60 years ago this question was on the International Mathematical Olympiad
Вставка
- Опубліковано 24 вер 2023
- Thanks to Jacques for the suggestion! The International Mathematical Olympiad (IMO) began in 1959 with only 7 countries. It has expanded to over 100 countries. This is problem 2 from the 1959 IMO.
AoPS 1959 IMO, problem 2
artofproblemsolving.com/wiki/...
Public Domain, commons.wikimedia.org/w/index...
FireCracker Math for quote
firecrackermath.org/a-challen...
@mathmemo2691 video
• [Very first IMO in his...
IMO pages
www.maa.org/math-competitions...
www.imo-official.org/
www.imo-official.org/year_cou...
www.imo-official.org/countrie...
imo2023.jp/en/countries/
Mathematics contests history
mathoverflow.net/questions/24...
Subscribe: ua-cam.com/users/MindYour...
Send me suggestions by email (address at end of many videos). I may not reply but I do consider all ideas!
If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.
If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.
Book ratings are from January 2023.
My Books (worldwide links)
mindyourdecisions.com/blog/my...
My Books (US links)
Mind Your Decisions: Five Book Compilation
amzn.to/2pbJ4wR
A collection of 5 books:
"The Joy of Game Theory" rated 4.3/5 stars on 290 reviews
amzn.to/1uQvA20
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 4.1/5 stars on 33 reviews
amzn.to/1o3FaAg
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4.2/5 stars on 54 reviews
amzn.to/1LOCI4U
"The Best Mental Math Tricks" rated 4.3/5 stars on 116 reviews
amzn.to/18maAdo
"Multiply Numbers By Drawing Lines" rated 4.4/5 stars on 37 reviews
amzn.to/XRm7M4
Mind Your Puzzles: Collection Of Volumes 1 To 3
amzn.to/2mMdrJr
A collection of 3 books:
"Math Puzzles Volume 1" rated 4.4/5 stars on 112 reviews
amzn.to/1GhUUSH
"Math Puzzles Volume 2" rated 4.2/5 stars on 33 reviews
amzn.to/1NKbyCs
"Math Puzzles Volume 3" rated 4.2/5 stars on 29 reviews
amzn.to/1NKbGlp
2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter.
My Blog
mindyourdecisions.com/blog/
Twitter
/ preshtalwalkar
Instagram
/ preshtalwalkar
Merch
teespring.com/stores/mind-you...
Patreon
/ mindyourdecisions
Press
mindyourdecisions.com/blog/press - Наука та технологія
not as hard as what i would expect for an IMO problem, but still challenging!
The early IMOs were not that hard at all.
Chewy, more than complicated.
that's true for any exam during that time@@AntonioLasoGonzalez
what did you expect from 1959 ? Permutations and combinations ?
I think it's easier to solve if you notice that the expressions under the big square root signs are full squares divided by 2: x+sqrt(2x-1) = (sqrt(2x-1) + 1)^2/2. Then we have (sqrt(2x-1) +1) + |(sqrt(2x-1) -1)| = A*sqrt(2) which comes to either 2x - 1 = A^2/2 or 2 = A*sqrt(2) depending on where your x is.
I was part of the first Math Olympia USA team in 1973-74. We had a summer training "camp" at Rutgers University in New Brunswick NJ lead by a great mathematics professor, and we used the earlier IMO problems and other fun problems for practice, and did daily classes in math theory, number theory, trigonometry, etc. One final team went to Hungary, if I remember correctly, and we had part of the playoffs in DC.
Wow! nice 😎
Whats your age
This... is one way to solve the problem, but in Vietnam it would be called "butchering" since there is a much better way to solve:
Let t = sqrt(2x-1), we would have x = (t^2 + 1)/2.
The equation becomes: sqrt((t^2 + 2t + 1)/2) + sqrt(t^2 - 2t + 1)/2) = A. From here it's a cakewalk.
i did it exactly like this and solved it under 2 min
It becomes much easier if you substitute the roots as a and b then you get a+b =A and a^2+b^2=2x
Just futzing around, I was able to figure x = 1/2 or 1 results in A=√2. I failed to figure out that it was endpoints of a range, or of the other solutions. Presh's solution is excellent. :-)
I've solved this problem many times and it stills hard😅
Thanks❤
This problem was published in an Indian mathematical magazine named 'mathematica' approx 18 or 19 years ago.
Mathematics as a subject serves as a basics to all subjects which is generally accepted at all levels of educational ladder & it plays a unique role in the development of each individual. My passion!!
TDS ONLINE MATHS
Nice problem with an impressive amount of mathematical magic that happens when you square the expression. A minor flaw in the graph shown near the end is that the curved part doesn't approach a vertical slope near x=1.
I did the problem in the same way up until this point:
A² = 2x + 2|x−1|
Then I considered the different cases:
1/2 ≤ x < 1 → A² = 2x + 2(1−x) = 2 → A = √2
x = 1 → A² = 2 + 2|1−1| = 2 = √2
x > 1 → A² = 2x + 2(x−1) = 4x−2 → x = (A²+2)/4 > 1 → A > √2
Then I can find solutions without graphing (which seems rather time consuming for a math contest)
A = √2 → 1/2 ≤ x ≤ 1
A = 1 → No solutions (since minimum value of A is √2)
A = 2 → x = (A²+2)/4 = (4+2)/4 = 3/2
Very nice! Great formatting here, too! Although, it looks like you're missing "→ A = √2" for your case when x=1.
Great one
*When I clicked on the video, i genuenly taught it was going to be some ancient Math problem.*
You can make it easier if you assume √(2x-1) as y and x as (y^2+1)/2
Only if modern Olympiad papers had this much easy questions. 😭
They sometimes have very hard questions which takes me hours to solve even after returning home.
Interesting that this complex looking function is constant from [0.5,1.0].
Great
I like how this implies that after quite some time, the problems we find hard now are going to be very classical problems that even slightly competitive middle schoolers find elementary.
Saw this problem from PK Math not too long ago
Yeah it is a beautiful problem, but compared to the IMO problems today relatively easy. I could solve that problem by myself which honestly is rarely the case for more recent IMO problems.
Ouch. I've got the case A=1 completely wrong. Only a Calculator convinced me that A(3/4) = Sqrt(2). Hopefully i've learned something from this. Thank you
I was curious, so I decided to plug it into a graphing calculator. According to the graph, it never hits y=1, it only hits y=2 when x=1.5, and it hits y=sqrt(2) when x is less than or equal to 1.
Any reason you don't list the variable first in your inequalities? E.g., x>1 compared to 11) is typically read as, "X is greater than one." While this (1
It was pretty common, essentially universal, practice in my math classes to arrange inequalities that are being used to indicate limits on variables in the format "a < x < b" even if one side of the set was missing. That way the lowest value is always to the left and the highest is always to the right. It was never outright stated as a rule or convention or anything like that, but was common enough to seem like one.
No, (x > 1) is written as "x is greater than one." While (1 < x) is written as "One is less than x." Write the same variable in the
appropriate lower case, or upper case, as needed.
У меня мозг отказывается функционировать, когда я смотрю на эти записи в обратном порядке
Presh the type of guy to ask for receipts when shopping, just to tell the cashier what the total cost is going to be, before the receipt prints.
I search the IMO problem list in the forlorn hope that I will even understand what just one problem is asking . So far , no luck .
Just square both sides. Then a lot of things cancel out and you can go from there.
Where did you get the second set of squares from? When hi squatted the equations in the beginning it should have just canceled out the square rooting that was happening. What did i miss or forget?
I hate how Presh doesn't put the full question in the preview frame.
When we taking sqaure on both side... We can solve it easily.. But if anyone has any suggestions.. Please give me.
We will get x+√2x-1+x-√2x-1+2√x^2-2x+1=A
At around 5:20 when you make the cases why didn't you choose the first condition to be x>=1 instead of x>1
Brutal
This is too much work. You can just write sqrt(x+-sqrt(2x-1)) as |sqrt(x-1/2) +- sqrt(1/2)| (you can do this by the same method as literally the previous video, writing sqrt(3-2sqrt(2)) as sqrt(2)-1). So the sum is actually max(sqrt(4x-2), sqrt(2)). So the interval [1/2,1] obviously goes to sqrt(2), nothing goes to 1, and only 3/2 goes to 2.
making a squared binomial equal to an absolute value.
Which grade can answer this question?easily… cause I am in 9th grade and I can hardly answer it
The REAL question 2
How are you counting the continents? It should be six: Europe, Asia, Africa, Australia and both Americas. If you join some together - which ones? Americas? Europe and Asia into Eurasia? And if so - why only one joint, and not both? I don't really get your nomenclature :/
I came here to ask this question. Either he has a math channel but can't count, or he's trying to say that Eurasia is only one continent. Please tell me that this isn't going to be another Pythagorean Theorem thing.
There are seven continents: Asia, Africa, North America, South America, Antarctica, Europe, and Australia.
the delineation of continents isn't standardized worldwide-it depends on who's counting (quick example: how many olympic rings?)
The biggest difficulty is the arbitrarily constrained problem. "A fool can ask more questions than a wise man can answer."
Your logic in minute 3 is wrong unless you explain that all the steps are reversible.
finally i got my first presh question correct 😭
This looks wrong to me. Substitute 3/2 into the original expression and you get sqrt(2) not 2. Similarly you can put say 13 into the original expression and you also get sqrt(2). I think it should be no solution for A=1 or 2 and x>=1/2 for A=sqrt(2).
Nice and easy
Problem 1 is even more ridiculous: You have to prove that gcd(21n+4,14n+3)=1.
Do I need to prove that the Euqlid algorithm works to get a full score? :)
@@wafemand Yes, but first you need to prove that the integers are closed under addition and multiplication.
This problem is weirdly simple. Use the fact that if two numbers share a common factor, so must their difference. (21n+4)-(14n+3)=7n+1. Apply again (14n+3)-(7n+1)=7n+2. And one last time (7n+2)-(7n+1)=1 Therefore the largest and only common factor of the original expressions is 1
Six continents.
I use desmos to solve this problem 😂😂
This solution is wrong.
For A=sqrt(2) x=1
For A=1 , x=3/4
For A=2, x=3/2
This is done by completion of square of x+sqrt(2*x-1) etc.
Please check
Mininmum value of A is sqrt(2), it cant be 3/4
The “International” Mathematics Olympiad in 1959?
If you consider the former Soviet Union and six of its satellites as “international” then I guess you’re right.
Math (and science, chess, etc.) should be politics free.
@@hrayzI agree, which is why I shared this comment, something I rarely do.
In what way is it not international? It involved multiple nations.
In 1959 none of these so-called “countries” were free and autonomous, independent of Soviet domination.
@@illinois_b and USSR was second last, while Romania (the one Soviet puppet that was always a problematic ally both politically and ethnically the most different of all others) overwhelmingly won. Do you call that MO rigged?
I almost solved the problem but lost a factor of 2 because of my dreaded scrawniness 😝
I don't know what to say except in early. And yes nobody cares
Great