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It'd be more useful if you gave us the exact ranking of your school last year so we can have a look at the performance trend over the years. But based off what you gave us, and assuming last year's performance is a good reflection of your school's general performance, I would agree with the...

Re: thoughts on independent 4u 2014? Were there any hard questions in the paper?

lmao why does eyeseeyou keep asking people how to state rank in maths lol. Judging by his abilities, shouldn't he be asking how to get at least a band 3?

Sounds different to what i was expecting, but it seems like legit progress so can you upload your solution? But what did you mean by T?

Ah, sorry about that, that was my mistake. All euclidean geometry is welcome and fine. No vectors though

Yeah dw, im also used to people who cant solve a problem and try to cover it up by making a load of excuses such as claiming to merely be "leaving a hole" when they are actually stuck but need to maintain their ego. Dont feel too bad about not being able to do it though, even some ruse and...

Lol, you lack understanding. The whole point of my last post wasnt because i specifically wanted a hsc proof for this question, i merely suggested that you present your "solution" in a clear and logical manner (presenting a solution using hsc tools was just one way i suggested in order to...

Lol, even now you still make claims without any proof (you are still yet to prove your claim), instead of just making vague references to spiral similarity To make yourself clear, I suggest you prove your claims via using hsc tools

I've interpreted your claim as: "Suppose we have two similar quadrilaterals, ABCD and EFGH. Call the midpoint of AE as X, and CG as Y. Also, label the midpoint of CD as P and FH as Q. Then, XYPQ is similar to both these quadrilaterals." If this is what you mean, there are counterexamples to this...

Spiral similarity isn't a theorem

As expected, on that page there is nothing to support your claim that if "Each of the corresponding vertices are the midpoints of the exterior vertices" then the figures are similar If you disagree, provide your full solution

What do you interpret spiral similarity as?

Yeah, there's a few steps which I thought were very difficult to see, so I'm trying to see whether other people could see (I personally could not crack this problem by myself, had to peek at the solutions)

Also please explain the part "Since the vertices of the inside quadrilateral are all exactly halfway between the corresponding vertices (p=0.5), the quadrilateral formed is similar to the two corresponding quadrilaterals." This has no clear justification behind it

Yes please do have a look again tomorrow. I believe that this problem will remain unsolved for quite some time though, (The solution is actually very tricky)

Unfortunately for you, the two exterior kites are not congruent, if by congruent you mean that they are identical. For some reason I believe you assumed that the lengths of GX and GY are equal Also, care to explain the part "Since the vertices of the inside quadrilateral are all exactly halfway...

Paradoxica, you are WRONG! Spiral similarity does not necessarily imply that EXFY is a kite, or if you meant something else by this statement you need to provide some form of justification.

Surely there must be someone capable enough on BoS to be able to do this question? No?

Re: HSC 2016 Complex Numbers Marathon The complex numbers method by considering the arg of (1+(x-y)i)(1+(x+y)i)(1+2x/(x^2-y^2-1)i) is more elegant, there are a few tricks to handle the expansion quickly Edit: Lol beaten by the above two people

No, is there some detail of the question that's confusing you?