Math Jams

copeland 2011-02-24 19:29:57
Welcome to the 2011 AMC 10B/12B Math Jam!

copeland 2011-02-24 19:30:07
I'm Jeremy Copeland, and I'll be leading our discussion tonight.

willwang123 2011-02-24 19:30:10
Yay!

mathcountsloser 2011-02-24 19:30:10
It's 6:30.

ctmusicgirl 2011-02-24 19:30:10
YAY :D

alex31415 2011-02-24 19:30:10
Start already!

willwang123 2011-02-24 19:30:10
Hi!

Joe10112 2011-02-24 19:30:10
Yay!

minirafa 2011-02-24 19:30:10
hi

chaos763 2011-02-24 19:30:10
Great!

gmarch89 2011-02-24 19:30:10
yay lol

copeland 2011-02-24 19:30:19
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.

copeland 2011-02-24 19:30:22
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.

copeland 2011-02-24 19:30:30
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.

copeland 2011-02-24 19:30:34
There are a lot of students here! As I said, only (a very small fraction of the) well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!

copeland 2011-02-24 19:30:41
Hmm.

copeland 2011-02-24 19:30:50
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the prerequisite material for every problem as we go. Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We usually do in our classes, but we have a large number of students tonight! So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!

copeland 2011-02-24 19:30:53
We do have two assistants tonight who can help answer some of the questions.

copeland 2011-02-24 19:30:57
The first is Patrick Hulin (worthawholebean). Patrick started doing math competitions in elementary school and attended National MATHCOUNTS in 2006. In 2010, he received an honorable mention on the USA Math Olympiad. He currently studies math at MIT.

copeland 2011-02-24 19:31:11
We also have Vincent Le who is a sophomore at MIT. He has participated in MATHCOUNTS, ARML, USAMO, and also has been to Canada/USA Mathcamp and AwesomeMath. He's been at MIT for one more year than Patrick, and is thus one year more awesome.

copeland 2011-02-24 19:31:33
They can answer questions by whispering to you or by opening a window with you to chat 1-on-1.

copeland 2011-02-24 19:31:42
Please also remember that the purpose of this Math Jam is to work through the solutions to AMC problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the final answer. That's not why we're here. We're going to work through the problems step-by-step, and comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, won't be posted.

copeland 2011-02-24 19:31:46
Also notice that there will be several cases where we actually find the answer or almost the answer but instead wander off. The goal is always to find a proof that our answer is correct and not just find the answers. Of course on the AMC you should aim to do much less work than this.

copeland 2011-02-24 19:31:55
Oh.

copeland 2011-02-24 19:31:59
Vincent = chessspro.

copeland 2011-02-24 19:32:06
The Math Jam will proceed as follows:

minirafa 2011-02-24 19:32:38
what about the other moderators?

copeland 2011-02-24 19:32:40
They'll be grading each of your performances. tlcruickshank is especially concerned with your "style."

copeland 2011-02-24 19:32:47
The Math Jam will proceed as follows:

copeland 2011-02-24 19:32:50
We will work the last 5 problems from the AMC 10A, then the last 5 problems from the AMC 12A. After that, time permitting, I will take requests for some other problems for discussion.

Doink 2011-02-24 19:33:08
what is style?

egbdf44 2011-02-24 19:33:08
What is "style"?

copeland 2011-02-24 19:33:10
I don't think you're off to a good start. :)

copeland 2011-02-24 19:33:20

copeland 2011-02-24 19:33:34
Notice that I will always put the current problem at the top of the window. You can resize the area at the top by dragging the horizontal gray bar.

copeland 2011-02-24 19:33:49
OK, which of the differences jump out at you?

Tribefan 2011-02-24 19:34:28
9

Relativity1618 2011-02-24 19:34:28
9

jeff10 2011-02-24 19:34:28
The difference between w and z is 9.

AkshajKadaveru 2011-02-24 19:34:28
1

chaos763 2011-02-24 19:34:28
The 9.

PowerOfPi 2011-02-24 19:34:28
w-z=9

RisingMathStar 2011-02-24 19:34:28
w - z = 9

tekgeek 2011-02-24 19:34:28
w-z must equal 9

.cpp 2011-02-24 19:34:28
The biggest difference, 9.

jxihong 2011-02-24 19:34:28
1

davidx 2011-02-24 19:34:28
1

anonymous0 2011-02-24 19:34:29
diff. between w and z =9

curethekitten 2011-02-24 19:34:29
1

KKNC123 2011-02-24 19:34:29
1

copeland 2011-02-24 19:34:34
Since 1 is the smallest difference we know that two of the numbers must differ by 1.

copeland 2011-02-24 19:34:37
Since 9 is the largest difference, we know that w-z=9.

copeland 2011-02-24 19:34:39
Now what?

professordad 2011-02-24 19:35:48
casework on values

AlphaMath1 2011-02-24 19:35:48
casework

TruffleMaster 2011-02-24 19:35:48
the sum of the individual differences must equal 9

jeff10 2011-02-24 19:35:48
The sum of the difference between w and x, x and y, and t and z is also 9.

NoWayHaze 2011-02-24 19:35:48
the sum of two of the differences is another

chaos763 2011-02-24 19:35:48
Some of the differences add up.

TrigMan 2011-02-24 19:35:48
You could start by guessing that x = w-1

capu 2011-02-24 19:35:48
1+3+5=9

dragonx 2011-02-24 19:35:48
We note that 3 differences must add up to 9 leaving us with two direct diffrrences being 3 and 5.

gravel 2011-02-24 19:35:48
Casework?

copeland 2011-02-24 19:35:52
Let's draw a number line.

copeland 2011-02-24 19:35:56

copeland 2011-02-24 19:35:58
Let's try to label the differences between the points on this line.

copeland 2011-02-24 19:36:11
Anything can be 1 so far.

copeland 2011-02-24 19:36:15

copeland 2011-02-24 19:36:18

copeland 2011-02-24 19:36:20

copeland 2011-02-24 19:36:22
What else do we know?

chaos763 2011-02-24 19:37:29
3 + 1 = 4.

capu 2011-02-24 19:37:29
the differences in consecutive numbers are 1, 3, and 5

TruffleMaster 2011-02-24 19:37:29
no sum can equal 3 so one of the smaller differences is 3

Spring 2011-02-24 19:37:29
The first and third do not work because one of the differences would be 8.

.cpp 2011-02-24 19:37:29
No difference of 8 exists, so cases 1 and 3 above are impossible.

dragonx 2011-02-24 19:37:29
Other two add up to 8: 3 and 5.

TrigMan 2011-02-24 19:37:29
The total is 9

Jasmine8925 2011-02-24 19:37:29
the other differences should add up to 8

smallpeoples343 2011-02-24 19:37:29

willwang123 2011-02-24 19:37:29
w and z are 9 apart

greatwhiteshark98 2011-02-24 19:37:29
3 and 5 are the other differences?

TruffleMaster 2011-02-24 19:37:29
so two of them must be 1 and 3 and the sum must be 9 so the third is 5

AkshajKadaveru 2011-02-24 19:37:29
the other two add to 8

copeland 2011-02-24 19:37:36
Since 3 is the next smallest, there must be an edge of length 3.

copeland 2011-02-24 19:37:40
Since the total length is 9 we know that the other length must be 5.

copeland 2011-02-24 19:37:44
Since 1+3 = 4 is the only way we can get 4, we must have the 3 next to the 1.

copeland 2011-02-24 19:37:48
This gives us exactly four possibilities:

copeland 2011-02-24 19:37:51

copeland 2011-02-24 19:37:53

copeland 2011-02-24 19:37:55

copeland 2011-02-24 19:37:58

copeland 2011-02-24 19:38:01
Are these all the answers?

jeff10 2011-02-24 19:38:39
NO only two are

theone142857 2011-02-24 19:38:39
5+3=8 bad

sparkle123 2011-02-24 19:38:39
first and last not okay, must have diff of 6

NoWayHaze 2011-02-24 19:38:39
no, some of them don't add up to other differences

professordad 2011-02-24 19:38:39
no because 8 is not a choice

JSGandora 2011-02-24 19:38:39
no, we can't get 6 with the first or last onews

Jasmine8925 2011-02-24 19:38:39
5 and 1 should be next to each other so we get 6 as a difference

copeland 2011-02-24 19:38:43
Actually that's too many! We've narrowed it down to these possibilities, but two of these are bogus. Notice that the top and bottom options have lengths of 8. This is unacceptable.

copeland 2011-02-24 19:38:47
Fortunately the middle two have lengths 1, 3, 5, 4, 6, and 9, which is exactly what we wanted.

copeland 2011-02-24 19:38:49

copeland 2011-02-24 19:38:52

copeland 2011-02-24 19:38:55
How do we finish?

jeff10 2011-02-24 19:39:36
Express everything in terms of z.

hsm174 2011-02-24 19:39:36
find w

Joe10112 2011-02-24 19:39:36
write equations!

sindennisz 2011-02-24 19:39:36
find w

RisingMathStar 2011-02-24 19:39:36
Their sum is 44

Iggy Iguana 2011-02-24 19:39:36
Now just plug them in to w+x+y+z=44

leinad47x 2011-02-24 19:39:36
add up possible values

andrewjjiang97 2011-02-24 19:39:36
find the values of w

Relativity1618 2011-02-24 19:39:36
w+x+y+z=44

AkshajKadaveru 2011-02-24 19:39:36
set variables

Jasmine8925 2011-02-24 19:39:36
we know the sum is 44, we should use that info now

theone142857 2011-02-24 19:39:36
sum=44

smallpeoples343 2011-02-24 19:39:36

TruffleMaster 2011-02-24 19:39:36
x+x+3+x+3+1+x+3+1+5=44

copeland 2011-02-24 19:39:44

copeland 2011-02-24 19:39:48

copeland 2011-02-24 19:39:58
What about the second case?

ahaanomegas 2011-02-24 19:40:23
We can express our sets as {w, w - 5, w - 6, w - 9} or {w, w - 3, w - 4, w - 9}

AlphaMath1 2011-02-24 19:40:23
w=15

theone142857 2011-02-24 19:40:23
15

Tribefan 2011-02-24 19:40:23
w=15

Iggy Iguana 2011-02-24 19:40:23
W=15

AstralMeson 2011-02-24 19:40:23
w = 15

airplanes1 2011-02-24 19:40:23
w=15

luminate 2011-02-24 19:40:23
w = 15

ahaanomegas 2011-02-24 19:40:23
In the second case, we have 4w - 16 = 44 --> w = 15

alikimlynn 2011-02-24 19:40:23
w = 15

TrigMan 2011-02-24 19:40:23
w=15, right?

.cpp 2011-02-24 19:40:23
Do the same thing for w=15.

copeland 2011-02-24 19:40:27

copeland 2011-02-24 19:40:30

copeland 2011-02-24 19:40:33
And the solution?

anonymous0 2011-02-24 19:40:58
The answer is B

krmathcounts 2011-02-24 19:40:58
b

jeff10 2011-02-24 19:40:58
16+15=31

danielguo94 2011-02-24 19:40:58
B

davidx 2011-02-24 19:40:58
31

alikimlynn 2011-02-24 19:40:58
31

andrewjjiang97 2011-02-24 19:40:58
sum is 31, B

professordad 2011-02-24 19:40:58
16+15=31.

calvinhobbesliker 2011-02-24 19:40:58
31

ooooo1 2011-02-24 19:40:58
b 31

luppleAOPS 2011-02-24 19:40:58
15+16=31

TrigMan 2011-02-24 19:40:58
B) 31 :)

Iggy Iguana 2011-02-24 19:40:58
16=15=31

j_f_c_w 2011-02-24 19:40:58
And the answer is 31=15+16

erica539 2011-02-24 19:40:58
31 the sum B

christerson 2011-02-24 19:40:58
33

kangchangood 2011-02-24 19:40:58
(b)

aleCTY 2011-02-24 19:40:58
31

googol.plex 2011-02-24 19:40:58
16+15=31

LightningStreak 2011-02-24 19:40:58
31

alex31415 2011-02-24 19:40:58
so, 15+16=31 :)

NoWayHaze 2011-02-24 19:40:58
15+16=31

sayeramesh 2011-02-24 19:40:58
B => 31

ooooo1 2011-02-24 19:40:58
B) 31

TemurAmir 2011-02-24 19:40:58
The solution is 16+15=31 B

theone142857 2011-02-24 19:40:58
31

kimmy12345 2011-02-24 19:40:58
15+16 = 31B

LightningStreak 2011-02-24 19:40:58
B. 31

TruffleMaster 2011-02-24 19:40:58
15+16=31

chengp 2011-02-24 19:40:58
31

infinity1 2011-02-24 19:40:58
15 + 16 = 31

chaos763 2011-02-24 19:40:58
31.

RogLiu26 2011-02-24 19:40:58
B

Gosav3122 2011-02-24 19:40:58
31

.cpp 2011-02-24 19:40:58
16+15 = 31 => B.

ahaanomegas 2011-02-24 19:40:58
Sum of all possible values of w = 16 + 15 = 31 --> (B)

christerson 2011-02-24 19:40:58
31

spaceguy524 2011-02-24 19:40:58
B

mathcool2009 2011-02-24 19:40:58
16+15=31

supermathman 2011-02-24 19:40:58
16

mathletepower 2011-02-24 19:40:58
(b) 31

vwu9 2011-02-24 19:40:58
31

Tribefan 2011-02-24 19:40:58
(B) 31

copeland 2011-02-24 19:41:01
The solution is 16+15=31, answer (B).

copeland 2011-02-24 19:41:26
MORE! That was fun.

copeland 2011-02-24 19:41:30

copeland 2011-02-24 19:41:41
First step?

AlphaMath1 2011-02-24 19:42:03
FIrst we need to try to visualize it

PowerOfPi 2011-02-24 19:42:03
diagram

Iggy Iguana 2011-02-24 19:42:03
draw a diagram

minirafa 2011-02-24 19:42:03
draw a diagram?

luminate 2011-02-24 19:42:03
draw a picture

googol.plex 2011-02-24 19:42:03
draw it

AkshajKadaveru 2011-02-24 19:42:03
PICTURE TIME!

jxihong 2011-02-24 19:42:03
draw a picture

TemurAmir 2011-02-24 19:42:03
Draw a diagram?

kimmy12345 2011-02-24 19:42:03
draw

erica539 2011-02-24 19:42:03
diagram.

sindennisz 2011-02-24 19:42:03
Draw a diagram.

ilovepink 2011-02-24 19:42:03
picture!

copeland 2011-02-24 19:42:07
Draw a picture:

copeland 2011-02-24 19:42:11

copeland 2011-02-24 19:42:14
And we need our cube inside:

copeland 2011-02-24 19:42:17

mathcountsloser 2011-02-24 19:42:25
What if you are a fail drawer.

copeland 2011-02-24 19:42:27
Good thing I'm not.

copeland 2011-02-24 19:42:30
Awesome. Now what?

copeland 2011-02-24 19:43:14
Wow that's a lot of suggestions.

copeland 2011-02-24 19:43:16
Coordinates.

copeland 2011-02-24 19:43:17
Labels

copeland 2011-02-24 19:43:19
. . .

mhy123 2011-02-24 19:43:26
we must take a cross section

erica539 2011-02-24 19:43:26
do a cross-section

.cpp 2011-02-24 19:43:26
Take an appropriate cross section.

aleCTY 2011-02-24 19:43:26
Right triangles

gmarch89 2011-02-24 19:43:26
calculate the volume of the pyramid?

LightningStreak 2011-02-24 19:43:26
draw an altitude?

luminate 2011-02-24 19:43:26
take a slice out of the pyramid and cube part

copeland 2011-02-24 19:43:35
OK.

copeland 2011-02-24 19:43:40
What cross-section?

Jasmine8925 2011-02-24 19:44:46
through the middle of the pyramid and cube

krmathcounts 2011-02-24 19:44:46
Down the middle of the pyramid, so you can find the height

dafjbomb 2011-02-24 19:44:46
from the top to the base so that the cube is cut into a perfect square cross section

ChipDale 2011-02-24 19:44:46
Through the middle

RisingMathStar 2011-02-24 19:44:46
center, laterally

TrigMan 2011-02-24 19:44:46
Right along the middle, so it hits the apex and the midpoints of two of the opposite base edges.

calvinhobbesliker 2011-02-24 19:44:46
The triangle whose altitude is the altidude of the pyramid

mathcool2009 2011-02-24 19:44:46
I think he means cut the pyramid in half vertically

copeland 2011-02-24 19:44:53
OK, I like this, but that's not enough info.

copeland 2011-02-24 19:44:58
There are lots of such slices.

copeland 2011-02-24 19:45:01
Which one do we want?

theone142857 2011-02-24 19:45:43
diagnal

sparkle123 2011-02-24 19:45:43
along the diagonal of the base

gravel 2011-02-24 19:45:43
To the midpoint of the opposing base side

ChipDale 2011-02-24 19:45:43
Through the diagonal of the cube

briantix 2011-02-24 19:45:43
opposite corners of the base of the pyramid

Candokevin 2011-02-24 19:45:43
slice it along the bottom diagnonal

copeland 2011-02-24 19:45:48
We want to try to find some 2D geometry somewhere that we can actually understand.

copeland 2011-02-24 19:45:52
If we look at one of the faces, that will give us some insight but it won't actually let us use the cube nicely.

copeland 2011-02-24 19:45:54
Instead we ought to slice diagonally across the pyramid.

copeland 2011-02-24 19:45:57

copeland 2011-02-24 19:46:01
OK, that's a little confusing so we'll push it into 2 dimensions:

copeland 2011-02-24 19:46:06

copeland 2011-02-24 19:46:08
What are the dimensions of this triangle?

theone142857 2011-02-24 19:46:38
1,1,xsqrt{2}

AlphaMath1 2011-02-24 19:46:38
sqrt 2 1 1

leinad47x 2011-02-24 19:46:38
1 by 1 by sqrt 2

luppleAOPS 2011-02-24 19:46:38
1,1, and sqrt2 for the base

andrewjjiang97 2011-02-24 19:46:38
1, 1, sqrt{2}

Iggy Iguana 2011-02-24 19:46:38
sqrt 2, 1, and 1

briantix 2011-02-24 19:46:38
1,1 sqrt2

willwang123 2011-02-24 19:46:38
1-1-sqrt2

PowerOfPi 2011-02-24 19:46:38
base is sqrt(2), legs are 1

STan321 2011-02-24 19:46:38
1-1-rt2

copeland 2011-02-24 19:46:43
Since the pyramid has equilateral faces of length 1, the legs of this triangle are length 1. The base is the diagonal of a square (the base of the pyramid) with side length 1. Therefore the base has length sqrt(2).

copeland 2011-02-24 19:46:47

copeland 2011-02-24 19:46:49
What kind of triangle is this?

AkshajKadaveru 2011-02-24 19:47:12
Right

dafjbomb 2011-02-24 19:47:12
right

sindennisz 2011-02-24 19:47:12
45-45-90

MrWhatWhen 2011-02-24 19:47:12
a right triangle

RisingMathStar 2011-02-24 19:47:12
isosceles right

briantix 2011-02-24 19:47:12
45-45-90

professordad 2011-02-24 19:47:12
its right triangle

usernamehi 2011-02-24 19:47:12
45, 45,90

airplanes1 2011-02-24 19:47:12
right isoceles

Tribefan 2011-02-24 19:47:12
45,45,90

copeland 2011-02-24 19:47:16
It's an isosceles right triangle! We can use that later.

copeland 2011-02-24 19:47:18
And the rectangle? (Of course the answer depends on the unknown side length, s, of the cube.)

Spring 2011-02-24 19:48:05
s*sqrt2 by s

aleCTY 2011-02-24 19:48:05
s, ssqrt2

hsm174 2011-02-24 19:48:05
s,srt2

.cpp 2011-02-24 19:48:05
h sqrt(2) and h are the sides of the rectangle.

aunch 2011-02-24 19:48:05
has height s and width s*sqrt2

erica539 2011-02-24 19:48:05
s by rt2s

thedrummer 2011-02-24 19:48:05
s by srt2

sindennisz 2011-02-24 19:48:05
s by s*sqrt2

copeland 2011-02-24 19:48:07

copeland 2011-02-24 19:48:10
Now what do you see?

PowerOfPi 2011-02-24 19:48:41
top triangle is similar to the whole because of parallel lines

NoWayHaze 2011-02-24 19:48:41
similar triangles!

letrangere 2011-02-24 19:48:41
similar triangles

Iggy Iguana 2011-02-24 19:48:41
similar triangles

thedrummer 2011-02-24 19:48:41
similar triangles

az94566 2011-02-24 19:48:41
a shape

mthcz11 2011-02-24 19:48:41
similar triangles

luppleAOPS 2011-02-24 19:48:41
we see more similar triangles

andrewjjiang97 2011-02-24 19:48:41
the two triangles on the bottom are 45-45-90

aleCTY 2011-02-24 19:48:41
similar triangles

copeland 2011-02-24 19:48:45
On the left and right we have two isosceles right triangles. We know this because the large triangle is isosceles (so, for example, the lines have slope 1 and -1).

copeland 2011-02-24 19:48:49

copeland 2011-02-24 19:48:53
OK, now the easy part. How do we finish?

Moldytape 2011-02-24 19:49:39
2s + s rt2 = rt2

Iggy Iguana 2011-02-24 19:49:39
WE know that 2s+ssqrt2 = sqrt2

hsm174 2011-02-24 19:49:39
2s+srt2=rt2

Doink 2011-02-24 19:49:39
s(2+sqrt2)=sqrt2

NoWayHaze 2011-02-24 19:49:39
2s+s*sqrt(2)=sqrt(2)

sindennisz 2011-02-24 19:49:39
2s+s*sqrt2=sqrt2

willwang123 2011-02-24 19:49:39
2s+s*sqrt2=sqrt2

calvinhobbesliker 2011-02-24 19:49:39
sqrt2=2s+ssqrt2?

awesomeusername 2011-02-24 19:49:39

RisingMathStar 2011-02-24 19:49:39

copeland 2011-02-24 19:49:44

copeland 2011-02-24 19:49:48
And the volume is. . .

professordad 2011-02-24 19:50:45

TruffleMaster 2011-02-24 19:50:45
5sqrt2-7

briantix 2011-02-24 19:50:45
5sqrt2-7

Iggy Iguana 2011-02-24 19:50:45
cube that to get 5sqrt2-7

NoWayHaze 2011-02-24 19:50:45
5sqrt2-7

erica539 2011-02-24 19:50:45
5rt2 - 7

.cpp 2011-02-24 19:50:45
5sqrt(2) - 7 => A.

ahaanomegas 2011-02-24 19:50:45

jxihong 2011-02-24 19:50:45
A 5 sq rt 2 -7

Relativity1618 2011-02-24 19:50:45
(sqrt2)-1 cubed

calvinhobbesliker 2011-02-24 19:50:45
Cube it to get 5sqrt2-7

theone142857 2011-02-24 19:50:45
cube it to get 5sqrt{2}-7

kangchangood 2011-02-24 19:50:45
(a)..

andrewjjiang97 2011-02-24 19:50:45
5sqrt{2}-7

Spring 2011-02-24 19:50:45

Relativity1618 2011-02-24 19:50:45
(5sqrt2)-7

Doink 2011-02-24 19:50:45
5sqrt2-7 (A)

MaggieKim 2011-02-24 19:50:45
5sqrt2-7

sparkle123 2011-02-24 19:50:45
A 5sqrt2 -7

dragonx 2011-02-24 19:50:45
A) 5sqrt(2)-7

copeland 2011-02-24 19:50:47

copeland 2011-02-24 19:50:57
I like that problem. It's fun to visualize.

mathcountsloser 2011-02-24 19:51:17

copeland 2011-02-24 19:51:19
(root(2)-1)^3 is prettier.

copeland 2011-02-24 19:51:31

andrewjjiang97 2011-02-24 19:51:31
I think it's better when you look from one face, and use similar triangles

alikimlynn 2011-02-24 19:51:33
Since you only need the hundreds’ digit and you can’t use a calculator (wah), use 11 instead of 2011. You’ll find that there is a cycle of 1, 3, 6, 0, 5 for the hundreds digit. (Found by 11^1, 11^2, 11^3, 11^4, 11^5, respectively.) Then, 2011 – 1 = 2010. 2010/5 = 402 with no remainder. Therefore, 2011^2011 has a hundreds’ digit of 5, or (C).

ahaanomegas 2011-02-24 19:51:36
Nothing is ugly in math if you know how to do it

copeland 2011-02-24 19:51:47
Dave did that.

copeland 2011-02-24 19:51:53
Dave is passing things without permission.

copeland 2011-02-24 19:52:00
How do we feel about Dave?

willwang123 2011-02-24 19:52:18
Gasp!!!

mathcountsloser 2011-02-24 19:52:18
Good job Dave.

davidx 2011-02-24 19:52:18
Dave is cool

willwang123 2011-02-24 19:52:18
Boo!!!!

alikimlynn 2011-02-24 19:52:18
Badly?

ChipDale 2011-02-24 19:52:18
Bad

sayeramesh 2011-02-24 19:52:18
bad

DVA6102 2011-02-24 19:52:18
good

dragonx 2011-02-24 19:52:18
Banhammer TIME

erica539 2011-02-24 19:52:18
shame on him!

Joe10112 2011-02-24 19:52:18
BAD DAVE!

usernamehi 2011-02-24 19:52:18
MAD!!!!!

smallpeoples343 2011-02-24 19:52:18
He's a bad person.

copeland 2011-02-24 19:52:21
Yes.

copeland 2011-02-24 19:52:24

copeland 2011-02-24 19:52:35
What is it that we need to compute?

AlphaMath1 2011-02-24 19:53:31
First reduce to mod 1000

VIPMaster 2011-02-24 19:53:31
the first digit of 2011^2011 (mod 1000)

aleCTY 2011-02-24 19:53:31
2011^2011 mod 1000

mthcz11 2011-02-24 19:53:31
2011^2011 in mod 1000

Iggy Iguana 2011-02-24 19:53:31
the thing mod 1000

AlphaBetaTheta 2011-02-24 19:53:31
remainder in mod 1000, binomial theorem to the rescue

copeland 2011-02-24 19:53:34

copeland 2011-02-24 19:53:46
Whenever you see one of these "hundreds" digits you know you should be reducing.

copeland 2011-02-24 19:53:50
What's the first thing we notice?

spaceguy524 2011-02-24 19:54:21
its the same as 11^(2011)

hi how are you doing toda 2011-02-24 19:54:21
get rid of 2000

luminate 2011-02-24 19:54:21
11^2011

capu 2011-02-24 19:54:21
compute 11^2011

xxrxxhxx 2011-02-24 19:54:21
2011 == 11 mod 1000

calculus321 2011-02-24 19:54:21
2011=2000+11

hi how are you doing toda 2011-02-24 19:54:21
2000 is unneccessary

briantix 2011-02-24 19:54:21
we can ignore the thoousands place

TruffleMaster 2011-02-24 19:54:21
2011 is 11 in mod 1000

mthcz11 2011-02-24 19:54:21
11^2011

googol.plex 2011-02-24 19:54:21
2011^2011=11^2011

AlphaBetaTheta 2011-02-24 19:54:21
2011^2011 = 11^2011 (mod 1000)

copeland 2011-02-24 19:54:26

copeland 2011-02-24 19:54:30
Now how do we compute this?

AlphaBetaTheta 2011-02-24 19:55:14
binomial theorem

AlphaMath1 2011-02-24 19:55:14
11=10+1 then BINOMIAL THEOREM!!!!!!

NoWayHaze 2011-02-24 19:55:14
binomial theorem

Spring 2011-02-24 19:55:15

.cpp 2011-02-24 19:55:15
(10+1)^2011

morpheus44 2011-02-24 19:55:15
binomial theorem?

Tribefan 2011-02-24 19:55:15
binomial theroem

copeland 2011-02-24 19:55:18
We use the binomial theorem on 11 = 10 + 1.

copeland 2011-02-24 19:55:24

copeland 2011-02-24 19:55:26
Why did I stop there?

VIPMaster 2011-02-24 19:55:59
because everything else cancels out mod 1000

calvinhobbesliker 2011-02-24 19:55:59
All else is 0 mod 1000

xxrxxhxx 2011-02-24 19:55:59
rest all have hundreds digit of 0 due to # of 10s as factors

PowerOfPi 2011-02-24 19:55:59
everything after that is divisble by 10^3=1000

AstralMeson 2011-02-24 19:55:59
that's the hundreds place

aleCTY 2011-02-24 19:55:59
10^3 is 0 mod 1000

.cpp 2011-02-24 19:55:59
All the other terms are 0 mod 1000, so they are unnecessary.

Spring 2011-02-24 19:55:59
Everything else is divisible by 1000 so they don't affect the hundreds digit.

Luminescence 2011-02-24 19:55:59
the others are 0mod1000

mathcountsloser 2011-02-24 19:55:59
10^3 = thousand, you only need the hundereds difit

copeland 2011-02-24 19:56:03
After that everything has a large power of 10. The next term, for example, is divisible by 10^3=1000, so is 0 mod 1000.

copeland 2011-02-24 19:56:05
Therefore removing all those ones gives

copeland 2011-02-24 19:56:09

copeland 2011-02-24 19:56:10
What's the answer?

quantammaths1234 2011-02-24 19:56:50
6.

TruffleMaster 2011-02-24 19:56:50
6

vwu9 2011-02-24 19:56:50
6

willwang123 2011-02-24 19:56:50
6!

soulspeedy 2011-02-24 19:56:50
6

NoWayHaze 2011-02-24 19:56:50
6

xxrxxhxx 2011-02-24 19:56:50
6 (D)

TemurAmir 2011-02-24 19:56:50
The answer is 6, D

Luminescence 2011-02-24 19:56:50
6

DavidTong 2011-02-24 19:56:50
6 D

infinity1 2011-02-24 19:56:50
D 6

ahaanomegas 2011-02-24 19:56:50

Relativity1618 2011-02-24 19:56:50
6

CRICKET229 2011-02-24 19:56:50
6

knowmath 2011-02-24 19:56:50
d

j_f_c_w 2011-02-24 19:56:50
(d) 6

CRICKET229 2011-02-24 19:56:50
(D) 6

girishvar12 2011-02-24 19:56:50
D. 6

Tribefan 2011-02-24 19:56:50
(D) 6

mthcz11 2011-02-24 19:56:50
D, 6

AlphaBetaTheta 2011-02-24 19:56:50
6.

capu 2011-02-24 19:56:50
D6

LightningStreak 2011-02-24 19:56:50
d.6

spaceguy524 2011-02-24 19:56:50
D

jeff10 2011-02-24 19:56:50
the last three digits are 611, so the answer is 6.

Lalagato 2011-02-24 19:56:50
6

sincostanseccsccot 2011-02-24 19:56:50
6?

agejiageji 2011-02-24 19:56:50
6

anonymous0 2011-02-24 19:56:50
D, 6

copeland 2011-02-24 19:56:54
Since I'm lazy, I'll reduce before multiplying:

copeland 2011-02-24 19:56:57

copeland 2011-02-24 19:57:03
The answer is (D) 6.

aleCTY 2011-02-24 19:57:17
Couldn't we just use Euler's extension of FlT?

copeland 2011-02-24 19:57:25
Many people suggested Euler at the begnning.

copeland 2011-02-24 19:57:31
As an aside (and ignore this or come back to it if you don't understand), I assumed this problem was going to be much harder, so I started by computing:

copeland 2011-02-24 19:57:36

copeland 2011-02-24 19:57:38

copeland 2011-02-24 19:57:48
This turns out to lead to fewer mistakes but in all I think it was the slower approach for this problem.

ahaanomegas 2011-02-24 19:57:57
And what is Euler's Extension of FIT?

copeland 2011-02-24 19:58:04
It's a little "l." Get it?

mathletepower 2011-02-24 19:58:21
NO

mthcz11 2011-02-24 19:58:21
no

willwang123 2011-02-24 19:58:21
No.

usernamehi 2011-02-24 19:58:21
I dont get it.

ChipDale 2011-02-24 19:58:21
No

davidx 2011-02-24 19:58:21
No

TemurAmir 2011-02-24 19:58:21
Nope

leinad47x 2011-02-24 19:58:21
no..

MrWhatWhen 2011-02-24 19:58:21
What does phi mean in that

Joe10112 2011-02-24 19:58:21
No. Please explain? XD

ctmusicgirl 2011-02-24 19:58:21
NO.

Relativity1618 2011-02-24 19:58:21
no

chaos763 2011-02-24 19:58:21
Not really.

sparkle123 2011-02-24 19:58:21
no :(

copeland 2011-02-24 19:58:24
Hmm.

copeland 2011-02-24 19:58:40
Check out Fermat's little theorem and Euler's totient.

copeland 2011-02-24 19:58:57
But for now we should move on. That wasn't actually necessary or even useful for this problem.

copeland 2011-02-24 19:59:04

copeland 2011-02-24 19:59:13
Where should we start?

lg5293 2011-02-24 19:59:44
graph?

erica539 2011-02-24 19:59:44
a diagram perhaps?

Iggy Iguana 2011-02-24 19:59:44
diagram?

googol.plex 2011-02-24 19:59:44
graph it

supermathman 2011-02-24 19:59:44
draw it!!!!1

ahaanomegas 2011-02-24 19:59:44
Draw an x-y plane?

mthcz11 2011-02-24 19:59:44
draw a diagram

mthcz11 2011-02-24 19:59:44
draw a graph

copeland 2011-02-24 19:59:53
You just like my drawins. I know you.

copeland 2011-02-24 19:59:57
We could begin by subtracting 2 because it's irrelevant, but we won't bother. It turns out not to matter too much and only makes the computations a little easier.

copeland 2011-02-24 20:00:00
Let's draw a graph:

copeland 2011-02-24 20:00:03

copeland 2011-02-24 20:00:09
What happens as we increase m?

Spring 2011-02-24 20:00:37
the slope increases

agejiageji 2011-02-24 20:00:37
slope increases

ChipDale 2011-02-24 20:00:37
The line is steeper

leinad47x 2011-02-24 20:00:37
slope goes up

Iggy Iguana 2011-02-24 20:00:37
the line gets steeper

luminate 2011-02-24 20:00:37
it becomes stteper

ahaanomegas 2011-02-24 20:00:37
The line becomes steeper

mathcool2009 2011-02-24 20:00:37
the grafh becomes steeper

professordad 2011-02-24 20:00:37
line gets steeper

Doink 2011-02-24 20:00:37
it gets steeper

pgmath 2011-02-24 20:00:37
The line becomes steeper.

copeland 2011-02-24 20:00:41
As we increase m the slope increases:

copeland 2011-02-24 20:00:43

copeland 2011-02-24 20:00:47

lg5293 2011-02-24 20:01:41
the lowest lattice point we can hit

Spring 2011-02-24 20:01:41
The maximum sloope the line can have and still not pass through any lattice points with x<100

MrWhatWhen 2011-02-24 20:01:41
the maximum slope that won't pass through any lattice point

TrigMan 2011-02-24 20:01:41
The highest you can go without hitting more dots.

Jasmine8925 2011-02-24 20:01:41
the next highest slope so that the line will go through a lattice point

RisingMathStar 2011-02-24 20:01:41
The minimum slope at which the line touches the next lattice point

.cpp 2011-02-24 20:01:41
The slope where y=mx+2 meets a lattice point first after m=1/2.

xxrxxhxx 2011-02-24 20:01:41
when the line will be steep enough to hit a lattice point

AlphaMath1 2011-02-24 20:01:41
the next slope wheres there's an integral solution (x, y)

gh625 2011-02-24 20:01:41
The slope that goes through the next highest lattice point

JSGandora 2011-02-24 20:01:41
The slope at which the line first concides with another lattice point

copeland 2011-02-24 20:01:43
The number a is the first slope larger than m=1/2 that hits another lattice point.

copeland 2011-02-24 20:01:46
Where will this collision occur?

Spring 2011-02-24 20:02:39
(99,52)

AlphaMath1 2011-02-24 20:02:39
at x=99?

calvinhobbesliker 2011-02-24 20:02:39
99, 52

RisingMathStar 2011-02-24 20:02:39
(99, 52)

JSGandora 2011-02-24 20:02:39
At the upper right hand corner of the 100x52 rectangle

copeland 2011-02-24 20:02:43
It'll occur somewhere at the end, so let me draw that for you:

copeland 2011-02-24 20:02:45

copeland 2011-02-24 20:02:48
For convenience we'll let L denote the line through (0,2) with slope 1/2.

copeland 2011-02-24 20:02:52
At this point we could compute the value of a right away, but it's only a tiny bit harder to get a nice, clean proof that our answer is correct, so instead we'll do everything the long way.

copeland 2011-02-24 20:02:58
For every integer x, there is a nearest lattice point above L with this x coordinate. What is this lattice point?

copeland 2011-02-24 20:03:31
First of all, it depends on whether x is even or odd. If x is even then what is the lattice point?

JSGandora 2011-02-24 20:04:18
(x,y+1)

Iggy Iguana 2011-02-24 20:04:18
(x,x+3)

.cpp 2011-02-24 20:04:18
1/2 x + 3.

Spring 2011-02-24 20:04:18
1/2x+3

RisingMathStar 2011-02-24 20:04:18

ahaanomegas 2011-02-24 20:04:18

copeland 2011-02-24 20:04:28

copeland 2011-02-24 20:04:28
What about if x is odd?

.cpp 2011-02-24 20:05:29
If x is odd, this is just (x,1/2x + 5/2).

PowerOfPi 2011-02-24 20:05:29
(x, x/2+2.5)

Spring 2011-02-24 20:05:29
1/2x+2 rounded up to the nearest integer

ahaanomegas 2011-02-24 20:05:29

VIPMaster 2011-02-24 20:05:29
(x, x/2 + 2.5)

Iggy Iguana 2011-02-24 20:05:29
(x,x/2+2.5)

JSGandora 2011-02-24 20:05:29
(x,y+1/2)

copeland 2011-02-24 20:05:33

copeland 2011-02-24 20:05:38

Spring 2011-02-24 20:06:26
(x/2+1)/x

ahaanomegas 2011-02-24 20:06:26

calvinhobbesliker 2011-02-24 20:06:26
0.5+1/x

Iggy Iguana 2011-02-24 20:06:26
(x/2+1)/x

luppleAOPS 2011-02-24 20:06:26
(x/2+1)/(x)

spaceguy524 2011-02-24 20:06:26

.cpp 2011-02-24 20:06:27
(x/2 + 1)/x

JSGandora 2011-02-24 20:06:27
(x/2+1)/x

TruffleMaster 2011-02-24 20:06:27
1/2+1/x

infinity1 2011-02-24 20:06:27
1/2 + 1/x

ChipDale 2011-02-24 20:06:27
1/2+1/x

sindennisz 2011-02-24 20:06:27
1/2+1/x

copeland 2011-02-24 20:06:29

copeland 2011-02-24 20:06:31
Do we want to maximize or minimize that?

ChipDale 2011-02-24 20:07:09
minimize

PowerOfPi 2011-02-24 20:07:09
minimize

erica539 2011-02-24 20:07:09
minimize

calvinhobbesliker 2011-02-24 20:07:09
Minimize

Gosav3122 2011-02-24 20:07:09
min

mstoenescu 2011-02-24 20:07:09
minimize

Spring 2011-02-24 20:07:09
Minimize because we want the first collision with a lattice point

Gosav3122 2011-02-24 20:07:09
biggest slope, so minimize x

jeff10 2011-02-24 20:07:09
minimize

mathcool2009 2011-02-24 20:07:09
minimize

copeland 2011-02-24 20:07:11
We want the line to be under all of these points so we want the minimum. How small can that be for even x from 1 to 100?

willwang123 2011-02-24 20:07:50
51/100

PowerOfPi 2011-02-24 20:07:50
1/2+1/100=51/100

Iggy Iguana 2011-02-24 20:07:50
1/2 + 1/100 = 51/100

calvinhobbesliker 2011-02-24 20:07:50
1/2+1/100=51/100

Spring 2011-02-24 20:07:50
1/2+1/100=51/100

TrigMan 2011-02-24 20:07:50
51/100

RisingMathStar 2011-02-24 20:07:50
x should be 100

infinity1 2011-02-24 20:07:50
x = 100 ==> 0.51

professordad 2011-02-24 20:07:50
1/2+1/100 or 51/100

copeland 2011-02-24 20:07:54

copeland 2011-02-24 20:07:55

Spring 2011-02-24 20:08:31
(x/2+1/2).x

calvinhobbesliker 2011-02-24 20:08:31
1/2+1/2x

Iggy Iguana 2011-02-24 20:08:31
(x/2+.5)/x

VIPMaster 2011-02-24 20:08:31
1/2 + (1/2x)

spaceguy524 2011-02-24 20:08:31

PowerOfPi 2011-02-24 20:08:31
1/2+1/(2x)

.cpp 2011-02-24 20:08:31
1/2 + 1/(2x)

MrWhatWhen 2011-02-24 20:08:31
(x/2-1/2)/x

ahaanomegas 2011-02-24 20:08:31

sindennisz 2011-02-24 20:08:31
1/2+1/2x

copeland 2011-02-24 20:08:37

copeland 2011-02-24 20:08:39
What is the minimum over odd x between 1 and 100?

LiBoy 2011-02-24 20:09:12
50/99

jeff10 2011-02-24 20:09:12
50/99

Iggy Iguana 2011-02-24 20:09:12
1/2 + 1/198 = 50/99

calvinhobbesliker 2011-02-24 20:09:12
1/2+1/198=100/198=50/99

ChipDale 2011-02-24 20:09:12
50/99

Spring 2011-02-24 20:09:12
1/2+1/(2*99)=50/99

Gosav3122 2011-02-24 20:09:12
99

PowerOfPi 2011-02-24 20:09:12
1/2+1/198=50/99

mstoenescu 2011-02-24 20:09:12
50/99

danielguo94 2011-02-24 20:09:12
50/99

AlphaMath1 2011-02-24 20:09:12
x=99 --->50/99

copeland 2011-02-24 20:09:16

copeland 2011-02-24 20:09:19
And the answer?

LightningStreak 2011-02-24 20:10:08
B

AlphaMath1 2011-02-24 20:10:08
50/99

agejiageji 2011-02-24 20:10:08
50/99

PowerOfPi 2011-02-24 20:10:08
B

gh625 2011-02-24 20:10:08
B 50/99

professordad 2011-02-24 20:10:08
50/99 or B

Jasmine8925 2011-02-24 20:10:08
50/99

minirafa 2011-02-24 20:10:08
b

supermathman 2011-02-24 20:10:08
50/99

Luminescence 2011-02-24 20:10:08
the smaller of the two, 50/99

ahaanomegas 2011-02-24 20:10:08

.cpp 2011-02-24 20:10:08
50/99 = (B).

girishvar12 2011-02-24 20:10:08
B

willwang123 2011-02-24 20:10:08
B?

ChipDale 2011-02-24 20:10:08
B

ctmusicgirl 2011-02-24 20:10:08
B

danielguo94 2011-02-24 20:10:08
B

TruffleMaster 2011-02-24 20:10:08
50/99

ooooo1 2011-02-24 20:10:08
50/99

mathletepower 2011-02-24 20:10:08
B for bee

sayeramesh 2011-02-24 20:10:08
B

ctmusicgirl 2011-02-24 20:10:08
B

az94566 2011-02-24 20:10:08
50/99

ooooo1 2011-02-24 20:10:08
B) 50/99

mathcounter7 2011-02-24 20:10:08
B

Spring 2011-02-24 20:10:08
50/99-5000/9900 and 51/100=5001/9900 so 50/99 is smaller

LiBoy 2011-02-24 20:10:08
50/99!

VIPMaster 2011-02-24 20:10:08
50/99

timelessmath 2011-02-24 20:10:08
50/99

knowmath 2011-02-24 20:10:08
b

mathletepower 2011-02-24 20:10:08
(B) 50/99

copeland 2011-02-24 20:10:12
Since 50/99 < 51/100, the answer must be 50/99, (B).

copeland 2011-02-24 20:10:18
Alright!

copeland 2011-02-24 20:10:33
Now it's time for the last problem on the 10.

copeland 2011-02-24 20:10:46

copeland 2011-02-24 20:10:55
OK, what first?

Joe10112 2011-02-24 20:11:18
Draw a diagram for this one first!

Iggy Iguana 2011-02-24 20:11:18
another diagram

calvinhobbesliker 2011-02-24 20:11:18
Diagram

mthcz11 2011-02-24 20:11:18
diagram

professordad 2011-02-24 20:11:18
draw the diagram :)

chaos763 2011-02-24 20:11:18
Diagram?

ChipDale 2011-02-24 20:11:18
Draw a picture

hoper190 2011-02-24 20:11:18
draw something

theone142857 2011-02-24 20:11:18
diagram

.cpp 2011-02-24 20:11:18
Diagram!

Luminescence 2011-02-24 20:11:18
draw a picture...?

az94566 2011-02-24 20:11:18
picture

VIPMaster 2011-02-24 20:11:18
We love your diagrams!!

ahaanomegas 2011-02-24 20:11:18
Diagram

copeland 2011-02-24 20:11:23
Well, drawing a picture has been useful for us so far. Let's draw another.

copeland 2011-02-24 20:11:27

copeland 2011-02-24 20:11:31
At every step we start with a triangle, we draw the incircle and we look at the three edge lengths above. Let's start by giving them variables.

copeland 2011-02-24 20:11:45

copeland 2011-02-24 20:11:52
Alright, what next? What else does geometry tell us?

AlphaMath1 2011-02-24 20:12:49
Congruent tangents

timelessmath 2011-02-24 20:12:49
tangent segments are congruent

Joe10112 2011-02-24 20:12:49
we note that z equals CE, and x equals FA, and y equals BD!

PowerOfPi 2011-02-24 20:12:49
CE=z, BD=y, AF=x because they are tangents from the same point

Iggy Iguana 2011-02-24 20:12:49
AB - x = y, and same thiing for the other sides.

professordad 2011-02-24 20:12:49
Two Tangent Theorem, so CE=CF, BE=BD, and AD=AF

ChipDale 2011-02-24 20:12:49
equidistant from the sides

copeland 2011-02-24 20:12:54
Since we're dealing with the incenter, we also know the other three edge lengths!

copeland 2011-02-24 20:12:57
Watch what happens if we draw the incenter and a couple of arcs:

copeland 2011-02-24 20:13:00

copeland 2011-02-24 20:13:04
Since the two triangles AOF and AOD are congruent (since these are right triangles with one equal edge and common hypotenuse) we know that AF = AD. (There's also a cute symmetry argument if you prefer, from flipping over OA.)

copeland 2011-02-24 20:13:17
The picture fleshes out like this:

copeland 2011-02-24 20:13:19

copeland 2011-02-24 20:13:24
OK, so back to the problem. Let's see how we include the initial conditions.

copeland 2011-02-24 20:13:27

copeland 2011-02-24 20:13:29
Can we solve this system or is it under- or overdetermined?

DVA6102 2011-02-24 20:14:11
it is solvable

TruffleMaster 2011-02-24 20:14:11
you can solve it if you add them

Jasmine8925 2011-02-24 20:14:11
i think we can solve it

bbgun34 2011-02-24 20:14:11
you can solve it

timelessmath 2011-02-24 20:14:11
add all equations

LightningStreak 2011-02-24 20:14:11
you can solve it

erica539 2011-02-24 20:14:11
we can solve

AlphaMath1 2011-02-24 20:14:11
Add them up, divide by 2, then start subtracting each equation

girishvar12 2011-02-24 20:14:11
solve

PowerOfPi 2011-02-24 20:14:11
add 2 of them, subtract one, then divide by 2

kangchangood 2011-02-24 20:14:11
we can. by sum/2 and subtraction

TrigMan 2011-02-24 20:14:11
2x + 2y + 2z = 6036

calvinhobbesliker 2011-02-24 20:14:11
We can solve it

TrigMan 2011-02-24 20:14:11
x + y + z = 3018

willwang123 2011-02-24 20:14:11
Add all the equations

copeland 2011-02-24 20:14:16
We can solve it! First add the three equations:

copeland 2011-02-24 20:14:19

copeland 2011-02-24 20:14:22

copeland 2011-02-24 20:14:24
What is x?

jeff10 2011-02-24 20:14:55
x+y=2011, and x+y+z=3018

.cpp 2011-02-24 20:14:55
1006.

airplanes1 2011-02-24 20:14:55
1006

calvinhobbesliker 2011-02-24 20:14:55
1006

ooooo1 2011-02-24 20:14:55
1006

kangchangood 2011-02-24 20:14:55
1006

Spring 2011-02-24 20:14:55
x=3018-(y+z)=3018-2012=1006

TruffleMaster 2011-02-24 20:14:55
1006

PowerOfPi 2011-02-24 20:14:55
3018-2012=1006

luppleAOPS 2011-02-24 20:14:55
3018-2012=1006

copeland 2011-02-24 20:15:00

copeland 2011-02-24 20:15:05

copeland 2011-02-24 20:15:18
Let's try one more iteration to see if we can't get an idea of what's going on.

copeland 2011-02-24 20:15:22

copeland 2011-02-24 20:15:29
(Notice that I'm using subscripts so I don't get confused. You don't have to care about them if you don't want. Also notice that x_2 is a partial edge of T_2 and a total edge of T_3. This is unfortunate but unavoidable.)

copeland 2011-02-24 20:15:37
Now what?

AlphaMath1 2011-02-24 20:16:15
Do the same thing

professordad 2011-02-24 20:16:15
add and divide by 2

Iggy Iguana 2011-02-24 20:16:15
same thing

Relativity1618 2011-02-24 20:16:15
do the same thing

jxihong 2011-02-24 20:16:15
add the equations

soulspeedy 2011-02-24 20:16:15
solve

mthcz11 2011-02-24 20:16:15
add them all together again

kangchangood 2011-02-24 20:16:15
same thing again..

ooooo1 2011-02-24 20:16:15
add them and subtract again

mthcz11 2011-02-24 20:16:15
do the same as before

jeff10 2011-02-24 20:16:15
x(2)+y(2)+z(2)=1509

girishvar12 2011-02-24 20:16:15
add all three then solve like before

Doink 2011-02-24 20:16:15
x+y+z=(1005+1006+1007)/2

copeland 2011-02-24 20:16:17

copeland 2011-02-24 20:16:20

copeland 2011-02-24 20:16:26

copeland 2011-02-24 20:16:33
I smell a pattern. What do you see?

PowerOfPi 2011-02-24 20:17:20
divide the middle by 2

RisingMathStar 2011-02-24 20:17:20
consecutive numbers

TruffleMaster 2011-02-24 20:17:20
the middle term divides by 2

Iggy Iguana 2011-02-24 20:17:20
just divide middle term by 2 to get the next middle term.

soulspeedy 2011-02-24 20:17:20
x,y,z increasing by 1

AlphaMath1 2011-02-24 20:17:20
Subtract y/2

BarbieRocks 2011-02-24 20:17:20
divide middle by 2, then plus 1 and minus 1

usernamehi 2011-02-24 20:17:20
increases by 1

briantix 2011-02-24 20:17:20
each triangle has half the perimeter of the one before it

Spring 2011-02-24 20:17:20
The middle numbers are dividing by 2 each time

qwerty323 2011-02-24 20:17:20
division by 2

mathcool2009 2011-02-24 20:17:20
I notice that the middle term is halved each time

lg5293 2011-02-24 20:17:20
middle term divides by two, and right and left terms are less than one and greater than one respectively

MrWhatWhen 2011-02-24 20:17:20
y is halved, x is 1 less than y, z is 1 more than y

ilovepink 2011-02-24 20:17:20
middle term is halving

kangchangood 2011-02-24 20:17:20
divide and the very right one down, middle one stays, and left one up

soulspeedy 2011-02-24 20:17:20
y,x,z increasing by 1

copeland 2011-02-24 20:17:24
It looks like the middle numbers are decreasing by half and the differences are always 1. Let's try to prove this.

copeland 2011-02-24 20:17:30

copeland 2011-02-24 20:17:38
Again we set things up:

copeland 2011-02-24 20:17:42

copeland 2011-02-24 20:17:50
What is the sum?

AlphaMath1 2011-02-24 20:18:12
3m

ooooo1 2011-02-24 20:18:12
3m

morpheus44 2011-02-24 20:18:12
3m

chimka 2011-02-24 20:18:12
3m

briantix 2011-02-24 20:18:12
3m

soulspeedy 2011-02-24 20:18:12
3m

TrigMan 2011-02-24 20:18:12
3m

qwerty323 2011-02-24 20:18:12
3m

ChipDale 2011-02-24 20:18:12
3m

mstoenescu 2011-02-24 20:18:12
3m

BarbieRocks 2011-02-24 20:18:12
3m

professordad 2011-02-24 20:18:12

ChipDale 2011-02-24 20:18:12
3m

copeland 2011-02-24 20:18:16

copeland 2011-02-24 20:18:25
And what does that make x, y, and z?

PowerOfPi 2011-02-24 20:19:47
1/2m+1, 1/2m, 1/2m-1

DVA6102 2011-02-24 20:19:47
1/2m-1,1/2m,1/2,+1

briantix 2011-02-24 20:19:47
m/2-1,m/2,m/2+1

TrigMan 2011-02-24 20:19:47
m/2-1,m/2,m/2+1

math243 2011-02-24 20:19:47
consecutive

NoWayHaze 2011-02-24 20:19:47
(m/2-1,m/2,m/2+1)

ChipDale 2011-02-24 20:19:47
m/2,m-1/2,m+1/2

agejiageji 2011-02-24 20:19:47
y = m/2, x = m/2-1, z = m/2 + 1

theone142857 2011-02-24 20:19:47
m/2,m/2-1,m/2+1

mathletepower 2011-02-24 20:19:47
m/2,m-2

BarbieRocks 2011-02-24 20:19:47
m/2-1, m/2, m/2+1

calvinhobbesliker 2011-02-24 20:19:47
m/2, m+2/2, m-2/2

mathcool2009 2011-02-24 20:19:47
x_n=m/2,y_n=m/2-1,z_n=m/2+1

professordad 2011-02-24 20:19:47
m/2-1,m/2,m/2+1

CRICKET229 2011-02-24 20:19:47
(m/2-1,m/2,m/2+1)

JSGandora 2011-02-24 20:19:47
x=m/2+1, y=m/2, z=m/2+1

girishvar12 2011-02-24 20:19:47
(m/2)-1=x y=m/2 z=(m/2)+1

copeland 2011-02-24 20:19:49

copeland 2011-02-24 20:19:53
(Notice that the order "changes," but that was also happening above.)

copeland 2011-02-24 20:19:57

copeland 2011-02-24 20:20:01

copeland 2011-02-24 20:20:09
Give me the middle term if you're feeling lazy.

LiBoy 2011-02-24 20:21:02
2012/2^N-1

qwerty323 2011-02-24 20:21:06
2012/2^n-1

BarbieRocks 2011-02-24 20:21:06
2012/2^(n-1)

AlphaMath1 2011-02-24 20:21:06
2012/2^n-1

ilovepink 2011-02-24 20:21:06
2012/2^(n-1)*

mathcool2009 2011-02-24 20:21:06
2012/2^(n-1) I think

CRICKET229 2011-02-24 20:21:06
2012/2^n-1

PowerOfPi 2011-02-24 20:21:06
2012/(2^(n-1))

willwang123 2011-02-24 20:21:06
2012/(2^(n-1))

copeland 2011-02-24 20:21:08

copeland 2011-02-24 20:21:12
Back to the problem. How do we know when we stop having a triangle?

Iggy Iguana 2011-02-24 20:22:04
We are trying to find the greatest value of T_n so that the triangle inequality still holds (x+y>z)

andrewjjiang97 2011-02-24 20:22:04
triangle inequality

erica539 2011-02-24 20:22:04
triangle inequality?

CRICKET229 2011-02-24 20:22:04
use triangle inequality

NoWayHaze 2011-02-24 20:22:04
triangle inequality

willwang123 2011-02-24 20:22:04
sides do not satisfy triangle inequality

sindennisz 2011-02-24 20:22:04
Triangle Inequality doesn't hold anymore.

adhya 2011-02-24 20:22:04
when the trianlge inequality isnt satisfied

LiBoy 2011-02-24 20:22:04
when sides do not conform to triangle inequality!

Spring 2011-02-24 20:22:04
When one of the sides is >= the sum of the other two

ahaanomegas 2011-02-24 20:22:04
Triangle Inequality

jxihong 2011-02-24 20:22:04
when two sides have a sum less than the third

TemurAmir 2011-02-24 20:22:04
When the two smaller sides added together, is less than the largest

supermathman 2011-02-24 20:22:04
when a term < 0

briantix 2011-02-24 20:22:04
when triangle inequalitie doesnt hold

professordad 2011-02-24 20:22:04
when it doesn't satisfy the triangle inequality

mathcool2009 2011-02-24 20:22:04
When the two shorter sides add up to not greater than the largest side.

copeland 2011-02-24 20:22:06
We stop having a triangle when the triangle inequality fails (or when the smallest edge has negative length). The triangle inequality fails if the two smallest lengths add to less than the largest length.

copeland 2011-02-24 20:22:13

copeland 2011-02-24 20:22:16
Gross. Let me simplify that for you.

copeland 2011-02-24 20:22:19

copeland 2011-02-24 20:22:27
or

copeland 2011-02-24 20:22:30

copeland 2011-02-24 20:22:45
What first power of 2 satisfies this first inequality with 2^{n-1}?

calvinhobbesliker 2011-02-24 20:23:23
n>10

NoWayHaze 2011-02-24 20:23:23
11

professordad 2011-02-24 20:23:23
n=11

BarbieRocks 2011-02-24 20:23:23
n=11

erica539 2011-02-24 20:23:23
11

CRICKET229 2011-02-24 20:23:23
11

theone142857 2011-02-24 20:23:23
1024

RisingMathStar 2011-02-24 20:23:23

ChipDale 2011-02-24 20:23:23
11

math243 2011-02-24 20:23:23
11

copeland 2011-02-24 20:23:28

copeland 2011-02-24 20:23:31

copeland 2011-02-24 20:23:35
So what's the answer to this problem?

mthcz11 2011-02-24 20:24:42
d

TemurAmir 2011-02-24 20:24:42
D, 1509/128

ksun48 2011-02-24 20:24:42
D

LiBoy 2011-02-24 20:24:42
2012/512=1509/128!

gh625 2011-02-24 20:24:42
D 1509/128

MrWhatWhen 2011-02-24 20:24:42
1509/128

AlphaMath1 2011-02-24 20:24:42
(D) 1509/128

theone142857 2011-02-24 20:24:42
1509/128

knowmath 2011-02-24 20:24:42
d

Iggy Iguana 2011-02-24 20:24:42
3x2012/512 = D (1509/128)

mthcz11 2011-02-24 20:24:42
(D) 1509/128

professordad 2011-02-24 20:24:42
so its D

Spring 2011-02-24 20:24:42
2012/512*3=1509/128, D

kangchangood 2011-02-24 20:24:42
(d)

infinity1 2011-02-24 20:24:42
D

az94566 2011-02-24 20:24:42
2

ahaanomegas 2011-02-24 20:24:42

danielguo94 2011-02-24 20:24:42
D

Relativity1618 2011-02-24 20:24:42
1509/128

ChipDale 2011-02-24 20:24:42
D

girishvar12 2011-02-24 20:24:42
1509/128 or D

mathletepower 2011-02-24 20:24:42
D

supermathman 2011-02-24 20:24:42
D

airplanes1 2011-02-24 20:24:42
1509/128

willwang123 2011-02-24 20:24:42
D

airplanes1 2011-02-24 20:24:42
so D

jeff10 2011-02-24 20:24:42
D

math_fun 2011-02-24 20:24:42
1509/128 or D

ooooo1 2011-02-24 20:24:42
D) 1509/128

PowerOfPi 2011-02-24 20:24:42
D

pgmath 2011-02-24 20:24:42
D) 1509/128

jxihong 2011-02-24 20:24:42
D

fortenforge 2011-02-24 20:24:42
D

copeland 2011-02-24 20:24:46

copeland 2011-02-24 20:24:52
(Notice that T_12, whatever that means, has a negative edge length. Therefore the negative edge length fear happens one term after the non-triangle case.)

copeland 2011-02-24 20:25:16
OK, crazy dance time. Three minute break then I want you back in your seats and exhausted.

davidx 2011-02-24 20:28:17
*dances*

DVA6102 2011-02-24 20:28:17

ahaanomegas 2011-02-24 20:28:17
Btw, I like that "crazy dance time".

1337Skills 2011-02-24 20:28:17
Time! Let's do this!

copeland 2011-02-24 20:28:22
Phew. Me too.

copeland 2011-02-24 20:28:32
Time for the 12.

copeland 2011-02-24 20:28:40
We might call this "12 o-clock."

copeland 2011-02-24 20:28:49
Or not. They can't all be winners.

copeland 2011-02-24 20:29:02

copeland 2011-02-24 20:29:14
When you see the word "digits," what does that hint you should do?

DVA6102 2011-02-24 20:29:51
expanded notation

profmusic 2011-02-24 20:29:51
Write it as 10a + b

danielguo94 2011-02-24 20:29:51
set variables as the digits

Gosav3122 2011-02-24 20:29:51
10a+b?

kangchangood 2011-02-24 20:29:51
set variables for each digit

bbgun34 2011-02-24 20:29:51
use 10a+b

calvinhobbesliker 2011-02-24 20:29:51
10a+b

lg5293 2011-02-24 20:29:51
separate into 10a + b

TrigMan 2011-02-24 20:29:51
Write numbers as 10a + b

.cpp 2011-02-24 20:29:51
10A+B notation.

gh625 2011-02-24 20:29:51
let it equal 10a+b

soulspeedy 2011-02-24 20:29:51
write as 10a+b

RisingMathStar 2011-02-24 20:29:51
write it as 10a + b

copeland 2011-02-24 20:29:58
clearly the tens digit is called a.

copeland 2011-02-24 20:30:02

copeland 2011-02-24 20:30:07

copeland 2011-02-24 20:30:14

copeland 2011-02-24 20:30:16
How do we define the geometric mean?

morpheus44 2011-02-24 20:30:50
sqrtxy

jeff10 2011-02-24 20:30:50
sqrt(xy)

AlphaMath1 2011-02-24 20:30:50
square root of product

lg5293 2011-02-24 20:30:50
sqrt{xy}

DVA6102 2011-02-24 20:30:50
root(xy)

luppleAOPS 2011-02-24 20:30:50
sqrt(xy)

erica539 2011-02-24 20:30:50
sqrt of xy

sindennisz 2011-02-24 20:30:50
sqrt(ab)

adhya 2011-02-24 20:30:50

mstoenescu 2011-02-24 20:30:50
sqrt (xy)

gqiao 2011-02-24 20:30:50
sqrt of x times y

RisingMathStar 2011-02-24 20:30:50

copeland 2011-02-24 20:30:55

copeland 2011-02-24 20:30:58
What can we do to talk about |x-y|?

calvinhobbesliker 2011-02-24 20:31:45
Square it?

.cpp 2011-02-24 20:31:45
Find (x-y)^2.

profmusic 2011-02-24 20:31:45
square both of those equations

TrigMan 2011-02-24 20:31:45
Square the equation, to start.

soulspeedy 2011-02-24 20:31:45
square

bbgun34 2011-02-24 20:31:45
find (x-y)^2

copeland 2011-02-24 20:31:51
If we square everything in sight we get good stuff.

copeland 2011-02-24 20:31:54

copeland 2011-02-24 20:32:00
So how do we get |x-y|^2?

VIPMaster 2011-02-24 20:32:53
(x+y)^2 - 4xy

erica539 2011-02-24 20:32:53
(2M)^2 -4G^2

bbgun34 2011-02-24 20:32:53
(2M)^2-4G^2=|x-y|^2

.cpp 2011-02-24 20:32:53
(x+y)^2 - 4xy = (x-y)^2.

PowerOfPi 2011-02-24 20:32:53
(x+y)^2-4xy

theone142857 2011-02-24 20:32:53
4M^2-4G^2

TrigMan 2011-02-24 20:32:53
(2M)^2 - 4(G)^2

profmusic 2011-02-24 20:32:53
subtract 4G^2, and we know a>b

luppleAOPS 2011-02-24 20:32:53
(2M)^2+4G^2

copeland 2011-02-24 20:32:57

copeland 2011-02-24 20:33:15
That was a bunch of steps. I'll give you a moment.

copeland 2011-02-24 20:33:35
What do we know about this number?

AlphaMath1 2011-02-24 20:34:18
It's a perfect square

PowerOfPi 2011-02-24 20:34:18
perfect square

lg5293 2011-02-24 20:34:18
its a perfect square

AlphaMath1 2011-02-24 20:34:18
(a+b)(a-b) must be divisible by 11

bbgun34 2011-02-24 20:34:18
it is a square, so 11|(a+b)(a-b)

pi37 2011-02-24 20:34:18
it is a perfect square

professordad 2011-02-24 20:34:18
its a perfect square.

.cpp 2011-02-24 20:34:18
Therefore, (a+b)(a-b) must be a square times 11.

calvinhobbesliker 2011-02-24 20:34:18
It's a perfect square, so either a+b or a-bis 11

kangchangood 2011-02-24 20:34:18
and one of them should be multiple of 11

mstoenescu 2011-02-24 20:34:18
it is a square

gqiao 2011-02-24 20:34:18
it's a perfect square?

VIPMaster 2011-02-24 20:34:18
it's a perfect square!

copeland 2011-02-24 20:34:26
We know it has to be a perfect square.

copeland 2011-02-24 20:34:30
Since this is a perfect square and a and b are both less than 10, we know that 11 must divide a+b. In fact, since a and b are small we must have that a+b=11.

copeland 2011-02-24 20:34:42
What about a-b?

dragoneye776 2011-02-24 20:35:29
a-b=1

willwang123 2011-02-24 20:35:29
1?

bbgun34 2011-02-24 20:35:29
a-b=1

mstoenescu 2011-02-24 20:35:29
it is 1

ahaanomegas 2011-02-24 20:35:29

VIPMaster 2011-02-24 20:35:29
either 1, 4 or 9

.cpp 2011-02-24 20:35:29
a-b=1.

RisingMathStar 2011-02-24 20:35:29
either 1 or 9

libiamo 2011-02-24 20:35:29
1

kangchangood 2011-02-24 20:35:29
can be 1,4?

mstoenescu 2011-02-24 20:35:29
1 or 4 or 9

danielguo94 2011-02-24 20:35:29
a-b = 1 or 9

TrigMan 2011-02-24 20:35:29
It equals 1

NoWayHaze 2011-02-24 20:35:29
=1

kangchangood 2011-02-24 20:35:29
only 1,

copeland 2011-02-24 20:35:38
That must be a perfect square? Which one?

copeland 2011-02-24 20:35:40
Since a and b are between 1 and 9, we know that a-b must be either 1 or 4. However, since a+b is odd we know a-b is also odd and a-b=1.

copeland 2011-02-24 20:35:46
So a and b are. . . .

ahaanomegas 2011-02-24 20:36:18
a = 6; b = 5

kthxbai 2011-02-24 20:36:18
6,5

mstoenescu 2011-02-24 20:36:18
6 and 5

airplanes1 2011-02-24 20:36:18
(a,b)=(6,5)

bbgun34 2011-02-24 20:36:18
a=6, b=5

danielguo94 2011-02-24 20:36:18
6 and 5

luppleAOPS 2011-02-24 20:36:18
a=6,b=5

PowerOfPi 2011-02-24 20:36:18
a=6, b=5

jeff10 2011-02-24 20:36:18
a=6, b=5

ooooo1 2011-02-24 20:36:18
6 and 5!!!???

copeland 2011-02-24 20:36:22
a=6 and b=5. The arithmetic mean is 65 and the geometric mean is 56. What is |x-y|?

VIPMaster 2011-02-24 20:37:36
66

Jongy 2011-02-24 20:37:36
66

bbgun34 2011-02-24 20:37:36
66

AlphaMath1 2011-02-24 20:37:36
66!

kangchangood 2011-02-24 20:37:36
66 (d)

.cpp 2011-02-24 20:37:36
66 => D.

willwin 2011-02-24 20:37:36
D

carmelninja 2011-02-24 20:37:36
66 --> D

professordad 2011-02-24 20:37:36
its 66.

VIPMaster 2011-02-24 20:37:36
sqrt(4*9*11*11*1)

luppleAOPS 2011-02-24 20:37:36
sqrt(4*9*11*11*1)=66

ooooo1 2011-02-24 20:37:36
66

.cpp 2011-02-24 20:37:36
sqrt(4 x 9 x 11 x 11) = |x-y| =66 =>D.

ksun48 2011-02-24 20:37:36
D66

DavidTong 2011-02-24 20:37:36
66

knowmath 2011-02-24 20:37:36
d

copeland 2011-02-24 20:37:43

copeland 2011-02-24 20:37:44
Therefore |x-y|=66 and the answer is (D).

copeland 2011-02-24 20:37:49

copeland 2011-02-24 20:38:11
Also notice that we were probably done a long time ago since we knew that |x-y|^2 was divisible by 2, 3, and 11.

copeland 2011-02-24 20:38:17
Not a lot of the choices have that property.

copeland 2011-02-24 20:38:22

copeland 2011-02-24 20:38:34
From this formula:

copeland 2011-02-24 20:38:51

davidx 2011-02-24 20:38:58
That'll be helpful

copeland 2011-02-24 20:39:00
I bet.

copeland 2011-02-24 20:39:21
I thought this was the trick when I first did this problem. That was a waste of 5 sure minutes.

copeland 2011-02-24 20:39:54
12B Problem 22 is the same as 10B Problem 25. It was fun, but let's skip it.

copeland 2011-02-24 20:40:00

copeland 2011-02-24 20:40:10
What first?

calvinhobbesliker 2011-02-24 20:40:41
Diagram

danielguo94 2011-02-24 20:40:41
diagram

AlphaMath1 2011-02-24 20:40:41
Your awesome diagrams pl0x

jeff10 2011-02-24 20:40:41
Draw a diagram to visualize it.

willwin 2011-02-24 20:40:41
diagram

willwang123 2011-02-24 20:40:41
Draw!!!!!!!!!!!!!1

sindennisz 2011-02-24 20:40:41
Picture?

mthcz11 2011-02-24 20:40:41
draw the graph :D

erica539 2011-02-24 20:40:41
draw a diagram once again?

copeland 2011-02-24 20:40:45
Let's draw another diagram. (I hope they gave you a lot of paper!)

copeland 2011-02-24 20:40:48

copeland 2011-02-24 20:40:52
And a path looks something like this:

copeland 2011-02-24 20:40:59

copeland 2011-02-24 20:41:01
What are we trying to find here?

calvinhobbesliker 2011-02-24 20:42:02
All dots on all possible paths

gqiao 2011-02-24 20:42:02
number of lattice points total in the paths

mthcz11 2011-02-24 20:42:02
which points lie on the red path

.cpp 2011-02-24 20:42:02
The number of lattice points the bug can reach.

erica539 2011-02-24 20:42:02
lattice points that lie on these paths

ahaanomegas 2011-02-24 20:42:02
The number of paths which follow the rules given (one of which is what you drew)

willwang123 2011-02-24 20:42:02
number of points within reach of a path.

ChipDale 2011-02-24 20:42:02
number of paths

carmelninja 2011-02-24 20:42:02
number of lattice points that could be included in such a path

Gosav3122 2011-02-24 20:42:02
farthest point away from the two that can be covered

copeland 2011-02-24 20:42:04
We're trying to find all the points on all the paths of length at most 20 from (-3,2) to (3,-2). That path above has length 12.

copeland 2011-02-24 20:42:07
What points do we absolutely know we can reach?

Jasmine8925 2011-02-24 20:43:05
anything within the rectangle that has those 2 points as corners

ahaanomegas 2011-02-24 20:43:06
Those that are in the bounded rectangle

jeff10 2011-02-24 20:43:06
the points inside the rectangle, brcause the perimeter is 20 itself

victorzhouaops 2011-02-24 20:43:06
the rectangle between the two points

erica539 2011-02-24 20:43:06
anything inside the rectangle.

mathletepower 2011-02-24 20:43:06
ones that are between two paths

.cpp 2011-02-24 20:43:06
All in rectangle with diagonal AB.

professordad 2011-02-24 20:43:06
all points in the rectangle bounded by (-3,2),(3,2),(3,-2),(-3,-2)

ooooo1 2011-02-24 20:43:06
the points in the rectangle: (x=-3, x=3, y=-2, y=2)

willwang123 2011-02-24 20:43:06
Inside a rectangle w/ opposite corners A and B

jeff10 2011-02-24 20:43:06
with vertices (-2,3) amd (2, -3)

VIPMaster 2011-02-24 20:43:06
all points around A and B

copeland 2011-02-24 20:43:06
We can reach any point in the rectangle with our vertices as corners:

copeland 2011-02-24 20:43:11

copeland 2011-02-24 20:43:17
These points are all reachable by a path that only travels down and right, and such a path has length 10. Let's figure out where else we can go.

copeland 2011-02-24 20:43:21
How can we efficiently describe a path?

kangchangood 2011-02-24 20:44:24
the.. letters?

erica539 2011-02-24 20:44:24
steps up/down and steps left/right

kangchangood 2011-02-24 20:44:24
move = letters?

dragoneye776 2011-02-24 20:44:24
list of directions

ChipDale 2011-02-24 20:44:24
A line where you have to go down or right except 10 instances

adhya 2011-02-24 20:44:24
choose what directions the bug is going in

copeland 2011-02-24 20:44:29
We can write a path as a sequence of U (up), D (down), L (left), and R (right).

copeland 2011-02-24 20:44:34
What do we know about each sequence?

VIPMaster 2011-02-24 20:45:40
it must have at most 20 letters

carmelninja 2011-02-24 20:45:40
at most 20 terms

luppleAOPS 2011-02-24 20:45:40
no more than twenty letters

mathtyro 2011-02-24 20:45:40
Each sequence must containt at least 6R (in some order) and 4D(in some order)

dragoneye776 2011-02-24 20:45:40
20 terms, D - U = 4 and R - L = 6

tan90 2011-02-24 20:45:40
It must end with 4 more D's than U's and 6 more R's than L's

ooooo1 2011-02-24 20:45:40
D-U=4 and L-R=6

BarbieRocks 2011-02-24 20:45:40
L-R=6, U-=4

VIPMaster 2011-02-24 20:45:40
D = -U, L = -R

kangchangood 2011-02-24 20:45:40
(D-U) = 4 (R - L) = 6

professordad 2011-02-24 20:45:40
its 20 units long

kangchangood 2011-02-24 20:45:40
and U + D + L + R <= 20

RisingMathStar 2011-02-24 20:45:40
the maximum sum of the numbers of U and L is 5

carmelninja 2011-02-24 20:45:40
there must be 6 more downs than ups, and 4 more rights than lefts

erica539 2011-02-24 20:45:40
U+D+L+R is equal to or less than 20

copeland 2011-02-24 20:45:42
Each sequence has at most 20 terms.

copeland 2011-02-24 20:45:45
The number of times D appears is exactly 4 more than the number of times U appears.

copeland 2011-02-24 20:45:48
The number of times R appears is exactly 6 more than the number of times L appears.

copeland 2011-02-24 20:45:52
How far to the left of the rectangle can we go?

VIPMaster 2011-02-24 20:46:27
5 steps

dragoneye776 2011-02-24 20:46:27
5

RisingMathStar 2011-02-24 20:46:27
5 units

bbgun34 2011-02-24 20:46:27
5

jeff10 2011-02-24 20:46:27
5 units

mathcool2009 2011-02-24 20:46:27
5 units

carmelninja 2011-02-24 20:46:27
5 units

willwang123 2011-02-24 20:46:27
5

calvinhobbesliker 2011-02-24 20:46:27
5

mathtyro 2011-02-24 20:46:27
5 (since we must use the 5 more to come back)

copeland 2011-02-24 20:46:29
We can always go 5 units to the left of (-3,2) by the path LLLLLRRRRR*, where * goes from green to green.

copeland 2011-02-24 20:46:33
We can go 5 units to the left of (-3,1) by a path DLLLLLRRRRR*.

copeland 2011-02-24 20:46:42
(reading left-to-right.)

copeland 2011-02-24 20:46:50
etc.

copeland 2011-02-24 20:46:51
So we can get any point 5 units to the left of the rectangle.

copeland 2011-02-24 20:46:55
What other points can we reach?

dragoneye776 2011-02-24 20:47:44
5 units up and down and right

ChipDale 2011-02-24 20:47:44
Down five more units

jeff10 2011-02-24 20:47:44
if you do not go left, you go 5 units up

willwang123 2011-02-24 20:47:44
5 up

ChipDale 2011-02-24 20:47:44
Up five more units

RisingMathStar 2011-02-24 20:47:44
5 units from any edge of the rectangle

hsm174 2011-02-24 20:47:44
5 units right

calvinhobbesliker 2011-02-24 20:47:44
5 above the rectangle

carmelninja 2011-02-24 20:47:44
points 5 units below the above/below the rectangle

Jasmine8925 2011-02-24 20:47:44
5 units up

.cpp 2011-02-24 20:47:44
5 units above rectangle, 5 units below rectangle, 5 units to the right of the rectangle.

gqiao 2011-02-24 20:47:44
any point 5 to the right

copeland 2011-02-24 20:47:49
We can also reach 5 units above, 5 units right and 5 units below.

copeland 2011-02-24 20:47:52
(We're going to need a bigger picture!)

copeland 2011-02-24 20:47:54

copeland 2011-02-24 20:47:57
Anywhere else?

jeff10 2011-02-24 20:48:53
If you took n units to the left, you take 5-n units up

VIPMaster 2011-02-24 20:48:53
anything with 4 Ls and 1 U, 3Ls and 2Us, ...

RisingMathStar 2011-02-24 20:48:53
and any combination thereof

kangchangood 2011-02-24 20:48:53
make some stairs..

theone142857 2011-02-24 20:48:53
or 2 and three or 1 and 4

dragoneye776 2011-02-24 20:48:53
in triangles of length and width 5, on each of the 4 corners

theone142857 2011-02-24 20:48:53
in between

professordad 2011-02-24 20:48:53
in the middle between the extensions, such as (5,5)

pikachoo 2011-02-24 20:48:53
units diagnal to the points?

carmelninja 2011-02-24 20:48:53
10 additional points in the corners

VIPMaster 2011-02-24 20:48:53
The lines between the vertices of the figure

.cpp 2011-02-24 20:48:53
Four triangles to make an octagon.

willwang123 2011-02-24 20:48:53
up and left

jeff10 2011-02-24 20:49:00
Yes, the points 5 unit steps away from the other green dot

copeland 2011-02-24 20:49:03
We can also venture into the top left region by traveling a path like UULLLRRRDD* This gives us any point in the upper left triangle. We can similarly hit any point in any of the other three triangles.

copeland 2011-02-24 20:49:07

copeland 2011-02-24 20:49:10
Can we go any further than this?

copeland 2011-02-24 20:49:28
(This is the hard question so I'll give you a minute to write good arguments.)

math243 2011-02-24 20:51:28
no, because when we go up or down, we need to come back

adhya 2011-02-24 20:51:28
no because we wont have enough moves left to come back

jeff10 2011-02-24 20:51:28
No, because we will either go 6 or more units farther from our start, or 6 or more units past our finish point, which leads us to a perimeter more than 20.

calvinhobbesliker 2011-02-24 20:51:28
No, because all outside the region have x+y>5, so you can't get back.

ChipDale 2011-02-24 20:51:28
No, if you go any further you can't get back to the other points

dragoneye776 2011-02-24 20:51:28
we can only go a maximum of 5 lattice points from the center rectangle, and you have all possible points 5 lattice points away

agejiageji 2011-02-24 20:51:28
no, because you can only go 5 steps out of the rectangle defined by the green points

luppleAOPS 2011-02-24 20:51:28
no. any move either gets you one step closer to (-3,2), or one step further away. We need ten steps in total, and these are the only ways to do it.

carmelninja 2011-02-24 20:51:28
any path that goes from a green point to a boundary point to the other green point maxes out at 20 units, so if a path goes to a point not already included, it would be more than 20 units in length

willwang123 2011-02-24 20:51:28
No, because we can only take 5 steps away from the starting point before we have to head towards the end. If we use more than 5 steps away (Up or left) we will not have enough steps to finish.

.cpp 2011-02-24 20:51:28
No, because otherwise we would stray more than 6 units from the rectangle; 6 more to get back + 10 extra moves for 7R and 3U gives more than 20 moves.

adhya 2011-02-24 20:51:28
no because at simplest it takes 10 moves to get from dot1 to dot 2, then we can move 5 more ahead and then come back with 5 but thats it because we can only have 20

gh625 2011-02-24 20:51:28
If you go more than five units up or to the left of A, then you would need more than 5 units to get back; combined wth the 10 needed to go from A to B, this would make more than 20 steps.

copeland 2011-02-24 20:51:33
No. The number of times we write U and L gives us a bound on how far away from the rectangle we can go. Therefore we can't possible go out of this octagon. (This isn't quite rigorous, but it encapsulates all of the aspects of a rigorous proof.)

copeland 2011-02-24 20:51:35
How many dots are there?

copeland 2011-02-24 20:51:38

wl2f 2011-02-24 20:53:20
195

lg5293 2011-02-24 20:53:20
195, use symmetry to make counting easier

dragoneye776 2011-02-24 20:53:20
195

professordad 2011-02-24 20:53:20
there are 195 dots.

VIPMaster 2011-02-24 20:53:20
195

jeff10 2011-02-24 20:53:20
195, lost of counting

mathcool2009 2011-02-24 20:53:20
2(7+9+11+13+15)+5(17)

VIPMaster 2011-02-24 20:53:20
24*6 + 17*3 = 195

ahaanomegas 2011-02-24 20:53:20
195

ahaanomegas 2011-02-24 20:53:20

.cpp 2011-02-24 20:53:20
4x15 + 4x25 + 7x5 = 60+100+35 = 195 => C.

kangchangood 2011-02-24 20:53:20
195

danielguo94 2011-02-24 20:53:20
195

theone142857 2011-02-24 20:53:20
(C)195!!!:whistle:

copeland 2011-02-24 20:53:24
We have 7+9+11+13+15+17+17+17+17+17+15+13+11+9+7 dots.

copeland 2011-02-24 20:53:27

copeland 2011-02-24 20:53:40
This is where you hope that your arithmetic errors aren't actually possible answers.

copeland 2011-02-24 20:53:52
If you're curious, the concept of distance we're talking about here is the "taxicab metric." The taxicab metric is built by imagining a grid as Chicago or your favorite other metropolitan city and measuring how far a cab travels to get from one intersection to another.

copeland 2011-02-24 20:54:03
The set of points satisfying d(P,X)+d(X,Q)<20 for fixed points P and Q (where d means distance) is just the interior of an ellipse. Therefore this question is asking us to compute the area of a taxicab ellipse.

copeland 2011-02-24 20:54:12
The cool thing we learned about a taxicab ellipse is that even though the major axis was not parallel to the x-axis, we still have a horizontal line of symmetry. That's pretty absurd and incredibly counter-intuitive.

copeland 2011-02-24 20:54:25
Our argument in general shows that taxicab ellipses are nothing more than octagons. Later you should think about what shape a taxicab ellipse is when you start with foci whose x-coordinates are equal. Also think about what happens when the foci are the same point (a "taxicab circle").

copeland 2011-02-24 20:55:13

copeland 2011-02-24 20:55:25
Alright, octics are hard. Give me a hint here.

jeff10 2011-02-24 20:55:36
What are octics?

ahaanomegas 2011-02-24 20:55:36
What are octics?

copeland 2011-02-24 20:55:40
octo = 8

copeland 2011-02-24 20:55:45
ics means, uh, polynomials.

calvinhobbesliker 2011-02-24 20:56:11
Let x=z^4 and factor

adhya 2011-02-24 20:56:11
make x = z^4

lg5293 2011-02-24 20:56:11
let u = z^4

.cpp 2011-02-24 20:56:11
Let x=z^4, solve a quadratci

gqiao 2011-02-24 20:56:11
set z^4 as x?

willwang123 2011-02-24 20:56:11
set x to z^4

copeland 2011-02-24 20:56:16
This polynomial is quadratic in z^4. Let's write w=z^4.

copeland 2011-02-24 20:56:39
(I prefer x, too, but I've been told not to use x for complex coordinates, so we'll say w for now.)

copeland 2011-02-24 20:56:45

copeland 2011-02-24 20:56:53
Let's find the roots. Do you see them?

airplanes1 2011-02-24 20:57:56
(z^4+4sqrt3+7)(z^4-1)

NoWayHaze 2011-02-24 20:57:56
(w-1)(w+(4sqrt3+7))

professordad 2011-02-24 20:57:56

ilovepink 2011-02-24 20:57:56
(w+(4sqrt3+7))(w-1)

professordad 2011-02-24 20:57:56
so w=1 or w=-4sqrt{3}-7

AlphaMath1 2011-02-24 20:57:56
Note that w=1 is a solution, so we already know that the fourth roots of unity are solutions to the octic.

wl2f 2011-02-24 20:57:56
z^4=1 and -4sqrt{3}-7

hsm174 2011-02-24 20:57:56
uh 1 woruks

airplanes1 2011-02-24 20:57:56
w=1, -(4sqrt3+7)

calvinhobbesliker 2011-02-24 20:57:56
Factor into (w-1)(w+4rad3+7) Roots are thus 4th roots of unity and 4th roots of -4sqrt3-7

.cpp 2011-02-24 20:57:56
w^2+(k)w-(k+1) = k+1 and -1 are roots.

bbgun34 2011-02-24 20:57:56
1 and -4rt3-7

copeland 2011-02-24 20:57:59
There are two ways to proceed here.

copeland 2011-02-24 20:58:01

copeland 2011-02-24 20:58:10

copeland 2011-02-24 20:58:36
Otherwise you might be skipping this one.

copeland 2011-02-24 20:58:39
Therefore we can factor P a little bit.

copeland 2011-02-24 20:58:44

copeland 2011-02-24 20:58:46
I know 4 roots instantly. What are they?

airplanes1 2011-02-24 20:59:11
1, -1, i, -i

bbgun34 2011-02-24 20:59:11
4th roots of unity

willwang123 2011-02-24 20:59:11
1 -1 -i i

kangchangood 2011-02-24 20:59:11
1 ,-1, i ,-i

lg5293 2011-02-24 20:59:11
1, i, -i, -1

Jongy 2011-02-24 20:59:11
+-1 and +-i

CRICKET229 2011-02-24 20:59:11
1, -1, i, -i

Aallisotsto 2011-02-24 20:59:11
1, -1, 1, -1

RisingMathStar 2011-02-24 20:59:11

aleCTY 2011-02-24 20:59:11
1,-1,i,-i

.cpp 2011-02-24 20:59:11
z = i, -i, 1, or -1 work.

bbgun34 2011-02-24 20:59:11
1,-1,i,-i

copeland 2011-02-24 20:59:14
The first quartic has roots 1, -1, i, and -i.

copeland 2011-02-24 20:59:18
Now they want us to take a fourth root of -(4sqrt(3)+7). That's just mean.

copeland 2011-02-24 20:59:25
First off, what's the argument (the angle it makes with the positive real axis) of -(4sqrt(3)+7)?

calvinhobbesliker 2011-02-24 21:01:00
180 degrees

mstoenescu 2011-02-24 21:01:00
pi

carmelninja 2011-02-24 21:01:00
pi, or 180 degrees

Jasmine8925 2011-02-24 21:01:00
pi

wl2f 2011-02-24 21:01:00
pi

.cpp 2011-02-24 21:01:00
Or 180 degrees, if you consider it to be negative.

mstoenescu 2011-02-24 21:01:00
180°

copeland 2011-02-24 21:01:06
This is a negative real number, so the argument is pi.

copeland 2011-02-24 21:01:08
What are the arguments of the four fourth roots of this number?

AlphaMath1 2011-02-24 21:02:06
180/4

gh625 2011-02-24 21:02:06
pi/4,3pi/4,5pi/4,7pi/4

RisingMathStar 2011-02-24 21:02:06

theone142857 2011-02-24 21:02:06
45,135,225,315

professordad 2011-02-24 21:02:06
45 degrees,135 degrees,225 degrees,315 degrees

nikeballa96 2011-02-24 21:02:06
by demoivre's theorem, they'll be 45 degrees

copeland 2011-02-24 21:02:10
Taking the fourth roots divides the argument by 4. However there are four fourth roots and they differ by multiples of 2pi/4.

copeland 2011-02-24 21:02:13
Let's draw a picture. We know four of the roots. We also know the other four lie on the rays of slope 1 and -1.

copeland 2011-02-24 21:02:17

copeland 2011-02-24 21:02:22
The next thing we need to compute is the magnitude of the fourth root. Notice that we can forget the minus sign when we do this since we already know about the argument.

copeland 2011-02-24 21:02:25
But fourth roots are hard. . . .

copeland 2011-02-24 21:02:28
Any ideas?

profmusic 2011-02-24 21:03:09
square root of square root

carmelninja 2011-02-24 21:03:09
square root of square root

ilovepink 2011-02-24 21:03:09
square root twice

jeff10 2011-02-24 21:03:09
sqrt(sqrtn)

mathman98 2011-02-24 21:03:09
take the square root, then take another square root?

jeff10 2011-02-24 21:03:09
sqrt(sqrt(n))

professordad 2011-02-24 21:03:09
take the square root and then square root again?

allicspteam 2011-02-24 21:03:09
make two sqrts

copeland 2011-02-24 21:03:13
Let's try taking a square root first. That still a little daunting. What guess can we make about the square root that might make this a little simpler?

copeland 2011-02-24 21:03:41
We want the square root of 4sqrt(3)+7.

Jasmine8925 2011-02-24 21:04:15
let the fourth root be a+bsqrt(c)

calvinhobbesliker 2011-02-24 21:04:15
a+bsqrtc?

Jasmine8925 2011-02-24 21:04:15
it'll have a sqrt(3) term

ilovepink 2011-02-24 21:04:15
(a+bsqrt3)

profmusic 2011-02-24 21:04:15
it can be written as a + bsqrt(3)

professordad 2011-02-24 21:04:15
its going to be in the form a+bsqrt{3}

dragoneye776 2011-02-24 21:04:15
set it equal to a + bsqrt(c) and then square both sides

mathcool2009 2011-02-24 21:04:16
has form a(fourthrt(3))+b(sqrt(3))+c

aleCTY 2011-02-24 21:04:16
(2+rt{3})^2

AlphaMath1 2011-02-24 21:04:16
(sqrt{3}+2)^2

copeland 2011-02-24 21:04:18

copeland 2011-02-24 21:04:20
This is a little bit naive. We don't have any reason to think that a and b here will be in any way nice.

copeland 2011-02-24 21:04:25
The only reason we have to even hope this will work is that we'd like to believe this problem is solvable and we're probably running short of other ideas.

copeland 2011-02-24 21:04:39
What is the square of a*sqrt(3)+b?

gh625 2011-02-24 21:05:20
3a^2+b^2+2absqrt{3}

ilovepink 2011-02-24 21:05:20
3a^2 + b^2 + 2absqrt3

nikeballa96 2011-02-24 21:05:20
3a^2+2absqrt3+b^2

CRICKET229 2011-02-24 21:05:20
3a^2+b^2+2absqrt3

professordad 2011-02-24 21:05:20

AlphaMath1 2011-02-24 21:05:20
3a^2+b^2+2ab*sqrt(3)

Jasmine8925 2011-02-24 21:05:20
$3a^2+2absqrt(3)+b^2$

dragoneye776 2011-02-24 21:05:20
3a^2 + b^2 + 2absqrt(3)

carmelninja 2011-02-24 21:05:20
(3a^2 + b^2) + (2ab)sqrt(3)

copeland 2011-02-24 21:05:26

copeland 2011-02-24 21:05:27
Now IF a and b are rational then we should be able to solve this by equating the coefficients:

copeland 2011-02-24 21:05:28

copeland 2011-02-24 21:05:31
Any ideas?

mstoenescu 2011-02-24 21:06:04
2+sqrt3

carmelninja 2011-02-24 21:06:04
lucky guess!! --> 2+sqrt(3)

trigonometry456103 2011-02-24 21:06:04
2+sqrt3

theone142857 2011-02-24 21:06:04
sqrt{3}+2

wl2f 2011-02-24 21:06:04
2+sqrt3

kthxbai 2011-02-24 21:06:04
1,2

jeff10 2011-02-24 21:06:04
a=1, b=2

carmelninja 2011-02-24 21:06:04
a=1, b=2

.cpp 2011-02-24 21:06:04
a=1, b=2.

professordad 2011-02-24 21:06:04
a=1,b=2

mstoenescu 2011-02-24 21:06:04
a=1 b=2

ChipDale 2011-02-24 21:06:04
a=1 b=2

copeland 2011-02-24 21:06:08
Neat! A solution: a = 1, b=2.

copeland 2011-02-24 21:06:11

copeland 2011-02-24 21:06:13
What next?

ahaanomegas 2011-02-24 21:06:37
Oooooh, nice!

nikeballa96 2011-02-24 21:06:37
square root againnn

profmusic 2011-02-24 21:06:37
now we do it again...

dragoneye776 2011-02-24 21:06:37
do it again

CRICKET229 2011-02-24 21:06:37
take square root of sqrt3+2

hsm174 2011-02-24 21:06:37
sqrt(sqrt3+2)

ChipDale 2011-02-24 21:06:37
Square root again

gh625 2011-02-24 21:06:37
take the square root of 2+sqrt{3}

calvinhobbesliker 2011-02-24 21:06:37
Take the square root of 2+sqrt3 in a similar way

Jasmine8925 2011-02-24 21:06:37
we need the square root of ;$sqrt(3)+2$

mathman98 2011-02-24 21:06:37
take the square root again'

copeland 2011-02-24 21:06:39
That was so nice that we have to try it twice.

copeland 2011-02-24 21:06:43

copeland 2011-02-24 21:06:45

copeland 2011-02-24 21:06:49
Hmm. That's pretty disappointing.

copeland 2011-02-24 21:06:59
I don't see any nice solutions.

copeland 2011-02-24 21:07:02
Any ideas?

copeland 2011-02-24 21:07:19
Does that look a little bit like a set of equations we COULD solve?

hsm174 2011-02-24 21:08:23
same thing?

profmusic 2011-02-24 21:08:23
add a sqrt(2) term in place of b

carmelninja 2011-02-24 21:08:23
instead, look for square root of 2(sqrt(3) + 2) = 4 + 2sqrt(3)

theone142857 2011-02-24 21:08:23
multiply sqrt{2} to both

professordad 2011-02-24 21:08:23
3a^2+b^2=4 and 2ab=2 is solvable

copeland 2011-02-24 21:08:30
I do see an equation that we can solve! Suppose we multiply the right sides only by 2, giving a new system of equations:

copeland 2011-02-24 21:08:32

copeland 2011-02-24 21:08:36
What are the solutions to this?

calvinhobbesliker 2011-02-24 21:08:58
1 and 1

theone142857 2011-02-24 21:08:58
1,1

CRICKET229 2011-02-24 21:08:58
1,1

professordad 2011-02-24 21:08:58
a=b=1

.cpp 2011-02-24 21:08:58
a=1, b=1.

willwang123 2011-02-24 21:08:58
a=b=1

christerson 2011-02-24 21:08:58
1,1

eb8368 2011-02-24 21:08:58
1,1

jeff10 2011-02-24 21:08:58
1 and 1

airplanes1 2011-02-24 21:08:58
a=b=1

carmelninja 2011-02-24 21:08:58
a=b=1

ooooo1 2011-02-24 21:08:58
1 and

evanteng1997 2011-02-24 21:08:58
b=1 a=1

danielguo94 2011-02-24 21:08:58
(1,1)

copeland 2011-02-24 21:09:02
a=b=1. Therefore we know that sqrt(3)+1 might be an interesting number. What is its square?

BarbieRocks 2011-02-24 21:09:25
4+2sqrt(3)

nikeballa96 2011-02-24 21:09:25
4 + 2sqrt3

gqiao 2011-02-24 21:09:25
4+2root3

professordad 2011-02-24 21:09:25
4+2sqrt3

calvinhobbesliker 2011-02-24 21:09:25
4+2sqrt3

jeff10 2011-02-24 21:09:25
2sqrt3+4

fries4guys 2011-02-24 21:09:25
4+2sqrt3

LiBoy 2011-02-24 21:09:25
4+2sqrt3

sindennisz 2011-02-24 21:09:25
4+2sqrt3

mstoenescu 2011-02-24 21:09:25
4+2sqrt3

ahaanomegas 2011-02-24 21:09:25

copeland 2011-02-24 21:09:34

copeland 2011-02-24 21:09:34
So what is the fourth root of 4sqrt(3)+7?

theone142857 2011-02-24 21:10:33
doulble so divide sqrt 2

AlphaMath1 2011-02-24 21:10:33
[sqrt(3)+1]/sqrt(2)

theone142857 2011-02-24 21:10:33
sqrt{6}/2+sqrt{2}/2

ilovepink 2011-02-24 21:10:33
sqrt(3/2) + (sqrt2)/2

mathcool2009 2011-02-24 21:10:33
(sqrt(3)+1)/sqrt(2)

gh625 2011-02-24 21:10:33
(1+sqrt{3})/sqrt{2}

copeland 2011-02-24 21:10:38

copeland 2011-02-24 21:10:45

copeland 2011-02-24 21:10:52
I'll sticky those:

copeland 2011-02-24 21:10:57

copeland 2011-02-24 21:11:05
And on the picture:

copeland 2011-02-24 21:11:07

copeland 2011-02-24 21:11:10
What we've just discovered is that the blue dots are distance d from the origin.

copeland 2011-02-24 21:11:14
And the polygon we want?

calvinhobbesliker 2011-02-24 21:11:54
Connect the dots wo intersection

professordad 2011-02-24 21:11:54
its an octagon (connect red with blue)

hsm174 2011-02-24 21:11:54
its a ninja star!!!

AlphaMath1 2011-02-24 21:11:54
A star

ooooo1 2011-02-24 21:11:54
is the dots connected

mathcool2009 2011-02-24 21:11:54
a concave star

calvinhobbesliker 2011-02-24 21:11:54
4-sided star

.cpp 2011-02-24 21:11:54
The obvious, shuriken shaped one.

copeland 2011-02-24 21:11:56
. . . . ninja star! Do you think it's a secret message?

copeland 2011-02-24 21:11:59
We just connect the dots consecutively.

copeland 2011-02-24 21:12:04

copeland 2011-02-24 21:12:06
Now how do we compute those distances?

professordad 2011-02-24 21:12:29
Distance Formula

adhya 2011-02-24 21:12:29
distance formula?

calvinhobbesliker 2011-02-24 21:12:29
8 times distance between (0,1) and point d

mathcool2009 2011-02-24 21:12:29
pythagoreabn theoroem

theone142857 2011-02-24 21:12:29
pythag

copeland 2011-02-24 21:12:32
yuck.

RisingMathStar 2011-02-24 21:12:54
How do you prove that that has the smallest perimeter?

Jasmine8925 2011-02-24 21:12:54
why did the problem ask for the minimum perimeter? isn't there only one octagon?

copeland 2011-02-24 21:13:00
This is an interesting question.

copeland 2011-02-24 21:13:15
However, once we do some computing we can actually show that these distances are even LOCALLY smallest

copeland 2011-02-24 21:13:26
Therefore the blue neighbors are the closest points to the red dots.

copeland 2011-02-24 21:13:33
And the red points are the closest neighbors to the blue dots.

copeland 2011-02-24 21:13:49
What's more awesome than the distance formula?

akscanb 2011-02-24 21:14:03
you

copeland 2011-02-24 21:14:06
and. . . .

tan90 2011-02-24 21:14:20
law of cosines

NoWayHaze 2011-02-24 21:14:20
law of cosines

carmelninja 2011-02-24 21:14:20
law of cosines

delta1 2011-02-24 21:14:20
law of cosines?

RisingMathStar 2011-02-24 21:14:20
law of cosines

copeland 2011-02-24 21:14:25
Let's use the law of cosines

copeland 2011-02-24 21:14:28

copeland 2011-02-24 21:14:41
I'll just compute for you.

copeland 2011-02-24 21:14:47

Jasmine8925 2011-02-24 21:15:11
can you give us another diagram with line segments labeled?

copeland 2011-02-24 21:15:26
We're using the origin and two neighboring dots.

copeland 2011-02-24 21:15:43
For example, 1 and de^{2pi i/8}.

copeland 2011-02-24 21:15:51
Oh. Well. That's pretty. What's the total perimeter?

.cpp 2011-02-24 21:16:39
Wow! The answer is just B!

calvinhobbesliker 2011-02-24 21:16:39
8sqrt2 is the answer

nikeballa96 2011-02-24 21:16:39
multiply sqrt2 by 8 and our answer is (B) 8sqrt2

professordad 2011-02-24 21:16:39
so the side is sqrt2 and the answer is 8sqrt2.

mstoenescu 2011-02-24 21:16:39
8sqrt2

nikeballa96 2011-02-24 21:16:39
8sqrt2, or answer choice B

ilovepink 2011-02-24 21:16:39
B 8*sqrt2

AlphaMath1 2011-02-24 21:16:39
8sqrt{2}

calculus321 2011-02-24 21:16:39
8sqrt2

lg5293 2011-02-24 21:16:39
8sqrt{2}, B

RisingMathStar 2011-02-24 21:16:39
8*sqrt{2}

.cpp 2011-02-24 21:16:39
8 sqrt 2 => B.

professordad 2011-02-24 21:16:39
8sqrt2. or B.

ChipDale 2011-02-24 21:16:39
B

NoWayHaze 2011-02-24 21:16:39
8sqrt2

carmelninja 2011-02-24 21:16:39
8sqrt(2) --> (B)

eb8368 2011-02-24 21:16:39
(B) 8Root2

copeland 2011-02-24 21:16:43
The lengths of the edges are all sqrt(2) by symmetry.

copeland 2011-02-24 21:16:48
Since there are 8 identical edges, the total perimeter must be 8sqrt(2), (B).

copeland 2011-02-24 21:17:09
This problem was fun because none of the things we did should have worked.

copeland 2011-02-24 21:17:20
You might have noticed cos(pi/12) in there somewhere. I have no idea how this forces the solution to be so clean, but if anybody sees the magic here you should definitely post it to the forums.

willwang123 2011-02-24 21:17:37
didn't they do that on purpose?

copeland 2011-02-24 21:17:40
Of course.

copeland 2011-02-24 21:17:53
That doesn't explain why those choices work, just what they achieve.

copeland 2011-02-24 21:18:02

copeland 2011-02-24 21:18:15
Whew.

copeland 2011-02-24 21:18:30
First thing's first. What is up with 99!? That number is what we in the business call "large."

danielguo94 2011-02-24 21:19:02
what about 5000!?

copeland 2011-02-24 21:19:05

dragoneye776 2011-02-24 21:19:13
It's over 9000

copeland 2011-02-24 21:19:16
Yes it is.

copeland 2011-02-24 21:19:41
No ideas?

copeland 2011-02-24 21:19:57
What possible property does 99! have?

narik 2011-02-24 21:20:23
something to do with divisible by the first 99 numbers

Goldey 2011-02-24 21:20:23
its just a number divisible by every number 1-99. Since everything repeats every k and k can be any number 1-99

calvinhobbesliker 2011-02-24 21:20:23
It's divisible by all possible k

edrill4000 2011-02-24 21:20:23
divisible by numbers 1-99

airplanes1 2011-02-24 21:20:23
its divisible by every integer less than and equal to 99

tan90 2011-02-24 21:20:23
It is divisible by all possible values of k

Gosav3122 2011-02-24 21:20:23
divisible by all numbers 1-99?

copeland 2011-02-24 21:20:27
Super.

copeland 2011-02-24 21:20:53
There's another clue here, but I guess we might just leave it by the wayside until we find it later.

copeland 2011-02-24 21:21:00
OK. There's no obvious way to start on this one. What should we do?

calvinhobbesliker 2011-02-24 21:21:41
Test a few k's?

Jasmine8925 2011-02-24 21:21:41
try some values of k and n

narik 2011-02-24 21:21:41
test the first few iterations out to see how it behaves

.cpp 2011-02-24 21:21:41
Try small examples of k.

1nf1n1tyP1 2011-02-24 21:21:41
cry?

dragoneye776 2011-02-24 21:21:41
work with small numbers to find a pattern

theone142857 2011-02-24 21:21:41
k=1,2,.........

delta1 2011-02-24 21:21:41
experiment with small values?

ahaanomegas 2011-02-24 21:21:41
Try a few numbers and hope we see something interesting.

copeland 2011-02-24 21:21:43
Let's try some cases. Any idea which cases we should try?

akscanb 2011-02-24 21:22:13
1

ooooo1 2011-02-24 21:22:13
try 1 first

ahaanomegas 2011-02-24 21:22:13
k = 1, 2, ...

wavelet 2011-02-24 21:22:13
5?

willwang123 2011-02-24 21:22:13
Easy ones.

adhya 2011-02-24 21:22:13
1,3

copeland 2011-02-24 21:22:15
We could try starting with k=1 or k=2. I have a better idea, though. What is it?

copeland 2011-02-24 21:22:27
Well. . .not 2.

copeland 2011-02-24 21:22:30
since k is odd.

BarbieRocks 2011-02-24 21:23:14
k=67

calvinhobbesliker 2011-02-24 21:23:14
The denoms of the answer choices?

copeland 2011-02-24 21:23:18
Looking at the answers we notice that the numbers 67 and 87 might be a good place to start.

copeland 2011-02-24 21:23:25
Notice that these are even quite likely to be relevant numbers since 67/99 ~2/3 and 87/99~7/8.

copeland 2011-02-24 21:23:28
Let's try 67 first.

copeland 2011-02-24 21:23:31

copeland 2011-02-24 21:23:38
(This is the point in the problem where I realized that right side is a constant.) What is the right side?

calvinhobbesliker 2011-02-24 21:24:04
1

professordad 2011-02-24 21:24:04
its 1

jeff10 2011-02-24 21:24:04
the right side is 1.

ooooo1 2011-02-24 21:24:04
1?

tan90 2011-02-24 21:24:04
1

RisingMathStar 2011-02-24 21:24:04
1

nikeballa96 2011-02-24 21:24:04
oh just kidding, it's 1

adhya 2011-02-24 21:24:06
1

snail2 2011-02-24 21:24:06
1 apparently

copeland 2011-02-24 21:24:10
The integer closest to 100/67 is 1. Therefore we have

copeland 2011-02-24 21:24:13

copeland 2011-02-24 21:24:16
What other simplification is there?

copeland 2011-02-24 21:24:43
(Notice that this isn't an "equality." It's a logical doodad that's either true or false.)

copeland 2011-02-24 21:25:14
Tell me specifically about [(100-n)/67].

copeland 2011-02-24 21:25:23
That looks pretty complicated.

lg5293 2011-02-24 21:25:34
split it up?

copeland 2011-02-24 21:25:43
yes, but NOT 100/67 - n/67.

MathAndKnowledge 2011-02-24 21:26:11
1 + (33 - n)/67

tan90 2011-02-24 21:26:11
you could split up the 67 and 33

Jasmine8925 2011-02-24 21:26:11
67/67+33/67-n/67?

.cpp 2011-02-24 21:26:11
1+ 33/67 - n/67.

AlphaMath1 2011-02-24 21:26:11
1-(n-33)/67

nikeballa96 2011-02-24 21:26:11
1+(33-n/67)

copeland 2011-02-24 21:26:16
We notice [a+1]=[a]+1, so we can remove 67/67 from the middle term.

copeland 2011-02-24 21:26:18

copeland 2011-02-24 21:26:27
When n ranges from 0 to 66, what happens to [n/67]?

.cpp 2011-02-24 21:27:28
It goes from 0 to 1.

delta1 2011-02-24 21:27:28
0 or 1

ChipDale 2011-02-24 21:27:28
It becomes 1

calvinhobbesliker 2011-02-24 21:27:28
It is 0 until n is 34, after which it is 1

BarbieRocks 2011-02-24 21:27:28
=0 from n=1 to 33 and 1 from n=34 to 67

Yongyi781 2011-02-24 21:27:28
It's 0 for n <= 33, and 1 for n >= 34

RisingMathStar 2011-02-24 21:27:28
changes from 0 to 1 at n=34

mathcool2009 2011-02-24 21:27:28
for n=0 to 33

professordad 2011-02-24 21:27:28
it is first 0 and then becomes 1

mstoenescu 2011-02-24 21:27:28
goes from 0 to 1

copeland 2011-02-24 21:27:31

copeland 2011-02-24 21:27:37
What about [(33-n)/67] on this interval?

agejiageji 2011-02-24 21:28:13
0

calvinhobbesliker 2011-02-24 21:28:13
It is always 0

ilovepink 2011-02-24 21:28:13
0.

Goldey 2011-02-24 21:28:13
always 0.

kad2361 2011-02-24 21:28:13
always 0

RisingMathStar 2011-02-24 21:28:13
always 0

jeff10 2011-02-24 21:28:13
it is always 0

ChipDale 2011-02-24 21:28:13
0

willwang123 2011-02-24 21:28:13
Always 0

dragoneye776 2011-02-24 21:28:13
from 0 to 66, it is 0

jeff10 2011-02-24 21:28:13
(33-0(/67=33/67 which is closer to 0

copeland 2011-02-24 21:28:18
Magically 33 is right in the middle of the interval, so 33-n ranges from 33 to -33. This ALWAYS gives zero. Therefore

copeland 2011-02-24 21:28:22

copeland 2011-02-24 21:28:29
What if n>=67? How can we deal with that?

kangchangood 2011-02-24 21:28:55
Well continuously I mistook it as floor function..

copeland 2011-02-24 21:28:57
Yes. Everyone has done this at least 3 times so far.

RisingMathStar 2011-02-24 21:29:26
modulo 67

Yongyi781 2011-02-24 21:29:26
Subtract 67 from n

.cpp 2011-02-24 21:29:26
This can be tackled mod 67 like we did for the middle term.

carmelninja 2011-02-24 21:29:26
n = 67x + y --> focus on y

kangchangood 2011-02-24 21:29:26
n = 67a + b

delta1 2011-02-24 21:29:26
split it up again

copeland 2011-02-24 21:29:30
If n>=67 we can write n=67+m.

copeland 2011-02-24 21:29:34

copeland 2011-02-24 21:29:45
Did that fall off the screen?

copeland 2011-02-24 21:29:56

copeland 2011-02-24 21:30:05

copeland 2011-02-24 21:30:09

copeland 2011-02-24 21:30:17
And. . .

lg5293 2011-02-24 21:31:01
back at the same place!

kangchangood 2011-02-24 21:31:01
so that means it applies for every number..

mstoenescu 2011-02-24 21:31:01
once again !

carmelninja 2011-02-24 21:31:01
its the exact same as before

dragoneye776 2011-02-24 21:31:01
Doesn't change, so just take mod 67

mstoenescu 2011-02-24 21:31:01
the "same" !

ilovepink 2011-02-24 21:31:01
it repeats!

copeland 2011-02-24 21:31:29
So. . . .

copeland 2011-02-24 21:31:40
Let's go back to an old question.

copeland 2011-02-24 21:31:47
Why do I care about 99!?

maxw3ll 2011-02-24 21:32:24
everything comes in nice multiples

dragoneye776 2011-02-24 21:32:24
its a factor of every value of k

snail2 2011-02-24 21:32:24
it is a multiple of everything from 1 to 99

ilovepink 2011-02-24 21:32:24
it divides everything evenly and nothing is left over

adhya 2011-02-24 21:32:24
its the case for all numbers from 1-99

jeff10 2011-02-24 21:32:24
99! is divisible by all numbers from 1-99

mathcool2009 2011-02-24 21:32:24
its 0(mod67)

.cpp 2011-02-24 21:32:24
Because it has dividers for all the range of k.

ahaanomegas 2011-02-24 21:32:24
Because our upper limit for k = 99

carmelninja 2011-02-24 21:32:24
its divisible by 67

copeland 2011-02-24 21:32:32
It's divisible by 67 and. . ..

RisingMathStar 2011-02-24 21:33:07
therefore each case is equally likely to happen

RisingMathStar 2011-02-24 21:33:07
making the probability 34/67

copeland 2011-02-24 21:33:08
Every residue class is represented equally!

copeland 2011-02-24 21:33:15
Does that make sense?

copeland 2011-02-24 21:33:26
Therefore we really only care about the residue classes, by induction.

copeland 2011-02-24 21:33:30
This gives us P(67)=34/67.

copeland 2011-02-24 21:33:33
That's nice. We at least now know that D could be the answer and E cannot. However maybe 87 does the same thing, so we certainly couldn't rule out C (or either of the smaller ones).

adhya 2011-02-24 21:33:49
no where diid the 34 come from

copeland 2011-02-24 21:33:56
We wandered too far from that argument.

copeland 2011-02-24 21:33:59

copeland 2011-02-24 21:34:06
We want to know the probability that this is not 0.

copeland 2011-02-24 21:34:11
er, that this is zer.

copeland 2011-02-24 21:34:19
That happens in 34 of the 67 cases.

copeland 2011-02-24 21:34:33
At this point I tried the same thing with 87 to see what happens. For the sake of everyone's sanity, though, I think we should go on to the general case.

copeland 2011-02-24 21:34:43
zer = 0.

copeland 2011-02-24 21:34:49
We fix k now, and consider the expression. What values of n concern us?

nikeballa96 2011-02-24 21:35:22
the values from 0 to k-1

calvinhobbesliker 2011-02-24 21:35:22
n between 0 and k

Goldey 2011-02-24 21:35:22
0<=n<k

copeland 2011-02-24 21:35:27
We only care about n such that 0<= n < k for the same inductive reason as above.

copeland 2011-02-24 21:35:36
(Oh, and there are 7 numbers from 0 to 6.)

copeland 2011-02-24 21:35:41
Furthermore we only care about the residue class of n as well.

copeland 2011-02-24 21:35:44
We can write 100=qk+r for for some integers q and r. How should we choose q and r?

nikeballa96 2011-02-24 21:36:57
maximize q and make sure r<k

carmelninja 2011-02-24 21:36:57
so r is minimized while still being positive

RisingMathStar 2011-02-24 21:36:57
q should be the floor of 100/k

Goldey 2011-02-24 21:36:57
0<=r<k

copeland 2011-02-24 21:36:59
This is almost always the right choice.

copeland 2011-02-24 21:37:01
However.

copeland 2011-02-24 21:37:21
In this case there's a better option. Remember all that "Confused it with the floor," stuff from before?

copeland 2011-02-24 21:37:38
We have a better option for the residue of 100.

Yongyi781 2011-02-24 21:38:22
-k/2 < r < k/2?

RisingMathStar 2011-02-24 21:38:22
-k/2 < r < k/2

Goldey 2011-02-24 21:38:22
-k/2<r<k/2? or something like that

calvinhobbesliker 2011-02-24 21:38:22
Use the "closest integer" finction?

copeland 2011-02-24 21:38:27
Excellent!

copeland 2011-02-24 21:38:27
We could force r to be between 0 an k, but I don't like that. Since we have [100/k] in our equation, let's instead let that be q. Therefore -k/2< r < k/2 is probably better.

copeland 2011-02-24 21:38:37

nikeballa96 2011-02-24 21:38:45
tricky tricky

copeland 2011-02-24 21:39:01
Yeah. Actually we would have balled up our paper in about 3 minutes and done this anyway.

copeland 2011-02-24 21:39:05
(And don't forget k is odd.)

copeland 2011-02-24 21:39:08
It's now or never, I guess.

copeland 2011-02-24 21:39:11

copeland 2011-02-24 21:39:29
Now how do we simplify that?

ahaanomegas 2011-02-24 21:39:54
Can we?

copeland 2011-02-24 21:40:01
We do want to compare this to something, right?

copeland 2011-02-24 21:40:13
Maybe "simplify" is not the best term.

ahaanomegas 2011-02-24 21:40:27
What do we want to compare it with?

copeland 2011-02-24 21:40:29
[100/k]

copeland 2011-02-24 21:40:33
Where do we see that?

.cpp 2011-02-24 21:40:46
For the original equation, we'll get [n/k] + [(r-n)/k] =0 again.

calvinhobbesliker 2011-02-24 21:40:48
That's q

Gosav3122 2011-02-24 21:40:50
q

copeland 2011-02-24 21:40:54
We've assumed that q=[100/k], so

copeland 2011-02-24 21:40:57

copeland 2011-02-24 21:41:07
(Deleting the middle stuff:)

copeland 2011-02-24 21:41:24

copeland 2011-02-24 21:41:30
Sorry. accidental smilie.

copeland 2011-02-24 21:41:40
Gaar.

copeland 2011-02-24 21:41:51

copeland 2011-02-24 21:42:17
Now what do we want?

jeff10 2011-02-24 21:42:48
we need q.

copeland 2011-02-24 21:42:59
We need the right hand side to equal [100/k].

copeland 2011-02-24 21:43:03
So what's that tell us?

ctmusicgirl 2011-02-24 21:43:41
isolate the term

narik 2011-02-24 21:43:41
that we need the probability that [n/k] + [r-n / k] = 0

mstoenescu 2011-02-24 21:43:41
that the other are zero !

calvinhobbesliker 2011-02-24 21:43:41
(n/k)+(r-n/k)=0

ooooo1 2011-02-24 21:43:41
the other two things on the right side equal to 0?

copeland 2011-02-24 21:43:45

copeland 2011-02-24 21:43:49
We can simplify that further. How does the negative sign affect [x/k]?

ahaanomegas 2011-02-24 21:44:29
It opposes the direction or rounding

jeff10 2011-02-24 21:44:29
we will go the other direction

copeland 2011-02-24 21:44:31
Right, [-x/k]=-[x/k].

copeland 2011-02-24 21:44:34

copeland 2011-02-24 21:44:39
OK, so we've algebrad ourselves out into the abyss. This equation is equivalent to the equation from the problem, except that we have introduced this new variable r.

copeland 2011-02-24 21:44:54
Thanks for enduring that. I promise it will pay off soon.

copeland 2011-02-24 21:44:55
Let's go back to the problem to see what this equation has to do with anything. First off, if r=0, what happens?

calvinhobbesliker 2011-02-24 21:45:19
It's true for all n

delta1 2011-02-24 21:45:19
always equal

RisingMathStar 2011-02-24 21:45:19
always true

mathcool2009 2011-02-24 21:45:19
its always treu

.cpp 2011-02-24 21:45:19
[n/k] = [n/k].

jeff10 2011-02-24 21:45:19
the two sides are equal by the reflexive property

Gosav3122 2011-02-24 21:45:19
the statement is true

copeland 2011-02-24 21:45:21
The equation is always true if r=0. So what is P(k) when r=0?

Gosav3122 2011-02-24 21:45:42
1

mathcool2009 2011-02-24 21:45:42
1

mstoenescu 2011-02-24 21:45:42
1

narik 2011-02-24 21:45:42
1

ahaanomegas 2011-02-24 21:45:42

RisingMathStar 2011-02-24 21:45:42
1

jeff10 2011-02-24 21:45:46
1.

copeland 2011-02-24 21:45:48
P=1. This isn't helpful since we want to minimize P.

copeland 2011-02-24 21:45:52
To extend our little example further, when is r=0?

RisingMathStar 2011-02-24 21:46:41
when k = 1, 5, or 25

ilovepink 2011-02-24 21:46:41
when k divides 100

.cpp 2011-02-24 21:46:41
k|100.

jeff10 2011-02-24 21:46:41
k=1

tan90 2011-02-24 21:46:41
when k=1

carmelninja 2011-02-24 21:46:41
k=1

copeland 2011-02-24 21:46:43
Since we defined r by 100 = qk+r, r is zero exactly when k divides 100. Therefore P(20)=1, P(50)=1, etc.

copeland 2011-02-24 21:46:50
OK, back to the task at hand. What values does [n/k] take?

copeland 2011-02-24 21:47:14
on 0<= n <k.

ahaanomegas 2011-02-24 21:47:38
0 and 1

calvinhobbesliker 2011-02-24 21:47:38
0 and 1

delta1 2011-02-24 21:47:38
0 or 1

.cpp 2011-02-24 21:47:38
0 or 1 only.

mstoenescu 2011-02-24 21:47:38
0 and 1

copeland 2011-02-24 21:47:52
For 0 <= n < k/2 we have [n/k]=0.

copeland 2011-02-24 21:47:55
For k/2 < n < k we have [n/k]=1.

copeland 2011-02-24 21:48:00
What are the values of [(n-r)/k]?

copeland 2011-02-24 21:48:14
on 0<=n < k

tan90 2011-02-24 21:48:19
0 or 1

jeff10 2011-02-24 21:48:19
0 and 1

copeland 2011-02-24 21:48:24
When does it flip?

Gosav3122 2011-02-24 21:49:21
n-r>k/2

copeland 2011-02-24 21:49:27
Since r is not that small, [(0-r)/k] must still be 0. The value continues to be zero until n-r>k/2 and then we get 1 from there on out.

copeland 2011-02-24 21:49:38
When does equality fail then?

copeland 2011-02-24 21:49:54

RisingMathStar 2011-02-24 21:50:56
between k/2 and k/2 +r

Gosav3122 2011-02-24 21:50:56
when n<k/2 but n-r>k/2

Gosav3122 2011-02-24 21:50:56
and vice versa

copeland 2011-02-24 21:51:01
As n crosses k/2, the value of [n/k] flips from 0 to 1.

copeland 2011-02-24 21:51:04
As n crosses k/2 + r, the value of [(n-r)/k] flips from 0 to 1.

copeland 2011-02-24 21:51:08
So the values are only different between k/2 and k/2 + r.

copeland 2011-02-24 21:51:11
Note that we don't know which is bigger, so we don't know which flip happens first. Or do we?

copeland 2011-02-24 21:51:14
. . . .

carmelninja 2011-02-24 21:52:16
r could be positive or negative

copeland 2011-02-24 21:52:18
If r>0 then [n/k] flips first. If r<0 then [n-r/k] flips first.

copeland 2011-02-24 21:52:23

copeland 2011-02-24 21:52:33

copeland 2011-02-24 21:52:43
What is P, as a function of r, in the first case?

RisingMathStar 2011-02-24 21:54:04
1 - r/k

carmelninja 2011-02-24 21:54:04
r/k

copeland 2011-02-24 21:54:09
There are exactly r integers in this interval so P=1-r/k.

copeland 2011-02-24 21:54:12
Likewise, if r<0 then the interval contains exactly -r integers and P=1+r/k.

copeland 2011-02-24 21:54:15

copeland 2011-02-24 21:54:18
We want to minimize P so we want to maximize |r|/k.

copeland 2011-02-24 21:54:25
What is the naive bound on |r|?

copeland 2011-02-24 21:54:39
(It's stickied.)

.cpp 2011-02-24 21:54:59
k/2?

carmelninja 2011-02-24 21:54:59
k/2

nikeballa96 2011-02-24 21:54:59
-k/2 to k/2

Gosav3122 2011-02-24 21:54:59
k/2

snail2 2011-02-24 21:54:59
-k/2<r<k/2

calvinhobbesliker 2011-02-24 21:54:59
k/2

copeland 2011-02-24 21:55:03
Actually k is odd.

copeland 2011-02-24 21:55:07

copeland 2011-02-24 21:55:14
Let's run with that. If 2r=k-1 then what does that say about the equation for 100?

copeland 2011-02-24 21:56:13
Recall 100 = qk+r

calvinhobbesliker 2011-02-24 21:56:57
100=(q+1/2)k-1/2

LawrenceHan 2011-02-24 21:56:57
2qk+k-1

copeland 2011-02-24 21:57:01

copeland 2011-02-24 21:57:07
And what do we know about 201?

RisingMathStar 2011-02-24 21:57:34
3 * 67

tan90 2011-02-24 21:57:34
3 * 67

.cpp 2011-02-24 21:57:34
Divisible by 67 and 3.

ilovepink 2011-02-24 21:57:34
it's divisible by 3 and 67

ChipDale 2011-02-24 21:57:34
It is 3*67

calvinhobbesliker 2011-02-24 21:57:34
It's div by 67

AlphaMath1 2011-02-24 21:57:34
it's divisible by 67

willwang123 2011-02-24 21:57:34
3*67

copeland 2011-02-24 21:57:38
201=67*3. Therefore we care about k=3 and k=67. (I'm feeling pretty good about 67 right now. I was starting to worry that you were leading me astray!)

copeland 2011-02-24 21:57:44
If k=3 then 100=3*33+1, so r=1. What is P(3)?

RisingMathStar 2011-02-24 21:58:46
2/3

calvinhobbesliker 2011-02-24 21:58:46
2/3

ilovepink 2011-02-24 21:58:46
2/3

.cpp 2011-02-24 21:58:46
2/3.

copeland 2011-02-24 21:58:55

copeland 2011-02-24 21:58:57
If k=67 we know P(67)=34/67, which is much less.

copeland 2011-02-24 21:59:00

copeland 2011-02-24 21:59:09

copeland 2011-02-24 21:59:23
Finally we need to knock out the non-edge cases. We assume that |r| is at most (k-3)/2. What does this make P?

ahaanomegas 2011-02-24 22:00:18
What are "non-edge cases"?

copeland 2011-02-24 22:00:19
I mean those r which aren't "maximal" in the sense that they bump into the k/2 bound.

calvinhobbesliker 2011-02-24 22:01:07
1/2-3/(2k)

ilovepink 2011-02-24 22:01:07
(k+3)/2k*

copeland 2011-02-24 22:01:11

copeland 2011-02-24 22:01:38
Since k is at most 99, 1/k is at least 1/99. Therefore

copeland 2011-02-24 22:01:42

copeland 2011-02-24 22:01:44
Tell me about 34/66.

RisingMathStar 2011-02-24 22:02:49
greater than 34/67

snail2 2011-02-24 22:02:49
it is bigger than 34/67

airplanes1 2011-02-24 22:02:49
it's greater than 34/67

.cpp 2011-02-24 22:02:49
Greater than 34/67; also not an answer choice.

Yongyi781 2011-02-24 22:02:49
34/66 > 34/67

copeland 2011-02-24 22:02:54
This is MORE than 34/67, so we can't do any better than 34/67.

copeland 2011-02-24 22:02:58
Therefore. . ..

danielguo94 2011-02-24 22:03:42
THE ANSWER IS D

nikeballa96 2011-02-24 22:03:42
so we should've given up waaay back then and guessed.

hyunjoon123 2011-02-24 22:03:42
it's d

jeff10 2011-02-24 22:03:42
our answer is D

airplanes1 2011-02-24 22:03:42
D

.cpp 2011-02-24 22:03:42
Answer is D.

snail2 2011-02-24 22:03:42
D!!!!!!!!!!!

ChipDale 2011-02-24 22:03:42
The answer is 34/67 or D

calvinhobbesliker 2011-02-24 22:03:42
D is the answer

delta1 2011-02-24 22:03:42
D!!!

copeland 2011-02-24 22:03:46
The answer, finally, is 34/67, (D).

tan90 2011-02-24 22:03:54
Was there a quicker way to do this problem?

RisingMathStar 2011-02-24 22:03:54
Moral of the story: don't attempt to solve #25

copeland 2011-02-24 22:04:17
Actually, I tried 87, realized that the computation that gives 44/87 is wrong and picked D.

willwang123 2011-02-24 22:04:24
that problem took almost an hour

hsm174 2011-02-24 22:04:24
Yes, that took around 15 mins

jeff10 2011-02-24 22:04:24
unless you solve the rest really quiclky

copeland 2011-02-24 22:04:33
Yes. that problem took 15minute hours.

copeland 2011-02-24 22:04:53
Alright, that's it for tonight.

copeland 2011-02-24 22:05:09
We've gone 2 hours and 36 minutes.

2011 AMC 10/12 B Discussion

Facilitator: Jeremy Copeland