HSC 2013 MX2 Marathon (archive) (5 Viewers)

JJ345 · Aug 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$An$ \ n \ $sided regular polygon is such that no side is vertical in gradient$

$Prove that if$ \ \ m_k \ \ $is the gradient of the$ \ k^{th} \ $ side of the polygon then$

$m_1m_2 + m_2m_3 + m_3 m_4 + \dots + m_{n-1}m_n + m_n m_1 = -n$

tanx= m(1) -m(2)/1+m(1)m(2)
Where x= (n-2)*180/n
Rearranging m(1)m(2) = (m(1)-m(2) /tanx )-1 ....(1)
Rewrite (1) for all combinations m(2)m(3).....m(n)m(1)
Then add side by side. I think?

Sy123 · Aug 3, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
tanx= m(1) -m(2)/1+m(1)m(2)
Where x= (n-2)*180/n
Rearranging m(1)m(2) = (m(1)-m(2) /tanx )-1 ....(1)
Rewrite (1) for all combinations m(2)m(3).....m(n)m(1)
Then add side by side. I think?

Yep that is correct.

============

$$The product of positive real numbers$ \ a_1, a_2, a_3, \dots , a_n \ $is equal to$ \ 1$

$Prove that$

$(1+a_1)(1+a_2)\dots (1+a_n) \geq 2^n$

HeroicPandas · Aug 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep that is correct.

============

$$The product of positive real numbers$ \ a_1, a_2, a_3, \dots , a_n \ $is equal to$ \ 1$

$Prove that$

$(1+a_1)(1+a_2)\dots (1+a_n) \geq 2^n$

$\rightarrow \\ (\sqrt{1} - \sqrt{a_i})^2 \geq 0 ($where i = 1, 2, 3, \dots n) \\\\ $1+ a_i\geq 2\sqrt{a_i} \\\\ $Let i = 1, 2, 3 \dots n and multiply all of them\\\\ $(1+a_1)(1+a_2)\dots (1+a_n) \geq 2^n \sqrt{a_1 \cdot a_2 \dots \cdot a_n} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 2^n ($since a_1 \cdot a_2 \dots \cdot a_n = 1)$

Sy123 · Aug 3, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$\rightarrow \\ (\sqrt{1} - \sqrt{a_i})^2 \geq 0 ($where i = 1, 2, 3, \dots n) \\\\ $1+ a_i\geq 2\sqrt{a_i} \\\\ $Let i = 1, 2, 3 \dots n and multiply all of them\\\\ $(1+a_1)(1+a_2)\dots (1+a_n) \geq 2^n \sqrt{a_1 \cdot a_2 \dots \cdot a_n} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 2^n ($since a_1 \cdot a_2 \dots \cdot a_n = 1)$

Yep nice work

===========

$You may use the following identity$

$\frac{a_1 + a_2 + a_3 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 a_3 \dots a_n}$

$$The sum of the positive real numbers$ \ x_1, x_2, x_3, \dots, x_n \ $is$ \ 1/2$

$Prove that$

$\frac{1-x_1}{1+x_1} \cdot \frac{1-x_2}{1+x_2} \cdot \frac{1-x_3}{1+x_3} \dots \frac{1-x_n}{1+x_n} \geq \frac{1}{3}$

JJ345 · Aug 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep nice work

===========

$You may use the following identity$

$\frac{a_1 + a_2 + a_3 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 a_3 \dots a_n}$

$$The sum of the positive real numbers$ \ x_1, x_2, x_3, \dots, x_n \ $is$ \ 1/2$

$Prove that$

$\frac{1-x_1}{1+x_1} \cdot \frac{1-x_2}{1+x_2} \cdot \frac{1-x_3}{1+x_3} \dots \frac{1-x_n}{1+x_n} \geq \frac{1}{3}$

Not too confident with my solution here. But i'm getting something along the lines of Let a(1)=1-x(1)....continue this for 1,2,3...n
So on the LHS we get (n-1/2)/n >= nth root of what we want on the LHS numerator in our final expression. ....(1)
Similarly, Let a(1)= 1+x(1)...continue for 1,2,3..n so on LHS we get (n+1/2)/n >=nth root of what we want on the LHS numerator in our final expression ....(2)

Divide (1) by (2) and let n=1 because we know that there must be atleast one term for this to happen so (n-1/2)/ (n+1/2) will become 1/3 and nth root will disappear.

Or something along the lines of that dodgy solution...sorry don't know how to use latex yet

Sy123 · Aug 3, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
Not too confident with my solution here. But i'm getting something along the lines of Let a(1)=1-x(1)....continue this for 1,2,3...n
So on the LHS we get (n-1/2)/n >= nth root of what we want on the LHS numerator in our final expression. ....(1)
Similarly, Let a(1)= 1+x(1)...continue for 1,2,3..n so on LHS we get (n+1/2)/n >=nth root of what we want on the LHS numerator in our final expression ....(2)

Divide (1) by (2) and let n=1 because we know that there must be atleast one term for this to happen so (n-1/2)/ (n+1/2) will become 1/3 and nth root will disappear.

Or something along the lines of that dodgy solution...sorry don't know how to use latex yet

Solution is correct until the 'let n=1' argument.

$f(n) = \left(\frac{n - 1/2}{n+1/2} \right )^n$ which is what you got as the lower bound

By subbing in n=1 we do get 1/3, so the next problem then is:

$$Prove$ \ \left( \frac{2n-1}{2n+1} \right )^n > \frac{1}{3}$

Or rather, Prove f(n) > f(1)

Which can be done then via.......

(an outline of the method is needed only, not the actual solution)

Sy123 · Aug 3, 2013

Re: HSC 2013 4U Marathon

$$Prove for some integer$ \ n$

$\binom{n}{0} - \frac{1}{3} \binom{n}{1} + \frac{1}{5} \binom{n}{2} - \dots + \frac{1}{2n+1} \binom{n}{n} = \frac{2^{2n}}{\binom{2n}{n}}$

JJ345 · Aug 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Solution is correct until the 'let n=1' argument.

$f(n) = \left(\frac{n - 1/2}{n+1/2} \right )^n$ which is what you got as the lower bound

By subbing in n=1 we do get 1/3, so the next problem then is:

$$Prove$ \ \left( \frac{2n-1}{2n+1} \right )^n > \frac{1}{3}$

Or rather, Prove f(n) > f(1)

Which can be done then via.......

(an outline of the method is needed only, not the actual solution)

Ok, that makes sense, so proving that f(n) is increasing for n>1 should suffice?
I'm thinking of differentiating using logarithms---> but maybe this method is going a bit overboard.

Sy123 · Aug 4, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
Ok, that makes sense, so proving that f(n) is increasing for n>1 should suffice?
I'm thinking of differentiating using logarithms---> but maybe this method is going a bit overboard.

Yep proving f is increasing is enough, and differentiating with logarithms is the most direct method, but its a little long and you might need to differentiate twice, alternatively:

$\frac{f(n+1)}{f(n)} = \frac{(2n+1)^{2n+1}}{(2n-1)^n (2n+3)^{n+1}} = \frac{2n+1}{2n+3} \left(\frac{(2n+1)^2}{(2n-1)(2n+3)} \right )^n$

$= \frac{2n+1}{2n+3} \left(\frac{4n^2 + 4n + 1}{4n^2+4n - 3} \right ) \times \left( \frac{4n^2+4n+1}{4n^2+4n-3} \right )^{n-1}$

$= \frac{8n^3+12n^2+6n+1}{8n^3+12n^2+6n-9} \left( \frac{4n^2 + 4n+1}{4n^2+4n-3} \right )^{n-1} > 1$

$\therefore f(n+1) > f(n)$

Hence n=1 is the lowest, hence f(n) > 1/3

JJ345 · Aug 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Prove for some integer$ \ n$

$\binom{n}{0} - \frac{1}{3} \binom{n}{1} + \frac{1}{5} \binom{n}{2} - \dots + \frac{1}{2n+1} \binom{n}{n} = \frac{2^{2n}}{\binom{2n}{n}}$

Binomial expansion of (1-x^2)^n, integrate both sides between limits of x=1 and x=0.
Weird thing is i'm getting an extra 2n+1 on the denominator on the RHS of the expression, maybe I integrated incorrectly. Isn't integral of (1-x^2)^n from 0 to 1 [2^(2n)][(n!)^2])/(2n+1)!?

seanieg89 · Aug 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Solution is correct until the 'let n=1' argument.

Is it?

How does finding an upper bound for the numerator and an upper bound for the denominator help us find a lower bound for a fraction?

Sy123 · Aug 4, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
Binomial expansion of (1-x^2)^n, integrate both sides between limits of x=1 and x=0, then let x=1.
Weird thing is i'm getting an extra 2n+1 on the denominator on the RHS of the expression, maybe I integrated incorrectly. Isn't integral of (1-x^2)^n from 0 to 1 [2^(2n)][(n!)^2])/(2n+1)!?

Yeah I did it again and you are correct

seanieg89 said:
Is it?

How does finding an upper bound for the numerator and an upper bound for the denominator help us find a lower bound for a fraction?

Ah yes reading the solution again it is incorrect, what I had in mind was

$(n- \frac{1}{2} \right )^n \geq (1-x_1)(1-x_2) \dots (1-x_n)$ ... (1)

$(n+1/2)^n \geq (1+x_1)(1+x_2) \dots (1+x_n)$ .... (2)

This is done by subbing in the AM-GM inequality, all substitutions are positive since x_k <= 1/2

Dividing (2) by (1), which we can do since all things are positive.

We get $\frac{(n+1/2)^n}{(n-1/2)^n} \geq 1/P$ (P is what we want)

Then flip everything, when we do that the inequality sign switches, yeilding P > (that)

But I was skimming through her solution and wasn't really paying attention, which is why I couldn't see that.

=====

I'm making way too many mistakes

(there is no mistake in this one)

$$For some real numbers$ \ x >y > 0 \ $and integer$ \ n$

$Prove$

$(n+1)y^n < \frac{x^{n+1} - y^{n+1}}{x-y} < (n+1) x^n$

seanieg89 · Aug 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yeah I did it again and you are correct

Ah yes reading the solution again it is incorrect, what I had in mind was

$(n- \frac{1}{2} \right )^n \geq (1-x_1)(1-x_2) \dots (1-x_n)$ ... (1)

$(n+1/2)^n \geq (1+x_1)(1+x_2) \dots (1+x_n)$ .... (2)

This is done by subbing in the AM-GM inequality, all substitutions are positive since x_k <= 1/2

Dividing (2) by (1), which we can do since all things are positive.

We get $\frac{(n+1/2)^n}{(n-1/2)^n} \geq 1/P$ (P is what we want)

Then flip everything, when we do that the inequality sign switches, yeilding P > (that)

But I was skimming through her solution and wasn't really paying attention, which is why I couldn't see that.

=====

I'm making way too many mistakes

(there is no mistake in this one)

$$For some real numbers$ \ x >y > 0 \ $and integer$ \ n$

$Prove$

$(n+1)y^n < \frac{x^{n+1} - y^{n+1}}{x-y} < (n+1) x^n$

Uhhh same issue, division will not preserve inequality in general.

Eg.
2 > 1 & 5 > 2 does not imply that 2/5 > 1/2.

Sy123 · Aug 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Uhhh same issue, division will not preserve inequality in general.

Eg.
2 > 1 & 5 > 2 does not imply that 2/5 > 1/2.

Yeah that's true :/

Also for anybody wanting to attempt that question, it is a correct result since I got it from a book, so the problem still stands if someone wants to attempt it.

seanieg89 · Aug 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yeah that's true :/

Also for anybody wanting to attempt that question, it is a correct result since I got it from a book, so the problem still stands if someone wants to attempt it.

Which book? Lagrange multipliers slay it pretty easily, but don't have the time to play around with MX2 tools right now. If it is something that can be done with elementary methods (I suspect it can, but nothing immediately jumps out at me) and no-one answers it by tomorrow evening I will have more time to look at it.

Sy123 · Aug 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Which book? Lagrange multipliers slay it pretty easily, but don't have the time to play around with MX2 tools right now. If it is something that can be done with elementary methods (I suspect it can, but nothing immediately jumps out at me) and no-one answers it by tomorrow evening I will have more time to look at it.

Book my teacher lent me:

'Mathematical Circles (Russian Experience)' by Dimitri Fomin, Sergey Genkin and Ilia Itenberg

Its a problem solving book for training and teaching olympiad students. The inequalities chapter uses only uses elementary inequalities such as AM-GM.
The non-MX2 chapters preceding it (just in case they might want you to use past knowledge) is Pigeon-hole Principle, Divisibility, Games, Graphs, Invariants. None of which I would imagine could be used to solve an inequality like this.

JJ345 · Aug 4, 2013

Re: HSC 2013 4U Marathon

Dividing doesn't preserve the inequality, but can we get away with using logs then subtracting?
Like ln(1-x)>=M
ln(1+x)>=N
Then saying ln(1-x/1+x) >= M-N
Either way I wouldn't know how to use the fact that x(1)+x(2)+....+x(n)= 1/2 in this method.

RealiseNothing · Aug 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Book my teacher lent me:

'Mathematical Circles (Russian Experience)' by Dimitri Fomin, Sergey Genkin and Ilia Itenberg

Its a problem solving book for training and teaching olympiad students. The inequalities chapter uses only uses elementary inequalities such as AM-GM.
The non-MX2 chapters preceding it (just in case they might want you to use past knowledge) is Pigeon-hole Principle, Divisibility, Games, Graphs, Invariants. None of which I would imagine could be used to solve an inequality like this.

Is the book any good? I've been meaning to find a good problem solving book.

seanieg89 · Aug 4, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
Dividing doesn't preserve the inequality, but can we get away with using logs then subtracting?
Like ln(1-x)>=M
ln(1+x)>=N
Then saying ln(1-x/1+x) >= M-N
Either way I wouldn't know how to use the fact that x(1)+x(2)+....+x(n)= 1/2 in this method.

No, that's just rewriting the same fallacy in a different form.

Sy123 · Aug 4, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
Dividing doesn't preserve the inequality, but can we get away with using logs then subtracting?
Like ln(1-x)>=M
ln(1+x)>=N
Then saying ln(1-x/1+x) >= M-N
Either way I wouldn't know how to use the fact that x(1)+x(2)+....+x(n)= 1/2 in this method.

Subtraction is essentially adding the negative version, so when we minus N, we are first multiplying by -1 then adding side by side.
Because for even with subtracting logs, we get stuff like

2 > 1, 5 > 2, ln(2/5) > ln(1/2) which doesn't work.

It seems that the only operations that one can work with inequalities side by side are addition and multiplication

RealiseNothing said:
Is the book any good? I've been meaning to find a good problem solving book.

I'd say its pretty good, there are only solutions for half of them though

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