HSC 2012 Marathon :)

**GoldyOrNugget** · 21 Oct 2012, 12:57 AM

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 80^o. I don't know how to simplify it manually though.

**Sy123** · 21 Oct 2012, 1:06 AM

$$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}$

**Carrotsticks** · 21 Oct 2012, 1:08 AM

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.

**seanieg89** · 21 Oct 2012, 1:16 AM

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

**Sy123** · 21 Oct 2012, 1:22 AM

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.

**seanieg89** · 21 Oct 2012, 1:25 AM

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

**Sy123** · 21 Oct 2012, 1:34 AM

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?

**seanieg89** · 21 Oct 2012, 1:37 AM

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

.

**seanieg89** · 21 Oct 2012, 1:37 AM

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

.

**Sy123** · 21 Oct 2012, 1:38 AM

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)

**seanieg89** · 21 Oct 2012, 1:42 AM

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.

**Sy123** · 22 Oct 2012, 5:56 PM

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?

**seanieg89** · 22 Oct 2012, 6:01 PM

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

**lgnorance** · 28 Oct 2012, 1:27 PM

Originally Posted by eskimogenius

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

GL.

PS. Answer is 80deg.

Fwark, this is a good question.

Just did it using complex numbers.

**Sy123** · 28 Oct 2012, 6:58 PM

Originally Posted by Sy123

Another question I made up:

$$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