Math Jams

copeland 2017-02-16 19:04:31
Are we ready to start?

solzonmars 2017-02-16 19:04:34
Are we doing AMC 10 or 12 first?

copeland 2017-02-16 19:04:41
10. We sort lexicographically.

copeland 2017-02-16 19:05:24
Welcome to the 2016 AMC 10A/12A Math Jam!

mumpu2k16 2017-02-16 19:05:26
I AM SO READY FOR THIS!

copeland 2017-02-16 19:05:34
No, wai! Me too!

ilovemath04 2017-02-16 19:05:38
Let's start!

MathTechFire 2017-02-16 19:05:38
Same

Vfire 2017-02-16 19:05:38
Same

mathchampion1 2017-02-16 19:05:45
2017 B!

jk23541 2017-02-16 19:05:45
i thought it was 2017

copeland 2017-02-16 19:05:51
I try to give you nice things.

copeland 2017-02-16 19:06:00
Welcome to the 2017 AMC 10A/12A Math Jam!

copeland 2017-02-16 19:06:05
OK, one more try.

copeland 2017-02-16 19:06:12
Welcome to the 2017 AMC 10B/12B Math Jam!

copeland 2017-02-16 19:06:21
So, we can edit all that out in the transcript.

copeland 2017-02-16 19:06:25
I'm Jeremy Copeland, and I'll be leading our discussion tonight.

copeland 2017-02-16 19:06:37
You'll notice devenware. He's responsible for that 2017 typo.

copeland 2017-02-16 19:06:45
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.

copeland 2017-02-16 19:06:56
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 2017-02-16 19:07:03
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 2017-02-16 19:07:04
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 2017-02-16 19:07:10
There are 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 2017-02-16 19:07:18
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 2017-02-16 19:07:53
We also have 4 assistants here helping out!

a1b2 2017-02-16 19:07:59
How do you type that fast?

copeland 2017-02-16 19:08:10
(The assistants and I all take turns typing the letters.)

copeland 2017-02-16 19:08:29
Your assistants for today will be Henrik Boecken (henrikjb), William Wang (willwang123), Joshua Morris (QuantumCat), and Kathryn Lesh (duilka).

copeland 2017-02-16 19:08:38
Henrik is an undergraduate at the Massachusetts Institute of Technology studying economics and, of course, math. After college, he plans on teaching at his old high school for a few years. Math contests were the backbone of his high school career, and now he hopes to give back. In his free time, Henrik enjoys Ultimate Frisbee, card games, and reading the Japanese manga One Piece.

copeland 2017-02-16 19:08:41
William is a freshman at the University of Pennsylvania. He is a 4-time USA(J)MO qualifier and a USA Physics Olympiad qualifier. In his spare time, he enjoys playing tennis, clarinet, and StarCraft.

copeland 2017-02-16 19:08:42
Josh, after getting his Bachelor's and Master's in Physics from UT El Paso and teaching high school physics, decided he loved science and working with students so much he went back to school to work pursue a career in medicine. While he's applying to medical school and going through the interview process, he's helping out here at Art of Problem Solving and keeping his math skills sharp. In his spare time, he loves cooking, running, reading and is (still) slowly teaching himself how to play the piano.

QuantumCat 2017-02-16 19:08:45
Hello everyone!

henrikjb 2017-02-16 19:08:50
Hi guys!

willwang123 2017-02-16 19:08:52
Hello!

copeland 2017-02-16 19:09:15
Kathryn decided to be a mathematician when she was twelve, and now teach mathematics in upstate New York. Her son and daughter are in college. Her son studies linguistics, and her daughter is finishing her degree this year and going to grad school in---guess what?--yes, math! In my spare time, she reads, watches Star Trek, and learns Japanese.

duilka 2017-02-16 19:09:19
Hi everyone!

copeland 2017-02-16 19:09:20
They will be sending you messages to answer questions or offer other help. However, due to the incredibly large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over may give you warm feelings, but just makes their poor lives harder, so please, only ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.

copeland 2017-02-16 19:09:32
Please also remember that the purpose of this Math Jam is to work through the solutions to AMC problems and learn! "Working through the solutions" includes discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the final answer (since I'll be ignoring those anyway). That's not what we're doing tonight. We're going to work through the problems step-by-step, and comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, won't be posted.

copeland 2017-02-16 19:09:36
Also notice that there will be several cases where we actually find the answer or almost the answer but instead wander off. The goal is always to find a proof that our answer is correct and not just find the answers. Of course on the AMC you should aim to do much less work than this.

copeland 2017-02-16 19:09:54
We will work the last 5 problems from the AMC 10A, then the last 5 problems from the AMC 12A. Two of these problems are the same, 10A Problem 25 and 12A Problem 21. We'll only solve that problem once.

copeland 2017-02-16 19:10:06
Should we, get started?

mathchampion1 2017-02-16 19:10:19
You wrote "in my spare time, she reads..."

copeland 2017-02-16 19:10:20
She only reads when I have free time.

zac15SCASD 2017-02-16 19:10:30
YEs!

Awesomekid05 2017-02-16 19:10:30
yes!

shootingstar8 2017-02-16 19:10:30
Yes!

Cardinals2014 2017-02-16 19:10:30
Yes

ILoveChess108 2017-02-16 19:10:30
YES

GeronimoStilton 2017-02-16 19:10:30
Yes!!!!

Slacker 2017-02-16 19:10:30
yes!

techguy2 2017-02-16 19:10:33
AGAIN WITH THE TYPOS

AthenasEye 2017-02-16 19:10:33
typo again

copeland 2017-02-16 19:10:39
deven. I'm sick of him.

devenware 2017-02-16 19:10:45
Yeah, my bad.

copeland 2017-02-16 19:10:53
21. In $\triangle ABC$, $AB = 6$, $AC = 8$, $BC = 10$, and $D$ is the midpoint of $BC$. What is the sum of the radii of the circles inscribed in $\triangle ADB$ and $\triangle ADC$ ?
$\textbf{(A)}\quad \sqrt{5} \qquad \qquad \textbf{(B)}\quad \dfrac{11}{4} \qquad\qquad \textbf{(C)}\quad 2\sqrt{2} \qquad\qquad \textbf{(D)}\quad \dfrac{17}{6} \qquad\qquad\textbf{(E)}\quad 3$

copeland 2017-02-16 19:11:03
What's always our first step in a geometry problem?

JJShan26 2017-02-16 19:11:22
diagram

summitwei 2017-02-16 19:11:22
diagram pl0x

reedmj 2017-02-16 19:11:22
Diagram!

letsgomath 2017-02-16 19:11:22
draw a picture!

a000 2017-02-16 19:11:22
draw a diagram

RayThroughSpace 2017-02-16 19:11:22
Draw a diagram

Skittlesftw 2017-02-16 19:11:22
diagram?

Ani10 2017-02-16 19:11:22
diagram

Reef334 2017-02-16 19:11:22
draw a diagram

iks92 2017-02-16 19:11:22
draw a picture!

rspr2001 2017-02-16 19:11:22
draw a diagram

naman12 2017-02-16 19:11:22
Draw the figure!!!

zhengyf 2017-02-16 19:11:22
diagrams

copeland 2017-02-16 19:11:23
Our first step is always to draw a diagram.

copeland 2017-02-16 19:11:30
But before we even do that, I want to make sure we do it right. What do we know about $\triangle ABC?$

ilovemath04 2017-02-16 19:11:59
right triangle

tootyfail 2017-02-16 19:11:59
its a right triangle

carlos8 2017-02-16 19:11:59
3-4-5 right triangle

blitzkrieg21 2017-02-16 19:11:59
it is a right triangle!

mathwiz0803 2017-02-16 19:11:59
right angle triangle (3-4-5)

000libyaclawdruse000 2017-02-16 19:11:59
It's a 3-4-5 right triangle

akaashp11 2017-02-16 19:11:59
3-4-5 Right triangle

yanyu2002 2017-02-16 19:11:59
It's a 3-4-5 Right triangle

copeland 2017-02-16 19:12:00
We know that $\triangle ABC$ is similar to a 3-4-5 right triangle, and $D$ is the midpoint of the hypotenuse. Okay, let's draw that.

copeland 2017-02-16 19:12:02

copeland 2017-02-16 19:12:03
(Remember to always draw a large, clear diagram.)

copeland 2017-02-16 19:12:05
What else should we add to this diagram immediately?

solzonmars 2017-02-16 19:12:28
AD

GeronimoStilton 2017-02-16 19:12:28
$AD$!

Mathaddict11 2017-02-16 19:12:28
segment AD

awesome_weisur 2017-02-16 19:12:28
draw AD

a1b2 2017-02-16 19:12:28
$AD$

ImpossibleCube 2017-02-16 19:12:28
median AD

mathwiz0803 2017-02-16 19:12:28
AD

ninjataco 2017-02-16 19:12:28
AD

copeland 2017-02-16 19:12:36
It makes sense to draw in $AD$.

copeland 2017-02-16 19:12:37
What else?

Slacker 2017-02-16 19:12:59
incircle

tdeng 2017-02-16 19:12:59
The circles

GeronimoStilton 2017-02-16 19:12:59
The circles!

Winston123 2017-02-16 19:12:59
The circles

BetaCentauri 2017-02-16 19:12:59
maybe the two circles could help

Derive_Foiler 2017-02-16 19:12:59
the circles?

copeland 2017-02-16 19:13:05
Let's also add the incircles of $\triangle ABD$ and $\triangle ADC.$

copeland 2017-02-16 19:13:07

copeland 2017-02-16 19:13:16
Okay. What else should we add to this diagram?

fractal161 2017-02-16 19:13:53
draw radii to the tangents?

InYourWildestDreams 2017-02-16 19:13:53
radii

letsgomath 2017-02-16 19:13:53
radii!

chessapple9 2017-02-16 19:13:53
The radii of the circles that touch the points of tangency

naman12 2017-02-16 19:13:53
Radii?

thinmint 2017-02-16 19:13:53
The radii

rick101 2017-02-16 19:13:53
Draw the radii

PiGuy3141592 2017-02-16 19:13:53
And the radii from each of the sides

carfan25 2017-02-16 19:13:53
the radii?

copeland 2017-02-16 19:13:56
We care about the radii of the circles, so it makes sense to draw those in. It also always makes sense to draw in radii to tangent points because they give us right angles. I'll draw them all in now, and we'll figure out which ones are useful later.

copeland 2017-02-16 19:13:59

copeland 2017-02-16 19:14:05
Okay, finally, let's add some lengths to our diagram. What lengths do we know?

Vfire 2017-02-16 19:14:42
6-8-10

zac15SCASD 2017-02-16 19:14:42
Ab,bc,ac

BetaCentauri 2017-02-16 19:14:42
the median is half the length of the hypotenuse so AD=5

a1b2 2017-02-16 19:14:42
$CD=5$

awesome_weisur 2017-02-16 19:14:42
AD = CD = BD = 5

jk23541 2017-02-16 19:14:42
AD=CD=BD=5

lfwei 2017-02-16 19:14:42
AB = 6, AC = 8, BC = 10

vishwathganesan 2017-02-16 19:14:42
AD = CD = BD = 5, AB = 6, AC = 8

MathTechFire 2017-02-16 19:14:42
Ab =6 Ac = 8 Bc =10

Magicsm 2017-02-16 19:14:42
AD = 5, CD = 5, DB = 5, AB = 6, AC = 8

copeland 2017-02-16 19:14:45
We know $AB = 6$ and $AC = 8$ because those were given to us.

copeland 2017-02-16 19:14:46
We also know $DC = DB = 5$ because $D$ is the midpoint of $BC$.

copeland 2017-02-16 19:14:50
Finally, we know $AD = 5$ because it's the median to the hypotenuse of a right triangle.

copeland 2017-02-16 19:14:54
If you didn't already know that, the hypotenuse of a right triangle is the diameter of the circumcircle. Therefore the midpoint of the hypotenuse is the center of the circumcircle, so it is equidistant from all three vertices.

copeland 2017-02-16 19:15:03

copeland 2017-02-16 19:15:15
Alright, finally, we have a beautiful, labeled diagram. I think it's time to get to work.

copeland 2017-02-16 19:15:24
Let's focus on triangle $ADB.$ We have three side lengths, and want to find the inradius. How do those two quantities relate to each other?

solzonmars 2017-02-16 19:15:47
Use [Area] = rs

Derive_Foiler 2017-02-16 19:15:47
Now we can use A=rs?

reaganchoi 2017-02-16 19:15:47
$A=rs$

Awesomekid05 2017-02-16 19:15:47
Now, A=rs?

Magicsm 2017-02-16 19:15:47
use the formula A = rs

Cardinals2014 2017-02-16 19:15:47
A = r*s

checkmatetang 2017-02-16 19:15:47
A=rs

geogirl08 2017-02-16 19:15:47
ABC = rs

mathwizard666 2017-02-16 19:15:47
A=s*r

thedoge 2017-02-16 19:15:47
Area = inradius * semipermeter

copeland 2017-02-16 19:15:50
We have the formula for the area of a triangle: $[ADB] = rs$ where $r$ is the inradius and $s$ is the semiperimeter.

copeland 2017-02-16 19:15:58
If you don't know this one, you can see this quickly in $\triangle ADB$:

copeland 2017-02-16 19:15:59

copeland 2017-02-16 19:16:01
This is three triangles with altitude $r$. The total area is \[\frac{6r}2+\frac{5r}2+\frac{5r}2=\frac{6+5+5}2\cdot r = sr.\]

copeland 2017-02-16 19:16:11
So the semiperimeter, like I just said, is $s=\dfrac{6+5+5}2=8.$

copeland 2017-02-16 19:16:12
What about the area?

copeland 2017-02-16 19:16:36
(And Heron is not the answer.)

awesome_weisur 2017-02-16 19:17:10
IT IS HALF THE AREA OF THE RIGHT TRIANGLE

reaganchoi 2017-02-16 19:17:10
Both $\frac{8 \cdot 6}{4}=12$

thedoge 2017-02-16 19:17:10
it's half of 24, or 12

geogirl08 2017-02-16 19:17:10
6*4/2 = 12

Magicsm 2017-02-16 19:17:10
area is 12, 1/2 the original

letsgomath 2017-02-16 19:17:10
half of ABC so 12

EulerMacaroni 2017-02-16 19:17:10
$1/2*6*4=12$

kiwitrader123 2017-02-16 19:17:10
12

Emathmaster 2017-02-16 19:17:10
12 for both

JJShan26 2017-02-16 19:17:10
draw the altitudes to get 4 3-4-5 triangles

mathman2048 2017-02-16 19:17:10
half the big triangle's area=12

copeland 2017-02-16 19:17:12
The area of $ADB$ is half the area of $ABC.$

copeland 2017-02-16 19:17:33
Everything in sight is either a 3-4-5 triangle or a pair of 3-4-5 triangles stuck together.

copeland 2017-02-16 19:17:47
Heron is a huge pain when you can just multiply lengths.

copeland 2017-02-16 19:18:01
Since the area of $ABC$ is $\dfrac{1}{2} \cdot 8 \cdot 6 = 24$, we know $[ADB] = 12.$

copeland 2017-02-16 19:18:04
So what is the inradius of $ADB$?

mathislife16 2017-02-16 19:18:35
3/2

PiGuy3141592 2017-02-16 19:18:35
3/2

ninjataco 2017-02-16 19:18:35
3/2

vishwathganesan 2017-02-16 19:18:35
12/8 = 3/2

Awesomekid05 2017-02-16 19:18:35
12/8 = 3/2

mathchampion1 2017-02-16 19:18:35
$12/8=3/2$

ilovemath04 2017-02-16 19:18:35
3/2

High 2017-02-16 19:18:35
3/2

solzonmars 2017-02-16 19:18:35
4/3

copeland 2017-02-16 19:18:37
That's $\dfrac{12}{8} = \dfrac{3}{2}$.

copeland 2017-02-16 19:18:38
Okay. And what about the inradius of $ADC$?

mathman2048 2017-02-16 19:19:07
4/3

ilovemath04 2017-02-16 19:19:07
4/3

warrenwangtennis 2017-02-16 19:19:07
4/3

GeronimoStilton 2017-02-16 19:19:07
$\frac{4}{3}$

Slacker 2017-02-16 19:19:07
do the same think

ImpossibleCube 2017-02-16 19:19:07
4/3

a000 2017-02-16 19:19:07
12/9 = 4/3

lfwei 2017-02-16 19:19:07
12/9 = 4/3

copeland 2017-02-16 19:19:09
By the exact same argument, that's \[\frac{[\triangle ADC]}{s}=\dfrac{\frac{24}2}{\frac{5+5+8}2}=\dfrac{12}{9}=\dfrac 43.\]

copeland 2017-02-16 19:19:11
So what is our answer?

Magicsm 2017-02-16 19:19:32
17/6

blitzkrieg21 2017-02-16 19:19:32
17/6

Metal_Bender19 2017-02-16 19:19:32
17/6

shootingstar8 2017-02-16 19:19:32
17/6

JJShan26 2017-02-16 19:19:32
D. 17/6

GeronimoStilton 2017-02-16 19:19:32
$(D)$

Awesomekid05 2017-02-16 19:19:32
4/3 + 3/2 = 17/6 , D

tdeng 2017-02-16 19:19:32
17/6

Ani10 2017-02-16 19:19:32
D

blitzkrieg21 2017-02-16 19:19:32
17/ 6

copeland 2017-02-16 19:19:34
The answer is $\dfrac 43 + \dfrac 32 = \boxed{\dfrac{17}{6}}$, (D).

copeland 2017-02-16 19:19:43
Alright. Everyone warmed up now?

Ani10 2017-02-16 19:20:19
wait but why are the two triangles ADC and ADB equal

copeland 2017-02-16 19:20:20
That's a great question. Did you see that they're both just a pair of 3-4-5 triangles joined along either the 3-sides or the 4-sides?

copeland 2017-02-16 19:20:47
Now ready for more?

Winston123 2017-02-16 19:20:56
Yes

rspr2001 2017-02-16 19:20:56
yup

AMC10Perfect 2017-02-16 19:20:56
yup

Fives 2017-02-16 19:20:56
yes

techguy2 2017-02-16 19:20:56
YEAAAA BOOOOI

yanyu2002 2017-02-16 19:20:56
Yes!

awesomemaths 2017-02-16 19:20:56
but SURE

ILoveChess108 2017-02-16 19:20:56
yes

kunsun 2017-02-16 19:20:56
yup

awesomemaths 2017-02-16 19:20:56
yes

rick101 2017-02-16 19:20:56
Yes!

copeland 2017-02-16 19:21:03
22. The diameter $\overline{AB}$ of a circle of radius 2 is extended to a point $D$ outside the circle so that $BD = 3$. Point $E$ is chosen so that $ED =5$ and line $ED$ is perpendicular to line $AD$. Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E.$ What is the area of $\triangle ABC?$
$\textbf{(A)}\quad \dfrac{120}{37} \qquad \qquad \textbf{(B)}\quad \dfrac{140}{39} \qquad\qquad \textbf{(C)}\quad \dfrac{145}{39} \qquad\qquad \textbf{(D)}\quad \dfrac{140}{37} \qquad\qquad\textbf{(E)}\quad \dfrac{120}{31}$

copeland 2017-02-16 19:21:05
Wait, where do we get started again?

Mathaddict11 2017-02-16 19:21:20
Draw a Diagram

StellarG 2017-02-16 19:21:20
Draw a diagram

HighQXMoney 2017-02-16 19:21:20
Draw the diagram

reedmj 2017-02-16 19:21:20
DIAGRAM!!!

wangymin 2017-02-16 19:21:20
diagram

Reef334 2017-02-16 19:21:20
diagram

fractal161 2017-02-16 19:21:20
DYE-A-GRAM!

copeland 2017-02-16 19:21:23
A diagram of course. How's this one?

copeland 2017-02-16 19:21:23

shootingstar8 2017-02-16 19:21:42
Bigger

lfwei 2017-02-16 19:21:42
too small

tootyfail 2017-02-16 19:21:42
NOOO bigger

pie314159265 2017-02-16 19:21:42
too small ._.

bogstop320 2017-02-16 19:21:42
too small

zac15SCASD 2017-02-16 19:21:42
small

BetaCentauri 2017-02-16 19:21:42
uh oh too small

dr3463 2017-02-16 19:21:42
make it bigger

AthenasEye 2017-02-16 19:21:42
too small

copeland 2017-02-16 19:21:44
TOO SMALL! We won't be able to do any work on this diagram. Remember to draw a large, clear diagram. In fact, I generally suggest you draw a full page diagram if you can.

copeland 2017-02-16 19:21:46

copeland 2017-02-16 19:21:49
What pops out at you in this diagram?

Vfire 2017-02-16 19:22:03
Nice

copeland 2017-02-16 19:22:04
Why thanks.

InYourWildestDreams 2017-02-16 19:22:27
ACB is a right triangle

JJShan26 2017-02-16 19:22:27
ACB is a right triangle and so is ADE

quartzgirl 2017-02-16 19:22:27
ABC must be a right triangle

ImpossibleCube 2017-02-16 19:22:27
ABC is a right triangle

akaashp11 2017-02-16 19:22:27
<ACB is a right angle!

letsgomath 2017-02-16 19:22:27
AC and BC are perpendicular

copeland 2017-02-16 19:22:30
We see that $\angle ACB$ is 90 degrees because $AB$ is the diameter of the circle. What else?

EulerMacaroni 2017-02-16 19:23:05
$\triangle ACB$ and $\triangle ADE$ are siilar

Naren12 2017-02-16 19:23:05
triangle ACB and ADE are similar

000libyaclawdruse000 2017-02-16 19:23:05
ABC is similar to AED

reedmj 2017-02-16 19:23:05
5-7-8 right triangle, ACB and ADE are similar

Ani10 2017-02-16 19:23:05
ACB and ADE similar kek

EulerMacaroni 2017-02-16 19:23:05
$\triangle ACB$ and $\triangle ADE$ are similar and oppositely oriented

Mathaddict11 2017-02-16 19:23:05
ABC and AED are similar triangles!

copeland 2017-02-16 19:23:06
Since $\angle ACB = 90^{\circ} = \angle ADE$, and $ACB$ and $ADE$ share $\angle A$, we know these two triangles are similar.

copeland 2017-02-16 19:23:19
(Don't forget to make the vertices match up when writing similarities.!)

copeland 2017-02-16 19:23:22
So what do we need to do to find the area of $ACB?$

a000 2017-02-16 19:24:13
ratio of lengths

a1b2 2017-02-16 19:24:13
Proportion

bluequest 2017-02-16 19:24:13
side ratio^2=area rati

quartzgirl 2017-02-16 19:24:13
Find the sides using the ratios first

ilovemath04 2017-02-16 19:24:13
find the ratio

PiGuy3141592 2017-02-16 19:24:13
Find the similarity ratio

Emathmaster 2017-02-16 19:24:13
Find the side ratio.

copeland 2017-02-16 19:24:16
All we have to do is find the similarity ratio and the area of $ADE.$

copeland 2017-02-16 19:24:18
One of those is a little easier. What's the area of $ADE?$

lfwei 2017-02-16 19:24:48
35/2

shootingstar8 2017-02-16 19:24:48
35/2

letsgomath 2017-02-16 19:24:48
35/2

reedmj 2017-02-16 19:24:48
35/2

EulerMacaroni 2017-02-16 19:24:48
$\frac{35}{2}$

Awesomekid05 2017-02-16 19:24:48
35/2

CornSaltButter 2017-02-16 19:24:48
35/2

mumpu2k16 2017-02-16 19:24:48
17.5

blitzkrieg21 2017-02-16 19:24:48
$35/2$

vishwathganesan 2017-02-16 19:24:48
35/2

Haphazard 2017-02-16 19:24:48
35/2

copeland 2017-02-16 19:24:53
That's $\dfrac 12 \cdot 7 \cdot 5 = \dfrac{35}{2}.$

copeland 2017-02-16 19:24:54
Okay, how can we find the similarity ratio?

GeronimoStilton 2017-02-16 19:25:46
Find $AE$ so we know $\frac{AB}{AE}$

dr3463 2017-02-16 19:25:46
we know the hypotenuse of ABC and ADE

quartzgirl 2017-02-16 19:25:46
Find the hypotenuse, and we already know the hypotenuse of ABC

copeland 2017-02-16 19:25:55
The similarity ratio is $\dfrac{AB}{AE}$.

copeland 2017-02-16 19:25:57
What is this ratio?

ninjataco 2017-02-16 19:26:35
4/sqrt(74)

MSTang 2017-02-16 19:26:35
4 / sqrt(74)

GeronimoStilton 2017-02-16 19:26:35
$\frac{4}{\sqrt{74}}$

quartzgirl 2017-02-16 19:26:35
4/(sqrt(74))

vishwathganesan 2017-02-16 19:26:35
4:sqrt(74)

abishek99 2017-02-16 19:26:35
4/sqrt(74)

ILoveChess108 2017-02-16 19:26:35
4/sqrt74

Skittlesftw 2017-02-16 19:26:35
4/\sqrt74

mathmagician 2017-02-16 19:26:35
$$4/sqrt(74)$$

copeland 2017-02-16 19:26:36
Since $AE = \sqrt{5^2 + 7^2} = \sqrt{74}$ and $AB = 4$, we know the similarity ratio is $\dfrac{4}{\sqrt{74}}.$

copeland 2017-02-16 19:26:37
So what is the area of $\triangle ACB?$

Reef334 2017-02-16 19:27:24
(D) 140/37

Emathmaster 2017-02-16 19:27:24
140/37

GeronimoStilton 2017-02-16 19:27:24
$\frac{140}{37}$, D

warrenwangtennis 2017-02-16 19:27:24
140/37

JJShan26 2017-02-16 19:27:24
35/2 * 16/74 = 140/37

tdeng 2017-02-16 19:27:24
8/37*35/2=140/37

Derive_Foiler 2017-02-16 19:27:24
35/2 * 16/74= 140/37

MSTang 2017-02-16 19:27:24
16/74 * 35/2 = 140/37

summitwei 2017-02-16 19:27:24
140/37

burunduchok 2017-02-16 19:27:24
140/37

copeland 2017-02-16 19:27:28
The area is \[ \frac{35}{2} \cdot \left( \frac{4}{\sqrt{74}} \right)^2 = \frac{35 \cdot 16}{148} = \frac{35\cdot 4}{37} = \boxed{\dfrac{140}{37}}.\]

copeland 2017-02-16 19:27:31
Our answer is (D).

copeland 2017-02-16 19:27:40
Any questions or should we soldier on?

JennyWenDE 2017-02-16 19:27:57
next question

Cardinals2014 2017-02-16 19:27:57
solider

First 2017-02-16 19:27:57
Solider on

MathTechFire 2017-02-16 19:27:57
soldier on

HighQXMoney 2017-02-16 19:27:57
Soldier on

zew 2017-02-16 19:27:57
keep going!

dr3463 2017-02-16 19:27:57
soldier on

Fives 2017-02-16 19:28:16
Can we pls have somthing easiar?

copeland 2017-02-16 19:28:20
OK, can do!

copeland 2017-02-16 19:28:24
23. Let \[N=123456789101112\ldots4344\] be the 79-digit number that is formed by writing the integers from 1 to 44 in order, one after the other. What is the remainder when $N$ is divided by 45?
$\textbf{(A)}\quad 1 \qquad \qquad \textbf{(B)}\quad 4 \qquad\qquad \textbf{(C)}\quad 9 \qquad\qquad \textbf{(D)}\quad 18 \qquad\qquad\textbf{(E)}\quad 44$

copeland 2017-02-16 19:28:32
Ooh, ooh! What are we going to use for this problem?

Cardinals2014 2017-02-16 19:29:05
CRT

First 2017-02-16 19:29:05
CRT

checkmatetang 2017-02-16 19:29:05
CRT

CornSaltButter 2017-02-16 19:29:05
Chinese remainder theorem

Metal_Bender19 2017-02-16 19:29:05
Chinese remainder theorem

burunduchok 2017-02-16 19:29:05
CRT

dipenm 2017-02-16 19:29:05
Chinese remainder theorem

summitwei 2017-02-16 19:29:05
chinese remainder theorem

MSTang 2017-02-16 19:29:05
CRT!!!

copeland 2017-02-16 19:29:11
This is a classic Chinese Remainder Theorem problem. If we want to find the remainder after dividing by 45, we can find the remainder after dividing by 5 and the remainder when dividing by 9 and work from there.

copeland 2017-02-16 19:29:12
What's the remainder when we divide $N$ by 5?

BetaCentauri 2017-02-16 19:29:34
4

geogirl08 2017-02-16 19:29:34
4

techguy2 2017-02-16 19:29:34
4

Mathcat1234 2017-02-16 19:29:34
4

jeffshen 2017-02-16 19:29:34
4

zhengyf 2017-02-16 19:29:34
4

Winston123 2017-02-16 19:29:34
4

Mathaddict11 2017-02-16 19:29:34
4

mathchampion1 2017-02-16 19:29:34
4

mathislife16 2017-02-16 19:29:34
4

copeland 2017-02-16 19:29:36
The last digit is 4, so

copeland 2017-02-16 19:29:37
\[N\equiv 4\pmod5.\]

copeland 2017-02-16 19:29:38
What's the remainder when we divide $N$ by 9?

Cardinals2014 2017-02-16 19:29:56
0

quanhui868 2017-02-16 19:29:56
0

PiGuy3141592 2017-02-16 19:29:56
0

Emathmaster 2017-02-16 19:29:56
0

tree3 2017-02-16 19:29:56
0

lfwei 2017-02-16 19:29:56
0

zhengyf 2017-02-16 19:29:56
0

ilovemath04 2017-02-16 19:29:56
0

shootingstar8 2017-02-16 19:29:56
0

belieber 2017-02-16 19:29:56
0

bgu 2017-02-16 19:29:56
0

bluequest 2017-02-16 19:29:56
0

akaashp11 2017-02-16 19:29:56
0 (mod 9)

summitwei 2017-02-16 19:29:56
0

copeland 2017-02-16 19:29:59
We can compute the sum of the digits trickily here. Instead of taking all the digits and adding them, we can take all the numbers and add them:\[N\equiv 12345\cdots4344\equiv1+2+3+\cdots+43+44\pmod{9}.\]

copeland 2017-02-16 19:30:09
This is an arithmetic progression, so we know the sum is $\dfrac{(1+44)\cdot44}2$, which, whatever it is, is definitely a multiple of 9. So. .

copeland 2017-02-16 19:30:13
\[N\equiv0\pmod9.\]

copeland 2017-02-16 19:30:17
What's the remainder when we divide $N$ by 45?

fractal161 2017-02-16 19:30:48
Has to be $9$

prasenjit_hazra 2017-02-16 19:30:48
9

Awesomekid05 2017-02-16 19:30:48
9

azmath333 2017-02-16 19:30:48
9

islander7 2017-02-16 19:30:48
9

thetank 2017-02-16 19:30:48
9

checkmatetang 2017-02-16 19:30:48
9

copeland 2017-02-16 19:30:56
We want to solve these two congruences above. That's best done using guess-and-check.

copeland 2017-02-16 19:31:00
We want a multiple of 9 that ends in 4 or 9. If you don't see it right away, you should just look at the answer choices!

copeland 2017-02-16 19:31:02
The answer is (C), $\boxed9$.

copeland 2017-02-16 19:31:36
OK, now back to harder?

The_Turtle 2017-02-16 19:31:58
YES!

awesomemaths 2017-02-16 19:31:58
SURE

zac15SCASD 2017-02-16 19:31:58
Sure

mumpu2k16 2017-02-16 19:31:58
yep

mathchampion1 2017-02-16 19:31:58
hyperbolas!!!

copeland 2017-02-16 19:32:06
Hyperbolas? Have you been looking ahead?

copeland 2017-02-16 19:32:09
24. The vertices of an equilateral triangle lie on the hyperbola $xy=1,$ and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\quad 48 \qquad \qquad \textbf{(B)}\quad 60 \qquad\qquad \textbf{(C)}\quad 108 \qquad\qquad \textbf{(D)}\quad 120 \qquad\qquad\textbf{(E)}\quad 169$

copeland 2017-02-16 19:32:12
Here's a hyperbola.

copeland 2017-02-16 19:32:13

copeland 2017-02-16 19:32:14
Where are the vertices of the hyperbola?

kunsun 2017-02-16 19:32:49
1,1 and -1,-1

InYourWildestDreams 2017-02-16 19:32:49
(1, 1) (-1, -1)

a000 2017-02-16 19:32:49
(1,1) and (-1,-1)

joslin94 2017-02-16 19:32:49
(1,1) and (-1,-1)

a1b2 2017-02-16 19:32:49
$(\pm 1, \pm 1)$

PiGuy3141592 2017-02-16 19:32:49
(1,1) and (-1,-1)

reedmj 2017-02-16 19:32:49
$(1,1) \text{ and } (-1,-1)$

quartzgirl 2017-02-16 19:32:49
(1, 1), (-1, -1)

mathwrite 2017-02-16 19:32:49
(1,1) and (-1,-1)

kiwitrader123 2017-02-16 19:32:49
(!,1), (-1,-1)

copeland 2017-02-16 19:32:51
The vertices are the closest points on the branches. They're the points where the line through the foci intersect the parabola. Since the hyperbola is symmetric across $y=x$, the foci are on $y=x$, so the vertices are at $(1,1)$, and $(-1,-1)$.

copeland 2017-02-16 19:33:02
Now we want an equilateral triangle with center at $(1,1)$. They use the word "centroid", but which center is more useful to think about?

mathfun5 2017-02-16 19:33:37
Circumcenter

zac15SCASD 2017-02-16 19:33:37
circumcenter

a1b2 2017-02-16 19:33:37
Circumcenter

letsgomath 2017-02-16 19:33:37
circumcenter?

mathchampion1 2017-02-16 19:33:37
circumcenter!

tdeng 2017-02-16 19:33:37
circumcenter

ninjataco 2017-02-16 19:33:37
circumcenter

copeland 2017-02-16 19:33:39
Why?

MSTang 2017-02-16 19:35:06
the circumcircle must hit the other vertex of the hyperbola

mathfun5 2017-02-16 19:35:06
contrsuct the circumcircle! It then passes through the hyperbola

copeland 2017-02-16 19:35:07
The circumcenter contains the vertices, so the circumcenter will intersect the hyperbola at the vertices (and maybe elsewhere).

copeland 2017-02-16 19:35:24
In an equilateral triangle, the centroid is also the circumcenter so we can use that. Let's think about circles with circumcenter $(1,1)$.

copeland 2017-02-16 19:35:33
Now we just draw a circle with center $(1,1)$ that intersects the hyperbola. Here are a few:

copeland 2017-02-16 19:35:34

copeland 2017-02-16 19:35:43
What do you notice?

shootingstar8 2017-02-16 19:35:58
Woah!

Emathmaster 2017-02-16 19:35:58
Yellow.

copeland 2017-02-16 19:36:00
Go go hsv.

copeland 2017-02-16 19:36:38
(hsv is a way to define colors. Nevermind.)

mathislife16 2017-02-16 19:36:50
the yellow one hits the other vertex

ninjataco 2017-02-16 19:36:50
intersects in 2, 3, or 4 points

The_Turtle 2017-02-16 19:36:50
The red circle intersects the graph twice, the yellow one intersects three times, and the rest intersect 4 times

zac15SCASD 2017-02-16 19:36:50
only one hits exactly 3 points (yellow)

techguy2 2017-02-16 19:36:50
we want the one that touches in only three places

tree3 2017-02-16 19:36:50
the yellow circle works

Zekrom 2017-02-16 19:36:50
it must intersect the other vertex because otherwise, there are four or two vertices on the hyperbola

copeland 2017-02-16 19:37:03
So, it's not a priori bad if the circle intersects 4 times instead of 3.

copeland 2017-02-16 19:37:13
However, why are those 4-intersection guys unsatisfying?

rafayaashary1 2017-02-16 19:37:58
Symmetry

Derive_Foiler 2017-02-16 19:37:58
not equilateral

vishwathganesan 2017-02-16 19:37:58
not symmetrical

dipenm 2017-02-16 19:37:58
they can't be equilateral

techguy2 2017-02-16 19:37:58
no three points are equidistant

mathwiz0803 2017-02-16 19:37:58
they don't form equilateral triangles

GeronimoStilton 2017-02-16 19:37:58
Because they give us two possible equilateral triangles, and they're not symmetric.

The_Turtle 2017-02-16 19:37:58
No subset of their points form an equilateral triangle

AlcumusGuy 2017-02-16 19:37:58
cannot be equilateral

copeland 2017-02-16 19:38:01
These are usually quadrilaterals.

copeland 2017-02-16 19:38:03
These all have $y=x$ as an axis of symmetry, so they're actually isosceles trapezoids.

copeland 2017-02-16 19:38:22
Three vertices of an isosceles trapezoid can't be the vertices of an equilateral triangle. The axis of symmetry would go through one of the sides of the triangle and through the other vertex. That would give us a degenerate trapezoid.

copeland 2017-02-16 19:38:33
So the only triangle we could get has to come from the circle that's tangent to the other branch:

copeland 2017-02-16 19:38:34

copeland 2017-02-16 19:38:46
I sure hope that's equilateral.

The_Turtle 2017-02-16 19:39:27
How can we tell?

pie314159265 2017-02-16 19:39:27
we can assume

yayups 2017-02-16 19:39:27
by the problem statement, it must be

BLCRAFT 2017-02-16 19:39:27
xD how do we know?

yanyu2002 2017-02-16 19:39:27
How can you just hope it's equilateral?

GeronimoStilton 2017-02-16 19:39:27
It doesn't matter because it's the only possible triangle.

copeland 2017-02-16 19:39:34
These are all fine emotions to be having right now. I have them all, too.

copeland 2017-02-16 19:39:46
What say we take a moment to check, actually? You should definitely not do that on the test, but we have the luxury of doing it now.

copeland 2017-02-16 19:40:01
I'm going to spew math at you for a moment, because this is pretty hard.

Ani10 2017-02-16 19:40:05
use coordinate geo?

copeland 2017-02-16 19:40:12
What's a generic point on the hyperbola look like?

Derive_Foiler 2017-02-16 19:40:32
(x,1/x)

no_name 2017-02-16 19:40:32
(a, 1/a)

akaashp11 2017-02-16 19:40:32
(x,1/x)

mathfun5 2017-02-16 19:40:32
(a,1/a)

AlcumusGuy 2017-02-16 19:40:32
(x, 1/x)

abishek99 2017-02-16 19:40:32
(a,1/a)

zac15SCASD 2017-02-16 19:40:32
(x, 1/x)

PiGuy3141592 2017-02-16 19:40:32
(x,1/x)

summitwei 2017-02-16 19:40:32
(x,1/x)

burunduchok 2017-02-16 19:40:32
$(x, \frac{1}{x})$

000libyaclawdruse000 2017-02-16 19:40:32
(x, 1/x)

copeland 2017-02-16 19:40:37
A generic point looks like $P=\left(x,\frac1x\right)$.

copeland 2017-02-16 19:40:38
The red circle has squared radius $r^2=8$. So a point on the circle looks like \[(x-1)^2+(y-1)^2=8.\]

copeland 2017-02-16 19:40:43
The intersection points of the circle with the hyperbola solve\[\left(x-1\right)^2+\left(\frac1x-1\right)^2=8.\]

copeland 2017-02-16 19:40:48
We can expand that to get \[x^2+\frac1{x^2}-2\left(x+\frac1x\right)+2=8.\]

copeland 2017-02-16 19:40:50
Now what should we always do with an expression like this?

Derive_Foiler 2017-02-16 19:41:24
x-1/x=a, and solve for the desired x^2+1/x^2

ninjataco 2017-02-16 19:41:24
substitute a = x + 1/x

awesome_weisur 2017-02-16 19:41:24
set x + 1/x = y

mtarun 2017-02-16 19:41:24
Let x+1/x=a

First 2017-02-16 19:41:24
make $(x+\frac{1}{x}$$=y$

copeland 2017-02-16 19:41:27
We should write it as a polynomial in $x+\dfrac1x$. Since $\left(x+\dfrac1x\right)^2=x^2+\dfrac1{x^2}+2$, we write

\[\left(x+\frac1{x}\right)^2-2\left(x+\frac1x\right)=8.\]

copeland 2017-02-16 19:41:39
The left side is almost a square! If we add 1 we get \[\left(x+\frac1{x}-1\right)^2=9,\]and given that we're in the first quadrant we get \[\frac{x+\frac1x-1}3=1.\]

copeland 2017-02-16 19:42:09
Now what?

mathfun5 2017-02-16 19:42:52
solve quadratic??

\

divijleisha 2017-02-16 19:42:52
x+1/x=4

naman12 2017-02-16 19:42:52
$x + \dfrac{1}{x} = 4$

tdeng 2017-02-16 19:42:52
x+1/x=4, which we can solve as a quadratic

copeland 2017-02-16 19:42:54
We could.

copeland 2017-02-16 19:42:57
Actually it's not necessary.

copeland 2017-02-16 19:43:02
Let's look more at what we have.

copeland 2017-02-16 19:43:22
What's the centroid of the triangle with vertices $\left(x,\frac1x\right)$, $\left(\frac1x,x\right)$, and $\left(-1,-1\right)$?

zac15SCASD 2017-02-16 19:43:49
(1, 1)

warrenwangtennis 2017-02-16 19:43:49
(1, 1)

donot 2017-02-16 19:43:49
1,1

Derive_Foiler 2017-02-16 19:43:49
oh. take the average of them all

a1b2 2017-02-16 19:43:49
$(1,1)$

HighQXMoney 2017-02-16 19:43:49
(1, 1)

First 2017-02-16 19:43:49
average the x coordinates and y coordinates

mathislife16 2017-02-16 19:43:49
(1,1)

GeronimoStilton 2017-02-16 19:43:51
The average of the sum of those points?

copeland 2017-02-16 19:43:53
Right!

copeland 2017-02-16 19:43:54
Now we're done. The centroid of the triangle $\left(x,\frac1x\right)$, $\left(\frac1x,x\right)$, $\left(-1,-1\right)$ is indeed \[\left(\dfrac{x+\frac1x-1}3,\dfrac{\frac1x+x-1}3\right)=(1,1).\]

copeland 2017-02-16 19:44:00
The centroid is the circumcenter. So the medians are angle bisectors. The Angle Bisector Theorem then gives that the edge lengths are equal and we're done.

copeland 2017-02-16 19:44:19
And that was a huge waste of time, testwise speaking, but at least we got to do some algebra, huh?

lfwei 2017-02-16 19:44:25
wow, magical

copeland 2017-02-16 19:44:31
It's not magic. It's math.

copeland 2017-02-16 19:44:40
OK, back to business. Let's compute its area. How can we find those other two vertices?

tdeng 2017-02-16 19:45:11
We don't need to

Derive_Foiler 2017-02-16 19:45:11
We don't need to . . .

MSTang 2017-02-16 19:45:11
we don't need to!

donot 2017-02-16 19:45:11
we don't need it

GeronimoStilton 2017-02-16 19:45:11
We don't need to know where they are.

Derive_Foiler 2017-02-16 19:45:11
We don't

tdeng 2017-02-16 19:45:11
We don't need to; we can just find the side length!

62861 2017-02-16 19:45:11
you don't need to

copeland 2017-02-16 19:45:15
Wait, that's a terrible idea. We can find the area much easier than that! How?

EpicDragonSlayr 2017-02-16 19:46:04
side length

dipenm 2017-02-16 19:46:04
The radius?

GeronimoStilton 2017-02-16 19:46:04
We have an equilateral triangle with circumradius $2\sqrt{2}$. We can solve.

letsgomath 2017-02-16 19:46:04
find the side length

speulers_theorem 2017-02-16 19:46:04
distance to centroid is $2\sqrt{2}$

maxplanck 2017-02-16 19:46:04
you know the circumradius

copeland 2017-02-16 19:46:06
We know the distance from the center of the equilateral triangle to one of the vertices. That determines the triangle completely. How far is it from the center to one vertex?

copeland 2017-02-16 19:46:16
Oh, wait. You just told me.

awesome_weisur 2017-02-16 19:46:25
2 sqrt 2

copeland 2017-02-16 19:46:27
The distance from the center to a vertex is the distance from $(1,1)$ to $(-1,-1)$. That's $2\sqrt2$.

copeland 2017-02-16 19:46:28
So what's the altitude?

First 2017-02-16 19:47:10
$3 \sqrt2$

CornSaltButter 2017-02-16 19:47:10
3sqrt2

abishek99 2017-02-16 19:47:10
3sqrt(2)

qrohn 2017-02-16 19:47:10
3sqrt2

bluequest 2017-02-16 19:47:10
3 sqrt 2

Devo123 2017-02-16 19:47:10
3sqrt2

Reef334 2017-02-16 19:47:10
3sqrt(2)

CaptainGeo 2017-02-16 19:47:10
3sqrt2

vonnan 2017-02-16 19:47:10
3root2

ImpossibleCube 2017-02-16 19:47:10
$3\sqrt{2}$

Emathmaster 2017-02-16 19:47:10
3sqrt2

a000 2017-02-16 19:47:10
3sqrt2

kunsun 2017-02-16 19:47:10
3sqrt2

copeland 2017-02-16 19:47:12
The altitude is $\dfrac32\cdot2\sqrt2=3\sqrt2$.

copeland 2017-02-16 19:47:15
What's the base length?

mathfun5 2017-02-16 19:47:44
2sqrt6

ilovemath04 2017-02-16 19:47:44
2sqrt6

Zekrom 2017-02-16 19:47:44
2*sqrt(6)

GeronimoStilton 2017-02-16 19:47:44
$2\sqrt{6}$

a000 2017-02-16 19:47:44
2sqrt6

Devo123 2017-02-16 19:47:44
2 sqrt 6

tree3 2017-02-16 19:47:44
2sqrt6

copeland 2017-02-16 19:47:46
The length of the base is $\dfrac2{\sqrt3}\cdot 3\sqrt2=2\sqrt6$.

copeland 2017-02-16 19:47:48
So what's the square of the area?

ilovemath04 2017-02-16 19:48:21
108

abishek99 2017-02-16 19:48:21
108

CaptainGeo 2017-02-16 19:48:21
$\text{(C) } 108$

ImpossibleCube 2017-02-16 19:48:21
108

prasenjit_hazra 2017-02-16 19:48:21
108

copeland 2017-02-16 19:48:22
The square of the area is \[\left(\frac12\cdot3\sqrt2\cdot2\sqrt6\right)^2=\left(3\sqrt{12}\right)^2=\boxed{108}.\] The answer is (C).

copeland 2017-02-16 19:48:27
Good going.

copeland 2017-02-16 19:49:01
Should we knock the rest of this "AMC 10" out now?

Mathaddict11 2017-02-16 19:49:16
yes!

vishwathganesan 2017-02-16 19:49:16
yeah!

sas4 2017-02-16 19:49:16
YES

PiGuy3141592 2017-02-16 19:49:16
Yes

Cardinals2014 2017-02-16 19:49:16
yeah sure

Mathcat1234 2017-02-16 19:49:16
sure!

divijleisha 2017-02-16 19:49:16
YES

mathfun5 2017-02-16 19:49:16
yes

jk23541 2017-02-16 19:49:16
yes

Slacker 2017-02-16 19:49:16
last one

Ani10 2017-02-16 19:49:29
why is that in quotes ;-;

Emathmaster 2017-02-16 19:49:29
lol, you put quotes

copeland 2017-02-16 19:49:30
I love "quotes".

copeland 2017-02-16 19:50:10
25. Last year Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\quad 92 \qquad \qquad \textbf{(B)}\quad 94 \qquad\qquad \textbf{(C)}\quad 96 \qquad\qquad \textbf{(D)}\quad 98 \qquad\qquad\textbf{(E)}\quad 100$

copeland 2017-02-16 19:50:12
Let's set up a bunch of variables that we probably won't need. I'll call $s_1, s_2, s_3, s_4, s_5, s_6,$ and $s_7$ the scores from Isabella's seven tests.

copeland 2017-02-16 19:50:13
Are there any other quantities we care about?

shootingstar8 2017-02-16 19:50:29

deven!!

copeland 2017-02-16 19:50:31
I know, right? Jerk.

islander7 2017-02-16 19:51:22
their sum

naman12 2017-02-16 19:51:22
Average

shootingstar8 2017-02-16 19:51:22
average

HighQXMoney 2017-02-16 19:51:22
Average

lsh0589 2017-02-16 19:51:22
average

geogirl08 2017-02-16 19:51:22
averages

Devo123 2017-02-16 19:51:22
the sum of the seven scores

MathTechFire 2017-02-16 19:51:22
average?

copeland 2017-02-16 19:51:25
Averages or sums?

copeland 2017-02-16 19:51:28
Which is better?

mumpu2k16 2017-02-16 19:51:58
Sums.

ninjataco 2017-02-16 19:51:58
sums

mathwiz0803 2017-02-16 19:51:58
no fractions = sum

kiwitrader123 2017-02-16 19:51:58
sum

alp_snow7 2017-02-16 19:51:58
Sum

mathmagician 2017-02-16 19:51:58
sums

mathchampion1 2017-02-16 19:51:58
sums

naman12 2017-02-16 19:51:58
Sum

Daniel2003 2017-02-16 19:51:58
sum

ilovemath04 2017-02-16 19:51:58
sum

quanhui868 2017-02-16 19:51:58
Sum

000libyaclawdruse000 2017-02-16 19:51:58
sums

copeland 2017-02-16 19:52:00
"The average is an integer" is way worse than "The sum is divisible by. . ."

copeland 2017-02-16 19:52:08
Since we're given information about the averages, we might also care about the totals, we'll let $T_1, T_2, T_3, T_4, T_5, T_6,$ and $T_7$ be Isabella's total score after each test.

copeland 2017-02-16 19:52:10
Since they gave us information about the seventh test, it makes sense to start there. What do we know?

SomethingNeutral 2017-02-16 19:52:34
s7 = 95

pie314159265 2017-02-16 19:52:34
the score was 95

a1b2 2017-02-16 19:52:34
$s_7=95$

awesomemaths 2017-02-16 19:52:34
the sevnth test score is a 95

copeland 2017-02-16 19:52:37
We know she scored a 95 on the seventh test. So, $s_7 = 95.$

copeland 2017-02-16 19:52:40
Additionally, we know that $T_7$ is a multiple of 7.

copeland 2017-02-16 19:52:43
There are lots of multiples of 7... can we narrow it down at all?

islander7 2017-02-16 19:54:09
between 7*94 and 7*97

Fridaychimp 2017-02-16 19:54:09
94*7-97*7

SomethingNeutral 2017-02-16 19:54:09
91+...97 to 94+..100

Ani10 2017-02-16 19:54:09
91+92+93+94+95+96+97 to 94+95+96+97+98+99+100 multiples of 7 inbetween

mathislife16 2017-02-16 19:54:09
94*7 through 97*7

copeland 2017-02-16 19:54:12
We know the lowest she could have possibly scored is $91 + 92 + 93 + 94 + 95 + 96 + 97 = 658.$

copeland 2017-02-16 19:54:14
The highest she could have possibly scored is $100 + 99 + 98 + 97 +96 + 95 + 94 = 679.$

copeland 2017-02-16 19:54:14
What are the multiples of 7 in that range?

kunsun 2017-02-16 19:54:55
658 665 672 679

vishwathganesan 2017-02-16 19:54:55
658, 665, 672, 679

weilunsun28 2017-02-16 19:54:55
658 665 672 679

mathwrite 2017-02-16 19:54:55
658, 665, 672, 679

letsgomath 2017-02-16 19:54:55
658, 665, 672, 679

SomethingNeutral 2017-02-16 19:54:55
658, 665, 672, 679

azmath333 2017-02-16 19:54:55
658, 665, 672, 679

copeland 2017-02-16 19:54:57
We know 658 and 679 are multiples of 7 since they're the sum of 7 consecutive integers, so we just list the others in between. Our possibilities are 658, 665, 672, and 679.

copeland 2017-02-16 19:55:04
Now we're in business. How can we narrow that down even further?

First 2017-02-16 19:55:48
Remove 95 and see if it's divisble by 6

Derive_Foiler 2017-02-16 19:55:48
subtract 95 gives a multiple of 6

islander7 2017-02-16 19:55:48
-95 becomes a multiple of 6

GeronimoStilton 2017-02-16 19:55:48
If we subtract $95$, we will get a multiple of $6$

SomethingNeutral 2017-02-16 19:55:48
-95 is multiple of 6

mathmagician 2017-02-16 19:55:48
the possibilities - 95 is divisible by 6

lsh0589 2017-02-16 19:55:48
-95 and see if that's a multiple of 6

mathwizard666 2017-02-16 19:55:48
95 less than the number is a multiple of 6

carfan25 2017-02-16 19:55:48
take away 95, the result has to be a multiple of 6

DaAsianPotatoe 2017-02-16 19:55:50
T_7 - 95 = 6*x

copeland 2017-02-16 19:55:54
We know that $T_6 = T_7 - 95$ is a multiple of 6. Do any of these satisfy that?

SomethingNeutral 2017-02-16 19:56:42
665.

tnalluri 2017-02-16 19:56:42
665

mathfun5 2017-02-16 19:56:42
665

Reef334 2017-02-16 19:56:42
665

tdeng 2017-02-16 19:56:42
665

mathfun5 2017-02-16 19:56:42
only 665

Cardinals2014 2017-02-16 19:56:42
665

abishek99 2017-02-16 19:56:42
only 665

divijleisha 2017-02-16 19:56:42
665

AAANNNMMMIII 2017-02-16 19:56:42
665

Slacker 2017-02-16 19:56:42
665

AOPS12142015 2017-02-16 19:56:42
665 works

Happytycho 2017-02-16 19:56:42
665

copeland 2017-02-16 19:56:43
Only $665 - 95 = 570$ is a multiple of 6. So, $T_7 = 665$ and $T_6 = 570.$

copeland 2017-02-16 19:56:46
Alright. Let's just keep working backwards. We know $T_5$ is a multiple of 5. How does $T_5$ relate to $T_6$?

vishwathganesan 2017-02-16 19:57:39
the difference is the answer

jeffshen 2017-02-16 19:57:39
100

tdeng 2017-02-16 19:57:39
They are both multiples of 5.

Piazolla13 2017-02-16 19:57:39
T6 - s6

hinna 2017-02-16 19:57:39
$T_5+s_6=T_6$

SomethingNeutral 2017-02-16 19:57:39
t6-t5 is a multiple of 5.

Reef334 2017-02-16 19:57:39
T_6 - S_6 = T_5

Slacker 2017-02-16 19:57:39
$T_5+s_6=T_6$

copeland 2017-02-16 19:57:42
We know $T_5 = T_6 - s_6$. Does that tell us anything about $s_6$?

ninjataco 2017-02-16 19:58:21
s6 is a mulitple of 5

belieber 2017-02-16 19:58:21
divisible by 5

shootingstar8 2017-02-16 19:58:21
Is a multiple of 5

Zekrom 2017-02-16 19:58:21
s_6 is a multiple of 5

Devo123 2017-02-16 19:58:21
multiple of 5

Antidifferentiation108 2017-02-16 19:58:21
is multiple of 5

AlcumusGuy 2017-02-16 19:58:21
needs to be multiple of 5

Smoothfang 2017-02-16 19:58:21
Multiple of 5

copeland 2017-02-16 19:58:23
Since $T_6$ is a multiple of 5, we know that $s_6$ must also be a multiple of 5. So?

SomethingNeutral 2017-02-16 19:58:49
s6 = 95 or 100 but 95 is taken

tdeng 2017-02-16 19:58:49
Must be 0(mod5), 95 is taken, so it must be 100

Piazolla13 2017-02-16 19:58:49
multiple of 5 so 100

jfmath04 2017-02-16 19:58:49
it has to be 100 in order for T_5 to be a multiple of 5

islander7 2017-02-16 19:58:49
s6=100

letsgomath 2017-02-16 19:58:49
not 95 so E 100

kiwitrader123 2017-02-16 19:58:49
only 100 is multipel

dilworthpenguins 2017-02-16 19:58:49
it must be 100 since it can't be 95

Cidkip 2017-02-16 19:58:49
100 is a multiple of 5, none of the other answer choices are

rspr2001 2017-02-16 19:58:49
The answer is E!

ilovemath04 2017-02-16 19:58:49
s6 must be 100

summitwei 2017-02-16 19:58:49
it has to be 100

mathmagician 2017-02-16 19:58:49
s6 is 100, since 95 is s7

copeland 2017-02-16 19:58:51
The only multiple of 5 available is 100. We conclude that $s_6 = \boxed{100}.$

copeland 2017-02-16 19:58:52
Our answer is (E).

copeland 2017-02-16 19:58:58
I think there are 4 sequences that work.

copeland 2017-02-16 19:59:00
$92+96+91+97+94+100+95.$

copeland 2017-02-16 19:59:00
$91+93+92+96+98+100+95$

copeland 2017-02-16 19:59:05
And swap the first 2 elements of each of those.

mumpu2k16 2017-02-16 19:59:30
AMC 12 now?

copeland 2017-02-16 19:59:31
Eh. What do you think?

Derive_Foiler 2017-02-16 19:59:58
yea!

awesomemaths 2017-02-16 19:59:58
YEAH

JP-GoVikes 2017-02-16 19:59:58
yessss

techguy2 2017-02-16 19:59:58
10 better

First 2017-02-16 19:59:58
Sure

mumpu2k16 2017-02-16 19:59:58
Of course!

SaltySnail2 2017-02-16 19:59:58
i wish to do amc 12, if possible

speulers_theorem 2017-02-16 19:59:58
Yeah!

MathTechFire 2017-02-16 19:59:58
Fine

copeland 2017-02-16 20:00:09
That's the kind of mostly-unfettered enthusiasm I love.

copeland 2017-02-16 20:00:20
But we won't start with Problem 21.

copeland 2017-02-16 20:00:24
AMC 10B Problem 25 and AMC 12B Problem 21 were the same. Let's move on to Problem 22.

copeland 2017-02-16 20:00:42
(I found it weird that 25 got demoted to 21 when it crossed over. Usually it lands later.)

copeland 2017-02-16 20:00:48
22. Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?
$\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}$

copeland 2017-02-16 20:01:00
This sounds like a boring game.

Slacker 2017-02-16 20:01:09
ikr

letsgomath 2017-02-16 20:01:12
i know right

copeland 2017-02-16 20:01:14
OK, how many total ways are there for the 4 players to draw these balls?

Ani10 2017-02-16 20:01:30
i got this problem right by playing it 400 times

copeland 2017-02-16 20:01:31
Me too!

vishwathganesan 2017-02-16 20:02:15
4!/2 = 12

speulers_theorem 2017-02-16 20:02:15
12

yrnsmurf 2017-02-16 20:02:15
12

Superwiz 2017-02-16 20:02:15
12

GeronimoStilton 2017-02-16 20:02:15
$12$

rspr2001 2017-02-16 20:02:15
12

copeland 2017-02-16 20:02:20
On each turn, there are 4 ways to hand out the green ball then three ways to hand out the red one. That's $4\cdot3=12$ ways to draw each turn.

copeland 2017-02-16 20:02:30
Four turns gives $12^4$. Great. That's our denominator.

copeland 2017-02-16 20:02:33
Now how can we work out the numerator?

ilovemath04 2017-02-16 20:03:18
casework

SomethingNeutral 2017-02-16 20:03:18
brute force or casework

yrnsmurf 2017-02-16 20:03:18
casework and graph theory?

Smoothfang 2017-02-16 20:03:18
Casework?

summitwei 2017-02-16 20:03:18
casework

copeland 2017-02-16 20:03:20
I guess we use casework. We haven't seen any casework yet and it is the AMC.

copeland 2017-02-16 20:03:21
How should we organize the cases?

copeland 2017-02-16 20:03:24
First, how many times can a person draw the green ball?

Derive_Foiler 2017-02-16 20:04:06
they can draw it once or twice

mathislife16 2017-02-16 20:04:06
twice or it will fail

Superwiz 2017-02-16 20:04:06
0,1or 2 times

GeronimoStilton 2017-02-16 20:04:06
$2$ for the event to happen

geogirl08 2017-02-16 20:04:06
2, 1, or 0

copeland 2017-02-16 20:04:10
A person can draw the green ball twice if he draws the red ball the other 2 times.

copeland 2017-02-16 20:04:18
He could draw it less, of course.

copeland 2017-02-16 20:04:21
Let's first consider the cases where someone draws the green ball twice.

copeland 2017-02-16 20:04:28
Say Abby draws the green ball twice and the red ball twice. That means she gets two coins from other players. Either those players are the same or different:

copeland 2017-02-16 20:04:33

copeland 2017-02-16 20:04:33

copeland 2017-02-16 20:04:40
Look how big those diagrams are!

copeland 2017-02-16 20:04:44
You know who's fault that is?

Reef334 2017-02-16 20:04:57
deven

shakeNbake 2017-02-16 20:04:57
deven

tdeng 2017-02-16 20:04:57
Deven!

SomethingNeutral 2017-02-16 20:04:57
deven

kiwitrader123 2017-02-16 20:04:57
DEVEN

naman12 2017-02-16 20:04:57
devenware

000libyaclawdruse000 2017-02-16 20:04:57
Deven!

copeland 2017-02-16 20:05:00
Right.

copeland 2017-02-16 20:05:04
If Abby draws two greens and passes to Bernardo both times then what has to happen on the other two draws?

tdeng 2017-02-16 20:06:10
Bernardo has to pass to Abby both times.

vishwathganesan 2017-02-16 20:06:10
bernardo draws both greens next

DaAsianPotatoe 2017-02-16 20:06:10
Bernardo passes them back

geogirl08 2017-02-16 20:06:10
abby draws red twice and bernardo draws green twice

oumarrbah_clf 2017-02-16 20:06:10
bernando has to draw two greens and abby has to draw two reds

mathislife16 2017-02-16 20:06:10
B passes to A

Happytycho 2017-02-16 20:06:10
pass them back

summitwei 2017-02-16 20:06:10
Bernardo passes to Abby twice

islander7 2017-02-16 20:06:10
bernardo passes to abby other 2 times

copeland 2017-02-16 20:06:12
If Abby passes twice to Bernard, then Bernardo must also pass twice to Abby.

copeland 2017-02-16 20:06:13

copeland 2017-02-16 20:06:19
Now THAT is a diagram.

copeland 2017-02-16 20:06:24
How many total different draws have two people exchanging coins in this way? (Any two people, not just Abby and Bernardo.)

sas4 2017-02-16 20:07:16
36

ilovemath04 2017-02-16 20:07:16
36

vishwathganesan 2017-02-16 20:07:16
4 choose 2 * 4 choose 2 = 36?

SomethingNeutral 2017-02-16 20:07:16
4 choose 2 times 4 choose 2 equals 36.

mathfun5 2017-02-16 20:07:16
36

Superwiz 2017-02-16 20:07:16
6* 6=36

mathfun5 2017-02-16 20:07:16
6*6

Tribefan 2017-02-16 20:07:16
4C2*4C2 = 36

geogirl08 2017-02-16 20:07:16
6*6

copeland 2017-02-16 20:07:18
There are 6 ways to choose a pair of people. Once we choose them, there are $\binom42=6$ ways to choose which turns the coins move one way and which turns the coins move the other. There are 36 total draws that have this form.

copeland 2017-02-16 20:07:34
What happens if Abby draws two greens and passes once to Bernardo and once to Carl? What can happen on the other two turns?

Superwiz 2017-02-16 20:08:16
C to A, B to A

GeronimoStilton 2017-02-16 20:08:16
Bernardo and Carl pass back.

tdeng 2017-02-16 20:08:16
Bernardo has to pass to Abby and Carl has to pass to Abby

First 2017-02-16 20:08:16
They each give a coin to Abby

vishwathganesan 2017-02-16 20:08:16
Abby draws 2 reds and carl and barnardo each draw a green

yrnsmurf 2017-02-16 20:08:16
C to A and B to A

ninjataco 2017-02-16 20:08:16
bernardo and carl pass back

oumarrbah_clf 2017-02-16 20:08:16
Abby has to draw two reds and bernando and carl each have to draw one green

copeland 2017-02-16 20:08:20

copeland 2017-02-16 20:08:20
On the other two turns, Bernardo and Carl have to pass once each and Abby needs to receive two coins. Therefore Bernardo and Carl must both pass to Abby on the other two turns.

copeland 2017-02-16 20:08:23

copeland 2017-02-16 20:08:27
How many total different draws have this L-shape? (Among any set of 3 players.)

mathfun5 2017-02-16 20:09:45
12*24 = 288

linqaszayi 2017-02-16 20:09:45
4*3*4*3*2=288

mathfun5 2017-02-16 20:09:45
288

GeronimoStilton 2017-02-16 20:09:45
4 ways to choose the first person times three ways to choose the person who doesn't participate times 24 ways to choose the order = 288$

copeland 2017-02-16 20:09:48
There are 4 people to choose as the node of the $L$ and then 3 choices for the lonely player. Then all four edges are distinct so there are $4!=24$ orders in which to perform the swaps. That gives all of $4\cdot3\cdot24=288$ ways for this type of swap to happen.

copeland 2017-02-16 20:09:50
OK, otherwise nobody draws two greens. Then what happens?

ninjataco 2017-02-16 20:10:54
everyone draws 1

vishwathganesan 2017-02-16 20:10:54
each person draws one green and one red

letsgomath 2017-02-16 20:10:54
they all draw one?

GeronimoStilton 2017-02-16 20:10:54
Everybody draws a green.

First 2017-02-16 20:10:54
Everyone draws a green

ilovemath04 2017-02-16 20:10:54
everyone draws one green

duck_master 2017-02-16 20:10:54
one green one red two white

copeland 2017-02-16 20:11:08
Since there are 4 greens drawn, everyone draws one green. Likewise for red. What are all the configurations where each person draws a single green and a single red?

copeland 2017-02-16 20:11:55
This is a question about graphs, not a question about numbers.

copeland 2017-02-16 20:12:10
Say Abby passes to Bernardo again. If Bernardo passes back to Abby then what?

copeland 2017-02-16 20:12:12

Derive_Foiler 2017-02-16 20:12:38
CD do a swap

geogirl08 2017-02-16 20:12:38
Carl passes to Debra and back

mathislife16 2017-02-16 20:12:38
the other two do a swap

yrnsmurf 2017-02-16 20:12:38
C and D have to trade

Reef334 2017-02-16 20:12:38
C and D pass to each other

copeland 2017-02-16 20:12:41
That means Carl and Debra have to swap as well:

copeland 2017-02-16 20:12:42

copeland 2017-02-16 20:12:49
Therefore we get a pair of swaps.

copeland 2017-02-16 20:12:53
However, if Abby passes to Bernardo and Bernardo then passes to Debra:

copeland 2017-02-16 20:12:54

copeland 2017-02-16 20:12:56
Then what?

SomethingNeutral 2017-02-16 20:13:34
cycle 4

vishwathganesan 2017-02-16 20:13:34
debra to carl and carl to abby

mathislife16 2017-02-16 20:13:34
D to C and C to A

Reef334 2017-02-16 20:13:34
D passes to C then to A

Derive_Foiler 2017-02-16 20:13:34
D to C, C to A?

Math-101 2017-02-16 20:13:34
A circle

GeronimoStilton 2017-02-16 20:13:34
Debra passes to Carl and Carl passes to Abby

MathTechFire 2017-02-16 20:13:34
A square cycle

Rushn 2017-02-16 20:13:34
D passes to C who passes to A

summitwei 2017-02-16 20:13:34
D->C and C->A

abishek99 2017-02-16 20:13:34
go around the square

copeland 2017-02-16 20:13:37
Then Bernardo is done playing. Now Carl still needs to pass to someone but the only one left to receive is Abby. So Carl passes to Abby and Debra must pass to Carl.

copeland 2017-02-16 20:13:40

copeland 2017-02-16 20:13:41
So, we have to have either a cycle or a pair of swaps.

copeland 2017-02-16 20:13:43
How many draws produce this type of configuration?

copeland 2017-02-16 20:13:43

linqaszayi 2017-02-16 20:14:54
3*4!=72

geogirl08 2017-02-16 20:14:54
3*24

vishwathganesan 2017-02-16 20:14:54
3*24 = 72

summitwei 2017-02-16 20:14:54
3*24

copeland 2017-02-16 20:14:57
There are 3 ways to pair off the players (pick Abby's partner). Once that's decided, the draws are all distinct so there are $4!=24$ ways to order the draws. There are $3\cdot24=72$ sets of draws in this case.

copeland 2017-02-16 20:14:58
What about if it's a cycle? How many ways can the players draw a cycle?

copeland 2017-02-16 20:14:59

summitwei 2017-02-16 20:16:22
6*24

linqaszayi 2017-02-16 20:16:22
3*2*4!=144

vishwathganesan 2017-02-16 20:16:22
no 144!

islander7 2017-02-16 20:16:22
6*4!

tdeng 2017-02-16 20:16:22
144

copeland 2017-02-16 20:16:24
There are 3 choices for who Abby passes to and then 2 choices for who passes to Abby. That gives 6 total ways to form a cycle. There are again $4!=24$ orders to draw the cycle, so there are $6\cdot24=144$ total sets draws in this case.

copeland 2017-02-16 20:16:25
What's the final answer?

duck_master 2017-02-16 20:18:47
540/12^4

Tribefan 2017-02-16 20:18:47
540/(12^4) = 5/192 (B)

duck_master 2017-02-16 20:18:47
B) 5/192

abishek99 2017-02-16 20:18:47
5/192 --> B

GeronimoStilton 2017-02-16 20:18:47
B

rt03 2017-02-16 20:18:47
5/192

copeland 2017-02-16 20:18:49
The total probability is

\begin{align*}

P

&=\frac{36+288+72+144}{12^4}\\

&=\frac{3+24+6+12}{12^3}\\

&=\frac{45}{12^3}\\

&=\frac{5}{4\cdot4\cdot12}\\

&=\boxed{\dfrac{5}{192}}.

\end{align*}

copeland 2017-02-16 20:18:50
The answer is (B).

copeland 2017-02-16 20:18:54
Ready to move on?

copeland 2017-02-16 20:19:05
I sense that was not everyone's favorite problem.

tfz4629 2017-02-16 20:19:11
Sure

mathchampion1 2017-02-16 20:19:11
yes!

HighQXMoney 2017-02-16 20:19:11
YES

HighQXMoney 2017-02-16 20:19:11
YES

YES

YES

copeland 2017-02-16 20:19:13
Great.

copeland 2017-02-16 20:19:27
I "solved" that problem at least a dozen times before I got the right answer.

copeland 2017-02-16 20:19:36
23. The graph of $y=f(x)$, where $f(x)$ is a polynomial of degree 3, contains points $A(2,4)$, $B(3,9)$, and $C(4,16)$. Lines $AB$, $AC$, and $BC$ intersect the graph again at points $D$, $E$, and $F$, respectively, and the sum of the $x$-coordinates of $D$, $E$, and $F$ is 24. What is $f(0)$?
$\textbf{(A)}\quad {-2} \qquad \qquad \textbf{(B)}\quad 0 \qquad\qquad \textbf{(C)}\quad 2 \qquad\qquad \textbf{(D)}\quad \dfrac{24}5 \qquad\qquad\textbf{(E)}\quad 8$

copeland 2017-02-16 20:19:39
OK, so what do you notice about those points?

ninjataco 2017-02-16 20:20:20
satisfy y = x^2

Derive_Foiler 2017-02-16 20:20:20
hmm . . . (x, x^2)

Reef334 2017-02-16 20:20:20
(x, x^2)

Tribefan 2017-02-16 20:20:20
Of the form (x,x^2)

SmartGuy101 2017-02-16 20:20:20
Squares

JP-GoVikes 2017-02-16 20:20:20
squares

SKundu13 2017-02-16 20:20:20
y = x^2

pie314159265 2017-02-16 20:20:20
they all lie on y=x^2

DaAsianPotatoe 2017-02-16 20:20:20
y = x^2

SmartGuy101 2017-02-16 20:20:20
y = x^2

copeland 2017-02-16 20:20:22
They're all of the form $(x,x^2)$.

copeland 2017-02-16 20:20:24
So how should we "encode" that in $f$?

acegikmoqsuwy2000 2017-02-16 20:21:04
$f(x)-x^2$ equals zero at those three values (and hence has them as roots)

yrnsmurf 2017-02-16 20:21:04
f(x)-x^2 has roots 2,3,4

rt03 2017-02-16 20:21:04
add a cubic polynomial with roots 2,3,4 to x^2

copeland 2017-02-16 20:21:07
What if we write $f(x)=x^2+\text{something}$ What do we know about the something?

ninjataco 2017-02-16 20:21:54
f(x) - x^2 = a(x-2)(x-3)(x-4)

pie314159265 2017-02-16 20:21:54
has roots of 2, 3, 4

Derive_Foiler 2017-02-16 20:21:54
it has zeros at 2, 3, 4

GeronimoStilton 2017-02-16 20:21:54
$f(x) = x^2 + a(x-2)(x-3)(x-4)$

abishek99 2017-02-16 20:21:54
0 for x = 2,3,4

Derive_Foiler 2017-02-16 20:21:54
it's a(x-2)(x-3)(x-4)

ilovemath04 2017-02-16 20:21:54
has roots 2,3, and 4

copeland 2017-02-16 20:21:56
The something has roots at 2, 3, and 4. It is also a cubic polynomial, so it is of the form $a(x-2)(x-3)(x-4)$.

copeland 2017-02-16 20:21:58
\[f(x)=x^2+a(x-2)(x-3)(x-4).\]

copeland 2017-02-16 20:22:00
OK, so about the line through $AB$? What line is that?

flyrain 2017-02-16 20:22:59
5x-6

Derive_Foiler 2017-02-16 20:22:59
that's y=5x-6

Tribefan 2017-02-16 20:22:59
5x-6

vishwathganesan 2017-02-16 20:22:59
y=5x-6

Reef334 2017-02-16 20:22:59
y=5x-6

HighQXMoney 2017-02-16 20:22:59
Line is $y=5x-6$

shakeNbake 2017-02-16 20:22:59
y=5x-6

thedoge 2017-02-16 20:22:59
$y = 5x-6$

copeland 2017-02-16 20:23:01
That line has slope $\dfrac{9-4}{3-2}=5$, so it is $y=5x-6$.

copeland 2017-02-16 20:23:04
How do we find the $x$-value of the other root?

geogirl08 2017-02-16 20:24:01
=

copeland 2017-02-16 20:24:08
One brilliant character. What's it mean?

geogirl08 2017-02-16 20:25:02
set them equal

brainiac1 2017-02-16 20:25:02
equate the expressions

Derive_Foiler 2017-02-16 20:25:02
set them equal?

islander7 2017-02-16 20:25:02
set them equal

copeland 2017-02-16 20:25:05
We set them equal:\[x^2+a(x-2)(x-3)(x-4)=5x-6.\]Bringing it all to the right-hand side gives\[x^2-5x+6+a(x-2)(x-3)(x-4)=0.\]

copeland 2017-02-16 20:25:06
See anything?

duck_master 2017-02-16 20:25:35
factoring!

abishek99 2017-02-16 20:25:35
factor first three terms

Antidifferentiation108 2017-02-16 20:25:35
can factor it

ninjataco 2017-02-16 20:25:35
factor x^2-5x+6 as (x-2)(x-3)

geogirl08 2017-02-16 20:25:39
(x-2)*(x-3)

PiAreSquared 2017-02-16 20:25:39
Quadratic

brainiac1 2017-02-16 20:25:39
the first few terms equal $(x-2)(x-3)$

copeland 2017-02-16 20:25:41
The "quadratic part" factors!

copeland 2017-02-16 20:25:42
\[(x-2)(x-3)+a(x-2)(x-3)(x-4)=0.\]

copeland 2017-02-16 20:25:44
By the way, why should we have guessed that the quadratic part would factor like this?

brainiac1 2017-02-16 20:26:34
2 and 3 are roots

ninjataco 2017-02-16 20:26:34
the cubic has roots 2 and 3

copeland 2017-02-16 20:26:36
The cubic part has roots 2 and 3. The whole thing also has roots 2 and 3. Therefore the quadratic part has roots 2 and 3. We also know its leading term is $x^2$, so that's the only thing it could have been.

copeland 2017-02-16 20:26:38
So what is the third root of this polynomial?

copeland 2017-02-16 20:27:04
Actually, just give me an equation that this root solves.

copeland 2017-02-16 20:27:08
That'll be just as helpful.

duck_master 2017-02-16 20:27:30
4-1/a

yrnsmurf 2017-02-16 20:27:30
4-1/a

Derive_Foiler 2017-02-16 20:27:30
1+a(x-4)?

ninjataco 2017-02-16 20:27:30
1+a(x-4)=0

duck_master 2017-02-16 20:27:30
1+a(x-4)=0

brainiac1 2017-02-16 20:27:30
4-1/a

tycooper 2017-02-16 20:27:30
a*(x-4)=-1

copeland 2017-02-16 20:27:32
Canceling $(x-2)(x-3)$ gives

copeland 2017-02-16 20:27:32
\[1+a(x_{AB}-4)=0.\]

copeland 2017-02-16 20:27:35
That tells us the $x$-value of the third root in terms of $a$. (Remember, our goal here is to use what we know about the roots to find $a$.)

copeland 2017-02-16 20:27:42
What similar equation will we get for the $x$-value of the third intersection of the line through $AC$?

copeland 2017-02-16 20:28:13
I'll give you a moment.

geogirl08 2017-02-16 20:29:17
1+a(xAC - 3) = 0

fractal161 2017-02-16 20:29:17
$1+a(x_{AC}-3)=0$

duck_master 2017-02-16 20:29:17
1+a(x-3)=0

kunsun 2017-02-16 20:29:17
1+a(x_AC-3) = 0

GeronimoStilton 2017-02-16 20:29:17
$1 + a(x_{AC} - 3) = 0$

yrnsmurf 2017-02-16 20:29:17
3-1/a

Tribefan 2017-02-16 20:29:17
1+a(x-3)=0

copeland 2017-02-16 20:29:19
Going the long way about it, we know that line has slope $\dfrac{16-4}{4-2}=6$, so must be $y=6x-8$.

copeland 2017-02-16 20:29:21
Substituting gives \[x^2+a(x-2)(x-3)(x-4)=6x-8\] so \[(x-2)(x-4)+a(x-2)(x-3)(x-4)=0,\] and

copeland 2017-02-16 20:29:23
\[1+a(x_{AC}-3)=0.\]

copeland 2017-02-16 20:29:26
Doing the same for $BC$ gives

copeland 2017-02-16 20:29:30
\[1+a(x_{BC}-2)=0.\]

copeland 2017-02-16 20:29:37
Now what?

GeronimoStilton 2017-02-16 20:30:48
Sum them!!!

Derive_Foiler 2017-02-16 20:30:48
use the fact that those x's add to 24? (Add the equations together)

fractal161 2017-02-16 20:30:48
Smash them all together?

geogirl08 2017-02-16 20:30:48
xAB + xAC + xBC = 24

yrnsmurf 2017-02-16 20:30:48
9-3/a=24

abishek99 2017-02-16 20:30:48
add them up

Reef334 2017-02-16 20:30:48
We know the sum of the x-coordinates, so solve for a

copeland 2017-02-16 20:30:55
When we add these we get \[3+a(x_{AB}+x_{AC}+x_{BC}-9)=0.\]

copeland 2017-02-16 20:31:04
We also know that $x_{AB}+x_{AC}+x_{BC}=24$.

copeland 2017-02-16 20:31:06
What is $a$?

letsgomath 2017-02-16 20:31:33
-1/5

DaAsianPotatoe 2017-02-16 20:31:33
-1/5

SomethingNeutral 2017-02-16 20:31:33
-1/5

ilovemath04 2017-02-16 20:31:33
-1/5

MathTechFire 2017-02-16 20:31:33
-1/5

copeland 2017-02-16 20:31:35
\[3+15a=0,\] so $a=-\dfrac15$.

copeland 2017-02-16 20:31:35
And what is $f(0)$?

ninjataco 2017-02-16 20:32:27
24/5

acegikmoqsuwy2000 2017-02-16 20:32:27
$\dfrac {24}5$

summitwei 2017-02-16 20:32:27
24/5

gradysocool 2017-02-16 20:32:27
24/5

rt03 2017-02-16 20:32:27
24/5

ilovemath04 2017-02-16 20:32:27
24/5

abishek99 2017-02-16 20:32:27
-1/5(-2)(-3)(-4) = 24/5 --> D

tdeng 2017-02-16 20:32:27
24/5

GeronimoStilton 2017-02-16 20:32:27
$D, \frac{24}{5}$

yrnsmurf 2017-02-16 20:32:27
24/5=D

Reef334 2017-02-16 20:32:27
-24a = (D) 24/5

Derive_Foiler 2017-02-16 20:32:27
24/5, D

geogirl08 2017-02-16 20:32:27
24/5

duck_master 2017-02-16 20:32:27
D) 24/5

anonymous0 2017-02-16 20:32:27
-24a

copeland 2017-02-16 20:32:29
$f(0)=0^2-\dfrac15(0-2)(0-3)(0-4)=\boxed{\dfrac{24}5}$. (D).

copeland 2017-02-16 20:32:42
Ooh, we're getting close. I'm getting all tingly.

copeland 2017-02-16 20:32:48
Shall we rock on?

SomethingNeutral 2017-02-16 20:33:12
yes!!!!!!

HighQXMoney 2017-02-16 20:33:12
I'm hungry, but sure.

copeland 2017-02-16 20:33:15
24. Quadrilateral $ABCD$ has right angles at $B$ and $C$, $\triangle ABC \sim \triangle BCD$, and $AB > BC$. There is a point $E$ in the interior of $ABCD$ such that $\triangle ABC \sim \triangle CEB$ and the area of $\triangle AED$ is $17$ times the area of $\triangle CEB$. What is $\dfrac{AB}{BC}$?
$\textbf{(A)}\quad 1 + \sqrt{2} \qquad \qquad \textbf{(B)}\quad 2 + \sqrt{2} \qquad\qquad \textbf{(C)}\quad \sqrt{17} \qquad\qquad \textbf{(D)}\quad 2 + \sqrt{5} \qquad\qquad\textbf{(E)}\quad 1 + 2\sqrt{3}$

summitwei 2017-02-16 20:33:35
diagram pl0x

GeronimoStilton 2017-02-16 20:33:35
Diagram!!!

CaptainGeo 2017-02-16 20:33:35
draw a diagram

yrnsmurf 2017-02-16 20:33:38
draw diagram

copeland 2017-02-16 20:33:44
OK, we start with a right trapezoid where the side lengths decrease by something that looks like a common ratio.

copeland 2017-02-16 20:33:45

copeland 2017-02-16 20:33:54
Let's drop some lengths in there. What should we label as our lengths?

gradysocool 2017-02-16 20:34:39
BC=1?

First 2017-02-16 20:34:39
WLOG assume $BC=1$

LaTeX_turtle 2017-02-16 20:34:39
BC=1?

copeland 2017-02-16 20:34:41
That's not bad. Then what should we call $AB$?

ninjataco 2017-02-16 20:35:06
x

LaTeX_turtle 2017-02-16 20:35:06
eggs

LaTeX_turtle 2017-02-16 20:35:06
AB=x

CaptainGeo 2017-02-16 20:35:06
x

Derive_Foiler 2017-02-16 20:35:06
AB=x. That's what we want to solve for

anonymous0 2017-02-16 20:35:06
x

gradysocool 2017-02-16 20:35:06
x of course

SomethingNeutral 2017-02-16 20:35:06
x

copeland 2017-02-16 20:35:07
OK, then if $AB=x$, what is $DC$?

Derive_Foiler 2017-02-16 20:35:27
DC=1/x

abishek99 2017-02-16 20:35:27
1/x

yrnsmurf 2017-02-16 20:35:27
1/x

gradysocool 2017-02-16 20:35:27
$\frac1x

GeronimoStilton 2017-02-16 20:35:27
$\frac{1}{x}$ by similar triangles

Tribefan 2017-02-16 20:35:27
1/x

islander7 2017-02-16 20:35:27
1/x

copeland 2017-02-16 20:35:28
Oh.

copeland 2017-02-16 20:35:30
That's terrible.

copeland 2017-02-16 20:35:33
Let's start over.

copeland 2017-02-16 20:35:37
New paper.

copeland 2017-02-16 20:35:41

copeland 2017-02-16 20:35:59
Give me better lengths, you fraction-wielding psychopaths.

abishek99 2017-02-16 20:36:24
CD = 1

Derive_Foiler 2017-02-16 20:36:24
Make BC = x, AB = x^2, and DC = 1?

SomethingNeutral 2017-02-16 20:36:24
dc = 1, bc = x, ab = x^2

geogirl08 2017-02-16 20:36:24
CD = 1

LaTeX_turtle 2017-02-16 20:36:24
DC=1, to get rid of remainders.

akaashp11 2017-02-16 20:36:24
$AB = x^2$

LaTeX_turtle 2017-02-16 20:36:24
BC=x

LaTeX_turtle 2017-02-16 20:36:24
AB=x^2

gradysocool 2017-02-16 20:36:31
why is that terrible? Fractions are nicer than remembering to divide AB by BC at the end. This way it's just "Find x"

copeland 2017-02-16 20:36:41
If $AB=x^2$ and $BC=x$ then what are we finding?

SomethingNeutral 2017-02-16 20:36:54
x

geogirl08 2017-02-16 20:36:54
x

CaptainGeo 2017-02-16 20:36:54
x

duck_master 2017-02-16 20:36:54
x lol

anonymous0 2017-02-16 20:36:54
x

HighQXMoney 2017-02-16 20:36:54
$x$

copeland 2017-02-16 20:36:55
We're still finding $\dfrac{AB}{BC}=x$.

copeland 2017-02-16 20:37:06
By $\triangle ABC\sim\triangle BCD$, we know that $\dfrac{AB}{BC}=\dfrac{BC}{CD}$. In order to avoid (at least for now) having denominators, let's let $CD=1$ and $BC=x$.

copeland 2017-02-16 20:37:15
Let's drop $E$ in there as well. What do we know that will help in placing $E$?

tdeng 2017-02-16 20:38:00
ABC is similar to CEB

Derive_Foiler 2017-02-16 20:38:00
Right angles!

duck_master 2017-02-16 20:38:00
CEB and ABC are similar triangles

summitwei 2017-02-16 20:38:00
on the semicircle with diameter BC

LaTeX_turtle 2017-02-16 20:38:00
ABC~CEB

GeronimoStilton 2017-02-16 20:38:00
$\triangle ABC \sim \triangle CEB$

tdeng 2017-02-16 20:38:00
It is on the semicircle with BC as its diameter

LaTeX_turtle 2017-02-16 20:38:00
CEB has right angle at E.

copeland 2017-02-16 20:38:05
Great. Since $\triangle ABC\sim\triangle CEB$, we know that $\angle CEB$ is a right angle. We also know that $\dfrac{EC}{BE}=x>1$.

copeland 2017-02-16 20:38:06
We'll put it right about here:

copeland 2017-02-16 20:38:07

copeland 2017-02-16 20:38:21
What other ratio do we know?

geogirl08 2017-02-16 20:38:57
AED/CEB = 17

LaTeX_turtle 2017-02-16 20:38:57
[AED] = 17[CEB]

duck_master 2017-02-16 20:38:57
the area of AED compared to CEB

tdeng 2017-02-16 20:38:57
$\frac{[\triangle AED]}{[\triangle CEB]}$

copeland 2017-02-16 20:39:05
Too soon. We only have a few lengths.

copeland 2017-02-16 20:39:19
That does tell us that our diagram is awful, but let's forget about that for now.

ninjataco 2017-02-16 20:39:56
CE/BE = x

kpatel2000 2017-02-16 20:39:56
AB/BC = CE/BE

cyborg108 2017-02-16 20:39:56
EC/EB = x

brainiac1 2017-02-16 20:39:56
we can figure out the side lengths of CEB

yrnsmurf 2017-02-16 20:39:56
find BE and CE

copeland 2017-02-16 20:40:00
Since $\triangle ABC\sim\triangle CEB$, we know $\dfrac{CE}{EB}=\dfrac{AB}{BC}=x$.

copeland 2017-02-16 20:40:18
We actually know exactly what $EB$ and $EC$ are now. We'll get there in a moment.

copeland 2017-02-16 20:40:22
I want to set up another variable here (it depends on $x$, but the diagram will be clearer with a second variable). Let's let $EB=t$.

copeland 2017-02-16 20:40:23
That makes $CE=xt$.

copeland 2017-02-16 20:40:24

copeland 2017-02-16 20:40:31
What else should we add to the diagram now so that we can better study $\triangle BEC$?

math9990 2017-02-16 20:41:25
altitude

fractal161 2017-02-16 20:41:25
Altitude from $E$ to $BC$?

copeland 2017-02-16 20:41:28
If we drop the altitude from $E$ to $BC$ we get more similar triangles.

copeland 2017-02-16 20:41:29

copeland 2017-02-16 20:41:30
And what is $EP$?

summitwei 2017-02-16 20:42:37
t^2

ninjataco 2017-02-16 20:42:37
t^2

brainiac1 2017-02-16 20:42:37
t^2

letsgomath 2017-02-16 20:42:37
T^2

geogirl08 2017-02-16 20:42:37
t^2

cyborg108 2017-02-16 20:42:37
t^2

letsgomath 2017-02-16 20:42:37
t^2

abishek99 2017-02-16 20:42:37
t^2

rt03 2017-02-16 20:42:37
t^2

DarkPikachu 2017-02-16 20:42:37
t^2

copeland 2017-02-16 20:42:40
We can compute the area of $\triangle CEB$ in two ways. first, since there's a right angle at $E$, we get \[[\triangle CEB]=\dfrac12\cdot t\cdot xt=\dfrac{xt^2}2.\] However, $EP$ is the altitude to $BC$, so \[[\triangle CEB]=\dfrac12\cdot x\cdot EP,\] so $EP=t^2$.

copeland 2017-02-16 20:42:57

copeland 2017-02-16 20:43:04
We have two variables in our diagram but they're dependent, so let's figure out what $t$ is. What equation does $t$ solve?

cyborg108 2017-02-16 20:43:56
Pythag on BEC

GeronimoStilton 2017-02-16 20:43:56
$t^2 + x^2t^2 = x^2$

LaTeX_turtle 2017-02-16 20:43:56
Pythagorean?

yrnsmurf 2017-02-16 20:43:56
$t^2(1+x^2)=x^2$

duck_master 2017-02-16 20:43:56
$t^2(1+x^2)=x^2$ by pythag and stuff

CaptainGeo 2017-02-16 20:43:56
t^2+x^2*t^2=x^2

abishek99 2017-02-16 20:43:56
pythaogrean theorem on BEC

Derive_Foiler 2017-02-16 20:43:56
t^2+x^2t^2=x^2 by pythag

LaTeX_turtle 2017-02-16 20:43:56
t^2+(xt)^2=x^2

kpatel2000 2017-02-16 20:43:56
pythagorean theorem on EBC

geogirl08 2017-02-16 20:43:56
t^2 + x^2t^2 = x^2

copeland 2017-02-16 20:43:59
The Pythagorean Theorem on $\triangle BEC$ gives \[t^2+x^2t^2=x^2,\]so

copeland 2017-02-16 20:44:01
\[t^2=\frac{x^2}{1+x^2}.\]

copeland 2017-02-16 20:44:04
That's cute.

copeland 2017-02-16 20:44:14
OK, so the area of $\triangle CEB$ is $\dfrac{xt^2}2$. How can we compute the area of $\triangle ABE$?

Royalreter1 2017-02-16 20:45:17
AB*BP/2

CaptainGeo 2017-02-16 20:45:17
figuring out BP

copeland 2017-02-16 20:45:36
Let's find $BP$ since that is the altitude to $E$.

GeronimoStilton 2017-02-16 20:45:39
Why do we need to know the area of $\triangle ABE$?

copeland 2017-02-16 20:45:44
Oh, that's a great question.

copeland 2017-02-16 20:45:48
What should our strategy be here?

copeland 2017-02-16 20:45:55
We want to find the ratio of some sides.

copeland 2017-02-16 20:46:02
We know something about the areas of two of the triangles.

copeland 2017-02-16 20:46:19
It feels like we need to invoke areas. But why would we want to think about the areas of those other two triangles?

BillZhuo 2017-02-16 20:47:11
just find the area of AED by subtracting pieces out of the trapezoid

GeronimoStilton 2017-02-16 20:47:11
We want to find the area of $\triangle ADE$ relative to the area of $\triangle BEC$.

cyborg108 2017-02-16 20:47:11
For a final equation for the trapezoid?

abishek99 2017-02-16 20:47:11
this helps us find the area of AED

eswa2000 2017-02-16 20:47:11
subtract the 3 triangles' areas from the quadrilateral to get [AED]

copeland 2017-02-16 20:47:13
We can easily compute the area of the full trapezoid and it depends only on $x$, so maybe we'll get two expressions for that area and that will take us home.

copeland 2017-02-16 20:47:26
Now we're finding $BP$. What similarity gives us $BP$?

gradysocool 2017-02-16 20:48:53
BEP~BEC

Derive_Foiler 2017-02-16 20:48:53
use triangle BEP and BCE or CBA

LaTeX_turtle 2017-02-16 20:48:53
BPE~BEC

copeland 2017-02-16 20:49:03
There are a billion similarities now: $\triangle BPE\sim\triangle BEC\sim\triangle CBA$, etc. And what is $BP$?

GeronimoStilton 2017-02-16 20:49:59
$\frac{t^2}{x}$

gradysocool 2017-02-16 20:49:59
$\frac{t^2}{x}$

Reef334 2017-02-16 20:49:59
t^2/x

LaTeX_turtle 2017-02-16 20:49:59
Gives $\frac{BP}{t} = \frac{t}{x}$

copeland 2017-02-16 20:50:01
Since $\dfrac{BP}{PE}=\dfrac1x$ and $PE=t^2$, we get $BP=\dfrac{t^2}x$.

copeland 2017-02-16 20:50:02
And what is $PC$?

BXU65 2017-02-16 20:52:02
(x^2-t^2)/x

yrnsmurf 2017-02-16 20:52:02
xt^2

Skittlesftw 2017-02-16 20:52:02
$ xt^2 $

GeronimoStilton 2017-02-16 20:52:02
$x - \frac{t^2}{x}$

letsgomath 2017-02-16 20:52:02
(x^2-t^2)/x

mathfever 2017-02-16 20:52:02
xt^2

Smoothfang 2017-02-16 20:52:02
xt^2

mathman3880 2017-02-16 20:52:02
xt^2

copeland 2017-02-16 20:52:20
We can subtract to get \[PC=BC-BP=x-\frac{t^2}x=\frac{x^2-t^2}{x}=\frac{x^2t^2}{x}=xt^2.\]

copeland 2017-02-16 20:52:24
We can also use similarity to get $\dfrac{PC}{t^2}=x$, so $PC=xt^2$.

copeland 2017-02-16 20:52:33

copeland 2017-02-16 20:52:37
OK, so what is the area of $\triangle ABE$ now?

Derive_Foiler 2017-02-16 20:53:18
t^2x/2

ninjataco 2017-02-16 20:53:18
xt^2/2

Smoothfang 2017-02-16 20:53:18
xt^2/2

geogirl08 2017-02-16 20:53:18
xt^2/2

Skittlesftw 2017-02-16 20:53:18
$ \frac{xt^2}{2} $

BXU65 2017-02-16 20:53:18
xt^2/2

copeland 2017-02-16 20:53:21
The base is $AB=x^2$ and the height is $\dfrac{t^2}x$, so the area is $\dfrac{xt^2}2$.

copeland 2017-02-16 20:53:22
What other area do we get?

mathman3880 2017-02-16 20:54:41
[BEC]=[ECD] = xt^2/2

duck_master 2017-02-16 20:54:41
CEB

tarzanjunior 2017-02-16 20:54:41
ECD

MP2016p42 2017-02-16 20:54:41
EDC

Derive_Foiler 2017-02-16 20:54:41
CED, which is xt^2/2 hmmmm . . .

BXU65 2017-02-16 20:54:41
Triangle DEC

summitwei 2017-02-16 20:54:41
[CDE]=xt^2/2

MP2016p42 2017-02-16 20:54:41
EDC area = x^2t/2

cyborg108 2017-02-16 20:54:41
EBC = x*t^2 / 2

copeland 2017-02-16 20:54:44
Triangle $DCE$ has base $CD=1$ and height $xt^2$, so has area $\dfrac{xt^2}2$.

copeland 2017-02-16 20:54:48
Huh, that looks familiar. Do we know the area of $\triangle ADE$?

BXU65 2017-02-16 20:56:38
17xt^2/2

letsgomath 2017-02-16 20:56:38
17 times the area of CEB

Skittlesftw 2017-02-16 20:56:38
$ 17 \cdot triangle CEB $

copeland 2017-02-16 20:56:40
It's given to us that it has area $\dfrac{17xt^2}2$.

copeland 2017-02-16 20:56:41
Let's put those areas in there.

copeland 2017-02-16 20:56:42
I'm going to write $\dfrac{xt^2}2$ as $a$:

copeland 2017-02-16 20:56:47

copeland 2017-02-16 20:56:52
The total area is $20a=10xt^2$. Do we have another expression for the total area?

duck_master 2017-02-16 20:57:51
$\frac{(x^2+1)x}{2}$

ninjataco 2017-02-16 20:57:51
x(1+x^2)/2

Reef334 2017-02-16 20:57:51
x(x^2+1)/2

SomethingNeutral 2017-02-16 20:57:51
(1+x^2)x/2

summitwei 2017-02-16 20:57:51
(1+x^2)*x/2

GeronimoStilton 2017-02-16 20:57:51
$\frac{x^3 + x}{2}$

BXU65 2017-02-16 20:57:51
(x^3+x)/2

geogirl08 2017-02-16 20:57:51
(x^2+1)x/2

CrystalEye 2017-02-16 20:57:51
(1+x^2)x/2

brainiac1 2017-02-16 20:57:51
(1+x^2)x/2

Tribefan 2017-02-16 20:57:51
(x^2+1)/2x

copeland 2017-02-16 20:57:53
The region is a right trapezoid so it has area $\dfrac{1+x^2}2\cdot x$.

copeland 2017-02-16 20:57:57
Equating these gives \[10xt^2=\frac{x(1+x^2)}2,\]and substituting $t^2=\dfrac{x^2}{1+x^2}$ and simplifying, we get

copeland 2017-02-16 20:58:04
\[20x^2=x^4+2x^2+1.\]

brainiac1 2017-02-16 20:58:20
it's a quadratic!

copeland 2017-02-16 20:58:21
Go us!

copeland 2017-02-16 20:59:47
What's $x^2$?

duck_master 2017-02-16 21:00:41
$9+\sqrt{80}$, $9-\sqrt{80}$

BXU65 2017-02-16 21:00:41
9+4sqrt(5)

celestialphoenix3768 2017-02-16 21:00:41
quadratic formula

Skittlesftw 2017-02-16 21:00:41
$ 9+4\sqrt5 $????

BXU65 2017-02-16 21:00:41
9 + 4sqrt(5)

letsgomath 2017-02-16 21:00:41
9 plus/minus 4sqrt5

Reef334 2017-02-16 21:00:41
9+4sqrt(5)

SomethingNeutral 2017-02-16 21:00:41
9$\pm$4sqrt5

copeland 2017-02-16 21:00:51
Wow, there are two solutions.

copeland 2017-02-16 21:01:03
Interestingly, they are reciprocals of one another.

copeland 2017-02-16 21:01:10
What happens if we choose the little one?

cyborg108 2017-02-16 21:02:11
AB < BC

cyborg108 2017-02-16 21:02:11
contradiction with the givens

summitwei 2017-02-16 21:02:11
then x<1 which is obviously wrong

GeronimoStilton 2017-02-16 21:02:11
Then $x < 1$

BXU65 2017-02-16 21:02:11
x would be 2- sqrt(5)

howie2000 2017-02-16 21:02:11
x<1 so AB<BC

yrnsmurf 2017-02-16 21:02:11
you get the smaller side

copeland 2017-02-16 21:02:13
If we choose the little one, then $AB<1$ and the diagram reflects. Cool. The little value is wrong, but it tells us about a symmetry to the problem that we knew about (and that the problem statement works to break).

copeland 2017-02-16 21:02:22
By the quadratic formula, we get $x^2=\dfrac{18\pm\sqrt{18^2-4}}2=9\pm4\sqrt5$.

copeland 2017-02-16 21:02:35
We need $x>1$ so $x^2=9\pm4\sqrt5$.

copeland 2017-02-16 21:02:37
So what must the answer be?

ninjataco 2017-02-16 21:03:17
D

BXU65 2017-02-16 21:03:17
2 + sqrt(5) or D

tdeng 2017-02-16 21:03:17
D

GeronimoStilton 2017-02-16 21:03:17
$2+\sqrt{5}$, D

mathman3880 2017-02-16 21:03:17
2+sqrt5

celestialphoenix3768 2017-02-16 21:03:17
D

nukelauncher 2017-02-16 21:03:17
2+sqrt5 so D

MathTechFire 2017-02-16 21:03:17
D?

yrnsmurf 2017-02-16 21:03:17
2+sqrt5=

cyborg108 2017-02-16 21:03:17
2+sqrt(5)

gradysocool 2017-02-16 21:03:19
well, it's gotta have a $\sqrt{5}$ so...

copeland 2017-02-16 21:03:22
The only answer choice that is friends with $\sqrt5$ is (D), $\boxed{2+\sqrt5}$.

copeland 2017-02-16 21:03:31
Indeed, if we square this value we get \[4+5+2\cdot2\sqrt5=9+4\sqrt5,\] so D is indeed the answer.

copeland 2017-02-16 21:03:41
If you're curious, $\dfrac{BC}{BP}=1+x^2\approx19$, so $E$ is really close to $\overline{AB}$ and diagram to scale looks like this:

copeland 2017-02-16 21:03:48

GeronimoStilton 2017-02-16 21:04:16
That's a harder diagram to work with.

copeland 2017-02-16 21:04:25
It's hard to believe those triangles have the same area, too.

copeland 2017-02-16 21:04:39
You know those experiments with the little kids and the water? I feel like that.

copeland 2017-02-16 21:04:43
(I'm the kid in the analogy.)

GeronimoStilton 2017-02-16 21:04:51
Final question!!!

gradysocool 2017-02-16 21:04:55
Why is [ABE]=[BCE]=[CDE] ? Is there an easy way to get this?

copeland 2017-02-16 21:04:59
Oh, yeah. I don't now.

copeland 2017-02-16 21:05:04
It's cool, though.

copeland 2017-02-16 21:05:14
Also, you're saying what we did wasn't easy?

copeland 2017-02-16 21:05:43
Here we go.

copeland 2017-02-16 21:05:46
25. A set of $n$ people participate in an online video basketball tournament. Each person may be a member of any number of 5-player teams, but no two teams may have exactly the same 5 members. The site statistics show a curious fact: The average, over all subsets of size 9 of the set of $n$ participants, of the number of complete teams whose members are among those 9 people is equal to the reciprocal of the average, over all subsets of size 8 of the set of $n$ participants, of the number of complete teams whose members are among those 8 people. How many values $n$, $9\leq n\leq 2017$, can be the number of participants?
$\textbf{(A)}\quad477 \qquad \qquad \textbf{(B)}\quad 482 \qquad\qquad \textbf{(C)}\quad 487 \qquad\qquad \textbf{(D)}\quad 557 \qquad\qquad\textbf{(E)}\quad 562$

copeland 2017-02-16 21:06:02
Alright, so that's a bucketful of words. Great. Usually I budget extra time to read the geometry problems.

copeland 2017-02-16 21:06:31
Let's try to get a grip on this problem. We have a bunch of teams. We want to compute some average of something. We do that twice. Let's figure out what that is.

copeland 2017-02-16 21:06:37
An average is just a sum of something divided by the number of things we're summing over.

copeland 2017-02-16 21:06:39
Let's unpack this statement:

"The average, over all subsets of size 9 of the set of $n$ participants, of the number of complete teams whose members are among those 9 people"

copeland 2017-02-16 21:06:57
So we're taking an average over all the 9-element subsets. How many of those are there?

BXU65 2017-02-16 21:08:06
N choose 9

First 2017-02-16 21:08:06
$n C 9$

geogirl08 2017-02-16 21:08:06
nC9

Jfault 2017-02-16 21:08:06
$\binom{n}{9}$?

nukelauncher 2017-02-16 21:08:06
nchoose9

ninjataco 2017-02-16 21:08:06
nC9

duck_master 2017-02-16 21:08:06
(n choose 9)

summitwei 2017-02-16 21:08:06
nC9

CrystalEye 2017-02-16 21:08:06
nC9

copeland 2017-02-16 21:08:09
There are $\dbinom n9$ subsets of size 9. That's the denominator.

copeland 2017-02-16 21:08:14
Now we want to average the number of complete teams that each of these subsets contains. That means, the numerator is the sum of all these numbers as the size-9 subset varies.

copeland 2017-02-16 21:08:42
How many subsets is a team contained in?

copeland 2017-02-16 21:09:54
For example, let $n=100$. You, me, your aunt Tracy, Ned from around the block, and some dude dressed like a banana form a team. How many subsets have the five of us in them?

geogirl08 2017-02-16 21:10:45
n-5C4?

GeronimoStilton 2017-02-16 21:10:45
$\binom{n-5}{4}$

az1326 2017-02-16 21:10:45
((n-5)choose 4)

brainiac1 2017-02-16 21:10:45
n-5 choose 4

geogirl08 2017-02-16 21:10:45
95C4

GeronimoStilton 2017-02-16 21:10:45
$\binom{95}{4}$

Reef334 2017-02-16 21:10:45
95 C 4

abishek99 2017-02-16 21:10:45
95C4

az1326 2017-02-16 21:10:45
(95 choose 4)

gradysocool 2017-02-16 21:10:45
95C4

First 2017-02-16 21:10:45
95C5?

Jfault 2017-02-16 21:10:45
$\binom{95}{4}$

CrystalEye 2017-02-16 21:10:45
(n-5)C4

copeland 2017-02-16 21:10:47
We just have to pick 4 more people to make a subset. There are 95 people to choose from.

copeland 2017-02-16 21:11:02
Each team contributes 1 to the value for each subset it is contained in. A team of 5 can be filled out to a subset of size 9 by choosing $\dbinom{n-5}4$ other members. Therefore each team contributes exactly $\dbinom{n-5}4$ to the numerator.

copeland 2017-02-16 21:11:10
If there are $t$ teams, then what is the average?

geogirl08 2017-02-16 21:13:10
t(n-5C4)/(nC9)

BXU65 2017-02-16 21:13:10
t * n-5 choose 4 / n choose 5

GeronimoStilton 2017-02-16 21:13:10
$\frac{t\binom{n-5}{4}}{\binom{n}{9}}$

BXU65 2017-02-16 21:13:10
t * n-5 choose 4 / n choose 9

az1326 2017-02-16 21:13:10
t ((n-5) choose 4) / (n choose 9)

duck_master 2017-02-16 21:13:10
$t \frac{\binom{n-5}{4}}{\binom{n}{9}}$

ninjataco 2017-02-16 21:13:10
t*(n-5)C4 / nC9

copeland 2017-02-16 21:13:13
The average blah blah size 9 blah blah blah is then\[\frac{t\dbinom{n-5}{4}}{\dbinom n9}.\]

ilikepie2003 2017-02-16 21:13:17
how do you know that the dude isn't a real banana

copeland 2017-02-16 21:13:21
He may be. Either way, he's an online video basketball phenom.

copeland 2017-02-16 21:13:43
Oh, by the way, you have 15 seconds to count the number of times "of" appears in that one long sentence. Go!

Jfault 2017-02-16 21:14:41
7?

geogirl08 2017-02-16 21:14:41
8

letsgomath 2017-02-16 21:14:41
7

leonlzg 2017-02-16 21:14:41
11?

Derive_Foiler 2017-02-16 21:14:41
11?

GeronimoStilton 2017-02-16 21:14:41
$13$

Jfault 2017-02-16 21:14:41
\boxed{9}!

copeland 2017-02-16 21:14:47
Yeah. Who knows?!!?

Reef334 2017-02-16 21:14:54
15?

duck_master 2017-02-16 21:14:54
12

copeland 2017-02-16 21:14:57
We think it's 11.

copeland 2017-02-16 21:15:04
But counting them is nondeterministic.

copeland 2017-02-16 21:15:09
What is the reciprocal of the average, over all subsets of size 8 of the set of $n$ participants, of the number of complete teams whose members are among those 8 people?

shakeNbake 2017-02-16 21:16:49
16

ProGameXD 2017-02-16 21:16:49
21

amburger66 2017-02-16 21:16:49
eleven!!11!11

BXU65 2017-02-16 21:16:49
7?

mathwiz0803 2017-02-16 21:16:49
18

shakeNbake 2017-02-16 21:16:49
oh its 15 just ctrl F

Haphazard 2017-02-16 21:16:49
there are 16 'of'

BXU65 2017-02-16 21:16:53
t * (n-5 choose 3) / n choose 8

gradysocool 2017-02-16 21:16:53
$\frac{\binom{n}{8}}{t\binom{n-5}{3}}$

duck_master 2017-02-16 21:16:53
$\frac{\binom{n}{8}}{t \binom{n-5}{3}}$

GeronimoStilton 2017-02-16 21:16:53
$$\frac{\binom{n}{8}}{t\binom{n-5}{3}}.$$

BXU65 2017-02-16 21:16:53
n choose 8 / t * n - 5 choose 3

copeland 2017-02-16 21:16:55
To fill out a group of 8 beginning with a team of size 5 we need to pick 3 more members from the remaining $n-5$.

copeland 2017-02-16 21:16:56
The reciprocal of blah average blah blah 8 blah is \[\frac{\dbinom n8}{t\dbinom{n-5}{3}}.\]

copeland 2017-02-16 21:17:01
Setting them equal gives

copeland 2017-02-16 21:17:06
\[\frac{t\dbinom{n-5}{4}}{\dbinom n9}=\frac{\dbinom n8}{t\dbinom{n-5}{3}}.\]

copeland 2017-02-16 21:17:17
Now what?

duck_master 2017-02-16 21:17:57
cross multiply

BXU65 2017-02-16 21:17:57
cross multiply?

Derive_Foiler 2017-02-16 21:17:57
expand . . . ?

celestialphoenix3768 2017-02-16 21:17:57
bash bash bash

SaltySnail2 2017-02-16 21:17:57
cross multiply?

EulerMacaroni 2017-02-16 21:17:57
cancel like crazy

ninjataco 2017-02-16 21:17:57
cross multiply

brainiac1 2017-02-16 21:17:57
now we expand and cancel all the factorials and pray it looks pretty

copeland 2017-02-16 21:18:05
Maybe I've got something wrong with me, but I kinda feel like expanding those binomial coefficients.

copeland 2017-02-16 21:18:15
Actually, this does make sense. When we see a quotient of two similar binomial coefficients, we can get a lot of cancellation by expanding both of them. So let's expand all our binomial coefficients. What happens to the left when we expand?

copeland 2017-02-16 21:18:34
(We'll cross-multiply in a moment. I love cross-multiplication.)

BXU65 2017-02-16 21:20:54
we get 9 * 8 * 7 * 6 * 5 * t / n * n-1 * n-2 * n-3 * n-4

duck_master 2017-02-16 21:20:54
t$\frac{9!(n-5)!}{4!n!}$

brainiac1 2017-02-16 21:20:54
$\frac{15120}{n(n-1)(n-2)(n-3)(n-4)}$

copeland 2017-02-16 21:20:59
On the left we get \[\frac{t\dbinom{n-5}{4}}{\dbinom n9}=\frac{t(n-5)!9!(n-9)!}{n!4!(n-9)!}=\frac{t(n-5)!9!}{n!4!}.\]

copeland 2017-02-16 21:21:02
On the right we get \[\frac{n!3!}{t(n-5)!8!}.\]

copeland 2017-02-16 21:21:03
Now what are we looking for?

GeronimoStilton 2017-02-16 21:21:38
The possible values of $n$

Derive_Foiler 2017-02-16 21:21:38
all possible n?

xayy 2017-02-16 21:21:38
n

copeland 2017-02-16 21:21:39
Which $n$? What makes $n$ special?

copeland 2017-02-16 21:23:53
We have\[\frac{t(n-5)!9!}{n!4!}=\frac{n!3!}{t(n-5)!8!}.\]Isolating $t^2$ gives\[t^2=\frac{n!^23!4!}{(n-5)!^28!9!}.\]

copeland 2017-02-16 21:23:56
So what makes $n$ special?

linqaszayi 2017-02-16 21:24:47
t is an integer so n must satisfy some division conditions

GeronimoStilton 2017-02-16 21:24:47
There exists an integer value of $t$

yrnsmurf 2017-02-16 21:24:47
the ones that make t an integer

copeland 2017-02-16 21:24:54
We're looking for all the values of $n$ such that $t$ is an integer.

copeland 2017-02-16 21:25:04
Now let's just go crazy and totally expand that. It really isn't that many terms after we cancel:

copeland 2017-02-16 21:25:05
\begin{align*}

t^2

&=\frac{\left[n(n-1)(n-2)(n-3)(n-4)\right]^2}{9\cdot8^2\cdot7^2\cdot6^2\cdot5^2\cdot4}\\

&=\left(\frac{n(n-1)(n-2)(n-3)(n-4)}{8\cdot7\cdot6\cdot5\cdot3\cdot2}\right)^2.\end{align*}

copeland 2017-02-16 21:25:12
Oh my. That's a perfect square. So we're looking for every value of $n$ that makes \[t=\frac{n(n-1)(n-2)(n-3)(n-4)}{8\cdot7\cdot6\cdot5\cdot3\cdot2}\]an integer.

copeland 2017-02-16 21:25:55
Guess what? We don't know that the right side is an integer. That's what we're exploring! Let's start simple. What do we know the numerator is divisible by?

nukelauncher 2017-02-16 21:26:39
2, 3, 4, and 5

daniellionyang 2017-02-16 21:26:39
5

tdeng 2017-02-16 21:26:39
5*3*2

brainiac1 2017-02-16 21:26:39
it's divisible by 3, 4, and 5

EulerMacaroni 2017-02-16 21:26:39
5

copeland 2017-02-16 21:26:41
We know for sure that the numerator is divisible by 5 since it's the product of 5 consecutive integers.

copeland 2017-02-16 21:26:42
Is it divisible by 7?

ninjataco 2017-02-16 21:26:58
maybe

summitwei 2017-02-16 21:26:58
not necessarily

BXU65 2017-02-16 21:26:58
not necessarily

copeland 2017-02-16 21:27:00
Not necessarily. For what $n$ is this divisible by 7?

EulerMacaroni 2017-02-16 21:27:44
slso $7$ is $n\equiv 0,1,2,3,4 \pmod 7$

nukelauncher 2017-02-16 21:27:44
n = 0, 1, 2, 3, 4 mod 7

brainiac1 2017-02-16 21:27:44
n = 0, 1, 2, 3, 4 mod 7

xayy 2017-02-16 21:27:44
0,1,2,3,4 mod 7

duck_master 2017-02-16 21:27:44
n is 0,1,2,3,4 mod 7

summitwei 2017-02-16 21:27:44
n = 4, 3, 2, 1, 0 mod 7

copeland 2017-02-16 21:27:47
The numerator is a multiple if 7 when $n\equiv 0,1,2,3,4\pmod7$.

copeland 2017-02-16 21:27:53
We probably want to factor that denominator better. Since we know it's always divisible by 5, let's get that out of there.

copeland 2017-02-16 21:27:55
We want to find when\[5t=\frac{n(n-1)(n-2)(n-3)(n-4)}{7\cdot3^2\cdot2^5}\]is an integer.

copeland 2017-02-16 21:28:09
The numerator is a multiple if 7 when $n\equiv 0,1,2,3,4\pmod7$.

copeland 2017-02-16 21:28:11
Now before we go too far down this path, what tool are we going to need to invoke on this problem?

Derive_Foiler 2017-02-16 21:28:48
CRT seems nice and bashy

linqaszayi 2017-02-16 21:28:48
CRT

brainiac1 2017-02-16 21:28:48
crt?

duck_master 2017-02-16 21:28:48
Chinese Remainder Theorem!

nukelauncher 2017-02-16 21:28:48
Chinese Remainder Theorem

summitwei 2017-02-16 21:28:48
chinese remainder theorem

GeronimoStilton 2017-02-16 21:28:48
CRT?

copeland 2017-02-16 21:28:50
A number is divisible by the product of coprime numbers if and only if it is divisible by each of the factors. That's a trivial case of the Chinese Remainder Theorem. It also follows directly from the Fundamental Theorem of Arithmetic. It's also just painfully obvious.

copeland 2017-02-16 21:28:58
So what's the product of those things in the denominator?

brainiac1 2017-02-16 21:30:03
2016!

tdeng 2017-02-16 21:30:03
2016

Slacker 2017-02-16 21:30:03
2016

brainiac1 2017-02-16 21:30:03
2016

abishek99 2017-02-16 21:30:03
2016

duck_master 2017-02-16 21:30:03
2016 which was last year

GeronimoStilton 2017-02-16 21:30:03
2016!

BXU65 2017-02-16 21:30:03
2016!

copeland 2017-02-16 21:30:05
\begin{align*}7\cdot3^2\cdot2^5

&=7\cdot9\cdot2^5\\

&=63\cdot2^5\\

&=(2^6-1)2^5\\

&=2^{11}-2^5\\

&=2048-32\\

&=2016.

\end{align*}

copeland 2017-02-16 21:30:08
Wow, unbelievable. Anyone feel like we're on the right track here?

GeronimoStilton 2017-02-16 21:30:27
Is that intentional?

tdeng 2017-02-16 21:30:27
Ye.

mssmath 2017-02-16 21:30:27
yup

BXU65 2017-02-16 21:30:27
This is nostalgic

Jfault 2017-02-16 21:30:27
no, we got 2016 instead of 2017 /s

copeland 2017-02-16 21:30:28
Yeah, retro.

copeland 2017-02-16 21:30:33
We can figure out the exact fraction of the numbers from 2 to 2017 that make this product divisible by all of these factors!

copeland 2017-02-16 21:30:35
Now that we're reinvigorated, when is the numerator divisible by 9?

Jfault 2017-02-16 21:31:57
whenever there are two factors of 3 in it

copeland 2017-02-16 21:32:03
When's that?

yrnsmurf 2017-02-16 21:32:58
when n is not 5 or 8 mod 9

summitwei 2017-02-16 21:32:58
n is not 5 or 8 mod 9

brainiac1 2017-02-16 21:32:58
n = 0, 1, 2, 3, 4, 6, 7 mod 9

BXU65 2017-02-16 21:32:58
n mod 9 is 0, 1, 2, 3, 4, 6, 7

copeland 2017-02-16 21:33:04
The product of 5 consecutive numbers is divisible by 9 when there are two threes or one nine. Therefore a product of 5 consecutive numbers is not divisible by 9 when the middle number is a multiple of 3 but not a multiple of nine. The middle number is $n-2$, so $n-2$ can't be 3 or 6.

copeland 2017-02-16 21:33:11
The numerator is divisible by 9 when $n\equiv0,1,2,3,4,6,7\pmod9$.

copeland 2017-02-16 21:33:19
Now let's figure out when the numerator is divisible by 32. (Yeah, I know - I didn't write the problem.)

copeland 2017-02-16 21:33:36
Let's make a table of $n\pmod{32}$ and the highest power of two that must divide $n(n-1)(n-2)(n-3)(n-4)$. We'll call that $V(n)$.

copeland 2017-02-16 21:33:38
Actually, let's start with $v(n)$, just the number of 2s dividing $n$.

copeland 2017-02-16 21:34:05
\[\begin{array}{c|c}

n\pmod{32}&-4&-3&-2&-1&0&1&2&3&4&5\\

\hline

v(n)&2&0&1&0&5&0&1&0&2&0\\

V(n)& & & & &\color{green}{8}&

\end{array}\]

copeland 2017-02-16 21:34:21
$V(0)=v(-4)+v(-3)+v(-2)+v(-1)+v(0).$

copeland 2017-02-16 21:34:30
We don't care about the values of $V$ for negative $n$. What are the entries for $n=1$ through $n=5$?

brainiac1 2017-02-16 21:36:23
6, 7, 6, 8, 3

BXU65 2017-02-16 21:36:23
6, 7, 6, 8, 3

xayy 2017-02-16 21:36:23
6,7,6,8,3

yrnsmurf 2017-02-16 21:36:23
67683

copeland 2017-02-16 21:36:26
\begin{align*}

V(1)&=v(-3)+v(-2)+v(-1)+v(0)+v(1)\\

V(2)&=v(-2)+v(-1)+v(0)+v(1)+v(2)\\

V(3)&=v(-1)+v(0)+v(1)+v(2)+v(3)\\

V(4)&=v(0)+v(1)+v(2)+v(3)+v(4)\\

V(5)&=v(1)+v(2)+v(3)+v(4)+v(5)

\end{align*}

copeland 2017-02-16 21:36:27
\[\begin{array}{c|c}

n\pmod{32}&-4&-3&-2&-1&0&1&2&3&4&5\\

\hline

v(n)&2&0&1&0&5&0&1&0&2&0\\

V(n)& & & & &\color{green}{8}&\color{green}{6}&\color{green}{7}&\color{green}{6}&\color{green}{8}&3\\

\end{array}\]

Jfault 2017-02-16 21:36:32
why is 8 in green

copeland 2017-02-16 21:36:38
Yeah, what is that green stuff about?

duck_master 2017-02-16 21:37:18
greater than 5

brainiac1 2017-02-16 21:37:18
it is greater than 5, so the numerator is divisible by 32

BXU65 2017-02-16 21:37:18
the values that are divisible by 32

GeronimoStilton 2017-02-16 21:37:18
Because it works for a possible value of $n$.

Derive_Foiler 2017-02-16 21:37:18
all above 5. good

copeland 2017-02-16 21:37:19
I marked all the numbers where $V(n)\geq5$ since those are the values of $n$ where the numerator is divisible by 35 (contains 5 powers of 2).

copeland 2017-02-16 21:37:21
What are the next two values of $V(n)$?

copeland 2017-02-16 21:37:23
\[\begin{array}{c|c}

n\pmod{32}&-4&-3&-2&-1&0&1&2&3&4&5&6&7\\

\hline

v(n)&2&0&1&0&5&0&1&0&2&0&1&0\\

V(n)& & & & &\color{green}{8}&\color{green}{6}&\color{green}{7}&\color{green}{6}&\color{green}{8}&3\\

\end{array}\]

BXU65 2017-02-16 21:38:07
4, 3

brainiac1 2017-02-16 21:38:07
4, 3

yrnsmurf 2017-02-16 21:38:07
43

GeronimoStilton 2017-02-16 21:38:07
$4, 3$

xayy 2017-02-16 21:38:07
4,3

copeland 2017-02-16 21:38:09
Next we have

copeland 2017-02-16 21:38:10
\[\begin{array}{c|c}

n\pmod{32}&-4&-3&-2&-1&0&1&2&3&4&5&6&7&8&9&10&11&12&13\\

\hline

v(n)&2&0&1&0&5&0&1&0&2&0&1&0&3&0&1&0&2&0\\

V(n)& & & & &\color{green}{8}&\color{green}{6}&\color{green}{7}&\color{green}{6}&\color{green}{8}&3&4&3\\

\end{array}\]

copeland 2017-02-16 21:38:11
And the next 6 terms?

BXU65 2017-02-16 21:39:25
6, 4, 5, 4, 6, 3

xayy 2017-02-16 21:39:25
6,4,5,4,6,3

brainiac1 2017-02-16 21:39:25
6, 4, 5, 4, 6, 3

duck_master 2017-02-16 21:39:25
6,4,5,4,6,3

yrnsmurf 2017-02-16 21:39:25
645463

Reef334 2017-02-16 21:39:25
6,4,5,4,6,3

rt03 2017-02-16 21:39:25
6,4,5,4,6,3

copeland 2017-02-16 21:39:28
\[\begin{array}{c|c}

n\pmod{32}&-4&-3&-2&-1&0&1&2&3&4&5&6&7&8&9&10&11&12&13\\

\hline

v(n)&2&0&1&0&5&0&1&0&2&0&1&0&3&0&1&0&2&0\\

V(n)& & & & &\color{green}{8}&\color{green}{6}&\color{green}{7}&\color{green}{6}&\color{green}{8}&3&4&3&\color{green}{6}&4&\color{green}{5}&4&\color{green}{6}&3\\

\end{array}\]

copeland 2017-02-16 21:39:29
OK, I think we get the idea. I'll give you a few more, clipping off those negatives for now:

copeland 2017-02-16 21:39:30
\[\begin{array}{c|c}

n\pmod{32}&0&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20\\

\hline

v(n)&5&0&1&0&2&0&1&0&3&0&1&0&2&0&1&0&4&0&1&0&2\\

V(n)&\color{green}{8}&\color{green}{6}&\color{green}{7}&\color{green}{6}&\color{green}{8}&3&4&3&\color{green}{6}&4&\color{green}{5}&4&\color{green}{6}&3&4&3&\color{green}{7}&\color{green}{5}&\color{green}{6}&\color{green}{5}&\color{green}{7}\\

\end{array}\]

copeland 2017-02-16 21:39:37
This sure isn't getting any more fun.

copeland 2017-02-16 21:39:39
Actually, I think we needed to do all that work, but maybe we can stop now, why?

brainiac1 2017-02-16 21:40:29
there's a pattern?

copeland 2017-02-16 21:40:32
It's patternish. How can we talk about the top half of the numbers?

Derive_Foiler 2017-02-16 21:40:35
it'll repeat to some extent

copeland 2017-02-16 21:40:46
Kinda. Does $v(n)$ repeat?

brainiac1 2017-02-16 21:41:11
every 32...

GeronimoStilton 2017-02-16 21:41:11
Yes, modulo 32.

copeland 2017-02-16 21:41:15
Yeah, that's true.

copeland 2017-02-16 21:41:24
Does it have some smaller similarity?

Derive_Foiler 2017-02-16 21:41:52
and doesn't it mirror at 16?

copeland 2017-02-16 21:41:53
Oh, it does "mirror"! There's reflection symmetry. Actually, the reflection at 2 is a little easier to see.

BXU65 2017-02-16 21:41:59
the number of factors of two has a pattern

linqaszayi 2017-02-16 21:41:59
mod 16?

copeland 2017-02-16 21:42:03
Also, $v(n)$ mostly repeats mod 16 except $v(16)$ and $v(0)$ are different.

copeland 2017-02-16 21:42:10
So there are two things that work here.

copeland 2017-02-16 21:42:12
We could notice that $k$ and $-k$ have the same number of factors so the number of factors of 2 in $n(n-1)(n-2)(n-3)(n-4)$ needs to be symmetric about $n=2$ and $n=18$. So we can look at $n=2$ and $n=18$ and we can double all the rest to get our final count.

copeland 2017-02-16 21:42:21
Alternatively, we could notice that everything repeats modulo 16 except that $n=0$ through $n=4$ get an extra power of 2. Therefore, again we can "double" the number of successes up to edge cases.

copeland 2017-02-16 21:42:30
How many of the residues modulo 32 solve $n(n-1)(n-2)(n-3)(n-4)\equiv0\pmod{32}$?

mathman2048 2017-02-16 21:43:28
16

yrnsmurf 2017-02-16 21:43:28
16

BXU65 2017-02-16 21:43:28
16

abishek99 2017-02-16 21:43:28
16

GeronimoStilton 2017-02-16 21:43:28
$16$?

copeland 2017-02-16 21:43:30
From 3 to 17 there are 7 green numbers. Doubling that and adding 2 for $n=2$ and $n=18$ gives us $\dfrac{16}{32}$ of the numbers are multiples of 32.

copeland 2017-02-16 21:43:36
What fraction of the numbers from 1 to 2017 make $\dfrac{n(n-1)(n-2)(n-3)(n-4)}{7\cdot9\cdot32}$ and integer?

copeland 2017-02-16 21:44:11
Or at least, how do we compute the fraction?

ninjataco 2017-02-16 21:45:23
multiply all the fractions from all of the mods

duck_master 2017-02-16 21:45:23
More Chinese Remainder Theorem!

BXU65 2017-02-16 21:45:23
1/2 * 7/9 * 5/7 which is 5/18

GeronimoStilton 2017-02-16 21:45:23
$\frac{7 \cdot 16 \cdot 5}{2016}$

Reef334 2017-02-16 21:45:23
5/7*7/9*16/32

yrnsmurf 2017-02-16 21:45:23
16*7*5

copeland 2017-02-16 21:45:26
Well, $\dfrac57$ of them are divisible by 7, $\dfrac79$ of them are divisible by 9, and $\dfrac{16}{32}$ of them are divisible by 32.

copeland 2017-02-16 21:45:29
Those are independent by the Painfully Obvious Lemma (though it feels less obvious than painful now), so the number of integers is\[\dfrac57\cdot\dfrac79\cdot\dfrac{20}{32}=\dfrac{35\cdot16}{2016}=\dfrac{560}{2016}.\]

copeland 2017-02-16 21:45:37
560 is not one of the answers!

brainiac1 2017-02-16 21:46:24
it's from 9 to 2017...

linqaszayi 2017-02-16 21:46:24
$n\ge 9$

celestialphoenix3768 2017-02-16 21:46:24
must subtract 3 for working 2-9

shakeNbake 2017-02-16 21:46:24
then assume u miscounted by 3 and get (D)

duck_master 2017-02-16 21:46:24
did you talk about 1,2,3,4,5,6,7,8,9,2017?

copeland 2017-02-16 21:46:27
Oh, we're not allowed to have $n=2$ through $n=8$. Too bad they didn't save this problem until 2024. (In fact, we can tell because the denominator is also wrong.)

copeland 2017-02-16 21:46:30
Remember, we're looking at the 2016 numbers 2-2017.

copeland 2017-02-16 21:46:34
Those first 7 values of $n$ could have worked or not. If they all were "good", then the answer is $560-7=553$. If they all failed, the answer is 560.

copeland 2017-02-16 21:46:34
So. . . .

BXU65 2017-02-16 21:47:22
it needs to be 557 or D

GeronimoStilton 2017-02-16 21:47:22
The answer's 557, D.

summitwei 2017-02-16 21:47:22
somewhere in between, so d

duck_master 2017-02-16 21:47:22
*ahem* look at the answer choices!

copeland 2017-02-16 21:47:24
The only answer in that range is (D) $\boxed{557}$.

copeland 2017-02-16 21:48:18
So, I guess we're obligated to figure out which three of the first 8 gave bad integers. AND, we're probably obligated to argue that we can make all of these.

copeland 2017-02-16 21:48:22
First, can we pick a set of $t=\dfrac{n(n-1)(n-2)(n-3)(n-4)}{8\cdot7\cdot6\cdot5\cdot3\cdot2}$ different teams from a group of $n$ people?

copeland 2017-02-16 21:49:12
How many total possible teams are there?

GeronimoStilton 2017-02-16 21:49:49
Yes, if $n \ge 4$

BXU65 2017-02-16 21:49:49
n choose 5

duck_master 2017-02-16 21:49:49
$\binom{n}{5}$

copeland 2017-02-16 21:49:51
Since \[\dbinom n5=\dfrac{n(n-1)(n-2)(n-3)(n-4)}{5\cdot4\cdot3\cdot2\cdot1}\] is way larger than $t$, we can pick any set of $t$ teams and be done with it (when $t$ is an integer).

copeland 2017-02-16 21:49:56
Second, let's see how many of the first 8 succeed:

copeland 2017-02-16 21:49:57
\[\begin{array}{c|c}

n&2&3&4&5&6&7&8\\

\hline

V(n)&\color{green}{7}&\color{green}{6}&\color{green}{8}&3&4&3&\color{green}{6}\\

n\pmod{7}&\color{green}{2}&\color{green}{3}&\color{green}{4}&5&6&\color{green}{0}&\color{green}{1}\\

n\pmod{9}&\color{green}{2}&\color{green}{3}&\color{green}{4}&5&\color{green}{6}&\color{green}{7}&8\\

\end{array}\]

yrnsmurf 2017-02-16 21:50:14
1,2,3,4 work 5,6,7 don't

copeland 2017-02-16 21:50:15
The numbers 2, 3, and 4 are the omitted ones so $560-3=557$ is indeed the right answer.

copeland 2017-02-16 21:50:24
And we win?

brainiac1 2017-02-16 21:50:48
yup?

copeland 2017-02-16 21:50:56
Enthusiasm and delight.

copeland 2017-02-16 21:51:11
Well, That's it for the Math Jam.

copeland 2017-02-16 21:51:18
Thanks everyone for coming!

GeronimoStilton 2017-02-16 21:51:21
Bye!!!

brainiac1 2017-02-16 21:51:21
Thanks!

copeland 2017-02-16 21:51:23
AoPS is running our weekend Special AIME Problem Seminar next weekend, Feb 25 and 26, from 3:30-6:30 PM Eastern each day. More information is at http://artofproblemsolving.com/school/course/catalog/maa-aime-special.

copeland 2017-02-16 21:51:23
Please join us again on March 9 and 24 when we will be discussing the AIME I and II contests.

BXU65 2017-02-16 21:52:07
What's the best way to prepare for AIME

copeland 2017-02-16 21:52:09
Buy all the AoPS books and take classes, too.

copeland 2017-02-16 21:52:31
Also, check out our forums. This question is on there a billion times and a lot of the real answers are very good.

copeland 2017-02-16 21:52:49
Alright, I'm going to close everything down now. Thanks for coming!

BXU65 2017-02-16 21:53:04
Ok Thanks Bye!

swimmerstar 2017-02-16 21:53:04
Thanks

ImpossibleCube 2017-02-16 21:53:04
Thanks!

anixsarkar 2017-02-16 21:53:04
thank you

2017 AMC 10/12 B Discussion

Facilitator: AoPS Staff