Solving an IMO Problem in 10 Minutes!! | International Mathematical Olympiad 1992 Problem 1
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- Опубліковано 31 січ 2021
- #IMO #NumberTheory #MathOlympiad
Here is the solution to IMO 1992 Problem 1!!
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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.
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Regularly kills IMO questions, incorrectly multiplies to get 2 instead of 4 😆
I’m not making fun or criticizing. I think it’s beneficial for students to see that even the greatest mathematicians will make casual errors if they are not careful.
Please keep up the excellent work. I wish I had such instruction in my mathematical competition days as a high school and college student.
👍
Good morning!
I did it in a similar way, but quite differnt, as well.
Let k(a,b,c) = (abc-1)/(a-1)(b-1)(c-1) So for a, kmax needs that b=a+1 and c=a+2 (for paritity I coul be more restrictive; b=a+2 and c= a+4, but we would not be able to simplify)
let k*(a,b,c) = abc/((a-1)(b-1)(c-1)
2 a < 4. So a=2 or a=3.
For parity a,b,c are all even and k is odd or a,b,c are all odd and k can be also odd or even.
If a=2
K(2,b,c) is maximum for some b if c=b+1 as k is odd k(2,b,b+1)>=3; then K*(2,b,b+1) >3, so b a and b is even b=4.
kmax(2,4,c)=k(2,4,5)= 10/3. As k needs to be in integer and k is odd and k>1, k=3 , so c=8. Then (2,4,8) is a solution
Doing the same for a=3
We find that b = 5, k=2 and then c= 15. and (3,5,15) is the other solution. Only two solutions.
Note that in both solutions c=ab and the first solucion is a geometric progression (GP).
I liked that you added a term to then factor, with a prime number as a product. And we find b and c at once. I have to restict first for b, then for k and aftewards, solve for c.
Good afternoon!
When I realize, after solving the problem, that ab=c, I have tried to see it earlier, but I failed. If I had a way to prove that, it would let me to a amazing solution.
((ab)^2-1)/(a-1)(b-1)(ab-1) is integer. Simplifing ... (ab+1)/((a-1)(b-1)) ==> b-1 | ab+1 As b-1 | ab-a ==> b-1 | a+1 (i)
from (i) we have b-1 b=a+2.
(a-1)(a+1) | a^2 +2a +1, Simplifing a-1 | a+1. As a-1 | a-1 a-1| 2 and then a=2 or a=3.
if a=2; b=a+2=4; c=ab=8. (2,4,8) Checking, it is ok!
if a=3; b=a+2=5; c= ab=15 (3,5,15) Checking ok!
Unfortunately, i am not able to prove that c=ab. So it has no value.
EXPLAINING a solution to a problem in 10 minutes is not the same as SOLVING the problem in 10 minutes.
The amazing solution, it's easy to understand, you're my favorite math channel, I always watched your video and it's helped me improve my math, and help improve my critical thinking... I enjoy your video, thank you very much 🥺 please don't get tired of making videos like this
Are/were you an olympian yourself?
Most likely an adult. An adult will be able to solve and explain olympiad question easier than the actual highschooler
Based on what ? Anyone who can solve the problem will be able to explain it as well . Has nothing to do with being an adult or not .
@@xwtek3505not how it works😭there needs to be a very high level of specialized studying to reach this level of undertsnding of contest math
You make it so easy even I can understand it. It feels like an AIME question now.
amazing method of solving this ecuation. Please try more of these kind of problems with naturals or integers.
Excellent content. Can you tell me please which software do u use? Is it like an app running on ipad?
Great problem and great solution.
Thanks so much!
Amazing
Much the same but slightly differently ...
natural numbers s.t. 1n²-1)
and a≥2,b≥3,c≥4 so max value of ...(2) is for min, a,b,c
from ...(2)
k ac+bc+ab-a-b-c=0 but ab > a and bc > b and ac > c, so k≠1
looking at a, if a≥4
k=(a/(a-1))(b/(b-1))(c/(c-1))-1/P
≤ (4/3)(5/4)(6/5)-1/P < 2 so no solution for a≥4
if a=3
k≤(3/2)(4/3)(5/4)-1/P < 2.5
so k=2
first consider a=2
k≤(2/1)(3/2)(4/3)-1/P < 4
so k=2 or 3
if k=3,
3(2-1)(b-1)(c-1)=2bc-1
3bc-2bc-3b-3c+3+1=0
bc-3b-3c+4=0 add 5 and factorise
(b-3)(c-3)=5 factor pairs 1,5
solutions a,b,c = 2,4,8
if k=2
2(2-1)(b-1)(c-1)=2bc-1
-2b-2c+3=0
2b+2c=3 has no solutions for b,c > a=2
consider a=3 and k=2
2(3-1)(b-1)(c-1)=3bc-1
4bc-4b-4c+4-3bc+1=0 add 11 and factorise
(b-4)(c-4)=11 factor pairs 1,11
solutions a,b,c = 3,5,15
Excelente
Really awesome! Must watch videos
Can you pls tell us your secret for solving questions with such an effective strategy I have my maths olympiad coming up.
nice and neat
This way to solve it is fairly good, IMO😉
At 5:09, shouldn't it be (4/3)(5/4)(6/5). You wrote 6/3 instead of 6/5
I believe you are correct. I was thinking the same thing. I imagine this video is made by following a worked out solution, and hence was likely a transcription error.
Thanks for a simple and nice presentation. DrRahul Rohtak Haryana India
Ram Ram bhai
@@ashirwadgarg174 Ram Ram bhaiya ji
That was simple even as a Problem 1. However, thank you for the nice presentation!
Can you suggest me books/online resources to learn from so I can be as good as this guy?Pleaseeeeee
@@ariyanbista5837 there are many online resources. Two of them are AoPS and IMOmath. AoPS has numerous interesting contest problems which are usually solved by the AoPS community. IMOmath has nice preparation material and many past exams of national, regional and international competitions. As for the books that I'd recommend, well, there are many, many, many nice books covering a single Olympiad topic. For instance, there are at least 3 good books on Olympiad inequalities and the book of Vasc, although advanced, is one of them. However, a book that covers almost all the Olympiad topics at a rather satisfying level has to be Arthur Engel's book. I cannot recall its exact title now, but it's a must. Hope I helped a bit.
is the checking part is important at olymp, i mean we should write checking part on our answer sheet?
Yes, checking whether the solution is valid is important.
Good video bro , I want to ask you from where you learnt that when to use which approach.?
As a former Olympiad competitor, my answer would be "problem solving practice", "trying to figure out a motivation for a solution", and "learning about new techniques".
Usually what you do is try around a bunch of stuff and one of them is a hit. There are some pointers you may get from say "playing around with small cases".
In this example you could've played around by saying "what if I set a = 5?" Then you get 4(b-1)(c-1) divides 5bc-1. Subtracting 4(b-1)(c-1) from the LHS you get 4(b-1)(c-1) divides bc +4b + 4c - 5. Now there's a "feeling" that this won't ever work because the RHS is "too big". Next comes the question "how do I formalize this feeling"? Oh I know, let me use the fact that LHS is k times the RHS where k is greater than or equal to 1. So now from bc + 4b + 4c - 5 >= 4bc -4b - 4c+4 you get 0>= 3bc - 8b - 8b +9 = b(c-8) + c(2b-8) + 9 which looks very close to a contradiction.
Now from here you see the basic idea was that "the LHS isn't too much bigger than the RHS". You try to solve the problem by generalizing the idea, ie looking at the equation k*(a-1)(b-1)(c-1) = abc - 1 and looking whether or not k = 3 or 4 would work here. There are many approaches to finish off but that's the basic process through which ideas comes from.
@@ShefsofProblemSolving Hi bro thanks for giving such a wonderful explanation to my question .Means basic idea is to play around with the equations and get a feel then tackle it .
@@divyanshtripathi7867 Yup pretty much. Unless you have an idea right away you wanna try out, you just play with the equations, explore the problem to get a sense of where it might be going
Says, "And this is equal to two." Contemplates life for two seconds after he contradicts every rule learnes in elemantary school. "No, I'll just stick to regular mathematics for this video."
Best
Love
Hi, you are really excellent. May I ask where you learnt all of these? And also maybe how you recognize when to use some approach or another? Thank you very much for your content
It's simply ALOT of practice and a fair amount of intelligence. There's no recipe in solving such problems as opposed to high school / undergrad level university math. You just do a bunch of them and when you're stuck, you think about them for days or weeks or even months and try to be creative without looking up the solution.
But there are a few hints, maybe purposefully given: If you're given integer variables you often need to use factorizations, prime properties, modular arithmetic. For real value variables you may need to use known inequalities in some clever form, mean-value theorem and so on. But it's almost never as easy. Many times you need to first notice stuff about the problem structure and even after rewriting and manipulations it may still be hard to see this theorem or that inequality is going to be useful.
But exactly this intuition is the thing you learn by hammering your brain with these kind of problems.
@@ZeonLP ohh thank you very much
@@gastoncastillo9946 Though there are also a couple of pointers you develop as you expose yourself to different problems.
For example, in combinatorics problems the direction you take your problem is usually informed by playing around with small cases. This "playing around with small cases" becomes a pointer for where to take the problem.
Most important is to try to figure out what the motivation for a certain solution was. This is much easier to do once you yourself actually try to solve the problem.
Turn on postifications
👏👏👏👏👏👏👏👏👏🤩
Wish I was smart . But we can't have everything .
were you an IMO candidate?
你講野咁似香港人嘅