HSC 2013 MX2 Marathon (archive) (1 Viewer)

Carrotsticks · Nov 21, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Hmmm...I got 2*log(2)-1. Are you sure your answer is correct?

My mistake Sean, the question was supposed to be an ACUTE ANGLED triangle, but I didn't type out the 'acute angled' part.

Your answer 2ln(2) - 1 is however correct. My method was to determine a set of inequalities which made a 'hyperbolic triangle', then found the area of the triangle either through a non-linear transformation or playing around with a bit of geometry and integration etc.

jyu said:
What you have worked out is not the answer to the question. Your answer is for forming a right angled triangle. Now I see why the probability is so low.

Read above. Fus Ro Dah is not mistaken, I sent him the correct question privately via PM whereas I typed down the wrong one here on the thread, which is why he acquired his answer. My fault for the typo.

RealiseNothing said:
This is going to be good.

My typos strike again!

Carrotsticks · Nov 21, 2012

Re: HSC 2013 4U Marathon

Here's a question more accessible to the majority of MX2 students.

$\\ $Suppose $ x + y + z = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ where $ x,y,z $ are non-zero real numbers.$ \\\\ $Prove that $ \frac{x^6 + y^6 + z^6}{x^3 + y^3 + z^3} = xyz$

RealiseNothing · Nov 21, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
Here's a question more accessible to the majority of MX2 students.

You think so? I thought your last question was heaps do-able by most MX2 students.

Carrotsticks · Nov 21, 2012

Re: HSC 2013 4U Marathon

Here's a slightly different question to what you would normally see. It is a fairly well-known problem.

Some pirates wish to bury their treasure on an island. They find a tree T and two rocks U and V. Starting at T, they pace off the distance from T to U, then turn right and pace off an equal distance from U to a point P, which they mark. Returning to T, they pace off the distance from T to V, then turn left and pace off an equal distance (to TV) to a point Q, which they mark. The treasure is then buried at the midpoint of PQ.

Years later a treasure hunter, who has the exact instructions for the location of the treasure, visits the island but finds that the tree T has disappeared! Being rash, he finds a random rock and decides to pretend that is the location of T, then proceeds to follow the instructions as usual. Use Complex Numbers and an Argand Diagram to determine if this treasure hunter get rich in the end.

Carrotsticks · Nov 21, 2012

Re: HSC 2013 4U Marathon

$\\ $Solve the polynomial $ (x-2)(x-3)(x-4)(x-5) = 360$

jyu · Nov 21, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
$\\ $Solve the polynomial $ (x-2)(x-3)(x-4)(x-5) = 360$

3x4x5x6=360, work from there

RealiseNothing · Nov 21, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
$\\ $Solve the polynomial $ (x-2)(x-3)(x-4)(x-5) = 360$

By inspection, since the four factors are all consecutive numbers, and knowing that $\frac{6!}{2} = 360$ , then $(x-2)=6$

$x=8$ is one solution.

Also since 4 negatives make a positive, $(x-5)=-6$

$x = -1$ is another solution.

Carrotsticks · Nov 21, 2012

Re: HSC 2013 4U Marathon

$\\ $Find the n'th roots of $ -1 $ and factor the polynomial $ P(t) = t^n + 1 $ over $ \mathbb{C}, \mathbb{R}, \mathbb{Q}$

Carrotsticks · Nov 21, 2012

Re: HSC 2013 4U Marathon

$\\ $Find values of $ a $ and $ b $ such that $ P(x) = (ax+b)(x^5+1)- (5x+1) $ is divisible by $ x^2 + 1$

jyu · Nov 21, 2012

Re: HSC 2013 4U Marathon

(x-1)(x-3)(x-6)(x-9)=252, find x

Carrotsticks · Nov 21, 2012

Re: HSC 2013 4U Marathon

$\\ $For what integer value(s) of $ a $ is the polynomial $ P(x) = x^{13} + x + 90 $ divisible by $ x^2 - x + a ?$

RealiseNothing · Nov 21, 2012

Re: HSC 2013 4U Marathon

jyu said:
(x-1)(x-3)(x-6)(x-9)=252, find x

x=10 is one solution from inspection, similar to that of the previous question.

Sy123 · Nov 21, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
$\\ $Solve the polynomial $ (x-2)(x-3)(x-4)(x-5) = 360$

Is the answer:

x=8
x=-1

$x=\frac{7\pm i \sqrt{71}}{2}$

Sy123 · Nov 21, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
$\\ $Find the n'th roots of $ -1 $ and factor the polynomial $ P(t) = t^n + 1 $ over $ \mathbb{C}, \mathbb{R}, \mathbb{Q}$

For C, we just do $(t+1)(t-cis 2\pi/n)...(t-t_n)=0$

t_k (k in N) are the roots.

For R, we simply add the conjugates and express as quadratic factors.

$(t+1)(t^2-2t\cos \frac{2\pi}{n}+1)(t^2-2t\cos \frac{4\pi}{n}+1)...$

For Q, am I right in saying:

$P(t)=(t+1)(t^{n-1}-t^{n-2}+t^{n-3}-...+1)$

bleakarcher · Nov 21, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
For C, we just do $(t+1)(t-cis 2\pi/n)...(t-t_n)=0$

t_k (k in N) are the roots.

For R, we simply add the conjugates and express as quadratic factors.

$(t+1)(t^2-2t\cos \frac{2\pi}{n}+1)(t^2-2t\cos \frac{4\pi}{n}+1)...$

For Q, am I right in saying:

$P(t)=(t+1)(t^{n-1}-t^{n-2}+t^{n-3}-...+1)$

I don't think the factorisation over Q is correct. You are not given whether n is even or odd.

Sy123 · Nov 21, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
I don't think the factorisation over Q is correct. You are not given whether n is even or odd.

Ah yes, my factorisation only works if n is odd. My bad.

Trebla · Nov 21, 2012

Re: HSC 2013 4U Marathon

$$Define $\omega_j=\dfrac{2\pi j}{n}$ for some fixed integers $j$ and $n$. Show that\\\\(a) $\displaystyle\sum_{t=1}^n\cos(t \omega_j) = \displaystyle\sum_{t=1}^n\sin(t \omega_j) = 0\\\\\\$(b) Hence show that for $j\neq k\\\\$(i) $\displaystyle\sum_{t=1}^n\cos(t \omega_j)\cos(t \omega_k) = 0\\\\$(ii) $ \displaystyle\sum_{t=1}^n\sin(t\omega_j)\sin(t \omega_k) = 0\\\\$(iii) $\displaystyle\sum_{t=1}^n\sin(t \omega_j)\cos(t \omega_k) = 0$

RealiseNothing · Nov 21, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
For Q, am I right in saying:

$P(t)=(t+1)(t^{n-1}-t^{n-2}+t^{n-3}-...+1)$

$t^n + 1 = (t - 1)(t^{n-1} + t^{n-2} + ... + t + 1)$

bleakarcher · Nov 21, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
$t^n + 1 = (t - 1)(t^{n-1} + t^{n-2} + ... + t + 1)$

That's t^n-1.

seanieg89 · Nov 21, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
$t^n + 1 = (t - 1)(t^{n-1} + t^{n-2} + ... + t + 1)$

nope.

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