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[Skyrunner]

More of a sort of venting than actual advice question :<

As I write this, I am taking a break from excessively difficult math problems that seem to exist juat to push the limits of mathematical puzzle-solving. In my short 1 year stay in the US, these types of problems don't exist in a typical school setting. But here is where they all go to live, and it's driving me crazy. Most problems hinge on finding a crucial hint that is hidden amongst the details and then using a weird method to follow that to the answer. To me, the finding of the crucial hint is too hard. :/ It never feels like I can improve myself because the harder problems are usually like that. And my school is one of the most competitive public high schools in Korea, which means the hardest problems occassional appear on te math tests. Because grading is relative, those who do manage to get the problems right get As (1 or 2) while I have to struggle to not go below a 4 (B, top 40%). It hugely stresses me and sometimes I imagine how better I would feel if I just stopped trying, even though that would ruin my chances of going to a good foreign university. =_=

Uhh.. guess that's it. I probably have more to say but I can't think of it right now. I could totes provide example problems If you wish.

[vagel7]

Aren't South-Korean universities ranked quite high internationally?

Also, if you're talking about math word problems then I agree with you. Those things are horrible

[Solymr]

Are them the type of problems where you have to think outside the box, but the box is a maze worthy of a minotaur?

[Skyrunner]

Aren't South-Korean universities ranked quite high internationally?

Also, if you're talking about math word problems then I agree with you. Those things are horrible

Not really. The highest one is maybe in the 50s? :/

Are them the type of problems where you have to think outside the box, but the box is a maze worthy of a minotaur?

Here's a few problems I posted to another topic

Define F_n(x) = Integrate for dx: (x³ⁿ-1)/(x²+x+1)

F_n(1) = 1/(3n-1) - 1/(3n-2) + 1/(3n-4) - 1/(3n-5) + ... + 1/2 - 1

Represent F_n(0) in terms of n.

f(x) is a differentiable function with the domain 0 <x <pi/2.
For all valid x, f'(x) = cos^2 (f (x)), and f(1)=pi/4. if g (x) is the inverse function of f (x), find g(pi/3) * g'(pi/3).

Assume f'(x) means 'the first order differential of f(x)'.

Note that calculators are not allowed, and that for the picture, originally there was no graph in the problem.

I don't know about you, but those sure are hard to me :<

[Solymr]

They look tedious but not that hard to me :<
Then again I got a medal on math olympics so I'm an outlier.

[vagel7]

Yeah, they don't look that horrible. My Achilles Heel are the bloody word problems. For things like that I just suggest you to take a math book, learn the theory and different ways of solving them and just start solving.

[Jelle]

Those don't seem to hard. Fair bit of numbers crunching, especially without calculator, but they seem rather obvious.

Is it possible you are putting to much focus in recognizing patterns in the given information and looking for a matching methodology to use? In my opinion the most logical way to solving such problems is to start with the answer to arrive to and work your way down from there. I would for example always read the latter part of the question for what is asked first, then read the given information.