Can VCE'ers compete with HSC Maths?

[pi]

What did you do here?

Looked a bit fluffy

Probably should have used arg(ab) = arg(a) + arg(b), hence arg((1+i)((1+2i)(1+3i)) = arg(-10) = pi

[cosine]

Looked a bit fluffy

Probably should have used arg(ab) = arg(a) + arg(b), hence arg((1+i)((1+2i)(1+3i)) = arg(-10) = pi

I simply just added the arguments together, why do you multiply them may I ask?

[keltingmeith]

I simply just added the arguments together

You did - you didn't actually show that they were pi, though. You showed that computationally, they would approximate pi, but not that they were exactly pi.

By doing as pi has done, you show that they are exactly pi.

[pi]

By doing as pi has done, you show that they are exactly pi.

I had to read this 5 times LOL

[keltingmeith]

I had to read this 5 times LOL

#metaaf

[nerdgasm]

Mm, I'd just like to say that judging by some of the content posts in the HSC maths threads on this forum, HSC maths definitely seems to go more into classical or Euclidean geometry than VCE maths does. I hadn't come across things like the 'alternate segment theorem' until I saw it being used in a maths competition paper some time ago. We don't really go into induction at the VCE level either.

As someone who enjoys recreational maths problems (of pretty much all levels), I'll leave the following ones here. Hopefully, there's something to challenge everybody:

Real numbers a, b and c satisfy the following two equations:
a + b + c = 0
a^2 + b^2 + c^2 = 1.
Prove that a^4 + b^4 +c^4 = 1/2. (This question I got from a guy's Youtube comment).

Consider the set {1, 2, 3, ... n}, where n is a positive integer. Your task is to place each of the numbers into one of two sets, such that the sum of the numbers in one set is equal to the product of the numbers in the other. You are not allowed to leave any numbers unallocated. For which values of n is this possible? (AMOC Senior Contest 2014, Problem 2)

A certain robot on the #atarnotes IRC channel randomly chooses one word in every sentence and replaces it with 'butt'. Thus, in a sentence of n words, every word has a 1/n chance of being replaced with 'butt' (even the word 'butt' can be replaced with 'butt', for no net effect). I take a sentence of n words (none of which are originally 'butt'), and run it through this robot (so guaranteed, on this first run, one word will be replaced with 'butt'). I then run this new sentence through the robot. I keep doing this until all of the words in the sentence are 'butt'. For a given sentence of n words (none of which are originally 'butt'), what is the expected number of times I will need to run the sentence through the robot to get an all-'butt' sentence? (Unashamedly inspired by some late-night IRC chats)

Evaluate 1/2 + 1/6 + 1/12 + 1/20 + ... + 1/n(n+1) + ... (the infinite series).

I have an isosceles, right angled triangle. In it, I place a square such that one side of the square is exactly the middle third of the hypotenuse. The other two corners of the square touch the other sides of the triangle. Is such a shape possible?

I have an infinitely thin (but not infinitely long) straight stick, and choose two distinct points along it. I cut the stick at those points, to form three pieces. What is the probability that I can form a triangle with these pieces?