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HSC 2012 MX2 Marathon (archive) (3 Viewers)

Carrotsticks

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Re: 2012 HSC MX2 Marathon

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 plus sign is redundant.
 

SpiralFlex

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Re: 2012 HSC MX2 Marathon

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 plus sign is redundant.
Ah sorry I could have sworn typed the integer symbol as "Z". o_O
 

Trebla

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Re: 2012 HSC MX2 Marathon

Integers can be negative. Natural numbers are always non-negative.

So you can write or for {1, 2, 3, ....}
 
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Drongoski

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Re: 2012 HSC MX2 Marathon

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.
 

SpiralFlex

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Re: 2012 HSC MX2 Marathon

Integers can be negative. Natural numbers are always non-negative.

So you can write or 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.
 

largarithmic

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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!
 

Carrotsticks

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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.

 

Carrotsticks

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Re: 2012 HSC MX2 Marathon

Sorry, I should have made some parts that lead to the final answer.

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.
 

tywebb

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Re: 2012 HSC MX2 Marathon

This took an unbelievable amount of time to type. Better be correct.
It's not as bad if you use eiθ 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.
 
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tywebb

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Re: 2012 HSC MX2 Marathon

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




or



For positive integers though it's more common to use
 

largarithmic

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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.

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 <OAP = 30 as it is equilateral, so AO/OP = 1/sin30 = 2. So AO = 2OP, but AO is the radius of the bigger circle, OP the radius of the smaller. So each circle has half the radius of the one before it. (this can also be proven by invoking Euler's formula for triangles: 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

 

lolcakes52

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Re: 2012 HSC MX2 Marathon

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 <oap =="" 30="" as="" it="" is="" equilateral,="" so="" ao="" op="1/sin30" 2.="" but="" the="" radius="" of="" bigger="" circle,="" smaller.="" each="" circle="" has="" half="" one="" before="" it.="" (this="" can="" also="" be="" proven="" by="" invoking="" euler's="" formula="" for="" triangles:=""
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.</oap>
 
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largarithmic

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Re: 2012 HSC MX2 Marathon

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.</oap>
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.
 

lolcakes52

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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.
 

largarithmic

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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.
You dont actually need to use an infinite series. You can use the following fact (prove it for yourself its easy):

If , then both these are also equal to .

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.

Post your whole method?
 

Carrotsticks

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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.
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?
 

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