60 years ago this question was on the International Mathematical Olympiad

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).

Just square both sides. Then a lot of things cancel out and you can go from there.

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

I hate how Presh doesn't put the full question in the preview frame.

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 .

You can make it easier if you assume √(2x-1) as y and x as (y^2+1)/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?

making a squared binomial equal to an absolute value.

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.

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

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.

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.

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

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.

When I clicked on the video, i genuenly taught it was going to be some ancient Math problem.

u + v = A ---------------------------------------------------> known
u^2 + v^2 = 2x
u+2 - v^2 = 2 * sqrt(2x-1)

(u + v)^2 = A^2 = 2x + 2uv ------------------------> known
u+2 - v^2 = 2 * sqrt(2x-1) = A(u-v)
2 * sqrt(2x-1) / A = u-v
4 * (2x-1) / A^2 = A^2 - 4uv = 2x - 2uv

u^2 * v^2 = (x-1)^2
uv = +- x-1

Then it can be substituted.

I use desmos to solve this problem 😂😂

The biggest difficulty is the arbitrarily constrained problem. "A fool can ask more questions than a wise man can answer."

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.

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.

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.

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.

Which grade can answer this question?easily… cause I am in 9th grade and I can hardly answer it

The REAL question 2

Your logic in minute 3 is wrong unless you explain that all the steps are reversible.

Saw this problem from PK Math not too long ago

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. :-)

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.

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 :/

This problem was published in an Indian mathematical magazine named 'mathematica' approx 18 or 19 years ago.

Any reason you don't list the variable first in your inequalities? E.g., x>1 compared to 1<x. I found it a bit confusing when we generally read from left to right. This(x>1) is typically read as, "X is greater than one." While this (1<x) is typically read as, "One is less than X."

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 almost solved the problem but lost a factor of 2 because of my dreaded scrawniness 😝

Problem 1 is even more ridiculous: You have to prove that gcd(21n+4,14n+3)=1.

Interesting that this complex looking function is constant from [0.5,1.0].

Great one

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.

Nice and easy

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

not as hard as what i would expect for an IMO problem, but still challenging!

Great

Great

I've solved this problem many times and it stills hard😅
Thanks❤

I don't know what to say except in early. And yes nobody cares