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trying to figure out how to activate it..

testing $x_1$

x_1 is just a number... so basically you plug numbers into those weird looking symbols eg if x_1=2, x_2=5, y_1=3, y_2=7 then cauchy-schwarz inequality says that (2^2+5^2)*(3^2+7^2)>=(2*3+5*7)^2. Which if you expand out you can see is true. PS x^2 means x squared...

Cauchy-Schwarz is an inequality (C-S) for short. It goes, for positive integer n and positive (can make it non-negative, but positive makes things easy to understand) x_i and y_i... {(x_1)^2+(x_2)^2+....+(x_n)^2}{(y_1)^2+(y_2)^2+....+(y_n)^2}>={(x_1.y_1)^2+(x_2.y_2)^2+....+(x_n.y_n)^2}...

You can't usually integrate random expressions... Including this one... If you're masochistic and want to figure out random stuff then go to http://integrals.wolfram.com. It comes out with an extremely ugly expression.. you have been warned... basically the techniques you're taught in 4U are...

UNSW comp you're allowed to bring in books anyway :-P

Just thinking on the run, but... you could always do a summation thing.. ie case 1: none of the 31 cards are A's. case 2: there's 1 A. Then for each case, you've determined how many non-A cards you need to choose (eg 30 non-A's for case 2) then just 198C30... Actually, so then generalising...

The thing is, if you consider their relative motion it gets a lot easier... ie let the first particle be defined as the origin (your euclidean plane is then a "camera" capturing what goes on about that point). Since both particles have equal acceleration, you can deem them to have zero...

Although, I reckon mclake has _actually_ done finite whereas I haven't... hence the "apparently"... but algebra is fun :-) Well I have an assignment due now so maybe I should actually stop wasting time..

yeah... well I don't do that much volume but it's good stuff :-) BTW apparently finite is a relative walk in the park compared to discrete..

lol I can't read..

yep, sounds wrong to me too..

I'd like to help but I have no idea of what the proof is..

Ahh this brings back fond memories...

There's always newton and lagrange to go on..... and you thought soccer rivalry was bad..

Scammed it from IMO2001 :-) Also got another cool poster about reimann zeta fn... both on my bedroom wall... nerdy isn't it.. but one of them is getting tattered so coming off soon. I did mean half a something.. let me go check Oh. Yeah, quartic isn't excessively long, you're right affinity...

GRRR my third time writing this: This had better work Probability: method 1. Let 2 people be A and B. Let the group A belongs to be the first group. B is then randomly allocated to one of the other slots. There are 4 slots left in first group, 5 in second group, so prob of them being...

EDIT: dbl post that didn't work.

In reply to what you said before... Let the intersection of the arc with the y-axis be Y. Then Y, -1, 1 form an equilateral triangle, you can work out Y from that, and the centre of the triangle is the same as the centre of the circle..

I reckon it's right...