IMO 2024 Problem 6 - the *FINAL BOSS* is always tough! Functional equation leaves few survivors
ฝัง
- เผยแพร่เมื่อ 14 ต.ค. 2024
- #mathematics #olympiad #math
International Mathematical Olympiad (IMO) 2024 Day 2
Solutions and discussion of problem 6
65th International Mathematical Olympiad Bath UK
Problem 6 - Algebra / Functional equation
That's a wrap! Thanks for tuning in to the IMO 2024 problems. Let me know what you think of this year's IMO problems!
I checked other solutions in AOPS and they did use the fact that f: Q->Q (to prove that the function is Cauchy) while yours did not at all. If this were changed to R I think this deserves to be the hardest functional equation ever appeared on IMO 😭
Very good point, I totally didnt realise that this solution doesnt need to use the f:Q->Q information and even works for f:R->R
Thank you for your work! There is very little content like this on TH-cam.
My pleasure!
Also a *huge* disclaimer that coming up with the example that works (at the end) is non-trivial. If anyone can share some motivation for others to learn, that would be helpful!
Yeah I would love to hear about motivation! When I tried it, it seemed like magic that it satisfied all the properties.
Maybe some motivation: in P1 of this year, I realized that floor(x) + floor(-x) was 0 if x is integer and -1 if not, so maybe that is why one starts trying with the floor function on P6
Hint:If {x}-{y}>0, use the front; otherwise use the latter
You can motivate the construction as follows. It's natural to start by proving that x + f(x) is a fixed point, so you might like to follow up by investigating the set of fixed points. If a and b are fixed points, then plug in a and b to get that f(a+b) = a+b, so a+b is a fixed point. If a is a fixed point, then plug in a and -a to get that f(a + f(-a)) = 0 or a + f(-a) = f(0), in either case -a is a fixed point. So the fixed points form a group.
From here you might guess that the set of fixed points is the set of integers, and then try to come up with something. Just thinking about what f will do to the interval (0, 1), you know it has to sort of reflect that interval (because f(x) + x has to be an integer). So maybe you try something like f(x) = -x on the interval, but you know that f(n) = n so maybe you want your function to be something that reflects the interval [n, n+1) about n for any n. (And this becomes floor(x) - {x} as was in the video.)
I think it's quite hard to motivate the construction if you don't know that the fixed points form a group - you can prove more stuff about the fixed points to help though (for example you can actually prove that f(x+a) = f(x)+a for any fixed point a, in the above I effectively just guessed that that was true. And of course you don't need to prove this to solve the problem)
You can prove the set of r such that f(r)+f(-r)=0 is a group, and then that every orbit in Q has the same value for f(r)+f(-r) (and that it is consistent within the orbits too ofcourse). From there one might decide to pick a subgroup of Q and try, so for example , and from there we take the easiest route: we use f(x)=x on and then use that f(x)+x is a set point for the rest of Q and that f is bijective to choose these set points as 2*floor x, and then we check if it works.
I think whoever came up with the example was hunting for a function that had a discontinuity in x and -x such that f(-x)=/=f(x) for some x, but f(x)=f(-x) for some other x.
22:43 most astonishing part of the solution. i have no idea of how to find out that actual example
Agree. Finding the example is also hard. Imagine proving all the way till here and you are so close to that 7 marks but yet so far
Hi, you have competed in IMO as a contestant, right? It would be great if you could share your experiences and what you have learned, this will help many aspiring people who want to come to IMO, I think.
You can make a video about that, but ofc, this is just my opinion, feel free to do whatever.
It's a great suggestion!
@@dedekindcuts3589Down for it
You explained the problem really well, and I learned a lot from this video via the techniques you used along with your thought process. Thank you for posting this! ❤
Glad it was helpful!
Note that the fixed point of f forms a subgroup of Q
And that f(r) +f(-r) is in that sub-group.
Please make long lectures of number theory cover each theory and nice problems please 🙏🏻
I have been thinking of that for a while!
Thank you soo mch
Could you recommend some good problem solving books that would help with viewing problems differently than usual
The Art and Craft of Problem Solving by Paul Zeitz is great!
Great explanation!
Glad it was helpful!
Is this posted by Singaporean?
hi i watch ur videos a lot can u recommend me books that u personally used while u were preparing for imo ? please
It's good to start with The Art and Craft of Problem Solving by Paul Zeitz. Personally I learned number theory from Elementary Number Theory by David Burton and geometry from Notes on Euclidean Geometry by Kiran Kedlaya
Perfect
Once you find that f must be biyective and that f(0)=0.
Why can't you say that
f(f(x))=f(f(x)+0)=x+f(0)=x and so f=f^(-1)?
@@pequenoduende7891 it is an "or" statement, so "f(f(x)+0)=x+f(0) or f(x+f(0)) = f(x)+0" is a true statement, and the right one leads to f(x)=f(x) which is always true, so we don't know at all if the left one is true
Ohh, ok, thx! I didn't understand the problem correctly sry, but now I do
how in the hell would you think of that example for c=2
I feel you. Same sentiments
After 2017, here comes a worthy opponent😅
Still 2.5 times more full score :p
Good explanation
Thanks!
Hello brother, I love olympiad styled math, but when facing college 's fixed dull curriculum and exam, I find myslef not able to freely explore math anymore, pls share how would u deal with it? thx u so much
After some foundation classes, you should be able to explore classes and subareas that interest you more! Also eventually you will be doing research which will be more open-ended. You should talk to your college professors - one of them might give you some research problem to work on!
@@dedekindcuts3589 Thank you so much for your reply!
Imo series has come to an end so what's next?🙂 putnam?
Usually I cover the first few problems of Putnam! There's some Putnam 2022 and 2023 videos on this channel that you can check out! (Unfortunately this year I have some other commitments during Putnam period so it will be 1-2 weeks later)
I was just asking I am sure you'll come up with something amazing! 🙌
@@dedekindcuts3589The upcoming IMC(International mathematics competition for undergraduate students) may have some interesting problems :)
@@dedekindcuts3589 why only the first few problems?
I didn't take the exam, so take my opinion with a grain of salt. But P3 seems a lot more mind-bending than P6! I only heard about P5 being a troll question, but did people also get tricked by this low answer?
Part of the difficulty is firstly people who get trolled by P5 already lost a lot of precious time and cannot really have enough time to work on P6. I fully agree with you that P3 is a lot more abstract and difficult to wrap your head around than P6 which is more about a very long series of exploring a web of equations - though without knowing that you are on the right path a priori, the web of equations can be overwhelming. We'll know when we see the stats!
To show that there exists an integer \( c \) such that for any aquaesulian function \( f \), there are at most \( c \) different rational numbers of the form \( f(r) + f(-r) \) for some rational number \( r \), and to find the smallest possible value of \( c \), let's first analyze the properties and constraints provided by the definition of an aquaesulian function.
Given:
\[ f: \mathbb{Q} \to \mathbb{Q} \]
is an aquaesulian function if for every \( x, y \in \mathbb{Q} \):
\[ f(x+f(y)) = f(x) + y \quad \text{or} \quad f(f(x)+y) = x + f(y). \]
### Step-by-Step Analysis
1. **Key Property Usage**:
Consider the first property \( f(x + f(y)) = f(x) + y \). Set \( x = 0 \):
\[ f(f(y)) = f(0) + y \]
This equation implies that \( f(f(y)) = y + f(0) \).
2. **Second Property Analysis**:
Now consider the second property \( f(f(x) + y) = x + f(y) \). Set \( y = 0 \):
\[ f(f(x)) = x + f(0) \]
3. **Equating Both Properties**:
From the above two equations, we get:
\[ f(f(y)) = y + f(0) \quad \text{and} \quad f(f(x)) = x + f(0) \]
Thus:
\[ f(f(r)) = r + f(0) \]
for any rational number \( r \).
4. **Key Observation for \( f(r) + f(-r) \)**:
Let \( y = -r \):
\[ f(f(r) - r) = r - r + f(0) = f(0) \]
Thus:
\[ f(f(r) - r) = f(0) \]
5. **Identifying \( f(r) + f(-r) \)**:
Consider the expression \( f(r) + f(-r) \):
\[ f(f(r) + (-r)) = f(0) = f(f(-r) + r) \]
Using the definition of an aquaesulian function, observe:
\[ f(f(r) + (-r)) = r + f(-r) \]
Therefore:
\[ r + f(-r) = f(0) \]
Solving this for \( f(-r) \):
\[ f(-r) = f(0) - r \]
Hence:
\[ f(r) + f(-r) = f(r) + (f(0) - r) = f(r) + f(0) - r \]
6. **Conclusion**:
Since \( f(r) \) can be any rational number depending on the specific function, observe that:
\[ f(r) + f(-r) = f(r) + f(0) - r \]
The number of different values this can take depends on \( f(0) \).
Since \( f(0) \) is fixed for any specific function \( f \) and \( f(r) \) and \( r \) can vary over all rational numbers, the number of different values of \( f(r) + f(-r) \) will be influenced by the variability introduced by \( r \) and the fixed \( f(0) \).
### Finding \( c \):
Considering the observations:
- \( f(r) + f(-r) = f(r) + f(0) - r \)
- Since \( f(r) \) maps rational numbers to rational numbers,
- The distinct values \( f(r) + f(-r) \) form a finite set.
Conclusively, the values of \( f(r) + f(-r) \) depend only on the structure and form of \( f \) as defined, and thus:
The smallest possible value of \( c \) such that there are at most \( c \) different rational numbers of the form \( f(r) + f(-r) \) for some rational number \( r \) is:
\[ c = 1 \]
This follows because the expression \( f(r) + f(-r) = f(r) + f(0) - r \) can be controlled by the constraints of \( f \) over \( \mathbb{Q} \).
Perfeito!
can you post google gemini proof.
Maybe its the hardest IMO i have ever seen.
Can't beat 2017 (: Highest score that year is a 35 and P3 that year had only 2 solves
2021 also had a killer p2
why is the explanation easy but the problem difficult?
because P probably != NP
Good analogy. Easier to understand a solution than to come up with it
FUN FACT: The city of Bath should be well known to cryptic crossword addicts. For instance a valid clue would be “Foreign capital in hte old city (4)” with the answer BAHT 😊
you think that a person has taken a perfect exam?
a chinese dude goT a perfect score
@@justmath.1533where can I see the results
Haojia Shi
And how did the United States fare?
@@MoyGomez-rp3ks
I believe they will excel at this year's IMO.
hi is there any way to connect with you ,line,telegram, discord,email, whatsapp, i want to compile a list of awesome maths papers and books,competitons for github, and get your level, as a mentor for advice and be better please thanks 🙏🙏🙏
can you teach me or guide me, if is too expensive classes just giving me advices and asking you questions