HSC 2013 MX2 Marathon (archive) (1 Viewer)

seanieg89 · Sep 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Oh I see, my answer was way off.
Pretty amazing how that strategy increases probability by a lot though....

=====

$$A regular polygon of unit length with points$ \ \ X_1, X_2, X_3, \dots , X_n$

$$Let$ \ \ d_k = X_n X_k$

$Evaluate$

$\lim_{n \to \infty} \frac{n^2}{\sum_{k=1}^n d_k}$

Yeah, definitely one of my favourite puzzle type questions. It's been proven relatively recently that that strategy is actually optimal, but that is MUCH harder I think.

Sy123 · Sep 21, 2013

Re: HSC 2013 4U Marathon

This is a good inequality:

$$Prove for real$ \ \ \{ a_k \} \ \ $and$ \ \ \{b_k \}$

$\sum_{k=1}^{n} \sqrt{a_k^2+ b_k^2} \geq \sqrt{ \left(\sum_{k=1}^n a_k \right)^2 + \left(\sum_{k=1}^n b_k \right )^2}$

Sy123 · Sep 21, 2013

Re: HSC 2013 4U Marathon

$Show$

$\sum_{k=1}^{\infty} \frac{6^k}{(3^{k+1} - 2^{k+1})(3^{k} - 2^k)} = 2$

spyris · Sep 22, 2013

Re: HSC 2013 4U Marathon

Hi, I have a general question, I heard somewhere that 1/z = z(bar) but im not sure if its true because i am unable to prove it...so is it true and if so how do i show it.

HeroicPandas · Sep 22, 2013

Re: HSC 2013 4U Marathon

spyris said:
Hi, I have a general question, I heard somewhere that 1/z = z(bar) but im not sure if its true because i am unable to prove it...so is it true and if so how do i show it.

That is true for roots of unity, where |z| = 1

So if |z| = 1, then 1/z = z(bar)

RealiseNothing · Sep 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$A regular polygon of unit length with points$ \ \ X_1, X_2, X_3, \dots , X_n$

$$Let$ \ \ d_k = X_n X_k$

$Evaluate$

$\lim_{n \to \infty} \frac{n^2}{\sum_{k=1}^n d_k}$

Is it 2?

Sy123 · Sep 22, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Is it 2?

The answer should be $\frac{\pi^2}{2}$

seanieg89 · Sep 22, 2013

Re: HSC 2013 4U Marathon

$P=\frac{n^2}{\sum_{k=1}^n d_k}=\frac{n^2\sin\left(\frac{\pi}{n}\right)}{\sum_{k=1}^n \sin\left(\frac{ k\pi}{n}\right)}=\frac{n^2\sin\left(\frac{\pi}{2n}\right)\sin\left(\frac{\pi}{n}\right)}{\sin \left(\frac{(n-1)\pi}{2n}\right)}\rightarrow \frac{\pi^2}{2}.$

Sy123 · Sep 22, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$P=\frac{n^2}{\sum_{k=1}^n d_k}=\frac{n^2\sin\left(\frac{\pi}{n}\right)}{\sum_{k=1}^n \sin\left(\frac{ k\pi}{n}\right)}=\frac{n^2\sin\left(\frac{\pi}{2n}\right)\sin\left(\frac{\pi}{n}\right)}{\sin \left(\frac{(n-1)\pi}{2n}\right)}\rightarrow \frac{\pi^2}{2}.$

Practising for the BOS trial?

(btw that was HSC 2007 4U 13 marks compressed).

$You may use the fact that$

$\sum_{k=1}^{\infty} \frac{\sin(k\theta)}{k} = \frac{\pi}{2} - \frac{\theta}{2}$

$Prove that$

$\pi = 3 + \frac{4}{2 \cdot 3 \cdot 4} - \frac{4}{4\cdot 5 \cdot 6} + \frac{4}{6 \cdot 7 \cdot 8} - \frac{4}{8\cdot 9 \cdot 10} + \dots$

seanieg89 · Sep 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Practising for the BOS trial? (btw that was HSC 2007 4U 13 marks compressed).

$You may use the fact that$

$\sum_{k=1}^{\infty} \frac{\sin(k\theta)}{k} = \frac{\pi}{2} - \frac{\theta}{2}$

$Prove that$

$\pi = 3 + \frac{4}{2 \cdot 3 \cdot 4} - \frac{4}{4\cdot 5 \cdot 6} + \frac{4}{6 \cdot 7 \cdot 8} - \frac{4}{8\cdot 9 \cdot 10} + \dots$

I don't want to be shown up by any current MX2 students

.

Haha yeah, I recognised it, although I forgot the way the question guided you through it (I did my HSC in 07).

P.S. Obviously, I would supply some intermediate reasoning in an actual exam haha.

seanieg89 · Sep 23, 2013

Re: HSC 2013 4U Marathon

$Find (with careful proof) \\ \\ $\lim_{n\rightarrow \infty}\sum_{k=0}^n {n\choose k}^{-1}$.$

RealiseNothing · Sep 23, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$Find (with careful proof) \\ \\ $\lim_{n\rightarrow \infty}\sum_{k=0}^n {n\choose k}^{-1}$.$

My first thought would be just 2 since the first and last terms are 0!, and the rest all $\to 0$ as there will be an 'n' in the denominator.

seanieg89 · Sep 23, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
My first thought would be just 2 since the first and last terms are 0!, and the rest all $\to 0$ as there will be an 'n' in the denominator.

But there are an increasing number of terms, so that isn't quite rigorous.

Eg.

$\lim_{n\rightarrow\infty}\sum_{j=0}^n \left(\frac{1}{\sqrt{n+j}}\right)=\infty$

You can check this counterexample yourself by mx2 level methods, but heuristically it is obvious...we are adding n objects, each of which is roughly 1/root(n) (roughly means bounded between two constant multiples of) so we expect the sum to be roughly root(n), which tends to infinity.

Sy123 · Sep 23, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$Find (with careful proof) \\ \\ $\lim_{n\rightarrow \infty}\sum_{k=0}^n {n\choose k}^{-1}$.$

$S_n = \sum_{k=0}^{n} \binom{n}{k}^{-1}$

$S_{n+1} = 1 + \sum_{k=0}^n \binom{n+1}{k}^{-1}$

$\binom{n+1}{k} = \binom{n}{k} \times \frac{n+1}{(n+1) - k}$

$\therefore \ \binom{n+1}{k}^{-1} = \left(1- \frac{k}{n+1} \right ) \binom{n}{k}^{-1}$

$S_{n+1} = 1 + \sum_{k=0}^n \left(1- \frac{k}{n+1} \right ) \binom{n}{k}^{-1}$

$S_{n+1} = 1 + S_{n} - \frac{1}{n+1} \sum_{k=0}^{n} k\binom{n}{k}^{-1}$

$$Since$ \ \ \sum_{k=0}^{n} k \binom{n}{k}^{-1} = \sum_{k=0}^n (n-k) \binom{n}{k}^{-1}$

$\therefore \ \ \sum_{k=0}^{n} k \binom{n}{k}^{-1} = \frac{n}{2} \sum_{k=0}^n \binom{n}{k}^{-1}$

$\therefore \ \ S_{n+1} = S_n + 1- \frac{1}{n+1} \left(\frac{n}{2} S_n \right )$

$\therefore \ \ S_{n+1} = \frac{n+2}{2n+2} S_{n} + 1$

-----

Now, we must prove that:

$\lim_{n \to \infty} S_n = L \ \ $for finite$ \ \ L$

Yeah........so I'm not too sure yet on how to do this within the scope of the HSC.

But lets complete the solution, assuming that the limit converges:

$\lim_{n \to \infty} S_{n+1} = \lim_{n \to \infty} S_n = L$

$L = L \lim_{n \to \infty} \frac{n+2}{2n+2} + 1$

$L = \frac{1}{2} L + 1$

$\therefore \ \ L = \lim_{n \to \infty} \sum_{k=0}^n \binom{n}{k}^{-1} = 2$

seanieg89 · Sep 23, 2013

Re: HSC 2013 4U Marathon

Hmm yeah, cannot think of any immediate way to show that the limit is finite.

My method was to assume n was large, (which is all we care about anyway) and use that binomial coefficients get bigger towards the middle.

So:

$0\leq \sum_{k=0}^n {n\choose k}^{-1} - 2 - \frac{2}{n} = \sum_{k=2}^{n-2} {n\choose k}^{-1} \leq (n-3)\cdot {n \choose 2}^{-1}=\frac{2(n-3)}{n(n-1)}\\ \\ $whence letting $n\rightarrow\infty$ completes the proof.$$

Sy123 · Sep 23, 2013

Re: HSC 2013 4U Marathon

$We establish$

$\frac{1}{2} < \frac{n+2}{2n+2} < \frac{3}{4} \ \ \ (*)$

$$for$ \ \ n \geq 1$

$$Consider the sequence$ \ \ b_{n+1} = \frac{3}{4} b_{n} + 1 \ \ \ b_0 = 1$

$By repeated iteration, we find$

$b_{n+1} = 1 + \frac{3}{4} + \dots + \frac{3^n}{4^n} + \frac{3^{n+1}}{4^{n+1}}$

$\lim_{n \to \infty} b_n = 4$

$Now consider the sequence$

$a_{n+1} = \frac{1}{2} a_n + 1$

$By repeated iterations we arrive at a geometric sequence, and take the limit to arrive at$

$\lim_{n \to \infty} a_n = 2$

$Therefore since we have 3 sequences$

$s_{n+1} = \frac{n+2}{2n+2} s_n + 1$

$a_{n+1} = \frac{1}{2} a_n + 1$

$b_{n+1} = \frac{3}{4} b_n + 1$

$$Due to$ \ \ (*)$

$a_n < s_n < b_n$

$\therefore \ \ 2 \leq \lim_{n \to \infty}s_n \leq 4$

---- (proof not yet complete)

One I think would then have to subsequently prove that s_(n+1) < s_n (which is indeed true), problem is this is only true for n =>4 (the partial sums increase until 3 and 4)
So I don't know of any way to do this....

EDIT: Or you can just do your way which is much easier

seanieg89 · Sep 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$We establish$

$\frac{1}{2} < \frac{n+2}{2n+2} < \frac{3}{4} \ \ \ (*)$

$$for$ \ \ n \geq 1$

$$Consider the sequence$ \ \ b_{n+1} = \frac{3}{4} b_{n} + 1 \ \ \ b_0 = 1$

$By repeated iteration, we find$

$b_{n+1} = 1 + \frac{3}{4} + \dots + \frac{3^n}{4^n} + \frac{3^{n+1}}{4^{n+1}}$

$\lim_{n \to \infty} b_n = 4$

$Now consider the sequence$

$a_{n+1} = \frac{1}{2} a_n + 1$

$By repeated iterations we arrive at a geometric sequence, and take the limit to arrive at$

$\lim_{n \to \infty} a_n = 2$

$Therefore since we have 3 sequences$

$s_{n+1} = \frac{n+2}{2n+2} s_n + 1$

$a_{n+1} = \frac{1}{2} a_n + 1$

$b_{n+1} = \frac{3}{4} b_n + 1$

$$Due to$ \ \ (*)$

$a_n < s_n < b_n$

$\therefore \ \ 2 \leq \lim_{n \to \infty}s_n \leq 4$

---- (proof not yet complete)

One I think would then have to subsequently prove that s_(n+1) < s_n (which is indeed true), problem is this is only true for n =>4 (the partial sums increase until 3 and 4)
So I don't know of any way to do this....

EDIT: Or you can just do your way which is much easier

Well we only need EVENTUAL monotonicity to use the theorem you are thinking of, but that theorem isn't really MX2 assumed knowledge. But yeah, syllabus aside...if your working is all correct then the method is valid. Well done for bashing it out

.

Sy123 · Sep 23, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Well we only need EVENTUAL monotonicity to use the theorem you are thinking of, but that theorem isn't really MX2 assumed knowledge. But yeah, syllabus aside...if your working is all correct then the method is valid. Well done for bashing it out .

Ahhh ok

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

Something a little easier:

For a triangle ABC, with point P, Q, R on AB, BC, CA respectively
Such that, AQ bisects BC, CP bisects AB, and BR bisects CA. These 3 lines intersect at 1 point (a property which you may assume), at a point K, this point is called the centroid of a triangle.

For a general hyperbola, when the tangent at a point S intersects the asymptotes at points M and N.
Show that the locus of the centroid of the triangle OMN (O being origin) is a hyperbola.

Sy123 · Sep 27, 2013

Re: HSC 2013 4U Marathon

$0<p , \ 0 < q$

$p+q < 1$

$Prove$

$(px+qy)^2 \leq px^2 + qy^2$

harryharper · Sep 27, 2013

Re: HSC 2013 4U Marathon

Is it $p+q\le1$ ?

$x^2+y^2\ge2xy$

$x^2(\frac{1-p}{q})+y^2(\frac{1-q}{p})\ge2xy$

$x^2(1-p)p+y^2(1-q)q\ge2pqxy$

$px^2+qy^2\ge p^2x^2+q^2y^2+2pqxy$

$px^2+qy^2\ge (px+qy)^2$

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