HSC 2012 MX2 Marathon (archive) (2 Viewers)

Carrotsticks · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

largarithmic said:
Hey your thing says HSC 2011, are you doing science / advanced maths this year and if so at which uni?

Also may as well post a question. Dunno how suitable thing one is but its neat:

Part (A). Trapezium ABCD is given with AB parallel to CD. Let diagonals AC and BD intersect at X. Prove that triangles ADX and BCX have the same area.

Part (B). A triangle and any point on one of its sides are given. Show how to construct, with straight edge and compass, a line through this given point that bisects the triangle into two regions of equal area.

(A) Using alternate angle theorem and vertically opposite angles, you can prove that the two triangles ADX and BCX are similar. Then use the expression $Area=\frac{1}{2} a b \sin C$ to prove equality of areas using identities from the ratios between the lengths of similar triangles.

(B) Using basic straight edge and compass techniques (the ones taught in Year 8 Maths), draw the median of the opposite side from given point. The median splits the triangle into two equal segments.

asianese · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Parvee said:
Why does this thread make me feel way behind in 4u?

THIS THIS THIS ^^^^

Getting very insecure..

Carrotsticks · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

asianese said:
THIS THIS THIS ^^^^

Getting very insecure..

Like Spiral said, there is little need for feeling bad about it. Here are my reasons.

1. The majority of the questions here won't actually appear in exams. They are simply questions that test our thinking abilities, not necessarily a reflection of what the HSC will be like.

2. The majority of the posts from this thread have been from 2012'ers who most likely have Mathematics as their strongest subject, hence why they are able to do the questions.

3. It is likely that the people posting there are being/have been tutored (perhaps even ahead of time) in Extension 2, hence why they are able to do other topics usually taught later.

4. Some of the questions here were posted by people who have either graduated or are currently studying at University. Many of these questions (like tywebb's proof for the Generalised Wallis Product) are very difficult and perhaps even beyond what is expected of even the best of Extension 2 students.

5. You will be able to do the majority of these questions by the time you have your Trials.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Haha, don't worry guys, most of questions are just random stuff we find/ think of, even if it is beyond the 4u course.

rolpsy · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

$\\\textup{Let } z = r\, \mathrm{cis}(\theta).\\\textup{Show that } |z+1|\times|z-1| = 1 \textup{ implies } r^2 = 2\cos(2\theta).$

Carrotsticks · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

rolpsy said:
$\\\textup{Let } z = r\, \mathrm{cis}(\theta).\\\textup{Show that } |z+1|\times|z-1| = 1 \textup{ implies } r^2 = 2\cos(2\theta).$

Rolpsy, I very much like your questions.

May I ask from where do you acquire them, unless you make them yourself?

Carrotsticks · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

^^^^^ I had a very similar proof for that question. The main difficulty was finding the necessary substitution in order to acquire the telescoping sum. I also got stuck during the calculations for Part (A). I got the exact same as you for line 3, except I had 8x^2 instead of 4x^2, and I could not proceed (similarly to the way that $(a+b)^{2}=a^{2}+b^{2}+2ab$

However, I do not think many students will be able to recognise:

1. The substitution to be used.

2. The use of Partial fractions in your solutions line 3

Here were my solutions, which I had written a while ago, but never got round to uploading them. I included explanations for 2012'ers who may not understand all the concepts there.

seanieg89 · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
(B) Using basic straight edge and compass techniques (the ones taught in Year 8 Maths), draw the median of the opposite side from given point. The median splits the triangle into two equal segments.

The point is an arbitrary point on one of the triangles sides, not a vertex.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

I don't think anyone did this question, so I'll repost it

$$Using $ sinx = \frac{e^{ix}-e^{-ix}}{2i} $, show sin^{-1}x = -i\ln(ix+\sqrt{1-x^2})$

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Let $y=\sin x=\frac{e^{ix}-e^{-ix}}{2i}$

Noting: $\sin^{-1}y=x$

$\frac{e^{ix}-e^{-ix}}{2i}$

Let $z=cis x =e^{ix}$

$\frac{1}{z}=\frac{1}{cis x}=e^{-ix}$

$\frac{z-\frac{1}{z}}{2i}=y$

$z^2-2iyz-1=0$

$z=iy \pm \sqrt{1-y^2}$

$\therefore iy \pm \sqrt{1-y^2}=e^{ix}$

Solve for $x$

$\ln (iy \pm \sqrt{1-y^2})=ix$

$\therefore x=-i\ln (iy \pm \sqrt{1-y^2})$

Taking the positive solution,

$\therefore \sin^{-1}{y}=-i\ln (iy + \sqrt{1-y^2})$

Editing

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

I'm not sure why you added the $z=cisx$ part for, but that's right.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Show the following:

$1. |z+w|^2 = |z|^2 +2Re(z\overline{w}) +|w|^2$

$2. |z-w|^2 = |z|^2 -2Re(z\overline{w}) +|w|^2$

$3. |z+w|^2 +|z-w|^2= 2(|z|^2 +|w|^2)$

Drongoski · Jan 20, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Show the following:

$1. |z+w|^2 = |z|^2 +2Re(z\overline{w}) +|w|^2$

$2. |z-w|^2 = |z|^2 -2Re(z\overline{w}) +|w|^2$

$3. |z+w|^2 +|z-w|^2= 2(|z|^2 +|w|^2)$

$1) |z+w|^2 = (z+w) \overline {(z+w)} =(z+w)(\overline {z} +\overline {w} ) = z \overline{z} + z \overline {w} + \overline {z} w + w \overline {w} \\ \\ = |z|^2 + (z\overline {w} + \overline {z\overline{w}}) + |w|^2 = |z|^2 +2 Re (z \overline {w}) + |w|^2$

2) Similar to 1)

3) Add 1) and 2)

seanieg89 · Jan 20, 2012

Re: 2012 HSC MX2 Marathon

$|z\pm w|^2=(z\pm w)\overline{(z\pm w)}=(z\pm w)(\overline{z}\pm\overline{w})=(z\overline{z}\pm(z\overline{w}+\overline{z\overline{w}})+w\overline{w})\\=|z|^2\pm2\Re(z\overline{w})+|w|^2.\\ \\$Add the two equations together to arrive at c.$$

SpiralFlex · Jan 20, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
I'm not sure why you added the $z=cisx$ part for, but that's right.

Reaction haha.

Carrotsticks · Jan 21, 2012

Re: 2012 HSC MX2 Marathon

tywebb said:
$\text{Sketch }y=e^{f(x)}$

Does it have inflections?

Yes. They occur when you e^f(x) the left and right branches.

deswa1 · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

Using the fact that <a href="http://www.codecogs.com/eqnedit.php?latex=arg\frac{z_{1}}{z_{2}}=argz_{1}-argz_{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" title="arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" /></a>

Prove that <a href="http://www.codecogs.com/eqnedit.php?latex=tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" title="tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" /></a>

Carrotsticks · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
Using the fact that <a href="http://www.codecogs.com/eqnedit.php?latex=arg\frac{z_{1}}{z_{2}}=argz_{1}-argz_{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" title="arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" /></a>

Prove that <a href="http://www.codecogs.com/eqnedit.php?latex=tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" title="tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" /></a>

Haha I made a very similar question to this a couple of months ago, but the first question was:

Show that:

$(2+i)(3+i)=5+5i$

Then I got them to deduce a similar identity to yours.

seanieg89 · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

An easy polynomials question:

$$Let $x_1,\ldots,x_{n+1}$ be $n+1$ distinct real numbers. Let $P(x)$ and $Q(x)$ be real polynomials of degree $n$ such that $P(x_j)=Q(x_j)$ for $j=1,\ldots,n+1.$ Prove that $P(x)=Q(x)$ for all real $x.$

kingkong123 · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
Using the fact that $arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}$

Prove that $tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}$

let z1 = 1 + 2i and z2 = 3 +i
therefore arg(z1) = arctan(2) and arg(z2)=arctan(1/3)
we want to prove arctan(2) - arctan(1/) = pi/4 ie arg(z1)-arg(z2)=pi/4

arg(z1)-arg(z2)=arg(z1/z2) where z1/z2 = (1+i)/2

arg( (1+i)/2 ) = pi/4.

therefore $tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}$ as required. were u asking this because u didnt know how or just as a forum question? nice question

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