HSC 2013 MX2 Marathon (archive) (6 Viewers)

dunjaaa · Jan 8, 2014

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.

dunjaaa · Jan 8, 2014

Re: HSC 2014 4U Marathon

why doesnt it edit wtf

dunjaaa · Jan 8, 2014

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 · Jan 8, 2014

Re: HSC 2014 4U Marathon

Wtf why doesnt it post the whole text?!

dunjaaa · Jan 8, 2014

Re: HSC 2014 4U Marathon

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

RealiseNothing · Jan 8, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
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 said:
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 · Jan 8, 2014

Re: HSC 2014 4U Marathon

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

seanieg89 · Jan 8, 2014

Re: HSC 2014 4U Marathon

$Prove that:\\ \\ $(1+m)^{-1/n}+(1+n)^{-1/m}>1$\\ \\ for positive integers $m,n>1.$$

dunjaaa · Jan 8, 2014

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 · Jan 8, 2014

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 · Jan 8, 2014

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.

asianese · Jan 8, 2014

Re: HSC 2014 4U Marathon

$Prove that the area enclosed by the ellipse $ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1 $ is $ \pi ab $ units.$

$$Consider the two hyperbolae $ xy=c^2 $ and $ \left(\dfrac{x}{a}-\dfrac{y}{b}\right)\left(\dfrac{x}{a}+\dfrac{y}{b}\right)=1 $ for positive constants $ a, b, c. $ Prove that the line going through the two points of intersection of the two hyperbolae always go through the origin. That is, prove that the line between the points of intersection is in the form $ kx + jy =0 $ for constants $ k $ and $ j.$

Sy123 · Jan 8, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
$Prove that:\\ \\ $(1+m)^{-1/n}+(1+n)^{-1/m}>1$\\ \\ for positive integers $m,n>1.$$

$\\ \sqrt[n]{m+1} < \frac{(1+1+...+1) + (m+1)}{n} = \frac{n-1 + m+1}{n} = \frac{m+n}{n} \ \ \ \ \ \fbox{1} \\ \\ \sqrt[m]{m+1} < \frac{m-1 + n+1}{m} = \frac{n+m}{m} \ \ \ \ \ \fbox{2} \\ \\ \frac{1}{\fbox{1}} + \frac{1}{\fbox{2}} \Rightarrow \ \ \frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} > \frac{n}{n+m} + \frac{n}{n+m} = 1 \\ \\ $There is no equality since not all terms in the use of AM-GM inequality are equal$$

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$\\ $i) Show that$ \ \ \ 1-x^n = (1-x)(1+x+\dots + x^{n-1}) \\ \\ $ii) Hence find the value of the sum$ \\ \\ \cos \theta + \cos 2\theta + \dots + \cos(n\theta) \\ \\ $in terms of$ \ \theta$

Sy123 · Jan 8, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
Given the limited information, I had a crack at it. This is what I manage to salvage View attachment 29614

Yea something like that

seanieg89 · Jan 8, 2014

Re: HSC 2014 4U Marathon

$For $a,b,c >1$, show that:\\ \\ $\frac{3}{1+abc}\leq \frac{1}{1+a^3}+\frac{1}{1+b^3}+\frac{1}{1+c^3}.$$

dunjaaa · Jan 9, 2014

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 · Jan 9, 2014

Re: HSC 2014 4U Marathon

Now try this

$\\ $find$ \ \ \ \cos^2 \theta + \cos^2 2\theta + \dots + \cos^2 n \theta$

dunjaaa · Jan 9, 2014

Re: HSC 2014 4U Marathon

Nice question yet again! Here's what I did

Screen shot 2014-01-09 at 3.18.17 AM.png

Sy123 · Jan 9, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
Nice question yet again! Here's what I did View attachment 29622

Well done, I don't know exactly what you did but the general idea was just doing double angle rule (which you did)

dunjaaa · Jan 10, 2014

Re: HSC 2014 4U Marathon

Had a go at making a question

Screen shot 2014-01-10 at 8.22.45 PM.png

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