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gah got it. Take 2 squared out as a common factor. *was blind* i didnt see it at first, coz when i divided 64/4, i managed to wind up with 18... its too early in the morning.

hmm. *guesses ramdomly* perhaps... Try to write the 2 squared + 4 squared + 6 squared pattern as sums of squares... eg 2 squared = 1 squared + 1 squared... 4 squared = 2 squared x 4 o.0 or maybe not...

Hey I don't really understand proofs unless they're algebraic when I'm reading online, so what I'll do is print yours of, Xayma, and take a look at it tonight so I have a hard copy with me, (and there's also the bonus of being able to look at circles, not ellipses) and then I'll get back to...

ahh genius. =) thanks!

err... ive got a diagram made up but dont know how to attach.. 0.o

Hmm *kicks self* -.- ive been doing the question WRONG! Blah. Sorry, ND, about that. :( My bad. Thanks for the proof, Xayma. :) *gives cake to all* Well, now I know what circumscribes truly means- and considering i wasted hours of my life on this, I wont be forgetting it any time soon...

Sorry, I dont understand why it can't be correct. I'm not really following. -.-' Perhaps we have to prove that a = (a + b)/2, then?

Why is it obviously not right? circumscribes = Makes a circle going through points A, B, C

Well, we're s'posed to use c/g skill s from 3u to do them, so Im guessing yes. But as to if well get asked them in tests Q3 was asked.. =D But I think the 1st one is a bit too complicated. Not sure about the 2nd.

Mine popped up first, though... =D Oh, and I'll hunt for my past papers book for that, thanks, CM. :)

both. :)

Hey guys- These are out of the 4u Fitzpatrick book, ex 37(a) If you haven't had a go at them you probably should try taking a look at them some time soon because they take aaagggesss to see.. (or it could've just been me). Well, anyway, there are three that I can't get out, any help would be...

uhmm... would this work?: subst x = [tanu]^2 and then after simplifying, subs v = sinu ?

3a. Let angle XAB = angle XAC = x let angle AXC = a, therefore angle AXB = 180 - a in triangle ACX: sinx / CX = sina / AC ....1 in triangle ABX: sinx / BX = sin(180-a) / AB sinx / BX = sina / AB ....2 1 / 2 : BX/CX = AC/BC which means that BX:CX = AC:BC as required...

3c. Let the lines/rays from q's 3a,b, intersect at point R. RO perpendicular PQ (from 3a) Since RO, PQ are the diagonals of QRPO, the QRPO must be a rhombus (since QO and PO are equal, it cannot be a kite). Therefore, PR = OQ and R = vector OP + vector OQ. that is, R = z(0) + i(z(0)) R...

-.-' i bet you just want some cake :P

Hmm... Grey_C, I don't really get what you're saying... perhaps i'll get you to explain it to me over msn later on... -.-' Affinity, what you're saying's true, but then what if the equilateral trianlge formed isnt actually formed with the origin as center? Because for part (ii), one of the...

Yup, I'm stuck, yet again: (i) State a condition for z, w, v (all complex) to form the vertices of an equilateral triangle on the Argand Diagram. (ii) If z = 1 + 3i, find a complex number w, so that z, w and the origin O form the vertices of an equilateral triangle. Thanks for taking a...

3a. line = perpendicular to the line joining P and Q, going through (0,0) 3b. Line parallel to vector OQ going through P. ?

Haha... awwww.... *gives you some lamingtons to make up for the stolen cakes* geez... this thread has become so .... cake-oriented...