HSC 2012 MX2 Marathon (archive) (2 Viewers)

nightweaver066 · Mar 6, 2012

Re: 2012 HSC MX2 Marathon

zeebobDD said:
View attachment 24537

$C: x^2 + y^2 = R^2$

$L: ax + by + c = 0$

$i) For L to be tangent to C, the perpendicular distance from the centre of the circle (0,0) to L must equal to R$

$\therefore |\frac{c}{\sqrt{a^2 + b^2}}| = R$

$Squaring both sides$

$\frac{c^2}{a^2 + b^2} = R^2$

$\therefore R^2(a^2 + b^2) = c^2$

$$ii) $ |\frac{2a + c}{\sqrt{a^2 + b^2}}| + |\frac{-2a + c}{\sqrt{a^2 + b^2}}| = 6$

$Squaring both sides$

$\frac{4a^2 + 4ac + c^2}{a^2 + b^2} + \frac{2}{a^2 + b^2}|c^2 - 4a^2| + \frac{4a^2 - 4ac + c^2}{a^2 + b^2} = 36$

$\frac{8a^2 + 2c^2}{a^2 + b^2} + \frac{2}{a^2 + b^2}|c^2 - 4a^2| = 36$

$$As $ c > 2a => c^2 > 4a^2, \; c^2 - 4a^2 > 0$

$\frac{8a^2 + 2c^2}{a^2 + b^2} + \frac{2c^2 - 8a^2}{a^2 + b^2} = 36$

$\frac{4c^2}{a^2 + b^2} = 36$

$c^2 = 9(a^2 + b^2)$

$$Similar to part (i), L must be tangent to the circle $ x^2 + y^2 = 9$

SpiralFlex · Mar 6, 2012

Re: 2012 HSC MX2 Marathon

zeebobDD said:
View attachment 24537

Ewww maffs.

By perpendicular distance formula,

$\frac{|c|}{\sqrt{a^2+b^2}}=R$

$R^2(a^2+b^2)=c^2$

$\frac{|c-2a|}{\sqrt{a^2+b^2}}+\frac{|c+2a|}{\sqrt{a^2+b^2}}=6$

Squaring all sides,

$\frac{(c-2a)^2+(c+2a)^2+2(c-2a)(c+2a)}{\sqrt{a^2+b^2}}=36$

$=\frac{4c^2}{a^2+b^2}=36$

$9(a^2+b^2)=c^2$

As proven before, the line is a tangent if the condition holds true. Clearly the form is of equivalence to, what we have shown before.

Your equation of your circle is,

$x^2+y^2=9$

IamBread · Mar 8, 2012

Re: 2012 HSC MX2 Marathon

Looks like we need more questions!

$z=1+i\sqrt{3}, w =1+i \\\\ $Show $ cos\frac{7\pi}{12} = \frac{1-\sqrt{3}}{2\sqrt{2}}$

Aesytic · Mar 8, 2012

Re: 2012 HSC MX2 Marathon

the complex number z and w in mod-arg form are 2cis(pi/3) and root2 cis(pi/4) respectively
.'. zw = 2root2 cis(pi/3 + pi/4)
=2root2cis(7pi/12)
= 2root2*cos(7pi/12) + i2root2sin(7pi/12)

also, zw = (1+iroot3)(1+i)
= 1 - root3 + i(root3 + 1)

equating real parts,
2root2*cos(7pi/12) = 1 - root3
.'.cos(7pi/12) = [1-root3]/[2root2]

zeebobDD · Mar 8, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Looks like we need more questions!

$z=1+i\sqrt{3}, w =1+i \\\\ $Show $ cos\frac{7\pi}{12} = \frac{1-\sqrt{3}}{2\sqrt{2}}$

LOL this came in our maths exam today, except it was show sin5pi/12:L

rolpsy · Mar 8, 2012

Re: 2012 HSC MX2 Marathon

rolpsy said:
$\\\textup{Let } z = r\, \mathrm{cis}(\theta).\\\textup{Show that } |z+1|\times|z-1| = 1 \textup{ implies } r^2 = 2\cos(2\theta).$

Nooblet94 · Mar 8, 2012

Re: 2012 HSC MX2 Marathon

rolpsy said:

You sure that question's right?

Drongoski · Mar 8, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
You sure that question's right?

Drongoski · Mar 9, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Looks like we need more questions!

$z=1+i\sqrt{3}, w =1+i \\\\ $Show $ cos\frac{7\pi}{12} = \frac{1-\sqrt{3}}{2\sqrt{2}}$

$cos\frac {7\pi}{12} = cos (\frac {\pi}{3} + \frac {\pi}{4}) = cos \frac{\pi}{3} cos \frac {\pi}{4} - sin \frac{\pi}{3} sin \frac {\pi}{4} = \frac {1}{2} \cdot \frac {1}{\sqrt 2} - \frac {\sqrt 3}{2} \cdot \frac {1}{\sqrt 2} = \frac {1-\sqrt 3}{2\sqrt 2}$

rolpsy · Mar 9, 2012

Re: 2012 HSC MX2 Marathon

Drongoski said:
rolpsy's question

$|z+1| \times |z-1| =1 \implies |z^2-1| = 1 \implies |z^2-1|^2 = 1$

$\implies |r^2 cis2 \theta -1|^2 = 1\implies (r^2 cos2\theta -1)^2 +( r^2sin 2\theta)^2 =1$

$\implies r^4(sin^2 2\theta + cos^2 2 \theta) -2r^2 cos 2 \theta = 0 \implies r^2 (r^2 - 2cos 2\theta ) = 0$

$\implies r^2 = 2cos 2 \theta$

correcto

$\\\textup{By considering the area under the curve,}\\\\y=\frac{1}{1+x^2},\\\\\textup{between 0 and 1 as the limit of a sum of rectangles,}\\\textup{evalutate,}\\\\\lim_{n\to \infty} \sum_{k=1}^{n} \frac{n}{n^2 + k^2}.$

seanieg89 · Mar 9, 2012

Re: 2012 HSC MX2 Marathon

$$Let $f(x)=\frac{1}{1+x^2}.$ From the definition of the Riemann integral:\\$\pi/4=\int_0^1\frac{dx}{1+x^2}=\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n f(k/n)\cdot n^{-1}\right)=\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n\frac{n}{n^2+k^2}\right).$

mirakon · Mar 9, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
$$Let $f(x)=\frac{1}{1+x^2}.$ From the definition of the Riemann integral:\\$\pi/4=\int_0^1\frac{dx}{1+x^2}=\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n f(k/n)\cdot n^{-1}\right)=\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n\frac{n}{n^2+k^2}\right).$

is the riemann hypothesis thing actually syllabus?

largarithmic · Mar 9, 2012

Re: 2012 HSC MX2 Marathon

mirakon said:
is the riemann hypothesis thing actually syllabus?

Uh thats not the riemann hypothesis...

Also you're usman, right?

seanieg89 · Mar 9, 2012

Re: 2012 HSC MX2 Marathon

I would say that the most important unsolved problem in mathematics is beyond syllabus

.

Carrotsticks · Mar 9, 2012

Re: 2012 HSC MX2 Marathon

haha very cute

IamBread · Mar 13, 2012

Re: 2012 HSC MX2 Marathon

$$Show $ |Re(z \overline{w})| \le|z||w|$

nightweaver066 · Mar 13, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$Show $ |Re(z \overline{w})| \le|z||w|$

$Let z = rcis\theta, w = Rcis\phi$

$|Re(Rrcis(\theta + \phi)| = Rrcos(\theta + \phi)$

$Rrcos(\theta + \phi) = Rr $ when \theta + \phi = $ \frac{\pi}{2}$

$cos(\theta + \phi) = 1 $ which is its maximum only when \theta + \phi = $ \pm \frac{n\pi}{2}, n \in \mathbb{Z}$

$\therefore Rrcos(\theta + \phi) \leq Rr$

$i.e. |Re(z\overline{w})| \leq |z||w|$

Aesytic · Mar 13, 2012

Re: 2012 HSC MX2 Marathon

could it also be done this way:
if Q was the complex number z*conj(w), and OP was |Re(z*conj(w))|,
angle OPQ would be a right angle, and therefore OQ > OP because OQ is the hypotenuse of triangle OPQ.
since OQ is also the modulus of z*conj(w), and |conj(w)| = |w|,
.'. |z*conj(w)| = |zw|
=|z||w|
.'. OQ = |z||w| and OP = |Re(z*conj(w)|
.'. |z||w| > |Re(z*conj(w))|

lolcakes52 · Mar 13, 2012

Re: 2012 HSC MX2 Marathon

Since we are on important mathematical problems, prove P is not NP.

Carrotsticks · Mar 13, 2012

Re: 2012 HSC MX2 Marathon

lolcakes52 said:
Since we are on important mathematical problems, prove P is not NP.

Very funny haha.

For those who don't know what's going on: http://en.wikipedia.org/wiki/P_versus_NP_problem

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