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HSC 2013 MX2 Marathon (archive) (1 Viewer)

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dunjaaa

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Re: HSC 2014 4U Marathon

I don't know how to latex but, for the second question, is the area overlapping = area of square - area of triangle = (√2)^2 - 1/2(√3)(3/2) = 2 - (3√3)/4units. In regards to the first question, I have no idea where to start but I was thinking if we consider a complex number z such that |z|=1 and arg(z)=θ, 0<θ<pi/2. The locus formed from arg(z+1) - arg(z-1) = A is a semicircle and then showing that A (i.e. angle at circumference) worth pi/2 will suffice a valid proof.
 
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dunjaaa

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Re: HSC 2014 4U Marathon

This was what I was intending to post "I don't know how to latex but, for the second question, is the area overlapping = area of square - area of triangle = (√2)^2 - 1/2(√3)(3/2) = 2 - (3√3)/4units. In regards to the first question, I have no idea where to start but I was thinking if we consider a complex number z such that |z|=1 and arg(z)=θ, 0<θ<pi/2. The locus formed from arg(z+1) - arg(z-1) = A is a semicircle and then showing that A (i.e. angle at circumference) worth pi/2 will suffice a valid proof"
 

dunjaaa

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Re: HSC 2014 4U Marathon

Wtf why doesnt it post the whole text?!
 

RealiseNothing

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Re: HSC 2014 4U Marathon

This was what I was intending to post "I don't know how to latex but, for the second question, is the area overlapping = area of square - area of triangle = (√2)^2 - 1/2(√3)(3/2) = 2 - (3√3)/4units. In regards to the first question, I have no idea where to start but I was thinking if we consider a complex number z such that |z|=1 and arg(z)=θ, 0 < θ< pi/2. The locus formed from arg(z+1) - arg(z-1) = A is a semicircle and then showing that A (i.e. angle at circumference) worth pi/2 will suffice a valid proof"
Screen shot 2014-01-08 at 2.32.53 AM.png download - 2shared what i intended on posting be4 ^ not sure if chat bugged cause i can only see parts of my posts and cant edit nor upload
Thank me later.
 

dunjaaa

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Re: HSC 2014 4U Marathon

oh idk but for some reason forums bugged on my screen when i posted
 

dunjaaa

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Re: HSC 2014 4U Marathon

Given the limited information, I had a crack at it. This is what I manage to salvage Screen shot 2014-01-08 at 3.43.22 PM.png
 

dunjaaa

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Re: HSC 2014 4U Marathon

The roots all have modulus 1 and by plotting them on a unit circle forms a rectangle, width 1 and length √3. Area spanned by roots is √3 x 1 = √3 units.
 

dunjaaa

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Re: HSC 2014 4U Marathon

Screen shot 2014-01-08 at 4.20.38 PM.png and for the area enclosed by the 5th, 6th, ... roots of any complex number, does the area approach the area of the circle dependent on what the modulus is or does that only happen to complex numbers with modulus 1, only generalisation I've come up with.
 

dunjaaa

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Re: HSC 2014 4U Marathon

Very nice question! Screen shot 2014-01-09 at 12.58.53 AM.png, part (iii) should be proving the result by induction haha
 

Sy123

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Re: HSC 2014 4U Marathon

Now try this

 
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