HSC 2014 MX2 Marathon (archive) (3 Viewers)

FrankXie · Nov 22, 2014

Re: HSC 2014 4U Marathon

for part (ii):

$$Consider the geometric series of complex numbers$ \sum_{n=0}^\infty[(\cos\theta+i\sin\theta)\cos\theta]^n$

Because the modulus of common ratio is less than 1, the limiting sum exists and

$S_\infty=\frac{1}{1-\cos\theta(\cos\theta+i\sin\theta)}$

$=\frac{1}{\sin\theta(\sin\theta-i\cos\theta)}$

$=\frac{\sin\theta+i\cos\theta}{\sin\theta}$

Finally equate the real part of both sides.

Axio · Nov 23, 2014

Re: HSC 2014 4U Marathon

In how many ways can the word:

$CARROTSTICKS$

be arranged in a line so that the two Rs (which are indistinguishable) are separated by at least one other letter?

integral95 · Nov 23, 2014

Re: HSC 2014 4U Marathon

Axio said:
In how many ways can the word:

$CARROTSTICKS$

be arranged in a line so that the two Rs (which are indistinguishable) are separated by at least one other letter?

$10(\frac{10!}{2!2!2!})$

i'm not sure hehehe

dunjaaa · Nov 23, 2014

Re: HSC 2014 4U Marathon

Insertion method; 11C2 x 10!/(2!2!2!)

integral95 · Nov 23, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
Insertion method; 11C2 x 10!/(2!2!2!)

11C2 could include 2 Rs being together lol

Axio · Nov 23, 2014

Re: HSC 2014 4U Marathon

I got (assuming same letters are always undistinguishable; I could be wrong lol):

Number of total arrangements - number of arrangements with 2 Rs together
= 12!/(2!2!2!) - (2!11!)/(2!2!2!)

FrankXie · Nov 24, 2014

Re: HSC 2014 4U Marathon

Axio said:
I got (assuming same letters are always undistinguishable; I could be wrong lol):

Number of total arrangements - number of arrangements with 2 Rs together
= 12!/(2!2!2!) - (2!11!)/(2!2!2!)

I would also do complementary. But do you mean two R's are different while 2C's, 2S's, 2T's are each pair identical? lol

Sy123 · Nov 24, 2014

Re: HSC 2014 4U Marathon

$\\ $Using complex numbers, show that the angle subtended by a semi-circle is a right angle$$

Chlee1998 · Nov 24, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Using complex numbers, show that the angle subtended by a semi-circle is a right angle$$

using the converse of pythagoras' theorem was one way i did it

FrankXie · Nov 24, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Using complex numbers, show that the angle subtended by a semi-circle is a right angle$$

$Without loss of generality, let the circle be $ |z|=1, $ and the diameter be on the real axis.$

$$ Let the points $ P, A $ and $ B $ represent complex numbers $ \cos\theta+i\sin\theta, -1 $ and $ 1.$

$$ Now $ \overrightarrow{AP}=\cos\theta+1+i\sin\theta, \overrightarrow{PB}=1-\cos\theta-i\sin\theta$

$\frac{\cos\theta+1+i\sin\theta}{1-\cos\theta-i\sin\theta}=\frac{2\cos^2\frac\theta2+2i\sin\frac\theta2\cos\frac\theta2}{2\sin^2\frac\theta2-2i\sin\frac\theta2\cos\frac\theta2}=i\cot \frac{\theta}{2}$

$therefore, the vectors $ \overrightarrow{AP} $ and $ \overrightarrow{PB} $ make a right angle.$

Sy123 · Nov 25, 2014

Re: HSC 2014 4U Marathon

FrankXie said:
$Without loss of generality, let the circle be $ |z|=1, $ and the diameter be on the real axis.$

$$ Let the points $ P, A $ and $ B $ represent complex numbers $ \cos\theta+i\sin\theta, -1 $ and $ 1.$

$$ Now $ \overrightarrow{AP}=\cos\theta+1+i\sin\theta, \overrightarrow{PB}=1-\cos\theta-i\sin\theta$

$\frac{\cos\theta+1+i\sin\theta}{1-\cos\theta-i\sin\theta}=\frac{2\cos^2\frac\theta2+2i\sin\frac\theta2\cos\frac\theta2}{2\sin^2\frac\theta2-2i\sin\frac\theta2\cos\frac\theta2}=i\cot \frac{\theta}{2}$

$therefore, the vectors $ \overrightarrow{AP} $ and $ \overrightarrow{PB} $ make a right angle.$

Similar to what I have in mind, I just translate the problem into showing:

$$arg$ \left(\frac{z-1}{z+1} \right) = \frac{\pi}{2} \ $for$ \ |z| = 1 \\ \\ \frac{z-1}{z+1} = \frac{(z-1)(\bar{z} + 1)}{|z+1|^2} = \frac{|z|^2 - 1 + (z - \bar{z})}{K} \\ \\ = C(z - \bar{z}) \ $for some constant$ \ C > 0$

So it avoids mod-arg form but its essentially the same thing

Chlee1998 · Nov 26, 2014

Re: HSC 2014 4U Marathon

integral from (6^(1/2) + 2^(1/2))/2 to 1: (1+x^2)/(1+x^4) using the substitution u= x-(1/x)

you cannot use any 4u specific integration formulas such as IBP, partial fractions, inverse tan integral, etc. Only 3u concepts such as u substitution.

FrankXie · Nov 26, 2014

Re: HSC 2014 4U Marathon

Chlee1998 said:
integral from (6^(1/2) + 2^(1/2))/2 to 1: (1+x^2)/(1+x^4) using the substitution u= x-(1/x)

you cannot use any 4u specific integration formulas such as IBP, partial fractions, inverse tan integral, etc. Only 3u concepts such as u substitution.

Is it allowed to use the standard table of integrals to evaluate $\int\frac{du}{u^2+2}$ ?

Chlee1998 · Nov 26, 2014

Re: HSC 2014 4U Marathon

FrankXie said:
Is it allowed to use the standard table of integrals to evaluate $\int\frac{du}{u^2+2}$ ?

please ignore my previous post.

Chlee1998 · Nov 27, 2014

Re: HSC 2014 4U Marathon

sort of difficult

FrankXie · Nov 27, 2014

Re: HSC 2014 4U Marathon

Chlee1998 said:
View attachment 31376

sort of difficult

Here is my solution using trigonometry. Let AD, DC and CB touch the semicircle at points E, F and G respectively. Let the radius of the circle be r.

$$ Let $ \angle{EOA}=\angle{GOB}=\alpha (\triangle{AOE}\equiv\triangle{BOG})$

$$ Let $ \angle{EOD}=\angle{DOF}=\beta (\triangle{DOE}\equiv\triangle{DOF})$

$$ Thus $ \angle{FOC}=\angle{COG}=\frac{\pi-2\alpha-2\beta}{2}=\frac{\pi}{2}-(\alpha+\beta) (\triangle{FOC}\equiv\triangle{COG})$

$$ Now, $ AO=r\sec\alpha, AD=r\tan\alpha+r\tan\beta, BC=r\tan\alpha+r\tan[\frac{\pi}{2}-(\alpha+\beta)]=r\tan\alpha+r\cot(\alpha+\beta)=r\tan\alpha+r \frac{1-\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}$

$\therefore AD\times BC=r^2(\tan\alpha+\tan\beta)(\tan\alpha+\frac{1-\tan\alpha\tan\beta}{\tan\alpha+\tan\beta})$

$=r^2(\tan^2\alpha+\tan\beta\tan\alpha+1-\tan\alpha\tan\beta)$

$=r^2(\tan^2\alpha+1)=r^2\sec^2\alpha=AO^2$

Chlee1998 · Nov 27, 2014

Re: HSC 2014 4U Marathon

FrankXie said:
Here is my solution using trigonometry. Let AD, DC and CB touch the semicircle at points E, F and G respectively. Let the radius of the circle be r.

$$ Let $ \angle{EOA}=\angle{GOB}=\alpha (\triangle{AOE}\equiv\triangle{BOG})$

$$ Let $ \angle{EOD}=\angle{DOF}=\beta (\triangle{DOE}\equiv\triangle{DOF})$

$$ Thus $ \angle{FOC}=\angle{COG}=\frac{\pi-2\alpha-2\beta}{2}=\frac{\pi}{2}-(\alpha+\beta) (\triangle{FOC}\equiv\triangle{COG})$

$$ Now, $ AO=r\sec\alpha, AD=r\tan\alpha+r\tan\beta, BC=r\tan\alpha+r\tan[\frac{\pi}{2}-(\alpha+\beta)]=r\tan\alpha+r\cot(\alpha+\beta)=r\tan\alpha+r \frac{1-\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}$

$\therefore AD\times BC=r^2(\tan\alpha+\tan\beta)(\tan\alpha+\frac{1-\tan\alpha\tan\beta}{\tan\alpha+\tan\beta})$

can u find another way without using so much trig

FrankXie · Nov 27, 2014

Re: HSC 2014 4U Marathon

hm thinking

FrankXie · Nov 28, 2014

Re: HSC 2014 4U Marathon

oops, how can i forget similar triangles

$\triangle{ADO}|||\triangle{BOC} (\angle{DAO}=\angle{OBC}, \angle{AOD}=\angle{BCO})$

the reason why these pairs of angles are equal is already in my first solution

Chlee1998 · Nov 28, 2014

Re: HSC 2014 4U Marathon

FrankXie said:
oops, how can i forget similar triangles

$\triangle{ADO}|||\triangle{BOC} (\angle{DAO}=\angle{OBC}, \angle{AOD}=\angle{BCO})$

the reason why these pairs of angles are equal is already in my first solution

yep good job this was what I did lol. Hence why I told u to find an alternatuve solution which hinted that a very simple did exist

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