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Are you saying that beating 80% of the students who sit the HSC exams would equate to a 96ish ATAR? Do you have stats to back this up? I really don't believe the effect of including students who don't complete the HSC is close to this significant. (I would expect it to be more like 86. A bit...

It is important to remember here that "raising to a complex power" is an undefined notion in high school. So e^(i*pi) has no meaning a priori. Without rigorously defining the objects you are working with, it is hard to pin down exactly what your "why?" question is asking. Given any way of...

Re: HSC 2014 4U Marathon - Advanced Level Yep, good. It should be noted that this question is just a rearrangement of Schur's inequality, which is why the method is somewhat similar to how Schur's is proven.

Re: HSC 2014 4U Marathon - Advanced Level $Let $a,b,c\geq 0.\\ \\$Prove that$: \\ \\ abc\geq (a+b-c)(a+c-b)(b+c-a).

A leading mathematician who was born in Australia and works at UCLA. He was a child prodigy and has lived up to his potential. He is conversant with a ridiculously broad array of topics and continues to output very original and sometimes technical research (his specialty being harmonic analysis...

Why are you still trying to solve this? Carrot showed that the integral does not converge.

There were 1,236 all rounders last year. With 75,168 students taking at least one HSC course, this is about 1.6% of students getting all-rounders. (These numbers may be slightly off, different sources gave different numbers.) There were ~800 state ranks, a slightly lower number...but this is...

Re: HSC 2014 4U Marathon - Advanced Level Well by symmetry, no strategy can assure you better than a 50% result. (Here a draw counts as half of a win). I stated a trivial optimal solution for (i), but if we have no information about our opponents action then the theory behind this game is...

Re: HSC 2014 4U Marathon - Advanced Level Presumably a "strategy" is just a set of rules determining whether or not to say stop after a given sequence of cards (and for (i) also with the knowledge of when an opponent says stop.) Care to define "optimal"? In the game theoretic sense of...

Any ambiguity as to what the question is asking is terrible practice imo. Especially hurting those with high aims. Such questions really hinder the exams potential as a means for fairly gauging the relative levels of students.

For the record, I agree with 90atarpls and Realise...but questions like these are the reason I would probably get a lower raw mark in general than MX2.

How I feel about teachers / courses that assess students with questions like the OP:

To stay what I would call "clean shaven" I would pretty much have to do it daily, I definitely get an afternoon shadow if I shave in the morning. That said, I actually only shave 2-3 times a week because I think a bit of stubble suits me (two days growth being optimal imo).

If you want to prove this property, it comes out quickly from considering similar triangles.

Re: HSC 2014 4U Marathon - Advanced Level Well, the colour of the square the knight is on after the 4000th move is fixed, so we can rule out half of the board. Your idea basically shows that it can be any of the squares of the correct colour though.

Re: HSC 2014 4U Marathon - Advanced Level Just a fun bit of morning procrastination :p.

Re: HSC 2014 4U Marathon - Advanced Level Suppose to the contrary that f IS reducible over Z and f = gh is one such expression with deg(g) > n/2 > deg(h). Note that we can assume strict inequality of degrees because their sum is deg(f) = n which is odd. This turns out to be key. g(aj)h(aj) =...

Re: HSC 2014 4U Marathon - Advanced Level For x < 0, define f(x) = af(-x) + bf(-2x)+cf(-3x) (*) for some undetermined constants a,b,c. It is clear that this will satisfy the differentiability and boundedness assumptions of f for positive x, just by the laws of how differentiation behaves...

Re: HSC 2014 4U Marathon - Advanced Level It's just similar triangles: Alternate angles tells us: \triangle AXF \sim \triangle CXB and \triangle AXD \sim \triangle EXB This implies: \frac{EX}{CX}=\frac{EX}{BX}\cdot\frac{BX}{CX}= \frac{AX}{DX} \cdot\frac{FX}{AX}=\frac{FX}{DX} Therefore...

June 10-12.