1. ## HSC 2017-2019 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/...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 $cos\ 36$ (degrees)

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

2. ## HSC 2017 MX2 Marathon ADVANCED

A nice question

3. ## Re: HSC 2017 MX2 Marathon

Originally Posted by Kingom
A nice question
What paper is that from? Looks interesting

4. ## Re: HSC 2017 MX2 Marathon

Originally Posted by wu345
What paper is that from? Looks interesting
It's a surprise!

5. ## Re: HSC 2017 MX2 Marathon

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

6. ## Re: HSC 2017 MX2 Marathon

Originally Posted by seanieg89
$Partial fractions can be used to decompose the LHS into \\ \\ 7+\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}\\ \\ where \omega=\textrm{cis}(2\pi/7)\\ \\ Now let H(z)=(z^7-1)\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}.\\ \\ We have H(\omega^k)=\omega^k\prod_{i\neq k}(\omega^k-\omega^i)=\omega^{nk}\prod_{i\neq 0}(1-\omega^i)\\ \\ = \frac{z^7-1}{z-1}|_{z=1}=7. Since H(z) has degree at most 6, we conclude that H(z) is identically equal to 7.\\ \\ Hence LHS\cdot (z^7-1)=7(z^7-1)+2H(z)=7(z^7+1)+2(H(z)-7)=7(z^7+1) as required.\\ \\ \\ Then we can rearrange to\\ \\ 2\sum_{k=1}^3 \frac{1}{z^2-2z\cos(2k\pi/7)+1}=\frac{7(z^7+1)}{(z^7-1)(z^2-1)}-\frac{1}{(z-1)^2}=\frac{6z^4+4z^3+8z^2+4z+6}{z^6+z^5+z^4+z^3+z ^2+z+1}.\\ \\ Setting z=1 and using the double angle formula yields \\ \\ \sum_{k=1}^3 \csc^2(k\pi/7)=8. \\ \\ Where was it from?$
IMO Longlist 1988

7. ## Re: HSC 2017 MX2 Marathon

Haha that was my suspicion.

8. ## Re: HSC 2017 MX2 Marathon

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

9. ## Re: HSC 2017 MX2 Marathon

Originally Posted by Kingom
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).

10. ## Re: HSC 2017 MX2 Marathon

Originally Posted by seanieg89
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

11. ## Re: HSC 2017 MX2 Marathon ADVANCED

Originally Posted by dan964
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/...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 $(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})^5=-1$. Consider $z^5=-1$ whose roots are $z_1=cis\frac{\pi}{5}, z_2 = cis\frac{-\pi}{5}, z_3 = cis\frac{3\pi}{5}, z_4 = cis\frac{-3\pi}{5}, z_5 =-1$. $z^5 + 1 = (z+1)(z^4-z^3+z^2-z+1) = 0$ Thus the roots of $z^4 - z^3 + z^2 -z +1=0$ are $z_1, z_2, z_3, z_4$. Note that $z_2, z_4$ are the conjugates of $z_1, z_3$ respectively.
$z^4 -z^3 +z^2-z+1 = (z-z_1)(z-z_2)(z-z_3)(z-z_4) = (z^2 - 2\cos\frac{\pi}{5}z + 1)(z^2 - 2\cos\frac{3\pi}{5}z+1)$
By comparison of coefficients for $z^3, z^2$ we get:
$\cos\frac{\pi}{5} + \cos\frac{3\pi}{5}=\frac{1}{2}$ and $\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=\frac{-1}{4}$
These are the sum of roots and product of roots of the quadratic equation; $4x^2-2x+1=0$.
$x=\frac{1\pm \sqrt{5}}{4}$, but $\cos\frac{\pi}{5}>0$. Thus $\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}$

12. ## Re: HSC 2017 MX2 Marathon

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

13. ## Re: HSC 2017 MX2 Marathon

Originally Posted by Mahan1
$\textrm{for} x,y,z \in \mathbb{R}\ xyz(x+y+z) =1$
$\textrm{Find the minimum of the following expression}\ (x+y)(y+z)(x+z)$
Lagrange multipliers because why not

14. ## Re: HSC 2017 MX2 Marathon

Originally Posted by Mahan1
$\textrm{for} x,y,z \in \mathbb{R}\ xyz(x+y+z) =1$
$\textrm{Find the minimum of the following expression}\ (x+y)(y+z)(x+z)$
By inspection, the limit is -∞

15. ## Re: HSC 2017 MX2 Marathon ADVANCED

$Time for a new question$

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

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

16. ## Re: HSC 2017 MX2 Marathon ADVANCED

Originally Posted by dan964
$Time for a new question$

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

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

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

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

fixed sorry.

18. ## Re: HSC 2017 MX2 Marathon ADVANCED

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

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

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

19. ## Re: HSC 2017 MX2 Marathon ADVANCED

Originally Posted by frog0101
Find the Cartesian coordinates of a complex number that has a modulus of 3 and argument pi/8
$\noindent \cos \frac{\pi}{4}=\frac{\sqrt{2}}{2} \\ \\ 2 \cos^2 \frac{\pi}{8}-1=\frac{\sqrt{2}}{2} \\ \\ \Rightarrow \cos \frac{\pi}{8}=\frac{1}{2}\sqrt{2+\sqrt{2}} \\ \\ Similarly, \sin \frac{\pi}{8}=\frac{1}{2}\sqrt{2-\sqrt{2}} \\ \\ \therefore 3 cis\frac{\pi}{8} = \frac{3}{2}\sqrt{2+\sqrt{2}}+\frac{3}{2}\sqrt{2-\sqrt{2}} i$

20. ## Re: HSC 2017 MX2 Marathon ADVANCED

A square pyramid has height exactly half it's base side length, a. Find the volume of the pyramid.

21. ## Re: HSC 2017 MX2 Marathon ADVANCED

A square pyramid has height exactly half it's base side length, a. Find the volume of the pyramid.
It's just $\frac { { a }^{ 3 } }{ 6 }$.
Consider a cube with side length a. If you cut along its diagonals u get 6 identical pyramids with height half the base.

22. ## Re: HSC 2017 MX2 Marathon ADVANCED

why can the standard V=1/3Ah be using used
where A is the area of the base = a^2
and h = a/2

23. ## Re: HSC 2017 MX2 Marathon ADVANCED

Originally Posted by dan964
why can the standard V=1/3Ah be using used
where A is the area of the base = a^2
and h = a/2
ewwww formula pleb jkjk

24. ## Re: HSC 2017 MX2 Marathon ADVANCED

Find the maximal possible angle that a circle centred at the origin can meet the ellipse x^2/a^2 + y^2/b^2 = 1.

25. ## Re: HSC 2017 MX2 Marathon ADVANCED

(The above Q isn't too hard, just something to get this thread rolling again .)

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