Mathcamp 2024 Qualifying Quiz Part 1 Math Jam, Problems 1-3
Go back to the Math Jam ArchiveThis Math Jam will discuss solutions to the first half of the 2024 Mathcamp Qualifying Quiz. Mathcamp is an intensive 5-week-long summer program for mathematically talented high school students. More than just a summer camp, Mathcamp is a vibrant community, made up of a wide variety of people who share a common love of learning and passion for mathematics. Join us tomorrow for the second half!
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Facilitator: AoPS Staff
jwelsh
2024-05-15 18:51:12
Welcome to the Canada/USA Mathcamp Qualifying Quiz Math Jam! The Math Jam will begin at 7:00 pm ET (4:00 pm PT). We will cover problems 1-3 today and be back tomorrow for problems 4-6.
Welcome to the Canada/USA Mathcamp Qualifying Quiz Math Jam! The Math Jam will begin at 7:00 pm ET (4:00 pm PT). We will cover problems 1-3 today and be back tomorrow for problems 4-6.
jwelsh
2024-05-15 18:51:25
The classroom is moderated: students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. So, when you send a message, it will not appear immediately, and may not appear at all.
The classroom is moderated: students can type into the classroom, but only the moderators can choose a comment to drop into the classroom. So, when you send a message, it will not appear immediately, and may not appear at all.
jwelsh
2024-05-15 18:52:22
This Math Jam will be led by our friends from Mathcamp!
This Math Jam will be led by our friends from Mathcamp!
jwelsh
2024-05-15 19:00:25
Hello and welcome to the Canada/USA Mathcamp Qualifying Quiz Math Jam!
Hello and welcome to the Canada/USA Mathcamp Qualifying Quiz Math Jam!
jwelsh
2024-05-15 19:00:36
Before I introduce our guests, let me briefly explain how our online classroom works.
Before I introduce our guests, let me briefly explain how our online classroom works.
jwelsh
2024-05-15 19:00:47
This room is moderated, which means that all your questions and comments come to the moderators. We may share your comments with the whole room if we so choose.
This room is moderated, which means that all your questions and comments come to the moderators. We may share your comments with the whole room if we so choose.
jwelsh
2024-05-15 19:00:56
Also, you'll find that you can adjust the classroom windows in a variety of ways by clicking on the user dropdown menu in the top-right corner of the classroom.
Also, you'll find that you can adjust the classroom windows in a variety of ways by clicking on the user dropdown menu in the top-right corner of the classroom.
jwelsh
2024-05-15 19:01:05
Canada/USA Mathcamp is an intensive five-week-long summer program for high-school students interested in mathematics, designed to expose students to the beauty of advanced mathematical ideas and to new ways of thinking. You can learn more about Canada/USA Mathcamp here: www.mathcamp.org
Canada/USA Mathcamp is an intensive five-week-long summer program for high-school students interested in mathematics, designed to expose students to the beauty of advanced mathematical ideas and to new ways of thinking. You can learn more about Canada/USA Mathcamp here: www.mathcamp.org
jwelsh
2024-05-15 19:01:13
Many AoPS instructors, assistants, and students are alumni of this outstanding program!
Many AoPS instructors, assistants, and students are alumni of this outstanding program!
jwelsh
2024-05-15 19:01:32
Each year, Mathcamp creates a Qualifying Quiz, which is the main component of the application process. If you haven't already seen it, you can find the 2024 Quiz problems at https://www.mathcamp.org/qualifying_quiz/
In this Math Jam, the following Mathcamp staff (all members of this year's Quiz committee) will discuss the problems and their solutions:
Each year, Mathcamp creates a Qualifying Quiz, which is the main component of the application process. If you haven't already seen it, you can find the 2024 Quiz problems at https://www.mathcamp.org/qualifying_quiz/
In this Math Jam, the following Mathcamp staff (all members of this year's Quiz committee) will discuss the problems and their solutions:
jwelsh
2024-05-15 19:01:57
Tim Black (timblack) has been a camper, JC, mentor, and faculty at Mathcamp. This summer — his 15th at camp — he will be an academic coordinator. He received his Ph.D. in math and computer science from the University of Chicago, where he studied applications of group theory to complexity theory and error correcting codes.
Tim Black (timblack) has been a camper, JC, mentor, and faculty at Mathcamp. This summer — his 15th at camp — he will be an academic coordinator. He received his Ph.D. in math and computer science from the University of Chicago, where he studied applications of group theory to complexity theory and error correcting codes.
jwelsh
2024-05-15 19:02:11
Yasha Berchenko-Kogan (Yasha) has been a camper and a JC at Mathcamp. He is currently an assistant professor at the Florida Institute of Technology, studying differential geometry and numerical analysis.
Yasha Berchenko-Kogan (Yasha) has been a camper and a JC at Mathcamp. He is currently an assistant professor at the Florida Institute of Technology, studying differential geometry and numerical analysis.
jwelsh
2024-05-15 19:02:20
Let's turn the room over to Tim and Yasha now to get us started!
Let's turn the room over to Tim and Yasha now to get us started!
timblack
2024-05-15 19:02:25
Hi all! Thanks, Jo, for the introduction. I'm Tim!. I'm excited to get to talk about the qualifying quiz with you all today.
Hi all! Thanks, Jo, for the introduction. I'm Tim!. I'm excited to get to talk about the qualifying quiz with you all today.
timblack
2024-05-15 19:02:43
You are welcome and encouraged to put questions in the chat at any time!
You are welcome and encouraged to put questions in the chat at any time!
timblack
2024-05-15 19:02:47
We're covering problems 1–3 today, but we're actually going to start with problem 3.
We're covering problems 1–3 today, but we're actually going to start with problem 3.
timblack
2024-05-15 19:02:56
Problem 3
Problem 3
timblack
2024-05-15 19:03:03
3. Here is a curious fact you may not have known about mathematicians: when asked a yes-or-no question, number theorists will always tell the truth, while analysts will always lie. Be careful: if you ask someone a yes-or-no question, and they cannot answer either "yes" or "no" to it, then the universe explodes in paradox. For example, this happens if you ask an number theorist, "Is your answer to this question 'no'?"
3. Here is a curious fact you may not have known about mathematicians: when asked a yes-or-no question, number theorists will always tell the truth, while analysts will always lie. Be careful: if you ask someone a yes-or-no question, and they cannot answer either "yes" or "no" to it, then the universe explodes in paradox. For example, this happens if you ask an number theorist, "Is your answer to this question 'no'?"
timblack
2024-05-15 19:03:06
a. What yes-or-no question can be asked to both a number theorist and an analyst to cause the universe to explode in both cases?
a. What yes-or-no question can be asked to both a number theorist and an analyst to cause the universe to explode in both cases?
timblack
2024-05-15 19:03:22
Wow, this hardly looks like a math problem!
Wow, this hardly looks like a math problem!
timblack
2024-05-15 19:03:30
But it actually does require some logical reasoning that is suspiciously mathematical, so we'll allow it!
But it actually does require some logical reasoning that is suspiciously mathematical, so we'll allow it!
timblack
2024-05-15 19:03:42
Let's try to solve this problem together.
Let's try to solve this problem together.
CheetahLover
2024-05-15 19:03:48
sus
sus
timblack
2024-05-15 19:03:54
A good place to start on a problem like this is to just start coming up with possible questions. Don't worry yet about whether the question you come up with "works" for the problem. The point is to just try something and see what happens. Think up some questions — I'll ask for them in a minute.
A good place to start on a problem like this is to just start coming up with possible questions. Don't worry yet about whether the question you come up with "works" for the problem. The point is to just try something and see what happens. Think up some questions — I'll ask for them in a minute.
timblack
2024-05-15 19:04:05
We can start with the the example question: "Is your answer to this question 'no'?". First, let's see what happens if we ask this question to a number theorist, who is supposed to always tell the truth.
We can start with the the example question: "Is your answer to this question 'no'?". First, let's see what happens if we ask this question to a number theorist, who is supposed to always tell the truth.
timblack
2024-05-15 19:04:11
If a number theorist answers "yes" to this question, is it a lie or the truth?
If a number theorist answers "yes" to this question, is it a lie or the truth?
timblack
2024-05-15 19:05:06
timblack
2024-05-15 19:05:10
It would be a lie! The answer is not "no", but by saying "yes" a number theorist would be claiming that it is.
It would be a lie! The answer is not "no", but by saying "yes" a number theorist would be claiming that it is.
timblack
2024-05-15 19:05:25
If a number theorist answers "no" to this question, is it a lie or the truth?
If a number theorist answers "no" to this question, is it a lie or the truth?
timblack
2024-05-15 19:06:07
timblack
2024-05-15 19:06:10
It's also a lie! This time, the answer is "no", but by saying "no", a number theorist is claiming it's not.
It's also a lie! This time, the answer is "no", but by saying "no", a number theorist is claiming it's not.
Shadow_Sniper
2024-05-15 19:06:31
That's one that would make the universe explode!
That's one that would make the universe explode!
timblack
2024-05-15 19:06:33
These are the two things we need to check to see if the universe explodes: if the mathematician can't answer "yes" and can't answer "no", then the universe explodes.
These are the two things we need to check to see if the universe explodes: if the mathematician can't answer "yes" and can't answer "no", then the universe explodes.
timblack
2024-05-15 19:06:37
In this case, a number theorist is supposed to supposed to tell the truth, but both "yes" and "no" would be a lie. So a number theorist can't give either answer, and the universe explodes.
In this case, a number theorist is supposed to supposed to tell the truth, but both "yes" and "no" would be a lie. So a number theorist can't give either answer, and the universe explodes.
timblack
2024-05-15 19:06:56
Let's keep using the same example question: "Is your answer to this question 'no'?". Now, let's try asking it to an analyst, who is supposed to always lie.
Let's keep using the same example question: "Is your answer to this question 'no'?". Now, let's try asking it to an analyst, who is supposed to always lie.
timblack
2024-05-15 19:07:05
If an analyst answers "yes" to this question, is it a lie or the truth?
If an analyst answers "yes" to this question, is it a lie or the truth?
timblack
2024-05-15 19:07:54
timblack
2024-05-15 19:07:59
It's a lie! (Which is good: the analyst is supposed to lie.)
It's a lie! (Which is good: the analyst is supposed to lie.)
timblack
2024-05-15 19:08:01
If an analyst answers "no" to this question, is it a lie or the truth?
If an analyst answers "no" to this question, is it a lie or the truth?
timblack
2024-05-15 19:08:30
timblack
2024-05-15 19:08:31
Again, it's a lie!
Again, it's a lie!
timblack
2024-05-15 19:08:38
So, actually, an analyst has a choice in how to answer this question: they can say "yes" or "no", and either way they'll be telling a lie, which is exactly what they are supposed to do. So the universe does not explode if we ask this to an analyst.
So, actually, an analyst has a choice in how to answer this question: they can say "yes" or "no", and either way they'll be telling a lie, which is exactly what they are supposed to do. So the universe does not explode if we ask this to an analyst.
JordanZhang2013
2024-05-15 19:08:53
It's all lies!
It's all lies!
CheetahLover
2024-05-15 19:08:57
no explosion
no explosion
timblack
2024-05-15 19:09:03
The problem wants us to find a question that will cause the universe to explode no matter which mathematician you ask it to. The example question causes the universe to explode when asked to a number theorist, but not to an analyst. What are some other questions we could ask? I don't care right now whether the questions actually cause the universe to explode or not; the point is just to explore.
The problem wants us to find a question that will cause the universe to explode no matter which mathematician you ask it to. The example question causes the universe to explode when asked to a number theorist, but not to an analyst. What are some other questions we could ask? I don't care right now whether the questions actually cause the universe to explode or not; the point is just to explore.
timblack
2024-05-15 19:10:05
Keep putting thoughts in the chat!
Keep putting thoughts in the chat!
TJmit
2024-05-15 19:10:09
Are you lying?
Are you lying?
timblack
2024-05-15 19:10:25
Can either mathematician answer "Are you lying?"
Can either mathematician answer "Are you lying?"
phoenixxxxxx
2024-05-15 19:10:58
yes
yes
aryajb2013
2024-05-15 19:10:58
yes
yes
shsiliveri
2024-05-15 19:10:58
yes
yes
RubyJaguar10
2024-05-15 19:10:58
yes
yes
ryankimse
2024-05-15 19:10:58
Yes?
Yes?
snow52
2024-05-15 19:10:58
yes
yes
timblack
2024-05-15 19:11:04
Yes!
Yes!
timblack
2024-05-15 19:11:21
The number theorist can truthfully answer "no", while the analyst can lie by saying "no"
The number theorist can truthfully answer "no", while the analyst can lie by saying "no"
nikola_tesla-roadster
2024-05-15 19:11:37
Are you hungry
Are you hungry
timblack
2024-05-15 19:11:46
What do you think about this? Can they answer "are you hungry?"
What do you think about this? Can they answer "are you hungry?"
BrianZhao
2024-05-15 19:12:04
yes
yes
Super_AA
2024-05-15 19:12:04
yes
yes
ewang17
2024-05-15 19:12:04
yes
yes
baboo12
2024-05-15 19:12:04
yes
yes
Kai321
2024-05-15 19:12:04
yes
yes
NAAUBE
2024-05-15 19:12:04
Yes
Yes
kjialili
2024-05-15 19:12:04
yes
yes
timblack
2024-05-15 19:12:08
Sure!
Sure!
timblack
2024-05-15 19:12:31
I don't know if they are hungry, but presumably they know, and can simply tell the truth about it if they are a number theorist, or say the opposite if they are an analyst
I don't know if they are hungry, but presumably they know, and can simply tell the truth about it if they are a number theorist, or say the opposite if they are an analyst
jmath10
2024-05-15 19:13:06
If lying is telling the truth, have you ever told a lie?
If lying is telling the truth, have you ever told a lie?
timblack
2024-05-15 19:13:12
What do we think of this question?
What do we think of this question?
Andrewwang828
2024-05-15 19:13:41
ngl pretty fire
ngl pretty fire
ChaitraliKA
2024-05-15 19:13:41
Interesting
Interesting
timblack
2024-05-15 19:13:53
I also think it's interesting, but I'm not sure what it means!
I also think it's interesting, but I'm not sure what it means!
timblack
2024-05-15 19:14:59
One could say that lying inherently isn't telling the truth, so the truthful answer is always "yes"
One could say that lying inherently isn't telling the truth, so the truthful answer is always "yes"
timblack
2024-05-15 19:15:10
Or one could say the question doesn't make sense
Or one could say the question doesn't make sense
timblack
2024-05-15 19:15:31
In any case, it doesn't feel like a very satisfying answer to the question, and indeed, the FAQ for this problem disallows it
In any case, it doesn't feel like a very satisfying answer to the question, and indeed, the FAQ for this problem disallows it
timblack
2024-05-15 19:16:10
The FAQ for the problem says: "Some questions just don't have clear answers, or don't have known answers, and these will not cause the universe to explode." I think this is fair; and in this particular case, the question really does have a true answer, even if a number theorist doesn't know what it is.
The FAQ for the problem says: "Some questions just don't have clear answers, or don't have known answers, and these will not cause the universe to explode." I think this is fair; and in this particular case, the question really does have a true answer, even if a number theorist doesn't know what it is.
timblack
2024-05-15 19:16:49
Okay, how about this question: "Am I thinking of a boat?". Does this cause the universe to explode when asked to a number theorist?
Okay, how about this question: "Am I thinking of a boat?". Does this cause the universe to explode when asked to a number theorist?
timblack
2024-05-15 19:17:23
Oops I got ahead of myself
Oops I got ahead of myself
timblack
2024-05-15 19:18:03
Here's the one where I think it is fair to disqualify it. The question really does have a true answer, even if a number theorist doesn't know what it is.
Here's the one where I think it is fair to disqualify it. The question really does have a true answer, even if a number theorist doesn't know what it is.
timblack
2024-05-15 19:18:08
(By the way, yes, I was thinking of a boat)
(By the way, yes, I was thinking of a boat)
timblack
2024-05-15 19:18:13
timblack
2024-05-15 19:18:39
We've learned something here: If a question is going to cause the universe to explode, it needs to depend on the answer. In other words, giving a different answer to the question needs to change the true answer to the question. The question needs to be self referential
We've learned something here: If a question is going to cause the universe to explode, it needs to depend on the answer. In other words, giving a different answer to the question needs to change the true answer to the question. The question needs to be self referential
AxoLuna
2024-05-15 19:18:43
TAU BOAT!
TAU BOAT!
gorigo
2024-05-15 19:18:51
nice boat!
nice boat!
timblack
2024-05-15 19:18:56
Can we come up with a question that causes the universe to explode when asked to an analyst? (Remember an analyst always lies.)
Can we come up with a question that causes the universe to explode when asked to an analyst? (Remember an analyst always lies.)
sr2022
2024-05-15 19:19:06
Are they selling cookies?
Are they selling cookies?
timblack
2024-05-15 19:19:25
They are giving out cookies to campers doing math at TAU (Mathcamp's worktime and office hours)!
They are giving out cookies to campers doing math at TAU (Mathcamp's worktime and office hours)!
ardentiat
2024-05-15 19:19:32
Is your answer to this question "yes"?
Is your answer to this question "yes"?
admathcamp353
2024-05-15 19:19:32
Is your answer to this question "yes?"
Is your answer to this question "yes?"
snow52
2024-05-15 19:19:32
is your answer to this question "yes"?
is your answer to this question "yes"?
Kai321
2024-05-15 19:19:32
is your answer to this question "yes"?
is your answer to this question "yes"?
timblack
2024-05-15 19:19:38
Sure, we can ask "Is your answer to this question 'yes'?".
Sure, we can ask "Is your answer to this question 'yes'?".
timblack
2024-05-15 19:19:49
If an analyst says "yes", they are telling the truth, because their answer to the question is indeed "yes". If they say "no", they are also telling the truth, because in this case the answer is not "yes". Either answer is the truth, but they are supposed to lie, so they can't give either answer. So the universe explodes.
If an analyst says "yes", they are telling the truth, because their answer to the question is indeed "yes". If they say "no", they are also telling the truth, because in this case the answer is not "yes". Either answer is the truth, but they are supposed to lie, so they can't give either answer. So the universe explodes.
ScramJack
2024-05-15 19:20:07
Boom
Boom
CheetahLover
2024-05-15 19:20:07
boom
boom
timblack
2024-05-15 19:20:11
I think we're getting closer to a final answer. We have one question that causes the universe to explode when asked to a number theorist: "Is your answer to this question 'no'?". And we have one question that causes the universe to explode when asked to an analyst: "Is your answer to this question 'yes'?". Can we somehow combine these two questions into a single question that works for both?
I think we're getting closer to a final answer. We have one question that causes the universe to explode when asked to a number theorist: "Is your answer to this question 'no'?". And we have one question that causes the universe to explode when asked to an analyst: "Is your answer to this question 'yes'?". Can we somehow combine these two questions into a single question that works for both?
stevecabbage
2024-05-15 19:20:52
can I have a cookie?
can I have a cookie?
timblack
2024-05-15 19:20:54
Hmm, I think I can truthfully answer "Can I have a cookie" with "no"p
Hmm, I think I can truthfully answer "Can I have a cookie" with "no"p
Mathdreams
2024-05-15 19:21:03
timblack
2024-05-15 19:21:12
Going back to the task at hand: The two questions are identical except for the last word ("'no'" vs "'yes'"). Can we come up with a phrase that is equivalent to "'no'" for a number theorist but equivalent to "'yes'" for an analyst? If so, we could substitute that in to the question.
Going back to the task at hand: The two questions are identical except for the last word ("'no'" vs "'yes'"). Can we come up with a phrase that is equivalent to "'no'" for a number theorist but equivalent to "'yes'" for an analyst? If so, we could substitute that in to the question.
joeym2011
2024-05-15 19:22:11
Is your answer to the question the same as the answer to "Are you an analyst"?
Is your answer to the question the same as the answer to "Are you an analyst"?
timblack
2024-05-15 19:22:15
Right: "the true answer to whether you are an analyst" is "no" for a number theorist and "yes" for an analyst.
Right: "the true answer to whether you are an analyst" is "no" for a number theorist and "yes" for an analyst.
timblack
2024-05-15 19:22:25
Putting it all together, we can ask the question "Is your answer to this question the same as the true answer to whether you are an analyst?"
Putting it all together, we can ask the question "Is your answer to this question the same as the true answer to whether you are an analyst?"
timblack
2024-05-15 19:23:04
For good measure, let's check what happens if you ask this question to a number theorist or analyst.
For good measure, let's check what happens if you ask this question to a number theorist or analyst.
timblack
2024-05-15 19:23:10
If a number theorist says "yes", then "your answer to this question" is "yes", while the "true answer to whether you are an analyst" is "no". These are not the same, so the number theorist lied by answering "yes".
If a number theorist says "yes", then "your answer to this question" is "yes", while the "true answer to whether you are an analyst" is "no". These are not the same, so the number theorist lied by answering "yes".
timblack
2024-05-15 19:23:23
If a number theorist says "no", then "your answer to this question" is "no", while the "true answer to whether you are an analyst" is "no". These are the same, so the number theorist lied by answering "no".
If a number theorist says "no", then "your answer to this question" is "no", while the "true answer to whether you are an analyst" is "no". These are the same, so the number theorist lied by answering "no".
timblack
2024-05-15 19:23:29
In summary, if you ask the question to a number theorist, answering either "yes" or "no" would be a lie, but a number theorist is supposed to tell the truth, so the universe explodes.
In summary, if you ask the question to a number theorist, answering either "yes" or "no" would be a lie, but a number theorist is supposed to tell the truth, so the universe explodes.
Matthew.C
2024-05-15 19:23:33
boom
boom
hiloo
2024-05-15 19:23:51
my brain goes boom
my brain goes boom
Andrewwang828
2024-05-15 19:23:51
is this math or just thinking really hard
is this math or just thinking really hard
timblack
2024-05-15 19:23:57
timblack
2024-05-15 19:24:03
If an analyst says "yes", then "your answer to this question" is "yes", while the "true answer to whether you are an analyst" is "yes". These are the same, so the analyst told the truth by answering "yes".
If an analyst says "yes", then "your answer to this question" is "yes", while the "true answer to whether you are an analyst" is "yes". These are the same, so the analyst told the truth by answering "yes".
timblack
2024-05-15 19:24:13
If an analyst says "no", then "your answer to this question" is "no", while the "true answer to whether you are an analyst" is "yes". These are not the same, so the analyst told the truth by answering "no".
If an analyst says "no", then "your answer to this question" is "no", while the "true answer to whether you are an analyst" is "yes". These are not the same, so the analyst told the truth by answering "no".
timblack
2024-05-15 19:24:24
To summarize, if you ask the question to an analyst, answering either "yes" or "no" is the truth, but an analyst is supposed to lie so the universe explodes.
To summarize, if you ask the question to an analyst, answering either "yes" or "no" is the truth, but an analyst is supposed to lie so the universe explodes.
RubyJaguar10
2024-05-15 19:24:40
yay universe go BOOM!!!
yay universe go BOOM!!!
nikola_tesla-roadster
2024-05-15 19:24:40
KABOOM
KABOOM
baboo12
2024-05-15 19:24:40
wowza
wowza
hiloo
2024-05-15 19:24:55
isn't math just thinking very hard @andrewwang?
isn't math just thinking very hard @andrewwang?
timblack
2024-05-15 19:24:58
So, we've come up with a question that satisfies the problem: "Is your answer to this question the same as the true answer to whether you are an analyst?"
So, we've come up with a question that satisfies the problem: "Is your answer to this question the same as the true answer to whether you are an analyst?"
timblack
2024-05-15 19:25:12
Here are a couple of other questions that are equivalent to this one (they are essentially rephrasings):
- "Are you a number theorist xor is your answer to this question 'yes'?"
- "Is your answer to this question 'no' if and only if you are a number theorist?"
Here are a couple of other questions that are equivalent to this one (they are essentially rephrasings):
- "Are you a number theorist xor is your answer to this question 'yes'?"
- "Is your answer to this question 'no' if and only if you are a number theorist?"
timblack
2024-05-15 19:25:29
There's also a different question that I frequently came across while reading applications:
There's also a different question that I frequently came across while reading applications:
timblack
2024-05-15 19:25:41
"Would the other type of mathematician answer 'yes' to this question?"
"Would the other type of mathematician answer 'yes' to this question?"
timblack
2024-05-15 19:25:50
This question can work, but you have to be really careful about how you write about it. Actually, we have to be really careful about any question that refers to what a different mathematician would do. Here's why.
This question can work, but you have to be really careful about how you write about it. Actually, we have to be really careful about any question that refers to what a different mathematician would do. Here's why.
timblack
2024-05-15 19:25:59
Let's say we ask this question to a number theorist. They, in their head, hypothesize about what would happen if they asked the question to an analyst. The analyst could say "yes" or "no", or they could cause the universe to explode. If, in their head, the number theorist decides that the analyst causes the universe to explode, then the number theorist can answer "no", in the sense of "no, the other type of mathematician doesn't answer 'yes' to this question, because they don't say anything, because their universe has exploded."
Let's say we ask this question to a number theorist. They, in their head, hypothesize about what would happen if they asked the question to an analyst. The analyst could say "yes" or "no", or they could cause the universe to explode. If, in their head, the number theorist decides that the analyst causes the universe to explode, then the number theorist can answer "no", in the sense of "no, the other type of mathematician doesn't answer 'yes' to this question, because they don't say anything, because their universe has exploded."
timblack
2024-05-15 19:26:24
You can make the argument that this question works, but you need to be precise in describing how you are modeling the universe.
You can make the argument that this question works, but you need to be precise in describing how you are modeling the universe.
timblack
2024-05-15 19:26:36
Any other thoughts on part (a)?
Any other thoughts on part (a)?
jmath10
2024-05-15 19:27:48
Would a question such as this work: Is your answer to this question neither a lie nor a truth? Or would that be going against the FAQ?
Would a question such as this work: Is your answer to this question neither a lie nor a truth? Or would that be going against the FAQ?
timblack
2024-05-15 19:27:51
I think you just have to be very careful if you veer into unusual definitions
I think you just have to be very careful if you veer into unusual definitions
AxoLuna
2024-05-15 19:28:24
How might you go about laying down that precision?
How might you go about laying down that precision?
timblack
2024-05-15 19:28:26
We'll get an example of some careful reasoning in part (b)
We'll get an example of some careful reasoning in part (b)
timblack
2024-05-15 19:29:13
On that note, let's move onto part (b)! You can still keep putting your questions in the chat.
On that note, let's move onto part (b)! You can still keep putting your questions in the chat.
timblack
2024-05-15 19:29:26
b. To try to prevent the universe from exploding in paradox, logicians have decided to act as paradox detectors. When you ask a logician a yes-or-no question, the logician will answer "yes" if asking the question to a number theorist would cause the universe to explode, and "no" otherwise. For example, if you ask a logician, "Is your answer to this question 'no'?" the logician will answer "yes".
What yes-or-no question can be asked to a logician to cause the universe to explode?
b. To try to prevent the universe from exploding in paradox, logicians have decided to act as paradox detectors. When you ask a logician a yes-or-no question, the logician will answer "yes" if asking the question to a number theorist would cause the universe to explode, and "no" otherwise. For example, if you ask a logician, "Is your answer to this question 'no'?" the logician will answer "yes".
What yes-or-no question can be asked to a logician to cause the universe to explode?
timblack
2024-05-15 19:29:43
This is getting pretty meta
This is getting pretty meta
timblack
2024-05-15 19:29:48
This problem is also weird because when you ask a logician a question, the logician doesn't actually try to answer the question themself. They still reply with "yes" or "no", but that reply is not intended to be an answer to the question. Instead, it indicates whether the universe explodes if you ask the question to a number theorist.
This problem is also weird because when you ask a logician a question, the logician doesn't actually try to answer the question themself. They still reply with "yes" or "no", but that reply is not intended to be an answer to the question. Instead, it indicates whether the universe explodes if you ask the question to a number theorist.
timblack
2024-05-15 19:30:09
Also, remember the universe explodes when someone can't say "yes" or "no". So we want a question where the logician can't respond "yes" or "no".
Also, remember the universe explodes when someone can't say "yes" or "no". So we want a question where the logician can't respond "yes" or "no".
Liontiger
2024-05-15 19:30:38
but why are we trying to explode the universe?
but why are we trying to explode the universe?
timblack
2024-05-15 19:30:39
Maybe we're trying to figure out what not to do to avoid the universe from exploding?
Maybe we're trying to figure out what not to do to avoid the universe from exploding?
timblack
2024-05-15 19:30:45
Let's get our bearings. Just like in part (a), we can expect that the question we ask will again have to be self-referential: We saw in part (a) that a question that doesn't reference to its own answer can always be answered "yes" or "no". So, a the logician can just say "no" to a non-self-referential question.
Let's get our bearings. Just like in part (a), we can expect that the question we ask will again have to be self-referential: We saw in part (a) that a question that doesn't reference to its own answer can always be answered "yes" or "no". So, a the logician can just say "no" to a non-self-referential question.
timblack
2024-05-15 19:31:03
Is it possible that one of the questions we considered in part (a) might also work for part (b)?
Is it possible that one of the questions we considered in part (a) might also work for part (b)?
snow52
2024-05-15 19:31:56
no
no
smartowl2020
2024-05-15 19:31:56
No
No
baboo12
2024-05-15 19:31:56
no
no
timblack
2024-05-15 19:31:58
No? Why not?
No? Why not?
eibc
2024-05-15 19:33:09
no, the logician would just answer "yes"
no, the logician would just answer "yes"
Kai321
2024-05-15 19:33:09
No; Those are all "first level" paradoxes. We need to reference the paridox of the question in the question
No; Those are all "first level" paradoxes. We need to reference the paridox of the question in the question
AxoLuna
2024-05-15 19:33:09
we were always able to determine whether they would explode the universe for the number theorist and analyst
we were always able to determine whether they would explode the universe for the number theorist and analyst
admathcamp353
2024-05-15 19:33:09
The questions either cause the universe to explode when asking the number theorist, or not
The questions either cause the universe to explode when asking the number theorist, or not
timblack
2024-05-15 19:33:11
Right. The questions in part (a) came in two flavors: there were the questions that caused the universe to explode when asked to a number theorist, and those that didn't. A logician says "yes" to the first flavor of question, and "no" to the second flavor. But either way, the logician can say something.
Right. The questions in part (a) came in two flavors: there were the questions that caused the universe to explode when asked to a number theorist, and those that didn't. A logician says "yes" to the first flavor of question, and "no" to the second flavor. But either way, the logician can say something.
timblack
2024-05-15 19:33:27
So we need a question where a logician's answer to the question affects whether a number theorist can answer the question. We need a question that refers to a logician's answer.
So we need a question where a logician's answer to the question affects whether a number theorist can answer the question. We need a question that refers to a logician's answer.
timblack
2024-05-15 19:33:41
And we already established that the question needs to refer to a number theorist's answer. So at this point, we could start playing around with questions that reference both a logician's answer and a number theorist's answer.
And we already established that the question needs to refer to a number theorist's answer. So at this point, we could start playing around with questions that reference both a logician's answer and a number theorist's answer.
timblack
2024-05-15 19:33:53
Here's another way to think of it. From part (a), we already know some questions that cause the universe to explode, and some that don't. For instance, asking a number theorist "Is your answer to this question 'no'?" causes the universe to explode, but asking a number theorist "Is your answer to this question 'yes'?" does not.
Here's another way to think of it. From part (a), we already know some questions that cause the universe to explode, and some that don't. For instance, asking a number theorist "Is your answer to this question 'no'?" causes the universe to explode, but asking a number theorist "Is your answer to this question 'yes'?" does not.
timblack
2024-05-15 19:34:07
Can we come up with a question that is equivalent to "Is your answer to this question 'no'?" if the logician says "no" but equivalent to "Is your answer to this question 'yes'?" if the logician says "yes"?
Can we come up with a question that is equivalent to "Is your answer to this question 'no'?" if the logician says "no" but equivalent to "Is your answer to this question 'yes'?" if the logician says "yes"?
timblack
2024-05-15 19:35:07
If we could come up with such a question, then the number theorist does cause the universe to explode if the logician says "no", but the number theorist does not cause the universe to explode if the logician says "yes". Either way, the logician said the wrong thing!
If we could come up with such a question, then the number theorist does cause the universe to explode if the logician says "no", but the number theorist does not cause the universe to explode if the logician says "yes". Either way, the logician said the wrong thing!
timblack
2024-05-15 19:35:16
Any ideas for what our final question could be?
Any ideas for what our final question could be?
NAAUBE
2024-05-15 19:36:23
Can both a theorist and a logician answer this question with the same answer?
Can both a theorist and a logician answer this question with the same answer?
timblack
2024-05-15 19:36:35
Nice! The solution to this problem is to ask the question "Is your answer to this question the same as a logician's answer to this question?"
Nice! The solution to this problem is to ask the question "Is your answer to this question the same as a logician's answer to this question?"
timblack
2024-05-15 19:36:47
(Or another equivalent phrasing)
(Or another equivalent phrasing)
timblack
2024-05-15 19:37:26
The idea is that we're again taking the questions "Is your answer to this question 'no'?" and "Is your answer to this question 'yes'?" and replacing the last word with something
The idea is that we're again taking the questions "Is your answer to this question 'no'?" and "Is your answer to this question 'yes'?" and replacing the last word with something
timblack
2024-05-15 19:37:57
"the logician's answer to this question" fits the bill
"the logician's answer to this question" fits the bill
timblack
2024-05-15 19:38:19
Let's check that this question works.
Let's check that this question works.
timblack
2024-05-15 19:38:30
Here's an example of us being really careful with our casework.
Here's an example of us being really careful with our casework.
timblack
2024-05-15 19:38:38
To verify that this question works, we need to consider two possible universes: The universe where a logician says "yes" and the universe where a logician says "no".
To verify that this question works, we need to consider two possible universes: The universe where a logician says "yes" and the universe where a logician says "no".
timblack
2024-05-15 19:38:53
In the universe where a logician says "yes":
In the universe where a logician says "yes":
timblack
2024-05-15 19:38:55
If a number theorist says "yes", then "your answer to this question" is "yes" and "a logician's answer to this question" is "yes". These are the same, so the number theorist has told the truth by answering "yes".
If a number theorist says "yes", then "your answer to this question" is "yes" and "a logician's answer to this question" is "yes". These are the same, so the number theorist has told the truth by answering "yes".
timblack
2024-05-15 19:39:10
It turns out the number theorist also has the option to say "no". But regardless, the number theorist can give a truthful answer, so the universe does not explode when this question is asked to the number theorist, so it was incorrect for the logician to say "yes".
It turns out the number theorist also has the option to say "no". But regardless, the number theorist can give a truthful answer, so the universe does not explode when this question is asked to the number theorist, so it was incorrect for the logician to say "yes".
timblack
2024-05-15 19:39:29
In the universe where a logician says "no":
In the universe where a logician says "no":
timblack
2024-05-15 19:39:35
If a number theorist says "yes", then "your answer to this question" is "yes" while "a logician's answer to this question" is "no". These are not the same, so the number theorist has lied by answering "yes".
If a number theorist says "yes", then "your answer to this question" is "yes" while "a logician's answer to this question" is "no". These are not the same, so the number theorist has lied by answering "yes".
timblack
2024-05-15 19:39:51
If a number theorist says "no", then "your answer to this question" is "no" and "a logician's answer to this question" is also "no". These are the same, so the number theorist has lied by answering "no".
If a number theorist says "no", then "your answer to this question" is "no" and "a logician's answer to this question" is also "no". These are the same, so the number theorist has lied by answering "no".
timblack
2024-05-15 19:40:07
The number theorist cannot answer "yes" or "no", so this universe explodes. But that means it was incorrect for the logician to say "no".
The number theorist cannot answer "yes" or "no", so this universe explodes. But that means it was incorrect for the logician to say "no".
timblack
2024-05-15 19:40:15
Putting it all together, the logician cannot correctly say "yes" or "no", so the universe explodes when this question is presented to a logician.
Putting it all together, the logician cannot correctly say "yes" or "no", so the universe explodes when this question is presented to a logician.
aryajb2013
2024-05-15 19:40:18
BOOM
BOOM
ek-nalshtheguy
2024-05-15 19:40:23
ahh my brain is exploding
ahh my brain is exploding
timblack
2024-05-15 19:40:27
Thematic!
Thematic!
CheetahLover
2024-05-15 19:40:38
Kaboom
Kaboom
ailiuda30
2024-05-15 19:40:38
kaboom
kaboom
Matthew.C
2024-05-15 19:40:38
boom
boom
Liontiger
2024-05-15 19:40:45
what did the universe ever do
what did the universe ever do
timblack
2024-05-15 19:40:50
AxoLuna
2024-05-15 19:41:02
this is so complicated haha! your answers are way more clear than mine were
this is so complicated haha! your answers are way more clear than mine were
timblack
2024-05-15 19:41:09
It's supposed to be a challenge!
It's supposed to be a challenge!
timblack
2024-05-15 19:41:31
We hope you all had fun thinking about it, despite — or because — it's a challenge
We hope you all had fun thinking about it, despite — or because — it's a challenge
Matthew.C
2024-05-15 19:41:35
now may i have a cookie?
now may i have a cookie?
timblack
2024-05-15 19:41:37
Okay.
Okay.
timblack
2024-05-15 19:41:44
And that's it!
And that's it!
timblack
2024-05-15 19:41:50
Once we came up with the questions, we could try to methodically step through the various cases to check that it worked, but coming up with the questions, and even just thinking about setup of the problem in the right way, can be challenging.
Once we came up with the questions, we could try to methodically step through the various cases to check that it worked, but coming up with the questions, and even just thinking about setup of the problem in the right way, can be challenging.
timblack
2024-05-15 19:42:04
Any last questions or thoughts on this problem? We'll pause here another minute before we move on to the next problem.
Any last questions or thoughts on this problem? We'll pause here another minute before we move on to the next problem.
baboo12
2024-05-15 19:42:06
can i have one too?
can i have one too?
timblack
2024-05-15 19:42:09
Sure, why not
Sure, why not
Mathdreams
2024-05-15 19:42:19
Can I have a virtual cookie
Can I have a virtual cookie
RubyJaguar10
2024-05-15 19:42:19
and me?
and me?
timblack
2024-05-15 19:42:26
Cookies for all (some assembly required)
Cookies for all (some assembly required)
NAAUBE
2024-05-15 19:42:30
Are you answering them as an analyst?
Are you answering them as an analyst?
timblack
2024-05-15 19:42:35
hmmm
hmmm
timblack
2024-05-15 19:42:43
With that, I'm going to hand things off to my friend (and former Mathcamp roommate) Yasha!
With that, I'm going to hand things off to my friend (and former Mathcamp roommate) Yasha!
Yasha
2024-05-15 19:42:49
Hi everyone!
Hi everyone!
Yasha
2024-05-15 19:43:23
I'll present problems 1 and 2. For problem 1, this solution and the pretty pictures were written/made by Molly, our most prolific grader.
I'll present problems 1 and 2. For problem 1, this solution and the pretty pictures were written/made by Molly, our most prolific grader.
Yasha
2024-05-15 19:43:41
Problem 1
Problem 1
Yasha
2024-05-15 19:43:47
The picture below shows a variant of a famous paradoxical puzzle. On the left, we take two rectangles of area 60, and cut each one into two pieces. On the right, we rearrange the four pieces, and put them together into a single rectangle of area 119. How could this be?
The picture below shows a variant of a famous paradoxical puzzle. On the left, we take two rectangles of area 60, and cut each one into two pieces. On the right, we rearrange the four pieces, and put them together into a single rectangle of area 119. How could this be?
Yasha
2024-05-15 19:43:51
Yasha
2024-05-15 19:43:54
a. Explain what's wrong.
a. Explain what's wrong.
Yasha
2024-05-15 19:43:58
Clearly, 60 + 60 = 120 = 119 + 1, so one unit of area is getting lost somewhere. There must be a place on the large rectangle where the four pieces overlap. Where is it?
Clearly, 60 + 60 = 120 = 119 + 1, so one unit of area is getting lost somewhere. There must be a place on the large rectangle where the four pieces overlap. Where is it?
happypi31415
2024-05-15 19:44:36
consider the slopes
consider the slopes
joeym2011
2024-05-15 19:44:36
The slope of the diagonal line on the right is not constant.
The slope of the diagonal line on the right is not constant.
admathcamp353
2024-05-15 19:44:36
The diagonal
The diagonal
hiloo
2024-05-15 19:44:36
along the diagonoal?
along the diagonoal?
phia4
2024-05-15 19:44:36
The diagonal
The diagonal
TallyKitten123
2024-05-15 19:44:36
somewhere on the hypotenuses
somewhere on the hypotenuses
Kai321
2024-05-15 19:44:36
The diagonal, the slopes don't match
The diagonal, the slopes don't match
AxoLuna
2024-05-15 19:44:36
in those suspiciously thick black lines!
in those suspiciously thick black lines!
smartowl2020
2024-05-15 19:44:36
The slope of the purple shape and the pink shape don't line up
The slope of the purple shape and the pink shape don't line up
jeffrey2024
2024-05-15 19:44:36
in the middle
in the middle
MathIsFun286
2024-05-15 19:44:36
In the center
In the center
ek-nalshtheguy
2024-05-15 19:44:36
yeah the traingle slopes arent the same
yeah the traingle slopes arent the same
Yasha
2024-05-15 19:44:44
Let's take a closer look at the diagonal.
Let's take a closer look at the diagonal.
Yasha
2024-05-15 19:44:46
Yasha
2024-05-15 19:44:52
The slope of the pink section is 5/12, and the slope of the purple section is 2/5. They're not the same, so the "diagonal" is not actually a straight line.
The slope of the pink section is 5/12, and the slope of the purple section is 2/5. They're not the same, so the "diagonal" is not actually a straight line.
Yasha
2024-05-15 19:45:00
The shape made up by the pink and purple pieces combined is not a triangle - it is a quadrilateral that bulges slightly outward. Same with the shape made up by the blue and yellow pieces. These bulges overlap each other in a tiny sliver of area 1, accounting for the missing unit.
The shape made up by the pink and purple pieces combined is not a triangle - it is a quadrilateral that bulges slightly outward. Same with the shape made up by the blue and yellow pieces. These bulges overlap each other in a tiny sliver of area 1, accounting for the missing unit.
ek-nalshtheguy
2024-05-15 19:45:16
2.5 and 2.4 so close!!
2.5 and 2.4 so close!!
jeffrey2024
2024-05-15 19:45:16
12/5=2.4,5/2=2.5
12/5=2.4,5/2=2.5
Yasha
2024-05-15 19:45:54
Yup, the slopes being almost but not exactly equal is why it looks deceiving.
Yup, the slopes being almost but not exactly equal is why it looks deceiving.
Yasha
2024-05-15 19:46:15
b. Although two 5 × 12 rectangles cannot really be rearranged into a 7 × 17 rectangle, it is possible to take two a × b rectangles and cut them as shown in the picture above to make a c × d rectangle, with no paradox. What should the lengths a, b, c, d be (up to scaling, of course)?
b. Although two 5 × 12 rectangles cannot really be rearranged into a 7 × 17 rectangle, it is possible to take two a × b rectangles and cut them as shown in the picture above to make a c × d rectangle, with no paradox. What should the lengths a, b, c, d be (up to scaling, of course)?
Yasha
2024-05-15 19:46:29
Without loss of generality, let's say a < b and c < d. Then we can label the rectangles as follows (ignoring the grid, we're just using the original diagram as a visual aid):
Without loss of generality, let's say a < b and c < d. Then we can label the rectangles as follows (ignoring the grid, we're just using the original diagram as a visual aid):
Yasha
2024-05-15 19:46:34
Yasha
2024-05-15 19:46:44
Can you express the value of d in terms of a and b?
Can you express the value of d in terms of a and b?
CheetahLover
2024-05-15 19:46:59
a+b
a+b
NAAUBE
2024-05-15 19:46:59
a+b
a+b
joeym2011
2024-05-15 19:46:59
$a+b$
$a+b$
Kai321
2024-05-15 19:46:59
a+b
a+b
eibc
2024-05-15 19:46:59
$a+b$
$a+b$
MathIsFun286
2024-05-15 19:46:59
a+b
a+b
MichelleWu09
2024-05-15 19:47:04
a+b
a+b
AxoLuna
2024-05-15 19:47:04
should be $a+b$
should be $a+b$
snow52
2024-05-15 19:47:04
a+b
a+b
aliabidi
2024-05-15 19:47:04
a + b
a + b
K124659
2024-05-15 19:47:04
a+b
a+b
Yasha
2024-05-15 19:47:09
Right - the bottom edge of the large rectangle has length d, and it's made up of a purple edge of length a and a pink edge of length b. So d = a+b.
Right - the bottom edge of the large rectangle has length d, and it's made up of a purple edge of length a and a pink edge of length b. So d = a+b.
Yasha
2024-05-15 19:47:14
Can you express the value of c in terms of a and b?
Can you express the value of c in terms of a and b?
MichelleWu09
2024-05-15 19:47:39
b-a
b-a
MathIsFun286
2024-05-15 19:47:39
b-a
b-a
EpicBird08
2024-05-15 19:47:39
b-a
b-a
anAOPS_2022
2024-05-15 19:47:39
b-a
b-a
Purple8cat
2024-05-15 19:47:39
b-a
b-a
Yasha
2024-05-15 19:47:45
We can see from the large rectangle that the longer base of the purple trapezoid has length c. The shorter base matches up with the short leg of the pink triangle, so it has length a. Since the second small rectangle is cut into two congruent trapezoids, the lengths of the two bases must add up to b. b = c+a, so c = b-a.
We can see from the large rectangle that the longer base of the purple trapezoid has length c. The shorter base matches up with the short leg of the pink triangle, so it has length a. Since the second small rectangle is cut into two congruent trapezoids, the lengths of the two bases must add up to b. b = c+a, so c = b-a.
Yasha
2024-05-15 19:47:54
Knowing this, we can label the large rectangle like so:
Knowing this, we can label the large rectangle like so:
Yasha
2024-05-15 19:47:56
Yasha
2024-05-15 19:48:11
To prevent the problem from part (a), we have to pick values of a and b such that the slopes of the purple and pink sections are the same.
To prevent the problem from part (a), we have to pick values of a and b such that the slopes of the purple and pink sections are the same.
Yasha
2024-05-15 19:48:15
What equation does this give us?
What equation does this give us?
EpicBird08
2024-05-15 19:49:17
so b/a = b+a/b-a
so b/a = b+a/b-a
joeym2011
2024-05-15 19:49:17
$\frac{b-2a}a=\frac ab$
$\frac{b-2a}a=\frac ab$
eibc
2024-05-15 19:49:17
$\frac{b-a}{a+b}=\frac{a}{b}$
$\frac{b-a}{a+b}=\frac{a}{b}$
smartowl2020
2024-05-15 19:49:17
a/b = (b-a)/(a+b)
a/b = (b-a)/(a+b)
snow52
2024-05-15 19:49:17
(b-2a)/a=a/b
(b-2a)/a=a/b
MichelleWu09
2024-05-15 19:49:17
a/b = (b-a)/(a+b)
a/b = (b-a)/(a+b)
baboo12
2024-05-15 19:49:17
(b-2a)/(a) = a/b
(b-2a)/(a) = a/b
AxoLuna
2024-05-15 19:49:17
$\frac{b-a}{a+b}= \frac{a}{b}$
$\frac{b-a}{a+b}= \frac{a}{b}$
CrunchyCucumber
2024-05-15 19:49:17
$\dfrac{b-a}{a+b}=\dfrac{a}{b}$
$\dfrac{b-a}{a+b}=\dfrac{a}{b}$
ChaitraliKA
2024-05-15 19:49:17
(b-a)/(a+b)=a/b
(b-a)/(a+b)=a/b
NAAUBE
2024-05-15 19:49:17
$a/b$ = $(b-a)/(a+b)$
$a/b$ = $(b-a)/(a+b)$
Yasha
2024-05-15 19:49:22
We get $\frac ab = \frac{b-2a}a$.
We get $\frac ab = \frac{b-2a}a$.
Yasha
2024-05-15 19:49:25
Alternatively and equivalently, using similar triangles, we can get the equation $\frac ab=\frac{b-a}{a+b}$.
Alternatively and equivalently, using similar triangles, we can get the equation $\frac ab=\frac{b-a}{a+b}$.
EpicBird08
2024-05-15 19:49:57
this gives b^2 - ab = ab + a^2 -> a^2 + 2ab - b^2 = 0 which is a quadratic equation!!!!!
this gives b^2 - ab = ab + a^2 -> a^2 + 2ab - b^2 = 0 which is a quadratic equation!!!!!
MathIsFun286
2024-05-15 19:49:57
$b^2-a^2=2ab$
$b^2-a^2=2ab$
NAAUBE
2024-05-15 19:49:57
$b^2 - 2ba - a^2 = 0$
$b^2 - 2ba - a^2 = 0$
cwc28
2024-05-15 19:49:57
(a+b)(b-a)=2ab
(a+b)(b-a)=2ab
admathcamp353
2024-05-15 19:49:57
(a+b)(a-b) = 2ab?
(a+b)(a-b) = 2ab?
NAAUBE
2024-05-15 19:50:12
Make a quadratic and solve!
Make a quadratic and solve!
Yasha
2024-05-15 19:50:34
Can we solve for $b$ in terms of $a$?
Can we solve for $b$ in terms of $a$?
joeym2011
2024-05-15 19:51:42
$a^2+2ab-b^2=0$ so $\frac ba=1+\sqrt2$
$a^2+2ab-b^2=0$ so $\frac ba=1+\sqrt2$
NAAUBE
2024-05-15 19:51:42
yielding $\frac{b-a}{a}=\sqrt{2}$
yielding $\frac{b-a}{a}=\sqrt{2}$
MathIsFun286
2024-05-15 19:51:42
$b=a(1+\sqrt{2})$
$b=a(1+\sqrt{2})$
EpicBird08
2024-05-15 19:51:42
b = (sqrt2 + 1) x a
b = (sqrt2 + 1) x a
EpicBird08
2024-05-15 19:51:42
b = (sqrt2 + 1)a
b = (sqrt2 + 1)a
Kai321
2024-05-15 19:51:42
a/(sqrt(2) - 1) = b
a/(sqrt(2) - 1) = b
MichelleWu09
2024-05-15 19:51:42
b=(√2+1)a
b=(√2+1)a
AxoLuna
2024-05-15 19:51:42
yes! $b = a(1+\sqrt2)$
yes! $b = a(1+\sqrt2)$
Yasha
2024-05-15 19:52:16
We get $b = (1+\sqrt2)a$. (One way is to clear denominators and then use the quadratic formula.)
We get $b = (1+\sqrt2)a$. (One way is to clear denominators and then use the quadratic formula.)
joeym2011
2024-05-15 19:52:30
$(a,b,c,d)=(x,x+\sqrt2x,\sqrt2x,2x+\sqrt2x)$
$(a,b,c,d)=(x,x+\sqrt2x,\sqrt2x,2x+\sqrt2x)$
Yasha
2024-05-15 19:53:15
From there, we get $c = b-a = \sqrt2 a$, and $d = a+b = (2+\sqrt2)a.$ For any positive value of a, these numbers will work, since the diagram can scale without a problem.
From there, we get $c = b-a = \sqrt2 a$, and $d = a+b = (2+\sqrt2)a.$ For any positive value of a, these numbers will work, since the diagram can scale without a problem.
Yasha
2024-05-15 19:53:46
Note that when a = 5, b ≈ 12.07, c ≈ 7.07, and d ≈ 17.07. These are very close to the numbers in the paradoxical puzzle - thus why the illusion works so well.
Note that when a = 5, b ≈ 12.07, c ≈ 7.07, and d ≈ 17.07. These are very close to the numbers in the paradoxical puzzle - thus why the illusion works so well.
EpicBird08
2024-05-15 19:54:27
and we can actually generalize!
and we can actually generalize!
EpicBird08
2024-05-15 19:54:27
29x70 and 41x99
29x70 and 41x99
Yasha
2024-05-15 19:54:44
Yes, now you can make lots of illusory triangles!
Yes, now you can make lots of illusory triangles!
Yasha
2024-05-15 19:55:13
c. It is also possible to take three congruent rectangles, cut each one into two pieces, and rearrange them to form a single rectangle similar to the original three. How can we do this?
Try to find an answer that lets you create a paradoxical decomposition of your own!
c. It is also possible to take three congruent rectangles, cut each one into two pieces, and rearrange them to form a single rectangle similar to the original three. How can we do this?
Try to find an answer that lets you create a paradoxical decomposition of your own!
Yasha
2024-05-15 19:55:30
Collectively, MathCamp's applicants this year came up with 7+ different correct decompositions for this problem! We don't have time to show all of them, but we encourage you to look for other valid solutions on your own time.
Collectively, MathCamp's applicants this year came up with 7+ different correct decompositions for this problem! We don't have time to show all of them, but we encourage you to look for other valid solutions on your own time.
Yasha
2024-05-15 19:55:46
The simple answer is to stack three $1 \times \sqrt3$ rectangles to create a $3 \times \sqrt3$ rectangle... but that's boring and can't be used to create a paradoxical decomposition. So let's try something else.
The simple answer is to stack three $1 \times \sqrt3$ rectangles to create a $3 \times \sqrt3$ rectangle... but that's boring and can't be used to create a paradoxical decomposition. So let's try something else.
Yasha
2024-05-15 19:55:54
By "extending" the decomposition from parts (a) and (b) and squeezing in a second pair of trapezoids, we can cut and rearrange three a x b rectangles like this:
By "extending" the decomposition from parts (a) and (b) and squeezing in a second pair of trapezoids, we can cut and rearrange three a x b rectangles like this:
Yasha
2024-05-15 19:55:56
Yasha
2024-05-15 19:56:17
Since the large rectangle is rotationally symmetric, it's clear that the second and third small triangles must be cut in the same way. Let's label the bases of the trapezoids like this:
Since the large rectangle is rotationally symmetric, it's clear that the second and third small triangles must be cut in the same way. Let's label the bases of the trapezoids like this:
Yasha
2024-05-15 19:56:22
Yasha
2024-05-15 19:56:31
What can you say about the values of x and y, in terms of a and b?
What can you say about the values of x and y, in terms of a and b?
Pommukutti
2024-05-15 19:57:17
What is a and b?
What is a and b?
Yasha
2024-05-15 19:57:36
a and b are the side lenghts of the small rectangles, with a being the smaller one
a and b are the side lenghts of the small rectangles, with a being the smaller one
MathIsFun286
2024-05-15 19:58:18
x+y=b
x+y=b
Kai321
2024-05-15 19:58:20
x = y = 1/2b
x = y = 1/2b
CrunchyCucumber
2024-05-15 19:58:21
$x=y=\dfrac{b}{2}$
$x=y=\dfrac{b}{2}$
aliabidi
2024-05-15 19:58:25
x + y = b
x + y = b
Yasha
2024-05-15 19:58:33
In the large rectangle, x and y match up to each other, so they must be equal. Also, they add up to b.
In the large rectangle, x and y match up to each other, so they must be equal. Also, they add up to b.
Yasha
2024-05-15 19:58:42
Thus, x = y = b/2.
Thus, x = y = b/2.
Yasha
2024-05-15 19:58:51
What can you say about the values of z and w, in terms of a and b?
What can you say about the values of z and w, in terms of a and b?
RubyJaguar10
2024-05-15 19:59:39
b=w+z?
b=w+z?
EpicBird08
2024-05-15 19:59:39
$z = a, w = b - a, x = y = b/2$
$z = a, w = b - a, x = y = b/2$
MathIsFun286
2024-05-15 19:59:39
w+z=b
w+z=b
cwc28
2024-05-15 19:59:39
z=a
z=a
Pommukutti
2024-05-15 19:59:39
z + w = b
z + w = b
baboo12
2024-05-15 19:59:42
z+w=b
z+w=b
TallyKitten123
2024-05-15 19:59:42
z+w=b
z+w=b
Yasha
2024-05-15 19:59:51
In the large rectangle, z matches up with the short leg of the pink triangle, so z = a. Also, z and w add up to b, so w = b-a.
In the large rectangle, z matches up with the short leg of the pink triangle, so z = a. Also, z and w add up to b, so w = b-a.
Yasha
2024-05-15 19:59:57
So we can label the large rectangle like this:
So we can label the large rectangle like this:
Yasha
2024-05-15 19:59:58
EpicBird08
2024-05-15 20:00:51
so by similar triangles a/b = (b-a)/(b+2a), another quadratic equation!
so by similar triangles a/b = (b-a)/(b+2a), another quadratic equation!
EpicBird08
2024-05-15 20:01:11
solving this one gives $b = (1+\sqrt{3})a$
solving this one gives $b = (1+\sqrt{3})a$
Yasha
2024-05-15 20:01:33
Yup, in the interests of time I'm going to skip a lot of the steps here, but it's pretty similar to what we did in part (b).
Yup, in the interests of time I'm going to skip a lot of the steps here, but it's pretty similar to what we did in part (b).
Yasha
2024-05-15 20:01:50
We equate the slopes and do the algebra. In the end, we get $b = (1+\sqrt3)a$.
We equate the slopes and do the algebra. In the end, we get $b = (1+\sqrt3)a$.
Yasha
2024-05-15 20:01:54
To get a paradoxical decomposition, we can use integers close to this $1:1+\sqrt3$ ratio. For example, take three 3x8 rectangles, cut one along the diagonal, and cut the other two so that z=3 and w=5. By rearranging them "as in the diagram above", we appear to make a 5x14 rectangle of area 70 out of three rectangles of area 24, and 2 units of area mysteriously go missing.
To get a paradoxical decomposition, we can use integers close to this $1:1+\sqrt3$ ratio. For example, take three 3x8 rectangles, cut one along the diagonal, and cut the other two so that z=3 and w=5. By rearranging them "as in the diagram above", we appear to make a 5x14 rectangle of area 70 out of three rectangles of area 24, and 2 units of area mysteriously go missing.
EpicBird08
2024-05-15 20:02:58
and this motivated me to use a $1 \times (1 + \sqrt{n})$ rectangle to make something similar for $n$ rectangles
and this motivated me to use a $1 \times (1 + \sqrt{n})$ rectangle to make something similar for $n$ rectangles
aliabidi
2024-05-15 20:02:58
Is it also possible to take n congruent rectangles, cut each one into two pieces and rearrange to form a single similar rectangle, with b = (1+sqrt(n))a?
Is it also possible to take n congruent rectangles, cut each one into two pieces and rearrange to form a single similar rectangle, with b = (1+sqrt(n))a?
Yasha
2024-05-15 20:03:01
I would guess that it generalizes that way, but I'd have to check.
I would guess that it generalizes that way, but I'd have to check.
Yasha
2024-05-15 20:03:23
Fun fact: you can find integers close to a $1:1+\sqrt3$ ratio by using the continued fraction expression for $1+\sqrt3$.
Fun fact: you can find integers close to a $1:1+\sqrt3$ ratio by using the continued fraction expression for $1+\sqrt3$.
EpicBird08
2024-05-15 20:03:33
it does, i worked out the details
it does, i worked out the details
Yasha
2024-05-15 20:03:36
Awesome!
Awesome!
Yasha
2024-05-15 20:04:29
OK, onwards to problem 2. I actually wrote the initial version of this problem.
OK, onwards to problem 2. I actually wrote the initial version of this problem.
Yasha
2024-05-15 20:04:33
Problem 2
Problem 2
Yasha
2024-05-15 20:04:35
Kayla has two red boxes, two green boxes, and two blue boxes. Each of the six boxes contains a secret number. Kayla hands the boxes to Leo and asks him to write the letters A, a, B, b, C, and c, on the boxes. We will write #A, #a, #B, #b, #C, and #c to refer to the secret numbers inside these boxes.
Kayla has two red boxes, two green boxes, and two blue boxes. Each of the six boxes contains a secret number. Kayla hands the boxes to Leo and asks him to write the letters A, a, B, b, C, and c, on the boxes. We will write #A, #a, #B, #b, #C, and #c to refer to the secret numbers inside these boxes.
Yasha
2024-05-15 20:04:47
From there, the boxes go to Maya, who opens the boxes and reports the differences #A−#a, #B−#b, and #C−#c, (in that order). Next, the boxes go to Nathan, who opens the boxes and reports the sum of the numbers in the red boxes, the sum of the numbers in the green boxes, and the sum of the numbers in the blue boxes (in that order). The resealed boxes, along with Maya's and Nathan's reports, are then handed back to Kayla, who must determine the numbers in each of the boxes.
From there, the boxes go to Maya, who opens the boxes and reports the differences #A−#a, #B−#b, and #C−#c, (in that order). Next, the boxes go to Nathan, who opens the boxes and reports the sum of the numbers in the red boxes, the sum of the numbers in the green boxes, and the sum of the numbers in the blue boxes (in that order). The resealed boxes, along with Maya's and Nathan's reports, are then handed back to Kayla, who must determine the numbers in each of the boxes.
Yasha
2024-05-15 20:04:54
Kayla expected Leo to label same-color boxes with the same letter, which would have made it easy for Kayla to figure the numbers: for example, knowing #A−#a from Maya and #A+#a from Nathan, Kayla could solve for #A and #a. However, to Kayla's surprise, Leo is color blind, so his labeling had nothing to do with the colors.
Kayla expected Leo to label same-color boxes with the same letter, which would have made it easy for Kayla to figure the numbers: for example, knowing #A−#a from Maya and #A+#a from Nathan, Kayla could solve for #A and #a. However, to Kayla's surprise, Leo is color blind, so his labeling had nothing to do with the colors.
Yasha
2024-05-15 20:05:08
a. For which of the labelings that Leo used is it still possible for Kayla to determine the six secret numbers? (Kayla can see which colors have which labels on them.)
a. For which of the labelings that Leo used is it still possible for Kayla to determine the six secret numbers? (Kayla can see which colors have which labels on them.)
Yasha
2024-05-15 20:06:15
There's a lot to unpack here. It can help to make a diagram.
There's a lot to unpack here. It can help to make a diagram.
Yasha
2024-05-15 20:06:18
Yasha
2024-05-15 20:06:27
I've linked the red boxes with a red line segment, so, in this example, $A$ and $b$ are red.
I've linked the red boxes with a red line segment, so, in this example, $A$ and $b$ are red.
Yasha
2024-05-15 20:06:33
So, in this example, Kayla knows:
* $A-a$, $B-b$, and $C-c$ from Maya's report, and
* $A+b$, $B+C$, and $c+a$ from Nathan's report.
(Note that I'm just writing $A$ instead of #A, and so forth.)
So, in this example, Kayla knows:
* $A-a$, $B-b$, and $C-c$ from Maya's report, and
* $A+b$, $B+C$, and $c+a$ from Nathan's report.
(Note that I'm just writing $A$ instead of #A, and so forth.)
Yasha
2024-05-15 20:06:58
Kayla's task is to figure out all of the numbers from this information.
Kayla's task is to figure out all of the numbers from this information.
Yasha
2024-05-15 20:07:06
For a moment, let's just focus on this part of the diagram:
For a moment, let's just focus on this part of the diagram:
Yasha
2024-05-15 20:07:10
Yasha
2024-05-15 20:07:20
Given that we know $A-a$, $B-b$, and $A+b$, what else can we figure out? For example, can we figure out $A+B$? If so, how?
Given that we know $A-a$, $B-b$, and $A+b$, what else can we figure out? For example, can we figure out $A+B$? If so, how?
joeym2011
2024-05-15 20:07:48
Add the last two.
Add the last two.
aliabidi
2024-05-15 20:07:54
Yes, just add B - b and A + b.
Yes, just add B - b and A + b.
Yasha
2024-05-15 20:07:56
Yes, we can figure $A+B$ because $A+B=(A+b)+(B-b)$.
Yes, we can figure $A+B$ because $A+B=(A+b)+(B-b)$.
baboo12
2024-05-15 20:08:36
how did you determine that the red boxes were A and b ???
how did you determine that the red boxes were A and b ???
Yasha
2024-05-15 20:08:38
We're doing an example to get the hang of the problem.
We're doing an example to get the hang of the problem.
Yasha
2024-05-15 20:08:56
By the end of it, we're going to have to be able to deal with all possible diagrams.
By the end of it, we're going to have to be able to deal with all possible diagrams.
Yasha
2024-05-15 20:09:26
What about $a+b$? How can we figure that out from $A-a$, $B-b$, and $A+b$?
What about $a+b$? How can we figure that out from $A-a$, $B-b$, and $A+b$?
tiangang
2024-05-15 20:09:51
subtract first and last
subtract first and last
aliabidi
2024-05-15 20:09:51
Subtract A - a from A + b
Subtract A - a from A + b
joeym2011
2024-05-15 20:09:51
We have $a+b=(A+b)-(A-a)$.
We have $a+b=(A+b)-(A-a)$.
Yasha
2024-05-15 20:09:59
We can figure $a+b$ because $a+b=(A+b)-(A-a)$.
We can figure $a+b$ because $a+b=(A+b)-(A-a)$.
Yasha
2024-05-15 20:10:11
What about $a+B$?
What about $a+B$?
aliabidi
2024-05-15 20:11:02
Add a + b and B - b
Add a + b and B - b
tiangang
2024-05-15 20:11:02
add second and third and subtract first
add second and third and subtract first
smartowl2020
2024-05-15 20:11:02
second - first + third
second - first + third
joeym2011
2024-05-15 20:11:02
$A+b+B-b-(A-a)$
$A+b+B-b-(A-a)$
MichelleWu09
2024-05-15 20:11:02
(A+b)+(B-b)-(A-a)
(A+b)+(B-b)-(A-a)
Yasha
2024-05-15 20:11:08
Yes, using $a+B=(A+b)-(A-a)+(B-b)$.
Yes, using $a+B=(A+b)-(A-a)+(B-b)$.
Yasha
2024-05-15 20:11:21
(We can also get it from the stuff we've figured out already.)
(We can also get it from the stuff we've figured out already.)
Yasha
2024-05-15 20:11:29
Likewise, we can reverse this process: Assuming we know $A-a$ and $B-b$, then, given $A+B$, $a+b$, or $a+B$, we can figure out $A+b$.
Likewise, we can reverse this process: Assuming we know $A-a$ and $B-b$, then, given $A+B$, $a+b$, or $a+B$, we can figure out $A+b$.
Yasha
2024-05-15 20:11:40
We can conclude that it doesn't matter which of the $Aa$ boxes are linked to which of the $Bb$ boxes: The following situations all provide the same information as the original:
We can conclude that it doesn't matter which of the $Aa$ boxes are linked to which of the $Bb$ boxes: The following situations all provide the same information as the original:
Yasha
2024-05-15 20:11:45
Yasha
2024-05-15 20:11:56
So, we can simplify our diagram by drawing it like this:
So, we can simplify our diagram by drawing it like this:
Yasha
2024-05-15 20:11:58
Yasha
2024-05-15 20:13:14
With this simplification, the entire diagram looks like this:
With this simplification, the entire diagram looks like this:
Yasha
2024-05-15 20:13:15
Yasha
2024-05-15 20:14:02
So, this actually helps us a lot with regards to the earlier question about why I labeled specific boxes red. With the insight we just had, we can cover a lot of different cases with just one diagram.
So, this actually helps us a lot with regards to the earlier question about why I labeled specific boxes red. With the insight we just had, we can cover a lot of different cases with just one diagram.
Yasha
2024-05-15 20:14:51
As long as one of the $Aa$ boxes has the same color as one of the $Bb$ boxes, and one of the $Bb$ boxes has the same color as one of the $Cc$ boxes, and one of the $Cc$ boxes has the same color as one of the $Aa$ boxes, this diagram, and hence our reasoning for it, will apply.
As long as one of the $Aa$ boxes has the same color as one of the $Bb$ boxes, and one of the $Bb$ boxes has the same color as one of the $Cc$ boxes, and one of the $Cc$ boxes has the same color as one of the $Aa$ boxes, this diagram, and hence our reasoning for it, will apply.
Yasha
2024-05-15 20:15:09
But it doesn't matter exactly which boxes are the same color as which other boxes.
But it doesn't matter exactly which boxes are the same color as which other boxes.
Yasha
2024-05-15 20:15:18
Even though this diagram no longer captures exactly which boxes are which color, it still completely captures what information is available to Kayla:
* The differences $A-a$, $B-b$, and $C-c$.
* The sums of any two boxes linked by a colored line.
Even though this diagram no longer captures exactly which boxes are which color, it still completely captures what information is available to Kayla:
* The differences $A-a$, $B-b$, and $C-c$.
* The sums of any two boxes linked by a colored line.
Yasha
2024-05-15 20:15:41
Is it enough information for Kayla to figure out what all the numbers are? Let's, for a moment, just focus on the capital letters:
Is it enough information for Kayla to figure out what all the numbers are? Let's, for a moment, just focus on the capital letters:
Yasha
2024-05-15 20:15:46
Yasha
2024-05-15 20:15:50
Kayla knows $A+B$, $B+C$, and $C+A$. Can she solve for $A$? If so, how?
Kayla knows $A+B$, $B+C$, and $C+A$. Can she solve for $A$? If so, how?
snow52
2024-05-15 20:16:56
add all 3 divide by 2 and subtract the second number
add all 3 divide by 2 and subtract the second number
MathIsFun286
2024-05-15 20:16:56
Add the three numbers, divide the sum by 2, then subtract (B+C)
Add the three numbers, divide the sum by 2, then subtract (B+C)
MichelleWu09
2024-05-15 20:16:56
half of the entire sum and minus B+C
half of the entire sum and minus B+C
NAAUBE
2024-05-15 20:16:56
Subtract the first two, then add that to the 3rd
Subtract the first two, then add that to the 3rd
snow52
2024-05-15 20:16:56
2A+2B+2C from adding them all then divide by 2 and subtract B+C
2A+2B+2C from adding them all then divide by 2 and subtract B+C
ProblemSolving704
2024-05-15 20:16:56
((A+B)-(B+C)+(C+A))/2
((A+B)-(B+C)+(C+A))/2
kikimaths
2024-05-15 20:16:56
(A+B) + (C+A) - (B+C) = 2A
(A+B) + (C+A) - (B+C) = 2A
Oshawoot
2024-05-15 20:16:56
subtract B+C from A+B, add C+A, to get 2A=?, divide by 2
subtract B+C from A+B, add C+A, to get 2A=?, divide by 2
Yasha
2024-05-15 20:16:59
Yes, if we add $A+B$ and $C+A$, we get $2A+B+C$. Subtracting $B+C$, we get $2A$. Dividing by $2$, we get $A$. In summary:
$A=\frac12((A+B)+(C+A)-(B+C)).$
Yes, if we add $A+B$ and $C+A$, we get $2A+B+C$. Subtracting $B+C$, we get $2A$. Dividing by $2$, we get $A$. In summary:
$A=\frac12((A+B)+(C+A)-(B+C)).$
Yasha
2024-05-15 20:17:36
Adding everything to get $2(A+B+C)$ is also a great way to do it.
Adding everything to get $2(A+B+C)$ is also a great way to do it.
Yasha
2024-05-15 20:17:47
Then you divide by $2$ to get $A+B+C$, and then you subtract $B+C$ to get $A$.
Then you divide by $2$ to get $A+B+C$, and then you subtract $B+C$ to get $A$.
Yasha
2024-05-15 20:18:17
She can similarly find $B$ and $C$. And, knowing $A-a$, and $A$, she can of course find $a$, and similarly $b$ and $c$.
She can similarly find $B$ and $C$. And, knowing $A-a$, and $A$, she can of course find $a$, and similarly $b$ and $c$.
NAAUBE
2024-05-15 20:18:38
Can we solve this with matrixes?
Can we solve this with matrixes?
Yasha
2024-05-15 20:18:40
Yes, at its core, this is just a linear algebra problem.
Yes, at its core, this is just a linear algebra problem.
Yasha
2024-05-15 20:19:11
Of course, we're not done yet: There are other possible diagrams. For example, here's what Kayla expected to happen:
Of course, we're not done yet: There are other possible diagrams. For example, here's what Kayla expected to happen:
Yasha
2024-05-15 20:19:12
Yasha
2024-05-15 20:19:17
This diagram represents both $A$ and $a$ being red, both $B$ and $b$ being green, and both $C$ and $c$ being blue.
This diagram represents both $A$ and $a$ being red, both $B$ and $b$ being green, and both $C$ and $c$ being blue.
Yasha
2024-05-15 20:19:21
As the problem states, in this case, it's easy for Kayla to figure out $A$, since she knows both $A+a$ and $A-a$, so she can solve
$A=\frac12((A+a)+(A-a)).$
She can figure out the other variables similarly.
As the problem states, in this case, it's easy for Kayla to figure out $A$, since she knows both $A+a$ and $A-a$, so she can solve
$A=\frac12((A+a)+(A-a)).$
She can figure out the other variables similarly.
Yasha
2024-05-15 20:19:28
Are there any other kinds of diagrams that are possible?
Are there any other kinds of diagrams that are possible?
joeym2011
2024-05-15 20:19:56
Could we have a situation like $AB,ab,Cc$?
Could we have a situation like $AB,ab,Cc$?
snow52
2024-05-15 20:19:56
2 go to each other and the third is a loop
2 go to each other and the third is a loop
MichelleWu09
2024-05-15 20:20:08
(A, a) and (B, c), (C, b)
(A, a) and (B, c), (C, b)
Kai321
2024-05-15 20:20:10
1 more: Ab, Ba, Cc
1 more: Ab, Ba, Cc
Yasha
2024-05-15 20:20:12
In the first case we considered, $Aa$, $Bb$, and $Cc$ were all linked together. In the second case, they were all separate. But we can also have two linked together and one separate:
In the first case we considered, $Aa$, $Bb$, and $Cc$ were all linked together. In the second case, they were all separate. But we can also have two linked together and one separate:
Yasha
2024-05-15 20:20:17
Yasha
2024-05-15 20:20:22
Here, $C$ and $c$ are both blue, whereas one of $Aa$ is red and the other is green, and likewise for $Bb$.
Here, $C$ and $c$ are both blue, whereas one of $Aa$ is red and the other is green, and likewise for $Bb$.
Yasha
2024-05-15 20:20:30
Of course, with this diagram, Kayla can immediately figure out $C$ and $c$. So we really only need to focus on this part:
Of course, with this diagram, Kayla can immediately figure out $C$ and $c$. So we really only need to focus on this part:
Yasha
2024-05-15 20:20:36
Yasha
2024-05-15 20:20:45
Can Kayla figure out what $A$ is?
Can Kayla figure out what $A$ is?
Yasha
2024-05-15 20:20:55
As before, we can focus just on the capital letters:
As before, we can focus just on the capital letters:
Yasha
2024-05-15 20:20:56
Yasha
2024-05-15 20:21:03
Given that Kayla knows $A+B$ and $B+A$, can she figure out $A$?
Given that Kayla knows $A+B$ and $B+A$, can she figure out $A$?
MathIsFun286
2024-05-15 20:21:42
no
no
joeym2011
2024-05-15 20:21:42
NO
NO
NAAUBE
2024-05-15 20:21:42
no
no
MichelleWu09
2024-05-15 20:21:42
no, A+B is B+A
no, A+B is B+A
admathcamp353
2024-05-15 20:21:42
No
No
Kai321
2024-05-15 20:21:42
no
no
smartowl2020
2024-05-15 20:21:42
No
No
sr2022
2024-05-15 20:21:42
no
no
Yasha
2024-05-15 20:21:46
No. For example, we could have $A=1$ and $B=-1$, or we could have $A=0$ and $B=0$. Kayla can't tell the difference.
No. For example, we could have $A=1$ and $B=-1$, or we could have $A=0$ and $B=0$. Kayla can't tell the difference.
Yasha
2024-05-15 20:21:50
The problem is that $B+A=A+B$, so the information she got is redundant.
The problem is that $B+A=A+B$, so the information she got is redundant.
Yasha
2024-05-15 20:21:54
We can build out a full example like so:
We can build out a full example like so:
Yasha
2024-05-15 20:21:55
Yasha
2024-05-15 20:22:04
In this case, all of the numbers Kayla receives from Maya and Nathan are zeros, so Kayla can't tell the difference between the above situation and the one where all the numbers are zeros:
In this case, all of the numbers Kayla receives from Maya and Nathan are zeros, so Kayla can't tell the difference between the above situation and the one where all the numbers are zeros:
Yasha
2024-05-15 20:22:08
Yasha
2024-05-15 20:22:14
Kayla can't figure out what the numbers are in this case.
Kayla can't figure out what the numbers are in this case.
Yasha
2024-05-15 20:22:42
Are any other diagrams possible?
Are any other diagrams possible?
MathIsFun286
2024-05-15 20:23:19
nope
nope
joeym2011
2024-05-15 20:23:19
No, except for $A,B,C$ permutated.
No, except for $A,B,C$ permutated.
A_P_11
2024-05-15 20:23:19
AC, Bb? ac
AC, Bb? ac
Yasha
2024-05-15 20:23:23
Well, we could have a diagram like the previous one except where $Aa$ is by itself and $Bb$ and $Cc$ are linked, or where $Bb$ is by itself and $Aa$ and $Cc$ are linked. But these are essentially the same situation.
Well, we could have a diagram like the previous one except where $Aa$ is by itself and $Bb$ and $Cc$ are linked, or where $Bb$ is by itself and $Aa$ and $Cc$ are linked. But these are essentially the same situation.
Yasha
2024-05-15 20:23:31
Other than that, there are no other possibilities. The colors link the letters into groups. They can either be
* one big group with all three letters,
* three small groups with just one letter, or
* two groups, one with two letters and one with just one letter.
Other than that, there are no other possibilities. The colors link the letters into groups. They can either be
* one big group with all three letters,
* three small groups with just one letter, or
* two groups, one with two letters and one with just one letter.
Yasha
2024-05-15 20:23:44
In the first two cases, Kayla can solve for all of the numbers. In the third case, she cannot. That's it for part a.
In the first two cases, Kayla can solve for all of the numbers. In the third case, she cannot. That's it for part a.
Yasha
2024-05-15 20:24:19
b. Generalize your answer to a problem with 2n boxes of n different colors, labeled with n different uppercase and lowercase letters.
b. Generalize your answer to a problem with 2n boxes of n different colors, labeled with n different uppercase and lowercase letters.
Yasha
2024-05-15 20:24:28
Any guesses on how we might generalize to more letters? Under what circumstances can Kayla figure out all the numbers? Under what circumstances is figuring out all the numbers impossible?
Any guesses on how we might generalize to more letters? Under what circumstances can Kayla figure out all the numbers? Under what circumstances is figuring out all the numbers impossible?
Yasha
2024-05-15 20:24:53
As before, the colors link the letters into groups. The groups are independent of each other: solving one group doesn't help us with the others.
As before, the colors link the letters into groups. The groups are independent of each other: solving one group doesn't help us with the others.
Yasha
2024-05-15 20:25:06
So, if we can solve all of the groups, Kayla succeeds. If there's one group that cannot be solved, Kayla fails.
So, if we can solve all of the groups, Kayla succeeds. If there's one group that cannot be solved, Kayla fails.
Yasha
2024-05-15 20:25:09
So, we must determine which linked groups can be solved, and which cannot.
So, we must determine which linked groups can be solved, and which cannot.
Yasha
2024-05-15 20:25:21
In the previous part, we saw that a group of size three can be solved, a group of size one can be solved, but a group of size two cannot be solved.
In the previous part, we saw that a group of size three can be solved, a group of size one can be solved, but a group of size two cannot be solved.
admathcamp353
2024-05-15 20:26:12
My idea was when numbers are switched between odd numbers of pairs, she can solve it, and with even numbers of pairs she cannot
My idea was when numbers are switched between odd numbers of pairs, she can solve it, and with even numbers of pairs she cannot
joeym2011
2024-05-15 20:26:12
groups of odd size can be solved and groups of even size cannot
groups of odd size can be solved and groups of even size cannot
NAAUBE
2024-05-15 20:26:12
I would guess odd groups can be solved, and even cannot
I would guess odd groups can be solved, and even cannot
MichelleWu09
2024-05-15 20:26:12
it can't be even?
it can't be even?
Yasha
2024-05-15 20:26:28
One guess could be that odd-size groups can be solved, but even-size groups cannot. This guess turns out to be correct.
One guess could be that odd-size groups can be solved, but even-size groups cannot. This guess turns out to be correct.
Yasha
2024-05-15 20:26:55
But some of you guessed that groups of size 2 can't be solved, and others can. That's also a reasonable guess, but turns out to be not correct.
But some of you guessed that groups of size 2 can't be solved, and others can. That's also a reasonable guess, but turns out to be not correct.
Yasha
2024-05-15 20:27:18
That's why we need to prove things! There are lots of reasonable guesses, but only some of them are right.
That's why we need to prove things! There are lots of reasonable guesses, but only some of them are right.
Yasha
2024-05-15 20:27:35
So, let's make a diagram. Each group looks like this:
So, let's make a diagram. Each group looks like this:
Yasha
2024-05-15 20:27:37
Yasha
2024-05-15 20:27:57
As in the previous part, we can figure things out just by looking at the capital letters.
As in the previous part, we can figure things out just by looking at the capital letters.
Yasha
2024-05-15 20:27:58
Yasha
2024-05-15 20:28:04
We know $A_1+A_2, A_2+A_3,\dotsc,A_n+A_1$.
We know $A_1+A_2, A_2+A_3,\dotsc,A_n+A_1$.
Yasha
2024-05-15 20:28:16
Assuming $n$ is odd, how can we figure out $A_1$?
Assuming $n$ is odd, how can we figure out $A_1$?
MathIsFun286
2024-05-15 20:28:55
So the adding all numbers strategy only works when n is odd
So the adding all numbers strategy only works when n is odd
joeym2011
2024-05-15 20:28:55
Take half the sum and subtract $A_2+A_3+\dots+A_n$.
Take half the sum and subtract $A_2+A_3+\dots+A_n$.
Yasha
2024-05-15 20:29:20
Sure, we add everything together, which gets us $2(A_1+A_2+\dotsb+A_n)$, since every number appears twice.
Sure, we add everything together, which gets us $2(A_1+A_2+\dotsb+A_n)$, since every number appears twice.
Yasha
2024-05-15 20:29:29
Then divide by two, getting $A_1+\dotsb+A_n$.
Then divide by two, getting $A_1+\dotsb+A_n$.
Yasha
2024-05-15 20:30:06
And then we subtract $(A_2+A_3)+(A_4+A_5)+\dotsb+(A_{n-1}+A_n)$, leaving us with $A_1$.
And then we subtract $(A_2+A_3)+(A_4+A_5)+\dotsb+(A_{n-1}+A_n)$, leaving us with $A_1$.
Yasha
2024-05-15 20:30:26
Note that for this to work the list $A_2,A_3,\dotsc,A_n$ has to have an even number of things in it, which works only because $n$ is odd.
Note that for this to work the list $A_2,A_3,\dotsc,A_n$ has to have an even number of things in it, which works only because $n$ is odd.
Yasha
2024-05-15 20:30:56
I actually thought about this a different way.
I actually thought about this a different way.
Yasha
2024-05-15 20:31:00
Starting with $A_1+A_2$, we subtract $A_2+A_3$ to cancel out the $A_2$, obtaining $A_1-A_3$. Then we add $A_3+A_4$ to cancel out the $-A_3$, obtaining $A_1+A_4$. Then we subtract $A_4+A_5$, and so forth.
Starting with $A_1+A_2$, we subtract $A_2+A_3$ to cancel out the $A_2$, obtaining $A_1-A_3$. Then we add $A_3+A_4$ to cancel out the $-A_3$, obtaining $A_1+A_4$. Then we subtract $A_4+A_5$, and so forth.
Yasha
2024-05-15 20:31:19
When we get to our last step, do we add $A_n+A_1$ or do we subtract $A_n+A_1$?
When we get to our last step, do we add $A_n+A_1$ or do we subtract $A_n+A_1$?
Oshawoot
2024-05-15 20:31:43
add?
add?
MathIsFun286
2024-05-15 20:31:43
depends on the parity of n
depends on the parity of n
NAAUBE
2024-05-15 20:31:43
add
add
Yasha
2024-05-15 20:31:46
Well, we added $A_1+A_2$, subtracted $A_2+A_3$, and added $A_3+A_4$. So, when the first summand is $A_i$ for $i$ odd, we add, and when the first summand is $A_i$ for $i$ even, we subtract. Since $n$ is odd, we add $A_n+A_1$.
Well, we added $A_1+A_2$, subtracted $A_2+A_3$, and added $A_3+A_4$. So, when the first summand is $A_i$ for $i$ odd, we add, and when the first summand is $A_i$ for $i$ even, we subtract. Since $n$ is odd, we add $A_n+A_1$.
Yasha
2024-05-15 20:31:58
So, we have
$(A_1+A_2)-(A_2+A_3)+(A_3+A_4)\pm\dotsb+(A_n+A_1)$.
So, we have
$(A_1+A_2)-(A_2+A_3)+(A_3+A_4)\pm\dotsb+(A_n+A_1)$.
Yasha
2024-05-15 20:32:02
Almost everything cancels, and we are left with
$(A_1+A_2)-(A_2+A_3)+(A_3+A_4)\pm\dotsb+(A_n+A_1)=2A_1$.
Almost everything cancels, and we are left with
$(A_1+A_2)-(A_2+A_3)+(A_3+A_4)\pm\dotsb+(A_n+A_1)=2A_1$.
Yasha
2024-05-15 20:32:16
Dividing by two, we can solve for $A_1$.
Dividing by two, we can solve for $A_1$.
MathIsFun286
2024-05-15 20:32:24
If n is odd, we add
If n is even, we subtract and get null information
If n is odd, we add
If n is even, we subtract and get null information
Yasha
2024-05-15 20:32:38
Yes, for this solution method, too, it is crucial that $n$ is odd.
Yes, for this solution method, too, it is crucial that $n$ is odd.
joeym2011
2024-05-15 20:32:50
For even $n$, everything cancels out.
For even $n$, everything cancels out.
NAAUBE
2024-05-15 20:32:50
but if n even you need to subtract getting null!
but if n even you need to subtract getting null!
Yasha
2024-05-15 20:33:50
But of course, just because our solution method fails when $n$ is even, doesn't necessarily mean that there's no solution when $n$ is even.
But of course, just because our solution method fails when $n$ is even, doesn't necessarily mean that there's no solution when $n$ is even.
Yasha
2024-05-15 20:34:05
We will have to show that Kayla fails when $n$ is even separately.
We will have to show that Kayla fails when $n$ is even separately.
Yasha
2024-05-15 20:34:25
To finish off the odd case, we just have one more remark:
To finish off the odd case, we just have one more remark:
Yasha
2024-05-15 20:34:30
As before, the same reasoning lets us solve for all of the $A_i$. And, since we know $A_i-a_i$, we can also solve for the $a_i$.
As before, the same reasoning lets us solve for all of the $A_i$. And, since we know $A_i-a_i$, we can also solve for the $a_i$.
Yasha
2024-05-15 20:34:39
So Kayla can figure out all the numbers.
So Kayla can figure out all the numbers.
Mathdreams
2024-05-15 20:34:58
Is it fine if you use matrices for this?
Is it fine if you use matrices for this?
Yasha
2024-05-15 20:34:59
Sure, at its core, this is a linear algebra problem, and you can use matrices to solve those.
Sure, at its core, this is a linear algebra problem, and you can use matrices to solve those.
NAAUBE
2024-05-15 20:37:14
Im happy i wasnt the only person to solve it with matrices
Im happy i wasnt the only person to solve it with matrices
smartowl2020
2024-05-15 20:37:14
How could you use matrices to solve this?
How could you use matrices to solve this?
NAAUBE
2024-05-15 20:37:14
I used matrices, so did you see a quicker general solution with those by someone else? Im just wondering because I'm sure there must have been an easier way
I used matrices, so did you see a quicker general solution with those by someone else? Im just wondering because I'm sure there must have been an easier way
Yasha
2024-05-15 20:37:45
So, even though I wrote the initial version of this problem, other volunteers did the grading, so I don't actually know what people did!
So, even though I wrote the initial version of this problem, other volunteers did the grading, so I don't actually know what people did!
Yasha
2024-05-15 20:38:29
At its core, this problem is a linear system of equations. So, once you learn how to solve those using matrices, you can apply that knowledge here.
At its core, this problem is a linear system of equations. So, once you learn how to solve those using matrices, you can apply that knowledge here.
Yasha
2024-05-15 20:39:06
But we don't expect Mathcamp applicants to have necessarily learned linear algebra, so the intended solution is definitely non-matrix.
But we don't expect Mathcamp applicants to have necessarily learned linear algebra, so the intended solution is definitely non-matrix.
Yasha
2024-05-15 20:39:11
But matrix solutions are great, too.
But matrix solutions are great, too.
Mathdreams
2024-05-15 20:40:03
But we can also show there are infinitely many solutions to the system if we want to show $n$ even doesn't work.
But we can also show there are infinitely many solutions to the system if we want to show $n$ even doesn't work.
ColorfulChameleon109
2024-05-15 20:40:03
wee caan dooo eeeveeen nooow?
wee caan dooo eeeveeen nooow?
Yasha
2024-05-15 20:40:41
Alright, let's move on to even. In fact, we don't need to find infinitely many solutions; just two different ones is enough.
Alright, let's move on to even. In fact, we don't need to find infinitely many solutions; just two different ones is enough.
Yasha
2024-05-15 20:40:54
Following the previous part, could we come up with an example where $A_1+A_2, A_2+A_3,\dotsc,A_n+A_1$ are all zero?
Following the previous part, could we come up with an example where $A_1+A_2, A_2+A_3,\dotsc,A_n+A_1$ are all zero?
MathIsFun286
2024-05-15 20:42:32
$A_k=(-1)^k$ and $n$ is even
$A_k=(-1)^k$ and $n$ is even
Mathdreams
2024-05-15 20:42:32
n, -n, n, -n, ..., n, -n
n, -n, n, -n, ..., n, -n
NAAUBE
2024-05-15 20:42:32
What about alternating + and - 1?
What about alternating + and - 1?
DogisSad
2024-05-15 20:42:32
Set them all equal to 0
Set them all equal to 0
Mathdreams
2024-05-15 20:42:32
$n, -n, n, -n, ..., n, -n$
$n, -n, n, -n, ..., n, -n$
NAAUBE
2024-05-15 20:42:32
alternating $\pm1$
alternating $\pm1$
ColorfulChameleon109
2024-05-15 20:42:32
yeah! just alternate $-1$s and $1$s!
yeah! just alternate $-1$s and $1$s!
Yasha
2024-05-15 20:42:37
Well, of course if we set all the $A_i$ to zero, then all these sums will be zero. But we can also make a nontrivial example by setting $A_i=(-1)^i$, so
$A_1=-1, A_2=1, A_3=-1,\dotsc$.
Well, of course if we set all the $A_i$ to zero, then all these sums will be zero. But we can also make a nontrivial example by setting $A_i=(-1)^i$, so
$A_1=-1, A_2=1, A_3=-1,\dotsc$.
Yasha
2024-05-15 20:42:54
Is $A_n$ equal to $1$ or $-1$ in this case?
Is $A_n$ equal to $1$ or $-1$ in this case?
A_P_11
2024-05-15 20:43:06
1
1
MathIsFun286
2024-05-15 20:43:06
1
1
Yasha
2024-05-15 20:43:09
Since $n$ is even, $A_n=(-1)^n=1$, so the full sequence is
$A_1=-1, A_2=1, A_3=-1,\dotsc,A_n=1$.
Since $n$ is even, $A_n=(-1)^n=1$, so the full sequence is
$A_1=-1, A_2=1, A_3=-1,\dotsc,A_n=1$.
anAOPS_2022
2024-05-15 20:43:15
1
1
Yasha
2024-05-15 20:43:19
Any two consecutive numbers sum to zero, including $A_n+A_1=0$. So, Kayla cannot tell the difference between this scenario and the one where all the numbers are zero.
Any two consecutive numbers sum to zero, including $A_n+A_1=0$. So, Kayla cannot tell the difference between this scenario and the one where all the numbers are zero.
Yasha
2024-05-15 20:43:24
To complete the example, we set the $a_i=(-1)^i$ as well.
To complete the example, we set the $a_i=(-1)^i$ as well.
Yasha
2024-05-15 20:43:25
Yasha
2024-05-15 20:43:30
In this example, Kayla will receive all zeros in her reports from Maya and Nathan, so she cannot tell apart the above scenario from the scenario where all of the secret numbers are zero.
In this example, Kayla will receive all zeros in her reports from Maya and Nathan, so she cannot tell apart the above scenario from the scenario where all of the secret numbers are zero.
Yasha
2024-05-15 20:44:23
So, to summarize: We group the boxes into uppercase/lowercase pairs. Then we link pairs of boxes if they share a color, thereby obtaining linked groups of uppercase/lowercase pairs. If each group contains an odd number of pairs, then Kayla succeeds. If any of the groups contains an even number of pairs, then Kayla fails.
So, to summarize: We group the boxes into uppercase/lowercase pairs. Then we link pairs of boxes if they share a color, thereby obtaining linked groups of uppercase/lowercase pairs. If each group contains an odd number of pairs, then Kayla succeeds. If any of the groups contains an even number of pairs, then Kayla fails.
NAAUBE
2024-05-15 20:45:13
Can there be a situation like that for n odd?
Can there be a situation like that for n odd?
Yasha
2024-05-15 20:45:16
Nope! In fact, we proved that when we did the $n$ odd case.
Nope! In fact, we proved that when we did the $n$ odd case.
Yasha
2024-05-15 20:45:46
When $n$ is odd, Kayla can figure things out. So if she gets reports of all zeros, she knows that all the secret numbers are, in fact, zero.
When $n$ is odd, Kayla can figure things out. So if she gets reports of all zeros, she knows that all the secret numbers are, in fact, zero.
Yasha
2024-05-15 20:45:54
There's no way to make an example like this.
There's no way to make an example like this.
Yasha
2024-05-15 20:46:47
So, that's it for problem 2, which is the last problem for today (since we started with problem 3). Tomorrow some other Mathcamp volunteers will present problems 4-6.
So, that's it for problem 2, which is the last problem for today (since we started with problem 3). Tomorrow some other Mathcamp volunteers will present problems 4-6.
Yasha
2024-05-15 20:46:55
I'll stick around for a little bit since there were some questions I didn't get to.
I'll stick around for a little bit since there were some questions I didn't get to.
zoeR123
2024-05-15 20:47:06
How do I solve using matrices and what is it
How do I solve using matrices and what is it
Yasha
2024-05-15 20:47:24
You can use matrices to solve systems of linear equations.
You can use matrices to solve systems of linear equations.
Yasha
2024-05-15 20:47:53
Systems of linear equations are talked about here: https://artofproblemsolving.com/wiki/index.php/System_of_equations#Solving_Linear_Systems
Systems of linear equations are talked about here: https://artofproblemsolving.com/wiki/index.php/System_of_equations#Solving_Linear_Systems
Yasha
2024-05-15 20:48:16
Unfortunately, they don't say too much about matrices there, other than that they are "Advanced Methods"
Unfortunately, they don't say too much about matrices there, other than that they are "Advanced Methods"
Yasha
2024-05-15 20:49:16
Here's another link I found Googling https://math.libretexts.org/Bookshelves/Algebra/Intermediate_Algebra_1e_(OpenStax)/04%3A_Systems_of_Linear_Equations/4.06%3A_Solve_Systems_of_Equations_Using_Matrices
Here's another link I found Googling https://math.libretexts.org/Bookshelves/Algebra/Intermediate_Algebra_1e_(OpenStax)/04%3A_Systems_of_Linear_Equations/4.06%3A_Solve_Systems_of_Equations_Using_Matrices
Yasha
2024-05-15 20:49:27
The key word you're looking for is "Linear Algebra"
The key word you're looking for is "Linear Algebra"
NAAUBE
2024-05-15 20:49:37
https://www.youtube.com/watch?v=uQhTuRlWMxw&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=7
https://www.youtube.com/watch?v=uQhTuRlWMxw&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=7
NAAUBE
2024-05-15 20:49:37
That video helps with this specific problem
That video helps with this specific problem
NAAUBE
2024-05-15 20:49:37
its 3blue1brown
its 3blue1brown
Yasha
2024-05-15 20:49:55
Haven't watched this one but I know 3blue1brown is good.
Haven't watched this one but I know 3blue1brown is good.
ColorfulChameleon109
2024-05-15 20:50:32
What's $-3!$?
What's $-3!$?
Yasha
2024-05-15 20:50:34
Well I'd say it's -6 since I think the factorial operation takes precedence over negation.
Well I'd say it's -6 since I think the factorial operation takes precedence over negation.
Yasha
2024-05-15 20:50:40
$(-3)!$ on the other hand...
$(-3)!$ on the other hand...
MathIsFun286
2024-05-15 20:50:56
Use the Gamma function!
Use the Gamma function!
Yasha
2024-05-15 20:51:15
Unfortunately, the Gamma function is defined on all complex numbers except for the non-positive integers.
Unfortunately, the Gamma function is defined on all complex numbers except for the non-positive integers.
Yasha
2024-05-15 20:51:31
So, sadly, we can define the factorial for pretty much everything except things like $(-3)!$.
So, sadly, we can define the factorial for pretty much everything except things like $(-3)!$.
A21
2024-05-15 20:51:36
3blue1brown is amazing
3blue1brown is amazing
ColorfulChameleon109
2024-05-15 20:52:12
Can I have a cookie?
Can I have a cookie?
Yasha
2024-05-15 20:52:14
Yes! But if it's from me, it'd only be virtual. I'd recommend your pantry instead.
Yes! But if it's from me, it'd only be virtual. I'd recommend your pantry instead.
paixiao
2024-05-15 20:53:33
Can we move onto P3
Can we move onto P3
Yasha
2024-05-15 20:53:39
We swapped the order and did that one first.
We swapped the order and did that one first.
Yasha
2024-05-15 20:53:44
I think there will be a transcript of this though?
I think there will be a transcript of this though?
jwelsh
2024-05-15 20:54:44
Yes! It'll be up soon after the Math Jam is over.
Yes! It'll be up soon after the Math Jam is over.
paixiao
2024-05-15 20:55:36
so all of them are done?
so all of them are done?
Yasha
2024-05-15 20:55:42
For today. 4-6 will be tomorrow.
For today. 4-6 will be tomorrow.
NAAUBE
2024-05-15 20:55:54
what do we use fractional derivatives for? is it just for certain vector spaces? I'm very curious abt this
what do we use fractional derivatives for? is it just for certain vector spaces? I'm very curious abt this
Yasha
2024-05-15 20:56:00
It comes up in partial differential equations.
It comes up in partial differential equations.
NAAUBE
2024-05-15 20:57:04
In what cases is a forier transform (or however you spell it) more useful than a laplace transform?
In what cases is a forier transform (or however you spell it) more useful than a laplace transform?
Yasha
2024-05-15 20:57:26
Fourier transform is useful for steady state situations. Laplace transform is useful for immediate effect situations (e.g. you hit something)
Fourier transform is useful for steady state situations. Laplace transform is useful for immediate effect situations (e.g. you hit something)
ColorfulChameleon109
2024-05-15 20:57:38
when tomorrow?
when tomorrow?
Yasha
2024-05-15 20:57:48
Same time, which is 7pm Eastern and 4pm Pacific
Same time, which is 7pm Eastern and 4pm Pacific
Oshawoot
2024-05-15 20:58:21
for finding percentages of populations, do we always round down or do we round to the nearest whole number, my teacher said always round down but i'm not entirely sure
for finding percentages of populations, do we always round down or do we round to the nearest whole number, my teacher said always round down but i'm not entirely sure
Yasha
2024-05-15 20:58:23
I don't think there's a universal rule. Which means that in your classroom your teacher's rule goes.
I don't think there's a universal rule. Which means that in your classroom your teacher's rule goes.
Yasha
2024-05-15 20:59:07
But, generally speaking, you might round not just to a whole number. For example, you might round 74.38% to 74.4%
But, generally speaking, you might round not just to a whole number. For example, you might round 74.38% to 74.4%
Yasha
2024-05-15 20:59:33
In science-type situations you want to figure out the number of significant digits.
In science-type situations you want to figure out the number of significant digits.
Yasha
2024-05-15 20:59:47
and round that way. Usually you round to the closest thing, whether it is down or up
and round that way. Usually you round to the closest thing, whether it is down or up
ColorfulChameleon109
2024-05-15 20:59:55
Where did you get your avatar?
Where did you get your avatar?
Yasha
2024-05-15 21:00:07
I was in high school when I set it, so I don't quite remember. But it's the escape key on the keyboard.
I was in high school when I set it, so I don't quite remember. But it's the escape key on the keyboard.
Oshawoot
2024-05-15 21:00:16
i meant like the amount of people given a percent and a population
i meant like the amount of people given a percent and a population
Yasha
2024-05-15 21:00:45
Oh I see. Still, probably significant figures is the way to go, because for most things you don't have precise enough information to know down to the last person.
Oh I see. Still, probably significant figures is the way to go, because for most things you don't have precise enough information to know down to the last person.
Yasha
2024-05-15 21:01:07
Like, it's not like we know exactly how many people are above 6 feet tall.
Like, it's not like we know exactly how many people are above 6 feet tall.
Yasha
2024-05-15 21:01:19
But we can estimate, but if so we'd probably say how many millions of people rather than exactly how many people.
But we can estimate, but if so we'd probably say how many millions of people rather than exactly how many people.
NAAUBE
2024-05-15 21:02:09
can you prove if a DiffEQ is unsolvable?
can you prove if a DiffEQ is unsolvable?
Yasha
2024-05-15 21:02:44
Yes, there are ways to do that. You might even learn of examples in a regular college DiffEQ class if it's challenging enough.
Yes, there are ways to do that. You might even learn of examples in a regular college DiffEQ class if it's challenging enough.
admathcamp353
2024-05-15 21:03:20
If they ask like, how many people can this many popsicles serve, or something like that, I think you're supposed to round down because you can't serve a person with half a popsicle
If they ask like, how many people can this many popsicles serve, or something like that, I think you're supposed to round down because you can't serve a person with half a popsicle
Yasha
2024-05-15 21:03:29
Yeah, maybe that's the idea with always rounding down.
Yeah, maybe that's the idea with always rounding down.
Yasha
2024-05-15 21:04:23
Anyways, that's all for today. Come back tomorrow for more Mathcamp volunteers and more Mathcamp problems!
Anyways, that's all for today. Come back tomorrow for more Mathcamp volunteers and more Mathcamp problems!
jwelsh
2024-05-15 21:05:01
Thank you all for coming! Over 500 students joined us today.
Thank you all for coming! Over 500 students joined us today.
jwelsh
2024-05-15 21:05:18
A special thanks to Yasha and Tim for leading us through problems 1-3!
A special thanks to Yasha and Tim for leading us through problems 1-3!
jwelsh
2024-05-15 21:05:35
Join us tomorrow for problems 4-6!
Join us tomorrow for problems 4-6!
jwelsh
2024-05-15 21:05:50
If you have questions for Mathcamp, you ask tomorrow or can contact them at www.mathcamp.org/contact.php
If you have questions for Mathcamp, you ask tomorrow or can contact them at www.mathcamp.org/contact.php
jwelsh
2024-05-15 21:06:10
Finally, a transcript of this Math Jam will be posted soon here: www.aops.com/school/mathjams-transcripts
Finally, a transcript of this Math Jam will be posted soon here: www.aops.com/school/mathjams-transcripts
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