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Just concerned that the second part has not been answered properly. Z^4+1=(z2-2cos3pi/4.z+1)(z2-2cospi/4.z+1) Divide both sides by z^2, to get the z+1/z =2cosΘ structure which would be useful. Seen a similar problem like this in the Blue Patel.

There are always two sides to an argument : the argument itself and the argument about the argument. Please excuse this Lewis Carroll type argument..:) Now for the argument itself : no there won't be any loss of generality by saying one side is size 12 unless you're talking about shoes! 12...

____________________________________________________ :: ryan.cck :: can u just simply say the greatest area is when a = 1... find the interesection between the two lines, which happens to be 1/3 of the distance away from the point x = 1... split up into two triangles and using area of one...

Idyll : Absolutely!

Well defended nike33. Another way is to draw CF // BE and DZ. Since this 3 parallel lines splits ZC equally, they must also split DC equally.

Idyll hasn't done too badly, he (or she) has the right approach. But in the name of elegance, lets follow ND's advise and more. Position the triangle with C at O, A at 12 and ... wait for it... B at 12z where z is a complex number. If Idyll (or anyone else) care to do it again with this...

just to comment that dropping a perpendicular from the midpoint of DZ and assuming that it will cut DC at O might be at best unnecessary or at worst wrong. But overall approach is good. Will give you a chance to revise that part or defend it.

nike33 : you're on the right track - but are you doing approach b) or a) ? At any rate, go on...

Some complex number problems could be done quiet elegantly using a geometric approach. How about the other way around? Let ABC be a triangle, D be the midpoint of AB and E a point on the side AC for which AE = 2EC. Prove that BE bisects the segment CD. a) by geometry b) by complex numbers

Ryan.cck : Method is right. Reflecting across y=x and y=0 seems to be an elegant approach. Nice careful work. Being careful is another requirement for doing well in Ext2. Practise your care on the following problem : Let 0 <= a <=4. Prove that the area of the bounded region enclosed by the...

Yes ryan.cck. Show us how, so we can see if its got class.:) *** also note, I've edited the storm problem***

In a similar vein to the other questions : a car travels at 2/3 km/min due east. A circular storm, w/ radius 51, starts with its center 110 kms due north of the car and travels southeast at 1/sqrt(2) km/min. The car enters the circle of the storm at time t1 and leaves at t2 (mins). Find...

Sorry guys, can't get classier than CM_tutor's.:o

Short, shorter, shortest. Maths' such a perverse world:D . Line through (8,6) parallel to OQ is 3x - 10y +36 = 0. This meets OP at (16/7, 30/7) which should also be the midpoint of OP. Thus P is (32/7,60/7). The distance of P to (8,6) is sqrt((24/7)^2+(18/7)^2) = 30/7. Hence PQ = 60/7.

Sorry for not being clear. Couldn't seem to download Xayma's. Though from the discussion it sounded correct.

__________________________________________________ for a) i got the answer... itll be a bit annoying to type on comp... the distance should be 36 and 36/49 units -ryan.cck _________________________________________________ your answer's too large. Would be interested in a brief...

I agree, McLake.

We had a thread last year on elegance -Spice Girl and ND were the main correspondents. To do well in Ext. 2, students need to develop their EQ elegance quotient. Here are three problems to practise on, the first two could easily be done by a Year 11 doing the preliminary, the third - well...

Sorry guys, it wasn't much of a Mother after all.:) Will try to make up for it later towards the exams. Couldn't see Affinity's answer (nor Xayma's), but imagine it follows from the expansion of (i-i)^10 and collect the 1's. Truth is I had this solution before I went looking for a problem to...

Who says OLDMAN can't attract attention?:D Played basketball with the young son this morning, and the court was filled with hoppers- yes the locusts have finally reached Mudgee. Even without a blade of grass, a grasshopper's orientation after it lands seem pretty random - so its next hopping...