HSC 2012 MX2 Marathon (archive) (1 Viewer)

IamBread · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

$\alpha $ and $ \beta $ are the roots of $ z^2 - 2z +2 = 0. \\\\ $If $ \cot\theta = x + 1, $ prove that$ \\\\ \frac{(x+\alpha)^n - (x + \beta)^n}{\alpha - \beta} = \frac{\sin n\theta}{(\sin\theta)^n}$

SpiralFlex · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

Lol Patel.

SpiralFlex · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

Let the roots be $\alpha, \beta$

$\cot \theta = x+1$

$x=\cot \theta - 1$

We need to prove,

$\frac{(\cot \theta - 1+\alpha)^n - (\cot \theta - 1+\beta)^}{\alpha - \beta}=\frac{\sin (n \theta)}{\sin^n \theta}$

By solving the equation,

$\alpha = 1+i, \beta =1-i$

$\therefore LHS=\frac{(\cot \theta - 1+1+i)^n - (\cot \theta - 1+1-i)^n}{\alpha - \beta}$

$=\frac{(\cot \theta +i)^n - (\cot \theta -i)^n}{2i}$

$=\frac{cis (n \theta)-cis(-n\theta)}{2i\sin (n\theta)}$

$=\frac{\sin (n \theta)}{\sin^n \theta}$

Drongoski · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$\alpha $ and $ \beta $ are the roots of $ z^2 - 2z +2 = 0. \\\\ $If $ \cot\theta = x + 1, $ prove that$ \\\\ \frac{(x+\alpha)^n - (x + \beta)^n}{\alpha - \beta} = \frac{\sin n\theta}{(\sin\theta)^n}$

$\therefore z^2-2z+1 = -1 \implies (z-1)^2 = -1 \implies z = 1 \pm i $ $ \implies \alpha = 1+i $ & $ \beta = i-i$

$\therefore x+\alpha = cot \theta -1 + 1+ i = cot \theta +i = \frac {cos \theta}{sin \theta} + i = \frac {cos \theta + isin \theta}{sin \theta} = \frac {cis \theta} {sin \theta}$

Similarly $x+\beta = \frac {cis(- \theta)}{sin \theta}$

$\therefore \frac {(x+\alpha)^n - (x+\beta)^n}{\alpha-\beta} = \frac {\frac {(cis \theta)^n}{sin^n \theta} - \frac {(cis(- \theta))^n}{(sin \theta)^n}}{2i} = \frac {cis (n\theta) - cis (-n\theta)}{2i(sin \theta)^n}$

$= \frac {cos(n\theta)+isin(n\theta) - cos(-n\theta)-isin(-n\theta)}{2i(sin \theta)^n}$

$= \frac{cos(n\theta)+isin(n\theta) - cos(n\theta)+isin(n\theta)}{ditto} = \frac {2isin(n\theta)}{2i(sin \theta)^n} = RHS$

IamBread · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

$$The polynomial $ 2x^3 - x^2 + 5 = 0 $ has roots $ \alpha, \beta, \gamma. \\\\ $Find the polynomial with integer coefficients whose roots are $ \alpha^3, \beta^3, \gamma^3$

nightweaver066 · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$The polynomial $ 2x^3 - x^2 + 5 = 0 $ has roots $ \alpha, \beta, \gamma. \\\\ $Find the polynomial with integer coefficients whose roots are $ \alpha^3, \beta^3, \gamma^3$

$2x^3 - x^2 + 5 = 0$

$$Related roots: y = x^3$

$x = ^3\sqrt{y}$

$\therefore 2y - y^{\frac{2}{3}} + 5 = 0$

$(2y + 5)^3 = y^2$

$8y^3 + 60y^2 + 150y + 125 = y^2$

$8y^3 + 59y^2 + 150y + 125 = 0$

gurmies · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

Prove that:

$3|z-1|^2 = |z+1|^2$ if and only if

$|z-2|^2 = 3$

IamBread · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

gurmies said:
Prove that:

$3|z-1|^2 = |z+1|^2$ if and only if

$|z-2|^2 = 3$

$$Let $ z = x+ iy, \\\\ 3|z-1|^2 = |z+1|^2$

$3(x-1)^2 + 3y^2 = (x+1)^2 + y^2 \\\\ 3x^2-6x+3+3y^2=x^2+2x+1+y^2 \\\\ 2x^2 -8x +2 +2y^2 = 0 \\\\ 2(x-2)^2+2y^2=6 \\\\ (x-2)^2+y^2 = 3 \\\\ |x +iy-2|^2=3 \\\\ $As $ x+iy =z, |z-2|^2=3$

$New question... $ p(x) = 2x^3 -19x^2+112x+d $ has root $5+6i.\\\\ $Find the other roots and the value of d.$

nightweaver066 · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$Let $ z = x+ iy, \\\\ 3|z-1|^2 = |z+1|^2$

$3(x-1)^2 + 3y^2 = (x+1)^2 + y^2 \\\\ 3x^2-6x+3+3y^2=x^2+2x+1+y^2 \\\\ 2x^2 -8x +2 +2y^2 = 0 \\\\ 2(x-2)^2+2y^2=6 \\\\ (x-2)^2+y^2 = 3 \\\\ |x +iy-2|^2=3 \\\\ $As $ x+iy =z, |z-2|^2=3$

$New question... $ p(x) = 2x^3 -19x^2+112x+d $ has root $5+6i.\\\\ $Find the other roots and the value of d.$

$As all the coefficients of p(x) are real, by conjugate root theorem, x = 5 - 6i is also a root$

$p(x) = (x - 5 + 6i)(x - 5 - 6i)(ax + b)$

$= (x^2 - 10x + 61)(ax + b) \equiv 2x^3 - 19x^2 + 112x + d$

$$Comparing the terms of x^3, $ a = 2$

$$Comparing the terms of x^2, $ b - 20 = - 19$

$b = 1$

$$Comparing the constants, $ d = 61$

$\therefore p(x) = (x^2 - 10x + 61)(2x + 1)$

$$The roots are x = 5 - 6i, 5 + 6i or $ -\frac{1}{2} $ and d = 61$

gurmies · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
$As all the coefficients of p(x) are real, by conjugate root theorem, x = 5 - 6i is also a root$

$p(x) = (x - 5 + 6i)(x - 5 - 6i)(ax + b)$

$= (x^2 - 10x + 61)(ax + b) \equiv 2x^3 - 19x^2 + 112x + d$

$$Comparing the terms of x^3, $ a = 2$

$$Comparing the terms of x^2, $ b - 20 = - 19$

$b = 1$

$$Comparing the constants, $ d = 61$

$\therefore p(x) = (x^2 - 10x + 61)(2x + 1)$

$$The roots are x = 5 - 6i, 5 + 6i or $ -\frac{1}{2} $ and d = 61$

Possibly a simpler way to do this would be to use the sum of roots to find the final root, and the product of roots to find d.

IamBread · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

gurmies said:
Possibly a simpler way to do this would be to use the sum of roots to find the final root, and the product of roots to find d.

I used sum of roots to find the final root, then subbed in x=-1/2 to find d. But either way works.

$P(a\cos\theta,b\sin\theta) $ and $ Q(a\cos\phi,b\sin\phi) $ lie on the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \\\\ $Find the equation of line $PQ$

mnmaa · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

gurmies · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

mnmaa said:

I suspect that it's sufficient to solve Z=Z^3 and discard -1 as a solution. My reasoning is that there clearly can't be an imaginary part present in Z. With this in mind, we can proceed noting that |Z| = Z, Z > 0 and |Z| = -Z, Z < 0. My method yields Z = 0 or/ 1. Could be wrong here though.

cutemouse · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

gurmies said:
Prove that:

$3|z-1|^2 = |z+1|^2$ if and only if

$|z-2|^2 = 3$

I asked that question in 2009, and you answered it oO.

cutemouse · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$Let $ z = x+ iy, \\\\ 3|z-1|^2 = |z+1|^2$

$3(x-1)^2 + 3y^2 = (x+1)^2 + y^2 \\\\ 3x^2-6x+3+3y^2=x^2+2x+1+y^2 \\\\ 2x^2 -8x +2 +2y^2 = 0 \\\\ 2(x-2)^2+2y^2=6 \\\\ (x-2)^2+y^2 = 3 \\\\ |x +iy-2|^2=3 \\\\ $As $ x+iy =z, |z-2|^2=3$

From memory, I think there's an easier way to do this question. Ie. using z z (bar) = |z|^2

IamBread · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

mnmaa said:

$z = 0, 1$

math man · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

cause this is a proof question i started with LHS:
$\!\!\!\!\!\!\!\!\!\!\!\!\!\! LHS= 3|z-1|^{2} \\ = 3(x-1)^{2} + 3y^{2} \\ = 3(x^{2}-2x+1) + 3y^{2} \\ = 3((x-2)^{2} +2x -3) +3y^{2} \\ = (x^{2}-4x+4 +6x -9)+y^{2} +2((x-2)^{2}+y^{2}) \\ = x^{2}+2x -5 +y^{2} +6 \\ =(x+1)^{2} + y^{2} \\ =|z+1|^{2}$

gurmies · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

$|z-2|^{2}=3 \\\\ (z-2)\overline{(z-2)}=3 \\\\ (z-2)(\overline{z}-2)=3 \\\\ z\overline{z}-2z-2\overline{z}+4=3 \\\\ z\overline{z}-2z-2\overline{z}+1=0 \\\\ 2z\overline{z}-4z-4\overline{z}+2=0 \\\\ 3z\overline{z}-3z-3\overline{z}+3=z\overline{z}+z+\overline{z}+1 \\\\ 3(z\overline{z}-z-\overline{z}+1)=z\overline{z}+z+\overline{z}+1 \\\\ 3(z-1)(\overline{z}-1)=(z+1)(\overline{z}+1) \\\\ 3(z-1)\overline{(z-1)}=(z+1)\overline{(z+1)} \\\\ 3|z-1|^{2}=|z+1|^{2}$

cutemouse · Feb 14, 2012

Re: 2012 HSC MX2 Marathon

gurmies said:
$|z-2|^{2}=3 \\\\ (z-2)\overline{(z-2)}=3 \\\\ (z-2)(\overline{z}-2)=3 \\\\ z\overline{z}-2z-2\overline{z}+4=3 \\\\ z\overline{z}-2z-2\overline{z}+1=0 \\\\ 2z\overline{z}-4z-4\overline{z}+2=0 \\\\ 3z\overline{z}-3z-3\overline{z}+3=z\overline{z}+z+\overline{z}+1 \\\\ 3(z\overline{z}-z-\overline{z}+1)=z\overline{z}+z+\overline{z}+1 \\\\ 3(z-1)(\overline{z}-1)=(z+1)(\overline{z}+1) \\\\ 3(z-1)\overline{(z-1)}=(z+1)\overline{(z+1)} \\\\ 3|z-1|^{2}=|z+1|^{2}$

Pro (Y).

largarithmic · Feb 14, 2012

Re: 2012 HSC MX2 Marathon

cutemouse said:
From memory, I think there's an easier way to do this question. Ie. using z z (bar) = |z|^2

It's also true because of

http://en.wikipedia.org/wiki/Circles_of_Apollonius

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