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HSC 2016 MX2 Marathon (archive) (3 Viewers)

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porcupinetree

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Re: HSC 2016 4U Marathon

Would have been faster to directly use the sine area formula.

i.e.

(sin120)(QB)(QC)/2 = (sin60)(QB)(QD)/2 + (sin60)(QC)(QD)/2

This immediately re-arranges to the desired result by inspection.
Thought there was something like that right under my nose.
 

InteGrand

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Re: HSC 2016 4U Marathon

you didn't think of chasing it? You dont have a HSC to worry about, you can prove it however you like so long as it's within the syllabus.
Oftentimes once one has found a solution, they type it up without first searching (too hard) for other ones, because there's no guarantee that another one will be easily found and they want to get their solution up.
 

porcupinetree

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Re: HSC 2016 4U Marathon

you didn't think of chasing it? You dont have a HSC to worry about, you can prove it however you like so long as it's within the syllabus.
Once it's been proven, it's been proven

Ceebs finding a more elegant proof for the sake of it lmao
 

Paradoxica

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Re: HSC 2016 4U Marathon

Oftentimes once one has found a solution, they type it up without first searching (too hard) for other ones, because there's no guarantee that another one will be easily found and they want to get their solution up.
true... I'm no different from that sometimes...

meh. moving on.

 

StudiousStudent

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Re: HSC 2016 4U Marathon

I have great admiration for everyone posting in this thread.

I'm moving on to a non-foreign land.
 

wu345

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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).
thanks, but what about algebraically? can you please give me a hint or two? I started with by squaring both sides, then writing them as the products complex numbers and conjugates but im not sure if thats on the right track
 

Paradoxica

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Re: HSC 2016 4U Marathon

thanks, but what about algebraically? can you please give me a hint or two? I started with by squaring both sides, then writing them as the products complex numbers and conjugates but im not sure if thats on the right track
you need to work with the real and imaginary parts of the complex numbers so make substitutions. Then complete squares. The algebra will work out at the end.
 

InteGrand

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Re: HSC 2016 4U Marathon

thanks, but what about algebraically? can you please give me a hint or two? I started with by squaring both sides, then writing them as the products complex numbers and conjugates but im not sure if thats on the right track
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 you what the locus is, so you know what to work towards.)
 
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