Recent content by alexandred

So I changed my UAC preference to combined med after the release of ATARs (have had no access to the internet for the last few weeks while on holiday, so no choice in the matter there). I'm probably only being neurotic, but does anyone know if (a) that will be a problem and (b) when those...

Two of my friends have science entry offers (and since neither are Olympiad people, probably not the four you have in mind), so it seems like 10 would be a minimum for entry (making the fairly safe assumption that between the two of us we haven't canvassed the entire group).

Got an offer for merit, conditional on >95 atar. Also, I want to do Science (Adv. Math)/Law; does a science merit scholarship cover the years spent on the law degree also (the fourth and fifth, as I understand it)?

There are two methods once you have realized that rt ~= 0.56. We have lnt = e^rt, but we also know e^rt = 1/rt. Solving lnt = e^(0.56) gives 0.0972, solving lnt = 1/0.56 gives 0.094 (rounded from 0.0939 if memory serves). Either is an acceptable approximation (I suspect they'll both get paid).

r ~= 0.56/(e^e^(0.56)) ~= 0.0972 was what most of those at my school who got (iii) out had.

Afaik there were people on 100% who didn't state rank last year. This year seems somewhat harder, though not by that much, so probably.

Your proof goes: qpt is a straight line => qpt collinear. You effectively assumed the result,

What vertically opposite angles were there? We don't know qpt are collinear so how did you do it? Curious to see your solution.

Thought the paper was reasonable overall, thought with the exception of part c q14 was very easy (if induction can't be too hard so as to not favour 4u students, why make it part of the tail end of the paper?). The circle geo in q13 was pretty good for 3u, though once you saw the construction it...

Note the (-1)^k

6 is b. Consider the graph of y=sinx; for any given vertical line y=a (|a|<=1) the solutions for intersections are: x = 2kpi + arcsin(a), x = (2k+1)pi - arcsin(a), or equivalently x = kpi + (-1)^k*arcsin(a). Divide this by 2 for 2x

These are right, but I assume you can (I did, anyway) also write it as the sum from 3 to 8 of the same expansion.

I had 4020 for similar reasons, think you miscounted. Obviously not d, since 4096 is the maximal value possible, and I had 76 from (4 ways to have them all in 1 room) + (6 ways to choose 1 outside, 4 rooms for the 5 left, 3 rooms for the other person). So 4096-76=4020 So I'm fairly sure q10 was...

Thought it was mostly fairly straightforward, but screwed up a bit - dropped 2 marks on 14 d (iv), and lost ~3 marks to sillies (calculator input wrong, and drew x^2 - y^2 = 4 because I missed the missed the <=). Certainly q16 was very easy this year - surely e4 cut off will be quite high?

Re: HSC 2013 4U Marathon Let the vertices of the inner quadrilateral on red be z1, z2, z3, z4 and let a = cis(pi/4)/sqrt(2). Then we can find p = z1 + a(z3-z1), etc; multiplying r-q by i and subtracting it from s-p gives 0 (if we note that a-ia = 1 and a+ia = i) so equal length and perpendicular.