HSC 2013 MX2 Marathon (archive) (3 Viewers)

asianese · Feb 27, 2013

Re: HSC 2013 4U Marathon

Think of the position of P - and the relation between PS and PS'

darkfenrir · Feb 27, 2013

Re: HSC 2013 4U Marathon

William Sidis said:
Very popular conics question (don't tell which paper it is from if you know, please):

for part iii), is doing sin of both angles a good way? wats quicker?

Lieutenant_21 · Feb 28, 2013

Re: HSC 2013 4U Marathon

darkfenrir said:
for part iii), is doing sin of both angles a good way? wats quicker?

Think about the triangles.

darkfenrir · Feb 28, 2013

Re: HSC 2013 4U Marathon

gah! TELL MEEE wats a quicker way plz!

or more elegant

Sine thing is super ugly...

Sy123 · Feb 28, 2013

Re: HSC 2013 4U Marathon

darkfenrir said:
gah! TELL MEEE wats a quicker way plz!

or more elegant

Sine thing is super ugly...

Prove that S'PS* is collinear
Then use vertically opposite angles (this is how I did it)

darkfenrir · Feb 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Prove that S'PS* is collinear
Then use vertically opposite angles (this is how I did it)

Wats your DP? Is that carl sagan or some guy like that?

asianese · Feb 28, 2013

Re: HSC 2013 4U Marathon

Example of latex:

\int \frac{\sin^3x}{\cos^5x} dx

Wrap that in [ t e x ] [ / t e x ] tags.

Sy123 · Feb 28, 2013

Re: HSC 2013 4U Marathon

darkfenrir said:
Wats your DP? Is that carl sagan or some guy like that?

pm'ed

=====================

$Prove that the angle between the tangent to an ellipse, and the line from the centre to the tangent passing through the point of intersection, is never a right angle unless the ellipse is a circle or if the tangent is horizontal or vertical$

petermik · Feb 28, 2013

Re: HSC 2013 4U Marathon

$\int \frac{\sin^3x}{\cos^5x} dx$

HeroicPandas · Feb 28, 2013

Re: HSC 2013 4U Marathon

petermik said:
$\int \frac{\sin^3x}{\cos^5x} dx$

int: tan^3 (x) sex^2 (x).dx

Let u = tanx

blah blah..

= (tanx)^4 / 4 + C

Sy123 · Feb 28, 2013

Re: HSC 2013 4U Marathon

petermik said:
$\int \frac{\sin^3x}{\cos^5x} dx$

$\int \frac{\sin^3 x}{\cos^5x} \ dx = \int \tan^{3}x \sec^2 x \ dx = \frac{1}{4} \tan^4 x + c$

seanieg89 · Feb 28, 2013

Re: HSC 2013 4U Marathon

Chad O Dude has 1000 women interested in him. He knows that exactly one of them has a vicious, contagious, and fatal (if he catches it) case of herpes.

He wants to sleep with all of the other 999 women, but needs to first determine which one is infected. The way he does this is by sending a number of friends to sleep with them, and observing whether they die by the next morning. Each friend can bed several women at once, and Chad can run these tests on separate nights (although no friend is willing to be used as a guinea pig on multiple nights).

What is the minimum number of friends Chad must have in order to successfully identify the infected woman?

Provide proof.

deswa1 · Feb 28, 2013

Re: HSC 2013 4U Marathon

^Ahh nice question. I know the answer but I won't spoil it- will be interesting to see who (if anyone) gets it

iBibah · Mar 1, 2013

seanieg89 said:
Chad O Dude has 1000 women interested in him. He knows that exactly one of them has a vicious, contagious, and fatal (if he catches it) case of herpes.

He wants to sleep with all of the other 999 women, but needs to first determine which one is infected. The way he does this is by sending a number of friends to sleep with them, and observing whether they die by the next morning. Each friend can bed several women at once, and Chad can run these tests on separate nights (although no friend is willing to be used as a guinea pig on multiple nights).

What is the minimum number of friends Chad must have in order to successfully identify the infected woman?

Provide proof.

Isn't this the wine problem in a different situation ?

deswa1 · Mar 1, 2013

Re: HSC 2013 4U Marathon

Yep its the same thing just with a new story haha.

erckle999 · Mar 1, 2013

Re: HSC 2013 4U Marathon

I get 10 friends, working on a proof now.

Sy123 · Mar 2, 2013

Re: HSC 2013 4U Marathon

$A circle$ \ \ c_0 \ \ $is placed inside the unit square$ \ \ s_0 \ \ $so that the edges of the squares are tangents to the circle$

$Inside the circle, square$ \ \ s_1 \ \ $is placed inside$ \ \ c_0 \ \ $such that the corners of the square touch the circle$

$A circle$ \ \ c_1 \ \ $is then placed within$ \ \ s_1 \ \ $such that the edges are tangents to the circle$

$This process is continued infinitely many times$

$$Find the radius of$ \ \ c_n$

RealiseNothing · Mar 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$A circle$ \ \ c_0 \ \ $is placed inside the unit square$ \ \ s_0 \ \ $so that the edges of the squares are tangents to the circle$

$Inside the circle, square$ \ \ s_1 \ \ $is placed inside$ \ \ c_0 \ \ $such that the corners of the square touch the circle$

$A circle$ \ \ c_1 \ \ $is then placed within$ \ \ s_1 \ \ $such that the edges are tangents to the circle$

$This process is continued infinitely many times$

$$Find the radius of$ \ \ c_n$

All squares are similar to eachother, and same with circles. Hence we only have to find the ratio between the the radius' of $c_0$ and $c_1$ to find the radius of $c_n$

The radius of:

$c_0 = \frac{1}{\sqrt{2}}$

$c_1 = 1$

So the radius has a ratio of $\sqrt{2}$ . Hence the radius of $c_n$ is $2^{\frac{n-1}{2}}$

Trebla · Mar 2, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Chad O Dude has 1000 women interested in him. He knows that exactly one of them has a vicious, contagious, and fatal (if he catches it) case of herpes.

He wants to sleep with all of the other 999 women, but needs to first determine which one is infected. The way he does this is by sending a number of friends to sleep with them, and observing whether they die by the next morning. Each friend can bed several women at once, and Chad can run these tests on separate nights (although no friend is willing to be used as a guinea pig on multiple nights).

What is the minimum number of friends Chad must have in order to successfully identify the infected woman?

Provide proof.

LOL

Sy123 · Mar 3, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
All squares are similar to eachother, and same with circles. Hence we only have to find the ratio between the the radius' of $c_0$ and $c_1$ to find the radius of $c_n$

The radius of:

$c_0 = \frac{1}{\sqrt{2}}$

$c_1 = 1$

So the radius has a ratio of $\sqrt{2}$ . Hence the radius of $c_n$ is $2^{\frac{n-1}{2}}$

Another elegant solution, nice work.

========

$The complex numbers$ \ \ z \ \ $and$ \ \ w \ \ $satisfy the conditions$

$z^{13}=w$

$w^{11} = z$

$$The imaginary part of$ \ \ z \ \ $ is$ \ \ \sin \left (\frac{m\pi}{n} \right )$

$\textbf{Where m and n are positive integers with NO common factors}$

$$Find$ \ \ n$

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