60 years ago this question was on the International Mathematical Olympiad
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- čas přidán 8. 09. 2024
- 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
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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
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.
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
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
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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.
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❤
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.
You can make it easier if you assume √(2x-1) as y and x as (y^2+1)/2
Interesting that this complex looking function is constant from [0.5,1.0].
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.
*When I clicked on the video, i genuenly taught it was going to be some ancient Math problem.*
This problem was published in an Indian mathematical magazine named 'mathematica' approx 18 or 19 years 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.
Saw this problem from PK Math not too long ago
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
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.
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
Just square both sides. Then a lot of things cancel out and you can go from there.
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.
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
Great
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
Great one
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 .
Your logic in minute 3 is wrong unless you explain that all the steps are reversible.
I hate how Presh doesn't put the full question in the preview frame.
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.
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).
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?
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?)
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 biggest difficulty is the arbitrarily constrained problem. "A fool can ask more questions than a wise man can answer."
Brutal
finally i got my first presh question correct 😭
The REAL question 2
Nice and easy
Six continents.
I use desmos to solve this problem 😂😂
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
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
It's so lengthy and boring, however there is an easy solution
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