HSC 2016 MX2 Marathon (archive) (1 Viewer)

KingOfActing · Dec 5, 2015

Re: HSC 2016 4U Marathon

leehuan said:
Never seen the vertical line before. What does it mean?

Divides. $a|b$ means "a divides b" or "b is divisible by a".

leehuan · Dec 5, 2015

Re: HSC 2016 4U Marathon

Ah, had no teacher use that notation before.

leehuan · Dec 5, 2015

Re: HSC 2016 4U Marathon

$\\ $In order to demonstrate divisibility by 7: \\ The case for $n=1$ yields 0, and the case for $n=2$ yields $7 \times 18$ \\ Assume true for $n=k$ to give: \\ ${ k }^{ 7 }-k=7M \quad (k, m \in \mathbb N) \\$ Then we must show ${ \left( k+1 \right) }^{ 7 }-\left( k+1 \right) $ is divisible by 42.$ \\{ \left( k+1 \right) }^{ 7 }-\left( k+1 \right) \\ ={ k }^{ 7 }+7{ k }^{ 6 }+21{ k }^{ 5 }+35{ k }^{ 4 }+35{ k }^{ 3 }+21{ k }^{ 2 }+6k\\ =42M+7{ k }^{ 6 }+21{ k }^{ 5 }+35{ k }^{ 4 }+35{ k }^{ 3 }+21{ k }^{ 2 }+7k\\ =7(6M+{ k }^{ 6 }+3{ k }^{ 5 }+5{ k }^{ 4 }+5{ k }^{ 3 }+3k+1)$

$\\ $Given that $M, k \in \mathbb N$, our statement is true. \\ Now, observe $42=7 \times 6$, so if the formula is also divisible by 6, it must be divisible by 42. $\\ { n }^{ 7 }-n\\ =n\left( { n }^{ 6 }-1 \right) \\ =n\left( { n }^{ 3 }-1 \right) \left( { n }^{ 3 }+1 \right) \\ =\left( n-1 \right) n\left( n+1 \right) \left( { n }^{ 2 }+n+1 \right) \left( { n }^{ 2 }-n+1 \right)$

$\\ \left( n-1 \right) n\left( n+1 \right) $ form three consecutive integers. It is possible to prove that this is divisible by 6 easily through induction as well, or alternatively by noting that the product of just two consecutive integers is always even, and that at least one of the three integers will be a multiple of 3. The divisibility test for a multiple of 6 is that it is a multiple of 2 and 3, and thus this is also satisfied.$$

$\\ $Ultimately, we arrive at our final result of divisibility by 42, because both that of 6 and 7 are satisfied.$

InteGrand · Dec 5, 2015

Re: HSC 2016 4U Marathon

leehuan said:
Ah, had no teacher use that notation before.

It's generally not used in the HSC or school (I guess because divisibility isn't a big topic in the HSC), but it is a standard mathematical notation.

Paradoxica · Dec 5, 2015

Re: HSC 2016 4U Marathon

$$1. It is given that $ A + B + C= \pi$

$$Define $p = \sin{A}\sin{B}\sin{C}, q = \cos{A}\cos{B}\cos{C}$

$$Evaluate $\tan^3{A} + \tan^3{B} + \tan^3{C}$ in terms of $p$ and $q.$

$\noindent 2. If $a,b,c$ are the sides of a triangle, show that$

$\sqrt[n]{a}, \sqrt[n]{b}, \sqrt[n]{c}, \; \forall n \in \mathbb N$ also form a triangle.$$

Paradoxica · Dec 6, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
$$1. It is given that $ A + B + C= \pi$

$$Define $p = \sin{A}\sin{B}\sin{C}, q = \cos{A}\cos{B}\cos{C}$

$$Evaluate $\tan^3{A} + \tan^3{B} + \tan^3{C}$ in terms of $p$ and $q.$

$\noindent 2. If $a,b,c$ are the sides of a triangle, show that$

$\sqrt[n]{a}, \sqrt[n]{b}, \sqrt[n]{c}, \; \forall n \in \mathbb N$ also form a triangle.$$

$$Hint for 1: Consider the triangle with angles $A,B,C.$

$$Hint for 2: Consider the limit as $n \rightarrow \infty.$

leehuan · Dec 6, 2015

Re: HSC 2016 4U Marathon

Admittedly, I used only algebra for 1. I'll post what I got up to incase anyone finishes it from there

$\\ \tan { \left( A+B+C \right) } =\tan { \pi } \\ \frac { \tan { \left( A+B \right) } +\tan { C } }{ 1-\tan { \left( A+B \right) \tan { C } } } =0\\ \frac { \tan { A } +\tan { B } }{ 1-\tan { A } \tan { B } } +\tan { C } =0\\ \tan { A } +\tan { B } +\tan { C } =\tan { A } \tan { B } \tan { C } \\ \therefore \tan { A } +\tan { B } +\tan { C } =\frac { p }{ q } =\tan { A } \tan { B } \tan { C }$

$\\ \therefore \frac { { p }^{ 3 } }{ { q }^{ 3 } } =\left( \tan ^{ 3 }{ A } +\tan ^{ 3 }{ B } +\tan ^{ 3 }{ C } \right) +6\tan { A } \tan { B } \tan { C } +3\tan { A } \tan { B } \left( \tan { A } +\tan { B } \right) +3\tan { B } \tan { C } \left( \tan { B } +\tan { C } \right) +3\tan { A } \tan { C } \left( \tan { A } +\tan { C } \right) \\$

$\\ \frac { { p }^{ 3 } }{ { q }^{ 3 } } -\frac { 6p }{ q } =\left( \tan ^{ 3 }{ A } +\tan ^{ 3 }{ B } +\tan ^{ 3 }{ C } \right) +\frac { 3p }{ q } \cot { C } \left( \frac { p }{ q } -\tan { C } \right) +\frac { 3p }{ q } \cot { A } \left( \frac { p }{ q } -\tan { A } \right) +\frac { 3p }{ q } \cot { B } \left( \frac { p }{ q } -\tan { B } \right) \\ \frac { { p }^{ 3 } }{ { q }^{ 3 } } -\frac { 6p }{ q } =\left( \tan ^{ 3 }{ A } +\tan ^{ 3 }{ B } +\tan ^{ 3 }{ C } \right) +\frac { 3p }{ q } \left( \frac { p }{ q } \left( \cot { A } +\cot { B } +\cot { C } \right) -3 \right)$

I knew that A+B+C=pi meant they formed the angles of a triangle but I was too lazy to consider if that would lead anywhere

Drsoccerball · Dec 7, 2015

Re: HSC 2016 4U Marathon

Im in japan and Latex isnt showing?

obliviousninja · Dec 7, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
Im in japan and Latex isnt showing?

Good

InteGrand · Dec 7, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
Im in japan and Latex isnt showing?

I think it's messing up on the site once again (it's not working for me either).

leehuan · Dec 7, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
Im in japan and Latex isnt showing?

It's not working.

omegadot · Dec 7, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
Im in japan and Latex isnt showing?

InteGrand said:
I think it's messing up on the site once again (it's not working for me either).

leehuan said:
It's not working.

It's working again (well, at least for me it's working again)

leehuan · Dec 7, 2015

Re: HSC 2016 4U Marathon

omegadot said:
It's working again (well, at least for me it's working again)

Yes it is now

omegadot · Dec 9, 2015

Re: HSC 2016 4U Marathon

Ambility said:
$$\textbf{NEW QUESTION}$\\\\$The complex numbers w and z with respective arguments $\phi$ and $\theta$ are two distinct points on the Argand diagram where $\phi < \theta$. The area of the triangle formed by the origin and these two points is A.$\\\\$Show that $z\bar{w}-w\bar{z}=4iA$

$$\noindent Writing the two complex numbers in modulus--argument form one has: $w = |w| \, \mbox{cis} \, \phi$ and $z = |z| \, \mbox{cis} \, \theta$. Now let $w$ form one side of the triangle and $z$ the other such that the included angle between these two sides is equal to $\theta - \phi.$

$$\noindent Note that $z \bar{w}$ is the conjugate of the complex number $w \bar{z}.$ Thus$\\\begin{align*}z\bar{w} - w\bar{z} &= 2 i \, \mbox{Im} \, (z \bar{w})\end{align*}\\$Now$\\\begin{align*}z\bar{w} &= |z| (\cos \theta + i \sin \theta) \cdot |w| (\cos \phi - i \sin \phi)\\&=|w| |z| \left (\cos \theta \cos \phi - i \cos \theta \sin \phi + i \sin \theta \cos \phi + \sin \theta \sin \phi \right )\\&= |w| |z| \left [\cos (\theta + \phi) + i \sin (\theta - \phi) \right ],\end{align*}$

$$\noindent giving $\mbox{Im} \, (z \bar{w}) = |w| |z| \sin (\theta - \phi).$ So one has$

$\begin{align}z\bar{w} - w\bar{z} &= 2 i |w| |z| \sin (\theta - \phi) = 4 i \left [\frac{1}{2}|w| |z| \sin (\theta - \phi) \right ].\end{align}$

$$\noindent For the triangle with sides of lengths $|w|$ and $|z|$ and included angle $\theta - \phi$, its area $A$ will be given by$\\A = \frac{1}{2} |w| |z| \sin (\theta - \phi).$

$$\noindent Thus Eq. (1) becomes $z\bar{w} - w\bar{z} = 4iA$, as required to show.$

omegadot · Dec 9, 2015

Re: HSC 2016 4U Marathon

$$\noindent \textbf{Next Question}$

$$\noindent If $v,w,z \in \mathbb{C}$ show that$\\\begin{align*}|v| + |v + w| + |w + z| + |2 + z| \geqslant 2.\end{align*}$

Paradoxica · Dec 9, 2015

Re: HSC 2016 4U Marathon

omegadot said:
$$\noindent \textbf{Next Question}$

$$\noindent If $v,w,z \in \mathbb{C}$ show that$\\\begin{align*}|v| + |v + w| + |w + z| + |2 + z| \geqslant 2.\end{align*}$

$|w+z| + |w| \geq |z|, |2+z| + |z| \geq 2, |v| + |v+w| \geq |w|$

$All are true by the triangle inequality. Add them together and cancel appropriate terms to yield the desired inequality$

Paradoxica said:
$$1. It is given that $ A + B + C= \pi$

$$Define $p = \sin{A}\sin{B}\sin{C}, q = \cos{A}\cos{B}\cos{C}$

$$Evaluate $\tan^3{A} + \tan^3{B} + \tan^3{C}$ in terms of $p$ and $q.$

$\noindent 2. If $a,b,c$ are the sides of a triangle, show that$

$\sqrt[n]{a}, \sqrt[n]{b}, \sqrt[n]{c}, \; \forall n \in \mathbb N$ also form a triangle.$$

Paradoxica said:
$$Hint for 1: Consider the triangle with angles $A,B,C.$

$$Hint for 2: Consider the limit as $n \rightarrow \infty.$

omegadot · Dec 11, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
$\noindent 2. If $a,b,c$ are the sides of a triangle, show that$

$\sqrt[n]{a}, \sqrt[n]{b}, \sqrt[n]{c}, \; \forall n \in \mathbb N$ also form a triangle.$$

$$\noindent To show $\sqrt[n]{a}, \sqrt[n]{b},$ and $\sqrt[n]{c}$ forms a triangle we will show that $\sqrt[n]{a}, \sqrt[n]{b},$ and $\sqrt[n]{c}$ satisfy the triangle inequality rule given that $a, b,$ and $c$ are positive and satisfy the triangle inequality rule, namely: $a + b > c, b + c > a,$ and $c + a > b$.$

$$\noindent Since $n \in \mathbb{N}$ from the binomial theorem one has$\\\begin{align*}\left (\sqrt[n]{a} + \sqrt[n]{b} \right )^n &= a + \binom{n}{1} \left (\sqrt[n]{a} \right )^{n - 1} \left (\sqrt[n]{b} \right )^1 + \cdots + \binom{n}{n - 1} \left (\sqrt[n]{a} \right )^1 \left (\sqrt[n]{b} \right )^{n - 1} + b > a + b.\end{align*}$

$$\noindent Since $a,b,$ and $c$ are positive and satisfy the triangle inequality rule, namely $a + b > c$, one therefore has$

$\left (\sqrt[n]{a} + \sqrt[n]{b} \right )^n > c \quad \Rightarrow \quad \sqrt[n]{a} + \sqrt[n]{b} > \sqrt[n]{c}.$

$$\noindent In a similar manner it can be shown that$\\\begin{align*}\sqrt[n]{b} + \sqrt[n]{c} > \sqrt[n]{a} \quad \mbox{and} \quad \sqrt[n]{c} + \sqrt[n]{a} > \sqrt[n]{b}.\end{align*}$

$$\noindent Thus $\sqrt[n]{a}, \sqrt[n]{b},$ and $\sqrt[n]{c}$ satisfy the triangle inequality rule. We therefore conclude $\sqrt[n]{a}, \sqrt[n]{b},$ and $\sqrt[n]{c}$ must also form a triangle.$

omegadot · Dec 11, 2015

Re: HSC 2016 4U Marathon

$$\noindent \textbf{Next Question}$

$$\noindent For all complex numbers $z,w$ satisfying $|z| = 12$ and $|w - 3 - 4i| = 5$, find the minimum and maximum value of $|z - w|.$

braintic · Dec 12, 2015

Re: HSC 2016 4U Marathon

omegadot said:
$$\noindent \textbf{Next Question}$

$$\noindent For all complex numbers $z,w$ satisfying $|z| = 12$ and $|w - 3 - 4i| = 5$, find the minimum and maximum value of $|z - w|.$

For anyone who wants to give this a go, the answers are 2 and 22. It's up to you to explain those answers.

KingOfActing · Dec 12, 2015

Re: HSC 2016 4U Marathon

braintic said:
For anyone who wants to give this a go, the answers are 2 and 22. It's up to you to explain those answers.

Note $|z-w|$ is the distance between a point on the circle centered on the origin with radius 12 and the circle with center $(3,4)$ and radius 5. Looking at this diagram:

We can see that $AB$ is the shortest distance whilst $BC$ is the longest.

$AB = OA - OB = 12 - 2\times5 = 12 - 10 = 2\\BC = OB + OC = 12 + 2\times5 = 12 + 10 = 22$

HSC 2016 MX2 Marathon (archive) (1 Viewer)

lukewarm mess

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

-insert title here-

Well-Known Member

Well-Known Member

(╯°□°）╯━︵ ┻━┻ - - - -

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Active Member

-insert title here-

Active Member

Active Member

Well-Known Member

lukewarm mess

Attachments

Users Who Are Viewing This Thread (Users: 0, Guests: 1)