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HSC 2016 MX2 Marathon ADVANCED (archive) (2 Viewers)

Thread starter Sy123
Start date Oct 24, 2015

Status: Not open for further replies.

L

lita1000

Member

#101

Re: HSC 2016 4U Marathon - Advanced Level

any interesting questions?

Sy123

This too shall pass

#102

Re: HSC 2016 4U Marathon - Advanced Level

$\\ $An integer polynomial is a polynomial where all the co-efficients are integers, let$ \ p(x) \ $be such an integer polynomial$ \\ \\ $i) Show that if$ \ p + \sqrt{q} \ $is a root of$ \ p(x) \ $then$ \ p - \sqrt{q} \ $is a root, where$ \ p \ $is rational and$ \ q \ $is rational, but not the square of a rational$ \\ \\ $ii) Find other roots of$ \ p(x) \ $given that$ \ \alpha + \sqrt[3]{\beta} \ $is a root of$ \ p(x) \ $where$ \ \alpha, \beta \ $are rational, and$ \ \beta \ $is not the cube of a rational$$

D

Drsoccerball

Well-Known Member

#103

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
$\\ $An integer polynomial is a polynomial where all the co-efficients are integers, let$ \ p(x) \ $be such an integer polynomial$ \\ \\ $i) Show that if$ \ p + \sqrt{q} \ $is a root of$ \ p(x) \ $then$ \ p - \sqrt{q} \ $is a root, where$ \ p \ $is rational and$ \ q \ $is rational, but not the square of a rational$ \\ \\ $ii) Find other roots of$ \ p(x) \ $given that$ \ \alpha + \sqrt[3]{\beta} \ $is a root of$ \ p(x) \ $where$ \ \alpha, \beta \ $are rational, and$ \ \beta \ $is not the cube of a rational$$

$i) x= p+\sqrt{q}$

$x^2-2px+p^2-q=0$

$x=p-\sqrt{q}$

$x^2-2px+p^2-q=0$

$$ Since Both equations are equal if $ p +\sqrt{q} $ is a root, then so is $ p=\sqrt{q}$

$ii)x=\alpha +\beta^{\frac{1}{3}}$

$x^3 -3x^2 \alpha +3x \alpha ^2 -\alpha ^3 -\beta =0$

$$ Factorise $ (x-\alpha -\beta ^{\frac{1}{3}})(x^2+(\beta ^{\frac{1}{3}}-2\alpha)x +\frac{\alpha ^3 +\beta}{\alpha + \beta ^{\frac{1}{3}}})$

$$ Use the quadratic equation to find a root and the root is in the form of (i)$

Paradoxica

-insert title here-

#104

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
$i) x= p+\sqrt{q}$

$x^2-2px+p^2-q=0$

$x=p-\sqrt{q}$

$x^2-2px+p^2-q=0$

$$ Since Both equations are equal if $ p +\sqrt{q} $ is a root, then so is $ p=\sqrt{q}$

$ii)x=\alpha +\beta^{\frac{1}{3}}$

$x^3 -3x^2 \alpha +3x \alpha ^2 -\alpha ^3 -\beta =0$

$$ Factorise $ (x-\alpha -\beta ^{\frac{1}{3}})(x^2+(\beta ^{\frac{1}{3}}-2\alpha)x +\frac{\alpha ^3 +\beta}{\alpha + \beta ^{\frac{1}{3}}})$

$$ Use the quadratic equation to find a root and the root is in the form of (i)$

Shouldn't you also assume that the quadratic is one of the factors of an arbitrary polynomial?

D

Drsoccerball

Well-Known Member

#105

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
Shouldn't you also assume that the quadratic is one of the factors of an arbitrary polynomial?

I guess so

Sy123

This too shall pass

#106

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
$i) x= p+\sqrt{q}$

$x^2-2px+p^2-q=0$

$x=p-\sqrt{q}$

$x^2-2px+p^2-q=0$

$$ Since Both equations are equal if $ p +\sqrt{q} $ is a root, then so is $ p=\sqrt{q}$

$ii)x=\alpha +\beta^{\frac{1}{3}}$

$x^3 -3x^2 \alpha +3x \alpha ^2 -\alpha ^3 -\beta =0$

$$ Factorise $ (x-\alpha -\beta ^{\frac{1}{3}})(x^2+(\beta ^{\frac{1}{3}}-2\alpha)x +\frac{\alpha ^3 +\beta}{\alpha + \beta ^{\frac{1}{3}}})$

$$ Use the quadratic equation to find a root and the root is in the form of (i)$

What do you mean by "both equations are equal"?
How does this allow us to prove that P(p + sqrt(q)) = 0 -> P(p - sqrt(q)) = 0? (for any integer polynomials P)

D

Drsoccerball

Well-Known Member

#107

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
What do you mean by "both equations are equal"?
How does this allow us to prove that P(p + sqrt(q)) = 0 -> P(p - sqrt(q)) = 0? (for any integer polynomials P)

Well I found the equations that satisfied each root separately and they happened to be equal and thus have the equation has both roots given.

Sy123

This too shall pass

#108

Re: HSC 2016 4U Marathon - Advanced Level

$\\ $On a perfect circular billiard table, the ball is hit at angle$ \ \theta \ $radians away from the line joining the starting point and the centre$$

$\\ $Find the average distance of the ball to the centre in terms of$ \ \theta$

Sy123

This too shall pass

#109

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
Well I found the equations that satisfied each root separately and they happened to be equal and thus have the equation has both roots given.

Your argument needs to be clearer for it to be a mathematical proof

I'm still not sure what you are trying to say here, the equations $(x-1)(x- (p+\sqrt(q)) = 0$ and $x - p - \sqrt{q} = 0$ both have a root in common, but why does tha tell us anything special about those equations?

Sy123

This too shall pass

#110

Re: HSC 2016 4U Marathon - Advanced Level

$\\ $An$ \ n \ $sided regular polygon is such that no side is vertical in gradient$ \\ $Prove that if$ \ m_k \ $is the gradient of the$ \ k^{\text{th}} \ $side, then:$ \\\\ m_1 m_2 + m_2 m_3 + \dots + m_{n-1}m_n + m_n m_1 = - n$

I

InteGrand

Well-Known Member

#111

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
Your argument needs to be clearer for it to be a mathematical proof

I'm still not sure what you are trying to say here, the equations $(x-1)(x- (p+\sqrt(q)) = 0$ and $x - p - \sqrt{q} = 0$ both have a root in common, but why does tha tell us anything special about those equations?

$I think he's pretty much saying that if $p+\sqrt{q}$ is a root of the polynomial he gave, so is $p-\sqrt{q}$.$

Sy123

This too shall pass

#112

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$I think he's pretty much saying that if $p+\sqrt{q}$ is a root of the polynomial he gave, so is $p-\sqrt{q}$.$

Right, then the next step is to show that any integer polynomial P has to have the same property

Paradoxica

-insert title here-

#113

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
$\\ $On a perfect circular billiard table, the ball is hit at angle$ \ \theta \ $radians away from the line joining the starting point and the centre$$

$\\ $Find the average distance of the ball to the centre in terms of$ \ \theta$

This question requires clarification before I can work on it.

Sy123

This too shall pass

#114

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
This question requires clarification before I can work on it.

O

outlier1

New Member

#115

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:

I got cos(theta) over 2

Paradoxica

-insert title here-

#116

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:

magnitude of force? speed of ball? I'm not sure how to work with a dimensionless vector.

I

InteGrand

Well-Known Member

#117

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
magnitude of force? speed of ball? I'm not sure how to work with a dimensionless vector.

I think it essentially can have any speed and the answer is not impacted, because the path traced out is the same (I think we are assuming a perfect mathematical universe with no friction or loss of energy in any way, and perfect reflection off the walls; so the ball moves for an infinite amount of time (and the speed is constant)).

Sy123

This too shall pass

#118

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
$\\ $An integer polynomial is a polynomial where all the co-efficients are integers, let$ \ p(x) \ $be such an integer polynomial$ \\ \\ $i) Show that if$ \ p + \sqrt{q} \ $is a root of$ \ p(x) \ $then$ \ p - \sqrt{q} \ $is a root, where$ \ p \ $is rational and$ \ q \ $is rational, but not the square of a rational$ \\ \\ $ii) Find other roots of$ \ p(x) \ $given that$ \ \alpha + \sqrt[3]{\beta} \ $is a root of$ \ p(x) \ $where$ \ \alpha, \beta \ $are rational, and$ \ \beta \ $is not the cube of a rational$$

$\\ $Let$ \ p + \sqrt{q} \ $be a root of the integer polynomial$ \ P$

$\\ $Consider the polynomial$ \ f(x) = ( x - (p + \sqrt{q}))(x - (p - \sqrt{q})) = (x-p)^2 - q$

$\\ f(x) \ $also has root$ \ p - \sqrt{q}$

$\\ $Divide then$ \ P(x) \ $by$ \ f(x) \ $and by the factor theorem$$

$\ P(x) = f(x) Q(x) + (Ax + B)$

$\\ $Where$ \ Q(x) \ $is a polynomial with rational co-efficients$$

$\\ $Thus$ \ A, B \ $are also rational, for if they were irrational, then$ \ P(x) - f(x)Q(x) \ $would be a polynomial with irrational co-efficients, which is impossible as$ \ P,Q,f \ $are all polynomials with rational co-efficients$$

$P( p + \sqrt{q}) = f(p + \sqrt{q}) Q(p + \sqrt{q}) + A(p + \sqrt{q}) + B$

$P(p + \sqrt{q}) = f(p + \sqrt{q}) = 0$

$\\ \Rightarrow \ 0 = (Ap + B) + \sqrt{q} (A)$

$\\ $Equating the rational and irrational parts, we have$ \ A = 0 \ \Rightarrow \ B = 0$

$\\ $Thus,$ \ P(x) = f(x) Q(x)$

$\\ $Thus,$ \ P(p- \sqrt{q}) = f(p - \sqrt{q}) Q(p - \sqrt{q}) = 0$

$\\ $Thus,$ \ p - \sqrt{q} \ $is a root$$

------

(ii) is very similar, where the other solutions are

$x = \alpha + (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})\sqrt[3]{\beta}$

$x = \alpha + (\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3})\sqrt[3]{\beta}$

Paradoxica

-insert title here-

#119

Re: HSC 2016 4U Marathon - Advanced Level

outlier1 said:
I got cos(theta) over 2

This does not match the answer I got for the special case theta = pi/4

Paradoxica

-insert title here-

#120

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
$\\ $On a perfect circular billiard table, the ball is hit at angle$ \ \theta \ $radians away from the line joining the starting point and the centre$$

$\\ $Find the average distance of the ball to the centre in terms of$ \ \theta$

Sy123 said:
$\\ $An$ \ n \ $sided regular polygon is such that no side is vertical in gradient$ \\ $Prove that if$ \ m_k \ $is the gradient of the$ \ k^{\text{th}} \ $side, then:$ \\\\ m_1 m_2 + m_2 m_3 + \dots + m_{n-1}m_n + m_n m_1 = - n$ a

I am flummoxed. Solutions?

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