# Thread: 2012 HSC MX2 Marathon

1. ## Re: 2012 HSC MX2 Marathon

Originally Posted by Carrotsticks
What?

The generally accepted definition of the field of Natural numbers are positive integers (some people also consider 0 to be a natural number, but that's just preference).
The set of natural numbers is not a field.

2. ## Re: 2012 HSC MX2 Marathon

Originally Posted by Drongoski
The set of natural numbers is not a field.
*set on the field of real numbers.

3. ## Re: 2012 HSC MX2 Marathon

Originally Posted by Trebla
Integers can be negative. Natural numbers are always non-negative.

So you can write $\mathbb{N}$ or $\mathbb{Z}^+$ for {1, 2, 3, ....}
Yes that is what I meant, but I mistook the N for a Z, that is why I added a N plus.

4. ## Re: 2012 HSC MX2 Marathon

Theres basically no agreed standard for what "N" means... I usually just write N_0 or N^+ (dunno how to do blackboard bold on this), in a way that would be pretty unambiguous.

Someone give a problem!

5. ## Re: 2012 HSC MX2 Marathon

Not really something that would appear in the HSC (perhaps a trial if some school likes weird questions).

Consider a unit circle. Inscribed within this unit circle is an equilateral triangle such that all 3 vertices touch the circle. Inscribed within this triangle is another circle such that all 3 sides of the triangle are in contact with the circle.

This process repeats infinitely.

Find the limiting ratio Red : Green as the number of iterations approaches infinity.

6. ## Re: 2012 HSC MX2 Marathon

But I love questions where you're just told what to find, and you can use any method to come to a solution. This is where you start seeing originality.

7. ## Re: 2012 HSC MX2 Marathon

Originally Posted by tywebb
Here's the solution to largarithmic's question:

Here's the solution to spiralfex's question:

Here's the next question:

$\ \newline(a)\text{ If }y=\textstyle{1\over x}+\sum_{n=1}^\infty{2x\over x^2-n^2}\text{ for }01\newline\ \newline (e)\text{ Let }k=2\text{ to prove Wallis' Product, }\prod_{n=1}^\infty{(2n)^2\over (2n)^2-1}={\pi\over2}$
Dude
Holy
Crap
Lots of writing!
How long did that take you?

8. ## Re: 2012 HSC MX2 Marathon

Originally Posted by nightweaver066
This took an unbelievable amount of time to type. Better be correct.
It's not as bad if you use e notation. (Here I'm referring to spiralflex's question).

Also, for part (iii) you are solving for n, not θ and you should get n=2k.

9. ## Re: 2012 HSC MX2 Marathon

Originally Posted by largarithmic
I usually just write N_0 or N^+ (dunno how to do blackboard bold on this)
Just write (without spaces)

[ tex ]\mathbb{N}_0[ /tex ]

or

[ tex ]\mathbb{N}^+[ /tex ]

and you should get

$\mathbb{N}_0$

or

$\mathbb{N}^+$

For positive integers though it's more common to use $\mathbb{Z}^+$

10. ## Re: 2012 HSC MX2 Marathon

Originally Posted by Carrotsticks
Not really something that would appear in the HSC (perhaps a trial if some school likes weird questions).

Consider a unit circle. Inscribed within this unit circle is an equilateral triangle such that all 3 vertices touch the circle. Inscribed within this triangle is another circle such that all 3 sides of the triangle are in contact with the circle.

This process repeats infinitely.

Find the limiting ratio Red : Green as the number of iterations approaches infinity.

cool problem!!!

Anyway here's how I'd do it.

Observation: each circle has half the radius as the one before it. Proof: Clearly all the circles, since the diagram has rotational symmetry, have the same centre; call this O. Now let a vertex of say the biggest triangle by A, and the second biggest circle be tangent to a side of the biggest triangle at say P where one of the endpoints of the side P is on is A. Then where O,I,R,r are the circumcentre, incentre, circumradius, inradius respectively.

We now find the ratio of the red area to the green area in just one annulus. So say take the biggest circle and triangle, and cut a hole where the next circle is. Say the radius of the big circle is one. Then the area of the triangle including the hole is 3*1/2*1*1*sin120 = (3root3)/4. Then, the red area is R = pi - (3root3)/4 and the green area is G = (3root3)/4 - pi/4.

We now find the final answer. All the red and green areas form a geometric series with ratio 1/4 since each circle is half the radius of the one before so areas are reduced to a quarter; so the total red area = R(1+1/4+1/16+...) = R/1-(1/4) = 4R/3. Similarly total green area = 4G/3. Then the final ratio is the same as R/G, i.e. it is

$R:G = \frac{4\pi - 3\sqrt{3}}{3\sqrt{3}-\pi}$

11. ## Re: 2012 HSC MX2 Marathon

Originally Posted by largarithmic
cool problem!!!

Anyway here's how I'd do it.

Observation: each circle has half the radius as the one before it. Proof: Clearly all the circles, since the diagram has rotational symmetry, have the same centre; call this O. Now let a vertex of say the biggest triangle by A, and the second biggest circle be tangent to a side of the biggest triangle at say P where one of the endpoints of the side P is on is A. Then
We now find the ratio of the red area to the green area in just one annulus. So say take the biggest circle and triangle, and cut a hole where the next circle is. Say the radius of the big circle is one. Then the area of the triangle including the hole is 3*1/2*1*1*sin120 = (3root3)/4. Then, the red area is R = pi - (3root3)/4 and the green area is G = (3root3)/4 - pi/4.

We now find the final answer. All the red and green areas form a geometric series with ratio 1/4 since each circle is half the radius of the one before so areas are reduced to a quarter; so the total red area = R(1+1/4+1/16+...) = R/1-(1/4) = 4R/3. Similarly total green area = 4G/3. Then the final ratio is the same as R/G, i.e. it is

$R:G = \frac{4\pi - 3\sqrt{3}}{3\sqrt{3}-\pi}$
What if the radius isn't one? When I did the question, I got a similar answer but I used a variable r and couldn't simplify past a point, I am sure I did something wrong but when I got a similar looking anwser.

12. ## Re: 2012 HSC MX2 Marathon

Originally Posted by lolcakes52
What if the radius isn't one? When I did the question, I got a similar answer but I used a variable r and couldn't simplify past a point, I am sure I did something wrong but when I got a similar looking anwser.
Doesn't matter because of dimensionality. Like, what does "radius one" mean? It means, "radius one UNIT" where that unit is something you chose. You could have, the outer radius is radius "one inch" for instance, or could be "two centimetres": but if its radius "two centimetres" you just let see, a wombiunit = two centimetres, and call is "radius one wombiunit". When dealing with areas you need to adjust because conversion factors are squared. But it should definitely work out if the radius is R instead of 1, you just literally add a factor of R onto every length and a factor of R^2 onto every area and its still correct.

13. ## Re: 2012 HSC MX2 Marathon

Right, so I did it again and I still get a different answer. Your method is definitely shorter as I am using infinite series and a lot more algebra.

14. ## Re: 2012 HSC MX2 Marathon

Originally Posted by lolcakes52
Right, so I did it again and I still get a different answer. Your method is definitely shorter as I am using infinite series and a lot more algebra.
You dont actually need to use an infinite series. You can use the following fact (prove it for yourself its easy):

If $\frac{a}{b} = \frac{c}{d}$, then both these are also equal to $\frac{a+c}{b+d}$.

Then if R_i and G_i denote the red area/green area within the ith annulus, clearly R1/G1 = R2/G2 = R3/G3 =... by similarity so applying the above identity the total ratio = R1+R2+.../G1+G2+... = R1/G1. I computed R1 and G1 above (and I'm pretty sure theyre both correct), giving another way to get my answer.

15. ## Re: 2012 HSC MX2 Marathon

Originally Posted by lolcakes52
Right, so I did it again and I still get a different answer. Your method is definitely shorter as I am using infinite series and a lot more algebra.
When you say 'infinite series', are you referring to the infinite geometric series? If the algebra is crazy, it implies that you are going down the wrong track. There is always a simpler solution.

Nice work larg on your solution. Perhaps you can post a similar 'fun' question?

16. ## Re: 2012 HSC MX2 Marathon

Yes, infinite geometric series. I think I am just confusing myself.
This is from the unsw school mathematics comp, it is interesting and I would consider it harder 3 unit.

Screen shot 2012-01-09 at 6.53.45 PM.png

17. ## Re: 2012 HSC MX2 Marathon

----------------------------------------------------------------------------------------------

I already posted up a question before but maybe people forgot, so here it is again:

$\ \newline(a)\text{ If }y=\textstyle{1\over x}+\sum_{n=1}^\infty{2x\over x^2-n^2}\text{ for }01\newline\ \newline (e)\text{ Let }k=2\text{ to prove Wallis' Product, }\prod_{n=1}^\infty{(2n)^2\over (2n)^2-1}={\pi\over2}$

18. ## Re: 2012 HSC MX2 Marathon

tywebb, I have already done it, but I didn't want to post a solution because I knew it would be a crazy amount of typing

I think somebody else should try posting a solution, this is a thread aimed at MX2 students after all.

19. ## Re: 2012 HSC MX2 Marathon

No. If you've got a solution, post it up. Don't wait for others to do it or we might be waiting forever.

You don't need to type it. Just scan and upload will do.

20. ## Re: 2012 HSC MX2 Marathon

Funny how this thread is titled "2012 HSC MX2 Marathon" yet there are hardly any current HSC students participating. How about toning down the difficulty and topic coverage of the questions to encourage participation from HSC students who've only done a term of the course? I'd imagine the questions above would be very intimidating for most of them.

21. ## Re: 2012 HSC MX2 Marathon

You are right.

Note: People who have posted in this thread are NOT ALLOWED to answer this. I want some fellow extension 2 students to feel welcomed.

$a) Solve in mod arg form: z^4+1=0$

$b) Plot the roots on an Argand diagram.$

$c) Find the area of the quadrilateral formed by these roots.$

22. ## Re: 2012 HSC MX2 Marathon

Originally Posted by largarithmic
$(ii) In the case when n is odd, \textbf{\underline{shower}} by considering (\sqrt{a+1} + \sqrt{a})T_m where m is even, or otherwise, that Tn can be written in the form C_n\sqrt{a+1} + D_n \sqrt{a} where Cn and Dn are integers depending on a and n and (a+1)C_n^2 = aD_n^2 + 1$
lol

23. ## Re: 2012 HSC MX2 Marathon

$a) z^4&=-1$
$&=cis(\pi +2k\pi)$
$&=cis(\pi/4), cis(3\pi/4), cis(-\pi/4), cis(-3\pi/4)$

I don't know how to plot them on here but the third part:
c) All diagonals are equal length (all have modulus 1). They are perpendicular to each other (separation of pi degrees). Therefore the quadrilateral is a square with diagonal length 2 units. Each side has length root 2 so the area of the quad is 2 units squared.

Hooray, first time using latex .
A few questions: How do you type text without it becoming bunched up and how do you align your equals signs? I tried following the guide but it wasn't working...

24. ## Re: 2012 HSC MX2 Marathon

Originally Posted by deswa1
$a) z^4&=-1$
$&=cis(\pi +2k\pi)$
$&=cis(\pi/4), cis(3\pi/4), cis(-\pi/4), cis(-3\pi/4)$

How do you type text without it becoming bunched up and how do you align your equals signs?

To align, type without spaces for [ tex ] and [ /tex ]:

[ tex ]\begin{aligned}a)\ z^4&=-1\\ &=\mathrm{c}i\mathrm{s}\; (\pi+2k\pi)\\ &=\mathrm{c}i\mathrm{s}\; (\pi/4),\ \mathrm{c}i\mathrm{s}\; (3\pi/4),\ \mathrm{c}i\mathrm{s}\; (-\pi/4),\ \mathrm{c}i\mathrm{s}\; (-3\pi/4)\end{aligned}[ /tex ]

to get

\begin{aligned}a)\ z^4&=-1\\ &=\mathrm{c}i\mathrm{s}\; (\pi+2k\pi)\\ &=\mathrm{c}i\mathrm{s}\; (\pi/4),\ \mathrm{c}i\mathrm{s}\; (3\pi/4),\ \mathrm{c}i\mathrm{s}\; (-\pi/4),\ \mathrm{c}i\mathrm{s}\; (-3\pi/4)\end{aligned}

Although your third line is actually incorrect and so it should be

\begin{aligned}a)\ z^4&=-1\\ &=\mathrm{c}i\mathrm{s}\; (\pi+2k\pi)\\ {\color{BrickRed}{\therefore z}}&=\mathrm{c}i\mathrm{s}\; (\pi/4),\ \mathrm{c}i\mathrm{s}\; (3\pi/4),\ \mathrm{c}i\mathrm{s}\; (-\pi/4),\ \mathrm{c}i\mathrm{s}\; (-3\pi/4)\end{aligned}

And suppose I want to type this sentence in latex without it being bunched up.

I'd do it like this:

[ tex ]\text{And suppose I want to type this sentence in latex without it being bunched up.}[ /tex ]

to get

$\text{And suppose I want to type this sentence in latex without it being bunched up.}$

25. ## Re: 2012 HSC MX2 Marathon

Thanks for that . How would you do this question:
complex.png

I could do it but the second part took me a bit of time whilst in the answers they just wrote it down with no working (past school paper). Am I missing something ridiculously simple?

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