60 years ago this question was on the International Mathematical Olympiad

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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[alinat.]

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.

[arjunraman]

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

[madelineveggie]

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.

[ucanhvungoc]

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.

[TDSONLINEMATHS]

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

[johnklinger]

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.

[MathsScienceandHinduism]

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

[mr.d]

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

[omaraldebs]

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

[FlatEarthMath]

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. :-)

[troys]

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.

[MarieAnne.]

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

[sanchitagrawal]

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

[charlesdbruce]

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."

[the-boy-who-lived]

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.

[jacobgoldman]

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

[XinLi]

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

[jaguar]

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

[SG]

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.

[MathsMadeSimple]

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.