2017 AMC 10/12 A Discussion
Go back to the Math Jam ArchiveAoPS Instructors will discuss problems from the AMC 10/12 A, administered February 7. We will discuss the last 5 problems on each test.
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Facilitator: AoPS Staff
devenware
2017-02-08 19:01:35
Welcome to the 2017 AMC 10A/12A Math Jam!
Welcome to the 2017 AMC 10A/12A Math Jam!
devenware
2017-02-08 19:01:42
I'm Deven Ware, and I'll be leading our discussion tonight.
I'm Deven Ware, and I'll be leading our discussion tonight.
devenware
2017-02-08 19:01:49
I've worked at AoPS since 2014. I left the Shire at an early age and honed my lethal spellcasting skills under the watchful guidance of some of the most famed wizards of Middle Earth. My hobbies include freestyle hopscotch, competitive jazz viola, and extreme grasshopper racing. I have a teenaged pet bonsai tree named Shigeru.
I've worked at AoPS since 2014. I left the Shire at an early age and honed my lethal spellcasting skills under the watchful guidance of some of the most famed wizards of Middle Earth. My hobbies include freestyle hopscotch, competitive jazz viola, and extreme grasshopper racing. I have a teenaged pet bonsai tree named Shigeru.
VGoma
2017-02-08 19:01:59
How does this work?
How does this work?
devenware
2017-02-08 19:02:02
Before we get started I would like to take a moment to explain our virtual classroom procedures to those who have not previously participated in a Math Jam or one of our online classes.
Before we get started I would like to take a moment to explain our virtual classroom procedures to those who have not previously participated in a Math Jam or one of our online classes.
devenware
2017-02-08 19:02:10
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.
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.
devenware
2017-02-08 19:02:17
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.
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.
devenware
2017-02-08 19:02:30
There are bunches and bunches of students here. As I said, only a 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!!
There are bunches and bunches of students here. As I said, only a 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!!
devenware
2017-02-08 19:02:39
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!
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!
devenware
2017-02-08 19:03:05
Your assistants for today will be Kaitlin Maile (kmaile32) and Kevin Chong An (projeceulerlover).
Your assistants for today will be Kaitlin Maile (kmaile32) and Kevin Chong An (projeceulerlover).
devenware
2017-02-08 19:03:09
Kaitlin has been working for AOPS since November 2014, when she was a senior in high school. While in high school, Kaitlin participated in math team, marching band, and various sports. Now, she is a junior at the University of Minnesota, majoring in Biomedical Engineering and minoring in Computer Science. When not working for AOPS, studying for classes, or working on research projects, she enjoys weight-lifting, listening to almost any genre of music, and going on adventures.
Kaitlin has been working for AOPS since November 2014, when she was a senior in high school. While in high school, Kaitlin participated in math team, marching band, and various sports. Now, she is a junior at the University of Minnesota, majoring in Biomedical Engineering and minoring in Computer Science. When not working for AOPS, studying for classes, or working on research projects, she enjoys weight-lifting, listening to almost any genre of music, and going on adventures.
devenware
2017-02-08 19:03:11
Kevin is currently a senior at California Institute of Technology majoring in Mathematics with a minor in Computer Science. He's done math competitions throughout his time at high school, and he really enjoys sharing his knowledge with others. Some of his non-problem solving interests are weightlifting, playing Hanabi, and video games.
Kevin is currently a senior at California Institute of Technology majoring in Mathematics with a minor in Computer Science. He's done math competitions throughout his time at high school, and he really enjoys sharing his knowledge with others. Some of his non-problem solving interests are weightlifting, playing Hanabi, and video games.
devenware
2017-02-08 19:03:18
They will be sending you messages to answer questions or offer other help. However, due to the incredibly large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over may give you warm feelings, but just makes their poor lives harder, so please, only ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.
They will be sending you messages to answer questions or offer other help. However, due to the incredibly large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over may give you warm feelings, but just makes their poor lives harder, so please, only ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.
devenware
2017-02-08 19:03:20
Please also remember that the purpose of this Math Jam is to work through the solutions to AMC problems and learn! "Working through the solutions" includes discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the final answer (since I'll be ignoring those anyway). That's not what we're doing tonight. 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.
Please also remember that the purpose of this Math Jam is to work through the solutions to AMC problems and learn! "Working through the solutions" includes discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the final answer (since I'll be ignoring those anyway). That's not what we're doing tonight. 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.
AmitLuke
2017-02-08 19:03:26
How do you type so fast?
How do you type so fast?
devenware
2017-02-08 19:03:35
I use several keyboards.
I use several keyboards.
AmitLuke
2017-02-08 19:04:00
Cool
Cool
hodori01
2017-02-08 19:04:00
jacksonhu
2017-02-08 19:04:00
cool
cool
Benjy450
2017-02-08 19:04:00
Impressive.
Impressive.
Wave-Particle
2017-02-08 19:04:00
padfoot6302
2017-02-08 19:04:00
interesting
interesting
IceParrot
2017-02-08 19:04:00
Really!?!
Really!?!
devenware
2017-02-08 19:04:05
devenware
2017-02-08 19:04:18
We will work the last 5 problems from the AMC 10A, then the last 5 problems from the AMC 12A. Two of these problems are the same, 10A Problem 24 and 12A Problem 23. We'll only solve that problem once.
We will work the last 5 problems from the AMC 10A, then the last 5 problems from the AMC 12A. Two of these problems are the same, 10A Problem 24 and 12A Problem 23. We'll only solve that problem once.
devenware
2017-02-08 19:04:23
Let's get started!
Let's get started!
devenware
2017-02-08 19:04:25
Oh, and there truly are a lot of people here tonight. I really don't like saying it, but we're probably going to miss some of the things that some of you say. (Especially during the geometry problems - sheesh.) Please forgive me in advance. That doesn't happen in our classes.
Oh, and there truly are a lot of people here tonight. I really don't like saying it, but we're probably going to miss some of the things that some of you say. (Especially during the geometry problems - sheesh.) Please forgive me in advance. That doesn't happen in our classes.
devenware
2017-02-08 19:04:41
OKAY. READY?
OKAY. READY?
mathchampion1
2017-02-08 19:05:12
yes!
yes!
mj434
2017-02-08 19:05:12
YES!
YES!
wiler5002
2017-02-08 19:05:12
IM READY
IM READY
XxkalleknightxXlelxd
2017-02-08 19:05:12
yes
yes
mathaction
2017-02-08 19:05:12
YEAH
YEAH
AmitLuke
2017-02-08 19:05:12
YES
YES
vlwilliam
2017-02-08 19:05:12
GO!\
GO!\
Auty
2017-02-08 19:05:12
YEAH!
YEAH!
B_Blake
2017-02-08 19:05:12
ya
ya
mepehe888
2017-02-08 19:05:12
yup
yup
padfoot6302
2017-02-08 19:05:12
yea1!!!
yea1!!!
challengetogo
2017-02-08 19:05:12
READY!!
READY!!
mathcrazymj
2017-02-08 19:05:12
YESH
YESH
ruchirk533
2017-02-08 19:05:12
yay
yay
Smoothfang
2017-02-08 19:05:12
YES
YES
Cardinals2014
2017-02-08 19:05:12
yeah
yeah
User2013
2017-02-08 19:05:12
yes!
yes!
PenguinJoe
2017-02-08 19:05:12
YES
YES
iks92
2017-02-08 19:05:12
yeah!!
yeah!!
GeronimoStilton
2017-02-08 19:05:12
Ready!
Ready!
AmitLuke
2017-02-08 19:05:12
IM READY
IM READY
bluephoenix
2017-02-08 19:05:12
YEAH!!!!
YEAH!!!!
JTMath
2017-02-08 19:05:12
YES
YES
kiwitrader123
2017-02-08 19:05:12
Yea
Yea
Bill9000
2017-02-08 19:05:12
YES! I'm excited!
YES!
I'm excited!
devenware
2017-02-08 19:05:18
OKAY, LET'S GO!
OKAY, LET'S GO!
devenware
2017-02-08 19:05:24
21. A square with side length $x$ is inscribed in a right triangle with sides of length 3, 4, and 5 so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides 3, 4, and 5 so that one side of the square lies on the hypotenuse of the triangle. What is $\dfrac{x}{y}$?
$\phantom{hi}$
$\text{(A) } \dfrac{12}{13} \quad
\text{(B) } \dfrac{35}{37} \quad
\text{(C) } 1 \quad
\text{(D) } \dfrac{37}{35} \quad
\text{(E) } \dfrac{13}{12}$
21. A square with side length $x$ is inscribed in a right triangle with sides of length 3, 4, and 5 so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides 3, 4, and 5 so that one side of the square lies on the hypotenuse of the triangle. What is $\dfrac{x}{y}$?
$\phantom{hi}$
$\text{(A) } \dfrac{12}{13} \quad
\text{(B) } \dfrac{35}{37} \quad
\text{(C) } 1 \quad
\text{(D) } \dfrac{37}{35} \quad
\text{(E) } \dfrac{13}{12}$
devenware
2017-02-08 19:05:32
What's the first step?
What's the first step?
GeneralCobra19
2017-02-08 19:05:52
Draw a diagram
Draw a diagram
hodori01
2017-02-08 19:05:52
draw diagram
draw diagram
quartzgirl
2017-02-08 19:05:52
First, draw a diagram
First, draw a diagram
Talker32332
2017-02-08 19:05:52
Diagram!
Diagram!
zjjc123
2017-02-08 19:05:52
diagram
diagram
mathforfun2016
2017-02-08 19:05:52
Draw it.
Draw it.
mt492
2017-02-08 19:05:52
diagram
diagram
Jessp
2017-02-08 19:05:52
draw a diagram
draw a diagram
mikhailgromov
2017-02-08 19:05:52
diagram
diagram
Brisingrfire
2017-02-08 19:05:52
draw a picture
draw a picture
theironcatfish
2017-02-08 19:05:52
Draw it out?
Draw it out?
uptownmath
2017-02-08 19:05:52
Draw a diagram.
Draw a diagram.
SomethingNeutral
2017-02-08 19:05:52
Draw a diagram
Draw a diagram
garretth
2017-02-08 19:05:52
Draw the diagram
Draw the diagram
devenware
2017-02-08 19:06:06
Draw a diagram. That's always my first step on a geometry problem.
Draw a diagram. That's always my first step on a geometry problem.
devenware
2017-02-08 19:06:11
pie314159265
2017-02-08 19:06:30
use similar triangles
use similar triangles
S0larPh03nix
2017-02-08 19:06:30
similar triangles
similar triangles
Liopleurodon
2017-02-08 19:06:30
Similar triangles!
Similar triangles!
devenware
2017-02-08 19:06:34
And what does similar triangles tell us?
And what does similar triangles tell us?
devenware
2017-02-08 19:07:43
Lots of great answers out there, I'm going with the one that I did when solving this:
Lots of great answers out there, I'm going with the one that I did when solving this:
Picroft
2017-02-08 19:07:49
(4-x)/x = 4/3
(4-x)/x = 4/3
amc8nov
2017-02-08 19:07:49
(4-x)/x=4/3
(4-x)/x=4/3
dragon6688
2017-02-08 19:07:49
$\frac{4-x}{x} = \frac{4}{3}$
$\frac{4-x}{x} = \frac{4}{3}$
devenware
2017-02-08 19:07:58
The lower left triangle is a 3-4-5 triangle so\[\frac{4-x}x=\frac43.\]
The lower left triangle is a 3-4-5 triangle so\[\frac{4-x}x=\frac43.\]
devenware
2017-02-08 19:08:00
Therefore\[12-3x=4x,\]so $x=\dfrac{12}7$.
Therefore\[12-3x=4x,\]so $x=\dfrac{12}7$.
devenware
2017-02-08 19:08:04
Now what?
Now what?
jybsmartguy2
2017-02-08 19:08:33
draw another diagram for y
draw another diagram for y
CrystalEye
2017-02-08 19:08:33
find y
find y
liant
2017-02-08 19:08:33
find y
find y
mathaction
2017-02-08 19:08:33
draw the other?
draw the other?
acegikmoqsuwy2000
2017-02-08 19:08:33
find $y$
find $y$
dudethefirst
2017-02-08 19:08:33
find y
find y
GeronimoStilton
2017-02-08 19:08:33
Solve for $y$.
Solve for $y$.
Hyun04
2017-02-08 19:08:33
draw another diagram
draw another diagram
tdeng
2017-02-08 19:08:33
Now draw the other square
Now draw the other square
mathforfun2016
2017-02-08 19:08:33
Draw diagram for y.
Draw diagram for y.
kiwitrader123
2017-02-08 19:08:33
draw the other diagram
draw the other diagram
skmc
2017-02-08 19:08:33
Draw the other diagram
Draw the other diagram
MathMan1234
2017-02-08 19:08:33
draw another diagram for y
draw another diagram for y
mikhailgromov
2017-02-08 19:08:33
diagram for y
diagram for y
CrystalEye
2017-02-08 19:08:33
draw a diagram to find y
draw a diagram to find y
oriduck.nk
2017-02-08 19:08:33
draw another diagram
draw another diagram
blacksheep2003
2017-02-08 19:08:33
Draw the second triangle, and mark the diagram
Draw the second triangle, and mark the diagram
Ani10
2017-02-08 19:08:33
find y using another diagram
find y using another diagram
devenware
2017-02-08 19:08:37
The other diagram:
The other diagram:
devenware
2017-02-08 19:08:39
devenware
2017-02-08 19:08:50
(The 3 and 4 are the long edge lengths.)
(The 3 and 4 are the long edge lengths.)
devenware
2017-02-08 19:08:56
Let's get some more numbers on here. What are these two values $a$ and $b$?
Let's get some more numbers on here. What are these two values $a$ and $b$?
devenware
2017-02-08 19:08:56
vlwilliam
2017-02-08 19:09:22
a=5-y-b
a=5-y-b
quanhui868
2017-02-08 19:09:22
5-y-b
5-y-b
yid
2017-02-08 19:09:22
a+y+b = 5
a+y+b = 5
ruchirk533
2017-02-08 19:09:22
a+b+y=5
a+b+y=5
devenware
2017-02-08 19:09:31
Definitely. We'll keep that in mind.
Definitely. We'll keep that in mind.
devenware
2017-02-08 19:09:39
Can we use similar triangles again to relate $a,b,$ and $y$?
Can we use similar triangles again to relate $a,b,$ and $y$?
CrystalEye
2017-02-08 19:10:17
a=4/3y b=3/4y
a=4/3y b=3/4y
dudethefirst
2017-02-08 19:10:17
4/3y and 3/4y
4/3y and 3/4y
acegikmoqsuwy2000
2017-02-08 19:10:17
$\dfrac 43y$ and $\dfrac 34y$
$\dfrac 43y$ and $\dfrac 34y$
uptownmath
2017-02-08 19:10:17
a=4y/3, b=3y/4
a=4y/3, b=3y/4
Arieoreos
2017-02-08 19:10:17
a=4/3y, b=3/4y
a=4/3y, b=3/4y
summitwei
2017-02-08 19:10:17
a=4/3*y, b=3/4*y
a=4/3*y, b=3/4*y
brainiac1
2017-02-08 19:10:17
a=4y/3, b=3y/4
a=4y/3, b=3y/4
sydney_z
2017-02-08 19:10:17
a = 4y/3
a = 4y/3
BunBulb
2017-02-08 19:10:17
a/y = 4/3
a/y = 4/3
sydney_z
2017-02-08 19:10:17
b=3y/4
b=3y/4
devenware
2017-02-08 19:10:22
Everything is a 3-4-5 triangle.
Everything is a 3-4-5 triangle.
devenware
2017-02-08 19:10:23
The left triangle gives $\dfrac{a}y=\dfrac43$, so $a=\dfrac{4y}3$.
The left triangle gives $\dfrac{a}y=\dfrac43$, so $a=\dfrac{4y}3$.
devenware
2017-02-08 19:10:25
The right triangle gives $\dfrac{b}y=\dfrac34$, so $b=\dfrac{3y}4$.
The right triangle gives $\dfrac{b}y=\dfrac34$, so $b=\dfrac{3y}4$.
devenware
2017-02-08 19:10:26
devenware
2017-02-08 19:10:38
Okay, now what?
Okay, now what?
amwmath
2017-02-08 19:11:07
Sum of the things on the bottom is 5
Sum of the things on the bottom is 5
tdeng
2017-02-08 19:11:07
Now 4y/3 + y + 3y/4=5
Now 4y/3 + y + 3y/4=5
MathMan1234
2017-02-08 19:11:07
now 4y/3 + 3y/4 + y = 5
now 4y/3 + 3y/4 + y = 5
yid
2017-02-08 19:11:07
a+b+y = 5
a+b+y = 5
wiler5002
2017-02-08 19:11:07
$\frac{4y}{3}+y+\frac{3y}{4}=5$
$\frac{4y}{3}+y+\frac{3y}{4}=5$
CrystalEye
2017-02-08 19:11:07
4/3y+y+3/4y=5
4/3y+y+3/4y=5
dragon6688
2017-02-08 19:11:07
$\frac{4y}{3} + y + \frac{3y}{4} = 5$
$\frac{4y}{3} + y + \frac{3y}{4} = 5$
ScienceSpirit
2017-02-08 19:11:07
Equate to 5
Equate to 5
devenware
2017-02-08 19:11:20
Now we use the fact that everything on the bottom adds to 5!
Now we use the fact that everything on the bottom adds to 5!
devenware
2017-02-08 19:11:24
What do you get for $y$?
What do you get for $y$?
mj434
2017-02-08 19:12:08
y=60/37
y=60/37
blacksheep2003
2017-02-08 19:12:08
$y=60/37$
$y=60/37$
gmdoss
2017-02-08 19:12:08
60/37
60/37
n2001
2017-02-08 19:12:08
60/37
60/37
mathusername
2017-02-08 19:12:08
y=60/37
y=60/37
sunnywoods
2017-02-08 19:12:08
y=60/37
y=60/37
Cardinals2014
2017-02-08 19:12:08
60/37
60/37
CharlesHong
2017-02-08 19:12:08
60/37
60/37
AAANNNMMMIII
2017-02-08 19:12:08
60/37
60/37
tfz4629
2017-02-08 19:12:08
60/37
60/37
CoolVincent122333
2017-02-08 19:12:08
60/37
60/37
evanhlu
2017-02-08 19:12:08
60/37
60/37
devenware
2017-02-08 19:12:20
The bottom edge has length 5, so we get the equation\[5=\frac{3y}4+y+\frac{4y}3=\frac{9y+12y+16y}{12}=\frac{37y}{12}.\] Therefore $y=\dfrac{60}{37}$.
The bottom edge has length 5, so we get the equation\[5=\frac{3y}4+y+\frac{4y}3=\frac{9y+12y+16y}{12}=\frac{37y}{12}.\] Therefore $y=\dfrac{60}{37}$.
devenware
2017-02-08 19:12:21
And the answer?
And the answer?
theartof
2017-02-08 19:12:54
D
D
MSTang
2017-02-08 19:12:54
D, 37/35
D, 37/35
Mathisfun04
2017-02-08 19:12:54
37/35
37/35
Wave-Particle
2017-02-08 19:12:54
D
D
bobjoe123
2017-02-08 19:12:54
So x/y = 37/35
So x/y = 37/35
garretth
2017-02-08 19:12:54
D
D
tooswagforyou
2017-02-08 19:12:54
37/35
37/35
uptownmath
2017-02-08 19:12:54
37/35
37/35
mikhailgromov
2017-02-08 19:12:54
D
D
Cardinals2014
2017-02-08 19:12:54
D
D
zhengyf
2017-02-08 19:12:54
D) 37/35
D) 37/35
FATRaichu
2017-02-08 19:12:54
37/35
37/35
AmitLuke
2017-02-08 19:12:54
D
D
Mario2357
2017-02-08 19:12:54
D) 37/35
D) 37/35
padfoot6302
2017-02-08 19:12:54
37/35
37/35
SomethingNeutral
2017-02-08 19:12:54
37/35
37/35
conniejeon1
2017-02-08 19:12:54
37/35
37/35
Liopleurodon
2017-02-08 19:12:54
$\frac{37}{35}$
$\frac{37}{35}$
devenware
2017-02-08 19:13:00
We want \[\dfrac xy=\frac{\frac{12}7}{\frac{60}{37}}=\frac{12\cdot37}{60\cdot7}=\boxed{\dfrac{37}{35}}.\]
We want \[\dfrac xy=\frac{\frac{12}7}{\frac{60}{37}}=\frac{12\cdot37}{60\cdot7}=\boxed{\dfrac{37}{35}}.\]
devenware
2017-02-08 19:13:05
The answer is (D).
The answer is (D).
dragon6688
2017-02-08 19:13:35
Makes it look so easy...
Makes it look so easy...
theironcatfish
2017-02-08 19:13:35
Well thats pretty neat
Well thats pretty neat
SmartGuy101
2017-02-08 19:13:35
YAY
YAY
Talker32332
2017-02-08 19:13:35
easier than i thought
easier than i thought
Brainiac2
2017-02-08 19:13:35
Yay!
Yay!
devenware
2017-02-08 19:13:40
devenware
2017-02-08 19:13:45
NEXT PROBLEM.
NEXT PROBLEM.
devenware
2017-02-08 19:14:00
22. Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$, respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?
$\phantom{hi}$
$\text{(A) } \dfrac{4\sqrt{3}\pi}{27} - \dfrac13\quad
\text{(B) } \dfrac{\sqrt{3}}{2} - \dfrac{\pi}8\quad
\text{(C) } \dfrac12 \quad
\text{(D) } \sqrt{3} - \dfrac{2\sqrt{3}\pi}{9} \quad
\text{(E) } \dfrac43-\dfrac{4\sqrt{3}\pi}{27}$
22. Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$, respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?
$\phantom{hi}$
$\text{(A) } \dfrac{4\sqrt{3}\pi}{27} - \dfrac13\quad
\text{(B) } \dfrac{\sqrt{3}}{2} - \dfrac{\pi}8\quad
\text{(C) } \dfrac12 \quad
\text{(D) } \sqrt{3} - \dfrac{2\sqrt{3}\pi}{9} \quad
\text{(E) } \dfrac43-\dfrac{4\sqrt{3}\pi}{27}$
devenware
2017-02-08 19:14:07
Okay, everyone better know the first step...
Okay, everyone better know the first step...
theartof
2017-02-08 19:14:41
Draw a diagram
Draw a diagram
quartzgirl
2017-02-08 19:14:41
Again, draw a diagram.
Again, draw a diagram.
Wave-Particle
2017-02-08 19:14:41
again draw a diagram
again draw a diagram
SomethingNeutral
2017-02-08 19:14:41
Draw another diagram
Draw another diagram
rplamkin
2017-02-08 19:14:41
draw a diagram
draw a diagram
N3RDBIRD
2017-02-08 19:14:41
draw
draw
mathchampion1
2017-02-08 19:14:41
diagram
diagram
chessapple9
2017-02-08 19:14:41
Draw a diagram
Draw a diagram
ShineBunny
2017-02-08 19:14:41
Diagram!
Diagram!
sbundlab1
2017-02-08 19:14:41
Diagram!
Diagram!
mathcrazymj
2017-02-08 19:14:41
draw diagram
draw diagram
Goliath
2017-02-08 19:14:41
diagram
diagram
jeffshen
2017-02-08 19:14:41
draw it
draw it
purpleapples
2017-02-08 19:14:41
draw a diagram
draw a diagram
TaffyQ
2017-02-08 19:14:41
draw a diagram!
draw a diagram!
challengetogo
2017-02-08 19:14:41
diagram!
diagram!
Yuny
2017-02-08 19:14:41
diagramm
diagramm
thinmint
2017-02-08 19:14:41
Draw a picture!!!
Draw a picture!!!
GM2B
2017-02-08 19:14:41
diagram
diagram
Auty
2017-02-08 19:14:41
Draw a diagram!
Draw a diagram!
Happy2020
2017-02-08 19:14:41
draw a diagram!
draw a diagram!
dzhou100
2017-02-08 19:14:41
DRAW A PIC/DIAGRAM!
DRAW A PIC/DIAGRAM!
GeneralCobra19
2017-02-08 19:14:41
DIAGRAm
DIAGRAm
kevindk
2017-02-08 19:14:41
diagram
diagram
PenguinJoe
2017-02-08 19:14:41
DIAGRAM
DIAGRAM
dipenm
2017-02-08 19:14:41
draw diagram
draw diagram
devenware
2017-02-08 19:14:47
DRAW A DIAGRAM!
DRAW A DIAGRAM!
devenware
2017-02-08 19:14:54
When sketching this diagram, where should you start?
When sketching this diagram, where should you start?
Arieoreos
2017-02-08 19:15:21
Draw the circle first
Draw the circle first
kiwitrader123
2017-02-08 19:15:21
Circle
Circle
jybsmartguy2
2017-02-08 19:15:21
the circle
the circle
WW92030
2017-02-08 19:15:21
circle
circle
Mario2357
2017-02-08 19:15:21
the circle
the circle
amc8nov
2017-02-08 19:15:21
with the circle
with the circle
rplamkin
2017-02-08 19:15:21
the circle
the circle
MSTang
2017-02-08 19:15:21
circle
circle
wiler5002
2017-02-08 19:15:21
With the circle
With the circle
mathchampion1
2017-02-08 19:15:21
the circle
the circle
theironcatfish
2017-02-08 19:15:21
circle then triangle
circle then triangle
mathforfun2016
2017-02-08 19:15:21
drawing the circle
drawing the circle
devenware
2017-02-08 19:15:26
It's hard to draw a circle tangent to two lines. It's a lot easier to draw the circle then the tangent lines.
It's hard to draw a circle tangent to two lines. It's a lot easier to draw the circle then the tangent lines.
devenware
2017-02-08 19:15:29
devenware
2017-02-08 19:15:32
Anything else we should add to the diagram before we get started?
Anything else we should add to the diagram before we get started?
uptownmath
2017-02-08 19:16:10
radii
radii
dipenm
2017-02-08 19:16:10
the center
the center
wiler5002
2017-02-08 19:16:10
Center of the circle and the right angles it makes
Center of the circle and the right angles it makes
purpleapples
2017-02-08 19:16:10
center of circle
center of circle
dipenm
2017-02-08 19:16:10
connect the center to the tangent points
connect the center to the tangent points
Arieoreos
2017-02-08 19:16:10
Two radii
Two radii
theartof
2017-02-08 19:16:10
Radii
Radii
Liopleurodon
2017-02-08 19:16:10
The radii to the points of tangency
The radii to the points of tangency
CrystalEye
2017-02-08 19:16:10
radii
radii
orchdork123
2017-02-08 19:16:10
points
points
amc8nov
2017-02-08 19:16:10
draw radii
draw radii
vbcnxm
2017-02-08 19:16:10
radii
radii
abvenkgoo
2017-02-08 19:16:10
Radii from the points of tangency to the center
Radii from the points of tangency to the center
Wave-Particle
2017-02-08 19:16:10
radii of circle
radii of circle
devenware
2017-02-08 19:16:14
Whenever we have tangents to a circle, it makes sense to draw the perpendicular radii.
Whenever we have tangents to a circle, it makes sense to draw the perpendicular radii.
devenware
2017-02-08 19:16:15
devenware
2017-02-08 19:16:18
What regions should we label?
What regions should we label?
dipenm
2017-02-08 19:17:26
the circular segment
the circular segment
amc8nov
2017-02-08 19:17:26
the triangle outside of the circle
the triangle outside of the circle
kiwitrader123
2017-02-08 19:17:26
region outside circle, sector
region outside circle, sector
GeronimoStilton
2017-02-08 19:17:26
The triangle, the bit of the sector not including the triangle
The triangle, the bit of the sector not including the triangle
Mario2357
2017-02-08 19:17:26
iscoceles triangle, missing arc, and other part of the equilateral
iscoceles triangle, missing arc, and other part of the equilateral
chessapple9
2017-02-08 19:17:26
We should label the region in the triangle outside of the circle and inside of the circle.
We should label the region in the triangle outside of the circle and inside of the circle.
WW92030
2017-02-08 19:17:26
outside the circle, inside the circle
outside the circle, inside the circle
Liopleurodon
2017-02-08 19:17:26
The area of the triangle outside the circle, the area of the circle that's outside the triangle but inside the circle, and the part that they share
The area of the triangle outside the circle, the area of the circle that's outside the triangle but inside the circle, and the part that they share
devenware
2017-02-08 19:17:30
Let's label the part of the triangle outside the circle, the part of the triangle inside the circle, and that other little triangle that just happened to appear.
Let's label the part of the triangle outside the circle, the part of the triangle inside the circle, and that other little triangle that just happened to appear.
devenware
2017-02-08 19:17:32
devenware
2017-02-08 19:17:33
And what value are we trying to compute?
And what value are we trying to compute?
devenware
2017-02-08 19:17:57
(READ THE PROBLEM CAREFULLY!)
(READ THE PROBLEM CAREFULLY!)
happypi3.14159265358979
2017-02-08 19:18:46
x/(x+y)
x/(x+y)
Mathisfun04
2017-02-08 19:18:46
x/(x + y)
x/(x + y)
Happy2020
2017-02-08 19:18:46
x/(x+y)
x/(x+y)
wiler5002
2017-02-08 19:18:46
$\frac{x}{x+y}$
$\frac{x}{x+y}$
User2013
2017-02-08 19:18:46
x/(x+y)
x/(x+y)
aragornmf
2017-02-08 19:18:46
x/(x+y)
x/(x+y)
AAANNNMMMIII
2017-02-08 19:18:46
x/(x+y)
x/(x+y)
iks92
2017-02-08 19:18:46
x/(x+y)
x/(x+y)
BunBulb
2017-02-08 19:18:46
x/(x+y)
x/(x+y)
ScienceSpirit
2017-02-08 19:18:46
x/(x+y)
x/(x+y)
CrystalEye
2017-02-08 19:18:46
x/(x+y)
x/(x+y)
mt492
2017-02-08 19:18:46
x/(x+y)
x/(x+y)
blacksheep2003
2017-02-08 19:18:46
$x/(x+y)$
$x/(x+y)$
devenware
2017-02-08 19:18:50
We want to find $\dfrac{x}{x+y}$.
We want to find $\dfrac{x}{x+y}$.
devenware
2017-02-08 19:18:52
Note that this doesn't have anything to do with the $z$ we labeled. So, even if we might use it to help, we'll want to get rid of it from our computations at the end.
Note that this doesn't have anything to do with the $z$ we labeled. So, even if we might use it to help, we'll want to get rid of it from our computations at the end.
devenware
2017-02-08 19:18:57
Before we get started, is there anything we can do to simplify our computations?
Before we get started, is there anything we can do to simplify our computations?
Arieoreos
2017-02-08 19:19:34
Make the circle a unit circke
Make the circle a unit circke
amwmath
2017-02-08 19:19:34
Set the radius to $1$
Set the radius to $1$
warrenwangtennis
2017-02-08 19:19:34
r = 1
r = 1
brainiac1
2017-02-08 19:19:34
assume a side length or radius or 1
assume a side length or radius or 1
challengetogo
2017-02-08 19:19:34
let the radius of the circle be 1
let the radius of the circle be 1
MSTang
2017-02-08 19:19:34
let the circle have radius 1
let the circle have radius 1
dragon6688
2017-02-08 19:19:34
Set the radius to 1 WLOG because the end result is a ratio
Set the radius to 1 WLOG because the end result is a ratio
mathchampion1
2017-02-08 19:19:34
let the radius be 1
let the radius be 1
wiler5002
2017-02-08 19:19:34
Let the radius be 1
Let the radius be 1
ScienceSpirit
2017-02-08 19:19:34
Make radius 1
Make radius 1
devenware
2017-02-08 19:19:41
Nothing in the problem depends on how big our diagram is -- we can scale it up or down and it doesn't matter. So, we can just pick some length to be whatever we want.
Nothing in the problem depends on how big our diagram is -- we can scale it up or down and it doesn't matter. So, we can just pick some length to be whatever we want.
devenware
2017-02-08 19:19:46
At first we might want the edge length of the equilateral triangle to be 1, but since the circle seems like it's going to be important in a couple of computations, let's set that to 1:
At first we might want the edge length of the equilateral triangle to be 1, but since the circle seems like it's going to be important in a couple of computations, let's set that to 1:
devenware
2017-02-08 19:19:48
devenware
2017-02-08 19:19:54
What is the edge-length of the equilateral triangle now?
What is the edge-length of the equilateral triangle now?
MSTang
2017-02-08 19:20:32
$\sqrt{3}$
$\sqrt{3}$
dudethefirst
2017-02-08 19:20:32
root 3
root 3
GeronimoStilton
2017-02-08 19:20:32
$\sqrt{3}$
$\sqrt{3}$
WW92030
2017-02-08 19:20:32
root 3
root 3
CrystalEye
2017-02-08 19:20:32
sqrt3
sqrt3
chessapple9
2017-02-08 19:20:32
$\sqrt 3$
$\sqrt 3$
MSTang
2017-02-08 19:20:32
dipenm
2017-02-08 19:20:32
$\sqrt{3}$
$\sqrt{3}$
throwhit
2017-02-08 19:20:32
sqrt(3)
sqrt(3)
mathchampion1
2017-02-08 19:20:32
sqrt(3)
sqrt(3)
fields123
2017-02-08 19:20:32
sqr3
sqr3
cakeguy
2017-02-08 19:20:32
$\sqrt{3}$
$\sqrt{3}$
purplecoat
2017-02-08 19:20:32
sqrt 3
sqrt 3
AidanNReilly
2017-02-08 19:20:32
sqrt 3
sqrt 3
adyj
2017-02-08 19:20:32
root 3
root 3
skmc
2017-02-08 19:20:32
sqrt(3)
sqrt(3)
devenware
2017-02-08 19:20:40
This triangle is a 30-60-90 triangle since the left vertex has measure $\dfrac{60^\circ}2$ and the top vertex is a right angle:
This triangle is a 30-60-90 triangle since the left vertex has measure $\dfrac{60^\circ}2$ and the top vertex is a right angle:
devenware
2017-02-08 19:20:41
devenware
2017-02-08 19:20:46
Therefore the equilateral triangle has edge-length $\sqrt3$:
Therefore the equilateral triangle has edge-length $\sqrt3$:
devenware
2017-02-08 19:20:47
devenware
2017-02-08 19:21:08
How do $x,y,$ and $z$ relate to the triangles we've drawn and to the circle?
How do $x,y,$ and $z$ relate to the triangles we've drawn and to the circle?
warrenwangtennis
2017-02-08 19:21:41
x+y=area of triangle
x+y=area of triangle
MSTang
2017-02-08 19:21:41
x+y is now the area of the triangle
x+y is now the area of the triangle
conniejeon1
2017-02-08 19:21:41
x+y= Area of the eqilateral triangle
x+y= Area of the eqilateral triangle
devenware
2017-02-08 19:21:47
We know that $x + y$ is the area of the equilateral triangle. So, \[x + y = (\sqrt{3})^2 \cdot \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}.\]
We know that $x + y$ is the area of the equilateral triangle. So, \[x + y = (\sqrt{3})^2 \cdot \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}.\]
mathchampion1
2017-02-08 19:22:11
y+z is a 120 degree sector
y+z is a 120 degree sector
SomethingNeutral
2017-02-08 19:22:11
x+y = equilateral triangle, y+z = sector (?) of circle
x+y = equilateral triangle, y+z = sector (?) of circle
Ani10
2017-02-08 19:22:11
y+z=sector of circle
y+z=sector of circle
wiler5002
2017-02-08 19:22:11
y+z is a sector of the cirlce with angle 120
y+z is a sector of the cirlce with angle 120
devenware
2017-02-08 19:22:16
We know that $y + z$ is the area of a sector of the circle with angle $120^{\circ}.$ So, $y + z = \dfrac{\pi}{3}.$
We know that $y + z$ is the area of a sector of the circle with angle $120^{\circ}.$ So, $y + z = \dfrac{\pi}{3}.$
abvenkgoo
2017-02-08 19:22:32
$x+y+z = /sqrt{3}$
$x+y+z = /sqrt{3}$
Arieoreos
2017-02-08 19:22:32
It's half of x+y+z, each triangle is
It's half of x+y+z, each triangle is
GeronimoStilton
2017-02-08 19:22:32
$x+y+z$ is $\sqrt{3}$
$x+y+z$ is $\sqrt{3}$
MSTang
2017-02-08 19:22:32
x+y+z=sqrt(3)
x+y+z=sqrt(3)
uptownmath
2017-02-08 19:22:33
x+y+z is the sum of the areas of the two triangles
x+y+z is the sum of the areas of the two triangles
devenware
2017-02-08 19:22:40
We know that $x + y + z$ is twice the area of one of our 30-60-90 triangles, or $x+y+z = \sqrt{3}.$
We know that $x + y + z$ is twice the area of one of our 30-60-90 triangles, or $x+y+z = \sqrt{3}.$
Ani10
2017-02-08 19:23:07
we have enough to find x
we have enough to find x
devenware
2017-02-08 19:23:08
So, what is $x$?
So, what is $x$?
SomethingNeutral
2017-02-08 19:23:41
x = sqrt(3)-pi/3
x = sqrt(3)-pi/3
chessapple9
2017-02-08 19:23:41
So $x = \sqrt 3 - (\pi/3)$
So $x = \sqrt 3 - (\pi/3)$
CaptainGeo
2017-02-08 19:23:41
sqrt3-pi/3
sqrt3-pi/3
uptownmath
2017-02-08 19:23:41
$sqrt3-\frac{pi}{3}$
$sqrt3-\frac{pi}{3}$
mathchampion1
2017-02-08 19:23:41
sqrt(3)-pi/3
sqrt(3)-pi/3
cooljoseph
2017-02-08 19:23:41
$\sqrt3-\frac\pi3$
$\sqrt3-\frac\pi3$
WW92030
2017-02-08 19:23:41
root 3 - pi/3
root 3 - pi/3
bomb427006
2017-02-08 19:23:41
$\sqrt3-\pi/3$
$\sqrt3-\pi/3$
LegoLdr
2017-02-08 19:23:41
root(3)-1/3pi
root(3)-1/3pi
seanwang2001
2017-02-08 19:23:45
sqrt(3)-pi/3
sqrt(3)-pi/3
devenware
2017-02-08 19:23:49
We have $x = (x+y+z) - (y+z) = \sqrt{3} - \dfrac{\pi}{3}.$
We have $x = (x+y+z) - (y+z) = \sqrt{3} - \dfrac{\pi}{3}.$
devenware
2017-02-08 19:23:50
So what is our answer?
So what is our answer?
SomethingNeutral
2017-02-08 19:24:23
Answer is E.
Answer is E.
Cardinals2014
2017-02-08 19:24:23
E
E
jybsmartguy2
2017-02-08 19:24:23
E
E
uptownmath
2017-02-08 19:24:23
$E$
$E$
AAANNNMMMIII
2017-02-08 19:24:23
E
E
FATRaichu
2017-02-08 19:24:23
E
E
CharlesHong
2017-02-08 19:24:23
E
E
Knight8
2017-02-08 19:24:23
E
E
Ani10
2017-02-08 19:24:23
is it E?
is it E?
gexin
2017-02-08 19:24:23
E
E
bomb427006
2017-02-08 19:24:23
E
E
hodori01
2017-02-08 19:24:23
E
E
devenware
2017-02-08 19:24:27
We have \begin{align*} \frac{x}{x+y} &= \frac{\sqrt{3} - \frac{\pi}{3}}{3\sqrt{3}/4} \\ &= \frac{4\sqrt{3} - \frac{4\pi}{3}}{3\sqrt{3}} \\ &= \frac{4}{3} - \frac{4\pi\sqrt{3}}{27}. \end{align*} Our answer is (E).
We have \begin{align*} \frac{x}{x+y} &= \frac{\sqrt{3} - \frac{\pi}{3}}{3\sqrt{3}/4} \\ &= \frac{4\sqrt{3} - \frac{4\pi}{3}}{3\sqrt{3}} \\ &= \frac{4}{3} - \frac{4\pi\sqrt{3}}{27}. \end{align*} Our answer is (E).
cooljoseph
2017-02-08 19:24:46
Yay!
Yay!
amwmath
2017-02-08 19:24:46
Yay
Yay
warrenwangtennis
2017-02-08 19:24:46
easy!
easy!
tfz4629
2017-02-08 19:24:46
Yay
Yay
mathchampion1
2017-02-08 19:24:46
so easy
so easy
devenware
2017-02-08 19:24:53
Or maybe we're just all really good.
Or maybe we're just all really good.
LegoLdr
2017-02-08 19:25:09
That's probably it
That's probably it
MSTang
2017-02-08 19:25:09
why not both?
why not both?
devenware
2017-02-08 19:25:19
We'll try the next one to test our hypotheses.
We'll try the next one to test our hypotheses.
devenware
2017-02-08 19:25:24
That's the scientific method or something.
That's the scientific method or something.
devenware
2017-02-08 19:25:42
23. How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$, inclusive?
$\phantom{hi}$
$\text{(A) } 2128 \quad
\text{(B) } 2148 \quad
\text{(C) } 2160 \quad
\text{(D) } 2200 \quad
\text{(E) } 2300$
23. How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$, inclusive?
$\phantom{hi}$
$\text{(A) } 2128 \quad
\text{(B) } 2148 \quad
\text{(C) } 2160 \quad
\text{(D) } 2200 \quad
\text{(E) } 2300$
devenware
2017-02-08 19:25:47
How should we approach this problem?
How should we approach this problem?
iNomOnCountdown
2017-02-08 19:26:31
Complementary counting
Complementary counting
Arieoreos
2017-02-08 19:26:31
Complementary counting
Complementary counting
amwmath
2017-02-08 19:26:31
Complementary counting
Complementary counting
dipenm
2017-02-08 19:26:31
Complementary counting?
Complementary counting?
Liopleurodon
2017-02-08 19:26:31
Complementary Counting
Complementary Counting
RedPhoenix
2017-02-08 19:26:31
complementary
complementary
challengetogo
2017-02-08 19:26:31
complementary counting?
complementary counting?
thedoge
2017-02-08 19:26:31
use complementary counting
use complementary counting
WW92030
2017-02-08 19:26:31
complementary counting
complementary counting
devenware
2017-02-08 19:26:40
We could try complementary counting. That is, let's count all the triples and subtract the triples that give degenerate triangles.
We could try complementary counting. That is, let's count all the triples and subtract the triples that give degenerate triangles.
devenware
2017-02-08 19:26:41
Why do we think complementary counting is useful here?
Why do we think complementary counting is useful here?
GeneralCobra19
2017-02-08 19:27:37
It would be hard to generate triangles "that work"
It would be hard to generate triangles "that work"
Liopleurodon
2017-02-08 19:27:37
Only have to count degenerate triangles
Only have to count degenerate triangles
WW92030
2017-02-08 19:27:37
the degenerate triples are collinear points
the degenerate triples are collinear points
garretth
2017-02-08 19:27:37
collinear subtraction
collinear subtraction
AAANNNMMMIII
2017-02-08 19:27:37
degenerate triangles (lines)
degenerate triangles (lines)
WW92030
2017-02-08 19:27:37
There are only a few degenerate triples
There are only a few degenerate triples
Brainiac2
2017-02-08 19:27:37
Not many degenerate answers.
Not many degenerate answers.
ItzVineeth
2017-02-08 19:27:37
Because we can take what we don't want subtract it from everything and recieve what we wan't
Because we can take what we don't want subtract it from everything and recieve what we wan't
adyj
2017-02-08 19:27:37
It's not as bashy as counting all possible triangles.
It's not as bashy as counting all possible triangles.
DarkPikachu
2017-02-08 19:27:37
there's a lot less cases of degenerate triangles than ones with positive area
there's a lot less cases of degenerate triangles than ones with positive area
brainiac1
2017-02-08 19:27:37
the complementary case only has one case: the points all lie on a line
the complementary case only has one case: the points all lie on a line
devenware
2017-02-08 19:27:42
Because degenerate triangles are pretty easy to count.
Because degenerate triangles are pretty easy to count.
devenware
2017-02-08 19:27:52
Degenerate triangles have all three vertices are on the same line.
Degenerate triangles have all three vertices are on the same line.
Happy2020
2017-02-08 19:28:30
Draw a diagram ...?
Draw a diagram ...?
GM2B
2017-02-08 19:28:30
diagram?
diagram?
mathchampion1
2017-02-08 19:28:30
diagram
diagram
ah123
2017-02-08 19:28:30
draw a picture
draw a picture
CaptainGeo
2017-02-08 19:28:30
draw a graph
draw a graph
alcumus1503
2017-02-08 19:28:30
diagram
diagram
it1023
2017-02-08 19:28:30
Draw a graph.
Draw a graph.
devenware
2017-02-08 19:28:32
Okay let's draw the diagram.
Okay let's draw the diagram.
devenware
2017-02-08 19:28:36
devenware
2017-02-08 19:28:43
OK, how many triangles are there total in this grid?
OK, how many triangles are there total in this grid?
rishis
2017-02-08 19:29:44
25 C3=2300
25 C3=2300
mathchampion1
2017-02-08 19:29:44
25C3 including degenerate
25C3 including degenerate
Jayjayliu
2017-02-08 19:29:44
25 choose 3
25 choose 3
DarkPikachu
2017-02-08 19:29:44
25 choose 3
25 choose 3
Bill9000
2017-02-08 19:29:44
2300
2300
conniejeon1
2017-02-08 19:29:44
2300
2300
tdeng
2017-02-08 19:29:44
$\dbinom{25}{3}$
$\dbinom{25}{3}$
bobjoe123
2017-02-08 19:29:44
$\binom{25}{3}$
$\binom{25}{3}$
mutinykids
2017-02-08 19:29:44
2300
2300
devenware
2017-02-08 19:29:48
There are 25 vertices and we need to pick 3, so there are \[\dbinom{25}3=\frac{25\cdot24\cdot23}{3\cdot2\cdot1}=25\cdot4\cdot23=2300.\]
There are 25 vertices and we need to pick 3, so there are \[\dbinom{25}3=\frac{25\cdot24\cdot23}{3\cdot2\cdot1}=25\cdot4\cdot23=2300.\]
mathchampion1
2017-02-08 19:30:30
we can eliminate choice (E)
we can eliminate choice (E)
devenware
2017-02-08 19:30:36
Yeah, so (E) is definitely wrong.
Yeah, so (E) is definitely wrong.
devenware
2017-02-08 19:30:40
Now, how can we organize our counting for all the degenerate triangles?
Now, how can we organize our counting for all the degenerate triangles?
amwmath
2017-02-08 19:31:17
Cases? By slope, perhaps?
Cases? By slope, perhaps?
3_14152
2017-02-08 19:31:17
by slope of line
by slope of line
amwmath
2017-02-08 19:31:17
Slope.
Slope.
PenguinJoe
2017-02-08 19:31:17
different line slopes?
different line slopes?
bomb427006
2017-02-08 19:31:17
slope of the line connecting the 3 points
slope of the line connecting the 3 points
CornSaltButter
2017-02-08 19:31:17
Line slopes
Line slopes
MathMan1234
2017-02-08 19:31:17
slope of the line formed by the points?
slope of the line formed by the points?
Brainiac2
2017-02-08 19:31:17
slopes?
slopes?
Ani10
2017-02-08 19:31:17
slope of lines
slope of lines
devenware
2017-02-08 19:31:23
Let's count by slope. What slope should we start with?
Let's count by slope. What slope should we start with?
SomethingNeutral
2017-02-08 19:31:51
0
0
Talker32332
2017-02-08 19:31:51
0?
0?
jeffshen
2017-02-08 19:31:51
0
0
challengetogo
2017-02-08 19:31:51
0
0
Brisingrfire
2017-02-08 19:31:51
0
0
CrystalEye
2017-02-08 19:31:51
0
0
FATRaichu
2017-02-08 19:31:51
0
0
spotty2012
2017-02-08 19:31:51
0
0
geomstar_1234
2017-02-08 19:31:51
0
0
orchdork123
2017-02-08 19:31:51
0?
0?
chris100
2017-02-08 19:31:51
0
0
devenware
2017-02-08 19:31:54
Slope 0:
Slope 0:
devenware
2017-02-08 19:31:56
devenware
2017-02-08 19:31:59
I'll get you started. The number of degenerate triangles here is 5 times, um, something. 5 times what?
I'll get you started. The number of degenerate triangles here is 5 times, um, something. 5 times what?
zjjc123
2017-02-08 19:32:46
5 choose 3
5 choose 3
SomethingNeutral
2017-02-08 19:32:46
5c3
5c3
dzhou100
2017-02-08 19:32:46
5 choose 3
5 choose 3
RedHawk
2017-02-08 19:32:46
5 C 3
5 C 3
pie314159265
2017-02-08 19:32:46
5C3
5C3
wiler5002
2017-02-08 19:32:46
$\binom{5}{3}$
$\binom{5}{3}$
chessapple9
2017-02-08 19:32:46
5 choose 3, which is 10.
5 choose 3, which is 10.
Smoothfang
2017-02-08 19:32:46
5C3=10
5C3=10
vbcnxm
2017-02-08 19:32:46
5 choose 3
5 choose 3
skmc
2017-02-08 19:32:46
5 choose 3
5 choose 3
Quaratinium
2017-02-08 19:32:46
5 choose 3 then that times 5
5 choose 3 then that times 5
devenware
2017-02-08 19:32:49
To get a degenerate triangle on one of those lines, we pick 3 of the five points.
To get a degenerate triangle on one of those lines, we pick 3 of the five points.
devenware
2017-02-08 19:32:51
There are $5\dbinom53=5\cdot10=50$ horizontal degenerate triangles.
There are $5\dbinom53=5\cdot10=50$ horizontal degenerate triangles.
devenware
2017-02-08 19:32:53
What should we do next?
What should we do next?
zjjc123
2017-02-08 19:33:26
slope undefined
slope undefined
RedHawk
2017-02-08 19:33:26
Undefined slope
Undefined slope
jeffshen
2017-02-08 19:33:26
undefined
undefined
GAN
2017-02-08 19:33:26
vertical
vertical
adyj
2017-02-08 19:33:26
slope is undefined
slope is undefined
geomstar_1234
2017-02-08 19:33:26
undefined slope?
undefined slope?
ShineBunny
2017-02-08 19:33:26
vertical slope
vertical slope
uptownmath
2017-02-08 19:33:26
slope undefined
slope undefined
ilovemath04
2017-02-08 19:33:26
undefined slopes
undefined slopes
devenware
2017-02-08 19:33:32
The number of vertical degenerate triangles should be the same, 50.
The number of vertical degenerate triangles should be the same, 50.
devenware
2017-02-08 19:33:34
devenware
2017-02-08 19:33:35
What other slopes are possible?
What other slopes are possible?
alcumus1503
2017-02-08 19:34:33
1
1
kiwitrader123
2017-02-08 19:34:33
1
1
NewbieGamer
2017-02-08 19:34:33
2
2
dzhou100
2017-02-08 19:34:33
1 and -1
1 and -1
awesome1028717
2017-02-08 19:34:33
slope 1, 1/2, -1, -1/2
slope 1, 1/2, -1, -1/2
iks92
2017-02-08 19:34:33
1/2, 2, 1
1/2, 2, 1
jybsmartguy2
2017-02-08 19:34:33
+ or - 1
+ or - 1
amwmath
2017-02-08 19:34:33
$\pm2^{\pm1}$ and $\pm1$
$\pm2^{\pm1}$ and $\pm1$
hodori01
2017-02-08 19:34:33
1,-1,1/2,-1/2,2,-2
1,-1,1/2,-1/2,2,-2
warrenwangtennis
2017-02-08 19:34:33
1/2, -1/2
1/2, -1/2
Picroft
2017-02-08 19:34:33
2, 0.5, -2, -0.5
2, 0.5, -2, -0.5
awesome1028717
2017-02-08 19:34:33
slope 1
slope 1
Benjy450
2017-02-08 19:34:33
1 and -1
1 and -1
xjeanniewx
2017-02-08 19:34:33
slope 1 and -1
slope 1 and -1
Arieoreos
2017-02-08 19:34:33
slope of 1, slope of 1/2, -1/2, 2, -2
slope of 1, slope of 1/2, -1/2, 2, -2
awesome1028717
2017-02-08 19:34:33
slope -1/2
slope -1/2
devenware
2017-02-08 19:34:37
We can only get slopes of $\pm1$, $\pm2$, and $\pm\frac 12$.
We can only get slopes of $\pm1$, $\pm2$, and $\pm\frac 12$.
devenware
2017-02-08 19:34:44
Specifically, we can look at any line with at least 3 points on it in this diagram. If the line isn't vertical or horizontal then the $x$-coordinates form part of an arithmetic progression inside $\{1,2,3,4,5\}$ and the same for the $y$-coordinates. Thus the difference between neighboring points is $\pm1$ or $\pm2$. Therefore the numerator and denominator of the slope are both at most 2.
Specifically, we can look at any line with at least 3 points on it in this diagram. If the line isn't vertical or horizontal then the $x$-coordinates form part of an arithmetic progression inside $\{1,2,3,4,5\}$ and the same for the $y$-coordinates. Thus the difference between neighboring points is $\pm1$ or $\pm2$. Therefore the numerator and denominator of the slope are both at most 2.
devenware
2017-02-08 19:34:55
How many lines of slope 2?
How many lines of slope 2?
devenware
2017-02-08 19:35:38
(I'll note that it'll probably help to be drawing along!)
(I'll note that it'll probably help to be drawing along!)
CornSaltButter
2017-02-08 19:35:42
3
3
GeneralCobra19
2017-02-08 19:35:42
There are 3 lines
There are 3 lines
kiwitrader123
2017-02-08 19:35:42
3
3
WW92030
2017-02-08 19:35:42
3
3
GeronimoStilton
2017-02-08 19:35:42
$3$
$3$
n2001
2017-02-08 19:35:42
3
3
AAANNNMMMIII
2017-02-08 19:35:42
3
3
Mario2357
2017-02-08 19:35:42
3
3
mafsalamander
2017-02-08 19:35:42
3
3
antmath2520
2017-02-08 19:35:42
3
3
evanhlu
2017-02-08 19:35:42
3
3
CaptainGeo
2017-02-08 19:35:42
3
3
mathcount2002
2017-02-08 19:35:42
3
3
Goliath
2017-02-08 19:35:42
nope, bleh, 3
nope, bleh, 3
acegikmoqsuwy2000
2017-02-08 19:35:42
$3$
$3$
devenware
2017-02-08 19:35:46
devenware
2017-02-08 19:35:50
Each of these lines contains one degenerate triangle.
Each of these lines contains one degenerate triangle.
devenware
2017-02-08 19:35:53
There are 3 degenerate triangles of slope 2.
There are 3 degenerate triangles of slope 2.
devenware
2017-02-08 19:35:56
What else?
What else?
Goliath
2017-02-08 19:36:20
another three for -2
another three for -2
jybsmartguy2
2017-02-08 19:36:20
-2
-2
uptownmath
2017-02-08 19:36:20
slope -2
slope -2
sbundlab1
2017-02-08 19:36:20
slope of -2
slope of -2
Ani10
2017-02-08 19:36:20
3 of slope -2
3 of slope -2
Arieoreos
2017-02-08 19:36:20
3 lines slope -2
3 lines slope -2
mutinykids
2017-02-08 19:36:20
and 3 of slope -2
and 3 of slope -2
WW92030
2017-02-08 19:36:20
3 lines for -2
3 lines for -2
jeffshen
2017-02-08 19:36:20
3 for -2 slope
3 for -2 slope
dzhou100
2017-02-08 19:36:20
-2 slope, also 3.
-2 slope, also 3.
ilovemath04
2017-02-08 19:36:20
3 for -2
3 for -2
oriduck.nk
2017-02-08 19:36:20
-2
-2
Bill9000
2017-02-08 19:36:20
slope of -2
slope of -2
devenware
2017-02-08 19:36:24
Symmetrically, there are 3 degenerate triangles of slope $-2$.
Symmetrically, there are 3 degenerate triangles of slope $-2$.
devenware
2017-02-08 19:36:28
tooswagforyou
2017-02-08 19:36:50
slope 1/2
slope 1/2
Mario2357
2017-02-08 19:36:50
also 3 of -2,1/2, -1/2
also 3 of -2,1/2, -1/2
LegoLdr
2017-02-08 19:36:50
3 lines with 1/2 slope too
3 lines with 1/2 slope too
acegikmoqsuwy2000
2017-02-08 19:36:50
slope -2, 1/2, -1/2 should all also yield 3
slope -2, 1/2, -1/2 should all also yield 3
amwmath
2017-02-08 19:36:50
$3$ of $-2$, $-\frac12$, and $\frac12$ each as well.
$3$ of $-2$, $-\frac12$, and $\frac12$ each as well.
warrenwangtennis
2017-02-08 19:36:50
1/2
1/2
mikhailgromov
2017-02-08 19:36:50
another 6 for \pm\frac12
another 6 for \pm\frac12
LegoLdr
2017-02-08 19:36:50
Same is true for 1/2 and -1/2
Same is true for 1/2 and -1/2
devenware
2017-02-08 19:36:59
There are also 3 degenerate triangles of slope $+\dfrac12$ and 3 of slope $-\dfrac12$.
There are also 3 degenerate triangles of slope $+\dfrac12$ and 3 of slope $-\dfrac12$.
devenware
2017-02-08 19:37:05
How many lines are there of slope 1?
How many lines are there of slope 1?
devenware
2017-02-08 19:37:34
(Look at your diagram carefully! Remember we need at least 3 of our dots on the line.)
(Look at your diagram carefully! Remember we need at least 3 of our dots on the line.)
mathmagician
2017-02-08 19:37:51
5
5
evanhlu
2017-02-08 19:37:51
5
5
jeffshen
2017-02-08 19:37:51
5
5
sbundlab1
2017-02-08 19:37:51
5
5
ilovemath04
2017-02-08 19:37:51
5
5
math101010
2017-02-08 19:37:51
5
5
uptownmath
2017-02-08 19:37:51
$5$
$5$
pratima4
2017-02-08 19:37:51
5
5
LegoLdr
2017-02-08 19:37:51
5?
5?
kenneo1
2017-02-08 19:37:51
5
5
dipenm
2017-02-08 19:37:51
no 5
no 5
pratima4
2017-02-08 19:37:51
5
5
kevindk
2017-02-08 19:37:51
5
5
Blocry
2017-02-08 19:37:51
5
5
devenware
2017-02-08 19:37:57
devenware
2017-02-08 19:37:59
That top and bottom lines each have 1 degenerate triangle, the second and fourth lines have $\dbinom43=4$, and the middle line has $\dbinom53=10$.
That top and bottom lines each have 1 degenerate triangle, the second and fourth lines have $\dbinom43=4$, and the middle line has $\dbinom53=10$.
devenware
2017-02-08 19:38:14
There are $1+4+10+4+1=20$ degenerate triangles of slope 1.
There are $1+4+10+4+1=20$ degenerate triangles of slope 1.
kiwitrader123
2017-02-08 19:38:43
same for slope -1
same for slope -1
spotty2012
2017-02-08 19:38:43
this is also true for a slope of -1
this is also true for a slope of -1
sbundlab1
2017-02-08 19:38:43
-1 also has 20
-1 also has 20
FATRaichu
2017-02-08 19:38:43
same for -1
same for -1
ilovemath04
2017-02-08 19:38:43
20 for -1
20 for -1
uptownmath
2017-02-08 19:38:43
same for slope -1
same for slope -1
fields123
2017-02-08 19:38:43
20 for slope -1
20 for slope -1
CornSaltButter
2017-02-08 19:38:43
Same for slope -1 by symmetry
Same for slope -1 by symmetry
devenware
2017-02-08 19:38:47
Likewise there are 20 degenerate triangles of slope $-1$.
Likewise there are 20 degenerate triangles of slope $-1$.
devenware
2017-02-08 19:38:50
What's the answer?
What's the answer?
GeronimoStilton
2017-02-08 19:39:38
It's 2148!
It's 2148!
Arieoreos
2017-02-08 19:39:38
2148
2148
CrystalEye
2017-02-08 19:39:38
2148
2148
SmartGuy101
2017-02-08 19:39:38
$\boxed B$
$\boxed B$
Benjy450
2017-02-08 19:39:38
2148, B
2148, B
FATRaichu
2017-02-08 19:39:38
2148, B
2148, B
SmartGuy101
2017-02-08 19:39:38
2148
2148
CrystalEye
2017-02-08 19:39:38
B
B
GeneralCobra19
2017-02-08 19:39:38
2300-252=2148, (B)
2300-252=2148, (B)
SnakeYu
2017-02-08 19:39:38
B
B
chessapple9
2017-02-08 19:39:38
I think it is B
I think it is B
amc8nov
2017-02-08 19:39:38
2148!!!
2148!!!
michigo
2017-02-08 19:39:38
2148
2148
amc8nov
2017-02-08 19:39:38
B
B
chris100
2017-02-08 19:39:38
B
B
orchdork123
2017-02-08 19:39:38
B
B
goodball
2017-02-08 19:39:38
B
B
devenware
2017-02-08 19:39:41
Complementary counting tells us the total number of triangles is \[2300-50-50-3-3-3-3-20-20=\boxed{2148}.\]
Complementary counting tells us the total number of triangles is \[2300-50-50-3-3-3-3-20-20=\boxed{2148}.\]
devenware
2017-02-08 19:39:42
The answer is (B).
The answer is (B).
rlzhang
2017-02-08 19:40:06
yay
yay
SmartGuy101
2017-02-08 19:40:06
YAY
YAY
devenware
2017-02-08 19:40:14
Well, we've scientifically proven it:
Well, we've scientifically proven it:
adyj
2017-02-08 19:40:18
So, the conclusion for easy or smartness is...
So, the conclusion for easy or smartness is...
AAANNNMMMIII
2017-02-08 19:40:18
so easy, we must be really good
so easy, we must be really good
pie314159265
2017-02-08 19:40:35
what about experimental error
what about experimental error
devenware
2017-02-08 19:40:38
Good point...
Good point...
devenware
2017-02-08 19:40:41
We better try a few more.
We better try a few more.
antmath2520
2017-02-08 19:41:30
Can't 'prove' anything scientifically, can only disprove
Can't 'prove' anything scientifically, can only disprove
devenware
2017-02-08 19:41:36
Too philosophical. Next problem.
Too philosophical. Next problem.
devenware
2017-02-08 19:41:42
24. For certain real numbers $a$, $b$, and $c$, the polynomial \[g(x) = x^3 +ax^2 + x+10\]has three distinct roots, and each root of $g(x)$ is also a root of the polynomial\[f(x) = x^4 +x^3 +bx^2 + 100x + c.\]What is $f(1)$?
$\phantom{hi}$
$\text{(A) } {-9009} \quad
\text{(B) } {-8008} \quad
\text{(C) } {-7007} \quad
\text{(D) } {-6006} \quad
\text{(E) } {-5005}$
24. For certain real numbers $a$, $b$, and $c$, the polynomial \[g(x) = x^3 +ax^2 + x+10\]has three distinct roots, and each root of $g(x)$ is also a root of the polynomial\[f(x) = x^4 +x^3 +bx^2 + 100x + c.\]What is $f(1)$?
$\phantom{hi}$
$\text{(A) } {-9009} \quad
\text{(B) } {-8008} \quad
\text{(C) } {-7007} \quad
\text{(D) } {-6006} \quad
\text{(E) } {-5005}$
devenware
2017-02-08 19:42:02
This problem is also Problem 23 from the 12A this year. We'll only solve the problem once.
This problem is also Problem 23 from the 12A this year. We'll only solve the problem once.
devenware
2017-02-08 19:42:15
The problem tells us that $g(x)$ has three distinct roots, so let's factor it as $g(x) = (x+q)(x+r)(x+s).$
The problem tells us that $g(x)$ has three distinct roots, so let's factor it as $g(x) = (x+q)(x+r)(x+s).$
devenware
2017-02-08 19:42:17
How can we factor $f(x)$?
How can we factor $f(x)$?
uptownmath
2017-02-08 19:43:06
(x+q)(x+r)(x+s)(x+t)
(x+q)(x+r)(x+s)(x+t)
warrenwangtennis
2017-02-08 19:43:06
g(x)(x-t)
g(x)(x-t)
Darth37
2017-02-08 19:43:06
(x + q)(x + r)(x + s)(x + u)
(x + q)(x + r)(x + s)(x + u)
challengetogo
2017-02-08 19:43:06
(x+q)(x+r)(x+s)(x+a)
(x+q)(x+r)(x+s)(x+a)
Wave-Particle
2017-02-08 19:43:06
$f(x) = (x+a)(x+q)(x+r)(x+s)$
$f(x) = (x+a)(x+q)(x+r)(x+s)$
Arieoreos
2017-02-08 19:43:06
(x+q)(x+r)(x+s)(x+t)
(x+q)(x+r)(x+s)(x+t)
tdeng
2017-02-08 19:43:06
(x+q)(x+r)(x+s)(x+t) for some t
(x+q)(x+r)(x+s)(x+t) for some t
raxu
2017-02-08 19:43:06
$(x+q)(x+r)(x+s)(x+t)$
$(x+q)(x+r)(x+s)(x+t)$
CaptainGeo
2017-02-08 19:43:06
(x+q)(x+r)(x+s)(x+m)
(x+q)(x+r)(x+s)(x+m)
SomethingNeutral
2017-02-08 19:43:06
or (x+q)(x+r)(x+s)(x+t)
or (x+q)(x+r)(x+s)(x+t)
brainiac1
2017-02-08 19:43:06
$(x+q)(x+r)(x+s)(x+t)$
$(x+q)(x+r)(x+s)(x+t)$
First
2017-02-08 19:43:06
(x+n)(x+q)(x+r)(x+s)
(x+n)(x+q)(x+r)(x+s)
devenware
2017-02-08 19:43:10
Since each of the roots of $g(x)$ is a root of $f(x)$, we have \[f(x) = (x+q)(x+r)(x+s)(x+t) = g(x)(x+t),\] for some real number $t.$
Since each of the roots of $g(x)$ is a root of $f(x)$, we have \[f(x) = (x+q)(x+r)(x+s)(x+t) = g(x)(x+t),\] for some real number $t.$
devenware
2017-02-08 19:43:11
That is,
That is,
devenware
2017-02-08 19:43:13
\[x^4+x^3+bx^2+100x+c=(x^3+ax^2+x+10)(x+t).\]
\[x^4+x^3+bx^2+100x+c=(x^3+ax^2+x+10)(x+t).\]
devenware
2017-02-08 19:43:17
Now what?
Now what?
warrenwangtennis
2017-02-08 19:43:51
expand and equate coefficients
expand and equate coefficients
Darth37
2017-02-08 19:43:51
Expand and equate like terms
Expand and equate like terms
skmc
2017-02-08 19:43:51
actually multiply through and match the coeficents?
actually multiply through and match the coeficents?
MathMan1234
2017-02-08 19:43:51
just write equations for the coefficients
just write equations for the coefficients
devenware
2017-02-08 19:43:56
Now we equate coefficients and proceed to rock.
Now we equate coefficients and proceed to rock.
devenware
2017-02-08 19:44:04
Now we equate coefficients and proceed to rock.
Now we equate coefficients and proceed to rock.
devenware
2017-02-08 19:44:06
What does the constant term tell us?
What does the constant term tell us?
AlisonH
2017-02-08 19:44:43
c=10t
c=10t
Bill9000
2017-02-08 19:44:43
$c=10t$
$c=10t$
amwmath
2017-02-08 19:44:43
$c=10t$
$c=10t$
mathchampion1
2017-02-08 19:44:43
10t=c
10t=c
summitwei
2017-02-08 19:44:43
10t = c
10t = c
jybsmartguy2
2017-02-08 19:44:43
c=10t
c=10t
CaptainGeo
2017-02-08 19:44:43
10t=c
10t=c
PenguinJoe
2017-02-08 19:44:43
awesome1028717
2017-02-08 19:44:43
c=10t
c=10t
devenware
2017-02-08 19:44:47
The constant term on the left is $c$ and the constant term on the right is $10t$. Thus \[c=10t.\]
The constant term on the left is $c$ and the constant term on the right is $10t$. Thus \[c=10t.\]
devenware
2017-02-08 19:44:59
What does the linear term tell us?
What does the linear term tell us?
Goliath
2017-02-08 19:45:38
t=90
t=90
Darth37
2017-02-08 19:45:38
t = 90
t = 90
jybsmartguy2
2017-02-08 19:45:38
100=t+10
100=t+10
summitwei
2017-02-08 19:45:38
10+t=100
10+t=100
FATRaichu
2017-02-08 19:45:38
10+t=100
10+t=100
tdeng
2017-02-08 19:45:38
100 = 10+t
100 = 10+t
abvenkgoo
2017-02-08 19:45:38
$10+t = 100$
$10+t = 100$
spotty2012
2017-02-08 19:45:38
t=90
t=90
wiler5002
2017-02-08 19:45:38
$10+t=100$
$10+t=100$
uptownmath
2017-02-08 19:45:38
10+t=100 or t=90
10+t=100 or t=90
mohanxue612
2017-02-08 19:45:38
t=90
t=90
ilovemath04
2017-02-08 19:45:38
100 = t+10
100 = t+10
abvenkgoo
2017-02-08 19:45:38
$t = 90$
$t = 90$
brainiac1
2017-02-08 19:45:38
$10+t=100$
$10+t=100$
Goliath
2017-02-08 19:45:38
10+t=100, t=90
10+t=100, t=90
math101010
2017-02-08 19:45:38
t=90
t=90
ScienceSpirit
2017-02-08 19:45:38
100=10+t so t=90
100=10+t so t=90
devenware
2017-02-08 19:45:43
The linear term on the left is $100$ and the linear term on the right is $10+t$, so \[100=10+t.\]
The linear term on the left is $100$ and the linear term on the right is $10+t$, so \[100=10+t.\]
devenware
2017-02-08 19:46:00
Quadratic?
Quadratic?
mathfun5
2017-02-08 19:46:40
at + 1 = b
at + 1 = b
Darth37
2017-02-08 19:46:40
at + 1 = b
at + 1 = b
tdeng
2017-02-08 19:46:40
b = at+1
b = at+1
antmath2520
2017-02-08 19:46:40
$b = at + 1$
$b = at + 1$
PenguinJoe
2017-02-08 19:46:40
at + 1 = b
at + 1 = b
FATRaichu
2017-02-08 19:46:40
at+1=b
at+1=b
brainiac1
2017-02-08 19:46:40
$b=at+1$
$b=at+1$
AlisonH
2017-02-08 19:46:40
at+1=b
at+1=b
devenware
2017-02-08 19:46:43
Equating the quadratic terms gives us \[b=at+1.\]
Equating the quadratic terms gives us \[b=at+1.\]
devenware
2017-02-08 19:46:44
Cubic?
Cubic?
wiler5002
2017-02-08 19:47:23
$a+t=1$
$a+t=1$
CrystalEye
2017-02-08 19:47:23
1=a+90
1=a+90
uptownmath
2017-02-08 19:47:23
1=t+a or 1=90+a or a=-89
1=t+a or 1=90+a or a=-89
SomethingNeutral
2017-02-08 19:47:23
t + a = 1
t + a = 1
Smoothfang
2017-02-08 19:47:23
a+t=1
a+t=1
amwmath
2017-02-08 19:47:23
$1=t+a$
$1=t+a$
quanhui868
2017-02-08 19:47:23
a+t=1
a+t=1
devenware
2017-02-08 19:47:28
Equating the cubic terms gives us \[1=a+t.\]
Equating the cubic terms gives us \[1=a+t.\]
devenware
2017-02-08 19:47:41
So we get the system
So we get the system
devenware
2017-02-08 19:47:42
\begin{align*}
c&=10t\\
100&=10+t\\
b&=at+1\\
1&=a+t\\
\end{align*}
\begin{align*}
c&=10t\\
100&=10+t\\
b&=at+1\\
1&=a+t\\
\end{align*}
oriduck.nk
2017-02-08 19:47:55
t=90
t=90
mathchampion1
2017-02-08 19:47:55
t=90 and find a
t=90 and find a
Bill9000
2017-02-08 19:48:04
We already know $t=90$
We already know $t=90$
spotty2012
2017-02-08 19:48:04
but we know t=90
but we know t=90
devenware
2017-02-08 19:48:07
The second equation tells us that $t=90$.
The second equation tells us that $t=90$.
Goliath
2017-02-08 19:48:17
a=-89
a=-89
CaptainGeo
2017-02-08 19:48:17
a=-89
a=-89
PenguinJoe
2017-02-08 19:48:17
a = -89
a = -89
LegoLdr
2017-02-08 19:48:17
a=-89
a=-89
Smoothfang
2017-02-08 19:48:17
a=-89
a=-89
devenware
2017-02-08 19:48:21
The fourth equation tells us that $1=a+90$. Therefore $a=-89$.
The fourth equation tells us that $1=a+90$. Therefore $a=-89$.
devenware
2017-02-08 19:49:02
We could figure out $b$ and $c$, too, but do we need to?
We could figure out $b$ and $c$, too, but do we need to?
mathforfun2016
2017-02-08 19:49:40
no
no
uptownmath
2017-02-08 19:49:40
No.
No.
EpicRuler101
2017-02-08 19:49:40
no
no
SomethingNeutral
2017-02-08 19:49:40
no
no
GeronimoStilton
2017-02-08 19:49:40
No.
No.
acegikmoqsuwy2000
2017-02-08 19:49:40
nope, just use the factored form
nope, just use the factored form
antmath2520
2017-02-08 19:49:40
No, use f(x) = g(x) * (x + t)!
No, use f(x) = g(x) * (x + t)!
amwmath
2017-02-08 19:49:40
$f(1)=g(1)(t+1)$\
$f(1)=g(1)(t+1)$\
mathfun5
2017-02-08 19:49:40
We can do instead f(1) = g(1)*(1+t)
We can do instead f(1) = g(1)*(1+t)
Benjy450
2017-02-08 19:49:40
no just solve for g(1) then multiply by 91
no just solve for g(1) then multiply by 91
devenware
2017-02-08 19:49:45
We want to find $f(1)$, which we also know is $g(1)(1+t)$. What does that equal?
We want to find $f(1)$, which we also know is $g(1)(1+t)$. What does that equal?
SomethingNeutral
2017-02-08 19:50:27
f(1) = (1-89+1+10)(1+90) = (-77)(91) = -7007 C
f(1) = (1-89+1+10)(1+90) = (-77)(91) = -7007 C
mathfun5
2017-02-08 19:50:27
-7007
-7007
quanhui868
2017-02-08 19:50:27
-7007,C
-7007,C
Darth37
2017-02-08 19:50:27
-7007
-7007
quartzgirl
2017-02-08 19:50:27
-7007
-7007
uptownmath
2017-02-08 19:50:27
-7007
-7007
CrystalEye
2017-02-08 19:50:27
-7007
-7007
acegikmoqsuwy2000
2017-02-08 19:50:27
$91(-77)=-7007$
$91(-77)=-7007$
XxkalleknightxXlelxd
2017-02-08 19:50:27
-7007
-7007
devenware
2017-02-08 19:50:31
\begin{align*}
f(1)&=g(1)(1+t)\\&=(1+a+1+10)(1+t)\\&=(12-89)(91)=-77\cdot91\\&=-7\cdot11\cdot91=-7\cdot1001\\&=\boxed{-7007}.
\end{align*}
\begin{align*}
f(1)&=g(1)(1+t)\\&=(1+a+1+10)(1+t)\\&=(12-89)(91)=-77\cdot91\\&=-7\cdot11\cdot91=-7\cdot1001\\&=\boxed{-7007}.
\end{align*}
devenware
2017-02-08 19:50:32
The answer is (C).
The answer is (C).
devenware
2017-02-08 19:50:55
(But I will admit, when I solved this problem, I didn't see that -- I found $b$ and $c$ and added everything up!)
(But I will admit, when I solved this problem, I didn't see that -- I found $b$ and $c$ and added everything up!
)
SomethingNeutral
2017-02-08 19:51:24
same here
same here
Goliath
2017-02-08 19:51:24
lol, same, i added it up
lol, same, i added it up
oriduck.nk
2017-02-08 19:51:24
me too
me too
goodball
2017-02-08 19:51:24
WOw!!!
WOw!!!
devenware
2017-02-08 19:51:34
Okay.
Okay.
devenware
2017-02-08 19:51:36
NEXT PROBLEM.
NEXT PROBLEM.
devenware
2017-02-08 19:52:17
25. How many integers between 100 and 999, inclusive, have the property that some permutation of its digits is a multiple of 11 between 100 and 999? For example, both 121 and 211 have this property.
$\phantom{hi}$
$\text{(A) } 226 \quad
\text{(B) } 243 \quad
\text{(C) } 270 \quad
\text{(D) } 469 \quad
\text{(E) } 486$
25. How many integers between 100 and 999, inclusive, have the property that some permutation of its digits is a multiple of 11 between 100 and 999? For example, both 121 and 211 have this property.
$\phantom{hi}$
$\text{(A) } 226 \quad
\text{(B) } 243 \quad
\text{(C) } 270 \quad
\text{(D) } 469 \quad
\text{(E) } 486$
devenware
2017-02-08 19:52:43
What's the divisibility rule for 11 again?
What's the divisibility rule for 11 again?
amc8nov
2017-02-08 19:53:44
alternating digits sum to multiples of11
alternating digits sum to multiples of11
summitwei
2017-02-08 19:53:44
for three-digit number abc, a-b+c
for three-digit number abc, a-b+c
kevindk
2017-02-08 19:53:44
even digits sum - odd digit sum is multiple of 11
even digits sum - odd digit sum is multiple of 11
Darth37
2017-02-08 19:53:44
alternate digit sum
alternate digit sum
FATRaichu
2017-02-08 19:53:44
alternating digits
alternating digits
BIGBUBBLE
2017-02-08 19:53:44
alternate digit sums differ by a multiple of 11
alternate digit sums differ by a multiple of 11
GeneralCobra19
2017-02-08 19:53:44
the alternating digits' sum have a difference that mod 11 is 0
the alternating digits' sum have a difference that mod 11 is 0
adyj
2017-02-08 19:53:44
Alternate adding and subtracting the digits. If result is divisible by 11, the number is.
Alternate adding and subtracting the digits. If result is divisible by 11, the number is.
tdeng
2017-02-08 19:53:44
Alternating sum is divisible by 11
Alternating sum is divisible by 11
garretth
2017-02-08 19:53:44
a+c = b mod 11
a+c = b mod 11
RedHawk
2017-02-08 19:53:44
alternating sum is congruent mod 11
alternating sum is congruent mod 11
vbcnxm
2017-02-08 19:53:44
alternating numbers add/subtract div by 11?!
alternating numbers add/subtract div by 11?!
devenware
2017-02-08 19:53:48
Since $10\equiv-1\pmod{11}$, a three-digit number $\overline{abc}$ is a multiple of 11 if \[100a+10b+c\equiv(-1)^2a+(-1)b+c\equiv a-b+c\equiv0\pmod{11}.\]
Since $10\equiv-1\pmod{11}$, a three-digit number $\overline{abc}$ is a multiple of 11 if \[100a+10b+c\equiv(-1)^2a+(-1)b+c\equiv a-b+c\equiv0\pmod{11}.\]
devenware
2017-02-08 19:53:51
That is, we test for divisibility by 11 by checking whether $a-b+c$ is a multiple of 11.
That is, we test for divisibility by 11 by checking whether $a-b+c$ is a multiple of 11.
devenware
2017-02-08 19:53:57
Furthermore, since all these are digits, we can only get $a+c=b$ or $a+c=b+11$.
Furthermore, since all these are digits, we can only get $a+c=b$ or $a+c=b+11$.
devenware
2017-02-08 19:54:35
So let's always use $b$ to be the number that's "subtracted", so we're looking for three types of numbers
\begin{align*}
\overline{abc}\\
\overline{bac}\\
\overline{acb}\\
\end{align*}
where $a+c\equiv b\pmod{11}$.
So let's always use $b$ to be the number that's "subtracted", so we're looking for three types of numbers
\begin{align*}
\overline{abc}\\
\overline{bac}\\
\overline{acb}\\
\end{align*}
where $a+c\equiv b\pmod{11}$.
devenware
2017-02-08 19:54:40
These will be our three types of numbers. Let's try listing out some examples and see if we find anything interesting.
These will be our three types of numbers. Let's try listing out some examples and see if we find anything interesting.
devenware
2017-02-08 19:54:42
$$\begin{array}{ |c|c|c|c| }
\hline
\overline{abc} & \overline{bac} & \overline{acb} \\
\hline
737 & 377 & 773 \\
484 & 844 & 448 \\
770 & 770 & 707 \\
\hline
\end{array}$$
$$\begin{array}{ |c|c|c|c| }
\hline
\overline{abc} & \overline{bac} & \overline{acb} \\
\hline
737 & 377 & 773 \\
484 & 844 & 448 \\
770 & 770 & 707 \\
\hline
\end{array}$$
devenware
2017-02-08 19:54:45
What's going on in that last row?
What's going on in that last row?
proztheboss
2017-02-08 19:55:31
first two numbers are the same
first two numbers are the same
chessapple9
2017-02-08 19:55:31
They're the same.
They're the same.
mathchampion1
2017-02-08 19:55:31
all are 770
all are 770
ilovemath04
2017-02-08 19:55:31
two are the same
two are the same
devenware
2017-02-08 19:55:42
If one of the digits of a number is 0, it might fall into two different categories. That's something to be careful about.
If one of the digits of a number is 0, it might fall into two different categories. That's something to be careful about.
devenware
2017-02-08 19:55:45
When exactly can a number be in two of these categories?
When exactly can a number be in two of these categories?
EpicRuler101
2017-02-08 19:56:47
it only counts as 1 permutation if there is a 0 in the number
it only counts as 1 permutation if there is a 0 in the number
jeffshen
2017-02-08 19:56:47
multiple of 10
multiple of 10
antmath2520
2017-02-08 19:56:47
Multiple of 110
Multiple of 110
Goliath
2017-02-08 19:56:47
when it has 0 as a digit
when it has 0 as a digit
SomethingNeutral
2017-02-08 19:56:47
when it is n-n-0
when it is n-n-0
Brisingrfire
2017-02-08 19:56:47
when it has repeating digits and a zero
when it has repeating digits and a zero
PenguinJoe
2017-02-08 19:56:47
multiple of 10 and 11?
multiple of 10 and 11?
devenware
2017-02-08 19:56:54
If $\overline{xyz}$ is of the form $\overline{abc}$ and also of the form $\overline{bac}$ then
\begin{align*}
x-y+z&\equiv0\pmod{11}\\
y-x+z&\equiv0\pmod{11}\\
\end{align*}
which when added tells us that one of the digits is zero. More generally, if a number fits more than one of these patterns then one of the digits has to be zero and the other 2 digits are equal.
If $\overline{xyz}$ is of the form $\overline{abc}$ and also of the form $\overline{bac}$ then
\begin{align*}
x-y+z&\equiv0\pmod{11}\\
y-x+z&\equiv0\pmod{11}\\
\end{align*}
which when added tells us that one of the digits is zero. More generally, if a number fits more than one of these patterns then one of the digits has to be zero and the other 2 digits are equal.
devenware
2017-02-08 19:57:08
Also, if all digits are nonzero then a number can only be in one of these forms:
770 has the form $\overline{abc}$ and $\overline{bac}$
707 has the form $\overline{acb}$ and $\overline{bac}$
Also, if all digits are nonzero then a number can only be in one of these forms:
770 has the form $\overline{abc}$ and $\overline{bac}$
707 has the form $\overline{acb}$ and $\overline{bac}$
devenware
2017-02-08 19:57:18
That's something we'll have to worry about along the way.
That's something we'll have to worry about along the way.
devenware
2017-02-08 19:57:19
What else is bad about the digit 0 here?
What else is bad about the digit 0 here?
v4913
2017-02-08 19:58:01
It cannot be the first digit.
It cannot be the first digit.
AlisonH
2017-02-08 19:58:01
it can't be the first digit
it can't be the first digit
uptownmath
2017-02-08 19:58:01
you can't start with it
you can't start with it
geomstar_1234
2017-02-08 19:58:01
it can't be 077
it can't be 077
adyj
2017-02-08 19:58:01
Cant be in the hundreds place
Cant be in the hundreds place
spotty2012
2017-02-08 19:58:01
it can't be in the first position
it can't be in the first position
bobjoe123
2017-02-08 19:58:01
Cant start with 0
Cant start with 0
sgadekar
2017-02-08 19:58:01
cant start the number
cant start the number
mathmagician
2017-02-08 19:58:01
077 is not a 3 digit number
077 is not a 3 digit number
mepehe888
2017-02-08 19:58:01
it can't go in the hundredths place
it can't go in the hundredths place
amwmath
2017-02-08 19:58:01
Can't be first digit
Can't be first digit
skmc
2017-02-08 19:58:01
it can't go as the first digit of the number
it can't go as the first digit of the number
mathfun5
2017-02-08 19:58:01
cant be first
cant be first
MSTang
2017-02-08 19:58:01
cant start a #
cant start a #
mathmagician
2017-02-08 19:58:01
cab doesn't exist
cab doesn't exist
Kirby703
2017-02-08 19:58:01
it can't start off numbers (either the ones we start with or the multiples of 11)
it can't start off numbers (either the ones we start with or the multiples of 11)
fields123
2017-02-08 19:58:01
it can't be the first digit
it can't be the first digit
ScienceSpirit
2017-02-08 19:58:01
can't have 0bc
can't have 0bc
challengetogo
2017-02-08 19:58:01
it can't be placed as the hundreds digit
it can't be placed as the hundreds digit
PenguinJoe
2017-02-08 19:58:01
it only has two permutations
it only has two permutations
Bill9000
2017-02-08 19:58:01
It can't be in the hundreds place!
It can't be in the hundreds place!
Brainiac2
2017-02-08 19:58:01
It can't be in front.
It can't be in front.
evanhlu
2017-02-08 19:58:01
only 2 possibilities
only 2 possibilities
devenware
2017-02-08 19:58:36
0 can't be the first digit since 077 is not a 3-digit number.
0 can't be the first digit since 077 is not a 3-digit number.
devenware
2017-02-08 19:58:51
So let's deal with the digit 0 first. Our strategy will be to find all the numbers that fit our patterns with one of the digits zero first. After that we'll count the numbers that fit our patterns using only nonzero digits.
So let's deal with the digit 0 first. Our strategy will be to find all the numbers that fit our patterns with one of the digits zero first. After that we'll count the numbers that fit our patterns using only nonzero digits.
devenware
2017-02-08 19:59:11
Are there any viable numbers with 2 digits equal to 0?
Are there any viable numbers with 2 digits equal to 0?
uptownmath
2017-02-08 19:59:37
no
no
XxkalleknightxXlelxd
2017-02-08 19:59:37
no
no
proztheboss
2017-02-08 19:59:37
no
no
CaptainGeo
2017-02-08 19:59:37
no
no
mathfun5
2017-02-08 19:59:37
no
no
goodball
2017-02-08 19:59:37
No.
No.
EpicRuler101
2017-02-08 19:59:37
no
no
evanhlu
2017-02-08 19:59:37
no
no
vlwilliam
2017-02-08 19:59:37
no
no
spotty2012
2017-02-08 19:59:37
no
no
antmath2520
2017-02-08 19:59:37
No
No
amc8nov
2017-02-08 19:59:37
NO
NO
edus1
2017-02-08 19:59:37
No
No
XxkalleknightxXlelxd
2017-02-08 19:59:37
none possible
none possible
3_14152
2017-02-08 19:59:37
no
no
adihaya
2017-02-08 19:59:37
no
no
devenware
2017-02-08 19:59:54
No. We can't solve $a+c\equiv b\pmod{11}$ when 2 of the digits are 0 (unless all three digits are zero).
No. We can't solve $a+c\equiv b\pmod{11}$ when 2 of the digits are 0 (unless all three digits are zero).
devenware
2017-02-08 19:59:56
Now we tackle the $\overline{abc}$ case with $b=0$.
Now we tackle the $\overline{abc}$ case with $b=0$.
devenware
2017-02-08 20:00:14
How many 3-digit numbers are there of the form $\overline {a0c}$ with $a+c=0$ or $a+c=11$?
How many 3-digit numbers are there of the form $\overline {a0c}$ with $a+c=0$ or $a+c=11$?
brainiac1
2017-02-08 20:00:49
8
8
uptownmath
2017-02-08 20:00:49
8
8
Kirby703
2017-02-08 20:00:49
8
8
PenguinJoe
2017-02-08 20:00:49
8
8
SomethingNeutral
2017-02-08 20:00:49
8
8
CrystalEye
2017-02-08 20:00:49
8
8
3_14152
2017-02-08 20:00:49
8
8
v4913
2017-02-08 20:00:49
8
8
mathfun5
2017-02-08 20:00:49
8
8
LearnAMC
2017-02-08 20:00:49
8
8
FATRaichu
2017-02-08 20:00:49
8
8
devenware
2017-02-08 20:00:59
Well $a+c=0$ is impossible, but for every choice of $a$ from 2 to 9, there's exactly one viable digit $c=11-a$. There are 8 of these.
Well $a+c=0$ is impossible, but for every choice of $a$ from 2 to 9, there's exactly one viable digit $c=11-a$. There are 8 of these.
XxkalleknightxXlelxd
2017-02-08 20:01:05
902 803
902 803
antmath2520
2017-02-08 20:01:05
8 (209, 308, 407, 506, 605, 704, 803, 902)
8 (209, 308, 407, 506, 605, 704, 803, 902)
mathfun5
2017-02-08 20:01:05
209 308 ... 902
209 308 ... 902
devenware
2017-02-08 20:01:10
There they are listed out.
There they are listed out.
devenware
2017-02-08 20:01:24
What about the $\overline{acb}$ case with $b=0$?
What about the $\overline{acb}$ case with $b=0$?
evanhlu
2017-02-08 20:02:20
8
8
PenguinJoe
2017-02-08 20:02:20
still 8
still 8
JoyceX
2017-02-08 20:02:20
also 8
also 8
XxkalleknightxXlelxd
2017-02-08 20:02:20
8
8
AlisonH
2017-02-08 20:02:20
8
8
LegoLdr
2017-02-08 20:02:20
same thing
same thing
jeffshen
2017-02-08 20:02:20
8 cases
8 cases
ilovemath04
2017-02-08 20:02:20
same
same
devenware
2017-02-08 20:02:24
This is the same. We're looking for 3-digit numbers of the form $\overline {ac0}$ with $a+c=0$ or $a+c=11$, which is in a different order but the same sets of numbers.
This is the same. We're looking for 3-digit numbers of the form $\overline {ac0}$ with $a+c=0$ or $a+c=11$, which is in a different order but the same sets of numbers.
devenware
2017-02-08 20:02:47
The answer is the same, 8. (290 is an example).
The answer is the same, 8. (290 is an example).
devenware
2017-02-08 20:02:55
Alright, we've handled $b=0$. Now we let $a$ or $c$ be zero.
Alright, we've handled $b=0$. Now we let $a$ or $c$ be zero.
devenware
2017-02-08 20:03:01
Let's look at the case, $\overline {ab0}$ with $a+0=b$ or $a+0=b+11$. How many numbers have this form?
Let's look at the case, $\overline {ab0}$ with $a+0=b$ or $a+0=b+11$. How many numbers have this form?
bao2022
2017-02-08 20:03:35
9.
9.
FATRaichu
2017-02-08 20:03:35
9
9
amwmath
2017-02-08 20:03:35
9.
9.
First
2017-02-08 20:03:35
9?
9?
XxkalleknightxXlelxd
2017-02-08 20:03:35
9
9
chessapple9
2017-02-08 20:03:35
There are 9 numbers.
There are 9 numbers.
ilovemath04
2017-02-08 20:03:35
9
9
fields123
2017-02-08 20:03:35
9
9
Bill9000
2017-02-08 20:03:35
$9$
$9$
goodball
2017-02-08 20:03:35
9
9
devenware
2017-02-08 20:03:40
Now $b+11$ is impossible, but for any $a$ from 1 to 9, we get a viable $b$ (these are the numbers 110, 220, etc.). There are 9 of these.
Now $b+11$ is impossible, but for any $a$ from 1 to 9, we get a viable $b$ (these are the numbers 110, 220, etc.). There are 9 of these.
devenware
2017-02-08 20:03:52
(Those are the ones a lot of you thought I meant earlier. Gotta work on my clarity.)
(Those are the ones a lot of you thought I meant earlier. Gotta work on my clarity.)
devenware
2017-02-08 20:04:03
How about $\overline {a0b}$ with $a+0=b$ or $a+0=b+11$?
How about $\overline {a0b}$ with $a+0=b$ or $a+0=b+11$?
ItzVineeth
2017-02-08 20:04:37
9 miscount
9 miscount
XxkalleknightxXlelxd
2017-02-08 20:04:37
same, 9
same, 9
bao2022
2017-02-08 20:04:37
9
9
LearnAMC
2017-02-08 20:04:37
9
9
bao2022
2017-02-08 20:04:37
9 still
9 still
LegoLdr
2017-02-08 20:04:37
also 9
also 9
ilovemath04
2017-02-08 20:04:37
same thing
same thing
brainiac1
2017-02-08 20:04:37
9 still
9 still
Bill9000
2017-02-08 20:04:37
$9$
$9$
SomethingNeutral
2017-02-08 20:04:37
9
9
Bill9000
2017-02-08 20:04:37
9
9
goodball
2017-02-08 20:04:37
9 as well.
9 as well.
amwmath
2017-02-08 20:04:37
9 again
9 again
evanhlu
2017-02-08 20:04:37
9??
9??
mathfun5
2017-02-08 20:04:37
9
9
dzhou100
2017-02-08 20:04:37
9
9
FATRaichu
2017-02-08 20:04:37
same
same
devenware
2017-02-08 20:04:41
There are also 9 of these (101, 202, etc.).
There are also 9 of these (101, 202, etc.).
devenware
2017-02-08 20:04:51
How about $\overline{b0c}$ and $\overline{ba0}$?
How about $\overline{b0c}$ and $\overline{ba0}$?
ItzVineeth
2017-02-08 20:05:14
9 again.....
9 again.....
mathchampion1
2017-02-08 20:05:14
9
9
jeffshen
2017-02-08 20:05:14
9
9
bao2022
2017-02-08 20:05:14
9 each
9 each
ilovemath04
2017-02-08 20:05:14
same again
same again
SomethingNeutral
2017-02-08 20:05:14
duplicates
duplicates
Bill9000
2017-02-08 20:05:14
SAME EXACT THING
SAME EXACT THING
evanhlu
2017-02-08 20:05:14
18
18
Kirby703
2017-02-08 20:05:14
9 for each?
9 for each?
LearnAMC
2017-02-08 20:05:20
18 together
18 together
devenware
2017-02-08 20:05:23
In these cases, we have to have $b=c$ or $b=a$. These are the numbers like 101, 202, etc., along with 110, 220, etc. That's 18 more, right?
In these cases, we have to have $b=c$ or $b=a$. These are the numbers like 101, 202, etc., along with 110, 220, etc. That's 18 more, right?
uptownmath
2017-02-08 20:06:09
No, they are duplicates!
No, they are duplicates!
SomethingNeutral
2017-02-08 20:06:09
no already counted them
no already counted them
goodball
2017-02-08 20:06:09
No, they repeat!!
No, they repeat!!
mt492
2017-02-08 20:06:09
no, duplicates
no, duplicates
mathfun5
2017-02-08 20:06:09
we already counted those!
we already counted those!
goodball
2017-02-08 20:06:09
No, repetitions.
No, repetitions.
brainiac1
2017-02-08 20:06:14
110 was repeated?
110 was repeated?
Kirby703
2017-02-08 20:06:14
wait a minute we counted these 18 didn't we :o
wait a minute we counted these 18 didn't we :o
devenware
2017-02-08 20:06:17
No! We've already counted these! The $\overline{b0c}$ numbers are also $\overline{a0b}$ numbers and the $\overline{ba0}$ numbers are also the $\overline{ab0}$ numbers. Weird, right?
No! We've already counted these! The $\overline{b0c}$ numbers are also $\overline{a0b}$ numbers and the $\overline{ba0}$ numbers are also the $\overline{ab0}$ numbers. Weird, right?
Bill9000
2017-02-08 20:06:31
No, not really.
No, not really.
devenware
2017-02-08 20:06:35
Not so weird. We noticed that there were going to be some overlaps when one of the digits is zero. We also pointed out that this can't happen in our casework once we move on to nonzero digits.
Not so weird. We noticed that there were going to be some overlaps when one of the digits is zero. We also pointed out that this can't happen in our casework once we move on to nonzero digits.
devenware
2017-02-08 20:06:48
Okay, that's it for the zeroes. How many total numbers have a digit of 0?
Okay, that's it for the zeroes. How many total numbers have a digit of 0?
mathfun5
2017-02-08 20:07:26
34
34
conniejeon1
2017-02-08 20:07:26
34
34
goodball
2017-02-08 20:07:26
34
34
uptownmath
2017-02-08 20:07:26
34
34
LegoLdr
2017-02-08 20:07:26
34?
34?
jeffshen
2017-02-08 20:07:26
34
34
goodball
2017-02-08 20:07:26
34 numbers
34 numbers
CrystalEye
2017-02-08 20:07:26
34
34
goodball
2017-02-08 20:07:26
There are 34 numbers.
There are 34 numbers.
devenware
2017-02-08 20:07:32
There are $8+8+9+9=34$ total numbers with a digit of zero.
There are $8+8+9+9=34$ total numbers with a digit of zero.
devenware
2017-02-08 20:07:46
Now we're restricted to thinking about numbers where none of the digits are 0.
Now we're restricted to thinking about numbers where none of the digits are 0.
devenware
2017-02-08 20:07:47
How could we construct such a number?
How could we construct such a number?
proztheboss
2017-02-08 20:08:31
go one digit at a time
go one digit at a time
goodball
2017-02-08 20:08:31
Casework?
Casework?
devenware
2017-02-08 20:08:38
Casework.. Yum.
Casework.. Yum.
devenware
2017-02-08 20:08:51
What should we use for cases?
What should we use for cases?
bobjoe123
2017-02-08 20:09:39
Middle Number
Middle Number
mathmagician
2017-02-08 20:09:39
the center number?
the center number?
wiler5002
2017-02-08 20:09:39
middle digit
middle digit
Ani10
2017-02-08 20:09:39
the minus plus rule
the minus plus rule
devenware
2017-02-08 20:09:44
Casework on the form of the number again: We will pick which of the three places to put $b$, the subtracted digit, then choose the three digits.
Casework on the form of the number again: We will pick which of the three places to put $b$, the subtracted digit, then choose the three digits.
devenware
2017-02-08 20:09:49
There's a dependence on the digits, though. If you pick 2, the third is mostly determined for you. Which 2 digits should we pick to make things possibly go most smoothly?
There's a dependence on the digits, though. If you pick 2, the third is mostly determined for you. Which 2 digits should we pick to make things possibly go most smoothly?
devenware
2017-02-08 20:10:13
(Remember, we're always going to call the subtracted number in our 11's trick $b$.)
(Remember, we're always going to call the subtracted number in our 11's trick $b$.)
devenware
2017-02-08 20:10:51
I don't want you to pick actual numbers -- I'm saying which two of $a,b,$ and $c$ should we specify?
I don't want you to pick actual numbers -- I'm saying which two of $a,b,$ and $c$ should we specify?
uptownmath
2017-02-08 20:11:12
a and c
a and c
goodball
2017-02-08 20:11:12
digits a and c
digits a and c
goodball
2017-02-08 20:11:12
Use digits a and c
Use digits a and c
XxkalleknightxXlelxd
2017-02-08 20:11:12
a and c
a and c
v4913
2017-02-08 20:11:12
a and c?
a and c?
mathislife16
2017-02-08 20:11:12
a and c
a and c
conniejeon1
2017-02-08 20:11:12
a and c
a and c
FATRaichu
2017-02-08 20:11:12
a,c
a,c
proztheboss
2017-02-08 20:11:12
a and c
a and c
goodball
2017-02-08 20:11:12
a and c
a and c
bao2022
2017-02-08 20:11:12
a and c
a and c
devenware
2017-02-08 20:11:17
Let's pick $a$ and $c$. That is, we pick two of the three places of the number and drop nonzero digits in those places. Then we hope that their sum, modulo 11, is a nonzero digit that we can put in the third place.
Let's pick $a$ and $c$. That is, we pick two of the three places of the number and drop nonzero digits in those places. Then we hope that their sum, modulo 11, is a nonzero digit that we can put in the third place.
devenware
2017-02-08 20:11:21
It's most natural to start with $b$ in the middle, so let's count the numbers of the form $\overline{abc}$ with all three digits nonzero and $a+c\equiv b\pmod{11}$.
It's most natural to start with $b$ in the middle, so let's count the numbers of the form $\overline{abc}$ with all three digits nonzero and $a+c\equiv b\pmod{11}$.
devenware
2017-02-08 20:11:22
For example, if we let $\overline{abc}=\overline{1b5}$, what are the choices for $b$?
For example, if we let $\overline{abc}=\overline{1b5}$, what are the choices for $b$?
SomethingNeutral
2017-02-08 20:12:10
6
6
mathmagician
2017-02-08 20:12:10
6
6
Brisingrfire
2017-02-08 20:12:10
6?
6?
mathfun5
2017-02-08 20:12:10
6
6
smishra
2017-02-08 20:12:10
6
6
AlisonH
2017-02-08 20:12:10
6
6
First
2017-02-08 20:12:10
6?
6?
shubhmit
2017-02-08 20:12:10
6
6
ilovemath04
2017-02-08 20:12:10
only 6
only 6
EpicRuler101
2017-02-08 20:12:10
6
6
bao2022
2017-02-08 20:12:10
6.
6.
v4913
2017-02-08 20:12:10
It has to be 6
It has to be 6
devenware
2017-02-08 20:12:19
The only choice for $b$ is 6.
The only choice for $b$ is 6.
devenware
2017-02-08 20:12:38
Because we need 1 - b + 5 = 0.
Because we need 1 - b + 5 = 0.
devenware
2017-02-08 20:12:41
Will there ever be two choices for $b$?
Will there ever be two choices for $b$?
GeneralCobra19
2017-02-08 20:13:16
No
No
ItzVineeth
2017-02-08 20:13:16
no
no
Kirby703
2017-02-08 20:13:16
no
no
goodball
2017-02-08 20:13:16
For each pair a and c we pick, b only has one choice.
For each pair a and c we pick, b only has one choice.
sahi
2017-02-08 20:13:16
no
no
LearnAMC
2017-02-08 20:13:16
no
no
it1023
2017-02-08 20:13:16
NO
NO
fields123
2017-02-08 20:13:16
NO
NO
devenware
2017-02-08 20:13:20
No. There is at most one nonzero digit $b$ that solves $b\equiv a+c\pmod{11}$ no matter what $a$ and $c$ are.
No. There is at most one nonzero digit $b$ that solves $b\equiv a+c\pmod{11}$ no matter what $a$ and $c$ are.
devenware
2017-02-08 20:13:38
Even if $a + c$ is more than 11 -- there's still only one choice.
Even if $a + c$ is more than 11 -- there's still only one choice.
devenware
2017-02-08 20:13:43
For example, if we had 7b7 what is b?
For example, if we had 7b7 what is b?
bobjoe123
2017-02-08 20:14:06
3
3
bao2022
2017-02-08 20:14:06
3
3
Bill9000
2017-02-08 20:14:06
$3$
$3$
smishra
2017-02-08 20:14:06
3
3
proztheboss
2017-02-08 20:14:06
3
3
mathfun5
2017-02-08 20:14:06
3
3
v4913
2017-02-08 20:14:06
3.
3.
jeffshen
2017-02-08 20:14:06
3
3
antmath2520
2017-02-08 20:14:06
3
3
mepehe888
2017-02-08 20:14:06
3
3
kevindk
2017-02-08 20:14:06
3
3
evanhlu
2017-02-08 20:14:06
3
3
geomstar_1234
2017-02-08 20:14:06
3
3
physangel
2017-02-08 20:14:08
3
3
amwmath
2017-02-08 20:14:08
$3$
$3$
devenware
2017-02-08 20:14:14
Yeah, $b$ would have to be 3 there.
Yeah, $b$ would have to be 3 there.
devenware
2017-02-08 20:14:32
So no matter what, we can't have two choices for $b$.
So no matter what, we can't have two choices for $b$.
devenware
2017-02-08 20:14:41
Okay what about 5b5, then what is b?
Okay what about 5b5, then what is b?
mathfun5
2017-02-08 20:15:19
none
none
jeffshen
2017-02-08 20:15:19
no
no
First
2017-02-08 20:15:19
can't happen
can't happen
uptownmath
2017-02-08 20:15:19
There is no possible choice.
There is no possible choice.
FATRaichu
2017-02-08 20:15:19
none
none
PenguinJoe
2017-02-08 20:15:19
doesn't work
doesn't work
smishra
2017-02-08 20:15:19
none exist
none exist
FATRaichu
2017-02-08 20:15:19
no solution
no solution
bobjoe123
2017-02-08 20:15:19
None??
None??
ilovemath04
2017-02-08 20:15:19
no value
no value
antmath2520
2017-02-08 20:15:19
None
None
Kirby703
2017-02-08 20:15:19
wait -1 and 10 aren't digits
wait -1 and 10 aren't digits
mt492
2017-02-08 20:15:19
not possible
not possible
brainiac1
2017-02-08 20:15:19
not possible
not possible
physangel
2017-02-08 20:15:19
10, so not possible
10, so not possible
AOPSmath68
2017-02-08 20:15:19
not possible
not possible
audbear
2017-02-08 20:15:19
None would work (it would be negative one...)
None would work (it would be negative one...)
LegoLdr
2017-02-08 20:15:19
it's not possible
it's not possible
devenware
2017-02-08 20:15:30
Yeah, we'd want to use b = 10 or b = -1, but those aren't digits...
Yeah, we'd want to use b = 10 or b = -1, but those aren't digits...
devenware
2017-02-08 20:15:40
So sometimes we don't have a possible $b$...
So sometimes we don't have a possible $b$...
devenware
2017-02-08 20:15:46
When do we not have a possible $b$?
When do we not have a possible $b$?
SomethingNeutral
2017-02-08 20:16:19
so if a+c = 10 no possible b.
so if a+c = 10 no possible b.
uptownmath
2017-02-08 20:16:19
When a+c=10
When a+c=10
Kirby703
2017-02-08 20:16:19
a+c=10
a+c=10
goodball
2017-02-08 20:16:19
a+c=10
a+c=10
bao2022
2017-02-08 20:16:19
When a + c = 10
When a + c = 10
jeffshen
2017-02-08 20:16:19
when outer digits add up to 10
when outer digits add up to 10
mathmagician
2017-02-08 20:16:19
when a+c = 10
when a+c = 10
evanhlu
2017-02-08 20:16:19
a+c=10
a+c=10
GeneralCobra19
2017-02-08 20:16:19
When a+c=10
When a+c=10
CrystalEye
2017-02-08 20:16:19
a+c=10
a+c=10
bobjoe123
2017-02-08 20:16:19
When the sum is 10(of a and c)
When the sum is 10(of a and c)
Brisingrfire
2017-02-08 20:16:19
when a+c=10
when a+c=10
shubhmit
2017-02-08 20:16:19
a+c =10
a+c =10
LegoLdr
2017-02-08 20:16:19
when a+c=10
when a+c=10
goodball
2017-02-08 20:16:19
When a+c=10
When a+c=10
mathfever
2017-02-08 20:16:19
when the first and last digits sum to 10?
when the first and last digits sum to 10?
summitwei
2017-02-08 20:16:21
a+c=10
a+c=10
adyj
2017-02-08 20:16:21
When the hundreds and ones add to 10, there is not a possible b.
When the hundreds and ones add to 10, there is not a possible b.
skywalker321
2017-02-08 20:16:24
when a+c = 10
when a+c = 10
devenware
2017-02-08 20:16:32
Right, if $a + c = 10$, there is no possible $b$.
Right, if $a + c = 10$, there is no possible $b$.
devenware
2017-02-08 20:16:35
But is that the only case?
But is that the only case?
mathfun5
2017-02-08 20:17:07
what about a+c = 11 then b cant b 0 cause we already counted that
what about a+c = 11 then b cant b 0 cause we already counted that
Ani10
2017-02-08 20:17:07
if a+c=11
if a+c=11
Archos
2017-02-08 20:17:07
11
11
SnakeYu
2017-02-08 20:17:07
what if it adds to 11?
what if it adds to 11?
mathislife16
2017-02-08 20:17:07
No, because we've eliminated 0s
No, because we've eliminated 0s
devenware
2017-02-08 20:17:18
Right, we can't have $a + c = 11$ either, because we're dealing with all nonzero digits now.
Right, we can't have $a + c = 11$ either, because we're dealing with all nonzero digits now.
devenware
2017-02-08 20:17:39
Okay, so we want to count how many pairs there are of $a,c$ that add up to something other than 10 or 11.
Okay, so we want to count how many pairs there are of $a,c$ that add up to something other than 10 or 11.
devenware
2017-02-08 20:17:45
Suggestions?
Suggestions?
SomethingNeutral
2017-02-08 20:18:06
complementary?
complementary?
Bill9000
2017-02-08 20:18:06
COMPLEMENTARY COUNTING!!!
COMPLEMENTARY COUNTING!!!
mathfever
2017-02-08 20:18:06
complementary counting?
complementary counting?
mt492
2017-02-08 20:18:06
complementary counting
complementary counting
RedHawk
2017-02-08 20:18:06
Complementary counting?
Complementary counting?
PenguinJoe
2017-02-08 20:18:06
complementary counting
complementary counting
antmath2520
2017-02-08 20:18:06
Complementary counting?
Complementary counting?
sgadekar
2017-02-08 20:18:14
complementary counting
complementary counting
devenware
2017-02-08 20:18:15
COMPLEMENTARY COUNTING!
COMPLEMENTARY COUNTING!
devenware
2017-02-08 20:18:21
Without the sum restriction, how many ordered pairs $a$ and $c$ can we pick?
Without the sum restriction, how many ordered pairs $a$ and $c$ can we pick?
conniejeon1
2017-02-08 20:18:51
81
81
SomethingNeutral
2017-02-08 20:18:51
9^2 = 81
9^2 = 81
mathmagician
2017-02-08 20:18:51
81
81
FATRaichu
2017-02-08 20:18:51
81
81
PenguinJoe
2017-02-08 20:18:51
81
81
adyj
2017-02-08 20:18:51
81
81
v4913
2017-02-08 20:18:51
81?
81?
jeffshen
2017-02-08 20:18:51
81
81
LearnAMC
2017-02-08 20:18:51
81
81
uptownmath
2017-02-08 20:18:51
81 cuz you can't have zeros!
81 cuz you can't have zeros!
Kirby703
2017-02-08 20:18:51
oh so that's what that's called... 81
oh so that's what that's called... 81
devenware
2017-02-08 20:18:54
There are 9 choices for $a$ and 9 for $c$, so 81 total pairs.
There are 9 choices for $a$ and 9 for $c$, so 81 total pairs.
devenware
2017-02-08 20:19:02
Okay, how many of those pairs add to 10?
Okay, how many of those pairs add to 10?
FATRaichu
2017-02-08 20:19:32
9
9
SomethingNeutral
2017-02-08 20:19:32
9.
9.
Kirby703
2017-02-08 20:19:32
9
9
jeffshen
2017-02-08 20:19:32
9
9
WW92030
2017-02-08 20:19:32
9
9
RedHawk
2017-02-08 20:19:32
9
9
uptownmath
2017-02-08 20:19:32
9
9
mathfun5
2017-02-08 20:19:32
9
9
ilovemath04
2017-02-08 20:19:32
9
9
v4913
2017-02-08 20:19:32
9
9
LearnAMC
2017-02-08 20:19:32
9
9
goodball
2017-02-08 20:19:32
9
9
mathfever
2017-02-08 20:19:32
9?
9?
physangel
2017-02-08 20:19:32
9
9
evanhlu
2017-02-08 20:19:32
9
9
mathfever
2017-02-08 20:19:32
9
9
LegoLdr
2017-02-08 20:19:32
9
9
Blocry
2017-02-08 20:19:32
9
9
devenware
2017-02-08 20:19:39
For any nonzero digit $a$, the value $c=10-a$ is also a nonzero digit, so there are 9 invalid pairs that add to 10.
For any nonzero digit $a$, the value $c=10-a$ is also a nonzero digit, so there are 9 invalid pairs that add to 10.
devenware
2017-02-08 20:19:40
How many pairs add to 11?
How many pairs add to 11?
fields123
2017-02-08 20:20:24
8
8
amburger66
2017-02-08 20:20:24
8
8
harry1234
2017-02-08 20:20:24
8
8
EpicRuler101
2017-02-08 20:20:24
8
8
brainiac1
2017-02-08 20:20:24
8
8
Brisingrfire
2017-02-08 20:20:24
8
8
sahi
2017-02-08 20:20:24
8
8
Blocry
2017-02-08 20:20:24
8
8
First
2017-02-08 20:20:24
8
8
AlisonH
2017-02-08 20:20:24
8
8
AnaT129
2017-02-08 20:20:24
8
8
PenguinJoe
2017-02-08 20:20:24
8
8
devenware
2017-02-08 20:20:31
For any digit $a>1$, the value $c=11-a$ is also a nonzero digit, so there are 8 invalid pairs that add to 11.
For any digit $a>1$, the value $c=11-a$ is also a nonzero digit, so there are 8 invalid pairs that add to 11.
devenware
2017-02-08 20:20:32
So how many valid pairs are there?
So how many valid pairs are there?
Kirby703
2017-02-08 20:21:24
64
64
v4913
2017-02-08 20:21:24
64
64
goodball
2017-02-08 20:21:24
64
64
harry1234
2017-02-08 20:21:24
64
64
mathfun5
2017-02-08 20:21:24
64
64
conniejeon1
2017-02-08 20:21:24
64
64
sgadekar
2017-02-08 20:21:24
64
64
Brisingrfire
2017-02-08 20:21:24
81-17=64
81-17=64
Blocry
2017-02-08 20:21:24
64
64
physangel
2017-02-08 20:21:24
64
64
adyj
2017-02-08 20:21:24
81-17=64
81-17=64
AlisonH
2017-02-08 20:21:24
64
64
mathmagician
2017-02-08 20:21:24
81-9-8 or 64
81-9-8 or 64
Bill9000
2017-02-08 20:21:24
64! (Not 64 factorial, just saying.)
64! (Not 64 factorial, just saying.)
bao2022
2017-02-08 20:21:24
64.
64.
mshanmugam
2017-02-08 20:21:24
64
64
Dolphin8
2017-02-08 20:21:24
64
64
devenware
2017-02-08 20:22:07
There are $81-9-8=64$ total valid pairs $a$ and $c$.
There are $81-9-8=64$ total valid pairs $a$ and $c$.
devenware
2017-02-08 20:22:44
What about numbers of the form $\overline{acb}$ with nonzero digits? How many of those are there?
What about numbers of the form $\overline{acb}$ with nonzero digits? How many of those are there?
mathfun5
2017-02-08 20:23:23
64 same!
64 same!
mathmagician
2017-02-08 20:23:23
also 64
also 64
FATRaichu
2017-02-08 20:23:23
64
64
goodball
2017-02-08 20:23:23
same.
same.
fields123
2017-02-08 20:23:23
same
same
conniejeon1
2017-02-08 20:23:23
64
64
goodball
2017-02-08 20:23:23
64 as well
64 as well
SomethingNeutral
2017-02-08 20:23:23
64.
64.
mathfun5
2017-02-08 20:23:23
a+c = b so same
a+c = b so same
Blocry
2017-02-08 20:23:23
same?
same?
mathislife16
2017-02-08 20:23:23
Still 64 right? It doesn't change, because there are no 0s
Still 64 right? It doesn't change, because there are no 0s
LearnAMC
2017-02-08 20:23:23
64
64
lemonpower
2017-02-08 20:23:23
same as abc?
same as abc?
devenware
2017-02-08 20:23:27
It's the same! We just take an $\overline{abc}$ number and swap the last two digits. There are 64 of these.
It's the same! We just take an $\overline{abc}$ number and swap the last two digits. There are 64 of these.
devenware
2017-02-08 20:23:28
What about $\overline{bac}$ with nonzero digits?
What about $\overline{bac}$ with nonzero digits?
devenware
2017-02-08 20:23:38
What about $\overline{bac}$ with nonzero digits?
What about $\overline{bac}$ with nonzero digits?
mathfun5
2017-02-08 20:24:01
64
64
mathmagician
2017-02-08 20:24:01
same thing! 64
same thing! 64
Kirby703
2017-02-08 20:24:01
another 64 for bac
another 64 for bac
SomethingNeutral
2017-02-08 20:24:01
64
64
mathchampion1
2017-02-08 20:24:01
64
64
mathchampion1
2017-02-08 20:24:01
same as well
same as well
ilovemath04
2017-02-08 20:24:01
SAME AGAIN
SAME AGAIN
jeffshen
2017-02-08 20:24:01
64
64
AnaT129
2017-02-08 20:24:03
still 64?
still 64?
LegoLdr
2017-02-08 20:24:04
same
same
devenware
2017-02-08 20:24:11
That's the same again. This time we take an $\overline{abc}$ number and swap the first two digits. There are still 64.
That's the same again. This time we take an $\overline{abc}$ number and swap the first two digits. There are still 64.
devenware
2017-02-08 20:24:14
How many total numbers do we have with all nonzero digits?
How many total numbers do we have with all nonzero digits?
bobjoe123
2017-02-08 20:25:03
192
192
Goliath
2017-02-08 20:25:03
192
192
CharlesHong
2017-02-08 20:25:03
192
192
shubhmit
2017-02-08 20:25:03
192
192
bao2022
2017-02-08 20:25:03
192
192
CrystalEye
2017-02-08 20:25:03
192
192
AOPSmath68
2017-02-08 20:25:03
192?
192?
devenware
2017-02-08 20:25:07
There are 3 places to put $b$ and for each of these 3 choices, there are 64 possible 3-digit numbers. We showed at the beginning that these cases don't overlap. Therefore there are $3\cdot64=192$ 3-digit numbers that are permutations of multiples of 11 with no digit zero.
There are 3 places to put $b$ and for each of these 3 choices, there are 64 possible 3-digit numbers. We showed at the beginning that these cases don't overlap. Therefore there are $3\cdot64=192$ 3-digit numbers that are permutations of multiples of 11 with no digit zero.
devenware
2017-02-08 20:25:08
And what is the final answer?
And what is the final answer?
guo_fang
2017-02-08 20:25:41
226
226
mathfun5
2017-02-08 20:25:41
226
226
FATRaichu
2017-02-08 20:25:41
226
226
bao2022
2017-02-08 20:25:41
(A)
(A)
LegoLdr
2017-02-08 20:25:41
A!!!
A!!!
goodball
2017-02-08 20:25:41
A
A
fields123
2017-02-08 20:25:41
226
226
jeffshen
2017-02-08 20:25:41
A
A
goodball
2017-02-08 20:25:41
226
226
mathfun5
2017-02-08 20:25:41
192+34 = 226
192+34 = 226
mathfun5
2017-02-08 20:25:41
226!
226!
Brisingrfire
2017-02-08 20:25:41
192+34= 226 (A!)
192+34= 226 (A!)
ilovemath04
2017-02-08 20:25:41
226, A
226, A
SnakeYu
2017-02-08 20:25:41
226, A
226, A
bao2022
2017-02-08 20:25:41
(A) 226
(A) 226
physangel
2017-02-08 20:25:41
A
A
CrystalEye
2017-02-08 20:25:41
226
226
CharlesHong
2017-02-08 20:25:41
226 A
226 A
First
2017-02-08 20:25:41
$\boxed{\textbf{A}}$!
$\boxed{\textbf{A}}$!
devenware
2017-02-08 20:25:44
Adding in the numbers with 0 as a digit we find $34+192=\boxed{226}$ total numbers that are permutations of multiples of 11. The answer is (A).
Adding in the numbers with 0 as a digit we find $34+192=\boxed{226}$ total numbers that are permutations of multiples of 11. The answer is (A).
devenware
2017-02-08 20:26:22
WOOHOO!
WOOHOO!
mathmagician
2017-02-08 20:26:43
WE DID IT
WE DID IT
Bill9000
2017-02-08 20:26:43
YAY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
YAY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
goodball
2017-02-08 20:26:43
That was GREAT!
That was GREAT!
RemiaFlower
2017-02-08 20:26:52
AWESOME!!!
AWESOME!!!
v4913
2017-02-08 20:26:52
IT WAS AWESOME!
IT WAS AWESOME!
devenware
2017-02-08 20:26:56
AWESOME!!
AWESOME!!
EpicRuler101
2017-02-08 20:27:05
YAY THANKS SO MUCH
YAY THANKS SO MUCH
ilovemath04
2017-02-08 20:27:05
AWESOME
AWESOME
devenware
2017-02-08 20:27:16
Alright, everyone ready for the AMC 12?
Alright, everyone ready for the AMC 12?
AnaT129
2017-02-08 20:27:39
YEAHHHHH
YEAHHHHH
amwmath
2017-02-08 20:27:39
YEAH!
YEAH!
First
2017-02-08 20:27:39
Yes!
Yes!
Bill9000
2017-02-08 20:27:39
GREAT!!!!
GREAT!!!!
PenguinJoe
2017-02-08 20:27:39
YAS
YAS
LegoLdr
2017-02-08 20:27:39
YEAH!!!
YEAH!!!
mathchampion1
2017-02-08 20:27:39
YES!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
YES!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
brainiac1
2017-02-08 20:27:39
Yep
Yep
goodball
2017-02-08 20:27:39
YES!!!!!!!!!!!!!!!!!!!
YES!!!!!!!!!!!!!!!!!!!
challengetogo
2017-02-08 20:27:39
YESSSS
YESSSS
amwmath
2017-02-08 20:27:39
Le''s DO this!
Le''s DO this!
devenware
2017-02-08 20:27:44
YEAH, LET'S DO THIS!
YEAH, LET'S DO THIS!
devenware
2017-02-08 20:27:56
21. A set $S$ is constructed as follows. To begin, $S=\{0,10\}$. Repeatedly, as long as possible, if $x$ is an integer root of some polynomial $a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0$ for some $n\ge 1$, all of whose coefficients $a_i$ are elements of $S$,then $x$ is put into $S$. When no more elements can be added to $S$, how many elements does $S$ have?
$\phantom{hi}$
$
\text{(A) } 4 \quad
\text{(B) } 5 \quad
\text{(C) } 7 \quad
\text{(D) } 9 \quad
\text{(E) } 11
$
21. A set $S$ is constructed as follows. To begin, $S=\{0,10\}$. Repeatedly, as long as possible, if $x$ is an integer root of some polynomial $a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0$ for some $n\ge 1$, all of whose coefficients $a_i$ are elements of $S$,then $x$ is put into $S$. When no more elements can be added to $S$, how many elements does $S$ have?
$\phantom{hi}$
$
\text{(A) } 4 \quad
\text{(B) } 5 \quad
\text{(C) } 7 \quad
\text{(D) } 9 \quad
\text{(E) } 11
$
devenware
2017-02-08 20:28:01
OK, important AMC moment. What relevant theorem haven't we used yet that we often get to see on the AMCs?
OK, important AMC moment. What relevant theorem haven't we used yet that we often get to see on the AMCs?
First
2017-02-08 20:29:07
Rational Root Theorem
Rational Root Theorem
mathcrazymj
2017-02-08 20:29:07
rational root thm
rational root thm
wiler5002
2017-02-08 20:29:07
Rational root theorem
Rational root theorem
summitwei
2017-02-08 20:29:07
rational root?
rational root?
devenware
2017-02-08 20:29:12
The Rational Root Theorem!
The Rational Root Theorem!
devenware
2017-02-08 20:29:13
If we were to build a polynomial here with coefficients 0 and 10 what do we know about its roots?
If we were to build a polynomial here with coefficients 0 and 10 what do we know about its roots?
wiler5002
2017-02-08 20:30:15
must be divisors of 10
must be divisors of 10
summitwei
2017-02-08 20:30:15
can only have roots of +- 1, 2, 5, 10
can only have roots of +- 1, 2, 5, 10
VinVinB
2017-02-08 20:30:15
Factors of 10
Factors of 10
mathfun5
2017-02-08 20:30:15
A_n = 10 and a_0 = 10 or 0
A_n = 10 and a_0 = 10 or 0
summitwei
2017-02-08 20:30:15
can only have roots of +/- 1, 2, 5, 10, 0
can only have roots of +/- 1, 2, 5, 10, 0
devenware
2017-02-08 20:30:21
Well, if $a_0=0$ then we can divide by $x$ and not change the nozero roots, so we should assume that $a_0=10$.
Well, if $a_0=0$ then we can divide by $x$ and not change the nozero roots, so we should assume that $a_0=10$.
devenware
2017-02-08 20:30:22
The Rational Root Theorem tells us that any integer root is a (positive or negative) factor of 10.
The Rational Root Theorem tells us that any integer root is a (positive or negative) factor of 10.
devenware
2017-02-08 20:30:28
What roots can we get with only coefficients of 10?
What roots can we get with only coefficients of 10?
devenware
2017-02-08 20:31:06
Someone give me a polynomial and a root we can get from it.
Someone give me a polynomial and a root we can get from it.
VinVinB
2017-02-08 20:31:56
10x + 10 gives -1
10x + 10 gives -1
CrystalEye
2017-02-08 20:31:56
10x+10=0, x=-1
10x+10=0, x=-1
Figpaste
2017-02-08 20:31:56
10x+10 , x=-1
10x+10 , x=-1
wiler5002
2017-02-08 20:31:56
$10x+10$ gives $-1$
$10x+10$ gives $-1$
summitwei
2017-02-08 20:31:56
10x+10, -1
10x+10, -1
challengetogo
2017-02-08 20:31:56
10x + 10 => x = -1
10x + 10 => x = -1
devenware
2017-02-08 20:32:00
$10x+10$ has root $-1$. (Actually that's the only new number we can add to $S$ so far.)
$10x+10$ has root $-1$. (Actually that's the only new number we can add to $S$ so far.)
devenware
2017-02-08 20:32:21
[For example, we can't use $x + 10$ or something like that -- because we don't have a coefficient 1 yet.]
[For example, we can't use $x + 10$ or something like that -- because we don't have a coefficient 1 yet.]
amwmath
2017-02-08 20:32:38
Wait, what about $-10$?
Wait, what about $-10$?
mathfun5
2017-02-08 20:32:38
Why not 1 or 2,5,10
Why not 1 or 2,5,10
devenware
2017-02-08 20:33:02
Since we can only use 10 as a coefficient at this point, we can't get -10. To get -10, you'd want something like x + 10.
Since we can only use 10 as a coefficient at this point, we can't get -10. To get -10, you'd want something like x + 10.
devenware
2017-02-08 20:33:22
So anyway, we can get $-1$ using $10x + 10$, so let's up our set $S$.
So anyway, we can get $-1$ using $10x + 10$, so let's up our set $S$.
devenware
2017-02-08 20:33:23
Now $S=\{-1,0,10\}$.
Now $S=\{-1,0,10\}$.
devenware
2017-02-08 20:34:49
What else can we add to $S$?
What else can we add to $S$?
LearnAMC
2017-02-08 20:35:22
1
1
SomethingNeutral
2017-02-08 20:35:22
1?
1?
devenware
2017-02-08 20:35:27
Can you find a polynomial with root 1?
Can you find a polynomial with root 1?
devenware
2017-02-08 20:36:00
Remember, we can only use -1 and 10 as our coefficients at this point.
Remember, we can only use -1 and 10 as our coefficients at this point.
mathfun5
2017-02-08 20:36:14
what about 0
what about 0
devenware
2017-02-08 20:36:18
And 0.
And 0.
Damalone
2017-02-08 20:36:39
10x^10-x^9-x^8...-x-1 gives x=1
10x^10-x^9-x^8...-x-1 gives x=1
FATRaichu
2017-02-08 20:36:39
-x^10-x^9-x^8.....x+10=0
-x^10-x^9-x^8.....x+10=0
summitwei
2017-02-08 20:36:39
-1x^10-1x^9-1x^8...-1x^2-1x+10, 1
-1x^10-1x^9-1x^8...-1x^2-1x+10, 1
devenware
2017-02-08 20:36:42
The number 1 is a root of \begin{align*}
&\phantom{+}\ \ (-1)x^{10}+(-1)x^{9}+(-1)x^{8}+(-1)x^{7}+(-1)x^{6}\\&+(-1)x^{5}+(-1)x^{4}+(-1)x^{3}+(-1)x^{2}+(-1)x^{1}+10.\end{align*}
The number 1 is a root of \begin{align*}
&\phantom{+}\ \ (-1)x^{10}+(-1)x^{9}+(-1)x^{8}+(-1)x^{7}+(-1)x^{6}\\&+(-1)x^{5}+(-1)x^{4}+(-1)x^{3}+(-1)x^{2}+(-1)x^{1}+10.\end{align*}
devenware
2017-02-08 20:36:44
More intuitively, to have a root 1, a polynomial needs its coefficients to sum to 0. If our coefficients come from $\{-1, 0, 10\}$, we can just choose ten $-1$s and one $10$.
More intuitively, to have a root 1, a polynomial needs its coefficients to sum to 0. If our coefficients come from $\{-1, 0, 10\}$, we can just choose ten $-1$s and one $10$.
devenware
2017-02-08 20:36:46
Now we have $S=\{-1,0,1,10\}$.
Now we have $S=\{-1,0,1,10\}$.
devenware
2017-02-08 20:37:03
I bet we can get $-10$. It's hard for me to imagine that $S$ won't be symmetric (containing only pairs $\pm n$) after all... Can you construct a polynomial that gives us $-10$?
I bet we can get $-10$. It's hard for me to imagine that $S$ won't be symmetric (containing only pairs $\pm n$) after all... Can you construct a polynomial that gives us $-10$?
amwmath
2017-02-08 20:37:50
$-10$ from $x+10$
$-10$ from $x+10$
mathman3880
2017-02-08 20:37:50
x+10
x+10
summitwei
2017-02-08 20:37:50
1x+10
1x+10
ToneLoke
2017-02-08 20:37:50
X+10
X+10
abvenkgoo
2017-02-08 20:37:50
$x+10$
$x+10$
acegikmoqsuwy2000
2017-02-08 20:37:50
$x+10$
$x+10$
xiongxiong
2017-02-08 20:37:50
X+10
X+10
FATRaichu
2017-02-08 20:37:50
x+10=0
x+10=0
CrystalEye
2017-02-08 20:37:50
x+10=0
x+10=0
wiler5002
2017-02-08 20:37:50
$x+10$
$x+10$
CharlesHong
2017-02-08 20:37:50
x+10
x+10
amburger66
2017-02-08 20:37:50
x+10 lol
x+10 lol
mathfun5
2017-02-08 20:37:50
x+10 = 0
x+10 = 0
Figpaste
2017-02-08 20:37:50
x+10
x+10
LegoLdr
2017-02-08 20:37:52
just x + 10
just x + 10
devenware
2017-02-08 20:37:54
$x+10$ has root $-10$.
$x+10$ has root $-10$.
devenware
2017-02-08 20:38:17
Indeed, since 1 is in $S$, if $s$ is any element of $S$ we can write the polynomial\[x+s\] with root $-s$.
Indeed, since 1 is in $S$, if $s$ is any element of $S$ we can write the polynomial\[x+s\] with root $-s$.
devenware
2017-02-08 20:38:25
Alright, so now we should see if we can include $\pm2$ and $\pm5$.
Alright, so now we should see if we can include $\pm2$ and $\pm5$.
devenware
2017-02-08 20:38:26
Is there a nice polynomial with 2 as a root?
Is there a nice polynomial with 2 as a root?
summitwei
2017-02-08 20:39:41
x^3+x-10
x^3+x-10
amwmath
2017-02-08 20:39:41
$x^3+x-10$.
$x^3+x-10$.
mathman3880
2017-02-08 20:39:41
x^3 + x - 10
x^3 + x - 10
Bill9000
2017-02-08 20:39:41
x^3+x-10=0
x^3+x-10=0
wiler5002
2017-02-08 20:39:41
$-x^3-x+10$
$-x^3-x+10$
Bill9000
2017-02-08 20:39:41
$x^3-x-10$
$x^3-x-10$
vbcnxm
2017-02-08 20:39:41
-x^3-x+10
-x^3-x+10
wiler5002
2017-02-08 20:39:41
or $x^3+x-10$
or $x^3+x-10$
challengetogo
2017-02-08 20:39:41
x^3 + x - 10
x^3 + x - 10
raxu
2017-02-08 20:39:41
$x^3+x-10=0$
$x^3+x-10=0$
mathcount2002
2017-02-08 20:39:41
x^3+0x^2+x-10=0
x^3+0x^2+x-10=0
FATRaichu
2017-02-08 20:39:41
x^3+x-10=0
x^3+x-10=0
devenware
2017-02-08 20:39:44
The binary expansion for 10 is $10=2+2^3$, so 2 must be a root of \[x^3+x-10.\]
The binary expansion for 10 is $10=2+2^3$, so 2 must be a root of \[x^3+x-10.\]
devenware
2017-02-08 20:39:46
Therefore $S$ contains $\pm2$.
Therefore $S$ contains $\pm2$.
devenware
2017-02-08 20:39:47
What about 5? Do you see a way to get 5?
What about 5? Do you see a way to get 5?
summitwei
2017-02-08 20:40:32
2x-10
2x-10
FATRaichu
2017-02-08 20:40:32
2x-10=0
2x-10=0
amwmath
2017-02-08 20:40:32
$2x-10$ now.
$2x-10$ now.
acegikmoqsuwy2000
2017-02-08 20:40:32
$2x-10$
$2x-10$
wiler5002
2017-02-08 20:40:32
$2x-10$
$2x-10$
drazn98
2017-02-08 20:40:32
2x-10
2x-10
VinVinB
2017-02-08 20:40:32
2x-10
2x-10
xiongxiong
2017-02-08 20:40:32
2x-10
2x-10
SomethingNeutral
2017-02-08 20:40:32
2x-10=0?
2x-10=0?
math-rules
2017-02-08 20:40:32
2x-10
2x-10
mathman3880
2017-02-08 20:40:32
2x-10
2x-10
Bill9000
2017-02-08 20:40:32
2x-10=0
2x-10=0
AidanNReilly
2017-02-08 20:40:32
2x-10
2x-10
Figpaste
2017-02-08 20:40:32
2x-10
2x-10
mathislife16
2017-02-08 20:40:32
2x-10
2x-10
arandomperson123
2017-02-08 20:40:32
2x-10 => x=5 works
2x-10 => x=5 works
Benjy450
2017-02-08 20:40:32
2x-10
2x-10
devenware
2017-02-08 20:40:36
Now that we have 2, getting 5 is easy:\[2x-10\] has 5 as a root.
Now that we have 2, getting 5 is easy:\[2x-10\] has 5 as a root.
devenware
2017-02-08 20:40:39
We're at the point where $S=\{0,\pm1,\pm2,\pm5,\pm10\}$. Now what elements can we add?
We're at the point where $S=\{0,\pm1,\pm2,\pm5,\pm10\}$. Now what elements can we add?
wiler5002
2017-02-08 20:41:13
none
none
summitwei
2017-02-08 20:41:13
no more by rrt
no more by rrt
LearnAMC
2017-02-08 20:41:13
none
none
Ani10
2017-02-08 20:41:13
nothing
nothing
mathfun5
2017-02-08 20:41:13
None cause of rational root theorem
None cause of rational root theorem
FATRaichu
2017-02-08 20:41:13
none
none
AnaT129
2017-02-08 20:41:13
none more because that is all the factors of 10
none more because that is all the factors of 10
First
2017-02-08 20:41:13
I think none
I think none
wiler5002
2017-02-08 20:41:13
because any new memebers must divide a current numnber and all the primes are in there
because any new memebers must divide a current numnber and all the primes are in there
Figpaste
2017-02-08 20:41:13
none
none
xiongxiong
2017-02-08 20:41:13
None
None
mathman3880
2017-02-08 20:41:13
Nothing
Nothing
arandomperson123
2017-02-08 20:41:16
by the RRT, we can not add any more...
by the RRT, we can not add any more...
RemiaFlower
2017-02-08 20:41:20
none
none
devenware
2017-02-08 20:41:23
We can't add anything! Since $a_0$ has to be an element of $S$--and we can assume it is nonzero--any new element we add to $S$ has to be a factor of one of these numbers (and thus a factor of 10). We've already found all the factors of 10 so we're done!
We can't add anything! Since $a_0$ has to be an element of $S$--and we can assume it is nonzero--any new element we add to $S$ has to be a factor of one of these numbers (and thus a factor of 10). We've already found all the factors of 10 so we're done!
devenware
2017-02-08 20:41:27
The answer?
The answer?
WW92030
2017-02-08 20:41:48
D
D
CrystalEye
2017-02-08 20:41:48
9
9
FATRaichu
2017-02-08 20:41:48
9
9
LearnAMC
2017-02-08 20:41:48
D
D
Figpaste
2017-02-08 20:41:48
D
D
goodball
2017-02-08 20:41:48
D
D
Benjy450
2017-02-08 20:41:48
9, D
9, D
Knight8
2017-02-08 20:41:48
D
D
brainiac1
2017-02-08 20:41:48
9
9
goodball
2017-02-08 20:41:48
9
9
ToneLoke
2017-02-08 20:41:48
D
D
arandomperson123
2017-02-08 20:41:48
9!!!
9!!!
devenware
2017-02-08 20:41:55
The answer is 9, (D).
The answer is 9, (D).
devenware
2017-02-08 20:42:02
devenware
2017-02-08 20:42:25
Okay, next problem.
Okay, next problem.
RemiaFlower
2017-02-08 20:42:33
YAY!!!!!
YAY!!!!!
goodball
2017-02-08 20:42:33
YAY!
YAY!
RemiaFlower
2017-02-08 20:42:33
YAY!!!!!!
YAY!!!!!!
devenware
2017-02-08 20:42:39
The next one was my favorite too.
The next one was my favorite too.
mathchampion1
2017-02-08 20:42:54
I love states!
I love states!
devenware
2017-02-08 20:43:02
California is my favorite.
California is my favorite.
mathchampion1
2017-02-08 20:43:08
Particles!
Particles!
devenware
2017-02-08 20:43:23
The Charm quark is my favorite.
The Charm quark is my favorite.
AnaT129
2017-02-08 20:43:40
ohhh that is a cool one
ohhh that is a cool one
devenware
2017-02-08 20:43:42
22. A square is drawn in the Cartesian coordinate plane with vertices at $(2,2)$, $(-2,2)$, $(-2,-2)$, and $(2,-2)$. A particle starts at $(0,0)$. Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is $\dfrac18$ that the particle will move from $(x,y)$ to each of $(x,y+1)$, $(x+1,y+1)$, $(x+1,y)$, $(x+1,y-1)$, $(x,y-1)$, $(x-1,y-1)$, $(x-1,y)$, or $(x-1,y+1)$. The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime integers. What is $m+n$?
$\phantom{hi}$
$\text{(A) } 4 \quad
\text{(B) } 5 \quad
\text{(C) } 7 \quad
\text{(D) } 15 \quad
\text{(E) } 39$
22. A square is drawn in the Cartesian coordinate plane with vertices at $(2,2)$, $(-2,2)$, $(-2,-2)$, and $(2,-2)$. A particle starts at $(0,0)$. Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is $\dfrac18$ that the particle will move from $(x,y)$ to each of $(x,y+1)$, $(x+1,y+1)$, $(x+1,y)$, $(x+1,y-1)$, $(x,y-1)$, $(x-1,y-1)$, $(x-1,y)$, or $(x-1,y+1)$. The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime integers. What is $m+n$?
$\phantom{hi}$
$\text{(A) } 4 \quad
\text{(B) } 5 \quad
\text{(C) } 7 \quad
\text{(D) } 15 \quad
\text{(E) } 39$
devenware
2017-02-08 20:43:47
Alright, what's the perfect tool for this problem?
Alright, what's the perfect tool for this problem?
Ani10
2017-02-08 20:44:20
graph paper
graph paper
quartzgirl
2017-02-08 20:44:20
graph paper
graph paper
mathfun5
2017-02-08 20:44:20
graph paper and a picture!
graph paper and a picture!
Eyed
2017-02-08 20:44:29
graph paper
graph paper
devenware
2017-02-08 20:44:31
You're right graph paper is super useful here!
You're right graph paper is super useful here!
devenware
2017-02-08 20:44:36
I hope you brought some to the test!
I hope you brought some to the test!
devenware
2017-02-08 20:44:52
Okay, now what's a good mathematical tool we can use?
Okay, now what's a good mathematical tool we can use?
wiler5002
2017-02-08 20:45:23
States
States
mathchampion1
2017-02-08 20:45:23
State diagrams!!!!
State diagrams!!!!
BobaFett101
2017-02-08 20:45:23
events with states
events with states
math-rules
2017-02-08 20:45:23
state diagram
state diagram
devenware
2017-02-08 20:45:33
This is a problem where we want to think about states. We want to think about all of the possible places that the particle could be at a given time. Let's draw the diagram:
This is a problem where we want to think about states. We want to think about all of the possible places that the particle could be at a given time. Let's draw the diagram:
devenware
2017-02-08 20:45:34
devenware
2017-02-08 20:45:39
So there are 25 different states: the particle could be at any one of these 25 locations.
So there are 25 different states: the particle could be at any one of these 25 locations.
devenware
2017-02-08 20:45:41
Gosh, 25 is a lot, huh?
Gosh, 25 is a lot, huh?
xiongxiong
2017-02-08 20:46:11
Symmetry
Symmetry
abvenkgoo
2017-02-08 20:46:11
symmetry though!
symmetry though!
BobaFett101
2017-02-08 20:46:11
symmetry
symmetry
mathislife16
2017-02-08 20:46:11
you can reduce it because of symmetry
you can reduce it because of symmetry
mathchampion1
2017-02-08 20:46:11
but it is symmetric around the 12 and 13th point!
but it is symmetric around the 12 and 13th point!
drazn98
2017-02-08 20:46:11
symmetry
symmetry
math-rules
2017-02-08 20:46:11
symmetry
symmetry
devenware
2017-02-08 20:46:17
A lot of the states are basically the same. The particle wins when it gets to an outer corner, it loses on the outer edge, the inner corners are all fundamentally the same and so are the inner edges (we'll define "same" in a moment).
A lot of the states are basically the same. The particle wins when it gets to an outer corner, it loses on the outer edge, the inner corners are all fundamentally the same and so are the inner edges (we'll define "same" in a moment).
devenware
2017-02-08 20:46:18
devenware
2017-02-08 20:46:23
Now in a state diagram, we usually label the states with some values. What values should we use to label these states?
Now in a state diagram, we usually label the states with some values. What values should we use to label these states?
First
2017-02-08 20:46:48
probabilities
probabilities
math-rules
2017-02-08 20:46:48
probability of winning from that point
probability of winning from that point
devenware
2017-02-08 20:46:59
Cool. Any probabilities we know for sure already?
Cool. Any probabilities we know for sure already?
xiongxiong
2017-02-08 20:47:36
0 red, 1 green
0 red, 1 green
tdeng
2017-02-08 20:47:36
Green = 1
Green = 1
wiler5002
2017-02-08 20:47:36
corners are 1
corners are 1
brainiac1
2017-02-08 20:47:36
the probability in the corner is 1
the probability in the corner is 1
wiler5002
2017-02-08 20:47:36
edges are 0
edges are 0
tdeng
2017-02-08 20:47:36
Red = 0, Green =1
Red = 0, Green =1
ToneLoke
2017-02-08 20:47:36
red = 0, green = 1
red = 0, green = 1
garretth
2017-02-08 20:47:36
corners and edges
corners and edges
xiongxiong
2017-02-08 20:47:36
Green is 1, red is 0
Green is 1, red is 0
devenware
2017-02-08 20:47:46
A particle at an outer corner has already won! Those get a probability of 1. Likewise a particle at an outer edge has already lost so we put 0 there.
A particle at an outer corner has already won! Those get a probability of 1. Likewise a particle at an outer edge has already lost so we put 0 there.
devenware
2017-02-08 20:47:48
devenware
2017-02-08 20:47:52
Oh, and by the way, we can give a good definition for "the same" now. We colored these dots the same color when we know, by symmetry, that they have the same probability of success.
Oh, and by the way, we can give a good definition for "the same" now. We colored these dots the same color when we know, by symmetry, that they have the same probability of success.
devenware
2017-02-08 20:47:57
It looks like we have three more colors. I don't see immediately what the probabilities at those points are, so let's give them variables:
It looks like we have three more colors. I don't see immediately what the probabilities at those points are, so let's give them variables:
devenware
2017-02-08 20:47:59
devenware
2017-02-08 20:48:05
I picked $o$ for origin, $c$ for corner, and $e$ for edge, since good variable names are good.
I picked $o$ for origin, $c$ for corner, and $e$ for edge, since good variable names are good.
devenware
2017-02-08 20:48:11
Do we have an expression for $o$?
Do we have an expression for $o$?
devenware
2017-02-08 20:48:40
Think: What type of state can you move to from $o$?
Think: What type of state can you move to from $o$?
BobaFett101
2017-02-08 20:48:59
e or c
e or c
garretth
2017-02-08 20:48:59
e or c
e or c
RemiaFlower
2017-02-08 20:49:08
e or c
e or c
devenware
2017-02-08 20:49:09
We can move to $e$ or $c$.
We can move to $e$ or $c$.
devenware
2017-02-08 20:49:16
Can you use that to write an expression for $o$?
Can you use that to write an expression for $o$?
tdeng
2017-02-08 20:49:31
o = 1/2e+1/2c
o = 1/2e+1/2c
ToneLoke
2017-02-08 20:49:31
c/2 + e/2
c/2 + e/2
math-rules
2017-02-08 20:49:31
o = (c+e)/2
o = (c+e)/2
stronto
2017-02-08 20:49:31
e/2 + c/2
e/2 + c/2
brainiac1
2017-02-08 20:49:31
$o=\frac{1}{2} c+\frac{1}{2} e$
$o=\frac{1}{2} c+\frac{1}{2} e$
wiler5002
2017-02-08 20:49:31
$\frac{e}{2}+\frac{c}{2}$
$\frac{e}{2}+\frac{c}{2}$
amwmath
2017-02-08 20:49:31
$\frac{c+e}2$
$\frac{c+e}2$
Blocry
2017-02-08 20:49:31
0.5e+0.5c
0.5e+0.5c
amwmath
2017-02-08 20:49:31
$\frac{4e+4c}8=\frac{e+c}2$
$\frac{4e+4c}8=\frac{e+c}2$
summitwei
2017-02-08 20:49:31
o=1/2*e+1/2*c
o=1/2*e+1/2*c
goodball
2017-02-08 20:49:31
so the expression is (c+e)/2
so the expression is (c+e)/2
devenware
2017-02-08 20:49:35
A particle standing at the middle vertex will go to a $c$ vertex half the time and an $e$ vertex the other half the time. Therefore \[o=\frac12c+\frac12e.\]
A particle standing at the middle vertex will go to a $c$ vertex half the time and an $e$ vertex the other half the time. Therefore \[o=\frac12c+\frac12e.\]
devenware
2017-02-08 20:49:38
I like to clear denominators, so \[2o=c+e.\]
I like to clear denominators, so \[2o=c+e.\]
devenware
2017-02-08 20:49:42
Can we get an expression for $e$?
Can we get an expression for $e$?
devenware
2017-02-08 20:50:31
Let's think about this. What types of states can $e$ move to?
Let's think about this. What types of states can $e$ move to?
LearnAMC
2017-02-08 20:51:05
0, c, or o
0, c, or o
mathmagician
2017-02-08 20:51:05
lose, c, and o
lose, c, and o
math-rules
2017-02-08 20:51:11
c, o, e, or 0
c, o, e, or 0
FATRaichu
2017-02-08 20:51:11
o, c, 0
o, c, 0
devenware
2017-02-08 20:51:26
Great. What's the probability that you move to a 0?
Great. What's the probability that you move to a 0?
CrystalEye
2017-02-08 20:51:41
3/8
3/8
WW92030
2017-02-08 20:51:41
3/8
3/8
dr3463
2017-02-08 20:51:41
3/8
3/8
mathcount2002
2017-02-08 20:51:41
3/8
3/8
mathfun5
2017-02-08 20:51:41
3/8
3/8
devenware
2017-02-08 20:51:49
Cool. Probability you move to another $e$?
Cool. Probability you move to another $e$?
CrystalEye
2017-02-08 20:52:03
1/4
1/4
FATRaichu
2017-02-08 20:52:03
1/4
1/4
mathcount2002
2017-02-08 20:52:03
1/4
1/4
goodball
2017-02-08 20:52:03
1/4
1/4
Eyed
2017-02-08 20:52:03
1/4
1/4
Bill9000
2017-02-08 20:52:03
1/4
1/4
adyj
2017-02-08 20:52:03
1/4
1/4
devenware
2017-02-08 20:52:14
Awesome. Probability you move to a $c$?
Awesome. Probability you move to a $c$?
CrystalEye
2017-02-08 20:52:33
1/4.
1/4.
mathfun5
2017-02-08 20:52:33
1/4
1/4
wiler5002
2017-02-08 20:52:33
also 1/4
also 1/4
dr3463
2017-02-08 20:52:33
2/8
2/8
mathfever
2017-02-08 20:52:33
1/4
1/4
CharlesHong
2017-02-08 20:52:33
1/4
1/4
PenguinJoe
2017-02-08 20:52:33
1/4
1/4
BooBooTM
2017-02-08 20:52:33
1/4
1/4
mathchampion1
2017-02-08 20:52:34
1/4 again
1/4 again
devenware
2017-02-08 20:52:45
And there's a 1/8 probability that we move to the $o$.
And there's a 1/8 probability that we move to the $o$.
devenware
2017-02-08 20:52:51
So can you write an expression for $e$?
So can you write an expression for $e$?
Figpaste
2017-02-08 20:53:21
e = 1/4 e + 1/8 o +1/4 c
e = 1/4 e + 1/8 o +1/4 c
garretth
2017-02-08 20:53:21
e=1/4c + 1/8o + 1/4e
e=1/4c + 1/8o + 1/4e
garretth
2017-02-08 20:53:21
e=1/4c + 1/8o + 1/4e
e=1/4c + 1/8o + 1/4e
garretth
2017-02-08 20:53:21
e=1/4c + 1/8o + 1/4e
e=1/4c + 1/8o + 1/4e
amwmath
2017-02-08 20:53:21
$\frac14c+\frac14e+\frac18\rm o$\
$\frac14c+\frac14e+\frac18\rm o$\
CrystalEye
2017-02-08 20:53:21
1/4c+1/4e+1/8o
1/4c+1/4e+1/8o
FATRaichu
2017-02-08 20:53:21
1/4e+1/4c+1/8o
1/4e+1/4c+1/8o
challengetogo
2017-02-08 20:53:21
e = e/4 + o/8 + c/4
e = e/4 + o/8 + c/4
mathfun5
2017-02-08 20:53:21
e = o/8 + c/4 + e/4
e = o/8 + c/4 + e/4
devenware
2017-02-08 20:53:26
A particle at $e$ will move each of the 8 directions with probability $\dfrac18$, so \[e=\frac18(3\cdot0+2\cdot c+2\cdot e+1\cdot o).\]
A particle at $e$ will move each of the 8 directions with probability $\dfrac18$, so \[e=\frac18(3\cdot0+2\cdot c+2\cdot e+1\cdot o).\]
devenware
2017-02-08 20:53:28
Clearing the denominator gives \[8e=2c+2e+o.\]
Clearing the denominator gives \[8e=2c+2e+o.\]
devenware
2017-02-08 20:53:35
And an expression for $c$?
And an expression for $c$?
mathfun5
2017-02-08 20:54:51
c = 1/8 + e/4 + o/8
c = 1/8 + e/4 + o/8
tdeng
2017-02-08 20:54:51
c=1/8+1/4e+1/8o
c=1/8+1/4e+1/8o
amwmath
2017-02-08 20:54:51
$\frac18+\frac14e+\frac18\bar{\rm o}$
$\frac18+\frac14e+\frac18\bar{\rm o}$
mathfun5
2017-02-08 20:54:51
c = 1/8 + e/4 + o/8
c = 1/8 + e/4 + o/8
FATRaichu
2017-02-08 20:54:51
1/8+1/8o+1/4e
1/8+1/8o+1/4e
LearnAMC
2017-02-08 20:54:51
c=1/8+1/4e+1/8o
c=1/8+1/4e+1/8o
CrystalEye
2017-02-08 20:54:51
1/8+1/4e+1/8o
1/8+1/4e+1/8o
mathcount2002
2017-02-08 20:54:51
c=1/8+e/4+o/8
c=1/8+e/4+o/8
devenware
2017-02-08 20:55:05
A particle at $c$ will move each of the 8 directions with probability $\dfrac18$, so \[c=\frac18(1\cdot1+4\cdot0+2\cdot e+1\cdot o).\]
A particle at $c$ will move each of the 8 directions with probability $\dfrac18$, so \[c=\frac18(1\cdot1+4\cdot0+2\cdot e+1\cdot o).\]
devenware
2017-02-08 20:55:07
Clearing the denominator gives \[8c=1+2e+o.\]
Clearing the denominator gives \[8c=1+2e+o.\]
wiler5002
2017-02-08 20:55:42
3 equations and 3 variables so just solve for o
3 equations and 3 variables so just solve for o
mathchampion1
2017-02-08 20:55:42
so we have a system of equations
so we have a system of equations
devenware
2017-02-08 20:55:47
We are given the system of equations
\begin{align*}
2o&=c+e\\
8e&=2c+2e+o\\
8c&=1+2e+o\\
\end{align*}
and we want to find $o$.
We are given the system of equations
\begin{align*}
2o&=c+e\\
8e&=2c+2e+o\\
8c&=1+2e+o\\
\end{align*}
and we want to find $o$.
devenware
2017-02-08 20:55:48
What should we do first?
What should we do first?
ToneLoke
2017-02-08 20:56:38
substitute o into the bottom two
substitute o into the bottom two
abvenkgoo
2017-02-08 20:56:38
subtract the bottom two equations
subtract the bottom two equations
mathfever
2017-02-08 20:56:38
subtract the 3rd equation from the 2nd
subtract the 3rd equation from the 2nd
devenware
2017-02-08 20:56:43
Funny enough, getting rid of $o$ is a decent first step, but do you see a way to simplify the above equations first, so that maybe we don't have to?
Funny enough, getting rid of $o$ is a decent first step, but do you see a way to simplify the above equations first, so that maybe we don't have to?
wiler5002
2017-02-08 20:57:50
sub o for c+e
sub o for c+e
stronto
2017-02-08 20:57:50
replace 2c+2e with 4o
replace 2c+2e with 4o
Octophi
2017-02-08 20:57:50
sub in $2c+2e=4o$ into second
sub in $2c+2e=4o$ into second
shakeNbake
2017-02-08 20:57:50
plug c + e = 2o into the 2nd equation
plug c + e = 2o into the 2nd equation
devenware
2017-02-08 20:57:55
In the second equation, we see $c+e$, but we know that's $2o$! Therefore\[8e=2(c+e)+o=5o.\]
In the second equation, we see $c+e$, but we know that's $2o$! Therefore\[8e=2(c+e)+o=5o.\]
devenware
2017-02-08 20:57:57
\begin{align*}
2o&=c+e\\
8e&=5o\\
8c&=1+2e+o\\
\end{align*}
\begin{align*}
2o&=c+e\\
8e&=5o\\
8c&=1+2e+o\\
\end{align*}
devenware
2017-02-08 20:58:03
Now we have $e$ in terms of $o$. What is $c$ in terms of $o$?
Now we have $e$ in terms of $o$. What is $c$ in terms of $o$?
xiongxiong
2017-02-08 20:59:29
11/8 o
11/8 o
CrystalEye
2017-02-08 20:59:29
11/8o
11/8o
stronto
2017-02-08 20:59:29
11o/8
11o/8
bobjoe123
2017-02-08 20:59:29
11o/8 = c
11o/8 = c
bobjoe123
2017-02-08 20:59:29
$\frac{11o}{8} = c$
$\frac{11o}{8} = c$
devenware
2017-02-08 20:59:34
Combining the first two equations gives us \[2o=c+\frac58o,\] so \[c=\frac{11}8o.\]
Combining the first two equations gives us \[2o=c+\frac58o,\] so \[c=\frac{11}8o.\]
devenware
2017-02-08 20:59:36
This makes sense: you're more likely to win from $c$ than from $e$, and $o$ is the average of the two.
This makes sense: you're more likely to win from $c$ than from $e$, and $o$ is the average of the two.
devenware
2017-02-08 20:59:38
Now we can use the last equation to find $o$. What is it?
Now we can use the last equation to find $o$. What is it?
xiongxiong
2017-02-08 21:00:53
4/35
4/35
Archos
2017-02-08 21:00:53
4/35
4/35
FATRaichu
2017-02-08 21:00:53
4/35
4/35
Acstar00
2017-02-08 21:00:53
4/35
4/35
bobjoe123
2017-02-08 21:00:53
4/35
4/35
brainiac1
2017-02-08 21:00:53
$o = \frac{4}{35}$
$o = \frac{4}{35}$
ToneLoke
2017-02-08 21:00:53
4/35
4/35
devenware
2017-02-08 21:00:57
Substituting gives\[11o=1+\frac54o+o,\]so \[44o=4+9o.\] Therefore $o=\dfrac{4}{35}$.
Substituting gives\[11o=1+\frac54o+o,\]so \[44o=4+9o.\] Therefore $o=\dfrac{4}{35}$.
devenware
2017-02-08 21:00:57
And the answer?
And the answer?
CrystalEye
2017-02-08 21:01:50
39
39
mathfun5
2017-02-08 21:01:50
39
39
LearnAMC
2017-02-08 21:01:50
E
E
mathchampion1
2017-02-08 21:01:50
39
39
garretth
2017-02-08 21:01:50
39
39
Acstar00
2017-02-08 21:01:50
(E) 39
(E) 39
xiongxiong
2017-02-08 21:01:50
E
E
garretth
2017-02-08 21:01:50
woops 39
woops 39
FATRaichu
2017-02-08 21:01:50
39, E
39, E
challengetogo
2017-02-08 21:01:50
39!
39!
amwmath
2017-02-08 21:01:50
39
39
First
2017-02-08 21:01:50
E
E
goodball
2017-02-08 21:01:50
39
39
AnaT129
2017-02-08 21:01:50
E-39!!!!!
E-39!!!!!
mathmagician
2017-02-08 21:01:50
39 of course!
39 of course!
ToneLoke
2017-02-08 21:01:50
E
E
brainiac1
2017-02-08 21:01:50
4+35=39
4+35=39
bobjoe123
2017-02-08 21:01:50
EEEEE
EEEEE
mathfun5
2017-02-08 21:01:50
(E) 39!
(E) 39!
PenguinJoe
2017-02-08 21:01:50
39
39
WW92030
2017-02-08 21:01:50
E
E
devenware
2017-02-08 21:01:55
The answer is $4+35=\boxed{39}$, (E).
The answer is $4+35=\boxed{39}$, (E).
RemiaFlower
2017-02-08 21:02:11
WE DID IT!!!
WE DID IT!!!
devenware
2017-02-08 21:02:24
Awesome!
Awesome!
devenware
2017-02-08 21:02:33
AMC 12 Problem 23 was a repeat of Problem 24 on the AMC 10, so we'll skip it now.
AMC 12 Problem 23 was a repeat of Problem 24 on the AMC 10, so we'll skip it now.
devenware
2017-02-08 21:02:36
That means. . .
That means. . .
devenware
2017-02-08 21:02:43
24. Quadrilateral $ABCD$ is inscribed in circle $O$ and has sides $AB=3$, $BC=2$, $CD = 6$, and $DA = 8$. Let $X$ and $Y$ be points on $\overline{BD}$ such that \[\frac{DX}{BD} = \frac14\qquad\text{and}\qquad\frac{BY}{BD} = \frac{11}{36}.\]Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$. Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$. Let $G$ be the point on circle $O$ other than $C$ that lies on $CX$. What is $XF\cdot XG$?
$\phantom{hi}$
$\text{(A) } 17 \quad
\text{(B) } \dfrac{59-5\sqrt{2}}{3} \quad
\text{(C) } \dfrac{91-12\sqrt{2}}{4} \quad
\text{(D) } \dfrac{67-10\sqrt{2}}{3} \quad
\text{(E) } 18$
24. Quadrilateral $ABCD$ is inscribed in circle $O$ and has sides $AB=3$, $BC=2$, $CD = 6$, and $DA = 8$. Let $X$ and $Y$ be points on $\overline{BD}$ such that \[\frac{DX}{BD} = \frac14\qquad\text{and}\qquad\frac{BY}{BD} = \frac{11}{36}.\]Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$. Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$. Let $G$ be the point on circle $O$ other than $C$ that lies on $CX$. What is $XF\cdot XG$?
$\phantom{hi}$
$\text{(A) } 17 \quad
\text{(B) } \dfrac{59-5\sqrt{2}}{3} \quad
\text{(C) } \dfrac{91-12\sqrt{2}}{4} \quad
\text{(D) } \dfrac{67-10\sqrt{2}}{3} \quad
\text{(E) } 18$
devenware
2017-02-08 21:02:57
Well 145.5 is a very respectable score. Problem 25 now?
Well 145.5 is a very respectable score. Problem 25 now?
amburger66
2017-02-08 21:03:26
noooo
noooo
challengetogo
2017-02-08 21:03:26
never!
never!
LearnAMC
2017-02-08 21:03:26
no
no
First
2017-02-08 21:03:26
No
No
RemiaFlower
2017-02-08 21:03:38
NO
NO
mathfun5
2017-02-08 21:03:38
NOOOOO
NOOOOO
devenware
2017-02-08 21:03:39
Fine, fine. Then tell me what we should do first?
Fine, fine. Then tell me what we should do first?
CrystalEye
2017-02-08 21:04:30
draw a diagram
draw a diagram
First
2017-02-08 21:04:30
Draw a diagram
Draw a diagram
challengetogo
2017-02-08 21:04:30
draw the diagram
draw the diagram
kevindk
2017-02-08 21:04:30
diagram
diagram
amburger66
2017-02-08 21:04:30
draw a diagram
draw a diagram
summitwei
2017-02-08 21:04:30
diagram pl0x
diagram pl0x
Bill9000
2017-02-08 21:04:30
DIAGRAM
DIAGRAM
ToneLoke
2017-02-08 21:04:30
something somesthing diagram
something somesthing diagram
Cytrew
2017-02-08 21:04:30
diagram!
diagram!
brainiac1
2017-02-08 21:04:30
this needs a well-drawn diagram
this needs a well-drawn diagram
raxu
2017-02-08 21:04:30
Draw a diagram.
Draw a diagram.
yid
2017-02-08 21:04:30
diagram?
diagram?
Jayjayliu
2017-02-08 21:04:30
draw something
draw something
Cytrew
2017-02-08 21:04:30
draw it!
draw it!
devenware
2017-02-08 21:04:37
devenware
2017-02-08 21:04:40
OK, that's where we start. Next we construct $E$ by extending $AX$ and intersecting the line through $Y$ parallel to $\overline{AD}$.
OK, that's where we start. Next we construct $E$ by extending $AX$ and intersecting the line through $Y$ parallel to $\overline{AD}$.
devenware
2017-02-08 21:04:41
devenware
2017-02-08 21:04:55
After that we construct $F$ by extending $CX$ and intersecting with the line through $E$ parallel to $\overline{AC}$.
After that we construct $F$ by extending $CX$ and intersecting with the line through $E$ parallel to $\overline{AC}$.
devenware
2017-02-08 21:04:56
devenware
2017-02-08 21:05:06
The point $G$ is the intersection of $CX$ and the circle.
The point $G$ is the intersection of $CX$ and the circle.
devenware
2017-02-08 21:05:13
Now this is a mess of lines, isn't it? We're trying to find $XF\cdot XG$.
Now this is a mess of lines, isn't it? We're trying to find $XF\cdot XG$.
First
2017-02-08 21:05:38
Looks scary
Looks scary
devenware
2017-02-08 21:05:40
Any ideas?
Any ideas?
AnaT129
2017-02-08 21:06:12
power of a point or similar triangles?
power of a point or similar triangles?
dr3463
2017-02-08 21:06:12
This is a wild guess but can we use power of a point theorem
This is a wild guess but can we use power of a point theorem
brainiac1
2017-02-08 21:06:12
looks like power of a point
looks like power of a point
pandadude
2017-02-08 21:06:12
power of point!
power of point!
Cytrew
2017-02-08 21:06:12
maybe a bit like power of a point??
maybe a bit like power of a point??
Jayjayliu
2017-02-08 21:06:12
power of a point
power of a point
goodball
2017-02-08 21:06:12
Power of a point?
Power of a point?
wiler5002
2017-02-08 21:06:12
Looks kinda like power of a point
Looks kinda like power of a point
devenware
2017-02-08 21:06:25
Definitely looks like power of a point or similar triangles are going to be relevant here.
Definitely looks like power of a point or similar triangles are going to be relevant here.
devenware
2017-02-08 21:06:50
That is, we want to rewrite $XF \cdot XG$ without using $F.$
That is, we want to rewrite $XF \cdot XG$ without using $F.$
devenware
2017-02-08 21:07:04
We can do this with power of a point or similar triangles, I'll try similar triangles.
We can do this with power of a point or similar triangles, I'll try similar triangles.
devenware
2017-02-08 21:07:25
One thing to note toward that goal, though. This problem has an interesting property: the configuration is constructed in a sequence of steps. The last thing we constructed was $F$, so it depends on the things we've constructed before. We should be able to "eliminate" $F$ from the problem by unwinding the construction.
One thing to note toward that goal, though. This problem has an interesting property: the configuration is constructed in a sequence of steps. The last thing we constructed was $F$, so it depends on the things we've constructed before. We should be able to "eliminate" $F$ from the problem by unwinding the construction.
devenware
2017-02-08 21:07:30
Let's highlight what we used to construct $F$:
Let's highlight what we used to construct $F$:
devenware
2017-02-08 21:07:32
devenware
2017-02-08 21:07:33
Now, what do we have here?
Now, what do we have here?
Bill9000
2017-02-08 21:07:54
two similar triangles!
two similar triangles!
mathfun5
2017-02-08 21:07:54
similar triangles!!!
similar triangles!!!
First
2017-02-08 21:07:54
A pair of similar triangles
A pair of similar triangles
shakeNbake
2017-02-08 21:07:54
similar triangles
similar triangles
brainiac1
2017-02-08 21:07:54
two similar triangles
two similar triangles
1023ong
2017-02-08 21:07:54
similar triangles
similar triangles
User2013
2017-02-08 21:07:54
similar triangles
similar triangles
challengetogo
2017-02-08 21:07:56
similar triangles!
similar triangles!
devenware
2017-02-08 21:07:59
Which triangles?
Which triangles?
bobjoe123
2017-02-08 21:08:57
ACX~EFX!!!!!!!!
ACX~EFX!!!!!!!!
mathfun5
2017-02-08 21:08:57
XFE and XCA
XFE and XCA
wiler5002
2017-02-08 21:08:57
FXE and CXA
FXE and CXA
Bill9000
2017-02-08 21:08:57
ACX and EFX
ACX and EFX
devenware
2017-02-08 21:09:05
Here we have parallel lines $\overline{FE}\parallel\overline{AC}$, so we get some similar triangles $\triangle FXE\sim\triangle CXA$.
Here we have parallel lines $\overline{FE}\parallel\overline{AC}$, so we get some similar triangles $\triangle FXE\sim\triangle CXA$.
devenware
2017-02-08 21:09:27
That's something to work with. However, we have an $XG$ in here, too...
That's something to work with. However, we have an $XG$ in here, too...
devenware
2017-02-08 21:09:54
If we let $H$ be the other intersection with the circle, we also get $\triangle FXE\sim\triangle HXG$, as these are triangles formed by intersecting chords.
If we let $H$ be the other intersection with the circle, we also get $\triangle FXE\sim\triangle HXG$, as these are triangles formed by intersecting chords.
devenware
2017-02-08 21:09:56
devenware
2017-02-08 21:09:58
So how can we rewrite $XF\cdot XG$?
So how can we rewrite $XF\cdot XG$?
devenware
2017-02-08 21:11:03
Hint: Use the similarity ratio from what we just got.
Hint: Use the similarity ratio from what we just got.
LearnAMC
2017-02-08 21:12:03
XE*XH
XE*XH
AnaT129
2017-02-08 21:12:03
XH*XE
XH*XE
brainiac1
2017-02-08 21:12:03
$XH \cdot XE$
$XH \cdot XE$
devenware
2017-02-08 21:12:06
By the similarity $\triangle FXE\sim\triangle HXG$, we get \[\frac{XF}{XH}=\frac{XE}{XG}.\] Now we can "replace" the line $CF$ with the line $AE$ by writing\[XF\cdot XG=XH\cdot XE.\]
By the similarity $\triangle FXE\sim\triangle HXG$, we get \[\frac{XF}{XH}=\frac{XE}{XG}.\] Now we can "replace" the line $CF$ with the line $AE$ by writing\[XF\cdot XG=XH\cdot XE.\]
devenware
2017-02-08 21:12:17
Notice that Power of a Point on $X$ is equivalent to invoking $\triangle CXA\sim\triangle HXG$ - that's a different way to visualize the same step.
Notice that Power of a Point on $X$ is equivalent to invoking $\triangle CXA\sim\triangle HXG$ - that's a different way to visualize the same step.
devenware
2017-02-08 21:12:23
(And it's how most of you wanted to do it.)
(And it's how most of you wanted to do it.)
dr3463
2017-02-08 21:12:29
but how does that help us?
but how does that help us?
devenware
2017-02-08 21:12:32
Good question!
Good question!
devenware
2017-02-08 21:12:35
Now we can completely forget the points $F$ and $G$:
Now we can completely forget the points $F$ and $G$:
devenware
2017-02-08 21:12:37
brainiac1
2017-02-08 21:12:50
good riddance
good riddance
devenware
2017-02-08 21:12:52
What should we focus on now?
What should we focus on now?
LearnAMC
2017-02-08 21:13:33
H and E
H and E
goodball
2017-02-08 21:13:33
Points H and E
Points H and E
AnaT129
2017-02-08 21:13:33
removing points E and H?
removing points E and H?
devenware
2017-02-08 21:13:43
Yeah, I don't like $E$ either. Let's try to get rid of it.
Yeah, I don't like $E$ either. Let's try to get rid of it.
devenware
2017-02-08 21:13:48
Let's throw down just those pieces that are relevant to the construction of $E$:
Let's throw down just those pieces that are relevant to the construction of $E$:
devenware
2017-02-08 21:13:51
devenware
2017-02-08 21:13:54
Let's play the same game again. What triangles are immediately similar given that $\overline{EY}\parallel\overline{AD}$?
Let's play the same game again. What triangles are immediately similar given that $\overline{EY}\parallel\overline{AD}$?
wiler5002
2017-02-08 21:15:17
EXY and AXD
EXY and AXD
LearnAMC
2017-02-08 21:15:17
XYE and XDA
XYE and XDA
pandadude
2017-02-08 21:15:17
XEY, XAD
XEY, XAD
brainiac1
2017-02-08 21:15:17
AXD and EXY
AXD and EXY
bobjoe123
2017-02-08 21:15:17
ADX and EYX
ADX and EYX
1023ong
2017-02-08 21:15:17
EYX, ADX
EYX, ADX
challengetogo
2017-02-08 21:15:17
ADX ~ EYX
ADX ~ EYX
Ani10
2017-02-08 21:15:17
ADX and EYX
ADX and EYX
bobjoe123
2017-02-08 21:15:17
ADX and EYX are similar!!!!
ADX and EYX are similar!!!!
ArcticWolf
2017-02-08 21:15:17
AXD and EXY?
AXD and EXY?
devenware
2017-02-08 21:15:22
Since those lines are parallel, we know $\triangle EXY\sim\triangle AXD$.
Since those lines are parallel, we know $\triangle EXY\sim\triangle AXD$.
devenware
2017-02-08 21:15:32
Now we also need $HX$ in here. How do we get that?
Now we also need $HX$ in here. How do we get that?
devenware
2017-02-08 21:17:04
Any other triangles that are similar to $AXD$?
Any other triangles that are similar to $AXD$?
devenware
2017-02-08 21:17:19
Let me add a little more to the diagram and see if we see anything.
Let me add a little more to the diagram and see if we see anything.
devenware
2017-02-08 21:17:22
adyj
2017-02-08 21:18:13
XHB
XHB
wiler5002
2017-02-08 21:18:13
HXB
HXB
LearnAMC
2017-02-08 21:18:13
BXH
BXH
dr3463
2017-02-08 21:18:13
I mean BHX
I mean BHX
Ani10
2017-02-08 21:18:13
ADX similar to BHX!
ADX similar to BHX!
devenware
2017-02-08 21:18:24
Since $DX$ and $AH$ intersect at $X$, we know $\triangle AXD\sim\triangle BXH$. Thus
\[\triangle EXY\sim\triangle BXH.\]
Since $DX$ and $AH$ intersect at $X$, we know $\triangle AXD\sim\triangle BXH$. Thus
\[\triangle EXY\sim\triangle BXH.\]
devenware
2017-02-08 21:18:28
What does that tell us about $XH\cdot XE$?
What does that tell us about $XH\cdot XE$?
drazn98
2017-02-08 21:19:44
$\frac{XE}{XY}=\frac{BX}{XH}$
$\frac{XE}{XY}=\frac{BX}{XH}$
wiler5002
2017-02-08 21:19:44
$XY*BX$
$XY*BX$
Ani10
2017-02-08 21:19:44
YX*BX
YX*BX
LearnAMC
2017-02-08 21:19:44
XY*XB
XY*XB
devenware
2017-02-08 21:19:49
By similarity,
\[XH\cdot XE = XY\cdot XB.\]
By similarity,
\[XH\cdot XE = XY\cdot XB.\]
devenware
2017-02-08 21:20:02
Okay, so we can finally ditch $E$.
Okay, so we can finally ditch $E$.
devenware
2017-02-08 21:20:09
We no longer have anything beyond our original cyclic quadrilateral. Let's delete all the extra nonsense and add in the numbers from the problem:
We no longer have anything beyond our original cyclic quadrilateral. Let's delete all the extra nonsense and add in the numbers from the problem:
devenware
2017-02-08 21:20:11
AnaT129
2017-02-08 21:20:20
yesssss the diagram is nice now
yesssss the diagram is nice now
devenware
2017-02-08 21:20:24
Agreed.
Agreed.
wiler5002
2017-02-08 21:20:54
find the total length of $BD$?
find the total length of $BD$?
brainiac1
2017-02-08 21:20:54
Looks like we now have to find BD
Looks like we now have to find BD
devenware
2017-02-08 21:21:00
Well, we have these ratios and they let us write everything in terms of $BD$.
Well, we have these ratios and they let us write everything in terms of $BD$.
devenware
2017-02-08 21:21:00
\begin{align*}
XF\cdot XG
&=XH\cdot XE\\
&=XY\cdot XB\\
&=(BD-BY-DX)\cdot (BD-DX)\\
&=\left(1-\frac{11}{36}-\frac14\right)BD\cdot \left(1-\frac14\right)BD\\
&=\frac{16}{36}\cdot\frac34\cdot BD^2\\
&=\frac{BD^2}3.
\end{align*}
\begin{align*}
XF\cdot XG
&=XH\cdot XE\\
&=XY\cdot XB\\
&=(BD-BY-DX)\cdot (BD-DX)\\
&=\left(1-\frac{11}{36}-\frac14\right)BD\cdot \left(1-\frac14\right)BD\\
&=\frac{16}{36}\cdot\frac34\cdot BD^2\\
&=\frac{BD^2}3.
\end{align*}
devenware
2017-02-08 21:21:09
So all we need now is $BD$. In fact, we don't need any of those other numbers. I wonder why they didn't just give us $BD$ to begin with. . .
So all we need now is $BD$. In fact, we don't need any of those other numbers. I wonder why they didn't just give us $BD$ to begin with. . .
devenware
2017-02-08 21:21:11
How can we find $BD$?
How can we find $BD$?
drazn98
2017-02-08 21:22:01
law of cosines on ABD and CBD
law of cosines on ABD and CBD
BobaFett101
2017-02-08 21:22:01
law of cosines
law of cosines
Alex2
2017-02-08 21:22:01
Use law of cosines twice, get system of equations and solve
Use law of cosines twice, get system of equations and solve
brainiac1
2017-02-08 21:22:01
The law of cosines on the two opposite angles
The law of cosines on the two opposite angles
devenware
2017-02-08 21:22:06
We can use the Law of Cosines. We know that $\angle A$ and $\angle C$ are supplementary, so $\cos A=-\cos C$. Let's call this value $c$.
We can use the Law of Cosines. We know that $\angle A$ and $\angle C$ are supplementary, so $\cos A=-\cos C$. Let's call this value $c$.
devenware
2017-02-08 21:22:09
The Law of Cosines on $\triangle DAB$ at angle $A$ gives
\begin{align*}
BD^2&=AD^2+AB^2-AD\cdot AB \cdot c\\
&=8^2+3^2-8\cdot3c\\
&=73-24c.
\end{align*}
The Law of Cosines on $\triangle DAB$ at angle $A$ gives
\begin{align*}
BD^2&=AD^2+AB^2-AD\cdot AB \cdot c\\
&=8^2+3^2-8\cdot3c\\
&=73-24c.
\end{align*}
devenware
2017-02-08 21:22:11
Likewise, the Law of Cosines on $\triangle DCB$ at angle $C$ gives
\begin{align*}
BD^2&=CD^2+CB^2+CD\cdot CB \cdot c\\
&=6^2+2^2+6\cdot2c\\
&=40+12c.
\end{align*}
Likewise, the Law of Cosines on $\triangle DCB$ at angle $C$ gives
\begin{align*}
BD^2&=CD^2+CB^2+CD\cdot CB \cdot c\\
&=6^2+2^2+6\cdot2c\\
&=40+12c.
\end{align*}
devenware
2017-02-08 21:22:17
(You're welcome for the algebra.)
(You're welcome for the algebra.
)
goodball
2017-02-08 21:22:30
Thank you.
Thank you.
pandadude
2017-02-08 21:22:37
Thanks!
Thanks!
devenware
2017-02-08 21:22:39
devenware
2017-02-08 21:22:55
So we're left with
\begin{align*}
BD^2&=73-24c\\
BD^2&=40+12c.
\end{align*}
So we're left with
\begin{align*}
BD^2&=73-24c\\
BD^2&=40+12c.
\end{align*}
devenware
2017-02-08 21:23:09
If we double the second and add we get
\[3BD^2=73+2\cdot40=153,\]so $BD=\sqrt{51}.$
If we double the second and add we get
\[3BD^2=73+2\cdot40=153,\]so $BD=\sqrt{51}.$
devenware
2017-02-08 21:23:12
What's the final answer?
What's the final answer?
bobjoe123
2017-02-08 21:24:04
17
17
goodball
2017-02-08 21:24:04
17
17
wiler5002
2017-02-08 21:24:04
A
A
First
2017-02-08 21:24:04
A
A
CharlesHong
2017-02-08 21:24:04
A
A
LearnAMC
2017-02-08 21:24:04
A
A
goodball
2017-02-08 21:24:04
A
A
FATRaichu
2017-02-08 21:24:04
17
17
bobjoe123
2017-02-08 21:24:04
!7!!!!!!!!!!17!!!!!!!!
!7!!!!!!!!!!17!!!!!!!!
dr3463
2017-02-08 21:24:04
17
17
AnaT129
2017-02-08 21:24:04
17 - A
17 - A
challengetogo
2017-02-08 21:24:04
A
A
brainiac1
2017-02-08 21:24:04
17
17
mathfun5
2017-02-08 21:24:04
51/3 = 17
51/3 = 17
summitwei
2017-02-08 21:24:04
17
17
mathfun5
2017-02-08 21:24:04
17!!!
17!!!
devenware
2017-02-08 21:24:09
We want \[XF\cdot XG=\frac{BD^2}3=17,\] so the answer is (A), $\boxed{17}$.
We want \[XF\cdot XG=\frac{BD^2}3=17,\] so the answer is (A), $\boxed{17}$.
RemiaFlower
2017-02-08 21:24:19
17!!
17!!
devenware
2017-02-08 21:24:26
17!!!
17!!!
RemiaFlower
2017-02-08 21:24:38
YAY we did it again!!!!
YAY we did it again!!!!
dr3463
2017-02-08 21:24:38
We did it!!
We did it!!
Cytrew
2017-02-08 21:24:38
YAY
YAY
goodball
2017-02-08 21:24:38
YAY,we finally did it!!!!!!!!!!!!!
YAY,we finally did it!!!!!!!!!!!!!
brainiac1
2017-02-08 21:24:38
One problem left
One problem left
adyj
2017-02-08 21:24:38
Let's finish this up.
Let's finish this up.
devenware
2017-02-08 21:24:46
Only one to go! Let's do this.
Only one to go! Let's do this.
devenware
2017-02-08 21:25:00
Everyone have their probleming scarves on?
Everyone have their probleming scarves on?
Lah-aops
2017-02-08 21:25:19
sure
sure
Bill9000
2017-02-08 21:25:19
Yesss
Yesss
goodball
2017-02-08 21:25:19
YESSSS!!!!!!!
YESSSS!!!!!!!
RemiaFlower
2017-02-08 21:25:19
Yes, and my matching thinking cap ;)
Yes, and my matching thinking cap ;)
Makorn
2017-02-08 21:25:19
I actually got a science bowl infinity scarf this weekend ;)
I actually got a science bowl infinity scarf this weekend ;)
challengetogo
2017-02-08 21:25:24
nope but I have my probleming hat on!
nope but I have my probleming hat on!
summitwei
2017-02-08 21:25:25
lemme get mine
lemme get mine
devenware
2017-02-08 21:25:30
Okay cool.
Okay cool.
devenware
2017-02-08 21:25:40
Any probleming gear is fine. I often use probleming socks instead.
Any probleming gear is fine. I often use probleming socks instead.
devenware
2017-02-08 21:26:28
25. The vertices $V$ of a centrally symmetric hexagon in the complex plane are given by \[ V = \left\{ \sqrt{2}i,-\sqrt{2}i,\frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i), \frac{1}{\sqrt{8}}(-1-i)\right\}.\]For each $j$, $1\le j\le12$, an element $z_j$ is chosen from $V$ at random, independently of the other choices. Let $P = \prod_{j=1}^{12}z_j$ be the product of the 12 numbers selected. What is the probability that $P=-1$?
$\phantom{hi}$
$\text{(A) } \dfrac{5\cdot 11}{3^{10}} \quad
\text{(B) } \dfrac{5^2\cdot 11}{2\cdot 3^{10}} \quad
\text{(C) } \dfrac{5\cdot 11}{3^9} \quad
\text{(D) } \dfrac{5\cdot 7\cdot 11}{2\cdot 3^{10}} \quad
\text{(E) } \dfrac{2^2 \cdot 5\cdot 11}{3^{10}}$
25. The vertices $V$ of a centrally symmetric hexagon in the complex plane are given by \[ V = \left\{ \sqrt{2}i,-\sqrt{2}i,\frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i), \frac{1}{\sqrt{8}}(-1-i)\right\}.\]For each $j$, $1\le j\le12$, an element $z_j$ is chosen from $V$ at random, independently of the other choices. Let $P = \prod_{j=1}^{12}z_j$ be the product of the 12 numbers selected. What is the probability that $P=-1$?
$\phantom{hi}$
$\text{(A) } \dfrac{5\cdot 11}{3^{10}} \quad
\text{(B) } \dfrac{5^2\cdot 11}{2\cdot 3^{10}} \quad
\text{(C) } \dfrac{5\cdot 11}{3^9} \quad
\text{(D) } \dfrac{5\cdot 7\cdot 11}{2\cdot 3^{10}} \quad
\text{(E) } \dfrac{2^2 \cdot 5\cdot 11}{3^{10}}$
devenware
2017-02-08 21:26:54
OK, I see $1+i$. What should you always think when you see $1+i$?
OK, I see $1+i$. What should you always think when you see $1+i$?
Jayjayliu
2017-02-08 21:27:50
45 degrees
45 degrees
Jayjayliu
2017-02-08 21:27:50
(1+i)^4=-4
(1+i)^4=-4
BobaFett101
2017-02-08 21:27:50
e^(i pi/4)
e^(i pi/4)
devenware
2017-02-08 21:28:03
The complex number $1+i$ should remind you of eighth roots of unity. That is, $(1+i)^2 = 1+2i-1 = 2i$, so \[(1+i)^8=(2i)^4=2^4i^4=2^4.\] Dividing by $\sqrt2$ gives\[\left(\frac{1+i}{\sqrt2}\right)^8=1.\]
The complex number $1+i$ should remind you of eighth roots of unity. That is, $(1+i)^2 = 1+2i-1 = 2i$, so \[(1+i)^8=(2i)^4=2^4i^4=2^4.\] Dividing by $\sqrt2$ gives\[\left(\frac{1+i}{\sqrt2}\right)^8=1.\]
devenware
2017-02-08 21:28:20
Let's write $\omega=\dfrac{1+i}{\sqrt2}$. Can we rewrite $V$ in terms of $\omega$?
Let's write $\omega=\dfrac{1+i}{\sqrt2}$. Can we rewrite $V$ in terms of $\omega$?
devenware
2017-02-08 21:29:13
What is $\sqrt{2}i$ in terms of $\omega$?
What is $\sqrt{2}i$ in terms of $\omega$?
devenware
2017-02-08 21:29:32
Okay, how about this, what is $i$ in terms of $\omega$?
Okay, how about this, what is $i$ in terms of $\omega$?
devenware
2017-02-08 21:30:04
(Think $\omega$ is an 8th root of unity -- what about $i$?)
(Think $\omega$ is an 8th root of unity -- what about $i$?)
mathman3880
2017-02-08 21:30:33
\omega^2
\omega^2
amburger66
2017-02-08 21:30:33
w^2
w^2
wiler5002
2017-02-08 21:30:33
%\omega^2$
%\omega^2$
BobaFett101
2017-02-08 21:30:33
w^2
w^2
xingxia_alexye
2017-02-08 21:30:33
w^2
w^2
devenware
2017-02-08 21:30:49
Right, $i$ is $\omega^2$.
Right, $i$ is $\omega^2$.
devenware
2017-02-08 21:30:55
So $\sqrt{2}i$ is?
So $\sqrt{2}i$ is?
brainiac1
2017-02-08 21:31:30
$\sqrt{2} \omega^2$
$\sqrt{2} \omega^2$
amwmath
2017-02-08 21:31:30
$\omega^2\sqrt2$
$\omega^2\sqrt2$
mathfun5
2017-02-08 21:31:30
sqrt(2)*w^2
sqrt(2)*w^2
wiler5002
2017-02-08 21:31:30
$\sqrt2\omega^2$
$\sqrt2\omega^2$
mathman3880
2017-02-08 21:31:30
$sqrt{2}*\omega^2$
$sqrt{2}*\omega^2$
First
2017-02-08 21:31:30
$\sqrt{2} w^2$
$\sqrt{2} w^2$
devenware
2017-02-08 21:31:46
Great. One down, 5 more to go.
Great. One down, 5 more to go.
devenware
2017-02-08 21:31:55
What is $-\sqrt{2}i$ in terms of $\omega$?
What is $-\sqrt{2}i$ in terms of $\omega$?
mathfun5
2017-02-08 21:32:17
-sqrt(2)*w^2
-sqrt(2)*w^2
wiler5002
2017-02-08 21:32:17
$-\sqrt2\omega^2$
$-\sqrt2\omega^2$
xingxia_alexye
2017-02-08 21:32:17
-sqrt(2)w^2
-sqrt(2)w^2
challengetogo
2017-02-08 21:32:17
-sqrt(2)*w^2
-sqrt(2)*w^2
devenware
2017-02-08 21:32:23
True, but can we write that without any negatives?
True, but can we write that without any negatives?
drazn98
2017-02-08 21:33:34
$sqrt{2}{\omega}^{6}$
$sqrt{2}{\omega}^{6}$
wiler5002
2017-02-08 21:33:34
$\sqrt2\omega^6$
$\sqrt2\omega^6$
brainiac1
2017-02-08 21:33:34
$\omega^6 \sqrt{2}$
$\omega^6 \sqrt{2}$
amwmath
2017-02-08 21:33:34
$\sqrt2\,\omega^6$
$\sqrt2\,\omega^6$
BobaFett101
2017-02-08 21:33:34
rt(2)w^6
rt(2)w^6
devenware
2017-02-08 21:33:35
Right, $-\omega^2 = -i = \omega^6$.
Right, $-\omega^2 = -i = \omega^6$.
devenware
2017-02-08 21:33:52
So $-\sqrt{2}i$ is $\sqrt{2}\omega^6$.
So $-\sqrt{2}i$ is $\sqrt{2}\omega^6$.
devenware
2017-02-08 21:34:20
Okay, the rest are a little easier. Can someone give them to me?
Okay, the rest are a little easier. Can someone give them to me?
drazn98
2017-02-08 21:35:04
$frac{\omega}{2}, \frac{{\omega}^3}{2}, frac{{\omega}^5}{2}, \frac{{\omega}^7}{2}$
$frac{\omega}{2}, \frac{{\omega}^3}{2}, frac{{\omega}^5}{2}, \frac{{\omega}^7}{2}$
pandadude
2017-02-08 21:35:04
w/2,w^3/2,w^5/2,w^7/2
w/2,w^3/2,w^5/2,w^7/2
Sparks29032
2017-02-08 21:35:04
the third one is w/2
the third one is w/2
devenware
2017-02-08 21:35:12
Right. Here's the whole set written out:
Right. Here's the whole set written out:
devenware
2017-02-08 21:35:18
\[V=\left\{
\sqrt2\omega^2,\sqrt2\omega^6,\frac{\omega}2,\frac{\omega^3}2,\frac{\omega^5}2,\frac{\omega^7}2
\right\}.\]
\[V=\left\{
\sqrt2\omega^2,\sqrt2\omega^6,\frac{\omega}2,\frac{\omega^3}2,\frac{\omega^5}2,\frac{\omega^7}2
\right\}.\]
goodball
2017-02-08 21:35:34
Nice.
Nice.
devenware
2017-02-08 21:35:40
Yeah, that looks a lot cleaner huh?
Yeah, that looks a lot cleaner huh?
warrenwangtennis
2017-02-08 21:36:03
why did we not just write them in polar form?
why did we not just write them in polar form?
devenware
2017-02-08 21:36:15
I chose to write this in this way since that's how I solved the problem and it's also good to try to become fluent with roots of unity.
I chose to write this in this way since that's how I solved the problem and it's also good to try to become fluent with roots of unity.
devenware
2017-02-08 21:36:21
It probably helps to see these in the complex plane with the unit circle:
It probably helps to see these in the complex plane with the unit circle:
devenware
2017-02-08 21:36:23
brainiac1
2017-02-08 21:36:31
now do we have to decipher that weird product in the second part?
now do we have to decipher that weird product in the second part?
devenware
2017-02-08 21:36:33
How many total ways can we construct the product $P$?
How many total ways can we construct the product $P$?
warrenwangtennis
2017-02-08 21:37:36
6^12
6^12
Jayjayliu
2017-02-08 21:37:36
6^12
6^12
mathfun5
2017-02-08 21:37:36
6^12
6^12
amwmath
2017-02-08 21:37:36
$6^{12}$
$6^{12}$
brainiac1
2017-02-08 21:37:36
6^12
6^12
Blocry
2017-02-08 21:37:36
6^12
6^12
drazn98
2017-02-08 21:37:36
$6^{12}$
$6^{12}$
yid
2017-02-08 21:37:38
6^12?
6^12?
devenware
2017-02-08 21:37:44
There are $6^{12}$ total possible strings
\[P=z_1z_2\cdots z_{12}.\]
There are $6^{12}$ total possible strings
\[P=z_1z_2\cdots z_{12}.\]
devenware
2017-02-08 21:37:54
Okay, there's our denominator. Unfortunately, every denominator in the answer choices for the problem is a factor of this, so we're no closer to solving the problem.
Okay, there's our denominator. Unfortunately, every denominator in the answer choices for the problem is a factor of this, so we're no closer to solving the problem.
devenware
2017-02-08 21:38:22
The product has to have the form $2^a\omega^b$. We want to find all the products that evaluate to $2^0\omega^4$.
The product has to have the form $2^a\omega^b$. We want to find all the products that evaluate to $2^0\omega^4$.
devenware
2017-02-08 21:38:35
Does the $2^0$ tell us anything interesting?
Does the $2^0$ tell us anything interesting?
pandadude
2017-02-08 21:39:28
so we need 8 of the first 2 and 4 of the last 4 so we can get a magnitude of 1
so we need 8 of the first 2 and 4 of the last 4 so we can get a magnitude of 1
warrenwangtennis
2017-02-08 21:39:28
take 8 $sqrt2$s and 4 1/2's
take 8 $sqrt2$s and 4 1/2's
15Pandabears
2017-02-08 21:39:28
You need 8 of the sqrt2 and 4 of the 1/2
You need 8 of the sqrt2 and 4 of the 1/2
drazn98
2017-02-08 21:39:28
the magnitude of the product has to be 1, i.e. pick 8 $\sqrt{2}$'s and 4 $\frac{1}{2}$
the magnitude of the product has to be 1, i.e. pick 8 $\sqrt{2}$'s and 4 $\frac{1}{2}$
wiler5002
2017-02-08 21:39:28
have to choose 8 from the first 2 and 4 from the last 4
have to choose 8 from the first 2 and 4 from the last 4
summitwei
2017-02-08 21:39:28
# of sqrt(2) is twice # of 1/2
# of sqrt(2) is twice # of 1/2
devenware
2017-02-08 21:39:36
In order to get a modulus of 1, we need to pick 8 "big" elements from
\[B=\left\{
\sqrt2\omega^2,\sqrt2\omega^6
\right\}
\] and four "little" elements from
\[
L=\left\{
\frac{\omega}2,\frac{\omega^3}2,\frac{\omega^5}2,\frac{\omega^7}2
\right\}.
\]
In order to get a modulus of 1, we need to pick 8 "big" elements from
\[B=\left\{
\sqrt2\omega^2,\sqrt2\omega^6
\right\}
\] and four "little" elements from
\[
L=\left\{
\frac{\omega}2,\frac{\omega^3}2,\frac{\omega^5}2,\frac{\omega^7}2
\right\}.
\]
devenware
2017-02-08 21:39:42
How many total products can we construct like that (ignoring the argument for now)?
How many total products can we construct like that (ignoring the argument for now)?
devenware
2017-02-08 21:40:32
Let's think about this. We have to have a product that looks like this:
Let's think about this. We have to have a product that looks like this:
devenware
2017-02-08 21:40:39
\[bb\ell b\ell bbb\ell \ell bb\]
\[bb\ell b\ell bbb\ell \ell bb\]
devenware
2017-02-08 21:40:45
With 8 b's and 4 l's.
With 8 b's and 4 l's.
devenware
2017-02-08 21:41:03
In how many ways can we place 8 b's and 4 l's?
In how many ways can we place 8 b's and 4 l's?
warrenwangtennis
2017-02-08 21:41:42
12 choose 4
12 choose 4
mathman3880
2017-02-08 21:41:42
12C4
12C4
brainiac1
2017-02-08 21:41:42
12 choose 4 or 495
12 choose 4 or 495
Sparks29032
2017-02-08 21:41:42
495?
495?
Jayjayliu
2017-02-08 21:41:43
12 choose 4
12 choose 4
xingxia_alexye
2017-02-08 21:42:01
C(12,4)
C(12,4)
goodball
2017-02-08 21:42:01
12 choose4?
12 choose4?
devenware
2017-02-08 21:42:03
There are 12 spaces in total, and we need to choose 4 of them to be $\ell$, so there are $\binom{12}{4}$ ways to do this.
There are 12 spaces in total, and we need to choose 4 of them to be $\ell$, so there are $\binom{12}{4}$ ways to do this.
devenware
2017-02-08 21:42:26
Once those are placed, in how many ways can we replace the $b$'s with elements from $B$?
Once those are placed, in how many ways can we replace the $b$'s with elements from $B$?
mathman3880
2017-02-08 21:43:44
256
256
15Pandabears
2017-02-08 21:43:44
2^8
2^8
amwmath
2017-02-08 21:43:44
$2^8$
$2^8$
goodball
2017-02-08 21:43:44
256?
256?
wiler5002
2017-02-08 21:43:44
$2^8$
$2^8$
ScienceSpirit
2017-02-08 21:43:46
2^8
2^8
devenware
2017-02-08 21:43:53
There are $2^8 = 256$ ways to do that.
There are $2^8 = 256$ ways to do that.
devenware
2017-02-08 21:44:02
In how many ways can we replace the $\ell$'s with elements from $L$?
In how many ways can we replace the $\ell$'s with elements from $L$?
warrenwangtennis
2017-02-08 21:44:35
$4^4$
$4^4$
xingxia_alexye
2017-02-08 21:44:35
2^8
2^8
pandadude
2017-02-08 21:44:35
256
256
ScienceSpirit
2017-02-08 21:44:35
4^4
4^4
brainiac1
2017-02-08 21:44:35
4^4=256
4^4=256
devenware
2017-02-08 21:44:43
There are $4^4 = 256$ ways to do that.
There are $4^4 = 256$ ways to do that.
devenware
2017-02-08 21:44:49
Therefore, ignoring the argument, we have \[N=\binom{12}82^84^4=\frac{12\cdot11\cdot10\cdot9}{24}\cdot2^{16}=2^{16}3^2\cdot5\cdot11\]strings that have the right modulus.
Therefore, ignoring the argument, we have \[N=\binom{12}82^84^4=\frac{12\cdot11\cdot10\cdot9}{24}\cdot2^{16}=2^{16}3^2\cdot5\cdot11\]strings that have the right modulus.
goodball
2017-02-08 21:45:25
Including the argument?
Including the argument?
devenware
2017-02-08 21:45:29
What are all the possible values of our product if it has modulus 1?
What are all the possible values of our product if it has modulus 1?
devenware
2017-02-08 21:47:06
Okay hold on.
Okay hold on.
devenware
2017-02-08 21:47:16
What can we get from the product of 8 elements in $B$?
What can we get from the product of 8 elements in $B$?
devenware
2017-02-08 21:47:27
What power of $\omega$ will we always get?
What power of $\omega$ will we always get?
mathman3880
2017-02-08 21:48:35
0mod4?
0mod4?
brainiac1
2017-02-08 21:48:35
a multiple of 4
a multiple of 4
BobaFett101
2017-02-08 21:48:35
multiple of 4
multiple of 4
devenware
2017-02-08 21:48:50
A multiple of 4 definitely.
A multiple of 4 definitely.
devenware
2017-02-08 21:49:04
Let's actually write this out and see what we find.
Let's actually write this out and see what we find.
devenware
2017-02-08 21:49:19
Eight copies of elements of $B$ gives us:\[(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)=\pm16\omega^{16}.\]
Eight copies of elements of $B$ gives us:\[(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)(\pm\sqrt2\omega^2)=\pm16\omega^{16}.\]
devenware
2017-02-08 21:49:24
How does that simplify?
How does that simplify?
warrenwangtennis
2017-02-08 21:50:02
plus or minus 16
plus or minus 16
Jayjayliu
2017-02-08 21:50:02
plusminus 16
plusminus 16
brainiac1
2017-02-08 21:50:02
just $\pm 16$
just $\pm 16$
BobaFett101
2017-02-08 21:50:02
w^16 = 1
w^16 = 1
mathman3880
2017-02-08 21:50:02
+-16
+-16
devenware
2017-02-08 21:50:19
Since $\omega^{16} = 1,$ that's just $\pm 16.$
Since $\omega^{16} = 1,$ that's just $\pm 16.$
devenware
2017-02-08 21:50:32
Multiplying 4 copies of elements from $L$ is a little harder. We get
\[
\left(\frac{\omega^a}2\right)
\left(\frac{\omega^b}2\right)
\left(\frac{\omega^c}2\right)
\left(\frac{\omega^d}2\right)
=\frac{\omega^{a+b+c+d}}{16}.
\]
Multiplying 4 copies of elements from $L$ is a little harder. We get
\[
\left(\frac{\omega^a}2\right)
\left(\frac{\omega^b}2\right)
\left(\frac{\omega^c}2\right)
\left(\frac{\omega^d}2\right)
=\frac{\omega^{a+b+c+d}}{16}.
\]
devenware
2017-02-08 21:50:39
What can we say about that exponent on $\omega$?
What can we say about that exponent on $\omega$?
warrenwangtennis
2017-02-08 21:51:52
even
even
brainiac1
2017-02-08 21:51:52
its even
its even
mathman3880
2017-02-08 21:51:52
0, 2 mod 4
0, 2 mod 4
Jayjayliu
2017-02-08 21:51:52
no, always even
no, always even
wiler5002
2017-02-08 21:51:52
0 mod 2
0 mod 2
ScienceSpirit
2017-02-08 21:51:52
it's even?
it's even?
devenware
2017-02-08 21:52:01
Thankfully we get the $\dfrac1{16}$ that we wanted. The exponent $a+b+c+d$ is a sum of four odd numbers. That means it is even. In fact, it can be any even residue modulo 8. Specifically, we could multiply
\[
\left(\frac{\omega^a}2\right)
\left(\frac{\omega^a}2\right)
\left(\frac{\omega^1}2\right)
\left(\frac{\omega^7}2\right)
=\frac{\omega^{2a}}{16}.
\]
Thankfully we get the $\dfrac1{16}$ that we wanted. The exponent $a+b+c+d$ is a sum of four odd numbers. That means it is even. In fact, it can be any even residue modulo 8. Specifically, we could multiply
\[
\left(\frac{\omega^a}2\right)
\left(\frac{\omega^a}2\right)
\left(\frac{\omega^1}2\right)
\left(\frac{\omega^7}2\right)
=\frac{\omega^{2a}}{16}.
\]
pandadude
2017-02-08 21:52:50
1,-1,i,-i
1,-1,i,-i
devenware
2017-02-08 21:52:54
Therefore in our set of products with the right modulus, we always get $P$ equal to $1$, $i$, $-1$, or $-i$.
Therefore in our set of products with the right modulus, we always get $P$ equal to $1$, $i$, $-1$, or $-i$.
devenware
2017-02-08 21:53:06
Here's a product equal to 1:
Here's a product equal to 1:
devenware
2017-02-08 21:53:07
\[
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
\left(\frac{\omega^1}2\right)
\left(\frac{\omega^7}2\right)
\left(\frac{\omega^1}2\right)
\left(\frac{\omega^7}2\right)
\]
\[
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
\left(\frac{\omega^1}2\right)
\left(\frac{\omega^7}2\right)
\left(\frac{\omega^1}2\right)
\left(\frac{\omega^7}2\right)
\]
devenware
2017-02-08 21:53:10
Here's a product equal to i:
Here's a product equal to i:
devenware
2017-02-08 21:53:11
\[
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
\left(\frac{\omega^3}2\right)
\left(\frac{\omega^7}2\right)
\left(\frac{\omega^1}2\right)
\left(\frac{\omega^7}2\right)
\]
\[
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
(\sqrt2\omega^4)
(\sqrt2\omega^2)
\left(\frac{\omega^3}2\right)
\left(\frac{\omega^7}2\right)
\left(\frac{\omega^1}2\right)
\left(\frac{\omega^7}2\right)
\]
devenware
2017-02-08 21:53:22
[Yikes those are long]
[Yikes those are long]
devenware
2017-02-08 21:53:32
This is a weird question, but which of the four products, $1$, $i$, $-1$, or $-i$, are we most likely to get?
This is a weird question, but which of the four products, $1$, $i$, $-1$, or $-i$, are we most likely to get?
wiler5002
2017-02-08 21:54:07
symmetric
symmetric
wiler5002
2017-02-08 21:54:07
all equally likely
all equally likely
warrenwangtennis
2017-02-08 21:54:07
all equally likely
all equally likely
letsgomath
2017-02-08 21:54:07
none, they all have probability 1/4
none, they all have probability 1/4
brainiac1
2017-02-08 21:54:07
all of them?
all of them?
goodball
2017-02-08 21:54:34
Equal chance
Equal chance
devenware
2017-02-08 21:54:39
We can turn a product that's equal to 1 into a product that's equal to $i$ by increasing the exponent on the first $\omega^a/2$ by 2 (like the pair of examples above).
We can turn a product that's equal to 1 into a product that's equal to $i$ by increasing the exponent on the first $\omega^a/2$ by 2 (like the pair of examples above).
devenware
2017-02-08 21:54:41
That's a bijection! There are the same number of strings that equal 1 as equal $i$. So?
That's a bijection! There are the same number of strings that equal 1 as equal $i$. So?
warrenwangtennis
2017-02-08 21:55:33
multiply the total number of ways by 1/4
multiply the total number of ways by 1/4
BobaFett101
2017-02-08 21:55:33
divide by 4
divide by 4
devenware
2017-02-08 21:55:48
So the number of products that are equal to $-1$ is $\dfrac N4$.
So the number of products that are equal to $-1$ is $\dfrac N4$.
devenware
2017-02-08 21:55:50
What's the final answer?
What's the final answer?
xingxia_alexye
2017-02-08 21:56:41
E
E
summitwei
2017-02-08 21:56:41
e
e
warrenwangtennis
2017-02-08 21:56:41
E
E
wiler5002
2017-02-08 21:56:41
E
E
devenware
2017-02-08 21:56:46
The probability of getting a product of $-1$ is \[\frac{\frac{N}4}{6^{12}}=\frac{2^{16}\cdot3^2\cdot5\cdot11}{4\cdot6^{12}}=\frac{2^{16}\cdot3^2\cdot5\cdot11}{2^{14}\cdot3^{12}}=\boxed{\dfrac{2^2\cdot5\cdot11}{3^{10}}}.\]
The probability of getting a product of $-1$ is \[\frac{\frac{N}4}{6^{12}}=\frac{2^{16}\cdot3^2\cdot5\cdot11}{4\cdot6^{12}}=\frac{2^{16}\cdot3^2\cdot5\cdot11}{2^{14}\cdot3^{12}}=\boxed{\dfrac{2^2\cdot5\cdot11}{3^{10}}}.\]
devenware
2017-02-08 21:56:47
The answer is (E).
The answer is (E).
devenware
2017-02-08 21:56:58
Please join us again on Thursday, February 16, when we will discuss the AMC 10B/12B contests and also again on March 9 and 24 when we will be discussing the AIME I and II contests.
Please join us again on Thursday, February 16, when we will discuss the AMC 10B/12B contests and also again on March 9 and 24 when we will be discussing the AIME I and II contests.
devenware
2017-02-08 21:57:08
Oh, and soon you will be able to find videos for the last 5 problems of each exam on the videos page:
http://www.artofproblemsolving.com/Videos/index.php?type=amc
We chose a few different tactics in their videos than we used today, so it might be worth checking them out.
Oh, and soon you will be able to find videos for the last 5 problems of each exam on the videos page:
http://www.artofproblemsolving.com/Videos/index.php?type=amc
We chose a few different tactics in their videos than we used today, so it might be worth checking them out.
Cytrew
2017-02-08 21:57:13
yay
yay
goodball
2017-02-08 21:57:13
Yay!!!!!
Yay!!!!!
devenware
2017-02-08 21:57:20
YAY!! WE'RE DONE!
YAY!! WE'RE DONE!
amburger66
2017-02-08 21:57:25
WOW thanks!!
WOW thanks!!
challengetogo
2017-02-08 21:57:25
Thank you!!
Thank you!!
dragonfly
2017-02-08 21:57:25
ty
ty
devenware
2017-02-08 21:57:32
THANK YOU FOR COMING!
THANK YOU FOR COMING!
wiler5002
2017-02-08 21:57:37
THANKS DEVEN
THANKS DEVEN
iks92
2017-02-08 21:57:37
Thank you!!
Thank you!!
amwmath
2017-02-08 21:57:44
$\huge\color{red}{\heartsuit}$
$\huge\color{red}{\heartsuit}$
letsgomath
2017-02-08 21:57:49
When will the transcript of this math jam be available?
When will the transcript of this math jam be available?
devenware
2017-02-08 21:58:03
Yes, you will be able to find it here: https://artofproblemsolving.com/school/mathjams-transcripts in a little bit.
Yes, you will be able to find it here: https://artofproblemsolving.com/school/mathjams-transcripts in a little bit.
devenware
2017-02-08 21:58:27
Have a nice night everyone!
Have a nice night everyone!
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