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Ah holy crap i realised where I went wrong. I had R = 10, rather than R = 10sqrt(3). I hate stupid shit like that. Thanks very much for the nicely typed solution, braintic :)

Patel exercise 9C, question 4. Looking for a worked solution. Thanks in advance A model plane of mass 5kg attached to the end of an inelastic wire of length 20m flies in a horizontal circle of elevation 30°, while the other end of the wire is held fixed. If the lift (force) L acts at right...

alright that's what i got, except my method for finding the area of the parabola was not as nice. bloody simpsons rule

as the title says, just a volumes question. not 100% sure on my solution for question 7. anyone wanna show me the best way?

there is no 0th second though.

4. When t = 1, we are at the end of the first second, 1 full second has passed. When t = 2, we are at the end of the second second (hehe), 2 full seconds have passed. etc. t = 4 is the end of the 4th second. stupid wording though.

This is what i was going to suggest Products <--> Sums = winning

ah holy crap didn't think about P and Q being vertical! wow, that simplifies things. I was going to find P by subbing in T=a/e but it was so horrid i figured i must be missing the easier way. thanks!

can someone give me a hand with the question attached? it's question 4 from the 2004 HSC. Got no clue how to go about part 2.

By constructing lines AD & CB, we get the ever-so familiar butterfly shape. The angle from the minor arc AC is ACD, & the angle from the minor arc BD is BAD. Since the exterior angle of a triangle is equal to the sum of the two opposite interior angles, AXC = ACD + BAD It's not a theorem, but...

Re: HSC 2014 4U Marathon \text {In order to avoid confusion (because I used alpha and beta as roots), I'm going to call a and b, p and q respectively} \\ \text {Let the roots of the quadratic be given by } \alpha \text { and }\beta \\ \alpha + \beta = -p \\ p^{2} = (\alpha + \beta)^{2} \\ q...

you do realise this thread is from 2010 right?

Re: HSC 2014 4U Marathon - Advanced Level By alternating colours though how can this be true? If A1 is black, surely any large odd number could only be white squares (or less if you don't allow moves back) than the whole board?

Re: HSC 2014 4U Marathon - Advanced Level Yep exactly what I was going to say. It always alternates colours so there is absolutely no way it can get from white to black in an even number of moves.

Re: HSC 2014 4U Marathon - Advanced Level Well thats exactly what I got I was just a minute late :(

Re: HSC 2014 4U Marathon - Advanced Level Following that logic, my solution is any square the same colour as A1

Re: HSC 2014 4U Marathon - Advanced Level what if on the 4000th move you end up on (x-1, y-2), rather than on(x,y)? EDIT: I did a drawing and tried it out a bit, and i'm fairly positive there is no combination of moves to remedy this. If you hit (x,y) on an odd move, i don't think you can get...

the basic parts of the 4U course aren't much of a step up but as soon as you get into some of the harder questions, the difficulty gap between 3U and 4U is enormous.

some 2012s

here are a few 2011 trials EDIT: i'll upload some 2012 ones in a bit