1. Re: HSC 2012 Marathon :)

If you represent the hexagon's vertices as points on a plane and do lots of coordinate geometry, you obtain the following expression for the angle:

$\theta = \tan^{-1}(\frac{\frac{3}{2}\tan10^{\circ}\tan50^{\circ} + \frac{\sqrt{3}}{2}\tan50^{\circ}}{\tan 50^{\circ} - \frac{\sqrt{3}}{2} - \frac{1}{2} \tan 10^{\circ}}})$

And if you type that into a calculator, it does indeed give 80o. I don't know how to simplify it manually though.

2. Re: HSC 2012 Marathon :)

$Using the definition of e \\ \\ e=\lim_{n \to \infty}(1+\frac{1}{n})^n \\ \\ Prove from first principles \frac{d}{dx}(\ln x)=\frac{1}{x} \\ \\ Use the substitution u=\frac{h}{x} \ \ \ for f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

3. Re: HSC 2012 Marathon :)

Okay worked out the hexagon Q.

HINT: Construct AC and it follows that angle BAC = angle BCA = 30 degrees (isosceles triangle). Work from there and focus purely on ABCP.

4. Re: HSC 2012 Marathon :)

First part of this question is a 3U induction if you replace the word complex with real .

http://community.boredofstudies.org/...d.php?t=292775

5. Re: HSC 2012 Marathon :)

Originally Posted by seanieg89
First part of this question is a 3U induction if you replace the word complex with real .

http://community.boredofstudies.org/...d.php?t=292775
When you define u' does it mean this:

$u=(u_1,u_2,u_3...) \\ \\ u'=(u_1',u_2',u_3'...)$

If it means this then I will have a go at the question first thing tomorrow.

6. Re: HSC 2012 Marathon :)

Yep, exactly. The question itself is a pretty standard induction...but it raises other interesting questions about this construction.

7. Re: HSC 2012 Marathon :)

Just going to check my step 1 here to see if I actually understand the question, Ill do the rest tomorrow too tired right now lol

$Define the polynomial u=a_0 x^n+a_1 x^{n-1}+...+a_n$

$Step 1 : n=0 \\ \\ u=a_n \\ \therefore u'=0 \ \ for all n in sequence \\ \\ \therefore u'=(0,0,0...)$

Correct or way off?

8. Re: HSC 2012 Marathon :)

Originally Posted by Sy123
Just going to check my step 1 here to see if I actually understand the question, Ill do the rest tomorrow too tired right now lol

$Define the polynomial u=a_0 x^n+a_1 x^{n-1}+...+a_n$

$Step 1 : n=0 \\ \\ u=a_n \\ \therefore u'=0 \ \ for all n in sequence \\ \\ \therefore u'=(0,0,0...)$

Correct or way off?
Yep, notation is a little funky but the idea is write, a constant sequence will get killed straight away by the operator '. Compare this to the behaviour of a constant function being differentiated .

9. Re: HSC 2012 Marathon :)

Originally Posted by Sy123
Just going to check my step 1 here to see if I actually understand the question, Ill do the rest tomorrow too tired right now lol

$Define the polynomial u=a_0 x^n+a_1 x^{n-1}+...+a_n$

$Step 1 : n=0 \\ \\ u=a_n \\ \therefore u'=0 \ \ for all n in sequence \\ \\ \therefore u'=(0,0,0...)$

Correct or way off?
Yep, notation is a little funky but the idea is right, a constant sequence will get killed straight away by the operator '. Compare this to the behaviour of a constant function being differentiated .

10. Re: HSC 2012 Marathon :)

Originally Posted by seanieg89
Yep, notation is a little funky but the idea is write, a constant sequence will get killed straight away by the operator '. Compare this to the behaviour of a constant function being differentiated .
Alright nice I will see if I can complete this tomorrow

(inb4 detroyed)

11. Re: HSC 2012 Marathon :)

Originally Posted by Sy123
Alright nice I will see if I can complete this tomorrow

(inb4 detroyed)
Honestly not that difficult, maybe think about how you would try to prove the analogue of this theorem for functions: f is killed by taking finitely many derivatives if and only if f is a polynomial . Because differentiation is more familiar this should be quite easy, then try to mimic this approach for the less familiar operation going on here.

12. Re: HSC 2012 Marathon :)

Progress

Thats just my progress on Step 3, but I keep running into that barrier, I want to be able to prove that
$v^{(k)}=0 \ \ \ OR \ \ \ C \ \ \ (constant C)$

But I cant because I just end up with:

$v^{(k)}=\sum_{i=0}^{k-1}u^{(k-1)}$

Am I missing something here?

13. Re: HSC 2012 Marathon :)

Hint: polynomial degree. Gtg now but will look closer at your work later if you don't get it yet.

14. Re: HSC 2012 Marathon :)

Originally Posted by eskimogenius
A point P is marked inside a regular hexagon ABCDEF so that angleBAP=angleDCP=50deg. Find angle ABP.

GL.

Fwark, this is a good question.

Just did it using complex numbers.

15. Re: HSC 2012 Marathon :)

Originally Posted by Sy123

$A projectile is fired from the origin with velocity V, under the conditions of gravity g and no air resistance. The projectile is fired such that the maximum height is h, and the horizontal distance from the origin to the maximum height is h. \\ \\ Show that V=\sqrt{\frac{5gh}{2}}$
$A projectile is fired from the origin with velocity V, under the conditions of gravity g and no air resistance. The projectile is fired such that the maximum height is half the range of the projectile. Taking maximum height as h. A particle is starting Simple Harmonic Motion where it's centre of motion is where the projectile hits the ground. It is initially at the centre of motion, with period of 2\pi . The projectile hits the particle while it is at its centre of motion, right when the particle has completed a Period. \\ \\ Show that h=\frac{g \pi^2}{2}$

Use the results from the previous question

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