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Re: 2012 HSC MX2 Marathon You dont actually need to use an infinite series. You can use the following fact (prove it for yourself its easy): If \frac{a}{b} = \frac{c}{d}, then both these are also equal to \frac{a+c}{b+d}. Then if R_i and G_i denote the red area/green area within the ith...

Re: 2012 HSC MX2 Marathon Doesn't matter because of dimensionality. Like, what does "radius one" mean? It means, "radius one UNIT" where that unit is something you chose. You could have, the outer radius is radius "one inch" for instance, or could be "two centimetres": but if its radius "two...

Re: 2012 HSC MX2 Marathon cool problem!!! Anyway here's how I'd do it. Observation: each circle has half the radius as the one before it. Proof: Clearly all the circles, since the diagram has rotational symmetry, have the same centre; call this O. Now let a vertex of say the biggest...

Re: 2012 HSC MX2 Marathon Theres basically no agreed standard for what "N" means... I usually just write N_0 or N^+ (dunno how to do blackboard bold on this), in a way that would be pretty unambiguous. Someone give a problem!

wrarrr tesseract

Alright, that quantity i/-k represents the vector CA divided by the vector CB, right? Now if A,B,C is isosceles right angled at C, that means that the length CA = the length CB, right? I.e. |CA| = |CB|, i.e. |CA/CB| = 1 right? So you have to get |i/-k| = 1, coz CA/CB = i/-k if CA and CB are...

Let A represent z1, B represent z2, C represent the other thing. Consider this: \frac{z_2 - \frac{z_1-ikz_2}{1-ik}}{z1 - \frac{z_1-ikz_2}{1-ik}} = \frac{z_2 - ikz_2 - z_1 + ikz_2}{z_1 - ikz_1 - z_1 + ikz_2} = \frac{z_2 - z_1}{ik(z_2 - z_1)} = \frac{1}{ik} = \frac{1}{-k}i This number is...

Re: 2012 HSC MX1 Marathon ^^ now try this question, and explain why it directly implies that problem (B) (i.e. show how to solve problem (B) without angle chasing: Given any points A,B,C,D on the plane with no three collinear, explain how to find the spiral centre that sends AB to CD. I.e...

Re: 2012 HSC MX1 Marathon I dont think anyone would complain if you just drew a diagram and drew in an angle chase :P

Re: 2012 HSC MX1 Marathon That was what I was looking for ^^ Anyway, geometry problem: $(a) Triangle $ABC$ is given, with arbitrary points $D, E, F$ chosen on sides $BC, CA, AB$ respectively. Prove that the three circles $AEF, BDF, CDE$ all go through a common point.$ $(b)Four distinct...

Re: 2012 HSC MX2 Marathon yup, STEP III 2011. STEP is a pretty good source of questions, I know grammar occasionally takes questions from it :p

Re: 2012 HSC MX2 Marathon You might've, but I kinda doubt it? Ive never seen it posted on these forums before, and it was written for a UK test so not in the "HSC circulation" per se.

Re: 2012 HSC MX1 Marathon 8 choose 3 = something... How about this gem: Prove, WITHOUT USING THE BINOMIAL THEOREM OR ANYTHING LIEK THAT, that ^nC_0^2 + ^nC_1^2 + ^nC_2^2 + ... + ^nC_n^2 = ^{2n}C_n

Re: 2012 HSC MX2 Marathon I don't really know how tex works on this forum... I think Ive fixed it now though!

Re: 2012 HSC MX2 Marathon Here's one I just found (havent done it yet) but it looks decently cool: $Let $T_n = (\sqrt{a+1} + \sqrt{a})^n$ where n is a positive integer and a is a given positive integer.$ $(i) When n is even, prove by induction that $T_n$ can be written in the form $A_n +...

Re: 2012 HSC MX2 Marathon Didn't one of these exist ages ago? Anyway let P(x) = x^3(ax^2 + bx + c). Now by the remainder theorem, P(1) = 1. Differentiate P(x) - 1, and you find that P'(x) is has a double root at x = 1, so P'(1) = 0 and P''(1) = 0. Put that into the above to get: a+b+c = 1...

I'm pretty sure you can get away with some not "mathematically correct" things at school... like full analytical rigour is not really part of the course, and I'm not convinced every marker would pick really subtle things up either.

I was under the impression 'beyond the syllabus' methods like calculusing complex wasnt allowed... I was told to completely avoid doing certain stuff I knew for inequalities and geometry, for instance. In any case, can one of the uni guys explain to me the rigorous basis behind the calculus of...

Re: 2012 HSC MX1 Marathon Isnt it now just L'Hopital's again?

The cool geometric thing is this, and this is the only way I know to prove muirhead. In three variables you can think of each sequence (x1,x2,x3) which has sum say k as a barycentric coordinate, determining a point within an equilateral triangle of "weight" k. If you haven't come across this...