HSC 2013 MX2 Marathon (archive) (11 Viewers)

rumbleroar · Dec 12, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
R is the centroid of triangle OPQ. By symmetry, triangle PQS is a reflection of triangle OPQ. Consider R' is the centroid of triangle PQS. A property of all triangles is the centroid divides the median (i.e. OM, QN, MS, etc) into the ratio 2:1. Thus OR:RM is 2:1 and similarly SR':MR' is 2:1.

Thus since OM=MS, then it follows that SR:OR is 2:1, hence $z=\frac{1}{3}(u+v)$

Nice!!
However, I think one person tried doing that and got marked down, because the question specifically stated hence and that meant they needed to use the previous parts.

rumbleroar · Dec 12, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
I'm surprised VC didn't get it.

Me too, but it was a tough test

Apparently it was especially designed for/targeted towards a certain class lolol

RealiseNothing · Dec 12, 2013

Re: HSC 2014 4U Marathon

rumbleroar said:
Nice!!
However, I think one person tried doing that and got marked down, because the question specifically stated hence and that meant they needed to use the previous parts.

It stated deduce not hence, not sure if the same still applies.

rumbleroar · Dec 12, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
It stated deduce not hence, not sure if the same still applies.

Oh crap, sorry, my bad. But one of the teachers went through it and told us it was marked down because it didn't relate to the previous parts, so idk if they would be marks back or not :/ but apparently the q was quite hard, as some teachers couldn't do it on the first go and stuff

Fizzy_Cyst · Dec 12, 2013

Re: HSC 2014 4U Marathon

$\\$find$\ \sqrt{-8-6i} \\ \\ $and hence solve$\\ \\ z^2-(5-i)z+8-i= 0$

integral95 · Dec 12, 2013

Re: HSC 2014 4U Marathon

$i) Show that$ \ (1+ki)(k+i) \ $for real$ \ k \ $is purely imaginary$

$$ii) Hence describe the sketch of the function$ \ \ f(x) = \tan^{-1} (x) + \tan^{-1} \left(\frac{1}{x} \right)$

$\\ i) \ (1+ki)(k+i) = k +i +ki - k \\ = 0 + i(k+1) \\ ii) \therefore arg(1+ki)(k+i) = arg[(i)(k+1)] \\ \therefore arg(1+ki)+ arg (k+i) = \pm \frac{\pi}{2} \\ tan^{-1}(k)+ tan^{-1}(\frac{1}{k}) = \pm \frac{\pi}{2} \\ \therefore tan^{-1}x + tan^{1}\frac {1}{x} = \frac{\pi}{2} for x >0 \\ tan^{-1}x + tan^{1}\frac {1}{x} = -\frac{\pi}{2} for x<0$

Sy123 · Dec 12, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
$a,b,c > 1$

$\frac{1}{a^2 -1} + \frac{1}{b^2 -1} + \frac{1}{c^2-1} = 1$

$\\ $Prove that$ \\ \\ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} \leq 1$

My solution:

$$Let$ \ x = \frac{1}{a^2-1} \ \Rightarrow a = \sqrt{\frac{x+1}{x}} , \ b = \sqrt{\frac{y+1}{y}} , \ c = \sqrt{\frac{z+1}{z}}$

$\therefore \ x+y+z = 1 \\ \\ $We must prove that$ \ \ \ \frac{\sqrt{x}}{\sqrt{x} + \sqrt{x+1}} + \frac{\sqrt{y}}{\sqrt{y} + \sqrt{y+1} } + \frac{\sqrt{z}}{\sqrt{z} + \sqrt{z+1}} < 1$

$\\ \frac{\sqrt{x}}{\sqrt{x} + \sqrt{x+1}} + \frac{\sqrt{y}}{\sqrt{y} + \sqrt{y+1} } + \frac{\sqrt{z}}{\sqrt{z} + \sqrt{z+1}} = \sqrt{x(x+1)} - x + \sqrt{y(y+1)} - y + \sqrt{z(z+1)} - z \\ \\ \frac{2x+1}{2} \geq \sqrt{x(x+1)} \\ \\ \therefore \ \frac{2x+1}{2} + \frac{2y+1}{2} + \frac{2z+1}{2} - (x+y+z) \geq \sqrt{x(x+1)} - x + \sqrt{y(y+1)} - y + \sqrt{z(z+1)} - z \\ \\ $With Cauchy we can see, again easily proven$ \ \ (x+y+z)^{1/2}((x+1) + (y+1) + (z+1))^{1/2} \geq \sqrt{x(x+1)} + \sqrt{y(y+1)} + \sqrt{z(z+1)} \\ \\ $Therefore$ \ \ 2\geq \sqrt{x(x+1)} + \sqrt{y(y+1)} + \sqrt{z(z+1)} \\ \\ \therefore 1 \geq \sqrt{x(x+1)} + \sqrt{y(y+1)} + \sqrt{z(z+1)} - (x+y+z) \\ \\ \therefore \ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} \leq 1$

--------

Option 1 (Harder 3U):

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

-------

Option 2 (Complex Numbers)

$\\ $Using complex numbers, prove that the angle subtended by a semi-circle is always a right angle$$

Carrotsticks · Dec 13, 2013

Re: HSC 2014 4U Marathon

rumbleroar said:
View attachment 29383

This was from our term 1 complex numbers assessment

Much tear
It got ripped off Sydney Grammar though so some of you may have seen it before
No one got iii

Have fun )))

Sent from my iPhone using Tapatalk

I could be derping atm, but I don't see how (3-2l)v = 2(k-l)z.

Because l and k are real numbers so if we re-arrange it, we have v=(3-2l)/(2(k-l))z, so v = A*z for some real number A, meaning that v and z are collinear, which is not the case as per the construction?

Sy123 · Dec 13, 2013

Re: HSC 2014 4U Marathon

Carrotsticks said:
I could be derping atm, but I don't see how (3-2l)v = 2(k-l)z.

Because l and k are real numbers so if we re-arrange it, we have v=(3-2l)/(2(k-l))z, so v = A*z for some real number A, meaning that v and z are collinear, which is not the case as per the construction?

Take the 2 equations from part (i), and re-arrange both so that $\frac{1}{2} u$ is the subject, equate the equations, to yield an equation only in z,v which re-arranged gives you the required result, I did it and the algebra matches with the Show that

Carrotsticks · Dec 13, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Take the 2 equations from part (i), and re-arrange both so that $\frac{1}{2} u$ is the subject, equate the equations, to yield an equation only in z,v which re-arranged gives you the required result, I did it and the algebra matches with the Show that

I understand how to get the result, but what I meant was that I do not understand how the result is supposed to work geometrically.

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Take the 2 equations from part (i), and re-arrange both so that $\frac{1}{2} u$ is the subject, equate the equations, to yield an equation only in z,v which re-arranged gives you the required result, I did it and the algebra matches with the Show that

I just did equation 1 minus equation 2, though I guess they are essentially the same thing.

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

Carrotsticks said:
I could be derping atm, but I don't see how (3-2l)v = 2(k-l)z.

Because l and k are real numbers so if we re-arrange it, we have v=(3-2l)/(2(k-l))z, so v = A*z for some real number A, meaning that v and z are collinear, which is not the case as per the construction?

Have you re-arranged it right?

Carrotsticks · Dec 13, 2013

Re: HSC 2014 4U Marathon

$\\ (3-2l)v = 2(k-l)z \\\\ v = \frac{2(k-l)}{3-2l}z \\\\ $Since $ k $ and $ l $ are real numbers, we can say that $ v = Az $ where $ A $ is some real number. Geometrically, this means that $ z $ and $ v $ lie on the same line, except one is a scalar multiple of another, which cannot be true from the construction of $ z $ unless $ v $ and $ u $ are collinear.$

Sy123 · Dec 13, 2013

Re: HSC 2014 4U Marathon

Carrotsticks said:
$\\ (3-2l)v = 2(k-l)z \\\\ v = \frac{2(k-l)}{3-2l}z \\\\ $Since $ k $ and $ l $ are real numbers, we can say that $ v = Az $ where $ A $ is some real number. Geometrically, this means that $ z $ and $ v $ lie on the same line, except one is a scalar multiple of another, which cannot be true from the construction of $ z $ unless $ v $ and $ u $ are collinear.$

I think I've found the problem, when we consider the final result, $z = \frac{1}{3} (u+v)$

Substituting this into the first 2 equations, we get:

$k= l = \frac{3}{2}$

Meaning the division or multiplication by $(k-l)$ cannot be done, since its zero.

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

Carrotsticks said:
$\\ (3-2l)v = 2(k-l)z \\\\ v = \frac{2(k-l)}{3-2l}z \\\\ $Since $ k $ and $ l $ are real numbers, we can say that $ v = Az $ where $ A $ is some real number. Geometrically, this means that $ z $ and $ v $ lie on the same line, except one is a scalar multiple of another, which cannot be true from the construction of $ z $ unless $ v $ and $ u $ are collinear.$

The other problem is that from (iii), if:

$z = \frac{1}{3}(u+v)$

and $kz = \frac{1}{2}(u+v)$

then it implies $k=\frac{3}{2}$

Also from part (ii) we get:

$(3-2l)v=(3-2l)z$ when we let $k=\frac{3}{2}$

Thus $3-2l=0$ and so $k=l=\frac{3}{2}$

But we chose arbitrary positive real numbers for (k,l) and so that would imply $z=v$ which is definitely not true.

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
I think I've found the problem, when we consider the final result, $z = \frac{1}{3} (u+v)$

Substituting this into the first 2 equations, we get:

$k= l = \frac{3}{2}$

Meaning the division or multiplication by $(k-l)$ cannot be done, since its zero.

ok b10

Carrotsticks · Dec 13, 2013

Re: HSC 2014 4U Marathon

Yep I get it now, I was derping, it's part of the proof.

Since z cannot be v, the only alternative is that (3-2l) and (k-l) = 0, so k = l = 3/2, so using (i) that instantly yields (iii).

Carrotsticks · Dec 13, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
The other problem is that from (iii), if:

$z = \frac{1}{3}(u+v)$

and $kz = \frac{1}{2}(u+v)$

then it implies $k=\frac{3}{2}$

Also from part (ii) we get:

$(3-2l)v=(3-2l)z$ when we let $k=\frac{3}{2}$

Thus $3-2l=0$ and so $k=l=\frac{3}{2}$

But we chose arbitrary positive real numbers for (k,l) and so that would imply $z=v$ which is definitely not true.

We aren't choosing arbitrary positive reals.

The construction is enough to uniquely define one value each of K and L, and it just so happens that the values are the same (3/2).

RealiseNothing · Dec 13, 2013

Re: HSC 2014 4U Marathon

Carrotsticks said:
We aren't choosing arbitrary positive reals.

The construction is enough to uniquely define one value each of K and L, and it just so happens that the values are the same (3/2).

Yep I mistook it as arbitrary values for k and l.

Carrotsticks · Dec 13, 2013

Re: HSC 2014 4U Marathon

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?

HSC 2013 MX2 Marathon (archive) (11 Viewers)

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