Summer schedule is now available! Enroll today to secure your spot!

2017 AIME I Discussion

Go back to the Math Jam Archive

AoPS Instructors discuss all 15 problems on the 2017 AIME I.

Copyright © 2024 AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this file with others.

Facilitator: Jeremy Copeland

copeland 2017-03-09 19:02:47
Welcome to the 2017 AIME I Math Jam!
copeland 2017-03-09 19:02:48
I'm Jeremy Copeland, and I'll be leading our discussion tonight.
copeland 2017-03-09 19:02:56
Before we get started I would like to take a moment to explain our virtual classroom to those who have not previously participated in a Math Jam or one of our online classes.
copeland 2017-03-09 19:02:58
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 2017-03-09 19:03:04
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 2017-03-09 19:03:17
Who doesn't want to be kept organized, amirite?
copeland 2017-03-09 19:03:24
Notice that this is a lot like one of our classes except there are a lot more of you and the same number of me. I won't be able to post all of your comments all of the time. It's not personal! Here's a secret though: I prefer to pass clear, well-written comments to the room.
copeland 2017-03-09 19:03:34
And of course, a lot of mathematics appears on the AIME. We don't have the chance to teach all the techniques, but you should take notes of the things you think are cool and want to research more and look into those things later. There are some great conversations going on about this year's AIME over in our community:
https://artofproblemsolving.com/community/c5t183f5_aime
If you want to know more about any of the problems, try there.
copeland 2017-03-09 19:03:43
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the necessary 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 always to try do so in our regular online 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 2017-03-09 19:03:54
We do have two teaching assistants with us tonight to help answer your questions:
copeland 2017-03-09 19:03:55
Henrik Boecken (henrikjb): 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 2017-03-09 19:03:56
William Wang (willwang123): 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.
henrikjb 2017-03-09 19:04:03
Hi guys!
willwang123 2017-03-09 19:04:04
Hi everyone!
zac15SCASD 2017-03-09 19:04:21
PA ftw!!!
quanhui868 2017-03-09 19:04:21
Hi
Radio2 2017-03-09 19:04:21
Hi!
tdeng 2017-03-09 19:04:21
Hey!
swagger 2017-03-09 19:04:21
hi
samuel 2017-03-09 19:04:21
Hi!
GeneralCobra19 2017-03-09 19:04:21
Hi. Glad to be here.
Liopleurodon 2017-03-09 19:04:21
Hello!
MrMXS 2017-03-09 19:04:21
hello henrikjb and willwang123
copeland 2017-03-09 19:04:24
They can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.
copeland 2017-03-09 19:04:30
Please also remember that the purpose of this Math Jam is to work through the solutions to AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. Also on occasion we may stop to prove things that you wouldn't necessary need to prove while doing the contest. So please, when a question is posted, do not simply respond with the final answer. That's not why we're here. We're going to work through the problems step-by-step, and comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, won't be acknowledged.
copeland 2017-03-09 19:04:42
Before we get started, here's a question for those of you who took the test: What was your favorite question on the test?
MSTang 2017-03-09 19:05:21
13
swagger 2017-03-09 19:05:21
#2
EasyAs_Pi 2017-03-09 19:05:21
question number 16!
pieater314159 2017-03-09 19:05:21
Problem 14. I like logarithms.
samuel 2017-03-09 19:05:21
#15
linqaszayi 2017-03-09 19:05:21
#14 maybe
zihang 2017-03-09 19:05:21
#11
mossie 2017-03-09 19:05:21
The 13th!
GeronimoStilton 2017-03-09 19:05:21
Problem 13, although I didn't get to it in time.
amzhao 2017-03-09 19:05:21
2, it was pretty simple
ilovemath04 2017-03-09 19:05:21
#1
NewbieGamer 2017-03-09 19:05:21
#13
hjia17 2017-03-09 19:05:21
1
reelmathematician 2017-03-09 19:05:21
Problem 15 because 15 is prime. jk
ethanliu247 2017-03-09 19:05:21
number 8,even if i got it wrong
Jyzhang12 2017-03-09 19:05:21
1
copeland 2017-03-09 19:05:27
Two of you said 14, huh?
copeland 2017-03-09 19:05:29
OK, then.
copeland 2017-03-09 19:05:32
Let's get started! We're going to work through all 15 problems from the 2017 AIME I, in order.
copeland 2017-03-09 19:05:37
1. Fifteen distinct points are designated on $\triangle ABC{:}$ the 3 vertices $A,$ $B,$ and $C;$ 3 other point on side $\overline{AB};$ 4 other points on side $\overline{BC};$ and 5 other points on side $\overline{CA}$. Find the number of triangles with positive area whose vertices are among these 15 points.
copeland 2017-03-09 19:05:43
I lorve semicolons.
tdeng 2017-03-09 19:06:06
Lorve?
Smileyklaws 2017-03-09 19:06:06
lorve
Liopleurodon 2017-03-09 19:06:06
lorve
copeland 2017-03-09 19:06:07
It's French.
copeland 2017-03-09 19:06:17
(A romance language.)
copeland 2017-03-09 19:06:26
What's the trick here?
celestialphoenix3768 2017-03-09 19:06:51
complementary counting
GeneralCobra19 2017-03-09 19:06:51
Complementary Counting!
SomethingNeutral 2017-03-09 19:06:51
complementary counting
strategos21 2017-03-09 19:06:51
Complimentary counting
ilikepie2003 2017-03-09 19:06:51
use complementary counting!
espeon12 2017-03-09 19:06:51
complementary counting
sxu 2017-03-09 19:06:51
complementary counting: subtract degenerate triangles from total
copeland 2017-03-09 19:06:54
If we want to see how many triangles there are with positive area we can find the number of triangles that have area zero and subtract.
richuw 2017-03-09 19:07:09
Draw a diagram!
oiler8 2017-03-09 19:07:09
A diagram
copeland 2017-03-09 19:07:11
copeland 2017-03-09 19:07:13
How many total triples are there?
MaxTplusAMSP 2017-03-09 19:07:37
15 choose 3
espeon12 2017-03-09 19:07:37
15C3
EulerMacaroni 2017-03-09 19:07:37
$\binom{15}{3}$
DemonPlat4 2017-03-09 19:07:37
15 choose 3
islander7 2017-03-09 19:07:37
15C3
ninjataco 2017-03-09 19:07:37
15C3
Deathranger999 2017-03-09 19:07:37
15 choose 3
W.Sun 2017-03-09 19:07:37
15 choose 3
letsgomath 2017-03-09 19:07:37
15 choose 3 = 455
fdas 2017-03-09 19:07:37
15 choose 3
gabrielsui 2017-03-09 19:07:37
15 nCr 3
copeland 2017-03-09 19:07:40
There are $\dbinom{15}3$ total triples of points.
copeland 2017-03-09 19:07:41
How many of those points give us "triangles" of area zero?
mossie 2017-03-09 19:08:19
$\binom{15}{3}-\binom{7}{3}-\binom{6}{3}-\binom{5}{3}=390$
zihang 2017-03-09 19:08:19
7C3+6C3+5C3
sxu 2017-03-09 19:08:19
5C3+6C3+7C3
abvenkgoo 2017-03-09 19:08:19
7C3+6C3+5C3
EulerMacaroni 2017-03-09 19:08:19
$\binom{5}{3}+\binom{6}{3}+\binom{7}{3}$
jonzli123 2017-03-09 19:08:19
5C3 + 6C3 + 7C3
vvluo 2017-03-09 19:08:19
7 choose 3 plus 5 choose 3 plus 6 choose 3
tdeng 2017-03-09 19:08:19
$\binom{5}{3}+\binom{6}{3}+\binom{7}{3}$
cwechsz 2017-03-09 19:08:19
6C3 + 5C3 + 7C3
copeland 2017-03-09 19:08:22
The points must be collinear to have area zero. There are $\dbinom53$ triples along $\overline{AB}$, $\dbinom63$ triples along $\overline{BC},$ and $\dbinom73$ triples along $\overline{CA}$.
copeland 2017-03-09 19:08:22
And the final answer?
math101010 2017-03-09 19:08:40
390
Pandasareamazing. 2017-03-09 19:08:40
390
nukelauncher 2017-03-09 19:08:40
390!
fireflame241 2017-03-09 19:08:40
390
DarthRen 2017-03-09 19:08:40
390
dhruv 2017-03-09 19:08:40
390
Mrkiller 2017-03-09 19:08:40
390
jonzli123 2017-03-09 19:08:40
390!!!!!!
ethanliu247 2017-03-09 19:08:40
390
copeland 2017-03-09 19:08:42
There are

\begin{align*}

\binom{15}3-\binom53-\binom63-\binom73

&=5\cdot7\cdot13-5\cdot2-5\cdot4-7\cdot5\\

&=5(91-2-4-7)\\

&=5\cdot78=\boxed{390}

\end{align*} total triangles of positive area.
copeland 2017-03-09 19:08:44
Great.
copeland 2017-03-09 19:08:47
Everyone warmed up?
First 2017-03-09 19:09:01
Yes!
BLCRAFT 2017-03-09 19:09:01
ya...
SomethingNeutral 2017-03-09 19:09:01
yes
Dude03 2017-03-09 19:09:01
yes
Jyzhang12 2017-03-09 19:09:01
yea
MountainHeight 2017-03-09 19:09:01
yes
cooleybz2013 2017-03-09 19:09:01
yep
lsh0589 2017-03-09 19:09:01
yeah
EasyAs_Pi 2017-03-09 19:09:01
Yes!
tdeng 2017-03-09 19:09:01
Yeah!
copeland 2017-03-09 19:09:02
Cool.
copeland 2017-03-09 19:09:14
Incidentally, we used the same trick on the AMC10A this year:
copeland 2017-03-09 19:09:19
23. How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$, inclusive?
copeland 2017-03-09 19:09:29
(We're not solving that problem.)
copeland 2017-03-09 19:09:35
We are going to solve. . .
ethanliu247 2017-03-09 19:09:42
THisone
SomethingNeutral 2017-03-09 19:09:42
I remember that one...
lego101 2017-03-09 19:09:42
oh i remember that question -_-
gabrielsui 2017-03-09 19:09:42
i was there!
Funnybunny5246 2017-03-09 19:09:48
On to #2
Liopleurodon 2017-03-09 19:09:48
NUMBER 2
W.Sun 2017-03-09 19:09:48
Number 2!
copeland 2017-03-09 19:09:56
2. When each of $702,$ $787,$ and $855$ is divided by the positive integer $m,$ the remainder is always the positive integer $r.$ When each of $412,$ $722,$ and $815$ is divided by the positive integer $n,$ the remainder is always the positive integer $s\neq r.$ Find $m+n+r+s.$
copeland 2017-03-09 19:10:00
Where should we start?
noodlemaster 2017-03-09 19:10:37
analyze differences
quanhui868 2017-03-09 19:10:37
find the difference between the numbers
jonzli123 2017-03-09 19:10:37
find the difference between the numbers
DemonPlat4 2017-03-09 19:10:37
differences of the numbers
smartpgp 2017-03-09 19:10:37
find the gcf of the differences
Liopleurodon 2017-03-09 19:10:37
subtract each number from each other, and find their GCD
rapturt9 2017-03-09 19:10:37
subtract each pair of the 3 numbers
vvluo 2017-03-09 19:10:37
differences
copeland 2017-03-09 19:10:40
If two numbers have the same remainder when divided by $m$ then their difference is a multiple of $m$. Here we have
copeland 2017-03-09 19:10:41
\begin{align*}

702&=am+r\\

787&=bm+r\\

855&=cm+r

\end{align*}
copeland 2017-03-09 19:10:42
When we subtract them pairwise we get

\begin{align*}

m&\mid 787-702=85\\

m&\mid 855-787=68\\

\end{align*}
copeland 2017-03-09 19:10:47
[Remember that we say $m\mid n$ if $n$ is a multiple of $m$, so $3\mid 6$ and $71\mid71$ and $431\mid0.$]
copeland 2017-03-09 19:10:53
So what do we know about $m?$
tdeng 2017-03-09 19:11:39
Only common factor is 17
MSTang 2017-03-09 19:11:39
m = 17
legolego 2017-03-09 19:11:39
factor of 17
espeon12 2017-03-09 19:11:39
it's 17
swagger 2017-03-09 19:11:39
17
nukelauncher 2017-03-09 19:11:39
m ≠ 1, so m = 17
DarthRen 2017-03-09 19:11:39
m|17
Liopleurodon 2017-03-09 19:11:39
m is larger than 1
MrMXS 2017-03-09 19:11:39
$m=17$
Pandasareamazing. 2017-03-09 19:11:39
so it is 17
amzhao 2017-03-09 19:11:39
m is 17
MathTechFire 2017-03-09 19:11:39
it is 17
copeland 2017-03-09 19:11:42
Since $m$ divides both of these numbers, it divides the GCD, which is 17. (Nicely, the first step in the Euclidean Algorithm here is $85-68=17$.)
copeland 2017-03-09 19:11:44
In order for $r$ to be positive we need $m>1,$ so $m=17$.
copeland 2017-03-09 19:11:45
So what is $r?$
tdeng 2017-03-09 19:12:06
5
swagger 2017-03-09 19:12:06
5
letsgomath 2017-03-09 19:12:06
5
legolego 2017-03-09 19:12:06
5
ilovemath04 2017-03-09 19:12:06
r=5
Funnybunny5246 2017-03-09 19:12:06
It is 5
copeland 2017-03-09 19:12:08
I like looking at 855. Since

\[855=850+5=17\cdot50+5,\] we see that $r=5$.
copeland 2017-03-09 19:12:14
OK, now for the other three numbers. Does anyone see what $n$ is?
Radio2 2017-03-09 19:12:37
31
GeronimoStilton 2017-03-09 19:12:37
$31$
mshanmugam 2017-03-09 19:12:37
31
Radio2 2017-03-09 19:12:37
n=31
SomethingNeutral 2017-03-09 19:12:37
n is 31
strategos21 2017-03-09 19:12:37
31
skiboy32 2017-03-09 19:12:37
31
copeland 2017-03-09 19:12:39
Subtracting pairwise again we get

\begin{align*}

n\mid722-412&=310\\

n\mid 815-722&=93.

\end{align*}
copeland 2017-03-09 19:12:47
Both of those numbers are obviously divisible by 31: we have $310=31\cdot10$ and $93=31\cdot3$. Since the quotients are coprime, the GCD is $n=31$.
copeland 2017-03-09 19:12:50
And what is $s?$
quanhui868 2017-03-09 19:13:26
9
stan23456 2017-03-09 19:13:26
9
rapturt9 2017-03-09 19:13:26
9
MrMXS 2017-03-09 19:13:26
$9$
DemonPlat4 2017-03-09 19:13:26
9
cooleybz2013 2017-03-09 19:13:26
9
SomethingNeutral 2017-03-09 19:13:26
actually 9
copeland 2017-03-09 19:13:29
Let's start with 722. We know 620 is a multiple of 31, so if we start by subtracting 620 we get

\[722\equiv 102\equiv71\equiv40\equiv9\pmod{31}.\]
copeland 2017-03-09 19:13:30
And the answer?
nukelauncher 2017-03-09 19:13:54
062
math_noob 2017-03-09 19:13:54
62
samuel 2017-03-09 19:13:54
062
ijava 2017-03-09 19:13:54
62
ilikepie2003 2017-03-09 19:13:54
62
Pandasareamazing. 2017-03-09 19:13:54
62=31+9+17+5
ilikepie2003 2017-03-09 19:13:54
062
samuel 2017-03-09 19:13:54
$\boxed{062}$
DemonPlat4 2017-03-09 19:13:54
31+9+17+5 = 062
zihang 2017-03-09 19:13:54
62
thinkinavi 2017-03-09 19:13:54
062
copeland 2017-03-09 19:13:56
\[m+r+n+s=17+5+31+9=\boxed{062}.\]
copeland 2017-03-09 19:14:01
I haven't figured out what $s\neq r$ does in this problem. My guess is that in the original version of the problem one of those GCDs was not prime and there was an extra step. If so, then they lightened the problem up by making them both prime and the $s\neq r$ constraint is vestigial.
copeland 2017-03-09 19:14:08
I once knew this kid, Willie, who had 11 toes. Who's ready for the next problem?
thomas9549 2017-03-09 19:14:26
i am
jam10307 2017-03-09 19:14:26
Me1!!!
swagger 2017-03-09 19:14:26
me!
Pandasareamazing. 2017-03-09 19:14:26
me!!
W.Sun 2017-03-09 19:14:26
ME
lego101 2017-03-09 19:14:26
okay! me!
W.Sun 2017-03-09 19:14:26
Moiii
GeronimoStilton 2017-03-09 19:14:26
Me!
amackenzie1 2017-03-09 19:14:26
I am!
copeland 2017-03-09 19:14:28
3. For a positive integer $n,$ let $d_n$ be the units digit of \[1+2+3+\cdots+n.\]Find the remainder when \[\sum_{n=1}^{2017}d_n\] is divided by 1000.
copeland 2017-03-09 19:14:37
Raise your hand if your first answer was way too big on this problem.
ilovetynker 2017-03-09 19:15:18
me!
First 2017-03-09 19:15:18
Me!
BLCRAFT 2017-03-09 19:15:18
*raise*
lego101 2017-03-09 19:15:18
*raises hand emphatically*
jam10307 2017-03-09 19:15:18
RAISE
W.Sun 2017-03-09 19:15:18
Me lol
quanhui868 2017-03-09 19:15:18
*raises hand*
sxu 2017-03-09 19:15:18
*raises hand
jonzli123 2017-03-09 19:15:18
*raises hand*
thomas9549 2017-03-09 19:15:18
me!
ilovetynker 2017-03-09 19:15:18
me
W.Sun 2017-03-09 19:15:18
*raises hand*
cooleybz2013 2017-03-09 19:15:18
me
jack74 2017-03-09 19:15:18
hand raise
samuel 2017-03-09 19:15:18
me me me
strategos21 2017-03-09 19:15:20
my hand is raised
copeland 2017-03-09 19:15:21
Yeah, me too. I had that, "Jeremy, what are you doing?" moment later than I should admit.
copeland 2017-03-09 19:15:25
Every $d_n$ is a digit. That's important. Half of the mistakes on this problem are from forgetting that. We're adding 2017 digits. If you ever add a number that's more than 9, you're doing it wrong.
copeland 2017-03-09 19:15:35
So we want the units digits of the triangle numbers. I don't think I've ever seen that. But what do you expect?
nukelauncher 2017-03-09 19:16:05
theres a repeating pattern
ilovemath04 2017-03-09 19:16:05
a pattern
DemonPlat4 2017-03-09 19:16:05
find the pattern
GeronimoStilton 2017-03-09 19:16:05
A pattern!
espeon12 2017-03-09 19:16:05
i expected it to be a cycle
Deathranger999 2017-03-09 19:16:05
Pattern
Liopleurodon 2017-03-09 19:16:05
A pattern
tdeng 2017-03-09 19:16:05
It will repeat
gabrielsui 2017-03-09 19:16:05
pattern
Mrkiller 2017-03-09 19:16:05
It to repeat
dhruv 2017-03-09 19:16:05
pattern?
Shri333 2017-03-09 19:16:05
a pattern
abvenkgoo 2017-03-09 19:16:05
a pattern in the units digits
copeland 2017-03-09 19:16:08
I expect the sequence $d_n$ to be periodic. This problem is going to be really hard if it's not.
copeland 2017-03-09 19:16:09
Do you see the period off-hand?
Mrkiller 2017-03-09 19:16:37
20
tdeng 2017-03-09 19:16:37
It has a period of 20
legolego 2017-03-09 19:16:37
period of 20
gabrielsui 2017-03-09 19:16:37
no
shenmaster88 2017-03-09 19:16:37
evey 20
atmchallenge 2017-03-09 19:16:37
no, so let's write out some terms
Pandasareamazing. 2017-03-09 19:16:37
not really
EulerMacaroni 2017-03-09 19:16:37
20
GeronimoStilton 2017-03-09 19:16:37
No. Let's write it out to find the pattern!
ilikepie2003 2017-03-09 19:16:37
nope
copeland 2017-03-09 19:16:40
You might see that it's 20. Since the units digits of $n$ repeat every 10, the period has to be a multiple of 10. Since $d_0=0$ and $d_{10}=5$ is the units digit of 55, the period isn't 10. But $d_{20}=0$ is the units digit of $210$. The period is 20.
copeland 2017-03-09 19:16:44
You might not see that. It's not even a big deal. Why?
LittleChimp 2017-03-09 19:17:35
We have to write out terms anyways
jonzli123 2017-03-09 19:17:35
you write out the terms
espeon12 2017-03-09 19:17:35
we can just write out some terms
LearningMath 2017-03-09 19:17:35
Because we want the last 3 digits
Pudentane 2017-03-09 19:17:35
cause we want the sum of the units digits
copeland 2017-03-09 19:17:37
Well, we actually need to sum these things so we need to figure out what they are. Once we list them we'll know exactly what the period is.
copeland 2017-03-09 19:17:43
OK, so let's list them. What are the first 10? (You can even just list 10 digits in a row!)
Liopleurodon 2017-03-09 19:18:24
$1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0$
SomethingNeutral 2017-03-09 19:18:24
1360518655681500
EasyAs_Pi 2017-03-09 19:18:24
yes, $1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,\cdots$
vishwathganesan 2017-03-09 19:18:24
1360518655
SomethingNeutral 2017-03-09 19:18:24
1360518655
pican 2017-03-09 19:18:24
1, 3, 6, 0, 5, 1, 8, 6, 5, 5
GeneralCobra19 2017-03-09 19:18:24
1 3 6 0 5 1 8 6 5 5
Funnybunny5246 2017-03-09 19:18:24
1360518655
KYang 2017-03-09 19:18:24
1 3 6 0 5 1 8 6 5 5
copeland 2017-03-09 19:18:28
\[\begin{array}{c|ccccc|ccccc}

n &1&2&3&4&5&6&7&8&9&10\\

\hline

d_n&1&3&6&0&5&1&8&6&5&5\\

\end{array}\]
copeland 2017-03-09 19:18:30
Here are the next 10:
copeland 2017-03-09 19:18:32
\[\begin{array}{c|ccccc|ccccc|ccccc|ccccc}

n &1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20\\

\hline

d_n&1&3&6&0&5&1&8&6&5&5&6&8&1&5&0&6&3&1&0&0\\

\end{array}\]
copeland 2017-03-09 19:18:33
Hm, it's symmetric. We don't care (except that it helps us double-check our arithmetic).
copeland 2017-03-09 19:18:36
What's the sum?
Liopleurodon 2017-03-09 19:19:16
$70$
nosaj 2017-03-09 19:19:16
70
Funnybunny5246 2017-03-09 19:19:16
70
LittleChimp 2017-03-09 19:19:16
70
rapturt9 2017-03-09 19:19:16
70
math_noob 2017-03-09 19:19:16
70
SomethingNeutral 2017-03-09 19:19:16
70
Radio2 2017-03-09 19:19:16
70
legolego 2017-03-09 19:19:16
70
letsgomath 2017-03-09 19:19:16
70
copeland 2017-03-09 19:19:19
The sum is $70$. That means the average value over any block of 20 consecutive numbers is $\dfrac72$.
copeland 2017-03-09 19:19:25
So how do we compute the sum of the first 2017 of these $d_n?$
LittleChimp 2017-03-09 19:20:31
the sum of the first 2000 is just 7000
rapturt9 2017-03-09 19:20:31
70*50 for the first 2000 and then sum the first 17
EasyAs_Pi 2017-03-09 19:20:31
compute the sum of the first 2000 then add the last 17...
vishwathganesan 2017-03-09 19:20:31
find the sum of the first 101 cycles and subtract 1+0+0
legolego 2017-03-09 19:20:31
first 2020 : 7070 - 0 - 0 - 1 = 7069 -> 69
ilovemath04 2017-03-09 19:20:31
the first 2000 is 70*100
spicyray 2017-03-09 19:20:31
20*101-1
amzhao 2017-03-09 19:20:31
2000*(7/2) then add first 17
owm 2017-03-09 19:20:31
100*7/2+69
MathTechFire 2017-03-09 19:20:31
7/2 x 100 + next 17
Funnybunny5246 2017-03-09 19:20:31
1000(70) + first 17 digits
copeland 2017-03-09 19:20:33
We can start with the first 2020 of them.
copeland 2017-03-09 19:20:36
Since the average value is $\dfrac72$, the sum of the first 2020 is 7070.
copeland 2017-03-09 19:20:42
Then we subtract $d_{20}+d_{19}+d_{18}.$ What's the final answer?
Radio2 2017-03-09 19:21:29
069
legolego 2017-03-09 19:21:29
69
sxu 2017-03-09 19:21:29
069
swagger 2017-03-09 19:21:29
069
MrMXS 2017-03-09 19:21:29
$069$
StellarG 2017-03-09 19:21:29
7069
MountainHeight 2017-03-09 19:21:29
069
amackenzie1 2017-03-09 19:21:29
$069$
GeneralCobra19 2017-03-09 19:21:29
69
copeland 2017-03-09 19:21:34
We want $7070-0-0-1=7069.$ The answer is $\boxed{069}$.
copeland 2017-03-09 19:21:37
Notice that there were two separate tactics suggested above. Tackling the first 2000 or the first 2020. Both work just fine.
copeland 2017-03-09 19:22:08
What next?
copeland 2017-03-09 19:22:12
Should we keep going?
First 2017-03-09 19:22:22
#4
GeneralCobra19 2017-03-09 19:22:22
#4
Shri333 2017-03-09 19:22:22
QUESTION 4!!!
nukelauncher 2017-03-09 19:22:22
YEs
stronto 2017-03-09 19:22:22
YEAH
quanhui868 2017-03-09 19:22:22
Yeah
owm 2017-03-09 19:22:22
Yep!
Shri333 2017-03-09 19:22:22
YES
GeorgeAvocados 2017-03-09 19:22:22
sure
Liopleurodon 2017-03-09 19:22:23
Number 4
copeland 2017-03-09 19:22:26
4. A pyramid has a triangular base with side lengths 20, 20, and 24. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length 25. The volume of the pyramid is $m\sqrt n,$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$
copeland 2017-03-09 19:22:29
Hm, OK. Do we like 20-20-24 triangles?
winnertakeover 2017-03-09 19:22:45
yes\
Deathranger999 2017-03-09 19:22:45
Yes.
cwechsz 2017-03-09 19:22:45
yes
NewbieGamer 2017-03-09 19:22:45
yes
copeland 2017-03-09 19:22:48
Why?
Deathranger999 2017-03-09 19:23:14
It's two 3-4-5 in disguise.
nosaj 2017-03-09 19:23:14
they're two 12-16-20's taped together
abvenkgoo 2017-03-09 19:23:14
it's 2 12-16-20 triangles together
Funnybunny5246 2017-03-09 19:23:14
turns into 12-16-20
ethanliu247 2017-03-09 19:23:14
two 20-16-12
jonzli123 2017-03-09 19:23:14
forms two 20-16-12 triangles
math101010 2017-03-09 19:23:14
12-16-20
Picroft 2017-03-09 19:23:14
3-4-5 triangles when you draw altitude
math.fever 2017-03-09 19:23:14
3-4-5 triangles!
Wuna 2017-03-09 19:23:14
3-4-5 based!
copeland 2017-03-09 19:23:19
It's isosceles anyway. We can even compute the altitude to the base:
copeland 2017-03-09 19:23:21
copeland 2017-03-09 19:23:21
Oh, right. That's 3-4-5:
copeland 2017-03-09 19:23:22
copeland 2017-03-09 19:23:24
How do we compute the volume?
vishwathganesan 2017-03-09 19:23:53
1/3Bh
GeronimoStilton 2017-03-09 19:23:53
We need the height.
EasyAs_Pi 2017-03-09 19:23:53
$\frac{bh}{3}$
tdeng 2017-03-09 19:23:53
Now we just need the height.
Pudentane 2017-03-09 19:23:53
1/3*B*h
gabrielsui 2017-03-09 19:23:53
1/3 * B * h
stronto 2017-03-09 19:23:53
1/3 [ABC] h
curry3030 2017-03-09 19:23:53
1/3 * b * h
young_3141 2017-03-09 19:23:53
1/3 * b * h
islander7 2017-03-09 19:23:53
find height to this face
espeon12 2017-03-09 19:23:53
bh/3
LearningMath 2017-03-09 19:23:53
The base's area times height divided by 3
copeland 2017-03-09 19:23:56
To compute the volume, we need the area of the base and the height. Toss in a $\dfrac13$ and stir. Let's save the variable $h$ for the height of the pyramid.
copeland 2017-03-09 19:23:57
What's the area of the base?
SDMKM 2017-03-09 19:24:15
192
ninjataco 2017-03-09 19:24:15
192
Pandasareamazing. 2017-03-09 19:24:15
192
FlamingDragon_9000 2017-03-09 19:24:15
192
ilovemath04 2017-03-09 19:24:15
192
vvluo 2017-03-09 19:24:15
192
SomethingNeutral 2017-03-09 19:24:15
16*12 = 192
dhruv 2017-03-09 19:24:15
192
copeland 2017-03-09 19:24:16
The area of the base is $12\cdot16.$
copeland 2017-03-09 19:24:24
Looks like you guys like multiplying more than I do.
copeland 2017-03-09 19:24:27
That's cool.
copeland 2017-03-09 19:24:31
And, qualitatively, what do we know about the altitude to this base?
a1b2 2017-03-09 19:25:15
It contains the circumcenter
richuw 2017-03-09 19:25:15
Touches the Circumcenter
sxu 2017-03-09 19:25:15
it's above the circumcenter
copeland 2017-03-09 19:25:28
The point we care about is the circumcenter of the base.
copeland 2017-03-09 19:25:35
That makes sense geometrically: every tetrahedron has a circumsphere. Since the points on the base are equidistant from the top vertex, the plane of their circumcircle is perpendicular to the diameter through the upper vertex, and, um, nevermind. . .
copeland 2017-03-09 19:25:41
Maybe you were thinking something else.
copeland 2017-03-09 19:25:48
Why is it the circumcenter again?
richuw 2017-03-09 19:26:09
Equi-distant from each vertex
jonzli123 2017-03-09 19:26:09
same distance from each vertex
stronto 2017-03-09 19:26:09
The top is equidistant from all 3 vertices
linqaszayi 2017-03-09 19:26:09
equidistance
First 2017-03-09 19:26:09
Center of equidistance?
copeland 2017-03-09 19:26:11
So?
sxu 2017-03-09 19:27:12
well since altitude is perp to base, it's perp to every line thru foot, so congruent triangles (that's how I convinced myself)
alifenix- 2017-03-09 19:27:12
Because all the lines upwards have the same length. The height is the same, thus the distance must be the same and so it is the definition of the circumcenter.
richuw 2017-03-09 19:27:12
well R^2 + Altitude^2 = 625
copeland 2017-03-09 19:27:17
Yeah! There are right triangles in here. We're just using HL similarity, or the Pythagorean Theorem or something.
copeland 2017-03-09 19:27:44
Where's the circumcenter of this triangle?
copeland 2017-03-09 19:27:48
Deathranger999 2017-03-09 19:28:12
On the altitude.
vvluo 2017-03-09 19:28:12
on altitude
MathTechFire 2017-03-09 19:28:12
on the 16 line
MrMXS 2017-03-09 19:28:12
on the altitute
Jyzhang12 2017-03-09 19:28:12
on the height 16
Radio2 2017-03-09 19:28:12
on the median with length 116
Radio2 2017-03-09 19:28:12
on the median with length 16
copeland 2017-03-09 19:28:23
We know that the distances from the vertices of this triangle to the fourth vertex are each 25.
copeland 2017-03-09 19:28:23
We also know by symmetry that the vertex lies over the triangle altitude that we drew in our picture. Let's plop the base of our altitude onto the picture.
copeland 2017-03-09 19:28:28
copeland 2017-03-09 19:28:37
If $a^2+h^2=25^2$, then $a$ is the distance from our point to any of the vertices.
copeland 2017-03-09 19:28:39
Let's plop $a$ on there, shall we?
copeland 2017-03-09 19:28:42
copeland 2017-03-09 19:28:49
How can we get $a$ from here?
tdeng 2017-03-09 19:29:40
(16-a)^2+12^2=a^2 by the Pythagorean Theorem
Jyzhang12 2017-03-09 19:29:40
pythagorean theorem
raxu 2017-03-09 19:29:40
$(16-a)^2+12^2=a^2$ using Pythagorean Theorem
legolego 2017-03-09 19:29:40
12^2 + (16-a)^2 = a^2
mathguy623 2017-03-09 19:29:40
$12^2+(16-a)^2=a^2$
Deathranger999 2017-03-09 19:29:40
(16 - a)^2 + 12^2 = a^2
Radio2 2017-03-09 19:29:40
Draw a perpendicular to a side with length 20 and use similarity
curry3030 2017-03-09 19:29:40
drop perp from altitude to side with length 20
copeland 2017-03-09 19:29:42
Oh, here are two solutions. We can either go with Pythagoras or similarity.
copeland 2017-03-09 19:29:44
Which do you want?
EasyAs_Pi 2017-03-09 19:30:42
similairiry
BuddyS 2017-03-09 19:30:42
similarity
leonlzg 2017-03-09 19:30:42
similarity
gabrielsui 2017-03-09 19:30:42
Similarity!!!
letsgomath 2017-03-09 19:30:42
Pythagoras
J1618 2017-03-09 19:30:42
I want to see how similarity can work here
QuestForKnowledge 2017-03-09 19:30:42
Pyhatgoras
Mathaddict11 2017-03-09 19:30:42
Pythagoras
nukelauncher 2017-03-09 19:30:42
pythagoras ftw
LearningMath 2017-03-09 19:30:42
Pythagoras
copeland 2017-03-09 19:30:47
Wow, OK.
copeland 2017-03-09 19:30:56
Similarity is a little faster. Let's do that.
copeland 2017-03-09 19:31:02
Tell me about this other altitude:
copeland 2017-03-09 19:31:04
GeronimoStilton 2017-03-09 19:31:59
It bisects because $a = a$.
GeneralCobra19 2017-03-09 19:31:59
The side it bisects is cut into two sides of 10
amackenzie1 2017-03-09 19:31:59
triangle formed by it is similar to half of the big triangle
EasyAs_Pi 2017-03-09 19:31:59
it creates similar triangles
amzhao 2017-03-09 19:31:59
similar trianlg
copeland 2017-03-09 19:32:01
It bisects that other side, so those edges have length 10:
copeland 2017-03-09 19:32:02
copeland 2017-03-09 19:32:02
So what is $a?$
AlisonH 2017-03-09 19:32:34
a/10=20/16
MrMXS 2017-03-09 19:32:34
using the new right triangle gives $\cfrac{a}{20}=\cfrac{10}{12}=\cfrac{5}{6}$
jonzli123 2017-03-09 19:32:34
25/2
math101010 2017-03-09 19:32:34
25/2
KYang 2017-03-09 19:32:34
25/2
MountainHeight 2017-03-09 19:32:34
25/2
Funnybunny5246 2017-03-09 19:32:34
25/2
copeland 2017-03-09 19:32:36
Now $a$ is the hypotenuse of a $3-4-5$ triangle. The long leg has length 10, so the hypotenuse solves $\dfrac a5=\dfrac{10}4.$ Therefore $a=\dfrac{25}2.$
copeland 2017-03-09 19:32:56
Incidentally, the Pythagoran approach, gone-plaid, looks like this:
copeland 2017-03-09 19:32:57
copeland 2017-03-09 19:33:01
$a+b=16.$
copeland 2017-03-09 19:33:07
And $12^2+b^2=a^2$.
copeland 2017-03-09 19:33:13
We have

\begin{align*}

12^2&=a^2-b^2\\

16&=a+b

\end{align*}

Dividing these gives \[a-b=\frac{12^2}{4^2}=9.\] Adding that to the second equation gives\[2a=16+9=25.\]
copeland 2017-03-09 19:33:26
Either way,
copeland 2017-03-09 19:33:27
\[a=\frac{25}2.\]
copeland 2017-03-09 19:33:31
What is the height?
nosaj 2017-03-09 19:34:03
so the height is 25*sqrt(3)/2
islander7 2017-03-09 19:34:03
25rt3/2
vishwathganesan 2017-03-09 19:34:03
25sqrt3/2
vvluo 2017-03-09 19:34:03
25 sqrt(3)/2
legolego 2017-03-09 19:34:03
25sqrt3/2
KYang 2017-03-09 19:34:03
{25/2}*sqrt{3}
NeeNeeMath 2017-03-09 19:34:03
25/2sqrt(3)
algebra_star1234 2017-03-09 19:34:03
$25\sqrt{3}/2$
rapturt9 2017-03-09 19:34:03
25sqrt(3)/2
copeland 2017-03-09 19:34:05
The height solves\[h^2=25^2-\left(\frac{25}2\right)^2=25^2\cdot\frac34.\] So $h=25\sqrt{\dfrac34}=\dfrac{25\sqrt3}2$.
copeland 2017-03-09 19:34:07
And the volume?
KYang 2017-03-09 19:34:37
800 sqrt{3}
algebra_star1234 2017-03-09 19:34:37
$800\sqrt{3}$
ilovemath04 2017-03-09 19:34:37
800\sqrt3
a1b2 2017-03-09 19:34:37
$800\sqrt3$
sxu 2017-03-09 19:34:37
$800\sqrt{3}$
stronto 2017-03-09 19:34:37
800\sqrt{3}
vishwathganesan 2017-03-09 19:34:37
800sqrt3
GeronimoStilton 2017-03-09 19:34:37
$800\sqrt{3}$
EasyAs_Pi 2017-03-09 19:34:37
$800\sqrt3$
rapturt9 2017-03-09 19:34:37
800sqrt3
copeland 2017-03-09 19:34:39
The volume is

\begin{align*}

V&=\frac13(12\cdot16)\cdot\dfrac{25\sqrt3}2\\

&=4\cdot8\cdot25\sqrt3=800\sqrt3.

\end{align*}
copeland 2017-03-09 19:34:41
And the final answer?
GeronimoStilton 2017-03-09 19:34:57
$803$
Picroft 2017-03-09 19:34:57
803
dhruv 2017-03-09 19:34:57
803
DylanX22 2017-03-09 19:34:57
803
swagger 2017-03-09 19:34:57
803
owm 2017-03-09 19:34:57
803
BooBooTM 2017-03-09 19:34:57
803
First 2017-03-09 19:34:57
803
winnertakeover 2017-03-09 19:34:57
803
alifenix- 2017-03-09 19:34:57
803
copeland 2017-03-09 19:34:59
The final answer is $800+3=\boxed{803}.$
copeland 2017-03-09 19:35:29
Incidentally, several of you suggested an even faster way to the circumradius:
pican 2017-03-09 19:35:32
We could use the fact that the circumradius is equal to $\dfrac{abc}{4A}$, where $a$, $b$, $c$ are the sides and $A$ is the area
DemonPlat4 2017-03-09 19:35:32
A = abc/4R
atmchallenge 2017-03-09 19:35:32
the circumradius, so $abc/4A$.
First 2017-03-09 19:35:32
We could have just used the formula $[ABC]=\frac{abc}{4R}$
copeland 2017-03-09 19:36:06
Often it's nice to work from first principles, but there just are a few formulas that you should know.
copeland 2017-03-09 19:36:12
Time for. . .
awesomemaths 2017-03-09 19:37:01
problem 5
lego101 2017-03-09 19:37:01
#5!!!!!
cooleybz2013 2017-03-09 19:37:01
5th problem!!!!
curry3030 2017-03-09 19:37:01
question 5
lego101 2017-03-09 19:37:01
Number five, it's alive!
Reef334 2017-03-09 19:37:01
#5
copeland 2017-03-09 19:37:04
5. A rational number written in base eight is $\underline a \underline b.\underline c\underline d,$ where all digits are nonzero. The same number in base twelve is $\underline b \underline b.\underline b\underline a.$ Find the base-ten number $\underline a \underline b\underline c.$
copeland 2017-03-09 19:37:27
OK, great. We have a rational number that's not an integer.
copeland 2017-03-09 19:37:28
Which of the two expressions is going to be more fun to look at?
richuw 2017-03-09 19:38:12
bb.ba
espeon12 2017-03-09 19:38:12
the second one?
a1b2 2017-03-09 19:38:12
$bb.ba_{12}$
ilovemath04 2017-03-09 19:38:12
bb.ba
MrMXS 2017-03-09 19:38:12
the second?
ethanliu247 2017-03-09 19:38:12
second one
legolego 2017-03-09 19:38:12
second one with only two distinct digits
rapturt9 2017-03-09 19:38:12
the second only 2 variables
BuddyS 2017-03-09 19:38:12
base 12
vvluo 2017-03-09 19:38:12
bb.ba becase lez pozibilideez
Pandasareamazing. 2017-03-09 19:38:12
bb.ba
copeland 2017-03-09 19:38:15
The base-12 number has a lot of repeated digits. Let's think about it first. Can you write $\underline b \underline b.\underline b\underline a$ in a more meaningful way?
vishwathganesan 2017-03-09 19:39:08
12b+b+b/12+a/144
pi_Plus_45x23 2017-03-09 19:39:08
$13b+\frac{b}{12}+\frac{a}{144}$
lego101 2017-03-09 19:39:08
12b + b + b/12 + a/144
DylanX22 2017-03-09 19:39:08
12b + b + b/12 + a/144
math101010 2017-03-09 19:39:08
12b+b+b/12+a/144
Deathranger999 2017-03-09 19:39:08
12b + b + b/12 + a/144
nukelauncher 2017-03-09 19:39:08
13b + b/12 + a/144
GeneralCobra19 2017-03-09 19:39:08
12b+b+b/12+a/144
WhaleVomit 2017-03-09 19:39:08
12b+b+b/12+a/144
J1618 2017-03-09 19:39:08
12b +b +b/12 +a/144
copeland 2017-03-09 19:39:11
We can kind-of turn it into a "mixed number":\[12b+b+\frac{12b+a}{12^2}.\]
copeland 2017-03-09 19:39:20
As a "mixed number" the base-8 dude becomes \[8a+b+\frac{8c+d}{8^2}.\]
copeland 2017-03-09 19:39:22
\[12b+b+\frac{12b+a}{12^2}=8a+b+\frac{8c+d}{8^2}\]
copeland 2017-03-09 19:39:23
What else can we say about this equation?
NeeNeeMath 2017-03-09 19:40:08
loook at the integere parts first
espeon12 2017-03-09 19:40:08
the fractions are equivalent
EulerMacaroni 2017-03-09 19:40:08
integer and fractional parts are respectively equal
hodori01 2017-03-09 19:40:08
use integer and fractional parts
vishwathganesan 2017-03-09 19:40:08
equate the integral parts and the fractional parts
mathguy623 2017-03-09 19:40:08
fractional parts and integer parts are equal
DemonPlat4 2017-03-09 19:40:08
12b = 8a
lego101 2017-03-09 19:40:08
we know the integer parts are equal and the fractional parts are equal too since a, b, c, d are all integers
rapturt9 2017-03-09 19:40:08
the fracs equal each other
copeland 2017-03-09 19:40:10
Since these are base-12 and base-8 numbers, we know that the fractions are less than 1. So the integer parts and fractional parts are equal.
copeland 2017-03-09 19:40:12
The fractional part is nasty. The integer part is cute:\[12b+b=8a+b.\]
copeland 2017-03-09 19:40:13
What's that say?
pican 2017-03-09 19:40:45
$3b=2a$
tdeng 2017-03-09 19:40:45
3b=2a
letsgomath 2017-03-09 19:40:45
3b=2a
treemath 2017-03-09 19:40:45
3b=2a
LearningMath 2017-03-09 19:40:45
3b = 2a
rapturt9 2017-03-09 19:40:45
a=2b/3
ninjataco 2017-03-09 19:40:45
3b = 2a
QuestForKnowledge 2017-03-09 19:40:45
3b=2a
islander7 2017-03-09 19:40:45
3b=2a
stronto 2017-03-09 19:40:45
a = 3/2 b
bomb427006 2017-03-09 19:40:45
$2a=3b$
KYang 2017-03-09 19:40:45
3b=2a
copeland 2017-03-09 19:40:47
This says that $a=\dfrac32b$.
copeland 2017-03-09 19:40:48
What are all the possible values for $\underline a\underline b?$
legolego 2017-03-09 19:41:34
32, 64
quanhui868 2017-03-09 19:41:34
32, 64, 96
J1618 2017-03-09 19:41:34
32 and 64
jonzli123 2017-03-09 19:41:34
32 & 64
cooljoseph 2017-03-09 19:41:34
64, 32
LearningMath 2017-03-09 19:41:34
64, 32
ezhao02 2017-03-09 19:41:34
32, 64
amackenzie1 2017-03-09 19:41:34
32, 64
copeland 2017-03-09 19:41:37
Since these are both base-8 digits, they're less than 8. Therefore $b$ is either 2 or 4 and the corresponding $a$ is 3 or 6.
copeland 2017-03-09 19:41:39
Therefore the base-12 number is either $\underline 2 \underline 2.\underline 2\underline 3$ or $\underline 4 \underline 4.\underline 4\underline 6.$
copeland 2017-03-09 19:41:43
Let's convert $\underline 2 \underline 2.\underline 2\underline 3_{12}$ to a base-8 number. We know the integer part is $\underline 3\underline 2_{8}$. What is $.\underline 2\underline3_{12}$ when we convert it to base 8?
Funnybunny5246 2017-03-09 19:43:00
14
Pandasareamazing. 2017-03-09 19:43:00
.14
letsgomath 2017-03-09 19:43:00
0.14
vvluo 2017-03-09 19:43:00
.14
mathdragon2000 2017-03-09 19:43:00
.14
jonzli123 2017-03-09 19:43:00
.14
ethanliu247 2017-03-09 19:43:00
.14
islander7 2017-03-09 19:43:00
.14
copeland 2017-03-09 19:43:03
\[.\underline 2\underline3_{12}=\frac{27}{144}=\frac3{16}=\frac{12}{64}=.\underline1\underline4_{8}.\]
copeland 2017-03-09 19:43:12
Therefore $\underline 2 \underline 2.\underline 2\underline 3_{12} = \underline 3 \underline 2.\underline 1\underline 4_{8}$.
copeland 2017-03-09 19:43:14
And what is $\underline 4 \underline 4.\underline 4\underline 6_{12}$ when converted to base 8?
Radio2 2017-03-09 19:44:11
64.28
gabrielsui 2017-03-09 19:44:11
64.28
ezhao02 2017-03-09 19:44:11
64.30
Funnybunny5246 2017-03-09 19:44:11
64.30
Mrkiller 2017-03-09 19:44:11
64.30
IsaacZ123 2017-03-09 19:44:11
64.3
amzhao 2017-03-09 19:44:11
64.28
MountainHeight 2017-03-09 19:44:11
64.3
DemonPlat4 2017-03-09 19:44:11
64.30
mathguy623 2017-03-09 19:44:11
64.30
First 2017-03-09 19:44:11
64.3
copeland 2017-03-09 19:44:19
This is twice the previous number!\[\underline 4 \underline 4.\underline 4\underline 6_{12}=2(\underline 2 \underline 2.\underline 2\underline 3_{12}) = 2(\underline 3 \underline 2.\underline 1\underline 4_{8})=\underline 6 \underline 4.\underline 3\underline 0_{8}.\]
copeland 2017-03-09 19:44:26
Notice that base 8 you need to carry the 8.
copeland 2017-03-09 19:44:30
So are there two answers?
sxu 2017-03-09 19:44:55
no since nonzero
BooBooTM 2017-03-09 19:44:55
No!!!!
legolego 2017-03-09 19:44:55
you can't have 0
DemonPlat4 2017-03-09 19:44:55
NO - nonzero digits
vvluo 2017-03-09 19:44:55
but there is a 0........!!!!!
Pandasareamazing. 2017-03-09 19:44:55
non zero digits
jonzli123 2017-03-09 19:44:55
NO! All the digits have to be nonzero
Funnybunny5246 2017-03-09 19:44:55
One has a 0
mingxu 2017-03-09 19:44:55
nonzero!
owm 2017-03-09 19:44:55
No, the digits are nonzero
AlisonH 2017-03-09 19:44:55
all digits nonzero
GeronimoStilton 2017-03-09 19:44:55
All the digits are nonzero.
copeland 2017-03-09 19:44:57
No! The digits have to be positive. So what's the answer?
legolego 2017-03-09 19:45:28
321
AlisonH 2017-03-09 19:45:28
321
Picroft 2017-03-09 19:45:28
321
SomethingNeutral 2017-03-09 19:45:28
321
jonzli123 2017-03-09 19:45:28
321
vvluo 2017-03-09 19:45:28
321
mingxu 2017-03-09 19:45:28
321
smartpgp 2017-03-09 19:45:28
521
MrMXS 2017-03-09 19:45:28
$321$
ilovemath04 2017-03-09 19:45:28
321
Deathranger999 2017-03-09 19:45:28
321
math101010 2017-03-09 19:45:28
321
smartpgp 2017-03-09 19:45:28
531
Gamabyte 2017-03-09 19:45:28
321
winnertakeover 2017-03-09 19:45:28
321
copeland 2017-03-09 19:45:30
The only viable pair is $\underline 2 \underline 2.\underline 2\underline 3_{12} = \underline 3 \underline 2.\underline 1\underline 4_{8}$. Therefore $\underline a\underline b\underline c=\boxed{321}.$
copeland 2017-03-09 19:45:35
Also, the second digit after the decimal point in a base-12 number, $\underline 0.\underline0\underline1_{12},$ is the "grossths" place.
copeland 2017-03-09 19:45:39
I just made that up.
copeland 2017-03-09 19:45:40
Next?
GeronimoStilton 2017-03-09 19:45:59
Problem 6!
espeon12 2017-03-09 19:45:59
number 6
IsaacZ123 2017-03-09 19:45:59
number 6
vishwathganesan 2017-03-09 19:45:59
number 6 yay!!
lego101 2017-03-09 19:45:59
Number six, pick up sticks, and the pace too.
amzhao 2017-03-09 19:45:59
numero 6
MathTechFire 2017-03-09 19:45:59
6!!!!!!!!
gabrielsui 2017-03-09 19:45:59
6!
copeland 2017-03-09 19:46:03
6. A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x.$ Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\dfrac{14}{25}.$ Find the difference between the largest and smallest possible values of $x.$
copeland 2017-03-09 19:46:06
OK, here's a picture:
copeland 2017-03-09 19:46:07
copeland 2017-03-09 19:46:08
When does the chord intersect the triangle?
IsaacZ123 2017-03-09 19:46:53
when the two end points are from two different arc segments
lego101 2017-03-09 19:46:53
When the two points are on different sides of the triangle
a1b2 2017-03-09 19:46:53
If the two chosen points are on different arcs
winnertakeover 2017-03-09 19:46:53
when 2 separte arcs are chosen
vishwathganesan 2017-03-09 19:46:53
when both points are in different arcs of the circle
J1618 2017-03-09 19:46:53
When the two endpoints are on two different arcs
DemonPlat4 2017-03-09 19:46:53
if the points are in 2 different of the 3 "regions" on the circle
QuestForKnowledge 2017-03-09 19:46:53
when not on same section
GeneralCobra19 2017-03-09 19:46:53
When the points are across different "parts" of the circle divided by the triangle.
alifenix- 2017-03-09 19:46:53
when the points are on two different arcs out of the three made by the points on the triangle
copeland 2017-03-09 19:46:56
The triangle cuts off three segments of the circle. A chord intersects the triangle when its vertices lie on the arcs of different segments.
copeland 2017-03-09 19:47:21
I think it's likely that we're going to want to use complementary probability on this problem, since the chords that do not intersect the triangle are easier to think about.
copeland 2017-03-09 19:47:23
Let's pick two points at random on the circle. What's the probability that the two points are in the top right (red) region below?
copeland 2017-03-09 19:47:24
ninjataco 2017-03-09 19:48:05
(x/180)^2
tdeng 2017-03-09 19:48:05
(2x/360)^2
rapturt9 2017-03-09 19:48:05
x^2/(180^2)
Reef334 2017-03-09 19:48:05
x^2/180^2
Funnybunny5246 2017-03-09 19:48:05
$(x/180)^2$
rapturt9 2017-03-09 19:48:05
(x/180)^2
copeland 2017-03-09 19:48:09
That region is $\dfrac{2x}{360}=\dfrac{x}{180}$ of the circle. The probability that both points are in that region is $\dfrac{x^2}{180^2}$. Let's call that $p^2$.
copeland 2017-03-09 19:48:10
What about the blue region?
IsaacZ123 2017-03-09 19:48:40
same as p^2
NeeNeeMath 2017-03-09 19:48:40
same
rapturt9 2017-03-09 19:48:40
(x/180)^2, the same thing
GeronimoStilton 2017-03-09 19:48:40
$\frac{x^2}{180^2}$
IsaacZ123 2017-03-09 19:48:40
same as the red region
Peggy 2017-03-09 19:48:40
same
smartpgp 2017-03-09 19:48:40
p^2
copeland 2017-03-09 19:48:42
Since the triangle is isosceles, the blue and red regions are congruent, so the probability that both points lie in the blue region is also $p^2$.
copeland 2017-03-09 19:48:43
What's the probability that both points are in the green region?
Picroft 2017-03-09 19:50:33
(1-2p)^2
legolego 2017-03-09 19:50:33
(1-2p)^2
ezhao02 2017-03-09 19:50:33
$(1-p)^2$
GeronimoStilton 2017-03-09 19:50:33
$\frac{(180-2x)^2}{180^2}$
ezhao02 2017-03-09 19:50:33
$(1-2p)^2$
amzhao 2017-03-09 19:50:33
((180-2x)/180)^2
rapturt9 2017-03-09 19:50:33
(180-2x)^2/(180)^2
MathTechFire 2017-03-09 19:50:33
(180-2x/180)^2
copeland 2017-03-09 19:50:45
The probability that a point is in the green region is $1-2p$, so the probability that both points are in the green region is $(1-2p)^2=1-4p+4p^2$.
copeland 2017-03-09 19:50:48
And in terms of $p$, what is the probability that the chord intersects the triangle?
cooljoseph 2017-03-09 19:52:08
$1-(6p^2-4p+1)$
dandyq 2017-03-09 19:52:08
1 - (1 - 4p + 4p^2 + 2p^2)
GeronimoStilton 2017-03-09 19:52:08
$4p - 6p^2$
alifenix- 2017-03-09 19:52:08
$1 - 2p^2 - 1 + 4p - 4p^2 = 4p - 6p^2$
QuestForKnowledge 2017-03-09 19:52:08
1-(6p^2-4p+1)
MathTechFire 2017-03-09 19:52:08
1-4p+6p^2
ezhao02 2017-03-09 19:52:08
$1-(1-2p)^2-p^2-p^2$
copeland 2017-03-09 19:52:11
The probability that the chord intersects the probability is complementary to the probability that both points are in one of the three regions above, so it is \[1-p^2-p^2-(1-4p+4p^2)=4p-6p^2.\]
copeland 2017-03-09 19:52:13
We are trying to solve\[4p-6p^2=\dfrac{14}{25}.\]
copeland 2017-03-09 19:52:15
What are the solutions to this equation?
GeneralCobra19 2017-03-09 19:52:40
Is there a nicer way without quadratic formula?
copeland 2017-03-09 19:52:47
For sure. Shall I?
MrMXS 2017-03-09 19:53:00
go ahead
MegaProblemSolver 2017-03-09 19:53:00
yes
MathTechFire 2017-03-09 19:53:00
definitely
awesomemaths 2017-03-09 19:53:00
yep
Pandasareamazing. 2017-03-09 19:53:00
yess!
Ani10 2017-03-09 19:53:00
yas
jfmath04 2017-03-09 19:53:00
yup
amackenzie1 2017-03-09 19:53:02
yes please!
winnertakeover 2017-03-09 19:53:02
yes
copeland 2017-03-09 19:53:04
When we multiply by 25 we get \[150p^2-100p+14=0.\]Dividing by 2 gives\[75p^2-50p+7=0.\]
copeland 2017-03-09 19:53:13
I see $5p$ in there.
copeland 2017-03-09 19:53:15
In terms of $5p$ we have \[3(5p)^2-10(5p)+7=0.\] Since $3+7=10$, we see $5p=1$ as an immediate solution, so we're set up to factor. This quadratic factors as

\[\left((3(5p)-7\right)\left((5p)-1\right)=0.\]
copeland 2017-03-09 19:53:24
The solutions are $p=\dfrac{7}{15}$ and $p=\dfrac15$.
copeland 2017-03-09 19:53:26
But what do we need now?
MrMXS 2017-03-09 19:53:52
$x$
jonzli123 2017-03-09 19:53:52
x
islander7 2017-03-09 19:53:52
value of x
Deathranger999 2017-03-09 19:53:52
The value of x
Atg 2017-03-09 19:53:52
Multiply p by 180
MathTechFire 2017-03-09 19:53:52
x
Funnybunny5246 2017-03-09 19:53:52
x
vishwathganesan 2017-03-09 19:53:52
we want x in degrees, not p's
fdas 2017-03-09 19:53:54
the angles
guoziyanglovemath 2017-03-09 19:53:54
work out x
copeland 2017-03-09 19:53:56
We want things in terms of $x$. What are the two possible values of $x?$
algebra_star1234 2017-03-09 19:54:20
84 and 36
vvluo 2017-03-09 19:54:20
36,84
Deathranger999 2017-03-09 19:54:20
x = 84 and x = 36.
cooljoseph 2017-03-09 19:54:20
36 and 84
Radio2 2017-03-09 19:54:20
36 and 84
Reef334 2017-03-09 19:54:20
84 and 36
GeronimoStilton 2017-03-09 19:54:20
$36$ and $84$
copeland 2017-03-09 19:54:22
And the answer?
mathman3880 2017-03-09 19:54:39
180(7/15 - 1/5) = 48
First 2017-03-09 19:54:39
48 is the answer
stronto 2017-03-09 19:54:39
48
ninjataco 2017-03-09 19:54:39
048
pythonsquared 2017-03-09 19:54:39
48
math101010 2017-03-09 19:54:39
048
owm 2017-03-09 19:54:39
48
a1b2 2017-03-09 19:54:39
$\boxed{048}$
DemonPlat4 2017-03-09 19:54:39
84 - 36 = 048 :<
cooleybz2013 2017-03-09 19:54:40
48
copeland 2017-03-09 19:54:43
Since $p=\dfrac{x}{180}$, these two probabilities correspond to $x=84$ and $x=36$.
copeland 2017-03-09 19:54:47
The difference between these two values of $x$ is $84-36=\boxed{048}.$
copeland 2017-03-09 19:54:55
Alright, not halfway there but almost. What's next?
letsgomath 2017-03-09 19:55:15
#7
alphamom 2017-03-09 19:55:15
7!
IsaacZ123 2017-03-09 19:55:15
number 7
GeorgeAvocados 2017-03-09 19:55:15
number 7 oh thank heaven
Jyzhang12 2017-03-09 19:55:15
numero 7
copeland 2017-03-09 19:55:18
7. For nonnegative integers $a$ and $b$ with $a+b\leq6,$ let $T(a,b)=\dbinom6a\dbinom6b\dbinom6{a+b}.$ Let $S$ denote the sum of all $T(a,b),$ where $a$ and $b$ are nonnegative integers with $a+b\leq6.$ Find the remainder when $S$ is divided by 1000.
copeland 2017-03-09 19:55:19
I like the smell of this problem. There just has to be something really elegant going on here.
copeland 2017-03-09 19:55:20
Let's think about how we construct a combinatorial argument. First, what is $\dbinom6a$?
legolego 2017-03-09 19:55:45
choose a objects out of 6 objects
IsaacZ123 2017-03-09 19:55:45
number of ways to choose a people from 6
First 2017-03-09 19:55:45
I have 6 people and I need to pick a committee of a people
Mathaddict11 2017-03-09 19:55:45
choosing a people from a group of 6 people
espeon12 2017-03-09 19:55:45
choosing a things from 6 things
MrMXS 2017-03-09 19:55:45
the number of ways to choose $a$ objects from $6$ objects
letsgomath 2017-03-09 19:55:45
number of ways to choose a people out of 6
WhaleVomit 2017-03-09 19:55:45
number of ways to choose a people to murder out of 6 people
amyhu910 2017-03-09 19:55:45
the number of ways to choose a items out of 6
copeland 2017-03-09 19:55:48
This is the number of ways to choose $a$ elements from a set of size 6. For example, we could be picking $a$ rabbits to put hats on.
copeland 2017-03-09 19:55:54
Of course there are a lot of equally nice stories that we can write, from block-walking to organizing strings of letters, etc., but they all boil down to the rabbit thing.
copeland 2017-03-09 19:55:56
So here we have a set of 6 things and we're picking $a$ of them.
copeland 2017-03-09 19:56:00
We also have another set of 6 things and we're picking $b$ of them. (Maybe it's the same set. Keep that option in mind.)
copeland 2017-03-09 19:56:04
Finally we have a third set and we're picking $a+b$ of them.
copeland 2017-03-09 19:56:07
So we have 6 rabbits and we're putting $a$ hats on them. We have 6 coyotes and we're putting $b$ hats on them. We have 6 snowmen and we're putting $a+b$ hats on them.
copeland 2017-03-09 19:56:15
Snowmen love hats.
copeland 2017-03-09 19:56:20
I don't see it yet. Is there some nice trick we can play with these sets?
lego101 2017-03-09 19:56:35
wouldnt the coyotes eat the rabbits
copeland 2017-03-09 19:56:36
It depends on $b-a$.
ninjataco 2017-03-09 19:57:18
6C(a+b) = 6C(6-a-b)
amackenzie1 2017-03-09 19:57:18
6 choose a + b = 6 choose 6 -(a + b)
GeronimoStilton 2017-03-09 19:57:18
We're choosing $6-a-b$ snowmen not to put hats on.
legolego 2017-03-09 19:57:18
6 choose (a+b) = 6 choose (6 - a - b)
vishwathganesan 2017-03-09 19:57:18
wait, we can also choose 6-a-b snowmen not to put hats on
copeland 2017-03-09 19:57:23
The ways to pick $a+b$ snowmen from a set of 6 are the same as the ways to pick $6-a-b$ snowmen from a set of 6.
copeland 2017-03-09 19:57:26
So this counts the ways to pick $a$ rabbits and then pick $b$ coyotes and then pick $6-a-b$ snowmen. How is that better?
IsaacZ123 2017-03-09 19:58:21
we are giving a total of 6 hats
vishwathganesan 2017-03-09 19:58:21
pick 6 out of 18 total
legolego 2017-03-09 19:58:21
our sum equals 18 choose 6
GeronimoStilton 2017-03-09 19:58:21
We're picking $6$ members out of a set with $18$ elements!
amackenzie1 2017-03-09 19:58:21
basically you're putting 6 hats on to 18 things
cooljoseph 2017-03-09 19:58:21
Well, this is the same as choosing 6 things out of 18 to put hats on.
IsaacZ123 2017-03-09 19:58:21
we are choosing a total of 6 out of 18
espeon12 2017-03-09 19:58:21
if you combine all the snowmen, coyotes, and rabbits, you're choosing 6 to put hats on
islander7 2017-03-09 19:58:21
18 objects choose 6 of them
sxu 2017-03-09 19:58:21
6 hats, give a to rabbits, b to coyotes, rest to snowmen
copeland 2017-03-09 19:58:34
Now we're counting ways to pick 6 critters out of a full set of 18 critters. Say that more precisely for me.
awesomemaths 2017-03-09 19:58:59
18 choose 6
MrMXS 2017-03-09 19:58:59
$\binom{18}{6}$
GeronimoStilton 2017-03-09 19:58:59
We want $\dbinom{18}{6}$
DemonPlat4 2017-03-09 19:58:59
18 choose 6
alifenix- 2017-03-09 19:58:59
thats $\binom{18}{6}$, wish I realized this lol
nukelauncher 2017-03-09 19:58:59
18 choose 6
IsaacZ123 2017-03-09 19:58:59
18C6
QuestForKnowledge 2017-03-09 19:58:59
18 chose 6
copeland 2017-03-09 19:59:01
Since we are summing over all possible $a$ and $b$, we are looking at all the ways to pick some number of bunnies, $a$, out of the first set, some other number of coyotes $b$ out of the second set, and then pulling whatever else is needed, $6-a-b$ snowmen, from the third set to create a set of 6 critters.
copeland 2017-03-09 19:59:07
That's just $\dbinom{18}6$.
copeland 2017-03-09 19:59:08
\[\binom{18}6=\frac{18\cdot17\cdot16\cdot15\cdot14\cdot13}{6\cdot5\cdot4\cdot3\cdot2\cdot1}.\]
copeland 2017-03-09 19:59:09
Anybody know what that is modulo 1000?
legolego 2017-03-09 19:59:46
564
thinkinavi 2017-03-09 19:59:46
564
IsaacZ123 2017-03-09 19:59:46
564
samuel 2017-03-09 19:59:46
564?
math101010 2017-03-09 19:59:46
564
awesomemaths 2017-03-09 19:59:46
too lazy to compute
algebra_star1234 2017-03-09 19:59:46
564!!!!
letsgomath 2017-03-09 19:59:46
564
GeronimoStilton 2017-03-09 19:59:46
It's congruent to $564$ modulo 1000.
GeneralCobra19 2017-03-09 19:59:46
My calculator says 564
ilikepie2003 2017-03-09 19:59:46
no.
DemonPlat4 2017-03-09 19:59:46
*cricket-chirp*
brainiac1 2017-03-09 19:59:46
564, clearly
Lance57 2017-03-09 19:59:46
564
copeland 2017-03-09 19:59:49
\begin{align*}

\frac{18\cdot17\cdot16\cdot15\cdot14\cdot13}{6\cdot5\cdot4\cdot3\cdot2\cdot1}

&=17\cdot2\cdot3\cdot14\cdot13\\

&=102\cdot182\\

&=18{,}200+364\\

&=18{,}564.

\end{align*}
copeland 2017-03-09 19:59:56
The answer is $\boxed{564}$.
copeland 2017-03-09 20:00:01
This is a case of the Vandermonde's Identity, which is the general name for all of the identities that you get by writing $\dbinom nm$ as a sum by breaking $n$ into subsets like this.
copeland 2017-03-09 20:00:13
Now we're definitely almost halfway there maybe.
First 2017-03-09 20:00:36
#8!
GeronimoStilton 2017-03-09 20:00:36
Problem 8!
MathTechFire 2017-03-09 20:00:36
POUND 8
ethanliu247 2017-03-09 20:00:36
Great! number 8 next!
IsaacZ123 2017-03-09 20:00:36
number 8 is not half of 15
jkittykitkat 2017-03-09 20:00:36
number 8
Pandasareamazing. 2017-03-09 20:00:36
On to number 8 we go!
copeland 2017-03-09 20:00:46
8. Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0,75).$ Let $O$ and $P$ be two points in the plane with $OP=200.$ Let $Q$ and $R$ be points on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b,$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR\leq 100$ is equal to $\dfrac mn,$ where $m$ and $N$ are relatively prime positive integers. Find $m+n$.
copeland 2017-03-09 20:00:49
Where should we start?
IsaacZ123 2017-03-09 20:01:10
drawing a picture
NewbieGamer 2017-03-09 20:01:10
Draw
mathman3880 2017-03-09 20:01:10
diagram
jonzli123 2017-03-09 20:01:10
draw a diagram!!!
awesomemaths 2017-03-09 20:01:10
diagram
Pandasareamazing. 2017-03-09 20:01:10
draw a diagram
ninjataco 2017-03-09 20:01:10
diagram
stronto 2017-03-09 20:01:10
DIAGRAM
copeland 2017-03-09 20:01:16
Let's draw a picture! What does the right angle constraint on $\angle OQP$ and $\angle ORP$ tell us?
fdas 2017-03-09 20:02:11
Draw a circle with OP as a diameter
jkittykitkat 2017-03-09 20:02:11
It makes a circle
DemonPlat4 2017-03-09 20:02:11
on a semicircle
ethanliu247 2017-03-09 20:02:11
in a circle
tdeng 2017-03-09 20:02:11
They are on a semicircle
WhaleVomit 2017-03-09 20:02:11
they are on circle with diameter OP
islander7 2017-03-09 20:02:11
on semicircle with diameter pq
fdas 2017-03-09 20:02:11
they are in a circle
GeronimoStilton 2017-03-09 20:02:11
$Q$ and $R$ are on a semicircle.
copeland 2017-03-09 20:02:13
If $\angle OQP=90^\circ$ then $Q$ lies on the circle with diameter $OP.$ So we're picking two points on this arc at random (for some notion of "random").
copeland 2017-03-09 20:02:15
copeland 2017-03-09 20:02:16
So, the center is $\heartsuit$. It's not my fault; the problem-writer already took $O.$
copeland 2017-03-09 20:02:21
OK, so when is $QR<100?$
dt800298 2017-03-09 20:03:31
when the minor arc QR is less than 60 degrees
AlisonH 2017-03-09 20:03:31
angle Q(heart)R is less than 60
Root01 2017-03-09 20:03:31
When arc formed by QR is less than 60 degrees
tdeng 2017-03-09 20:03:31
When $\angle Q\heart R \leq 60$
Celebrated 2017-03-09 20:03:31
arc QR is less than or equal to 60
WhaleVomit 2017-03-09 20:03:31
when <Q<3R < 60 degrees
rapturt9 2017-03-09 20:03:31
When angle QheartR is <60 degrees
Radio2 2017-03-09 20:03:31
When $\angle \heartsuit > 60^\circ$
copeland 2017-03-09 20:03:33
(Props for <Q<3R<60.)
copeland 2017-03-09 20:03:36
The radius of the circle is $100$, so if $QR=100$ then $\triangle\heartsuit QR$ is equilateral.
copeland 2017-03-09 20:03:38
What is $\angle Q\heartsuit R$ in terms of $a$ and $b?$
vishwathganesan 2017-03-09 20:04:52
when |a-b| < 30
stronto 2017-03-09 20:04:52
2a-2b
WhaleVomit 2017-03-09 20:04:52
|2a-2b|
ninjataco 2017-03-09 20:04:52
|2a-2b|
guoziyanglovemath 2017-03-09 20:04:52
2|a-b|
vvluo 2017-03-09 20:04:52
a-b<=30
jonzli123 2017-03-09 20:04:52
|2b-2a|
rapturt9 2017-03-09 20:04:52
The absolute value of 2a-2b
copeland 2017-03-09 20:04:54
Since $\angle POQ=a$ the central angle is $\angle P\heartsuit Q=2a$. Likewise, $\angle P\heartsuit R=2b.$ Therefore $\angle Q\heartsuit R=|2a-2b|.$
copeland 2017-03-09 20:04:57
We want to find the probability that two angles $a$ and $b$ chosen uniformly from $[0,75]$ satisfy $|2a-2b|<60.$
copeland 2017-03-09 20:05:00
This is a tool for. . .
IsaacZ123 2017-03-09 20:05:43
geometric probability
ninjataco 2017-03-09 20:05:43
geometric probability
First 2017-03-09 20:05:43
geometric probability
stronto 2017-03-09 20:05:43
geometric probability
a1b2 2017-03-09 20:05:43
Geometric probability
pythonsquared 2017-03-09 20:05:43
geometric probability
tdeng 2017-03-09 20:05:43
Geometric probability
WhaleVomit 2017-03-09 20:05:43
geometric probability
fdas 2017-03-09 20:05:43
geometric probability
Radio2 2017-03-09 20:05:43
geometric probability
copeland 2017-03-09 20:05:45
Geometric probability! First we draw a rectangle representing the possible values for $a$ and $b$:
copeland 2017-03-09 20:05:46
copeland 2017-03-09 20:05:47
And what does the success region look like?
IsaacZ123 2017-03-09 20:06:38
two triangles in the corners
tdeng 2017-03-09 20:06:38
A hexagon
vishwathganesan 2017-03-09 20:06:38
it is a hexagon
andsun19 2017-03-09 20:06:38
diagonal
brainiac1 2017-03-09 20:06:38
a strip along the long diagonal
NewbieGamer 2017-03-09 20:06:38
Square minus two triangles
rapturt9 2017-03-09 20:06:38
a wide diagonal line
Blue_Whale 2017-03-09 20:06:38
diagonal stripe from bottomleft to topright?
J1618 2017-03-09 20:06:38
A diagonal strip with horizontal width 30
ninjataco 2017-03-09 20:06:38
chop off two triangles in the upper left and lower right corners
legolego 2017-03-09 20:06:38
a stripe
vvluo 2017-03-09 20:06:38
stripe with 2 isosceles right triangles being unsuccessful
J1618 2017-03-09 20:06:38
A diagonal strip from b=30 and a=30
copeland 2017-03-09 20:06:41
The success region is where $a$ and $b$ are within 30 of one another. We want to draw the lines $b=a+30$ and $b=a-30,$ since those are the boundaries of the successful region.
copeland 2017-03-09 20:06:43
The whole diagonal is in the success region, so we want to color between the boundary lines.
copeland 2017-03-09 20:06:44
copeland 2017-03-09 20:06:49
What do we want to compute?
IsaacZ123 2017-03-09 20:07:23
the area of the shaded region/total area
cooleybz2013 2017-03-09 20:07:23
the area ratio
Celebrated 2017-03-09 20:07:23
shaded region in terms of whole thig
math129 2017-03-09 20:07:23
area of stripe/area of square
Radio2 2017-03-09 20:07:23
the shaded divided by the total
mathboy4 2017-03-09 20:07:23
the shaded place
stronto 2017-03-09 20:07:23
shaded region/total area
copeland 2017-03-09 20:07:30
We want to compute the proportion of the square that is gray.
copeland 2017-03-09 20:07:32
What should we do first?
copeland 2017-03-09 20:07:40
Eh. That's a bad question.
vishwathganesan 2017-03-09 20:07:50
scale it!!!!!!!1
copeland 2017-03-09 20:07:56
But that's a good answer:
copeland 2017-03-09 20:07:57
It's fine to rescale everything in sight by 15, since that won't change the proportions. Here's a better picture:
copeland 2017-03-09 20:07:58
copeland 2017-03-09 20:08:22
What's the probability that $QR$ is less than 100?
letsgomath 2017-03-09 20:08:54
16/25
sxu 2017-03-09 20:08:54
16/25
jkittykitkat 2017-03-09 20:08:54
25-9=16. 16/25
bogstop320 2017-03-09 20:08:54
16/25
mathman3880 2017-03-09 20:08:54
16/25
Reef334 2017-03-09 20:08:54
16/25
islander7 2017-03-09 20:08:54
16/25
Ani10 2017-03-09 20:08:54
16/25
MountainHeight 2017-03-09 20:08:54
16/25
owm 2017-03-09 20:08:54
16/25
copeland 2017-03-09 20:08:58
The white area can be shoved together to get a $3\times3$ square, so its area is 9.
copeland 2017-03-09 20:08:59
The full square has area 25 and the grey region has area $25-9=16$. Therefore the probability of success is $\dfrac{16}{25}.$ The answer is $16+25=\boxed{041}.$
copeland 2017-03-09 20:09:02
Theorem: the probability that a geometric probability problem uses some version of this diagram is also $\dfrac{16}{25}.$
copeland 2017-03-09 20:09:04
copeland 2017-03-09 20:09:14
Ready for the next one?
Celebrated 2017-03-09 20:09:34
prob 9
Celebrated 2017-03-09 20:09:34
yea
IsaacZ123 2017-03-09 20:09:34
wait what
MegaProblemSolver 2017-03-09 20:09:34
yeah!
jonzli123 2017-03-09 20:09:34
yeah!
jfmath04 2017-03-09 20:09:34
yas
yrnsmurf 2017-03-09 20:09:34
yes #9
IsaacZ123 2017-03-09 20:09:34
what was the theorem u just stated
lego101 2017-03-09 20:09:34
Number nine, lookin fine.
jkittykitkat 2017-03-09 20:09:34
Time to get another one wrong
copeland 2017-03-09 20:09:52
9. Let $a_{10}=10,$ and for each integer $n>10$ let $a_n=100a_{n-1}+n.$ Find the least $n>10$ such that $a_n$ is a multiple of 99.
copeland 2017-03-09 20:09:58
So we have "is a multiple of 99", which means. . .
dhruv 2017-03-09 20:10:41
mods
sxu 2017-03-09 20:10:41
mod 99
First 2017-03-09 20:10:41
$\pmod{99}$
GeronimoStilton 2017-03-09 20:10:41
Express it modulo $99$!
ninjataco 2017-03-09 20:10:41
a_n == 0 (mod 99)
Mrkiller 2017-03-09 20:10:41
congruent to 0 mod 99
yojan_sushi 2017-03-09 20:10:41
0 mod 99
jkittykitkat 2017-03-09 20:10:41
0 mod 99
pythonsquared 2017-03-09 20:10:41
modular arithmetic
jonzli123 2017-03-09 20:10:41
mod 99
copeland 2017-03-09 20:10:49
Consider everything modulo 99. What's the first simplification of the problem?
copeland 2017-03-09 20:11:00
(Also, nobody loves the CRT more than I do, but let's hold off until we need it.)
Mrkiller 2017-03-09 20:11:41
so 100 becomes 1
QuestForKnowledge 2017-03-09 20:11:41
an=a(n-1)+n MUCH nicer
Reef334 2017-03-09 20:11:41
a_n = a_{n-1)+n
LittleChimp 2017-03-09 20:11:41
change the 100 to a 1
sxu 2017-03-09 20:11:41
$a_n=a_{n-1}+n$
amackenzie1 2017-03-09 20:11:41
$100{a_{n-1}} = {a_{n-1}}$.
jkittykitkat 2017-03-09 20:11:41
an=a(n-1)+n
MountainHeight 2017-03-09 20:11:41
a_n = a_(n-1) + n
copeland 2017-03-09 20:11:43
Well, $100\equiv1\pmod{99},$ so we should think of the recursion $a_n=a_{n-1}+n$.
copeland 2017-03-09 20:11:45
What's that the recursion for?
yrnsmurf 2017-03-09 20:12:17
triangular numbers
letsgomath 2017-03-09 20:12:17
triangular numbers
First 2017-03-09 20:12:17
Triangular numbers
DemonPlat4 2017-03-09 20:12:17
triangular numbers
amackenzie1 2017-03-09 20:12:17
triangle numbers
tdeng 2017-03-09 20:12:17
Triangular numbers
ethanliu247 2017-03-09 20:12:17
triangular numbers
copeland 2017-03-09 20:12:19
That's the recursion for the triangle numbers. What's the relationship between $a_n$ and $\dfrac{n(n+1)}2?$
yrnsmurf 2017-03-09 20:13:02
an=n(n+1)/2-45
letsgomath 2017-03-09 20:13:02
missing 1+2+3+4+5+6+7+8+9
tdeng 2017-03-09 20:13:02
a_n=n(n+1)/2-45
IsaacZ123 2017-03-09 20:13:02
we have to take out 1+2+3+4+5+6+7+8+9
vishwathganesan 2017-03-09 20:13:02
the first is 45 more than the seconf
islander7 2017-03-09 20:13:02
a_n=n(n+1)/2-45
vishwathganesan 2017-03-09 20:13:05
i mean 45 less
copeland 2017-03-09 20:13:08
The difference is that we forgot to add in the first $1+2+\cdots+9=\dfrac{9(10)}2$.
copeland 2017-03-09 20:13:12
\[a_n=\dfrac{n(n+1)}2-\dfrac{9\cdot10}2.\]
copeland 2017-03-09 20:13:14
This is a polynomial in $n$. Check this trick out: Do you see any roots of this polynomial?
ninjataco 2017-03-09 20:13:39
9
math129 2017-03-09 20:13:39
n=9
amackenzie1 2017-03-09 20:13:39
9 is a root
IsaacZ123 2017-03-09 20:13:39
n=9
DemonPlat4 2017-03-09 20:13:39
n=9
brainiac1 2017-03-09 20:13:39
9
copeland 2017-03-09 20:13:43
If you set $n=9$ you get \[\dfrac{n(n+1)}2-\dfrac{9\cdot10}2=\dfrac{9\cdot10}2-\dfrac{9\cdot10}2=0.\] That makes sense since $a_9=0$ must be true if $a_{10}=10$.
copeland 2017-03-09 20:13:43
See the other root?
nukelauncher 2017-03-09 20:13:58
9 and -10
brainiac1 2017-03-09 20:13:58
9 and -10
brainiac1 2017-03-09 20:13:58
-10
QuestForKnowledge 2017-03-09 20:13:58
-10
DemonPlat4 2017-03-09 20:13:58
n=-10
Funnybunny5246 2017-03-09 20:13:58
-10
NeeNeeMath 2017-03-09 20:13:58
-10
fdas 2017-03-09 20:13:58
-10
copeland 2017-03-09 20:13:59
If you set $n=-10$ you get \[\dfrac{n(n+1)}2-\dfrac{9\cdot10}2=\dfrac{(-10)(-9)}2-\dfrac{9\cdot10}2=0.\]
copeland 2017-03-09 20:14:00
Therefore $a_n=\dfrac12(n-9)(n+10).$
copeland 2017-03-09 20:14:03
Incidentally, difference of rectangles is a thing. Try to find a geometric proof that $a(a+x)-b(b+x)=(a-b)(a+b+x)$. Difference of squares is $x=0$, and we just used the $x=1$ case.
copeland 2017-03-09 20:14:11
But not now.
copeland 2017-03-09 20:14:13
What are we looking for?
letsgomath 2017-03-09 20:15:10
0 mod 99
QuestForKnowledge 2017-03-09 20:15:10
a n such that the equation is 0 mod 99
jkittykitkat 2017-03-09 20:15:10
it being a multiple of 99
a1b2 2017-03-09 20:15:10
$\equiv 0 \mod 99$
amackenzie1 2017-03-09 20:15:10
a_n to be a multiple of $99$.
IsaacZ123 2017-03-09 20:15:10
looking for the next time it is 0 mod 99
guoziyanglovemath 2017-03-09 20:15:10
multiple of 99
DjokerNole 2017-03-09 20:15:10
find the least a_n is a multiple of 99
First 2017-03-09 20:15:10
0 \mod 99
vvluo 2017-03-09 20:15:10
multiple of 99
copeland 2017-03-09 20:15:14
Since 99 is odd, we're looking for the first $n>10$ such that $99\mid (n-9)(n+10)$.
copeland 2017-03-09 20:15:17
Can we simplify this a little?
copeland 2017-03-09 20:16:29
We could take a swing at CRT. Do we need to?
letsgomath 2017-03-09 20:16:46
9 and 11
NeeNeeMath 2017-03-09 20:16:46
yeah crt
IsaacZ123 2017-03-09 20:16:46
99=9*11
NeeNeeMath 2017-03-09 20:16:46
divisible b 9 and 11
jkittykitkat 2017-03-09 20:16:46
do the case for 11 and 9 seperately
Radio2 2017-03-09 20:16:46
no
IsaacZ123 2017-03-09 20:16:46
no
NeeNeeMath 2017-03-09 20:16:46
noooo
DjokerNole 2017-03-09 20:16:46
no
copeland 2017-03-09 20:17:10
Let's just start searching for the answer.
copeland 2017-03-09 20:17:22
I want to make life easier by letting $m=n-9$ then we want the first $m>1$ such that $99\mid m(m+19)$.
copeland 2017-03-09 20:17:30
Now what?
Dr4gon39 2017-03-09 20:18:00
BASH
copeland 2017-03-09 20:18:02
Yeah, there had to be a point where we bash. This is the AIME after all.
fdas 2017-03-09 20:18:08
Start listing ms that create a multiple of 11
legolego 2017-03-09 20:18:12
either factor must have a factor of 11
copeland 2017-03-09 20:18:18
One of $m$ and $m+19$ is a multiple of 11. Let's start listing multiples of 11:
copeland 2017-03-09 20:18:20
\[\begin{array}{c|c|c}

x&x-19&x+19\\

\hline

11&-8&30\\

22&3&41\\

33&14&52\\

44&25&63\\

55&36&74\\

66&47&85\\

77&58&96\\

\end{array}\]
copeland 2017-03-09 20:18:23
See any winners on that list?
jkittykitkat 2017-03-09 20:19:10
55
GeronimoStilton 2017-03-09 20:19:10
$x = 44$
islander7 2017-03-09 20:19:10
55
legolego 2017-03-09 20:19:10
55!
GeronimoStilton 2017-03-09 20:19:10
$x = 55$
amackenzie1 2017-03-09 20:19:10
55 - 36
yrnsmurf 2017-03-09 20:19:10
63 and 36 for 44 and 55
GeronimoStilton 2017-03-09 20:19:10
$x = 44,55$
QuestForKnowledge 2017-03-09 20:19:10
x=44
Funnybunny5246 2017-03-09 20:19:10
36
vishwathganesan 2017-03-09 20:19:10
63?
amyhu910 2017-03-09 20:19:10
x=44
IsaacZ123 2017-03-09 20:19:10
55
copeland 2017-03-09 20:19:12
The pairs $44\cdot63$ and $36\cdot55$ both work.
copeland 2017-03-09 20:19:13
So what's $m?$
copeland 2017-03-09 20:20:16
Remember that $m$ is the smallest number such that $m(m+19)$ is a multiple of 99.
legolego 2017-03-09 20:20:28
36
GeronimoStilton 2017-03-09 20:20:28
$36$
fdas 2017-03-09 20:20:28
36 because it is lower
Pandasareamazing. 2017-03-09 20:20:28
36
stronto 2017-03-09 20:20:28
36
guoziyanglovemath 2017-03-09 20:20:28
36
yrnsmurf 2017-03-09 20:20:28
36
Reef334 2017-03-09 20:20:28
36
islander7 2017-03-09 20:20:28
36
Jyzhang12 2017-03-09 20:20:28
36
copeland 2017-03-09 20:20:32
Since there aren't going to be any numbers less than 36 "lower" on the table, the first multiple of 99 comes with $m=36.$
copeland 2017-03-09 20:20:32
And what's the final answer?
nukelauncher 2017-03-09 20:21:01
045
brainiac1 2017-03-09 20:21:01
45
brainiac1 2017-03-09 20:21:01
45
strategos21 2017-03-09 20:21:01
045
rapturt9 2017-03-09 20:21:01
$45$
fdas 2017-03-09 20:21:01
45
mshanmugam 2017-03-09 20:21:01
45
stronto 2017-03-09 20:21:01
45
LittleChimp 2017-03-09 20:21:01
045
Pandasareamazing. 2017-03-09 20:21:01
36+9=45
thinkinavi 2017-03-09 20:21:01
045
copeland 2017-03-09 20:21:04
The first multiple of 99 comes when $n=9+36=\boxed{045}$, when $a_{45}=\dfrac{36\cdot55}2.$
copeland 2017-03-09 20:21:10
Great.
copeland 2017-03-09 20:21:37
I'm sad we didn't get to CRT, but maybe we'll get to later.
Lance57 2017-03-09 20:22:13
Question 10!
Pandasareamazing. 2017-03-09 20:22:13
now we go to #10
lego101 2017-03-09 20:22:13
next problem?
MathTechFire 2017-03-09 20:22:13
zen 10
copeland 2017-03-09 20:22:22
CRT is the Chinese Remainder Theorem.
copeland 2017-03-09 20:22:27
10. Let $z_1=18+83i,$ $z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that \[\frac{z_3-z_1}{z_2-z_1}\cdot\frac{z-z_2}{z-z_3}\] is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$.
copeland 2017-03-09 20:22:30
OK, so that's a lot of fancy complex numbers stuff. I think this is probably a geometry problem. Think for a second where those three points are in the complex plane and I'll draw them for you.
copeland 2017-03-09 20:22:45
copeland 2017-03-09 20:22:49
Let's focus on the hard part.
copeland 2017-03-09 20:22:51
We have this crazy expression, \[\frac{z_3-z_1}{z_2-z_1}\cdot\frac{z-z_2}{z-z_3},\]and we want to know when it's real. What does being real tell us about the geometry here?
ninjataco 2017-03-09 20:23:28
on the x-axis
a1b2 2017-03-09 20:23:28
It is on the real axis
espeon12 2017-03-09 20:23:28
it's on teh x axis?
Ani10 2017-03-09 20:23:28
lies on the x axis
tdeng 2017-03-09 20:23:28
It's on the real axis
JJShan26 2017-03-09 20:23:28
it's on the x axis
NeeNeeMath 2017-03-09 20:23:28
x axis
copeland 2017-03-09 20:23:29
So a number is real when it's on the "$x$-axis".
copeland 2017-03-09 20:23:37
When have we seen a product be real?
IsaacZ123 2017-03-09 20:24:14
two conjugates
First 2017-03-09 20:24:14
cojugates
vishwathganesan 2017-03-09 20:24:14
conjugates
ethanliu247 2017-03-09 20:24:14
when the complex number are conjugates
sxu 2017-03-09 20:24:14
complex conjugates?
GeneralCobra19 2017-03-09 20:24:14
conjugates
copeland 2017-03-09 20:24:15
Fabulous! Multiplying conjugates gives us a real number.
copeland 2017-03-09 20:24:43
Complex conjugates have an angle relationship and a length relationship. Which one does reality care about?
richuw 2017-03-09 20:25:34
Degree measures add up to be 180n
islander7 2017-03-09 20:25:34
angle
tree3 2017-03-09 20:25:34
angle
a1b2 2017-03-09 20:25:34
Angle
IsaacZ123 2017-03-09 20:25:34
angle relationshiop
fishy15 2017-03-09 20:25:34
angle
samuel 2017-03-09 20:25:34
angle!
dr3463 2017-03-09 20:25:34
angle
samuel 2017-03-09 20:25:36
length doesn't matter
copeland 2017-03-09 20:25:37
If we change the lengths, it doesn't change reality. This is a statement about angles.
copeland 2017-03-09 20:26:11
We've been talking about angles that add to zero. How can we convert that to a statement like, "angles are equal?"
dr3463 2017-03-09 20:26:50
reflection
copeland 2017-03-09 20:26:58
Cool, we could reflect one of the numbers.
copeland 2017-03-09 20:27:10
How do you reflect a number across the real axis?
a1b2 2017-03-09 20:27:50
Complex Conjugate
MountainHeight 2017-03-09 20:27:50
take its conjugate
ninjataco 2017-03-09 20:27:50
take its conjugate
espeon12 2017-03-09 20:27:50
graph its conjugate
copeland 2017-03-09 20:27:56
That's good. Anything else?
yrnsmurf 2017-03-09 20:28:09
take its reciprocal
MSTang 2017-03-09 20:28:09
Reciprocate
copeland 2017-03-09 20:28:24
Great. The argument of $\dfrac1z$ is the negative of the argument of $z.$
copeland 2017-03-09 20:28:26
If we have a pair of complex numbers $u$ and $v$ then $\dfrac u v$ is real when $u$ and $v$ can be written with the same argument. That is, if $u=re^{i\theta}$ and $v=se^{i\phi}$ then in order for $\dfrac uv$ to be real, we have to have $\theta=\phi$ (or possibly $\theta=\phi+\pi$).
copeland 2017-03-09 20:28:34
So in order for $\dfrac{z_3-z_1}{z_2-z_1}\cdot\dfrac{z-z_2}{z-z_3}$ to be real, we would need $\dfrac{z_3-z_1}{z_2-z_1}$ and $\dfrac{z-z_3}{z-z_2}$ to have the same argument.
copeland 2017-03-09 20:28:44
Where can I find the argument of $\dfrac{z_3-z_1}{z_2-z_1}$ on this diagram?
copeland 2017-03-09 20:28:53
copeland 2017-03-09 20:30:03
Where can I find $z_3-z_1?$
IsaacZ123 2017-03-09 20:30:28
the line connecting z_3 and z_1
GeronimoStilton 2017-03-09 20:30:28
The line from $z_3$ to $z_1$.
copeland 2017-03-09 20:30:30
That's the "vector" that points from $z_1$ to $z_3.$
copeland 2017-03-09 20:31:19
And $z_2-z_1$ is the "vector" pointing from $z_1$ to $z_2$.
copeland 2017-03-09 20:31:20
What is the angle of their quotient?
IsaacZ123 2017-03-09 20:31:35
draw $z_3z_1z_2$
KYang 2017-03-09 20:31:35
the angle z2 z1 z3 ??
IsaacZ123 2017-03-09 20:31:35
the angle of $z_2z_1z_3$
yrnsmurf 2017-03-09 20:31:35
the angle of <z3z1z2
GeronimoStilton 2017-03-09 20:31:35
The angle $Z_2Z_1Z_3$?
brainiac1 2017-03-09 20:31:35
the angle from z2 to z3 through z1
copeland 2017-03-09 20:31:37
$z_3-z_1$ is the "vector" that points from $z_1$ to $z_3$ and $z_2-z_1$ is the "vector" that points from $z_1$ to $z_2$. The argument of their quotient is the angle at $z_1$ from $z_3$ to $z_2$.
copeland 2017-03-09 20:31:39
copeland 2017-03-09 20:31:51
And if we have a $z$ such that $\dfrac{z-z_3}{z-z_2}$ has the same argument, what does that say?
islander7 2017-03-09 20:32:12
same angle
vishwathganesan 2017-03-09 20:32:12
similar triangles?
copeland 2017-03-09 20:32:13
So?
cjquines0 2017-03-09 20:32:44
it's concyclic with $z1, z2, z3$!
yrnsmurf 2017-03-09 20:32:44
it lies on the circle with z1z2z3
MSTang 2017-03-09 20:32:44
cyclic quad!!
sxu 2017-03-09 20:32:44
z lies on the same arc?
mathman3880 2017-03-09 20:32:44
the quadrilateral is cyclic
Root01 2017-03-09 20:32:44
On the circumcircle of z1,z2,z3
Mrkiller 2017-03-09 20:32:44
a semicircle?
copeland 2017-03-09 20:32:51
We can rewrite $\dfrac{z_3-z}{z_2-z}=\dfrac{z-z_3}{z-z_2}$. This is the same expression that we had for $z_1$ above. That means that the angle at $z$ from $z_3$ to $z_2$ is the same as the corresponding angle at $z_1$, so $\angle z_3z_1z_2=\angle z_3zz_2$.
copeland 2017-03-09 20:32:57
The set of $z$ that we care about are the points on the circumcircle of $\triangle z_3z_1z_2$.
copeland 2017-03-09 20:32:58
copeland 2017-03-09 20:33:02
[If you want to space out at this picture more later, think about this: The function $f(z)=\dfrac{z_3-z_1}{z_2-z_1}\cdot\dfrac{z-z_2}{z-z_3}$ takes points on this circle and spits out real numbers. That means the circle is identified with the real line. Where is 0? Where is $\infty?$ What is the difference between the positive and negatives? Where are the integers?]
copeland 2017-03-09 20:33:13
Back to the problem. What are we trying to find?
richuw 2017-03-09 20:34:01
when the imaginary part is as great as possible (The peak of the circle)
First 2017-03-09 20:34:01
the highest from the real axis
Dr4gon39 2017-03-09 20:34:01
mazimize the imaginary part and then give the real part of the complex number with the optimized imaginary part
Mrkiller 2017-03-09 20:34:01
the z with the biggest imaginary part
yrnsmurf 2017-03-09 20:34:01
the point with the greatest y
brainiac1 2017-03-09 20:34:01
the maximum possible imaginary part for z (basically the top of the circle)
gabrielsui 2017-03-09 20:34:01
greatest possible imaginary part
GeronimoStilton 2017-03-09 20:34:01
The $z$ such that $\operatorname{Im} z$ is the greatest possible.
yrnsmurf 2017-03-09 20:34:01
the x coordinate of the point with greatest y
IsaacZ123 2017-03-09 20:34:01
the greatest imaginary part of z
copeland 2017-03-09 20:34:03
We want the point with largest imaginary part and we want to figure out what its real part is. That point is definitely way up at the top:
copeland 2017-03-09 20:34:05
copeland 2017-03-09 20:34:09
Now I think it's time for some numbers. What should our goal be now?
yrnsmurf 2017-03-09 20:34:59
the circumcenter of z1z2z3
ninjataco 2017-03-09 20:34:59
find circumradius and center of the circle
gabrielsui 2017-03-09 20:34:59
find the center of the circle
brainiac1 2017-03-09 20:34:59
find the center of this circle, since it will have the same x-value as our desired point
vishwathganesan 2017-03-09 20:34:59
find center and radius
fdas 2017-03-09 20:34:59
Find the center of the circle
copeland 2017-03-09 20:35:01
The top of the circle has the same real part as the center of the circle. So if we can find the center of this circle then we're done.
copeland 2017-03-09 20:35:06
Let's just turn these complex coordinates into vectors. (That's the first and last time in my life I'll ever say that.)
copeland 2017-03-09 20:35:08
We want to find the $x$-coordinate of the center of the circle. What's its $y$-coordinate?
copeland 2017-03-09 20:35:57
vishwathganesan 2017-03-09 20:36:38
61
brainiac1 2017-03-09 20:36:38
61
cjquines0 2017-03-09 20:36:38
midway between $39$ and $83$, so $61$
sxu 2017-03-09 20:36:38
61
strategos21 2017-03-09 20:36:38
61
copeland 2017-03-09 20:36:41
Since there is a vertical chord between $z_1=(18,83)$ and $z_2=(18,39)$, we know that the horizontal diameter bisects this chord. So the horizontal diameter of the circle is $y=\dfrac{83+39}2=61$.
copeland 2017-03-09 20:36:42
This is our circle:
copeland 2017-03-09 20:36:43
copeland 2017-03-09 20:36:52
Now what?
liuh008 2017-03-09 20:37:34
Distance between 2 points?
IsaacZ123 2017-03-09 20:37:34
we can create a set of equations using the radius of the circle
fdas 2017-03-09 20:37:34
Draw right triangles
J1618 2017-03-09 20:37:34
Find x using a right triangle
copeland 2017-03-09 20:37:37
We could use the distance formula. That would be nice. Stare at the coordinates of those three points, though. Do you see anything nice?
copeland 2017-03-09 20:38:24
We have $z_2=(18,39)$ and $z_3=(78,99)$. Those have the same ones digits. . .
brainiac1 2017-03-09 20:38:37
find the perpendicular bisector of z2z3, which happens to have slope -1
yrnsmurf 2017-03-09 20:38:37
line through z2 and z3 has slope 1
sxu 2017-03-09 20:38:37
oh the x and y coords both differ by 60
duck_master 2017-03-09 20:38:37
z1 and z3 are 45 degrees to the x-axis!
vishwathganesan 2017-03-09 20:38:37
oops z_2z_3 slope is 1
copeland 2017-03-09 20:38:39
The difference between $z_3$ and $z_2$ is $z_3-z_2=(78,99)-(18,39)=(60,60).$
copeland 2017-03-09 20:38:42
That means the line through $z_2$ and $z_3$ has slope $+1$.
copeland 2017-03-09 20:38:45
So its perpendicular bisector has slope $-1!$ It's $x+y=k$. What is $k?$
brainiac1 2017-03-09 20:39:53
117
samuel 2017-03-09 20:39:53
$k=117$
vishwathganesan 2017-03-09 20:39:53
117
sxu 2017-03-09 20:39:53
117
duck_master 2017-03-09 20:39:53
117?
cjquines0 2017-03-09 20:39:53
it passes through the midpoint $(48, 69)$, so it's $48+69=117$
islander7 2017-03-09 20:39:53
117
copeland 2017-03-09 20:39:55
The perpendicular bisector contains the midpoint $(48,69)$, so its equation is $x+y=117.$
copeland 2017-03-09 20:39:56
What is the $x$-coordinate of our center?
Buddy03 2017-03-09 20:40:37
56
IsaacZ123 2017-03-09 20:40:37
56
Reef334 2017-03-09 20:40:37
56
legolego 2017-03-09 20:40:37
56
strategos21 2017-03-09 20:40:37
56
pythonsquared 2017-03-09 20:40:37
56
DemonPlat4 2017-03-09 20:40:37
56
vishwathganesan 2017-03-09 20:40:37
56
brainiac1 2017-03-09 20:40:37
56
copeland 2017-03-09 20:40:38
The center lies on the lines

\begin{align*}

x+y&=117\\

y&=61.

\end{align*}
copeland 2017-03-09 20:40:41
That makes $x=117-61=\boxed{056}.$
copeland 2017-03-09 20:40:48
Cool.
copeland 2017-03-09 20:40:58
When I was a kid, people used to say "Cool beans."
copeland 2017-03-09 20:41:03
I wonder why that went out of style.
copeland 2017-03-09 20:41:04
What's next?
QuestForKnowledge 2017-03-09 20:41:22
palindrome .... 11
letsgomath 2017-03-09 20:41:22
#11!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
lego101 2017-03-09 20:41:22
Number eleven, we're in heaven bc we're doing math
NewbieGamer 2017-03-09 20:41:22
q 11
sxu 2017-03-09 20:41:22
11
MathTechFire 2017-03-09 20:41:22
#11
copeland 2017-03-09 20:41:25
Palindrome, huh?
copeland 2017-03-09 20:41:50
(You know most of them have been palindromes already.)
brainiac1 2017-03-09 20:42:05
the only palindromic prime with an even number of digits
copeland 2017-03-09 20:42:10
OK, I did not know that.
copeland 2017-03-09 20:42:11
11. Consider arrangements of the 9 numbers $1,2,3,\ldots,9$ in a $3\times3$ array. For each such arrangement, let $a_1,$ $a_2,$ and $a_3$ be the medians of the numbers in rows 1, 2, and 3, respectively, and let $m$ be the median of $\{a_1,a_2,a_3\}$. Let $Q$ be the number of arrangements for which $m=5.$ Find the remainder when $Q$ is divided by 1000.
copeland 2017-03-09 20:42:20
What does it mean for 5 to be the median of the medians?
checkmatetang 2017-03-09 20:43:08
m is median of row
checkmatetang 2017-03-09 20:43:08
5 is median of a row
IsaacZ123 2017-03-09 20:43:08
one median is larger and the other is smaller
vishwathganesan 2017-03-09 20:43:08
it is the median of its row
GeronimoStilton 2017-03-09 20:43:08
It means that one of the medians is greater than $5$ and one is less.
mshanmugam 2017-03-09 20:43:08
one of the medians is 5
copeland 2017-03-09 20:43:12
5 must be one of the medians. One of the medians is smaller than 5 and the other median is larger than 5.
copeland 2017-03-09 20:43:14
Well, 5 is definitely in some row. What's the probability that it's the median of its own row?
brainiac1 2017-03-09 20:44:59
4/7
JJShan26 2017-03-09 20:44:59
4/7
DemonPlat4 2017-03-09 20:44:59
$\frac{4}{7}$?
strategos21 2017-03-09 20:44:59
4/7
IsaacZ123 2017-03-09 20:44:59
4/7
copeland 2017-03-09 20:45:03
5 is the median of its row when one of its neighbors is less than 5 and the other neighbor is more than 5. We have 4 little numbers and 4 big numbers in the array.
copeland 2017-03-09 20:45:06
We need the probability that 5 is in a row with one little number and one big number. Without loss of generality, assume 5 is in the middle. If the left number is little, then $\dfrac47$ of the time the right number will be big. If the left number is big then $\dfrac47$ of the time the right number will be little.
copeland 2017-03-09 20:45:10
Either way, there is a $\dfrac47$ chance that 5 is the median of one of the rows.
copeland 2017-03-09 20:45:18
Or this:
owm 2017-03-09 20:45:20
(8C2-4C2-4C2)/8C2?
copeland 2017-03-09 20:45:26
If 5 is the median of one of the rows then we have one little and one big number in the row with 5. How can we fill the array such that 5 is the median of the medians?
copeland 2017-03-09 20:45:49
How can we fill in the rest of such an array, that is?
IsaacZ123 2017-03-09 20:47:39
wait that's what the problem is asking for basically because if it is the median of its own row it has to be the median of the set
a1b2 2017-03-09 20:47:39
Any arrangement
yrnsmurf 2017-03-09 20:47:39
any way you want
strategos21 2017-03-09 20:47:39
The rest of the array doesn't matter; just calculate the number of arrangements.
brainiac1 2017-03-09 20:47:39
every other arrangement has to work
zihang 2017-03-09 20:47:39
6!=720
a1b2 2017-03-09 20:47:39
It is impossible for both other medians to be more than 5 so any arrangement is valid.
Dr4gon39 2017-03-09 20:47:39
ANY WAY works as long as five is already the median of its own row
copeland 2017-03-09 20:47:42
We have 3 little numbers and 3 big numbers to put in the grid.
copeland 2017-03-09 20:47:43
If all 3 little numbers go in the same row then that row's median will be smaller than 5 and the other row will have larger median and 5 is the median of the medians.
copeland 2017-03-09 20:47:44
If one row has 2 little numbers and 1 big then the median of that row will be little. Symmetrically, the other row will have a big median and 5 is still the median of the medians.
copeland 2017-03-09 20:47:50
So if 5 is the median of its row, it's the median of the medians of all the rows.
copeland 2017-03-09 20:47:53
No matter what, $\dfrac47$ of the time 5 is the median of the medians.
copeland 2017-03-09 20:47:55
How many total ways are there to fill the array?
Buddy03 2017-03-09 20:48:49
9!
vishwathganesan 2017-03-09 20:48:49
9!
brainiac1 2017-03-09 20:48:49
9!
IsaacZ123 2017-03-09 20:48:49
362880 or 9!
GeronimoStilton 2017-03-09 20:48:49
$362880 = 9!$
Kiola 2017-03-09 20:48:49
9!
checkmatetang 2017-03-09 20:48:49
9!
Einsteinhead 2017-03-09 20:48:49
9!
yrnsmurf 2017-03-09 20:48:49
9!
mathcrazymar 2017-03-09 20:48:49
9!
copeland 2017-03-09 20:48:56
Who knows 9!? You guys are crazy.
copeland 2017-03-09 20:48:59
There are $9!$ total ways to fill the array.
copeland 2017-03-09 20:49:00
How many arrays have 5 as the median of the medians?
Buddy03 2017-03-09 20:49:41
9!*(4/7)
a1b2 2017-03-09 20:49:41
$9!*\frac{4}{7}$
IsaacZ123 2017-03-09 20:49:41
9!*4/7
awesomemaths 2017-03-09 20:49:41
4/7*9!
Mrkiller 2017-03-09 20:49:41
4/7 of them so
Einsteinhead 2017-03-09 20:49:41
(4/7)*9!
checkmatetang 2017-03-09 20:49:41
207360
a1b2 2017-03-09 20:49:41
$\frac{4}{7}*9!=207360$
letsgomath 2017-03-09 20:49:41
207360
strategos21 2017-03-09 20:49:41
4/7 of them
mathcrazymar 2017-03-09 20:49:41
4/7 * 9!
Buddy03 2017-03-09 20:49:41
207360=9!*(4/7)
sxu 2017-03-09 20:49:41
9*8*6*5*4*4*3*2
Pot 2017-03-09 20:49:41
207360
Mrkiller 2017-03-09 20:49:41
207360
DemonPlat4 2017-03-09 20:49:41
$\frac{4}{7}\cdot362880$
ninjataco 2017-03-09 20:49:41
4*9!/7
cjquines0 2017-03-09 20:49:41
$\frac47$ of those, so $207360$
copeland 2017-03-09 20:49:45
Since $\dfrac47$ of the arrays have this property, there are $\dfrac47\cdot9!$ total arrays.
copeland 2017-03-09 20:49:47
\[\frac47\cdot9!=9\cdot8\cdot7\cdot6!=9\cdot8\cdot4\cdot720.\]
copeland 2017-03-09 20:49:53
And what's the final answer?
letsgomath 2017-03-09 20:50:12
360
Pot 2017-03-09 20:50:12
360
IsaacZ123 2017-03-09 20:50:12
360
stronto 2017-03-09 20:50:12
360
KYang 2017-03-09 20:50:12
360
GeronimoStilton 2017-03-09 20:50:12
$360$
QuestForKnowledge 2017-03-09 20:50:12
360
AlisonH 2017-03-09 20:50:12
360
MountainHeight 2017-03-09 20:50:12
360
legolego 2017-03-09 20:50:12
360
amgmflannigan 2017-03-09 20:50:12
360
math.fever 2017-03-09 20:50:12
360
Damalone 2017-03-09 20:50:12
360
copeland 2017-03-09 20:50:13
\begin{align*}

9\cdot8\cdot4\cdot720

&\equiv9\cdot8\cdot880\\

&\equiv9\cdot40\\

&\equiv\boxed{360}\\

\end{align*}
copeland 2017-03-09 20:50:18
OK, great.
copeland 2017-03-09 20:50:44
11 down. 4 to go. I bet the rest are easy. Should we skip them?
awesomemaths 2017-03-09 20:51:05
problem 12
awesomemaths 2017-03-09 20:51:05
problem 12 yay!!!
cjquines0 2017-03-09 20:51:05
totally
legolego 2017-03-09 20:51:05
noo
DemonPlat4 2017-03-09 20:51:05
no
First 2017-03-09 20:51:05
No!
MountainHeight 2017-03-09 20:51:05
NO
amackenzie1 2017-03-09 20:51:05
Haha no!
tdeng 2017-03-09 20:51:05
No.
awesomemaths 2017-03-09 20:51:05
nope
strategos21 2017-03-09 20:51:05
nah, you should just humor us
yrnsmurf 2017-03-09 20:51:05
nah
lego101 2017-03-09 20:51:10
Number twelve, as deeper into the world of mathematica shall we delve.
lego101 2017-03-09 20:51:10
NO POWER THROUGH
copeland 2017-03-09 20:51:16
POWER THROUGH, in caps. I like it.
copeland 2017-03-09 20:51:18
12. Call a set $S$ product-free if there do not exist $a,b,c\in S$ (not necessarily distinct) such that $ab=c.$ For example, the empty set and the set $\{16,20\}$ are product-free, whereas the sets $\{4,16\}$ and $\{2,8,16\}$ are not product-free. Find the number of product-free subsets of the set $\{1,2,3,4,5,6,7,8,9,10\}.$
copeland 2017-03-09 20:51:32
I'm thinking more people should use caps. It would make the classroom much more interesting.
copeland 2017-03-09 20:51:36
Also, that's a joke.
copeland 2017-03-09 20:51:40
Don't do that.
Pot 2017-03-09 20:51:46
OKAY
reelmathematician 2017-03-09 20:51:46
CAPS FOR THE WIN
IsaacZ123 2017-03-09 20:51:46
ARE YOU SURE IT WAS A JOKE
MSTang 2017-03-09 20:51:46
I WON'T!
copeland 2017-03-09 20:51:48
Alright, what do you see immediately?
vishwathganesan 2017-03-09 20:52:46
anything with 1 in it fails
GeronimoStilton 2017-03-09 20:52:46
$1$ Can't be a member of any of the sets!
Mrkiller 2017-03-09 20:52:46
1 can be in any of them
ninjataco 2017-03-09 20:52:46
1 can't be in the subset
Einsteinhead 2017-03-09 20:52:46
Set cannot contain 1
tdeng 2017-03-09 20:52:46
We can't have 1
duck_master 2017-03-09 20:52:46
DON'T PUT 1 THERE OR U WON'T BE ABLE TO PUT ANYTHING ELSE
mathfever 2017-03-09 20:52:46
that they can't have 1
amackenzie1 2017-03-09 20:52:46
can;t have a 1
strategos21 2017-03-09 20:52:46
1 doesn't fit anywhere
amyhu910 2017-03-09 20:52:46
1 cannot be in the set
KYang 2017-03-09 20:52:46
you can't have 1
MountainHeight 2017-03-09 20:52:46
we can ignore 1
copeland 2017-03-09 20:52:49
I see that 1 can't be in the set since $1\cdot1=1$.
copeland 2017-03-09 20:52:53
Are there a lot of products to avoid or just a few?
IsaacZ123 2017-03-09 20:54:11
a few
EdwinNational 2017-03-09 20:54:11
Just a few
amgmflannigan 2017-03-09 20:54:11
2 and 3
QuestForKnowledge 2017-03-09 20:54:11
few
DemonPlat4 2017-03-09 20:54:11
few-ish
andsun19 2017-03-09 20:54:11
2*2=4, 2*3=6, 2*4=8, 2*5=10, 3*3=9
Mathaddict11 2017-03-09 20:54:11
really only a few and we can quickly list them out
rraj 2017-03-09 20:54:11
just a few
Metal_Bender19 2017-03-09 20:54:11
a few
itxxc12 2017-03-09 20:54:11
just a few
ilikepie2003 2017-03-09 20:54:11
a few
copeland 2017-03-09 20:54:14
There are only a few products we have to avoid. Every product we could make (omitting 1) has either a 2 or a 3 in it.
copeland 2017-03-09 20:54:16
Actually, here are all the products just so we see them:
copeland 2017-03-09 20:54:16
\[\begin{array}{c}
2\cdot2=4&2\cdot3=6&2\cdot4=8&2\cdot5=10\\
3\cdot2=6&3\cdot3=9.\\
\end{array}\]
copeland 2017-03-09 20:54:18
I put $2\cdot3=3\cdot2$ on there twice because it kinda fits twice.
copeland 2017-03-09 20:54:22
So how should we organize our count?
a1b2 2017-03-09 20:55:24
By twos and threes
legolego 2017-03-09 20:55:24
has a 2, has a 3, both, or neither
tdeng 2017-03-09 20:55:24
{Has 2, no 3}, {has 3, no 2}, {has 2 and 3}, {no 2, no 3}
copeland 2017-03-09 20:55:26
Let's use casework on whether we have 2 or 3 in the set.
copeland 2017-03-09 20:55:28
If we don't have 2 or 3 in the set, how many possible sets can we make?
IsaacZ123 2017-03-09 20:55:56
2^7
tdeng 2017-03-09 20:55:56
2^7
legolego 2017-03-09 20:55:56
2^7
IsaacZ123 2017-03-09 20:55:56
128
letsgomath 2017-03-09 20:55:56
128
Ericaops 2017-03-09 20:55:56
128
DemonPlat4 2017-03-09 20:55:56
128
GeronimoStilton 2017-03-09 20:55:56
$2^7 = 128$
brainiac1 2017-03-09 20:55:56
2^7=128
copeland 2017-03-09 20:55:58
If we don't have 2 or 3 in the set then we want a subset of $\{4,5,6,7,8,9,10\}.$ Any numbers can be in our set, so there are $2^7$ total sets that miss 2 and 3.
copeland 2017-03-09 20:56:00
What if 3 is in the set and 2 is not? (Three is in fewer products, so putting it in the set is less restrictive. That's why it's a better case to tackle next.)
legolego 2017-03-09 20:56:54
2^6
owm 2017-03-09 20:56:54
2^6
a1b2 2017-03-09 20:56:54
9 can't be in it
IsaacZ123 2017-03-09 20:56:54
64
mathcrazymar 2017-03-09 20:56:54
2^6
IsaacZ123 2017-03-09 20:56:54
2^6
cjquines0 2017-03-09 20:56:54
any subset of $\{4, 5, 6, 7, 8, 10\}$ so $2^6$
MountainHeight 2017-03-09 20:56:54
2^6
Ericaops 2017-03-09 20:56:54
cant have 9, 2^6
Mathaddict11 2017-03-09 20:56:54
we can't have 3 and 9 so there are $2^6=64$ ways
Archimedes15 2017-03-09 20:56:54
2^6 = 64
brainiac1 2017-03-09 20:56:54
2^6=64
copeland 2017-03-09 20:56:56
We can add anything but 9 now.
copeland 2017-03-09 20:56:57
If 3 is in the set and 2 is not then we finish the set by adding some subset of $\{4,5,6,7,8,10\}$ to the set. There are $2^6$ subsets like this.
copeland 2017-03-09 20:57:00
What if 2 is in the set and 3 is not? This is harder. What numbers can we add or not, without fear?
Inco 2017-03-09 20:57:51
6, 7, 9
owm 2017-03-09 20:57:51
we can add 6,7,and 9
copeland 2017-03-09 20:57:54
We can add any of $\{6,7,9\}$ to the set without worrying about making products with any of the other elements.
copeland 2017-03-09 20:57:55
What about 5 and 10?
Ericaops 2017-03-09 20:58:28
cannot have both
legolego 2017-03-09 20:58:28
choose one or neither
yrnsmurf 2017-03-09 20:58:28
5, 10, or neither
letsgomath 2017-03-09 20:58:28
only one of the 2
QuestForKnowledge 2017-03-09 20:58:28
u cant have em both
Archimedes15 2017-03-09 20:58:28
2*5 = 10, meaning we can't have 5 and 10
cooljoseph 2017-03-09 20:58:28
You can have one, but not the other
cjquines0 2017-03-09 20:58:28
we have to pick between $5$, $10$, or neither
Peggy 2017-03-09 20:58:28
only one of them
mishai 2017-03-09 20:58:28
only one of them or none
DemonPlat4 2017-03-09 20:58:28
one or the other, but not both
BuddyS 2017-03-09 20:58:28
only one of 5 or 10
IsaacZ123 2017-03-09 20:58:28
we just can't have both
MountainHeight 2017-03-09 20:58:28
they can't be in at the same time
copeland 2017-03-09 20:58:31
We can have either of them or none. How many ways is that to add 5 or 10?
brainiac1 2017-03-09 20:58:48
3
amackenzie1 2017-03-09 20:58:48
3
Doink 2017-03-09 20:58:48
3
QuestForKnowledge 2017-03-09 20:58:48
3
thetank 2017-03-09 20:58:48
3
letsgomath 2017-03-09 20:58:48
3
vishwathganesan 2017-03-09 20:58:48
3
mishai 2017-03-09 20:58:48
3
copeland 2017-03-09 20:58:51
There are three choices: neither, 5, or 10.
copeland 2017-03-09 20:58:52
What about 4 and 8?
IsaacZ123 2017-03-09 20:59:24
also can't we add 8 safely as well since we can't have 4 in there
tdeng 2017-03-09 20:59:24
We can't have 4=2*2, so we can just add 8
letsgomath 2017-03-09 20:59:24
no 4, 8 works
DjokerNole 2017-03-09 20:59:24
only 8
Reef334 2017-03-09 20:59:24
neither or only 8
IsaacZ123 2017-03-09 20:59:24
we can only have 8 because 2*2=4
mishai 2017-03-09 20:59:24
no 4 but 8 can appear
cooljoseph 2017-03-09 20:59:24
You can never have 4, so you can always have 8.
owm 2017-03-09 20:59:24
there are 2 possibilities
a1b2 2017-03-09 20:59:24
The 8, or nothing!
EdwinNational 2017-03-09 20:59:24
No 4
GeronimoStilton 2017-03-09 20:59:24
We can have $8$ either way, but we can't have $4$ since $2 \cdot 2 = 4$
MountainHeight 2017-03-09 20:59:24
4 can't be in but 8 can
copeland 2017-03-09 20:59:26
We can't have 4 since $2\cdot2=4$, so we can either have 8 or not.
copeland 2017-03-09 20:59:29
How many sets have 2 and not 3?
GeronimoStilton 2017-03-09 21:00:32
$48$
mishai 2017-03-09 21:00:32
48
IsaacZ123 2017-03-09 21:00:32
48
yrnsmurf 2017-03-09 21:00:32
2*2*2*2*3=48
Tribefan 2017-03-09 21:00:32
48
legolego 2017-03-09 21:00:32
48
letsgomath 2017-03-09 21:00:32
48
MountainHeight 2017-03-09 21:00:32
48
vishwathganesan 2017-03-09 21:00:32
48
cjquines0 2017-03-09 21:00:32
$2^3$ ways to pick from $\{6, 7, 9\}$, $3$ ways to pick $5$ or $10$ or neither, and $2$ ways to pick either $8$ or nothing, so $48$
islander7 2017-03-09 21:00:32
48
copeland 2017-03-09 21:00:33
The sets with 2 and not 3 are made by choosing from the $2^4$ subests of $\{6,7,8,9\}$ and then choosing either 5, 10, or neither (3 choices). There are $2^4\cdot3$ product-free sets that contain 2 and not 3.
copeland 2017-03-09 21:00:36
Now let's look at all sets with 2 and 3. What numbers can we throw in without fear?
GeronimoStilton 2017-03-09 21:01:23
$7$
a1b2 2017-03-09 21:01:23
7
MountainHeight 2017-03-09 21:01:23
7
owm 2017-03-09 21:01:23
7
Z_Math404 2017-03-09 21:01:23
7
letsgomath 2017-03-09 21:01:23
7
copeland 2017-03-09 21:01:25
Only 7. All the other numbers appear in products with 2 or 3.
copeland 2017-03-09 21:01:34
Oh, wait, there's one more we just learned goes well with 2.
IsaacZ123 2017-03-09 21:01:48
7 and 8
vishwathganesan 2017-03-09 21:01:48
7, 8
GeronimoStilton 2017-03-09 21:01:48
$7$ and $8$
Funnybunny5246 2017-03-09 21:01:48
7,8
yrnsmurf 2017-03-09 21:01:48
7 and 8
tdeng 2017-03-09 21:01:48
7,8
vishwathganesan 2017-03-09 21:01:48
also 8 though
IsaacZ123 2017-03-09 21:01:48
wait but 8 can be in there by the same reasoning as the last case
Tribefan 2017-03-09 21:01:48
8
amackenzie1 2017-03-09 21:01:48
8
letsgomath 2017-03-09 21:01:48
8
copeland 2017-03-09 21:02:00
We can add 7 or 8 without any issues.
copeland 2017-03-09 21:02:03
What other numbers can we easily make sense of?
copeland 2017-03-09 21:02:11
Are there any that are forbidden?
letsgomath 2017-03-09 21:02:33
no 4 or 9
Z_Math404 2017-03-09 21:02:33
no 6
pianoman24 2017-03-09 21:02:33
6
islander7 2017-03-09 21:02:33
4,6,9
vishwathganesan 2017-03-09 21:02:33
4,6,9
IsaacZ123 2017-03-09 21:02:33
6 9
checkmatetang 2017-03-09 21:02:33
4,6,9
rockyisi 2017-03-09 21:02:33
4,6,9
wrc400 2017-03-09 21:02:33
4, 6, 9
treemath 2017-03-09 21:02:33
4,9
andsun19 2017-03-09 21:02:33
4,6,9
copeland 2017-03-09 21:02:38
We definitely can't have 6 if we already have 2 or 3. We also can't have 4 since we have 2 or 9 since we have 3.
copeland 2017-03-09 21:02:40
What about 5?
mishai 2017-03-09 21:03:24
either 5 or 10
Z_Math404 2017-03-09 21:03:24
either 5 or 10
yrnsmurf 2017-03-09 21:03:24
either 5, 10, or neither
mshanmugam 2017-03-09 21:03:24
same as last time; 5 or 10
GeronimoStilton 2017-03-09 21:03:24
We can have $5$, $10$, or neither.
brainiac1 2017-03-09 21:03:24
if we have 5, we can't have 10 and vice versa
MountainHeight 2017-03-09 21:03:24
we can either have 5 or 10 but not both
Archimedes15 2017-03-09 21:03:24
we can only have 5 if there is no 10
owm 2017-03-09 21:03:24
we can have 5 or 10 but not both
treemath 2017-03-09 21:03:24
5 or 10, but not both
andsun19 2017-03-09 21:03:24
5 or 10 or neither
copeland 2017-03-09 21:03:26
We can have 5, 10, or neither.
copeland 2017-03-09 21:03:26
I think that's all. How many sets have 2 and 3?
IsaacZ123 2017-03-09 21:03:57
12
mishai 2017-03-09 21:03:57
12
yrnsmurf 2017-03-09 21:03:57
12
vishwathganesan 2017-03-09 21:03:57
12
andsun19 2017-03-09 21:03:57
12
samuel 2017-03-09 21:03:57
12
copeland 2017-03-09 21:03:59
In a set with 2 and 3, we can't have any of 4, 6, or 9. We can have 7 or 8 whenever we want ($2^2$ choices). We can have either 5, 10, or neither (3 choices). There are $3\cdot2^2$ total sets with both 2 and 3.
copeland 2017-03-09 21:04:00
How many total product-free sets are there?
GeronimoStilton 2017-03-09 21:04:24
$252$
vishwathganesan 2017-03-09 21:04:24
252
IsaacZ123 2017-03-09 21:04:24
252
brainiac1 2017-03-09 21:04:24
252
a1b2 2017-03-09 21:04:24
$\boxed{252}$
stronto 2017-03-09 21:04:24
252
tdeng 2017-03-09 21:04:24
252
copeland 2017-03-09 21:04:27
There are \[2^7+2^6+2^4\cdot3+2^2\cdot3=128+64+48+12=\boxed{252}\]total sets.
copeland 2017-03-09 21:04:39
Twelve down.
lego101 2017-03-09 21:04:49
Number thirteen, by this time our scrap paper is far from clean. #POWERTHROUGH
copeland 2017-03-09 21:04:57
Yeah. Maybe we'll get to another palindrome soon, too!
copeland 2017-03-09 21:05:04
13. For every $m\geq2,$ let $Q(m)$ be the least positive integer with the following property: For every $n\geq Q(m),$ there is always a perfect cube $k^3$ in the range $n< k^3\leq m\cdot n.$ Find the reminader when
\[\sum_{m=2}^{2017} Q(m)\]is divided by 1000.
copeland 2017-03-09 21:05:14
This is an alphabet soup of a problem. I don't really know what's going on. What should we do to get a better grasp on the problem?
Doink 2017-03-09 21:05:41
try small cases
liuh008 2017-03-09 21:05:41
Small examples!
letsgomath 2017-03-09 21:05:41
plug in a few numbers
sxu 2017-03-09 21:05:41
try out some stuff
First 2017-03-09 21:05:41
Try an example
amackenzie1 2017-03-09 21:05:41
try some simpler cases?
Ericaops 2017-03-09 21:05:41
try small cases
copeland 2017-03-09 21:05:44
Let's experiment! We probably want to start with small numbers, but I don't even know which variable to make small. There's $m$ and there's $Q$, of course, but there's also $n$ and $k$.
copeland 2017-03-09 21:05:49
Let's focus on $m$ since we know for sure that $m=2$ is a small and interesting case.
copeland 2017-03-09 21:05:53
Now we look at all these intervals

\begin{align*}

&(2,4]\\

&(3,6]\\

&(4,8]\\

&(5,10]\\

&(6,12]\\

&(7,14]\\

\end{align*}

and eventually these intervals should all have cubes in them.
copeland 2017-03-09 21:05:54
Which one of those first six have cubes in them?
vishwathganesan 2017-03-09 21:07:10
third onwards
IYN 2017-03-09 21:07:10
(4, 8] and the ones after
Doink 2017-03-09 21:07:10
(4,8], (5,10], (6,12], (7,14]
ninjataco 2017-03-09 21:07:10
all except the first two
andsun19 2017-03-09 21:07:10
3rd, 4th, 5th, 6th
letsgomath 2017-03-09 21:07:10
(4,8],(5,10],(6,12],(7,14]
alifenix- 2017-03-09 21:07:10
the last four
DemonPlat4 2017-03-09 21:07:10
(4,8], (5,10], (6,12], (7,14]
a1b2 2017-03-09 21:07:23
The last 4, but that doesn't matter as $(8,16]$ doesn't have any.
vishwathganesan 2017-03-09 21:07:23
but then there's (8, 16] which doesn't have any
copeland 2017-03-09 21:07:25
These red ones have cubes in them:

\begin{align*}

&(2,4]\\

&(3,6]\\

&\color{red}{(4,8]}\\

&\color{red}{(5,10]}\\

&\color{red}{(6,12]}\\

&\color{red}{(7,14]}\\

\end{align*}

but the next interval $(8,16]$ does not contain a cube.
copeland 2017-03-09 21:07:28
What's the next interval $(n,2n]$ that has a cube in it?
Picroft 2017-03-09 21:07:55
14,28
cooljoseph 2017-03-09 21:07:55
(14,28]
Reef334 2017-03-09 21:07:55
(14,28]
GeronimoStilton 2017-03-09 21:07:55
$(14,28]$
liuh008 2017-03-09 21:07:55
(14, 28]
dhruv 2017-03-09 21:07:55
(14, 28]
letsgomath 2017-03-09 21:07:55
(14,28]
QuestForKnowledge 2017-03-09 21:07:55
the 27 one, (14,28]
Inco 2017-03-09 21:07:55
(14, 28]
thetank 2017-03-09 21:07:55
(14, 28]
copeland 2017-03-09 21:07:57
The next available cube is 27, and the first interval that contains 27 is $(14,28]$.
copeland 2017-03-09 21:08:00
And what's the last interval that contains 27?
DemonPlat4 2017-03-09 21:08:34
(26,52]
stronto 2017-03-09 21:08:34
(26,52]
sxu 2017-03-09 21:08:34
(26, 52]
yrnsmurf 2017-03-09 21:08:34
(26, 52]
Doink 2017-03-09 21:08:34
(26,52]
mishai 2017-03-09 21:08:34
(26,52]
awesomemaths 2017-03-09 21:08:34
(26, 52]
dantaxyz 2017-03-09 21:08:34
(26,52]
letsgomath 2017-03-09 21:08:34
(26,52]
grant.wilkins 2017-03-09 21:08:34
(26, 52]
Mrkiller 2017-03-09 21:08:34
(26,52]
EpicCarrotMaster 2017-03-09 21:08:34
(26, 52]
copeland 2017-03-09 21:08:36
The last interval that contains 27 is $(26,52]$. Does the next interval $(27,54]$ contain a cube?
curry3030 2017-03-09 21:08:55
no
grant.wilkins 2017-03-09 21:08:55
No
checkmatetang 2017-03-09 21:08:55
no
fdas 2017-03-09 21:08:55
No
cooleybz2013 2017-03-09 21:08:55
no
copeland 2017-03-09 21:08:57
No. What's the next interval $(n,2n]$ that does contain a cube?
NewbieGamer 2017-03-09 21:09:21
32,64
mishai 2017-03-09 21:09:21
(32,64]
cooljoseph 2017-03-09 21:09:21
32,64
Reef334 2017-03-09 21:09:21
(32,64]
owm 2017-03-09 21:09:21
32,64
dhruv 2017-03-09 21:09:21
(32, 64]
letsgomath 2017-03-09 21:09:21
(32,64]
varun555 2017-03-09 21:09:21
(32, 64]
liuh008 2017-03-09 21:09:21
(32, 64]
EpicCarrotMaster 2017-03-09 21:09:21
(32, 64]
islander7 2017-03-09 21:09:21
(32,64]
copeland 2017-03-09 21:09:23
The next cube is 64, so the next interval that contains a cube is $(32,64]$.
copeland 2017-03-09 21:10:00
And the last interval that contains 64 is $(63,126]$.
copeland 2017-03-09 21:10:01
What's the next interval $(n,2n]$ that doesn't contain a cube?
Z_Math404 2017-03-09 21:10:13
none
vishwathganesan 2017-03-09 21:10:13
there is none
NewbieGamer 2017-03-09 21:10:13
none
DemonPlat4 2017-03-09 21:10:13
NONE
Reef334 2017-03-09 21:10:13
none of them
cooljoseph 2017-03-09 21:10:13
there isn't one
GeronimoStilton 2017-03-09 21:10:13
There aren't any more!
MountainHeight 2017-03-09 21:10:13
no more
amackenzie1 2017-03-09 21:10:13
it seems that from now on they all contain cubes
Funnybunny5246 2017-03-09 21:10:13
none
copeland 2017-03-09 21:10:16
Hm. This is a harder question. Since 125 is in $(64,128]$, all the intervals from $n=64$ to $n=124$ contain $5^3=125.$ That last interval is $(124,248]$.
copeland 2017-03-09 21:10:19
Alright, I think it's time for a picture.
copeland 2017-03-09 21:10:20
Here are the cubes along the top. Below them I've graphed all of the intervals $(n,2n]$. An interval contains a cube if it crosses one of the vertical lines. Those are the red ones. The problem suggests that there is a first $n$ such that every line below the line $(n,2n]$ crosses one of the vertical lines (so will be red).
copeland 2017-03-09 21:10:22
copeland 2017-03-09 21:10:31
Why can't we fit an interval in between 125 and 216?
a1b2 2017-03-09 21:11:02
The lines are longer than the intervals
varun555 2017-03-09 21:11:02
There is no interval short enough
tdeng 2017-03-09 21:11:08
2*125=250
mishai 2017-03-09 21:11:08
216/125<2
curry3030 2017-03-09 21:11:08
125*2 = 250
samuel 2017-03-09 21:11:08
because 216/125<2
GeronimoStilton 2017-03-09 21:11:08
Because $2(125-1) > 216$
EpicCarrotMaster 2017-03-09 21:11:08
216-125 < 125 * 2
liuh008 2017-03-09 21:11:14
What a pretty picture
copeland 2017-03-09 21:11:15
Why thanks!
copeland 2017-03-09 21:11:16
Since $\dfrac{216}{125}<2$, there can't be a pair $n$ and $2n$ that are both between these bars.
copeland 2017-03-09 21:11:17
What about between 216 and 343?
Z_Math404 2017-03-09 21:11:31
did deven make that
copeland 2017-03-09 21:11:32
I made that. You can tell because it's so pro.
DemonPlat4 2017-03-09 21:11:42
216 < 125$\cdot$2
sxu 2017-03-09 21:11:42
also <2
mishai 2017-03-09 21:11:42
343/216<216/125<2
vishwathganesan 2017-03-09 21:11:42
343/216 is less that 2 too
copeland 2017-03-09 21:11:44
Since $\dfrac{343}{216}<2$, there isn't a good interval in here either.
copeland 2017-03-09 21:11:45
What about further to the right?
IsaacZ123 2017-03-09 21:12:27
the further we go, the less the ratios become
IYN 2017-03-09 21:12:27
The fractions get smaller and smaller
vishwathganesan 2017-03-09 21:12:27
(n+1)^3/n^3 tends to one
brainiac1 2017-03-09 21:12:27
the quotient (n+1/n)^3 only decreases as n increases, there will never be another place where a cube is more than twice the previous cube after 64
islander7 2017-03-09 21:12:27
ratios get smaller
mishai 2017-03-09 21:12:27
(n+2)^3/(n+1)^3<(n+1)^3/n^3<2
Archimedes15 2017-03-09 21:12:27
It is the same issue... it will always be less than 2.
QuestForKnowledge 2017-03-09 21:12:27
(x/(x-1))^3 decreases
copeland 2017-03-09 21:12:28
We're looking at \[\dfrac{(k+1)^3}{k^3}=\frac{1+3k^2+3k+1}{k^3}=1+\frac3k+\frac3{k^2}+\frac1{k^3}.\] When $k$ gets bigger, this value gets smaller, so the ratio is always going to be less than 2.
copeland 2017-03-09 21:12:31
Just so we are clear what happened, what is $Q(2)?$
mishai 2017-03-09 21:13:06
32
vishwathganesan 2017-03-09 21:13:06
32
legolego 2017-03-09 21:13:06
32
a1b2 2017-03-09 21:13:06
$Q(2)=32$
Z_Math404 2017-03-09 21:13:06
32
Makorn 2017-03-09 21:13:06
It would be $32$
DemonPlat4 2017-03-09 21:13:06
Q(2) = 32
yrnsmurf 2017-03-09 21:13:06
32
IsaacZ123 2017-03-09 21:13:06
32
IYN 2017-03-09 21:13:06
32
copeland 2017-03-09 21:13:07
The last interval that doesn't contain a cube is $(31,62],$ so the final set of red blocks begins at $(32,64]$. That makes
copeland 2017-03-09 21:13:10
$Q(2)=32$.
copeland 2017-03-09 21:13:12
Now on to $Q(3)$. Hopefully this is faster.
copeland 2017-03-09 21:13:14
There ought to be an inequality between $Q(2)$ and $Q(3)$. What is it?
yrnsmurf 2017-03-09 21:14:05
Q(3)=<Q(2)
DemonPlat4 2017-03-09 21:14:05
Q(2)$\geq$Q(3)
a1b2 2017-03-09 21:14:05
$Q(x)>=Q(x+1)$
vishwathganesan 2017-03-09 21:14:05
oops I meant Q(n) >= Q(n + 1)
copeland 2017-03-09 21:14:11
If there's a cube in $(n,2n]$ then there's definitely a cube in $(n,3n]$ so $Q(3)\leq Q(2)$. In fact, $Q$ is going to be a decreasing function. Let's compute $Q(3)$ by hand again to see what we can see.
copeland 2017-03-09 21:14:19
First of all for $m=2$ we found the last interval that doesn't contain a cube lies between 27 and 64. Is there an interval of the form $(n,3n]$ between 27 and 64?
thetank 2017-03-09 21:14:48
no
bomb427006 2017-03-09 21:14:48
no
Doink 2017-03-09 21:14:48
no
First 2017-03-09 21:14:48
No
Mrkiller 2017-03-09 21:14:48
no
copeland 2017-03-09 21:14:50
Why not?
curry3030 2017-03-09 21:15:02
no, 27*3 = 81
Makorn 2017-03-09 21:15:02
$27\cdot3=81>64$
tdeng 2017-03-09 21:15:02
27*3=81>64
vishwathganesan 2017-03-09 21:15:02
64/27 < 3
varun555 2017-03-09 21:15:02
64/27<3
copeland 2017-03-09 21:15:06
Since $\dfrac{64}{27}<3$, there aren't any intervals inside here.
copeland 2017-03-09 21:15:07
Oh good! $Q(3)<Q(2)$.
copeland 2017-03-09 21:15:09
Are there any intervals $(n,3n]$ that fit between 8 and 27?
DemonPlat4 2017-03-09 21:15:36
yes (8,24]
a1b2 2017-03-09 21:15:36
$(8,24]$
Makorn 2017-03-09 21:15:36
Yes, $(8,24]$
GeronimoStilton 2017-03-09 21:15:36
$(8,24]$
vishwathganesan 2017-03-09 21:15:36
(8, 24]
cooljoseph 2017-03-09 21:15:36
Yes, 8,24
IsaacZ123 2017-03-09 21:15:36
(8,24] should be the last one
checkmatetang 2017-03-09 21:15:36
(8,24]
IYN 2017-03-09 21:15:36
(8, 24]
amackenzie1 2017-03-09 21:15:36
8, 24
First 2017-03-09 21:15:36
Yes,8, 24
copeland 2017-03-09 21:15:38
Yes, but just barely! $(8,24]$ fits, but $(9,27]$ does not!
copeland 2017-03-09 21:15:39
Do we know what $Q(3)$ is yet?
GeronimoStilton 2017-03-09 21:16:09
$Q(3) = 9$
Makorn 2017-03-09 21:16:09
$Q(3)=9$
a1b2 2017-03-09 21:16:09
$Q(3)=9$
stronto 2017-03-09 21:16:09
9
samuel 2017-03-09 21:16:09
yes 9
islander7 2017-03-09 21:16:09
9
Mrkiller 2017-03-09 21:16:09
9?
cooljoseph 2017-03-09 21:16:09
yes: 9
ninjataco 2017-03-09 21:16:11
9
copeland 2017-03-09 21:16:13
Yes, we know that after $n=8$ the intervals all contain cubes. The interval $(8,24]$ does not contain a cube, so the intervals all turn red at $(9,27]$. That makes
copeland 2017-03-09 21:16:14
$Q(3)=9.$
copeland 2017-03-09 21:16:15
copeland 2017-03-09 21:16:20
Oh, wow. That graph tells us even more, right? Can you tell me what $Q(4)$ is going to be right away?
mishai 2017-03-09 21:16:52
2
GeronimoStilton 2017-03-09 21:16:52
$2$.
ninjataco 2017-03-09 21:16:52
2
NewbieGamer 2017-03-09 21:16:52
2
tdeng 2017-03-09 21:16:52
2
cooljoseph 2017-03-09 21:16:52
2
brainiac1 2017-03-09 21:16:52
2
SomethingNeutral 2017-03-09 21:16:52
actually 2
yrnsmurf 2017-03-09 21:16:52
2
Ericaops 2017-03-09 21:16:52
2
stronto 2017-03-09 21:16:52
2
copeland 2017-03-09 21:16:56
When we draw the graph for $m=4$, it's going to be just like this graph but with longer bars. For example, instead of $(5,15]$ we'll have $(5,20]$. The new bars will always be red when the old bars were.
copeland 2017-03-09 21:16:58
However, the bar at 8 is going to be $(8,32]$, which contains 27 so it will turn red.
copeland 2017-03-09 21:17:02
What about the bar at 2?
QuestForKnowledge 2017-03-09 21:17:30
red
DemonPlat4 2017-03-09 21:17:30
it reaches 8
SomethingNeutral 2017-03-09 21:17:30
2,8 contains 8
stronto 2017-03-09 21:17:30
itll be red just barely
IsaacZ123 2017-03-09 21:17:30
(2,8] contains a cube
owm 2017-03-09 21:17:30
it will also be red
copeland 2017-03-09 21:17:32
The bar at 2 will be $(2,8]$, which also contains a cube so will also turn red.
copeland 2017-03-09 21:17:36
However, the bar at 1 will be $(1,4]$ which does not contain a cube.
copeland 2017-03-09 21:17:39
That makes $Q(4)=2$.
copeland 2017-03-09 21:17:40
Every bar after the first bar is red.
copeland 2017-03-09 21:17:41
copeland 2017-03-09 21:17:44
For what other values of $m$ is $Q(m)=2$?
vishwathganesan 2017-03-09 21:18:07
Q(4) = Q(5) = Q(6) = Q(7) = 2
Makorn 2017-03-09 21:18:07
$Q(4)=Q(5)=Q(6)=Q(7)=2$
SomethingNeutral 2017-03-09 21:18:07
5,6,7
GeronimoStilton 2017-03-09 21:18:07
$m = 5,6,7$
vishwathganesan 2017-03-09 21:18:07
5,6,7
legolego 2017-03-09 21:18:07
5, 6, 7
Kiola 2017-03-09 21:18:07
5 6 7
MountainHeight 2017-03-09 21:18:07
5-7
DemonPlat4 2017-03-09 21:18:07
Q(5) = Q(6) = Q(7) = 2
IsaacZ123 2017-03-09 21:18:07
5,6,7
IYN 2017-03-09 21:18:07
5, 6, 7
liuh008 2017-03-09 21:18:07
m <= 7
copeland 2017-03-09 21:18:10
Since $(1,5]$, $(1,6]$, and $(1,7]$ do not contain cubes, $Q(5)=Q(6)=Q(7)=2.$
copeland 2017-03-09 21:18:10
Then what?
pianoman24 2017-03-09 21:18:32
Wait as $m$ goes up, it looks like $Q(m)$ will eventually always be $1$
IsaacZ123 2017-03-09 21:18:32
and then the rest are just 1
Doink 2017-03-09 21:18:32
it's all 1 from there
liuh008 2017-03-09 21:18:32
The rest are all 1.
lego101 2017-03-09 21:18:32
then 1?
owm 2017-03-09 21:18:32
Q(8)=1
vishwathganesan 2017-03-09 21:18:32
Q(n) for n >= 8 equals 1
tdeng 2017-03-09 21:18:32
AFter that it's always 1
yrnsmurf 2017-03-09 21:18:32
so Q(x)=1 if x>7
mishai 2017-03-09 21:18:32
Q(8)=1
Reef334 2017-03-09 21:18:32
they all equal 1
Mrkiller 2017-03-09 21:18:32
then they are all 1!!!
copeland 2017-03-09 21:18:34
After 7, every bar is red. Therefore $Q(m)=1$ for $m>7$.
copeland 2017-03-09 21:18:35
So what is our final sum?
vishwathganesan 2017-03-09 21:19:15
so total is 2059
Z_Math404 2017-03-09 21:19:15
2059
vishwathganesan 2017-03-09 21:19:15
so answer is 059
mishai 2017-03-09 21:19:15
1*2010+4*2+9+32=2059
legolego 2017-03-09 21:19:15
59
DemonPlat4 2017-03-09 21:19:15
32+9+4*2+2010*1=2059 -> 059
owm 2017-03-09 21:19:15
32+9+2+2+2+2+1+1....
IsaacZ123 2017-03-09 21:19:15
59
IYN 2017-03-09 21:19:15
32+8+2+2+2+2+1*(2017-7)
IsaacZ123 2017-03-09 21:19:15
2059
Makorn 2017-03-09 21:19:15
$32+9+2+2+2+2+2+2010(1)=2059$
copeland 2017-03-09 21:19:17
The final sum is

\[32+9+4\cdot2+1\cdot2010\equiv32+9+8+10\equiv\boxed{059}.\]
copeland 2017-03-09 21:19:24
Great work.
copeland 2017-03-09 21:19:28
I'm really feeling it now.
NewbieGamer 2017-03-09 21:19:41
14!
W.Sun 2017-03-09 21:19:41
Yay! 2 More!
mishai 2017-03-09 21:19:41
yes my favorite problem
lego101 2017-03-09 21:19:41
Number fourteen…I need some caffeine…
GeronimoStilton 2017-03-09 21:19:41
Problem 14!
copeland 2017-03-09 21:19:44
Favorite?
copeland 2017-03-09 21:19:46
14. Let $a>1$ and $x>1$ satisfy\[\log_a\left(\log_a(\log_a 2)+\log_a24-128\right)=128\]and \[\log_a(\log_ax)=256.\] Find the remainder when $x$ is divided by 1000.
copeland 2017-03-09 21:19:48
There's a rule of thumb about AMC and AIME problems that when you see a logarithm, that means the problem is really scary but way less hard than it looks.
copeland 2017-03-09 21:19:53
This problem proves that rule doesn't always hold.
copeland 2017-03-09 21:20:00
There are lots of different avenues to take with this problem. Most of them fail. There are a couple of nice first steps, though. Any ideas?
Doink 2017-03-09 21:20:36
Take equation 1 to the power of a
potatosensei 2017-03-09 21:20:36
undo the logs
stronto 2017-03-09 21:20:36
remove the first nested log
amackenzie1 2017-03-09 21:20:36
exp?
nukelauncher 2017-03-09 21:20:36
convert to exponential form
ppu 2017-03-09 21:20:36
rewrite in exponential form?
pianoman24 2017-03-09 21:20:36
Write in exponent form instead of logarithm form
AwesomeAaron 2017-03-09 21:20:36
reverse the logs
claserken 2017-03-09 21:20:36
Turn logs into exponentials?
copeland 2017-03-09 21:20:37
Let's exponentiate both sides with respect to base $a$:
copeland 2017-03-09 21:20:38
\[\log_a(\log_a 2)+\log_a24-128=a^{128}\]
copeland 2017-03-09 21:20:41
That looks nicer. Now we can push the 128 to the other side since I'm not really in the mood for negatives (or fractions) yet.
copeland 2017-03-09 21:20:43
\[\log_a(\log_a 2)+\log_a24=a^{128}+128\]
copeland 2017-03-09 21:20:45
And now what?
Z_Math404 2017-03-09 21:21:06
do it again
sxu 2017-03-09 21:21:06
a again
ucap 2017-03-09 21:21:06
more exponents
Makorn 2017-03-09 21:21:06
Combine the two logs
amackenzie1 2017-03-09 21:21:06
exponentiate again
brainiac1 2017-03-09 21:21:06
sum the two logs
nosaj 2017-03-09 21:21:06
combine the loggs
DemonPlat4 2017-03-09 21:21:06
combine the logarithms
stronto 2017-03-09 21:21:06
combine logs on left side
cooljoseph 2017-03-09 21:21:06
exponentiate again
duck_master 2017-03-09 21:21:06
raise it to a again!
legolego 2017-03-09 21:21:06
combine logs
copeland 2017-03-09 21:21:08
We have $\log u+\log v =\log uv$, so
copeland 2017-03-09 21:21:09
\[\log_a(24\log_a2)=a^{128}+128.\]
copeland 2017-03-09 21:21:10
Let's ditch those logs by exponentiating:
copeland 2017-03-09 21:21:13
\[24\log_a2=a^{a^{128}+128}.\]
copeland 2017-03-09 21:21:14
And what's $a^{24\log_a2}$ equal?
mishai 2017-03-09 21:21:35
2^24
IsaacZ123 2017-03-09 21:21:35
2^24
yrnsmurf 2017-03-09 21:21:35
2^24
Z_Math404 2017-03-09 21:21:35
2^24
monkeybanana 2017-03-09 21:21:35
2^24
cooleybz2013 2017-03-09 21:21:35
2^24
tdeng 2017-03-09 21:21:35
2^24
copeland 2017-03-09 21:21:37
Next we have
copeland 2017-03-09 21:21:38
\[2^{24}=a^{a^{a^{128}+128}}.\]
copeland 2017-03-09 21:21:42
Those aren't the same base. Can we fix that?
mishai 2017-03-09 21:22:08
a=2^n
math101010 2017-03-09 21:22:08
a=2^b
stronto 2017-03-09 21:22:08
let a=2^k
yrnsmurf 2017-03-09 21:22:08
set a=2^b
brainiac1 2017-03-09 21:22:08
assume a = 2^b
yrnsmurf 2017-03-09 21:22:08
take the log base 2
copeland 2017-03-09 21:22:10
Let's write $a=2^b$. Then we have
copeland 2017-03-09 21:22:12
\[2^{24}=(2^b)^{(2^b)^{(2^b)^{128}+128}}=2^{b2^{b2^{128b}+128b}},\]so
copeland 2017-03-09 21:22:16
\[24=b2^{b2^{128b}+128b}.\]
copeland 2017-03-09 21:22:25
And do you see the magical simplification on the right side?
samuel 2017-03-09 21:23:16
a little
potatosensei 2017-03-09 21:23:16
not really lol
vishwathganesan 2017-03-09 21:23:16
no not really
copeland 2017-03-09 21:23:21
Alright, great.
liuh008 2017-03-09 21:23:24
b * 2^(128b)=k
copeland 2017-03-09 21:23:31
Wait, what's that about?
copeland 2017-03-09 21:23:42
It looks like there might be some symmetry on the right hand side.
copeland 2017-03-09 21:23:45
Almost.
copeland 2017-03-09 21:23:56
There's all this $2^{128b}$ nonsense.
copeland 2017-03-09 21:24:00
But it's not quite the same.
copeland 2017-03-09 21:24:26
Can we invent another $b2^{128b}$ somewhere?
a1b2 2017-03-09 21:24:32
split up the sum in the exponent
copeland 2017-03-09 21:24:35
Then what happens?
Makorn 2017-03-09 21:25:49
$24=b2^{b2^{128b}}\cdotb^{128}$
rt03 2017-03-09 21:25:49
the RHS is $b2^{128b} \cdot 2^{b2^{128b}}$
tdeng 2017-03-09 21:25:49
$b\cdot2^{128b}\cdot2^{b2^{128b}}$
duck_master 2017-03-09 21:25:49
$b2^{128b} 2^{b2^{128b}}$
copeland 2017-03-09 21:25:52
There is a magical step in every solution to this problem. Here the magical step is
copeland 2017-03-09 21:25:53
\[2^{24}=b2^{b2^{128b}+128b}=(b2^{128b})2^{b2^{128b}}.\]
copeland 2017-03-09 21:25:54
That right side is $t2^t$ for some $t$. Can we write 24 in the same way?
mishai 2017-03-09 21:26:16
3*2^3
liuh008 2017-03-09 21:26:16
t = 3
brainiac1 2017-03-09 21:26:16
24 = 3*2^3
SomethingNeutral 2017-03-09 21:26:16
3*2^3
tdeng 2017-03-09 21:26:16
3*2^3
rt03 2017-03-09 21:26:16
3 \cdot 2^3
copeland 2017-03-09 21:26:18
$24=3\cdot2^3!$
copeland 2017-03-09 21:26:19
Therefore
copeland 2017-03-09 21:26:21
\[3\cdot2^3=(b2^{128b})2^{b2^{128b}}.\]
copeland 2017-03-09 21:26:25
So?
stronto 2017-03-09 21:26:47
b*2^128b = 3
mishai 2017-03-09 21:26:47
b*2^(128b)=3
DemonPlat4 2017-03-09 21:26:47
b2^128b=3
randomsolver 2017-03-09 21:26:47
3=b2^(128b)
SeanGee 2017-03-09 21:26:47
b2^128b = 3
copeland 2017-03-09 21:26:49
Since $f(t)=t2^t$ is monotonic, we have to have $b2^{128b}=3.$
copeland 2017-03-09 21:26:50
We're still not done. We need another substitution. What should we make $b?$
mishai 2017-03-09 21:27:45
b=3/2^k
liuh008 2017-03-09 21:27:45
3/2^c could work
rt03 2017-03-09 21:27:45
$3 \cdot 2^z$
copeland 2017-03-09 21:27:48
$b=3\cdot2^c$ is nice, but let's set $b=3\cdot2^{-c}.$ I want to get $b$ over to the right side of the equation. This gives
copeland 2017-03-09 21:27:52
\[3\cdot 2^{-c}2^{128\cdot3\cdot 2^{-c}}=3,\]so\[2^{3\cdot2^{7-c}}=2^c\] and
copeland 2017-03-09 21:28:05
\[3\cdot 2^{7-c}=c.\]
copeland 2017-03-09 21:28:09
Do you see a $c$ that solves this equation?
mishai 2017-03-09 21:28:39
c=6
Makorn 2017-03-09 21:28:39
$c=6$
EpicCarrotMaster 2017-03-09 21:28:39
6
cooleybz2013 2017-03-09 21:28:39
6
pianoman24 2017-03-09 21:28:39
6.
yrnsmurf 2017-03-09 21:28:39
6
math.fever 2017-03-09 21:28:39
6
nukelauncher 2017-03-09 21:28:39
6
BIGBUBBLE 2017-03-09 21:28:39
6
QuestForKnowledge 2017-03-09 21:28:39
shamelessly gueass 6
mathcrazymar 2017-03-09 21:28:39
6
fangmu 2017-03-09 21:28:39
6
Funnybunny5246 2017-03-09 21:28:39
6
NewbieGamer 2017-03-09 21:28:39
6
Pudentane 2017-03-09 21:28:39
6
MrMXS 2017-03-09 21:28:39
$6$
lego101 2017-03-09 21:28:39
6
copeland 2017-03-09 21:28:41
$c=6$ solves this equation. If $c=6,$ what is $a?$
vishwathganesan 2017-03-09 21:29:26
2^3/64
liuh008 2017-03-09 21:29:26
2^(3/64)
rt03 2017-03-09 21:29:26
$2^{\frac{3}{64}}$
Makorn 2017-03-09 21:29:26
$a=2^{\frac{6}{64}}$?
QuestForKnowledge 2017-03-09 21:29:26
2^(3/64)
copeland 2017-03-09 21:29:28
If $c=6$ then $b=3\cdot2^{-6},$ and $a=2^{3\cdot 2^{-6}}.$
copeland 2017-03-09 21:29:56
\[a=8^b=8^{2^{-6}}\]
copeland 2017-03-09 21:30:05
If you got this far, then congrats. You're still not done. We need to find the $x$ that solves $\log_a\log_a x=256.$
copeland 2017-03-09 21:30:09
So we need some logs. What do we do?
IsaacZ123 2017-03-09 21:31:30
unlock the logarithms
math.fever 2017-03-09 21:31:30
convert a to logs
potatosensei 2017-03-09 21:31:30
undo the logs
DemonPlat4 2017-03-09 21:31:30
change logarithms into exponentation
copeland 2017-03-09 21:31:50
We have two paths. We could turn one equation into an exponential equation or turn the other equation into a logs equation.
copeland 2017-03-09 21:32:03
Let's simplify our equation for $a$ a tiny bit first.
copeland 2017-03-09 21:32:10
Let's take the log of our equation:\[1=\log_a8^{2^{-6}}=2^{-6}\log_a8,\] and \[\log_a8=2^6.\]
copeland 2017-03-09 21:32:12
Can we get 256 in there?
vishwathganesan 2017-03-09 21:33:09
multiply by 4
Makorn 2017-03-09 21:33:09
$4log_a8=256$
owm 2017-03-09 21:33:09
multiply by 4
GeronimoStilton 2017-03-09 21:33:09
By multiplying both sides by $4$, yes.
tdeng 2017-03-09 21:33:09
256/4=2^6
math.fever 2017-03-09 21:33:09
2^2log_a(8)=2^8
copeland 2017-03-09 21:33:11
Yeah, multiply by $2^2$:\[256=2^2\cdot2^6=4\log_a8=\log_a8^4=\log_a2^{12}.\]
copeland 2017-03-09 21:33:12
This is progress! We've gotten rid of one of the layers of log. Since $256=\log_a\log_a x$ we need to solve
copeland 2017-03-09 21:33:13
\[\log_a x=2^{12}.\]
copeland 2017-03-09 21:33:17
Congrats. Now we're here. We're still not done.
copeland 2017-03-09 21:33:21
I want to exponentiate:\[x=a^{2^{12}}.\] Now what?
QuestForKnowledge 2017-03-09 21:34:00
plug in that a
nukelauncher 2017-03-09 21:34:00
substitute the value of a
amackenzie1 2017-03-09 21:34:00
we know $a$!
liuh008 2017-03-09 21:34:00
a = 2^(3/2^6)
duck_master 2017-03-09 21:34:00
a=2^(3/64)
stronto 2017-03-09 21:34:00
a = 2^(3/64)
copeland 2017-03-09 21:34:02
We do know that $a=8^{2^{-6}}$, so \[x=(8^{2^{-6}})^{2^{12}}=8^{2^6}=2^{3\cdot2^6}=2^{192}.\]
copeland 2017-03-09 21:34:09
Oh great! We're still not done, but we're close. We just need to compute $2^{192}$ modulo 1000.
copeland 2017-03-09 21:34:11
What do we need?
duck_master 2017-03-09 21:34:34
CRT
brainiac1 2017-03-09 21:34:34
CRT and some luck
GeronimoStilton 2017-03-09 21:34:34
CRT!
liuh008 2017-03-09 21:34:34
mod 8 and mod 125
stronto 2017-03-09 21:34:34
CRT!
QuestForKnowledge 2017-03-09 21:34:34
take mod and reduce
S0larPh03nix 2017-03-09 21:34:34
crt and euler
Z_Math404 2017-03-09 21:34:34
chinese remainder theroesm
Funnybunny5246 2017-03-09 21:34:34
CRT
nukelauncher 2017-03-09 21:34:34
crt
copeland 2017-03-09 21:34:36
Oh, finally. I was worried this wasn't an AIME at all.
copeland 2017-03-09 21:34:43
We've made it this far without the Chinese Remainder Theorem. Now's the time. I'll get you started. $1000=8\cdot125$.
copeland 2017-03-09 21:34:45
What is $2^{192}\pmod8?$
SomethingNeutral 2017-03-09 21:34:54
0
ninjataco 2017-03-09 21:34:54
0
Makorn 2017-03-09 21:34:54
$0$
ppu 2017-03-09 21:34:54
0
owm 2017-03-09 21:34:54
0
Z_Math404 2017-03-09 21:34:54
0
copeland 2017-03-09 21:34:56
It's zero. All day long.
copeland 2017-03-09 21:34:57
Now we need to know $2^{192}\pmod{125}.$ First, what is the order of an element modulo $125?$ That is, what is $\phi(125)?$
legolego 2017-03-09 21:35:15
100
yrnsmurf 2017-03-09 21:35:15
100
nosaj 2017-03-09 21:35:15
100
ppu 2017-03-09 21:35:15
100
varun555 2017-03-09 21:35:15
100
SnakeYu 2017-03-09 21:35:15
100
tdeng 2017-03-09 21:35:15
100
copeland 2017-03-09 21:36:02
$\phi(p^n)=p^n-p^{n-1}$, so $\phi(125)=125-25=100.$
copeland 2017-03-09 21:36:24
So what should we compute instead of $2^{192}\pmod{125}?$
andsun19 2017-03-09 21:36:31
2^-8 MOD 125
amackenzie1 2017-03-09 21:36:31
$2^{-8}$
SnakeYu 2017-03-09 21:36:31
2^-8
rt03 2017-03-09 21:36:31
$2^{-8}$
cjquines0 2017-03-09 21:36:31
$2^{-8}$ seems easier
celestialphoenix3768 2017-03-09 21:36:31
2^-8 mod 125
copeland 2017-03-09 21:36:35
We can compute $2^{-8}\pmod{125}$, which (another freebie) is \[256^{-1}\equiv6^{-1}\pmod{125}.\]
copeland 2017-03-09 21:36:37
Can you see the inverse of 6 modulo 125?
copeland 2017-03-09 21:36:41
Can you think of any multiples of 6 that are one more than a multiple of 125?
ppu 2017-03-09 21:37:01
126 XD
rt03 2017-03-09 21:37:01
126
a1b2 2017-03-09 21:37:01
126
Makorn 2017-03-09 21:37:01
$126=6\cdot21$
lego101 2017-03-09 21:37:01
126
amackenzie1 2017-03-09 21:37:01
126
liuh008 2017-03-09 21:37:01
126
brainiac1 2017-03-09 21:37:01
126
owm 2017-03-09 21:37:01
126
copeland 2017-03-09 21:37:04
Oh, 126 is a multiple of 6. So what is $6^{-1}\pmod{125}?$
S0larPh03nix 2017-03-09 21:37:18
21
celestialphoenix3768 2017-03-09 21:37:18
21
tdeng 2017-03-09 21:37:18
21
GeronimoStilton 2017-03-09 21:37:18
$21$
S0larPh03nix 2017-03-09 21:37:18
21
dhruv 2017-03-09 21:37:18
21
SomethingNeutral 2017-03-09 21:37:18
21
stronto 2017-03-09 21:37:18
21
MountainHeight 2017-03-09 21:37:18
21
rt03 2017-03-09 21:37:18
21
copeland 2017-03-09 21:37:20
Since $6\cdot21=126$ we know that $6^{-1}\equiv21\pmod{1000}.$
copeland 2017-03-09 21:37:22
Great, but we're still not done. Now we need to find a solution to the congruence:

\begin{align*}

x&\equiv0\pmod8\\

x&\equiv21\pmod{125}.

\end{align*}
copeland 2017-03-09 21:37:23
Anyone got it?
celestialphoenix3768 2017-03-09 21:38:01
896
GeronimoStilton 2017-03-09 21:38:01
$896$!
liuh008 2017-03-09 21:38:01
896
celestialphoenix3768 2017-03-09 21:38:01
896
brainiac1 2017-03-09 21:38:01
896
rd123 2017-03-09 21:38:01
896
S0larPh03nix 2017-03-09 21:38:01
896
islander7 2017-03-09 21:38:01
896
a1b2 2017-03-09 21:38:01
$\boxed{896}$
Z_Math404 2017-03-09 21:38:01
896
copeland 2017-03-09 21:38:03
The value $21-125=-104$ is actually a multiple of 8, so that's the right residue class. The answer is $1000+(-104)=\boxed{896}.$
copeland 2017-03-09 21:38:08
Great work, team. Uno mas.
copeland 2017-03-09 21:38:13
(That's Spanish.)
lego101 2017-03-09 21:38:33
Number fifteen. THE FINAL BATTLE. It's you vs. the math, and no one can intervene.
QuestForKnowledge 2017-03-09 21:38:33
one more
IsaacZ123 2017-03-09 21:38:33
oh my
Makorn 2017-03-09 21:38:33
I only do Esperanto; it's logically equivalent to math
pie314159265 2017-03-09 21:38:33
i thought it was chinese jkjk
poremon 2017-03-09 21:38:33
Yo entiendo
copeland 2017-03-09 21:38:38
15. The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt3,$ $5,$ and $\sqrt{37},$ as shown, is $\dfrac{m\sqrt p}n,$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.
copeland 2017-03-09 21:38:40
copeland 2017-03-09 21:38:52
Let's pick some variables. There are lots of choices, but I have a personal favorite. What are the options that you see?
math101010 2017-03-09 21:39:23
s=side length
pianoman24 2017-03-09 21:39:23
How about $s$ for side length?
Regardy 2017-03-09 21:39:23
the area?
tdeng 2017-03-09 21:39:23
Sidelength of the equilaterel triangle
copeland 2017-03-09 21:39:48
We could use side length or area. Those are options, but I don't like side length because it might not give us a unique triangle.
QuestForKnowledge 2017-03-09 21:39:55
alpha and omega and beta and ...
copeland 2017-03-09 21:39:59
Those are lots of letters.
copeland 2017-03-09 21:40:05
Where else could we put them?
yrnsmurf 2017-03-09 21:40:18
the coordinates of the vertices of the triangle
QuestForKnowledge 2017-03-09 21:40:21
for angles maybe, thats there oringinal purpose
nosaj 2017-03-09 21:40:23
we should have a variable for the horizontal and vertical positions of the two vertices on the legs
copeland 2017-03-09 21:40:29
We could label $\theta$ as the variable below and work entirely in terms of that. We could also work with $\phi=90^\circ-\theta$ just as easily. We can also work with either of the lengths, $a$, or $b$, of that small right triangle.
copeland 2017-03-09 21:40:30
copeland 2017-03-09 21:40:34
My inclination is to work with both of the variables $a$ and $b$. There are two reasons for this.
copeland 2017-03-09 21:40:36
First, I like symmetry, and I expect that the symmetrical relationship between $a$ and $b$ will help us deal with the variables together.
copeland 2017-03-09 21:40:40
Second, I have a hunch that if we use only $a$, then whatever ugly expression gives us $b$ is going to be carried around through a lot of computations trying to make things harder on us.
lego101 2017-03-09 21:40:47
Why wouldn't side length give a unique triangle?
copeland 2017-03-09 21:40:49
Good question.
copeland 2017-03-09 21:41:06
If there's a unique minimum, then moving either direction from it should give you the numbers just larger than it.
copeland 2017-03-09 21:41:15
Let's think in terms of $a$ and $b$. There is a relation between them that makes the equilateral fit inside the big triangle. What's it time for now?
GeronimoStilton 2017-03-09 21:41:42
Please not coordinate bash!
amgmflannigan 2017-03-09 21:41:42
coordinates
GeronimoStilton 2017-03-09 21:41:42
Complex bash!
nosaj 2017-03-09 21:41:42
uhh coordinate geometry?
copeland 2017-03-09 21:41:55
It's time for some analytic techniques. Let's skip complex numbers on this, though that works nicely, too. Here are some coordinates:
copeland 2017-03-09 21:41:56
copeland 2017-03-09 21:41:58
Now, if we randomly pick $a$ and $b$, we won't have $(x,y)$ on that line.
copeland 2017-03-09 21:42:01
In terms of $a$ and $b$, let's figure out $(x,y)$. Synthetically, how would you get to $(x,y)?$
nosaj 2017-03-09 21:42:52
rotatate 60 degrees!
yrnsmurf 2017-03-09 21:42:52
rotate (a,0) 60 degrees around (0,b)
ninjataco 2017-03-09 21:42:52
rotate (a,0) around (0,b)
cooljoseph 2017-03-09 21:42:52
rotating 60 degrees
IsaacZ123 2017-03-09 21:42:52
rotate the line with (0,b) and (a,0) by -60 degrees
copeland 2017-03-09 21:43:01
That's one idea. Here's the other:
owm 2017-03-09 21:43:07
Find the midpoint and slope of (0,b)-(a,0)
liuh008 2017-03-09 21:43:07
Perpendicular bisector of (0,b) and (a,0)
mathfever 2017-03-09 21:43:07
perpendicular bisector of the line
Funnybunny5246 2017-03-09 21:43:07
Perpendicular bisector of ab
copeland 2017-03-09 21:43:11
Let's go with the second.
copeland 2017-03-09 21:43:20
One way to get there is to go to the midpoint of the segment from $(a,0)$ to $(0,b)$ and then walk along the altitude. The slope of the that base is $-\frac ba$. What's the slope of the altitude?
raxu 2017-03-09 21:43:37
$\frac{a}{b}$
IsaacZ123 2017-03-09 21:43:37
a/b
math101010 2017-03-09 21:43:37
a/b
islander7 2017-03-09 21:43:37
a/b
nosaj 2017-03-09 21:43:37
a/b
copeland 2017-03-09 21:43:39
The slope of the altitude is $\dfrac ab$.
copeland 2017-03-09 21:43:45
If the edge length is $s$, what's the altitude?
swagger21 2017-03-09 21:44:24
s*sqrt3/2
math101010 2017-03-09 21:44:24
s*sqrt(3)/2
ThorJames 2017-03-09 21:44:24
s sqrt(3) / 2
raxu 2017-03-09 21:44:24
$\frac{sqrt{3}}{2}s$
tdeng 2017-03-09 21:44:24
$\frac{s\sqrt{3}}{2}$
Funnybunny5246 2017-03-09 21:44:24
s sqrt(3)/2
vishwathganesan 2017-03-09 21:44:24
s*sqrt3/2
GeronimoStilton 2017-03-09 21:44:24
$\frac{s\sqrt{3}}{2}$
MountainHeight 2017-03-09 21:44:24
s√3/2
MrMXS 2017-03-09 21:44:24
$\cfrac{s\sqrt{3}}{2}$
copeland 2017-03-09 21:44:27
The altitude is $s\dfrac{\sqrt3}2$.
copeland 2017-03-09 21:44:37
So if we start at $\left(\dfrac a2,\dfrac b2\right)$ and move $s\dfrac{\sqrt3}2$ in the $\dfrac ab$ direction where do we end up?
raxu 2017-03-09 21:45:21
$(x,y)=(\frac{a+\sqrt{3}b}{2},\frac{b+\sqrt{3}a}{2})$
IsaacZ123 2017-03-09 21:45:24
(a/2+a/(sqrt(a^2+b^2)*s*sqrt(3)/2,b/2+b/(sqrt(a^2+b^2)*s*sqrt(3)/2)
copeland 2017-03-09 21:45:26
The vector we want to travel along points toward $(b,a)$. The vector $(b,a)$ of course has length $s$, since $(a,-b)$ does. So the vector along the altitude is $\dfrac{\sqrt3}2(b,a)$ from the center of the base.
copeland 2017-03-09 21:45:27
The third vertex is \[\left(\dfrac a2,\dfrac b2\right)+\left(\dfrac {b\sqrt3}2,\dfrac{a\sqrt3}2\right),\]which is
copeland 2017-03-09 21:45:29
\[\left(\dfrac{a+b\sqrt3}2,\dfrac{b+a\sqrt3}2\right).\]
copeland 2017-03-09 21:45:33
This is a nice expression since when we swap $a$ and $b,$ the coordinates swap (because the equilateral triangle flips over $y=x$).
copeland 2017-03-09 21:45:34
What's the equation for the hypotenuse?
raxu 2017-03-09 21:46:48
$y=5-\frac{5}{2\sqrt{3}}x$
duck_master 2017-03-09 21:46:48
$\frac{x}{5}+\frac{y}{2\sqrt{3}}=1$
math101010 2017-03-09 21:46:48
y=(-2sqrt3/5)*x+5
Funnybunny5246 2017-03-09 21:46:48
ay+bx=ab
vishwathganesan 2017-03-09 21:46:48
5y+2sqrt3x=something
liuh008 2017-03-09 21:46:48
-5/(2sqrt{3})x + 5 = y
copeland 2017-03-09 21:46:51
If a line has intercepts $(p,0)$ and $(0,q)$, then its equation is $qx+py=pq$.
copeland 2017-03-09 21:46:52
The equation for the hypotenuse is \[5x+2\sqrt3 y=10\sqrt3.\]
copeland 2017-03-09 21:47:00
We can combine these to get the relationship between $a$ and $b$. Awesome. I won't torture you with the work.
copeland 2017-03-09 21:47:04
Combining $(x,y)=\left(\dfrac{a+b\sqrt3}2,\dfrac{b+a\sqrt3}2\right)$ with $5x+2\sqrt3 y=10\sqrt3$ gives

\[

5\left(\dfrac{a+b\sqrt3}2\right)+2\sqrt3\left(\dfrac{b+a\sqrt3}2\right)=10\sqrt3.

\]Multiplying by 2 and expanding gives\[

11a+7\sqrt3 b=20\sqrt3.

\]
copeland 2017-03-09 21:47:14
What are we trying to minimize?
swagger21 2017-03-09 21:47:43
a^2 + b^2
math101010 2017-03-09 21:47:43
a^2+b^2
pie314159265 2017-03-09 21:47:43
a^2+b^2
ninjataco 2017-03-09 21:47:43
a^2 + b^2
raxu 2017-03-09 21:47:43
$\frac{\sqrt{3}}{4}(a^2+b^2)$
cooljoseph 2017-03-09 21:47:43
$a^2+b^2$
treemath 2017-03-09 21:47:43
a^2+b^2
liuh008 2017-03-09 21:47:43
a^2 + b^2
owm 2017-03-09 21:47:43
a^2+b^2
copeland 2017-03-09 21:47:48
We're trying to minimize the area of that gray triangle.
copeland 2017-03-09 21:47:52
The area of an equilateral triangle with side $s$ is $\dfrac{s^2\sqrt3}4$. Since $s^2=a^2+b^2$. . .
copeland 2017-03-09 21:47:53
We want to minimize\[\dfrac{\sqrt3}4(a^2+b^2)\] given that \[11a+7\sqrt3 b=20\sqrt3.\]
copeland 2017-03-09 21:47:57
What tool do we need?
CornSaltButter 2017-03-09 21:48:31
Cauchy-schwarz inequality
randomsolver 2017-03-09 21:48:31
cauchy
liuh008 2017-03-09 21:48:31
CauchyS?
tdeng 2017-03-09 21:48:31
Cauchy Schwarz?
cooljoseph 2017-03-09 21:48:31
C-S?
raxu 2017-03-09 21:48:31
Solve for $b$ in terms of $a$, or use Cauchy-Schwarz Inequality.
copeland 2017-03-09 21:48:33
This is the perfect time for Cauchy-Schwartz. ( http://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality )
copeland 2017-03-09 21:48:34
We need two lists of numbers to mush together. What are our lists?
IsaacZ123 2017-03-09 21:49:23
a^2+b^2 and 11a+7sqrt(3)b
GeronimoStilton 2017-03-09 21:49:23
$a$ and $b$, $11$ and $7\sqrt{3}$
mishai 2017-03-09 21:49:23
(a^2+b^2)(121+147)>=(11a+7bsqrt3)^2
raxu 2017-03-09 21:49:23
$11,7\sqrt{3};a,b$
brainiac1 2017-03-09 21:49:23
(a^2+b^2)(121+147)
swagger21 2017-03-09 21:49:23
(a^2 + b^2)(121 + 147) ≥ (11a + 7root3b)^2
lego101 2017-03-09 21:49:29
i knew there was going to be a cauchy schwarz problem on aime this year
copeland 2017-03-09 21:49:37
Shocker, right?
copeland 2017-03-09 21:49:39
Certainly one of the lists is $a,b$. The other list is $7,\sqrt3$.
copeland 2017-03-09 21:49:47
Cauchy-Schwarz tells us that \[(11a+7\sqrt3b)^2\leq(a^2+b^2)(11^2+(7\sqrt3)^2).\]
copeland 2017-03-09 21:49:51
Expanding all that out and substituting what we know about the left side gives

\begin{align*}

a^2+b^2

\geq\quad&\frac{(11a+7\sqrt3b)^2}{(11^2+(7\sqrt3)^2)}\\

&=\frac{(20\sqrt3)^2}{(121+147)}\\

&=\frac{1200}{268}.\\

&=\frac{300}{67}.\\

\end{align*}

We even know that equality occurs when $(a,b)$ is proportional to $(11,7\sqrt3)$, so we are going to get a minimum from this.
copeland 2017-03-09 21:50:01
That's nice. And what is the minimum area?
GeronimoStilton 2017-03-09 21:50:22
You mean $11$ and $7\sqrt{3}$ is the second list. @copeland
brainiac1 2017-03-09 21:51:00
75sqrt3/67
yrnsmurf 2017-03-09 21:51:00
75/67 sqrt 3
IsaacZ123 2017-03-09 21:51:00
75sqrt(3)/67
NewbieGamer 2017-03-09 21:51:00
75*rt3/67
Mudkipswims42 2017-03-09 21:51:00
$\dfrac{75\sqrt{3}}{67}$
amgmflannigan 2017-03-09 21:51:00
300/67*\sqrt{3}/4
ninjataco 2017-03-09 21:51:00
300sqrt(3)/268
liuh008 2017-03-09 21:51:00
75sqrt{3}
GeronimoStilton 2017-03-09 21:51:00
$\frac{75\sqrt{3}}{67}$
mishai 2017-03-09 21:51:00
75sqrt3/67
swagger21 2017-03-09 21:51:00
75root3/67
liuh008 2017-03-09 21:51:00
75sqrt{3]/67
copeland 2017-03-09 21:51:02
The area is $\dfrac{s^2\sqrt3}4$ so the minimum area is $\dfrac{\sqrt3}4\cdot\dfrac{300}{67}=\dfrac{75\sqrt3}{67}.$
copeland 2017-03-09 21:51:02
The final answer is. . .
raxu 2017-03-09 21:51:33
$\frac{75\sqrt{3}}{67}$, giving us $\boxed{145}$.
mishai 2017-03-09 21:51:33
145
Metal_Bender19 2017-03-09 21:51:33
145
IsaacZ123 2017-03-09 21:51:33
145
raxu 2017-03-09 21:51:33
$\boxed{145}$
yrnsmurf 2017-03-09 21:51:33
145
Z_Math404 2017-03-09 21:51:33
145
GeronimoStilton 2017-03-09 21:51:33
$145$
MountainHeight 2017-03-09 21:51:33
145
Mudkipswims42 2017-03-09 21:51:33
145?
ninjataco 2017-03-09 21:51:33
145
nature 2017-03-09 21:51:33
145
duck_master 2017-03-09 21:51:33
145
liuh008 2017-03-09 21:51:33
145
vishwathganesan 2017-03-09 21:51:33
oops 145
randomsolver 2017-03-09 21:51:33
145
Jyzhang12 2017-03-09 21:51:33
145
curry3030 2017-03-09 21:51:33
145
amackenzie1 2017-03-09 21:51:33
145
illumination 2017-03-09 21:51:33
145
copeland 2017-03-09 21:51:36
\[75+3+67=\boxed{145}.\]
copeland 2017-03-09 21:51:45
Guess what?
Z_Math404 2017-03-09 21:52:12
145 YAS WE DEFEATED ALL OF THEM WITH OUR MATH SWORDS!
strategos21 2017-03-09 21:52:12
"and we are done"
amackenzie1 2017-03-09 21:52:12
We did it!
brainiac1 2017-03-09 21:52:12
and we're done
awesomemaths 2017-03-09 21:52:12
partly spleeping now
IsaacZ123 2017-03-09 21:52:12
that was easier than 14
lego101 2017-03-09 21:52:12
We did it. We beat the problems. WE POWERED THROUGH
amackenzie1 2017-03-09 21:52:12
Yay!
vishwathganesan 2017-03-09 21:52:12
yay
raxu 2017-03-09 21:52:12
We're done!
IsaacZ123 2017-03-09 21:52:12
we're done
awesomemaths 2017-03-09 21:52:12
we are done
GeronimoStilton 2017-03-09 21:52:12
We finished the test in the required time limit!
samuel 2017-03-09 21:52:12
We all got a 15 today!
lego101 2017-03-09 21:52:12
WE POWERED THROUGH
vishwathganesan 2017-03-09 21:52:12
we're not done though
xxu110 2017-03-09 21:52:12
what
cooleybz2013 2017-03-09 21:52:12
the end
samuel 2017-03-09 21:52:12
What?
copeland 2017-03-09 21:52:22
Yeah, I think 14 was epically hard.
copeland 2017-03-09 21:52:34
That's all for the 2017 AIME A. Thanks for spending your evening with us.
copeland 2017-03-09 21:52:34
There will be a shindig just like this in a couple of weeks on March 24 when the inimitable Dave Patrick will work with you through the 2017 AIME II.
lego101 2017-03-09 21:52:58
Thank you so much!!!! This was amazing!
copeland 2017-03-09 21:53:00
No, thank you guys.

Copyright © 2024 AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this file with others.