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Re: HSC 2014 4U Marathon Could you also do this: Assume all of the eqns have non-real complex roots. Then by conjugate root theorem and summing product of roots you get a sum of real squares = 0. Therefore, all the roots must be zero, a contradiction of the assumption.

Look at the angles, cointerior angles are supplementary. Those ARE parallel lines, just drawn to not look like they are.

Once you realise it's a trapezium it's fairly straightforward. Just drop perpendiculars to find the height and you can use trig to find the difference between top and bottom.

LOL er guys, that shape is a trapezium, the diagram is dodgy

State rankings are a bit weird. In 2012: First in state: 98/97 Second: 98/98 Third: 97/98 (i think...) All the other state rankers got ~97 so lowest HSC (aligned) mark may have been 97 or 96.

http://www.imo-official.org/country_individual_r.aspx?code=AUS 1/6 chance ;) But 'imo guy' said he skipped q14 and 15 and went straight to 16 so there's that...

if you let z=x+iy and then use tan definition of arg it works out

inb4 3 hours

okay nvm just read the first post :L You change the polynomial so that the roots are c_q-i etc. i.e. replace z with z+i then consider product of roots, which would simply be simply P(i). So now we take modulus of everything and we get: |prod.ofroots|<1 but from part i), the real roots (or...

Are we allowed to spoil answers in here or...?

I knew how to do the birthday paradox MC, but my calculator couldn't handle >___>

I think you can, because of perpendicular bisector thingo from centre. I didn't get it anyway LOL, left like 15 mins for q16 :S... also @chlee, it's not that easy of a proof to see, you could be underestimating your inability or just happened to have had the clarity to see it on the day :P it's...

Falconpride lol what a silly name hehe

Put me down for 4u and 3u thanks! ^_^

Re: HSC 2014 4U Marathon Did it a different way by letting w=x+iy and simplifying z, to get z=i(y/(1-x))=+-i(sqrt((1-x)/(1+x)), which for -1(lessthan)x(lessthan)1 ,draws out the imaginary axis with a hole at origin. Like your solution better though :P edit: how do you use less than signs...

Re: HSC 2014 4U Marathon Equality holds when a+b=b+c=a+c=sqrt(3)

I think its easier to let z=rcis(theta) rather than a+ib. Using DMT gives r-(1/r)=0 which clearly means r=1 (since r>0)

Haha hopefully! English is the weakest subject at NSB though, so there's that to consider... Anyway thanks for the feedback guys, might make a new thread with confirmed ranks, but right now full steam ahead to station fix-up-english!

Thanks!

Hey sorry guys, my friend actually took my phone and posted this... but I suppose since this thread's already started, those aren't exactly correct: English would be more like 30/160 (stuffed up trials) Maths ext 1 - 1 (equal)/115 Maths ext 2 - 2 or 3/69 Physics - 2/70-80 Chem - 1 or 2/60-70...