fuck math it ruined me throughout high school
Given are five line segments (or equivalently any five positive numbers aka lengths) so that any three of them can be put together to make a (nondegenerate) triangle. Prove that at least one of those triangles must be acute.
triangle inequality
what does the law of cosines tell you about acute triangles?
I forget most of the non-simple math I learned back in school already and I only graduated 2 years ago 
5
Over for retardcels
Math sucks, hated my bitch of a math teacher and It was difficult for me
Assuming psychological theories are even statistically falsifiable
Answer requires pretty much no physics.
a) Note that L/a and N must have the same parity, for L/a + N = 2N_+ (*). If L/a = 2k and N = 2m, then we must have that N_+ = m + k and N_- = m - k (solve (*) for N_+). \Omega is therefore (2m choose m + k) = (2m choose m - k) = (2m)! / ( (m + k)! * (m - k)! ).
Answer requires pretty much no physics.
I'm too lazy to deal with the case where both L/a and N are odd.
b) first of all, it's Stirling with two i's. Using log properties:
ln(\Omega) \simeq 2m*ln(2m) - 2m - (m + k)*ln(m + k) + (m + k) - (m - k)*ln(m - k) + (m - k)
= 2m*ln(2) - m*ln( (m + k) * (m - k) / m^2 ) + k*ln( (m - k) / (m + k) ) = N*ln(2) - m*ln( 1 - k^2 / m^2) + k*ln( 1 - 2k / (m + k) )
now using that ln(1 - x) \simeq -x for small x and the fact that N >> L/a AKA m >> k
ln(\Omega) \simeq N*ln(2) - m*( k^2 / m^2 ) - 2k^2 / m = N*ln(2) - m*( 3k^2 / m^2 ) = N*( ln(2) - 3k^2 / (2m^2) )
which is not quite the desideratum
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Assuming psychological theories are even statistically falsifiable
If a>b> c>d>e
In this case I've reached de > 1/3(D2+e2). Must be some other piece of info I'm not using
I remember everyone always mogging me with maths in school
I can't help you if I don't know how you got there
You should have greater than or equal to signs (right triangles are not acute). Try adding them up to make 2(d^2 + e^2) ≤ a^2 < (d + e)^2.
Isn't that why I shouldn't have >= signs? None of these are acute.
so we have 2de > d^2 + e^2. I guess this removes the possibility that any triangle involving those two is right angled
Exactly. By way of contradiction you assume none of the triangles are acute.
Rearranging the inequality yields that d^2 − 2de + e^2 < 0. Can you factor the expression on the left-hand side?
Ah. Ic. Noice. I was this close jfl
Yup you had the right idea from the get-go.
Fuark, low IQ shitskin genes won't let me master mathematical shit.
What 2 + 2?
Interesting watch. Based on the video I get the sense that Newtonian mechanics is a collection of mathematical ideas rather than a proper mathematical model. Differential equation can be quite finicky, however, so determinism of Newtonian mechanics hinges on details too finicky for a mere set of ideas. In short, I get the sense that determinism of Newtonian mechanics is a fundamentally ill-posed problem.
It's like asking whether God exists without specifying what God ought to be up front. I think it's contentious precisely because the terms ain't clear-cut.
No AI.
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this problem looks like tedium more than anything else
This was asked in a college entrance examination. With no calculator allowed and only pen and paper.
ew. I guess I'd write tan as sin over cos and start using prosthaphaeresis, but I don't feel like it.
Lazy boy. Poor, starving chinese students had to solve this in a dingy examination hall with feather-ink pens and sitting cross-legged with no shoes and socks in hope of putting food on their families table.
Nigga I'm doing a PhD in math. Let me have time off.
No. You have to solve this now. No one cares about your useless math PhD.
Billons must calculate
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