HSC 2013 MX2 Marathon (archive) (4 Viewers)

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

That's basically the same as my centroid proof then. k=l=3/2 just says that the centroid divides the median into the ratio 2:1.

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

Carrotsticks said:
Haha I just realised, k=3/2 was on the photo of the question.

Rumbleroar, surely you would have seen what to do from here? Or did you write that in afterwards?

I think she realised she essentially just had to prove that k=3/2 to do the question, by observing the equation given and the result needed.

Carrotsticks · Dec 13, 2013

Re: HSC 2014 4U Marathon

Oh btw for those interested, this question touches on a recurring theme in Linear Algebra called 'Linear Independence', and this is a pretty common technique for killing off 'ratios within triangles' problems (ie: centroid is 2/3 along the median)

Very roughly, it is that if we have vectors such that $\alpha Z + \beta W = 0$ (and you can clearly see that if we re-arrange (ii), we get that form), where Z and W are 'linearly independent' (meaning that one cannot be constructed from a 'combination' of the other), then $\alpha = \beta = 0$

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Option 1 (Harder 3U):

$\\ $Evaluate$ \\ \\ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{\sqrt{2nk - k^2}}$

I'm getting that it $\to \infty$ ?

Edit: nope working out had a plus instead of a minus :/

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

Hmm does it $\to \infty$ ?

I'm thinking it should.

Edit: wait no, idk, I should sleep soon.

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

Let's hope I haven't completely derped...

Consider the GM-HM inequality:

$\sqrt[n]{x_1...x_n} \geq \frac{n}{\frac{1}{x_1}+...+\frac{1}{x_n}}$

Flipping this gives:

$\frac{\sum_{k=1}^{n} \frac{1}{x_k}}{n} \geq \frac{1}{\sqrt[n]{x_1...x_n}}$

Now let:

$x_k = \sqrt{k(2n-k)}$

$\frac{\sum_{k=1}^{n} \frac{1}{\sqrt{k(2n-k)}}}{n} \geq \sqrt[n]{\prod_{k=1}^{n} \sqrt{k(2n-k)}}$

$\frac{\sum_{k=1}^{n} \frac{1}{\sqrt{k(2n-k)}}}{n} \geq \sqrt[n]{n(2n-1)!}$

The RHS $\to \infty$ and thus the LHS (our sum) also $\to \infty$

Edit: Yep I derped, I think I'll go to sleep now...

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

Though one thing I have noticed is that:

$\frac{1}{\sqrt{2nk-k^2}} = \frac{1}{\sqrt{n^2-(n-k)^2}}$

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Though one thing I have noticed is that:

$\frac{1}{\sqrt{2nk-k^2}} = \frac{1}{\sqrt{n^2-(n-k)^2}}$

$\sum_{k=1}^{n} \frac{1}{\sqrt{n^2-(n-k)^2}} = \int_{1}^{n} \frac{dk}{\sqrt{n^2-(n-k)^2}}$

If we do a substitution and let $n-k=u$ we get:

$sin^{-1}\frac{u}{n}$ with limits from 0 to (n-1).

$\lim_{n \to \infty}{sin^{-1}\frac{n-1}{n}} = \frac{\pi}{2}$

Carrotsticks · Dec 13, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
$\sum_{k=1}^{n} \frac{1}{\sqrt{n^2-(n-k)^2}} = \int_{1}^{n} \frac{dk}{\sqrt{n^2-(n-k)^2}}$

If we do a substitution and let $n-k=u$ we get:

$sin^{-1}\frac{u}{n}$ with limits from 0 to (n-1).

$\lim_{n \to \infty}{sin^{-1}\frac{n-1}{n}} = \frac{\pi}{2}$

Eh how does the partial sum become a definite integral? Did you mean that the LHS is a Riemann Sum, then take the limit as N -> infinity?

Sy123 · Dec 13, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
$\sum_{k=1}^{n} \frac{1}{\sqrt{n^2-(n-k)^2}} = \int_{1}^{n} \frac{dk}{\sqrt{n^2-(n-k)^2}}$

If we do a substitution and let $n-k=u$ we get:

$sin^{-1}\frac{u}{n}$ with limits from 0 to (n-1).

$\lim_{n \to \infty}{sin^{-1}\frac{n-1}{n}} = \frac{\pi}{2}$

That is the correct answer, but I was able to do it without resorting to Riemann Sums (indeed a rather crude version of it, try lower and upper rectangles of fixed length)

seanieg89 · Dec 13, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
That is the correct answer, but I was able to do it without resorting to Riemann Sums (indeed a rather crude version of it, try lower and upper rectangles of fixed length)

Huh, what do you mean by this?

My solution is also based on Riemann sums.

$S=\sum_{k=1}^n \frac{1}{\sqrt{2nk-k^2}}=\sum_{k=1}^n \frac{1}{n}\frac{1}{\sqrt{2k/n-(k/n)^2}}\rightarrow \int_0^1 \frac{dx}{\sqrt{2x-x^2}}=\frac{\pi}{2}.$

By the way, using Riemann sums shouldn't be thought of as outside MX2. Otherwise we shouldn't be able to use integration at all in MX2 because that is how the Riemann integral is defined. (Although the rigorous details of this construction are beyond the scope of MX2, so too are the rigorous details of much simpler things like the very notion of a limit or what it means to raise a real number to a real power etc.)

Sy123 · Dec 13, 2013

Re: HSC 2014 4U Marathon

seanieg89 said:
Huh, what do you mean by this?

My solution is also based on Riemann sums.

$S=\sum_{k=1}^n \frac{1}{\sqrt{2nk-k^2}}=\sum_{k=1}^n \frac{1}{n}\frac{1}{\sqrt{2k/n-(k/n)^2}}\rightarrow \int_0^1 \frac{dx}{\sqrt{2x-x^2}}=\frac{\pi}{2}.$

By the way, using Riemann sums shouldn't be thought of as outside MX2. Otherwise we shouldn't be able to use integration at all in MX2 because that is how the Riemann integral is defined. (Although the rigorous details of this construction are beyond the scope of MX2, so too are the rigorous details of much simpler things like the very notion of a limit or what it means to raise a real number to a real power etc.)

Sketch the graph of:

$y = \frac{1}{\sqrt{2nx - x^2}} \ \ $for integer$ \ n$

Its easy enough to graph, then consider the upper and lower rectangles at, $x=1,2,3, \dots , n$

Considering the lower rectangles first

$\sum_{k=2}^n \frac{1}{\sqrt{2nk - k^2}} < \int_1^n \frac{dx}{\sqrt{2nx-x^2}}$

Then the upper rectangles

$\sum_{k=1}^{(n-1)} \frac{1}{\sqrt{2nk-k^2}} > \int_1^n \frac{dx}{\sqrt{2nx-x^2}}$

Add something to both sides of the inequality, make the sum the middle term in the inequality, find the integrals and apply squeeze theorem.

seanieg89 · Dec 13, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Sketch the graph of:

$y = \frac{1}{\sqrt{2nx - x^2}} \ \ $for integer$ \ n$

Its easy enough to graph, then consider the upper and lower rectangles at, $x=1,2,3, \dots , n$

Considering the lower rectangles first

$\sum_{k=2}^n \frac{1}{\sqrt{2nk - k^2}} < \int_1^n \frac{dx}{\sqrt{2nx-x^2}}$

Then the upper rectangles

$\sum_{k=1}^{(n-1)} \frac{1}{\sqrt{2nk-k^2}} > \int_1^n \frac{dx}{\sqrt{2nx-x^2}}$

Add something to both sides of the inequality, make the sum the middle term in the inequality, find the integrals and apply squeeze theorem.

Ah okay, I thought it might be this. This is actually a kind of similar idea to how one proves that the definition of a Riemann integral as a limit of Riemann sums makes sense and is unambiguous.

But yeah, any argument you construct using integrals is implicitly using Riemann sums anyway. (Otherwise what does the integral mean?)

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

Carrotsticks said:
Eh how does the partial sum become a definite integral? Did you mean that the LHS is a Riemann Sum, then take the limit as N -> infinity?

That's what I was trying to get at, but like I all nightered it and was ceebs explaining properly. When posting it I knew it wasn't very rigorous.

asianese · Dec 13, 2013

Re: HSC 2014 4U Marathon

Sure realise

bottleofyarn · Dec 14, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Make sure to say so if this one is too hard/open

$\\ $Using complex numbers find the sum$ \\ \\ \cos(\theta) + \cos(\theta + \alpha) + \cos(\theta + 2\alpha) + \dots + \cos(\theta + (n-1)\alpha)$

This isn't too bad actually and a rather nice question, once you realise that $cis\theta\times cis\alpha=cis(\theta+\alpha)$ . Taking the real part of the geometric sum you get $S_n=Re\left(cis\theta\left(\frac{cis^n\alpha-1}{cis\alpha-1}\right)\right)$ .
Make the denominator real and then using half angle sets up the numerator nicely for sums to products. Simplifying it down, I got
$S_n=\csc\frac{\alpha}{2}\cos\left(\theta+\frac{n-1}{2}\alpha\right)\sin\frac{n}{2}\alpha}$

==

Reminds me of a similar question.
$$Show that $ \sin nt+\sin(nt+\alpha)+\sin(nt+2\alpha)+\sin(nt+3\alpha)+\dots+\sin(nt+(k-1)\alpha)=0,$ where $\alpha=\frac{2\pi}{k}$

There are parts before it but they lead away from the really nice solution.

Sy123 · Dec 16, 2013

Re: HSC 2014 4U Marathon

bottleofyarn said:
This isn't too bad actually and a rather nice question, once you realise that $cis\theta\times cis\alpha=cis(\theta+\alpha)$ . Taking the real part of the geometric sum you get $S_n=Re\left(cis\theta\left(\frac{cis^n\alpha-1}{cis\alpha-1}\right)\right)$ .
Make the denominator real and then using half angle sets up the numerator nicely for sums to products. Simplifying it down, I got
$S_n=\csc\frac{\alpha}{2}\cos\left(\theta+\frac{n-1}{2}\alpha\right)\sin\frac{n}{2}\alpha}$

==

Reminds me of a similar question.
$$Show that $ \sin nt+\sin(nt+\alpha)+\sin(nt+2\alpha)+\sin(nt+3\alpha)+\dots+\sin(nt+(k-1)\alpha)=0,$ where $\alpha=\frac{2\pi}{k}$

There are parts before it but they lead away from the really nice solution.

Yea something like that

$\\ $The area of a triangle with sides$ \ a,b,c \ $is given by the formula$ \ A = \sqrt{s(s-a)(s-b)(s-c)} \ $where$ \ s= \frac{1}{2} (a+b+c) \\ \\ $Show that$ \ \ A \leq \frac{a^2+b^2+c^2}{4}$

Something wrong with latex?

Sy123 · Dec 16, 2013

Re: HSC 2014 4U Marathon

$\\ $The area of a triangle with sides$ \ a,b,c \ $is given by the formula$ \ A = \sqrt{s(s-a)(s-b)(s-c)} \ $where$ \ s= \frac{1}{2} (a+b+c) \\ \\ $Show that$ \ \ A \leq \frac{a^2+b^2+c^2}{4}$

Why is latex not working?

asianese · Dec 16, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $The area of a triangle with sides$ \ a,b,c \ $is given by the formula$ \ A = \sqrt{s(s-a)(s-b)(s-c)} \ $where$ \ s= \frac{1}{2} (a+b+c) \\ \\ $Show that$ \ \ A \leq \frac{a^2+b^2+c^2}{4}$

Why is latex not working?

It hatttesss you!

$$The area of a triangle with sides$ \ a,b,c \ $is given by the formula$ \ A = \sqrt{s(s-a)(s-b)(s-c)} \ $where$ \ s= \frac{1}{2} (a+b+c) \\ \\ $Show that$ \ \ A \leq \frac{a^2+b^2+c^2}{4}$

integral95 · Dec 16, 2013

Re: HSC 2014 4U Marathon

asianese said:
It hatttesss you!

$$The area of a triangle with sides$ \ a,b,c \ $is given by the formula$ \ A = \sqrt{s(s-a)(s-b)(s-c)} \ $where$ \ s= \frac{1}{2} (a+b+c) \\ \\ $Show that$ \ \ A \leq \frac{a^2+b^2+c^2}{4}$

I used the Identity $\\ (a+b)(b+c)(c+a))\geq 8abc$

and got

$\\ s(s-a)(s-b)(s-c) \leq \frac{(a+b+c)(abc)}{16}$

but then I"m stuck lol

HSC 2013 MX2 Marathon (archive) (4 Viewers)

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