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Not necessarily undefined. It basically means that there's no value you could define for the function there that would make it continuous there (as opposed to a 'removable discontinuity', which is one where you can define a function value appropriately and make the function continuous there)...

Re: HSC 2016 MX2 Combinatorics Marathon Monty knows what is behind each door, right?

I just don't think it'll be the spectacle people are hoping for. But OK sure, I'll play sometime hopefully (can't play right now though).

Lol, in terms of lightning, usually between 30 seconds and 2 min. In terms of "longest", atm I'd probably play 10-15 min. In terms of variants, I think the only one atm I like is Crazyhouse. But I think by playing these and super short lightning games a lot, it deteriorates your regular chess...

I'm actually fairly bad (and rusty). Normally when I play nowadays I just play fun variant games or lightning haha (and haven't even done that in some time). I think there are some other pretty good chess players on BOS though.

$\noindent Well if we wanted to do it like this, we'd get $a+b = \left(p+r\right)+\left(q+s\right)i$, so $|a+b| = \sqrt{\left(p+r\right)^2 + \left(q+s\right)^2}= \sqrt{p^2 + q^2 + r^2 + s^2 + 2pr + 2qs}$. So the triangle inequality would give us $\sqrt{p^2 + q^2}+ \sqrt{r^2 + s^2} \geq \sqrt{p^2...

$\noindent Doing this shows that $\left | \sqrt{a} \right | + \left |\sqrt{b} \right | \geq \left | \sqrt{a} +\sqrt{b}\right |$, which is just equivalent to an equality as both sides are equal, so this doesn't prove the claim.$ $\noindent The claim is easily proved though by showing $LHS^2 \geq...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread It says that he goes down the other side of the hill, so he doesn't come back to where he starts; his final displacement should be 2 m.

If they're "15 years old" now, they were born in either 2000 or 2001.

Re: HSC 2016 Complex Numbers Marathon To save people having to go search up the paper, it can be found in the Dropbox link in this post: http://community.boredofstudies.org/1094/general-discussion-2015-hsc/343889/extension-2-bos-trial-2015-results-documents.html#post7051009 .

Re: HSC 2016 Complex Numbers Marathon It can be proved using HSC geometry (iirc this was done in the 2015 BOS 4U Trial).

Re: HSC 2016 4U Marathon It's essentially just tedious algebra I think. You need to show two things: 1) any point satisfying that relationship between the moduli lies on the given circle, and 2) any point on that circle satisfies the relationship between the moduli. (Luckily the question tells...

I think the topic of discussion at the time was people doing the HSC younger than usual.

Re: HSC 2016 Complex Numbers Marathon Also note that the circle they ask us to prove is the locus isn't just any old circle, it's actually the circle that has as diameter the points that internally and externally divide the line segment joining the points z1 and z2 in the complex plane in ratio...

Re: HSC 2016 Complex Numbers Marathon If I recall correctly, this result was proved geometrically in one of the Q's in the 2015 BOS HSC 4U trial (the proof was based on the angle bisector theorem (https://en.wikipedia.org/wiki/Angle_bisector_theorem) which the paper got you to prove as an...

Re: HSC 2016 4U Marathon If I recall correctly, this result was proved geometrically in one of the Q's in the 2015 BOS HSC 4U trial (the proof was based on the angle bisector theorem (https://en.wikipedia.org/wiki/Angle_bisector_theorem) which the paper got you to prove as an earlier part).

True

Re: HSC 2016 Complex Numbers Marathon I think it's a sort of BOS tradition (the old marathons were more popular I think, maybe more high-schoolers posted maths things on BOS then).

Re: HSC 2016 Complex Numbers Marathon I think the main reason the Combinatorics Marathon is rarely active is that most people hate/dislike that topic.

Well it's a wave moving at a constant speed v (whose value we find at the end). So if it travels one-eighth of the wavelength in 0.35 seconds, it travels the full wavelength in 8 times this, i.e. 8*0.35 s. This value is then by definition the period T.