2015 AIME I Math Jam
Go back to the Math Jam ArchiveAoPS instructors discuss all 15 problems of the 2015 AIME I.
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Facilitator: Dave Patrick
DPatrick
2015-03-21 19:00:28
Welcome to the 2015 AIME I Math Jam!
Welcome to the 2015 AIME I Math Jam!
DPatrick
2015-03-21 19:00:35
I'm Dave Patrick, and I'll be leading our discussion tonight.
I'm Dave Patrick, and I'll be leading our discussion tonight.
DPatrick
2015-03-21 19:00:41
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.
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.
DPatrick
2015-03-21 19:00:51
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.
DPatrick
2015-03-21 19:01:01
This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick
2015-03-21 19:01:16
There are a lot of students here! As I said, only a relatively small fraction of the well-written comments will be passed to the entire group. Please do not take it personally if your responses do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!
There are a lot of students here! As I said, only a relatively small fraction of the well-written comments will be passed to the entire group. Please do not take it personally if your responses do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!
DPatrick
2015-03-21 19:01:31
Also, we won't always 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!
Also, we won't always 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!
DPatrick
2015-03-21 19:01:59
We do have two teaching assistants with us tonight to help answer your questions: Luis (Duelist) and Alex (BOGTRO). (Or at least we're supposed to have 2 -- Alex must be running late.)
We do have two teaching assistants with us tonight to help answer your questions: Luis (Duelist) and Alex (BOGTRO). (Or at least we're supposed to have 2 -- Alex must be running late.)
DPatrick
2015-03-21 19:02:14
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.
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.
DPatrick
2015-03-21 19:02:36
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.
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.
DPatrick
2015-03-21 19:02:48
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.
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.
AlcumusGuy
2015-03-21 19:02:51
will rrusczyk also be teaching tonight?
will rrusczyk also be teaching tonight?
mathboxboro
2015-03-21 19:02:51
how about rruczyk?
how about rruczyk?
DPatrick
2015-03-21 19:03:01
I think he's here just to keep an eye on me...
I think he's here just to keep an eye on me...
DPatrick
2015-03-21 19:03:19
Before we get started, I have a question for those of you who took the test: What was your favorite question on the test? What was your least favorite?
Before we get started, I have a question for those of you who took the test: What was your favorite question on the test? What was your least favorite?
DPatrick
2015-03-21 19:04:19
A lot of you liked 12, which was pretty nice too I agree. My favorites were 5 and 11 because they were a little different.
A lot of you liked 12, which was pretty nice too I agree. My favorites were 5 and 11 because they were a little different.
DPatrick
2015-03-21 19:04:30
My least favorite was #9, which also seems to be a popular choice for least favorite.
My least favorite was #9, which also seems to be a popular choice for least favorite.
DPatrick
2015-03-21 19:04:47
But we'll get to them all...let's get started! We're going to do all of 1 through 15, in order.
But we'll get to them all...let's get started! We're going to do all of 1 through 15, in order.
DPatrick
2015-03-21 19:04:55
1. The expressions \[A=1\times2+3\times4+5\times6+\cdots+37\times38+39\] and \[B=1+2\times3+4\times5+\cdots+38\times39\]are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers $A$ and $B$.
1. The expressions \[A=1\times2+3\times4+5\times6+\cdots+37\times38+39\] and \[B=1+2\times3+4\times5+\cdots+38\times39\]are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers $A$ and $B$.
DPatrick
2015-03-21 19:05:15
You'll notice the current problem will also be at the top of the window. You can resize that top area by dragging the horizontal bar.
You'll notice the current problem will also be at the top of the window. You can resize that top area by dragging the horizontal bar.
akim99
2015-03-21 19:05:40
Group into terms
Group into terms
Tommy2000
2015-03-21 19:05:40
Group up products in the subtraction!
Group up products in the subtraction!
Darn
2015-03-21 19:05:40
Group corresponding terms in the two sums seems like a good idea (find a pattern)
Group corresponding terms in the two sums seems like a good idea (find a pattern)
AKAL3
2015-03-21 19:05:40
make some nice groupings for the differences
make some nice groupings for the differences
math_boy
2015-03-21 19:05:40
B is obviously greater, so we want B - A
B is obviously greater, so we want B - A
urmilla
2015-03-21 19:05:40
Do B-A.
Do B-A.
DPatrick
2015-03-21 19:05:48
$B$ looks bigger because it has that $38 \times 39$ term at the end.
$B$ looks bigger because it has that $38 \times 39$ term at the end.
DPatrick
2015-03-21 19:06:03
So let's compute $B-A$. (If we're wrong about it being bigger, we'll just get a negative number.)
So let's compute $B-A$. (If we're wrong about it being bigger, we'll just get a negative number.)
DPatrick
2015-03-21 19:06:22
As some of you mentioned, there are a lot of common factors. So let's pair them up to make it easy to subtract, like so:
As some of you mentioned, there are a lot of common factors. So let's pair them up to make it easy to subtract, like so:
DPatrick
2015-03-21 19:06:27
\[\begin{array}{ccccccccccccc}
\phantom{-}(B&=&1&+&2\times3&+&4\times5&+&6\times7&+&\cdots&+&38\times39 &&\phantom{39})\\
-(A&=& & &1\times2&+&3\times4&+&5\times6&+&\cdots&+&37\times38&+&39)\\
\end{array}\]
\[\begin{array}{ccccccccccccc}
\phantom{-}(B&=&1&+&2\times3&+&4\times5&+&6\times7&+&\cdots&+&38\times39 &&\phantom{39})\\
-(A&=& & &1\times2&+&3\times4&+&5\times6&+&\cdots&+&37\times38&+&39)\\
\end{array}\]
DPatrick
2015-03-21 19:06:49
Then what happens?
Then what happens?
Eugenis
2015-03-21 19:07:13
$B-A=-38+(2\times 2)+(2\times 4)+(2\times 6)+\cdots +(2\times 36)+(2\times 38)$
$B-A=-38+(2\times 2)+(2\times 4)+(2\times 6)+\cdots +(2\times 36)+(2\times 38)$
IMGUNNA
2015-03-21 19:07:13
common factors
common factors
cumo99
2015-03-21 19:07:13
factor
factor
MSTang
2015-03-21 19:07:13
-38 + 2*(2+4+6+...+38)
-38 + 2*(2+4+6+...+38)
pieslinger
2015-03-21 19:07:13
stuff cancels nicely
stuff cancels nicely
summitwei
2015-03-21 19:07:13
The middle terms are all multiples of 4
The middle terms are all multiples of 4
DPatrick
2015-03-21 19:07:20
Subtracting gives
\[\begin{array}{ccccccccccccc}
\phantom{-}(B&=&1&+&2\times3&+&4\times5&+&6\times7&+&\cdots&+&38\times39 &&\phantom{39})\\
-(A&=& & &1\times2&+&3\times4&+&5\times6&+&\cdots&+&37\times38&+&39)\\
\hline
B-A&=&1&+&2\times2&+&4\times2&+&6\times2&+&\cdots&+&38\times2 &-&39\\
\end{array}\]
Subtracting gives
\[\begin{array}{ccccccccccccc}
\phantom{-}(B&=&1&+&2\times3&+&4\times5&+&6\times7&+&\cdots&+&38\times39 &&\phantom{39})\\
-(A&=& & &1\times2&+&3\times4&+&5\times6&+&\cdots&+&37\times38&+&39)\\
\hline
B-A&=&1&+&2\times2&+&4\times2&+&6\times2&+&\cdots&+&38\times2 &-&39\\
\end{array}\]
DPatrick
2015-03-21 19:07:32
That's nice. Except for the first and the last terms, the middle terms are all multiples of 4.
That's nice. Except for the first and the last terms, the middle terms are all multiples of 4.
supercomputer
2015-03-21 19:07:43
-38+4(sum from 1 to 19)
-38+4(sum from 1 to 19)
jam10307
2015-03-21 19:07:48
factor out 4
factor out 4
nosaj
2015-03-21 19:07:48
So factor out a 4 from middle terms
So factor out a 4 from middle terms
akim99
2015-03-21 19:07:51
which is $ 4(1+2+...+19) -38$
which is $ 4(1+2+...+19) -38$
DPatrick
2015-03-21 19:07:56
Exactly:
\[
B-A = 1 + 4(1+2+3+\cdots+19)-39.
\]
Exactly:
\[
B-A = 1 + 4(1+2+3+\cdots+19)-39.
\]
ryanyz10
2015-03-21 19:08:26
4(190) - 38
4(190) - 38
Tommy2000
2015-03-21 19:08:26
$760-38=\boxed{722}$
$760-38=\boxed{722}$
csmath
2015-03-21 19:08:26
which is thus 190*4-38
which is thus 190*4-38
C-bass
2015-03-21 19:08:26
=4(19 x 20 / 2) -38
=4(19 x 20 / 2) -38
MiamiHeat88
2015-03-21 19:08:32
4x190 - 38 = 760 - 38 = 722
4x190 - 38 = 760 - 38 = 722
urmilla
2015-03-21 19:08:32
722
722
stan23456
2015-03-21 19:08:32
which is 722.
which is 722.
DPatrick
2015-03-21 19:08:40
Right, now we're done:
\[
B-A = 1 + 4 \cdot \frac{19(20)}{2} - 39 = 1 + 760 - 39 = \boxed{722}.
\]
Right, now we're done:
\[
B-A = 1 + 4 \cdot \frac{19(20)}{2} - 39 = 1 + 760 - 39 = \boxed{722}.
\]
DPatrick
2015-03-21 19:08:59
Pretty nice #1 to get us warmed up, I thought.
Pretty nice #1 to get us warmed up, I thought.
DPatrick
2015-03-21 19:09:10
On to #2:
On to #2:
DPatrick
2015-03-21 19:09:15
2. The nine delegates to the Economic Cooperation Conference include 2 officials from Mexico, 3 officials from Canada, and 4 officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
2. The nine delegates to the Economic Cooperation Conference include 2 officials from Mexico, 3 officials from Canada, and 4 officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
simon1221
2015-03-21 19:09:48
total ways is 9C3
total ways is 9C3
MathDog0809
2015-03-21 19:09:48
9 choose three
9 choose three
urmilla
2015-03-21 19:09:48
first off 9C3 ways in total to pick the sleepers
first off 9C3 ways in total to pick the sleepers
DPatrick
2015-03-21 19:10:07
I like starting with the denominator in probability problems, since it's usually easy and on occasion provides insight.
I like starting with the denominator in probability problems, since it's usually easy and on occasion provides insight.
DPatrick
2015-03-21 19:10:14
We must choose 3 people from the 9, so $\binom93 = \frac{9 \cdot 8 \cdot 7}{3!} = 84$ ways possible.
We must choose 3 people from the 9, so $\binom93 = \frac{9 \cdot 8 \cdot 7}{3!} = 84$ ways possible.
DPatrick
2015-03-21 19:10:34
And now we have to decide how to count the numerator.
And now we have to decide how to count the numerator.
spartan168
2015-03-21 19:11:03
Complimentary Counting!
Complimentary Counting!
fluffyanimal
2015-03-21 19:11:03
Casework or complementary counting both work
Casework or complementary counting both work
Sciolympian
2015-03-21 19:11:03
could we also do complementary counting?
could we also do complementary counting?
DPatrick
2015-03-21 19:11:14
You could count this directly by casework, but I found complementary counting to be easier.
You could count this directly by casework, but I found complementary counting to be easier.
DPatrick
2015-03-21 19:11:19
What has to happen if it's not exactly two sleepers from the same country?
What has to happen if it's not exactly two sleepers from the same country?
ninjataco
2015-03-21 19:11:59
1 sleeper from each country, or 3 from the same country
1 sleeper from each country, or 3 from the same country
bli1999
2015-03-21 19:11:59
One from each country, or three from one country
One from each country, or three from one country
AlcumusGuy
2015-03-21 19:11:59
one sleeper from each country or all three from same country
one sleeper from each country or all three from same country
summitwei
2015-03-21 19:11:59
Either all from different countries or all from the same country
Either all from different countries or all from the same country
distortedwalrus
2015-03-21 19:11:59
either three from the same country, or all from different countries
either three from the same country, or all from different countries
willabc
2015-03-21 19:11:59
all three are from the same country or they are all from different countries
all three are from the same country or they are all from different countries
math_boy
2015-03-21 19:11:59
all three are from different countries, or all three the same
all three are from different countries, or all three the same
hesa57
2015-03-21 19:11:59
All the sleepers are from the same country or all are from different countries
All the sleepers are from the same country or all are from different countries
DPatrick
2015-03-21 19:12:05
Right: either there's 1 sleeper from each country, or all 3 are from the same country.
Right: either there's 1 sleeper from each country, or all 3 are from the same country.
DPatrick
2015-03-21 19:12:12
And both of these are really easy to count.
And both of these are really easy to count.
DPatrick
2015-03-21 19:12:15
In how many ways can we select 1 person from each country?
In how many ways can we select 1 person from each country?
amburger66
2015-03-21 19:12:32
1 from each country: 2*3*4
1 from each country: 2*3*4
AKAL3
2015-03-21 19:12:32
2*3*4
2*3*4
FTW
2015-03-21 19:12:32
2*3*4
2*3*4
swirlykick
2015-03-21 19:12:32
2*3*4=24
2*3*4=24
rocket13jg
2015-03-21 19:12:32
2*3*4=24
2*3*4=24
BFYSharks
2015-03-21 19:12:32
2*3*4
2*3*4
Darn
2015-03-21 19:12:32
First count cases where all are different: $2\times3\times4=24$
First count cases where all are different: $2\times3\times4=24$
PencilPal
2015-03-21 19:12:32
2*3*4 = 24
2*3*4 = 24
football017
2015-03-21 19:12:32
2*3*4=24?
2*3*4=24?
DPatrick
2015-03-21 19:12:39
We just multiply the choices for each person: $2 \cdot 3 \cdot 4 = 24$.
We just multiply the choices for each person: $2 \cdot 3 \cdot 4 = 24$.
DPatrick
2015-03-21 19:12:45
In how many ways can we have all 3 sleepers from the same country?
In how many ways can we have all 3 sleepers from the same country?
morninglight
2015-03-21 19:13:08
1+4=5
1+4=5
brightknight
2015-03-21 19:13:08
4+1=5
4+1=5
eyux
2015-03-21 19:13:08
3C3+4C3=5
3C3+4C3=5
whatshisbucket
2015-03-21 19:13:08
4C3+3C3
4C3+3C3
william122
2015-03-21 19:13:08
1+4=5
1+4=5
xyuqing
2015-03-21 19:13:08
1+4=5
1+4=5
cumo99
2015-03-21 19:13:08
3C3 + 4C3 = 5
3C3 + 4C3 = 5
DPatrick
2015-03-21 19:13:12
Either we have all 3 Canadians, or we have 3 of the 4 Americans (which amounts to choosing an American to stay awake).
Either we have all 3 Canadians, or we have 3 of the 4 Americans (which amounts to choosing an American to stay awake).
DPatrick
2015-03-21 19:13:21
Thus there are $1+4 = 5$ ways all 3 can be from the same country.
Thus there are $1+4 = 5$ ways all 3 can be from the same country.
spartan168
2015-03-21 19:13:36
So 24 + 5 = 29
So 24 + 5 = 29
akim99
2015-03-21 19:13:36
24+5=29, which is what we DON'T want
24+5=29, which is what we DON'T want
DPatrick
2015-03-21 19:13:43
Right, this makes $24+5 = 29$ ways to not have what we want.
Right, this makes $24+5 = 29$ ways to not have what we want.
ryanyz10
2015-03-21 19:13:58
84 - 29 = 55... 55/84 so answer is 139?
84 - 29 = 55... 55/84 so answer is 139?
danusv
2015-03-21 19:13:58
SO we have 1-29/84 = 55/84
SO we have 1-29/84 = 55/84
hesa57
2015-03-21 19:13:58
$\frac{84-29}{84}$
$\frac{84-29}{84}$
princess-of-mathland
2015-03-21 19:13:58
1-29/84=55/84
1-29/84=55/84
DPatrick
2015-03-21 19:14:04
Thus there are $84 - 29 = 55$ ways to have what we want, and the probability is $\dfrac{55}{84}$.
Thus there are $84 - 29 = 55$ ways to have what we want, and the probability is $\dfrac{55}{84}$.
nosaj
2015-03-21 19:14:13
so the probability is $\frac{55}{84}$, giving us an answer of 139
so the probability is $\frac{55}{84}$, giving us an answer of 139
DPatrick
2015-03-21 19:14:16
This is already in lowest terms, so our answer is $55 + 84 = \boxed{139}$.
This is already in lowest terms, so our answer is $55 + 84 = \boxed{139}$.
DPatrick
2015-03-21 19:14:42
Ok, onwards:
Ok, onwards:
DPatrick
2015-03-21 19:14:47
3. There is a prime number $p$ such that $16p + 1$ is the cube of a positive integer. Find $p$.
3. There is a prime number $p$ such that $16p + 1$ is the cube of a positive integer. Find $p$.
Tommy2000
2015-03-21 19:15:06
Convert words to algebra!
Convert words to algebra!
p2pcmlp
2015-03-21 19:15:06
16p+1=n^3
16p+1=n^3
MSTang
2015-03-21 19:15:06
16p + 1 = n^3
16p + 1 = n^3
nosyarg
2015-03-21 19:15:06
16p+1=x^3
16p+1=x^3
Eugenis
2015-03-21 19:15:06
Well we can start by making equations
Well we can start by making equations
stan23456
2015-03-21 19:15:06
let 16p+1=n^3, for some integer n.
let 16p+1=n^3, for some integer n.
DPatrick
2015-03-21 19:15:11
We can let $16p + 1 = n^3$, where $n$ is an integer. What's next?
We can let $16p + 1 = n^3$, where $n$ is an integer. What's next?
jam10307
2015-03-21 19:15:33
difference o cubes
difference o cubes
C-bass
2015-03-21 19:15:33
subtract 1 from both sides
subtract 1 from both sides
bli1999
2015-03-21 19:15:33
Difference of cubes
Difference of cubes
AlcumusGuy
2015-03-21 19:15:33
Rewrite as 16p = n^3 - 1 = (n-1)(n^2 + n + 1)
Rewrite as 16p = n^3 - 1 = (n-1)(n^2 + n + 1)
hlasker1
2015-03-21 19:15:33
subtract the 1 and simplify the difference of cubes
subtract the 1 and simplify the difference of cubes
azmath333
2015-03-21 19:15:33
16p = n^3 - 1^3
16p = n^3 - 1^3
csmath
2015-03-21 19:15:33
Then we can do 16p = n^3-1
Then we can do 16p = n^3-1
SilverPersian1
2015-03-21 19:15:33
-1 from both sides
-1 from both sides
DPatrick
2015-03-21 19:15:40
Good idea. We can write this as $16p = n^3 - 1$, which is nice because we can factor $n^3 - 1$ as a difference of cubes: $$16p = (n-1)(n^2 + n + 1).$$
Good idea. We can write this as $16p = n^3 - 1$, which is nice because we can factor $n^3 - 1$ as a difference of cubes: $$16p = (n-1)(n^2 + n + 1).$$
DPatrick
2015-03-21 19:15:47
Hmm, we'd like to say that one of these factors is equal to $p$. Any observations that might help us in that direction?
Hmm, we'd like to say that one of these factors is equal to $p$. Any observations that might help us in that direction?
whatshisbucket
2015-03-21 19:16:24
n^2+n+1 is always odd
n^2+n+1 is always odd
brightknight
2015-03-21 19:16:24
n^2+n+1 is odd
n^2+n+1 is odd
MSTang
2015-03-21 19:16:24
n^2 + n + 1 is always odd
n^2 + n + 1 is always odd
stan23456
2015-03-21 19:16:24
n^2+n+1 is odd
n^2+n+1 is odd
Tommy2000
2015-03-21 19:16:24
n^2+n+1 is always odd so n-1=16
n^2+n+1 is always odd so n-1=16
zebrae
2015-03-21 19:16:24
n^2+n+1 is always odd, so n has to be 17
n^2+n+1 is always odd, so n has to be 17
mjoshi
2015-03-21 19:16:24
n-1 = 16, so that n^2+n+1 is prime
n-1 = 16, so that n^2+n+1 is prime
mssmath
2015-03-21 19:16:24
n^2+n+1 is always odd
n^2+n+1 is always odd
donot
2015-03-21 19:16:24
The last term is always odd
The last term is always odd
DPatrick
2015-03-21 19:16:37
Right: note that $n^2+n+1 = n(n+1) + 1$, so $n(n+1)$ is the product of an odd and an even, regardless of what $n$ is. Thus the second term is always odd.
Right: note that $n^2+n+1 = n(n+1) + 1$, so $n(n+1)$ is the product of an odd and an even, regardless of what $n$ is. Thus the second term is always odd.
DPatrick
2015-03-21 19:16:50
Therefore, since $p \not= 2$ (because 33 is certainly not the cube of a positive integer!), $p$ must be odd, so we must have $16 = n-1$ and $p = n^2 + n + 1$.
Therefore, since $p \not= 2$ (because 33 is certainly not the cube of a positive integer!), $p$ must be odd, so we must have $16 = n-1$ and $p = n^2 + n + 1$.
raxu
2015-03-21 19:17:21
n=17, p=307
n=17, p=307
cdl031apos
2015-03-21 19:17:21
n=17
n=17
Epic777
2015-03-21 19:17:21
n=17, p=307
n=17, p=307
brainiac1
2015-03-21 19:17:21
and 307=17^2+17+1 is prime
and 307=17^2+17+1 is prime
DPatrick
2015-03-21 19:17:33
Yes. Thus $n=17$ and $$p = n^2 + n + 1 = 17^2 + 17 + 1 = 289 + 17 + 1 = \boxed{307}.$$
Yes. Thus $n=17$ and $$p = n^2 + n + 1 = 17^2 + 17 + 1 = 289 + 17 + 1 = \boxed{307}.$$
DPatrick
2015-03-21 19:17:46
(You might check that 307 is prime as a check of your answer. Happily, it is!)
(You might check that 307 is prime as a check of your answer. Happily, it is!)
DPatrick
2015-03-21 19:18:07
Next, our first geometry problem:
Next, our first geometry problem:
DPatrick
2015-03-21 19:18:12
4. Point $B$ lies on line segment $\overline{AC}$ with $AB = 16$ and $BC = 4$. Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$. Let $M$ be the midpoint of $\overline{AE}$, and $N$ be the midpoint of $\overline{CD}$. The area of $\triangle BMN$ is $x$. Find $x^2$.
4. Point $B$ lies on line segment $\overline{AC}$ with $AB = 16$ and $BC = 4$. Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$. Let $M$ be the midpoint of $\overline{AE}$, and $N$ be the midpoint of $\overline{CD}$. The area of $\triangle BMN$ is $x$. Find $x^2$.
maverick8
2015-03-21 19:18:25
Draw a diagram!
Draw a diagram!
rlz
2015-03-21 19:18:25
An accurate diagram is important!
An accurate diagram is important!
simon1221
2015-03-21 19:18:25
make a diagram
make a diagram
tdeng
2015-03-21 19:18:25
diagram
diagram
MiamiHeat88
2015-03-21 19:18:25
draw it
draw it
DPatrick
2015-03-21 19:18:29
We can try to sketch a picture of all this. (Having a ruler on contest day really helps!)
We can try to sketch a picture of all this. (Having a ruler on contest day really helps!)
DPatrick
2015-03-21 19:18:33
DPatrick
2015-03-21 19:18:54
A fairly good diagram might lead to a conjecture using this picture...
A fairly good diagram might lead to a conjecture using this picture...
WhaleVomit
2015-03-21 19:19:08
MNB looks like equilateral!
MNB looks like equilateral!
C-bass
2015-03-21 19:19:08
mbn is equilateral?????
mbn is equilateral?????
whatshisbucket
2015-03-21 19:19:08
BMN looks equilateral
BMN looks equilateral
akim99
2015-03-21 19:19:08
Hmmm $NMB$ looks equilateral
Hmmm $NMB$ looks equilateral
william122
2015-03-21 19:19:08
its an equilateral triangle!
its an equilateral triangle!
Darn
2015-03-21 19:19:08
$\triangle MNB$ is equilateral
$\triangle MNB$ is equilateral
fluffyanimal
2015-03-21 19:19:08
looks equilateral
looks equilateral
PencilPal
2015-03-21 19:19:08
equilateral
equilateral
Jz2435
2015-03-21 19:19:08
MNB is equilateral
MNB is equilateral
NikhilP
2015-03-21 19:19:08
MNB is equalateral
MNB is equalateral
DPatrick
2015-03-21 19:19:15
One thing a really good sketch helps with is that you might notice that $BMN$ looks equilateral. That might help us later, if we can prove it.
One thing a really good sketch helps with is that you might notice that $BMN$ looks equilateral. That might help us later, if we can prove it.
DPatrick
2015-03-21 19:19:25
How are we going to keep track of where all these points are?
How are we going to keep track of where all these points are?
AKAL3
2015-03-21 19:19:46
coord bash
coord bash
BFYSharks
2015-03-21 19:19:46
Coordinates
Coordinates
howie2000
2015-03-21 19:19:46
coordinate plane
coordinate plane
Sciolympian
2015-03-21 19:19:46
using coordinates
using coordinates
temp8909
2015-03-21 19:19:46
analytic geometry
analytic geometry
RedPhoenix
2015-03-21 19:19:46
Could we use coordinates?
Could we use coordinates?
morninglight
2015-03-21 19:19:46
coordinates!
coordinates!
math_boy
2015-03-21 19:19:46
coordinate plane
coordinate plane
spartan168
2015-03-21 19:19:46
coordinate bashing lol
coordinate bashing lol
xyuqing
2015-03-21 19:19:46
Coordinates
Coordinates
summitwei
2015-03-21 19:19:46
Coordinates!
Coordinates!
DPatrick
2015-03-21 19:19:54
I agree; I used coordinates. Coordinates are good of keeping track of collinear points (like $A$, $B$, $C$), especially when the lengths are known, and they're good for computing midpoints.
I agree; I used coordinates. Coordinates are good of keeping track of collinear points (like $A$, $B$, $C$), especially when the lengths are known, and they're good for computing midpoints.
DPatrick
2015-03-21 19:20:08
So let's set $A = (0,0)$, $B = (16,0)$, and $C = (20,0)$.
So let's set $A = (0,0)$, $B = (16,0)$, and $C = (20,0)$.
DPatrick
2015-03-21 19:20:23
What are $D$ and $E$?
What are $D$ and $E$?
MSTang
2015-03-21 19:21:03
D = (8, 8sqrt(3)) and E = (18, 2sqrt(3))
D = (8, 8sqrt(3)) and E = (18, 2sqrt(3))
urmilla
2015-03-21 19:21:03
D(8,8sqrt3); E(18,2sqrt3)
D(8,8sqrt3); E(18,2sqrt3)
ryanyz10
2015-03-21 19:21:03
8, 8sqrt(3) is D
8, 8sqrt(3) is D
rocket13jg
2015-03-21 19:21:03
D(8, 8root3), E(18, 2root3)
D(8, 8root3), E(18, 2root3)
ryanyz10
2015-03-21 19:21:03
E is 18, 2sqrt(3)
E is 18, 2sqrt(3)
ninjataco
2015-03-21 19:21:03
D (8, 8sqrt3) and E (18, 2sqrt3)
D (8, 8sqrt3) and E (18, 2sqrt3)
DPatrick
2015-03-21 19:21:08
$D$'s $x$-coordinate is halfway between $A$ and $B$, and $D$'s $y$-coordinate is the height of $ABD$, which is $\sqrt3/2$ times $AB$.
$D$'s $x$-coordinate is halfway between $A$ and $B$, and $D$'s $y$-coordinate is the height of $ABD$, which is $\sqrt3/2$ times $AB$.
DPatrick
2015-03-21 19:21:16
So $D = (8,8\sqrt3)$.
So $D = (8,8\sqrt3)$.
DPatrick
2015-03-21 19:21:23
By similar logic, $E = (18,2\sqrt3)$.
By similar logic, $E = (18,2\sqrt3)$.
akim99
2015-03-21 19:21:35
Now find the midpoints
Now find the midpoints
spartan168
2015-03-21 19:21:39
Midpoint Formula!
Midpoint Formula!
jam10307
2015-03-21 19:21:39
midpt formula
midpt formula
DPatrick
2015-03-21 19:21:43
Right. To compute a midpoint, we average the coordinates.
Right. To compute a midpoint, we average the coordinates.
24iam24
2015-03-21 19:22:07
so M(9, √3) and N(14, 4√3)
so M(9, √3) and N(14, 4√3)
Eugenis
2015-03-21 19:22:07
m= $(9,\sqrt{3})$
m= $(9,\sqrt{3})$
DPatrick
2015-03-21 19:22:11
So, $M$ is the average of $E = (18,2\sqrt3)$ and $A = (0,0)$. Hence $M = (9,\sqrt3)$.
Similarly, $N$ is the average of $D = (8,8\sqrt3)$ and $C=(20,0)$. Hence $N = (14,4\sqrt3)$.
So, $M$ is the average of $E = (18,2\sqrt3)$ and $A = (0,0)$. Hence $M = (9,\sqrt3)$.
Similarly, $N$ is the average of $D = (8,8\sqrt3)$ and $C=(20,0)$. Hence $N = (14,4\sqrt3)$.
DPatrick
2015-03-21 19:22:19
So our triangle $BMN$ has vertices $(16,0)$, $(9,\sqrt3)$, and $(14,4\sqrt3)$.
So our triangle $BMN$ has vertices $(16,0)$, $(9,\sqrt3)$, and $(14,4\sqrt3)$.
hesa57
2015-03-21 19:22:47
Distance formula
Distance formula
brightknight
2015-03-21 19:22:47
distance formula for side lengths
distance formula for side lengths
bli1999
2015-03-21 19:22:47
Distance formula
Distance formula
amburger66
2015-03-21 19:22:47
using distance formula, we can confirm that it's equilateral
using distance formula, we can confirm that it's equilateral
stan23456
2015-03-21 19:22:47
we can prove that its equilateral
we can prove that its equilateral
Sciolympian
2015-03-21 19:22:47
calculate side lengths now
calculate side lengths now
princess-of-mathland
2015-03-21 19:22:47
By distance formula, each side is sqrt52
By distance formula, each side is sqrt52
raxu
2015-03-21 19:22:47
And by distance formula, BMN is equilateral with side length sqrt(52).
And by distance formula, BMN is equilateral with side length sqrt(52).
DPatrick
2015-03-21 19:23:01
Right. We suspected it was equilateral...
Right. We suspected it was equilateral...
DPatrick
2015-03-21 19:23:07
...and we can verify from directly computation that each side has length $\sqrt{52}$:
\begin{align*}
BM &= \sqrt{7^2 + (\sqrt3)^2} = \sqrt{49+3} = \sqrt{52}, \\
BN &= \sqrt{2^2 + (4\sqrt3)^2} = \sqrt{4+8} = \sqrt{52}, \\
MN &= \sqrt{5^2 + (3\sqrt3)^2} = \sqrt{25+27} = \sqrt{52}.
\end{align*}
...and we can verify from directly computation that each side has length $\sqrt{52}$:
\begin{align*}
BM &= \sqrt{7^2 + (\sqrt3)^2} = \sqrt{49+3} = \sqrt{52}, \\
BN &= \sqrt{2^2 + (4\sqrt3)^2} = \sqrt{4+8} = \sqrt{52}, \\
MN &= \sqrt{5^2 + (3\sqrt3)^2} = \sqrt{25+27} = \sqrt{52}.
\end{align*}
DPatrick
2015-03-21 19:23:16
So $BMN$ is an equilateral triangle with side length $\sqrt{52}$. What is its area?
So $BMN$ is an equilateral triangle with side length $\sqrt{52}$. What is its area?
MSTang
2015-03-21 19:23:35
52sqrt(3)/4 = 13sqrt(3)
52sqrt(3)/4 = 13sqrt(3)
legolego
2015-03-21 19:23:35
13sqrt3
13sqrt3
mathboxboro
2015-03-21 19:23:35
13root3
13root3
MathDog0809
2015-03-21 19:23:35
13sqrt3?
13sqrt3?
cumo99
2015-03-21 19:23:35
13\sqrt{3}
13\sqrt{3}
csmath
2015-03-21 19:23:35
13sqrt(3)
13sqrt(3)
IMGUNNA
2015-03-21 19:23:35
13sqrt3
13sqrt3
DPatrick
2015-03-21 19:23:39
The area is $\dfrac{s^2\sqrt3}{4} = \dfrac{52\sqrt3}{4} = 13\sqrt3$.
The area is $\dfrac{s^2\sqrt3}{4} = \dfrac{52\sqrt3}{4} = 13\sqrt3$.
DPatrick
2015-03-21 19:23:45
And the final answer is this area squared, or $13^2 \cdot 3 = 169 \cdot 3 = \boxed{507}$.
And the final answer is this area squared, or $13^2 \cdot 3 = 169 \cdot 3 = \boxed{507}$.
DPatrick
2015-03-21 19:24:11
I should mention, as many of you did, that you could use the Shoelace Theorem to compute the area from the coordinates; for example:
\[
\left|\frac12\begin{vmatrix} -7 & \sqrt3 \\ -2 & 4\sqrt3 \end{vmatrix}\right| = \left|\frac12(-28\sqrt3 + 2\sqrt3)\right| = 13\sqrt3.
\]
I should mention, as many of you did, that you could use the Shoelace Theorem to compute the area from the coordinates; for example:
\[
\left|\frac12\begin{vmatrix} -7 & \sqrt3 \\ -2 & 4\sqrt3 \end{vmatrix}\right| = \left|\frac12(-28\sqrt3 + 2\sqrt3)\right| = 13\sqrt3.
\]
DPatrick
2015-03-21 19:24:30
(If you don't know the Shoelace Theorem, I encourage you to look it up when we're done!)
(If you don't know the Shoelace Theorem, I encourage you to look it up when we're done!)
DPatrick
2015-03-21 19:25:00
But since we suspected the triangle was equilateral, and it was pretty easy to verify that it was, I found that to be a bit easier and less computational.
But since we suspected the triangle was equilateral, and it was pretty easy to verify that it was, I found that to be a bit easier and less computational.
DPatrick
2015-03-21 19:25:19
On to #5:
On to #5:
DPatrick
2015-03-21 19:25:27
5. In a drawer Sandy has 5 pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the 10 socks in the drawer. On Tuesday Sandy selects 2 of the remaining 8 socks at random and on Wednesday two of the remaining 6 socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
5. In a drawer Sandy has 5 pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the 10 socks in the drawer. On Tuesday Sandy selects 2 of the remaining 8 socks at random and on Wednesday two of the remaining 6 socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
DPatrick
2015-03-21 19:25:49
You could certainly solve this problem with some careful casework, but there's a clever trick we can use to make this problem a lot easier.
You could certainly solve this problem with some careful casework, but there's a clever trick we can use to make this problem a lot easier.
ryanyz10
2015-03-21 19:26:07
work backwards?
work backwards?
nosaj
2015-03-21 19:26:07
It is easier to do this problem backwards (pick Wednesday first)
It is easier to do this problem backwards (pick Wednesday first)
AlcumusGuy
2015-03-21 19:26:07
you can rearrange the days
you can rearrange the days
william122
2015-03-21 19:26:07
work backwards
work backwards
xyuqing
2015-03-21 19:26:07
Do the last day first
Do the last day first
DPatrick
2015-03-21 19:26:11
Right!
Right!
DPatrick
2015-03-21 19:26:14
Generally in counting (or probability, which is really just counting most of the time), we want to deal with the most restrictive condition first.
Generally in counting (or probability, which is really just counting most of the time), we want to deal with the most restrictive condition first.
DPatrick
2015-03-21 19:26:22
Here, the most restrictive condition is that the socks match on Wednesday.
Here, the most restrictive condition is that the socks match on Wednesday.
DPatrick
2015-03-21 19:26:29
So let's pretend that Wednesday comes first!
So let's pretend that Wednesday comes first!
DPatrick
2015-03-21 19:26:36
Now if you're thinking "Wait a minute, we can't do that -- Monday clearly comes first!": imagine that Sandy lays out all of her socks in a row for the week, then just takes the first 2 socks on Monday, the next two socks on Tuesday, and so on. It doesn't matter in what order she places socks in her row -- all 10 socks will eventually get placed.
Now if you're thinking "Wait a minute, we can't do that -- Monday clearly comes first!": imagine that Sandy lays out all of her socks in a row for the week, then just takes the first 2 socks on Monday, the next two socks on Tuesday, and so on. It doesn't matter in what order she places socks in her row -- all 10 socks will eventually get placed.
DPatrick
2015-03-21 19:26:54
So, pretending that Wednesday is first, what's the probability that she gets a match?
So, pretending that Wednesday is first, what's the probability that she gets a match?
BFYSharks
2015-03-21 19:27:12
Probability 1/9
Probability 1/9
jigglypuff
2015-03-21 19:27:12
1/9
1/9
summitwei
2015-03-21 19:27:12
1/9
1/9
RedPhoenix
2015-03-21 19:27:12
1/9
1/9
Studiosa
2015-03-21 19:27:12
1/9
1/9
temp8909
2015-03-21 19:27:12
1/9
1/9
whatshisbucket
2015-03-21 19:27:12
1/9
1/9
doodlemaster7
2015-03-21 19:27:12
1/9
1/9
NikhilP
2015-03-21 19:27:12
1/9
1/9
DPatrick
2015-03-21 19:27:19
Sure: no matter what first socks she picks, exactly 1 of the remaining 9 socks will match. So the probability of a match is $\frac19$.
Sure: no matter what first socks she picks, exactly 1 of the remaining 9 socks will match. So the probability of a match is $\frac19$.
DPatrick
2015-03-21 19:27:29
With the 8 remaining socks, what's the probability that she does NOT get a match on Monday?
With the 8 remaining socks, what's the probability that she does NOT get a match on Monday?
ninjataco
2015-03-21 19:27:52
6/7
6/7
Matt17
2015-03-21 19:27:52
6/7
6/7
raxu
2015-03-21 19:27:52
6/7
6/7
jigglypuff
2015-03-21 19:27:52
6/7
6/7
nosyarg
2015-03-21 19:27:52
6/7
6/7
dtxiong
2015-03-21 19:27:52
6/7
6/7
mjoshi
2015-03-21 19:27:52
6/7
6/7
meteor88
2015-03-21 19:27:52
6/7
6/7
DPatrick
2015-03-21 19:28:10
Right, it's the reverse of Wednesday: no matter what first sock she picks, there are 1 matching sock and 6 non-matching socks remaining.
Right, it's the reverse of Wednesday: no matter what first sock she picks, there are 1 matching sock and 6 non-matching socks remaining.
DPatrick
2015-03-21 19:28:14
So the probability of a non-match on Monday (given that we already have a match on Wednesday) is $\frac67$.
So the probability of a non-match on Monday (given that we already have a match on Wednesday) is $\frac67$.
DPatrick
2015-03-21 19:28:22
Now what's the probability of a non-match on Tuesday?
Now what's the probability of a non-match on Tuesday?
DPatrick
2015-03-21 19:28:36
Note there are 6 socks left: 2 matching pairs, and 2 non-matching extras (the two that match Monday's socks).
Note there are 6 socks left: 2 matching pairs, and 2 non-matching extras (the two that match Monday's socks).
fluffyanimal
2015-03-21 19:28:58
13/15
13/15
xyuqing
2015-03-21 19:28:58
13/15
13/15
RedPhoenix
2015-03-21 19:28:58
13/15
13/15
raxu
2015-03-21 19:28:58
1-2/(6C2)=13/15
1-2/(6C2)=13/15
bli1999
2015-03-21 19:28:58
1-2/15=13/15
1-2/15=13/15
whatshisbucket
2015-03-21 19:28:58
13/15
13/15
Studiosa
2015-03-21 19:28:58
13/15
13/15
DPatrick
2015-03-21 19:29:05
Exactly. Out of the $\binom62 = 15$ pairs that she could choose, 2 match and 13 don't.
Exactly. Out of the $\binom62 = 15$ pairs that she could choose, 2 match and 13 don't.
DPatrick
2015-03-21 19:29:10
Thus the probability of a non-match on Tuesday is $\frac{13}{15}$.
Thus the probability of a non-match on Tuesday is $\frac{13}{15}$.
ryanyz10
2015-03-21 19:29:30
multiply them
multiply them
csmath
2015-03-21 19:29:30
then we can just multiply these all
then we can just multiply these all
akim99
2015-03-21 19:29:30
Time to multiply
Time to multiply
DPatrick
2015-03-21 19:29:38
Right: the overall probability is
\[
\frac{1}{9} \cdot \frac{6}{7} \cdot \frac{13}{15} = \frac{26}{315}.
\]
(A factor of 3 cancels when we convert to lowest terms.)
Right: the overall probability is
\[
\frac{1}{9} \cdot \frac{6}{7} \cdot \frac{13}{15} = \frac{26}{315}.
\]
(A factor of 3 cancels when we convert to lowest terms.)
DPatrick
2015-03-21 19:29:48
And the final answer is $26 + 315 = \boxed{341}$.
And the final answer is $26 + 315 = \boxed{341}$.
DPatrick
2015-03-21 19:30:19
Again, the moral of the story: generally in counting (or probability, which is really just counting most of the time), we want to deal with the most restrictive condition first.
Again, the moral of the story: generally in counting (or probability, which is really just counting most of the time), we want to deal with the most restrictive condition first.
DPatrick
2015-03-21 19:30:37
OK, we're one-third of the way home! On to #6:
OK, we're one-third of the way home! On to #6:
DPatrick
2015-03-21 19:30:42
6. Points $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I,$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^{\circ}$. Find the degree measure of $\angle BAG$.
6. Points $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I,$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^{\circ}$. Find the degree measure of $\angle BAG$.
DPatrick
2015-03-21 19:30:45
mathtastic
2015-03-21 19:31:15
on the actual contest the points weren't dotted
on the actual contest the points weren't dotted
nosaj
2015-03-21 19:31:15
Wait; on the actual test the points didn't have dots.
Wait; on the actual test the points didn't have dots.
DPatrick
2015-03-21 19:31:33
That's weird, because the copy that the AMC sent me does have dots!
That's weird, because the copy that the AMC sent me does have dots!
AlcumusGuy
2015-03-21 19:31:59
Let x = arc AB, y = arc AI
Let x = arc AB, y = arc AI
summitwei
2015-03-21 19:31:59
Call angle ABC $\alpha$ and AIH $\beta$
Call angle ABC $\alpha$ and AIH $\beta$
DPatrick
2015-03-21 19:32:09
Sure, let's set up some variables. Let's make the 4 top arcs each be $x$ (degrees), and the bottom 5 arcs each be $y$ (degrees).
Sure, let's set up some variables. Let's make the 4 top arcs each be $x$ (degrees), and the bottom 5 arcs each be $y$ (degrees).
DPatrick
2015-03-21 19:32:22
How can we relate $x$ and $y$?
How can we relate $x$ and $y$?
Tommy2000
2015-03-21 19:32:51
First relate ABD and AHG
First relate ABD and AHG
math_boy
2015-03-21 19:32:51
<ABD = <AHG + 12
<ABD = <AHG + 12
xyuqing
2015-03-21 19:32:51
C is a center.
C is a center.
p2pcmlp
2015-03-21 19:32:51
find ABD and AHG
find ABD and AHG
Tommy2000
2015-03-21 19:32:51
Express ABD iand AHG in terms of x and y
Express ABD iand AHG in terms of x and y
DPatrick
2015-03-21 19:32:57
There seem to be two ways:
- using the data given in the problem that $C$ is the center of the bottom arc's circle
- using the data given in the problem about the two angles that differ by $12^\circ$.
There seem to be two ways:
- using the data given in the problem that $C$ is the center of the bottom arc's circle
- using the data given in the problem about the two angles that differ by $12^\circ$.
DPatrick
2015-03-21 19:33:08
The hope is: each of these should give us an equation relating $x$ and $y$ (hopefully) -- then we'll have 2 equations in 2 variables, so we'll be able to solve the system!
The hope is: each of these should give us an equation relating $x$ and $y$ (hopefully) -- then we'll have 2 equations in 2 variables, so we'll be able to solve the system!
DPatrick
2015-03-21 19:33:18
What does $C$ being the center of the bottom arc's circle tell us?
What does $C$ being the center of the bottom arc's circle tell us?
nosaj
2015-03-21 19:33:46
ECA=5y
ECA=5y
BFYSharks
2015-03-21 19:33:46
Angle ECA = 5y
Angle ECA = 5y
DPatrick
2015-03-21 19:33:51
Right. $\angle ACE$ is a central angle that intercepts the entire bottom arc.
Right. $\angle ACE$ is a central angle that intercepts the entire bottom arc.
DPatrick
2015-03-21 19:33:57
So $\angle ACE = 5y$.
So $\angle ACE = 5y$.
DPatrick
2015-03-21 19:34:06
But how can we use the top arc to compute $\angle ACE$ in terms of $x$?
But how can we use the top arc to compute $\angle ACE$ in terms of $x$?
nosaj
2015-03-21 19:34:37
However, $\angle ECA$ is also $\frac{360-4x}{2}$ it inscribes that arc.
However, $\angle ECA$ is also $\frac{360-4x}{2}$ it inscribes that arc.
wangth100
2015-03-21 19:34:37
$ /angle{ACE} = 180-2x $
$ /angle{ACE} = 180-2x $
ryanyz10
2015-03-21 19:34:37
inscribed angle
inscribed angle
Tommy2000
2015-03-21 19:34:37
$180-2x$
$180-2x$
viv7000
2015-03-21 19:34:37
ACE is (360-4x)/2
ACE is (360-4x)/2
rlz
2015-03-21 19:34:37
(360-4x)/2
(360-4x)/2
maverick8
2015-03-21 19:34:37
360-4x divided by 2
360-4x divided by 2
stan23456
2015-03-21 19:34:37
180-2x
180-2x
mathboxboro
2015-03-21 19:34:37
180-2x
180-2x
DPatrick
2015-03-21 19:34:44
Arc ${AE}$ is 4x, so $\angle ACE$ is inscribed in the top circle, intercepting an arc of size $360 - 4x$.
Arc ${AE}$ is 4x, so $\angle ACE$ is inscribed in the top circle, intercepting an arc of size $360 - 4x$.
DPatrick
2015-03-21 19:34:54
Thus $\angle ACE = \frac12(360 - 4x) = 180 - 2x$.
Thus $\angle ACE = \frac12(360 - 4x) = 180 - 2x$.
DPatrick
2015-03-21 19:35:08
Great: this gives us the equation \[5y = 180 - 2x.\]
Great: this gives us the equation \[5y = 180 - 2x.\]
DPatrick
2015-03-21 19:35:18
Now on to the other data: $\angle ABD = \angle AHG + 12$.
Now on to the other data: $\angle ABD = \angle AHG + 12$.
DPatrick
2015-03-21 19:35:23
What's $\angle ABD$?
What's $\angle ABD$?
mathboxboro
2015-03-21 19:35:53
(360-3x)/2
(360-3x)/2
ryanyz10
2015-03-21 19:35:53
(360 - 3x)/2
(360 - 3x)/2
rlz
2015-03-21 19:35:53
(360-3x)/2
(360-3x)/2
RedPhoenix
2015-03-21 19:35:53
180 - 3x/2
180 - 3x/2
brainiac1
2015-03-21 19:35:53
180-3x/2
180-3x/2
william122
2015-03-21 19:35:53
1/2(360-3x)
1/2(360-3x)
azmath333
2015-03-21 19:35:53
180 degrees - 3x/2
180 degrees - 3x/2
DPatrick
2015-03-21 19:35:59
Yes. It intercepts an arc of size $360 - 3x$ from the top circle, so $\angle ABD = \frac12(360-3x) = 180 - \frac32x$.
Yes. It intercepts an arc of size $360 - 3x$ from the top circle, so $\angle ABD = \frac12(360-3x) = 180 - \frac32x$.
DPatrick
2015-03-21 19:36:12
What's $\angle AHG$?
What's $\angle AHG$?
ninjataco
2015-03-21 19:36:34
(360-3y)/2
(360-3y)/2
dtxiong
2015-03-21 19:36:34
1/2(360-3y)
1/2(360-3y)
swe1
2015-03-21 19:36:34
(360-3y)/2
(360-3y)/2
bli1999
2015-03-21 19:36:34
180-3y/2
180-3y/2
whatshisbucket
2015-03-21 19:36:34
(360-3y)/2
(360-3y)/2
DPatrick
2015-03-21 19:36:41
Basically the same idea: It intercepts an arc of $360 - 3y$ of the bottom circle, so $\angle AHG = \frac12(360 - 3y) = 180 - \frac32y$.
Basically the same idea: It intercepts an arc of $360 - 3y$ of the bottom circle, so $\angle AHG = \frac12(360 - 3y) = 180 - \frac32y$.
DPatrick
2015-03-21 19:36:49
So we have the equation
\[
180 - \frac32x = 180 - \frac32y + 12.
\]
So we have the equation
\[
180 - \frac32x = 180 - \frac32y + 12.
\]
MathDog0809
2015-03-21 19:37:01
find x and y!
find x and y!
C-bass
2015-03-21 19:37:01
solve system?
solve system?
DPatrick
2015-03-21 19:37:14
Right, our plan worked! We've got two linear equations in $x$ and $y$, so we can solve for them.
Right, our plan worked! We've got two linear equations in $x$ and $y$, so we can solve for them.
jigglypuff
2015-03-21 19:37:27
difference of 8
difference of 8
dragon21
2015-03-21 19:37:34
simplified is y-x=8
simplified is y-x=8
DPatrick
2015-03-21 19:37:42
The second equation simplifies to $3x = 3y - 24$, or $x = y - 8$.
The second equation simplifies to $3x = 3y - 24$, or $x = y - 8$.
Studiosa
2015-03-21 19:38:01
x=20 y=28
x=20 y=28
rocket13jg
2015-03-21 19:38:01
x=20, y=28
x=20, y=28
ninjataco
2015-03-21 19:38:01
x=20, y=28
x=20, y=28
DPatrick
2015-03-21 19:38:06
Substituting this back into the first equation gives
\[
5y = 180 - 2(y-8) = 196 - 2y.
\]
Substituting this back into the first equation gives
\[
5y = 180 - 2(y-8) = 196 - 2y.
\]
DPatrick
2015-03-21 19:38:13
So $7y = 196$, and $y = 28$. And then $x = y-8 = 20$.
So $7y = 196$, and $y = 28$. And then $x = y-8 = 20$.
DPatrick
2015-03-21 19:38:26
To finish, we want to compute $\angle BAG$:
To finish, we want to compute $\angle BAG$:
DPatrick
2015-03-21 19:38:29
DPatrick
2015-03-21 19:38:36
That angle doesn't nicely intercept one circle or the other, so what can we do?
That angle doesn't nicely intercept one circle or the other, so what can we do?
nosaj
2015-03-21 19:38:52
Note that $\angle BAG = \angle BAE + \angle EAG$.
Note that $\angle BAG = \angle BAE + \angle EAG$.
Matt17
2015-03-21 19:38:52
Make line from A to E to split angle
Make line from A to E to split angle
temp8909
2015-03-21 19:38:52
BAE+EAG
BAE+EAG
csmath
2015-03-21 19:38:52
Make it the sum of 2 angles
Make it the sum of 2 angles
mathtastic
2015-03-21 19:38:52
$\angle BAE+\angle GAE$
$\angle BAE+\angle GAE$
AKAL3
2015-03-21 19:38:52
draw AE
draw AE
nsd
2015-03-21 19:38:52
sum if BAE and EAG
sum if BAE and EAG
DPatrick
2015-03-21 19:39:00
No problem -- we can draw in $\overline{AE}$ so that we break up the angle into two angles, each of which intercepts an arc of a single circle:
No problem -- we can draw in $\overline{AE}$ so that we break up the angle into two angles, each of which intercepts an arc of a single circle:
DPatrick
2015-03-21 19:39:05
maverick8
2015-03-21 19:39:33
BAG=BAE+EAG=3x/2+2y/2
BAG=BAE+EAG=3x/2+2y/2
Sciolympian
2015-03-21 19:39:33
inscribed angle formula
inscribed angle formula
csmath
2015-03-21 19:39:33
So it's just y + 3x/2
So it's just y + 3x/2
mathtastic
2015-03-21 19:39:33
$\frac{3x+2y}{2}$
$\frac{3x+2y}{2}$
DPatrick
2015-03-21 19:39:40
The top part intercepts an arc of $3x$, so $\angle BAE = \frac32(x) = \frac32(20) = 30$.
The top part intercepts an arc of $3x$, so $\angle BAE = \frac32(x) = \frac32(20) = 30$.
DPatrick
2015-03-21 19:39:47
The bottom part intercepts an arc of $2y$, so $\angle EAG = \frac12(2y) = y = 28$.
The bottom part intercepts an arc of $2y$, so $\angle EAG = \frac12(2y) = y = 28$.
xyuqing
2015-03-21 19:40:05
30+28=58
30+28=58
yuvi18
2015-03-21 19:40:05
58
58
summitwei
2015-03-21 19:40:05
058
058
DPatrick
2015-03-21 19:40:06
And thus,
\[
\angle BAG = \angle BAE + \angle EAG = 30 + 28 = \boxed{058}.
\]
And thus,
\[
\angle BAG = \angle BAE + \angle EAG = 30 + 28 = \boxed{058}.
\]
DPatrick
2015-03-21 19:40:43
From one algebra-laced geometry problem to another, here's #7:
From one algebra-laced geometry problem to another, here's #7:
DPatrick
2015-03-21 19:40:47
7. In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$ and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$.
7. In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$ and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$.
DPatrick
2015-03-21 19:40:51
DPatrick
2015-03-21 19:41:02
What strikes you about this picture?
What strikes you about this picture?
AlcumusGuy
2015-03-21 19:41:25
Similar triangles!
Similar triangles!
Darn
2015-03-21 19:41:25
Similar triangles
Similar triangles
trumpeter
2015-03-21 19:41:25
similar triangles!
similar triangles!
fluffyanimal
2015-03-21 19:41:25
similar triangles everywhere
similar triangles everywhere
ninjataco
2015-03-21 19:41:25
a bunch of similar $1-2-\sqrt{5}$ right triangles!
a bunch of similar $1-2-\sqrt{5}$ right triangles!
princess-of-mathland
2015-03-21 19:41:25
Similar triangles!!!
Similar triangles!!!
raxu
2015-03-21 19:41:25
Note that all the triangles we see are similar.
Note that all the triangles we see are similar.
eyux
2015-03-21 19:41:25
theres a lot of similar triangles
theres a lot of similar triangles
torquoiseworld
2015-03-21 19:41:25
similar triangles
similar triangles
math_boy
2015-03-21 19:41:25
SIMILAR TRIANGLES
SIMILAR TRIANGLES
cumo99
2015-03-21 19:41:25
similar triangles
similar triangles
WhaleVomit
2015-03-21 19:41:25
similar triangles
similar triangles
gaberen
2015-03-21 19:41:25
So many similar triangles
So many similar triangles
MiamiHeat88
2015-03-21 19:41:25
it has a lot of similar triangles
it has a lot of similar triangles
brainiac1
2015-03-21 19:41:25
the similar triangles
the similar triangles
maverick8
2015-03-21 19:41:25
Similar Triangles
Similar Triangles
Epic777
2015-03-21 19:41:25
Lots of similar triangles
Lots of similar triangles
hlasker1
2015-03-21 19:41:25
similar triangles
similar triangles
BFYSharks
2015-03-21 19:41:25
Similar triangles
Similar triangles
morninglight
2015-03-21 19:41:25
Similar triangles
Similar triangles
stan23456
2015-03-21 19:41:25
so many 1-2-sqrt5 triangles.
so many 1-2-sqrt5 triangles.
DPatrick
2015-03-21 19:41:36
Yeah, there are tons of similar triangles lying about.
Yeah, there are tons of similar triangles lying about.
DPatrick
2015-03-21 19:41:46
$ED$ is half of $DC$, so all of the triangles have sides in the ratio $1 : 2 : \sqrt{5}$.
$ED$ is half of $DC$, so all of the triangles have sides in the ratio $1 : 2 : \sqrt{5}$.
DPatrick
2015-03-21 19:42:08
And, conveniently, sides $AB$ and $BC$ of the big square are made up of pieces of smaller triangles, each of which also has a side of one of our smaller squares. So we can relate these pieces to the sides of the squares, and use $AB = BC$ to set up an equation.
And, conveniently, sides $AB$ and $BC$ of the big square are made up of pieces of smaller triangles, each of which also has a side of one of our smaller squares. So we can relate these pieces to the sides of the squares, and use $AB = BC$ to set up an equation.
DPatrick
2015-03-21 19:42:30
In particular, let's set $x$ to be the be the side of the square $FGHJ$. We'll try to use $AB = BC$ to set up an equation for $x$.
In particular, let's set $x$ to be the be the side of the square $FGHJ$. We'll try to use $AB = BC$ to set up an equation for $x$.
DPatrick
2015-03-21 19:42:41
What's $AB$ in terms of $x$?
What's $AB$ in terms of $x$?
DPatrick
2015-03-21 19:43:42
We have to break up $AB = AN + NH + HB$ in order to compute it.
We have to break up $AB = AN + NH + HB$ in order to compute it.
DPatrick
2015-03-21 19:43:46
What's $AN$?
What's $AN$?
Studiosa
2015-03-21 19:44:11
3 sqrt(11)/sqrt(5)
3 sqrt(11)/sqrt(5)
MSTang
2015-03-21 19:44:11
sqrt(99)/5
sqrt(99)/5
ninjataco
2015-03-21 19:44:11
$\sqrt{99}/\sqrt{5}$
$\sqrt{99}/\sqrt{5}$
summitwei
2015-03-21 19:44:11
sqrt 99/sqrt 5
sqrt 99/sqrt 5
MSTang
2015-03-21 19:44:11
sqrt(99) / sqrt(5)
sqrt(99) / sqrt(5)
MiamiHeat88
2015-03-21 19:44:11
(sqrt99)/(sqrt5)
(sqrt99)/(sqrt5)
bli1999
2015-03-21 19:44:11
sqrt99/sqrt5
sqrt99/sqrt5
DPatrick
2015-03-21 19:44:15
$AN$ is the "1" side of $AMN$, where the "$\sqrt5$" side is $MN = \sqrt{99}$. (Let's leave it as $\sqrt{99}$ rather than $3\sqrt{11}$ until we have a compelling reason to change it.)
$AN$ is the "1" side of $AMN$, where the "$\sqrt5$" side is $MN = \sqrt{99}$. (Let's leave it as $\sqrt{99}$ rather than $3\sqrt{11}$ until we have a compelling reason to change it.)
DPatrick
2015-03-21 19:44:22
Thus $AN = \dfrac{\sqrt{99}}{\sqrt5}$.
Thus $AN = \dfrac{\sqrt{99}}{\sqrt5}$.
DPatrick
2015-03-21 19:44:30
How about $NH$?
How about $NH$?
MiamiHeat88
2015-03-21 19:44:38
we should use a variable for MN and plug in the value when we are done computing
we should use a variable for MN and plug in the value when we are done computing
DPatrick
2015-03-21 19:45:02
We could certainly have done that...it's not a bad idea at all. If it were any more complicated than $\sqrt{99}$ I certainly would have done that.
We could certainly have done that...it's not a bad idea at all. If it were any more complicated than $\sqrt{99}$ I certainly would have done that.
rlz
2015-03-21 19:45:21
$\frac{\sqrt{5*99}}{2}$
$\frac{\sqrt{5*99}}{2}$
hwl0304
2015-03-21 19:45:21
sqrt99*sqrt5/2
sqrt99*sqrt5/2
IMGUNNA
2015-03-21 19:45:21
sqrt99 * sqrt 5 / 2
sqrt99 * sqrt 5 / 2
azmath333
2015-03-21 19:45:21
$\frac{\sqrt{99\cdot 5}}{2}$
$\frac{\sqrt{99\cdot 5}}{2}$
chaoshadow37
2015-03-21 19:45:28
NH= sqrt(99)*sqrt(5)/2
NH= sqrt(99)*sqrt(5)/2
DPatrick
2015-03-21 19:45:29
$NH$ is the "$\sqrt5$" side of $NHK$, where the "2" side is $\sqrt{99}$.
$NH$ is the "$\sqrt5$" side of $NHK$, where the "2" side is $\sqrt{99}$.
DPatrick
2015-03-21 19:45:37
Thus $NH = \dfrac{\sqrt{5}\cdot\sqrt{99}}{2}$.
Thus $NH = \dfrac{\sqrt{5}\cdot\sqrt{99}}{2}$.
DPatrick
2015-03-21 19:45:45
Finally, what's $HB$?
Finally, what's $HB$?
ninjataco
2015-03-21 19:46:10
$2x/\sqrt{5}$
$2x/\sqrt{5}$
rocket13jg
2015-03-21 19:46:10
2x/sqrt5
2x/sqrt5
SHARKYBOY
2015-03-21 19:46:10
FTW
2015-03-21 19:46:10
2x/sqrt5
2x/sqrt5
nosaj
2015-03-21 19:46:10
$x \times \frac{2}{\sqrt5}$
$x \times \frac{2}{\sqrt5}$
whatshisbucket
2015-03-21 19:46:10
2x/sqrt5
2x/sqrt5
eswa2000
2015-03-21 19:46:10
2x/sqrt{5}
2x/sqrt{5}
24iam24
2015-03-21 19:46:10
x*2/√5
x*2/√5
DPatrick
2015-03-21 19:46:15
$HB$ is the "2" side of $BHJ$, where the "$\sqrt{5}$" side is $x$.
$HB$ is the "2" side of $BHJ$, where the "$\sqrt{5}$" side is $x$.
DPatrick
2015-03-21 19:46:20
Thus $HB = \dfrac{2x}{\sqrt5}$.
Thus $HB = \dfrac{2x}{\sqrt5}$.
DPatrick
2015-03-21 19:46:26
Adding these up gives
\[
AB = AN + NH + HB = \frac{\sqrt{99}}{\sqrt5} + \frac{\sqrt5 \cdot \sqrt{99}}{2} + \frac{2x}{\sqrt5}.
\]
Adding these up gives
\[
AB = AN + NH + HB = \frac{\sqrt{99}}{\sqrt5} + \frac{\sqrt5 \cdot \sqrt{99}}{2} + \frac{2x}{\sqrt5}.
\]
DPatrick
2015-03-21 19:46:40
Looks messy, but maybe nice things will happen later.
Looks messy, but maybe nice things will happen later.
DPatrick
2015-03-21 19:46:44
OK, how about $BC = BJ + JC$?
OK, how about $BC = BJ + JC$?
FTW
2015-03-21 19:47:10
Bj = x/sqrt5
Bj = x/sqrt5
nosaj
2015-03-21 19:47:10
$BJ=\frac{x}{\sqrt5}$
$BJ=\frac{x}{\sqrt5}$
csmath
2015-03-21 19:47:10
We can do BJ = x/sqrt(5)
We can do BJ = x/sqrt(5)
whatshisbucket
2015-03-21 19:47:10
BJ=x/sqrt5
BJ=x/sqrt5
MSTang
2015-03-21 19:47:29
BJ = x/sqrt(5) and JC = x * sqrt(5)/2
BJ = x/sqrt(5) and JC = x * sqrt(5)/2
ryanyz10
2015-03-21 19:47:29
x/sqrt(5) + xsqrt(5)/2
x/sqrt(5) + xsqrt(5)/2
maverick8
2015-03-21 19:47:29
$BC=BJ+JC=x/sqrt(5)+xsqrt(5)/2 $
$BC=BJ+JC=x/sqrt(5)+xsqrt(5)/2 $
azmath333
2015-03-21 19:47:29
$\frac{x}{\sqrt{5}} + \frac{x\sqrt{5}}{2}$
$\frac{x}{\sqrt{5}} + \frac{x\sqrt{5}}{2}$
maverick8
2015-03-21 19:47:29
BC=BJ+JC=x/sqrt(5)+xsqrt(5)/2
BC=BJ+JC=x/sqrt(5)+xsqrt(5)/2
Darn
2015-03-21 19:47:29
$\frac{x\sqrt{5}}{2}+\frac{x}{\sqrt{5}}$
$\frac{x\sqrt{5}}{2}+\frac{x}{\sqrt{5}}$
DPatrick
2015-03-21 19:47:33
$BJ$ is the "1" side of $BJH$ where the "$\sqrt5$" side is $x$. So $BJ = \dfrac{x}{\sqrt5}$.
$BJ$ is the "1" side of $BJH$ where the "$\sqrt5$" side is $x$. So $BJ = \dfrac{x}{\sqrt5}$.
DPatrick
2015-03-21 19:47:38
$JC$ is the "$\sqrt5$" side of $CJF$ where the "2" side is $x$. So $JC = \dfrac{\sqrt5 \cdot x}{2}$.
$JC$ is the "$\sqrt5$" side of $CJF$ where the "2" side is $x$. So $JC = \dfrac{\sqrt5 \cdot x}{2}$.
DPatrick
2015-03-21 19:47:44
Thus
\[
BC = BJ + JC = \dfrac{x}{\sqrt5} + \dfrac{\sqrt5 \cdot x}{2}.
\]
Thus
\[
BC = BJ + JC = \dfrac{x}{\sqrt5} + \dfrac{\sqrt5 \cdot x}{2}.
\]
ninjataco
2015-03-21 19:47:59
now we can set AB=BC and solve for x
now we can set AB=BC and solve for x
nsd
2015-03-21 19:47:59
AB = BC
AB = BC
24iam24
2015-03-21 19:47:59
then equate them because AB=BC
then equate them because AB=BC
DPatrick
2015-03-21 19:48:03
So using $AB = BC$, we get the equation:
\[
\frac{\sqrt{99}}{\sqrt5} + \frac{\sqrt5 \cdot \sqrt{99}}{2} + \frac{2x}{\sqrt5} = \dfrac{x}{\sqrt5} + \dfrac{\sqrt5 \cdot x}{2}.
\]
So using $AB = BC$, we get the equation:
\[
\frac{\sqrt{99}}{\sqrt5} + \frac{\sqrt5 \cdot \sqrt{99}}{2} + \frac{2x}{\sqrt5} = \dfrac{x}{\sqrt5} + \dfrac{\sqrt5 \cdot x}{2}.
\]
DPatrick
2015-03-21 19:48:21
How do we solve this?
How do we solve this?
whatshisbucket
2015-03-21 19:48:30
multiply by 2sqrt5
multiply by 2sqrt5
nosaj
2015-03-21 19:48:30
multiply both sides by $2\sqrt5$ cause fractions eww
multiply both sides by $2\sqrt5$ cause fractions eww
stan23456
2015-03-21 19:48:30
multiply by 2sqrt5
multiply by 2sqrt5
nosyarg
2015-03-21 19:48:30
clear the denominators?
clear the denominators?
danusv
2015-03-21 19:48:30
multiply by 2 root 5
multiply by 2 root 5
DPatrick
2015-03-21 19:48:35
I'd multiply through by $2\sqrt5$ to clear the denominators. Happily, that gets rid of the $\sqrt5$'s as well:
\[
2\sqrt{99} + 5\sqrt{99} + 4x = 2x + 5x.
\]
I'd multiply through by $2\sqrt5$ to clear the denominators. Happily, that gets rid of the $\sqrt5$'s as well:
\[
2\sqrt{99} + 5\sqrt{99} + 4x = 2x + 5x.
\]
DPatrick
2015-03-21 19:48:45
Wow, that turned out nice!
Wow, that turned out nice!
csmath
2015-03-21 19:49:16
3x = 7sqrt (99)
3x = 7sqrt (99)
C-bass
2015-03-21 19:49:16
x = 7sqrt(11)
x = 7sqrt(11)
summitwei
2015-03-21 19:49:16
So x=7sqrt11
So x=7sqrt11
ryanyz10
2015-03-21 19:49:16
x = 7sqrt(99)/3
x = 7sqrt(99)/3
IMGUNNA
2015-03-21 19:49:16
7sqrt99 = 3x
7sqrt99 = 3x
DPatrick
2015-03-21 19:49:30
Right. We have just $7\sqrt{99} = 3x$. That gives $x = 7\sqrt{11}$.
Right. We have just $7\sqrt{99} = 3x$. That gives $x = 7\sqrt{11}$.
maverick8
2015-03-21 19:49:45
x^2=539
x^2=539
RedPhoenix
2015-03-21 19:49:45
x = 7 rt 11, so x^2 is 539
x = 7 rt 11, so x^2 is 539
SHARKYBOY
2015-03-21 19:49:45
DPatrick
2015-03-21 19:49:53
Thus $x^2 = 49 \cdot 11 = \boxed{539}$ is our answer.
Thus $x^2 = 49 \cdot 11 = \boxed{539}$ is our answer.
DPatrick
2015-03-21 19:50:24
Enough geometry for a little while...on to #8:
Enough geometry for a little while...on to #8:
DPatrick
2015-03-21 19:50:28
8. For a positive integer $n$, let $s(n)$ denote the sum of the digits of $n$. Find the smallest positive integer $n$ satisfying $s(n)=s(n+864)=20$.
8. For a positive integer $n$, let $s(n)$ denote the sum of the digits of $n$. Find the smallest positive integer $n$ satisfying $s(n)=s(n+864)=20$.
DPatrick
2015-03-21 19:50:43
How many digits does the target number have?
How many digits does the target number have?
doodlemaster7
2015-03-21 19:51:01
we know that n has to be a 3 digit number
we know that n has to be a 3 digit number
spartan168
2015-03-21 19:51:01
At least 3
At least 3
summitwei
2015-03-21 19:51:01
3 because this is AIME
3 because this is AIME
hwl0304
2015-03-21 19:51:01
0<n<999 so 3
0<n<999 so 3
brainiac1
2015-03-21 19:51:01
3 digits
3 digits
DPatrick
2015-03-21 19:51:06
Since each digit of $n$ contributes at most 9 to $s(n)$, we know $n$ must have at least 3 digits.
Since each digit of $n$ contributes at most 9 to $s(n)$, we know $n$ must have at least 3 digits.
DPatrick
2015-03-21 19:51:12
And since it's the AIME and $n$ is our final answer, we know that $n$ has exactly 3 digits!
And since it's the AIME and $n$ is our final answer, we know that $n$ has exactly 3 digits!
DPatrick
2015-03-21 19:51:26
What does $s(n)=s(n+864)$ tell us about what happens when we add? What must happen when we add in order for the digit sum not to change?
What does $s(n)=s(n+864)$ tell us about what happens when we add? What must happen when we add in order for the digit sum not to change?
MSTang
2015-03-21 19:51:59
carrying
carrying
chaoshadow37
2015-03-21 19:51:59
it has to regroup
it has to regroup
rlz
2015-03-21 19:51:59
Carrying
Carrying
eswa2000
2015-03-21 19:51:59
there's carry over's
there's carry over's
ooooo1
2015-03-21 19:51:59
exactly two digits carry over when we add
exactly two digits carry over when we add
xyuqing
2015-03-21 19:51:59
2 carries
2 carries
Sciolympian
2015-03-21 19:51:59
2 carries since each carry subtracts 9 from digit sum and 864 has digit sum of 18
2 carries since each carry subtracts 9 from digit sum and 864 has digit sum of 18
kuilin
2015-03-21 19:51:59
carrying
carrying
DPatrick
2015-03-21 19:52:06
We have to carry. In fact, every time we carry, we reduce one digit by 10 and increase the next digit by 1. Therefore, carrying reduces the digit sum by 9.
We have to carry. In fact, every time we carry, we reduce one digit by 10 and increase the next digit by 1. Therefore, carrying reduces the digit sum by 9.
DPatrick
2015-03-21 19:52:27
Since 8+6+4=18, we have to carry exactly twice when we add 864 to $n$, in order to subtract 18 from the digit sum and leave it overall unchanged.
Since 8+6+4=18, we have to carry exactly twice when we add 864 to $n$, in order to subtract 18 from the digit sum and leave it overall unchanged.
nosaj
2015-03-21 19:52:39
Do casework based on which place values carry.
Do casework based on which place values carry.
DPatrick
2015-03-21 19:52:49
Indeed -- is there any digits that we know must carry?
Indeed -- is there any digits that we know must carry?
dragon21
2015-03-21 19:53:05
The smallest possible is 299
The smallest possible is 299
zacchro
2015-03-21 19:53:05
hundreds digit has to carry
hundreds digit has to carry
MSTang
2015-03-21 19:53:05
Hundreds - since n >= 299
Hundreds - since n >= 299
nosaj
2015-03-21 19:53:05
Yes hundreds
Yes hundreds
AlcumusGuy
2015-03-21 19:53:05
hundreds
hundreds
mathbeida
2015-03-21 19:53:05
hundreds
hundreds
fluffyanimal
2015-03-21 19:53:05
hundreds
hundreds
willabc
2015-03-21 19:53:05
hundreds digit must carry
hundreds digit must carry
DPatrick
2015-03-21 19:53:13
Right. Since $20=9+9+2$, the smallest the hundreds digit can possibly be is 2, so the hundreds digit must carry.
Right. Since $20=9+9+2$, the smallest the hundreds digit can possibly be is 2, so the hundreds digit must carry.
DPatrick
2015-03-21 19:53:19
We have two cases now: we carry the hundreds and tens digits, or the hundreds and ones digit.
We have two cases now: we carry the hundreds and tens digits, or the hundreds and ones digit.
DPatrick
2015-03-21 19:53:30
Which one is likely to be more fruitful?
Which one is likely to be more fruitful?
Studiosa
2015-03-21 19:53:59
hundreds and tens
hundreds and tens
mathperson9
2015-03-21 19:53:59
Hundreds and tehns
Hundreds and tehns
ninjataco
2015-03-21 19:53:59
hundreds and tens
hundreds and tens
temp8909
2015-03-21 19:53:59
hundreds and tens
hundreds and tens
Sciolympian
2015-03-21 19:53:59
hundreds and tens
hundreds and tens
ryanyz10
2015-03-21 19:53:59
hundreds and tens?
hundreds and tens?
chaoshadow37
2015-03-21 19:53:59
hundreds and tens
hundreds and tens
Ramanan369
2015-03-21 19:53:59
hundreds and tens
hundreds and tens
DPatrick
2015-03-21 19:54:13
The ones digit of $n$ can be much larger without carrying than the tens digit. Therefore we can make the sum of the tens and ones digit largest (and thus the entire number smallest, which is our goal) by carrying from the tens digit.
The ones digit of $n$ can be much larger without carrying than the tens digit. Therefore we can make the sum of the tens and ones digit largest (and thus the entire number smallest, which is our goal) by carrying from the tens digit.
DPatrick
2015-03-21 19:54:20
So what's the smallest $n$ with $s(n)=20$ and such that $n+864$ carries in the tens and hundreds digits?
So what's the smallest $n$ with $s(n)=20$ and such that $n+864$ carries in the tens and hundreds digits?
bli1999
2015-03-21 19:54:45
695
695
whatshisbucket
2015-03-21 19:54:45
695
695
xyuqing
2015-03-21 19:54:45
695!
695!
hlasker1
2015-03-21 19:54:45
695
695
Royalreter1
2015-03-21 19:54:45
695
695
maverick8
2015-03-21 19:54:45
The last digit should be 5
The last digit should be 5
PencilPal
2015-03-21 19:54:45
695
695
DPatrick
2015-03-21 19:55:10
Exactly. We want the units digit as large as possible (so that the tens and hundreds digits can be small). The largest it can be without carrying when we add is 5.
Exactly. We want the units digit as large as possible (so that the tens and hundreds digits can be small). The largest it can be without carrying when we add is 5.
DPatrick
2015-03-21 19:55:26
That leaves 15 left, and to make the number smaller, we want as much as possible in the tens digits.
That leaves 15 left, and to make the number smaller, we want as much as possible in the tens digits.
DPatrick
2015-03-21 19:55:31
So that gives the number 695.
So that gives the number 695.
DPatrick
2015-03-21 19:55:52
(To check, 695 + 864 = 1559 has digit sum of 20 too.)
(To check, 695 + 864 = 1559 has digit sum of 20 too.)
DPatrick
2015-03-21 19:56:01
Just to check that our intuition was correct: what's the smallest $n$ with $s(n)=20$ and such that $n+864$ carries in the ones and hundreds digits?
Just to check that our intuition was correct: what's the smallest $n$ with $s(n)=20$ and such that $n+864$ carries in the ones and hundreds digits?
xyuqing
2015-03-21 19:56:35
929
929
nosyarg
2015-03-21 19:56:35
929
929
MSTang
2015-03-21 19:56:35
929
929
zacchro
2015-03-21 19:56:35
929
929
tenniskidperson3
2015-03-21 19:56:35
929
929
kingsave3166
2015-03-21 19:56:35
929
929
DPatrick
2015-03-21 19:57:03
Right. 2 is the smallest tens digit we can have to avoid carrying (remember we'll get an extra +1 from the carry in the units digit). So 929 is the only such number.
Right. 2 is the smallest tens digit we can have to avoid carrying (remember we'll get an extra +1 from the carry in the units digit). So 929 is the only such number.
Studiosa
2015-03-21 19:57:15
bigger than 695
bigger than 695
MathDog0809
2015-03-21 19:57:15
bigger than 695
bigger than 695
DPatrick
2015-03-21 19:57:29
Right. Our intuition for which case to look in was correct, and the answer is $\boxed{695}$.
Right. Our intuition for which case to look in was correct, and the answer is $\boxed{695}$.
DPatrick
2015-03-21 19:57:51
Well, we can't avoid it any longer...on to #9:
Well, we can't avoid it any longer...on to #9:
DPatrick
2015-03-21 19:57:55
9. Let $S$ be the set of all ordered triples of integers $(a_1,a_2,a_3)$ with $1\leq a_1,a_2,a_3\leq10$. Each ordered triple in $S$ generates a sequence according to the rule \[a_n=a_{n-1} \cdot |a_{n-2}-a_{n-3}|\] for $n\geq4$. Find the number of such sequences for which $a_n=0$ for some $n$.
9. Let $S$ be the set of all ordered triples of integers $(a_1,a_2,a_3)$ with $1\leq a_1,a_2,a_3\leq10$. Each ordered triple in $S$ generates a sequence according to the rule \[a_n=a_{n-1} \cdot |a_{n-2}-a_{n-3}|\] for $n\geq4$. Find the number of such sequences for which $a_n=0$ for some $n$.
AlcumusGuy
2015-03-21 19:58:38
We need $a_{n-2} = a_{n-3}$ at some point.
We need $a_{n-2} = a_{n-3}$ at some point.
DPatrick
2015-03-21 19:59:13
Right: the value of $a_n$ is zero whenever $a_{n-2}=a_{n-3}$. That's the only way we can get our first zero: getting two consecutive terms equal.
Right: the value of $a_n$ is zero whenever $a_{n-2}=a_{n-3}$. That's the only way we can get our first zero: getting two consecutive terms equal.
nosyarg
2015-03-21 19:59:55
if 2 consecutive terms differ by one that will happen
if 2 consecutive terms differ by one that will happen
raxu
2015-03-21 20:00:01
We can start listing the possibilities: a_1=a_2, a_2=a_3, ...
We can start listing the possibilities: a_1=a_2, a_2=a_3, ...
maverick8
2015-03-21 20:00:01
The difference between 2 consecutive terms can be 1
The difference between 2 consecutive terms can be 1
DPatrick
2015-03-21 20:00:25
Right. One possibility is we start out that way from the beginning -- we definitely hit zero quickly if $a_2$ is equal to either $a_1$ or $a_3$:
Right. One possibility is we start out that way from the beginning -- we definitely hit zero quickly if $a_2$ is equal to either $a_1$ or $a_3$:
DPatrick
2015-03-21 20:00:30
\begin{align*}
&1,1,5,0,0\ldots\\
&6,6,2,0,0\ldots\\
&3,8,8,40,0,\ldots\\
&5,2,2,6,0,\ldots
\end{align*}
\begin{align*}
&1,1,5,0,0\ldots\\
&6,6,2,0,0\ldots\\
&3,8,8,40,0,\ldots\\
&5,2,2,6,0,\ldots
\end{align*}
DPatrick
2015-03-21 20:00:55
But we can also get two equal terms if we have two consecutive terms that differ by 1.
But we can also get two equal terms if we have two consecutive terms that differ by 1.
DPatrick
2015-03-21 20:01:03
When $|a_{n-2}-a_{n-3}|=1$ that forces
\[a_n=a_{n-1}|a_{n-2}-a_{n-3}|=a_{n-1}\cdot1=a_{n-1}\]
When $|a_{n-2}-a_{n-3}|=1$ that forces
\[a_n=a_{n-1}|a_{n-2}-a_{n-3}|=a_{n-1}\cdot1=a_{n-1}\]
DPatrick
2015-03-21 20:01:13
So we get two consecutive equal values, and that forces a zero value down the line.
So we get two consecutive equal values, and that forces a zero value down the line.
DPatrick
2015-03-21 20:01:21
We can see some examples of this:
We can see some examples of this:
DPatrick
2015-03-21 20:01:24
\begin{align*}
&1,2,3,3,3,0,\ldots\\
&1,2,5,5,15,0,\ldots\\
&9,1,2,16,16,224,0\ldots\\
&5,3,2,4,4,8,0\ldots
\end{align*}
\begin{align*}
&1,2,3,3,3,0,\ldots\\
&1,2,5,5,15,0,\ldots\\
&9,1,2,16,16,224,0\ldots\\
&5,3,2,4,4,8,0\ldots
\end{align*}
DPatrick
2015-03-21 20:01:47
So perhaps we have a conjecture:
There are no zeros if and only if $|a_2 - a_1| \ge 2$ and $|a_3 - a_2| \ge 2$.
So perhaps we have a conjecture:
There are no zeros if and only if $|a_2 - a_1| \ge 2$ and $|a_3 - a_2| \ge 2$.
rlz
2015-03-21 20:02:20
There are more sneaky cases
There are more sneaky cases
supercomputer
2015-03-21 20:02:20
7, 5, 1
7, 5, 1
mathtastic
2015-03-21 20:02:20
try $1,3,5$
try $1,3,5$
mssmath
2015-03-21 20:02:27
1,3,1
1,3,1
brightknight
2015-03-21 20:02:27
531 doesnt work
531 doesnt work
DPatrick
2015-03-21 20:02:36
Yeah... wait a second. What about this example:\[1,3,1,2,4,4,8,0,\ldots\]That one fails!
Yeah... wait a second. What about this example:\[1,3,1,2,4,4,8,0,\ldots\]That one fails!
Studiosa
2015-03-21 20:02:49
how do you find the sneaky cases systematically?
how do you find the sneaky cases systematically?
DPatrick
2015-03-21 20:03:06
That's a really good question. The way I found them is to try to prove my earlier "conjecture".
That's a really good question. The way I found them is to try to prove my earlier "conjecture".
DPatrick
2015-03-21 20:03:22
I made the leap to try to show that if
\begin{align*}
|a_2-a_1|&\geq2\\
|a_3-a_2|&\geq2
\end{align*}
then $|a_n-a_{n-1}|\geq2$ for all $n$.
I made the leap to try to show that if
\begin{align*}
|a_2-a_1|&\geq2\\
|a_3-a_2|&\geq2
\end{align*}
then $|a_n-a_{n-1}|\geq2$ for all $n$.
DPatrick
2015-03-21 20:03:57
...and if this happens, then we can never hit 0.
...and if this happens, then we can never hit 0.
DPatrick
2015-03-21 20:04:07
If you were to try to prove this, how would you likely proceed?
If you were to try to prove this, how would you likely proceed?
MSTang
2015-03-21 20:04:25
induction
induction
danusv
2015-03-21 20:04:25
Induction!
Induction!
xyuqing
2015-03-21 20:04:25
Induction?
Induction?
ws5188
2015-03-21 20:04:25
induction
induction
chaoshadow37
2015-03-21 20:04:25
induction?
induction?
kingsave3166
2015-03-21 20:04:29
itd probably be induction
itd probably be induction
DPatrick
2015-03-21 20:04:33
We can try to prove this by induction. We've already got two base cases. So suppose it's true that $|a_k - a_{k-1}| \ge 2$ for all $k < n$ for some $n$.
We can try to prove this by induction. We've already got two base cases. So suppose it's true that $|a_k - a_{k-1}| \ge 2$ for all $k < n$ for some $n$.
DPatrick
2015-03-21 20:04:45
Then
\[
a_n = a_{n-1}|a_{n-2} - a_{n-3}| \ge 2a_{n-1}.
\]
Then
\[
a_n = a_{n-1}|a_{n-2} - a_{n-3}| \ge 2a_{n-1}.
\]
DPatrick
2015-03-21 20:05:08
This means that $a_n - a_{n-1} \ge 2a_{n-1} - a_{n-1} = a_{n-1}$. But how do we know that that's greater than 2?
This means that $a_n - a_{n-1} \ge 2a_{n-1} - a_{n-1} = a_{n-1}$. But how do we know that that's greater than 2?
summitwei
2015-03-21 20:05:24
WE DONT
WE DONT
MSTang
2015-03-21 20:05:24
we don't
we don't
DPatrick
2015-03-21 20:05:32
Well, what if we go back one more step:
\[
a_{n-1} = a_{n-2} \cdot |a_{n-3} - a_{n-4}| \ge 2a_{n-2} \ge 2,
\]
since all the $a$'s are positive.
Well, what if we go back one more step:
\[
a_{n-1} = a_{n-2} \cdot |a_{n-3} - a_{n-4}| \ge 2a_{n-2} \ge 2,
\]
since all the $a$'s are positive.
DPatrick
2015-03-21 20:05:42
So we're done...right?
So we're done...right?
AlcumusGuy
2015-03-21 20:06:20
no, you can't go back further because of base case
no, you can't go back further because of base case
summitwei
2015-03-21 20:06:20
What if we can't go back a step?
What if we can't go back a step?
mssmath
2015-03-21 20:06:20
For n sufficiently large
For n sufficiently large
DPatrick
2015-03-21 20:06:44
That's the problem. What I did doesn't work for $n=4$ in particular, since then the $a_{n-4}$ that I tried to use in my last line doesn't exist!
That's the problem. What I did doesn't work for $n=4$ in particular, since then the $a_{n-4}$ that I tried to use in my last line doesn't exist!
DPatrick
2015-03-21 20:07:30
And if we back up a step, we see the root of the flaw. For $n=4$, we needed $a_3 \ge 2$ (that was the point of the last step).
And if we back up a step, we see the root of the flaw. For $n=4$, we needed $a_3 \ge 2$ (that was the point of the last step).
DPatrick
2015-03-21 20:07:35
But what if $a_3 = 1$?
But what if $a_3 = 1$?
Studiosa
2015-03-21 20:07:57
conjecture doesn't work then
conjecture doesn't work then
AlcumusGuy
2015-03-21 20:08:05
then you get the sneaky case
then you get the sneaky case
nosaj
2015-03-21 20:08:05
Then if the first 2 terms differ by 2 it goes to zero
Then if the first 2 terms differ by 2 it goes to zero
rlz
2015-03-21 20:08:05
That's the sneaky case
That's the sneaky case
DPatrick
2015-03-21 20:08:45
Exactly. My proof fails when $a_3 = 1$ and $|a_2 - a_1| = 2$, because we get $a_4 = 2$.
Exactly. My proof fails when $a_3 = 1$ and $|a_2 - a_1| = 2$, because we get $a_4 = 2$.
eyux
2015-03-21 20:08:49
so in all other cases, the conjecture holds?
so in all other cases, the conjecture holds?
csmath
2015-03-21 20:08:49
How do you know that it is the only exception?
How do you know that it is the only exception?
DPatrick
2015-03-21 20:08:59
Because that's the only place where my proof fails!
Because that's the only place where my proof fails!
DPatrick
2015-03-21 20:09:13
Indeed, when we fix that flaw, we have a theorem: there are no zeros if and only if all of the following inequalities hold:
\begin{align*}
|a_2-a_1|&\geq2,\\
|a_3-a_2|&\geq2,\\
|a_4-a_3|&\geq2.
\end{align*}
Indeed, when we fix that flaw, we have a theorem: there are no zeros if and only if all of the following inequalities hold:
\begin{align*}
|a_2-a_1|&\geq2,\\
|a_3-a_2|&\geq2,\\
|a_4-a_3|&\geq2.
\end{align*}
Sciolympian
2015-03-21 20:09:34
now the counting!
now the counting!
DPatrick
2015-03-21 20:09:40
Right. Now we need to count. Ugh.
Right. Now we need to count. Ugh.
DPatrick
2015-03-21 20:10:02
Let's try to count all the triples that satisfy all three inequalities above. Then, at the end, we have to remember to subtract this count from 1000 to get the number of triples that generate a sequence with a zero term.
Let's try to count all the triples that satisfy all three inequalities above. Then, at the end, we have to remember to subtract this count from 1000 to get the number of triples that generate a sequence with a zero term.
mssmath
2015-03-21 20:10:30
The first two cases are easy then bash
The first two cases are easy then bash
DPatrick
2015-03-21 20:10:40
Right. Let's use casework on the value of $a_2$ and count how many triples $(a_1,a_2,a_3)$ satisfy the first two inequalities. Then we'll worry about the which might fail on the third. (Remember the third inequality is pretty hard to fail if the first two work.)
Right. Let's use casework on the value of $a_2$ and count how many triples $(a_1,a_2,a_3)$ satisfy the first two inequalities. Then we'll worry about the which might fail on the third. (Remember the third inequality is pretty hard to fail if the first two work.)
DPatrick
2015-03-21 20:10:52
If $a_2=1$, how many triples satisfy the first two inequalities?
If $a_2=1$, how many triples satisfy the first two inequalities?
BFYSharks
2015-03-21 20:11:24
64
64
mssmath
2015-03-21 20:11:24
64
64
summitwei
2015-03-21 20:11:24
64
64
TheStrangeCharm
2015-03-21 20:11:24
64
64
CornSaltButter
2015-03-21 20:11:31
8^2=64
8^2=64
DPatrick
2015-03-21 20:11:32
If $a_2=1$ then we need to have $a_1,a_3\geq3$. That gives $8\cdot8=64$ good triples.
If $a_2=1$ then we need to have $a_1,a_3\geq3$. That gives $8\cdot8=64$ good triples.
DPatrick
2015-03-21 20:11:39
Symmetrically, if $a_2=10$ we need to have $a_1,a_3\leq8$. That gives $8 \times 8 = 64$ more good triples.
Symmetrically, if $a_2=10$ we need to have $a_1,a_3\leq8$. That gives $8 \times 8 = 64$ more good triples.
DPatrick
2015-03-21 20:12:28
And for each $2\leq a_2\leq9$, how many triples satisfy the first two inequalities?
And for each $2\leq a_2\leq9$, how many triples satisfy the first two inequalities?
xyuqing
2015-03-21 20:12:39
49
49
kingsave3166
2015-03-21 20:12:39
49
49
ohmcfifth
2015-03-21 20:12:39
49?
49?
thequantumguy
2015-03-21 20:12:50
49
49
DPatrick
2015-03-21 20:12:53
If $2\leq a_2\leq9$ then we need $a_1$ and $a_3$ to avoid $\{a_2-1,a_2,a_2+1\}$. That gives us $7\cdot7 = 49$ total triples for each $a_2$.
If $2\leq a_2\leq9$ then we need $a_1$ and $a_3$ to avoid $\{a_2-1,a_2,a_2+1\}$. That gives us $7\cdot7 = 49$ total triples for each $a_2$.
DPatrick
2015-03-21 20:13:06
So summing these cases up, we've got $2 \cdot 64 + 8 \cdot 49 = 520$ triples satisfying the first two inequalities.
So summing these cases up, we've got $2 \cdot 64 + 8 \cdot 49 = 520$ triples satisfying the first two inequalities.
DPatrick
2015-03-21 20:13:14
But how many of these fail the third?
But how many of these fail the third?
C-bass
2015-03-21 20:13:33
not very many
not very many
DPatrick
2015-03-21 20:13:36
Right. Recall that $a_4 = a_3 \cdot |a_2 - a_1|$, so $|a_4 - a_3| = a_3 \cdot \left(|a_2 - a_1| - 1\right)$.
Right. Recall that $a_4 = a_3 \cdot |a_2 - a_1|$, so $|a_4 - a_3| = a_3 \cdot \left(|a_2 - a_1| - 1\right)$.
DPatrick
2015-03-21 20:13:58
So $|a_4 - a_3| = 0$ only where $|a_2 - a_1| = 1$, but this isn't true from any of our 520 triples.
So $|a_4 - a_3| = 0$ only where $|a_2 - a_1| = 1$, but this isn't true from any of our 520 triples.
DPatrick
2015-03-21 20:14:01
How many, though, have $|a_4 - a_3| = 1$?
How many, though, have $|a_4 - a_3| = 1$?
DPatrick
2015-03-21 20:14:08
For this, we need $|a_2 - a_1| = 2$ and $a_3 = 1$. How many are there?
For this, we need $|a_2 - a_1| = 2$ and $a_3 = 1$. How many are there?
Sciolympian
2015-03-21 20:14:45
14 triples
14 triples
dantx5
2015-03-21 20:14:45
14
14
DPatrick
2015-03-21 20:15:00
There are two for each $3 \le a_2 \le 8$ (setting $a_1 = a_2 \pm 2$), but just one for $a_2 = 9 \text{ or } 10$ (setting $a_1 = a_2 - 2$). (Note none of our 520 triples have $a_2 = 1$ or $2$ because $|a_2 - a_3| \ge 2$.)
There are two for each $3 \le a_2 \le 8$ (setting $a_1 = a_2 \pm 2$), but just one for $a_2 = 9 \text{ or } 10$ (setting $a_1 = a_2 - 2$). (Note none of our 520 triples have $a_2 = 1$ or $2$ because $|a_2 - a_3| \ge 2$.)
DPatrick
2015-03-21 20:15:22
So we have to exclude $2 \cdot 6 + 2 = 14$ triples from our count, leaving $520 - 14 = 506$ triples satisfying all three inequalities.
So we have to exclude $2 \cdot 6 + 2 = 14$ triples from our count, leaving $520 - 14 = 506$ triples satisfying all three inequalities.
DPatrick
2015-03-21 20:15:31
Is that our final answer?
Is that our final answer?
mathtastic
2015-03-21 20:15:44
take complement
take complement
csmath
2015-03-21 20:15:44
So the answer is 494
So the answer is 494
RedPhoenix
2015-03-21 20:15:44
answer is 494
answer is 494
csmath
2015-03-21 20:15:44
1000-506
1000-506
Tommy2000
2015-03-21 20:15:47
no 1000-506=494 is the answer
no 1000-506=494 is the answer
DPatrick
2015-03-21 20:15:58
Remember, we counted the sequences that don't contain zero. We need the sequences that do contain 0. There are $10^3$ total sequences so there are $1000-506=\boxed{494}$ sequences that contain 0.
Remember, we counted the sequences that don't contain zero. We need the sequences that do contain 0. There are $10^3$ total sequences so there are $1000-506=\boxed{494}$ sequences that contain 0.
DPatrick
2015-03-21 20:16:34
That problem was pretty ugly. I think it was the hardest problem of all 15 on the contest. It's too easy to miscount and/or miss cases.
That problem was pretty ugly. I think it was the hardest problem of all 15 on the contest. It's too easy to miscount and/or miss cases.
DPatrick
2015-03-21 20:16:58
We played with it a lot around the office trying to find a nicer solution. We did not succeed.
We played with it a lot around the office trying to find a nicer solution. We did not succeed.
DPatrick
2015-03-21 20:17:19
I think after that problem, we should take a brief break. Let's resume in about 4 minutes at :21 past!
I think after that problem, we should take a brief break. Let's resume in about 4 minutes at :21 past!
DPatrick
2015-03-21 20:21:10
OK, on to the double-digit problems!
OK, on to the double-digit problems!
DPatrick
2015-03-21 20:21:14
10. Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[ \lvert f(1)\rvert = \lvert f(2) \rvert = \lvert f(3)\rvert = \lvert f(5)\rvert = \lvert f(6)\rvert = \lvert f(7)\rvert = 12.\] Find $\lvert f(0)\rvert$.
10. Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[ \lvert f(1)\rvert = \lvert f(2) \rvert = \lvert f(3)\rvert = \lvert f(5)\rvert = \lvert f(6)\rvert = \lvert f(7)\rvert = 12.\] Find $\lvert f(0)\rvert$.
mathtastic
2015-03-21 20:21:35
GRAPH IT
GRAPH IT
nosyarg
2015-03-21 20:21:35
draw a curve?
draw a curve?
nosaj
2015-03-21 20:21:35
Draw the graph of the cubic polynomial.
Draw the graph of the cubic polynomial.
DPatrick
2015-03-21 20:21:39
A good step here is to draw a generic graph of a cubic polynomial.
A good step here is to draw a generic graph of a cubic polynomial.
DPatrick
2015-03-21 20:21:43
DPatrick
2015-03-21 20:21:55
(Note that this cubic has leading coefficient positive. But because of the absolute values, we could just multiply by -1 if necessary to make it look like this.)
(Note that this cubic has leading coefficient positive. But because of the absolute values, we could just multiply by -1 if necessary to make it look like this.)
rlz
2015-03-21 20:22:04
It can be 12 in at most 3 places
It can be 12 in at most 3 places
mathtastic
2015-03-21 20:22:04
draw lines $y=\pm 12$
draw lines $y=\pm 12$
jam10307
2015-03-21 20:22:04
i drew 2 lines
i drew 2 lines
DPatrick
2015-03-21 20:22:09
Yes, let's draw in two horizontal lines, to represent $y=12$ and $y=-12$:
Yes, let's draw in two horizontal lines, to represent $y=12$ and $y=-12$:
DPatrick
2015-03-21 20:22:13
DPatrick
2015-03-21 20:22:31
This is a very generic picture, but how does this help us?
This is a very generic picture, but how does this help us?
AKAL3
2015-03-21 20:22:51
1,5,6 are equal
1,5,6 are equal
rlz
2015-03-21 20:22:51
There's a - + + - - + pattern
There's a - + + - - + pattern
summitwei
2015-03-21 20:22:51
3 of the roots are 12 and 3 are -12
3 of the roots are 12 and 3 are -12
dantx5
2015-03-21 20:22:51
determines order
determines order
Studiosa
2015-03-21 20:22:51
know the alternation of positives and negatives
know the alternation of positives and negatives
FTW
2015-03-21 20:22:51
it shows us what sign each solution is
it shows us what sign each solution is
CornSaltButter
2015-03-21 20:22:51
So f(1)=f(5)=f(6)=-12 and f(2)=f(3)=f(7)=12
So f(1)=f(5)=f(6)=-12 and f(2)=f(3)=f(7)=12
DPatrick
2015-03-21 20:22:55
It looks like that, going from left-to-right, we must have -12, 12, 12, -12, -12, 12.
It looks like that, going from left-to-right, we must have -12, 12, 12, -12, -12, 12.
DPatrick
2015-03-21 20:23:09
In fact this is the case. It actually requires some calculus-ness to prove this rigorously. But (I hope) it's clear enough from the diagram.
In fact this is the case. It actually requires some calculus-ness to prove this rigorously. But (I hope) it's clear enough from the diagram.
DPatrick
2015-03-21 20:23:15
And we don't actually have to prove it anyway: we can just assume it, and if we can find an $f$ that works, we're done.
And we don't actually have to prove it anyway: we can just assume it, and if we can find an $f$ that works, we're done.
DPatrick
2015-03-21 20:23:33
So let's assume, based on very good evidence, that
\[
f(2) = f(3) = f(7) = 12
\]
and
\[
f(1) = f(5) = f(6) = -12.
\]
So let's assume, based on very good evidence, that
\[
f(2) = f(3) = f(7) = 12
\]
and
\[
f(1) = f(5) = f(6) = -12.
\]
akim99
2015-03-21 20:23:36
We want $|f(0)|$ so it won't matter if it's flipped or not
We want $|f(0)|$ so it won't matter if it's flipped or not
DPatrick
2015-03-21 20:23:49
Right: the top three as I wrote them might be -12 and the bottom three 12, but it doesn't matter.
Right: the top three as I wrote them might be -12 and the bottom three 12, but it doesn't matter.
DPatrick
2015-03-21 20:23:58
How do we proceed from here?
How do we proceed from here?
ryanyz10
2015-03-21 20:24:24
say g(x) = f(x) + 12
say g(x) = f(x) + 12
ninjataco
2015-03-21 20:24:24
f(x)+12=c(x-1)(x-5)(x-6) and f(x)-12=c(x-2)(x-3)(x-7)
f(x)+12=c(x-1)(x-5)(x-6) and f(x)-12=c(x-2)(x-3)(x-7)
Darn
2015-03-21 20:24:24
Create a new polynomial with those roots, then subtract/add
Create a new polynomial with those roots, then subtract/add
Ramanan369
2015-03-21 20:24:30
f(x)=a(x-2)(x-3)(x-7)+12
f(x)=a(x-2)(x-3)(x-7)+12
DPatrick
2015-03-21 20:24:48
Right. Consider the cubic $g(x) = f(x) - 12$. We know this has its roots at 2, 3, and 7.
Right. Consider the cubic $g(x) = f(x) - 12$. We know this has its roots at 2, 3, and 7.
DPatrick
2015-03-21 20:24:56
So $g(x) = c(x-2)(x-3)(x-7)$ for some constant $c$. Thus
\[
f(x) = c(x-2)(x-3)(x-7) + 12.
\]
So $g(x) = c(x-2)(x-3)(x-7)$ for some constant $c$. Thus
\[
f(x) = c(x-2)(x-3)(x-7) + 12.
\]
DPatrick
2015-03-21 20:25:22
This is super-important: once you know all $n$ roots of a degree $n$ polynomial, then (up to a constant) you know the polynomial!
This is super-important: once you know all $n$ roots of a degree $n$ polynomial, then (up to a constant) you know the polynomial!
DPatrick
2015-03-21 20:25:44
And using the other three values, we similarly get
\[
f(x) = c(x-1)(x-5)(x-6) - 12.
\]
And using the other three values, we similarly get
\[
f(x) = c(x-1)(x-5)(x-6) - 12.
\]
DPatrick
2015-03-21 20:26:01
It's the same $c$ in both equations, because it's the coefficient of the $x^3$ term of $f(x)$.
It's the same $c$ in both equations, because it's the coefficient of the $x^3$ term of $f(x)$.
DPatrick
2015-03-21 20:26:15
Now what?
Now what?
pieslinger
2015-03-21 20:26:39
plug in 0
plug in 0
Ramanan369
2015-03-21 20:26:39
plug in 0
plug in 0
MSTang
2015-03-21 20:26:39
Compare constant terms to get $c$
Compare constant terms to get $c$
24iam24
2015-03-21 20:26:39
plug in 0 and solve for c
plug in 0 and solve for c
swe1
2015-03-21 20:26:39
equate constant terms
equate constant terms
Sciolympian
2015-03-21 20:26:39
solve for c
solve for c
DPatrick
2015-03-21 20:26:46
We can compute $f(0)$ using either expression:
\[
f(0) = c(-2)(-3)(-7) + 12 = -42c + 12,
\]
and
\[
f(0) = c(-1)(-5)(-6) - 12 = -30c - 12.
\]
We can compute $f(0)$ using either expression:
\[
f(0) = c(-2)(-3)(-7) + 12 = -42c + 12,
\]
and
\[
f(0) = c(-1)(-5)(-6) - 12 = -30c - 12.
\]
dtxiong
2015-03-21 20:27:11
-30c-12=-42c-12
-30c-12=-42c-12
ninjataco
2015-03-21 20:27:11
c=2
c=2
BFYSharks
2015-03-21 20:27:11
c=2
c=2
william122
2015-03-21 20:27:11
c=2
c=2
DPatrick
2015-03-21 20:27:12
But of course, these have to be equal!
So $-42c+12 = -30c -12$.
This gives $24 = 12c$, so $c=2$.
But of course, these have to be equal!
So $-42c+12 = -30c -12$.
This gives $24 = 12c$, so $c=2$.
nosaj
2015-03-21 20:27:35
so $c=2$ and $|f(0)|=\boxed{072}$, and we are done.
so $c=2$ and $|f(0)|=\boxed{072}$, and we are done.
maverick8
2015-03-21 20:27:35
Answer is 72
Answer is 72
xyuqing
2015-03-21 20:27:35
072
072
nosyarg
2015-03-21 20:27:35
072
072
DPatrick
2015-03-21 20:27:38
And thus $f(0) = -42(2) + 12 = -72$, so $|f(0)| = \boxed{072}$ is our answer.
And thus $f(0) = -42(2) + 12 = -72$, so $|f(0)| = \boxed{072}$ is our answer.
DPatrick
2015-03-21 20:27:50
(As a check, you can verify that $2(x-2)(x-3)(x-7)+12 = 2(x-1)(x-5)(x-6)-12 = 2x^3 - 24x^2 + 82x - 72$ is the cubic that works.)
(As a check, you can verify that $2(x-2)(x-3)(x-7)+12 = 2(x-1)(x-5)(x-6)-12 = 2x^3 - 24x^2 + 82x - 72$ is the cubic that works.)
DPatrick
2015-03-21 20:28:10
On to #11:
On to #11:
DPatrick
2015-03-21 20:28:14
11. Triangle $ABC$ has positive integer side lengths with $AB = AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI = 8$. Find the smallest possible perimeter of $\triangle ABC$.
11. Triangle $ABC$ has positive integer side lengths with $AB = AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI = 8$. Find the smallest possible perimeter of $\triangle ABC$.
DPatrick
2015-03-21 20:28:29
Of course, let's start with a diagram.
Of course, let's start with a diagram.
DPatrick
2015-03-21 20:28:33
DPatrick
2015-03-21 20:28:39
I've labeled $\angle IBC = x$ and the midpoint $M$ of the base $\overline{BC}$. Note that $I$ lies on $\overline{AM}$ because the triangle is isosceles.
I've labeled $\angle IBC = x$ and the midpoint $M$ of the base $\overline{BC}$. Note that $I$ lies on $\overline{AM}$ because the triangle is isosceles.
DPatrick
2015-03-21 20:29:00
What's $BC$ in terms of $x$?
What's $BC$ in terms of $x$?
brightknight
2015-03-21 20:29:25
16cos(x)
16cos(x)
ninjataco
2015-03-21 20:29:25
16cos x
16cos x
ryanyz10
2015-03-21 20:29:25
16cosx
16cosx
william122
2015-03-21 20:29:25
BC=16cosx
BC=16cosx
yimingz89
2015-03-21 20:29:25
$16\cos(x)$
$16\cos(x)$
C-bass
2015-03-21 20:29:25
16cosx
16cosx
dyang
2015-03-21 20:29:25
16cosx
16cosx
DPatrick
2015-03-21 20:29:30
We know that $\cos x = \frac{BM}{8}$ from right triangle $BMI$, so $BC = 2BM = 16 \cos x$.
We know that $\cos x = \frac{BM}{8}$ from right triangle $BMI$, so $BC = 2BM = 16 \cos x$.
crastybow
2015-03-21 20:30:00
since BC is an integer, cos x = k/16, with k an integer
since BC is an integer, cos x = k/16, with k an integer
DPatrick
2015-03-21 20:30:09
Exactly. Since $BC$ is a positive integer, we must have $\cos x = \frac{k}{16}$ for some $1 \le k \le 16$. (Then $BC = k$.)
Exactly. Since $BC$ is a positive integer, we must have $\cos x = \frac{k}{16}$ for some $1 \le k \le 16$. (Then $BC = k$.)
DPatrick
2015-03-21 20:30:22
But there's more! $0^\circ < x < 45^\circ$, so what bounds does that put on $k$?
But there's more! $0^\circ < x < 45^\circ$, so what bounds does that put on $k$?
crastybow
2015-03-21 20:31:02
12 <= k < 16
12 <= k < 16
wangth100
2015-03-21 20:31:02
$ k \ge 12 $ (since $ k $ is an integer)
$ k \ge 12 $ (since $ k $ is an integer)
DPatrick
2015-03-21 20:31:12
We must have $1 > \cos x > \frac{1}{\sqrt2} \approx 0.707$, so $k$ must be 12, 13, 14, or 15.
We must have $1 > \cos x > \frac{1}{\sqrt2} \approx 0.707$, so $k$ must be 12, 13, 14, or 15.
DPatrick
2015-03-21 20:31:27
Next: what is $AB$ is terms of $k$?
Next: what is $AB$ is terms of $k$?
DPatrick
2015-03-21 20:31:56
There are a few steps to this...maybe getting $AB$ in terms of $x$ is a start.
There are a few steps to this...maybe getting $AB$ in terms of $x$ is a start.
BFYSharks
2015-03-21 20:32:12
k/(2cos2x)
k/(2cos2x)
DPatrick
2015-03-21 20:32:20
Using triangle $ABM$, we have $\cos 2x = \dfrac{BM}{AB}$, so $AB = \dfrac{BM}{\cos 2x} = \dfrac{k}{2\cos 2x}$.
Using triangle $ABM$, we have $\cos 2x = \dfrac{BM}{AB}$, so $AB = \dfrac{BM}{\cos 2x} = \dfrac{k}{2\cos 2x}$.
xyuqing
2015-03-21 20:32:46
use cos (2x) = 2cos(x)^2-1
use cos (2x) = 2cos(x)^2-1
nosaj
2015-03-21 20:32:46
Use double angle formula
Use double angle formula
brainiac1
2015-03-21 20:32:46
and use the double angle formula
and use the double angle formula
C-bass
2015-03-21 20:32:46
double angle identity?
double angle identity?
brightknight
2015-03-21 20:32:46
cos2x=2cos^2-1
cos2x=2cos^2-1
william122
2015-03-21 20:32:46
cos2x=cos^2x-1!
cos2x=cos^2x-1!
DPatrick
2015-03-21 20:32:53
And then $\cos 2x = 2\cos^2 x - 1$ by the double-angle formula, so
\[
\cos 2x = 2\left(\frac{k}{16}\right)^2 - 1 = \frac{k^2 - 128}{128}.
\]
And then $\cos 2x = 2\cos^2 x - 1$ by the double-angle formula, so
\[
\cos 2x = 2\left(\frac{k}{16}\right)^2 - 1 = \frac{k^2 - 128}{128}.
\]
DPatrick
2015-03-21 20:33:08
This gives $AB = \dfrac{k}{2\cos 2x} = \dfrac{64k}{k^2 - 128}$.
This gives $AB = \dfrac{k}{2\cos 2x} = \dfrac{64k}{k^2 - 128}$.
DPatrick
2015-03-21 20:33:17
For which $k$ in 12, 13, 14, 15 is this an integer?
For which $k$ in 12, 13, 14, 15 is this an integer?
Darn
2015-03-21 20:33:35
12
12
zachman99323
2015-03-21 20:33:35
only $12$ works, which is silly
only $12$ works, which is silly
RedPhoenix
2015-03-21 20:33:35
12
12
akim99
2015-03-21 20:33:35
12
12
bli1999
2015-03-21 20:33:35
12
12
AkshajK
2015-03-21 20:33:35
12
12
DPatrick
2015-03-21 20:33:40
$k=12$ works! We get $AB = \dfrac{64(12)}{16} = 48$.
$k=12$ works! We get $AB = \dfrac{64(12)}{16} = 48$.
DPatrick
2015-03-21 20:33:50
I'll save us time and tell you that none of the other $k$'s give an integer for $AB$.
I'll save us time and tell you that none of the other $k$'s give an integer for $AB$.
DPatrick
2015-03-21 20:33:58
So not only does $k=12$ give the minimum perimeter -- it's actually the only triangle that works!
So not only does $k=12$ give the minimum perimeter -- it's actually the only triangle that works!
william122
2015-03-21 20:34:06
48*2+12=108
48*2+12=108
wangth100
2015-03-21 20:34:06
so the answer is $ \boxed{108} $.
so the answer is $ \boxed{108} $.
DPatrick
2015-03-21 20:34:10
The perimeter is $48 + 48 + 12 = \boxed{108}$.
The perimeter is $48 + 48 + 12 = \boxed{108}$.
xyuqing
2015-03-21 20:34:33
Why would they ask minimum?
Why would they ask minimum?
DPatrick
2015-03-21 20:34:54
Not sure why they phrased it that way. Their solution is essentially identical to what we just did.
Not sure why they phrased it that way. Their solution is essentially identical to what we just did.
DPatrick
2015-03-21 20:35:09
On to #12:
On to #12:
DPatrick
2015-03-21 20:35:12
12. Consider all 1000-element subsets of the set $\{1,2,3,\ldots,2015\}$. From each such subset choose the least element. The arithmetic mean of all of these least elements if $\frac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
12. Consider all 1000-element subsets of the set $\{1,2,3,\ldots,2015\}$. From each such subset choose the least element. The arithmetic mean of all of these least elements if $\frac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
AlcumusGuy
2015-03-21 20:35:44
Look at how many times each number is the minimum element.
Look at how many times each number is the minimum element.
nosaj
2015-03-21 20:35:44
Find the number of subsets with smallest element 1,2,3,...,1016.
Find the number of subsets with smallest element 1,2,3,...,1016.
summitwei
2015-03-21 20:35:44
Calculate # of subsets where min is 1, 2, 3, etc.
Calculate # of subsets where min is 1, 2, 3, etc.
swe1
2015-03-21 20:35:44
2015C1000 subsets total
2015C1000 subsets total
DPatrick
2015-03-21 20:35:52
Right. To start, there are $\dbinom{2015}{1000}$ ways to choose an 1000-element subset from a set of 2015 elements.
Right. To start, there are $\dbinom{2015}{1000}$ ways to choose an 1000-element subset from a set of 2015 elements.
DPatrick
2015-03-21 20:35:56
Let's hope we don't have to compute that number.
Let's hope we don't have to compute that number.
DPatrick
2015-03-21 20:36:08
We can count how many of them there are with each minimum element.
We can count how many of them there are with each minimum element.
DPatrick
2015-03-21 20:36:12
How many of them have 1 as the minimum element?
How many of them have 1 as the minimum element?
dyang
2015-03-21 20:36:30
2014 C 999
2014 C 999
blueberry7
2015-03-21 20:36:30
(2014 C 999)
(2014 C 999)
mjoshi
2015-03-21 20:36:30
2014C999
2014C999
ninjataco
2015-03-21 20:36:30
2014C999
2014C999
kingsave3166
2015-03-21 20:36:30
2014 choose 999
2014 choose 999
rocket13jg
2015-03-21 20:36:30
2014 choose 999
2014 choose 999
flyrain
2015-03-21 20:36:30
2014 choose 999
2014 choose 999
mssmath
2015-03-21 20:36:30
(2004 choose 999)
(2004 choose 999)
Epic777
2015-03-21 20:36:30
2014 choose 999
2014 choose 999
DPatrick
2015-03-21 20:36:35
We have to choose 999 more elements from $\{2,3,\ldots,2015\}$. So there are $\dbinom{2014}{999}$ of these sets.
We have to choose 999 more elements from $\{2,3,\ldots,2015\}$. So there are $\dbinom{2014}{999}$ of these sets.
DPatrick
2015-03-21 20:36:39
How many of them have 2 as the minimum element?
How many of them have 2 as the minimum element?
chaoshadow37
2015-03-21 20:36:56
2 as minimum there are 2013C999
2 as minimum there are 2013C999
ryanyz10
2015-03-21 20:36:56
2013C999
2013C999
morninglight
2015-03-21 20:36:56
2013C999
2013C999
dtxiong
2015-03-21 20:36:56
2013C999
2013C999
amburger66
2015-03-21 20:36:56
(2013C999)
(2013C999)
SHARKYBOY
2015-03-21 20:36:56
2013 choose 999
2013 choose 999
rlz
2015-03-21 20:36:56
$2013 \choose 999$
$2013 \choose 999$
nosyarg
2015-03-21 20:36:56
2013C999
2013C999
DPatrick
2015-03-21 20:37:01
We have to choose 999 more elements from $\{3,4,\ldots,2015\}$. So there are $\dbinom{2013}{999}$ of these sets.
We have to choose 999 more elements from $\{3,4,\ldots,2015\}$. So there are $\dbinom{2013}{999}$ of these sets.
akim99
2015-03-21 20:37:20
And so on all the way to $999 \choose 999$ for 1016
And so on all the way to $999 \choose 999$ for 1016
dtxiong
2015-03-21 20:37:20
pattern continues until 1016
pattern continues until 1016
DPatrick
2015-03-21 20:37:22
Similarly there are $\dbinom{2012}{999}$ subsets with 3 as the least element, $\dbinom{2011}{999}$ with 4 as the least element, and so on...
...down to just $\dbinom{999}{999} = 1$ with 1016 as the least element.
Similarly there are $\dbinom{2012}{999}$ subsets with 3 as the least element, $\dbinom{2011}{999}$ with 4 as the least element, and so on...
...down to just $\dbinom{999}{999} = 1$ with 1016 as the least element.
DPatrick
2015-03-21 20:37:44
In general, each least element $k$ gets counted $\dbinom{2015-k}{999}$ times, so the average is
\[
\frac{\dbinom{2014}{999}\cdot 1 + \dbinom{2013}{999} \cdot 2 + \dbinom{2012}{999} \cdot 3 + \cdots + \dbinom{999}{999} \cdot 1016}{\dbinom{2015}{1000}}.
\]
In general, each least element $k$ gets counted $\dbinom{2015-k}{999}$ times, so the average is
\[
\frac{\dbinom{2014}{999}\cdot 1 + \dbinom{2013}{999} \cdot 2 + \dbinom{2012}{999} \cdot 3 + \cdots + \dbinom{999}{999} \cdot 1016}{\dbinom{2015}{1000}}.
\]
DPatrick
2015-03-21 20:37:55
How can we compute that numerator? (And, hopefully, what we get will cancel with a lot of the denominator!)
How can we compute that numerator? (And, hopefully, what we get will cancel with a lot of the denominator!)
RedPhoenix
2015-03-21 20:38:25
We can use hockey-stick!
We can use hockey-stick!
blueberry7
2015-03-21 20:38:25
hockey stick
hockey stick
mjoshi
2015-03-21 20:38:25
Hockey Stick
Hockey Stick
BFYSharks
2015-03-21 20:38:25
Hockey Stick
Hockey Stick
summitwei
2015-03-21 20:38:25
Use hockey stick repeatedly
Use hockey stick repeatedly
AlcumusGuy
2015-03-21 20:38:25
Rewrite the sum in a clever way to exploit Hockey Stick.
Rewrite the sum in a clever way to exploit Hockey Stick.
brightknight
2015-03-21 20:38:25
Hockey-Stick Identity
Hockey-Stick Identity
eyux
2015-03-21 20:38:25
hockeystick!
hockeystick!
willabc
2015-03-21 20:38:25
hockey stick
hockey stick
morninglight
2015-03-21 20:38:25
Hockey Stick!
Hockey Stick!
ooooo1
2015-03-21 20:38:25
hockeystick
hockeystick
kingsave3166
2015-03-21 20:38:25
hockey stick identity
hockey stick identity
FTW
2015-03-21 20:38:25
hockey stick identity
hockey stick identity
Royalreter1
2015-03-21 20:38:25
Hockey Stick Theorem!
Hockey Stick Theorem!
DPatrick
2015-03-21 20:38:40
It's the coefficients that makes it tricky, because if we didn't have the 1, 2, 3, etc., we could use the Hockey Stick Identity!
It's the coefficients that makes it tricky, because if we didn't have the 1, 2, 3, etc., we could use the Hockey Stick Identity!
DPatrick
2015-03-21 20:38:45
\[
\dbinom{2014}{999} + \dbinom{2013}{999} + \dbinom{2012}{999} + \cdots + \dbinom{999}{999} = \dbinom{2015}{1000}.
\]
\[
\dbinom{2014}{999} + \dbinom{2013}{999} + \dbinom{2012}{999} + \cdots + \dbinom{999}{999} = \dbinom{2015}{1000}.
\]
DPatrick
2015-03-21 20:38:56
(If you don't know this identity, look it up after class, or try to figure it out on your own.)
(If you don't know this identity, look it up after class, or try to figure it out on your own.)
DPatrick
2015-03-21 20:39:05
But those coefficients are a problem...
But those coefficients are a problem...
DPatrick
2015-03-21 20:39:09
...or are they?
...or are they?
24iam24
2015-03-21 20:39:36
then you can repeatedly do it
then you can repeatedly do it
Jz2435
2015-03-21 20:39:36
no they are not
no they are not
C-bass
2015-03-21 20:39:36
no use it again
no use it again
csmath
2015-03-21 20:39:36
WAIT apply repeatedly taking one out each time
WAIT apply repeatedly taking one out each time
Wave-Particle
2015-03-21 20:39:36
use it over and over
use it over and over
william122
2015-03-21 20:39:36
we can make it into 1016 seperate hockey sticks whose sums combine into a larger hockey stick
we can make it into 1016 seperate hockey sticks whose sums combine into a larger hockey stick
brainiac1
2015-03-21 20:39:36
just subtract and reuse the identity
just subtract and reuse the identity
acegikmoqsuwy2000
2015-03-21 20:39:36
Just subtract that from the numerator, and then use hockey stick again
Just subtract that from the numerator, and then use hockey stick again
tongzhao
2015-03-21 20:39:36
hockey stick over and over
hockey stick over and over
Darn
2015-03-21 20:39:36
no just account for decrease of 1
no just account for decrease of 1
DPatrick
2015-03-21 20:39:48
Right! We could certainly take one term from each and use the Hockey Stick Identity; that leaves us with:
Right! We could certainly take one term from each and use the Hockey Stick Identity; that leaves us with:
DPatrick
2015-03-21 20:39:52
\[
\frac{\dbinom{2015}{1000} + \left(\dbinom{2014}{999}\cdot 0 + \dbinom{2013}{999} \cdot 1 + \dbinom{2012}{999} \cdot 2 + \cdots + \dbinom{999}{999} \cdot 1015\right)}{\dbinom{2015}{1000}}.
\]
\[
\frac{\dbinom{2015}{1000} + \left(\dbinom{2014}{999}\cdot 0 + \dbinom{2013}{999} \cdot 1 + \dbinom{2012}{999} \cdot 2 + \cdots + \dbinom{999}{999} \cdot 1015\right)}{\dbinom{2015}{1000}}.
\]
ryanyz10
2015-03-21 20:40:08
keep doing that
keep doing that
ninjataco
2015-03-21 20:40:08
hockey stick again!
hockey stick again!
william122
2015-03-21 20:40:08
DO IT AGAIN!
DO IT AGAIN!
DPatrick
2015-03-21 20:40:12
We peel one more term off of each (ignoring the first term that's 0), and that's another hockey stick:
\[
\dbinom{2013}{999} + \dbinom{2012}{999} + \cdots + \dbinom{999}{999} = \dbinom{2014}{2010}.
\]
We peel one more term off of each (ignoring the first term that's 0), and that's another hockey stick:
\[
\dbinom{2013}{999} + \dbinom{2012}{999} + \cdots + \dbinom{999}{999} = \dbinom{2014}{2010}.
\]
DPatrick
2015-03-21 20:40:20
So now we have (dropping the terms that become 0):
\[
\frac{\dbinom{2015}{1000} + \dbinom{2014}{1000} + \left(\dbinom{2012}{999} \cdot 1 + \dbinom{2011}{999} \cdot 2 + \cdots + \dbinom{999}{999} \cdot 1014\right)}{\dbinom{2015}{1000}}.
\]
So now we have (dropping the terms that become 0):
\[
\frac{\dbinom{2015}{1000} + \dbinom{2014}{1000} + \left(\dbinom{2012}{999} \cdot 1 + \dbinom{2011}{999} \cdot 2 + \cdots + \dbinom{999}{999} \cdot 1014\right)}{\dbinom{2015}{1000}}.
\]
chaoshadow37
2015-03-21 20:40:36
Just keep using hockey stick
Just keep using hockey stick
fluffyanimal
2015-03-21 20:40:36
moar
moar
crastybow
2015-03-21 20:40:36
do it again
do it again
MathPro2816
2015-03-21 20:40:36
Keep doing that.
Keep doing that.
DPatrick
2015-03-21 20:40:38
And we just keep turning the crank: we peel off one binomial coefficient from each term inside the big parentheses, and apply the Hockey Stick Identity to sum them.
And we just keep turning the crank: we peel off one binomial coefficient from each term inside the big parentheses, and apply the Hockey Stick Identity to sum them.
DPatrick
2015-03-21 20:40:45
Eventually, after applying all these hockey sticks, we end up with
\[
\frac{\dbinom{2015}{1000} + \dbinom{2014}{1000} + \dbinom{2013}{1000} + \cdots + \dbinom{1000}{1000}}{\dbinom{2015}{1000}}.
\]
Eventually, after applying all these hockey sticks, we end up with
\[
\frac{\dbinom{2015}{1000} + \dbinom{2014}{1000} + \dbinom{2013}{1000} + \cdots + \dbinom{1000}{1000}}{\dbinom{2015}{1000}}.
\]
summitwei
2015-03-21 20:40:59
And then you get one (gigantic) hockey stick
And then you get one (gigantic) hockey stick
AlcumusGuy
2015-03-21 20:40:59
One more Hockey Stick!
One more Hockey Stick!
BFYSharks
2015-03-21 20:40:59
One last time!
One last time!
hwl0304
2015-03-21 20:40:59
Hockeystick AGAIN
Hockeystick AGAIN
fz0718
2015-03-21 20:40:59
again
again
DPatrick
2015-03-21 20:41:04
It's yet one last hockey stick! We end up with $\dfrac{\dbinom{2016}{1001}}{\dbinom{2015}{1000}}$.
It's yet one last hockey stick! We end up with $\dfrac{\dbinom{2016}{1001}}{\dbinom{2015}{1000}}$.
chaoshadow37
2015-03-21 20:41:28
expand and cancel
expand and cancel
flyrain
2015-03-21 20:41:28
and that's 2016/1001
and that's 2016/1001
Epic777
2015-03-21 20:41:28
2016/1001
2016/1001
mssmath
2015-03-21 20:41:28
2016/1001
2016/1001
brainiac1
2015-03-21 20:41:28
almost everything cancels
almost everything cancels
Darn
2015-03-21 20:41:28
so it's 2016/1001
so it's 2016/1001
Reverse_Osmosis
2015-03-21 20:41:28
now you simplify
now you simplify
mjoshi
2015-03-21 20:41:28
Now rewrite as factorials
Now rewrite as factorials
DPatrick
2015-03-21 20:41:35
One safe way to simplify is to write it in terms of factorials: $\dfrac{\frac{2016!}{1001!1015!}}{\frac{2015!}{1000!1015!}}$.
One safe way to simplify is to write it in terms of factorials: $\dfrac{\frac{2016!}{1001!1015!}}{\frac{2015!}{1000!1015!}}$.
DPatrick
2015-03-21 20:41:43
The 1005!'s cancel, and most of the other factorials cancel, and we're left with just $\dfrac{2016}{1001}$.
The 1005!'s cancel, and most of the other factorials cancel, and we're left with just $\dfrac{2016}{1001}$.
lucasxia01
2015-03-21 20:42:07
simplify to 288/143
simplify to 288/143
blueberry7
2015-03-21 20:42:07
288/143
288/143
rocket13jg
2015-03-21 20:42:07
288/143
288/143
lolcreative883
2015-03-21 20:42:07
divide by 7?
divide by 7?
DPatrick
2015-03-21 20:42:15
Canceling a factor of 7 leaves us with $\dfrac{288}{143}$ in lowest terms, so our answer is $288 + 143 = \boxed{431}$.
Canceling a factor of 7 leaves us with $\dfrac{288}{143}$ in lowest terms, so our answer is $288 + 143 = \boxed{431}$.
donot
2015-03-21 20:42:38
If you forgot Hockey Stick, is it even possible to solve this problem?
If you forgot Hockey Stick, is it even possible to solve this problem?
DPatrick
2015-03-21 20:43:06
There is a really clever trick to sum it in one swoop. Hint: we end up with $\dbinom{2016}{1001}$, so what's a counting argument for where that number comes from?
There is a really clever trick to sum it in one swoop. Hint: we end up with $\dbinom{2016}{1001}$, so what's a counting argument for where that number comes from?
DPatrick
2015-03-21 20:43:14
I'll leave that for you to ponder.
I'll leave that for you to ponder.
DPatrick
2015-03-21 20:43:38
Although I like the problem, it is a pretty well-known problem. It was #2 on the 1981 IMO (but with general $n$ and $r$ instead of 2015 and 1000).
Although I like the problem, it is a pretty well-known problem. It was #2 on the 1981 IMO (but with general $n$ and $r$ instead of 2015 and 1000).
DPatrick
2015-03-21 20:43:56
Speaking of which, on to #13:
Speaking of which, on to #13:
DPatrick
2015-03-21 20:44:02
13. With all angles measured in degrees, the product $\displaystyle\prod_{k=1}^{45} \csc^2(2k-1)^{\circ} = m^n$, where $m$ and $n$ are integers greater than 1. Find $m + n$.
13. With all angles measured in degrees, the product $\displaystyle\prod_{k=1}^{45} \csc^2(2k-1)^{\circ} = m^n$, where $m$ and $n$ are integers greater than 1. Find $m + n$.
brainiac1
2015-03-21 20:44:14
this is slightly conceptually similar to problem number 3.32 in the Aops Precalculus
this is slightly conceptually similar to problem number 3.32 in the Aops Precalculus
nosaj
2015-03-21 20:44:16
Wasn't this in AoPS Precalc?
Wasn't this in AoPS Precalc?
spartan168
2015-03-21 20:44:20
omg the one from the precalc book
omg the one from the precalc book
DPatrick
2015-03-21 20:44:35
Yes, this is almost identical to Problem 3.32 from our Precalculus textbook.
Yes, this is almost identical to Problem 3.32 from our Precalculus textbook.
DPatrick
2015-03-21 20:44:50
So naturally, the solution I'm going to lead us along is pretty much the same as the one from the book.
So naturally, the solution I'm going to lead us along is pretty much the same as the one from the book.
chezbgone
2015-03-21 20:45:06
rewrite as product of reciprocals of sines
rewrite as product of reciprocals of sines
DPatrick
2015-03-21 20:45:18
Yes, plus the product notation is a little hard to deal with, so let's write out what this really is:
\[
\frac{1}{(\sin^2 1)(\sin^2 3)(\sin^2 5)\cdots(\sin^2 87)(\sin^2 89)}.
\]
(I'm not going to write all the degree signs in this problem. Just remember that we're working in degrees.)
Yes, plus the product notation is a little hard to deal with, so let's write out what this really is:
\[
\frac{1}{(\sin^2 1)(\sin^2 3)(\sin^2 5)\cdots(\sin^2 87)(\sin^2 89)}.
\]
(I'm not going to write all the degree signs in this problem. Just remember that we're working in degrees.)
DPatrick
2015-03-21 20:45:37
There are some false trails you can wander down in this problem, but it's the type of problem that if you apply enough trig identities, you'll hopefully end up at the answer.
There are some false trails you can wander down in this problem, but it's the type of problem that if you apply enough trig identities, you'll hopefully end up at the answer.
DPatrick
2015-03-21 20:45:46
We eventually want to try to get things to cancel out so that we're left with just an integer. One place to start is to try to get rid of the squares.
We eventually want to try to get things to cancel out so that we're left with just an integer. One place to start is to try to get rid of the squares.
C-bass
2015-03-21 20:46:09
sinx = cos(90-x)
sinx = cos(90-x)
High
2015-03-21 20:46:09
sin1 = cos 89
sin1 = cos 89
chaoshadow37
2015-03-21 20:46:15
will changing sin 89 to cos 1 help (and sin97 to cos 3 and sin 85 to cos 5)?
will changing sin 89 to cos 1 help (and sin97 to cos 3 and sin 85 to cos 5)?
DPatrick
2015-03-21 20:46:23
Let's do that: let's replace one sine in each $\sin^2 \theta$ term with $\cos(90-\theta)$:
Let's do that: let's replace one sine in each $\sin^2 \theta$ term with $\cos(90-\theta)$:
DPatrick
2015-03-21 20:46:27
\[
\frac{1}{(\sin 1)(\cos 89)(\sin 3)(\cos 87)\cdots(\sin 87)(\cos 3)(\sin 89)(\cos 1)}.
\]
\[
\frac{1}{(\sin 1)(\cos 89)(\sin 3)(\cos 87)\cdots(\sin 87)(\cos 3)(\sin 89)(\cos 1)}.
\]
nosyarg
2015-03-21 20:46:54
double angles?
double angles?
Sciolympian
2015-03-21 20:46:54
sin 89 cos 89!
sin 89 cos 89!
mathtastic
2015-03-21 20:46:54
double angle that
double angle that
mjoshi
2015-03-21 20:46:54
pair them up now
pair them up now
xyuqing
2015-03-21 20:46:54
Double angles?
Double angles?
AlcumusGuy
2015-03-21 20:46:54
now pair up the terms and use double-angle formula
now pair up the terms and use double-angle formula
DPatrick
2015-03-21 20:47:04
Good idea. The double-angle formula for sine is $\sin 2x = 2(\sin x)(\cos x)$. So $\dfrac{1}{(\sin x)(\cos x)} = \dfrac{2}{\sin 2x}$.
Good idea. The double-angle formula for sine is $\sin 2x = 2(\sin x)(\cos x)$. So $\dfrac{1}{(\sin x)(\cos x)} = \dfrac{2}{\sin 2x}$.
DPatrick
2015-03-21 20:47:14
That's helpful. We do this for all 45 pairs and we get that our expression is equal to
\[
\frac{2^{45}}{(\sin 2)(\sin 6)\cdots(\sin 174)(\sin 178)}.
\]
That's helpful. We do this for all 45 pairs and we get that our expression is equal to
\[
\frac{2^{45}}{(\sin 2)(\sin 6)\cdots(\sin 174)(\sin 178)}.
\]
DPatrick
2015-03-21 20:47:27
The $2^{45}$ is probably a good sign (no pun intended) -- we need a power of an integer at the end.
The $2^{45}$ is probably a good sign (no pun intended) -- we need a power of an integer at the end.
crastybow
2015-03-21 20:47:44
sin(178) = sin(2)
sin(178) = sin(2)
MSTang
2015-03-21 20:47:44
sin 178 = sin 2, so we get squares again
sin 178 = sin 2, so we get squares again
chrislee777
2015-03-21 20:47:44
sin x = sin (180-x)
sin x = sin (180-x)
DPatrick
2015-03-21 20:47:48
Angles bigger than 90 seem unnecessary -- let's replace $\sin 178$ with $\sin 2$, $\sin 174$ with $\sin 6$, and so on.
Angles bigger than 90 seem unnecessary -- let's replace $\sin 178$ with $\sin 2$, $\sin 174$ with $\sin 6$, and so on.
DPatrick
2015-03-21 20:48:10
Note that the $\sin 90 = 1$ in the middle just goes away, and now our expression is
\[
\frac{2^{45}}{(\sin^2 2)(\sin^2 6)\cdots(\sin^2 82)(\sin^2 86)}.
\]
(Note there are 22 squares in the denominator now.)
Note that the $\sin 90 = 1$ in the middle just goes away, and now our expression is
\[
\frac{2^{45}}{(\sin^2 2)(\sin^2 6)\cdots(\sin^2 82)(\sin^2 86)}.
\]
(Note there are 22 squares in the denominator now.)
DPatrick
2015-03-21 20:48:18
This is pretty similar to what we started with, but we've only got 22 terms now instead of 45, so this seems like progress.
This is pretty similar to what we started with, but we've only got 22 terms now instead of 45, so this seems like progress.
Wave-Particle
2015-03-21 20:48:29
we can do it again!
we can do it again!
tongzhao
2015-03-21 20:48:29
replace with cosine again?
replace with cosine again?
blueberry7
2015-03-21 20:48:29
do it again
do it again
acegikmoqsuwy2000
2015-03-21 20:48:29
Do it again
Do it again
blueberry7
2015-03-21 20:48:29
do the same thing
do the same thing
ryanyz10
2015-03-21 20:48:29
same thing again?
same thing again?
DPatrick
2015-03-21 20:48:39
Good idea, but the same "convert one of the sines to a cosine" trick we had before won't work in the same way as it did before, because $\sin 86$ converts to $\cos 4$, and there's no $\sin 4$ term to match it up to.
Good idea, but the same "convert one of the sines to a cosine" trick we had before won't work in the same way as it did before, because $\sin 86$ converts to $\cos 4$, and there's no $\sin 4$ term to match it up to.
DPatrick
2015-03-21 20:49:01
But it still feels like we want to use the $\sin 2x = 2 \sin x \cos x$ identity somehow to create more cancellation.
But it still feels like we want to use the $\sin 2x = 2 \sin x \cos x$ identity somehow to create more cancellation.
DPatrick
2015-03-21 20:49:35
One idea is to just apply it to all the squares.
One idea is to just apply it to all the squares.
ninjataco
2015-03-21 20:49:41
sin x = sin 2x /(2cosx)
sin x = sin 2x /(2cosx)
DPatrick
2015-03-21 20:49:47
Exactly. Let's try it as
\[
\dfrac{1}{\sin x} = \dfrac{2 \cos x}{\sin 2x},
\]
so that
\[
\dfrac{1}{\sin^2 x} = \dfrac{2^2 \cos^2 x}{\sin^2 2x}.
\]
Exactly. Let's try it as
\[
\dfrac{1}{\sin x} = \dfrac{2 \cos x}{\sin 2x},
\]
so that
\[
\dfrac{1}{\sin^2 x} = \dfrac{2^2 \cos^2 x}{\sin^2 2x}.
\]
DPatrick
2015-03-21 20:50:00
We apply this 22 times, so this makes our expression
\[
\frac{2^{89}(\cos^2 2)(\cos^2 6)\cdots(\cos^2 82)(\cos^2 86)}{(\sin^2 4)(\sin^2 12)\cdots(\sin^2 164)(\sin^2 172)}.
\]
We apply this 22 times, so this makes our expression
\[
\frac{2^{89}(\cos^2 2)(\cos^2 6)\cdots(\cos^2 82)(\cos^2 86)}{(\sin^2 4)(\sin^2 12)\cdots(\sin^2 164)(\sin^2 172)}.
\]
DPatrick
2015-03-21 20:50:20
Did this help?
Did this help?
ryanyz10
2015-03-21 20:50:32
replace angles greater than 90
replace angles greater than 90
mjoshi
2015-03-21 20:50:32
change the 172 to 8 and so on
change the 172 to 8 and so on
acegikmoqsuwy2000
2015-03-21 20:50:32
sin^2(172) = sin^2(8), etc
sin^2(172) = sin^2(8), etc
Sciolympian
2015-03-21 20:50:32
and convert the angles greater than 90 again
and convert the angles greater than 90 again
swe1
2015-03-21 20:50:32
get rid of sines bigger than 90
get rid of sines bigger than 90
william122
2015-03-21 20:50:32
replace sins above 90
replace sins above 90
DPatrick
2015-03-21 20:50:41
Again, angles bigger than 90 are harder to work with, so let's make all the angles acute. Note that $\sin^2 172 = \sin^2 8$, $\sin^2 164 = \sin^2 16$, etc., so that we fill in all the missing multiples of 4 between 0 and 90.
Again, angles bigger than 90 are harder to work with, so let's make all the angles acute. Note that $\sin^2 172 = \sin^2 8$, $\sin^2 164 = \sin^2 16$, etc., so that we fill in all the missing multiples of 4 between 0 and 90.
DPatrick
2015-03-21 20:50:47
We now have:
\[
\frac{2^{89}(\cos^2 2)(\cos^2 6)\cdots(\cos^2 82)(\cos^2 86)}{(\sin^2 4)(\sin^2 8)\cdots(\sin^2 84)(\sin^2 88)}.
\]
We now have:
\[
\frac{2^{89}(\cos^2 2)(\cos^2 6)\cdots(\cos^2 82)(\cos^2 86)}{(\sin^2 4)(\sin^2 8)\cdots(\sin^2 84)(\sin^2 88)}.
\]
HYP135peppers
2015-03-21 20:51:13
Now use sin(x)=cos(90-x) on numerator
Now use sin(x)=cos(90-x) on numerator
bluephoenix
2015-03-21 20:51:13
They cancel because cos(90-x) = sinx
They cancel because cos(90-x) = sinx
Studiosa
2015-03-21 20:51:13
things can cancel now if you do sin becomes cos thing
things can cancel now if you do sin becomes cos thing
AlcumusGuy
2015-03-21 20:51:13
Now all the sin/cos's cancel!
Now all the sin/cos's cancel!
crastybow
2015-03-21 20:51:13
hey look sin(88) = cos(2)
hey look sin(88) = cos(2)
zxz
2015-03-21 20:51:13
it cancels!
it cancels!
crastybow
2015-03-21 20:51:13
so everything cancels
so everything cancels
kingsave3166
2015-03-21 20:51:13
they match up!
they match up!
chaoshadow37
2015-03-21 20:51:13
cos 86=sin4 so we can cancel
cos 86=sin4 so we can cancel
ooooo1
2015-03-21 20:51:13
sin88=cos2
sin88=cos2
jam10307
2015-03-21 20:51:13
they cancel
they cancel
High
2015-03-21 20:51:13
cos 2 = sin 88, cos 6 = sin 84 etc. they all cancel
cos 2 = sin 88, cos 6 = sin 84 etc. they all cancel
ninjataco
2015-03-21 20:51:13
rewrite the sines as cosines, or the cosines as sines, and they all cancel!
rewrite the sines as cosines, or the cosines as sines, and they all cancel!
DPatrick
2015-03-21 20:51:21
We win: everything cancels now! $\cos^2 2 = \sin^2 88$, and $\cos^2 6 = \sin^2 84$, and so on, all the way down through to $\cos^2 86 = \sin^2 4$.
We win: everything cancels now! $\cos^2 2 = \sin^2 88$, and $\cos^2 6 = \sin^2 84$, and so on, all the way down through to $\cos^2 86 = \sin^2 4$.
DPatrick
2015-03-21 20:51:30
So our expression is just $2^{89}$, and the final answer is $2 + 89 = \boxed{091}$.
So our expression is just $2^{89}$, and the final answer is $2 + 89 = \boxed{091}$.
DPatrick
2015-03-21 20:52:08
OK, on to #14:
OK, on to #14:
DPatrick
2015-03-21 20:52:13
14. For each integer $n\ge 2$, let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\leq x < n$ and $0 \le y \le x \lfloor \sqrt{x} \rfloor$, where $\lfloor \sqrt{x} \rfloor$ is the greatest integer not exceeding $\sqrt{x}$. Find the number of values of $n$ with $2 \le n \le 1000$ for which $A(n)$ is an integer.
14. For each integer $n\ge 2$, let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\leq x < n$ and $0 \le y \le x \lfloor \sqrt{x} \rfloor$, where $\lfloor \sqrt{x} \rfloor$ is the greatest integer not exceeding $\sqrt{x}$. Find the number of values of $n$ with $2 \le n \le 1000$ for which $A(n)$ is an integer.
DPatrick
2015-03-21 20:52:37
Ick. This is a classic AIME ugly bookkeeping problem.
Ick. This is a classic AIME ugly bookkeeping problem.
Sciolympian
2015-03-21 20:52:55
draw a graph first!
draw a graph first!
brainiac1
2015-03-21 20:52:55
graph it
graph it
xyuqing
2015-03-21 20:52:55
Draw a picture?
Draw a picture?
summitwei
2015-03-21 20:52:55
Graph it?
Graph it?
DPatrick
2015-03-21 20:52:59
Maybe sketching a graph of the region is a good place to start.
Maybe sketching a graph of the region is a good place to start.
DPatrick
2015-03-21 20:53:03
What does the graph of $y = x\lfloor\sqrt{x}\rfloor$ look like?
What does the graph of $y = x\lfloor\sqrt{x}\rfloor$ look like?
MSTang
2015-03-21 20:53:33
Bunch of line segments with increasing slopes
Bunch of line segments with increasing slopes
zxz
2015-03-21 20:53:33
bunch of line segments
bunch of line segments
flyrain
2015-03-21 20:53:33
a graph with a bunch of gradually increasing steps
a graph with a bunch of gradually increasing steps
summitwei
2015-03-21 20:53:33
A bunch of lines, gradually increasing in slope
A bunch of lines, gradually increasing in slope
RedPhoenix
2015-03-21 20:53:33
several trapezoids
several trapezoids
AlcumusGuy
2015-03-21 20:53:33
A collection of trapezoids
A collection of trapezoids
C-bass
2015-03-21 20:53:33
broken lines
broken lines
DPatrick
2015-03-21 20:53:37
It's $y = x$ for $1 \le x < 4$.
It's $y = 2x$ for $4 \le x < 9$.
It's $y = 3x$ for $9 \le x < 16$.
And so on.
It's $y = x$ for $1 \le x < 4$.
It's $y = 2x$ for $4 \le x < 9$.
It's $y = 3x$ for $9 \le x < 16$.
And so on.
DPatrick
2015-03-21 20:53:44
So it's a bunch of line segments:
So it's a bunch of line segments:
DPatrick
2015-03-21 20:53:47
DPatrick
2015-03-21 20:54:02
How does this help us compute $A(n)$?
How does this help us compute $A(n)$?
raxu
2015-03-21 20:54:19
The area is a bunch of trapezoids.
The area is a bunch of trapezoids.
kingsave3166
2015-03-21 20:54:19
trapezoid formula
trapezoid formula
Studiosa
2015-03-21 20:54:19
trapezoids
trapezoids
Tommy2000
2015-03-21 20:54:19
Find sum of area of trapezoids
Find sum of area of trapezoids
Sciolympian
2015-03-21 20:54:19
we need the integer areas
we need the integer areas
zxz
2015-03-21 20:54:19
trapezoids
trapezoids
DPatrick
2015-03-21 20:54:30
Well...keep in mind we don't actually need to compute $A(n)$ -- we just need to know when it's an integer and when it's not.
Well...keep in mind we don't actually need to compute $A(n)$ -- we just need to know when it's an integer and when it's not.
DPatrick
2015-03-21 20:54:36
Let's zoom in a little bit in our picture:
Let's zoom in a little bit in our picture:
DPatrick
2015-03-21 20:54:40
DPatrick
2015-03-21 20:55:13
Let's abbreviate $s(x) = \lfloor \sqrt{x} \rfloor$ so that it's easier to type. What's the area between $x = n$ and $x=n+1$ (for any integer $n \ge 1$)?
Let's abbreviate $s(x) = \lfloor \sqrt{x} \rfloor$ so that it's easier to type. What's the area between $x = n$ and $x=n+1$ (for any integer $n \ge 1$)?
DPatrick
2015-03-21 20:55:41
Note that this region is a trapezoid of "height" 1 (in the $x$-direction) and parallel sides of length $ns(n)$ and $(n+1)s(n)$. For example, I've shaded the regions for $n=3$ (in pink) and $n=6$ (in blue) below:
Note that this region is a trapezoid of "height" 1 (in the $x$-direction) and parallel sides of length $ns(n)$ and $(n+1)s(n)$. For example, I've shaded the regions for $n=3$ (in pink) and $n=6$ (in blue) below:
DPatrick
2015-03-21 20:55:44
raxu
2015-03-21 20:56:27
(ns(n)+(n+1)s(n))/2
(ns(n)+(n+1)s(n))/2
swe1
2015-03-21 20:56:27
s(n) * (n+1/2)
s(n) * (n+1/2)
DPatrick
2015-03-21 20:56:37
Right, the area in one column is
\[
\frac12(ns(n)+(n+1)s(n)) = \frac12(2n+1)s(n).
\]
Right, the area in one column is
\[
\frac12(ns(n)+(n+1)s(n)) = \frac12(2n+1)s(n).
\]
DPatrick
2015-03-21 20:57:02
Again, that's the area of the trapezoidal column between $x=n$ and $x=n+1$.
Again, that's the area of the trapezoidal column between $x=n$ and $x=n+1$.
DPatrick
2015-03-21 20:57:06
What do we know about this quantity?
What do we know about this quantity?
morninglight
2015-03-21 20:57:19
2n+1 is odd
2n+1 is odd
MSTang
2015-03-21 20:57:22
Either int or int + 1/2
Either int or int + 1/2
raxu
2015-03-21 20:57:27
Whether that area is integer or not depends on the parity of s(n), since (2n+1) is odd
Whether that area is integer or not depends on the parity of s(n), since (2n+1) is odd
vinayak-kumar
2015-03-21 20:57:27
Integer when s(n) is even, and a half integer otherwise
Integer when s(n) is even, and a half integer otherwise
akim99
2015-03-21 20:57:27
To be an integer, s(n) is even
To be an integer, s(n) is even
DPatrick
2015-03-21 20:57:36
$(2n+1)$ is always an odd integer, so this is an integer if $s(n)$ is even, and it's a half-integer if $s(n)$ is odd.
$(2n+1)$ is always an odd integer, so this is an integer if $s(n)$ is even, and it's a half-integer if $s(n)$ is odd.
DPatrick
2015-03-21 20:58:01
Let's introduce some helpful terminology: let's say that $n$ is good if $A(n)$ is an integer, and bad if it's not. We're trying to count all the good integers.
Let's introduce some helpful terminology: let's say that $n$ is good if $A(n)$ is an integer, and bad if it's not. We're trying to count all the good integers.
DPatrick
2015-03-21 20:58:13
So now we have the formula
\[
A(n+1) = A(n) + \frac12(2n+1)s(n).
\]
So now we have the formula
\[
A(n+1) = A(n) + \frac12(2n+1)s(n).
\]
DPatrick
2015-03-21 20:58:38
So the conclusion so far is: $A(n+1)$'s "goodness" stays the same as $A(n)$ if $s(n)$ is even, and flips (good to bad or vice versa) if $s(n)$ is odd.
So the conclusion so far is: $A(n+1)$'s "goodness" stays the same as $A(n)$ if $s(n)$ is even, and flips (good to bad or vice versa) if $s(n)$ is odd.
summitwei
2015-03-21 20:58:50
Casework on the value of s(n)
Casework on the value of s(n)
DPatrick
2015-03-21 20:58:56
Right. We've made the key observation. From here on out, it's just bookkeeping.
Right. We've made the key observation. From here on out, it's just bookkeeping.
DPatrick
2015-03-21 20:59:06
Let's start making a chart.
Let's start making a chart.
DPatrick
2015-03-21 20:59:18
For what values of $n$ is $s(n) = 1$?
For what values of $n$ is $s(n) = 1$?
akim99
2015-03-21 20:59:33
What's bookkepping?
What's bookkepping?
william122
2015-03-21 20:59:33
what do you mean by BookKeeping?
what do you mean by BookKeeping?
DPatrick
2015-03-21 20:59:57
By "bookkeeping" I just mean carefully collecting a lot of data, and keeping track of it carefully to get our answer.
By "bookkeeping" I just mean carefully collecting a lot of data, and keeping track of it carefully to get our answer.
MSTang
2015-03-21 21:00:16
1, 2, 3
1, 2, 3
BFYSharks
2015-03-21 21:00:16
1-3
1-3
summitwei
2015-03-21 21:00:16
1,2,3
1,2,3
brainiac1
2015-03-21 21:00:16
1, 2, 3
1, 2, 3
C-bass
2015-03-21 21:00:16
1,2,3
1,2,3
jameswangisb
2015-03-21 21:00:16
1,2,3
1,2,3
mssmath
2015-03-21 21:00:16
1,2,3
1,2,3
nosyarg
2015-03-21 21:00:16
1,2,3
1,2,3
Th3Numb3rThr33
2015-03-21 21:00:16
1,2,3?
1,2,3?
DPatrick
2015-03-21 21:00:24
Right. $s(n) = 1$ for $n=1,2,3$.
Right. $s(n) = 1$ for $n=1,2,3$.
DPatrick
2015-03-21 21:00:41
Since $s(n)$ is odd, the goodness flips for every $n$. That is:
$A(2)$ is bad
$A(3)$ is good
$A(4)$ is bad.
Since $s(n)$ is odd, the goodness flips for every $n$. That is:
$A(2)$ is bad
$A(3)$ is good
$A(4)$ is bad.
DPatrick
2015-03-21 21:01:04
\[\begin{array}{c|c|c}
s(n) & n & \text{number of good } A(n) \\ \hline
1 & 1\ldots3 & 1
\end{array}\]
\[\begin{array}{c|c|c}
s(n) & n & \text{number of good } A(n) \\ \hline
1 & 1\ldots3 & 1
\end{array}\]
raxu
2015-03-21 21:01:14
A(5) to A(9) are all bad.
A(5) to A(9) are all bad.
FractalMathHistory
2015-03-21 21:01:14
$A(5) \rightarrow A(9)$ is bad
$A(5) \rightarrow A(9)$ is bad
DPatrick
2015-03-21 21:01:48
Right. For $n=4\ldots 8$, we have $s(n) =2$, so the "goodness" of the $A$'s doesn't change.
Right. For $n=4\ldots 8$, we have $s(n) =2$, so the "goodness" of the $A$'s doesn't change.
DPatrick
2015-03-21 21:01:59
Since $A(4)$ is bad, all of $A(5)\ldots A(9)$ are bad too.
Since $A(4)$ is bad, all of $A(5)\ldots A(9)$ are bad too.
DPatrick
2015-03-21 21:02:03
\[\begin{array}{c|c|c}
s(n) & n & \text{number of good } A(n) \\ \hline
1 & 1\ldots3 & 1 \\
2 & 4\ldots8 & 0
\end{array}\]
\[\begin{array}{c|c|c}
s(n) & n & \text{number of good } A(n) \\ \hline
1 & 1\ldots3 & 1 \\
2 & 4\ldots8 & 0
\end{array}\]
DPatrick
2015-03-21 21:02:47
It's a little confusing, since it's the parity of $s(n)$ that determines whether $A(n+1)$ "flips" or not. That +1 can be confusing, especially when we get to the end.
It's a little confusing, since it's the parity of $s(n)$ that determines whether $A(n+1)$ "flips" or not. That +1 can be confusing, especially when we get to the end.
brainiac1
2015-03-21 21:03:01
A(10, 12, 14, 16) are all good
A(10, 12, 14, 16) are all good
xyuqing
2015-03-21 21:03:01
10 good 11 bad 12 good ...
10 good 11 bad 12 good ...
summitwei
2015-03-21 21:03:01
Then from 9 to 15 it alternates again
Then from 9 to 15 it alternates again
DPatrick
2015-03-21 21:03:12
$n = 9\ldots15$ satisfy $s(n) = 3$, and the $A'$s start flipping again.
$n = 9\ldots15$ satisfy $s(n) = 3$, and the $A'$s start flipping again.
DPatrick
2015-03-21 21:03:20
We're coming in to this block with a bad $A(9)$, so $A(10),A(12),A(14),A(16)$ are all good.
We're coming in to this block with a bad $A(9)$, so $A(10),A(12),A(14),A(16)$ are all good.
DPatrick
2015-03-21 21:03:24
\[\begin{array}{c|c|c}
s(n) & n & \text{number of good } A(n) \\ \hline
1 & 1\ldots3 & 1 \\
2 & 4\ldots8 & 0 \\
3 & 9\ldots15 & 4
\end{array}\]
\[\begin{array}{c|c|c}
s(n) & n & \text{number of good } A(n) \\ \hline
1 & 1\ldots3 & 1 \\
2 & 4\ldots8 & 0 \\
3 & 9\ldots15 & 4
\end{array}\]
DPatrick
2015-03-21 21:03:35
How about $s(n) = 4$?
How about $s(n) = 4$?
RedPhoenix
2015-03-21 21:03:54
then 16-24 are all good
then 16-24 are all good
ryanyz10
2015-03-21 21:03:54
n = 16 - 24 are all good
n = 16 - 24 are all good
dantx5
2015-03-21 21:03:54
all good
all good
raxu
2015-03-21 21:03:54
And $A(17)\to A(25)$ are all good!
And $A(17)\to A(25)$ are all good!
nosaj
2015-03-21 21:03:54
all good
all good
Sciolympian
2015-03-21 21:03:54
all good
all good
bli1999
2015-03-21 21:03:54
A(16)->A(25) are all good
A(16)->A(25) are all good
mathbeida
2015-03-21 21:03:54
16 to 24 not changing pairity
16 to 24 not changing pairity
DPatrick
2015-03-21 21:04:08
$n = 16\ldots24$, and since $A(16)$ is good, all the $A$'s are good in this block are good too:
$n = 16\ldots24$, and since $A(16)$ is good, all the $A$'s are good in this block are good too:
DPatrick
2015-03-21 21:04:12
\[\begin{array}{c|c|c}
s(n) & n & \text{number of good } A(n) \\ \hline
1 & 1\ldots3 & 1 \\
2 & 4\ldots8 & 0 \\
3 & 9\ldots15 & 4 \\
4 & 16\ldots24 & 9
\end{array}\]
\[\begin{array}{c|c|c}
s(n) & n & \text{number of good } A(n) \\ \hline
1 & 1\ldots3 & 1 \\
2 & 4\ldots8 & 0 \\
3 & 9\ldots15 & 4 \\
4 & 16\ldots24 & 9
\end{array}\]
DPatrick
2015-03-21 21:04:23
So now we see the pattern.
In an odd $s(n)$ block, every other $A(n)$ is good. If we enter on a good $A(n)$, then we leave on a bad $A(n)$, and vice versa.
In an even block, all the $A(n)$ are either good or bad.
So now we see the pattern.
In an odd $s(n)$ block, every other $A(n)$ is good. If we enter on a good $A(n)$, then we leave on a bad $A(n)$, and vice versa.
In an even block, all the $A(n)$ are either good or bad.
DPatrick
2015-03-21 21:04:38
More specifically, there's a cycle of 4:
If $n \equiv 1 \pmod{4}$, then we start on good, so half (rounded DOWN) are good, and we leave on bad.
If $n \equiv 2 \pmod{4}$, then we're all bad
If $n \equiv 3 \pmod{4}$, then we start on bad, so half (rounded UP) are good, and we leave on good.
If $n \equiv 0 \pmod{4}$, then all are good.
More specifically, there's a cycle of 4:
If $n \equiv 1 \pmod{4}$, then we start on good, so half (rounded DOWN) are good, and we leave on bad.
If $n \equiv 2 \pmod{4}$, then we're all bad
If $n \equiv 3 \pmod{4}$, then we start on bad, so half (rounded UP) are good, and we leave on good.
If $n \equiv 0 \pmod{4}$, then all are good.
DPatrick
2015-03-21 21:05:07
So now it's just a matter of extending our chart all the way up to $n=1000$. Since $s(1000) = 31$ we'll need 31 rows unless we see a more clever pattern.
So now it's just a matter of extending our chart all the way up to $n=1000$. Since $s(1000) = 31$ we'll need 31 rows unless we see a more clever pattern.
akim99
2015-03-21 21:05:31
Squares!
Squares!
tongzhao
2015-03-21 21:05:31
squares?
squares?
nosyarg
2015-03-21 21:05:31
the squares?
the squares?
xyuqing
2015-03-21 21:05:31
Wait, 1^2=1, 2^2=4, 3^2=9?
Wait, 1^2=1, 2^2=4, 3^2=9?
BFYSharks
2015-03-21 21:05:31
Perfect squares?
Perfect squares?
DPatrick
2015-03-21 21:05:37
Good guess, but sadly no.
Good guess, but sadly no.
DPatrick
2015-03-21 21:05:41
Let's at least do the next 4 rows. I'm going to add the method of counting to our chart:
Let's at least do the next 4 rows. I'm going to add the method of counting to our chart:
DPatrick
2015-03-21 21:05:55
\[\begin{array}{c|c|c|c}
s(n) & n & \text{count} & \text{number of good } A(n) \\ \hline
1 & 1\ldots3 & \text{half rounded DOWN} & 1 \\
2 & 4\ldots8 & \text{none} & 0 \\
3 & 9\ldots15 & \text{half rounded UP} & 4 \\
4 & 16\ldots24 & \text{all} & 9 \\
5 & 25\ldots35 & \text{half rounded DOWN} & 5 \\
6 & 36\ldots48 & \text{none} & 0 \\
7 & 49\ldots63 & \text{half rounded UP} & 8 \\
8 & 64\ldots80 & \text{all} & 17
\end{array}\]
\[\begin{array}{c|c|c|c}
s(n) & n & \text{count} & \text{number of good } A(n) \\ \hline
1 & 1\ldots3 & \text{half rounded DOWN} & 1 \\
2 & 4\ldots8 & \text{none} & 0 \\
3 & 9\ldots15 & \text{half rounded UP} & 4 \\
4 & 16\ldots24 & \text{all} & 9 \\
5 & 25\ldots35 & \text{half rounded DOWN} & 5 \\
6 & 36\ldots48 & \text{none} & 0 \\
7 & 49\ldots63 & \text{half rounded UP} & 8 \\
8 & 64\ldots80 & \text{all} & 17
\end{array}\]
DPatrick
2015-03-21 21:06:06
Notice anything?
Notice anything?
Eugenis
2015-03-21 21:06:28
It looks like we have some kind of recursion
It looks like we have some kind of recursion
chezbgone
2015-03-21 21:06:28
pattern of cycle 4
pattern of cycle 4
Tommy2000
2015-03-21 21:06:28
increases by 4 everytime
increases by 4 everytime
AlcumusGuy
2015-03-21 21:06:28
multiples of 4
multiples of 4
bli1999
2015-03-21 21:06:28
Adds 8 to each term when it cycles through
Adds 8 to each term when it cycles through
DPatrick
2015-03-21 21:06:31
Each block of $s(n)$ counts exactly 4 more (for the ``half'' blocks) or 8 more (for the ``all'' block) than $s(n+4)$. Is that a coincidence?
Each block of $s(n)$ counts exactly 4 more (for the ``half'' blocks) or 8 more (for the ``all'' block) than $s(n+4)$. Is that a coincidence?
temp8909
2015-03-21 21:06:49
no
no
mathwizard888
2015-03-21 21:06:49
no
no
dyang
2015-03-21 21:06:49
nope
nope
csmath
2015-03-21 21:06:49
No!
No!
DPatrick
2015-03-21 21:06:51
It's not, because
\begin{align*}
(n+1)^2 - n^2 &= 2n+1, \\
(n+5)^2 - (n+4)^2 &= 2n + 9,
\end{align*}
so there are always 8 more terms in the $s(n+4)$ block than in the $s(n)$ block.
It's not, because
\begin{align*}
(n+1)^2 - n^2 &= 2n+1, \\
(n+5)^2 - (n+4)^2 &= 2n + 9,
\end{align*}
so there are always 8 more terms in the $s(n+4)$ block than in the $s(n)$ block.
DPatrick
2015-03-21 21:07:06
That simplifies things quite a bit. Each set of 4 blocks counts 16 more than the block before.
That simplifies things quite a bit. Each set of 4 blocks counts 16 more than the block before.
summitwei
2015-03-21 21:07:24
Arithmetic series
Arithmetic series
ryanyz10
2015-03-21 21:07:24
arithmetic sequence
arithmetic sequence
csmath
2015-03-21 21:07:24
So we can just sum these as arithmetic series?
So we can just sum these as arithmetic series?
DPatrick
2015-03-21 21:07:33
Indeed.
$s(1)\ldots s(4)$ counts $1+4+9 = 14$.
$s(5)\ldots s(8)$ counts $14+16 = 30$.
etc.
$s(25)\ldots s(28)$ counts $14+6(16) = 110$.
Indeed.
$s(1)\ldots s(4)$ counts $1+4+9 = 14$.
$s(5)\ldots s(8)$ counts $14+16 = 30$.
etc.
$s(25)\ldots s(28)$ counts $14+6(16) = 110$.
DPatrick
2015-03-21 21:07:39
So $s(1)$ through $s(28)$ have the sum of the 7-term arithmetic series
\[
14 + 30 + 46 + \cdots + 110 = \frac72 \cdot 124 = 434.
\]
So $s(1)$ through $s(28)$ have the sum of the 7-term arithmetic series
\[
14 + 30 + 46 + \cdots + 110 = \frac72 \cdot 124 = 434.
\]
AlcumusGuy
2015-03-21 21:07:56
We have to be careful about the upper end since $n \le 1000$.
We have to be careful about the upper end since $n \le 1000$.
DPatrick
2015-03-21 21:08:00
Absolutely.
Absolutely.
DPatrick
2015-03-21 21:08:11
We'll have to do the end by hand.
We'll have to do the end by hand.
Th3Numb3rThr33
2015-03-21 21:08:19
Then just add s(29), s(30), s(31)?
Then just add s(29), s(30), s(31)?
DPatrick
2015-03-21 21:08:23
$s(29)$ runs from $29^2 = 841$ to $30^2-1 = 899$, and we take half rounded up.
$s(29)$ runs from $29^2 = 841$ to $30^2-1 = 899$, and we take half rounded up.
DPatrick
2015-03-21 21:08:34
That's another 29 good ones, for a total so far of $434 + 29 = 463$.
That's another 29 good ones, for a total so far of $434 + 29 = 463$.
DPatrick
2015-03-21 21:08:48
$s(30)$ runs from $30^2 = 900$ to $31^2-1 = 960$, but we don't take any of them.
$s(30)$ runs from $30^2 = 900$ to $31^2-1 = 960$, but we don't take any of them.
DPatrick
2015-03-21 21:09:00
And finally, $s(31)$ runs from 961 up to 999 (since $s(n) = 999$ determines $A(1000)$). That's 39 terms. We take half rounded up, so that's another 20.
And finally, $s(31)$ runs from 961 up to 999 (since $s(n) = 999$ determines $A(1000)$). That's 39 terms. We take half rounded up, so that's another 20.
DPatrick
2015-03-21 21:09:14
This gives us a final count of $463 + 20 = \boxed{483}$.
This gives us a final count of $463 + 20 = \boxed{483}$.
DPatrick
2015-03-21 21:09:38
Very, very careful bookkeeping was the key to solving this problem.
Very, very careful bookkeeping was the key to solving this problem.
akim99
2015-03-21 21:09:43
HEre we go #15
HEre we go #15
spartan168
2015-03-21 21:09:43
And now for the grand finale
And now for the grand finale
DPatrick
2015-03-21 21:09:54
#15 is actually a lot shorter compared to the one we just did:
#15 is actually a lot shorter compared to the one we just did:
DPatrick
2015-03-21 21:09:57
15. A block of wood has the shape of a right circular cylinder with radius 6 and height 8, and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overset{\frown}{AB}$ on that face measures $120^{\circ}$. The block is then sliced in half along the plane that passes through point $A$, point $B$, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces if $a\pi + b\sqrt{c}$, where $a, b,$ and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a + b + c$.
15. A block of wood has the shape of a right circular cylinder with radius 6 and height 8, and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overset{\frown}{AB}$ on that face measures $120^{\circ}$. The block is then sliced in half along the plane that passes through point $A$, point $B$, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces if $a\pi + b\sqrt{c}$, where $a, b,$ and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a + b + c$.
DPatrick
2015-03-21 21:10:05
DPatrick
2015-03-21 21:10:12
Everyone should thank AoPS community members Royalreter1 and chezbgone2 for posting this excellent diagram!
Everyone should thank AoPS community members Royalreter1 and chezbgone2 for posting this excellent diagram!
DPatrick
2015-03-21 21:10:29
My only contribution was to color it blue.
My only contribution was to color it blue.
Royalreter1
2015-03-21 21:10:36
Wait chezbgone2 did everything, I just pasted what he told me to
Wait chezbgone2 did everything, I just pasted what he told me to
DPatrick
2015-03-21 21:10:53
Most 3-D geometry problems are 2-D geometry problems in disguise, and this problem is no exception.
Most 3-D geometry problems are 2-D geometry problems in disguise, and this problem is no exception.
DPatrick
2015-03-21 21:11:01
If we smush our flat unpainted face down onto the base, what does it look like?
If we smush our flat unpainted face down onto the base, what does it look like?
Tuxianeer
2015-03-21 21:11:24
a circle with two parallel chords
a circle with two parallel chords
nosaj
2015-03-21 21:11:24
Circle with 2 120 degree chords cut off.
Circle with 2 120 degree chords cut off.
C-bass
2015-03-21 21:11:24
A rectangle with curved sides!
A rectangle with curved sides!
DPatrick
2015-03-21 21:11:30
It's the region of the circle that lies between two 120-degree arcs:
It's the region of the circle that lies between two 120-degree arcs:
DPatrick
2015-03-21 21:11:39
DPatrick
2015-03-21 21:11:53
So, what's the area of the gray region that results when the white region is projected onto the base of the cylinder?
So, what's the area of the gray region that results when the white region is projected onto the base of the cylinder?
DPatrick
2015-03-21 21:12:06
How can we chop it up to find its area?
How can we chop it up to find its area?
cumo99
2015-03-21 21:12:35
draw diagonals
draw diagonals
High
2015-03-21 21:12:35
Connect A and B to its center
Connect A and B to its center
Darn
2015-03-21 21:12:35
Connect endpoints of arcs to center
Connect endpoints of arcs to center
RedPhoenix
2015-03-21 21:12:35
draw lines to the center?
draw lines to the center?
Tommy2000
2015-03-21 21:12:35
draw diagonals
draw diagonals
dantx5
2015-03-21 21:12:35
draw 4 radii to form triangles and sectors
draw 4 radii to form triangles and sectors
morninglight
2015-03-21 21:12:35
4 30-60-90s and 2 equilateral triangles
4 30-60-90s and 2 equilateral triangles
ninjataco
2015-03-21 21:12:35
draw radii to A and B
draw radii to A and B
sophiazhi
2015-03-21 21:12:35
connect center to A and B
connect center to A and B
brightknight
2015-03-21 21:12:35
Connect the center to A and B
Connect the center to A and B
DPatrick
2015-03-21 21:12:50
It's 2 copies of triangle $OAB$ (where $O$ is the center of the circle), plus 2 60-degree wedges of the circle, as shown below:
It's 2 copies of triangle $OAB$ (where $O$ is the center of the circle), plus 2 60-degree wedges of the circle, as shown below:
DPatrick
2015-03-21 21:12:53
DPatrick
2015-03-21 21:13:11
What's the area of $OAB$?
What's the area of $OAB$?
xyuqing
2015-03-21 21:13:38
9sqrt3
9sqrt3
Lee541
2015-03-21 21:13:38
9sqrt3
9sqrt3
dantx5
2015-03-21 21:13:38
9sqrt3
9sqrt3
SHARKYBOY
2015-03-21 21:13:38
9sqrt3
9sqrt3
summitwei
2015-03-21 21:13:38
9sqrt3
9sqrt3
theoldaops
2015-03-21 21:13:38
9root3
9root3
swirlykick
2015-03-21 21:13:38
9sqrt3
9sqrt3
DPatrick
2015-03-21 21:13:42
$OAB$ has height 3 and base $6\sqrt3$, so its area is $9\sqrt3$.
$OAB$ has height 3 and base $6\sqrt3$, so its area is $9\sqrt3$.
DPatrick
2015-03-21 21:13:56
And what's the area of one of those circular-wedge pieces?
And what's the area of one of those circular-wedge pieces?
bli1999
2015-03-21 21:14:16
6pi
6pi
brainiac1
2015-03-21 21:14:16
6pi
6pi
Sciolympian
2015-03-21 21:14:16
$6\pi$
$6\pi$
DPatrick
2015-03-21 21:14:21
Each wedge is 1/6 of the circle, so each wedge has area $6\pi$.
Each wedge is 1/6 of the circle, so each wedge has area $6\pi$.
DPatrick
2015-03-21 21:14:28
Thus the gray area is $2(6\pi + 9\sqrt3) = 12\pi + 18\sqrt3$.
Thus the gray area is $2(6\pi + 9\sqrt3) = 12\pi + 18\sqrt3$.
DPatrick
2015-03-21 21:14:34
Now what? How does this flattened grey area relate to the white area that we really want?
Now what? How does this flattened grey area relate to the white area that we really want?
xyuqing
2015-03-21 21:14:59
stretch it
stretch it
va2010
2015-03-21 21:14:59
multiply by 5/3
multiply by 5/3
nosaj
2015-03-21 21:14:59
stretch factor of 5/3
stretch factor of 5/3
Tuxianeer
2015-03-21 21:14:59
it's $\dfrac{3}{5}$ of the white area
it's $\dfrac{3}{5}$ of the white area
pieslinger
2015-03-21 21:14:59
it's a projection
it's a projection
csmath
2015-03-21 21:14:59
It's dilated by a factor
It's dilated by a factor
raxu
2015-03-21 21:14:59
The white one is 5/3 the grey one.
The white one is 5/3 the grey one.
Th3Numb3rThr33
2015-03-21 21:14:59
the white is a stretched version of the gray
the white is a stretched version of the gray
DivideBy0
2015-03-21 21:14:59
its just a rescaling
its just a rescaling
DPatrick
2015-03-21 21:15:08
It's the same basic shape, but stretched in one direction: the "horizontal" direction as we look at it above.
It's the same basic shape, but stretched in one direction: the "horizontal" direction as we look at it above.
DPatrick
2015-03-21 21:15:19
Right now, the gray region has width 6, but what width does it have when we stretch it back to its original shape?
Right now, the gray region has width 6, but what width does it have when we stretch it back to its original shape?
Tuxianeer
2015-03-21 21:15:38
$10$
$10$
akim99
2015-03-21 21:15:38
10
10
chaoshadow37
2015-03-21 21:15:38
10
10
temp8909
2015-03-21 21:15:38
10
10
DPatrick
2015-03-21 21:15:43
Right. It's "width" is the hypotenuse of a triangle with legs 6 (the grey width) and 8 (the height of the cylinder), as shown below:
Right. It's "width" is the hypotenuse of a triangle with legs 6 (the grey width) and 8 (the height of the cylinder), as shown below:
DPatrick
2015-03-21 21:15:46
DPatrick
2015-03-21 21:15:57
So the width of the white region is 10 (the hypotenuse of the 6-8-10 right triangle shown above).
So the width of the white region is 10 (the hypotenuse of the 6-8-10 right triangle shown above).
DPatrick
2015-03-21 21:16:08
Thus the gray region gets stretched by a factor of $\frac{10}{6} = \frac53$ to form the white region.
Thus the gray region gets stretched by a factor of $\frac{10}{6} = \frac53$ to form the white region.
DPatrick
2015-03-21 21:16:20
Since this stretching is only along one dimension, it multiplies the area by that same factor.
Since this stretching is only along one dimension, it multiplies the area by that same factor.
nosaj
2015-03-21 21:16:37
\[\tfrac{5}{3}(12\pi + 18\sqrt{3})=20\pi+30\sqrt{3}\]
\[\tfrac{5}{3}(12\pi + 18\sqrt{3})=20\pi+30\sqrt{3}\]
DPatrick
2015-03-21 21:16:39
So the white region's area is $\frac53(12\pi + 18\sqrt3) = 20\pi + 30\sqrt3$.
So the white region's area is $\frac53(12\pi + 18\sqrt3) = 20\pi + 30\sqrt3$.
Darn
2015-03-21 21:16:57
So our answer is 20+30+3=53
So our answer is 20+30+3=53
akim99
2015-03-21 21:16:57
20+30+3 = 53
20+30+3 = 53
Th3Numb3rThr33
2015-03-21 21:16:57
20 + 30 + 3 = 53
20 + 30 + 3 = 53
DPatrick
2015-03-21 21:16:58
This is in the format requested by the problem, so the final answer is $20 + 30 + 3 = \boxed{053}$.
This is in the format requested by the problem, so the final answer is $20 + 30 + 3 = \boxed{053}$.
csmath
2015-03-21 21:17:13
That was quite a straightforward 15.
That was quite a straightforward 15.
spartan168
2015-03-21 21:17:13
that was fast
that was fast
Darn
2015-03-21 21:17:13
wait that was easy for a #15
wait that was easy for a #15
DPatrick
2015-03-21 21:17:33
It was fairly easy for a #15, if you just remember the usual advice: most 3D geometry problems are just 2D geometry problems in disguise!
It was fairly easy for a #15, if you just remember the usual advice: most 3D geometry problems are just 2D geometry problems in disguise!
DPatrick
2015-03-21 21:17:42
And that's it for the Math Jam! Thanks for coming.
And that's it for the Math Jam! Thanks for coming.
DPatrick
2015-03-21 21:17:49
Join us again in just 6 days, on Friday, March 27, for the AIME II Math Jam!
Join us again in just 6 days, on Friday, March 27, for the AIME II Math Jam!
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