HSC 2013 MX2 Marathon (archive) (5 Viewers)

Carrotsticks · Nov 20, 2012

Re: HSC 2013 4U Marathon

A stick of some length L is split into two smaller sticks (not necessarily equal). The larger of these two are split into two more sticks at some arbitrary point. As a result, there are three sticks in total.

Show that the probability that the three sticks form a triangle is 3ln(2) - 2.

God damn it Sy123, your dp confuses me (but my Santa hat > your top hat).

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
There you go Sy123

$$Suppose $k\left | z-z_1 \right |=l\left | z-z_2 \right |,$ where $k\neq l$ and both are positive real numbers.$\\$(a) Show that the locus of $z$ in the Argand Diagram is a circle with centre $\frac{k^2z_1-l^2z_2}{k^2-l^2}$ and radius $\frac{kl|z_2-z_1|}{|k^2-l^2|}$ by (i) letting $z=x+iy$ (ii) geometric means\\(b)What happens in the limit as $k$ approaches $l?$

Well for i, I managed to simply brute force my way through it by:

$z=x+iy, z_1=a+ib, z_2=c+id$

Then just doing loads of algebra eventually yielding me the answer. However if this is the only way to do this question (by letting z=x+iy), then I will be a bit disappointed heh. So is there a good/clever way to do this?

I will try the geometric means later.

But as for part b. As k approaches l, the circle will become the perpendicular bisector of z_1 and z_2.

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
A stick of some length L is split into two smaller sticks (not necessarily equal). The larger of these two are split into two more sticks at some arbitrary point. As a result, there are three sticks in total.

Show that the probability that the three sticks form a triangle is 3ln(2) - 2.

God damn it Sy123, your dp confuses me (but my Santa hat > your top hat).

Can the sticks cross over each other, or must the sticks make a perfect triangle without any extra bits sticking out?

(Also my slowpoke is more classy (and manly)) than yours :s)

Carrotsticks · Nov 20, 2012

Re: HSC 2013 4U Marathon

No extra bits sticking out.

RealiseNothing · Nov 20, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
A stick of some length L is split into two smaller sticks (not necessarily equal). The larger of these two are split into two more sticks at some arbitrary point. As a result, there are three sticks in total.

Show that the probability that the three sticks form a triangle is 3ln(2) - 2.

God damn it Sy123, your dp confuses me (but my Santa hat > your top hat).

Nice question, I'm giving it a go and have an idea what to do, but let me make a variation question:

A stick of some length L is split into two smaller sticks (not necessarily equal). The smaller of these two are split into two more sticks at some arbitrary point. As a result, there are three sticks in total.

Find the probability that the sticks form a triangle.

jyu · Nov 20, 2012

Re: HSC 2013 4U Marathon

0

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

jyu said:
0

By Triangle Inequality yes?

RealiseNothing · Nov 20, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
By Triangle Inequality yes?

Yes. It's how you do carrot's question I think. Basically what's the probability that the sum of the two shorter sides cut is larger than the longest side.

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Yes. It's how you do carrot's question I think. Basically what's the probability that the sum of the two shorter sides cut is larger than the longest side.

That is what I originally thought, however I was not able to do much with the 3 inequalities that I had and no numbers.

RealiseNothing · Nov 20, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
That is what I originally thought, however I was not able to do much with the 3 inequalities that I had and no numbers.

Same, I got stuck doing it that way. I'll try it again tomorrow if no one has got it by then, I really need to get back to my chemistry assignment lol.

Fus Ro Dah · Nov 20, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
That is what I originally thought, however I was not able to do much with the 3 inequalities that I had and no numbers.

RealiseNothing said:
Same, I got stuck doing it that way. I'll try it again tomorrow if no one has got it by then, I really need to get back to my chemistry assignment lol.

A clue: We simplify the problem by considering the triangle OAB, where O is the origin and A and B are the x and y intercepts, respectively, of the line x+y=k. The set of all possible trisections into triangles exists in triangle OAB. Generally, questions such as these are done using geometric probability, which usually involves finding the area between a curve and a line or possibly the ratio between the area of a shape that lies entirely within another area representing the total set.

jyu · Nov 20, 2012

Re: HSC 2013 4U Marathon

3ln2-2 is small. One would think the prob is higher.

Fus Ro Dah · Nov 20, 2012

Re: HSC 2013 4U Marathon

jyu said:
3ln2-2 is small. One would think the prob is higher.

3ln2-2 is correct. I worked it out just then.

Trebla · Nov 21, 2012

Re: HSC 2013 4U Marathon

Damn these Slowpokes everywhere...

cutemouse · Nov 21, 2012

Re: HSC 2013 4U Marathon

Trebla said:
Damn these Slowpokes everywhere...

Long time no see. Out of the closet?

seanieg89 · Nov 21, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
A stick of some length L is split into two smaller sticks (not necessarily equal). The larger of these two are split into two more sticks at some arbitrary point. As a result, there are three sticks in total.

Show that the probability that the three sticks form a triangle is 3ln(2) - 2.

God damn it Sy123, your dp confuses me (but my Santa hat > your top hat).

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

Sy123 · Nov 21, 2012

Re: HSC 2013 4U Marathon

$$Using the identity $ z^n+\frac{1}{z^n}=2 \cos n\theta $ if $ z= \cos \theta + i \sin \theta$

$$Show that $ \int_0^{\frac{\pi}{2}} \cos^{2n} \theta \ d\theta = \frac{\pi (2n)!}{2^{2n}(n!)^2}$

Who needs integration by parts when you have this :s

cutemouse · Nov 21, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Well for i, I managed to simply brute force my way through it by:

$z=x+iy, z_1=a+ib, z_2=c+id$

Then just doing loads of algebra eventually yielding me the answer. However if this is the only way to do this question (by letting z=x+iy), then I will be a bit disappointed heh. So is there a good/clever way to do this?

Maybe you can use the fact that |z|^2 = z zbar...

jyu · Nov 21, 2012

Re: HSC 2013 4U Marathon

Fus Ro Dah said:
3ln2-2 is correct. I worked it out just then.

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.

RealiseNothing · Nov 21, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
A stick of some length L is split into two smaller sticks (not necessarily equal). The larger of these two are split into two more sticks at some arbitrary point. As a result, there are three sticks in total.

Show that the probability that the three sticks form a triangle is 3ln(2) - 2.

God damn it Sy123, your dp confuses me (but my Santa hat > your top hat).

Fus Ro Dah said:
3ln2-2 is correct. I worked it out just then.

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

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.

This is going to be good.

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