HSC 2013 MX2 Marathon (archive) (6 Viewers)

seanieg89 · Oct 20, 2013

Re: HSC 2013 4U Marathon

SpiralFlex said:
I was informed that if you use a method that is not within the bounds of the syllabus they will still mark it. Could I be wrong?

Informed by whom?

What if the method completely circumvents the concept they are trying to test?

Where do you draw the line? You certainly shouldn't be allowed to use out of syllabus theorems in "Prove" questions, as that would let you write zero-effort proofs using really high powered machinery.

You might be right, but I have never heard this...and it doesn't seem to make sense to me.

seanieg89 · Oct 20, 2013

Re: HSC 2013 4U Marathon

VBN2470 said:
Yeah I understand the area of the slice etc. it's just the limits that's confusing me...

With these sorts of things, try to reduce to problem to a two dimensional one. In this case, we want to find the length OP in terms of r and theta. Try this by drawing a 2-d diagram and using things you know about circles and triangles.

SpiralFlex · Oct 20, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Informed by whom?

What if the method completely circumvents the concept they are trying to test?

Where do you draw the line? You certainly shouldn't be allowed to use out of syllabus theorems in "Prove" questions, as that would let you write zero-effort proofs using really high powered machinery.

You might be right, but I have never heard this...and it doesn't seem to make sense to me.

I don't know really my teachers from high school told me this. But yeah what if they are trying to test a certain concept.

alexandred · Oct 20, 2013

Re: HSC 2013 4U Marathon

SpiralFlex said:
:O Back to complex numbers are we?

$Suppose we have$\; 4\;$squares$\;A, B, C, D\; $with centres$\;p, q, r, s \;$respectively arranged as shown in the diagram below. Prove that the line segments joining the centres of the opposite squares are equal in length and also perpendicular. [Hint: consider the inside edges of the highlighted in red and use your knowledge of vectors and complex numbers]$

Let the vertices of the inner quadrilateral on red be z1, z2, z3, z4 and let a = cis(pi/4)/sqrt(2). Then we can find p = z1 + a(z3-z1), etc; multiplying r-q by i and subtracting it from s-p gives 0 (if we note that a-ia = 1 and a+ia = i) so equal length and perpendicular.

Carrotsticks · Oct 20, 2013

Re: HSC 2013 4U Marathon

The idea just popped in my head. Haven't attempted it but here's an interesting thought.

$\\ $The goat from Taylor Swift's song 'Trouble' is tied to a water tank of radius $ r $ by a string of length $ l = 2 \pi r. $ Find the area of grass that it can munch on.$$

Sy123 · Oct 20, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
The idea just popped in my head. Haven't attempted it but here's an interesting thought.

$\\ $The goat from Taylor Swift's song 'Trouble' is tied to a water tank of radius $ r $ by a string of length $ l = 2 \pi r. $ Find the area of grass that it can munch on.$$

Its just the area of circle with radius l minus circle radius r isn't it?

---

$Consider the sequence$

$u_{n} = 4u_{n-1} - 4u_{n-2} \ \ u_1=2 \ \ u_2 = 8$

$$Find a formula for$ \ \ u_n$

$And then hence without induction prove$

$\sum_{r=1}^n u_n = 2 + (n-1)2^{n+1}$

seanieg89 · Oct 20, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Its just the area of circle with radius l minus circle radius r isn't it?

No, the string wraps around the water tank as you walk around, the shape is not circular. From a quick glance it seems the integrals involved may not have nice closed expressions (especially since r is variable).

Sy123 · Oct 20, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
No, the string wraps around the water tank as you walk around, the shape is not circular. From a quick glance it seems the integrals involved may not have nice closed expressions (especially since r is variable).

Ah ok

I'm getting $2\pi^2 r$ now,

seanieg89 · Oct 20, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Ah ok

I'm getting $2\pi^2 r$ now,

Working?

RealiseNothing · Oct 20, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
The idea just popped in my head. Haven't attempted it but here's an interesting thought.

$\\ $The goat from Taylor Swift's song 'Trouble' is tied to a water tank of radius $ r $ by a string of length $ l = 2 \pi r. $ Find the area of grass that it can munch on.$$

Don't you mean a string of length $\pi r^2$ ?

Sy123 · Oct 20, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Working?

Sorry?

seanieg89 · Oct 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Sorry?

Meaning could you explain your method? I don't think it can possibly be this as your answer only grows linearly with r. If we just look at the semicircular part of the region, this scales as r^2.

RealiseNothing · Oct 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Its just the area of circle with radius l minus circle radius r isn't it?

Even if the string didn't wrap around, the radius of the larger circle would be $l+r$

Sy123 · Oct 21, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Meaning could you explain your method?

Yep

http://i.imgur.com/wRLNGGU.jpg

Hopefully its right

seanieg89 · Oct 21, 2013

Re: HSC 2013 4U Marathon

It's not as simple as that, think of what happens if you stand at an arbitrary point holding the string and then pull it taut, it doesn't wrap around the post in the way you have shown in your diagram. For instance, on one side of the pole a taut string won't make any contact with the pole except for the point to which it is tied.

Sy123 · Oct 21, 2013

Re: HSC 2013 4U Marathon

Is there an elementary solution to Carrot's problem then?

If not can someone post another question

Lanxal · Oct 21, 2013

Re: HSC 2013 4U Marathon

I worked out a rudimentary (and probably flawed) way of solving it, but it requires solving $\theta - \tan \theta =2\pi$ , which I cannot do. So

.

Here is basically the way I was doing it:

http://i.imgur.com/voDXmVw.png

The dark blue is a semicircle. The green is the area swept out if the string is tangential to the circle, which I reasoned (possibly incorrectly, someone check me on this) is the furthest the goat can go and so forms the perimeter of the area. Reaching the grass on the left side of the goat is best done by going around the left, and similar with the right, so there is no need for the goat to go all the way around. This simplifies solving for the area because it prevents area overlap from affecting the answer. The light blue bit is the area within this perimeter that isn't found by solving the integral.

The problem is that you have to find the angle $a$ that occurs when the goat is directly on the other side of the tank from where the rope is fixed. That requires solving the aforementioned equation, which I can't do. Thus I give up. /flips table.

If you were to find $a$ though, solving becomes simple. The semicircle is easy to find, the green area can be solved with an integral like Sy did, and the light blue area is just geometry (the area of a right-angle triangle minus the area of the sector, all multiplied by two).

But yes. Let's move on.

Let's just do Sy's question:

Sy123 said:
$Consider the sequence$

$u_{n} = 4u_{n-1} - 4u_{n-2} \ \ u_1=2 \ \ u_2 = 8$

$$Find a formula for$ \ \ u_n$

$And then hence without induction prove$

$\sum_{r=1}^n u_n = 2 + (n-1)2^{n+1}$

Edit: Ok, so, I worked out the formula $u_n = n2^n$ by working backwards from the second part, which obviously you can't do in a proof. I'm really not quite sure how much proving I need to do in this kind of question: after I figure it out in my head using bastard backwards mathematics, can I just state it? Do I have to derive it from the recursive formula? And if I just state it, should I then prove it by induction? Not really sure on these points.

The second part is simple enough using a telescoping sum:

$u_n = 4u_{n-1} - 4u_{n-2}$

$u_n = 4u_{n-2} - 4u_{n-3}$

$...$

$u_3 = 4u_{2} - 4u_{1}$

$\therefore \sum_{r=3}^n u_r = 4u_{n-1} - 4u_{n-2} + 4u_{n-2} - 4u_{n-3} + ... + 4u_{2} - 4u_{1}$

$= 4u_{n-1} - 8$

$\therefore \sum_{r=1}^n u_r = 4u_{n-1} - 8 + u_2 + u_1$

$= 4u_{n-1} - 8 + 8 + 2$

$= 4(n-1)2^{n-1} + 2$

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

bottleofyarn · Oct 21, 2013

Re: HSC 2013 4U Marathon

Alright, I got a funky answer building on Lanxal and Sy's stuff.
http://i.imgur.com/8tUqegy.jpg
I was happy when the top part (light blue) cancelled, but less so when the integral (green) sucked.
Again, how to solve $x - \tan x = 2\pi$ :/

Lanxal · Oct 21, 2013

Re: HSC 2013 4U Marathon

Dear god what did you draw.

Also I have no question to give and apologise for that. ;-;

bottleofyarn · Oct 21, 2013

Re: HSC 2013 4U Marathon

Lanxal said:
Dear god what did you draw.

Also I have no question to give and apologise for that. ;-;

It's terribly inelegant IMO. Here's one:
$$For a monic cubic, the discriminant is $ {(\alpha - \beta)^2 (\beta - \gamma)^2 (\gamma - \alpha)^2$
$$Prove that there are three distinct real zeroes iff $ \Delta > 0$

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