# HSC 2017-2019 MX2 Marathon ADVANCED (1 Viewer)

#### dan964

##### MOD
Moderator
Post any questions within the scope and level of Mathematics Extension 2 mainly targeting Q16 difficulty in the HSC.

Any questions beyond the scope of the HSC syllabus should be posted in the Extracurricular Topics forum:
http://community.boredofstudies.org/14/mathematics-extension-2/360463/14/mathematics-extension-2/360463/14/mathematics-extension-2/360463/14/mathematics-extension-2/360463/14/mathematics-extension-2/360463/14/mathematics-extension-2/360463/...icular-topics/

Once a question is posted, it needs to be answered before the next question is raised.

I encourage all current students in particular to participate in this marathon.

To get the ball-rolling

Question:
Using MX2 methods & complex numbers, find an expression for $\bg_white cos\ 36$ (degrees)

Edit 6/3/17: I have moved some of the advanced questions to this forum...

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A nice question

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#### wu345

##### Member
Re: HSC 2017 MX2 Marathon

A nice question
What paper is that from? Looks interesting

#### Kingom

##### Member
Re: HSC 2017 MX2 Marathon

What paper is that from? Looks interesting
It's a surprise!

#### seanieg89

##### Well-Known Member
Re: HSC 2017 MX2 Marathon

A nice question
$\bg_white Partial fractions can be used to decompose the LHS into \\ \\ 7+\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}\\ \\ where \omega=\textrm{cis}(2\pi/7)\\ \\ Now let H(z)=(z^7-1)\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}.\\ \\ We have H(\omega^k)=\omega^k\prod_{i\neq k}(\omega^k-\omega^i)=\omega^{nk}\prod_{i\neq 0}(1-\omega^i)\\ \\ = \frac{z^7-1}{z-1}|_{z=1}=7. Since H(z) has degree at most 6, we conclude that H(z) is identically equal to 7.\\ \\ Hence LHS\cdot (z^7-1)=7(z^7-1)+2H(z)=7(z^7+1)+2(H(z)-7)=7(z^7+1) as required.\\ \\ \\ Then we can rearrange to\\ \\ 2\sum_{k=1}^3 \frac{1}{z^2-2z\cos(2k\pi/7)+1}=\frac{7(z^7+1)}{(z^7-1)(z^2-1)}-\frac{1}{(z-1)^2}=\frac{6z^4+4z^3+8z^2+4z+6}{z^6+z^5+z^4+z^3+z^2+z+1}.\\ \\ Setting z=1 and using the double angle formula yields \\ \\ \sum_{k=1}^3 \csc^2(k\pi/7)=8. \\ \\ Where was it from?$

#### Kingom

##### Member
Re: HSC 2017 MX2 Marathon

$\bg_white Partial fractions can be used to decompose the LHS into \\ \\ 7+\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}\\ \\ where \omega=\textrm{cis}(2\pi/7)\\ \\ Now let H(z)=(z^7-1)\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}.\\ \\ We have H(\omega^k)=\omega^k\prod_{i\neq k}(\omega^k-\omega^i)=\omega^{nk}\prod_{i\neq 0}(1-\omega^i)\\ \\ = \frac{z^7-1}{z-1}|_{z=1}=7. Since H(z) has degree at most 6, we conclude that H(z) is identically equal to 7.\\ \\ Hence LHS\cdot (z^7-1)=7(z^7-1)+2H(z)=7(z^7+1)+2(H(z)-7)=7(z^7+1) as required.\\ \\ \\ Then we can rearrange to\\ \\ 2\sum_{k=1}^3 \frac{1}{z^2-2z\cos(2k\pi/7)+1}=\frac{7(z^7+1)}{(z^7-1)(z^2-1)}-\frac{1}{(z-1)^2}=\frac{6z^4+4z^3+8z^2+4z+6}{z^6+z^5+z^4+z^3+z^2+z+1}.\\ \\ Setting z=1 and using the double angle formula yields \\ \\ \sum_{k=1}^3 \csc^2(k\pi/7)=8. \\ \\ Where was it from?$
IMO Longlist 1988

#### seanieg89

##### Well-Known Member
Re: HSC 2017 MX2 Marathon

Haha that was my suspicion.

#### Kingom

##### Member
Re: HSC 2017 MX2 Marathon

I believe it's accessible to the extension 2 student so i put it on the marathon

#### seanieg89

##### Well-Known Member
Re: HSC 2017 MX2 Marathon

I believe it's accessible to the extension 2 student so i put it on the marathon
Yeah fair enough, they were a lot easier back in the day.

Still should probably put questions like it in the advanced thread (Dan only created it today I believe, but for the last couple of years we have usually split the questions into two marathons, one for routine stuff and one for questions at HSC Q16 difficulty or above. As long as it is still within syllabus of course).

#### Kingom

##### Member
Re: HSC 2017 MX2 Marathon

Yeah fair enough, they were a lot easier back in the day.

Still should probably put questions like it in the advanced thread (Dan only created it today I believe, but for the last couple of years we have usually split the questions into two marathons, one for routine stuff and one for questions at HSC Q16 difficulty or above. As long as it is still within syllabus of course).
Yeh i wanted to put this in the advanced marathon, but none existed at the time it was posted so

#### frog1944

##### Member
Re: HSC 2017 MX2 Marathon ADVANCED

Post any questions within the scope and level of Mathematics Extension 2 mainly targeting Q16 difficulty in the HSC.

Any questions beyond the scope of the HSC syllabus should be posted in the Extracurricular Topics forum:
http://community.boredofstudies.org/14/mathematics-extension-2/360463/14/mathematics-extension-2/360463/...icular-topics/

Once a question is posted, it needs to be answered before the next question is raised.

I encourage all current students in particular to participate in this marathon.

To get the ball-rolling

Question:
Using MX2 methods & complex numbers, find an expression for the cos 36.
Note that $\bg_white (\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})^5=-1$. Consider $\bg_white z^5=-1$ whose roots are $\bg_white z_1=cis\frac{\pi}{5}, z_2 = cis\frac{-\pi}{5}, z_3 = cis\frac{3\pi}{5}, z_4 = cis\frac{-3\pi}{5}, z_5 =-1$. $\bg_white z^5 + 1 = (z+1)(z^4-z^3+z^2-z+1) = 0$ Thus the roots of $\bg_white z^4 - z^3 + z^2 -z +1=0$ are $\bg_white z_1, z_2, z_3, z_4$. Note that $\bg_white z_2, z_4$ are the conjugates of $\bg_white z_1, z_3$ respectively.
$\bg_white z^4 -z^3 +z^2-z+1 = (z-z_1)(z-z_2)(z-z_3)(z-z_4) = (z^2 - 2\cos\frac{\pi}{5}z + 1)(z^2 - 2\cos\frac{3\pi}{5}z+1)$
By comparison of coefficients for $\bg_white z^3, z^2$ we get:
$\bg_white \cos\frac{\pi}{5} + \cos\frac{3\pi}{5}=\frac{1}{2}$ and $\bg_white \cos\frac{\pi}{5}\cos\frac{3\pi}{5}=\frac{-1}{4}$
These are the sum of roots and product of roots of the quadratic equation; $\bg_white 4x^2-2x+1=0$.
$\bg_white x=\frac{1\pm \sqrt{5}}{4}$, but $\bg_white \cos\frac{\pi}{5}>0$. Thus $\bg_white \cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}$

#### Mahan1

##### Member
Re: HSC 2017 MX2 Marathon

$\bg_white \textrm{for} x,y,z \in \mathbb{R}\ xyz(x+y+z) =1$
$\bg_white \textrm{Find the minimum of the following expression}\ (x+y)(y+z)(x+z)$

#### Kingom

##### Member
Re: HSC 2017 MX2 Marathon

$\bg_white \textrm{for} x,y,z \in \mathbb{R}\ xyz(x+y+z) =1$
$\bg_white \textrm{Find the minimum of the following expression}\ (x+y)(y+z)(x+z)$
Lagrange multipliers because why not

##### -insert title here-
Re: HSC 2017 MX2 Marathon

$\bg_white \textrm{for} x,y,z \in \mathbb{R}\ xyz(x+y+z) =1$
$\bg_white \textrm{Find the minimum of the following expression}\ (x+y)(y+z)(x+z)$
By inspection, the limit is -∞

#### dan964

##### MOD
Moderator
Re: HSC 2017 MX2 Marathon ADVANCED

$\bg_white Time for a new question$

$\bg_white Q1: Prove by induction on n that$
$\bg_white (fg)^{(n)}=\sum_{i=0}^{n}\binom{n}{i}f^{(n-i)}g^{(i)}$

$\bg_white Q2: If P_{n}(x)=\frac{d^n}{d x^n}[(x^2-1)^n]. Find P_n(1) and P_n(-1)$
$\bg_white where f^{(n)} is the notation for the n-th derivative$

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#### calamebe

##### Active Member
Re: HSC 2017 MX2 Marathon ADVANCED

$\bg_white Time for a new question$

$\bg_white Q1: Prove by induction on n that$
$\bg_white (fg)^{(n)}=\sum_{i=0}^{n}\binom{n}{i}f^{(n-1)}g^{(i)}$

$\bg_white Q2: If P_{n}(x)=\frac{d^n}{d x^n}[(x^2-1)^n]. Find P_n(1) and P_n(-1)$
$\bg_white where f^{(n)} is the notation for the n-th derivative$
For Q1 instead of (n-1) at the top it should just be n-i.

$\bg_white Anyway for that one in the prove for k+1 part (the induction step is it called?) get (fg)^{(k+1)} which will be the derivative of (fg)^{(k)}, and this can value can be derived from the induction hypothesis. I derived by breaking the sum into two sums, and using a substitution r=i+1 on one of the substitutions. Then you can pretty easily get to the desired result.$

$\bg_white And for Q2 it is n!2^n and n!(-2)^n respectively right?$

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#### dan964

##### MOD
Moderator
Re: HSC 2017 MX2 Marathon ADVANCED

fixed sorry.

#### dan964

##### MOD
Moderator
Re: HSC 2017 MX2 Marathon ADVANCED

For Q1 instead of (n-1) at the top it should just be n-i.

$\bg_white Anyway for that one in the prove for k+1 part (the induction step is it called?) get (fg)^{(k+1)} which will be the derivative of (fg)^{(k)}, and this can value can be derived from the induction hypothesis. I derived by breaking the sum into two sums, and using a substitution r=i+1 on one of the substitutions. Then you can pretty easily get to the desired result.$

$\bg_white And for Q2 it is n!2^n and n!(-2)^n respectively right?$
Q2 is correct. Split it into (x+1)(x-1)

#### jathu123

##### Active Member
Re: HSC 2017 MX2 Marathon ADVANCED

Find the Cartesian coordinates of a complex number that has a modulus of 3 and argument pi/8
$\bg_white \noindent \cos \frac{\pi}{4}=\frac{\sqrt{2}}{2} \\ \\ 2 \cos^2 \frac{\pi}{8}-1=\frac{\sqrt{2}}{2} \\ \\ \Rightarrow \cos \frac{\pi}{8}=\frac{1}{2}\sqrt{2+\sqrt{2}} \\ \\ Similarly, \sin \frac{\pi}{8}=\frac{1}{2}\sqrt{2-\sqrt{2}} \\ \\ \therefore 3 cis\frac{\pi}{8} = \frac{3}{2}\sqrt{2+\sqrt{2}}+\frac{3}{2}\sqrt{2-\sqrt{2}} i$