:56
consider the factored for of x^13+x^12+.....+1
consider the factored for of x^13+x^12+.....+1
rrusczyk
2009-04-03 21:46:10
Let's see where that comes from:
Let's see where that comes from:
rrusczyk
2009-04-03 21:46:10
ThinkFlow
2009-04-03 21:46:37
it doesn't depend on w?
it doesn't depend on w?
rrusczyk
2009-04-03 21:46:57
funtwo
2009-04-03 21:47:00
roots of x^14=1
roots of x^14=1
xelalex1231030
2009-04-03 21:47:00
it's x^14 - 1!
it's x^14 - 1!
rrusczyk
2009-04-03 21:47:08
rrusczyk
2009-04-03 21:47:21
So, how will we get from this to the product we want:
So, how will we get from this to the product we want:
rrusczyk
2009-04-03 21:47:34
rrusczyk
2009-04-03 21:47:51
We want to jam in x = 1, but then we get . . . 0=0.
We want to jam in x = 1, but then we get . . . 0=0.
rrusczyk
2009-04-03 21:47:58
Hurm. That's no fun. What do we do?
Hurm. That's no fun. What do we do?
tinytim
2009-04-03 21:48:16
divide by (x-1) first.
divide by (x-1) first.
Yongyi781
2009-04-03 21:48:16
Divide by (x-1)
Divide by (x-1)
rrusczyk
2009-04-03 21:48:25
rrusczyk
2009-04-03 21:48:29
How does the right-hand side simplify?
How does the right-hand side simplify?
Yongyi781
2009-04-03 21:48:46
1+x+x^2+...+x^13
1+x+x^2+...+x^13
rrusczyk
2009-04-03 21:48:48
rrusczyk
2009-04-03 21:48:54
rrusczyk
2009-04-03 21:49:08
So, do we have our desired product?
So, do we have our desired product?
tinytim
2009-04-03 21:49:20
iremember to divide by 2
iremember to divide by 2
math154
2009-04-03 21:49:20
No, but w^7=-1, so 1-w^7=2 and we can divide this out to get the desired expression.
No, but w^7=-1, so 1-w^7=2 and we can divide this out to get the desired expression.
xelalex1231030
2009-04-03 21:49:20
but then divide by two because 1 - w^7 = 1 - (-1) = 2
but then divide by two because 1 - w^7 = 1 - (-1) = 2
rrusczyk
2009-04-03 21:49:28
rrusczyk
2009-04-03 21:49:32
rrusczyk
2009-04-03 21:49:39
rrusczyk
2009-04-03 21:49:48
And now we're finally ready to find:
And now we're finally ready to find:
tinytim
2009-04-03 21:50:05
So the answer is 2^12*7=672 (mod 1000)
So the answer is 2^12*7=672 (mod 1000)
funtwo
2009-04-03 21:50:06
7*96=672
7*96=672
rrusczyk
2009-04-03 21:50:11
rrusczyk
2009-04-03 21:50:34
Complex numbers are fun.
Complex numbers are fun.
rrusczyk
2009-04-03 21:50:44
And now it's time for number 14.
And now it's time for number 14.
wangsacl
2009-04-03 21:50:47
Is there any other approaches?
Is there any other approaches?
rrusczyk
2009-04-03 21:50:57
A wall of trig identities can get you there, I'm betting.
A wall of trig identities can get you there, I'm betting.
spelljack825
2009-04-03 21:51:04
much nicer than trig bashing XP
much nicer than trig bashing XP
rrusczyk
2009-04-03 21:51:18
Yeah, complex numbers hide a lot of the machinery of trig in very nice ways.
Yeah, complex numbers hide a lot of the machinery of trig in very nice ways.
dragon96
2009-04-03 21:51:24
its interesting how a geometry problem turns out to use the roots of unity
its interesting how a geometry problem turns out to use the roots of unity
rrusczyk
2009-04-03 21:51:39
Whenever you see a regular polygon in a hard problem, this is something you should consider.
Whenever you see a regular polygon in a hard problem, this is something you should consider.
rrusczyk
2009-04-03 21:51:45
rrusczyk
2009-04-03 21:51:59
Where would you naturally start with this problem to get a handle on it?
Where would you naturally start with this problem to get a handle on it?
MathAndKnowledge
2009-04-03 21:52:11
compute some terms
compute some terms
happysunnyshine
2009-04-03 21:52:11
to try some
to try some
ThinkFlow
2009-04-03 21:52:11
plug small values in
plug small values in
charles.du
2009-04-03 21:52:11
generate first few terms
generate first few terms
rrusczyk
2009-04-03 21:52:22
rrusczyk
2009-04-03 21:52:31
rrusczyk
2009-04-03 21:52:40
rrusczyk
2009-04-03 21:52:51
You may be able to see some patterns, but the most significant observation is that each of the terms we computed is rational.
You may be able to see some patterns, but the most significant observation is that each of the terms we computed is rational.
rrusczyk
2009-04-03 21:53:09
rrusczyk
2009-04-03 21:53:39
Any suggestions (assuming you don't see the magical trig substitution at this point)?
Any suggestions (assuming you don't see the magical trig substitution at this point)?
tinytim
2009-04-03 21:54:02
Take out the square root?
Take out the square root?
rrusczyk
2009-04-03 21:54:17
That would be nice, but squaring this looks ugly.
That would be nice, but squaring this looks ugly.
rrusczyk
2009-04-03 21:54:38
What's ugly about the expression in the square root?
What's ugly about the expression in the square root?
rrusczyk
2009-04-03 21:54:46
(That is, what's hard to deal with?)
(That is, what's hard to deal with?)
funtwo
2009-04-03 21:54:49
4^n
4^n
ThinkFlow
2009-04-03 21:54:49
n in an exponenet
n in an exponenet
xelalex1231030
2009-04-03 21:54:49
the 4^n
the 4^n
tinytim
2009-04-03 21:54:49
4^n
4^n
rrusczyk
2009-04-03 21:54:56
Yeah, the 4^n is no fun.
Yeah, the 4^n is no fun.
rrusczyk
2009-04-03 21:55:01
What might we do about that?
What might we do about that?
xelalex1231030
2009-04-03 21:55:10
we could divide out by it
we could divide out by it
rrusczyk
2009-04-03 21:55:19
Let's see what that looks like:
Let's see what that looks like:
rrusczyk
2009-04-03 21:55:33
What does the resulting equation look like?
What does the resulting equation look like?
PI-Dimension
2009-04-03 21:56:22
8/5 a_n+6*2^n/5sqrt(1-thingy)
8/5 a_n+6*2^n/5sqrt(1-thingy)
rrusczyk
2009-04-03 21:56:36
rrusczyk
2009-04-03 21:57:05
Hmm. Not beautiful. But not quite as scary. What might we do to make it even less scary?
Hmm. Not beautiful. But not quite as scary. What might we do to make it even less scary?
xelalex1231030
2009-04-03 21:57:15
let b_n = a_n / 2^n
let b_n = a_n / 2^n
rrusczyk
2009-04-03 21:57:21
And why would we do that?
And why would we do that?
PI-Dimension
2009-04-03 21:57:41
get rid of the denominator
get rid of the denominator
ThinkFlow
2009-04-03 21:57:41
get rid of n in the exponent
get rid of n in the exponent
xelalex1231030
2009-04-03 21:57:41
then we could get rid of all the 2^n terms
then we could get rid of all the 2^n terms
rrusczyk
2009-04-03 21:57:42
Our goal here is to make the radical expression something we are less afraid of.
Our goal here is to make the radical expression something we are less afraid of.
rrusczyk
2009-04-03 21:57:56
xelal's substitution gives:
xelal's substitution gives:
rrusczyk
2009-04-03 21:58:19
rrusczyk
2009-04-03 21:58:30
Ugh. Now we have a's and b's. What do we do?
Ugh. Now we have a's and b's. What do we do?
chessmaster7
2009-04-03 21:58:57
write everything in terms of b_n?
write everything in terms of b_n?
bbill955
2009-04-03 21:58:57
convert a's to b's
convert a's to b's
ThinkFlow
2009-04-03 21:59:02
express everything in b's. The 2^n disappears as well
express everything in b's. The 2^n disappears as well
facis
2009-04-03 21:59:02
divide by 2^n and substitute more b's in
divide by 2^n and substitute more b's in
rrusczyk
2009-04-03 21:59:04
We sub in for b for the other a's as well.
We sub in for b for the other a's as well.
rrusczyk
2009-04-03 21:59:12
rrusczyk
2009-04-03 21:59:18
rrusczyk
2009-04-03 21:59:27
This is an equation you can almost take home to mom and dad.
This is an equation you can almost take home to mom and dad.
Yongyi781
2009-04-03 21:59:30
Trig substitution?
Trig substitution?
funtwo
2009-04-03 21:59:30
looks like pythagorean identity
looks like pythagorean identity
rrusczyk
2009-04-03 21:59:58
Now, the trig substitution becomes a little more apparent. But we don't need it to finish. What might we do now that we have a simpler equation?
Now, the trig substitution becomes a little more apparent. But we don't need it to finish. What might we do now that we have a simpler equation?
benzi455
2009-04-03 22:00:15
crank it
crank it
rrusczyk
2009-04-03 22:00:34
We can start cranking out terms. We already have the first few:
We can start cranking out terms. We already have the first few:
rrusczyk
2009-04-03 22:00:38
rrusczyk
2009-04-03 22:00:51
What happens when we go after the next one?
What happens when we go after the next one?
ThinkFlow
2009-04-03 22:00:57
difference of squares!
difference of squares!
rrusczyk
2009-04-03 22:01:06
This observation is particularly helpful here.
This observation is particularly helpful here.
hello123
2009-04-03 22:02:20
what?
what?
rrusczyk
2009-04-03 22:02:34
I mean, using difference of squares makes computing b_4 particularly easy.
I mean, using difference of squares makes computing b_4 particularly easy.
rrusczyk
2009-04-03 22:02:40
What is it?
What is it?
RoFlLoLcOpT
2009-04-03 22:03:34
$\frac{24}{25}$
$\frac{24}{25}$
rrusczyk
2009-04-03 22:03:37
ThinkFlow
2009-04-03 22:03:56
it repeats???
it repeats???
funtwo
2009-04-03 22:03:56
epic.
epic.
rrusczyk
2009-04-03 22:04:03
What does this tell us?
What does this tell us?
Bl00d
2009-04-03 22:04:05
wow
wow
rrusczyk
2009-04-03 22:04:07
yeah.
yeah.
xelalex1231030
2009-04-03 22:04:22
b_10 = 24/25 also
b_10 = 24/25 also
myrealname
2009-04-03 22:04:28
follow the pattern, you get b_10 = 24/25
follow the pattern, you get b_10 = 24/25
rrusczyk
2009-04-03 22:04:33
It means that the terms b_2, b_3, b_4, ... alternate between 24/25 and 117/125.
It means that the terms b_2, b_3, b_4, ... alternate between 24/25 and 117/125.
rrusczyk
2009-04-03 22:04:42
Bl00d
2009-04-03 22:04:44
Could we have just cranked it out from the very beginning?
Could we have just cranked it out from the very beginning?
rrusczyk
2009-04-03 22:04:59
You'd have to catch the fact that the numerators are going up by factors of 4.
You'd have to catch the fact that the numerators are going up by factors of 4.
m1sterzer0
2009-04-03 22:05:02
it takes a bit of chutzpah to keep churning after that 117/125
it takes a bit of chutzpah to keep churning after that 117/125
rrusczyk
2009-04-03 22:05:10
True, but it's a recursion, and the AIME.
True, but it's a recursion, and the AIME.
rrusczyk
2009-04-03 22:05:18
But the more natural approach is this:
But the more natural approach is this:
charles.du
2009-04-03 22:05:20
let b_n = sin theta
let b_n = sin theta
rrusczyk
2009-04-03 22:05:35
Once you see sqrt(1-b^2), this substitution should come to mind.
Once you see sqrt(1-b^2), this substitution should come to mind.
rrusczyk
2009-04-03 22:06:00
(It makes finishing the problem a good bit easier if you're careful.)
(It makes finishing the problem a good bit easier if you're careful.)
rrusczyk
2009-04-03 22:06:18
(It's also a more powerful general technique that you'll use again and again.
(It's also a more powerful general technique that you'll use again and again.
Bl00d
2009-04-03 22:06:27
Coould we have just calculated a4
Coould we have just calculated a4
rrusczyk
2009-04-03 22:06:35
Sure, but the a_i don't repeat!
Sure, but the a_i don't repeat!
rrusczyk
2009-04-03 22:06:46
Remember, the b_i are a_i divided by 2^i.
Remember, the b_i are a_i divided by 2^i.
rrusczyk
2009-04-03 22:07:08
Work out the trig substitution approach -- there's still a little trickery to work through there!
Work out the trig substitution approach -- there's still a little trickery to work through there!
hello123
2009-04-03 22:07:15
so b_n+1=4/5*sintheta +3/5*costheta how does this help?
so b_n+1=4/5*sintheta +3/5*costheta how does this help?
rrusczyk
2009-04-03 22:07:20
How does that help?
How does that help?
xelalex1231030
2009-04-03 22:07:45
sin addition formula
sin addition formula
funtwo
2009-04-03 22:08:02
3-4-5 triangle
3-4-5 triangle
rrusczyk
2009-04-03 22:08:21
There's your hint. I'll let you sort out the details :)
There's your hint. I'll let you sort out the details :)
rrusczyk
2009-04-03 22:08:33
One more problem!
One more problem!
rrusczyk
2009-04-03 22:08:39
rrusczyk
2009-04-03 22:08:57
Sigh. All those words.
Sigh. All those words.
rrusczyk
2009-04-03 22:09:02
We start with a diagram.
We start with a diagram.
rrusczyk
2009-04-03 22:09:05
rrusczyk
2009-04-03 22:09:24
As with many good geometry problems, there are all kinds of first steps we can take here. There are many extra segments we can add, many angles and lengths we can find, and many angles and lengths we can relate to each other.
As with many good geometry problems, there are all kinds of first steps we can take here. There are many extra segments we can add, many angles and lengths we can find, and many angles and lengths we can relate to each other.
rrusczyk
2009-04-03 22:09:31
Before getting lost in a thicket of these observations, let's focus our efforts. Let's come up with a plan by focusing on what we need to find.
Before getting lost in a thicket of these observations, let's focus our efforts. Let's come up with a plan by focusing on what we need to find.
rrusczyk
2009-04-03 22:09:38
We want to learn about d, which is PQ. What are likely strategies for doing so?
We want to learn about d, which is PQ. What are likely strategies for doing so?
rrusczyk
2009-04-03 22:09:45
For example, are we likely to be able to hit triangle PQC with the Law of Cosines?
For example, are we likely to be able to hit triangle PQC with the Law of Cosines?
funtwo
2009-04-03 22:10:25
well angle PCQ is constant
well angle PCQ is constant
rrusczyk
2009-04-03 22:10:36
True. A and B are fixed, so <PCQ is fixed.
True. A and B are fixed, so <PCQ is fixed.
rrusczyk
2009-04-03 22:10:50
But why are we scared about law of cosines on PQC?
But why are we scared about law of cosines on PQC?
hello123
2009-04-03 22:11:07
PC and QC are variing
PC and QC are variing
benzi455
2009-04-03 22:11:07
that's the only part of PQC that we know
that's the only part of PQC that we know
rrusczyk
2009-04-03 22:11:11
We know that <ACB is fixed, which is promising, but we know very little about PC and QC, so the Law of Cosines doesn't look too promising. The Law of Sines suffers from the same problem.
We know that <ACB is fixed, which is promising, but we know very little about PC and QC, so the Law of Cosines doesn't look too promising. The Law of Sines suffers from the same problem.
rrusczyk
2009-04-03 22:11:29
What's likely to be our general strategy for getting at PQ? What are we likely to go after?
What's likely to be our general strategy for getting at PQ? What are we likely to go after?
m1sterzer0
2009-04-03 22:12:00
MN - MP - QN?
MN - MP - QN?
e714212835
2009-04-03 22:12:00
MP and QN?
MP and QN?
rrusczyk
2009-04-03 22:12:18
The most obvious remaining strategy is to go after MP and QN, and then subtract from 1.
The most obvious remaining strategy is to go after MP and QN, and then subtract from 1.
rrusczyk
2009-04-03 22:12:34
Are we going to be able to combine them in some magical way, or should we go after them separately?
Are we going to be able to combine them in some magical way, or should we go after them separately?
tinytim
2009-04-03 22:12:58
separately
separately
dgreenb801
2009-04-03 22:12:58
separately, then add
separately, then add
ThinkFlow
2009-04-03 22:13:04
separately
separately
all4math
2009-04-03 22:13:04
separate
separate
rrusczyk
2009-04-03 22:13:06
We probably aren't going to be able to get at both of them at once, so let's focus on just one of them:
We probably aren't going to be able to get at both of them at once, so let's focus on just one of them:
rrusczyk
2009-04-03 22:13:09
rrusczyk
2009-04-03 22:13:26
(In complicated geometry problems, redrawing a simpler diagram with just the pieces you're focusing on will sometimes help a great deal. Here, we know we can ignore AC, since MP does not in any way depend on A or C.)
(In complicated geometry problems, redrawing a simpler diagram with just the pieces you're focusing on will sometimes help a great deal. Here, we know we can ignore AC, since MP does not in any way depend on A or C.)
ThinkFlow
2009-04-03 22:13:40
should be able to determine based on location of C
should be able to determine based on location of C
rrusczyk
2009-04-03 22:13:53
That's the goal. But how? What observations can we make?
That's the goal. But how? What observations can we make?
myrealname
2009-04-03 22:14:04
cross-chord theorem?
cross-chord theorem?
tinytim
2009-04-03 22:14:04
power of a point
power of a point
rrusczyk
2009-04-03 22:14:24
We have (MP)(PN) = (BP)(CP).
We have (MP)(PN) = (BP)(CP).
rrusczyk
2009-04-03 22:14:45
It's not clear what we'll do with that.
It's not clear what we'll do with that.
rrusczyk
2009-04-03 22:15:00
We'll stick it up there anyway in case it becomes handy.
We'll stick it up there anyway in case it becomes handy.
xelalex1231030
2009-04-03 22:15:02
but PN = 1 - MP
but PN = 1 - MP
dgreenb801
2009-04-03 22:15:03
PN=1-MP
PN=1-MP
rrusczyk
2009-04-03 22:15:10
We know that MN = 1, so MP = 1 - PN.
We know that MN = 1, so MP = 1 - PN.
rrusczyk
2009-04-03 22:15:24
OK. We could combine that with the power of a point equation:
OK. We could combine that with the power of a point equation:
tinytim
2009-04-03 22:15:26
(1-MP)(MP)=(BP)(CP)
(1-MP)(MP)=(BP)(CP)
rrusczyk
2009-04-03 22:15:34
Not so obvious what to do with that...
Not so obvious what to do with that...
rrusczyk
2009-04-03 22:15:36
Hmmm.
Hmmm.
rrusczyk
2009-04-03 22:15:59
Stuck. On a geometry problem. What do we wonder when we are stuck on a geometry problem?
Stuck. On a geometry problem. What do we wonder when we are stuck on a geometry problem?
ThinkFlow
2009-04-03 22:16:13
What have we not used?
What have we not used?
MathAndKnowledge
2009-04-03 22:16:18
what haven't we used yet
what haven't we used yet
rrusczyk
2009-04-03 22:16:28
Exactly -- that's what we ask when we are stuck on a geometry problem.
Exactly -- that's what we ask when we are stuck on a geometry problem.
dgreenb801
2009-04-03 22:16:32
we havent used diameter
we havent used diameter
ThinkFlow
2009-04-03 22:16:32
MB=3/5
MB=3/5
m1sterzer0
2009-04-03 22:16:32
BM=3/5
BM=3/5
hello123
2009-04-03 22:16:32
location of B
location of B
tinytim
2009-04-03 22:16:32
MB=3/5
MB=3/5
rrusczyk
2009-04-03 22:16:51
Haven't used the fact that MN is a diameter, and we haven't touched the length of BM.
Haven't used the fact that MN is a diameter, and we haven't touched the length of BM.
rrusczyk
2009-04-03 22:16:56
How can we use these?
How can we use these?
hello123
2009-04-03 22:17:14
BN=4/5
BN=4/5
rrusczyk
2009-04-03 22:17:34
Since MN is a diameter, BMN is a right triangle, and BN = 4/5.
Since MN is a diameter, BMN is a right triangle, and BN = 4/5.
rrusczyk
2009-04-03 22:17:46
Hmm. What else does that suggest we look at?
Hmm. What else does that suggest we look at?
ThinkFlow
2009-04-03 22:18:10
MCN
MCN
rrusczyk
2009-04-03 22:18:16
And what do we know about that?
And what do we know about that?
tinytim
2009-04-03 22:18:37
also right triangle
also right triangle
e714212835
2009-04-03 22:18:37
its a right triangle, too.
its a right triangle, too.
spelljack825
2009-04-03 22:18:37
right angle triangle
right angle triangle
dgreenb801
2009-04-03 22:18:37
its right
its right
rrusczyk
2009-04-03 22:18:43
rrusczyk
2009-04-03 22:18:58
OK, we now have some right triangles, and we know BN.
OK, we now have some right triangles, and we know BN.
rrusczyk
2009-04-03 22:19:03
Um, now what?
Um, now what?
tinytim
2009-04-03 22:19:32
cyclic quad?
cyclic quad?
rrusczyk
2009-04-03 22:19:44
True. Are we excited about breaking out Ptolemy?
True. Are we excited about breaking out Ptolemy?
tinytim
2009-04-03 22:19:54
no...
no...
xelalex1231030
2009-04-03 22:19:54
not really
not really
rrusczyk
2009-04-03 22:20:01
Ptolemy: (MN)(BC) = (BN)(CM) + (BM)(CN)
Ptolemy: (MN)(BC) = (BN)(CM) + (BM)(CN)
rrusczyk
2009-04-03 22:20:04
This doesn't involve MP or PN. No good.
This doesn't involve MP or PN. No good.
rrusczyk
2009-04-03 22:20:12
Hurm.
Hurm.
rrusczyk
2009-04-03 22:20:34
Still stuck. The only useful relationships we have for MP so far are those we have above.
Still stuck. The only useful relationships we have for MP so far are those we have above.
rrusczyk
2009-04-03 22:20:40
Those with MP and PN.
Those with MP and PN.
rrusczyk
2009-04-03 22:20:47
MP and PN . . .
MP and PN . . .
rrusczyk
2009-04-03 22:20:50
MP and PN . . .
MP and PN . . .
rrusczyk
2009-04-03 22:20:56
Tell me about those two.
Tell me about those two.
ThinkFlow
2009-04-03 22:21:01
Which sum to 1
Which sum to 1
Yongyi781
2009-04-03 22:21:01
MP + PN = 1...
MP + PN = 1...
rrusczyk
2009-04-03 22:21:05
True. . .
True. . .
rrusczyk
2009-04-03 22:21:14
What else is interesting about those two segments?
What else is interesting about those two segments?
late20s
2009-04-03 22:21:37
the base of two triangles?
the base of two triangles?
rrusczyk
2009-04-03 22:21:54
And what does that make us think?
And what does that make us think?
tinytim
2009-04-03 22:22:10
area?
area?
funtwo
2009-04-03 22:22:10
areas?
areas?
Yongyi781
2009-04-03 22:22:10
Area?
Area?
rrusczyk
2009-04-03 22:22:35
. . . Interesting. And how will we relate these two segments to area?
. . . Interesting. And how will we relate these two segments to area?
dgreenb801
2009-04-03 22:22:51
part of similar triangles?
part of similar triangles?
rrusczyk
2009-04-03 22:22:57
They are part of similar triangles.
They are part of similar triangles.
rrusczyk
2009-04-03 22:23:02
So?
So?
rrusczyk
2009-04-03 22:23:08
What else are they part of ?
What else are they part of ?
ThinkFlow
2009-04-03 22:23:11
They are a diameter?
They are a diameter?
rrusczyk
2009-04-03 22:23:21
They are part of the same *line*.
They are part of the same *line*.
rrusczyk
2009-04-03 22:23:34
This leads me to one of my very favorite sneaky geometry strategies.
This leads me to one of my very favorite sneaky geometry strategies.
tinytim
2009-04-03 22:23:38
their ratio is equal to the ratio of the areas of the triangles
their ratio is equal to the ratio of the areas of the triangles
benzi455
2009-04-03 22:23:38
MP is to [BMC] as NP is to [BNC]
MP is to [BMC] as NP is to [BNC]
rrusczyk
2009-04-03 22:23:41
Whenever I have two important segment lengths along the same line in a tough geometry problem, I consider using area as a tool. We haveMP/PN = [BMP]/[BNP].
Whenever I have two important segment lengths along the same line in a tough geometry problem, I consider using area as a tool. We haveMP/PN = [BMP]/[BNP].
rrusczyk
2009-04-03 22:23:53
Do we know anything about the areas of these triangles that might be helpful?
Do we know anything about the areas of these triangles that might be helpful?
ThinkFlow
2009-04-03 22:24:11
They add up to 6/5
They add up to 6/5
rrusczyk
2009-04-03 22:24:16
True.
True.
rrusczyk
2009-04-03 22:24:21
Anything else?
Anything else?
tinytim
2009-04-03 22:24:40
6/25 you mean?
6/25 you mean?
rrusczyk
2009-04-03 22:24:43
Ya.
Ya.
rrusczyk
2009-04-03 22:24:45
(We know the sum PM + PN, too. . . )
(We know the sum PM + PN, too. . . )
rrusczyk
2009-04-03 22:24:50
What else?
What else?
ThinkFlow
2009-04-03 22:24:59
proportional to MCN
proportional to MCN
rrusczyk
2009-04-03 22:25:07
What do you mean?
What do you mean?
tinytim
2009-04-03 22:25:58
[BPC]/[PCN]=[BPM]/[MPC]
[BPC]/[PCN]=[BPM]/[MPC]
spelljack825
2009-04-03 22:25:59
MP/PN = [BMP]/[BNP] = [MPC]/[NPC]
MP/PN = [BMP]/[BNP] = [MPC]/[NPC]
rrusczyk
2009-04-03 22:26:02
We similarly have MP/PN = [PMC]/[PCN].
We similarly have MP/PN = [PMC]/[PCN].
rrusczyk
2009-04-03 22:26:27
We don't know a whole lot about those two triangles. Hurm.
We don't know a whole lot about those two triangles. Hurm.
rrusczyk
2009-04-03 22:26:44
But these triangle area equalities seem interesting. . .
But these triangle area equalities seem interesting. . .
dgreenb801
2009-04-03 22:27:12
a/b=c/d means a+b/c+d equals them?
a/b=c/d means a+b/c+d equals them?
rrusczyk
2009-04-03 22:27:22
True; and what does that give us here?
True; and what does that give us here?
dgreenb801
2009-04-03 22:27:47
[BMC][BNC]
[BMC][BNC]
rrusczyk
2009-04-03 22:27:53
We can combine the two area relationships we just found and note thatMP/PN = [BMC]/[BNC].
We can combine the two area relationships we just found and note thatMP/PN = [BMC]/[BNC].
rrusczyk
2009-04-03 22:28:01
Is there anything interesting about these two triangles?
Is there anything interesting about these two triangles?
mathq
2009-04-03 22:28:45
share BC
share BC
late20s
2009-04-03 22:28:45
same base: BC
same base: BC
rrusczyk
2009-04-03 22:28:47
What else?
What else?
BenM
2009-04-03 22:29:05
related angles
related angles
rrusczyk
2009-04-03 22:29:08
How?
How?
spelljack825
2009-04-03 22:29:10
BNC + BMC = 180..?
BNC + BMC = 180..?
ThinkFlow
2009-04-03 22:29:10
M is the supplement of N
M is the supplement of N
m1sterzer0
2009-04-03 22:29:13
angle M and angle N are supplementary
angle M and angle N are supplementary
rrusczyk
2009-04-03 22:29:21
Aha! Why is that interesting?
Aha! Why is that interesting?
dgreenb801
2009-04-03 22:29:33
area=(1/2)ab sin C?
area=(1/2)ab sin C?
benzi455
2009-04-03 22:29:33
same sine
same sine
m1sterzer0
2009-04-03 22:29:37
same sine
same sine
rrusczyk
2009-04-03 22:29:59
Ah, this might be an interesting tree to bark at; what do we find for MP/PN starting from this observation?
Ah, this might be an interesting tree to bark at; what do we find for MP/PN starting from this observation?
dgreenb801
2009-04-03 22:31:14
MP/PN=(3MC)/(4NC)
MP/PN=(3MC)/(4NC)
tinytim
2009-04-03 22:31:24
that is (BM)(MC)/(BN)(CN)
that is (BM)(MC)/(BN)(CN)
xelalex1231030
2009-04-03 22:31:24
MP/PN = (BM)(MC)/(BN)(NC)
MP/PN = (BM)(MC)/(BN)(NC)
rrusczyk
2009-04-03 22:31:54
rrusczyk
2009-04-03 22:32:11
Interesting
Interesting
rrusczyk
2009-04-03 22:32:14
Why is this *particularly interesting* in this problem?
Why is this *particularly interesting* in this problem?
rrusczyk
2009-04-03 22:33:23
What I'm getting at here is that we've related MP/PN to MC/NC. MC and NC have *nothing to do with B*.
What I'm getting at here is that we've related MP/PN to MC/NC. MC and NC have *nothing to do with B*.
late20s
2009-04-03 22:33:49
cuz the other line share the same property
cuz the other line share the same property
rrusczyk
2009-04-03 22:33:55
That means we can do essentially the same thing with point A to get an expression for MQ/QN in terms of C, not A or B. In other words, we can relate MP and QN, which is the goal in getting at PQ.
That means we can do essentially the same thing with point A to get an expression for MQ/QN in terms of C, not A or B. In other words, we can relate MP and QN, which is the goal in getting at PQ.
rrusczyk
2009-04-03 22:34:04
BenM
2009-04-03 22:35:36
Won't MQ/QN be equal to MC/CN
Won't MQ/QN be equal to MC/CN
rrusczyk
2009-04-03 22:35:45
We do the same thing with Q:
We do the same thing with Q:
rrusczyk
2009-04-03 22:35:46
rrusczyk
2009-04-03 22:36:02
ThinkFlow
2009-04-03 22:36:14
Because MAN is isoceles, right?
Because MAN is isoceles, right?
rrusczyk
2009-04-03 22:36:16
Right.
Right.
rrusczyk
2009-04-03 22:36:43
Now, what are we going to do.
Now, what are we going to do.
rrusczyk
2009-04-03 22:36:49
We keep our eye on the ball. We want an expression for PQ that we can use. We have PQ = 1 - MP - QN.
We keep our eye on the ball. We want an expression for PQ that we can use. We have PQ = 1 - MP - QN.
rrusczyk
2009-04-03 22:37:03
So, what do we have to extract from those two equations up there?
So, what do we have to extract from those two equations up there?
hello123
2009-04-03 22:37:14
find expressions for PN and QN?
find expressions for PN and QN?
myrealname
2009-04-03 22:37:35
isolate MP and QN
isolate MP and QN
rrusczyk
2009-04-03 22:37:37
We need expressions for MP and QN (PN and QN would be fine, too).
We need expressions for MP and QN (PN and QN would be fine, too).
rrusczyk
2009-04-03 22:37:41
And how will we get these?
And how will we get these?
ThinkFlow
2009-04-03 22:37:43
Pn = 1-Mp and MQ = 1- QN
Pn = 1-Mp and MQ = 1- QN
rrusczyk
2009-04-03 22:38:08
And what will we do about the right sides of those ratio equations to make this easier to deal with?
And what will we do about the right sides of those ratio equations to make this easier to deal with?
BenM
2009-04-03 22:38:25
make a new variable?
make a new variable?
rrusczyk
2009-04-03 22:38:31
We can express both MP/PN and MQ/QN in terms of MC/NC. Moreover, we have PN = 1-MP and MQ = 1-QN, so we can express MP and QN in terms of MC/NC. That means we can express PQ in terms of this ratio. Let's give it a try.
We can express both MP/PN and MQ/QN in terms of MC/NC. Moreover, we have PN = 1-MP and MQ = 1-QN, so we can express MP and QN in terms of MC/NC. That means we can express PQ in terms of this ratio. Let's give it a try.
spelljack825
2009-04-03 22:38:34
substitute a = MC/NC
substitute a = MC/NC
rrusczyk
2009-04-03 22:38:39
rrusczyk
2009-04-03 22:39:06
Now we're getting somewhere. We find MP and QN:
Now we're getting somewhere. We find MP and QN:
rrusczyk
2009-04-03 22:39:09
ThinkFlow
2009-04-03 22:39:11
Now we can solve for MP and QN in terms of t, right?
Now we can solve for MP and QN in terms of t, right?
rrusczyk
2009-04-03 22:39:14
Exactly.
Exactly.
rrusczyk
2009-04-03 22:39:19
And what do we do with these?
And what do we do with these?
BenM
2009-04-03 22:39:31
substitute into that expression we had for PQ
substitute into that expression we had for PQ
ThinkFlow
2009-04-03 22:39:34
Subtract from 1
Subtract from 1
rrusczyk
2009-04-03 22:39:43
rrusczyk
2009-04-03 22:39:49
What do we do with that?
What do we do with that?
BenM
2009-04-03 22:39:56
simplify
simplify
mathq
2009-04-03 22:39:56
simplify
simplify
rrusczyk
2009-04-03 22:40:03
rrusczyk
2009-04-03 22:40:07
rrusczyk
2009-04-03 22:40:21
This is promising. It's a quadratic. The AIME-ized form of the answer looks like the root of a quadratic equation. We're on the right track.
This is promising. It's a quadratic. The AIME-ized form of the answer looks like the root of a quadratic equation. We're on the right track.
rrusczyk
2009-04-03 22:40:26
But what do we do now?
But what do we do now?
BenM
2009-04-03 22:40:47
solve for t
solve for t
ThinkFlow
2009-04-03 22:40:49
maximize d
maximize d
rrusczyk
2009-04-03 22:40:51
How?
How?
rrusczyk
2009-04-03 22:40:58
What do we know about t?
What do we know about t?
late20s
2009-04-03 22:41:32
oh, t is positive
oh, t is positive
BenM
2009-04-03 22:41:37
t is positive
t is positive
spelljack825
2009-04-03 22:41:37
t is positive?
t is positive?
rrusczyk
2009-04-03 22:41:41
Do we know anything else about it?
Do we know anything else about it?
rrusczyk
2009-04-03 22:42:18
What is t defined as?
What is t defined as?
dgreenb801
2009-04-03 22:42:22
t is a root of the quadratic
t is a root of the quadratic
rrusczyk
2009-04-03 22:42:30
It is a root of the quadratic.
It is a root of the quadratic.
mathq
2009-04-03 22:42:33
MC/NC
MC/NC
tinytim
2009-04-03 22:42:33
MC/CN
MC/CN
ThinkFlow
2009-04-03 22:42:33
MC/NC
MC/NC
rrusczyk
2009-04-03 22:42:41
It is a ratio. This ratio is positive.
It is a ratio. This ratio is positive.
rrusczyk
2009-04-03 22:42:46
It is also . . .
It is also . . .
m1sterzer0
2009-04-03 22:42:48
it can range from 0+ --> infinity
it can range from 0+ --> infinity
rrusczyk
2009-04-03 22:42:50
real.
real.
rrusczyk
2009-04-03 22:42:56
Which tells us . . .
Which tells us . . .
dgreenb801
2009-04-03 22:43:16
discriminant is positive!
discriminant is positive!
MathAndKnowledge
2009-04-03 22:43:16
discriminant >= 0
discriminant >= 0
rrusczyk
2009-04-03 22:43:32
The discriminant of the quadratic is nonnegative.
The discriminant of the quadratic is nonnegative.
rrusczyk
2009-04-03 22:43:37
rrusczyk
2009-04-03 22:43:41
There was the equation we had.
There was the equation we had.
rrusczyk
2009-04-03 22:44:18
What do we do now?
What do we do now?
funtwo
2009-04-03 22:44:45
move the t over
move the t over
rrusczyk
2009-04-03 22:44:50
MathAndKnowledge
2009-04-03 22:45:17
d^2-14d+1>=0 i think by plugging in the discriminant
d^2-14d+1>=0 i think by plugging in the discriminant
tinytim
2009-04-03 22:45:17
D=d^2-14d+1
D=d^2-14d+1
funtwo
2009-04-03 22:45:23
compute the discriminant in terms of d
compute the discriminant in terms of d
rrusczyk
2009-04-03 22:45:29
This quadratic in t must have real solutions. That means that its discriminant must be nonnegative. We're set now.
This quadratic in t must have real solutions. That means that its discriminant must be nonnegative. We're set now.
rrusczyk
2009-04-03 22:45:33
rrusczyk
2009-04-03 22:46:08
And how do we finish this?
And how do we finish this?
ThinkFlow
2009-04-03 22:46:33
draops
2009-04-03 22:46:33
quadratic formula
quadratic formula
aggarwal
2009-04-03 22:46:33
complete the square
complete the square
funtwo
2009-04-03 22:46:33
14-rt192/2=7-4rt3 so 7+4+3=14
14-rt192/2=7-4rt3 so 7+4+3=14
rrusczyk
2009-04-03 22:46:43
Completing the square or the quadratic formula tell us:
Completing the square or the quadratic formula tell us:
rrusczyk
2009-04-03 22:46:46
rrusczyk
2009-04-03 22:47:02
What does this tell you about the quadratic d^2 - 14d +1 ?
What does this tell you about the quadratic d^2 - 14d +1 ?
rrusczyk
2009-04-03 22:47:43
Remember, we had the inequality d^2 - 14d + 1 >= 0
Remember, we had the inequality d^2 - 14d + 1 >= 0
rrusczyk
2009-04-03 22:47:58
We just found the roots. How can we finish?
We just found the roots. How can we finish?
funtwo
2009-04-03 22:48:00
it cant be more than 1 the diameter and it cant be between 7-4rt3 and 7+4rt3 so the maximum is 7-4rt3
it cant be more than 1 the diameter and it cant be between 7-4rt3 and 7+4rt3 so the maximum is 7-4rt3
rrusczyk
2009-04-03 22:48:14
mathq
2009-04-03 22:48:21
7-4sqrt3 because it's less than 1
7-4sqrt3 because it's less than 1
rrusczyk
2009-04-03 22:48:23
rrusczyk
2009-04-03 22:48:29
This was a hard, hard problem.
This was a hard, hard problem.
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