# Thread: HSC 2017 MX2 Marathon (archive)

1. ## Re: HSC 2017 MX2 Marathon

Not sure if its me or the book error but I had a complex number question

(6+i)(a+bi)=2

My solution for
$a= \frac{12}{37} \ and \ b =\frac{-2}{37}$

My book has a different solution to that.

2. ## Re: HSC 2017 MX2 Marathon

Originally Posted by davidgoes4wce
Not sure if its me or the book error but I had a complex number question

(6+i)(a+bi)=2

My solution for
$a= \frac{12}{37} \ and \ b =\frac{-2}{37}$

My book has a different solution to that.
I also got the same answer as you.

3. ## Re: HSC 2017 MX2 Marathon

I spent 5 minutes on this and couldn't solve it.

$By writing z=x+yi, solve the following equations:$

$z^2 = 5+i$

$Ans: \pm \sqrt{\frac{\sqrt{26}+5}{2}}+ i \sqrt{\frac{\sqrt{26}-5}{2}}$

4. ## Re: HSC 2017 MX2 Marathon

I think my book has done another mistake. (The book has made a few mistakes so I'm just checking with people on here)

5. ## Re: HSC 2017 MX2 Marathon

i got an

x+yi

my x-component was :

$x = \pm (\sqrt{{\frac{5}{4}}+\frac{\sqrt{26}}{2}})$

I haven't solved the y-component yet. Want to get 'x' right before I do the y.

6. ## Re: HSC 2017 MX2 Marathon

The book's answer is right, you can check it by squaring it to get 5 + i.

7. ## Re: HSC 2017 MX2 Marathon

Originally Posted by davidgoes4wce
I spent 5 minutes on this and couldn't solve it.

$By writing z=x+yi, solve the following equations:$

$z^2 = 5+i$

$Ans: \pm \sqrt{\frac{\sqrt{26}+5}{2}}+ i \sqrt{\frac{\sqrt{26}-5}{2}}$
x^2 - y^2 + 2xyi = 5+i

Comparing real and imaginary
2xy = 1
y = 1/2x

x^2 - y^2 = 5
x^2 - 1/4x^2 = 5
4x^4 - 20x^2 - 1 = 0
x^2 = (20 +- sqrt(20^2 - 4(4)(-1))/2(4)
x^2 = (20+- sqrt(416))/8
x^2 = (20 +- 4sqrt(26))/8
x^2 = (5 +- sqrt(26))/2

But x is real so take the positive case only
x = +- sqrt[(5+sqrt26)/2]

8. ## Re: HSC 2017 MX2 Marathon

Originally Posted by davidgoes4wce
I spent 5 minutes on this and couldn't solve it.

$By writing z=x+yi, solve the following equations:$

$z^2 = 5+i$

$Ans: \pm \sqrt{\frac{\sqrt{26}+5}{2}}+ i \sqrt{\frac{\sqrt{26}-5}{2}}$
http://i.imgur.com/xWfCJry.jpg

9. ## Re: HSC 2017 MX2 Marathon

$Let S_{\left ( n,k \right ) }= k+kk+kkk+... +\underbrace{kkkk...k}_{\text{n number of k's}} where {n,k} \in \mathbb{Z}: n>0 and 0

apologies if the question isn't clear (hence my two examples)

10. ## Re: HSC 2017 MX2 Marathon

Originally Posted by jathu123
$Let S_{\left ( n,k \right ) }= k+kk+kkk+... +\underbrace{kkkk...k}_{\text{n number of k's}} where {n,k} \in \mathbb{Z}: n>0 and 0

apologies if the question isn't clear (hence my two examples)
\noindent \begin{align*}S_{(n,k)} &= k + (k+10k) + (k +10k+100k)+ \ldots + (k+10k+\ldots+10^{n-1}k) \\ &= \frac{k(10^1-1)}{9}+\frac{k(10^2-1)}{9}+\frac{k(10^3-1)}{9}+\ldots+\frac{k(10^n-1)}{9} \\ 9S_{(n,k)} &= k(10^1+10^2+\ldots+10^n)-k(\underbrace{1+1+\ldots+1}_{n \text{ times}}) \\ &= \frac{10k(10^n-1)}{9}-kn \\ S_{(n,k)} &= \frac{10k(10^n-1)-9kn}{81}\end{align*}

11. ## Re: HSC 2017 MX2 Marathon

$Two circles \Gamma_1,\Gamma_2 of differing radii share a common point of tangency P and lie on the same side of the tangent line \ell. Let M be the point on \ell such that \angle Q_1MQ_2 is maximised, where Q_j is the point other than P on \Gamma_j with tangent passing through M. (Briefly explain why such a point M must exist, and why it is unique up to reflection in the line perpendicular to \ell through P.)\\ For this choice of M, find the angle \angle PMQ_1+\angle PMQ_2 .$

12. ## Re: HSC 2017 MX2 Marathon

Is there a formula for finding the sum of all points (angles where it's spiky) on an n pointed star? For example, for a 5 pointed star, it is 180 degrees

13. ## Re: HSC 2017 MX2 Marathon

Originally Posted by si2136
Is there a formula for finding the sum of all points (angles where it's spiky) on an n pointed star? For example, for a 5 pointed star, it is 180 degrees
For higher n, there becomes multiple ways to construct the star, each with their own angle sum. ie. the points of the star may create an n-gon, or another star, as shown for n = 7, 11. Note in each case, the first star creates an n-gon by joining every second vertex, while the others create other stars.

14. ## Re: HSC 2017 MX2 Marathon

Originally Posted by seanieg89
$Two circles \Gamma_1,\Gamma_2 of differing radii share a common point of tangency P and lie on the same side of the tangent line \ell. Let M be the point on \ell such that \angle Q_1MQ_2 is maximised, where Q_j is the point other than P on \Gamma_j with tangent passing through M. (Briefly explain why such a point M must exist, and why it is unique up to reflection in the line perpendicular to \ell through P.)\\ For this choice of M, find the angle \angle PMQ_1+\angle PMQ_2 .$
This may not be in the spirit of the question, but if you had something different in mind with a more geometric proof, I'd like to see it. Let $O_j$ be the centre of circle $\Gamma_j$. Consider the kite $MPO_j Q_j$, and apply the cosine rule to the side $PQ_j$ from both $\triangle MPQ_j$ and $\triangle O_j P Q_j$. Letting, $A = \angle PMQ_1, B = \angle PMQ_2, x = |PM| = |MQ_1| = |MQ_2|, r_j = \text{radius of } \ \Gamma_j$. We obtain that, $2r_1^2 (1 + \cos A) = PQ_1^2 = 2x^2 ( 1 - \cos A)$. A similar expression is obtained for $PQ_2^2$. From this we obtain,

$\tan \frac{A}{2} = \frac{r_1}{x}, \tan \frac{B}{2} = \frac{r_2}{x}$. As $0 < B - A < B < \pi$ we have that $\tan \frac{x}{2}$ is monotonic increasing, and so $B - A = \angle Q_1MQ_2$ is maximised if and only if $\tan \frac{B-A}{2}$ is maximised. Using the appropriate formulae we obtain,

$\tan \frac{B-A}{2} = (r_2 - r_1)\frac{x}{x^2 + r_1r_2x}$, through calculus or the AM-GM inequality we obtain a maxima when $x = \sqrt{r_1r_2}$ that is unique and so on. However this sends $\tan \frac{A+B}{2} = \frac{r_1/x + r_2/x}{1 - (r_1r_2)/x^2} \to \infty$ suggesting that $\frac{A+B}{2} = \frac{\pi}{2} \Rightarrow \angle PMQ_1 + \angle PMQ_2 = \pi$.

15. ## Re: HSC 2017 MX2 Marathon

$\\ Suppose you are analysing the decay of particles from a radioactive source, suppose you discover that the probability that the source emits \ k \ particles from your source in an hour is \\\\ p_k = \frac{e^{-\lambda} \lambda^k}{k!}, \ k \geq 0 \\\\ Where \ \lambda \ is a constant and a positive integer, the rate of emission for your source.$

$\\ (a) By considering the ratio \frac{p_{k+1}}{p_k} or otherwise, find the most likely number of particles to be emitted in an hour$

$\\ (b) Suppose you have a friend and she is analysing the decay from her own radioactive source, and that in fact the probability that \ n \ particles are emitted from her source in an hour is \\\\ q_n = \frac{e^{-\mu} \mu^n}{n!}, \ n \geq 0 \\\\ Where \ \mu \ is a constant and a positive integer, the rate of emission in her source.$

$\\ Show that the probability that the sum of your and her observations is \ m \ is given by, \\\\ r_m = \frac{e^{-(\lambda + \mu)} (\lambda + \mu)^m}{m!}, \ m \geq 0$

16. ## Re: HSC 2017 MX2 Marathon

In how many ways can seven identical cats be put into three identical pens so that all of the pens are occupied? You must state reasoning. (2 marks)

Why can we not use stars and bars? I.e. 9C2

17. ## Re: HSC 2017 MX2 Marathon

Originally Posted by pikachu975
In how many ways can seven identical cats be put into three identical pens so that all of the pens are occupied? You must state reasoning. (2 marks)

Why can we not use stars and bars? I.e. 9C2
$\noindent Because ''Stars and Bars'' involves placing n identical objects into r \textbf{distinct} bins.$

Here is a useful article for placing n identical objects into r identical bins: https://brilliant.org/wiki/identical...identical-bins .

18. ## Re: HSC 2017 MX2 Marathon

Originally Posted by Sy123
This may not be in the spirit of the question, but if you had something different in mind with a more geometric proof, I'd like to see it. Let $O_j$ be the centre of circle $\Gamma_j$. Consider the kite $MPO_j Q_j$, and apply the cosine rule to the side $PQ_j$ from both $\triangle MPQ_j$ and $\triangle O_j P Q_j$. Letting, $A = \angle PMQ_1, B = \angle PMQ_2, x = |PM| = |MQ_1| = |MQ_2|, r_j = \text{radius of } \ \Gamma_j$. We obtain that, $2r_1^2 (1 + \cos A) = PQ_1^2 = 2x^2 ( 1 - \cos A)$. A similar expression is obtained for $PQ_2^2$. From this we obtain,

$\tan \frac{A}{2} = \frac{r_1}{x}, \tan \frac{B}{2} = \frac{r_2}{x}$. As $0 < B - A < B < \pi$ we have that $\tan \frac{x}{2}$ is monotonic increasing, and so $B - A = \angle Q_1MQ_2$ is maximised if and only if $\tan \frac{B-A}{2}$ is maximised. Using the appropriate formulae we obtain,

$\tan \frac{B-A}{2} = (r_2 - r_1)\frac{x}{x^2 + r_1r_2x}$, through calculus or the AM-GM inequality we obtain a maxima when $x = \sqrt{r_1r_2}$ that is unique and so on. However this sends $\tan \frac{A+B}{2} = \frac{r_1/x + r_2/x}{1 - (r_1r_2)/x^2} \to \infty$ suggesting that $\frac{A+B}{2} = \frac{\pi}{2} \Rightarrow \angle PMQ_1 + \angle PMQ_2 = \pi$.
This is what I had in mind.

On the topic of a more geometric solution, I suspect there is a way to kill it quicker using nonsyllabus techniques (inversive geometry perhaps), but it is not immediately apparent to me, I might have a crack later this week.

19. ## Re: HSC 2017 MX2 Marathon

Originally Posted by seanieg89
$Two circles \Gamma_1,\Gamma_2 of differing radii share a common point of tangency P and lie on the same side of the tangent line \ell. Let M be the point on \ell such that \angle Q_1MQ_2 is maximised, where Q_j is the point other than P on \Gamma_j with tangent passing through M. (Briefly explain why such a point M must exist, and why it is unique up to reflection in the line perpendicular to \ell through P.)\\ For this choice of M, find the angle \angle PMQ_1+\angle PMQ_2 .$
I think LaTeX broke

20. ## Re: HSC 2017 MX2 Marathon (archive)

2017 marathon now locked.

New 2018 MX2 Marathon here:
HSC 2018 MX2 Marathon

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