HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)
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glittergal96
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Re: HSC 2014 4U Marathon - Advanced Level
Well sure lol. The power means inequalities in general are all helpful / easy to remember.
Well sure lol. The power means inequalities in general are all helpful / easy to remember.
Re: HSC 2014 4U Marathon - Advanced Level
The USSR Olympiad Problem book by Shklarsky has a small but deeply insightful chapter on the algebra of polynomials. Chapter 8, i think. All questions have worked solutions too.Does anyone know a book that has deeper complex numbers/polynomial theoretical styled problems like the 2011 Extension 2 HSC Exam Q8 (c). If so, please let us know.
Re: HSC 2014 4U Marathon - Advanced Level
http://prntscr.com/4fvye5
Any idea with part (c) (iii)
This is BOS hsc 1996 Q8 btw
http://prntscr.com/4fvye5
Any idea with part (c) (iii)
This is BOS hsc 1996 Q8 btw
sepseminar
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Re: HSC 2014 4U Marathon - Advanced Level
With regards to the circle geometry problem above, there are a number of ways to solve the problem. If you want a hint, just highlight the following text: Proof by contradiction is very helpful in existence problems like these. For instance, you could consider what would happen if you had a quadrilateral with only three sides tangent to a circle. You'll find that the given condition on the sides leads to a contradiction.
I believe the result is called Pitot's Theorem, which, although outside the syllabus, is definitely provable within the confines of the MX2 syllabus and is a nifty result.
With regards to the circle geometry problem above, there are a number of ways to solve the problem. If you want a hint, just highlight the following text: Proof by contradiction is very helpful in existence problems like these. For instance, you could consider what would happen if you had a quadrilateral with only three sides tangent to a circle. You'll find that the given condition on the sides leads to a contradiction.
I believe the result is called Pitot's Theorem, which, although outside the syllabus, is definitely provable within the confines of the MX2 syllabus and is a nifty result.
sepseminar
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Re: HSC 2014 4U Marathon - Advanced Level
Sorry, I should have been more clear when I said 'confines of the syllabus'. While I don't think it's explicitly mentioned in the syllabus, it is a technique that comes up in HSC 4u exams (generally in the harder questions, for instance in 2011). Moreover, it's a technique that's used very generally in proofs of the results that have come up in the 3u/4u syllabus (and in mathematics as a field of research). For instance, the proofs of the converses of many of the basic circle geometry theorems are all good examples of the logically intuitive idea of proof by contradiction, which has to be used subtly on most occasions.
I should note that the problem is still difficult with that hint and so kudos to anyone who solves it.
Sorry, I should have been more clear when I said 'confines of the syllabus'. While I don't think it's explicitly mentioned in the syllabus, it is a technique that comes up in HSC 4u exams (generally in the harder questions, for instance in 2011). Moreover, it's a technique that's used very generally in proofs of the results that have come up in the 3u/4u syllabus (and in mathematics as a field of research). For instance, the proofs of the converses of many of the basic circle geometry theorems are all good examples of the logically intuitive idea of proof by contradiction, which has to be used subtly on most occasions.
I should note that the problem is still difficult with that hint and so kudos to anyone who solves it.
RealiseNothing
what is that?It is Cowpea
Re: HSC 2014 4U Marathon - Advanced Level
Just because something isn't in the textbook doesn't mean it isn't a skill that can't be used.
Take for example a telescoping sum, which isn't in textbooks, but has come up plenty of times in HSC exams.
It has come up in many past HSC exams.
Just because something isn't in the textbook doesn't mean it isn't a skill that can't be used.
Take for example a telescoping sum, which isn't in textbooks, but has come up plenty of times in HSC exams.
RealiseNothing
what is that?It is Cowpea
RealiseNothing
what is that?It is Cowpea
Re: HSC 2014 4U Marathon - Advanced Level
Let
be the k'th Fibonacci number.
Show that:
Let
Show that:
RealiseNothing
what is that?It is Cowpea
Re: HSC 2014 4U Marathon - Advanced Level
Just in case there isn't, try this:
Prove that!+1 = kp)
for some integer 
if and only if 
is prime.
I'm not sure if there is a HSC method for this question, but feel free to try.Let
Show that:
Just in case there isn't, try this:
Prove that
RealiseNothing
what is that?It is Cowpea
Re: HSC 2014 4U Marathon - Advanced Level
2) Hence construct a tangent, say AE.
3) Combine the equation you are given that AB+CD=AD+BC with part (b).
4) You can arrive at a contradiction by messing around with this.
1) Assume that AD is not a tangent.
2) Hence construct a tangent, say AE.
3) Combine the equation you are given that AB+CD=AD+BC with part (b).
4) You can arrive at a contradiction by messing around with this.
RealiseNothing
what is that?It is Cowpea
Re: HSC 2014 4U Marathon - Advanced Level
I tried and got close, can't finish it off though.
Not even sure how doable this is with HSC methods.I'm not sure if there is a HSC method for this question, but feel free to try.
Just in case there isn't, try this:
Prove that
I tried and got close, can't finish it off though.
Carrotsticks
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Re: HSC 2014 4U Marathon - Advanced Level
I doubt there are any 'HSC method proofs' without resorting to some sort of modular arithmetic.
sepseminar
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Re: HSC 2014 4U Marathon - Advanced Level
Yeah, a proof of that result about the primes (I believe it's called Wilson's Theorem) in all probability will involve modular arithmetic, which isn't included in the syllabus. But, if you do happen to know how that works, it is possible to understand a proof. However, the ideas are pretty difficult to motivate and you'd need to lead people through the problem. Here's a probability question.
$)
Yeah, a proof of that result about the primes (I believe it's called Wilson's Theorem) in all probability will involve modular arithmetic, which isn't included in the syllabus. But, if you do happen to know how that works, it is possible to understand a proof. However, the ideas are pretty difficult to motivate and you'd need to lead people through the problem. Here's a probability question.
RealiseNothing
what is that?It is Cowpea
Re: HSC 2014 4U Marathon - Advanced Level
I ended up getting a 4U solution to it. It uses the ideas behind mod arith without directly using mod arith which I think a good 4U student is capable of.
RealiseNothing
what is that?It is Cowpea
Re: HSC 2014 4U Marathon - Advanced Level
I get 51.2%Yeah, a proof of that result about the primes (I believe it's called Wilson's Theorem) in all probability will involve modular arithmetic, which isn't included in the syllabus. But, if you do happen to know how that works, it is possible to understand a proof. However, the ideas are pretty difficult to motivate and you'd need to lead people through the problem. Here's a probability question.
Carrotsticks
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glittergal96
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Re: HSC 2014 4U Marathon - Advanced Level
For the probability question I got 3361/6561.
I just enumerated the colourings that don't contain a non-degenerate red triangle as follows:
(Colourings with less than 3 red points) + (Colourings with 3 red collinear points).
For Wilson's theorem, you could show that:
(~) If p is prime, then every nonzero a in {0,1,2,...,p-1} has a multiplicative inverse b (mod p), and that 1 and (p-1) are the only numbers that are self-inverse.
Once you have (~), the product (p-1)! cancels out in pairs except for the first and last term. This leaves a residue of 1*(p-1)=-1 mod p. This proves one direction of Wilson's and the converse is trivial.
To make this an extension two solution, I think we could just recast (~) as
For any prime p and any positive integer a there is a positive integer b such that ab-1 is divisible by p. Moreover, if b' is another such integer, then b-b' must be divisible by p. Also, if a-b is divisible by p and ab-1 is divisible by p, then either a-1 or a+1 is divisible by p.
Knowing about division/remainders is in the course, so this modified statement leads to a proof without mentioning anything out of syllabus I think.
For the probability question I got 3361/6561.
I just enumerated the colourings that don't contain a non-degenerate red triangle as follows:
(Colourings with less than 3 red points) + (Colourings with 3 red collinear points).
For Wilson's theorem, you could show that:
(~) If p is prime, then every nonzero a in {0,1,2,...,p-1} has a multiplicative inverse b (mod p), and that 1 and (p-1) are the only numbers that are self-inverse.
Once you have (~), the product (p-1)! cancels out in pairs except for the first and last term. This leaves a residue of 1*(p-1)=-1 mod p. This proves one direction of Wilson's and the converse is trivial.
To make this an extension two solution, I think we could just recast (~) as
For any prime p and any positive integer a there is a positive integer b such that ab-1 is divisible by p. Moreover, if b' is another such integer, then b-b' must be divisible by p. Also, if a-b is divisible by p and ab-1 is divisible by p, then either a-1 or a+1 is divisible by p.
Knowing about division/remainders is in the course, so this modified statement leads to a proof without mentioning anything out of syllabus I think.
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