HSC 2012 MX1 Marathon #1 (archive) (1 Viewer)

SpiralFlex

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Re: 2012 HSC MX1 Marathon

The function is integrable using methods taught in Extension 1.

Perhaps the sketching part is inappropriate for MX1. Everything else seems okay.
Of course it's integrable for MX1. But I am talking about the sketching part. Prehaps Zeebobby would like to alter the question.
 

largarithmic

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Re: 2012 HSC MX1 Marathon

Just realised that the value was half that of the golden ratio.
really cool fact (kinda unrelated but alright...)

you know every positive integer can be written as sum of distinct integer powers of phi?

e.g.





(where )

pretty hard question: prove it.
 

bleakarcher

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Re: 2012 HSC MX1 Marathon

Isnt that coroneos that loop question?
 

bleakarcher

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Re: 2012 HSC MX1 Marathon

Also, spiralflex i just realised you made an error when integrating 4cos^(x)
 

Carrotsticks

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Re: 2012 HSC MX1 Marathon

really cool fact (kinda unrelated but alright...)

you know every positive integer can be written as sum of distinct integer powers of phi?

e.g.





(where )

pretty hard question: prove it.
I'll let somebody else give it a try, but does this proof happen to use the following properties:







 
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zeebobDD

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Re: 2012 HSC MX1 Marathon

okay guys, actually i didnt get it from any of the books you mentioned:p, i got it from the cambridge 3u(yr11) though it was on the extension part, probz been in others too

aha i think the only difficulty in that question is the sketching:p alrite stuff that q, lets just moveee on
 

largarithmic

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Re: 2012 HSC MX1 Marathon

I'll let somebody else give it a try, but does this proof happen to use the following properties:







Uses the first property only - don't need anything to do with fibonacci numbers. Try it, its a fun question :)
 

Carrotsticks

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Re: 2012 HSC MX1 Marathon

Uses the first property only - don't need anything to do with fibonacci numbers. Try it, its a fun question :)
I have a proof in mind, but it uses two of the aforementioned identities. I'll give it a try once I get back from work.
 

Carrotsticks

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Re: 2012 HSC MX1 Marathon

Larg, regarding your question with the powers of phi.

Did you mean...



...instead of...



I managed to prove the following:

If n is odd, then:



If n is even, then:



Using a combination of the above, we can acquire the integers 1, 3, 4, 7, 11, 18 ...

For example:



This is a relationship using the same recursive formula as the Fibbonaci Sequence:



But with the conditions:



As opposed to:



Zeckendorf's Theorem (http://en.wikipedia.org/wiki/Zeckendorf's_theorem) states that all positive integers can be expressed as the sum of distinct terms of the Fibbonaci sequence.

Not quite sure about the validity of this, but since we are utilising the same recursive definition (but different starting conditions), all integers can be expressed as distinct sums (and subtractions) of powers of phi (though usually a positive power is paired with its equivalent negative power).

For example:

 

largarithmic

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Re: 2012 HSC MX1 Marathon

Larg, regarding your question with the powers of phi.

Did you mean...



...instead of...



I managed to prove the following:

If n is odd, then:



If n is even, then:



Using a combination of the above, we can acquire the integers 1, 3, 4, 7, 11, 18 ...

For example:



This is a relationship using the same recursive formula as the Fibbonaci Sequence:



But with the conditions:



As opposed to:



Zeckendorf's Theorem (http://en.wikipedia.org/wiki/Zeckendorf's_theorem) states that all positive integers can be expressed as the sum of distinct terms of the Fibbonaci sequence.

Not quite sure about the validity of this, but since we are utilising the same recursive definition (but different starting conditions), all integers can be expressed as distinct sums (and subtractions) of powers of phi (though usually a positive power is paired with its equivalent negative power).

For example:

You've answered a slightly different question - I said *sum* of distinct integer powers of phi - you can't have differences. But if you could, yeah that still works ^^ And I believe that theorem is actually pretty easy to prove (you can induct with a pretty simple bounding argument I think). But yeah the question without differences you can do without any knowlegde of the fibonacci sequence - and with barely any algebra actually.
 

cssftw

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Re: 2012 HSC MX1 Marathon

good heavens....i got 95 in the ext1 hsc - and now 3months of not touching a single page of maths - i've forgotten over 80% of the course.

G fucking G
 

king chopper

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Re: 2012 HSC MX1 Marathon

Okay - AB is a chord of a circle. P is the midpoint of the minor arc cut off by AB, and Q is the midpoint of the major arc. Show that PQ is a diameter (hint: include the centre of the circle in your diagram.)
 

largarithmic

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Re: 2012 HSC MX1 Marathon

Okay - AB is a chord of a circle. P is the midpoint of the minor arc cut off by AB, and Q is the midpoint of the major arc. Show that PQ is a diameter (hint: include the centre of the circle in your diagram.)
You dont even need to use the centre. here's a pretty simple proof:
Note that the combined length of arc PAQ is the same as arc PBQ (since arcs AP = PB, BQ = QA). Hence the angles they subtend are equal, so /_PAQ = /_PBQ, but by cyclic quadrilateral interior angle sum /_PAQ + /_PBQ = 180 so /_PAQ = 90. It follows that PQ is a diameter

Alternatively you can do a really neat argument from symmetry. Suppose we relabelled A as B, and B as A. Now the definition of a midpoint of an arc is fundamentally symmetric (you can relabel the arc endpoints and the actual midpoint remains the same). Hence any transformation that (1) preserves the circle and (2) flips A and B would not change the position of P and Q. Circles have an axis of symmetry in every line through their centre, so pick the one that flips A and B. Note that under reflexion about this axis, both P and Q remain at the same points - and when you reflect an entire plane about a point on it, the only "invariant" points are points on the line itself. So P and Q are on the axis of symmetry - which was through the centre in the first place, i.e. is a diameter. So PQ is a diameter. (a nice, subtle argument that doesnt use any equations or working out, just intuition and logic)
 

Trebla

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Re: 2012 HSC MX1 Marathon

Suppose that there are two angles subtended from a common interval AB (say angles ACB and ADB). If angle ACB is twice the size of angle ADB and AC = BC show that the points A, B and D form a circle with C as the centre.
 
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largarithmic

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Re: 2012 HSC MX1 Marathon

Suppose that there are two angles subtended from a common interval AB (say angles ACB and ADB). If angle ACB is twice the size of angle ADB, show that the points A, B and D are concyclic.
wahhhh... A, B and D are always going to be concyclic so long as <ADB is not zero (i.e. A,D,B collinear) since any three noncollinear points are concyclic. If you meant A, B, D are on a circle centre C, thats not true; although A, B, C and the centre of the circle through A, B and D are themselves concyclic
 

Carrotsticks

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Re: 2012 HSC MX1 Marathon

wahhhh... A, B and D are always going to be concyclic so long as
so long as... ? I don't see anything lol.

And to Trebla, any 3 points in 2D space are concyclic, therefore A, B and D are concyclic (assuming they are not collinear).
 

king chopper

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Re: 2012 HSC MX1 Marathon

Alright, another one ... PQ is the diameter of a circle, centre O. X is any point on the circle (not at P or Q). PX is extended to Y such that XY = OX. Find in simplest form the ratio of angle YOQ : anglee YOX
 

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