HSC 2014 MX2 Marathon ADVANCED (archive) (4 Viewers)

[mreditor16]

Re: HSC 2014 4U Marathon - Advanced Level

Advanced

OP most probably posted here, to get someone to do the question for them....

 

[faisalabdul16]

Re: HSC 2014 4U Marathon - Advanced Level

OP most probably posted here, to get someone to do the question for them....

Thanks captain obvious hahaha

 

[TL1998]

Re: HSC 2014 4U Marathon - Advanced Level

Prove that 1/a + 1/b + 1/c + 1/d >= 3/(b+c+d) + 3/(a+b+d) + 3/(a+c+d) + 3/(a+b+c)

 

[glittergal96]

Re: HSC 2014 4U Marathon - Advanced Level

Prove that 1/a + 1/b + 1/c + 1/d >= 3/(b+c+d) + 3/(a+b+d) + 3/(a+c+d) + 3/(a+b+c)

In an exam I would include a proof of 3-variable AM-GM, but hopefully you all know how to do that.

Applying AM-GM to 1/x, 1/y and 1/z and taking reciprocals we get AM(x,y,z) >= GM(x,y,z) >= 3/(1/x+1/y+1/z) (=HM(x,y,z)).

Applying the above to each triple gives:

3*LHS = (1/a+1/b+1/c)+(1/a+1/b+1/d) +(1/a+1/c+1/d) +(1/b+1/c+1/d) >= 9((a+b+c)+(a+b+d)+(a+c+d)+(b+c+d)) = 3*RHS.

 

[TL1998]

Re: HSC 2014 4U Marathon - Advanced Level

Just asking Glittergal, how have you performed in the AMC, UNSW maths comp and the austalian intermediate olympiads/ senior contests (if you've participated)

 

[dan964]

Re: HSC 2014 4U Marathon - Advanced Level

Wait, so what is the full question then? I don't see how this fits in anywhere that makes sense.

excluding pseudoprimes. Read amended comment above.
there is a way to remove Carmichael primes from showing up, unfortunately it also removes some other numbers.

Divide the LHS by n to remove factor of Carmichael primes.
Add the restriction of n not being equal to p. Or even better n is a prime that is not p. (e.g. 2, 3, 5).

 

[dan964]

Re: HSC 2014 4U Marathon - Advanced Level

http://community.boredofstudies.org...1d1411026981-complex-numbers-help-photo-1.jpg

Just some tips since you may not have learnt this yet:

The modulus |z| is given by sqrt(x^2 + y^2)
The argument (arg z) of a number is the angle between OZ and the x-axis (provided z is not the number 0)

Therefore use this to get (A=arg z)
x= |z|cos A
y= |z|sin A

Therefore z = |z|(cos A + i sin A) which is written as |z|(cis A) and is called the mod-arg form of a complex number

Question 2 use the substitution x^2+y^2 for |z|^2

Question 3 is not too hard. The proof for first case (n=1) is almost given let z=z2= z1. Assume the formula is true for n=k, then for n=k+1 write down the LHS, then use your assumption and prove true for n=k+1.....
Question 4 involves conjugates. Just sub x-iy for the conjugate (z bar).
last p of Question 4 use substitution and definition of |z|

 

[Sy123]

Re: HSC 2014 4U Marathon - Advanced Level

 

[Sy123]

Re: HSC 2014 4U Marathon - Advanced Level

 

[glittergal96]

Re: HSC 2014 4U Marathon - Advanced Level

Just asking Glittergal, how have you performed in the AMC, UNSW maths comp and the austalian intermediate olympiads/ senior contests (if you've participated)

Quite well, I won't be more specific for the sake of anonymity. I didn't do the AIMO or the UNSW though.

excluding pseudoprimes. Read amended comment above.
there is a way to remove Carmichael primes from showing up, unfortunately it also removes some other numbers.

Divide the LHS by n to remove factor of Carmichael primes.
Add the restriction of n not being equal to p. Or even better n is a prime that is not p. (e.g. 2, 3, 5).

Your last sentence is confusing to me, I will take the following interpretation though I am not 100% sure this is what you mean.

Show that the statement:
"n^(p-1)-1 is divisible by p for all n not divisible by p"
is a necessary and sufficient condition for p to be prime.

First assume p is prime. As discussed before, induction shows n^p-n is then divisible by p for ALL n. But n^p-n=n(n^(p-1)-1). As the RHS is divisible by p and it's first factor isn't, we must have n^(p-1)-1 divisible by p as required. (Note here the fundamental property of primes that if p|xy then p|x or p|y.)

Conversely, suppose p is such that a^(p-1)-1 is divisible by p for all a not divisible by p. Let d =/= p be a factor of p. Since d^(p-1)-1 is divisible by p, it must certainly also be divisible by d. But this would imply that d divides 1. As such, the only factor p can have other than itself is 1...ie p is prime.

 

[glittergal96]

Re: HSC 2014 4U Marathon - Advanced Level

It suffices to show that the polynomial

p(x)=x^2+(x+1)^2-b^4-(b+1)^4 has no positive integer roots for any positive integer b.

This polynomial is trivially increasing, so we must have a unique positive real root. But we can compute

p(b^2+b)=-2b^2-2b < 0,

and

p(b^2+b+1)=2b^2+2b+4 > 0.

So the unique positive real root lies strictly between two consecutive positive integers and cannot itself be a positive integer.

 

[TL1998]

Re: HSC 2014 4U Marathon - Advanced Level

Sy123, please do not answer this question as you have already done so before. Give the 2014ers a go at this thank you.

The lengths of the sides of a triangle form an arithmetic progression. The largest angle exceeds the smallest by 90 degrees. Find the ratios of the lengths of the sides.

 

[dunjaaa]

Re: HSC 2014 4U Marathon - Advanced Level

Just asking, are the numbers nice, because I got something really ugly.

 

[TL1998]

Re: HSC 2014 4U Marathon - Advanced Level

How ugly is it. They arent integers but theyre still pretty neat. What did you get

 

[dunjaaa]

Re: HSC 2014 4U Marathon - Advanced Level

Here's what I got

 

[glittergal96]

Re: HSC 2014 4U Marathon - Advanced Level

The sine rule (and a little trick involving supplementary angles when the triangle is obtuse) gives us us that the smallest side of a triangle is opposite its smallest angle. (And similarly for its largest side.)

Now we scale the triangle so it's smallest side is 1, for simplicity. This will of course not change the angles or the ratios of side lengths.

The sine rule gives us



Considering the first and second of these expressions gives



Considering the first and third gives:



Removing common factors, multiplying out and squaring gives a quadratic in d, whose only positive root is



So the two ratios of larger side lengths to the smallest one are 1+d,1+2d, which are equal to

 

[dunjaaa]

Re: HSC 2014 4U Marathon - Advanced Level

Nice solution, I don't know why my final expression turned out so ugly.

 

[glittergal96]

Re: HSC 2014 4U Marathon - Advanced Level

Nice solution, I don't know why my final expression turned out so ugly.

Note the minor correction, I originally wrote 2+d instead of 1+2d for the second lol.

 

[3ddementor]

Re: HSC 2014 4U Marathon - Advanced Level

Can anyone give me a heads up as to how to do Q16 (iii) , it's really doing my head in :S (from Dan964's set of questions)

 

[Sy123]

Re: HSC 2014 4U Marathon - Advanced Level

Does this question need clarification?