2008 AIME Math Jam
Go back to the Math Jam ArchiveAoPS instructors will lead a discussion on all 15 problems from the 2008 AIME.
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Facilitator: Dave Patrick
DPatrick19:00:01
Hello and welcome to the 2008 AIME I Math Jam!
Hello and welcome to the 2008 AIME I Math Jam!
DPatrick19:00:12
Before we get started, please note that this 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 discussion organized and on track.
Before we get started, please note that this 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 discussion organized and on track.
DPatrick19:00:23
This also means that only well-written comments will be dropped into the classroom, so please take time to write responses that are complete and easy to read.
This also means that only well-written comments will be dropped into the classroom, so please take time to write responses that are complete and easy to read.
DPatrick19:00:46
Due to the large size of this Math Jam, I won't be able to respond to every individual question, nor post a lot of responses. Please do not complain if your comments are not responded to.
Due to the large size of this Math Jam, I won't be able to respond to every individual question, nor post a lot of responses. Please do not complain if your comments are not responded to.
DPatrick19:00:58
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.
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.
DPatrick19:01:22
These solutions are the ones that (to me) are the most logical, the simplest, and/or the ones that you'd be most likely to think of on your own.
These solutions are the ones that (to me) are the most logical, the simplest, and/or the ones that you'd be most likely to think of on your own.
DPatrick19:01:30
Without further ado, let's get started!
Without further ado, let's get started!
DPatrick19:01:35
DPatrick19:01:39
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DPatrick19:01:44
(I will post a link to each problem after I post the problem statement. You should be able to click on the link to view the problem in a separate window. You may have to hold the Ctrl key, and depending on how your popup-blocker is configured it may not work at all for you.)
(I will post a link to each problem after I post the problem statement. You should be able to click on the link to view the problem in a separate window. You may have to hold the Ctrl key, and depending on how your popup-blocker is configured it may not work at all for you.)
107740019:02:27
make expressions on the # of girls before and after the boys come and set them equal
make expressions on the # of girls before and after the boys come and set them equal
syckls19:02:32
b + g = n, so g = .6n
b + g = n, so g = .6n
syckls19:02:32
also g = .58(n + 20)
also g = .58(n + 20)
DPatrick19:03:03
Right..most of the difficulty here is converting the word problem into mathematics in a nice way.
Right..most of the difficulty here is converting the word problem into mathematics in a nice way.
DPatrick19:03:19
Let n be the total number of students before the extra boys show up, and g be the number of girls.
Let n be the total number of students before the extra boys show up, and g be the number of girls.
unlimitd3K19:03:30
First find out how many total students were at the dance initially: .60n = .58(n+20)
First find out how many total students were at the dance initially: .60n = .58(n+20)
DPatrick19:03:41
Right, there are two equations for g in terms of n.
Right, there are two equations for g in terms of n.
DPatrick19:03:46
Before the extra boys show up, we have .6n = g.
Before the extra boys show up, we have .6n = g.
DPatrick19:03:53
After the extra boys show up, we have .58(n+20) = g.
After the extra boys show up, we have .58(n+20) = g.
DPatrick19:04:03
We set them equal and solve for n.
We set them equal and solve for n.
ebaudry1019:04:10
2n=1160
2n=1160
DPatrick19:04:20
Yes, I would multiply by 100 first to make life a little simpler.
Yes, I would multiply by 100 first to make life a little simpler.
BOGTRO19:04:27
n=580
n=580
ebaudry1019:04:27
so n = 580
so n = 580
DPatrick19:04:33
2n - 1160 = 0, so n = 580.
2n - 1160 = 0, so n = 580.
yjneb19:04:57
after we find n, we need .4*n+20=.4*580+20=252
after we find n, we need .4*n+20=.4*580+20=252
syckls19:04:57
then the number that want to dance = .4 * 580 + 20
then the number that want to dance = .4 * 580 + 20
ebaudry1019:04:59
then students who like do dance = .4n + 20, or 58 * 4 + 20, or 252
then students who like do dance = .4n + 20, or 58 * 4 + 20, or 252
DPatrick19:05:07
Right. Before the extra boys show up, .4n = .4(580) = 232 students like to dance.
Right. Before the extra boys show up, .4n = .4(580) = 232 students like to dance.
DPatrick19:05:12
Adding the 20 extra boys, we have a total of 232 + 20 = 252 students who like to dance.
Adding the 20 extra boys, we have a total of 232 + 20 = 252 students who like to dance.
DPatrick19:05:32
DPatrick19:05:37
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susanzduan19:05:52
diagram!
diagram!
davidyko19:05:52
Hah, draw a diagram.
Hah, draw a diagram.
DPatrick19:05:54
We'll want to start by drawing a picture...but it's important for the picture to be relatively accurate.
We'll want to start by drawing a picture...but it's important for the picture to be relatively accurate.
syckls19:06:10
80 > .5 * 10^2, so we know that the height of the triangle is greater than 10
80 > .5 * 10^2, so we know that the height of the triangle is greater than 10
unlimitd3K19:06:14
We know that G has to be outside of AIME because if it is inside, the max area of GEM is 50
We know that G has to be outside of AIME because if it is inside, the max area of GEM is 50
DPatrick19:06:20
Yes -- the key fact in our picture is that G must be outside of AIME. (If G is inside AIME, then the triangle only has area at most 50, but we need it to have at least 80.)
Yes -- the key fact in our picture is that G must be outside of AIME. (If G is inside AIME, then the triangle only has area at most 50, but we need it to have at least 80.)
DPatrick19:06:29
So here's our picture:
So here's our picture:
DPatrick19:06:36
DPatrick19:06:37
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/aa73ffd7b1f6322563afb4afb878ccec.png
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/aa73ffd7b1f6322563afb4afb878ccec.png
pythag01119:07:00
The are common to both figures is a trapezoid
The are common to both figures is a trapezoid
mchoi81519:07:00
The area in common is a trapezoid with larger base 10 and height 10.
The area in common is a trapezoid with larger base 10 and height 10.
Paboga19:07:00
The intersection is an isoceles trapezoid with h = 10 and bases 10 and 6
The intersection is an isoceles trapezoid with h = 10 and bases 10 and 6
DPatrick19:07:20
Right. The overlapping area is an isosceles trapezoid with height 10 and one base 10, and area 80.
Right. The overlapping area is an isosceles trapezoid with height 10 and one base 10, and area 80.
DPatrick19:07:46
Since the area of a trapezoid is its height times the average of the bases, the average of the bases has to be 8. Thus the top length of our trapezoid is 6.
Since the area of a trapezoid is its height times the average of the bases, the average of the bases has to be 8. Thus the top length of our trapezoid is 6.
DPatrick19:08:00
And to finish...
And to finish...
lingomaniac8819:08:09
similar triangles
similar triangles
BOGTRO19:08:09
And then we just get simple similar triangles.
And then we just get simple similar triangles.
susanzduan19:08:09
we can use similar triangles
we can use similar triangles
DPatrick19:08:19
Right -- the little triangle in the above picture is similar to the big triangle.
Right -- the little triangle in the above picture is similar to the big triangle.
DPatrick19:08:40
You might be able to "eyeball" the answer from here, but if you wanted to set up an equation for h...
You might be able to "eyeball" the answer from here, but if you wanted to set up an equation for h...
DPatrick19:08:56
jamieding5119:09:16
h = 25
h = 25
BOGTRO19:09:16
So h=025
So h=025
lingomaniac8819:09:16
therefore h=25
therefore h=25
mz9419:09:22
crossmultiply to get 4h=100; h=25
crossmultiply to get 4h=100; h=25
DPatrick19:09:34
Right: or, equivalently, 1 - 10/h = 6/10, so 10/h = 4/10, and h = 100/4 = 025.
Right: or, equivalently, 1 - 10/h = 6/10, so 10/h = 4/10, and h = 100/4 = 025.
DPatrick19:09:58
DPatrick19:10:02
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mathnerd31419:10:19
set up a system of equations
set up a system of equations
davidyko19:10:19
System of two equations in three variables.
System of two equations in three variables.
Ratclaw19:10:19
Set up the two equations with three variables
Set up the two equations with three variables
DPatrick19:10:27
Ok...we can let b, j, and s be their biking, jogging, and swimming rates.
Ok...we can let b, j, and s be their biking, jogging, and swimming rates.
unlimitd3K19:10:40
2b + 3j + 4s = 74
4b + 2j + 3s = 91
2b + 3j + 4s = 74
4b + 2j + 3s = 91
DPatrick19:10:51
Indeed, that's our system of equations.
Indeed, that's our system of equations.
DPatrick19:11:00
There are potentially lots of ways we can proceed.
There are potentially lots of ways we can proceed.
l Bob l19:11:10
multiply the top equation by 2 and subtract the two equations
multiply the top equation by 2 and subtract the two equations
mz9419:11:13
multiply the first equation by 2 to get rid of a variable
multiply the first equation by 2 to get rid of a variable
cyberspace19:11:16
double the first, subtract the second
double the first, subtract the second
DPatrick19:11:23
That's good...that will eliminate b.
That's good...that will eliminate b.
DPatrick19:11:27
4j + 5s = 57.
4j + 5s = 57.
DPatrick19:11:32
So what?
So what?
BOGTRO19:11:48
and now you have to guess and check
and now you have to guess and check
mathnerd31419:11:48
Guess and check?
Guess and check?
DPatrick19:11:52
You could...but you don't need to.
You could...but you don't need to.
Currylover19:12:05
mods?
mods?
l Bob l19:12:05
use modular arithmetic
use modular arithmetic
unlimitd3K19:12:05
we know that j and s are integers, and that 4j = 2 mod 5
we know that j and s are integers, and that 4j = 2 mod 5
DPatrick19:12:29
Right. Looking at 4j + 5s = 57, we see that 4j must be 2 (mod 5), so j must be 3 (mod 5).
Right. Looking at 4j + 5s = 57, we see that 4j must be 2 (mod 5), so j must be 3 (mod 5).
mchoi81519:12:48
j = 3,8,3+5k
j = 3,8,3+5k
mnmath19:12:48
j=3,8,13
j=3,8,13
tenniskidperson319:12:48
3, 8, 13...
3, 8, 13...
DPatrick19:12:54
You can narrow it even further.
You can narrow it even further.
DPatrick19:13:06
Our first equation was: 2b + 3j + 4s = 74
Our first equation was: 2b + 3j + 4s = 74
tenniskidperson319:13:09
has to be even
has to be even
DPatrick19:13:19
Right...j has to be even, because all the other terms are even.
Right...j has to be even, because all the other terms are even.
mnmath19:13:24
j=8
j=8
BOGTRO19:13:24
so it must be 8
so it must be 8
DPatrick19:13:37
Right. We no know j = 8 (mod 10).
Right. We no know j = 8 (mod 10).
DPatrick19:13:41
...now know...
...now know...
DPatrick19:13:47
But j=18 is too big for 4j + 5s = 57.
But j=18 is too big for 4j + 5s = 57.
DPatrick19:14:01
So j = 8 is the only possibilites (no guessing required!)
So j = 8 is the only possibilites (no guessing required!)
tenniskidperson319:14:08
use that to solve for b and s
use that to solve for b and s
mchoi81519:14:12
Therefore s = 5
Therefore s = 5
Kongster19:14:21
plug that back in and s = 5
plug that back in and s = 5
DPatrick19:14:28
Then we easily solve for s = 5 and b = 15. (At this point, I'd check these in our original equations, just to be sure.)
Then we easily solve for s = 5 and b = 15. (At this point, I'd check these in our original equations, just to be sure.)
DPatrick19:14:38
So the answer is 5^2 + 8^2 + 15^2 = 25 + 64 + 225 = 314.
So the answer is 5^2 + 8^2 + 15^2 = 25 + 64 + 225 = 314.
DPatrick19:15:06
DPatrick19:15:11
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davidyko19:15:30
Complete the square.
Complete the square.
ebaudry1019:15:30
Complete the square
Complete the square
yjneb19:15:30
complete the square
complete the square
Brut3Forc319:15:30
Complete the square on x.
Complete the square on x.
Kongster19:15:30
complete the square
complete the square
mchoi81519:15:30
Complete the square.
Complete the square.
mnmath19:15:30
complete the square
complete the square
LawOfSigns19:15:30
complete the square for x
complete the square for x
alkjash19:15:30
complete the square
complete the square
DPatrick19:15:43
Great minds think alike...the left side is a little hard to deal with, so let's complete the square.
Great minds think alike...the left side is a little hard to deal with, so let's complete the square.
DPatrick19:15:50
rowan19:16:02
difference of 2 squares
difference of 2 squares
davidyko19:16:02
Diff. of squares.
Diff. of squares.
james4l19:16:02
difference of squares
difference of squares
DPatrick19:16:16
First, to make things a little simpler, I'd let z = x+42. So z^2 + 244 = y^2.
First, to make things a little simpler, I'd let z = x+42. So z^2 + 244 = y^2.
DPatrick19:16:25
Thus 244 = y^2 - z^2 = (y+z)(y-z).
Thus 244 = y^2 - z^2 = (y+z)(y-z).
mchoi81519:16:33
244 = 2*2*61
244 = 2*2*61
zephyredx19:16:36
both must be even
both must be even
DPatrick19:17:04
Right...y+z and y-z have the same parity (they're either both odd or both even), so the only possibility is y+z = 122 and y-z = 2.
Right...y+z and y-z have the same parity (they're either both odd or both even), so the only possibility is y+z = 122 and y-z = 2.
karatemagic719:17:29
122 and 2 are y+z and y-z
122 and 2 are y+z and y-z
karatemagic719:17:32
62 and 60
62 and 60
DPatrick19:17:38
This gives y = 62 and z = 60.
This gives y = 62 and z = 60.
Kongster19:17:50
x=18
x=18
mathnerd31419:17:52
so x=18, y=62
so x=18, y=62
DPatrick19:18:02
Hence x = 18 and y = 62 (again, I'd check these before writing the final answer).
Hence x = 18 and y = 62 (again, I'd check these before writing the final answer).
DPatrick19:18:07
So the answer is 18+62 = 080.
So the answer is 18+62 = 080.
DPatrick19:18:32
DPatrick19:18:36
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DPatrick19:19:07
There's one thing you can do right away that makes life simpler.
There's one thing you can do right away that makes life simpler.
DPatrick19:19:37
Since everything scales in this problem, we might as well assume that r = 1.
Since everything scales in this problem, we might as well assume that r = 1.
DPatrick19:19:48
So all we're trying to do is find h.
So all we're trying to do is find h.
DPatrick19:20:00
What is the radius of the circle traced out by the cone?
What is the radius of the circle traced out by the cone?
Alumen19:20:17
the slant height
the slant height
Pavan19:20:17
the slant height
the slant height
Ratclaw19:20:19
the slant height of the cone
the slant height of the cone
MathPie19:20:19
The slant height of the cone
The slant height of the cone
DPatrick19:20:32
Right...the radius of the circle is the "slant height" of the cone, so it's sqrt(h^2 + 1).
Right...the radius of the circle is the "slant height" of the cone, so it's sqrt(h^2 + 1).
DPatrick19:20:41
It's the hypotenuse of a right triangle with base 1 (the radius of the base of the cone) and height h (the height of the cone).
It's the hypotenuse of a right triangle with base 1 (the radius of the base of the cone) and height h (the height of the cone).
DPatrick19:20:53
DPatrick19:21:04
On the other hand, how else can we express this?
On the other hand, how else can we express this?
yjneb19:21:23
17 times the circumference of the base of the cone
17 times the circumference of the base of the cone
susanzduan19:21:25
17*2pi
17*2pi
ebaudry1019:21:29
17*2*pi, or 17 times the circumference of the base of the cone
17*2*pi, or 17 times the circumference of the base of the cone
DPatrick19:21:33
The cone makes 17 rotations, so the circumference of the base of the cone gets traced out 17 times.
The cone makes 17 rotations, so the circumference of the base of the cone gets traced out 17 times.
DPatrick19:21:43
Thus this path has length 2*pi*17 = 34*pi.
Thus this path has length 2*pi*17 = 34*pi.
DPatrick19:21:50
james4l19:22:04
2pi cancels out
2pi cancels out
BOGTRO19:22:04
sqrt(h^2+1)=17
sqrt(h^2+1)=17
BOGTRO19:22:04
divide 2pi
divide 2pi
mchoi81519:22:04
divide by 2pi
divide by 2pi
karatemagic719:22:07
h^2+1=289
h^2+1=289
mnmath19:22:10
h=12sqrt(2)
h=12sqrt(2)
DPatrick19:22:18
williamtz19:22:43
have you seen the other interpretation of the problem on the forum..with 16 roatations?
have you seen the other interpretation of the problem on the forum..with 16 roatations?
DPatrick19:22:52
I have and I totally disagree with it, though I'm not going to debate it here.
I have and I totally disagree with it, though I'm not going to debate it here.
DPatrick19:23:04
DPatrick19:23:07
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unlimitd3K19:23:30
we can fill in a few more rows to try to find a pattern
we can fill in a few more rows to try to find a pattern
karatemagic719:23:30
Look for patterns.
Look for patterns.
DPatrick19:23:42
Indeed, let's just compute a few terms and see what happens.
Indeed, let's just compute a few terms and see what happens.
DPatrick19:23:48
DPatrick19:24:03
What do we notice?
What do we notice?
pythag01119:24:22
lots of multiples of 2..., divide first number in each row by 2^n
lots of multiples of 2..., divide first number in each row by 2^n
susanzduan19:24:22
they're all even
they're all even
chickendude19:24:22
each row is an arithmetic sequence
each row is an arithmetic sequence
yjneb19:24:31
factor a 2 out of the second row, a 4 out of the third, an 8 out of the forth, a 16 out of the 5th, etc
factor a 2 out of the second row, a 4 out of the third, an 8 out of the forth, a 16 out of the 5th, etc
DPatrick19:24:39
What I notice is:
Everything in the first row is odd, and increases by 2.
Everything in the second row is even, and increases by 4.
Everything in the third row is a multiple of 4, and increases by 8.
And so on.
What I notice is:
Everything in the first row is odd, and increases by 2.
Everything in the second row is even, and increases by 4.
Everything in the third row is a multiple of 4, and increases by 8.
And so on.
HoratioK19:24:51
67 is prime, that's useful
67 is prime, that's useful
DPatrick19:25:08
Indeed, all we care about is which entries are multiples of 67. So we can divide by all these powers of 2 as we go; dividing by 2 doesn't affect multiples of 67.
Indeed, all we care about is which entries are multiples of 67. So we can divide by all these powers of 2 as we go; dividing by 2 doesn't affect multiples of 67.
DPatrick19:25:18
In other words, let's change the rules a little bit. Instead of each entry being the sum of the two above, let's make it so that each entry is the average of the two above. This won't affect which entries are multiples of 67.
In other words, let's change the rules a little bit. Instead of each entry being the sum of the two above, let's make it so that each entry is the average of the two above. This won't affect which entries are multiples of 67.
DPatrick19:25:29
This has the effect of dividing out by 2 as we go.
This has the effect of dividing out by 2 as we go.
DPatrick19:25:36
If we do this, we get:
If we do this, we get:
DPatrick19:25:40
zephyredx19:26:05
only terms directly above of 67 are 67
only terms directly above of 67 are 67
kops72319:26:05
67 appears in all the odd rows... at least the ones that are long enough
67 appears in all the odd rows... at least the ones that are long enough
unlimitd3K19:26:07
every other row has 67 in it, until 67 becomes the last term in the row
every other row has 67 in it, until 67 becomes the last term in the row
tenniskidperson319:26:07
same in column
same in column
tenniskidperson319:26:10
same numbers in the columns
same numbers in the columns
mathlete10419:26:13
the numbers in columns are all the same
the numbers in columns are all the same
DPatrick19:26:18
Aha! We see that the columns are constant!
Aha! We see that the columns are constant!
DPatrick19:26:33
So all the multiples of 67 appear in the column directly below the 67 in the top row.
So all the multiples of 67 appear in the column directly below the 67 in the top row.
tenniskidperson319:26:42
67 is the 34th number in top row
67 is the 34th number in top row
DPatrick19:27:00
Right, the only multiple of 67 in the odd rows is in the 34th position (from the left) of the top row.
Right, the only multiple of 67 in the odd rows is in the 34th position (from the left) of the top row.
DPatrick19:27:09
So we need to count the number of odd rows that have that 34th position.
So we need to count the number of odd rows that have that 34th position.
silversheep19:27:13
or the 17th number counting from the right
or the 17th number counting from the right
DPatrick19:27:37
That's the easiest way to count it: 34th from the left is also 17th from the right. And every other row is one element shorter on each side.
That's the easiest way to count it: 34th from the left is also 17th from the right. And every other row is one element shorter on each side.
DPatrick19:27:43
So there are 017 rows with a multiple of 67.
So there are 017 rows with a multiple of 67.
DPatrick19:28:08
DPatrick19:28:12
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mnmath19:28:58
316^2<100000, 317^2>100000
316^2<100000, 317^2>100000
chickendude19:28:59
After some point, each set can only contain up to one perfect square.
After some point, each set can only contain up to one perfect square.
mathnerd31419:28:59
every 100 integers contains a perfect square, up to 2500...
every 100 integers contains a perfect square, up to 2500...
silversheep19:28:59
First find when consecutive perfect squares can differ by >100.
First find when consecutive perfect squares can differ by >100.
DPatrick19:29:14
One fact we'll need is 316^2 < 100000 < 317^2, so the highest square is 316^2.
One fact we'll need is 316^2 < 100000 < 317^2, so the highest square is 316^2.
cyberspace19:29:17
the gaps between squares are increasing odd numbers
the gaps between squares are increasing odd numbers
DPatrick19:29:28
DPatrick19:29:49
So for n<50, the distance between consecutive squares is less than 100, so no S's can get "skipped".
So for n<50, the distance between consecutive squares is less than 100, so no S's can get "skipped".
Paboga19:29:54
After 50^2 there is at most 1 square in each set
After 50^2 there is at most 1 square in each set
karatemagic719:29:57
at 50 the gap becomes >100
at 50 the gap becomes >100
DPatrick19:30:04
So if n >= 50, then we start to see the S's getting "skipped".
So if n >= 50, then we start to see the S's getting "skipped".
DPatrick19:30:11
For instance, 50^2 = 2500 is in S_25. Every S from S_0, S_1, up through S_25 will contain at least one perfect square. But after that, the squares are more than 100 apart, so some S's will get skipped.
For instance, 50^2 = 2500 is in S_25. Every S from S_0, S_1, up through S_25 will contain at least one perfect square. But after that, the squares are more than 100 apart, so some S's will get skipped.
DPatrick19:30:25
This means that there are 316-50 = 266 perfect squares in the S's above S_25.
This means that there are 316-50 = 266 perfect squares in the S's above S_25.
DPatrick19:30:39
Do we need to determine exactly where this skipping takes place?
Do we need to determine exactly where this skipping takes place?
lingomaniac8819:31:13
no, we just need to know how many skips occur
no, we just need to know how many skips occur
ebaudry1019:31:13
No because we are only worried about the number of sets, not which sets they are
No because we are only worried about the number of sets, not which sets they are
DPatrick19:31:19
No...we can just count!
No...we can just count!
DPatrick19:31:42
There are 999-25 = 974 S's above S_25, and 266 squares among them, and each S has at most 1 square.
There are 999-25 = 974 S's above S_25, and 266 squares among them, and each S has at most 1 square.
l Bob l19:32:14
so 974 - 266 = 708 sets without a perfect square
so 974 - 266 = 708 sets without a perfect square
williamtz19:32:14
then subtract and get 708
then subtract and get 708
tenniskidperson319:32:14
974-266=708
974-266=708
mathnerd31419:32:14
So 708 have no squares
So 708 have no squares
DPatrick19:32:19
Therefore, 974 - 266 = 708 of them get skipped and do not contain a perfect square.
Therefore, 974 - 266 = 708 of them get skipped and do not contain a perfect square.
DPatrick19:32:32
Note: this is an AIME problem that is really easy to be off by 1 on (for example, to mistakenly answer 707 or 709). In AIME counting problems, this is something that you really have to watch out for.
Note: this is an AIME problem that is really easy to be off by 1 on (for example, to mistakenly answer 707 or 709). In AIME counting problems, this is something that you really have to watch out for.
DPatrick19:32:58
DPatrick19:33:02
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Brut3Forc319:33:23
sum of tangents
sum of tangents
kops72319:33:23
we should probably use tangent sum formulas
we should probably use tangent sum formulas
tenniskidperson319:33:27
take tan of both sides
take tan of both sides
cyberspace19:33:27
big tan sum formula
big tan sum formula
DPatrick19:33:33
The key fact that we need for this problem is the tangent addition formula:
The key fact that we need for this problem is the tangent addition formula:
DPatrick19:33:41
DPatrick19:33:50
(If you don't have this formula memorized, you can easily derive it from the sine and cosine angle addition formulas. And if you don't have those formula memorized, you can easily derive them from Euler's formula for e^(ix).)
(If you don't have this formula memorized, you can easily derive it from the sine and cosine angle addition formulas. And if you don't have those formula memorized, you can easily derive them from Euler's formula for e^(ix).)
DPatrick19:34:31
(And if you don't know what any of this means, then you probably haven't taken trig yet and had no chance on this problem I'm afraid.)
(And if you don't know what any of this means, then you probably haven't taken trig yet and had no chance on this problem I'm afraid.)
DPatrick19:35:07
So now we just bash the algebra. I just did it straightforwardly; some people might perfer to move arctan(1/n) over to the right side first.
So now we just bash the algebra. I just did it straightforwardly; some people might perfer to move arctan(1/n) over to the right side first.
williamtz19:35:16
let arctan1/3=x
arctan1/4=y
and so on and then plug into formula
let arctan1/3=x
arctan1/4=y
and so on and then plug into formula
tenniskidperson319:35:16
(1/3+1/4)/(1-(1/3)(1/4))
(1/3+1/4)/(1-(1/3)(1/4))
Ed Z Hou19:35:16
I'd start by finding tan(arctan 1/3 + arctan 1/4).
I'd start by finding tan(arctan 1/3 + arctan 1/4).
DPatrick19:35:23
DPatrick19:35:43
DPatrick19:35:59
So now we can take tan of the whole left side:
So now we can take tan of the whole left side:
DPatrick19:36:07
DPatrick19:36:16
(The "1" on the right side is of course tan(pi/4))
(The "1" on the right side is of course tan(pi/4))
DPatrick19:36:25
Now we just solve.
Now we just solve.
james4l19:36:29
common denominator
common denominator
karatemagic719:36:35
multiply by the denominator
multiply by the denominator
DPatrick19:36:41
We multiply by 11(5n-1) to clear the little denominators:
We multiply by 11(5n-1) to clear the little denominators:
DPatrick19:36:46
DPatrick19:36:57
This simplifies to 46n + 48 = 48n - 46.
This simplifies to 46n + 48 = 48n - 46.
DPatrick19:37:01
So 2n = 94, hence n = 047.
So 2n = 94, hence n = 047.
DPatrick19:37:47
As I said, there are other ways we could have done the algebra (or we even could have converted the whole thing to complex numbers), but I think the straightforward way is pretty simple too.
As I said, there are other ways we could have done the algebra (or we even could have converted the whole thing to complex numbers), but I think the straightforward way is pretty simple too.
DPatrick19:38:09
DPatrick19:38:12
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DPatrick19:38:42
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Ed Z Hou19:38:55
Every ordered list of 3s, 4s, and 6s represents a way you can orient the crates, and each of these ways occurs with equal probability.
Every ordered list of 3s, 4s, and 6s represents a way you can orient the crates, and each of these ways occurs with equal probability.
HoratioK19:38:55
total of 3^10 ways to arrange crates
total of 3^10 ways to arrange crates
serialk11r19:38:57
3^10 is the denominator, find all ways to make 41 with 10 crates, add in permutations
3^10 is the denominator, find all ways to make 41 with 10 crates, add in permutations
DPatrick19:39:09
Each blocks can go 3 ways, so there are 3^10 possible stacks. Note: no need to expand this out!
Each blocks can go 3 ways, so there are 3^10 possible stacks. Note: no need to expand this out!
zephyredx19:39:23
we only need to find how many ways to make 41
we only need to find how many ways to make 41
MathPie19:39:29
If there are a crates at 3ft, b crates at 4 ft, and c crates at 6ft, then 3a+4b+6c=41 and a+b+c=10
If there are a crates at 3ft, b crates at 4 ft, and c crates at 6ft, then 3a+4b+6c=41 and a+b+c=10
tenniskidperson3_219:39:29
3a+4b+6b=41
3a+4b+6b=41
lingomaniac8819:39:33
a+b+c=10, 3a+4b+6c=41
a+b+c=10, 3a+4b+6c=41
DPatrick19:39:46
Right, if there are a blocks with height 3, b with height 4, and c with height 6, then we must have 3a + 4b + 6c = 41.
Right, if there are a blocks with height 3, b with height 4, and c with height 6, then we must have 3a + 4b + 6c = 41.
Kiran19:40:08
mod 3?
mod 3?
mnmath19:40:11
b=2,5,8
b=2,5,8
DPatrick19:40:28
I think the easiest way to solve this is to observe that 4b is the only term on the left side of the equation that is not a multiple of 3.
I think the easiest way to solve this is to observe that 4b is the only term on the left side of the equation that is not a multiple of 3.
DPatrick19:40:35
So 4b must be 2 mod 3, which means that b must be 2 mod 3.
So 4b must be 2 mod 3, which means that b must be 2 mod 3.
DPatrick19:40:45
So b must be 2, 5, or 8.
So b must be 2, 5, or 8.
DPatrick19:40:54
This gives the possible triples (a,b,c) = (5,2,3), (3,5,2), or (1,8,1).
This gives the possible triples (a,b,c) = (5,2,3), (3,5,2), or (1,8,1).
DPatrick19:41:00
How many possible stackings does this give?
How many possible stackings does this give?
Ed Z Hou19:41:34
10!/(2!3!5!) + 10!/(2!3!5!) + 10!/(1!1!8!)
10!/(2!3!5!) + 10!/(2!3!5!) + 10!/(1!1!8!)
mathnerd31419:41:34
10!/5!2!3!+10!/3!2!5!+10!/1!8!1!
10!/5!2!3!+10!/3!2!5!+10!/1!8!1!
lingomaniac8819:41:36
10! / (5!*2!*3!) + 10! / (3!*5!*2!) + 10! / (1!*8!*1!)
10! / (5!*2!*3!) + 10! / (3!*5!*2!) + 10! / (1!*8!*1!)
DPatrick19:41:42
DPatrick19:41:55
(Note that the first two are equal.)
(Note that the first two are equal.)
DPatrick19:42:03
So we need to divide this by 3^10, then simplify down to lowest terms.
So we need to divide this by 3^10, then simplify down to lowest terms.
DPatrick19:42:23
You could multiply it all out, but we can also cancel as we go...
You could multiply it all out, but we can also cancel as we go...
DPatrick19:42:36
Note that 10!/5! = 10*9*8*7*6, and the 6 will cancel with 3! in the denominators. The fact that there are 2 terms will cancel the 2!.
Note that 10!/5! = 10*9*8*7*6, and the 6 will cancel with 3! in the denominators. The fact that there are 2 terms will cancel the 2!.
DPatrick19:42:49
Also note that the last term is just 10*9.
Also note that the last term is just 10*9.
DPatrick19:42:55
BOGTRO19:43:10
we can get rid of the 9
we can get rid of the 9
HoratioK19:43:15
10*57/3^8
10*57/3^8
DPatrick19:43:24
9 will cancel with two powers of 3, and what's left in the numerator is 570.
9 will cancel with two powers of 3, and what's left in the numerator is 570.
BOGTRO19:43:31
then we are left iwth 570, which divided by 3 is 190
then we are left iwth 570, which divided by 3 is 190
james4l19:43:31
57 => 19*3
57 => 19*3
DPatrick19:43:39
570 will cancel another power of 3, and we'll be left with 190/3^7 in lowest terms.
570 will cancel another power of 3, and we'll be left with 190/3^7 in lowest terms.
DPatrick19:43:43
So the answer is 190.
So the answer is 190.
DPatrick19:44:10
DPatrick19:44:14
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kops72319:44:26
diagram
diagram
107620719:44:26
diagram!
diagram!
BOGTRO19:44:26
Diagram
Diagram
davidyko19:44:26
Draw a diagram, again
Draw a diagram, again
DPatrick19:44:39
We'll want a diagram, but note that the point E is somewhat ill-defined. It's just kind of floating out there somewhere. Let's keep that in mind.
We'll want a diagram, but note that the point E is somewhat ill-defined. It's just kind of floating out there somewhere. Let's keep that in mind.
DPatrick19:44:47
DPatrick19:44:49
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http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/2612af5fabfaeba8093fe1706f12b099.png
Kiran19:45:13
can we assume that B and C are the same point?
can we assume that B and C are the same point?
DPatrick19:45:19
Not necessarily...
Not necessarily...
syckls19:45:34
extend AB and CD to meet at a point, forming an equilateral triangle
extend AB and CD to meet at a point, forming an equilateral triangle
Brut3Forc319:45:34
extend into an equilateral triangle
extend into an equilateral triangle
karatemagic719:45:34
can we extend ab and cd to make an equilateral triangle
can we extend ab and cd to make an equilateral triangle
DPatrick19:45:47
That's one good idea, and it's often useful when you have 60 degree angles.
That's one good idea, and it's often useful when you have 60 degree angles.
DPatrick19:45:53
I solved this problem a little more naively.
I solved this problem a little more naively.
McDutchy19:45:59
It's possible to assume AB = BC = CD
It's possible to assume AB = BC = CD
DPatrick19:46:12
It turns out that this is true, so "assuming" this is actually a lucky guess.
It turns out that this is true, so "assuming" this is actually a lucky guess.
DPatrick19:46:22
But let's see how you would discover this.l
But let's see how you would discover this.l
Kongster19:46:34
there are infinite trapezoids that satisfies these requirements, so we can assume something
there are infinite trapezoids that satisfies these requirements, so we can assume something
DPatrick19:46:46
Right. The other thing that's a little weird is that other than the diagonal, we don't know the size of ABCD. And it's not uniquely determined.
Right. The other thing that's a little weird is that other than the diagonal, we don't know the size of ABCD. And it's not uniquely determined.
DPatrick19:47:05
Let's set up some variables. Let's set BC = p and AB = q.
Let's set up some variables. Let's set BC = p and AB = q.
DPatrick19:47:12
What's AD?
What's AD?
mnmath19:47:27
p+q
p+q
l Bob l19:47:31
2(BC)?
2(BC)?
isabella229619:47:32
p+q
p+q
MathPie19:47:38
p+q
p+q
karatemagic719:47:38
p+q
p+q
DPatrick19:47:49
CFD is 30-60-90, so FD = q/2. There's the same length on the other side.
So AD = p+q.
CFD is 30-60-90, so FD = q/2. There's the same length on the other side.
So AD = p+q.
DPatrick19:47:58
(It'll turn out later that it is 2(BC), but we don't know that yet.)
(It'll turn out later that it is 2(BC), but we don't know that yet.)
DPatrick19:48:06
DPatrick19:48:07
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DPatrick19:48:24
Any other relationship between p and q that we know?
Any other relationship between p and q that we know?
mathnerd31419:48:37
law of cosines
law of cosines
caffeineboy19:48:37
Law of cosines?
Law of cosines?
tenniskidperson3_219:48:46
law of cosines
law of cosines
DPatrick19:48:53
We can do Law of Cosines on triangle ABC:
We can do Law of Cosines on triangle ABC:
DPatrick19:48:57
DPatrick19:49:19
Now let's look for point E.
Now let's look for point E.
DPatrick19:49:50
AED is a (possibly degenerate) triangle. So it must satisfy the triangle inequality, and we know EA and ED, so that gives us bounds on AD.
AED is a (possibly degenerate) triangle. So it must satisfy the triangle inequality, and we know EA and ED, so that gives us bounds on AD.
HoratioK19:50:27
ad<=40rt7
ad<=40rt7
davidyko19:50:28
less than 40 sqrt (7), more than 20 sqrt (7)
less than 40 sqrt (7), more than 20 sqrt (7)
mathnerd31419:50:30
20sqrt(7) <= AD <= 40sqrt(7)
20sqrt(7) <= AD <= 40sqrt(7)
DPatrick19:50:34
DPatrick19:50:51
Paboga19:51:22
The symmetry suggests that p=q
The symmetry suggests that p=q
DPatrick19:51:36
Indeed, and we also have an expression involving both p+q and pq, and we want a bound.
Indeed, and we also have an expression involving both p+q and pq, and we want a bound.
DPatrick19:51:42
All of these things suggest AM-GM to me.
All of these things suggest AM-GM to me.
DPatrick19:51:48
DPatrick19:52:05
DPatrick19:52:25
DPatrick19:52:57
But our triangle inequality told us that p+q >= 20sqrt(7).
But our triangle inequality told us that p+q >= 20sqrt(7).
davidyko19:53:03
Implying p+q = 20 sqrt(7)
Implying p+q = 20 sqrt(7)
desperado19:53:03
p+q=20sqrt7
p+q=20sqrt7
tenniskidperson3_219:53:07
so p+q=20sqrt(7)
so p+q=20sqrt(7)
DPatrick19:53:15
We get that p+q = 20*sqrt(7).
We get that p+q = 20*sqrt(7).
mz9419:53:46
since p=q, p=q=10sqrt7
since p=q, p=q=10sqrt7
DPatrick19:53:52
Furthermore, E must lie on the line AD, to the left of A, with EA = 10sqrt(7) and ED = 30sqrt(7).
Furthermore, E must lie on the line AD, to the left of A, with EA = 10sqrt(7) and ED = 30sqrt(7).
DPatrick19:54:03
HoratioK19:54:12
since am-gm has equality, p=q, right?
since am-gm has equality, p=q, right?
DPatrick19:54:21
Right (I didn't explicitly mention that step).
Right (I didn't explicitly mention that step).
l Bob l19:54:37
just a question, back at the beginning, if you say that AB = BC = CD, and angle B = 120 isn't it just three equilateral triangles?
just a question, back at the beginning, if you say that AB = BC = CD, and angle B = 120 isn't it just three equilateral triangles?
DPatrick19:55:06
Right...but that involved assuming that AB = BC = CD. And that's a perfectly valid strategy: you might get lucky (and in this case you do).
Right...but that involved assuming that AB = BC = CD. And that's a perfectly valid strategy: you might get lucky (and in this case you do).
DPatrick19:55:15
But what we just did proves it.
But what we just did proves it.
kops72319:55:24
so in other words the trapezoid IS uniquely determined? just subtly
so in other words the trapezoid IS uniquely determined? just subtly
DPatrick19:55:35
The existence of point E uniquely determines it.
The existence of point E uniquely determines it.
DPatrick19:55:47
Without the data on point E, the trapezoid is not uniquely determined.
Without the data on point E, the trapezoid is not uniquely determined.
mathgirl_fractals19:56:08
How do you know to use inequalities? Is this a common technique in AIME geometry?
How do you know to use inequalities? Is this a common technique in AIME geometry?
DPatrick19:56:32
You actually don't specifically need it: there are other messier ways to do it (when I first did the problem, I solved it using --ick-- coordinates).
You actually don't specifically need it: there are other messier ways to do it (when I first did the problem, I solved it using --ick-- coordinates).
DPatrick19:56:49
Let's move on:
Let's move on:
DPatrick19:56:54
DPatrick19:56:58
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kops72319:57:21
definitely a recursion here
definitely a recursion here
mathnerd31419:57:21
recursion.
recursion.
james4l19:57:21
recursion
recursion
tenniskidperson3_219:57:21
recursion
recursion
Brut3Forc319:57:21
recursion
recursion
DPatrick19:57:39
We could do casework or try to come up with a formula based on the lengths of the runs.
We could do casework or try to come up with a formula based on the lengths of the runs.
DPatrick19:57:44
Or we could use recursion, which is my preference.
Or we could use recursion, which is my preference.
DPatrick19:58:01
Let a_n be the number of sequences of length n starting with A, and b_n be the number of sequences starting with B.
Let a_n be the number of sequences of length n starting with A, and b_n be the number of sequences starting with B.
DPatrick19:58:05
What are the recursions?
What are the recursions?
Ed Z Hou19:58:48
Then a_n = a_(n-2) + b_(n-2); and b_n = a_(n-1) + b_(n-2)
Then a_n = a_(n-2) + b_(n-2); and b_n = a_(n-1) + b_(n-2)
kops72319:58:48
a_n = a_(n-2) + b_(n-2) and b_n = a_(n-1) + b_(n-2)
a_n = a_(n-2) + b_(n-2) and b_n = a_(n-1) + b_(n-2)
DPatrick19:59:05
Any sequence that starts with A must start AA. Then what's left is a sequence of length n-2.
Any sequence that starts with A must start AA. Then what's left is a sequence of length n-2.
DPatrick19:59:10
DPatrick19:59:28
Sequences starting with B are slightly trickier:
If the B is followed by an A, then what's after the B is a sequence starting with A of length n-1.
If the B is followed by another B, then what's after the BB is a sequence starting with B of length n-2.
Sequences starting with B are slightly trickier:
If the B is followed by an A, then what's after the B is a sequence starting with A of length n-1.
If the B is followed by another B, then what's after the BB is a sequence starting with B of length n-2.
DPatrick19:59:42
DPatrick19:59:53
tenniskidperson3_220:00:08
a_1=0, a_2=1, b_1=1, b_2=0
a_1=0, a_2=1, b_1=1, b_2=0
lingomaniac8820:00:17
getting some smaller values: a_1=0, a_2=1, b_1=1, b_2=0
getting some smaller values: a_1=0, a_2=1, b_1=1, b_2=0
DPatrick20:00:26
Right...the only sequence of length 1 is B, and the only sequence of length 2 is AA.
Right...the only sequence of length 1 is B, and the only sequence of length 2 is AA.
kops72320:00:34
now we make a table
now we make a table
Ed Z Hou20:00:34
Then you can make a table and solve using addition.
Then you can make a table and solve using addition.
tenniskidperson3_220:00:34
now just plug and chug
now just plug and chug
DPatrick20:00:44
Sure...we could try to cleverly solve this for n=14, but since these recursions are just simple arithmetic, we can just crank them out by hand:
Sure...we could try to cleverly solve this for n=14, but since these recursions are just simple arithmetic, we can just crank them out by hand:
DPatrick20:00:53
As a check of the first step, note a_3 = 0+1 = 1 and b_3 = 1+1 = 2. Indeed there are 3 sequences of length 3: AAB, BAA, and BBB.
As a check of the first step, note a_3 = 0+1 = 1 and b_3 = 1+1 = 2. Indeed there are 3 sequences of length 3: AAB, BAA, and BBB.
DPatrick20:01:03
kops72320:01:23
so we want 80+92 = 172
so we want 80+92 = 172
tenniskidperson3_220:01:23
80+92=172
80+92=172
DPatrick20:01:33
So the answer is 80 + 92 = 172.
So the answer is 80 + 92 = 172.
mathnerd31420:01:42
any explicit way to do it?
any explicit way to do it?
DPatrick20:02:00
Certainly...there are techniques to solve the recursion to get a formula in terms of n.
Certainly...there are techniques to solve the recursion to get a formula in terms of n.
DPatrick20:02:19
Alternatively, you could forget about recursions altogether and do it directly as a "balls and urns" style problem.
Alternatively, you could forget about recursions altogether and do it directly as a "balls and urns" style problem.
DPatrick20:02:28
...but this latter is messy.
...but this latter is messy.
DPatrick20:02:43
Check the message board...I think at least one such solution is posted there.
Check the message board...I think at least one such solution is posted there.
DPatrick20:02:59
DPatrick20:03:05
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-96331086.gif
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-96331086.gif
DPatrick20:04:04
I've seen a lot of bogus, hand-wavy solutions to this problem...my goal is to present a valid solution.
I've seen a lot of bogus, hand-wavy solutions to this problem...my goal is to present a valid solution.
DPatrick20:04:18
A large part of the battle here is just deciphering the problem and figuring out how to best represent it mathematically.
A large part of the battle here is just deciphering the problem and figuring out how to best represent it mathematically.
DPatrick20:04:37
One simplification is that we can remove "time" from the problem.
One simplification is that we can remove "time" from the problem.
syckls20:04:44
imagine the whole stretch of road moving at a certain speed for an hour
imagine the whole stretch of road moving at a certain speed for an hour
DPatrick20:05:01
Instead of counting the number of cars that pass a certain spot in 1 hour, we can count the number of cars in a certain length at a fixed moment in time.
Instead of counting the number of cars that pass a certain spot in 1 hour, we can count the number of cars in a certain length at a fixed moment in time.
DPatrick20:05:12
In particular, let s be the speed of the cars (in km/h).
In particular, let s be the speed of the cars (in km/h).
DPatrick20:05:18
So instead of looking at cars/hour, we instead count how many whole cars fit inside an s kilometer stretch of road.
So instead of looking at cars/hour, we instead count how many whole cars fit inside an s kilometer stretch of road.
DPatrick20:05:41
I haven't really done anything, but for some reason I found it much easier to visualize this way.
I haven't really done anything, but for some reason I found it much easier to visualize this way.
kops72320:05:55
we need to first recognize that to fit a maximum number of cars in, they must be traveling at a multiple of 15 km/h
we need to first recognize that to fit a maximum number of cars in, they must be traveling at a multiple of 15 km/h
mathnerd31420:05:55
let s be a multiple of 15
let s be a multiple of 15
DPatrick20:06:24
It may not even be totally clear that we can do that, so let's put that off for a moment.
It may not even be totally clear that we can do that, so let's put that off for a moment.
ebaudry1020:06:35
so each car takes up (4+s/15*4 meters)
so each car takes up (4+s/15*4 meters)
DPatrick20:06:58
Every car need 4t meters of space, where t = (s/15) rounded up, plus 4 meters for the car itself.
Every car need 4t meters of space, where t = (s/15) rounded up, plus 4 meters for the car itself.
kops72320:07:00
the last one doesn't (or the first one depending on how you look at it)
the last one doesn't (or the first one depending on how you look at it)
Brut3Forc320:07:05
except for one car on the end
except for one car on the end
DPatrick20:07:30
Right. The first car takes up only 4 meters, then every subsequent car needs (4t+4) meters of space, where t = (s/15) rounded up.
Right. The first car takes up only 4 meters, then every subsequent car needs (4t+4) meters of space, where t = (s/15) rounded up.
DPatrick20:07:36
DPatrick20:08:04
The "1" out front is the first car, then the rest is the number of "space + car"s that can fit in the remaining 1000s-4 meters.
The "1" out front is the first car, then the rest is the number of "space + car"s that can fit in the remaining 1000s-4 meters.
DPatrick20:08:16
NOW it's clear that to maximize this number, we want s to be a multiple of 15 (since if it's less, we can move it up to the next highest multiple: the denominator doesn't change and the numerator gets bigger).
NOW it's clear that to maximize this number, we want s to be a multiple of 15 (since if it's less, we can move it up to the next highest multiple: the denominator doesn't change and the numerator gets bigger).
DPatrick20:08:46
So we can let s = 15k for some positive integer k.
So we can let s = 15k for some positive integer k.
DPatrick20:08:52
McDutchy20:09:09
Simplifiy the expression
Simplifiy the expression
DPatrick20:09:23
What's the easiest way to simplify it so that we can clearly see how it's maximized?
What's the easiest way to simplify it so that we can clearly see how it's maximized?
lingomaniac8820:10:23
1 + (3750k+3750)/(k+1) - 3751/(k+1)
1 + (3750k+3750)/(k+1) - 3751/(k+1)
DPatrick20:10:31
If we rewrite the numerator as 3750k - 1 = 3750k + 3750 - 3751, we get:
If we rewrite the numerator as 3750k - 1 = 3750k + 3750 - 3751, we get:
DPatrick20:10:36
DPatrick20:10:57
Now it's clear. The term inside the parentheses is always less than 1, but can be made arbitrarily close to 1 (by making k very large so that 1/(k+1) is very small).
Now it's clear. The term inside the parentheses is always less than 1, but can be made arbitrarily close to 1 (by making k very large so that 1/(k+1) is very small).
DPatrick20:11:07
In particular, if k >> 0 (this notation means k is very large), this number will be slightly less than 3751.
In particular, if k >> 0 (this notation means k is very large), this number will be slightly less than 3751.
DPatrick20:11:19
So we can achieve 3750 cars.
So we can achieve 3750 cars.
DPatrick20:11:25
Thus the answer is 375.
Thus the answer is 375.
mathnerd31420:11:48
The cars would have to be traveling ridiculously fast, though... ;)
The cars would have to be traveling ridiculously fast, though... ;)
Brut3Forc320:11:48
(Won't the cars be going 56,250 km/h?)
(Won't the cars be going 56,250 km/h?)
DPatrick20:11:57
Indeed. What I don't like about this problem is that in order to get the number above 3750, we need (1 - 1/k+1) >= 3750/3751.
Indeed. What I don't like about this problem is that in order to get the number above 3750, we need (1 - 1/k+1) >= 3750/3751.
DPatrick20:12:04
This means k/k+1 >= 3750/3751, or 3751k >= 3750k + 3750, or k >= 3750.
So the cars would have to be going 15*3750 = 56250 km/h. That's about Mach 45.7.
This means k/k+1 >= 3750/3751, or 3751k >= 3750k + 3750, or k >= 3750.
So the cars would have to be going 15*3750 = 56250 km/h. That's about Mach 45.7.
DPatrick20:12:17
But, in fairness, the problem did say the cars can be going any speed.
But, in fairness, the problem did say the cars can be going any speed.
DPatrick20:12:37
DPatrick20:12:43
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-240380574.gif
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-240380574.gif
yjneb20:12:57
plug in the pts
plug in the pts
james4l20:12:57
plug them in
plug them in
caffeineboy20:13:07
Plug the points in to find out about coefficients
Plug the points in to find out about coefficients
kops72320:13:09
well a0 can immediately be seen to be 0
well a0 can immediately be seen to be 0
DPatrick20:13:15
We can just start plugging in the given points. Rather than plug them all in at once, we'll do them systematically.
We can just start plugging in the given points. Rather than plug them all in at once, we'll do them systematically.
DPatrick20:13:20
p(0,0) = 0 just gives us a_0 = 0. So the constant term is 0.
p(0,0) = 0 just gives us a_0 = 0. So the constant term is 0.
DPatrick20:13:42
The next pair p(1,0) = p(-1,0) = 0 kills all the y terms and leave us with a cubic in x. Specifically:
The next pair p(1,0) = p(-1,0) = 0 kills all the y terms and leave us with a cubic in x. Specifically:
DPatrick20:13:46
karatemagic720:14:01
a_3=0
a_3=0
jamieding5120:14:01
a3 = 0
a3 = 0
kops72320:14:01
so adding, a_3 = 0
so adding, a_3 = 0
DPatrick20:14:09
Right, this immediately gives us a_3 = 0.
Right, this immediately gives us a_3 = 0.
frodo20:14:15
Brut3Forc320:14:19
then plugging back in, a_1+a_6=0
then plugging back in, a_1+a_6=0
DPatrick20:14:25
We also get a_1 + a_6 = 0, so a_6 = -a_1.
We also get a_1 + a_6 = 0, so a_6 = -a_1.
DPatrick20:14:34
We can do the same thing with the p(0,1) = p(0,-1) = 0 condition. This kills all the x terms and leaves:
We can do the same thing with the p(0,1) = p(0,-1) = 0 condition. This kills all the x terms and leaves:
DPatrick20:14:39
karatemagic720:14:51
a_5=0
a_5=0
Sea[Weed]20:14:51
a5 is zero
a5 is zero
lingomaniac8820:14:51
similarly a5=0 and a2=-a9
similarly a5=0 and a2=-a9
kops72320:14:51
so we get the same result... namely a_5 = 0, a_2 = -a_9
so we get the same result... namely a_5 = 0, a_2 = -a_9
DPatrick20:14:58
Right: a_5 = 0 and a_9 = -a_2.
Right: a_5 = 0 and a_9 = -a_2.
DPatrick20:15:07
DPatrick20:15:19
We haven't yet used p(1,1) = p(1,-1) = p(2,2) = 0.
We haven't yet used p(1,1) = p(1,-1) = p(2,2) = 0.
unlimitd3K20:15:40
use 1,1 and 1,-1 first
use 1,1 and 1,-1 first
DPatrick20:15:42
p(1,1) and p(1,-1) will have canceling terms, so let's look at them next.
p(1,1) and p(1,-1) will have canceling terms, so let's look at them next.
DPatrick20:15:49
The a_1 and a_2 terms will cancel, and we're left with:
The a_1 and a_2 terms will cancel, and we're left with:
DPatrick20:16:11
kops72320:16:33
and again, a8 = 0, a4 = -a7
and again, a8 = 0, a4 = -a7
HoratioK20:16:33
a8=0,a7=-a4
a8=0,a7=-a4
uclabb20:16:33
same as before
same as before
DPatrick20:16:39
So a_8 = 0 and a_4 = -a_7.
So a_8 = 0 and a_4 = -a_7.
DPatrick20:16:42
DPatrick20:16:54
At this stage, I don't like all those subscripts.
At this stage, I don't like all those subscripts.
DPatrick20:17:02
Let's replace them with r,s,t so it's easier to write.
Let's replace them with r,s,t so it's easier to write.
DPatrick20:17:06
mgao_220:17:17
sub p(2,2) now
sub p(2,2) now
kops72320:17:17
finally, use 2,2
finally, use 2,2
McDutchy20:17:17
now plug in (2,2)
now plug in (2,2)
DPatrick20:17:22
Our last given point is p(2,2) = 0. Let's plug it in:
p(2,2) = -6r -6s -4t = 0.
Our last given point is p(2,2) = 0. Let's plug it in:
p(2,2) = -6r -6s -4t = 0.
DPatrick20:17:47
We can solve for one in terms of the other two.
We can solve for one in terms of the other two.
Brut3Forc320:17:52
solve for t?
solve for t?
karatemagic720:17:52
3r+3s=-2t
3r+3s=-2t
yjneb20:17:52
so t=-3/2*(r+s)
so t=-3/2*(r+s)
DPatrick20:18:03
Probably t is best to solve for, so that we preserve some symmetry.
Probably t is best to solve for, so that we preserve some symmetry.
DPatrick20:18:10
We write t in terms of r and s, giving t = -3/2(r+s).
We write t in terms of r and s, giving t = -3/2(r+s).
DPatrick20:18:17
DPatrick20:18:32
How do we find our point (a/c,b/c) from here?
How do we find our point (a/c,b/c) from here?
yjneb20:18:52
put the r terms together and the s terms together
put the r terms together and the s terms together
davidyko20:18:52
Factor?
Factor?
DPatrick20:18:59
Let's factor the expression in r and s:
Let's factor the expression in r and s:
DPatrick20:19:06
DPatrick20:19:24
Any point we plug in has to work for all r and s, so we need both expressions in parentheses to be 0.
Any point we plug in has to work for all r and s, so we need both expressions in parentheses to be 0.
DPatrick20:19:43
kops72320:20:03
fortunately, we can divide out x from the left and y from the right (top and bottom in the new version)
fortunately, we can divide out x from the left and y from the right (top and bottom in the new version)
DPatrick20:20:11
There are a number of ways to finish from here.
There are a number of ways to finish from here.
DPatrick20:20:35
For example, the top one is linear in y, so we can write y in terms of x (we first divide by x):
For example, the top one is linear in y, so we can write y in terms of x (we first divide by x):
DPatrick20:20:41
DPatrick20:21:17
Then we divide the second equation by y, then plug this in:
Then we divide the second equation by y, then plug this in:
DPatrick20:21:30
DPatrick20:21:49
It's now just a quadratic in x that we can solve!
It's now just a quadratic in x that we can solve!
DPatrick20:22:03
We can make life a little simpler by multiplying through by 18 to get rid of fractions:
We can make life a little simpler by multiplying through by 18 to get rid of fractions:
DPatrick20:22:16
McDutchy20:22:50
you missed a sqrt sign
you missed a sqrt sign
DPatrick20:22:59
oops..that should be sqrt(1089) on the last line.
oops..that should be sqrt(1089) on the last line.
DPatrick20:23:19
(I realize I did the algebra quickly, but it's easy and boring.)
(I realize I did the algebra quickly, but it's easy and boring.)
karatemagic720:23:26
x=2 or 5/19
x=2 or 5/19
DPatrick20:23:34
Right. We get x=2 or x=5/19.
Right. We get x=2 or x=5/19.
DPatrick20:23:47
Taking x=2 gives y=2, but we already know that point.
Taking x=2 gives y=2, but we already know that point.
syckls20:23:58
c > 1
c > 1
Brut3Forc320:23:58
2 is invalid since c>1
2 is invalid since c>1
DPatrick20:24:07
Right, that's the other reason to take 5/19.
Right, that's the other reason to take 5/19.
DPatrick20:24:16
That gives y = (2/3)(1+x) = 2/3(24/19) = 16/19.
That gives y = (2/3)(1+x) = 2/3(24/19) = 16/19.
DPatrick20:24:25
The desired point is thus (5/19,16/19), and the answer is 5+16+19 = 040.
The desired point is thus (5/19,16/19), and the answer is 5+16+19 = 040.
DPatrick20:24:50
DPatrick20:24:54
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-197028526.gif
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-197028526.gif
DPatrick20:25:10
Naturally we start with a picture:
Naturally we start with a picture:
DPatrick20:25:18
DPatrick20:25:19
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/68f04152d700241db640f3a49e51b61f.png
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/68f04152d700241db640f3a49e51b61f.png
DPatrick20:25:28
O is the center of the circle. We're trying to maximize BP.
O is the center of the circle. We're trying to maximize BP.
tenniskidperson3_220:25:43
similar triangles
similar triangles
kops72320:25:43
similar triangles of course
similar triangles of course
mz9420:25:43
can u use similar triangles?
can u use similar triangles?
DPatrick20:26:07
You can, and that's how I did it at first (similar triangles and Law of Cosines and some algebra), but then I found a simpler way.
You can, and that's how I did it at first (similar triangles and Law of Cosines and some algebra), but then I found a simpler way.
DPatrick20:26:21
What in this diagram is just crying out to be added to the picture?
What in this diagram is just crying out to be added to the picture?
Ratclaw20:26:42
altitude from B to CT
altitude from B to CT
karatemagic720:26:45
line from b to ct
line from b to ct
abgs_220:26:48
perpendicular from B to the extension of CT
perpendicular from B to the extension of CT
DPatrick20:26:53
Let's draw BQ parallel to AP and OT.
Let's draw BQ parallel to AP and OT.
DPatrick20:26:59
DPatrick20:27:01
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/cfc23adf9a0280071411881080b1372d.png
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/cfc23adf9a0280071411881080b1372d.png
DPatrick20:27:12
What do we know about ABQP?
What do we know about ABQP?
jamieding5120:27:30
Trapezoid?
Trapezoid?
MathAndKnowledge20:27:30
trapezoid...
trapezoid...
HoratioK20:27:31
right trapezoid
right trapezoid
DPatrick20:27:42
It's a trapezoid with one leg perpendicular to the bases.
It's a trapezoid with one leg perpendicular to the bases.
DPatrick20:27:48
And what about OT?
And what about OT?
greydandelion20:27:59
median
median
Kongster20:28:05
midline
midline
davidyko20:28:05
average of AP and BQ
average of AP and BQ
BOGTRO20:28:05
the median
the median
ytrewq20:28:07
length 9
length 9
DPatrick20:28:23
Right! OT is the median, has length 9 (since OT is a radius), and is the average of the bases AP and BQ.
Right! OT is the median, has length 9 (since OT is a radius), and is the average of the bases AP and BQ.
DPatrick20:28:28
So we can let AP = 9-x and BQ = 9+x for some 0 < x < 9.
So we can let AP = 9-x and BQ = 9+x for some 0 < x < 9.
DPatrick20:28:37
Let's clean up the picture and show just the trapezoid:
Let's clean up the picture and show just the trapezoid:
DPatrick20:28:44
DPatrick20:28:46
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/d383e9673657a9cf0bbdbefdeb573fcc.png
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/d383e9673657a9cf0bbdbefdeb573fcc.png
DPatrick20:28:53
Now what? How do we determine BP?
Now what? How do we determine BP?
kops72320:29:13
well we need PQ first
well we need PQ first
PI-Dimension20:29:13
pythag?
pythag?
DPatrick20:29:23
Right, we can use the Pythagorean theorem on BPQ:
Right, we can use the Pythagorean theorem on BPQ:
DPatrick20:29:27
(BP)^2 = (PQ)^2 + (9+x)^2 by the Pythagorean Theorem.
(BP)^2 = (PQ)^2 + (9+x)^2 by the Pythagorean Theorem.
DPatrick20:29:31
So we need PQ.
So we need PQ.
DPatrick20:29:38
How do we get PQ?
How do we get PQ?
kops72320:29:44
extend perp from A to BQ
extend perp from A to BQ
DPatrick20:29:59
Bingo. Draw a perpendicular from A down to BQ. Call the foot R:
Bingo. Draw a perpendicular from A down to BQ. Call the foot R:
DPatrick20:30:07
DPatrick20:30:08
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/5fbc4528ba1ece2d9f2e58a14cdea200.png
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/5fbc4528ba1ece2d9f2e58a14cdea200.png
PI-Dimension20:30:20
BR=2x
BR=2x
107620720:30:20
BR is 2x
BR is 2x
HoratioK20:30:21
then BR = 2x and use pythag on ABR
then BR = 2x and use pythag on ABR
DPatrick20:30:26
Exactly.
Exactly.
DPatrick20:30:31
Note that BR = (BQ) - (AP) = (9+x)-(9-x) = 2x.
Note that BR = (BQ) - (AP) = (9+x)-(9-x) = 2x.
DPatrick20:30:43
So (PQ)^2 = (AR)^2 = (AB)^2 - (BR)^2 = 324-4x^2, using Pythagoras on ARB.
So (PQ)^2 = (AR)^2 = (AB)^2 - (BR)^2 = 324-4x^2, using Pythagoras on ARB.
DPatrick20:31:04
And now we can take this and apply Pythagoras on BPQ to find BP:
And now we can take this and apply Pythagoras on BPQ to find BP:
DPatrick20:31:13
DPatrick20:31:38
Aha, this is our expression for (BP)^2, and we just have to maximize it.
Aha, this is our expression for (BP)^2, and we just have to maximize it.
FloodFilter:20:31:45
-b/2a for the maximum distance?
-b/2a for the maximum distance?
karatemagic720:31:45
maximum is when x=3
maximum is when x=3
frodo20:31:45
-b/2a
-b/2a
lingomaniac8820:31:55
x = -b/2a
x = -b/2a
jeffdshen20:31:55
take x = -b/2a
take x = -b/2a
DPatrick20:31:57
This is maximized when x = 3 (in general, ax^2 + bx + c has its max/min at x = -b/2a).
This is maximized when x = 3 (in general, ax^2 + bx + c has its max/min at x = -b/2a).
DPatrick20:32:08
(And if you don't see this right away, you can always complete the square and get it that way.)
(And if you don't see this right away, you can always complete the square and get it that way.)
DPatrick20:32:18
When x=3, we get (BP)^2 = m^2 = -27 + 54 + 405 = 432.
When x=3, we get (BP)^2 = m^2 = -27 + 54 + 405 = 432.
DPatrick20:32:45
And finally...
And finally...
DPatrick20:32:51
DPatrick20:32:54
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-105381966.gif
//s3.amazonaws.com/classroom.artofproblemsolving.com/Classroom/cbe6/images/lx-105381966.gif
DPatrick20:33:02
DPatrick20:33:03
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/27988e45485b4ff23b4d233dfffda539.png
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/27988e45485b4ff23b4d233dfffda539.png
DPatrick20:33:37
Despite the fact that it looks somewhat scary, it's not really that hard a problem for a #15.
Despite the fact that it looks somewhat scary, it's not really that hard a problem for a #15.
rowan20:33:47
law of sine to find lengths of cut
law of sine to find lengths of cut
107620720:34:03
find the length of the cut, which is the slant diagonal
find the length of the cut, which is the slant diagonal
DPatrick20:34:05
We can first compute the length of the cut.
We can first compute the length of the cut.
DPatrick20:34:16
We can use Law of Sines on the triangles to find the cut length (which I'll call c):
We can use Law of Sines on the triangles to find the cut length (which I'll call c):
DPatrick20:34:20
frodo20:34:33
c=rt34
c=rt34
karatemagic720:34:38
c=sqqrt(34)
c=sqqrt(34)
DPatrick20:34:39
Since sin 30 = 1/2 and sin 45 = sqrt(2)/2, we have c = sqrt(34).
Since sin 30 = 1/2 and sin 45 = sqrt(2)/2, we have c = sqrt(34).
DPatrick20:34:49
Now what?
Now what?
rowan20:35:14
then we want to find how far the top corners of this objectstick out from the base
then we want to find how far the top corners of this objectstick out from the base
kops72320:35:16
the key observation here is that when youre folding up the bottom piece, the x coordinate of the end point is preserved
the key observation here is that when youre folding up the bottom piece, the x coordinate of the end point is preserved
DPatrick20:35:33
Right...you kind of have to picture what this is going to look like after it's folded.
Right...you kind of have to picture what this is going to look like after it's folded.
DPatrick20:35:41
(Or, you can rip a piece of paper and try to do it yourself!)
(Or, you can rip a piece of paper and try to do it yourself!)
DPatrick20:35:54
Let me label a few points:
Let me label a few points:
DPatrick20:36:01
DPatrick20:36:03
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/6ae36e5b3f7da29f2402410aefc2f867.png
http://s3.amazonaws.com/classroom.artofproblemsolving.com/Images/Transcripts/6ae36e5b3f7da29f2402410aefc2f867.png
DPatrick20:36:13
There are a number of ways to visualize it. I though of making O the origin of 3-D space, so that the "fold" lines in the original picture are the positive x and y axes.
There are a number of ways to visualize it. I though of making O the origin of 3-D space, so that the "fold" lines in the original picture are the positive x and y axes.
DPatrick20:36:53
As was mentioned, the horizontal length AP gets preserved as you fold, so the point A (after the fold) will have x-coordinate -(AP) relative to O, even after folding.
As was mentioned, the horizontal length AP gets preserved as you fold, so the point A (after the fold) will have x-coordinate -(AP) relative to O, even after folding.
DPatrick20:37:08
What is length AP?
What is length AP?
lingomaniac8820:37:32
AOP is a 15-degree angle, use right triangle trig from there
AOP is a 15-degree angle, use right triangle trig from there
DPatrick20:37:40
APO is a right triangle with hypotenuse AO and angle AOP = 15, so AP = sqrt(34) * sin(15).
APO is a right triangle with hypotenuse AO and angle AOP = 15, so AP = sqrt(34) * sin(15).
DPatrick20:37:53
(Remember AO = c = sqrt(34) from our first computation)
(Remember AO = c = sqrt(34) from our first computation)
DPatrick20:37:59
Let's not evaulate sin(15) just yet. No need to until we're sure we have to.
Let's not evaulate sin(15) just yet. No need to until we're sure we have to.
DPatrick20:38:23
So after we fold, the x-coordinate of A (with O as the origin) will be -sqrt(34)*sin(15).
So after we fold, the x-coordinate of A (with O as the origin) will be -sqrt(34)*sin(15).
DPatrick20:38:34
By symmetry, that's the y coordinate too.
By symmetry, that's the y coordinate too.
DPatrick20:38:39
And the z coordinate is what we're trying to find!
And the z coordinate is what we're trying to find!
DPatrick20:38:49
How do we tie it all together?
How do we tie it all together?
kops72320:39:07
we know the slant height AO
we know the slant height AO
mgao_220:39:07
3d pythagorean
3d pythagorean
mathnerd31420:39:07
distance formula
distance formula
lingomaniac8820:39:14
3D distance formula
3D distance formula
uclabb20:39:14
pythag in 3 dimensions
pythag in 3 dimensions
DPatrick20:39:21
Aha! We know that x^2+y^2+z^2 = c^2 (where c is the length of the cut).
Aha! We know that x^2+y^2+z^2 = c^2 (where c is the length of the cut).
DPatrick20:39:30
kops72320:39:55
we can get sin(15) from our half angle formula
we can get sin(15) from our half angle formula
mgao_220:39:55
now we need to know sin15
now we need to know sin15
DPatrick20:40:01
But we don't have to compute it!
But we don't have to compute it!
DPatrick20:40:06
jeffdshen20:40:09
use 1-2sin^2(15) = cos(30)
use 1-2sin^2(15) = cos(30)
DPatrick20:40:23
DPatrick20:40:34
Thus z^2 is just 17*sqrt(3).
Thus z^2 is just 17*sqrt(3).
DPatrick20:40:53
DPatrick20:41:18
I felt that 13 and 14 were harder than 15. It's a lot of personal preference too.
I felt that 13 and 14 were harder than 15. It's a lot of personal preference too.
DPatrick20:41:33
That's it for the Math Jam.
The AIME II is on April 2 and the AIME II Math Jam is currently scheduled for April 4.
That's it for the Math Jam.
The AIME II is on April 2 and the AIME II Math Jam is currently scheduled for April 4.
frodo20:41:46
How do they organize the problems?
How do they organize the problems?
DPatrick20:42:00
They roughly try to put them in order of difficulty. It's not an exact science.
They roughly try to put them in order of difficulty. It's not an exact science.
DPatrick20:42:15
3-D geometry is scary for a lot of people, so they maybe thought #15 was the hardest.
3-D geometry is scary for a lot of people, so they maybe thought #15 was the hardest.
serialk11r20:42:24
What date do you think the results will be released
What date do you think the results will be released
DPatrick20:42:38
let me look it up....
let me look it up....
kshweh_220:43:00
When will the cutoffs be announced?
When will the cutoffs be announced?
mcsquared20:43:00
when will we know if we qualify for the usamo?
when will we know if we qualify for the usamo?
DPatrick20:43:11
On or before April 11 (according to the AMC).
On or before April 11 (according to the AMC).
DPatrick20:43:17
So the week of April 7-11.
So the week of April 7-11.
Brut3Forc320:43:34
What will the (approximate) floor and index be?
What will the (approximate) floor and index be?
williamtz20:43:34
what do you think the floor and index is?
what do you think the floor and index is?
Kongster20:43:34
do you have a rough estimate for the aime I index or the floor for >11th?
do you have a rough estimate for the aime I index or the floor for >11th?
kostya20:43:34
do you know what the floor is for 10th grade and under?
do you know what the floor is for 10th grade and under?
DPatrick20:43:43
As a matter of policy, I do not speculate on such things.
As a matter of policy, I do not speculate on such things.
DPatrick20:44:10
AIME I scores itself should be available soon than that, prior to the AIME II date most likely.
AIME I scores itself should be available soon than that, prior to the AIME II date most likely.
mz9420:44:24
where do we get the transcript for this again?
where do we get the transcript for this again?
DPatrick20:44:49
Go to the Math Jam page on the website and click the "Transcript" button. It won't be up yet...it doesn't go up until a few minutes after we finish.
Go to the Math Jam page on the website and click the "Transcript" button. It won't be up yet...it doesn't go up until a few minutes after we finish.
uclabb20:45:00
is it an expanded field again like last year?
is it an expanded field again like last year?
DPatrick20:45:07
Yes: the target is 500 students for the USAMO.
Yes: the target is 500 students for the USAMO.
ebaudry1020:45:12
how does qualifying for the USAMO when you're an underclassman differ from when you're a junior or senior?
how does qualifying for the USAMO when you're an underclassman differ from when you're a junior or senior?
DPatrick20:45:21
Read the AMC website: it's too complicated for me to explain here.
Read the AMC website: it's too complicated for me to explain here.
DPatrick20:45:24
www.unl.edu/a,c
www.unl.edu/a,c
DPatrick20:45:25
oops
oops
DPatrick20:45:27
www.unl.edu/amc
www.unl.edu/amc
DPatrick20:45:36
Look on page 4 of the USAMO Teachers Manual.
Look on page 4 of the USAMO Teachers Manual.
kops72320:45:49
seniors can only advance past the USAMO by being in the top 12 in the nation, correct?
seniors can only advance past the USAMO by being in the top 12 in the nation, correct?
DPatrick20:46:03
Generally, yes: seniors only take the selection test and/or go to MOP if they're in the top 12 on the USAMO.
Generally, yes: seniors only take the selection test and/or go to MOP if they're in the top 12 on the USAMO.
kostya20:46:26
how many 9th graders are they planning to take to MOSP?
how many 9th graders are they planning to take to MOSP?
DPatrick20:46:28
Don't know.
Don't know.
DPatrick20:46:45
Actually. the Teachers Manual says 25.
Actually. the Teachers Manual says 25.
DPatrick20:46:47
(page 6)
(page 6)
cyberspace20:47:46
what score would a 9th grader have to get on usamo to go to MOP?
what score would a 9th grader have to get on usamo to go to MOP?
DPatrick20:47:53
Again, as policy, I do not speculate on such matters.
Again, as policy, I do not speculate on such matters.
MathAndKnowledge20:48:07
can 8th graders be taken to MOSP provided that their score on the AIME is relatively high?
can 8th graders be taken to MOSP provided that their score on the AIME is relatively high?
DPatrick20:48:17
Yes, if an 8th grader does well on the USAMO.
Yes, if an 8th grader does well on the USAMO.
Kongster20:48:21
once one makes it to the USAMO, do amc and aime scores still matter?
once one makes it to the USAMO, do amc and aime scores still matter?
DPatrick20:48:27
Yes: read the Teachers Manual for the details.
Yes: read the Teachers Manual for the details.
DPatrick20:49:21
(actually that last question doesn't seem to be addressed in there..they may have changed the policy)
(actually that last question doesn't seem to be addressed in there..they may have changed the policy)
DPatrick20:50:12
AoPS is not the AMC. We do not set their policy. Administrative questions about the AMC are best answered by looking at the AMC web site, specifically their teacher's manuals.
AoPS is not the AMC. We do not set their policy. Administrative questions about the AMC are best answered by looking at the AMC web site, specifically their teacher's manuals.
MathAndKnowledge20:50:16
is it likely that the AIME II Math Jam will be pushed back for similar reasons that the AIME I Math Jam was pushed back for?
is it likely that the AIME II Math Jam will be pushed back for similar reasons that the AIME I Math Jam was pushed back for?
DPatrick20:50:40
From what we know right now, no. But the AMC can always request a delay (as they did for this one), and if so we will accomodate their request.
From what we know right now, no. But the AMC can always request a delay (as they did for this one), and if so we will accomodate their request.
DPatrick20:51:20
I guess that's it...thanks for coming!
I guess that's it...thanks for coming!
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