2017 AIME II Discussion
Go back to the Math Jam ArchiveAoPS Instructors discuss all 15 problems on the 2017 AIME II .
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Facilitator: Dave Patrick
DPatrick
2017-03-24 19:00:26
Welcome to the 2017 AIME II Math Jam!
Welcome to the 2017 AIME II Math Jam!
DPatrick
2017-03-24 19:00:33
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
2017-03-24 19:00:42
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
2017-03-24 19:00:50
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.
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.
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.
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
2017-03-24 19:01:07
Notice that this is a lot like one of our classes except there are a lot more of you and the same number of me. I won't be able to post all of your comments all of the time. It's not personal! Here's a secret though: I prefer to pass clear, well-written comments to the room.
Notice that this is a lot like one of our classes except there are a lot more of you and the same number of me. I won't be able to post all of your comments all of the time. It's not personal! Here's a secret though: I prefer to pass clear, well-written comments to the room.
DPatrick
2017-03-24 19:01:26
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the necessary material for every problem as we go.
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the necessary material for every problem as we go.
DPatrick
2017-03-24 19:01:37
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!
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
2017-03-24 19:01:48
We do have two teaching assistants with us tonight to help answer your questions.
We do have two teaching assistants with us tonight to help answer your questions.
DPatrick
2017-03-24 19:01:58
Jessica Oehrlein (numberdance): Jessie graduated from Olin College of Engineering with a degree in mechanical engineering, and she now studies applied math and atmospheric science at Columbia University. She has attended Canada/USA Mathcamp, RSI (Research Science Institute), and BSM (Budapest Semesters in Mathematics). In her free time, she contra dances, rides roller coasters, and watches far too much ballet and modern dance.
Jessica Oehrlein (numberdance): Jessie graduated from Olin College of Engineering with a degree in mechanical engineering, and she now studies applied math and atmospheric science at Columbia University. She has attended Canada/USA Mathcamp, RSI (Research Science Institute), and BSM (Budapest Semesters in Mathematics). In her free time, she contra dances, rides roller coasters, and watches far too much ballet and modern dance.
numberdance
2017-03-24 19:02:02
Hi y'all!
Hi y'all!
DPatrick
2017-03-24 19:02:26
Our second TA is unfortunately having some technical issues right now, but I'll introduce him anyway:
Our second TA is unfortunately having some technical issues right now, but I'll introduce him anyway:
DPatrick
2017-03-24 19:02:31
Aakarsh Gottumukkala (numberman): Aakarsh majored in philosophy and minored in mathematics at The College of Wooster. Currently, he is a masters student in the liberal arts at St. John's College. He has been a Robinson Center Teaching Assistant for Mathematical Logic three summers in a row in Seattle, WA. He is really interested in algebra and number theory.
Aakarsh Gottumukkala (numberman): Aakarsh majored in philosophy and minored in mathematics at The College of Wooster. Currently, he is a masters student in the liberal arts at St. John's College. He has been a Robinson Center Teaching Assistant for Mathematical Logic three summers in a row in Seattle, WA. He is really interested in algebra and number theory.
DPatrick
2017-03-24 19:02:45
Hopefully he'll be with us soon.
Hopefully he'll be with us soon.
DPatrick
2017-03-24 19:02:59
Our TAs 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.
Our TAs 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
2017-03-24 19:03:23
Please also remember that the purpose of this Math Jam is to work through the solutions to AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. Also on occasion we may stop to prove things that you wouldn't necessary need to prove while doing the contest. So please, when a question is posted, do not simply respond with the final answer. That's not why we're here.
Please also remember that the purpose of this Math Jam is to work through the solutions to AIME problems, and not to merely present the answers. "Working through the solutions" includes discussing problem-solving tactics. Also on occasion we may stop to prove things that you wouldn't necessary need to prove while doing the contest. So please, when a question is posted, do not simply respond with the final answer. That's not why we're here.
DPatrick
2017-03-24 19:03:56
Probably about halfway through, I will get thirsty or hungry, or my fingers will get tired, and I'll take a 1-2 minute break. Other than that, we will simply plow through all 15 problems. It'll take 2-3 hours.
Probably about halfway through, I will get thirsty or hungry, or my fingers will get tired, and I'll take a 1-2 minute break. Other than that, we will simply plow through all 15 problems. It'll take 2-3 hours.
DPatrick
2017-03-24 19:04:23
Let's get started!
Let's get started!
DPatrick
2017-03-24 19:04:35
1. Find the number of subsets of $\{1,2,3,4,5,6,7,8\}$ that are subsets of neither $\{1,2,3,4,5\}$ nor $\{4,5,6,7,8\}$.
1. Find the number of subsets of $\{1,2,3,4,5,6,7,8\}$ that are subsets of neither $\{1,2,3,4,5\}$ nor $\{4,5,6,7,8\}$.
DPatrick
2017-03-24 19:04:55
(Note that I'll always pin the current problem at the top, so we can refer to it while we discuss the solution.)
(Note that I'll always pin the current problem at the top, so we can refer to it while we discuss the solution.)
lsh0589
2017-03-24 19:05:06
complementary counting?
complementary counting?
SomethingNeutral
2017-03-24 19:05:06
complementary counting!!
complementary counting!!
Ani10
2017-03-24 19:05:06
They're literally telling you to do complementary counting
They're literally telling you to do complementary counting
DPatrick
2017-03-24 19:05:38
Certainly one good option is to use complementary counting: count the subsets of the big set, and subtract the subsets of the smaller sets.
Certainly one good option is to use complementary counting: count the subsets of the big set, and subtract the subsets of the smaller sets.
DPatrick
2017-03-24 19:05:56
For whatever reason, though, that didn't occur to me in real time -- I just did it directly.
For whatever reason, though, that didn't occur to me in real time -- I just did it directly.
alifenix-
2017-03-24 19:06:02
I first used constructive counting, then checked with complementary counting.
I first used constructive counting, then checked with complementary counting.
DPatrick
2017-03-24 19:06:21
Indeed: you can just construct a valid subset directly. (Complementary counting becomes a nice check.)
Indeed: you can just construct a valid subset directly. (Complementary counting becomes a nice check.)
DPatrick
2017-03-24 19:06:32
What does it mean for a subset of the big set to not be a subset of $\{1,2,3,4,5\}$?
What does it mean for a subset of the big set to not be a subset of $\{1,2,3,4,5\}$?
baseballcat
2017-03-24 19:06:51
it has to contain 6,7, or 8
it has to contain 6,7, or 8
antmath2520
2017-03-24 19:06:51
It must contain at least one number from ${6, 7, 8}$.
It must contain at least one number from ${6, 7, 8}$.
vishwathganesan
2017-03-24 19:06:54
it has at least one of 6, 7, or 8
it has at least one of 6, 7, or 8
62861
2017-03-24 19:06:54
it contains at least one element of {6, 7, 8}
it contains at least one element of {6, 7, 8}
DPatrick
2017-03-24 19:07:00
Right: it must have a 6, a 7, or an 8.
Right: it must have a 6, a 7, or an 8.
DPatrick
2017-03-24 19:07:07
And similarly, to not be a subset of $\{4,5,6,7,8\}$ it must have a 1, a 2, or a 3.
And similarly, to not be a subset of $\{4,5,6,7,8\}$ it must have a 1, a 2, or a 3.
GeneralCobra19
2017-03-24 19:07:19
You can also do it another way, by using constructive... since you need to pick at least one from {1, 2, 3} and {6, 7, 8} and {4, 5} doesn't matter.
You can also do it another way, by using constructive... since you need to pick at least one from {1, 2, 3} and {6, 7, 8} and {4, 5} doesn't matter.
antmath2520
2017-03-24 19:07:22
It must contain a number from ${1, 2, 3}$ AND a number from ${6, 7, 8}$.
It must contain a number from ${1, 2, 3}$ AND a number from ${6, 7, 8}$.
DPatrick
2017-03-24 19:07:37
Exactly. So we can construct a valid subset by looking at the different pieces.
Exactly. So we can construct a valid subset by looking at the different pieces.
DPatrick
2017-03-24 19:07:43
We can take any subset of $\{1,2,3\}$ except the empty set.
We can take any subset of $\{4,5\}$ (including the empty set).
We can take any subset of $\{6,7,8\}$ except the empty set.
We can take any subset of $\{1,2,3\}$ except the empty set.
We can take any subset of $\{4,5\}$ (including the empty set).
We can take any subset of $\{6,7,8\}$ except the empty set.
DPatrick
2017-03-24 19:07:50
And then we mash them together.
And then we mash them together.
DPatrick
2017-03-24 19:08:06
How many choices does that give us?
How many choices does that give us?
GeneralCobra19
2017-03-24 19:08:34
That makes it (2^3-1)(2^3-1)*2^2=196
That makes it (2^3-1)(2^3-1)*2^2=196
SomethingNeutral
2017-03-24 19:08:34
so 7*4*7 = 196 !!! (not triple factorial)
so 7*4*7 = 196 !!! (not triple factorial)
arghh
2017-03-24 19:08:34
7 * 4 * 7 ways
7 * 4 * 7 ways
willmathxu
2017-03-24 19:08:34
7*7*4
7*7*4
antmath2520
2017-03-24 19:08:34
$\left(2^3 - 1\right) \times 2^2 \times \left(2^3 - 1\right) = 7 \times 4 \times 7 = \boxed{196}$
$\left(2^3 - 1\right) \times 2^2 \times \left(2^3 - 1\right) = 7 \times 4 \times 7 = \boxed{196}$
baseballcat
2017-03-24 19:08:34
7*7*4
7*7*4
DPatrick
2017-03-24 19:08:40
There are $2^3 - 1 = 7$ choices of nonempty subset of $\{1,2,3\}$.
There are $2^2 = 4$ choices of any subset of $\{4,5\}$.
There are $2^3 - 1 = 7$ choices of nonempty subset of $\{6,7,8\}$.
There are $2^3 - 1 = 7$ choices of nonempty subset of $\{1,2,3\}$.
There are $2^2 = 4$ choices of any subset of $\{4,5\}$.
There are $2^3 - 1 = 7$ choices of nonempty subset of $\{6,7,8\}$.
DPatrick
2017-03-24 19:08:48
So, altogether, there are $7 \cdot 4 \cdot 7 = \boxed{196}$ possible subsets of the big set.
So, altogether, there are $7 \cdot 4 \cdot 7 = \boxed{196}$ possible subsets of the big set.
DPatrick
2017-03-24 19:09:11
2. Teams $T_1$, $T_2$, $T_3$, and $T_4$ are in the playoffs. In the semifinal matches, $T_1$ plays $T_4$, and $T_2$ plays $T_3$. The winners of those two matches will play each other in the final match to determine the champion. When $T_i$ plays $T_j$, the probability that $T_i$ wins is $\dfrac{i}{i+j}$, and the outcomes of all matches are independent. The probability that $T_4$ will be the champion is $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
2. Teams $T_1$, $T_2$, $T_3$, and $T_4$ are in the playoffs. In the semifinal matches, $T_1$ plays $T_4$, and $T_2$ plays $T_3$. The winners of those two matches will play each other in the final match to determine the champion. When $T_i$ plays $T_j$, the probability that $T_i$ wins is $\dfrac{i}{i+j}$, and the outcomes of all matches are independent. The probability that $T_4$ will be the champion is $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
DPatrick
2017-03-24 19:09:33
What has to happen for $T_4$ to win it all?
What has to happen for $T_4$ to win it all?
SomethingNeutral
2017-03-24 19:10:05
t4 beats t1 and hwoever wins
t4 beats t1 and hwoever wins
baseballcat
2017-03-24 19:10:05
t4 beats t1 and beats the winner of t2 vs t3
t4 beats t1 and beats the winner of t2 vs t3
GeronimoStilton
2017-03-24 19:10:05
First $T_4$ must beat $T_1$, and then $T_2$ or $T_3$
First $T_4$ must beat $T_1$, and then $T_2$ or $T_3$
thedoge
2017-03-24 19:10:05
$T_4$ has to win the both games
$T_4$ has to win the both games
popcorn1
2017-03-24 19:10:05
He beats T1 and then the winner of T2 and T3's match.
He beats T1 and then the winner of T2 and T3's match.
DPatrick
2017-03-24 19:10:07
First, $T_4$ must defeat $T_1$ in the semifinal. Then $T_4$ must defeat the winner of $T_2$-vs-$T_3$.
First, $T_4$ must defeat $T_1$ in the semifinal. Then $T_4$ must defeat the winner of $T_2$-vs-$T_3$.
DPatrick
2017-03-24 19:10:18
What's the probability of $T_4$'s semifinal win?
What's the probability of $T_4$'s semifinal win?
mathrocks001
2017-03-24 19:10:38
4/5
4/5
phi_ftw1618
2017-03-24 19:10:38
$\frac{4}{5}$
$\frac{4}{5}$
mathmagician
2017-03-24 19:10:38
4/5
4/5
serd
2017-03-24 19:10:38
4/5
4/5
FieryEntei
2017-03-24 19:10:38
The probability of that happening is $\frac{4}{5}$ as given by the formula in the question
The probability of that happening is $\frac{4}{5}$ as given by the formula in the question
DPatrick
2017-03-24 19:10:43
We just use the formula given in the problem: it's $\dfrac{4}{4+1} = \dfrac45$.
We just use the formula given in the problem: it's $\dfrac{4}{4+1} = \dfrac45$.
DPatrick
2017-03-24 19:10:50
Now what?
Now what?
carzland
2017-03-24 19:11:08
We could do casework based on who wins the semifinal match that $T_4$ is not in.
We could do casework based on who wins the semifinal match that $T_4$ is not in.
legolego
2017-03-24 19:11:08
casework whether T_2 or T_3 wins
casework whether T_2 or T_3 wins
vishwathganesan
2017-03-24 19:11:08
casework on who wins t2 vs t3?
casework on who wins t2 vs t3?
DPatrick
2017-03-24 19:11:16
Right. We have two cases depending on whether $T_2$ or $T_3$ wins the other semifinal.
Right. We have two cases depending on whether $T_2$ or $T_3$ wins the other semifinal.
DPatrick
2017-03-24 19:11:28
Note that $T_2$ wins with probability $\dfrac{2}{2+3} = \dfrac25$ and $T_3$ wins with probability $\dfrac{3}{3+2} = \dfrac35$.
Note that $T_2$ wins with probability $\dfrac{2}{2+3} = \dfrac25$ and $T_3$ wins with probability $\dfrac{3}{3+2} = \dfrac35$.
DPatrick
2017-03-24 19:11:34
If $T_2$ wins the semi, then what's the probability that $T_4$ defeats $T_2$?
If $T_2$ wins the semi, then what's the probability that $T_4$ defeats $T_2$?
noob_user
2017-03-24 19:11:52
2/3
2/3
MathTechFire
2017-03-24 19:11:52
2/3
2/3
goodball
2017-03-24 19:11:52
2/3
2/3
DPatrick
2017-03-24 19:11:56
It's $\dfrac{4}{4+2} = \dfrac46 = \dfrac23$.
It's $\dfrac{4}{4+2} = \dfrac46 = \dfrac23$.
DPatrick
2017-03-24 19:12:02
And if $T_3$ wins the semi, what's the probability that $T_4$ defeats $T_3$?
And if $T_3$ wins the semi, what's the probability that $T_4$ defeats $T_3$?
vy1234
2017-03-24 19:12:26
4/7
4/7
QuestForKnowledge
2017-03-24 19:12:26
4/7
4/7
vishwathganesan
2017-03-24 19:12:26
4/(3+4)=4/7
4/(3+4)=4/7
Ani10
2017-03-24 19:12:26
4/7
4/7
DPatrick
2017-03-24 19:12:29
It's $\dfrac{4}{4+3} = \dfrac47$.
It's $\dfrac{4}{4+3} = \dfrac47$.
DPatrick
2017-03-24 19:12:40
So how do we combine all this data? What's the overall probability of $T_4$ winning the championship?
So how do we combine all this data? What's the overall probability of $T_4$ winning the championship?
Ani10
2017-03-24 19:12:52
so take a weighted average right?
so take a weighted average right?
DPatrick
2017-03-24 19:13:13
Right -- for the second game, we have to add the two cases, weighted by the probability of $T_2$ or $T_3$ winning.
Right -- for the second game, we have to add the two cases, weighted by the probability of $T_2$ or $T_3$ winning.
tdeng
2017-03-24 19:13:19
(4/5)(2/3*2/5+4/7*3/5)
(4/5)(2/3*2/5+4/7*3/5)
DPatrick
2017-03-24 19:13:30
Indeed: $T_4$ wins the semifinal with probability $\dfrac45$, then wins the final with probability $\dfrac25 \cdot \dfrac23 + \dfrac35 \cdot \dfrac47$.
Indeed: $T_4$ wins the semifinal with probability $\dfrac45$, then wins the final with probability $\dfrac25 \cdot \dfrac23 + \dfrac35 \cdot \dfrac47$.
DPatrick
2017-03-24 19:13:39
So the overall probability is \[\frac45 \cdot \left(\frac25 \cdot \frac23 + \frac35 \cdot \frac47\right).\]
So the overall probability is \[\frac45 \cdot \left(\frac25 \cdot \frac23 + \frac35 \cdot \frac47\right).\]
mathrocks001
2017-03-24 19:13:55
256/525
256/525
GeronimoStilton
2017-03-24 19:13:55
256/625
256/625
thedoge
2017-03-24 19:13:55
so $\frac{256}{525}$
so $\frac{256}{525}$
skiboy32
2017-03-24 19:13:55
so, the answer is $\frac{256}{525}$
so, the answer is $\frac{256}{525}$
DPatrick
2017-03-24 19:14:01
The common denominator inside the parentheses is $5 \cdot 3 \cdot 7 = 105$, so this is \[\frac45 \cdot \left(\frac{28}{105} + \frac{36}{105}\right) = \frac45 \cdot \frac{64}{105} = \frac{256}{525}.\]
The common denominator inside the parentheses is $5 \cdot 3 \cdot 7 = 105$, so this is \[\frac45 \cdot \left(\frac{28}{105} + \frac{36}{105}\right) = \frac45 \cdot \frac{64}{105} = \frac{256}{525}.\]
DPatrick
2017-03-24 19:14:13
And this is already in lowest terms, so the final answer is $256 + 525 = \boxed{781}$.
And this is already in lowest terms, so the final answer is $256 + 525 = \boxed{781}$.
DPatrick
2017-03-24 19:14:54
Don't get too hung up on the term "weighted average" -- that just means that when we combine the cases, each case doesn't have the same probability.
Don't get too hung up on the term "weighted average" -- that just means that when we combine the cases, each case doesn't have the same probability.
DPatrick
2017-03-24 19:15:04
We've "weighted" the cases by the probability that each one occurs.
We've "weighted" the cases by the probability that each one occurs.
DPatrick
2017-03-24 19:15:16
Anyway, onwards!
Anyway, onwards!
DPatrick
2017-03-24 19:15:20
3. A triangle has vertices $A(0,0)$, $B(12,0)$, and $C(8,10)$. The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
3. A triangle has vertices $A(0,0)$, $B(12,0)$, and $C(8,10)$. The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
GeronimoStilton
2017-03-24 19:15:37
Diagram?
Diagram?
SomethingNeutral
2017-03-24 19:15:37
DRAw a diagram
DRAw a diagram
QuestForKnowledge
2017-03-24 19:15:37
draw it
draw it
carzland
2017-03-24 19:15:37
Draw it!
Draw it!
serd
2017-03-24 19:15:37
graph it?
graph it?
DPatrick
2017-03-24 19:15:41
Let's start with a picture -- notice that I'll draw it on "graph paper" like I hope you did in real time:
Let's start with a picture -- notice that I'll draw it on "graph paper" like I hope you did in real time:
DPatrick
2017-03-24 19:15:44
DPatrick
2017-03-24 19:15:52
How do we compute the probability asked for in the problem?
How do we compute the probability asked for in the problem?
mathman3880
2017-03-24 19:16:13
find the area
find the area
Skywalker64
2017-03-24 19:16:13
by area
by area
legolego
2017-03-24 19:16:13
find desired region / total region
find desired region / total region
phi_ftw1618
2017-03-24 19:16:13
Find the set of points that are closest to B
Find the set of points that are closest to B
reelmathematician
2017-03-24 19:16:13
area of favorable region
area of favorable region
noob_user
2017-03-24 19:16:13
draw the feasible area
draw the feasible area
phi_ftw1618
2017-03-24 19:16:13
and then divide by area
and then divide by area
DPatrick
2017-03-24 19:16:21
Right. We need to determine the region inside in the triangle that's closest to $B$, compute its area, and compare to the area of $\triangle ABC$.
Right. We need to determine the region inside in the triangle that's closest to $B$, compute its area, and compare to the area of $\triangle ABC$.
DPatrick
2017-03-24 19:16:34
How do we determine the desired region?
How do we determine the desired region?
dtxiong
2017-03-24 19:16:50
draw perpendicular bisectors
draw perpendicular bisectors
fdas
2017-03-24 19:16:50
Draw perpendicular bisectors
Draw perpendicular bisectors
arghh
2017-03-24 19:16:50
Draw perpendicular bisectors of AB and BC
Draw perpendicular bisectors of AB and BC
thedoge
2017-03-24 19:16:50
we draw the perpendicular bisectors of AB and BC
we draw the perpendicular bisectors of AB and BC
Funnybunny5246
2017-03-24 19:16:50
Perpendicular bisectors
Perpendicular bisectors
reedmj
2017-03-24 19:16:50
First draw perpendicular bisectors of each side
First draw perpendicular bisectors of each side
spartan168
2017-03-24 19:16:50
Perpendicular bisectors
Perpendicular bisectors
DPatrick
2017-03-24 19:16:58
Aha -- perpendicular bisectors!
Aha -- perpendicular bisectors!
DPatrick
2017-03-24 19:17:16
Let's start with "points closer to $B$ than to $A$".
Let's start with "points closer to $B$ than to $A$".
goodball
2017-03-24 19:17:52
x>6
x>6
arghh
2017-03-24 19:17:52
That's the line x = 6, we want the region x > 6
That's the line x = 6, we want the region x > 6
Ani10
2017-03-24 19:17:52
perp bisector of AB
perp bisector of AB
serd
2017-03-24 19:17:52
the line x=6
the line x=6
DPatrick
2017-03-24 19:18:01
Right: it must have $x$-coordinate 6 or more.
Right: it must have $x$-coordinate 6 or more.
DPatrick
2017-03-24 19:18:05
So the blue region below is closer to $B$ than to $A$:
So the blue region below is closer to $B$ than to $A$:
DPatrick
2017-03-24 19:18:12
DPatrick
2017-03-24 19:18:27
And now for points closer to $B$ than to $C$...?
And now for points closer to $B$ than to $C$...?
reelmathematician
2017-03-24 19:18:43
perpendicular bisector of bc
perpendicular bisector of bc
QuestForKnowledge
2017-03-24 19:18:43
same thing
same thing
DPatrick
2017-03-24 19:19:00
Right. The points on the perpendicular bisector of $\overline{BC}$ are equally distant from $B$ and from $C$.
Right. The points on the perpendicular bisector of $\overline{BC}$ are equally distant from $B$ and from $C$.
DPatrick
2017-03-24 19:19:18
So the points "below" that bisector in our diagram will be closer to $B$ than to $C$.
So the points "below" that bisector in our diagram will be closer to $B$ than to $C$.
DPatrick
2017-03-24 19:19:22
DPatrick
2017-03-24 19:19:31
I've labeled the midpoint of $\overline{BC}$ as $Y$.
I've labeled the midpoint of $\overline{BC}$ as $Y$.
DPatrick
2017-03-24 19:19:39
How do we compute the area of $XBYZ$?
How do we compute the area of $XBYZ$?
arghh
2017-03-24 19:20:02
Find the coordinates of Z
Find the coordinates of Z
serd
2017-03-24 19:20:02
break it up into pieces or draw a big rectangle and subtract
break it up into pieces or draw a big rectangle and subtract
DPatrick
2017-03-24 19:20:57
I'd probably find the coordinates of $Z$ and then just break it up into nice right triangles.
I'd probably find the coordinates of $Z$ and then just break it up into nice right triangles.
DPatrick
2017-03-24 19:21:08
How do we find point $Z$?
How do we find point $Z$?
reelmathematician
2017-03-24 19:21:29
find the intersection of both bisectors
find the intersection of both bisectors
JJShan26
2017-03-24 19:21:29
set the two equations of the lines equal to each other
set the two equations of the lines equal to each other
a1b2
2017-03-24 19:21:29
Set up a system of equations
Set up a system of equations
serd
2017-03-24 19:21:29
system of equations of the perp. bisectors
system of equations of the perp. bisectors
mathrocks001
2017-03-24 19:21:33
slope of YZ
slope of YZ
DPatrick
2017-03-24 19:21:57
Well, we know the $x$-coordinate of $Z$ is 6. So we just can find the equation for $\overline{YZ}$ and plug in $x=6$.
Well, we know the $x$-coordinate of $Z$ is 6. So we just can find the equation for $\overline{YZ}$ and plug in $x=6$.
edward1000
2017-03-24 19:22:02
it's perpendicular to CB, and the slope of CB is -10/4, or -5/2, so the slope of ZY is 2/5
it's perpendicular to CB, and the slope of CB is -10/4, or -5/2, so the slope of ZY is 2/5
DPatrick
2017-03-24 19:22:16
Exactly. The slope of $\overline{BC}$ is $-\dfrac52$. So the slope of $\overline{YZ}$ is $\dfrac25$.
Exactly. The slope of $\overline{BC}$ is $-\dfrac52$. So the slope of $\overline{YZ}$ is $\dfrac25$.
DPatrick
2017-03-24 19:22:33
Therefore, the equation of the line containing $Y$ and $Z$ is \[y-5 = \frac25(x-10).\]
Therefore, the equation of the line containing $Y$ and $Z$ is \[y-5 = \frac25(x-10).\]
rainbow10
2017-03-24 19:22:51
so y=2x/5+1
so y=2x/5+1
giacomorizzo
2017-03-24 19:22:51
y=(2/5)x + 1
y=(2/5)x + 1
DPatrick
2017-03-24 19:23:08
...or you could pretty easily write the slope-intercept form, since $(10,5)$ is an "easy" point.
...or you could pretty easily write the slope-intercept form, since $(10,5)$ is an "easy" point.
reelmathematician
2017-03-24 19:23:13
plug in 6 for x
plug in 6 for x
Ani10
2017-03-24 19:23:13
We know x=6, so plug in to find y
We know x=6, so plug in to find y
spartan168
2017-03-24 19:23:16
Next, plug in x = 6 to get y = 17/5 or (6, 17/5) for point Z
Next, plug in x = 6 to get y = 17/5 or (6, 17/5) for point Z
DPatrick
2017-03-24 19:23:21
$Z$ has $x$-coordinate 6, so when we plug $x=6$ into the equation for $\overline{YZ}$, we get \[y-5=\frac25(6-10).\]
This solves to get $y = \dfrac{17}{5}$.
$Z$ has $x$-coordinate 6, so when we plug $x=6$ into the equation for $\overline{YZ}$, we get \[y-5=\frac25(6-10).\]
This solves to get $y = \dfrac{17}{5}$.
DPatrick
2017-03-24 19:23:31
So $Z$ has coordinates $\left(6,\dfrac{17}{5}\right)$.
So $Z$ has coordinates $\left(6,\dfrac{17}{5}\right)$.
DPatrick
2017-03-24 19:23:55
And so now to compute the area of $XBYZ$...
And so now to compute the area of $XBYZ$...
DjokerNole
2017-03-24 19:24:07
rectangle - triangles
rectangle - triangles
Ani10
2017-03-24 19:24:07
I would make a trapezoid and right triangle at this point by drawing x=10
I would make a trapezoid and right triangle at this point by drawing x=10
reelmathematician
2017-03-24 19:24:07
break into 2 right triangles
break into 2 right triangles
mathrocks001
2017-03-24 19:24:16
create rectangle around 4 points
create rectangle around 4 points
DPatrick
2017-03-24 19:24:19
All of these ideas are good.
All of these ideas are good.
DPatrick
2017-03-24 19:24:29
I did something even more basic:
I did something even more basic:
DPatrick
2017-03-24 19:24:33
DPatrick
2017-03-24 19:24:46
The bottom piece $XPQZ$ is a rectangle with sides $\dfrac{17}{5}$ and 4, so its area is $\dfrac{68}{5}$.
The top piece $ZQY$ is a right triangle with legs 4 and $\dfrac85$, so its area is $\dfrac{16}{5}$.
The right piece $YBP$ is a right triangle with legs 5 and 2, so its area is 5.
The bottom piece $XPQZ$ is a rectangle with sides $\dfrac{17}{5}$ and 4, so its area is $\dfrac{68}{5}$.
The top piece $ZQY$ is a right triangle with legs 4 and $\dfrac85$, so its area is $\dfrac{16}{5}$.
The right piece $YBP$ is a right triangle with legs 5 and 2, so its area is 5.
DPatrick
2017-03-24 19:25:00
There's probably a dozen different good ways to compute the area of this quadrilateral.
There's probably a dozen different good ways to compute the area of this quadrilateral.
DPatrick
2017-03-24 19:25:11
However you do it, the overall area of $XBYZ$ is $\dfrac{68}{5} + \dfrac{16}{5} + 5 = \dfrac{109}{5}$.
However you do it, the overall area of $XBYZ$ is $\dfrac{68}{5} + \dfrac{16}{5} + 5 = \dfrac{109}{5}$.
DPatrick
2017-03-24 19:25:28
And how do we finish?
And how do we finish?
DaAsianPotatoe
2017-03-24 19:25:41
divide by total area
divide by total area
legolego
2017-03-24 19:25:41
find area of whole triangle
find area of whole triangle
GeronimoStilton
2017-03-24 19:25:41
Find the area of $ABC$
Find the area of $ABC$
DPatrick
2017-03-24 19:25:46
Right. The original triangle $ABC$ has base $AB = 12$ and height 10, so its area is 60.
Right. The original triangle $ABC$ has base $AB = 12$ and height 10, so its area is 60.
DPatrick
2017-03-24 19:25:56
Therefore, the probability that we want is \[\frac{[XBYZ]}{[ABC]} = \frac{\frac{109}{5}}{60} = \frac{109}{300}.\]
Therefore, the probability that we want is \[\frac{[XBYZ]}{[ABC]} = \frac{\frac{109}{5}}{60} = \frac{109}{300}.\]
stardestroyer
2017-03-24 19:26:10
so 409
so 409
goodball
2017-03-24 19:26:10
Therefore, the answer is 409
Therefore, the answer is 409
DPatrick
2017-03-24 19:26:14
This cannot be further simplified, so our final answer is $109 + 300 = \boxed{409}$.
This cannot be further simplified, so our final answer is $109 + 300 = \boxed{409}$.
DPatrick
2017-03-24 19:26:34
This is a good argument for a "sanity check" of your answer at the end -- go re-read the problem and make sure you're answering what was asked for!
This is a good argument for a "sanity check" of your answer at the end -- go re-read the problem and make sure you're answering what was asked for!
DPatrick
2017-03-24 19:26:54
Several of you went to $109 + 5 = 114$ here...but if you re-read the problem, there's no way 109/5 can be a probability!
Several of you went to $109 + 5 = 114$ here...but if you re-read the problem, there's no way 109/5 can be a probability!
DPatrick
2017-03-24 19:27:17
4. Find the number of positive integers less than or equal to 2017 whose base-three representation contains no digit equal to 0.
4. Find the number of positive integers less than or equal to 2017 whose base-three representation contains no digit equal to 0.
DPatrick
2017-03-24 19:27:27
What data would probably help?
What data would probably help?
arghh
2017-03-24 19:27:46
Figure out what 2017 is in base 3
Figure out what 2017 is in base 3
serd
2017-03-24 19:27:46
write 2017 in base 3
write 2017 in base 3
to_chicken
2017-03-24 19:27:46
What $2017$ is in base $3$.
What $2017$ is in base $3$.
QuestForKnowledge
2017-03-24 19:27:46
what 2017 is in base 3
what 2017 is in base 3
Fibonachos
2017-03-24 19:27:46
The base-3 representation of 2017
The base-3 representation of 2017
DPatrick
2017-03-24 19:28:00
We're probably going to need to find what 2017 is in base 3.
We're probably going to need to find what 2017 is in base 3.
DPatrick
2017-03-24 19:28:06
You can easily work it out on your own, but to save time, here it is: \[ 2017 = 2 \cdot 3^6 + 2 \cdot 3^5 + 2 \cdot 3^3 + 2 \cdot 3^2 + 1 = 2202201_3.\]
You can easily work it out on your own, but to save time, here it is: \[ 2017 = 2 \cdot 3^6 + 2 \cdot 3^5 + 2 \cdot 3^3 + 2 \cdot 3^2 + 1 = 2202201_3.\]
DPatrick
2017-03-24 19:28:19
So what numbers in base 3 are less than this and don't contain a zero?
So what numbers in base 3 are less than this and don't contain a zero?
carzland
2017-03-24 19:28:51
Partial-Casework:
Partial-Casework:
DPatrick
2017-03-24 19:29:04
Right, there are a couple of different cases to consider.
Right, there are a couple of different cases to consider.
legolego
2017-03-24 19:29:07
have less than 7 digits or 7 digits and do not start with 22
have less than 7 digits or 7 digits and do not start with 22
DPatrick
2017-03-24 19:29:13
We can either have:
- a number with 6 digits or less, where each digit is 1 or 2; or
- a 7-digit number, where each digit is 1 or 2, provided it's less than $2202201_3$
We can either have:
- a number with 6 digits or less, where each digit is 1 or 2; or
- a 7-digit number, where each digit is 1 or 2, provided it's less than $2202201_3$
DPatrick
2017-03-24 19:29:29
How many of the first type are there? (6 digits or less)
How many of the first type are there? (6 digits or less)
reaganchoi
2017-03-24 19:29:42
1 digit: 2
2 digit: 4
3 digit: 8
4 digit: 16
5 digit: 32
6 digit: 64
1 digit: 2
2 digit: 4
3 digit: 8
4 digit: 16
5 digit: 32
6 digit: 64
thedoge
2017-03-24 19:29:42
2^6+2^5+2^4+2^3+2^2+2^1
2^6+2^5+2^4+2^3+2^2+2^1
DPatrick
2017-03-24 19:29:48
For each digit, we have two choices: 1 or 2.
For each digit, we have two choices: 1 or 2.
DPatrick
2017-03-24 19:29:54
So there are $2^1$ 1-digit numbers, $2^2$ 2-digit numbers, etc., up through $2^6$ 6-digit numbers.That is, there are $2^1 + 2^2 + \cdots + 2^6$ total.
So there are $2^1$ 1-digit numbers, $2^2$ 2-digit numbers, etc., up through $2^6$ 6-digit numbers.That is, there are $2^1 + 2^2 + \cdots + 2^6$ total.
Visconformity
2017-03-24 19:30:02
2^7-2
2^7-2
DPatrick
2017-03-24 19:30:15
Indeed, this is one less than $1 + 2 + \cdots + 2^6$, which itself is one less than $2^7$. So this quantity is $2^7 - 2 = 128 - 2 = 126$.
Indeed, this is one less than $1 + 2 + \cdots + 2^6$, which itself is one less than $2^7$. So this quantity is $2^7 - 2 = 128 - 2 = 126$.
DPatrick
2017-03-24 19:30:33
How about of the second type? (7-digit numbers less than $2202201_3$)
How about of the second type? (7-digit numbers less than $2202201_3$)
Ani10
2017-03-24 19:31:00
i would separate into numbers that start with 11, 12, 21, and 22
i would separate into numbers that start with 11, 12, 21, and 22
GeronimoStilton
2017-03-24 19:31:07
They are $2122222_3$ or less.
They are $2122222_3$ or less.
JJShan26
2017-03-24 19:31:09
numbers that start with 21,11,12
numbers that start with 21,11,12
DPatrick
2017-03-24 19:31:16
Good idea. The first two digits must be 11, 12, or 21. Then the last five digits can each either be 1 or 2.
Good idea. The first two digits must be 11, 12, or 21. Then the last five digits can each either be 1 or 2.
mathrocks001
2017-03-24 19:31:32
32 for each
32 for each
DPatrick
2017-03-24 19:31:37
This gives us 3 choices for the first two digits, then $2^5 = 32$ choices for the remaining five digits, for a total of $3 \cdot 32 = 96$ 7-digit numbers.
This gives us 3 choices for the first two digits, then $2^5 = 32$ choices for the remaining five digits, for a total of $3 \cdot 32 = 96$ 7-digit numbers.
SomethingNeutral
2017-03-24 19:31:55
126+96=$\boxed{222}$
126+96=$\boxed{222}$
Archimedes15
2017-03-24 19:32:00
96 + the first 126, so our answer is 222
96 + the first 126, so our answer is 222
DPatrick
2017-03-24 19:32:01
So altogether, there are $126 + 96 = \boxed{222}$ such numbers.
So altogether, there are $126 + 96 = \boxed{222}$ such numbers.
DPatrick
2017-03-24 19:32:17
5. A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are 189, 320, 287, 234, $x$, and $y$. Find the greatest possible value of $x+y$.
5. A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are 189, 320, 287, 234, $x$, and $y$. Find the greatest possible value of $x+y$.
DPatrick
2017-03-24 19:32:39
Before doing anything else, let's organize our data first so that the four given numbers are in order:
Before doing anything else, let's organize our data first so that the four given numbers are in order:
DPatrick
2017-03-24 19:32:44
5. A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are 189, 234, 287, 320, $x$, and $y$. Find the greatest possible value of $x+y$.
5. A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are 189, 234, 287, 320, $x$, and $y$. Find the greatest possible value of $x+y$.
DPatrick
2017-03-24 19:32:54
And let's say our four numbers are $a < b < c < d$. (This is just to give us common ground to discuss the problem.)
And let's say our four numbers are $a < b < c < d$. (This is just to give us common ground to discuss the problem.)
DPatrick
2017-03-24 19:33:01
What other quantity might be helpful to know?
What other quantity might be helpful to know?
Archimedes15
2017-03-24 19:33:22
It might be helpful to know the sum of all of the numbers
It might be helpful to know the sum of all of the numbers
skiboy32
2017-03-24 19:33:22
a+b+c+d
a+b+c+d
vishwathganesan
2017-03-24 19:33:22
their sum?
their sum?
DPatrick
2017-03-24 19:33:37
The sum of all four numbers is probably key. Let's set $s = a + b + c + d$.
The sum of all four numbers is probably key. Let's set $s = a + b + c + d$.
DPatrick
2017-03-24 19:33:43
But what else do we know about $s$?
But what else do we know about $s$?
edward1000
2017-03-24 19:33:57
3*s = the sum of all those numbers
3*s = the sum of all those numbers
SomethingNeutral
2017-03-24 19:34:00
189+234+287+320+x+y=3s
189+234+287+320+x+y=3s
Littlelachy
2017-03-24 19:34:07
We know that 3s is 189+234+287+320+z+y
We know that 3s is 189+234+287+320+z+y
reelmathematician
2017-03-24 19:34:07
3 times the pariwise sum
3 times the pariwise sum
shootingstar8
2017-03-24 19:34:09
189+234+287+320+x+y = 3s
189+234+287+320+x+y = 3s
DPatrick
2017-03-24 19:34:11
If we sum our six pairwise sums, we get $3s$, since each number in $\{a,b,c,d\}$ appears in three of the pairwise sums.
If we sum our six pairwise sums, we get $3s$, since each number in $\{a,b,c,d\}$ appears in three of the pairwise sums.
DPatrick
2017-03-24 19:34:20
So $189 + 234 + 287 + 320 + x + y = 3s$, and thus $x + y = 3s - 1030$.
So $189 + 234 + 287 + 320 + x + y = 3s$, and thus $x + y = 3s - 1030$.
DPatrick
2017-03-24 19:34:32
Thus, if we can maximize $s$, we'll automatically maximize $x+y$.
Thus, if we can maximize $s$, we'll automatically maximize $x+y$.
vishwathganesan
2017-03-24 19:34:43
s=(a+b)+(c+d)=(a+c)+(b+d)=(a+d)+(b+c)
s=(a+b)+(c+d)=(a+c)+(b+d)=(a+d)+(b+c)
DPatrick
2017-03-24 19:34:54
Aha, that's an good observation too.
Aha, that's an good observation too.
DPatrick
2017-03-24 19:35:15
The pairwise sums themselves come in 3 pairs that each sum to $s$.
The pairwise sums themselves come in 3 pairs that each sum to $s$.
arghh
2017-03-24 19:35:29
It's got to be the sum of two of the given numbers
It's got to be the sum of two of the given numbers
DPatrick
2017-03-24 19:35:40
Indeed, two of our four given sums must sum to $s$. That's because the pairwise sums come in three pairs of "opposites": $\{(a+b) \text{ and } (c+d)\}$, $\{(a+c) \text{ and } (b+d)\}$, and $\{(a+d) \text{ and } (b+c)\}$, so our four given sums must contain (at least) one of these pairs.
Indeed, two of our four given sums must sum to $s$. That's because the pairwise sums come in three pairs of "opposites": $\{(a+b) \text{ and } (c+d)\}$, $\{(a+c) \text{ and } (b+d)\}$, and $\{(a+d) \text{ and } (b+c)\}$, so our four given sums must contain (at least) one of these pairs.
DPatrick
2017-03-24 19:35:49
So what?
So what?
a1b2
2017-03-24 19:36:03
pick two largest and let x and y solve itself
pick two largest and let x and y solve itself
arghh
2017-03-24 19:36:06
Try 287 + 320, since we want maximum s?
Try 287 + 320, since we want maximum s?
rainbow10
2017-03-24 19:36:10
make 287 and 320 two of these pairs
make 287 and 320 two of these pairs
liyuxiao
2017-03-24 19:36:10
maximize the two numbers to get the greatest amount for s
maximize the two numbers to get the greatest amount for s
reelmathematician
2017-03-24 19:36:13
pick the 2 largest numbers
pick the 2 largest numbers
DPatrick
2017-03-24 19:36:16
Since we're trying to maximize $s$, we can hope that $s = 287 + 320 = 607$ is the sum of the two largest given numbers.
Since we're trying to maximize $s$, we can hope that $s = 287 + 320 = 607$ is the sum of the two largest given numbers.
willmathxu
2017-03-24 19:36:40
1821-1030=x+y
1821-1030=x+y
DPatrick
2017-03-24 19:36:45
This would make $x+y = 3(607) - 1030 = 1821 - 1030 = 791$.
This would make $x+y = 3(607) - 1030 = 1821 - 1030 = 791$.
GeronimoStilton
2017-03-24 19:36:59
How do we know this works?
How do we know this works?
DPatrick
2017-03-24 19:37:04
Good question -- is this actually possible?
Good question -- is this actually possible?
serd
2017-03-24 19:37:21
solve for a, b, c, and d
solve for a, b, c, and d
DPatrick
2017-03-24 19:37:35
Indeed it is: we can just try a configuration, such as $a+b = 189$, $a+c = 234$, $b+c = 287$, and $a+d = 320$.
Indeed it is: we can just try a configuration, such as $a+b = 189$, $a+c = 234$, $b+c = 287$, and $a+d = 320$.
vishwathganesan
2017-03-24 19:37:39
It says "distinct," so it may not work
It says "distinct," so it may not work
DPatrick
2017-03-24 19:38:12
Right, that's really the only worry. They could have played a really nasty trick on us and given us a "maximal" solution that has two numbers that are equal.
Right, that's really the only worry. They could have played a really nasty trick on us and given us a "maximal" solution that has two numbers that are equal.
GeneralCobra19
2017-03-24 19:38:23
You can just assign 189 to $a+b$ and onward...(234 to $a+c$, and onto the next two highest, which doesn't matter by the end) afterward, you can do some fun elimination(only three sums) to get desired b+c+2d.
You can just assign 189 to $a+b$ and onward...(234 to $a+c$, and onto the next two highest, which doesn't matter by the end) afterward, you can do some fun elimination(only three sums) to get desired b+c+2d.
DPatrick
2017-03-24 19:38:30
True. If you solve this (I won't take up our time doing so), you'll get $(a,b,c,d) = (68,121,166,252)$. (And for what it's worth, this gives $x = 121+252 = 373$ and $y = 166+252 = 418$ as the two missing sums, and indeed $373 + 418 = 791$, so that's a nice check.)
True. If you solve this (I won't take up our time doing so), you'll get $(a,b,c,d) = (68,121,166,252)$. (And for what it's worth, this gives $x = 121+252 = 373$ and $y = 166+252 = 418$ as the two missing sums, and indeed $373 + 418 = 791$, so that's a nice check.)
DPatrick
2017-03-24 19:38:49
So our answer is in fact $\boxed{791}$.
So our answer is in fact $\boxed{791}$.
DPatrick
2017-03-24 19:39:00
And in fact, it turns out this is not the only configuration -- we can swap the last two sums and get $(a,b,c,d) = \{51.5, 137.5, 182.5, 235.5\}$ works too, giving $(x,y) = (373,418)$ again.
And in fact, it turns out this is not the only configuration -- we can swap the last two sums and get $(a,b,c,d) = \{51.5, 137.5, 182.5, 235.5\}$ works too, giving $(x,y) = (373,418)$ again.
DPatrick
2017-03-24 19:39:25
One-third of the way through already!
One-third of the way through already!
DPatrick
2017-03-24 19:39:35
6. Find the sum of all positive integers $n$ such that $\sqrt{n^2 + 85n + 2017}$ is an integer.
6. Find the sum of all positive integers $n$ such that $\sqrt{n^2 + 85n + 2017}$ is an integer.
DPatrick
2017-03-24 19:39:49
Gee, it'd be nice if we could write what's under the radical in terms of a square.
Gee, it'd be nice if we could write what's under the radical in terms of a square.
thedoge
2017-03-24 19:40:03
Complete the square
Complete the square
legolego
2017-03-24 19:40:03
complete the square
complete the square
Funnybunny5246
2017-03-24 19:40:03
Complete square
Complete square
DPatrick
2017-03-24 19:40:10
Good idea!
Good idea!
DPatrick
2017-03-24 19:40:22
But first, it's a slight pain to complete the square when the middle coefficient is odd, so what can we do?
But first, it's a slight pain to complete the square when the middle coefficient is odd, so what can we do?
62861
2017-03-24 19:40:44
multiply radicand by 4
multiply radicand by 4
SomethingNeutral
2017-03-24 19:40:44
Multiply by 4/4
Multiply by 4/4
thedoge
2017-03-24 19:40:44
multiply the inside by 4
multiply the inside by 4
goodball
2017-03-24 19:40:44
Multiply by 2
Multiply by 2
awesomemaths
2017-03-24 19:40:44
we could times 2
we could times 2
letsgomath
2017-03-24 19:40:49
multiply by 4
multiply by 4
shark2018
2017-03-24 19:40:49
multiply by 2
multiply by 2
avy
2017-03-24 19:40:49
multiply the whole square root by 2 and so the inside by 4
multiply the whole square root by 2 and so the inside by 4
DPatrick
2017-03-24 19:40:54
I agree. Let's multiply under the radical by a 4, and put a $\dfrac12$ outside: \[ \sqrt{n^2 + 85n + 2017} = \frac12\sqrt{4n^2 + 4\cdot85n+ 8068}.\]
I agree. Let's multiply under the radical by a 4, and put a $\dfrac12$ outside: \[ \sqrt{n^2 + 85n + 2017} = \frac12\sqrt{4n^2 + 4\cdot85n+ 8068}.\]
DPatrick
2017-03-24 19:41:16
So now when we complete the square, we get $\dfrac12\sqrt{(2n+85)^2+843}$.
So now when we complete the square, we get $\dfrac12\sqrt{(2n+85)^2+843}$.
DPatrick
2017-03-24 19:41:33
(Trust me -- the 843 is correct.)
(Trust me -- the 843 is correct.)
DPatrick
2017-03-24 19:41:42
And if we want this to be an integer, let's give that integer a name: \[ \frac12\sqrt{(2n+85)^2 + 843} = m.\]
And if we want this to be an integer, let's give that integer a name: \[ \frac12\sqrt{(2n+85)^2 + 843} = m.\]
DPatrick
2017-03-24 19:41:46
Now what?
Now what?
letsgomath
2017-03-24 19:42:14
843 is the difference of squares
843 is the difference of squares
reedmj
2017-03-24 19:42:14
Get rid of the radical
Get rid of the radical
awesomemaths
2017-03-24 19:42:14
multiply 2 both sides
multiply 2 both sides
arghh
2017-03-24 19:42:14
Expand? (2n + 85)^2 + 843 = 4m^2
Expand? (2n + 85)^2 + 843 = 4m^2
DPatrick
2017-03-24 19:42:27
Right, if we multiply by 2 and then square, we get $843 = 4m^2 - (2n+85)^2$.
Right, if we multiply by 2 and then square, we get $843 = 4m^2 - (2n+85)^2$.
DPatrick
2017-03-24 19:42:33
Aha, that's a difference of squares!
Aha, that's a difference of squares!
SomethingNeutral
2017-03-24 19:42:46
Difference of squares so factor 843 (3*281)
Difference of squares so factor 843 (3*281)
shootingstar8
2017-03-24 19:42:50
Find the factors of 843
Find the factors of 843
DPatrick
2017-03-24 19:42:54
So it factors as $843 = (2m - 2n - 85)(2m + 2n + 85)$.
So it factors as $843 = (2m - 2n - 85)(2m + 2n + 85)$.
DPatrick
2017-03-24 19:43:04
Happily, the only ways we can factor $843$ into two positive integers are $843 = 1 \cdot 843 = 3 \cdot 281$ (note 281 is prime).
Happily, the only ways we can factor $843$ into two positive integers are $843 = 1 \cdot 843 = 3 \cdot 281$ (note 281 is prime).
DPatrick
2017-03-24 19:43:19
How do we use this info most efficiently to complete the problem?
How do we use this info most efficiently to complete the problem?
Ani10
2017-03-24 19:43:43
since n is positive, (phew) first factor is smaller than first, so only two cases
since n is positive, (phew) first factor is smaller than first, so only two cases
QuestForKnowledge
2017-03-24 19:43:43
second term must be bigger, solve system
second term must be bigger, solve system
DPatrick
2017-03-24 19:43:58
We could solve this completely, but we only care about $n$, right?
We could solve this completely, but we only care about $n$, right?
DPatrick
2017-03-24 19:44:14
How do we quickly isolate $n$?
How do we quickly isolate $n$?
fdas
2017-03-24 19:44:19
2n+85 is half of the difference
2n+85 is half of the difference
DPatrick
2017-03-24 19:44:34
Right! Or to say it another way: the difference in the two factors is $4n + 170$. This difference must equal $842$ or $278$.
Right! Or to say it another way: the difference in the two factors is $4n + 170$. This difference must equal $842$ or $278$.
DPatrick
2017-03-24 19:44:51
Solving $4n + 170 = 842$ gives $n = 168$.
Solving $4n + 170 = 278$ gives $n = 27$.
Solving $4n + 170 = 842$ gives $n = 168$.
Solving $4n + 170 = 278$ gives $n = 27$.
DPatrick
2017-03-24 19:45:18
To be fair, if we wanted to be complete, we should probably also check that $m$ is an integer. By summing the two factors we get $4m = 844$ or $4m = 284$, so $m=211$ or $m=71$ respectively.
To be fair, if we wanted to be complete, we should probably also check that $m$ is an integer. By summing the two factors we get $4m = 844$ or $4m = 284$, so $m=211$ or $m=71$ respectively.
celestialphoenix3768
2017-03-24 19:45:26
168+27=195
168+27=195
GeneralCobra19
2017-03-24 19:45:26
168+27 is 195
168+27 is 195
DPatrick
2017-03-24 19:45:35
And finally, the sum of the possible $n$'s is $168 + 27 = \boxed{195}$.
And finally, the sum of the possible $n$'s is $168 + 27 = \boxed{195}$.
DPatrick
2017-03-24 19:46:03
This might have been my favorite problem -- it tickled me for some reason.
This might have been my favorite problem -- it tickled me for some reason.
DPatrick
2017-03-24 19:46:17
7. Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx) = 2\log(x+2)$ has exactly one real solution.
7. Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx) = 2\log(x+2)$ has exactly one real solution.
arghh
2017-03-24 19:46:55
Get rid of the logs: kx = (x + 2)^2
Get rid of the logs: kx = (x + 2)^2
SomethingNeutral
2017-03-24 19:46:55
expand logs into exponents
expand logs into exponents
Archimedes15
2017-03-24 19:46:55
kx = (x+2)^2
kx = (x+2)^2
DPatrick
2017-03-24 19:47:02
Right -- we can use the log exponent rule to rewrite the equation as $\log(kx) = \log\left((x+2)^2\right)$.
Right -- we can use the log exponent rule to rewrite the equation as $\log(kx) = \log\left((x+2)^2\right)$.
DPatrick
2017-03-24 19:47:09
And we know that two logs are equal if and only if the numbers themselves are equal.
And we know that two logs are equal if and only if the numbers themselves are equal.
DPatrick
2017-03-24 19:47:16
In other words, we have $kx = (x+2)^2$.
In other words, we have $kx = (x+2)^2$.
DPatrick
2017-03-24 19:47:32
So we're just looking for values of $k$ in $[-500,500]$ such that $kx = (x+2)^2$ has exactly one real solution...right?
So we're just looking for values of $k$ in $[-500,500]$ such that $kx = (x+2)^2$ has exactly one real solution...right?
arghh
2017-03-24 19:47:57
Though we need to make sure that kx > 0 and x + 2 > 0
Though we need to make sure that kx > 0 and x + 2 > 0
gradysocool
2017-03-24 19:47:57
except no we must keep track of our original domains!!
except no we must keep track of our original domains!!
DPatrick
2017-03-24 19:48:17
Right, this is the trick to this problem. Not all solutions of $kx = (x+2)^2$ are necessarily solutions to our original equation!
Right, this is the trick to this problem. Not all solutions of $kx = (x+2)^2$ are necessarily solutions to our original equation!
DPatrick
2017-03-24 19:48:26
We can only take logs of positive numbers.
We can only take logs of positive numbers.
DPatrick
2017-03-24 19:48:32
That is, we need $kx > 0$ and $x+2 > 0$ (or $x > -2$) in order to have a valid solution.
That is, we need $kx > 0$ and $x+2 > 0$ (or $x > -2$) in order to have a valid solution.
DPatrick
2017-03-24 19:48:49
So we can rewrite the question without logs if we put this condition in:
So we can rewrite the question without logs if we put this condition in:
DPatrick
2017-03-24 19:48:53
7. Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $kx = (x+2)^2$, with the restrictions $kx > 0$ and $x > -2$, has exactly one real solution.
7. Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $kx = (x+2)^2$, with the restrictions $kx > 0$ and $x > -2$, has exactly one real solution.
DPatrick
2017-03-24 19:49:07
Now what?
Now what?
fdas
2017-03-24 19:49:30
Move the kx on the other side and use quadratic formula
Move the kx on the other side and use quadratic formula
Funnybunny5246
2017-03-24 19:49:30
Expand square and move to one side
Expand square and move to one side
thedoge
2017-03-24 19:49:30
expand and use quadratic formula
expand and use quadratic formula
DPatrick
2017-03-24 19:49:56
I like the "expand and move to one side" idea -- let's make this look like a "standard" quadratic -- but I'm not as fond of the quadratic formula here. Let's see...
I like the "expand and move to one side" idea -- let's make this look like a "standard" quadratic -- but I'm not as fond of the quadratic formula here. Let's see...
DPatrick
2017-03-24 19:50:03
7. Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $x^2 + (4-k)x + 4 = 0$, with the restrictions $kx > 0$ and $x > -2$, has exactly one real solution.
7. Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $x^2 + (4-k)x + 4 = 0$, with the restrictions $kx > 0$ and $x > -2$, has exactly one real solution.
DPatrick
2017-03-24 19:50:18
(All I did was rearrange the quadratic.)
(All I did was rearrange the quadratic.)
Fibonachos
2017-03-24 19:50:35
Look at the discriminant!!
Look at the discriminant!!
letsgomath
2017-03-24 19:50:35
use the discriminant
use the discriminant
DPatrick
2017-03-24 19:50:44
Sure, let's just look at the discriminant first.
Sure, let's just look at the discriminant first.
DPatrick
2017-03-24 19:50:52
The discriminant of this quadratic is $(4-k)^2 - 16$.
The discriminant of this quadratic is $(4-k)^2 - 16$.
DPatrick
2017-03-24 19:51:10
So there are no real roots if $0 < k < 8$, a double (real) root if $k=0$ or $k=8$, and two real roots if $k<0$ or $k>8$.
So there are no real roots if $0 < k < 8$, a double (real) root if $k=0$ or $k=8$, and two real roots if $k<0$ or $k>8$.
DPatrick
2017-03-24 19:51:19
We can't have $0 < k < 8$ -- there aren't any solutions at all.
We can't have $0 < k < 8$ -- there aren't any solutions at all.
vishwathganesan
2017-03-24 19:51:25
k=0 works, but it doesn't fit restrictions
k=0 works, but it doesn't fit restrictions
fdas
2017-03-24 19:51:25
k cant equal 0
k cant equal 0
DPatrick
2017-03-24 19:51:34
We can't have $k=0$, because of the $kx > 0$ inequality.
We can't have $k=0$, because of the $kx > 0$ inequality.
pslv
2017-03-24 19:51:45
k=8
k=8
BooBooTM
2017-03-24 19:51:45
8 works, right?
8 works, right?
DPatrick
2017-03-24 19:52:05
If $k=8$, then $x=2$ is a repeated root, and that's valid with our inequalities. So $k=8$ works.
If $k=8$, then $x=2$ is a repeated root, and that's valid with our inequalities. So $k=8$ works.
DPatrick
2017-03-24 19:52:10
What happens if $k>8$?
What happens if $k>8$?
Archimedes15
2017-03-24 19:52:32
two real roots... distinct
two real roots... distinct
legolego
2017-03-24 19:52:32
two distinct valid real roots
two distinct valid real roots
skiboy32
2017-03-24 19:52:32
then, we get two positive roots, so nope
then, we get two positive roots, so nope
DPatrick
2017-03-24 19:52:53
We get two real roots, and they're both positive (since their sum is $k-4$ and their product is $4$).
We get two real roots, and they're both positive (since their sum is $k-4$ and their product is $4$).
DPatrick
2017-03-24 19:53:14
So these roots are both valid for our original problem. That's no good, because we only want one root.
So these roots are both valid for our original problem. That's no good, because we only want one root.
DPatrick
2017-03-24 19:53:19
So none of $k>8$ work.
So none of $k>8$ work.
DPatrick
2017-03-24 19:53:23
What about $k<0$?
What about $k<0$?
SomethingNeutral
2017-03-24 19:53:40
2 roots too
2 roots too
DPatrick
2017-03-24 19:53:57
Right: there are also 2 real roots of our quadratic. But what do we know about them?
Right: there are also 2 real roots of our quadratic. But what do we know about them?
DPatrick
2017-03-24 19:54:11
Remember, they sum to $k-4$ and their product is 4.
Remember, they sum to $k-4$ and their product is 4.
DPatrick
2017-03-24 19:55:06
We have two roots whose sum is negative but whose product is positive. What does that tell us?
We have two roots whose sum is negative but whose product is positive. What does that tell us?
serd
2017-03-24 19:55:18
theyre both negative
theyre both negative
QuestForKnowledge
2017-03-24 19:55:18
2 negative roots
2 negative roots
arghh
2017-03-24 19:55:21
They are both negative
They are both negative
shootingstar8
2017-03-24 19:55:21
The roots are negative?
The roots are negative?
DPatrick
2017-03-24 19:55:46
Right. If $k<0$, then the quadratic has two negative roots for $x$. That's good because that will make $kx > 0$ true.
Right. If $k<0$, then the quadratic has two negative roots for $x$. That's good because that will make $kx > 0$ true.
DPatrick
2017-03-24 19:55:56
In which case will exactly one of these roots satisfy $x > -2$?
In which case will exactly one of these roots satisfy $x > -2$?
reelmathematician
2017-03-24 19:56:38
one root less than -2 and one more
one root less than -2 and one more
DPatrick
2017-03-24 19:56:46
Right! If two different negative numbers multiply to give 4, then one must be smaller than $-2$ and one must be larger than $-2$.
Right! If two different negative numbers multiply to give 4, then one must be smaller than $-2$ and one must be larger than $-2$.
DPatrick
2017-03-24 19:57:31
So when $k < 0$, our quadratic has two distinct negative roots: one less than $-2$ and one greater than $-2$. Only the second one satisfies $x > -2$, so only the second one is also a solution to our original problem.
So when $k < 0$, our quadratic has two distinct negative roots: one less than $-2$ and one greater than $-2$. Only the second one satisfies $x > -2$, so only the second one is also a solution to our original problem.
vishwathganesan
2017-03-24 19:57:37
so all k<0 work?
so all k<0 work?
DPatrick
2017-03-24 19:57:54
Exactly. All $k < 0$ will produce exactly one solution to our original problem.
Exactly. All $k < 0$ will produce exactly one solution to our original problem.
flamescarlet
2017-03-24 19:58:14
so 501!
so 501!
SomethingNeutral
2017-03-24 19:58:14
so answer $\boxed{501}?$
so answer $\boxed{501}?$
SimonSun
2017-03-24 19:58:14
so 500 negative values
so 500 negative values
GeronimoStilton
2017-03-24 19:58:14
$501$!
$501$!
DPatrick
2017-03-24 19:58:16
Therefore, in the given interval, all $-500 \le k \le -1$ work, along with $k=8$.
Therefore, in the given interval, all $-500 \le k \le -1$ work, along with $k=8$.
DPatrick
2017-03-24 19:58:31
So the answer is $\boxed{501}$.
So the answer is $\boxed{501}$.
DPatrick
2017-03-24 19:59:19
8. Find the number of positive integers $n$ less than 2017 such that \[ 1 + n + \frac{n^2}{2!} + \frac{n^3}{3!} + \frac{n^4}{4!} + \frac{n^5}{5!} + \frac{n^6}{6!} \] is an integer.
8. Find the number of positive integers $n$ less than 2017 such that \[ 1 + n + \frac{n^2}{2!} + \frac{n^3}{3!} + \frac{n^4}{4!} + \frac{n^5}{5!} + \frac{n^6}{6!} \] is an integer.
GeronimoStilton
2017-03-24 19:59:42
This looks like $e^n$.
This looks like $e^n$.
DPatrick
2017-03-24 20:00:03
Indeed, if you know some calculus, you'll recognize this as the first 8 terms of the Taylor series for $e^n$.
Indeed, if you know some calculus, you'll recognize this as the first 8 terms of the Taylor series for $e^n$.
DPatrick
2017-03-24 20:00:16
You can completely ignore that last sentence though, as it is of absolutely no help in solving the problem.
You can completely ignore that last sentence though, as it is of absolutely no help in solving the problem.
DPatrick
2017-03-24 20:00:27
(And I guess it's only the first 7 terms.)
(And I guess it's only the first 7 terms.)
Fibonachos
2017-03-24 20:00:33
We don't need to worry about $1+n$
We don't need to worry about $1+n$
DPatrick
2017-03-24 20:00:43
That's a start: clearly we don't have to worry about the $1+n$ part of the expression. So let's simplify the problem.
That's a start: clearly we don't have to worry about the $1+n$ part of the expression. So let's simplify the problem.
DPatrick
2017-03-24 20:00:47
8. Find the number of positive integers $n$ less than 2017 such that \[ \frac{n^2}{2!} + \frac{n^3}{3!} + \frac{n^4}{4!} + \frac{n^5}{5!} + \frac{n^6}{6!} \] is an integer.
8. Find the number of positive integers $n$ less than 2017 such that \[ \frac{n^2}{2!} + \frac{n^3}{3!} + \frac{n^4}{4!} + \frac{n^5}{5!} + \frac{n^6}{6!} \] is an integer.
flamescarlet
2017-03-24 20:01:00
we can make 6! common denominator
we can make 6! common denominator
hunter117
2017-03-24 20:01:11
we can factor out $n^2$
we can factor out $n^2$
DPatrick
2017-03-24 20:01:12
Sounds good. We can factor out $n^2$, and put the rest of the terms over a common denominator:
Sounds good. We can factor out $n^2$, and put the rest of the terms over a common denominator:
DPatrick
2017-03-24 20:01:16
8. Find the number of positive integers $n$ less than 2017 such that \[ \frac{n^2(360 + 120n + 30n^2 + 6n^3 + n^4)}{720} \] is an integer.
8. Find the number of positive integers $n$ less than 2017 such that \[ \frac{n^2(360 + 120n + 30n^2 + 6n^3 + n^4)}{720} \] is an integer.
Ani10
2017-03-24 20:01:27
prime factorize denominators?
prime factorize denominators?
giacomorizzo
2017-03-24 20:01:27
does n must have 2, 3, and 5 as factors
does n must have 2, 3, and 5 as factors
skiboy32
2017-03-24 20:01:27
it can be a multiple of $2\cdot 3\cdot 5=30$
it can be a multiple of $2\cdot 3\cdot 5=30$
DPatrick
2017-03-24 20:01:35
Right. We have to worry that the denominator "cancels out" with the numerator to leave an integer.
Right. We have to worry that the denominator "cancels out" with the numerator to leave an integer.
DPatrick
2017-03-24 20:01:40
So prime factors of these numbers might be more helpful.
So prime factors of these numbers might be more helpful.
DPatrick
2017-03-24 20:01:44
8. Find the number of positive integers $n$ less than 2017 such that \[ \frac{n^2\left((2^3\cdot3^2\cdot 5) + (2^3 \cdot 3 \cdot 5)n + (2 \cdot 3 \cdot 5)n^2 + (2 \cdot 3)n^3 + n^4)\right)}{2^4 \cdot 3^2 \cdot 5} \] is an integer.
8. Find the number of positive integers $n$ less than 2017 such that \[ \frac{n^2\left((2^3\cdot3^2\cdot 5) + (2^3 \cdot 3 \cdot 5)n + (2 \cdot 3 \cdot 5)n^2 + (2 \cdot 3)n^3 + n^4)\right)}{2^4 \cdot 3^2 \cdot 5} \] is an integer.
DPatrick
2017-03-24 20:01:59
So we've got to worry about 2's, 3's, and 5's.
So we've got to worry about 2's, 3's, and 5's.
DPatrick
2017-03-24 20:02:11
And the nice thing is: we can worry about them separately!
And the nice thing is: we can worry about them separately!
DPatrick
2017-03-24 20:02:23
What's the condition on $n$ that will cancel the $2^4$ in the denominator?
What's the condition on $n$ that will cancel the $2^4$ in the denominator?
legolego
2017-03-24 20:02:49
it is even
it is even
fdas
2017-03-24 20:02:49
multiple of 2
multiple of 2
shootingstar8
2017-03-24 20:02:49
Having 2 as a factor
Having 2 as a factor
skiboy32
2017-03-24 20:02:49
$n$ is divisible by $2.$
$n$ is divisible by $2.$
hunter117
2017-03-24 20:02:49
$n$ must have a factor of $2^1$
$n$ must have a factor of $2^1$
DPatrick
2017-03-24 20:03:16
Certainly if $n$ is odd, then the numerator is odd ($n^2$) times odd (every term in the big parentheses is even except for $n^4$, which is odd), so the numerator overall is odd. We don't even cancel a single 2, much less $2^4$!
Certainly if $n$ is odd, then the numerator is odd ($n^2$) times odd (every term in the big parentheses is even except for $n^4$, which is odd), so the numerator overall is odd. We don't even cancel a single 2, much less $2^4$!
DPatrick
2017-03-24 20:03:31
If $n$ is merely even, do we have enough powers of 2 in the numerator?
If $n$ is merely even, do we have enough powers of 2 in the numerator?
legolego
2017-03-24 20:03:48
yes
yes
Funnybunny5246
2017-03-24 20:03:48
Yes
Yes
avy
2017-03-24 20:03:48
yes
yes
DPatrick
2017-03-24 20:03:57
Yes, way more than enough: we get at least a $2^2$ from the outer $n^2$, and each term inside the parentheses has at least $2^3$.
Yes, way more than enough: we get at least a $2^2$ from the outer $n^2$, and each term inside the parentheses has at least $2^3$.
DPatrick
2017-03-24 20:04:12
So we cancel the 2's if and only if $n$ is even.
So we cancel the 2's if and only if $n$ is even.
DPatrick
2017-03-24 20:04:17
How about the 3's? What's the condition on $n$ to cancel $3^2$?
How about the 3's? What's the condition on $n$ to cancel $3^2$?
carzland
2017-03-24 20:04:40
$n$ must be divisible by 3.
$n$ must be divisible by 3.
FieryEntei
2017-03-24 20:04:40
multiples of 3?
multiples of 3?
SimonSun
2017-03-24 20:04:40
Similar logic about 3, n is divisible by 3
Similar logic about 3, n is divisible by 3
colinyao
2017-03-24 20:04:40
n is divisible by 3
n is divisible by 3
SomethingNeutral
2017-03-24 20:04:40
3|n
3|n
vishwathganesan
2017-03-24 20:04:40
div. by 3
div. by 3
DPatrick
2017-03-24 20:04:49
If $n$ is not a multiple of 3, then we're multiplying two non-multiples of 3 in the numerator, and we don't even cancel a single 3.
If $n$ is not a multiple of 3, then we're multiplying two non-multiples of 3 in the numerator, and we don't even cancel a single 3.
DPatrick
2017-03-24 20:05:05
If $n$ is a multiple of 3, then we get our $3^2$ just from the outer $n^2$ term, and don't need to worry about the 3's inside the big parentheses.
If $n$ is a multiple of 3, then we get our $3^2$ just from the outer $n^2$ term, and don't need to worry about the 3's inside the big parentheses.
DPatrick
2017-03-24 20:05:13
So we cancel the 3's if and only if $n$ is a multiple of 3.
So we cancel the 3's if and only if $n$ is a multiple of 3.
DPatrick
2017-03-24 20:05:18
How about the 5's? What's the condition on $n$ to cancel the 5 in the denominator?
How about the 5's? What's the condition on $n$ to cancel the 5 in the denominator?
flamescarlet
2017-03-24 20:05:46
n is a multiple of 5 or 6n^3 + n^4 is a multiple of 5
n is a multiple of 5 or 6n^3 + n^4 is a multiple of 5
bertasaurus
2017-03-24 20:05:51
n must be divisible by 5 or (n+1) must be divisible by 5
n must be divisible by 5 or (n+1) must be divisible by 5
DPatrick
2017-03-24 20:05:59
Right. First, if $n$ is a multiple of 5, we're set, because everything in sight is then a multiple of 5.
Right. First, if $n$ is a multiple of 5, we're set, because everything in sight is then a multiple of 5.
DPatrick
2017-03-24 20:06:24
Then, if $n$ is not a multiple of 5, we need $6n^3 + n^4$ to be a multiple of 5, because the other terms are already multiples of 5.
Then, if $n$ is not a multiple of 5, we need $6n^3 + n^4$ to be a multiple of 5, because the other terms are already multiples of 5.
DPatrick
2017-03-24 20:06:34
And this is a multiple of 5 if and only if $1 + n$ is a multiple of 5 (we can factor out $n^3$, and 6 is the same as 1 modulo 5).
And this is a multiple of 5 if and only if $1 + n$ is a multiple of 5 (we can factor out $n^3$, and 6 is the same as 1 modulo 5).
DPatrick
2017-03-24 20:06:53
So we cancel the 5's if and only if $n$ is either a multiple of 5 or one less than a multiple of 5.
So we cancel the 5's if and only if $n$ is either a multiple of 5 or one less than a multiple of 5.
DPatrick
2017-03-24 20:07:03
How do we combine all this info to count the $n$ that work?
How do we combine all this info to count the $n$ that work?
hunter117
2017-03-24 20:07:17
chinese remainder theorem
chinese remainder theorem
flamescarlet
2017-03-24 20:07:17
chinese remainder theorem!
chinese remainder theorem!
letsgomath
2017-03-24 20:07:20
has to satisfy all three mod congruences
has to satisfy all three mod congruences
DPatrick
2017-03-24 20:07:24
By the Chinese Remainder Theorem, we can combine our facts about 2's, 3's, and 5's into a single fact about what $n$ looks like in relationship to multiples of 30.
By the Chinese Remainder Theorem, we can combine our facts about 2's, 3's, and 5's into a single fact about what $n$ looks like in relationship to multiples of 30.
FTW
2017-03-24 20:07:46
0 or 24 mod 30
0 or 24 mod 30
legolego
2017-03-24 20:07:46
equivalent to 0 or 24 mod 30
equivalent to 0 or 24 mod 30
DPatrick
2017-03-24 20:08:04
Specifically, we need $n$ to be a multiple of 30 or 6 less than a multiple of 30. Or, to say it another way, we must have $n$ equivalent to 0 or 24 mod 30.
Specifically, we need $n$ to be a multiple of 30 or 6 less than a multiple of 30. Or, to say it another way, we must have $n$ equivalent to 0 or 24 mod 30.
Mathguy1492
2017-03-24 20:08:14
$30n$ or $30n-6$
$30n$ or $30n-6$
DPatrick
2017-03-24 20:08:21
Right.
Right.
DPatrick
2017-03-24 20:08:30
For every block of 30 integers, 2 of them will work.
For every block of 30 integers, 2 of them will work.
Fibonachos
2017-03-24 20:08:36
There are 67 multiples of 30
There are 67 multiples of 30
DPatrick
2017-03-24 20:08:43
Yes -- since $2017 = 67 \cdot 30 + 7$, we have 67 full blocks of 30, plus 7 extra.
Yes -- since $2017 = 67 \cdot 30 + 7$, we have 67 full blocks of 30, plus 7 extra.
DPatrick
2017-03-24 20:09:31
Actually only 6 extra since we want "less than 2017" from the problem statement.
Actually only 6 extra since we want "less than 2017" from the problem statement.
DPatrick
2017-03-24 20:09:57
But the "extra" are 2011, 2012, ..., 2016, and none of those work. (These are 1,2,...,6 mod 30, and the only numbers that work are either 0 or 24 mod 30.)
But the "extra" are 2011, 2012, ..., 2016, and none of those work. (These are 1,2,...,6 mod 30, and the only numbers that work are either 0 or 24 mod 30.)
shark2018
2017-03-24 20:10:07
67*2= 134
67*2= 134
flamescarlet
2017-03-24 20:10:07
so 67 * 2 = 134
so 67 * 2 = 134
arghh
2017-03-24 20:10:07
67 * 2
67 * 2
DPatrick
2017-03-24 20:10:19
Each of the 67 blocks of 30 have two numbers that work. So there are $2 \cdot 67 = \boxed{134}$ values of $n$ that work.
Each of the 67 blocks of 30 have two numbers that work. So there are $2 \cdot 67 = \boxed{134}$ values of $n$ that work.
DPatrick
2017-03-24 20:10:40
9. A special deck of cards contains 49 cards, each labeled with a number from 1 to 7 and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and still have at least one card of each color and at least one card with each number is $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
9. A special deck of cards contains 49 cards, each labeled with a number from 1 to 7 and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and still have at least one card of each color and at least one card with each number is $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
DPatrick
2017-03-24 20:11:03
Let's give these situations names, so we can more easily discuss them.
Let's give these situations names, so we can more easily discuss them.
DPatrick
2017-03-24 20:11:08
Let's say a "good" hand is a hand of 8 cards which contains all 7 numbers and all 7 colors.
Let's say a "great" hand is a good hand that has a discard that still leaves all 7 numbers and all 7 colors.
Let's say a "good" hand is a hand of 8 cards which contains all 7 numbers and all 7 colors.
Let's say a "great" hand is a good hand that has a discard that still leaves all 7 numbers and all 7 colors.
DPatrick
2017-03-24 20:11:15
So we want the probability that a good hand is also great.
So we want the probability that a good hand is also great.
DPatrick
2017-03-24 20:11:29
Any observations we can make before we start counting?
Any observations we can make before we start counting?
Mudkipswims42
2017-03-24 20:11:54
One duplicate
One duplicate
GeronimoStilton
2017-03-24 20:12:01
Well, by the Pigeonhole Principle, she has 2 of one color, and 2 of one number.
Well, by the Pigeonhole Principle, she has 2 of one color, and 2 of one number.
DPatrick
2017-03-24 20:12:14
Right: in any good hand, there are 2 cards with the same color, and 2 cards with the same number.
Right: in any good hand, there are 2 cards with the same color, and 2 cards with the same number.
DPatrick
2017-03-24 20:12:32
And what makes a great hand great?
And what makes a great hand great?
ESWN123456789
2017-03-24 20:12:50
If the hand is great, the double number and double color overlap
If the hand is great, the double number and double color overlap
vishwathganesan
2017-03-24 20:12:50
in a great hand, the "bonus" color/number are the same card
in a great hand, the "bonus" color/number are the same card
DPatrick
2017-03-24 20:13:09
Right! We know every good hand must have exactly 2 cards with the same number (let's call that the "special number"), and exactly 2 cards with the same color (the "special color").
Right! We know every good hand must have exactly 2 cards with the same number (let's call that the "special number"), and exactly 2 cards with the same color (the "special color").
DPatrick
2017-03-24 20:13:26
When the special number and the special color appear on the same card, then we can discard that special card and still be left will all 7 numbers and all 7 colors. So that's what makes a good hand great.
When the special number and the special color appear on the same card, then we can discard that special card and still be left will all 7 numbers and all 7 colors. So that's what makes a good hand great.
DPatrick
2017-03-24 20:13:46
The hands that are good but not great have the special number on non-special colors and the special color on non-special numbers.
The hands that are good but not great have the special number on non-special colors and the special color on non-special numbers.
DPatrick
2017-03-24 20:13:57
With these observations in place, let's try to count!
With these observations in place, let's try to count!
DPatrick
2017-03-24 20:14:01
How many great hands are there?
How many great hands are there?
JJShan26
2017-03-24 20:14:10
there are 7! ways to have all 7 numbers and 7 colors on 7 cards
there are 7! ways to have all 7 numbers and 7 colors on 7 cards
DPatrick
2017-03-24 20:14:29
Right. We could first choose the 7 cards that remain after the discard. These have to have 7 numbers and 7 colors on 7 cards.
Right. We could first choose the 7 cards that remain after the discard. These have to have 7 numbers and 7 colors on 7 cards.
DPatrick
2017-03-24 20:14:50
We can construct such a 7-card set.
We need to choose one of the 7 colors for the 1.
Then we need to choose one of the 6 remaining colors for the 2.
Then we need to choose one of the 5 remaining colors for the 3.
And so on.
We can construct such a 7-card set.
We need to choose one of the 7 colors for the 1.
Then we need to choose one of the 6 remaining colors for the 2.
Then we need to choose one of the 5 remaining colors for the 3.
And so on.
DPatrick
2017-03-24 20:15:01
So there are $7!$ ways to choose a perfect 7-card set.
So there are $7!$ ways to choose a perfect 7-card set.
DPatrick
2017-03-24 20:15:09
And then how many ways to choose an 8th card to make a great hand?
And then how many ways to choose an 8th card to make a great hand?
letsgomath
2017-03-24 20:15:25
42
42
legolego
2017-03-24 20:15:25
42
42
DPatrick
2017-03-24 20:15:40
Right. We can choose any card left in the deck! That card will be the "special card" in our 8-card great hand.
Right. We can choose any card left in the deck! That card will be the "special card" in our 8-card great hand.
DPatrick
2017-03-24 20:15:48
Since there are 42 cards left, there are $7! \cdot 42$ ways to make a great hand.
Since there are 42 cards left, there are $7! \cdot 42$ ways to make a great hand.
DPatrick
2017-03-24 20:16:02
A couple people are asking if we double-counted when we did this...did we?
A couple people are asking if we double-counted when we did this...did we?
SimonSun
2017-03-24 20:16:28
no
no
Metal_Bender19
2017-03-24 20:16:28
no
no
DPatrick
2017-03-24 20:16:59
I don't think we did. If we have a great hand, then the discarded card is uniquely determined, and the 7-card set it leaves behind is also uniquely determined.
I don't think we did. If we have a great hand, then the discarded card is uniquely determined, and the 7-card set it leaves behind is also uniquely determined.
DPatrick
2017-03-24 20:17:17
(If we discarded any other card, we'll be losing a non-special number or color.)
(If we discarded any other card, we'll be losing a non-special number or color.)
DPatrick
2017-03-24 20:17:27
So there are $7! \cdot 42$ great hands.
So there are $7! \cdot 42$ great hands.
DPatrick
2017-03-24 20:17:35
Next, how many "good but not great" hands are there?
Next, how many "good but not great" hands are there?
DPatrick
2017-03-24 20:17:46
Recall: the hands that are good but not great have the special number on non-special colors and the special color on non-special numbers.
Recall: the hands that are good but not great have the special number on non-special colors and the special color on non-special numbers.
DPatrick
2017-03-24 20:18:03
How can we construct such a hand?
How can we construct such a hand?
DPatrick
2017-03-24 20:18:56
Generally, when we constructively count, we try to count the most restrictive part of the construction first.
Generally, when we constructively count, we try to count the most restrictive part of the construction first.
DPatrick
2017-03-24 20:19:22
Here the restriction is that we need two cards with the same number, and we need two cards with the same color, but these four cards are all different.
Here the restriction is that we need two cards with the same number, and we need two cards with the same color, but these four cards are all different.
DPatrick
2017-03-24 20:19:49
There are probably different ways you could do it, but here's what I suggest:
There are probably different ways you could do it, but here's what I suggest:
DPatrick
2017-03-24 20:19:57
We have 7 choices for the special number.
We have 7 choices for the special number.
DPatrick
2017-03-24 20:20:03
Then, $\dbinom72$ choices for the two colors that get paired with the special number.
Then, $\dbinom72$ choices for the two colors that get paired with the special number.
DPatrick
2017-03-24 20:20:15
Then, 5 choices for the special color (we can't use either of the two colors from above, or else we'll end up making a special card, and we'll get a great hand).
Then, 5 choices for the special color (we can't use either of the two colors from above, or else we'll end up making a special card, and we'll get a great hand).
DPatrick
2017-03-24 20:20:26
Then, $\dbinom62$ choices for the two numbers that get paired with the special color (we can't use the special number, for the same reason).
Then, $\dbinom62$ choices for the two numbers that get paired with the special color (we can't use the special number, for the same reason).
DPatrick
2017-03-24 20:20:40
This leaves 4 more cards, 4 unused numbers, and 4 unused colors. How many ways to finish the hand from here?
This leaves 4 more cards, 4 unused numbers, and 4 unused colors. How many ways to finish the hand from here?
letsgomath
2017-03-24 20:20:58
24
24
Metal_Bender19
2017-03-24 20:20:58
4!
4!
GeronimoStilton
2017-03-24 20:20:58
We have $24$ choices for that.
We have $24$ choices for that.
shark2018
2017-03-24 20:20:58
4!
4!
Mathguy1492
2017-03-24 20:20:58
24
24
DPatrick
2017-03-24 20:21:20
We can pick these in $4!$ ways (we have to match up numbers with colors, like we did with our 7-card set from before).
We can pick these in $4!$ ways (we have to match up numbers with colors, like we did with our 7-card set from before).
DPatrick
2017-03-24 20:21:29
So there are a total of \[7 \cdot \binom72 \cdot 5 \cdot \binom62 \cdot 4!\] ways to make a good-not-great hand.
So there are a total of \[7 \cdot \binom72 \cdot 5 \cdot \binom62 \cdot 4!\] ways to make a good-not-great hand.
DPatrick
2017-03-24 20:21:57
If we write this out, this is \[\frac{7 \cdot 7 \cdot 6 \cdot 5 \cdot 6 \cdot 5 \cdot 4!}{4},\] so if we pull out a factor of $7!$ (to make it look more like our "great" count), we get $7! \cdot \dfrac{7 \cdot 6 \cdot 5}{4} = 7! \cdot \dfrac{105}{2}$.
If we write this out, this is \[\frac{7 \cdot 7 \cdot 6 \cdot 5 \cdot 6 \cdot 5 \cdot 4!}{4},\] so if we pull out a factor of $7!$ (to make it look more like our "great" count), we get $7! \cdot \dfrac{7 \cdot 6 \cdot 5}{4} = 7! \cdot \dfrac{105}{2}$.
DPatrick
2017-03-24 20:22:20
And how do we finish the problem from here?
And how do we finish the problem from here?
vishwathganesan
2017-03-24 20:22:46
find probability
find probability
DPatrick
2017-03-24 20:22:55
We want \[P = \frac{\text{great}}{\text{good}} = \frac{\text{great}}{\text{great}+\text{good-not-great}}.\]
We want \[P = \frac{\text{great}}{\text{good}} = \frac{\text{great}}{\text{great}+\text{good-not-great}}.\]
DPatrick
2017-03-24 20:23:10
Plugging in our counts, this is \[P = \frac{7! \cdot 42}{7! \cdot 42 + 7! \cdot \frac{105}{2}}.\]
Plugging in our counts, this is \[P = \frac{7! \cdot 42}{7! \cdot 42 + 7! \cdot \frac{105}{2}}.\]
letsgomath
2017-03-24 20:23:23
factor out 7!
factor out 7!
flamescarlet
2017-03-24 20:23:23
factor out 7!
factor out 7!
DPatrick
2017-03-24 20:23:33
The $7!$'s cancel, and multiplying through by 2 we're left with $\dfrac{84}{84 + 105}$.
The $7!$'s cancel, and multiplying through by 2 we're left with $\dfrac{84}{84 + 105}$.
Mathguy1492
2017-03-24 20:23:42
Then factor out another 7
Then factor out another 7
letsgomath
2017-03-24 20:23:52
21
21
DPatrick
2017-03-24 20:23:54
Indeed, there's a common factor of 21 here.
Indeed, there's a common factor of 21 here.
DPatrick
2017-03-24 20:24:05
This is just \[\frac{84}{84+105}=\frac4{4+5}=\frac49.\]
This is just \[\frac{84}{84+105}=\frac4{4+5}=\frac49.\]
DPatrick
2017-03-24 20:24:15
So the final answer is $4+9 = \boxed{013}$.
So the final answer is $4+9 = \boxed{013}$.
DPatrick
2017-03-24 20:24:47
Someone asked me why we didn't just could all the "good" hands. I didn't find a good way to do that except by casework into "great" and "good-not-great".
Someone asked me why we didn't just could all the "good" hands. I didn't find a good way to do that except by casework into "great" and "good-not-great".
DPatrick
2017-03-24 20:25:20
Now's a good time for a short break, before we get to the double-digit problems!
Now's a good time for a short break, before we get to the double-digit problems!
DPatrick
2017-03-24 20:25:24
Let's resume at :30 past.
Let's resume at :30 past.
DPatrick
2017-03-24 20:30:09
And we're back!
And we're back!
DPatrick
2017-03-24 20:30:13
10. Rectangle $ABCD$ has side lengths $AB = 84$ and $AD = 42$. Point $M$ is the midpoint of $\overline{AD}$, point $N$ is the trisection point of $\overline{AB}$ closer to $A$, and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$. Point $P$ lies on the quadrilateral $BCON$, and $\overline{BP}$ bisects the area of $BCON$. Find the area of $\triangle CDP$.
10. Rectangle $ABCD$ has side lengths $AB = 84$ and $AD = 42$. Point $M$ is the midpoint of $\overline{AD}$, point $N$ is the trisection point of $\overline{AB}$ closer to $A$, and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$. Point $P$ lies on the quadrilateral $BCON$, and $\overline{BP}$ bisects the area of $BCON$. Find the area of $\triangle CDP$.
DPatrick
2017-03-24 20:30:26
Mmmmmm......bacon.....
Mmmmmm......bacon.....
GeronimoStilton
2017-03-24 20:30:44
Diagram?
Diagram?
shootingstar8
2017-03-24 20:30:44
Diagram!
Diagram!
awesomemaths
2017-03-24 20:30:44
diagram
diagram
DPatrick
2017-03-24 20:30:53
Sure, let's draw a picture. It's also pretty easy to add coordinates in case we want them:
Sure, let's draw a picture. It's also pretty easy to add coordinates in case we want them:
DPatrick
2017-03-24 20:30:57
GeneralCobra19
2017-03-24 20:31:07
Can we coordinate bash this one?
Can we coordinate bash this one?
DPatrick
2017-03-24 20:31:20
That seems promising to me. What are the coordinates of point $O$?
That seems promising to me. What are the coordinates of point $O$?
legolego
2017-03-24 20:31:55
(12, 24)
(12, 24)
Skywalker64
2017-03-24 20:31:55
(12,24)
(12,24)
DPatrick
2017-03-24 20:32:02
Line $CM$ has slope $\dfrac14$ and $y$-intercept 21, so its equation is $y = \dfrac14x + 21$.
Line $DN$ has slope $-\dfrac32$ and $y$-intercept 42, so its equation is $y = -\dfrac32x + 42$.
Line $CM$ has slope $\dfrac14$ and $y$-intercept 21, so its equation is $y = \dfrac14x + 21$.
Line $DN$ has slope $-\dfrac32$ and $y$-intercept 42, so its equation is $y = -\dfrac32x + 42$.
DPatrick
2017-03-24 20:32:14
Equating these and solving gives $O = (12,24)$.
Equating these and solving gives $O = (12,24)$.
DPatrick
2017-03-24 20:32:30
(You can work it out on your own if you don't believe me.)
(You can work it out on your own if you don't believe me.)
DPatrick
2017-03-24 20:32:41
Now what?
Now what?
jonzli123
2017-03-24 20:32:53
find area of BCON
find area of BCON
carzland
2017-03-24 20:32:53
Find the area.
Find the area.
shootingstar8
2017-03-24 20:32:56
Find point P
Find point P
DPatrick
2017-03-24 20:33:09
We need to figure out where $P$ is. So that probably means finding the area $[BCON]$.
We need to figure out where $P$ is. So that probably means finding the area $[BCON]$.
DPatrick
2017-03-24 20:33:19
We can find the area of $BCON$ by drawing in $\overline{BO}$ and computing the areas of the two triangles $BON$ and $BOC$.
We can find the area of $BCON$ by drawing in $\overline{BO}$ and computing the areas of the two triangles $BON$ and $BOC$.
DPatrick
2017-03-24 20:33:21
DPatrick
2017-03-24 20:33:50
This is pretty straightforward since we can just read the bases and the heights right from the picture.
This is pretty straightforward since we can just read the bases and the heights right from the picture.
DPatrick
2017-03-24 20:34:09
We have $[BON] = \dfrac12 \cdot 24 \cdot 56 = 672$ and $[BOC] = \dfrac12 \cdot 42 \cdot 72 = 1512$.
We have $[BON] = \dfrac12 \cdot 24 \cdot 56 = 672$ and $[BOC] = \dfrac12 \cdot 42 \cdot 72 = 1512$.
DPatrick
2017-03-24 20:34:28
(None of these computations are particularly interesting so I'm just going to skip the steps in the interest of time.)
(None of these computations are particularly interesting so I'm just going to skip the steps in the interest of time.)
GeronimoStilton
2017-03-24 20:34:34
$2184$
$2184$
DPatrick
2017-03-24 20:34:38
Therefore, $[BCON] = 672 + 1512 = 2184$, and thus $\overline{BP}$ splits it into two regions of area $1092$.
Therefore, $[BCON] = 672 + 1512 = 2184$, and thus $\overline{BP}$ splits it into two regions of area $1092$.
DPatrick
2017-03-24 20:34:47
More specifically, where is $P$?
More specifically, where is $P$?
Ani10
2017-03-24 20:35:07
on OC
on OC
jonzli123
2017-03-24 20:35:07
on OC
on OC
GeronimoStilton
2017-03-24 20:35:07
On line $CO$
On line $CO$
DPatrick
2017-03-24 20:35:22
Right. It lies on $\overline{OC}$, and it's far enough down so that $[BOP] = 1512 - 1092 = 420$.
Right. It lies on $\overline{OC}$, and it's far enough down so that $[BOP] = 1512 - 1092 = 420$.
DPatrick
2017-03-24 20:35:28
DPatrick
2017-03-24 20:36:00
Is there a good way to describe how far down $P$ is on $\overline{OC}$?
Is there a good way to describe how far down $P$ is on $\overline{OC}$?
DPatrick
2017-03-24 20:36:35
What I mean is...sure we could find the coordinates of $P$, but do we really need to?
What I mean is...sure we could find the coordinates of $P$, but do we really need to?
carzland
2017-03-24 20:37:44
We just need the length of the base.
We just need the length of the base.
DPatrick
2017-03-24 20:38:00
In fact, $[OBC]$ and $[PBC]$ share the same base $BC$, right?
In fact, $[OBC]$ and $[PBC]$ share the same base $BC$, right?
DPatrick
2017-03-24 20:38:26
So $\dfrac{PC}{OC} = \dfrac{[PBC]}{[OBC]}$.
So $\dfrac{PC}{OC} = \dfrac{[PBC]}{[OBC]}$.
DPatrick
2017-03-24 20:38:58
And we already knew $[PBC] = 1092$ and $[OBC] = 1512$, so $\dfrac{PC}{OC} = \dfrac{1092}{1512}$.
And we already knew $[PBC] = 1092$ and $[OBC] = 1512$, so $\dfrac{PC}{OC} = \dfrac{1092}{1512}$.
DPatrick
2017-03-24 20:39:19
This simplifies to $\dfrac{PC}{OC} = \dfrac{13}{18}$.
This simplifies to $\dfrac{PC}{OC} = \dfrac{13}{18}$.
DPatrick
2017-03-24 20:39:32
And what does that tell us about $[CDP]$?
And what does that tell us about $[CDP]$?
GeronimoStilton
2017-03-24 20:40:14
It's $13 \cdot 42$
It's $13 \cdot 42$
shootingstar8
2017-03-24 20:40:14
18 * CDP = 13 * OBC
18 * CDP = 13 * OBC
fdas
2017-03-24 20:40:14
It has 13/18 the height of COD
It has 13/18 the height of COD
DPatrick
2017-03-24 20:40:37
Right! We don't need to compute the coordinates of $P$, since we know that $\dfrac{[CDP]}{[CDO]} = \dfrac{PC}{OC}$ (since they share the base $CD$).
Right! We don't need to compute the coordinates of $P$, since we know that $\dfrac{[CDP]}{[CDO]} = \dfrac{PC}{OC}$ (since they share the base $CD$).
DPatrick
2017-03-24 20:40:56
So we have that $[CDP] = \dfrac{13}{18}[CDO]$.
So we have that $[CDP] = \dfrac{13}{18}[CDO]$.
_--__--_
2017-03-24 20:41:15
84 * 13 / 2 = 546
84 * 13 / 2 = 546
GeronimoStilton
2017-03-24 20:41:22
$546$
$546$
DPatrick
2017-03-24 20:41:27
This gives $[CDP] = \dfrac{13}{18}\left(\dfrac12 \cdot 18 \cdot 84\right) = 13 \cdot 42 = \boxed{546}$.
This gives $[CDP] = \dfrac{13}{18}\left(\dfrac12 \cdot 18 \cdot 84\right) = 13 \cdot 42 = \boxed{546}$.
DPatrick
2017-03-24 20:41:58
It would have been perfectly fine to find the coordinates of $P$, but I always like using common bases and/or common heights when we have them.
It would have been perfectly fine to find the coordinates of $P$, but I always like using common bases and/or common heights when we have them.
DPatrick
2017-03-24 20:42:09
11. Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such as way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way).
11. Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such as way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way).
DPatrick
2017-03-24 20:42:49
Or... we can ask the opposite: when is it not possible to travel between any two towns?
Or... we can ask the opposite: when is it not possible to travel between any two towns?
a1b2
2017-03-24 20:43:09
They can't all point to/from a town
They can't all point to/from a town
DPatrick
2017-03-24 20:43:25
Right. If there's a town where all the roads leave town and none arrive into town, then there's no way to travel to that town. Such a town is called a source.
Right. If there's a town where all the roads leave town and none arrive into town, then there's no way to travel to that town. Such a town is called a source.
DPatrick
2017-03-24 20:43:38
Similarly, if there's a town where all the roads arrive into town and none leave town, then there's no way to travel away from that town. Such a town is called a sink.
Similarly, if there's a town where all the roads arrive into town and none leave town, then there's no way to travel away from that town. Such a town is called a sink.
DPatrick
2017-03-24 20:43:50
These clearly fail.
These clearly fail.
DPatrick
2017-03-24 20:43:57
Are those the only situations where the road system fails? That is, if we know that no town is a source or a sink, can we conclude that we can travel from any town to any other town?
Are those the only situations where the road system fails? That is, if we know that no town is a source or a sink, can we conclude that we can travel from any town to any other town?
a1b2
2017-03-24 20:44:31
All others work.
All others work.
DPatrick
2017-03-24 20:44:35
Can we prove this?
Can we prove this?
GeneralCobra19
2017-03-24 20:45:02
intuition?
intuition?
palatine
2017-03-24 20:45:02
yes, probably
yes, probably
DPatrick
2017-03-24 20:45:33
The proof has a couple of subtle points, and in contest conditions I might be tempted to believe "it's obvious" and move on with the counting.
The proof has a couple of subtle points, and in contest conditions I might be tempted to believe "it's obvious" and move on with the counting.
DPatrick
2017-03-24 20:45:47
But let's quickly glance at the proof.
But let's quickly glance at the proof.
DPatrick
2017-03-24 20:45:59
Specifically, suppose no town is a source or a sink. We'll prove that we can get from any town to any other town.
Specifically, suppose no town is a source or a sink. We'll prove that we can get from any town to any other town.
to_chicken
2017-03-24 20:46:15
There will always be a 'triangle' of towns that you can go to and from.
There will always be a 'triangle' of towns that you can go to and from.
DPatrick
2017-03-24 20:46:25
Indeed, one good method is to find the largest cycle that you can: that is, a sequence of roads that form a loop between 3 or more towns.
Indeed, one good method is to find the largest cycle that you can: that is, a sequence of roads that form a loop between 3 or more towns.
DPatrick
2017-03-24 20:46:34
Obviously if you can find a cycle that includes all 5 towns, we're done, because we can just travel around the loop to get from anywhere to anywhere.
Obviously if you can find a cycle that includes all 5 towns, we're done, because we can just travel around the loop to get from anywhere to anywhere.
DPatrick
2017-03-24 20:46:45
Suppose you can find a cycle between 4 of the towns:
Suppose you can find a cycle between 4 of the towns:
DPatrick
2017-03-24 20:46:48
DPatrick
2017-03-24 20:47:13
Can we always get to and from the 5th town?
Can we always get to and from the 5th town?
vishwathganesan
2017-03-24 20:47:25
yes
yes
anshrai
2017-03-24 20:47:25
yes
yes
a1b2
2017-03-24 20:47:27
Yes because it is neither source nor sink.
Yes because it is neither source nor sink.
tdeng
2017-03-24 20:47:32
Yes, because there is no sink or source
Yes, because there is no sink or source
DPatrick
2017-03-24 20:47:35
Right, that's no problem either, because one of the 4 towns on the cycle will have a road to the 5th town, and one of the 4 towns on the cycle will have a road from the 5th town. (Otherwise the 5th town would be a source or a sink.)
Right, that's no problem either, because one of the 4 towns on the cycle will have a road to the 5th town, and one of the 4 towns on the cycle will have a road from the 5th town. (Otherwise the 5th town would be a source or a sink.)
DPatrick
2017-03-24 20:47:50
So we can get to the town off the cycle from the cycle, and vice versa, and hence all 5 towns are connected to each other.
So we can get to the town off the cycle from the cycle, and vice versa, and hence all 5 towns are connected to each other.
DPatrick
2017-03-24 20:47:56
Finally, suppose you can only find a cycle with 3 towns:
Finally, suppose you can only find a cycle with 3 towns:
DPatrick
2017-03-24 20:47:59
DPatrick
2017-03-24 20:48:11
I've also drawn the road between the other two towns (called X and Y); it doesn't matter which way it goes.
I've also drawn the road between the other two towns (called X and Y); it doesn't matter which way it goes.
Mudkipswims42
2017-03-24 20:48:38
Y has to travel to at least one of the other towns, and have arrivals from at least one
Y has to travel to at least one of the other towns, and have arrivals from at least one
reaganchoi
2017-03-24 20:48:41
Y must lead into the loop; X must have an entrance from the loop.
Y must lead into the loop; X must have an entrance from the loop.
DPatrick
2017-03-24 20:48:43
Since Y isn't a sink, we have to be able to get from Y to the cycle.
Since Y isn't a sink, we have to be able to get from Y to the cycle.
DPatrick
2017-03-24 20:48:46
And since X isn't a source, we have to be able to get from the cycle to X.
And since X isn't a source, we have to be able to get from the cycle to X.
DPatrick
2017-03-24 20:48:49
DPatrick
2017-03-24 20:49:07
We don't know which town(s) the dotted roads connect to in the cycle, but it doesn't matter!
We don't know which town(s) the dotted roads connect to in the cycle, but it doesn't matter!
DPatrick
2017-03-24 20:49:24
So we can get from any town to any town.
So we can get from any town to any town.
DPatrick
2017-03-24 20:49:46
By the way, before we proceed with the counting: does this still work if there are more than 5 towns?
By the way, before we proceed with the counting: does this still work if there are more than 5 towns?
fdas
2017-03-24 20:50:02
No. 6 towns can form 2 loops
No. 6 towns can form 2 loops
ESWN123456789
2017-03-24 20:50:17
No with 6 there could be two cycles of 3
No with 6 there could be two cycles of 3
DPatrick
2017-03-24 20:50:42
Exactly. If we had 6 towns, then we could have a loop between 3 of them (call them A,B,C) and a different loop between the other 3 (call them X,Y,Z).
Exactly. If we had 6 towns, then we could have a loop between 3 of them (call them A,B,C) and a different loop between the other 3 (call them X,Y,Z).
DPatrick
2017-03-24 20:51:19
And then, if all the roads went from {A,B,C} to {X,Y,Z}, there'd be no way to get from the second loop back to the first. (For example, no way to travel from Y to C.)
And then, if all the roads went from {A,B,C} to {X,Y,Z}, there'd be no way to get from the second loop back to the first. (For example, no way to travel from Y to C.)
DPatrick
2017-03-24 20:51:37
So it's not quite so simple for 6 or more towns!
So it's not quite so simple for 6 or more towns!
DPatrick
2017-03-24 20:51:50
Happily, we only have 5 towns, so now we know that our answer is \[(\text{total number of road arrangements}) - (\text{arrangements with a source or a sink}).\]
Happily, we only have 5 towns, so now we know that our answer is \[(\text{total number of road arrangements}) - (\text{arrangements with a source or a sink}).\]
DPatrick
2017-03-24 20:51:58
How do we count this?
How do we count this?
Mudkipswims42
2017-03-24 20:52:12
PIE
PIE
GeronimoStilton
2017-03-24 20:52:12
PIE?
PIE?
RedPhoenix
2017-03-24 20:52:17
use PIE for the source or sink, subtract from 1024
use PIE for the source or sink, subtract from 1024
DPatrick
2017-03-24 20:52:52
Right: for the second term, we can count "source arrangements" and "sink arrangements" separately, but that will double-count arrangements that have both.
Right: for the second term, we can count "source arrangements" and "sink arrangements" separately, but that will double-count arrangements that have both.
DPatrick
2017-03-24 20:53:01
So to refine our answer a bit, using PIE, it's \[(\text{total number}) - (\text{with a source}) - (\text{with a sink}) + (\text{with both}).\]
So to refine our answer a bit, using PIE, it's \[(\text{total number}) - (\text{with a source}) - (\text{with a sink}) + (\text{with both}).\]
DPatrick
2017-03-24 20:53:13
What's the total number of possible road arrangements?
What's the total number of possible road arrangements?
to_chicken
2017-03-24 20:53:35
Total number of road arrangements is $2^{\binom{5}{2}}=1024$.
Total number of road arrangements is $2^{\binom{5}{2}}=1024$.
Diophantine33
2017-03-24 20:53:35
1024
1024
DPatrick
2017-03-24 20:53:41
There are $\dbinom52 = 10$ roads, and each can go in either of 2 directions. So there are $2^{10} = 1024$ total road arrangements.
There are $\dbinom52 = 10$ roads, and each can go in either of 2 directions. So there are $2^{10} = 1024$ total road arrangements.
DPatrick
2017-03-24 20:53:48
How many with a source?
How many with a source?
Metal_Bender19
2017-03-24 20:54:23
320
320
GeronimoStilton
2017-03-24 20:54:23
$5 \cdot 64$
$5 \cdot 64$
jonzli123
2017-03-24 20:54:23
320
320
DPatrick
2017-03-24 20:54:28
We have 5 choices for the town that's the source (note that we can't have more than one!), and that fixes those 4 roads to go out of that town.
We have 5 choices for the town that's the source (note that we can't have more than one!), and that fixes those 4 roads to go out of that town.
DPatrick
2017-03-24 20:54:36
The other 6 roads still have 2 choices each, so that $5 \cdot 2^6 = 5 \cdot 64 = 320$ arrangements with a source.
The other 6 roads still have 2 choices each, so that $5 \cdot 2^6 = 5 \cdot 64 = 320$ arrangements with a source.
DPatrick
2017-03-24 20:54:44
How many with a sink?
How many with a sink?
to_chicken
2017-03-24 20:54:58
same: 320
same: 320
Mudkipswims42
2017-03-24 20:54:58
same
same
tdeng
2017-03-24 20:54:58
Same number.
Same number.
colinyao
2017-03-24 20:54:58
same as with a source
same as with a source
DPatrick
2017-03-24 20:55:05
By symmetry, it's the same: 320. (Or you could do the same computation.)
By symmetry, it's the same: 320. (Or you could do the same computation.)
DPatrick
2017-03-24 20:55:10
How many with both?
How many with both?
jonzli123
2017-03-24 20:55:38
20*8
20*8
Metal_Bender19
2017-03-24 20:55:38
160
160
SomethingNeutral
2017-03-24 20:55:40
5*4*8 = 160
5*4*8 = 160
DPatrick
2017-03-24 20:55:44
We have 5 choices for the source and then 4 choices for the sink. This fixes 7 of the roads, leaving the other 3 (the roads running between the 3 remaining towns) free to choose their direction. So that's $5 \cdot 4 \cdot 2^3 = 20 \cdot 8 = 160$ with both.
We have 5 choices for the source and then 4 choices for the sink. This fixes 7 of the roads, leaving the other 3 (the roads running between the 3 remaining towns) free to choose their direction. So that's $5 \cdot 4 \cdot 2^3 = 20 \cdot 8 = 160$ with both.
soojoong
2017-03-24 20:55:59
$544$!!
$544$!!
DPatrick
2017-03-24 20:56:00
Therefore, our final answer is $1024 - 320 - 320 + 160 = \boxed{544}$.
Therefore, our final answer is $1024 - 320 - 320 + 160 = \boxed{544}$.
DPatrick
2017-03-24 20:56:13
12. Circle $C_0$ has radius 1, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r < 1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^\circ$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circle $C_1,C_2,C_3,\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^\circ$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$, the distance from the center of $C_0$ to $B$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
12. Circle $C_0$ has radius 1, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r < 1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^\circ$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circle $C_1,C_2,C_3,\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^\circ$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$, the distance from the center of $C_0$ to $B$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
DPatrick
2017-03-24 20:56:17
DPatrick
2017-03-24 20:57:11
This is not nearly as hard a problem as it looks, once you read it carefully and decipher what's going on.
This is not nearly as hard a problem as it looks, once you read it carefully and decipher what's going on.
Mudkipswims42
2017-03-24 20:57:25
Well... let's get bashing!
Well... let's get bashing!
DPatrick
2017-03-24 20:57:35
Yeah...we can set up a coordinate system so that the center of the original circle is the origin, and $A_0 = (1,0)$.
Yeah...we can set up a coordinate system so that the center of the original circle is the origin, and $A_0 = (1,0)$.
DPatrick
2017-03-24 20:57:47
In this coordinate system, what's point $A_1$?
In this coordinate system, what's point $A_1$?
DPatrick
2017-03-24 20:58:19
To figure this out, think about how we "move" from $A_0$ to $A_1$.
To figure this out, think about how we "move" from $A_0$ to $A_1$.
DPatrick
2017-03-24 20:58:50
And let's everything in terms of $r$ and not plug in the given fraction...that'll be a mess. Let's not use $\frac{11}{60}$ until we absolutely have to.
And let's everything in terms of $r$ and not plug in the given fraction...that'll be a mess. Let's not use $\frac{11}{60}$ until we absolutely have to.
to_chicken
2017-03-24 20:58:56
$(1-r,r)$
$(1-r,r)$
Piazolla13
2017-03-24 20:58:56
(1-r,r)
(1-r,r)
DPatrick
2017-03-24 20:59:07
Right! To get from $A_0$ to $A_1$, we move left $r$ units and up $r$ units.
Right! To get from $A_0$ to $A_1$, we move left $r$ units and up $r$ units.
DPatrick
2017-03-24 20:59:20
(Because the circle we're moving along has radius $r$.)
(Because the circle we're moving along has radius $r$.)
DPatrick
2017-03-24 20:59:27
So $A_1 = (1-r,r)$.
So $A_1 = (1-r,r)$.
DPatrick
2017-03-24 20:59:31
Now what's $A_2$?
Now what's $A_2$?
Mudkipswims42
2017-03-24 20:59:57
$(1-r-r^2,~r-r^2)$?
$(1-r-r^2,~r-r^2)$?
GeronimoStilton
2017-03-24 20:59:57
$A_2 = (1-r-r^2, r-r^2)$
$A_2 = (1-r-r^2, r-r^2)$
SomethingNeutral
2017-03-24 21:00:01
$A_2 = (1-r-r^2, r-r^2)
$A_2 = (1-r-r^2, r-r^2)
DPatrick
2017-03-24 21:00:04
To get from $A_1$ to $A_2$, we move left $r^2$ units and down $r^2$ units.
To get from $A_1$ to $A_2$, we move left $r^2$ units and down $r^2$ units.
DPatrick
2017-03-24 21:00:08
So $A_2 = (1-r-r^2,r-r^2)$.
So $A_2 = (1-r-r^2,r-r^2)$.
DPatrick
2017-03-24 21:00:18
What's $A_3$?
What's $A_3$?
RedPhoenix
2017-03-24 21:00:44
(1-r-r^2+r^3, r-r^2-r^3)
(1-r-r^2+r^3, r-r^2-r^3)
DPatrick
2017-03-24 21:00:52
To get from $A_2$ to $A_3$, we move right $r^3$ units and down $r^3$ units.
To get from $A_2$ to $A_3$, we move right $r^3$ units and down $r^3$ units.
DPatrick
2017-03-24 21:00:56
So $A_3 = (1-r-r^2+r^3,r-r^2-r^3)$.
So $A_3 = (1-r-r^2+r^3,r-r^2-r^3)$.
DPatrick
2017-03-24 21:01:01
See the pattern?
See the pattern?
DPatrick
2017-03-24 21:01:08
To get from $A_{n-1}$ to $A_n$, we move $r^n$ units left or right and $r^n$ units up or down.
To get from $A_{n-1}$ to $A_n$, we move $r^n$ units left or right and $r^n$ units up or down.
DPatrick
2017-03-24 21:01:22
Furthermore, for both the vertical and horizontal directions, we do two moves in the same direction, then two moves in the opposite direction, etc.
Furthermore, for both the vertical and horizontal directions, we do two moves in the same direction, then two moves in the opposite direction, etc.
DPatrick
2017-03-24 21:01:49
So $A_n = (1-r-r^2+r^3+r^4-r^5-r^6+r^7+\cdots\pm r^n,$
$r-r^2-r^3+r^4+r^5-r^6-r^7+\cdots\pm r^n)$.
So $A_n = (1-r-r^2+r^3+r^4-r^5-r^6+r^7+\cdots\pm r^n,$
$r-r^2-r^3+r^4+r^5-r^6-r^7+\cdots\pm r^n)$.
DPatrick
2017-03-24 21:02:02
And how does $B$ relate to all this?
And how does $B$ relate to all this?
Mudkipswims42
2017-03-24 21:02:11
Infinite!
Infinite!
SomethingNeutral
2017-03-24 21:02:15
B is when n approaches infinity
B is when n approaches infinity
DPatrick
2017-03-24 21:02:26
Right. Notice that $A_n$ is inside all of the previous circles $C_0,C_1,\ldots,C_{n-1}$.
Right. Notice that $A_n$ is inside all of the previous circles $C_0,C_1,\ldots,C_{n-1}$.
DPatrick
2017-03-24 21:02:37
So if we take the "limit" point of the $A_n$'s as $n$ goes to $\infty$, we'll get a point that's inside all of the $C$'s.
So if we take the "limit" point of the $A_n$'s as $n$ goes to $\infty$, we'll get a point that's inside all of the $C$'s.
DPatrick
2017-03-24 21:02:46
That is, $B$ is the point whose $x$-coordinate is the infinite series \[1-r-r^2+r^3+r^4-r^5-r^6+r^7+\cdots\] and whose $y$-coordinate is the infinite series \[r-r^2-r^3+r^4+r^5-r^6-r^7+\cdots\]where the terms are powers of $r$ and the signs change every other term.
That is, $B$ is the point whose $x$-coordinate is the infinite series \[1-r-r^2+r^3+r^4-r^5-r^6+r^7+\cdots\] and whose $y$-coordinate is the infinite series \[r-r^2-r^3+r^4+r^5-r^6-r^7+\cdots\]where the terms are powers of $r$ and the signs change every other term.
DPatrick
2017-03-24 21:03:03
How do we sum these series?
How do we sum these series?
carzland
2017-03-24 21:03:13
Geometric
Geometric
GeronimoStilton
2017-03-24 21:03:13
Geometric Series!
Geometric Series!
SomethingNeutral
2017-03-24 21:03:13
they are geometric!!
they are geometric!!
DPatrick
2017-03-24 21:03:32
Really? At first glance, they don't look geometric to me, because of the changing signs.
Really? At first glance, they don't look geometric to me, because of the changing signs.
DaAsianPotatoe
2017-03-24 21:03:49
each term has two terms
each term has two terms
DPatrick
2017-03-24 21:04:08
Aha! The first one is really a geometric series if we group the terms in pairs: \[(1-r) - r^2(1-r) + r^4(1-r) - r^6(1-r) + \cdots.\]
Aha! The first one is really a geometric series if we group the terms in pairs: \[(1-r) - r^2(1-r) + r^4(1-r) - r^6(1-r) + \cdots.\]
DPatrick
2017-03-24 21:04:27
That is, it's the infinite geometric series with first term $1-r$ and common ratio $-r^2$.
That is, it's the infinite geometric series with first term $1-r$ and common ratio $-r^2$.
DPatrick
2017-03-24 21:04:36
What does it sum to?
What does it sum to?
reaganchoi
2017-03-24 21:04:53
$\frac{1-r}{1-(-r^2)}$
$\frac{1-r}{1-(-r^2)}$
Mudkipswims42
2017-03-24 21:04:53
$\dfrac{1-r}{1+r^2}$
$\dfrac{1-r}{1+r^2}$
tdeng
2017-03-24 21:04:53
$\frac{1-r}{1+r^2}$
$\frac{1-r}{1+r^2}$
GeronimoStilton
2017-03-24 21:04:53
$\frac{1-r}{1+r^2}$
$\frac{1-r}{1+r^2}$
DPatrick
2017-03-24 21:05:02
Right: first term $1-r$ and common ratio $-r^2$, so its sum is $\dfrac{1-r}{1+r^2}$.
Right: first term $1-r$ and common ratio $-r^2$, so its sum is $\dfrac{1-r}{1+r^2}$.
DPatrick
2017-03-24 21:05:15
So that's the $x$-coordinate of $B$. What about the $y$-coordinate?
So that's the $x$-coordinate of $B$. What about the $y$-coordinate?
DPatrick
2017-03-24 21:05:31
(You might have to scroll up a little to see it.)
(You might have to scroll up a little to see it.)
DPatrick
2017-03-24 21:05:44
I'll repost it: $B$ is the point whose $x$-coordinate is the infinite series \[1-r-r^2+r^3+r^4-r^5-r^6+r^7+\cdots\] and whose $y$-coordinate is the infinite series \[r-r^2-r^3+r^4+r^5-r^6-r^7+\cdots\]where the terms are powers of $r$ and the signs change every other term.
I'll repost it: $B$ is the point whose $x$-coordinate is the infinite series \[1-r-r^2+r^3+r^4-r^5-r^6+r^7+\cdots\] and whose $y$-coordinate is the infinite series \[r-r^2-r^3+r^4+r^5-r^6-r^7+\cdots\]where the terms are powers of $r$ and the signs change every other term.
GeronimoStilton
2017-03-24 21:05:50
It's $\frac{r(1-r)}{1+r^2}$
It's $\frac{r(1-r)}{1+r^2}$
reaganchoi
2017-03-24 21:05:50
$\frac{r(1-r)}{1-(-r^2)}$
$\frac{r(1-r)}{1-(-r^2)}$
DPatrick
2017-03-24 21:06:03
Indeed. Notice that the second series (the $y$-coordinate expression) is just $r$ times the first one.
Indeed. Notice that the second series (the $y$-coordinate expression) is just $r$ times the first one.
DPatrick
2017-03-24 21:06:15
This means that $B = \left(\dfrac{1-r}{1+r^2},\dfrac{r(1-r)}{1+r^2}\right)$.
This means that $B = \left(\dfrac{1-r}{1+r^2},\dfrac{r(1-r)}{1+r^2}\right)$.
DPatrick
2017-03-24 21:06:22
And what is the distance from $B$ to the origin?
And what is the distance from $B$ to the origin?
Metal_Bender19
2017-03-24 21:06:57
(1-r)/($\sqrt{1+r^2}$)
(1-r)/($\sqrt{1+r^2}$)
DPatrick
2017-03-24 21:07:17
Right -- notice there's a big constant $\dfrac{1-r}{1+r^2}$ in common.
Right -- notice there's a big constant $\dfrac{1-r}{1+r^2}$ in common.
DPatrick
2017-03-24 21:07:26
So it's just $\dfrac{1-r}{1+r^2}\sqrt{1+r^2} = \dfrac{1-r}{\sqrt{1+r^2}}$.
So it's just $\dfrac{1-r}{1+r^2}\sqrt{1+r^2} = \dfrac{1-r}{\sqrt{1+r^2}}$.
SomethingNeutral
2017-03-24 21:07:40
plug in 11/60 now
plug in 11/60 now
a1b2
2017-03-24 21:07:40
$\frac{11}{60}
$\frac{11}{60}
DPatrick
2017-03-24 21:07:46
Right, we can't put it off any longer.
Right, we can't put it off any longer.
DPatrick
2017-03-24 21:07:51
Now we can plug in $r = \dfrac{11}{60}$ and see what we get.
Now we can plug in $r = \dfrac{11}{60}$ and see what we get.
DPatrick
2017-03-24 21:07:55
\[\frac{1-r}{\sqrt{1+r^2}} = \frac{\frac{49}{60}}{\sqrt{1+\frac{11^2}{60^2}}}.\]
\[\frac{1-r}{\sqrt{1+r^2}} = \frac{\frac{49}{60}}{\sqrt{1+\frac{11^2}{60^2}}}.\]
DPatrick
2017-03-24 21:08:10
Multiply by 60/60 and this simplifies to $\dfrac{49}{\sqrt{60^2 + 11^2}}$.
Multiply by 60/60 and this simplifies to $\dfrac{49}{\sqrt{60^2 + 11^2}}$.
letsgomath
2017-03-24 21:08:24
49/61
49/61
soojoong
2017-03-24 21:08:24
49/61!!!!!!!!!!!!
49/61!!!!!!!!!!!!
RedPhoenix
2017-03-24 21:08:24
the denominator is just 61
the denominator is just 61
DPatrick
2017-03-24 21:08:31
If you know your Pythagorean Triples well, and in particular notice that 11-60-61 is a triple, you can quickly see that the denominator is just 61. So the distance we want is $\dfrac{49}{61}$.
If you know your Pythagorean Triples well, and in particular notice that 11-60-61 is a triple, you can quickly see that the denominator is just 61. So the distance we want is $\dfrac{49}{61}$.
DPatrick
2017-03-24 21:09:03
Or you can compute it...but given the way the problem is phrased, we know it that $60^2 + 11^2$ has to be a perfect square.
Or you can compute it...but given the way the problem is phrased, we know it that $60^2 + 11^2$ has to be a perfect square.
skiboy32
2017-03-24 21:09:11
so the answer is $49+61=\boxed{110}.$
so the answer is $49+61=\boxed{110}.$
DPatrick
2017-03-24 21:09:18
This fraction is in lowest terms, so our answer is $49 + 61 = \boxed{110}$.
This fraction is in lowest terms, so our answer is $49 + 61 = \boxed{110}$.
DPatrick
2017-03-24 21:09:36
13. For each integer $n \ge 3$, let $f(n)$ be the number of 3-element subsets of the vertices of a regular $n$-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1) = f(n) + 78$.
13. For each integer $n \ge 3$, let $f(n)$ be the number of 3-element subsets of the vertices of a regular $n$-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1) = f(n) + 78$.
DPatrick
2017-03-24 21:09:58
It'd be nice if there were a super-easy formula for $f(n)$ to plug into, but there's probably not. (The problem would be too easy for a #13 if there were.)
It'd be nice if there were a super-easy formula for $f(n)$ to plug into, but there's probably not. (The problem would be too easy for a #13 if there were.)
DPatrick
2017-03-24 21:10:04
Indeed, what sorts of things do we have to worry about?
Indeed, what sorts of things do we have to worry about?
Mudkipswims42
2017-03-24 21:10:41
well there are 4 cases
well there are 4 cases
Mudkipswims42
2017-03-24 21:10:41
parity, and equilaterals
parity, and equilaterals
Tan
2017-03-24 21:10:41
divisible by 2 or 3
divisible by 2 or 3
High
2017-03-24 21:10:41
when n is a multiple of 3
when n is a multiple of 3
skiboy32
2017-03-24 21:10:41
if it's divisible by 3, because we then have equilateral triangles
if it's divisible by 3, because we then have equilateral triangles
RedPhoenix
2017-03-24 21:10:41
equilateral triangles, odd or even....
equilateral triangles, odd or even....
DPatrick
2017-03-24 21:10:52
Right. The calculation is slightly different for $n$ odd and $n$ even, and the calculation is slightly different if $n$ is a multiple of 3.
Right. The calculation is slightly different for $n$ odd and $n$ even, and the calculation is slightly different if $n$ is a multiple of 3.
DPatrick
2017-03-24 21:11:12
Let's first look at when $n$ not a multiple of 3, since this is easier: there are no equilateral triangles, so every isosceles triangle has a unique base side.
Let's first look at when $n$ not a multiple of 3, since this is easier: there are no equilateral triangles, so every isosceles triangle has a unique base side.
DPatrick
2017-03-24 21:11:18
An example might help. Let's look at $n=7$:
An example might help. Let's look at $n=7$:
DPatrick
2017-03-24 21:11:24
DPatrick
2017-03-24 21:11:32
If I draw any diagonal, can I make an isosceles triangle, and if so how many?
If I draw any diagonal, can I make an isosceles triangle, and if so how many?
DPatrick
2017-03-24 21:11:51
More specifically, can I make an isosceles triangle with my diagonal as its base?
More specifically, can I make an isosceles triangle with my diagonal as its base?
skiboy32
2017-03-24 21:12:08
yes, and 1
yes, and 1
RedPhoenix
2017-03-24 21:12:08
yes, exactly 1
yes, exactly 1
DPatrick
2017-03-24 21:12:13
DPatrick
2017-03-24 21:12:25
There's exactly one. I need to take the "midpoint" of the side of the blue line with an odd number of vertices, and use that as my third point:
There's exactly one. I need to take the "midpoint" of the side of the blue line with an odd number of vertices, and use that as my third point:
DPatrick
2017-03-24 21:12:29
DPatrick
2017-03-24 21:12:46
I can't do it on the other side of the blue segment, since there's no "midpoint".
I can't do it on the other side of the blue segment, since there's no "midpoint".
DPatrick
2017-03-24 21:12:55
Since $n-2$ is odd, only one side of the base will have an odd number of vertices, so we can only make one isosceles triangle with that base.
Since $n-2$ is odd, only one side of the base will have an odd number of vertices, so we can only make one isosceles triangle with that base.
DPatrick
2017-03-24 21:13:06
And this works even if we pick an edge of the $n$-gon to be our base, since there are $n-2$ (an odd number) of vertices left.
And this works even if we pick an edge of the $n$-gon to be our base, since there are $n-2$ (an odd number) of vertices left.
DPatrick
2017-03-24 21:13:09
DPatrick
2017-03-24 21:13:26
So if $n$ is odd (and not a multiple of 3), how many isosceles triangles are there?
So if $n$ is odd (and not a multiple of 3), how many isosceles triangles are there?
_--__--_
2017-03-24 21:13:49
n(n-1)/2
n(n-1)/2
DaAsianPotatoe
2017-03-24 21:13:49
nC2
nC2
fdas
2017-03-24 21:13:49
(n-1)/2
(n-1)/2
DPatrick
2017-03-24 21:13:56
Right: if $n$ is odd and not a multiple of 3, the number of isosceles triangles is just the number of pairs of vertices.
Right: if $n$ is odd and not a multiple of 3, the number of isosceles triangles is just the number of pairs of vertices.
DPatrick
2017-03-24 21:14:02
This is $\dbinom{n}{2} = \dfrac{n(n-1)}{2}$.
This is $\dbinom{n}{2} = \dfrac{n(n-1)}{2}$.
DPatrick
2017-03-24 21:14:40
To recap: if $n$ is odd, not a multiple of 3, we can pick any diagonal or edge to be our base, and there will be one isosceles triangle on the "odd" side of that segment.
To recap: if $n$ is odd, not a multiple of 3, we can pick any diagonal or edge to be our base, and there will be one isosceles triangle on the "odd" side of that segment.
DPatrick
2017-03-24 21:14:46
What if $n$ is even? Let's take $n=8$ as our example:
What if $n$ is even? Let's take $n=8$ as our example:
DPatrick
2017-03-24 21:14:49
DPatrick
2017-03-24 21:14:56
Can every diagonal be a base of an isosceles triangle?
Can every diagonal be a base of an isosceles triangle?
willmathxu
2017-03-24 21:15:09
no
no
vishwathganesan
2017-03-24 21:15:09
no
no
Metal_Bender19
2017-03-24 21:15:09
no
no
legolego
2017-03-24 21:15:09
no
no
DPatrick
2017-03-24 21:15:14
Why not?
Why not?
SomethingNeutral
2017-03-24 21:15:35
if you take the side that splits into a trapezoid
if you take the side that splits into a trapezoid
skiboy32
2017-03-24 21:15:35
Try when you choose vertices 3 over, for instance
Try when you choose vertices 3 over, for instance
a1b2
2017-03-24 21:15:38
Split into $2$ parts with an even parity of points
Split into $2$ parts with an even parity of points
DPatrick
2017-03-24 21:15:43
If the diagonal splits the rest of the vertices into two even groups (which is possible because $n-2$ is even), then we can't pick a "middle" vertex on either side to make an isosceles triangle:
If the diagonal splits the rest of the vertices into two even groups (which is possible because $n-2$ is even), then we can't pick a "middle" vertex on either side to make an isosceles triangle:
DPatrick
2017-03-24 21:15:49
DPatrick
2017-03-24 21:16:04
The above blue diagonal is not the base of any isosceles triangle.
The above blue diagonal is not the base of any isosceles triangle.
DPatrick
2017-03-24 21:16:14
But what if the diagonal splits the other $n-2$ vertices into two odd groups?
But what if the diagonal splits the other $n-2$ vertices into two odd groups?
legolego
2017-03-24 21:16:33
two triangles
two triangles
SomethingNeutral
2017-03-24 21:16:33
then 2 triangles
then 2 triangles
vishwathganesan
2017-03-24 21:16:33
then 2 triiangles
then 2 triiangles
skiboy32
2017-03-24 21:16:33
Then, it can make two.
Then, it can make two.
DPatrick
2017-03-24 21:16:43
If the diagonal splits the other $n-2$ vertices into two odd groups, then we get 2 isosceles triangles, one on each side:
If the diagonal splits the other $n-2$ vertices into two odd groups, then we get 2 isosceles triangles, one on each side:
DPatrick
2017-03-24 21:16:46
DPatrick
2017-03-24 21:16:52
So the number of triangles is $2 \cdot \text{(number of diagonals an even # of vertices apart)}$.
So the number of triangles is $2 \cdot \text{(number of diagonals an even # of vertices apart)}$.
DPatrick
2017-03-24 21:17:00
How many of those diagonals are there?
How many of those diagonals are there?
DPatrick
2017-03-24 21:18:03
Indeed, in the octagon $n=8$ there are 12.
Indeed, in the octagon $n=8$ there are 12.
DPatrick
2017-03-24 21:18:08
What's the formula for general even $n$?
What's the formula for general even $n$?
DPatrick
2017-03-24 21:18:58
Think of the vertices in two sets of $n/2$: the "even" vertices and the "odd" vertices,
Think of the vertices in two sets of $n/2$: the "even" vertices and the "odd" vertices,
DPatrick
2017-03-24 21:19:13
We can pick two of the evens and we're OK, or we can pick two of the odds and we're OK.
We can pick two of the evens and we're OK, or we can pick two of the odds and we're OK.
DPatrick
2017-03-24 21:19:28
So the number of such diagonals is $2\dbinom{n/2}{2}$.
So the number of such diagonals is $2\dbinom{n/2}{2}$.
DPatrick
2017-03-24 21:19:52
And remember that each of these diagonals gives us 2 isosceles triangles!
And remember that each of these diagonals gives us 2 isosceles triangles!
DPatrick
2017-03-24 21:19:59
So the number of isosceles triangles is $4\dbinom{n/2}{2} = 4\dfrac{\frac{n}{2}\cdot\frac{n-2}{2}}{2} = \dfrac{n(n-2)}{2}$.
So the number of isosceles triangles is $4\dbinom{n/2}{2} = 4\dfrac{\frac{n}{2}\cdot\frac{n-2}{2}}{2} = \dfrac{n(n-2)}{2}$.
DPatrick
2017-03-24 21:20:13
To recap so far: if $n$ is not a multiple of 3, then \[ f(n) = \left\{\begin{array}{ll} \dfrac{n(n-1)}{2} & \text{if $n$ is odd} \\ \dfrac{n(n-2)}{2} & \text{if $n$ is even}\end{array}\right.\]
To recap so far: if $n$ is not a multiple of 3, then \[ f(n) = \left\{\begin{array}{ll} \dfrac{n(n-1)}{2} & \text{if $n$ is odd} \\ \dfrac{n(n-2)}{2} & \text{if $n$ is even}\end{array}\right.\]
DPatrick
2017-03-24 21:20:32
And now, what happens when $n$ is a multiple of 3?
And now, what happens when $n$ is a multiple of 3?
skiboy32
2017-03-24 21:20:55
Then, the equilateral triangle make us overcount.
Then, the equilateral triangle make us overcount.
DPatrick
2017-03-24 21:21:11
Right. Our count above hinged on the fact that every isosceles triangle had a unique base.
Right. Our count above hinged on the fact that every isosceles triangle had a unique base.
DPatrick
2017-03-24 21:21:32
But when $n$ is a multiple of 3, we get some equilateral triangles by taking 3 evenly-spaced vertices of the $n$-gon.
But when $n$ is a multiple of 3, we get some equilateral triangles by taking 3 evenly-spaced vertices of the $n$-gon.
DPatrick
2017-03-24 21:21:45
How many equilateral triangles are there?
How many equilateral triangles are there?
SomethingNeutral
2017-03-24 21:22:09
n/3
n/3
legolego
2017-03-24 21:22:09
n/3
n/3
jonzli123
2017-03-24 21:22:09
n/3
n/3
tdeng
2017-03-24 21:22:09
n/3
n/3
DPatrick
2017-03-24 21:22:17
Right. The equilateral triangle can be rotated into $n/3$ different positions, so there are $n/3$ of them.
Right. The equilateral triangle can be rotated into $n/3$ different positions, so there are $n/3$ of them.
DPatrick
2017-03-24 21:22:26
How many times has each one been counted in our $f(n)$ formula above?
How many times has each one been counted in our $f(n)$ formula above?
skiboy32
2017-03-24 21:22:37
3.
3.
legolego
2017-03-24 21:22:37
three times
three times
flamescarlet
2017-03-24 21:22:37
3
3
abishek99
2017-03-24 21:22:37
3
3
DPatrick
2017-03-24 21:22:41
3 times: once for each "base".
3 times: once for each "base".
DPatrick
2017-03-24 21:22:49
So they've been triple-counted, which means they've been double-overcounted.
So they've been triple-counted, which means they've been double-overcounted.
DPatrick
2017-03-24 21:23:07
So, how do we correct our $f(n)$ formula when $n$ is a multiple of 3?
So, how do we correct our $f(n)$ formula when $n$ is a multiple of 3?
vishwathganesan
2017-03-24 21:23:25
-2n/3
-2n/3
Diophantine33
2017-03-24 21:23:27
Subtract 2n/3
Subtract 2n/3
SomethingNeutral
2017-03-24 21:23:27
subtract $\dfrac{2n}3$
subtract $\dfrac{2n}3$
DPatrick
2017-03-24 21:23:47
Right. There are $\dfrac{n}{3}$ equilaterals, and they got doubly-overcounted. So we have to subtract $\dfrac{2n}{3}$ whenever $n$ is a multiple of 3.
Right. There are $\dfrac{n}{3}$ equilaterals, and they got doubly-overcounted. So we have to subtract $\dfrac{2n}{3}$ whenever $n$ is a multiple of 3.
DPatrick
2017-03-24 21:23:53
Thus, with a little abuse of notation, we can write: \[ f(n) = \left\{\begin{array}{ll} \dfrac{n(n-1)}{2} & \text{if $n$ is odd} \\ \dfrac{n(n-2)}{2} & \text{if $n$ is even}\end{array}\right\} - \left\{\begin{array}{ll} 0 & \text{if $n$ is not a multiple of 3} \\ \dfrac{2n}{3} & \text{if $n$ is a multiple of 3}\end{array}\right\}\]
Thus, with a little abuse of notation, we can write: \[ f(n) = \left\{\begin{array}{ll} \dfrac{n(n-1)}{2} & \text{if $n$ is odd} \\ \dfrac{n(n-2)}{2} & \text{if $n$ is even}\end{array}\right\} - \left\{\begin{array}{ll} 0 & \text{if $n$ is not a multiple of 3} \\ \dfrac{2n}{3} & \text{if $n$ is a multiple of 3}\end{array}\right\}\]
DPatrick
2017-03-24 21:24:02
Ick. This seems like a lot to keep track of!
Ick. This seems like a lot to keep track of!
DPatrick
2017-03-24 21:24:19
What would make this simpler to work with?
What would make this simpler to work with?
azmath333
2017-03-24 21:24:26
now bash out mod 6
now bash out mod 6
DPatrick
2017-03-24 21:24:32
Sure. We could look at the different residues mod 6 and get a formula for each one.
Sure. We could look at the different residues mod 6 and get a formula for each one.
DPatrick
2017-03-24 21:24:40
For example, what's $f(6k)$?
For example, what's $f(6k)$?
DPatrick
2017-03-24 21:24:58
That's the "even, is a multiple of 3" case.
That's the "even, is a multiple of 3" case.
DPatrick
2017-03-24 21:25:19
So it works out to \[f(6k) = \dfrac{(6k)(6k-2)}{2} - \dfrac{2(6k)}{3}.\]
So it works out to \[f(6k) = \dfrac{(6k)(6k-2)}{2} - \dfrac{2(6k)}{3}.\]
DPatrick
2017-03-24 21:25:38
This simplifies to $f(6k) = 18k^2 - 10k$.
This simplifies to $f(6k) = 18k^2 - 10k$.
DPatrick
2017-03-24 21:25:59
I'll run through the other calculations, since they're not that interested to spend time on.
I'll run through the other calculations, since they're not that interested to spend time on.
DPatrick
2017-03-24 21:26:03
$f(6k+1) = \dfrac{(6k+1)(6k+1-1)}{2} = 18k^2 + 3k$.
$f(6k+1) = \dfrac{(6k+1)(6k+1-1)}{2} = 18k^2 + 3k$.
DPatrick
2017-03-24 21:26:09
$f(6k+2) = \dfrac{(6k+2)(6k+2-2)}{2} = 18k^2 + 6k$.
$f(6k+2) = \dfrac{(6k+2)(6k+2-2)}{2} = 18k^2 + 6k$.
DPatrick
2017-03-24 21:26:13
$f(6k+3) = \dfrac{(6k+3)(6k+3-1)}{2} - \dfrac{2(6k+3)}{3} = 18k^2 + 11k + 1$.
$f(6k+3) = \dfrac{(6k+3)(6k+3-1)}{2} - \dfrac{2(6k+3)}{3} = 18k^2 + 11k + 1$.
DPatrick
2017-03-24 21:26:17
$f(6k+4) = \dfrac{(6k+4)(6k+4-2)}{2} = 18k^2 + 18k + 4$.
$f(6k+4) = \dfrac{(6k+4)(6k+4-2)}{2} = 18k^2 + 18k + 4$.
DPatrick
2017-03-24 21:26:20
$f(6k+5) = \dfrac{(6k+5)(6k+5-1)}{1} = 18k^2 + 27k + 10$.
$f(6k+5) = \dfrac{(6k+5)(6k+5-1)}{1} = 18k^2 + 27k + 10$.
DPatrick
2017-03-24 21:26:30
That's a lot of data. Now what?
That's a lot of data. Now what?
DPatrick
2017-03-24 21:27:05
By now we may have forgotten the original problem, so a glance back might help...
By now we may have forgotten the original problem, so a glance back might help...
GeronimoStilton
2017-03-24 21:27:08
We look at the difference between these.
We look at the difference between these.
carzland
2017-03-24 21:27:10
Find the one that is 78 apart.
Find the one that is 78 apart.
DPatrick
2017-03-24 21:27:18
Right. We care about when $f(n+1) - f(n) = 78$. So let's define $g(n) = f(n+1)-f(n)$ and compute it for our mod 6 residues.
Right. We care about when $f(n+1) - f(n) = 78$. So let's define $g(n) = f(n+1)-f(n)$ and compute it for our mod 6 residues.
DPatrick
2017-03-24 21:27:36
\begin{align*}
g(6k) &= f(6k+1)-f(6k) = 13k \\
g(6k+1) &= f(6k+2)-f(6k+1) = 3k \\
g(6k+2) &= f(6k+3)-f(6k+2) = 5k+1 \\
g(6k+3) &= f(6k+4)-f(6k+3) = 7k+3 \\
g(6k+4) &= f(6k+5)-f(6k+4) = 9k+6 \\
g(6k+5) &= f(6k+6)-f(6k+5) = -k-2
\end{align*}
\begin{align*}
g(6k) &= f(6k+1)-f(6k) = 13k \\
g(6k+1) &= f(6k+2)-f(6k+1) = 3k \\
g(6k+2) &= f(6k+3)-f(6k+2) = 5k+1 \\
g(6k+3) &= f(6k+4)-f(6k+3) = 7k+3 \\
g(6k+4) &= f(6k+5)-f(6k+4) = 9k+6 \\
g(6k+5) &= f(6k+6)-f(6k+5) = -k-2
\end{align*}
DPatrick
2017-03-24 21:27:46
(Again, the calculations are not that interesting.)
(Again, the calculations are not that interesting.)
DPatrick
2017-03-24 21:27:55
Which of these $g$'s could possibly equal 78?
Which of these $g$'s could possibly equal 78?
QuestForKnowledge
2017-03-24 21:28:24
the 5th one or the first
the 5th one or the first
carzland
2017-03-24 21:28:24
The second.
The second.
DPatrick
2017-03-24 21:28:29
We just have to run through them.
We just have to run through them.
DPatrick
2017-03-24 21:28:34
$g(6k) = 13k = 78$ is true when $k=6$. So $n = 6k = 36$ is a solution.
$g(6k+1) = 3k = 78$ is true when $k=26$. So $n = 6k+1 = 6(26)+1 = 157$ is a solution.
$g(6k+2) = 5k+1 = 78$ simplifies to $5k = 77$. This has no solution.
$g(6k+3) = 7k+3 = 78$ simplifies to $7k = 75$. This has no solution.
$g(6k+4) = 9k+6 = 78$ simplifies to $9k = 72$, so $k=8$ works. So $n = 6k+4 = 6(8)+4 = 52$ is a solution.
$g(6k+5) = -k-2 = 78$ has no solution (the left side is negative!).
$g(6k) = 13k = 78$ is true when $k=6$. So $n = 6k = 36$ is a solution.
$g(6k+1) = 3k = 78$ is true when $k=26$. So $n = 6k+1 = 6(26)+1 = 157$ is a solution.
$g(6k+2) = 5k+1 = 78$ simplifies to $5k = 77$. This has no solution.
$g(6k+3) = 7k+3 = 78$ simplifies to $7k = 75$. This has no solution.
$g(6k+4) = 9k+6 = 78$ simplifies to $9k = 72$, so $k=8$ works. So $n = 6k+4 = 6(8)+4 = 52$ is a solution.
$g(6k+5) = -k-2 = 78$ has no solution (the left side is negative!).
DPatrick
2017-03-24 21:28:58
So the solutions are $n=36$, $n=157$, and $n=52$.
So the solutions are $n=36$, $n=157$, and $n=52$.
DPatrick
2017-03-24 21:29:11
Our final answer is $36 + 157 + 52 = \boxed{245}$.
Our final answer is $36 + 157 + 52 = \boxed{245}$.
DPatrick
2017-03-24 21:29:45
This problem was mainly a lot of bookkeeping.
This problem was mainly a lot of bookkeeping.
GeronimoStilton
2017-03-24 21:30:04
What do you mean by 'bookkeeping'?
What do you mean by 'bookkeeping'?
DPatrick
2017-03-24 21:30:12
Just a lot of data to accurately collect.
Just a lot of data to accurately collect.
DPatrick
2017-03-24 21:30:20
Two to go!
Two to go!
DPatrick
2017-03-24 21:30:23
14. A $10 \times 10 \times 10$ grid of points consists of all points in space of the form $(i,j,k)$, where $i$, $j$, and $k$ are integers between 1 and 10, inclusive. Find the number of different lines that contain exactly 8 of these points.
14. A $10 \times 10 \times 10$ grid of points consists of all points in space of the form $(i,j,k)$, where $i$, $j$, and $k$ are integers between 1 and 10, inclusive. Find the number of different lines that contain exactly 8 of these points.
DPatrick
2017-03-24 21:30:53
There are lots of ways to approach this problem. Most involve (again) careful bookkeeping.
There are lots of ways to approach this problem. Most involve (again) careful bookkeeping.
SomethingNeutral
2017-03-24 21:31:08
Diagram
Diagram
DPatrick
2017-03-24 21:31:31
We can try to draw a diagram. I'm not going to try to draw a cube of 100 points, though.
We can try to draw a diagram. I'm not going to try to draw a cube of 100 points, though.
DPatrick
2017-03-24 21:31:42
In fact, I like 2-D better than 3-D, so I'm going to project my line down to the $xy$-plane (that is, set all the $z$-coordinates to 0):
In fact, I like 2-D better than 3-D, so I'm going to project my line down to the $xy$-plane (that is, set all the $z$-coordinates to 0):
DPatrick
2017-03-24 21:31:45
DPatrick
2017-03-24 21:32:24
If we have a line that works, and we project it down to the $xy$-plane, we still have 8 points on a line segment in this plane.
If we have a line that works, and we project it down to the $xy$-plane, we still have 8 points on a line segment in this plane.
DPatrick
2017-03-24 21:33:01
So we can try to count the 8-point segments in this 10x10 grid, and count how many different ways we could "unproject" it up into our cube.
So we can try to count the 8-point segments in this 10x10 grid, and count how many different ways we could "unproject" it up into our cube.
DPatrick
2017-03-24 21:33:18
Could I have an 8-point segment in my 10x10 grid that doesn't hit the boundary, like so?
Could I have an 8-point segment in my 10x10 grid that doesn't hit the boundary, like so?
DPatrick
2017-03-24 21:33:21
legolego
2017-03-24 21:33:47
no
no
GeronimoStilton
2017-03-24 21:33:47
No.
No.
DPatrick
2017-03-24 21:33:50
Why not?
Why not?
abishek99
2017-03-24 21:34:24
contains more than 8 points
contains more than 8 points
DPatrick
2017-03-24 21:34:58
Right. When I lift this up into the cube, I'm guaranteed to have enough room at one end or the other to extend to a 9th point!
Right. When I lift this up into the cube, I'm guaranteed to have enough room at one end or the other to extend to a 9th point!
DPatrick
2017-03-24 21:35:13
Maybe this is easier to see if we compare to this line segment:
Maybe this is easier to see if we compare to this line segment:
DPatrick
2017-03-24 21:35:16
DPatrick
2017-03-24 21:35:30
Can I lift this segment in the $xy$-plane up to the cube so that it only contains 8 points of the cube?
Can I lift this segment in the $xy$-plane up to the cube so that it only contains 8 points of the cube?
SomethingNeutral
2017-03-24 21:35:56
Yes
Yes
62861
2017-03-24 21:36:03
yes, if it contains lattice points from $z = 1, 2, \dots, 8$?
yes, if it contains lattice points from $z = 1, 2, \dots, 8$?
DPatrick
2017-03-24 21:36:11
Right!
Right!
DPatrick
2017-03-24 21:36:33
We're OK, provided that the non-boundary end of the segment (the point $(8,5)$ in my example) hits either $z=1$ or $z=10$, so that I can't extend it any further and pick up a 9th point.
We're OK, provided that the non-boundary end of the segment (the point $(8,5)$ in my example) hits either $z=1$ or $z=10$, so that I can't extend it any further and pick up a 9th point.
DPatrick
2017-03-24 21:36:50
For example, with the red segment shown running from $(1,5)$ to $(8,5)$, I could have the line in the cube passing through \[(1,5,8),(2,5,7),(3,5,6),\ldots,(8,5,1)\] or the line in the cube passing through \[(1,5,3),(2,5,4),(3,5,5),\ldots,(8,5,10)\] and we don't get a ninth point on that line in our cube.
For example, with the red segment shown running from $(1,5)$ to $(8,5)$, I could have the line in the cube passing through \[(1,5,8),(2,5,7),(3,5,6),\ldots,(8,5,1)\] or the line in the cube passing through \[(1,5,3),(2,5,4),(3,5,5),\ldots,(8,5,10)\] and we don't get a ninth point on that line in our cube.
DPatrick
2017-03-24 21:37:31
(Because there's no room in the $z$ direction to the ninth point at $(9,5)$ in the plane.)
(Because there's no room in the $z$ direction to the ninth point at $(9,5)$ in the plane.)
DPatrick
2017-03-24 21:38:08
So each 8-point segment in the $10 \times 10$ grid that has one endpoint on the boundary gives us 2 valid lines.
So each 8-point segment in the $10 \times 10$ grid that has one endpoint on the boundary gives us 2 valid lines.
DPatrick
2017-03-24 21:38:21
How many such segments are there?
How many such segments are there?
SomethingNeutral
2017-03-24 21:38:51
40
40
DPatrick
2017-03-24 21:38:55
Each row has 2, one at each end, so that's 20.
Each row has 2, one at each end, so that's 20.
DPatrick
2017-03-24 21:39:00
Each column has 2, one at each end, so that's another 20.
Each column has 2, one at each end, so that's another 20.
DPatrick
2017-03-24 21:39:09
But there are some diagonal ones too!
But there are some diagonal ones too!
DPatrick
2017-03-24 21:39:27
The three middle southwest-to-northeast diagonals each give us 2, one at each end, like so (I'm showing you the one at the southwest end below), so that's another 6:
The three middle southwest-to-northeast diagonals each give us 2, one at each end, like so (I'm showing you the one at the southwest end below), so that's another 6:
DPatrick
2017-03-24 21:39:32
DPatrick
2017-03-24 21:39:45
And the three middle northwest-to-southeast diagonals each gives us 2, one at each end, so that's another 6.
And the three middle northwest-to-southeast diagonals each gives us 2, one at each end, so that's another 6.
DPatrick
2017-03-24 21:40:01
So we have a total of $20+20+6+6 = 52$ 8-point segments in our grid that hits the boundary at one end of the grid.
So we have a total of $20+20+6+6 = 52$ 8-point segments in our grid that hits the boundary at one end of the grid.
DPatrick
2017-03-24 21:40:07
And each of these gave us 2 lines in the cube, so that's a total of $2 \cdot 52 = 104$ lines so far.
And each of these gave us 2 lines in the cube, so that's a total of $2 \cdot 52 = 104$ lines so far.
DPatrick
2017-03-24 21:40:29
And finally, what if we start with an 8-point segment that hits the boundary of the $xy$-plane at both ends?
And finally, what if we start with an 8-point segment that hits the boundary of the $xy$-plane at both ends?
DPatrick
2017-03-24 21:40:32
DPatrick
2017-03-24 21:40:45
In how many ways can each of these be lifted up to the cube to make a valid line?
In how many ways can each of these be lifted up to the cube to make a valid line?
vishwathganesan
2017-03-24 21:41:07
a lot
a lot
DPatrick
2017-03-24 21:41:31
Indeed, now there's very little restriction on the $z$-coordinates. We can't get a 9th point even in the $xy$-plane, so we don't have to worry about $z$ so much.
Indeed, now there's very little restriction on the $z$-coordinates. We can't get a 9th point even in the $xy$-plane, so we don't have to worry about $z$ so much.
GeronimoStilton
2017-03-24 21:41:36
$16$
$16$
DPatrick
2017-03-24 21:41:56
Right: there are 16 ways to lift each segment up to the cube, as follows.
Right: there are 16 ways to lift each segment up to the cube, as follows.
DPatrick
2017-03-24 21:42:03
$z$ could be constant. That's 10 possibilities.
$z$ could be constant. That's 10 possibilities.
DPatrick
2017-03-24 21:42:08
$z$ could start at 1, 2, or 3 and increase as we go left-to-right. That's 3 possibilities.
$z$ could start at 1, 2, or 3 and increase as we go left-to-right. That's 3 possibilities.
DPatrick
2017-03-24 21:42:16
$z$ could start at 8, 9, or 10 and decrease as we go left-to-right. That's 3 possibilities.
$z$ could start at 8, 9, or 10 and decrease as we go left-to-right. That's 3 possibilities.
DPatrick
2017-03-24 21:42:26
So each 8-point segment in the picture above gives us $10+3+3=16$ lines in the cube.
So each 8-point segment in the picture above gives us $10+3+3=16$ lines in the cube.
vishwathganesan
2017-03-24 21:42:52
so sixty4 total
so sixty4 total
DPatrick
2017-03-24 21:43:01
There were 4 such segments in the planar grid, so that's a total of $4 \cdot 16 = 64$ lines in the cube.
There were 4 such segments in the planar grid, so that's a total of $4 \cdot 16 = 64$ lines in the cube.
DPatrick
2017-03-24 21:43:05
Adding this to the $104$ lines we found earlier gives us a grand total of $64 + 104 = \boxed{168}$ total lines.
Adding this to the $104$ lines we found earlier gives us a grand total of $64 + 104 = \boxed{168}$ total lines.
DPatrick
2017-03-24 21:43:29
I thought that this problem, by far, was the hardest problem on the contest.
I thought that this problem, by far, was the hardest problem on the contest.
DPatrick
2017-03-24 21:43:53
#15 by comparison is a lot shorter.
#15 by comparison is a lot shorter.
DPatrick
2017-03-24 21:44:03
15. Tetrahedron $ABCD$ has $AD=BC=28$, $AC=BD=44$, and $AB=CD=52$. For any point $X$ in space, define $f(X) = AX + BX + CX + DX$. The least possible value of $f(X)$ can be expressed as $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.
15. Tetrahedron $ABCD$ has $AD=BC=28$, $AC=BD=44$, and $AB=CD=52$. For any point $X$ in space, define $f(X) = AX + BX + CX + DX$. The least possible value of $f(X)$ can be expressed as $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.
DPatrick
2017-03-24 21:44:31
Of course, we notice that this is a tetrahedron whose opposite sides are equal.
Of course, we notice that this is a tetrahedron whose opposite sides are equal.
DPatrick
2017-03-24 21:44:36
So there's probably some symmetry that we can use somewhere.
So there's probably some symmetry that we can use somewhere.
DPatrick
2017-03-24 21:45:02
Is there a clever way to embed this tetrahedron in 3-D Cartesian space?
Is there a clever way to embed this tetrahedron in 3-D Cartesian space?
flamescarlet
2017-03-24 21:45:27
as a rectangle?
as a rectangle?
carzland
2017-03-24 21:45:30
The middle is the origin.
The middle is the origin.
agbdmrbirdyface
2017-03-24 21:45:32
symmetric about the origin?
symmetric about the origin?
DPatrick
2017-03-24 21:45:42
How about as every other vertex of a rectangular box?
How about as every other vertex of a rectangular box?
DPatrick
2017-03-24 21:45:55
Specifically, $A,B,C,D$ are four of the eight corners of a rectangular box with face diagonals of 28, 44, and 52, and with sides $2r$, $2s$, and $2t$, and center at the origin.
Specifically, $A,B,C,D$ are four of the eight corners of a rectangular box with face diagonals of 28, 44, and 52, and with sides $2r$, $2s$, and $2t$, and center at the origin.
DPatrick
2017-03-24 21:45:59
DPatrick
2017-03-24 21:46:09
In coordinates, we can set $A = (r,s,t)$, $B = (-r,-s,t)$, $C = (-r,s,-t)$, and $D = (r,-s,-t)$ for some positive values of $r,s,t$.
In coordinates, we can set $A = (r,s,t)$, $B = (-r,-s,t)$, $C = (-r,s,-t)$, and $D = (r,-s,-t)$ for some positive values of $r,s,t$.
DPatrick
2017-03-24 21:46:24
So now our pairs of equal opposite sides are the face diagonals of our box.
So now our pairs of equal opposite sides are the face diagonals of our box.
agbdmrbirdyface
2017-03-24 21:46:41
and now we use distance formula?
and now we use distance formula?
DPatrick
2017-03-24 21:46:53
Yeah: the computation part of the solution is now pretty straightforward.
Yeah: the computation part of the solution is now pretty straightforward.
GeronimoStilton
2017-03-24 21:46:58
We want $4\sqrt{r^2 + s^2 + t^2}$!
We want $4\sqrt{r^2 + s^2 + t^2}$!
DPatrick
2017-03-24 21:47:09
Right, and we have \begin{align*}
28 = AD = BC &= 2\sqrt{s^2+t^2}, \\
44 = AC = BD &= 2\sqrt{r^2+t^2}, \\
52 = AB = CD &= 2\sqrt{r^2+s^2}.
\end{align*}
Right, and we have \begin{align*}
28 = AD = BC &= 2\sqrt{s^2+t^2}, \\
44 = AC = BD &= 2\sqrt{r^2+t^2}, \\
52 = AB = CD &= 2\sqrt{r^2+s^2}.
\end{align*}
DPatrick
2017-03-24 21:47:20
This simplifies to the system \begin{align*}
196 = 14^2 &= s^2 + t^2, \\
484 = 22^2 &= r^2 + t^2, \\
676 = 26^2 &= r^2 + s^2.
\end{align*}
This simplifies to the system \begin{align*}
196 = 14^2 &= s^2 + t^2, \\
484 = 22^2 &= r^2 + t^2, \\
676 = 26^2 &= r^2 + s^2.
\end{align*}
DPatrick
2017-03-24 21:47:35
Now, it's a pretty good guess that the origin $(0,0,0)$ is the point $X$ that we want. This point is the center of the tetrahedron.
Now, it's a pretty good guess that the origin $(0,0,0)$ is the point $X$ that we want. This point is the center of the tetrahedron.
GeronimoStilton
2017-03-24 21:47:53
Add them and then divide by 2
Add them and then divide by 2
vishwathganesan
2017-03-24 21:47:53
add everything together
add everything together
agbdmrbirdyface
2017-03-24 21:47:53
add them up, divide by 2, square root, multiply by 4
add them up, divide by 2, square root, multiply by 4
DPatrick
2017-03-24 21:48:22
So we see that $f((0,0,0)) = 4\sqrt{r^2+s^2+t^2}$, and we can get this quantity by summing our system of equations.
So we see that $f((0,0,0)) = 4\sqrt{r^2+s^2+t^2}$, and we can get this quantity by summing our system of equations.
DPatrick
2017-03-24 21:48:28
Summing our system gives $196 + 484 + 676 = 1356 = 2(r^2+s^2+t^2)$, so $r^2 + s^2 + t^2 = \dfrac{1356}{2} = 678$.
Summing our system gives $196 + 484 + 676 = 1356 = 2(r^2+s^2+t^2)$, so $r^2 + s^2 + t^2 = \dfrac{1356}{2} = 678$.
DPatrick
2017-03-24 21:48:40
And this makes $f((0,0,0)) = 4\sqrt{678}$.
And this makes $f((0,0,0)) = 4\sqrt{678}$.
DPatrick
2017-03-24 21:48:48
This doesn't simplify further, so it suggests that $4 + 678 = \boxed{682}$ is the answer.
This doesn't simplify further, so it suggests that $4 + 678 = \boxed{682}$ is the answer.
62861
2017-03-24 21:48:53
why is $X$ the origin
why is $X$ the origin
DPatrick
2017-03-24 21:48:58
Aha...there's the rub.
Aha...there's the rub.
DPatrick
2017-03-24 21:49:06
This isn't rigorous yet, because we guessed that $(0,0,0)$ is the point we want. Can we prove that?
This isn't rigorous yet, because we guessed that $(0,0,0)$ is the point we want. Can we prove that?
DPatrick
2017-03-24 21:49:23
Let's go back to the picture:
Let's go back to the picture:
DPatrick
2017-03-24 21:49:27
DPatrick
2017-03-24 21:49:38
Suppose $X$ is an arbitrary point. How can we show that $f((0,0,0)) \le f(X)$?
Suppose $X$ is an arbitrary point. How can we show that $f((0,0,0)) \le f(X)$?
flamescarlet
2017-03-24 21:49:54
triangle ineq?
triangle ineq?
DPatrick
2017-03-24 21:50:04
It's basically a Triangle Inequality argument.
It's basically a Triangle Inequality argument.
DPatrick
2017-03-24 21:50:12
We thought a lot about this here, and we couldn't find a good way to show it "all at once" using symmetry. Most of the solutions we came up with (and most of the solutions I've seen posted) first show that $X$ can be moved to one of the axes to decrease $f(X)$. And then, you do that twice (with two different axes) to move it to the origin.
We thought a lot about this here, and we couldn't find a good way to show it "all at once" using symmetry. Most of the solutions we came up with (and most of the solutions I've seen posted) first show that $X$ can be moved to one of the axes to decrease $f(X)$. And then, you do that twice (with two different axes) to move it to the origin.
DPatrick
2017-03-24 21:51:01
In particular, let's show that we can move $X$ "laterally" to the $z$-axis to decrease $f$. Suppose $X$ is an arbitrary point, and $Y$ is the point in the same $xy$-plane cross-section of $X$ that's on the $z$-axis. We'll try to show that $f(Y) \le f(X)$.
In particular, let's show that we can move $X$ "laterally" to the $z$-axis to decrease $f$. Suppose $X$ is an arbitrary point, and $Y$ is the point in the same $xy$-plane cross-section of $X$ that's on the $z$-axis. We'll try to show that $f(Y) \le f(X)$.
DPatrick
2017-03-24 21:51:08
Here's a "top-down" picture of what's going on, looking straight down from above:
Here's a "top-down" picture of what's going on, looking straight down from above:
DPatrick
2017-03-24 21:51:12
DPatrick
2017-03-24 21:51:22
Remember that these points are at different "heights": $A$ and $B$ are "on top" in the $z=t$ plane, $C$ and $D$ are "on the bottom" in the $z=-t$ plane, and $X$ and $Y$ are in some horizontal plane somewhere in the middle.
Remember that these points are at different "heights": $A$ and $B$ are "on top" in the $z=t$ plane, $C$ and $D$ are "on the bottom" in the $z=-t$ plane, and $X$ and $Y$ are in some horizontal plane somewhere in the middle.
DPatrick
2017-03-24 21:51:32
We want to show that $f(Y) \le f(X)$.
We want to show that $f(Y) \le f(X)$.
DPatrick
2017-03-24 21:51:56
One idea is to reflect $X$ in its plane across the origin, creating a new point $X'$:
One idea is to reflect $X$ in its plane across the origin, creating a new point $X'$:
DPatrick
2017-03-24 21:52:00
DPatrick
2017-03-24 21:52:05
One nice thing that happens is that $Y$ is the midpoint of $\overline{XX'}$.
One nice thing that happens is that $Y$ is the midpoint of $\overline{XX'}$.
DPatrick
2017-03-24 21:52:09
But we also see that $f(X') = f(X)$...why?
But we also see that $f(X') = f(X)$...why?
GeronimoStilton
2017-03-24 21:52:29
Symmetry.
Symmetry.
SomethingNeutral
2017-03-24 21:52:29
symmetry
symmetry
flamescarlet
2017-03-24 21:52:35
A and B at the same height, everything's rotated 180
A and B at the same height, everything's rotated 180
DPatrick
2017-03-24 21:52:42
Exactly. This same reflection across the $z$-axis moves $A$ to $B$ and vice versa, and moves $C$ to $D$ and vice versa. In particular, $AX = BX'$ and $BX = AX'$, and the same for $C$ and $D$.
Exactly. This same reflection across the $z$-axis moves $A$ to $B$ and vice versa, and moves $C$ to $D$ and vice versa. In particular, $AX = BX'$ and $BX = AX'$, and the same for $C$ and $D$.
DPatrick
2017-03-24 21:53:00
So now, if we can show that $AX + AX' \ge 2(AY)$ (and the same for the other points), we'd be done, because then when we sum over all four points, we'd have $f(X) + f(X') \ge 2f(Y)$, or $f(X) \ge f(Y)$.
So now, if we can show that $AX + AX' \ge 2(AY)$ (and the same for the other points), we'd be done, because then when we sum over all four points, we'd have $f(X) + f(X') \ge 2f(Y)$, or $f(X) \ge f(Y)$.
DPatrick
2017-03-24 21:53:19
So let's try to show that $AX + AX' \ge 2(AY)$. But this is much simpler since all these points lie in the same plane:
So let's try to show that $AX + AX' \ge 2(AY)$. But this is much simpler since all these points lie in the same plane:
DPatrick
2017-03-24 21:53:22
DPatrick
2017-03-24 21:53:29
This is a general statement about triangles: if $AXX'$ is a triangle and $Y$ is the midpoint of $\overline{XX'}$, then $AX + AX' \ge 2(AY)$. Why is this true?
This is a general statement about triangles: if $AXX'$ is a triangle and $Y$ is the midpoint of $\overline{XX'}$, then $AX + AX' \ge 2(AY)$. Why is this true?
FieryEntei
2017-03-24 21:53:47
triangle inequality
triangle inequality
mssmath
2017-03-24 21:53:47
Create a parallelogram
Create a parallelogram
DPatrick
2017-03-24 21:53:52
Sure. Just reflect the triangle across $\overline{XX'}$ to make a parallelogram to point $Z$:
Sure. Just reflect the triangle across $\overline{XX'}$ to make a parallelogram to point $Z$:
DPatrick
2017-03-24 21:53:55
DPatrick
2017-03-24 21:54:18
"Rotate" is probably a better word than "reflect" in my previous sentence.
"Rotate" is probably a better word than "reflect" in my previous sentence.
DPatrick
2017-03-24 21:54:32
So now $AX' = ZX$ and $AY = ZY$, so $AX + AX' = AX + XZ$, and $2AY = AZ$.
So now $AX' = ZX$ and $AY = ZY$, so $AX + AX' = AX + XZ$, and $2AY = AZ$.
DPatrick
2017-03-24 21:54:43
Since $AX + XZ \ge AZ$ by the Triangle Inequality, we have $AX + AX' \ge 2AY$, just like we wanted.
Since $AX + XZ \ge AZ$ by the Triangle Inequality, we have $AX + AX' \ge 2AY$, just like we wanted.
DPatrick
2017-03-24 21:54:58
To summarize: we've shown that we can take any point $X$ in our box and make $f(X)$ smaller by moving $X$ over to the $z$-axis.
To summarize: we've shown that we can take any point $X$ in our box and make $f(X)$ smaller by moving $X$ over to the $z$-axis.
DPatrick
2017-03-24 21:55:05
So doing this twice in a row (once with the $z$-axis and once with a different axis) shows that we can always make $f(X)$ smaller by moving it to the origin.
So doing this twice in a row (once with the $z$-axis and once with a different axis) shows that we can always make $f(X)$ smaller by moving it to the origin.
DPatrick
2017-03-24 21:55:13
Therefore, $f((0,0,0)) \le f(X)$ for any $X$, and indeed the origin is the best point.
Therefore, $f((0,0,0)) \le f(X)$ for any $X$, and indeed the origin is the best point.
DPatrick
2017-03-24 21:55:21
And thus, $f((0,0,0)) = 4\sqrt{678}$ is the least possible value of $f(X)$, and our final answer is $4 + 678 = \boxed{682}$.
And thus, $f((0,0,0)) = 4\sqrt{678}$ is the least possible value of $f(X)$, and our final answer is $4 + 678 = \boxed{682}$.
DPatrick
2017-03-24 21:55:58
I thought this was easier than #14, AIME-wise, because you can "guess" that the center point is the best point without having to prove it. And if you put the tetrahedron in a box, the computation is pretty straightforward.
I thought this was easier than #14, AIME-wise, because you can "guess" that the center point is the best point without having to prove it. And if you put the tetrahedron in a box, the computation is pretty straightforward.
DPatrick
2017-03-24 21:56:15
#14 is just a nasty 3-D visualization and counting problem where you have to be really really careful.
#14 is just a nasty 3-D visualization and counting problem where you have to be really really careful.
GeronimoStilton
2017-03-24 21:56:21
We did all 15 problems in less than 3:00 hours!
We did all 15 problems in less than 3:00 hours!
DPatrick
2017-03-24 21:56:28
Success! That's it for the 2017 AIME II!
Success! That's it for the 2017 AIME II!
DPatrick
2017-03-24 21:56:37
Thanks for coming. Have a great weekend!
Thanks for coming. Have a great weekend!
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