HSC 2012 MX1 Marathon #1 (archive) (2 Viewers)

SpiralFlex · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

Examine said:
So should I stay with the first 2 and attempt the extension sometimes then?

You won't have time to attempt all the questions.

SpiralFlex · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
No. You were right the first time. It's 7/4. I made a silly mistake. But here is a different method:

I like the solution haha!

Carrotsticks · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

Examine said:
For Cambridge 3u Year 11, should I do the extension part? I'm finding it quite difficult though will the syllabus need me to do it?

Try a few of them. Although they may not appear in exams (I have only seen a couple appear in exams), it's the experience and the thought processes that you develop by doing them, that you have to gain. You can later use these in your exams.

SpiralFlex said:
You won't have time to attempt all the questions.

SpiralFlex · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
Try a few of them. Although they may not appear in exams (I have only seen a couple appear in exams), it's the experience and the thought processes that you develop by doing them, that you have to gain. You can later use these in your exams.

I'll race you after the HSC. Post worked solutions on BOS.

Bored_of_HSC · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

SpiralFlex said:
You won't have time to attempt all the questions.

I remember in my old tutor they made us do some of the questions

.

So tiring haha. But there were some fun proofs.

cutemouse · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
No. You were right the first time. It's 7/4. I made a silly mistake. But here is a different method:

Nice, but i think it's worth noting that the first line is only valid if the terms of the series are absolutely convergent.

Trebla · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

SpiralFlex said:
I'll race you after the HSC. Post worked solutions on BOS.

Good luck.

This had been attempted a few years ago but failed miserably (it was even attempted as a team effort).

cutemouse · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

Trebla said:
Good luck. This had been attempted a few years ago but failed miserably (it was even attempted as a team effort).

Teachers (and other people in general) become lazy when it comes to writing solutions, even to their own examinations... So go figure.

SpiralFlex · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

Trebla said:
Good luck. This had been attempted a few years ago but failed miserably (it was even attempted as a team effort).

Do not worry, Spiral loves helping humans! I shall make work solutions and get them checked then post them.

tywebb · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

cutemouse said:
the first line is only valid if the terms of the series are absolutely convergent

And of course it is.

$\ \newline\textstyle|{3(k+1)-1\over3^{k+1}}\big/{3k-1\over3^k}|=|{1\over3}+{1\over{3k-1}}|\newline\rightarrow{1\over3}\text{ as }k\rightarrow\infty\newline<1\newline\therefore\sum_{k=1}^\infty{3k-1\over3^k}\text{ is absolutely convergent}$

I didn't bother to put this in my solution because it is not required thesedays in HSC exams.

Exam committees and HSC marking teams would be in a lot of trouble if it was required to prove convergence (or absolute convergence) for a lot of HSC questions asked since 1980.

nightweaver066 · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
No. You were right the first time. It's 7/4. I made a silly mistake. But here is a different method:

This was what i was trying to do last night but didn't end up correct..

Why do you square the term on the left in the 2nd line?

bleakarcher · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

What did you do in line 2? I dont understand you did with the first term.

tywebb · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

nightweaver066 said:
Why do you square the term on the left in the 2nd line?

$\ \newline\textstyle1+{2\over r}+{3\over r^2}+\cdots\newline\ \newline=(1+{1\over r}+{1\over r^2}+\cdots)+{1\over r}(1+{1\over r}+{1\over r^2}+\cdots)+{1\over r^2}(1+{1\over r}+{1\over r^2}+\cdots)+\cdots\newline\ \newline=(1+{1\over r}+{1\over r^2}+\cdots)^2$

bleakarcher · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

^ Damn. How did you realise that?

tywebb · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

It's fairly standard. Don't your teachers teach you this stuff?

nightweaver066 · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
It's fairly standard. Don't your teachers teach you this stuff?

I haven't learnt this at school yet so it's interesting to me haha.

Thanks for clearing that up.

Trebla said:
Alternatively, a relatively easier problem:

$\hspace{-1cm} $Show that the quadratic polynomial $kx^2 + 3kx + 6$ can never be negative definite.$$

$$The conditions for a quadratic polynomial to be negative definite are that $ k < 0 \; $and$ \; \Delta < 0$

$\Delta = 9k^2 - 4(k)(6)$

$= 9k^2 - 24k$

$= 3k(3k - 8) < 0$

$0 < k < \frac{8}{3}$

$However, this set of values for k is greater than 0. Hence it can never be negative definite$

New question..

largarithmic · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

nightweaver066 said:
I haven't learnt this at school yet so it's interesting to me haha.

Thanks for clearing that up.

$$The conditions for a quadratic polynomial to be negative definite are that $ k < 0 \; $and$ \; \Delta < 0$

$\Delta = 9k^2 - 4(k)(6)$

$= 9k^2 - 24k$

$= 3k(3k - 8) < 0$

$0 < k < \frac{8}{3}$

$However, this set of values for k is greater than 0. Hence it can never be negative definite$

New question..

8 choose 3 = something...

How about this gem: Prove, WITHOUT USING THE BINOMIAL THEOREM OR ANYTHING LIEK THAT, that
$^nC_0^2 + ^nC_1^2 + ^nC_2^2 + ... + ^nC_n^2 = ^{2n}C_n$

deterministic · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

largarithmic said:
8 choose 3 = something...

How about this gem: Prove, WITHOUT USING THE BINOMIAL THEOREM OR ANYTHING LIEK THAT, that
$^nC_0^2 + ^nC_1^2 + ^nC_2^2 + ... + ^nC_n^2 = ^{2n}C_n$

Is a combinatorial argument acceptable?

If so, then since:
$^{2n}C_n$ = number of ways to choose n objects from 2n objects.
One way to do so is to split 2n objects into 2 groups with n objects in each (Doesn't matter how its done)
Consider number of ways of choosing k objects from group 1 and n-k objects in group 2 for k=0,1,..,n.
Add them together and equate with RHS to get result.

largarithmic · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

deterministic said:
Is a combinatorial argument acceptable?

If so, then since:
$^{2n}C_n$ = number of ways to choose n objects from 2n objects.
One way to do so is to split 2n objects into 2 groups with n objects in each (Doesn't matter how its done)
Consider number of ways of choosing k objects from group 1 and n-k objects in group 2 for k=0,1,..,n.
Add them together and equate with RHS to get result.

That was what I was looking for ^^

Anyway, geometry problem:

$(a) Triangle $ABC$ is given, with arbitrary points $D, E, F$ chosen on sides $BC, CA, AB$ respectively. Prove that the three circles $AEF, BDF, CDE$ all go through a common point.$

$(b)Four distinct arbitrary lines $l_1, l_2, l_3, l_4$ are given in the plane, no pair parallel and no three concurrent. Let $\gamma_1$ be the circle going through the vertices of the triangle determined by $l_2, l_3, l_4$; $\gamma_2$ be the circle going through the vertices of the triangle determined by $l_1, l_3, l_4$; $\gamma_3$ be the circle going through the vertices of the triangle determined by $l_1, l_2, l_4$; and $\gamma_4$ be the circle going through the vertices of the triangle determined by $l_1, l_2, l_3$. Prove that $\gamma_1$, $\gamma_2$, $\gamma_3$ and $\gamma_4$ are concurrent.$

Carrotsticks · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

largarithmic said:
That was what I was looking for ^^

Anyway, geometry problem:

$(a) Triangle $ABC$ is given, with arbitrary points $D, E, F$ chosen on sides $BC, CA, AB$ respectively. Prove that the three circles $AEF, BDF, CDE$ all go through a common point.$

$(b)Four distinct arbitrary lines $l_1, l_2, l_3, l_4$ are given in the plane, no pair parallel and no three concurrent. Let $\gamma_1$ be the circle going through the vertices of the triangle determined by $l_2, l_3, l_4$; $\gamma_2$ be the circle going through the vertices of the triangle determined by $l_1, l_3, l_4$; $\gamma_3$ be the circle going through the vertices of the triangle determined by $l_1, l_2, l_4$; and $\gamma_4$ be the circle going through the vertices of the triangle determined by $l_1, l_2, l_3$. Prove that $\gamma_1$, $\gamma_2$, $\gamma_3$ and $\gamma_4$ are concurrent.$

Geometry problems are fun to do, but they are the absolute worst when it comes to typing them down.

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