HSC 2013 MX2 Marathon (archive) (4 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.

HSC 2013 MX2 Marathon (archive) (4 Viewers)

Retired

This too shall pass

This too shall pass

Retired

what is that?It is Cowpea

Member

This too shall pass

what is that?It is Cowpea

This too shall pass

what is that?It is Cowpea

Member

Member

Member

Administrator

Account Closed

Well-Known Member

This too shall pass

Account Closed

Member

what is that?It is Cowpea

Users Who Are Viewing This Thread (Users: 0, Guests: 4)