2016 AMC 10/12 A Discussion
Go back to the Math Jam ArchiveA discussion of problems from the AMC 10/12 A, which is administered February 2. We will cover the last 5 problems on each test, as well as requested earlier problems on the tests.
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Facilitator: Jeremy Copeland
copeland
2016-02-03 19:03:00
Welcome to the 2016 AMC 10A/12A Math Jam!
Welcome to the 2016 AMC 10A/12A Math Jam!
copeland
2016-02-03 19:03:02
I'm Jeremy Copeland, and I'll be leading our discussion tonight.
I'm Jeremy Copeland, and I'll be leading our discussion tonight.
leonlzg
2016-02-03 19:03:10
Hi jeremy
Hi jeremy
copeland
2016-02-03 19:03:12
Hi.
Hi.
copeland
2016-02-03 19:03:13
I'm the school director here at AoPS. That means when something goes wrong, I either get yelled at or have to yell at someone else. Before AoPS, I was an instructor at MIT, and before that I got my Ph.D. from the University of Chicago. Before that I was an undergrad at Reed College and going back even further, I can't really remember. I used to have hobbies, but I'm a parent now, so those days are all over.
I'm the school director here at AoPS. That means when something goes wrong, I either get yelled at or have to yell at someone else. Before AoPS, I was an instructor at MIT, and before that I got my Ph.D. from the University of Chicago. Before that I was an undergrad at Reed College and going back even further, I can't really remember. I used to have hobbies, but I'm a parent now, so those days are all over.
chessdude2015
2016-02-03 19:03:21
Wow! How do you type so fast?
Wow! How do you type so fast?
copeland
2016-02-03 19:03:22
I have two keyboards.
I have two keyboards.
copeland
2016-02-03 19:03:25
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.
copeland
2016-02-03 19:03:35
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.
copeland
2016-02-03 19:03:45
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.
copeland
2016-02-03 19:03:53
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!!
copeland
2016-02-03 19:04:09
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!
copeland
2016-02-03 19:04:35
We plan to have 3 assistants tonight: Nicholas Yang (nackster), Ian Osborn (Frostwriter111), and Nick Bravo (nbravo).
We plan to have 3 assistants tonight: Nicholas Yang (nackster), Ian Osborn (Frostwriter111), and Nick Bravo (nbravo).
copeland
2016-02-03 19:04:46
Right now Ian is the only one who did not get caught in traffic:
Right now Ian is the only one who did not get caught in traffic:
copeland
2016-02-03 19:04:48
Ian joined AoPS in 2015. He graduated from MIT in 2014 with a BS in Mathematics and Chemical Engineering. In high school, he was an active member of the math team. Ian's accomplishments include high individual rankings at competitions such as HMMT and PUMaC, and qualifying for the USAMO four times. His team also finished in 4th place in the 2010 Moody's Mega Math Challenge, a mathematical modeling competition. In his free time, Ian enjoys singing, logic puzzles, and board games.
Ian joined AoPS in 2015. He graduated from MIT in 2014 with a BS in Mathematics and Chemical Engineering. In high school, he was an active member of the math team. Ian's accomplishments include high individual rankings at competitions such as HMMT and PUMaC, and qualifying for the USAMO four times. His team also finished in 4th place in the 2010 Moody's Mega Math Challenge, a mathematical modeling competition. In his free time, Ian enjoys singing, logic puzzles, and board games.
AlanMTuring
2016-02-03 19:05:00
Mr. Copeland, how can you use two keyboards at once?
Mr. Copeland, how can you use two keyboards at once?
Frostwriter111
2016-02-03 19:05:00
Hello everyone!
Hello everyone!
copeland
2016-02-03 19:05:02
Very quickly.
Very quickly.
christbow
2016-02-03 19:05:15
Who's levans?
Who's levans?
copeland
2016-02-03 19:05:16
He doesn't exist. You imagined him.
He doesn't exist. You imagined him.
copeland
2016-02-03 19:05:29
Assistants can answer questions by whispering to you or by opening a window with you to chat 1-on-1. 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 is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.
Assistants can answer questions by whispering to you or by opening a window with you to chat 1-on-1. 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 is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.
chessdude2015
2016-02-03 19:05:47
Who is markchen?
Who is markchen?
BAYMAX_HAMADA
2016-02-03 19:05:47
whos mark chen?
whos mark chen?
shivao
2016-02-03 19:05:47
who's markchen
who's markchen
copeland
2016-02-03 19:05:48
He's another of our awesome instructors, just hanging out.
He's another of our awesome instructors, just hanging out.
markchen
2016-02-03 19:05:57
Hi!
Hi!
copeland
2016-02-03 19:06:28
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 Problems 24 and 25 are 12A Problems 21 and 22. We'll only solve these problems 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 Problems 24 and 25 are 12A Problems 21 and 22. We'll only solve these problems once.
copeland
2016-02-03 19:06:40
We usually promise to chat about other problems on the test, but honestly, I tend to meander and most of us will be exhausted by the end of these 8 problems.
We usually promise to chat about other problems on the test, but honestly, I tend to meander and most of us will be exhausted by the end of these 8 problems.
copeland
2016-02-03 19:06:55
Who's ready to rock?
Who's ready to rock?
ImpossibleSphere
2016-02-03 19:07:10
ME!
ME!
Natsu_Dragneel
2016-02-03 19:07:10
ME
ME
Lion7
2016-02-03 19:07:10
MathConquer2014
2016-02-03 19:07:10
ME!
ME!
song2sons
2016-02-03 19:07:10
me!
me!
claserken
2016-02-03 19:07:10
Me
Me
overninethousand
2016-02-03 19:07:10
me!
me!
Wh2002
2016-02-03 19:07:10
me
me
IceFireGold1
2016-02-03 19:07:10
Me
Me
jfurf
2016-02-03 19:07:10
Me!
Me!
goodbear
2016-02-03 19:07:10
me
me
stan23456
2016-02-03 19:07:10
Me!
Me!
TroyMathGuy
2016-02-03 19:07:10
Im so exited
Im so exited
happyribbon
2016-02-03 19:07:10
Me!
Me!
amcwu
2016-02-03 19:07:10
me
me
leonlzg
2016-02-03 19:07:10
I am!
I am!
MaYang
2016-02-03 19:07:10
fatcat36
2016-02-03 19:07:10
ME
ME
beanielove2
2016-02-03 19:07:10
Meeeee
Meeeee
Aspen
2016-02-03 19:07:10
not me
not me
AoPSuser056
2016-02-03 19:07:10
Me!
Me!
zardum
2016-02-03 19:07:10
space_space
2016-02-03 19:07:10
not me
not me
lcalvert99
2016-02-03 19:07:10
Yeah!!!!
Yeah!!!!
Luke Skywalker
2016-02-03 19:07:10
I am!
I am!
celestialphoenix3768
2016-02-03 19:07:10
me
me
Emathmaster
2016-02-03 19:07:10
Me!!!
Me!!!
shivao
2016-02-03 19:07:10
me!!!!!!!!!!!!!!!!!!!!!!!
me!!!!!!!!!!!!!!!!!!!!!!!
noobynoob
2016-02-03 19:07:10
Teams
Teams
jasonmath
2016-02-03 19:07:10
mes
mes
Nirnivna
2016-02-03 19:07:10
Me!
Me!
goldypeng
2016-02-03 19:07:10
Me
Me
CobbleHead
2016-02-03 19:07:10
me
me
copeland
2016-02-03 19:07:18
OK, let's get started!
OK, let's get started!
copeland
2016-02-03 19:07:32
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. 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. Please forgive me in advance. That doesn't happen in our classes.
copeland
2016-02-03 19:07:38
And on with the AMC10:
And on with the AMC10:
copeland
2016-02-03 19:07:43
21. Circles with centers $P, Q,$ and $R,$ having radii 1, 2, and 3, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q',$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of $\triangle PQR$?
$\phantom{10A:21}$
(A) $0 \qquad$ (B) $\sqrt{\dfrac{2}{3}} \qquad$ (C) $1 \qquad$ (D) $\sqrt{6} - \sqrt{2} \qquad$ (E) $\sqrt{\dfrac{3}{2}}$
21. Circles with centers $P, Q,$ and $R,$ having radii 1, 2, and 3, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q',$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of $\triangle PQR$?
$\phantom{10A:21}$
(A) $0 \qquad$ (B) $\sqrt{\dfrac{2}{3}} \qquad$ (C) $1 \qquad$ (D) $\sqrt{6} - \sqrt{2} \qquad$ (E) $\sqrt{\dfrac{3}{2}}$
copeland
2016-02-03 19:07:52
Step one in a problem like this?
Step one in a problem like this?
Alanshenkerman
2016-02-03 19:08:07
diagram
diagram
stan23456
2016-02-03 19:08:07
Diagram?
Diagram?
Mudkipswims42
2016-02-03 19:08:07
Picture!
Picture!
akaashp11
2016-02-03 19:08:07
Diagram
Diagram
jfurf
2016-02-03 19:08:07
Picture!
Picture!
fractal161
2016-02-03 19:08:07
Diagram!!!!
Diagram!!!!
thkim1011
2016-02-03 19:08:07
diagram
diagram
Radical_CC
2016-02-03 19:08:07
picture!
picture!
bigboss
2016-02-03 19:08:07
draw diagram
draw diagram
mathfan1256
2016-02-03 19:08:07
Draw a diagram
Draw a diagram
Ultimate_draco
2016-02-03 19:08:07
draw a picture
draw a picture
baseballcat
2016-02-03 19:08:07
draw a diagram
draw a diagram
copeland
2016-02-03 19:08:10
Let's draw a diagram!
Let's draw a diagram!
copeland
2016-02-03 19:08:13
copeland
2016-02-03 19:08:16
Now I don't yet know whether $PR$ passes above or below $Q$, but we can verify this in a second. It's important to start by drawing some diagram and we can always change it up when we get more information.
Now I don't yet know whether $PR$ passes above or below $Q$, but we can verify this in a second. It's important to start by drawing some diagram and we can always change it up when we get more information.
copeland
2016-02-03 19:08:27
What are $PQ$ and $QR$?
What are $PQ$ and $QR$?
swagchicken
2016-02-03 19:08:58
3 and 5
3 and 5
eswa2000
2016-02-03 19:08:58
3,5
3,5
thomas0115
2016-02-03 19:08:58
3 and 5
3 and 5
Alacarter
2016-02-03 19:08:58
3 and 5
3 and 5
GeorgCantor
2016-02-03 19:08:58
3 and 5
3 and 5
duruphi
2016-02-03 19:08:58
3 and 5
3 and 5
ChickenOnRage
2016-02-03 19:08:58
3 and 5 respectively
3 and 5 respectively
arvind2002
2016-02-03 19:08:58
3 and 5
3 and 5
FlakeLCR
2016-02-03 19:08:58
3,5
3,5
yrnsmurf
2016-02-03 19:08:58
pq is 3 and qr is 5
pq is 3 and qr is 5
xiaodongxi
2016-02-03 19:08:58
PQ=3, QR=5
PQ=3, QR=5
copeland
2016-02-03 19:09:00
Since $PQ$ and $QR$ are made from radii of the circles, $PQ=1+2=3$ and $QR=2+3=5$.
Since $PQ$ and $QR$ are made from radii of the circles, $PQ=1+2=3$ and $QR=2+3=5$.
copeland
2016-02-03 19:09:03
Now how should we compute the area of $\triangle PQR?$
Now how should we compute the area of $\triangle PQR?$
arvind2002
2016-02-03 19:09:20
herons?
herons?
Alacarter
2016-02-03 19:09:20
heron's formula
heron's formula
awesomethree
2016-02-03 19:09:20
herons
herons
ciao_potter
2016-02-03 19:09:20
herons?
herons?
lcalvert99
2016-02-03 19:09:20
Heron?
Heron?
copeland
2016-02-03 19:09:23
Crazy.
Crazy.
copeland
2016-02-03 19:09:25
Anything simpler?
Anything simpler?
tipro
2016-02-03 19:09:50
subtaction
subtaction
garyasho
2016-02-03 19:09:50
Subtract area of 2 trapezoids
Subtract area of 2 trapezoids
thomas0115
2016-02-03 19:09:50
subtract areas
subtract areas
eswa2000
2016-02-03 19:09:50
add two trapezoids and subtract one
add two trapezoids and subtract one
pedronr
2016-02-03 19:09:50
subtraction
subtraction
akaashp11
2016-02-03 19:09:50
Subtracting areas
Subtracting areas
tipro
2016-02-03 19:09:50
find the area of the big figure and subtract triangles
find the area of the big figure and subtract triangles
copeland
2016-02-03 19:09:54
We can subtract! There are a number of figures we can easily compute the areas of that we can use to find $\triangle PQR$. Let's drop a couple of vertical and horizontal lines on the page:
We can subtract! There are a number of figures we can easily compute the areas of that we can use to find $\triangle PQR$. Let's drop a couple of vertical and horizontal lines on the page:
copeland
2016-02-03 19:10:10
copeland
2016-02-03 19:10:12
Now what are these horizontal lengths?
Now what are these horizontal lengths?
spartan168
2016-02-03 19:10:33
Pythagorean Theorem!
Pythagorean Theorem!
copeland
2016-02-03 19:10:34
For sure!
For sure!
Alacarter
2016-02-03 19:11:12
rad 8 and rad 24
rad 8 and rad 24
vitriolhumor
2016-02-03 19:11:12
sqrt8 and sqrt24
sqrt8 and sqrt24
MaYang
2016-02-03 19:11:12
P'Q'=2root(2), Q'R'=2root(6)
P'Q'=2root(2), Q'R'=2root(6)
jclash
2016-02-03 19:11:12
2sqrt2 and 2sqrt6
2sqrt2 and 2sqrt6
MaYang
2016-02-03 19:11:12
P'Q'=2root(2), Q'R'=2root(6)
P'Q'=2root(2), Q'R'=2root(6)
bigboss
2016-02-03 19:11:12
2\sqrt2 and 2\sqrt6
2\sqrt2 and 2\sqrt6
yrnsmurf
2016-02-03 19:11:12
$2\sqrt{6}$ for qr and $2\sqrt2$ for pq
$2\sqrt{6}$ for qr and $2\sqrt2$ for pq
eswa2000
2016-02-03 19:11:12
copeland
2016-02-03 19:11:25
The vertical lengths are 1, so the Pythagorean Theorem gives us horizontal lengths of $\sqrt{9-1}=\sqrt8=2\sqrt2$ and $\sqrt{25-1}=\sqrt24=2\sqrt6$.
The vertical lengths are 1, so the Pythagorean Theorem gives us horizontal lengths of $\sqrt{9-1}=\sqrt8=2\sqrt2$ and $\sqrt{25-1}=\sqrt24=2\sqrt6$.
copeland
2016-02-03 19:11:32
copeland
2016-02-03 19:11:37
Incidentally, what's the slope of $PQ?$
Incidentally, what's the slope of $PQ?$
pedronr
2016-02-03 19:12:18
1/(2sqrt2)
1/(2sqrt2)
mathcool2009
2016-02-03 19:12:18
1/2\sqrt{2}
1/2\sqrt{2}
idomath12345
2016-02-03 19:12:18
sqrt2/4
sqrt2/4
beanielove2
2016-02-03 19:12:18
1/2\sqrt{2}.
1/2\sqrt{2}.
strategos21
2016-02-03 19:12:18
sqrt2/4
sqrt2/4
awesomethree
2016-02-03 19:12:18
sqrt(2)/4
sqrt(2)/4
myungsooglee
2016-02-03 19:12:18
sqrt2/4
sqrt2/4
PNPADU
2016-02-03 19:12:18
1/2root2
1/2root2
alexli2014
2016-02-03 19:12:18
sqrt(2)/4
sqrt(2)/4
ninjataco
2016-02-03 19:12:18
sqrt2 / 4
sqrt2 / 4
copeland
2016-02-03 19:12:22
The slope of $PQ$ is $\dfrac1{2\sqrt2}$. Likewise the slope of $QR$ is $\dfrac1{2\sqrt{6}}$. Since $\dfrac1{2\sqrt2}>\dfrac1{2\sqrt{6}}$ we now know $Q$ lies above $PR$.
The slope of $PQ$ is $\dfrac1{2\sqrt2}$. Likewise the slope of $QR$ is $\dfrac1{2\sqrt{6}}$. Since $\dfrac1{2\sqrt2}>\dfrac1{2\sqrt{6}}$ we now know $Q$ lies above $PR$.
copeland
2016-02-03 19:12:31
Finally we see that $\triangle PQR$ can be build by combining two triangles and a rectangle and then subtracting a triangle.
Finally we see that $\triangle PQR$ can be build by combining two triangles and a rectangle and then subtracting a triangle.
copeland
2016-02-03 19:12:32
We begin by adding the three regions:
We begin by adding the three regions:
copeland
2016-02-03 19:12:34
copeland
2016-02-03 19:12:34
copeland
2016-02-03 19:12:35
copeland
2016-02-03 19:12:41
and then subtracting the triangle
and then subtracting the triangle
copeland
2016-02-03 19:12:44
copeland
2016-02-03 19:12:54
What's the area of this region:
What's the area of this region:
copeland
2016-02-03 19:12:56
Myst_Hawk
2016-02-03 19:13:19
sqrt2
sqrt2
leafshadow
2016-02-03 19:13:19
sqrt 2
sqrt 2
bringerofawesomeness
2016-02-03 19:13:19
sqrt2
sqrt2
Natsu_Dragneel
2016-02-03 19:13:19
root2
root2
ImpossibleSphere
2016-02-03 19:13:19
$\sqrt{2}$
$\sqrt{2}$
Matthew2000
2016-02-03 19:13:19
sqrt2
sqrt2
Locust
2016-02-03 19:13:19
IceFireGold1
2016-02-03 19:13:19
\sqrt{2}
\sqrt{2}
CobbleHead
2016-02-03 19:13:19
sqrt(2)
sqrt(2)
Wh2002
2016-02-03 19:13:19
root 2
root 2
copeland
2016-02-03 19:13:22
This region has area $\dfrac12\cdot2\sqrt2\cdot1=\sqrt2$.
This region has area $\dfrac12\cdot2\sqrt2\cdot1=\sqrt2$.
copeland
2016-02-03 19:13:22
What's the area of this region:
What's the area of this region:
copeland
2016-02-03 19:13:23
amcwu
2016-02-03 19:13:51
2root6
2root6
shivao
2016-02-03 19:13:51
2sqrt(6)
2sqrt(6)
goodbear
2016-02-03 19:13:51
2√6
2√6
APR02
2016-02-03 19:13:51
2 rad 6
2 rad 6
rocket13jg
2016-02-03 19:13:51
2rt6
2rt6
axue
2016-02-03 19:13:51
2 sqrt6
2 sqrt6
RiverTam
2016-02-03 19:13:51
2 sqrt(6)
2 sqrt(6)
math1012
2016-02-03 19:13:51
$2\sqrt{6}$
$2\sqrt{6}$
PI.E_3.14
2016-02-03 19:13:51
2sqrt(6)
2sqrt(6)
Emathmaster
2016-02-03 19:13:51
2sqrt6
2sqrt6
chessdude2015
2016-02-03 19:13:51
2 root 6
2 root 6
goodbear
2016-02-03 19:13:51
2sqrt(6)
2sqrt(6)
celestialphoenix3768
2016-02-03 19:13:51
2sqrt6
2sqrt6
copeland
2016-02-03 19:13:55
This region has area $1\cdot2\sqrt6=2\sqrt6$.
This region has area $1\cdot2\sqrt6=2\sqrt6$.
copeland
2016-02-03 19:13:56
What's the area of this region:
What's the area of this region:
copeland
2016-02-03 19:13:58
jfurf
2016-02-03 19:14:22
$\sqrt{6}$
$\sqrt{6}$
BAYMAX_HAMADA
2016-02-03 19:14:22
sqrt6
sqrt6
Mudkipswims42
2016-02-03 19:14:22
$\sqrt{6}$
$\sqrt{6}$
space_space
2016-02-03 19:14:22
sqrt6
sqrt6
blueberry2
2016-02-03 19:14:22
sqrt6
sqrt6
BillytheChamp
2016-02-03 19:14:22
sqrt6
sqrt6
MathConquer2014
2016-02-03 19:14:22
sqrt6
sqrt6
buzhidao
2016-02-03 19:14:22
$\sqrt{6}$
$\sqrt{6}$
claserken
2016-02-03 19:14:22
root(6)
root(6)
TheRealDeal
2016-02-03 19:14:22
sqrt6
sqrt6
blizzard10
2016-02-03 19:14:22
$\sqrt{6}$
$\sqrt{6}$
Peggy
2016-02-03 19:14:22
$\sqrt{6}$
$\sqrt{6}$
copeland
2016-02-03 19:14:26
This region is half as large as the rectangle we just computed so it has area $\sqrt6$.
This region is half as large as the rectangle we just computed so it has area $\sqrt6$.
copeland
2016-02-03 19:14:28
The total area of the green regions is $\sqrt2+2\sqrt6+\sqrt6=\sqrt2+3\sqrt6.$
The total area of the green regions is $\sqrt2+2\sqrt6+\sqrt6=\sqrt2+3\sqrt6.$
copeland
2016-02-03 19:14:28
What's the area of the blue region:
What's the area of the blue region:
copeland
2016-02-03 19:14:30
amcwu
2016-02-03 19:15:13
2root2 + 2 root 6
2root2 + 2 root 6
bigboss
2016-02-03 19:15:13
2\sqrt2 + 2\sqrt6
2\sqrt2 + 2\sqrt6
alexli2014
2016-02-03 19:15:13
2sqrt(2) + 2sqrt(6)
2sqrt(2) + 2sqrt(6)
deltaepsilon6
2016-02-03 19:15:13
2root(6)+2root(2)
2root(6)+2root(2)
Shruige1
2016-02-03 19:15:13
2rt2 + 2rt6
2rt2 + 2rt6
Emathmaster
2016-02-03 19:15:13
2sqrt2+2sqrt6
2sqrt2+2sqrt6
Sophiarulz
2016-02-03 19:15:13
2sqrt2 + 2sqrt6
2sqrt2 + 2sqrt6
ethanproz
2016-02-03 19:15:13
$2\sqrt{2} + 2\sqrt{6}$
$2\sqrt{2} + 2\sqrt{6}$
goodbear
2016-02-03 19:15:13
2√2+2√6
2√2+2√6
Anno
2016-02-03 19:15:17
2\sqrt2 + 2\sqrt6
2\sqrt2 + 2\sqrt6
copeland
2016-02-03 19:15:22
This region has area $\dfrac12\cdot(2\sqrt2+2\sqrt6)\cdot2=2\sqrt2+2\sqrt6$.
This region has area $\dfrac12\cdot(2\sqrt2+2\sqrt6)\cdot2=2\sqrt2+2\sqrt6$.
copeland
2016-02-03 19:15:22
And what is the area of $\triangle PQR?$
And what is the area of $\triangle PQR?$
dchen04
2016-02-03 19:16:13
rad(2)+2*rad(6)+rad(6)-2*rad(2)-2*rad(6)=rad(6)-rad(2)
rad(2)+2*rad(6)+rad(6)-2*rad(2)-2*rad(6)=rad(6)-rad(2)
ciao_potter
2016-02-03 19:16:13
$\sqrt{6}-\sqrt{2}$
$\sqrt{6}-\sqrt{2}$
thomas0115
2016-02-03 19:16:13
sqrt(6)-sqrt(2)
sqrt(6)-sqrt(2)
happyribbon
2016-02-03 19:16:13
sqrt6 - sqrt2
sqrt6 - sqrt2
yuvi18
2016-02-03 19:16:13
sqrt(6)- sqrt(2)
sqrt(6)- sqrt(2)
arvind2002
2016-02-03 19:16:13
root 6 -root2
root 6 -root2
garyasho
2016-02-03 19:16:13
root 6 - root 2
root 6 - root 2
lasergazer
2016-02-03 19:16:13
sqrt6 - sqrt2
sqrt6 - sqrt2
ciao_potter
2016-02-03 19:16:13
$\sqrt{6}-\sqrt{2}$
$\sqrt{6}-\sqrt{2}$
Derrick.Liang
2016-02-03 19:16:13
sqrt(6)-sqrt(2)
sqrt(6)-sqrt(2)
copeland
2016-02-03 19:16:16
Triangle $PQR$ has area
\[(\sqrt2+3\sqrt6)-(2\sqrt2+2\sqrt6)=\boxed{\sqrt6-\sqrt2}.\]
Triangle $PQR$ has area
\[(\sqrt2+3\sqrt6)-(2\sqrt2+2\sqrt6)=\boxed{\sqrt6-\sqrt2}.\]
copeland
2016-02-03 19:16:20
The answer is $D$.
The answer is $D$.
skimisgod
2016-02-03 19:17:14
Yay!
Yay!
rocket13jg
2016-02-03 19:17:14
yay
yay
Aspen
2016-02-03 19:17:14
YAY~
YAY~
FarmGirl
2016-02-03 19:17:14
That is so cool! Thank you.
That is so cool! Thank you.
copeland
2016-02-03 19:17:20
Indeed. One down.
Indeed. One down.
copeland
2016-02-03 19:17:36
What's next?
What's next?
Nirnivna
2016-02-03 19:17:52
22
22
BillytheChamp
2016-02-03 19:17:52
22
22
yrnsmurf
2016-02-03 19:17:52
question 22?
question 22?
ImpossibleSphere
2016-02-03 19:17:52
22, right?
22, right?
akaashp11
2016-02-03 19:17:52
{22}
{22}
copeland
2016-02-03 19:17:57
OK, great.
OK, great.
copeland
2016-02-03 19:18:01
22. For some positive integer $n$, the number $110n^3$ has 110 positive integer divisors, including 1 and the number $110n^3$. How many positive integer divisors does $81n^4$ have?
$\phantom{10A:22}$
(A) $110\qquad$ (B) $191 \qquad$ (C) $261 \qquad$ (D) $325 \qquad$ (E) $425$
22. For some positive integer $n$, the number $110n^3$ has 110 positive integer divisors, including 1 and the number $110n^3$. How many positive integer divisors does $81n^4$ have?
$\phantom{10A:22}$
(A) $110\qquad$ (B) $191 \qquad$ (C) $261 \qquad$ (D) $325 \qquad$ (E) $425$
copeland
2016-02-03 19:18:05
This problem is about counting divisors. We're probably going to need that formula so remind me. How many divisors does $3^87^{12}17^4$ have?
This problem is about counting divisors. We're probably going to need that formula so remind me. How many divisors does $3^87^{12}17^4$ have?
Ericaops
2016-02-03 19:18:43
9*13*5
9*13*5
mathcool2009
2016-02-03 19:18:43
(8+1)(12+1)(4+1)
(8+1)(12+1)(4+1)
ChickenOnRage
2016-02-03 19:18:43
9*13*5
9*13*5
coconut_force
2016-02-03 19:18:43
9 * 13 * 5
9 * 13 * 5
eswa2000
2016-02-03 19:18:43
9*13*5
9*13*5
owm
2016-02-03 19:18:43
(8+1)(12+1)(4+1)
(8+1)(12+1)(4+1)
Alanshenkerman
2016-02-03 19:18:43
(8+1)(12+1)(4+1)
(8+1)(12+1)(4+1)
galaxy1234
2016-02-03 19:18:43
(9)(13)(5)
(9)(13)(5)
beanielove2
2016-02-03 19:18:43
$9\cdot 13\cdot 5.$
$9\cdot 13\cdot 5.$
spartan168
2016-02-03 19:18:43
(8+1) * (12+1) * (4+1) = 585
(8+1) * (12+1) * (4+1) = 585
Einsteinhead
2016-02-03 19:18:43
$(8+1)(12+1)(4+1)=595$
$(8+1)(12+1)(4+1)=595$
copeland
2016-02-03 19:18:45
In order to construct a divisor of $3^87^{12}17^4$ we need to pick some number of 3s between 0 and 8, we need to pick some number of 7s between 0 and 12 and we need to pick some number of 17s between 0 and 4.
In order to construct a divisor of $3^87^{12}17^4$ we need to pick some number of 3s between 0 and 8, we need to pick some number of 7s between 0 and 12 and we need to pick some number of 17s between 0 and 4.
copeland
2016-02-03 19:18:52
There are 9 ways to pick a number of threes, 13 ways to pick a number of 7s and 5 ways to pick a number of 17. The number of divisors of $3^87^{12}17^4$ is \[(8+1)(12+1)(4+1).\]
There are 9 ways to pick a number of threes, 13 ways to pick a number of 7s and 5 ways to pick a number of 17. The number of divisors of $3^87^{12}17^4$ is \[(8+1)(12+1)(4+1).\]
copeland
2016-02-03 19:18:55
More generally, if we have a number factored into a product of prime powers, we can count its divisors by adding one to each exponent and multiplying all the results.
More generally, if we have a number factored into a product of prime powers, we can count its divisors by adding one to each exponent and multiplying all the results.
copeland
2016-02-03 19:19:05
(Assuming the bases are all distinct.)
(Assuming the bases are all distinct.)
copeland
2016-02-03 19:19:13
So here, when we add one to all the exponents of the prime powers in $110n^3$ and multiply those together we get 110.
So here, when we add one to all the exponents of the prime powers in $110n^3$ and multiply those together we get 110.
copeland
2016-02-03 19:19:14
Does that tell us anything immediately about $n?$
Does that tell us anything immediately about $n?$
eisirrational
2016-02-03 19:20:20
so n can only have 2,5,11 as prime factors
so n can only have 2,5,11 as prime factors
eswa2000
2016-02-03 19:20:20
it only has prime factors 2,5,11
it only has prime factors 2,5,11
BitterGummy
2016-02-03 19:20:26
n has factors of 2, 5 or 11
n has factors of 2, 5 or 11
Ericaops
2016-02-03 19:20:26
has prime powers of 2,5, and/or 11
has prime powers of 2,5, and/or 11
copeland
2016-02-03 19:20:29
The number 110 is the product of three primes, $110=2\cdot5\cdot11.$ Therefore there are at most 3 nonzero prime powers in the factorization of $110n^3.$ Since we already know 2, 3, and 5 divide $110n^3$ there can be no other primes that divide $n$.
The number 110 is the product of three primes, $110=2\cdot5\cdot11.$ Therefore there are at most 3 nonzero prime powers in the factorization of $110n^3.$ Since we already know 2, 3, and 5 divide $110n^3$ there can be no other primes that divide $n$.
copeland
2016-02-03 19:21:12
So we can write $n=2^a5^b11^c$ for nonnegative $a$, $b$, and $c$.
So we can write $n=2^a5^b11^c$ for nonnegative $a$, $b$, and $c$.
copeland
2016-02-03 19:21:14
And what equation does $a$, $b$, and $c$ solve?
And what equation does $a$, $b$, and $c$ solve?
copeland
2016-02-03 19:22:03
Tough, question. Let's back it out a little. How do we factor $110n^3,$ first?
Tough, question. Let's back it out a little. How do we factor $110n^3,$ first?
Natsu_Dragneel
2016-02-03 19:22:36
2*5*11*n^3
2*5*11*n^3
Mudkipswims42
2016-02-03 19:22:36
$2*5*11*n^3$
$2*5*11*n^3$
claserken
2016-02-03 19:22:36
2*5*11*n*n*n
2*5*11*n*n*n
copeland
2016-02-03 19:22:47
And we know the factorization of $n$. . .
And we know the factorization of $n$. . .
ninjataco
2016-02-03 19:23:24
$2^{3a+1}5^{3b+1}11^{3c+1}$
$2^{3a+1}5^{3b+1}11^{3c+1}$
coconut_force
2016-02-03 19:23:24
2 * 5 * 11 * (2^a * 5^b * 11^c) ^ 3
2 * 5 * 11 * (2^a * 5^b * 11^c) ^ 3
goodbear
2016-02-03 19:23:24
2^(3a+1)*5^(3b+1)*11^(3c+1)
2^(3a+1)*5^(3b+1)*11^(3c+1)
copeland
2016-02-03 19:23:43
Great.
\[110n^3=(2\cdot5\cdot11)(2^a5^b11^c)^3=2^{3a+1}5^{3b+1}11^{3c+1}.\]
Great.
\[110n^3=(2\cdot5\cdot11)(2^a5^b11^c)^3=2^{3a+1}5^{3b+1}11^{3c+1}.\]
copeland
2016-02-03 19:23:49
What does our divisor formula say?
What does our divisor formula say?
eswa2000
2016-02-03 19:24:32
(3a+2)(3b+2)(3c+2)=110
(3a+2)(3b+2)(3c+2)=110
ninjataco
2016-02-03 19:24:32
(3a + 2)(3b + 2)(3c + 2) = 110
(3a + 2)(3b + 2)(3c + 2) = 110
Lion7
2016-02-03 19:24:32
(3a+2)(3b+2)(3c+2)=110
(3a+2)(3b+2)(3c+2)=110
blizzard10
2016-02-03 19:24:32
So $(3a+2)(3b+2)(3c+2) = 110$
So $(3a+2)(3b+2)(3c+2) = 110$
jfurf
2016-02-03 19:24:32
$(3a+2)\cdot{(3b+2)}\cdot{(3c+2)}=110$.
$(3a+2)\cdot{(3b+2)}\cdot{(3c+2)}=110$.
brisane
2016-02-03 19:24:32
(3a+2)(3b+2)(3c+2) = 110.
(3a+2)(3b+2)(3c+2) = 110.
skimisgod
2016-02-03 19:24:32
(3a+2)(3b+2)(3c+2)=110
(3a+2)(3b+2)(3c+2)=110
idomath12345
2016-02-03 19:24:32
(3a+2)(3b+2)(3c+2)=110?
(3a+2)(3b+2)(3c+2)=110?
copeland
2016-02-03 19:24:35
Our divisor formula says\[(3a+2)(3b+2)(3c+2)=2\cdot5\cdot11.\]
Our divisor formula says\[(3a+2)(3b+2)(3c+2)=2\cdot5\cdot11.\]
copeland
2016-02-03 19:24:37
So what are $a$, $b$, and $c$?
So what are $a$, $b$, and $c$?
fractal161
2016-02-03 19:25:36
0, 1, and 3, in some semi-random order
0, 1, and 3, in some semi-random order
SHARKYBOY
2016-02-03 19:25:36
3, 1, 0 in some order
3, 1, 0 in some order
Peggy
2016-02-03 19:25:36
0, 1, 3 in some order
0, 1, 3 in some order
bigboss
2016-02-03 19:25:36
0,1,3
0,1,3
GeorgCantor
2016-02-03 19:25:36
0, 1, 3
0, 1, 3
leafshadow
2016-02-03 19:25:36
0, 1, 3
0, 1, 3
copeland
2016-02-03 19:25:40
We don't know what they are exactly, but in some order \[\{3a+2,3b+2,3c+2\}=\{2,5,11\}\]so \[\{a,b,c\}=\{0,1,3\}.\]
We don't know what they are exactly, but in some order \[\{3a+2,3b+2,3c+2\}=\{2,5,11\}\]so \[\{a,b,c\}=\{0,1,3\}.\]
copeland
2016-02-03 19:25:43
Now how many positive divisors does $81n^4$ have?
Now how many positive divisors does $81n^4$ have?
thomas0115
2016-02-03 19:26:54
325
325
ryanyz10
2016-02-03 19:26:54
5 * 5 * 13 = 325
5 * 5 * 13 = 325
Emathmaster
2016-02-03 19:26:54
325
325
yrnsmurf
2016-02-03 19:26:54
5*5*13=325
5*5*13=325
Ericaops
2016-02-03 19:26:54
5*13*5
5*13*5
blueberry2
2016-02-03 19:26:54
5(5)(13)=325
5(5)(13)=325
SimonSun
2016-02-03 19:26:54
5(13)(5)
5(13)(5)
eswa2000
2016-02-03 19:26:54
5*5*13
5*5*13
geogirl08
2016-02-03 19:26:54
5*13*5 = 325
5*13*5 = 325
mathproash2
2016-02-03 19:26:54
13*5*1*5 = D) 325
13*5*1*5 = D) 325
MaYang
2016-02-03 19:26:54
5(4a+1)(4b+1)(4c+1)
5(4a+1)(4b+1)(4c+1)
Alacarter
2016-02-03 19:26:57
(4+1)(4a+1)(4b+1)(4c+1)
(4+1)(4a+1)(4b+1)(4c+1)
copeland
2016-02-03 19:27:01
Well \[81n^4=(3^4)(2^a5^b11^c)^4=2^{4a}3^45^{4b}11^{4c}\] so it has \[(4a+1)(4+1)(4b+1)(4c+1)\] total divisors.
Well \[81n^4=(3^4)(2^a5^b11^c)^4=2^{4a}3^45^{4b}11^{4c}\] so it has \[(4a+1)(4+1)(4b+1)(4c+1)\] total divisors.
copeland
2016-02-03 19:27:08
We still don't know what $a$, $b$, and $c$ are, but they again appear symmetrically in this product so we know $4a+1,$ $4b+1,$ and $4c+1$ are equal to $4\cdot0+1=4,$ $4\cdot1+1=5,$ and $4\cdot3+1=13$ in some order.
We still don't know what $a$, $b$, and $c$ are, but they again appear symmetrically in this product so we know $4a+1,$ $4b+1,$ and $4c+1$ are equal to $4\cdot0+1=4,$ $4\cdot1+1=5,$ and $4\cdot3+1=13$ in some order.
copeland
2016-02-03 19:27:10
The total number of divisors is \[5\cdot1\cdot5\cdot13=\boxed{325}.\]
The total number of divisors is \[5\cdot1\cdot5\cdot13=\boxed{325}.\]
copeland
2016-02-03 19:27:12
The answer is $D$.
The answer is $D$.
copeland
2016-02-03 19:27:29
Great. Two down. Everybody warmed up now?
Great. Two down. Everybody warmed up now?
overninethousand
2016-02-03 19:27:50
yes
yes
Nirnivna
2016-02-03 19:27:50
YES!!!
YES!!!
SmartYuRuo
2016-02-03 19:27:50
YEA
YEA
goldypeng
2016-02-03 19:27:50
yes
yes
dchen04
2016-02-03 19:27:50
yes
yes
mathmaniatoo
2016-02-03 19:27:50
yee
yee
copeland
2016-02-03 19:27:55
OK, on to. . . .
OK, on to. . . .
copeland
2016-02-03 19:27:58
23. A binary operation $\diamondsuit$ has the properties that $a\diamondsuit (b \diamondsuit c) = (a \diamondsuit b) \cdot c$ and that $a \diamondsuit a = 1$ for all nonzero real numbers $a,b,$ and $c$. (Here the dot $\cdot$ represents the usual multiplication operation.) The solution to the equation $2016 \diamondsuit(6\diamondsuit x) = 100$ can be written as $\frac pq$, where $p$ and $q$ are relatively prime positive integers. What is $p+q$?
$\phantom{10A:23}$
(A) $109 \qquad$ (B) $201 \qquad$ (C) $301 \qquad$ (D) $3049 \qquad$ (E) $33,601$
23. A binary operation $\diamondsuit$ has the properties that $a\diamondsuit (b \diamondsuit c) = (a \diamondsuit b) \cdot c$ and that $a \diamondsuit a = 1$ for all nonzero real numbers $a,b,$ and $c$. (Here the dot $\cdot$ represents the usual multiplication operation.) The solution to the equation $2016 \diamondsuit(6\diamondsuit x) = 100$ can be written as $\frac pq$, where $p$ and $q$ are relatively prime positive integers. What is $p+q$?
$\phantom{10A:23}$
(A) $109 \qquad$ (B) $201 \qquad$ (C) $301 \qquad$ (D) $3049 \qquad$ (E) $33,601$
copeland
2016-02-03 19:28:04
That's $\$$\diamondsuit$\$$ in TeX. Let's agree that you can use ":" to represent $\diamondsuit$, so the first identity is
a:(b:c)=(a:b)c
I'll keep using $\diamondsuit$ since it will be clearer.
That's $\$$\diamondsuit$\$$ in TeX. Let's agree that you can use ":" to represent $\diamondsuit$, so the first identity is
a:(b:c)=(a:b)c
I'll keep using $\diamondsuit$ since it will be clearer.
copeland
2016-02-03 19:28:07
OK, do you see any binary operations that satisfy both of these properties?
OK, do you see any binary operations that satisfy both of these properties?
ChickenOnRage
2016-02-03 19:28:41
division -_-
division -_-
idomath12345
2016-02-03 19:28:41
Diamond is /?
Diamond is /?
ciao_potter
2016-02-03 19:28:41
division
division
Zyanrhangca
2016-02-03 19:28:41
division
division
spin8
2016-02-03 19:28:41
division
division
beanielove2
2016-02-03 19:28:41
division
division
ChickenOnRage
2016-02-03 19:28:41
division
division
ciao_potter
2016-02-03 19:28:41
DIVISION
DIVISION
garyasho
2016-02-03 19:28:41
Division!!!
Division!!!
math1012
2016-02-03 19:28:41
division
division
ChickenOnRage
2016-02-03 19:28:41
division ;-;
division ;-;
thequantumguy
2016-02-03 19:28:41
DIVISION
DIVISION
awesomethree
2016-02-03 19:28:41
division
division
weilunsun28
2016-02-03 19:28:41
division
division
Wh2002
2016-02-03 19:28:41
division
division
Ultimate_draco
2016-02-03 19:28:41
division xd
division xd
alexli2014
2016-02-03 19:28:41
division
division
Quinn
2016-02-03 19:28:41
division
division
baseballcat
2016-02-03 19:28:41
divison
divison
spartan168
2016-02-03 19:28:41
division!
division!
copeland
2016-02-03 19:28:47
Division!\[a\div(b\div c)=\frac{\phantom aa\phantom a}{\frac bc}=\frac ab\cdot c=(a\div b)\cdot c\]and \[a\div a=1.\]
Division!\[a\div(b\div c)=\frac{\phantom aa\phantom a}{\frac bc}=\frac ab\cdot c=(a\div b)\cdot c\]and \[a\div a=1.\]
copeland
2016-02-03 19:28:48
So what's $x$?
So what's $x$?
ythomashu
2016-02-03 19:29:47
100/2016*6
100/2016*6
Quinn
2016-02-03 19:29:47
600/2016
600/2016
Anno
2016-02-03 19:29:47
600/2016
600/2016
lgbam
2016-02-03 19:29:47
25/84
25/84
Anno
2016-02-03 19:29:47
600/2016 = 25/84
600/2016 = 25/84
vitriolhumor
2016-02-03 19:29:47
100/336 or 25/84
100/336 or 25/84
Wh2002
2016-02-03 19:29:47
which is 25/84
which is 25/84
mathguy623
2016-02-03 19:29:47
25/84
25/84
beanielove2
2016-02-03 19:29:51
$\dfrac{2016}{\frac{6}{x}} = 100.$ $2016x = 600.$
$\dfrac{2016}{\frac{6}{x}} = 100.$ $2016x = 600.$
copeland
2016-02-03 19:29:53
This is the classic cheat: there could be a lot of different operations that satisfy these two properties. However, the problem is asked in such a way that we know that whichever binary operation we pick we're going to get the same answer. Therefore we can solve the problem assuming $\diamondsuit$ is $\div$.
This is the classic cheat: there could be a lot of different operations that satisfy these two properties. However, the problem is asked in such a way that we know that whichever binary operation we pick we're going to get the same answer. Therefore we can solve the problem assuming $\diamondsuit$ is $\div$.
copeland
2016-02-03 19:30:02
If \[2016\div(6\div x)=100\]then\[2016x=600.\]
If \[2016\div(6\div x)=100\]then\[2016x=600.\]
copeland
2016-02-03 19:30:06
Let's factor that a little. I see at least a 4 on both sides:\[504x=150.\]
Let's factor that a little. I see at least a 4 on both sides:\[504x=150.\]
copeland
2016-02-03 19:30:09
Oh, there's another 2,\[252x=75,\]and also a 3,\[84x=25.\]
Oh, there's another 2,\[252x=75,\]and also a 3,\[84x=25.\]
copeland
2016-02-03 19:30:15
This tells us $x=\dfrac{25}{84}.$ And what is the answer?
This tells us $x=\dfrac{25}{84}.$ And what is the answer?
Tenthgrade2015
2016-02-03 19:30:36
A
A
jfurf
2016-02-03 19:30:36
109!
109!
dyang
2016-02-03 19:30:36
$(A)$
$(A)$
smart99
2016-02-03 19:30:36
a
a
Lion7
2016-02-03 19:30:36
A: 109!
A: 109!
song2sons
2016-02-03 19:30:36
109
109
song2sons
2016-02-03 19:30:36
A
A
shivao
2016-02-03 19:30:36
A
A
goodbear
2016-02-03 19:30:36
109
109
A1012
2016-02-03 19:30:36
25+84=109
25+84=109
shivao
2016-02-03 19:30:36
A) 109
A) 109
Mudkipswims42
2016-02-03 19:30:36
$109\implies A$
$109\implies A$
amplreneo
2016-02-03 19:30:36
109
109
copeland
2016-02-03 19:30:38
Since these are coprime, the answer is $25+84=\boxed{109},$ $A.$
Since these are coprime, the answer is $25+84=\boxed{109},$ $A.$
copeland
2016-02-03 19:30:40
There's the answer, but it's pretty unsatisfying since we had to assume that every such binary operation gives the same value. Let's think a moment about that.
There's the answer, but it's pretty unsatisfying since we had to assume that every such binary operation gives the same value. Let's think a moment about that.
copeland
2016-02-03 19:30:45
We're faced with a couple of identities satisfied by our operation:
\begin{align*}
a\diamondsuit (b\diamondsuit c)&=(a\diamondsuit b)\cdot c\\
a\diamondsuit a&=1.
\end{align*}
We're faced with a couple of identities satisfied by our operation:
\begin{align*}
a\diamondsuit (b\diamondsuit c)&=(a\diamondsuit b)\cdot c\\
a\diamondsuit a&=1.
\end{align*}
copeland
2016-02-03 19:30:49
What are some standard tactics when we are faced with functional identities?
What are some standard tactics when we are faced with functional identities?
ninjataco
2016-02-03 19:31:26
substitute stuff
substitute stuff
amplreneo
2016-02-03 19:31:26
Plug in values
Plug in values
idomath12345
2016-02-03 19:31:26
Sub in different stuff.
Sub in different stuff
.
copeland
2016-02-03 19:31:37
Sure, like what? What are usually good choices. Speaking generally now.
Sure, like what? What are usually good choices. Speaking generally now.
MathWhizzz
2016-02-03 19:32:20
plug x in for a,b,c
plug x in for a,b,c
reddymathcounts
2016-02-03 19:32:20
0 or 1
0 or 1
shivao
2016-02-03 19:32:20
-1,0,1
-1,0,1
TigerLily14
2016-02-03 19:32:20
1 and 0 maybe
1 and 0 maybe
ninjataco
2016-02-03 19:32:20
substitute 1 or a
substitute 1 or a
idomath12345
2016-02-03 19:32:20
Like x,x,x
Like x,x,x
ChickenOnRage
2016-02-03 19:32:20
a = b or b = c
a = b or b = c
yrnsmurf
2016-02-03 19:32:20
1 and 0
1 and 0
copeland
2016-02-03 19:32:22
Some common tactics are:
$\bullet$ Set some variables in the identity to 1
$\bullet$ Set some variables in the identity to 0
$\bullet$ Set some variables in the identity equal each other
$\bullet$ Set some pair of variables to add to 0 or 1
$\bullet$ Set some pair of variables to multiply to 1
Some common tactics are:
$\bullet$ Set some variables in the identity to 1
$\bullet$ Set some variables in the identity to 0
$\bullet$ Set some variables in the identity equal each other
$\bullet$ Set some pair of variables to add to 0 or 1
$\bullet$ Set some pair of variables to multiply to 1
copeland
2016-02-03 19:32:37
Now we're omitting zero and we're only working with multiplication/division (not addition/subtraction) here so we should ditch several of these ideas, but we still have
$\bullet$ Set some variables in the identity to 1
$\bullet$ Set some variables in the identity equal each other
$\bullet$ Set some pair of variables in the identity to multiply to 1
Now we're omitting zero and we're only working with multiplication/division (not addition/subtraction) here so we should ditch several of these ideas, but we still have
$\bullet$ Set some variables in the identity to 1
$\bullet$ Set some variables in the identity equal each other
$\bullet$ Set some pair of variables in the identity to multiply to 1
copeland
2016-02-03 19:32:42
Let's try setting some variables equal to each other. What should we try?
Let's try setting some variables equal to each other. What should we try?
ninjataco
2016-02-03 19:33:24
set them all equal to a
set them all equal to a
Mudkipswims42
2016-02-03 19:33:24
$a,a,a$?
$a,a,a$?
copeland
2016-02-03 19:33:29
If we set $a=b=c$ in the first identity what do we get?
If we set $a=b=c$ in the first identity what do we get?
copeland
2016-02-03 19:34:14
It's an identity, so if you didn't type "=" in your response, maybe go back and check.
It's an identity, so if you didn't type "=" in your response, maybe go back and check.
maxplanck
2016-02-03 19:34:29
a : 1 = a
a : 1 = a
shivao
2016-02-03 19:34:29
a:1=1:a
a:1=1:a
fractal161
2016-02-03 19:34:29
a : 1 = a
a : 1 = a
math1012
2016-02-03 19:34:29
a:1=a
a:1=a
eswa2000
2016-02-03 19:34:29
a:1 = a
a:1 = a
thomas0115
2016-02-03 19:34:29
a:1 = a
a:1 = a
reddymathcounts
2016-02-03 19:34:29
a:1=a
a:1=a
brian22
2016-02-03 19:34:29
$a \Diamond 1 = a$
$a \Diamond 1 = a$
goodbear
2016-02-03 19:34:29
$a\diamondsuit 1 = a$
$a\diamondsuit 1 = a$
copeland
2016-02-03 19:34:33
If we set $a=b=c$ in the first identity then we get\[a\diamondsuit(a\diamondsuit a)=(a\diamondsuit a)\cdot a.\]Applying the second identity on each side gives
\[a\diamondsuit1=1\cdot a=a.\]
If we set $a=b=c$ in the first identity then we get\[a\diamondsuit(a\diamondsuit a)=(a\diamondsuit a)\cdot a.\]Applying the second identity on each side gives
\[a\diamondsuit1=1\cdot a=a.\]
copeland
2016-02-03 19:34:34
(Formally, this says that 1 is a right identity for $\diamondsuit$.)
(Formally, this says that 1 is a right identity for $\diamondsuit$.)
copeland
2016-02-03 19:35:06
Others suggested this substitution:
Others suggested this substitution:
yrnsmurf
2016-02-03 19:35:08
b and c
b and c
mathguy623
2016-02-03 19:35:08
b=c
b=c
alexli2014
2016-02-03 19:35:08
b and c
b and c
Mathfun01
2016-02-03 19:35:08
b=c
b=c
MaYang
2016-02-03 19:35:08
b=c
b=c
maxplanck
2016-02-03 19:35:08
b = c
b = c
copeland
2016-02-03 19:35:17
What happens if we just set $b=c$, since that will still let us use the second identity?
What happens if we just set $b=c$, since that will still let us use the second identity?
ninjataco
2016-02-03 19:36:15
a = (a:b) * b
a = (a:b) * b
math1012
2016-02-03 19:36:15
a:1= (a:c)c
a:1= (a:c)c
reddymathcounts
2016-02-03 19:36:15
a:1=(a:b)*b
a:1=(a:b)*b
idomath12345
2016-02-03 19:36:15
a:1=(a:x)*x
a:1=(a:x)*x
TigerLily14
2016-02-03 19:36:15
a:1=a=a:b * c
a:1=a=a:b * c
happyribbon
2016-02-03 19:36:15
a:1 = (a:b):b
a:1 = (a:b):b
copeland
2016-02-03 19:36:18
If we set $b=c$ then the first identity reads\[a\diamondsuit(b\diamondsuit b)=(a\diamondsuit b)\cdot b.\]The left side simplifies to $a\diamondsuit1.$
If we set $b=c$ then the first identity reads\[a\diamondsuit(b\diamondsuit b)=(a\diamondsuit b)\cdot b.\]The left side simplifies to $a\diamondsuit1.$
copeland
2016-02-03 19:36:20
What does that say?
What does that say?
garyasho
2016-02-03 19:37:14
a/b = a:b
a/b = a:b
brian22
2016-02-03 19:37:14
$\frac{a}{b} = a \Diamond b$, after dividing by $b$
$\frac{a}{b} = a \Diamond b$, after dividing by $b$
maxplanck
2016-02-03 19:37:14
a : b = a/b
a : b = a/b
thomas0115
2016-02-03 19:37:14
a:b = a/b
a:b = a/b
Wh2002
2016-02-03 19:37:14
a=ab/b
a=ab/b
geogirl08
2016-02-03 19:37:14
a : b = a / b
a : b = a / b
copeland
2016-02-03 19:37:18
We just saw that $a\diamondsuit 1=a$, so we get \[a=(a\diamondsuit b)\cdot b.\]We can divide this by $b$ to get \[\frac ab=a\diamondsuit b.\]So what?
We just saw that $a\diamondsuit 1=a$, so we get \[a=(a\diamondsuit b)\cdot b.\]We can divide this by $b$ to get \[\frac ab=a\diamondsuit b.\]So what?
elpers21
2016-02-03 19:37:52
division
division
amcwu
2016-02-03 19:37:52
diamond equals division
diamond equals division
ThorJames
2016-02-03 19:37:52
it is division
it is division
SimonSun
2016-02-03 19:37:52
it is division
it is division
thomas0115
2016-02-03 19:37:52
the operation is division
the operation is division
snapdragon1
2016-02-03 19:37:52
diamond is division
diamond is division
thequantumguy
2016-02-03 19:37:52
it is division
it is division
baseballcat
2016-02-03 19:37:52
so the operation is division?
so the operation is division?
Emathmaster
2016-02-03 19:37:52
It is divison.
It is divison.
BobThePotato
2016-02-03 19:37:52
the operation is equivalent to division
the operation is equivalent to division
Ericaops
2016-02-03 19:37:52
the operation is indeed division
the operation is indeed division
blueberry2
2016-02-03 19:37:55
the binary operation is division
the binary operation is division
copeland
2016-02-03 19:37:57
This tells us that the only possible thing that $\diamondsuit$ can be is division! These two properties completely define division.
This tells us that the only possible thing that $\diamondsuit$ can be is division! These two properties completely define division.
copeland
2016-02-03 19:38:08
You learn something every day, huh?
You learn something every day, huh?
happyribbon
2016-02-03 19:38:23
Yes.
Yes.
SimonSun
2016-02-03 19:38:23
YUP!!
YUP!!
goldypeng
2016-02-03 19:38:23
yes
yes
BitterGummy
2016-02-03 19:38:23
WHOA
WHOA
vijaym
2016-02-03 19:38:23
yup!
yup!
claserken
2016-02-03 19:38:43
We wouldn't do this part of the test, right?
We wouldn't do this part of the test, right?
copeland
2016-02-03 19:38:44
I sure wouldn't have unless I'd given up on (or finished) the rest.
I sure wouldn't have unless I'd given up on (or finished) the rest.
copeland
2016-02-03 19:38:57
OK, who wants to go on to the AMC12 and who wants to stick with the AMC10?
OK, who wants to go on to the AMC12 and who wants to stick with the AMC10?
haradica
2016-02-03 19:39:21
AMC 12 pls
AMC 12 pls
Mudkipswims42
2016-02-03 19:39:21
AMC10!
AMC10!
Ishaan
2016-02-03 19:39:21
AMC 10
AMC 10
dhuang26
2016-02-03 19:39:21
Stick with Amc 10
Stick with Amc 10
aryastark
2016-02-03 19:39:21
amc10
amc10
duruphi
2016-02-03 19:39:21
AMC10!!!
AMC10!!!
yuunderstand168
2016-02-03 19:39:21
AMC12
AMC12
Radical_CC
2016-02-03 19:39:21
12!
12!
GeorgCantor
2016-02-03 19:39:21
AMC10
AMC10
hliu70
2016-02-03 19:39:21
AMC 10
AMC 10
brisane
2016-02-03 19:39:21
Stick with AMC10.
Stick with AMC10.
deltaepsilon6
2016-02-03 19:39:21
AMC 12
AMC 12
copeland
2016-02-03 19:39:27
Actually everyone wins.
Actually everyone wins.
copeland
2016-02-03 19:39:28
Problems 24 and 25 of the AMC10 were also Problems 21 and 22 on the AMC12, so if you're here for the AMC10, there are two more problems to go, but if you're here for the 12 then we're about to begin!
Problems 24 and 25 of the AMC10 were also Problems 21 and 22 on the AMC12, so if you're here for the AMC10, there are two more problems to go, but if you're here for the 12 then we're about to begin!
copeland
2016-02-03 19:40:03
24.A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length 200. What is the length of its fourth side?
$\phantom{10A:24}$
(A) $200 \qquad$ (B) $200\sqrt{2} \qquad$ (C) $200\sqrt{3} \qquad$ (D) $300\sqrt{2} \qquad$ (E) $500$
24.A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length 200. What is the length of its fourth side?
$\phantom{10A:24}$
(A) $200 \qquad$ (B) $200\sqrt{2} \qquad$ (C) $200\sqrt{3} \qquad$ (D) $300\sqrt{2} \qquad$ (E) $500$
copeland
2016-02-03 19:40:06
[Also AMC12, Problem 21]
[Also AMC12, Problem 21]
copeland
2016-02-03 19:40:22
I don't know what's going on with these giant numbers in this problem. I'm just going to divide every length in sight by 100 and start over:
I don't know what's going on with these giant numbers in this problem. I'm just going to divide every length in sight by 100 and start over:
copeland
2016-02-03 19:40:24
24.A quadrilateral is inscribed in a circle of radius $2\sqrt{2}$. Three of the sides of this quadrilateral have length 2. What is the length of its fourth side?
$\phantom{10A:24}$
(A) $2 \qquad$ (B) $2\sqrt{2} \qquad$ (C) $2\sqrt{3} \qquad$ (D) $3\sqrt{2} \qquad$ (E) $5$
24.A quadrilateral is inscribed in a circle of radius $2\sqrt{2}$. Three of the sides of this quadrilateral have length 2. What is the length of its fourth side?
$\phantom{10A:24}$
(A) $2 \qquad$ (B) $2\sqrt{2} \qquad$ (C) $2\sqrt{3} \qquad$ (D) $3\sqrt{2} \qquad$ (E) $5$
copeland
2016-02-03 19:40:43
Does everyone agree those are the same problem?
Does everyone agree those are the same problem?
APR02
2016-02-03 19:40:58
that's better
that's better
lithium123
2016-02-03 19:40:58
yep
yep
Peggy
2016-02-03 19:40:58
yes
yes
beanielove2
2016-02-03 19:40:58
Yeah
Yeah
TheRealDeal
2016-02-03 19:40:58
yes
yes
spin8
2016-02-03 19:40:58
yes
yes
space_space
2016-02-03 19:40:58
yes
yes
ciao_potter
2016-02-03 19:40:58
Yes
Yes
bigboss
2016-02-03 19:40:58
yes
yes
yangyifei
2016-02-03 19:40:58
yes
yes
copeland
2016-02-03 19:41:00
Good.
Good.
coconut_force
2016-02-03 19:41:17
diagram
diagram
MathConquer2014
2016-02-03 19:41:17
diagram!
diagram!
yrnsmurf
2016-02-03 19:41:17
Start with drawing a diagram
Start with drawing a diagram
space_space
2016-02-03 19:41:17
diagram
diagram
patatoe
2016-02-03 19:41:17
it helps to visualize
it helps to visualize
copeland
2016-02-03 19:41:23
Diagrams are awesome.
Diagrams are awesome.
copeland
2016-02-03 19:41:27
Notice that throughout I'm going to write $\sqrt8$ on my diagrams instead of $2\sqrt2$ since it will make the diagrams easier to read. These are the same number.
Notice that throughout I'm going to write $\sqrt8$ on my diagrams instead of $2\sqrt2$ since it will make the diagrams easier to read. These are the same number.
copeland
2016-02-03 19:41:31
So we're given some quadrilateral in a circle. What kind of quadrilateral is it?
So we're given some quadrilateral in a circle. What kind of quadrilateral is it?
akaashp11
2016-02-03 19:41:47
$Cyclical Quadrilaterals$
$Cyclical Quadrilaterals$
Luke Skywalker
2016-02-03 19:41:47
cyclic!
cyclic!
SHARKYBOY
2016-02-03 19:41:47
cyclic
cyclic
axue
2016-02-03 19:41:47
cyclic
cyclic
Mathfun01
2016-02-03 19:41:47
cyclic
cyclic
leeandrew1029gmail.com
2016-02-03 19:41:47
cyclic
cyclic
copeland
2016-02-03 19:41:53
Yup. What else?
Yup. What else?
ryanyz10
2016-02-03 19:42:22
isosceles trapezoid
isosceles trapezoid
eswa2000
2016-02-03 19:42:22
isosceles trapezoid
isosceles trapezoid
AlcumusGuy
2016-02-03 19:42:22
isosceles trapezoid
isosceles trapezoid
thomas0115
2016-02-03 19:42:22
isosceles trapezoid
isosceles trapezoid
goldypeng
2016-02-03 19:42:22
isosceles
isosceles
hliu70
2016-02-03 19:42:22
isoceles trapezoid
isoceles trapezoid
math1012
2016-02-03 19:42:22
isosceles trapezoid
isosceles trapezoid
FTW
2016-02-03 19:42:22
isosceles trapezoid
isosceles trapezoid
brian22
2016-02-03 19:42:22
Isosceles trapezoid!
Isosceles trapezoid!
mathlete478
2016-02-03 19:42:24
Isosceles trapezoid
Isosceles trapezoid
copeland
2016-02-03 19:42:26
It must be an isosceles trapezoid. Any cyclic quadrilateral with two opposite edges of the same length is an isosceles trapezoid.
It must be an isosceles trapezoid. Any cyclic quadrilateral with two opposite edges of the same length is an isosceles trapezoid.
copeland
2016-02-03 19:42:29
I don't know quite how to draw it yet. It's either
I don't know quite how to draw it yet. It's either
copeland
2016-02-03 19:42:30
copeland
2016-02-03 19:42:30
or
or
copeland
2016-02-03 19:42:32
copeland
2016-02-03 19:42:35
What distance can we throw on there to help us figure it out?
What distance can we throw on there to help us figure it out?
copeland
2016-02-03 19:43:22
(All I want now is to know which picture we have. . .)
(All I want now is to know which picture we have. . .)
copeland
2016-02-03 19:43:52
What will help us determine which is which?
What will help us determine which is which?
spartan168
2016-02-03 19:44:09
root 7 height of each isosceles triangle
root 7 height of each isosceles triangle
Locust
2016-02-03 19:44:09
Heights?
Heights?
ThorJames
2016-02-03 19:44:10
perpendicular to the base?
perpendicular to the base?
copeland
2016-02-03 19:44:14
We have three of these $2-2-2\sqrt2$ triangles lying around. By the Pythagorean Theorem, these have altitude $\sqrt{8-1}=\sqrt7$.
We have three of these $2-2-2\sqrt2$ triangles lying around. By the Pythagorean Theorem, these have altitude $\sqrt{8-1}=\sqrt7$.
copeland
2016-02-03 19:44:14
Can this next picture be right?
Can this next picture be right?
copeland
2016-02-03 19:44:15
copeland
2016-02-03 19:44:30
And why or why not?
And why or why not?
mathwhiz16
2016-02-03 19:45:24
No! $2<\sqrt{7}$
No! $2<\sqrt{7}$
idomath12345
2016-02-03 19:45:24
No, since the hypo.>then leg.
No, since the hypo.>then leg.
exhonourated
2016-02-03 19:45:24
sqrt 7 >2 so no
sqrt 7 >2 so no
ciao_potter
2016-02-03 19:45:24
Triangle Inequality?
Triangle Inequality?
owm
2016-02-03 19:45:24
no because \sqrt{7}>2
no because \sqrt{7}>2
toper2
2016-02-03 19:45:24
it cannot since sqrt(7) > 2
it cannot since sqrt(7) > 2
bigboss
2016-02-03 19:45:24
because the sides of the triangle can for a right triangle with the altitude. However, \sqrt7>2
because the sides of the triangle can for a right triangle with the altitude. However, \sqrt7>2
A1012
2016-02-03 19:45:24
No, because root 7 is greater than 2
No, because root 7 is greater than 2
jfurf
2016-02-03 19:45:24
No. $\sqrt{7}$ is greater than $2$.
No. $\sqrt{7}$ is greater than $2$.
copeland
2016-02-03 19:45:27
No, that's a bit absurd. Notice that $\sqrt7>2$. However, the altitude of a trapezoid is always less than or equal to the lengths of a leg. Therefore the configuration is actually:
No, that's a bit absurd. Notice that $\sqrt7>2$. However, the altitude of a trapezoid is always less than or equal to the lengths of a leg. Therefore the configuration is actually:
copeland
2016-02-03 19:45:31
copeland
2016-02-03 19:45:32
I've thrown away the circle. We can resurrect if later if we need to.
I've thrown away the circle. We can resurrect if later if we need to.
copeland
2016-02-03 19:45:50
Now what do you see in this diagram? (I've labeled points for you.)
Now what do you see in this diagram? (I've labeled points for you.)
copeland
2016-02-03 19:45:51
ciao_potter
2016-02-03 19:46:31
Similar triagles
Similar triagles
Einsteinhead
2016-02-03 19:46:31
similar triangles!
similar triangles!
deltaepsilon6
2016-02-03 19:46:31
similar triangles
similar triangles
akaashp11
2016-02-03 19:46:31
Similar Triangles!!
Similar Triangles!!
yrnsmurf
2016-02-03 19:46:31
Similar triangles
Similar triangles
blizzard10
2016-02-03 19:46:31
Similar/congruent triangles
Similar/congruent triangles
alexli2014
2016-02-03 19:46:31
similar triangles
similar triangles
blueberry2
2016-02-03 19:46:31
similar triangles
similar triangles
lychnis3
2016-02-03 19:46:31
similar triangles
similar triangles
strategos21
2016-02-03 19:46:31
similar triangles
similar triangles
copeland
2016-02-03 19:46:40
Which triangles are similar?
Which triangles are similar?
GeorgCantor
2016-02-03 19:47:13
PQO ~ YZO
PQO ~ YZO
Einsteinhead
2016-02-03 19:47:13
PQO~YZO
PQO~YZO
A1012
2016-02-03 19:47:13
POQ and YOZ
POQ and YOZ
MaYang
2016-02-03 19:47:13
OPQ~OYZ
OPQ~OYZ
reddymathcounts
2016-02-03 19:47:13
POQ and YOZ
POQ and YOZ
smart99
2016-02-03 19:47:13
OPQ and OYZ
OPQ and OYZ
Emathmaster
2016-02-03 19:47:13
YOZ and POQ
YOZ and POQ
copeland
2016-02-03 19:47:16
Great, that's a useful pair.
Great, that's a useful pair.
copeland
2016-02-03 19:47:18
Since $XW$ is parallel to $YZ$, we have similar triangles $\triangle OYZ\sim\triangle OPQ$.
Since $XW$ is parallel to $YZ$, we have similar triangles $\triangle OYZ\sim\triangle OPQ$.
copeland
2016-02-03 19:47:19
Let's call the angle $\angle OYZ=\theta$ for simplicity and mark these equalities:
Let's call the angle $\angle OYZ=\theta$ for simplicity and mark these equalities:
copeland
2016-02-03 19:47:20
copeland
2016-02-03 19:47:23
Now what do you see?
Now what do you see?
dchen04
2016-02-03 19:48:13
xyp and opq
xyp and opq
lcalvert99
2016-02-03 19:48:13
XYP and OPX
XYP and OPX
Mudkipswims42
2016-02-03 19:48:13
$\triangle YXP$ is also similiar!
$\triangle YXP$ is also similiar!
memc38123
2016-02-03 19:48:13
XYP is isosceles
XYP is isosceles
kunsun
2016-02-03 19:48:13
ZQW ~ POQ and PXY ~ POQ
ZQW ~ POQ and PXY ~ POQ
copeland
2016-02-03 19:48:16
Ooh! Ooh! $\angle XPY=\theta$ by vertical angles, so $\triangle XYP\sim\triangle OYZ!$ What are the side lengths of $\triangle XYP?$
Ooh! Ooh! $\angle XPY=\theta$ by vertical angles, so $\triangle XYP\sim\triangle OYZ!$ What are the side lengths of $\triangle XYP?$
geogirl08
2016-02-03 19:49:10
2, 2, sqrt(2)
2, 2, sqrt(2)
FTW
2016-02-03 19:49:10
2,2,$\sqrt{2}$
2,2,$\sqrt{2}$
Interest
2016-02-03 19:49:10
2,2,sqrt(2)
2,2,sqrt(2)
TheRealDeal
2016-02-03 19:49:10
2 2 sqrt2
2 2 sqrt2
thomas0115
2016-02-03 19:49:10
2,2,sqrt(2)
2,2,sqrt(2)
ryanyz10
2016-02-03 19:49:10
2, 2, sqrt(2)
2, 2, sqrt(2)
math1012
2016-02-03 19:49:10
2,2,sqrt2
2,2,sqrt2
eswa2000
2016-02-03 19:49:10
2,2,sqrt{2}
2,2,sqrt{2}
goodbear
2016-02-03 19:49:10
2, 2,√2
2, 2,√2
hliu70
2016-02-03 19:49:10
2-2-sqrt(2)
2-2-sqrt(2)
copeland
2016-02-03 19:49:13
Since $\dfrac{XY}{OY}=\dfrac{2}{\sqrt8}=\dfrac1{\sqrt2}$, we get that $\triangle XYP$ is $2-2-\sqrt2$ isosceles triangle. In particular $XP=2$ and $YP=\sqrt2$.
Since $\dfrac{XY}{OY}=\dfrac{2}{\sqrt8}=\dfrac1{\sqrt2}$, we get that $\triangle XYP$ is $2-2-\sqrt2$ isosceles triangle. In particular $XP=2$ and $YP=\sqrt2$.
copeland
2016-02-03 19:49:16
What does knowing $YP$ tell us?
What does knowing $YP$ tell us?
memc38123
2016-02-03 19:50:02
OP
OP
cinamon
2016-02-03 19:50:02
PO
PO
memc38123
2016-02-03 19:50:02
length of OP
length of OP
strategos21
2016-02-03 19:50:02
OP
OP
yrnsmurf
2016-02-03 19:50:02
its half of YO
its half of YO
Einsteinhead
2016-02-03 19:50:02
PO
PO
eswa2000
2016-02-03 19:50:02
P is midpoint of YO
P is midpoint of YO
ssy899
2016-02-03 19:50:02
we know op
we know op
patatoe
2016-02-03 19:50:02
the length of PO
the length of PO
Mathfun01
2016-02-03 19:50:02
you know OP
you know OP
copeland
2016-02-03 19:50:05
Now we know $OP=OY-YP=2\sqrt2-\sqrt2=\sqrt2$.
Now we know $OP=OY-YP=2\sqrt2-\sqrt2=\sqrt2$.
copeland
2016-02-03 19:50:06
What can we determine from there?
What can we determine from there?
leafshadow
2016-02-03 19:50:33
PQ
PQ
mjoshi
2016-02-03 19:50:33
gives us PQ
gives us PQ
memc38123
2016-02-03 19:50:33
QP
QP
Interest
2016-02-03 19:50:33
PQ
PQ
Einsteinhead
2016-02-03 19:50:33
PQ
PQ
buzhidao
2016-02-03 19:50:33
PQ
PQ
blizzard10
2016-02-03 19:50:33
PQ
PQ
TigerLily14
2016-02-03 19:50:33
PQ
PQ
copeland
2016-02-03 19:50:34
Great. What is $PQ$?
Great. What is $PQ$?
eswa2000
2016-02-03 19:50:54
PQ is 1
PQ is 1
FTW
2016-02-03 19:50:54
$PQ=1$ from sim triangles
$PQ=1$ from sim triangles
Alacarter
2016-02-03 19:50:54
PQ = 1
PQ = 1
mjoshi
2016-02-03 19:50:54
1
1
SimonSun
2016-02-03 19:50:54
1
1
owm
2016-02-03 19:50:54
1
1
thatmathgeek
2016-02-03 19:50:54
1
1
goldypeng
2016-02-03 19:50:54
1
1
Locust
2016-02-03 19:50:54
1
1
brian22
2016-02-03 19:50:54
$1$
$1$
copeland
2016-02-03 19:51:01
Since $\triangle OYZ\sim\triangle OPQ$ and $\dfrac{OY}{OP}=\dfrac{2\sqrt2}{\sqrt2}=2$, we know $2=\dfrac{YZ}{PQ}=\dfrac2{PQ}$, so $PQ=1$.
Since $\triangle OYZ\sim\triangle OPQ$ and $\dfrac{OY}{OP}=\dfrac{2\sqrt2}{\sqrt2}=2$, we know $2=\dfrac{YZ}{PQ}=\dfrac2{PQ}$, so $PQ=1$.
copeland
2016-02-03 19:51:02
Now what?
Now what?
owm
2016-02-03 19:51:44
so XW=5
so XW=5
blizzard10
2016-02-03 19:51:44
Thus XW is 5
Thus XW is 5
Mathfun01
2016-02-03 19:51:44
1 + 2 + 2
1 + 2 + 2
amcwu
2016-02-03 19:51:44
2+2+1=5
2+2+1=5
Arginine
2016-02-03 19:51:44
add them up
add them up
blueberry2
2016-02-03 19:51:44
add them together
add them together
spartan168
2016-02-03 19:51:44
XP + PQ + QW = 2 + 1 + 2 = 5
XP + PQ + QW = 2 + 1 + 2 = 5
thatmathgeek
2016-02-03 19:51:44
add XW up to get 5
add XW up to get 5
Locust
2016-02-03 19:51:44
We know the total length is 2+1+2=5
We know the total length is 2+1+2=5
buzhidao
2016-02-03 19:51:44
Add together, get XW
Add together, get XW
FTW
2016-02-03 19:51:44
just add the lengths
just add the lengths
goldypeng
2016-02-03 19:51:44
xw is 5
xw is 5
claserken
2016-02-03 19:51:44
the answer is 5
the answer is 5
ciao_potter
2016-02-03 19:51:44
$XW=5$
$XW=5$
TigerLily14
2016-02-03 19:51:44
add up the parts of XW
add up the parts of XW
happyribbon
2016-02-03 19:51:44
XW is 5.
XW is 5.
thomas0115
2016-02-03 19:51:44
XW is 5, so the answer is (E)
XW is 5, so the answer is (E)
memc38123
2016-02-03 19:51:44
XP + PQ + QW = 5
XP + PQ + QW = 5
copeland
2016-02-03 19:51:48
Now we're done! $XW=XP+PQ+QW=2+1+2=\boxed{5}.$
Now we're done! $XW=XP+PQ+QW=2+1+2=\boxed{5}.$
copeland
2016-02-03 19:51:51
The answer to the original problem is $\boxed{500},$ $E.$
The answer to the original problem is $\boxed{500},$ $E.$
Mudkipswims42
2016-02-03 19:52:05
Wow that was super neat!
Wow that was super neat!
copeland
2016-02-03 19:52:08
Thanks.
Thanks.
copeland
2016-02-03 19:52:15
There are a lot of really fancy solutions, too.
There are a lot of really fancy solutions, too.
MathWhizzz
2016-02-03 19:52:23
why do you think they made the numbers so big?
why do you think they made the numbers so big?
copeland
2016-02-03 19:52:26
I have no idea.
I have no idea.
copeland
2016-02-03 19:52:49
My best guess is that it lets them use different numbers than appeared in 21, but that's kinda weird.
My best guess is that it lets them use different numbers than appeared in 21, but that's kinda weird.
stevenrgao
2016-02-03 19:53:13
I think they did it to confuse you
I think they did it to confuse you
copeland
2016-02-03 19:53:15
Were you confused?
Were you confused?
thomas0115
2016-02-03 19:53:31
no
no
lychnis3
2016-02-03 19:53:31
yes
yes
awesomethree
2016-02-03 19:53:31
No
No
amplreneo
2016-02-03 19:53:31
nope just annoyed
nope just annoyed
yrnsmurf
2016-02-03 19:53:31
no
no
ThorJames
2016-02-03 19:53:31
yes
yes
mathwhiz16
2016-02-03 19:53:31
Not at all
Not at all
chewbacca2
2016-02-03 19:53:31
a littel
a littel
Shruige1
2016-02-03 19:53:31
not really
not really
Sophiarulz
2016-02-03 19:53:31
very
very
copeland
2016-02-03 19:53:42
Alright, ready to move on?
Alright, ready to move on?
jfurf
2016-02-03 19:54:05
YEP!!
YEP!!
fatcat36
2016-02-03 19:54:05
yes
yes
xiaodongxi
2016-02-03 19:54:05
yes
yes
spin8
2016-02-03 19:54:05
yes
yes
leeandrew1029gmail.com
2016-02-03 19:54:05
yes
yes
pcptc
2016-02-03 19:54:05
yeah!
yeah!
leeandrew1029gmail.com
2016-02-03 19:54:05
lets go
lets go
Emathmaster
2016-02-03 19:54:05
Yes!!!
Yes!!!
lasergazer
2016-02-03 19:54:05
yes
yes
copeland
2016-02-03 19:54:07
25. How many ordered triples $(x,y,z)$ of positive integers satisfy $\operatorname{lcm}(x,y) = 72$, $\operatorname{lcm}(x,z) = 600$, and $\operatorname{lcm}(y,z) = 900$?
$\phantom{10A:25}$
(A) $15 \qquad$ (B) $16 \qquad$ (C) $24 \qquad$ (D) $27 \qquad$ (E) $64$
25. How many ordered triples $(x,y,z)$ of positive integers satisfy $\operatorname{lcm}(x,y) = 72$, $\operatorname{lcm}(x,z) = 600$, and $\operatorname{lcm}(y,z) = 900$?
$\phantom{10A:25}$
(A) $15 \qquad$ (B) $16 \qquad$ (C) $24 \qquad$ (D) $27 \qquad$ (E) $64$
copeland
2016-02-03 19:54:09
[Also AMC12, Problem 22]
[Also AMC12, Problem 22]
copeland
2016-02-03 19:54:16
We have three numbers and we have information about pairs of numbers. This implies a nice organizational structure for this problem. Does anyone see it?
We have three numbers and we have information about pairs of numbers. This implies a nice organizational structure for this problem. Does anyone see it?
idomath12345
2016-02-03 19:54:48
We set up a table.
We set up a table.
memc38123
2016-02-03 19:54:48
A chart!
A chart!
idomath12345
2016-02-03 19:54:48
Table for powers of 2,3,5 with respect to x,y,z
Table for powers of 2,3,5 with respect to x,y,z
brian22
2016-02-03 19:54:51
A graph?
A graph?
copeland
2016-02-03 19:54:56
I imagine a lot of people used tables and everything worked out fine, but I found it a lot easier to see what is going on by creating a graph with vertices $x$, $y$, and $z$.
I imagine a lot of people used tables and everything worked out fine, but I found it a lot easier to see what is going on by creating a graph with vertices $x$, $y$, and $z$.
copeland
2016-02-03 19:55:00
copeland
2016-02-03 19:55:04
Throwing our data on here gives us
Throwing our data on here gives us
copeland
2016-02-03 19:55:04
copeland
2016-02-03 19:55:17
What does $\text{lcm}(y,z)=900$ tell us about $y$ and $z$?
What does $\text{lcm}(y,z)=900$ tell us about $y$ and $z$?
jrexmo
2016-02-03 19:56:04
they are both factors of 900
they are both factors of 900
goodbear
2016-02-03 19:56:04
divide 900
divide 900
akaashp11
2016-02-03 19:56:04
Divisors of 900
Divisors of 900
ninjataco
2016-02-03 19:56:29
prime factorization is 2^a 3^b 5^c
prime factorization is 2^a 3^b 5^c
SimonSun
2016-02-03 19:56:29
The highest in both must be $3^3*2^2*5^2$
The highest in both must be $3^3*2^2*5^2$
spartan168
2016-02-03 19:56:29
One of y and z has 2^2, one of them has 3^2 and one of them has 5^2
One of y and z has 2^2, one of them has 3^2 and one of them has 5^2
Locust
2016-02-03 19:56:29
the maximum prime powers are 3^2, 2^2, and 5^2
the maximum prime powers are 3^2, 2^2, and 5^2
brisane
2016-02-03 19:56:29
At least one is divisible by 9, at least one is divisible by 4, and at least one is divisible by 25.
At least one is divisible by 9, at least one is divisible by 4, and at least one is divisible by 25.
Mathaddict11
2016-02-03 19:56:29
The max exponents for each of y and z is 3^2, 2^2, and 5^2
The max exponents for each of y and z is 3^2, 2^2, and 5^2
ryanyz10
2016-02-03 19:56:29
at most 2 factors of 2, 3, and 5
at most 2 factors of 2, 3, and 5
copeland
2016-02-03 19:56:32
Since $900=2^23^25^2,$ it tells us
$\bullet$ Either $y$ or $z$ has a largest power of 2 equal to 2
$\bullet$ Either $y$ or $z$ has a largest power of 3 equal to 2
$\bullet$ Either $y$ or $z$ has a largest power of 5 equal to 2
$\bullet$ $y$ and $z$ are not divisible by any primes other than $2$, $3$, and $5$.
Since $900=2^23^25^2,$ it tells us
$\bullet$ Either $y$ or $z$ has a largest power of 2 equal to 2
$\bullet$ Either $y$ or $z$ has a largest power of 3 equal to 2
$\bullet$ Either $y$ or $z$ has a largest power of 5 equal to 2
$\bullet$ $y$ and $z$ are not divisible by any primes other than $2$, $3$, and $5$.
copeland
2016-02-03 19:56:43
copeland
2016-02-03 19:56:51
These are a lot of different constraints that we need to satisfy all at once. Is there some easy simplification that we can make?
These are a lot of different constraints that we need to satisfy all at once. Is there some easy simplification that we can make?
ThorJames
2016-02-03 19:57:28
use cases
use cases
copeland
2016-02-03 19:57:43
Casework is viable. Anybody have a different approach?
Casework is viable. Anybody have a different approach?
beanielove2
2016-02-03 19:58:29
Just pick the powers for one number
Just pick the powers for one number
copeland
2016-02-03 19:58:33
Huh. What's that mean?
Huh. What's that mean?
Math_Magicians
2016-02-03 19:58:56
constructive counting
constructive counting
copeland
2016-02-03 19:59:04
Cool. How can we "construct"?
Cool. How can we "construct"?
leeandrew1029gmail.com
2016-02-03 19:59:28
do one prime at at time
do one prime at at time
akaashp11
2016-02-03 19:59:28
Limit to the 2s first
Limit to the 2s first
alexli2014
2016-02-03 19:59:28
chose where 5 goes to start
chose where 5 goes to start
copeland
2016-02-03 19:59:34
The powers of 2 in each of the numbers don't actually care about the powers of 3 and 5. We can figure out the ways to distribute 2s, 3s, and 5s separately and then multiply.
The powers of 2 in each of the numbers don't actually care about the powers of 3 and 5. We can figure out the ways to distribute 2s, 3s, and 5s separately and then multiply.
copeland
2016-02-03 19:59:39
How do we start with filling out this triangle?
How do we start with filling out this triangle?
copeland
2016-02-03 19:59:40
idomath12345
2016-02-03 20:00:08
x NEEDS the 8.
x NEEDS the 8.
Math1331Math
2016-02-03 20:00:08
x must be 2^3
x must be 2^3
elpers21
2016-02-03 20:00:08
x must have 2^3
x must have 2^3
eswa2000
2016-02-03 20:00:08
x must have 2^3
x must have 2^3
bigboss
2016-02-03 20:00:08
x is 2^3
x is 2^3
baseballcat
2016-02-03 20:00:08
x has 2^3
x has 2^3
Mudkipswims42
2016-02-03 20:00:08
$x$ must have a factor of $2^3$
$x$ must have a factor of $2^3$
vijaym
2016-02-03 20:00:08
x = 2^3
x = 2^3
yrnsmurf
2016-02-03 20:00:08
x is $2^3$
x is $2^3$
cinamon
2016-02-03 20:00:08
top is 2^3
top is 2^3
Mathfun01
2016-02-03 20:00:08
x = 2^3
x = 2^3
geogirl08
2016-02-03 20:00:08
x corner 2^3
x corner 2^3
copeland
2016-02-03 20:00:11
$y$ and $z$ both have at most 2 factors of 2. In order to satisfy the left and right edges, $x$ must have a factor of $2^3$:
$y$ and $z$ both have at most 2 factors of 2. In order to satisfy the left and right edges, $x$ must have a factor of $2^3$:
copeland
2016-02-03 20:00:14
copeland
2016-02-03 20:00:18
How many possible values are there for the number of twos in the pair $(y,z)?$
How many possible values are there for the number of twos in the pair $(y,z)?$
hliu70
2016-02-03 20:01:07
5
5
memc38123
2016-02-03 20:01:07
5
5
memc38123
2016-02-03 20:01:07
5 pairs
5 pairs
SimonSun
2016-02-03 20:01:07
5
5
Metal_Bender19
2016-02-03 20:01:07
5
5
rahulkk
2016-02-03 20:01:07
5
5
cinamon
2016-02-03 20:01:07
5?
5?
beanielove2
2016-02-03 20:01:07
$(2, 1), (2, 0), (0, 1), (0, 2), (2, 2)$
$(2, 1), (2, 0), (0, 1), (0, 2), (2, 2)$
Mudkipswims42
2016-02-03 20:01:07
wait $5$
wait $5$
copeland
2016-02-03 20:01:11
Remember that one of them must have a factor of $2^2$, so there are 5 pairs:\[(0,2)\quad(1,2)\quad(2,2)\quad(2,1)\quad(2,0).\]
Remember that one of them must have a factor of $2^2$, so there are 5 pairs:\[(0,2)\quad(1,2)\quad(2,2)\quad(2,1)\quad(2,0).\]
copeland
2016-02-03 20:01:13
Incidentally, what do you notice in the original problem?
Incidentally, what do you notice in the original problem?
thomas0115
2016-02-03 20:01:56
only one multiple of 5
only one multiple of 5
AlcumusGuy
2016-02-03 20:01:56
Only (A) is divisible by 5
Only (A) is divisible by 5
Ericaops
2016-02-03 20:01:56
A is the only answer that is a multiple of 5
A is the only answer that is a multiple of 5
mjoshi
2016-02-03 20:01:56
only one is a multiply of 5
only one is a multiply of 5
memc38123
2016-02-03 20:01:56
Only 1 answer has a multiple of 5
Only 1 answer has a multiple of 5
coconut_force
2016-02-03 20:01:56
only A is divisible by 5
only A is divisible by 5
leeandrew1029gmail.com
2016-02-03 20:01:56
one multiple of 5
one multiple of 5
goodbear
2016-02-03 20:01:56
A
A
Shruige1
2016-02-03 20:01:56
only one is a multiple of 5
only one is a multiple of 5
ssy899
2016-02-03 20:01:56
that the only answer choice divisible by 5 is a?
that the only answer choice divisible by 5 is a?
ninjataco
2016-02-03 20:01:56
only 1 choice is divisible by 5
only 1 choice is divisible by 5
brian22
2016-02-03 20:01:56
Only one answer is divisible by $5$
Only one answer is divisible by $5$
dchen04
2016-02-03 20:01:56
A) 15 is divisible by 5
A) 15 is divisible by 5
claserken
2016-02-03 20:01:56
There is a multiple of five.
There is a multiple of five.
kyleliu
2016-02-03 20:01:56
only (a) divisible by 5
only (a) divisible by 5
copeland
2016-02-03 20:02:02
There is only one answer that is a multiple of 5. Since there are 5 ways to distribute the powers of 2, the answer must be a multiple of 5 so the answer is $\boxed{15}$, $A$.
There is only one answer that is a multiple of 5. Since there are 5 ways to distribute the powers of 2, the answer must be a multiple of 5 so the answer is $\boxed{15}$, $A$.
copeland
2016-02-03 20:02:03
Let's keep going just as a check. What do we learn from this diagram?
Let's keep going just as a check. What do we learn from this diagram?
copeland
2016-02-03 20:02:07
mjoshi
2016-02-03 20:02:40
Y = 3^2
Y = 3^2
TigerLily14
2016-02-03 20:02:40
y is 2
y is 2
amcwu
2016-02-03 20:02:40
y has to be 9
y has to be 9
bringerofawesomeness
2016-02-03 20:02:40
y = 3^2
y = 3^2
Emathmaster
2016-02-03 20:02:40
y=3^2
y=3^2
blueberry2
2016-02-03 20:02:40
y is 9
y is 9
spartan168
2016-02-03 20:02:40
y is 3^2
y is 3^2
ryanyz10
2016-02-03 20:02:40
y has 2 3's
y has 2 3's
annasun19
2016-02-03 20:02:40
y = 9
y = 9
copeland
2016-02-03 20:02:43
We see that $y$ must have $3^2$ as its power of 3:
We see that $y$ must have $3^2$ as its power of 3:
copeland
2016-02-03 20:02:43
copeland
2016-02-03 20:02:46
How many ways are there to distribute the threes?
How many ways are there to distribute the threes?
cinamon
2016-02-03 20:03:26
3 possibilities
3 possibilities
rahulkk
2016-02-03 20:03:26
3
3
Mathfun01
2016-02-03 20:03:26
3
3
snapdragon1
2016-02-03 20:03:26
3
3
owm
2016-02-03 20:03:26
3
3
ThorJames
2016-02-03 20:03:26
$3$
$3$
space_space
2016-02-03 20:03:26
3
3
danusv
2016-02-03 20:03:26
3 ways
3 ways
leonlzg
2016-02-03 20:03:26
3
3
space_space
2016-02-03 20:03:26
3
3
copeland
2016-02-03 20:03:28
For the powers of 3 in the pair $(y,z),$ we must have one equal to 1 and the other is either 0 or 1. There are 3 ways to distribute the threes:\[(0,1)\quad(1,1)\quad(1,0).\]
For the powers of 3 in the pair $(y,z),$ we must have one equal to 1 and the other is either 0 or 1. There are 3 ways to distribute the threes:\[(0,1)\quad(1,1)\quad(1,0).\]
copeland
2016-02-03 20:03:34
And for the fives. How many ways are there to finish this triangle?
And for the fives. How many ways are there to finish this triangle?
copeland
2016-02-03 20:03:39
Emathmaster
2016-02-03 20:04:13
1
1
Mathaddict11
2016-02-03 20:04:13
1 way
1 way
FrostBlitz
2016-02-03 20:04:13
1
1
Metal_Bender19
2016-02-03 20:04:13
1
1
goodbear
2016-02-03 20:04:13
0,0,2
0,0,2
geogirl08
2016-02-03 20:04:13
Only 1 ( 0, 0, 2 )
Only 1 ( 0, 0, 2 )
space_space
2016-02-03 20:04:13
1
1
Alanshenkerman
2016-02-03 20:04:13
1
1
stevenrgao
2016-02-03 20:04:13
1 way
1 way
akaashp11
2016-02-03 20:04:13
1 way (x,y,z) = (0,0,2)
1 way (x,y,z) = (0,0,2)
axue
2016-02-03 20:04:13
one way: z gets 5^2
one way: z gets 5^2
copeland
2016-02-03 20:04:15
Since $\text{lcm}(x,y)$ contains no fives, $x$ and $y$ both contain no factors of 5. That also means $z$ must have $5^2$ as a prime power factor:
Since $\text{lcm}(x,y)$ contains no fives, $x$ and $y$ both contain no factors of 5. That also means $z$ must have $5^2$ as a prime power factor:
copeland
2016-02-03 20:04:20
copeland
2016-02-03 20:04:23
This is unique, so there is 1 way to distribute the fives.
This is unique, so there is 1 way to distribute the fives.
copeland
2016-02-03 20:04:24
And was our guess right?
And was our guess right?
snapdragon1
2016-02-03 20:04:46
yes
yes
Peggy
2016-02-03 20:04:46
yes it was
yes it was
MasterSnipes
2016-02-03 20:04:46
Yes
Yes
happyribbon
2016-02-03 20:04:46
Yes
Yes
FrostBlitz
2016-02-03 20:04:46
yes, because 3x5=15
yes, because 3x5=15
thomas0115
2016-02-03 20:04:46
of course
of course
deltaepsilon6
2016-02-03 20:04:46
yes
yes
lkarhat
2016-02-03 20:04:46
yes
yes
brisane
2016-02-03 20:04:46
Yes.
Yes.
copeland
2016-02-03 20:04:56
Yes. In all, there are $5\cdot3\cdot1=\boxed{15}$ ways to construct $(x,y,z).$ The answer is $A.$
Yes. In all, there are $5\cdot3\cdot1=\boxed{15}$ ways to construct $(x,y,z).$ The answer is $A.$
yiqun
2016-02-03 20:05:13
ANOTHER ONE PLZ
ANOTHER ONE PLZ
copeland
2016-02-03 20:05:17
Are you sure?
Are you sure?
ciao_potter
2016-02-03 20:05:27
yeah mann
yeah mann
MasterSnipes
2016-02-03 20:05:27
Another one please!
Another one please!
usharaj99
2016-02-03 20:05:27
yes
yes
copeland
2016-02-03 20:05:30
OK.
OK.
copeland
2016-02-03 20:05:32
As I said, Problems 24 and 25 on the AMC10 were also Problems 21 and 22 on the AMC12, so the next problem is. . .
As I said, Problems 24 and 25 on the AMC10 were also Problems 21 and 22 on the AMC12, so the next problem is. . .
goodbear
2016-02-03 20:05:51
23
23
Aspen
2016-02-03 20:05:51
23!
23!
owm
2016-02-03 20:05:51
23
23
lasergazer
2016-02-03 20:05:51
12-23
12-23
blizzard10
2016-02-03 20:05:51
Problem 23 of the AMC 12?
Problem 23 of the AMC 12?
reddymathcounts
2016-02-03 20:05:51
23 on AMC12!
23 on AMC12!
copeland
2016-02-03 20:05:53
23. Three numbers in the interval $[0,1]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\phantom{12A:23}$
(A) $\dfrac{1}{6} \qquad$ (B) $\dfrac{1}{3} \qquad$ (C) $\dfrac{1}{2} \qquad$ (D) $\dfrac{2}{3} \qquad$ (E) $\dfrac{5}{6}$
23. Three numbers in the interval $[0,1]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\phantom{12A:23}$
(A) $\dfrac{1}{6} \qquad$ (B) $\dfrac{1}{3} \qquad$ (C) $\dfrac{1}{2} \qquad$ (D) $\dfrac{2}{3} \qquad$ (E) $\dfrac{5}{6}$
copeland
2016-02-03 20:06:03
What kind of problem is this?
What kind of problem is this?
Locust
2016-02-03 20:06:24
3D Geometric Probability?
3D Geometric Probability?
amplreneo
2016-02-03 20:06:24
Geometric Probability?
Geometric Probability?
Math1331Math
2016-02-03 20:06:24
geometric probability
geometric probability
bigboss
2016-02-03 20:06:24
geometric prob
geometric prob
_--__--_
2016-02-03 20:06:24
Geometric probability
Geometric probability
strategos21
2016-02-03 20:06:24
geometry probability
geometry probability
alexli2014
2016-02-03 20:06:24
geometric probability
geometric probability
TheStrangeCharm
2016-02-03 20:06:24
geometric probability
geometric probability
muti66
2016-02-03 20:06:27
Geo probability
Geo probability
copeland
2016-02-03 20:06:29
This is a geometric probability problem. Arguably, it's one of the quintessential geometric probability problems. It is a very valuable problem to see if you haven't seen it before.
This is a geometric probability problem. Arguably, it's one of the quintessential geometric probability problems. It is a very valuable problem to see if you haven't seen it before.
copeland
2016-02-03 20:06:33
In geometric probability, we compute probabilities as $\dfrac{\text{success}}{\text{total}},$ however the success and total values are computed as sizes (lengths, areas, volumes, etc.) of geometric regions.
In geometric probability, we compute probabilities as $\dfrac{\text{success}}{\text{total}},$ however the success and total values are computed as sizes (lengths, areas, volumes, etc.) of geometric regions.
copeland
2016-02-03 20:06:40
You can get more info about geometric probability from Chapter 10 of our Introduction to Counting and Probability book:
https://artofproblemsolving.com/store/item/intro-counting
You can get more info about geometric probability from Chapter 10 of our Introduction to Counting and Probability book:
https://artofproblemsolving.com/store/item/intro-counting
copeland
2016-02-03 20:06:42
What does the set of all possible choices for $x$, $y$, and $z$ look like?
What does the set of all possible choices for $x$, $y$, and $z$ look like?
geogirl08
2016-02-03 20:07:07
Unit cube
Unit cube
Alanshenkerman
2016-02-03 20:07:07
a unit cube
a unit cube
eswa2000
2016-02-03 20:07:07
unit cube
unit cube
thomas0115
2016-02-03 20:07:07
a cube with side length 1
a cube with side length 1
idomath12345
2016-02-03 20:07:07
Cube.
Cube.
wjq8g6
2016-02-03 20:07:07
unit cube
unit cube
MathDog0809
2016-02-03 20:07:07
cube
cube
TheStrangeCharm
2016-02-03 20:07:07
a unit cube
a unit cube
math1012
2016-02-03 20:07:07
a cube
a cube
ethanproz
2016-02-03 20:07:07
a cube
a cube
copeland
2016-02-03 20:07:10
I can pick any value between 0 and 1 for each of my variables. The set of all possibilities is a unit cube in space:
I can pick any value between 0 and 1 for each of my variables. The set of all possibilities is a unit cube in space:
copeland
2016-02-03 20:07:11
copeland
2016-02-03 20:07:12
The "total" region has volume 1.
The "total" region has volume 1.
copeland
2016-02-03 20:07:13
Now when do I actually get a triangle?
Now when do I actually get a triangle?
A1012
2016-02-03 20:08:05
x+y>z, x+z>y, y+z>x
x+y>z, x+z>y, y+z>x
mjoshi
2016-02-03 20:08:05
x+y>z, x+z>y, y+z>x
x+y>z, x+z>y, y+z>x
MaYang
2016-02-03 20:08:05
a+b>c, b+c>a, a+c>b
a+b>c, b+c>a, a+c>b
A1012
2016-02-03 20:08:05
x+y>z, x+z>y, y+z>x
x+y>z, x+z>y, y+z>x
dchen04
2016-02-03 20:08:05
z+x>y x+y>z z+y>x
z+x>y x+y>z z+y>x
ninjataco
2016-02-03 20:08:05
x+y > z, x+z > y, y+z > x
x+y > z, x+z > y, y+z > x
copeland
2016-02-03 20:08:09
I get a triangle when the point $(x,y,z)$ satisfies the three triangle inequalities:
\begin{align*}
x+y> z\\
y+z> x\\
z+x> y.\\
\end{align*}
I get a triangle when the point $(x,y,z)$ satisfies the three triangle inequalities:
\begin{align*}
x+y> z\\
y+z> x\\
z+x> y.\\
\end{align*}
copeland
2016-02-03 20:08:13
Qualitatively, how do we compute the probability we seek?
Qualitatively, how do we compute the probability we seek?
memc38123
2016-02-03 20:08:59
Draw a shape within the cube
Draw a shape within the cube
TheStrangeCharm
2016-02-03 20:08:59
we want to find the volume of the region(s) of all points that satisfy these three inequalities
we want to find the volume of the region(s) of all points that satisfy these three inequalities
eswa2000
2016-02-03 20:08:59
find the volume corresponding to the intersection of these inequalities
find the volume corresponding to the intersection of these inequalities
AlcumusGuy
2016-02-03 20:08:59
successful volume divided by total volume
successful volume divided by total volume
mathguy623
2016-02-03 20:08:59
find the volume of the region satisfying these inequalities
find the volume of the region satisfying these inequalities
ExuberantGPACN
2016-02-03 20:08:59
successful volume over total cube volume
successful volume over total cube volume
copeland
2016-02-03 20:09:02
We need to find the subset of the unit cube where all of these inequalities hold and compute its volume. Then we divide by the total volume of the cube (which we saw is 1) and that is our probability.
We need to find the subset of the unit cube where all of these inequalities hold and compute its volume. Then we divide by the total volume of the cube (which we saw is 1) and that is our probability.
copeland
2016-02-03 20:09:05
Let's focus on just one of these inequalities. How do we find where $x+y> z$?
Let's focus on just one of these inequalities. How do we find where $x+y> z$?
toper2
2016-02-03 20:10:10
graph it?
graph it?
claserken
2016-02-03 20:10:10
Graph
Graph
copeland
2016-02-03 20:10:13
Wait, how?
Wait, how?
TheStrangeCharm
2016-02-03 20:10:52
draw in the plane z = x+ y
draw in the plane z = x+ y
amplreneo
2016-02-03 20:10:52
$x +y = z$ is a plane D:
$x +y = z$ is a plane D:
eswa2000
2016-02-03 20:10:52
graph x+y=z
graph x+y=z
brian22
2016-02-03 20:10:52
Graph $x+y=z$, and figure out which side we're bounding
Graph $x+y=z$, and figure out which side we're bounding
MasterSnipes
2016-02-03 20:10:52
When x+y = z
When x+y = z
Arginine
2016-02-03 20:10:52
graph z = x + y
graph z = x + y
copeland
2016-02-03 20:11:05
We begin by graphing $x+y=z$. Can you name 3 quick points on this plane?
We begin by graphing $x+y=z$. Can you name 3 quick points on this plane?
MathDog0809
2016-02-03 20:11:44
1,0,1
1,0,1
deltaepsilon6
2016-02-03 20:11:44
0,0,0
0,0,0
spartan168
2016-02-03 20:11:44
0,0,0 and 0,1,1 and 1,0,1
0,0,0 and 0,1,1 and 1,0,1
eswa2000
2016-02-03 20:11:44
(0,0,0), (1,0,1), (0,1,1)
(0,0,0), (1,0,1), (0,1,1)
thomas0115
2016-02-03 20:11:44
(0,0,0), (0,1,1),(1,0,1)
(0,0,0), (0,1,1),(1,0,1)
Locust
2016-02-03 20:11:44
(1,0,1) (0,0,0) (0,1,1)
(1,0,1) (0,0,0) (0,1,1)
TigerLily14
2016-02-03 20:11:44
(0,0,0)(1,0,1)(0,1,1)
(0,0,0)(1,0,1)(0,1,1)
axue
2016-02-03 20:11:44
(0,0,0)
(0,0,0)
jfurf
2016-02-03 20:11:44
$(1,0,1)$ $(0,0,0)$ and $(0,1,1)$
$(1,0,1)$ $(0,0,0)$ and $(0,1,1)$
copeland
2016-02-03 20:11:47
First off we have $0+0=0$ so $(0,0,0)$ is on the plane.
First off we have $0+0=0$ so $(0,0,0)$ is on the plane.
copeland
2016-02-03 20:11:48
We also have $1+0=1$ and $0+1=1$ so $(1,0,1)$ and $(0,1,1)$ are on the plane:
We also have $1+0=1$ and $0+1=1$ so $(1,0,1)$ and $(0,1,1)$ are on the plane:
copeland
2016-02-03 20:11:50
copeland
2016-02-03 20:11:51
We can connect those to define the intersection of the plane with the cube:
We can connect those to define the intersection of the plane with the cube:
copeland
2016-02-03 20:11:52
copeland
2016-02-03 20:12:01
What is the volume of the region where we don't get a triangle: when $x+y\leq z$?
What is the volume of the region where we don't get a triangle: when $x+y\leq z$?
alexli2014
2016-02-03 20:12:38
1/6
1/6
brian22
2016-02-03 20:12:38
$\frac{1}{6}$
$\frac{1}{6}$
wjq8g6
2016-02-03 20:12:38
1/6
1/6
mjoshi
2016-02-03 20:12:38
The left pyramid: 1/6
The left pyramid: 1/6
yrnsmurf
2016-02-03 20:12:38
$\dfrac16$
$\dfrac16$
loserboy
2016-02-03 20:12:38
1/6
1/6
spartan168
2016-02-03 20:12:38
since 0,0,1 works, it is the left side which is a tetrahedron of volume 1/6
since 0,0,1 works, it is the left side which is a tetrahedron of volume 1/6
ThorJames
2016-02-03 20:12:38
1/6
1/6
garyasho
2016-02-03 20:12:38
1/6
1/6
_--__--_
2016-02-03 20:12:38
1/6
1/6
strategos21
2016-02-03 20:12:38
1/6
1/6
Metal_Bender19
2016-02-03 20:12:38
1/6
1/6
copeland
2016-02-03 20:12:41
This is the region containing $(0,0,1)$ which is a tetrahedron:
This is the region containing $(0,0,1)$ which is a tetrahedron:
copeland
2016-02-03 20:12:44
copeland
2016-02-03 20:12:49
It is a pyramid with base of area $\dfrac12$ and height 1 so has total volume of $\dfrac{Bh}3=\dfrac16$.
It is a pyramid with base of area $\dfrac12$ and height 1 so has total volume of $\dfrac{Bh}3=\dfrac16$.
copeland
2016-02-03 20:12:50
What does this $\dfrac16$ mean probabilistically?
What does this $\dfrac16$ mean probabilistically?
copeland
2016-02-03 20:13:12
Gosh, that's a confusing question, huh?
Gosh, that's a confusing question, huh?
copeland
2016-02-03 20:13:24
What happens 1/6 of the time?
What happens 1/6 of the time?
Locust
2016-02-03 20:13:56
Any x,y,z between [0,1] have a 1/6 chance of not satisfying x+y>z
Any x,y,z between [0,1] have a 1/6 chance of not satisfying x+y>z
ryanyz10
2016-02-03 20:13:56
1/6 chance of getting a triangle that doesn't satisfy x + y > z
1/6 chance of getting a triangle that doesn't satisfy x + y > z
spartan168
2016-02-03 20:13:56
Probability of 1/6 that it satisfies the first equation of the sum of x and y is less than z
Probability of 1/6 that it satisfies the first equation of the sum of x and y is less than z
wjq8g6
2016-02-03 20:13:56
x+y<z
x+y<z
blizzard10
2016-02-03 20:13:56
$x+y<=z$
$x+y<=z$
alexli2014
2016-02-03 20:13:56
x+y <= z
x+y <= z
akaashp11
2016-02-03 20:13:56
x + y < z
x + y < z
copeland
2016-02-03 20:13:59
The probability that a randomly chosen triple $(x,y,z)$ satisfies $x+y\leq z$ is equal to $\dfrac16$. Therefore a random triple will fail this particular triangle inequality $\dfrac16$ of the time.
The probability that a randomly chosen triple $(x,y,z)$ satisfies $x+y\leq z$ is equal to $\dfrac16$. Therefore a random triple will fail this particular triangle inequality $\dfrac16$ of the time.
copeland
2016-02-03 20:14:10
There are other ways to fail, of course. 1/6 of the time we fail IN THIS WAY.
There are other ways to fail, of course. 1/6 of the time we fail IN THIS WAY.
copeland
2016-02-03 20:14:14
What is the probability that $z+x\leq y$?
What is the probability that $z+x\leq y$?
PNPADU
2016-02-03 20:14:31
1/6
1/6
fishy15
2016-02-03 20:14:31
1/6
1/6
MaYang
2016-02-03 20:14:31
1/6
1/6
BobThePotato
2016-02-03 20:14:31
1/6 again
1/6 again
goodbear
2016-02-03 20:14:31
1/6
1/6
dchen04
2016-02-03 20:14:31
1/6
1/6
owm
2016-02-03 20:14:31
1/6
1/6
Peggy
2016-02-03 20:14:31
also 1/6?
also 1/6?
XturtleX
2016-02-03 20:14:31
1/6
1/6
danusv
2016-02-03 20:14:31
1/6 as well
1/6 as well
copeland
2016-02-03 20:14:34
By symmetry, the probability that $z+x\leq y$ must also be $\dfrac16$. This is the volume of the tetrahedron below this plane:
By symmetry, the probability that $z+x\leq y$ must also be $\dfrac16$. This is the volume of the tetrahedron below this plane:
copeland
2016-02-03 20:14:36
copeland
2016-02-03 20:14:37
Likewise, the probability that $y+z\leq x$ is also $\dfrac16$. This is the volume of the tetrahedron below this plane:
Likewise, the probability that $y+z\leq x$ is also $\dfrac16$. This is the volume of the tetrahedron below this plane:
copeland
2016-02-03 20:14:44
copeland
2016-02-03 20:14:49
What do we need to do in order to find the probability we are not in any one of these regions?
What do we need to do in order to find the probability we are not in any one of these regions?
idomath12345
2016-02-03 20:15:14
Add.
Add.
fishy15
2016-02-03 20:15:14
add and then subtract from one
add and then subtract from one
duruphi
2016-02-03 20:15:14
So add the probablities together and get 1/2?
So add the probablities together and get 1/2?
idomath12345
2016-02-03 20:15:14
And subtract from 1.
And subtract from 1.
math1012
2016-02-03 20:15:14
subtract from 1
subtract from 1
garyasho
2016-02-03 20:15:14
Add: 1/6 * 3 = 1/2
Add: 1/6 * 3 = 1/2
copeland
2016-02-03 20:15:21
That doesn't sound like enough to me. . .
That doesn't sound like enough to me. . .
copeland
2016-02-03 20:15:32
We certainly need to chop off all 3 tetrahedra and compute the volume of the remaining region. However, what do we need to worry about?
We certainly need to chop off all 3 tetrahedra and compute the volume of the remaining region. However, what do we need to worry about?
ThorJames
2016-02-03 20:16:06
find the intersection of them
find the intersection of them
BobThePotato
2016-02-03 20:16:06
overlapping areas
overlapping areas
Locust
2016-02-03 20:16:06
If they overlap anywhere
If they overlap anywhere
claserken
2016-02-03 20:16:06
intersection
intersection
spin8
2016-02-03 20:16:06
intersections
intersections
amcwu
2016-02-03 20:16:06
overlap
overlap
XturtleX
2016-02-03 20:16:06
overlap
overlap
A1012
2016-02-03 20:16:06
overcounting
overcounting
coconut_force
2016-02-03 20:16:06
if they overlap
if they overlap
Math99
2016-02-03 20:16:06
overlap
overlap
yiqun
2016-02-03 20:16:06
overlapping
overlapping
copeland
2016-02-03 20:16:10
We will be in trouble if the regions overlap since that would make it easy to over- or undercount the regions of overlap. Is that a problem here?
We will be in trouble if the regions overlap since that would make it easy to over- or undercount the regions of overlap. Is that a problem here?
thomas0115
2016-02-03 20:16:39
the 3 tetrahedrons are disjoint, so we can just subtract
the 3 tetrahedrons are disjoint, so we can just subtract
TheStrangeCharm
2016-02-03 20:16:39
these three bad events are all disjoint clearly
these three bad events are all disjoint clearly
brian22
2016-02-03 20:16:39
Well they don't intersect, so we can just add and complement
Well they don't intersect, so we can just add and complement
geogirl08
2016-02-03 20:16:39
they don't intersect
they don't intersect
MathWhizzz
2016-02-03 20:16:39
wait htey obviously dont overlap
wait htey obviously dont overlap
copeland
2016-02-03 20:16:53
No! If $x+y\leq z$ then $z$ is the longest edge. Likewise if $y+z\leq x$ then $x$ is longest and if $z+x\leq y$ then $y$ is longest. The only places these regions can possibly overlap is when 2 of the edges are equal and the third is zero. The overlapping regions are lines so they have volume 0 and we can ignore them.
No! If $x+y\leq z$ then $z$ is the longest edge. Likewise if $y+z\leq x$ then $x$ is longest and if $z+x\leq y$ then $y$ is longest. The only places these regions can possibly overlap is when 2 of the edges are equal and the third is zero. The overlapping regions are lines so they have volume 0 and we can ignore them.
copeland
2016-02-03 20:17:16
So they actually DO overlap, but only trivially so.
So they actually DO overlap, but only trivially so.
copeland
2016-02-03 20:17:46
What is the probability that we don't form a triangle when we pick a triple $(x,y,z)?$
What is the probability that we don't form a triangle when we pick a triple $(x,y,z)?$
A1012
2016-02-03 20:18:13
So the answer is 1/2?
So the answer is 1/2?
goodbear
2016-02-03 20:18:13
1-1/6-1/6-1/6=1/2 (C)
1-1/6-1/6-1/6=1/2 (C)
amplreneo
2016-02-03 20:18:13
1/2
1/2
MasterSnipes
2016-02-03 20:18:13
1/2
1/2
happyribbon
2016-02-03 20:18:13
1/2
1/2
cinamon
2016-02-03 20:18:13
(C) 1/2
(C) 1/2
Hydrahead
2016-02-03 20:18:13
1/2
1/2
deltaepsilon6
2016-02-03 20:18:13
1-(1/6*3)
1-(1/6*3)
mathguy623
2016-02-03 20:18:13
1/2
1/2
FrostBlitz
2016-02-03 20:18:13
1/2
1/2
Emathmaster
2016-02-03 20:18:13
1/2
1/2
MathDog0809
2016-02-03 20:18:13
1/2
1/2
picoo
2016-02-03 20:18:13
1/2
1/2
deltaepsilon6
2016-02-03 20:18:13
1/2
1/2
mathproash2
2016-02-03 20:18:13
1/2
1/2
Interest
2016-02-03 20:18:13
1/2
1/2
Locust2
2016-02-03 20:18:13
1/2
1/2
copeland
2016-02-03 20:18:18
Since the overlaps have volume 0, the probability that our $(x,y,z)$ lands in one of the invalid tetrahedra is $\dfrac16+\dfrac16+\dfrac16=\dfrac12$.
Since the overlaps have volume 0, the probability that our $(x,y,z)$ lands in one of the invalid tetrahedra is $\dfrac16+\dfrac16+\dfrac16=\dfrac12$.
copeland
2016-02-03 20:18:19
The probability we can't construct a triangle is complementary to the probability we do pick a triangle so the final probability is $1-\dfrac12=\boxed{\dfrac12}.$
The probability we can't construct a triangle is complementary to the probability we do pick a triangle so the final probability is $1-\dfrac12=\boxed{\dfrac12}.$
copeland
2016-02-03 20:18:43
Note that the answer is somehow not $\dfrac12$, but instead it is $1-\dfrac12$. This is one thing I don't like about this problem.
Note that the answer is somehow not $\dfrac12$, but instead it is $1-\dfrac12$. This is one thing I don't like about this problem.
copeland
2016-02-03 20:18:49
So I should admit that I wanted to solve the problem that way first and led you along the harder path.
So I should admit that I wanted to solve the problem that way first and led you along the harder path.
copeland
2016-02-03 20:18:57
I know for sure that (at least) one of the numbers is largest. Let's say it is $z.$ Then what are the possible values for $x$ and $y?$
I know for sure that (at least) one of the numbers is largest. Let's say it is $z.$ Then what are the possible values for $x$ and $y?$
ninjataco
2016-02-03 20:19:47
0 through z
0 through z
fishy15
2016-02-03 20:19:47
z>x, z>y
z>x, z>y
deltaepsilon6
2016-02-03 20:19:47
0<x<z
0<x<z
deltaepsilon6
2016-02-03 20:19:47
0<y<z
0<y<z
copeland
2016-02-03 20:19:51
For any $z,$ if $z$ is the largest value then $(x,y)$ must be chosen in a square of size $z\times z$:
For any $z,$ if $z$ is the largest value then $(x,y)$ must be chosen in a square of size $z\times z$:
copeland
2016-02-03 20:19:52
copeland
2016-02-03 20:19:57
And which values of $(x,y)$ form a triangle with our $z$?
And which values of $(x,y)$ form a triangle with our $z$?
thomas0115
2016-02-03 20:20:36
x+y>=z
x+y>=z
yrnsmurf
2016-02-03 20:20:36
Top right half
Top right half
jfurf
2016-02-03 20:20:36
$x+y>z$
$x+y>z$
deltaepsilon6
2016-02-03 20:20:36
x+y>z
x+y>z
A1012
2016-02-03 20:20:36
x+y>z
x+y>z
mjoshi
2016-02-03 20:20:36
x+y>z
x+y>z
copeland
2016-02-03 20:20:39
If we know $z$ is the longest edge then $(x,y,z)$ forms a triangle whenever $x+y\geq z.$
If we know $z$ is the longest edge then $(x,y,z)$ forms a triangle whenever $x+y\geq z.$
copeland
2016-02-03 20:20:40
copeland
2016-02-03 20:20:43
So if $z$ is the biggest number, what is the probability that we can form a triangle?
So if $z$ is the biggest number, what is the probability that we can form a triangle?
strategos21
2016-02-03 20:21:05
1/2
1/2
mathwizard888
2016-02-03 20:21:05
1/2
1/2
blizzard10
2016-02-03 20:21:05
1/2
1/2
MaYang
2016-02-03 20:21:05
1/2
1/2
duruphi
2016-02-03 20:21:05
1/2
1/2
spin8
2016-02-03 20:21:05
1/2
1/2
dchen04
2016-02-03 20:21:05
1/2
1/2
bigboss
2016-02-03 20:21:05
1/2
1/2
happyribbon
2016-02-03 20:21:05
1/2
1/2
toper2
2016-02-03 20:21:05
1/2
1/2
aq1048576
2016-02-03 20:21:05
1/2
1/2
copeland
2016-02-03 20:21:08
For each $z$, if $z$ is the biggest number this probability is 1/2. Therefore the probability that we can form a triangle assuming $z$ is biggest is 1/2.
For each $z$, if $z$ is the biggest number this probability is 1/2. Therefore the probability that we can form a triangle assuming $z$ is biggest is 1/2.
copeland
2016-02-03 20:21:09
If $x$ is biggest, what is the probability we can form a triangle?
If $x$ is biggest, what is the probability we can form a triangle?
baseballcat
2016-02-03 20:21:26
1/2
1/2
MaYang
2016-02-03 20:21:26
1/2
1/2
Metal_Bender19
2016-02-03 20:21:26
1/2
1/2
MathDog0809
2016-02-03 20:21:26
1/2 too
1/2 too
math1012
2016-02-03 20:21:26
1/2
1/2
geogirl08
2016-02-03 20:21:26
still 1/2
still 1/2
spartan168
2016-02-03 20:21:26
1/2
1/2
ThorJames
2016-02-03 20:21:26
1/2
1/2
cinamon
2016-02-03 20:21:26
ditto
ditto
copeland
2016-02-03 20:21:29
If $x$ is biggest, the probability is still 1/2. Likewise if $y$ is biggest. So what?
If $x$ is biggest, the probability is still 1/2. Likewise if $y$ is biggest. So what?
math1012
2016-02-03 20:21:47
the answer is 1/2 !
the answer is 1/2 !
alexli2014
2016-02-03 20:21:47
its always 1/2
its always 1/2
thomas0115
2016-02-03 20:21:47
the overall probability is also 1/2
the overall probability is also 1/2
Emathmaster
2016-02-03 20:21:47
The answer is 1/2
The answer is 1/2
mssmath
2016-02-03 20:21:47
1/2
1/2
MaYang
2016-02-03 20:21:47
The answer is 1/2?
The answer is 1/2?
lkarhat
2016-02-03 20:21:47
the total proboality is 1/2
the total proboality is 1/2
goldypeng
2016-02-03 20:21:47
still 1/2
still 1/2
copeland
2016-02-03 20:21:49
Those are all the cases! No matter which variable is biggest, the probability we can form a triangle is 1/2.
Those are all the cases! No matter which variable is biggest, the probability we can form a triangle is 1/2.
copeland
2016-02-03 20:21:50
We can look back on our pictures and see this argument in action. The set of points where $z$ is biggest is a square pyramid:
We can look back on our pictures and see this argument in action. The set of points where $z$ is biggest is a square pyramid:
copeland
2016-02-03 20:21:54
copeland
2016-02-03 20:22:02
Picking a particular $z$ gives us a square cross-section:
Picking a particular $z$ gives us a square cross-section:
copeland
2016-02-03 20:22:03
copeland
2016-02-03 20:22:17
And we get a red and green triangular pyramid with base area 1/2 and height 1, so we're equally likely to be in the red or green pyramid.
And we get a red and green triangular pyramid with base area 1/2 and height 1, so we're equally likely to be in the red or green pyramid.
copeland
2016-02-03 20:22:23
copeland
2016-02-03 20:22:29
The whole cube breaks into three such pairs of equally likely pyramids to half the time we are in a "success" region and half the time we are in a "fail" region. The answer is again $\dfrac12.$
The whole cube breaks into three such pairs of equally likely pyramids to half the time we are in a "success" region and half the time we are in a "fail" region. The answer is again $\dfrac12.$
duruphi
2016-02-03 20:22:55
So the answer is C!
So the answer is C!
goldypeng
2016-02-03 20:22:55
So the answer is 1/2 (C)?
So the answer is 1/2 (C)?
copeland
2016-02-03 20:22:59
The answer is very much $C$.
The answer is very much $C$.
copeland
2016-02-03 20:23:01
Are we having fun yet?
Are we having fun yet?
XturtleX
2016-02-03 20:23:14
yes
yes
owm
2016-02-03 20:23:14
yep!!!
yep!!!
PNPADU
2016-02-03 20:23:14
yes
yes
spin8
2016-02-03 20:23:14
YES
YES
awesomethree
2016-02-03 20:23:14
YES
YES
Locust
2016-02-03 20:23:14
YEAH
YEAH
GrilledOnions
2016-02-03 20:23:14
Yes!
Yes!
akaashp11
2016-02-03 20:23:14
Next one please!!
Next one please!!
copeland
2016-02-03 20:23:17
Next up is. . .
Next up is. . .
akaashp11
2016-02-03 20:23:39
24!!!
24!!!
lcalvert99
2016-02-03 20:23:39
24
24
deltaepsilon6
2016-02-03 20:23:39
24!
24!
awesomethree
2016-02-03 20:23:39
24
24
owm
2016-02-03 20:23:39
24
24
Math99
2016-02-03 20:24:24
why don't you have to divide by three based on whether x is the biggest, y is the biggest, or z is the biggest?
why don't you have to divide by three based on whether x is the biggest, y is the biggest, or z is the biggest?
copeland
2016-02-03 20:24:32
Here's a good question. You don't want to divide by 3. You want to take some sort of (weighted) average of the halves. Since they're all 1/2, it doesn't even matter that the weights are equal.
Here's a good question. You don't want to divide by 3. You want to take some sort of (weighted) average of the halves. Since they're all 1/2, it doesn't even matter that the weights are equal.
copeland
2016-02-03 20:24:36
Alright, Problem 24:
Alright, Problem 24:
copeland
2016-02-03 20:24:38
24. There is a smallest positive real number $a$ such that there exists a positive number $b$ such that all the roots of the polynomial $x^3 - ax^2 + bx - a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is the value of $b$?
$\phantom{12A:24}$
(A) $8 \qquad$ (B) $9 \qquad$ (C) $10 \qquad$ (D) $11 \qquad$ (E) 12
24. There is a smallest positive real number $a$ such that there exists a positive number $b$ such that all the roots of the polynomial $x^3 - ax^2 + bx - a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is the value of $b$?
$\phantom{12A:24}$
(A) $8 \qquad$ (B) $9 \qquad$ (C) $10 \qquad$ (D) $11 \qquad$ (E) 12
copeland
2016-02-03 20:24:46
First, does having $a$ and $b$ positive tell us anything about the roots?
First, does having $a$ and $b$ positive tell us anything about the roots?
brian22
2016-02-03 20:25:29
all positive
all positive
ThorJames
2016-02-03 20:25:29
the roots are positive
the roots are positive
math1012
2016-02-03 20:25:29
all roots are positive
all roots are positive
deltaepsilon6
2016-02-03 20:25:29
roots are positive
roots are positive
ninjataco
2016-02-03 20:25:29
all positive
all positive
mathwizard888
2016-02-03 20:25:29
all positive
all positive
MaYang
2016-02-03 20:25:29
All of the roots are positive?
All of the roots are positive?
copeland
2016-02-03 20:25:30
Since $a$ and $b$ are positive, Descartes Rule of Signs tells us that all the roots must be positive. You can also see this directly since if you substitute any negative value for $x$ you get the sum of four negative real numbers (and substituting 0 gives $-a\neq0$).
Since $a$ and $b$ are positive, Descartes Rule of Signs tells us that all the roots must be positive. You can also see this directly since if you substitute any negative value for $x$ you get the sum of four negative real numbers (and substituting 0 gives $-a\neq0$).
copeland
2016-02-03 20:25:36
Now how should we approach this problem?
Now how should we approach this problem?
Locust
2016-02-03 20:26:02
Use Vietas
Use Vietas
jfurf
2016-02-03 20:26:02
VIETA'S!!
VIETA'S!!
MaYang
2016-02-03 20:26:02
Vietas
Vietas
deltaepsilon6
2016-02-03 20:26:02
vieta's formula
vieta's formula
SimonSun
2016-02-03 20:26:02
vietas
vietas
akaashp11
2016-02-03 20:26:02
Vieta??
Vieta??
mssmath
2016-02-03 20:26:02
Let u,v,w be the roots and then use vieat's formulas
Let u,v,w be the roots and then use vieat's formulas
ryanyz10
2016-02-03 20:26:02
vietas?
vietas?
owm
2016-02-03 20:26:02
Vieta
Vieta
math1012
2016-02-03 20:26:02
viete's
viete's
fishy15
2016-02-03 20:26:02
vieta
vieta
yrnsmurf
2016-02-03 20:26:02
Vieta's?
Vieta's?
copeland
2016-02-03 20:26:06
Since we have 3 real roots we can give them names and use Vieta. Let the roots be $r,$ $s,$ and $t.$ Then \begin{align*}x^3-ax^2+bx-a&=(x-r)(x-s)(x-t)\\&=x^3-(r+s+t)x^2+(rs+st+tr)x-(rst)x.\end{align*}
Since we have 3 real roots we can give them names and use Vieta. Let the roots be $r,$ $s,$ and $t.$ Then \begin{align*}x^3-ax^2+bx-a&=(x-r)(x-s)(x-t)\\&=x^3-(r+s+t)x^2+(rs+st+tr)x-(rst)x.\end{align*}
copeland
2016-02-03 20:26:08
Therefore \begin{align*}
a&=r+s+t,\\
b&=rs+st+tr,\\
a&=rst.
\end{align*}
Therefore \begin{align*}
a&=r+s+t,\\
b&=rs+st+tr,\\
a&=rst.
\end{align*}
copeland
2016-02-03 20:26:09
What can we do with the two expressions that are equal to $a$?
What can we do with the two expressions that are equal to $a$?
mssmath
2016-02-03 20:27:25
r+s+t=rst
r+s+t=rst
Interest
2016-02-03 20:27:25
set them equal
set them equal
amcwu
2016-02-03 20:27:25
set them equal
set them equal
fractal161
2016-02-03 20:27:25
equate them?
equate them?
owm
2016-02-03 20:27:25
rst=r+s+t
rst=r+s+t
yiqun
2016-02-03 20:27:25
rst=r+s+t
rst=r+s+t
Natsu_Dragneel
2016-02-03 20:27:25
set them equal
set them equal
MasterSnipes
2016-02-03 20:27:25
MAKE EM EQUAL
MAKE EM EQUAL
copeland
2016-02-03 20:27:26
We could do that. That would give us some way to eliminate variables, but it really kills the symmetry.
We could do that. That would give us some way to eliminate variables, but it really kills the symmetry.
copeland
2016-02-03 20:27:30
Any other ideas?
Any other ideas?
ychen
2016-02-03 20:28:16
amgm
amgm
bharatputra
2016-02-03 20:28:16
am-gm
am-gm
mssmath
2016-02-03 20:28:16
Use AM-GM, in particular $xyz=x+y+z\ge3(xyz)^{\frac{1}{3}}$
Use AM-GM, in particular $xyz=x+y+z\ge3(xyz)^{\frac{1}{3}}$
brian22
2016-02-03 20:28:16
AM-GM FTW
AM-GM FTW
copeland
2016-02-03 20:28:22
Those values are reminiscent of the AM-GM inequality. The AM-GM inequality states that the arithmetic mean of a set of nonnegative numbers is greater than or equal to the set's geometric mean. See our Wiki for more details:
https://www.artofproblemsolving.com/wiki/index.php?title=Arithmetic_Mean-Geometric_Mean_Inequality
Those values are reminiscent of the AM-GM inequality. The AM-GM inequality states that the arithmetic mean of a set of nonnegative numbers is greater than or equal to the set's geometric mean. See our Wiki for more details:
https://www.artofproblemsolving.com/wiki/index.php?title=Arithmetic_Mean-Geometric_Mean_Inequality
copeland
2016-02-03 20:28:26
What do we get from AM-GM?
What do we get from AM-GM?
ninjataco
2016-02-03 20:29:34
a/3 >= a^(1/3)
a/3 >= a^(1/3)
fractal161
2016-02-03 20:29:34
a >= 3sqrt3
a >= 3sqrt3
ychen
2016-02-03 20:29:34
a = 3 root 3
a = 3 root 3
blizzard10
2016-02-03 20:29:34
$r+s+t>=3\sqrt[3]{rst}$
$r+s+t>=3\sqrt[3]{rst}$
mjoshi
2016-02-03 20:29:34
a>=3a^{1/3}
a>=3a^{1/3}
goodbear
2016-02-03 20:29:34
xyz≥3xyz^(1/3)
xyz≥3xyz^(1/3)
scfliu5
2016-02-03 20:29:34
a/3>=a^(1/3)
a/3>=a^(1/3)
claserken
2016-02-03 20:29:34
(x+y+z)/3 = cube root of xyz
(x+y+z)/3 = cube root of xyz
copeland
2016-02-03 20:29:38
Since all three roots are positive, AM-GM tells us that \[\sqrt[3]{rst}\leq\frac{r+s+t}3.\](The left side is the geometric mean and the right side is the arithmetic mean.)
Since all three roots are positive, AM-GM tells us that \[\sqrt[3]{rst}\leq\frac{r+s+t}3.\](The left side is the geometric mean and the right side is the arithmetic mean.)
copeland
2016-02-03 20:29:44
In this case we have \[\sqrt[3]a\leq \frac a3.\]
In this case we have \[\sqrt[3]a\leq \frac a3.\]
copeland
2016-02-03 20:29:51
Cubing gives $a\leq \dfrac{a^3}{27}$ or $27\leq a^2$. Taking the square root gives $a\geq 3\sqrt3$.
Cubing gives $a\leq \dfrac{a^3}{27}$ or $27\leq a^2$. Taking the square root gives $a\geq 3\sqrt3$.
copeland
2016-02-03 20:30:15
What does that tell us?
What does that tell us?
Alanshenkerman
2016-02-03 20:31:06
smallest value of a = 3sqrt3
smallest value of a = 3sqrt3
wjq8g6
2016-02-03 20:31:06
a = 3sqrt3 since we want the smallest a
a = 3sqrt3 since we want the smallest a
Locust
2016-02-03 20:31:06
the smallest value of a is $3\sqrt{3}$
the smallest value of a is $3\sqrt{3}$
ethanproz
2016-02-03 20:31:06
Since $a$ must be minimum, $a=3\sqrt{3}$
Since $a$ must be minimum, $a=3\sqrt{3}$
blizzard10
2016-02-03 20:31:06
$3\sqrt{3}$ is the smallest possible value for a
$3\sqrt{3}$ is the smallest possible value for a
copeland
2016-02-03 20:31:08
Close! We do want the smallest value, but we're not quiet home yet, right?
Close! We do want the smallest value, but we're not quiet home yet, right?
akaashp11
2016-02-03 20:31:38
We need to make b unique
We need to make b unique
mssmath
2016-02-03 20:31:38
We now have to find b
We now have to find b
blizzard10
2016-02-03 20:31:38
The value of b must be unique
The value of b must be unique
psolver1
2016-02-03 20:31:38
We need to find b
We need to find b
shark0416
2016-02-03 20:31:38
we need to see when b is positive?
we need to see when b is positive?
copeland
2016-02-03 20:31:44
Since we're looking for the minimum value of $a$ with some certain properties, we know the target value for $a$ is at least $3\sqrt3$. Specifically, we know that if there are to be 3 real roots to any such equation, we must have $a$ at least this large. We still have not looked into $b$ so we still do not know whether such a $b$ exists.
Since we're looking for the minimum value of $a$ with some certain properties, we know the target value for $a$ is at least $3\sqrt3$. Specifically, we know that if there are to be 3 real roots to any such equation, we must have $a$ at least this large. We still have not looked into $b$ so we still do not know whether such a $b$ exists.
copeland
2016-02-03 20:31:55
Assuming that $a=3\sqrt3$, what are the roots?
Assuming that $a=3\sqrt3$, what are the roots?
brian22
2016-02-03 20:32:37
ALSO, we have the equality case of AM-GM if we want minimum value, so $r=s=t$ to get $a=3\sqrt{3}$
ALSO, we have the equality case of AM-GM if we want minimum value, so $r=s=t$ to get $a=3\sqrt{3}$
deltaepsilon6
2016-02-03 20:32:37
$x=y=z=\sqrt{3},P(x)=(x-\sqrt{3})^3$
$x=y=z=\sqrt{3},P(x)=(x-\sqrt{3})^3$
math1012
2016-02-03 20:32:37
$\sqrt{3}$
$\sqrt{3}$
yrnsmurf
2016-02-03 20:32:37
All of them are $\sqrt3$
All of them are $\sqrt3$
deltaepsilon6
2016-02-03 20:32:37
triple root at x-root(3)
triple root at x-root(3)
Peggy
2016-02-03 20:32:37
all roots are $\sqrt{3}$
all roots are $\sqrt{3}$
eswa2000
2016-02-03 20:32:37
all sqrt{3}
all sqrt{3}
fractal161
2016-02-03 20:32:37
all sqrt3
all sqrt3
copeland
2016-02-03 20:32:44
Since $a=3\sqrt3$ is the "equality case" of AM-GM, the only possible values of the roots are $r=s=t=\sqrt3$.
Since $a=3\sqrt3$ is the "equality case" of AM-GM, the only possible values of the roots are $r=s=t=\sqrt3$.
copeland
2016-02-03 20:32:45
And what is $b$?
And what is $b$?
SimonSun
2016-02-03 20:33:28
9
9
spartan168
2016-02-03 20:33:28
3 + 3 + 3 = 9
3 + 3 + 3 = 9
aq1048576
2016-02-03 20:33:28
9
9
MaYang
2016-02-03 20:33:28
9
9
MasterSnipes
2016-02-03 20:33:28
9
9
shark0416
2016-02-03 20:33:28
9
9
wjq8g6
2016-02-03 20:33:28
9
9
Peggy
2016-02-03 20:33:28
9
9
fishy15
2016-02-03 20:33:28
rs + rt +st =9
rs + rt +st =9
mjoshi
2016-02-03 20:33:28
9
9
TigerLily14
2016-02-03 20:33:28
9 (B)
9 (B)
aq1048576
2016-02-03 20:33:28
BBBB
BBBB
loserboy
2016-02-03 20:33:28
9
9
dchen04
2016-02-03 20:33:32
9
9
Sophiarulz
2016-02-03 20:33:32
9
9
MathDog0809
2016-02-03 20:33:32
rs+st+rt=9
rs+st+rt=9
ThorJames
2016-02-03 20:33:32
$9$
$9$
copeland
2016-02-03 20:33:34
The value of $b$ is \
The value of $b$ is \
copeland
2016-02-03 20:33:37
Is this the answer?
Is this the answer?
lkarhat
2016-02-03 20:34:11
Yes
Yes
Emathmaster
2016-02-03 20:34:11
Yes!!!
Yes!!!
TheBlackBeltGuy
2016-02-03 20:34:11
yes
yes
copeland
2016-02-03 20:34:13
This is the answer. We know that $a$ must be at least $3\sqrt3$ in order for $b$ to exist. If $a=3\sqrt3$, this value of $b$ exists and indeed is unique. The answer is $\boxed{9},$ $B$.
This is the answer. We know that $a$ must be at least $3\sqrt3$ in order for $b$ to exist. If $a=3\sqrt3$, this value of $b$ exists and indeed is unique. The answer is $\boxed{9},$ $B$.
copeland
2016-02-03 20:34:24
Alright, now what?
Alright, now what?
MaYang
2016-02-03 20:34:53
25
25
goodbear
2016-02-03 20:34:53
25
25
spartan168
2016-02-03 20:34:53
The final question
The final question
yrnsmurf
2016-02-03 20:34:53
Last question?
Last question?
goldypeng
2016-02-03 20:34:53
Last Problem!!!!!!!
Last Problem!!!!!!!
Quinn
2016-02-03 20:34:53
25!
25!
duruphi
2016-02-03 20:34:53
NUMBER 25!!! THE FINALE!!!
NUMBER 25!!! THE FINALE!!!
claserken
2016-02-03 20:34:53
Number 25!!!
Number 25!!!
deltaepsilon6
2016-02-03 20:34:53
25!
25!
copeland
2016-02-03 20:34:58
25. Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k + 1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k = 1$, then the numbers that Bernardo writes are 16, 25, 36, 49, and 64, and the numbers showing on the board after Silvia erases are 1, 2, 3, 4, and 6, and thus $f(1) = 5$. What is the sum of the digits of $f(2) + f(4) + f(6) + \cdots + f(2016)$?
$\phantom{12A:25}$
(A) $7986 \qquad$ (B) $8002 \qquad$ (C) $8030 \qquad$ (D) $8048 \qquad$ (E) 8064
25. Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k + 1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k = 1$, then the numbers that Bernardo writes are 16, 25, 36, 49, and 64, and the numbers showing on the board after Silvia erases are 1, 2, 3, 4, and 6, and thus $f(1) = 5$. What is the sum of the digits of $f(2) + f(4) + f(6) + \cdots + f(2016)$?
$\phantom{12A:25}$
(A) $7986 \qquad$ (B) $8002 \qquad$ (C) $8030 \qquad$ (D) $8048 \qquad$ (E) 8064
copeland
2016-02-03 20:35:04
OK, so it took me a lot of failing at this problem before I figured out exactly what was wanted here. First of all, Bernardo is making a list of squares, such as, for $k=2$:
OK, so it took me a lot of failing at this problem before I figured out exactly what was wanted here. First of all, Bernardo is making a list of squares, such as, for $k=2$:
copeland
2016-02-03 20:35:10
\begin{array}{c|ccccccccccccccc}
\text{Bernardo}&
100&
121&
144&
169&
196&
225&
256&
289&
324&
361&
400&
441&
484&
529&
576&
625&
676&
729&
785&
841&
900&
961
\end{array}
\begin{array}{c|ccccccccccccccc}
\text{Bernardo}&
100&
121&
144&
169&
196&
225&
256&
289&
324&
361&
400&
441&
484&
529&
576&
625&
676&
729&
785&
841&
900&
961
\end{array}
copeland
2016-02-03 20:35:17
(Scroll right. . . )
(Scroll right. . . )
copeland
2016-02-03 20:35:21
As he's doing this, Silvia is changing his list by striking off the last 2 digits:
As he's doing this, Silvia is changing his list by striking off the last 2 digits:
copeland
2016-02-03 20:35:22
\begin{array}{c|ccccccccccccccc}
\text{Bernardo}&
100&
121&
144&
169&
196&
225&
256&
289&
324&
361&
400&
441&
484&
529&
576&
625&
676&
729&
784&
841&
900&
961&
\cdots\\
\hline
\text{Silvia}&
1\phantom{00}&
1\phantom{21}&
1\phantom{44}&
1\phantom{69}&
1\phantom{96}&
2\phantom{25}&
2\phantom{56}&
2\phantom{89}&
3\phantom{24}&
3\phantom{61}&
4\phantom{00}&
4\phantom{41}&
4\phantom{84}&
5\phantom{29}&
5\phantom{76}&
6\phantom{25}&
6\phantom{76}&
7\phantom{29}&
7\phantom{84}&
8\phantom{41}&
9\phantom{00}&
9\phantom{61}&
\cdots\\
\end{array}
\begin{array}{c|ccccccccccccccc}
\text{Bernardo}&
100&
121&
144&
169&
196&
225&
256&
289&
324&
361&
400&
441&
484&
529&
576&
625&
676&
729&
784&
841&
900&
961&
\cdots\\
\hline
\text{Silvia}&
1\phantom{00}&
1\phantom{21}&
1\phantom{44}&
1\phantom{69}&
1\phantom{96}&
2\phantom{25}&
2\phantom{56}&
2\phantom{89}&
3\phantom{24}&
3\phantom{61}&
4\phantom{00}&
4\phantom{41}&
4\phantom{84}&
5\phantom{29}&
5\phantom{76}&
6\phantom{25}&
6\phantom{76}&
7\phantom{29}&
7\phantom{84}&
8\phantom{41}&
9\phantom{00}&
9\phantom{61}&
\cdots\\
\end{array}
copeland
2016-02-03 20:35:38
The next square 1024. What does Silvia end up with?
The next square 1024. What does Silvia end up with?
axue
2016-02-03 20:36:24
10
10
amcwu
2016-02-03 20:36:24
10
10
celestialphoenix3768
2016-02-03 20:36:24
10
10
scfliu5
2016-02-03 20:36:24
10
10
SimonSun
2016-02-03 20:36:24
10
10
Math99
2016-02-03 20:36:24
10
10
Emathmaster
2016-02-03 20:36:24
10
10
blizzard10
2016-02-03 20:36:24
10
10
ryanyz10
2016-02-03 20:36:24
10
10
stevenrgao
2016-02-03 20:36:24
10
10
copeland
2016-02-03 20:36:27
Killing the last 2 digits of 1024 leaves just 10 on the board. With $k=2$ still, the sequence continues as
Killing the last 2 digits of 1024 leaves just 10 on the board. With $k=2$ still, the sequence continues as
copeland
2016-02-03 20:36:28
\begin{array}{c|cccccccccccccccc}
\text{Bernardo}&
\cdots&
1024&
1089&
1156&
1225&
1369&
1444&
1521&
1600&
1681&
1764&
1849&
1936&
2025&
2116&
2209&
2304&
2401\\
\hline
\text{Silvia}&
\cdots&
10\phantom{24}&
10\phantom{89}&
11\phantom{56}&
12\phantom{25}&
13\phantom{69}&
14\phantom{44}&
15\phantom{21}&
16\phantom{00}&
16\phantom{81}&
17\phantom{64}&
18\phantom{49}&
19\phantom{36}&
20\phantom{25}&
21\phantom{16}&
22\phantom{09}&
23\phantom{04}&
24\phantom{01}\\
\end{array}
\begin{array}{c|cccccccccccccccc}
\text{Bernardo}&
\cdots&
1024&
1089&
1156&
1225&
1369&
1444&
1521&
1600&
1681&
1764&
1849&
1936&
2025&
2116&
2209&
2304&
2401\\
\hline
\text{Silvia}&
\cdots&
10\phantom{24}&
10\phantom{89}&
11\phantom{56}&
12\phantom{25}&
13\phantom{69}&
14\phantom{44}&
15\phantom{21}&
16\phantom{00}&
16\phantom{81}&
17\phantom{64}&
18\phantom{49}&
19\phantom{36}&
20\phantom{25}&
21\phantom{16}&
22\phantom{09}&
23\phantom{04}&
24\phantom{01}\\
\end{array}
copeland
2016-02-03 20:36:29
When are they supposed to stop writing?
When are they supposed to stop writing?
maxplanck
2016-02-03 20:37:08
when a first digit is skipped
when a first digit is skipped
PNPADU
2016-02-03 20:37:08
smallest positive integer
smallest positive integer
Emathmaster
2016-02-03 20:37:08
When it differs by 2
When it differs by 2
goodbear
2016-02-03 20:37:08
differ by 2
differ by 2
Peggy
2016-02-03 20:37:08
when silvia skips a number
when silvia skips a number
fishy15
2016-02-03 20:37:08
until numbers differ by 2
until numbers differ by 2
math1012
2016-02-03 20:37:08
when Silvia's differ by at least 2
when Silvia's differ by at least 2
hexagram
2016-02-03 20:37:08
When Silvia's numbers get >=2 apart
When Silvia's numbers get >=2 apart
copeland
2016-02-03 20:37:12
So far, the neighboring numbers in Silvia's list are all either equal or one greater. They stop writing the first time two consecutive numbers in Silvia's list jump by more than 1.
So far, the neighboring numbers in Silvia's list are all either equal or one greater. They stop writing the first time two consecutive numbers in Silvia's list jump by more than 1.
copeland
2016-02-03 20:37:16
That is, they stop writing when neighboring numbers are $n$ and $m$ with $m>n+1$. And what does that make $f(k)?$
That is, they stop writing when neighboring numbers are $n$ and $m$ with $m>n+1$. And what does that make $f(k)?$
goodbear
2016-02-03 20:38:21
n+1
n+1
blizzard10
2016-02-03 20:38:21
f(k)=n+1
f(k)=n+1
eswa2000
2016-02-03 20:38:21
n+1
n+1
Peggy
2016-02-03 20:38:21
n+1
n+1
baseballcat
2016-02-03 20:38:21
n+1?
n+1?
akaashp11
2016-02-03 20:38:21
n + 1
n + 1
ImpossibleSphere
2016-02-03 20:38:21
n + 2
n + 2
copeland
2016-02-03 20:38:24
When they stop writing, the first number skipped is $f(k)$. (So $n+1$ in our example.)
When they stop writing, the first number skipped is $f(k)$. (So $n+1$ in our example.)
copeland
2016-02-03 20:38:26
OK, now we have the setup. Let's do some math!
OK, now we have the setup. Let's do some math!
copeland
2016-02-03 20:38:27
First, let's agree to talk about a specific $k$ so for now $k$ is fixed.
First, let's agree to talk about a specific $k$ so for now $k$ is fixed.
copeland
2016-02-03 20:38:35
Let us start by setting up some variable. What is a nice thing to look at?
Let us start by setting up some variable. What is a nice thing to look at?
copeland
2016-02-03 20:39:13
What should our variable represent?
What should our variable represent?
aq1048576
2016-02-03 20:40:07
the root of the square?
the root of the square?
copeland
2016-02-03 20:40:13
Bernardo is writing a bunch of numbers.
Bernardo is writing a bunch of numbers.
copeland
2016-02-03 20:40:22
We want to be able to talk about those numbers.
We want to be able to talk about those numbers.
copeland
2016-02-03 20:40:27
It would be nice to talk about the terms of this sequence so let's let the number Bernardo writes $b^2.$
It would be nice to talk about the terms of this sequence so let's let the number Bernardo writes $b^2.$
copeland
2016-02-03 20:40:32
We should be able to figure out what number Silvia creates. If Bernardo writes $b^2$ and Silvia erases the last $k$ digits of $b^2$ then what number is left?
We should be able to figure out what number Silvia creates. If Bernardo writes $b^2$ and Silvia erases the last $k$ digits of $b^2$ then what number is left?
ninjataco
2016-02-03 20:41:57
floor(b^2 / 10^k) ?
floor(b^2 / 10^k) ?
goodbear
2016-02-03 20:41:57
floor of b^2/10^k
floor of b^2/10^k
goodbear
2016-02-03 20:41:57
floor(b^2/10^k)
floor(b^2/10^k)
atmchallenge
2016-02-03 20:41:57
$\left \lfloor \frac{b^2}{10^k} \right \rfloor$
$\left \lfloor \frac{b^2}{10^k} \right \rfloor$
deltaepsilon6
2016-02-03 20:41:57
$\left\lfloor \frac{b^2}{10^k} \right\rfloor$
$\left\lfloor \frac{b^2}{10^k} \right\rfloor$
_--__--_
2016-02-03 20:41:57
If // means integer division, b^2 // 10^k
If // means integer division, b^2 // 10^k
loserboy
2016-02-03 20:41:57
floor(b^2/10^k)
floor(b^2/10^k)
copeland
2016-02-03 20:42:00
We can find Silvia's number by rounding $b^2$ down to the nearest multiple of $10^k$ and then dividing by $10^k$. That is the same as dividing by $10^k$ and then rounding down to the nearest integer. Mathematically, this is written
\[s(b)=\left\lfloor
\frac{b^2}{10^{k}}
\right\rfloor.\]
We can find Silvia's number by rounding $b^2$ down to the nearest multiple of $10^k$ and then dividing by $10^k$. That is the same as dividing by $10^k$ and then rounding down to the nearest integer. Mathematically, this is written
\[s(b)=\left\lfloor
\frac{b^2}{10^{k}}
\right\rfloor.\]
copeland
2016-02-03 20:42:09
Now, as we increment $b$, the value of $s$ increases. Since it is quadratic in $b$, the value of $s(b)$ increases slowly at first and then grows faster.
Now, as we increment $b$, the value of $s$ increases. Since it is quadratic in $b$, the value of $s(b)$ increases slowly at first and then grows faster.
copeland
2016-02-03 20:42:18
For our $k=2$ example above, we had to start at $b=10$ and $s(10)=1$. Here are $s(b)$ for $b$ between 10 and 50 when $k=2$:
For our $k=2$ example above, we had to start at $b=10$ and $s(10)=1$. Here are $s(b)$ for $b$ between 10 and 50 when $k=2$:
copeland
2016-02-03 20:42:26
copeland
2016-02-03 20:42:32
Incidentally, this is the floor of the graph of $f(b)=\dfrac{b^2}{100}$:
Incidentally, this is the floor of the graph of $f(b)=\dfrac{b^2}{100}$:
copeland
2016-02-03 20:42:33
copeland
2016-02-03 20:42:41
We care about the first point $b$ where $s(b+1)-s(b)>1$. Let's write that down:
We care about the first point $b$ where $s(b+1)-s(b)>1$. Let's write that down:
copeland
2016-02-03 20:42:43
\[
\left\lfloor
\frac{(b+1)^2}{10^{k}}
\right\rfloor
-\left\lfloor
\frac{b^2}{10^{k}}
\right\rfloor>1.
\]
\[
\left\lfloor
\frac{(b+1)^2}{10^{k}}
\right\rfloor
-\left\lfloor
\frac{b^2}{10^{k}}
\right\rfloor>1.
\]
copeland
2016-02-03 20:42:55
When $b$ is little, this doesn't grow very much. Roughly when is the first chance we get for $s$ to skip an integer?
When $b$ is little, this doesn't grow very much. Roughly when is the first chance we get for $s$ to skip an integer?
mssmath
2016-02-03 20:44:22
$b=\sqrt{10^k}$
$b=\sqrt{10^k}$
goodbear
2016-02-03 20:44:22
s approx. 50
s approx. 50
copeland
2016-02-03 20:44:27
OK, why? What are we doing here?
OK, why? What are we doing here?
copeland
2016-02-03 20:46:04
Let's make this more abstract. What must be true in order for $\lfloor x\rfloor-\lfloor y\rfloor > 1?$
Let's make this more abstract. What must be true in order for $\lfloor x\rfloor-\lfloor y\rfloor > 1?$
aq1048576
2016-02-03 20:46:42
x must be at least 1 greater then y
x must be at least 1 greater then y
looyee2001
2016-02-03 20:46:42
X>=Y+1
X>=Y+1
aq1048576
2016-02-03 20:46:42
x must be 1+ more than y
x must be 1+ more than y
deltaepsilon6
2016-02-03 20:46:42
$x-y>1 and x,y\in\Bbb{Z}$
$x-y>1 and x,y\in\Bbb{Z}$
ychen
2016-02-03 20:46:42
x - y > 1
x - y > 1
copeland
2016-02-03 20:46:47
For that to be true, $x$ and $y$ must certainly be at least 1 apart.
For that to be true, $x$ and $y$ must certainly be at least 1 apart.
copeland
2016-02-03 20:46:56
That holds here as well.
That holds here as well.
copeland
2016-02-03 20:46:58
If
\[
\frac{(b+1)^2}{10^{k}}
-
\frac{b^2}{10^{k}}
\leq1
\]
then there is no chance for $s$ to skip an integer. That is, if two real numbers differ by less than 1, their floors must differ by 1 or 0.
If
\[
\frac{(b+1)^2}{10^{k}}
-
\frac{b^2}{10^{k}}
\leq1
\]
then there is no chance for $s$ to skip an integer. That is, if two real numbers differ by less than 1, their floors must differ by 1 or 0.
copeland
2016-02-03 20:47:07
So what do we know about $b$?
So what do we know about $b$?
copeland
2016-02-03 20:48:25
I asked you for too many steps at once, sorry.
I asked you for too many steps at once, sorry.
copeland
2016-02-03 20:48:41
First, we just argued that we need that difference to be at least 1:
First, we just argued that we need that difference to be at least 1:
copeland
2016-02-03 20:48:55
\[
\frac{(b+1)^2}{10^{k}}
-
\frac{b^2}{10^{k}}
>1
\]
\[
\frac{(b+1)^2}{10^{k}}
-
\frac{b^2}{10^{k}}
>1
\]
copeland
2016-02-03 20:49:02
What does that tell us?
What does that tell us?
mathbeida
2016-02-03 20:49:50
2b >= 10^k
2b >= 10^k
mjoshi
2016-02-03 20:49:50
2b+1>10^k
2b+1>10^k
deltaepsilon6
2016-02-03 20:49:50
$2b+1>10^k$
$2b+1>10^k$
ychen
2016-02-03 20:49:50
2b + 1 > 10^ k
2b + 1 > 10^ k
macandcheese
2016-02-03 20:49:50
2b+1>10^k
2b+1>10^k
mathbeida
2016-02-03 20:49:50
(2b+1)/10^k >1
(2b+1)/10^k >1
Peggy
2016-02-03 20:49:50
b > (10^k - 1)/2
b > (10^k - 1)/2
lkarhat
2016-02-03 20:49:50
2b + 1 > 10^k
2b + 1 > 10^k
subarudl87
2016-02-03 20:49:50
2b+1 > 10^k
2b+1 > 10^k
geogirl08
2016-02-03 20:49:50
2b+1 > 10^k
2b+1 > 10^k
copeland
2016-02-03 20:50:03
\[
\frac{b^2+2b+1-b^2}{10^{k}}
>1
\]
\[
\frac{b^2+2b+1-b^2}{10^{k}}
>1
\]
copeland
2016-02-03 20:50:06
That simplifies to
\[
2b+1> 10^{k}
\]
That simplifies to
\[
2b+1> 10^{k}
\]
copeland
2016-02-03 20:50:10
In fact, $2b$ is even so we really have
\[2b\geq10^k.\]
In fact, $2b$ is even so we really have
\[2b\geq10^k.\]
copeland
2016-02-03 20:50:15
The first $b$ for which $s$ skips an integer satisfies the ineqality\[b\geq5\cdot10^{k-1}\]
The first $b$ for which $s$ skips an integer satisfies the ineqality\[b\geq5\cdot10^{k-1}\]
copeland
2016-02-03 20:50:25
However, let's think a little about this. When $k=6$, say, we have $b\geq 500{,}000$. The square of $500{,}000$ is $250{,}000{,}000{,}000$. Silvia is going to drop the last 6 zeroes from this and get $250{,}000$. We didn't round down at all!
However, let's think a little about this. When $k=6$, say, we have $b\geq 500{,}000$. The square of $500{,}000$ is $250{,}000{,}000{,}000$. Silvia is going to drop the last 6 zeroes from this and get $250{,}000$. We didn't round down at all!
copeland
2016-02-03 20:50:57
Will we skip an integer going to $s(500{,}001)?$
Will we skip an integer going to $s(500{,}001)?$
goodbear
2016-02-03 20:51:27
no
no
mathwizard888
2016-02-03 20:51:27
no
no
ychen
2016-02-03 20:51:27
no?
no?
owm
2016-02-03 20:51:27
no
no
copeland
2016-02-03 20:51:29
Why not?
Why not?
ychen
2016-02-03 20:52:37
because you start with 0 mod 1000000
because you start with 0 mod 1000000
geogirl08
2016-02-03 20:52:37
no. 250001000001 => 250001.
no. 250001
goodbear
2016-02-03 20:52:37
all 0's
all 0's
owm
2016-02-03 20:52:37
because 500001^=250001000001
because 500001^=250001000001
deltaepsilon6
2016-02-03 20:52:37
250001000001
250001000001
yrnsmurf
2016-02-03 20:52:37
it is 250001000001
it is 250001000001
copeland
2016-02-03 20:52:50
No. We actually don't even have to compute this, though. We know that we are just now passing the point where $\dfrac{b^2}{10^6}$ increases by 1. Since $\dfrac{500{,}000^2}{10^6}$ is an integer and $\dfrac{500{,}001^2}{10^6}$ is very, very close to 1 more than it, we know $s(500{,}001)$ must be $250{,}001$.
No. We actually don't even have to compute this, though. We know that we are just now passing the point where $\dfrac{b^2}{10^6}$ increases by 1. Since $\dfrac{500{,}000^2}{10^6}$ is an integer and $\dfrac{500{,}001^2}{10^6}$ is very, very close to 1 more than it, we know $s(500{,}001)$ must be $250{,}001$.
copeland
2016-02-03 20:53:30
Those of you who did compute noticed that we only lost 0.000001 when we rounded down. Plenty of wiggle room left.
Those of you who did compute noticed that we only lost 0.000001 when we rounded down. Plenty of wiggle room left.
copeland
2016-02-03 20:53:35
In order for two numbers that differ by about 1 to "round down" to numbers that differ by two, the smaller number must be just under an integer and the larger number must be just past the next integer:
In order for two numbers that differ by about 1 to "round down" to numbers that differ by two, the smaller number must be just under an integer and the larger number must be just past the next integer:
copeland
2016-02-03 20:53:44
copeland
2016-02-03 20:53:48
So we're looking for a number that may even be a lot bigger than $5\cdot10^{k-1}$.
So we're looking for a number that may even be a lot bigger than $5\cdot10^{k-1}$.
copeland
2016-02-03 20:54:02
Any ideas what to do now?
Any ideas what to do now?
mssmath
2016-02-03 20:54:48
try 500000+x
try 500000+x
copeland
2016-02-03 20:54:55
This is the point in the problem where I gave up on actually thinking and just threw more arithmetic/variables/machinery at the problem.
This is the point in the problem where I gave up on actually thinking and just threw more arithmetic/variables/machinery at the problem.
copeland
2016-02-03 20:54:57
Let's write $b=5\cdot10^{k-1}+c.$ Then
\[
s(b)=\left\lfloor
\frac{(5\cdot10^{k-1}+c)^2}{10^{k}}
\right\rfloor\]
Let's write $b=5\cdot10^{k-1}+c.$ Then
\[
s(b)=\left\lfloor
\frac{(5\cdot10^{k-1}+c)^2}{10^{k}}
\right\rfloor\]
copeland
2016-02-03 20:55:06
There's a nice theoretical underpinning to writing $b$ like this. We should think of $5\cdot10^{k-1}$ as a "first-order approximation" to $b$. That is, we expect this to be a pretty good approximation for $b$, especially when $k$ is really large. But we also know it's not right, so we "correct" our approximation. We call $c$ the "second order correction" for our approximation. If all our guesses are right, $c$ should be a lot smaller than $b$, but it should be able to fix any of the problems with our approximation.
There's a nice theoretical underpinning to writing $b$ like this. We should think of $5\cdot10^{k-1}$ as a "first-order approximation" to $b$. That is, we expect this to be a pretty good approximation for $b$, especially when $k$ is really large. But we also know it's not right, so we "correct" our approximation. We call $c$ the "second order correction" for our approximation. If all our guesses are right, $c$ should be a lot smaller than $b$, but it should be able to fix any of the problems with our approximation.
mathguy5041
2016-02-03 20:55:31
expand
expand
copeland
2016-02-03 20:55:35
What happens when we expand that?
What happens when we expand that?
copeland
2016-02-03 20:55:43
I'll give you a few moments.
I'll give you a few moments.
copeland
2016-02-03 20:56:04
And remember that we care about $k\geq2$.
And remember that we care about $k\geq2$.
deltaepsilon6
2016-02-03 20:58:06
$s(b)=\left\lfloor\frac{25\cdot10^{2k-2}+10^k*c+c^2}{10^k}\right\rfloor$
$s(b)=\left\lfloor\frac{25\cdot10^{2k-2}+10^k*c+c^2}{10^k}\right\rfloor$
copeland
2016-02-03 20:58:53
\begin{align*}
s(5\cdot10^{k-1}+c)&=\left\lfloor
\frac{(5\cdot10^{k-1}+c)^2}{10^{k}}
\right\rfloor\\
&=
\left\lfloor
\frac{25\cdot10^{2k-2}+10\cdot10^{k-1}c+c^2}{10^{k}}
\right\rfloor\\
&=
\left\lfloor
25\cdot10^{k-2}+c+\frac{c^2}{10^{k}}
\right\rfloor
\end{align*}
\begin{align*}
s(5\cdot10^{k-1}+c)&=\left\lfloor
\frac{(5\cdot10^{k-1}+c)^2}{10^{k}}
\right\rfloor\\
&=
\left\lfloor
\frac{25\cdot10^{2k-2}+10\cdot10^{k-1}c+c^2}{10^{k}}
\right\rfloor\\
&=
\left\lfloor
25\cdot10^{k-2}+c+\frac{c^2}{10^{k}}
\right\rfloor
\end{align*}
copeland
2016-02-03 20:58:55
That's easier to read.
That's easier to read.
copeland
2016-02-03 20:59:28
Now what?
Now what?
Alanshenkerman
2016-02-03 20:59:39
take out the 25*10^k-2 and c as they are integers
take out the 25*10^k-2 and c as they are integers
owm
2016-02-03 20:59:39
the part with no fraction can be taken out of the floor function
the part with no fraction can be taken out of the floor function
goodbear
2016-02-03 20:59:41
$25\cdot10^{k-2}+c+\left\lfloor \frac{c^2}{10^{k}} \right\rfloor$
$25\cdot10^{k-2}+c+\left\lfloor \frac{c^2}{10^{k}} \right\rfloor$
copeland
2016-02-03 20:59:47
Now $k\geq2$ for all the terms in the sum we care about so the first two numbers are integers:
\[
s(5\cdot10^{k-1}+c)
=
\left\lfloor
25\cdot10^{k-2}+c+\frac{c^2}{10^{k}}
\right\rfloor
=
25\cdot10^{k-2}+c+
\left\lfloor
\frac{c^2}{10^{k}}
\right\rfloor.
\]
Now $k\geq2$ for all the terms in the sum we care about so the first two numbers are integers:
\[
s(5\cdot10^{k-1}+c)
=
\left\lfloor
25\cdot10^{k-2}+c+\frac{c^2}{10^{k}}
\right\rfloor
=
25\cdot10^{k-2}+c+
\left\lfloor
\frac{c^2}{10^{k}}
\right\rfloor.
\]
copeland
2016-02-03 20:59:48
Now tell me what happens to $s$ as we increase $c$ starting at 0.
Now tell me what happens to $s$ as we increase $c$ starting at 0.
mathwizard888
2016-02-03 21:01:26
keeps going up by 1 until the last term hits 1
keeps going up by 1 until the last term hits 1
mssmath
2016-02-03 21:01:26
We just need $\floor{c^2/10^k}=1$ instead of 0 now
We just need $\floor{c^2/10^k}=1$ instead of 0 now
yrnsmurf
2016-02-03 21:01:26
It stays 0 until it goes over $10^{\frac k2}$
It stays 0 until it goes over $10^{\frac k2}$
eswa2000
2016-02-03 21:01:26
s stays the same until c reaches sqrt{10^k}
s stays the same until c reaches sqrt{10^k}
ychen
2016-02-03 21:01:26
it increases by 1 until c^2 >= 10^k
it increases by 1 until c^2 >= 10^k
copeland
2016-02-03 21:01:31
When $c$ is small, that last floor is zero, and incrementing $c$ only increases the "$+c$" part, so $s$ increases by 1.
When $c$ is small, that last floor is zero, and incrementing $c$ only increases the "$+c$" part, so $s$ increases by 1.
copeland
2016-02-03 21:01:35
When does it skip an integer?
When does it skip an integer?
eswa2000
2016-02-03 21:02:30
c reaches sqrt{10^k}
c reaches sqrt{10^k}
jfurf
2016-02-03 21:02:30
When $c^2=10^k$
When $c^2=10^k$
deltaepsilon6
2016-02-03 21:02:30
$c^2\geq 10^k$
$c^2\geq 10^k$
TigerLily14
2016-02-03 21:02:30
when the last part is greater than 1 (rounds down to 1)
when the last part is greater than 1 (rounds down to 1)
mssmath
2016-02-03 21:02:30
$c\ge10^{\frac{k}{2}}$
$c\ge10^{\frac{k}{2}}$
aq1048576
2016-02-03 21:02:30
when the last term reaches 1?
when the last term reaches 1?
mathwizard888
2016-02-03 21:02:30
at $c=10^{k/2}$
at $c=10^{k/2}$
copeland
2016-02-03 21:02:34
It skips an integer when the floor goes from 0 to 1. That means we skip an integer the first time $\left\lfloor\dfrac{c^2}{10^k}\right\rfloor=1$.
It skips an integer when the floor goes from 0 to 1. That means we skip an integer the first time $\left\lfloor\dfrac{c^2}{10^k}\right\rfloor=1$.
copeland
2016-02-03 21:02:36
Great. Now we just need to think about when $\sqrt{\dfrac{c^2}{10^k}}\approx1$. What's going to help us a lot in finishing the problem?
Great. Now we just need to think about when $\sqrt{\dfrac{c^2}{10^k}}\approx1$. What's going to help us a lot in finishing the problem?
copeland
2016-02-03 21:02:41
Look back at the problem statement. . .
Look back at the problem statement. . .
SimonSun
2016-02-03 21:03:14
sum up from 2.. 2016
sum up from 2.. 2016
deltaepsilon6
2016-02-03 21:03:14
sum of digits
sum of digits
copeland
2016-02-03 21:03:16
What's particularly useful about that sum?
What's particularly useful about that sum?
mssmath
2016-02-03 21:04:23
k=2l
k=2l
mathwizard888
2016-02-03 21:04:23
k is even
k is even
akaashp11
2016-02-03 21:04:23
sum of even integers in x^2
sum of even integers in x^2
geogirl08
2016-02-03 21:04:23
Sum of evens n(n+1)
Sum of evens n(n+1)
copeland
2016-02-03 21:04:26
We only care when $k$ is even!
We only care when $k$ is even!
copeland
2016-02-03 21:04:27
So what is $c$?
So what is $c$?
yrnsmurf
2016-02-03 21:05:18
$10^{\frac k2}$
$10^{\frac k2}$
aq1048576
2016-02-03 21:05:18
10^{k/2}
10^{k/2}
deltaepsilon6
2016-02-03 21:05:18
10^(k/2)
10^(k/2)
owm
2016-02-03 21:05:18
c=10^{k/2}
c=10^{k/2}
copeland
2016-02-03 21:05:21
The value of $c$ where we skip our first integer is $c=\sqrt{10^k}=10^{k/2}.$
The value of $c$ where we skip our first integer is $c=\sqrt{10^k}=10^{k/2}.$
copeland
2016-02-03 21:05:28
When we plug this value of $c$ in we get
\[s(5\cdot10^{k-1}+10^{k/2})=25\cdot10^{k-2}+10^{k/2}+1.\]
When we plug this value of $c$ in we get
\[s(5\cdot10^{k-1}+10^{k/2})=25\cdot10^{k-2}+10^{k/2}+1.\]
copeland
2016-02-03 21:05:32
What number did we skip?
What number did we skip?
copeland
2016-02-03 21:07:21
So. . . we determined that this particular value of $c$ is the place where the floor flips from 0 to 1. This is the value at that $c$. What happened with Silvia's sequence?
So. . . we determined that this particular value of $c$ is the place where the floor flips from 0 to 1. This is the value at that $c$. What happened with Silvia's sequence?
yrnsmurf
2016-02-03 21:07:51
1 less than that number
1 less than that number
ThorJames
2016-02-03 21:07:51
it skipped a number
it skipped a number
copeland
2016-02-03 21:08:11
We finally skipped a number.
We finally skipped a number.
aq1048576
2016-02-03 21:08:20
25*10^{k-2}+10^{k/2}
25*10^{k-2}+10^{k/2}
copeland
2016-02-03 21:08:22
We skipped $25\cdot10^{k-2}+10^{k/2}$.
We skipped $25\cdot10^{k-2}+10^{k/2}$.
copeland
2016-02-03 21:08:29
Now we finish by adding these up, starting at $k=2$ and finishing at $k=2016$.
Now we finish by adding these up, starting at $k=2$ and finishing at $k=2016$.
copeland
2016-02-03 21:08:31
That looks a bit like:
\begin{array}{rr}
&25\cdot10^0+10^1\\
&25\cdot10^2+10^2\\
&25\cdot10^4+10^3\\
&25\cdot10^6+10^4\\
&25\cdot10^8+10^5\\
+&25\cdot10^{10}+10^6\\
\hline
\end{array}
That looks a bit like:
\begin{array}{rr}
&25\cdot10^0+10^1\\
&25\cdot10^2+10^2\\
&25\cdot10^4+10^3\\
&25\cdot10^6+10^4\\
&25\cdot10^8+10^5\\
+&25\cdot10^{10}+10^6\\
\hline
\end{array}
copeland
2016-02-03 21:08:32
This problem never ends, huh. . . .
This problem never ends, huh. . . .
copeland
2016-02-03 21:08:41
I can make those into for-real numbers, though:
\begin{array}{rr}
&35\\
&2600\\
&251000\\
&25010000\\
&2500100000\\
+&250001000000\\
\hline
\end{array}
I can make those into for-real numbers, though:
\begin{array}{rr}
&35\\
&2600\\
&251000\\
&25010000\\
&2500100000\\
+&250001000000\\
\hline
\end{array}
copeland
2016-02-03 21:08:44
And when we write this super-big number out and add its digits, what happens?
And when we write this super-big number out and add its digits, what happens?
mssmath
2016-02-03 21:09:56
Their is no carrying so just add the digits for the win
Their is no carrying so just add the digits for the win
geogirl08
2016-02-03 21:09:56
8+8+8+8+...... (1008 8's in all)
8+8+8+8+...... (1008 8's in all)
deltaepsilon6
2016-02-03 21:09:56
8n
8n
deltaepsilon6
2016-02-03 21:09:56
1008*8
1008*8
SmallKid
2016-02-03 21:09:56
It becomes 8064?
It becomes 8064?
yrnsmurf
2016-02-03 21:09:56
8064
8064
ychen
2016-02-03 21:09:56
each row sum up to 8
each row sum up to 8
copeland
2016-02-03 21:10:01
In that expression, the 25s move to the left very fast and the ones also move to the left. The 1s always appear in subsequently larger places. Likewise for the 2s and 5s.
Specifically, no place value will ever get more than $1+2+5$, so there will never be any carrying. Therefore each term contributes $1+2+5=8$ to the final digit sum.
In that expression, the 25s move to the left very fast and the ones also move to the left. The 1s always appear in subsequently larger places. Likewise for the 2s and 5s.
Specifically, no place value will ever get more than $1+2+5$, so there will never be any carrying. Therefore each term contributes $1+2+5=8$ to the final digit sum.
copeland
2016-02-03 21:10:13
And the answer?
And the answer?
goldypeng
2016-02-03 21:10:49
E is the answer
E is the answer
yiqun
2016-02-03 21:10:49
E
E
geogirl08
2016-02-03 21:10:49
E
E
SimonSun
2016-02-03 21:10:49
E
E
jfurf
2016-02-03 21:10:49
$\large{\boxed{8064}}$
$\large{\boxed{8064}}$
ychen
2016-02-03 21:10:49
E 8064
E 8064
akaashp11
2016-02-03 21:10:49
$E$
$E$
happyribbon
2016-02-03 21:10:49
E
E
SimonSun
2016-02-03 21:10:49
8064
8064
owm
2016-02-03 21:10:49
8*1008
8*1008
deltaepsilon6
2016-02-03 21:10:49
$\boxed{E}$
$\boxed{E}$
yrnsmurf
2016-02-03 21:10:49
E
E
yiqun
2016-02-03 21:10:49
E
E
owm
2016-02-03 21:10:49
E
E
copeland
2016-02-03 21:10:52
There are 1008 total terms so the final digit sum is $8\cdot1008=\boxed{8064}$. $E$.
There are 1008 total terms so the final digit sum is $8\cdot1008=\boxed{8064}$. $E$.
copeland
2016-02-03 21:11:10
Incidentally, at the end there Bernardo was squaring thousands of thousand-digit numbers.
Incidentally, at the end there Bernardo was squaring thousands of thousand-digit numbers.
copeland
2016-02-03 21:11:21
I imagine the time it would take just to erase everything would be substantial.
I imagine the time it would take just to erase everything would be substantial.
copeland
2016-02-03 21:11:53
Cool, we're done. We've gone a bit over 2 hours so I think I'm going to call it. There are a lot of really good discussions of all the problems on the forums.
Cool, we're done. We've gone a bit over 2 hours so I think I'm going to call it. There are a lot of really good discussions of all the problems on the forums.
copeland
2016-02-03 21:11:58
Please go join them!
Please go join them!
copeland
2016-02-03 21:12:15
That's all for tonight's Math Jam!
That's all for tonight's Math Jam!
copeland
2016-02-03 21:12:16
Please join us again on Thursday, February 18, when we will discuss the AMC 10B/12B contests and also again on March 5 and 18 when we will be discussing the AIME I and II contests.
Please join us again on Thursday, February 18, when we will discuss the AMC 10B/12B contests and also again on March 5 and 18 when we will be discussing the AIME I and II contests.
copeland
2016-02-03 21:12:22
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
I believe that Deven and Kyle might have chosen some 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
I believe that Deven and Kyle might have chosen some different tactics in their videos than we used today, so it might be worth checking them out.
yrnsmurf
2016-02-03 21:13:37
Wait should the amc 10 be discussed in the middle school or high school forum?
Wait should the amc 10 be discussed in the middle school or high school forum?
copeland
2016-02-03 21:13:39
In the contests forum:
http://www.artofproblemsolving.com/community/c3158_usa_contests
In the contests forum:
http://www.artofproblemsolving.com/community/c3158_usa_contests
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