HSC 2015 MX2 Marathon (archive) (5 Viewers)

Ekman · Feb 12, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
a little change would make a good question:

$$ Simplify $ \sin{x}+\sin2x+\sin3x+\cdots+\sin{nx}. $ Hence show that $ \sin\frac{\pi}{2015}+\sin\frac{2\pi}{2015}+ \sin \frac{3\pi}{2015}+\cdots+\sin\frac{\2014\pi}{2015}= \cot\frac{\pi}{4030}.$

I did something a little different. I made a geometric sum of this:

$cisx+ cis2x+ cis3x+cis4x+...+cisnx = cisx+(cisx)^2+(cisx)^3+...+(cisx)^n$

Then made the whole geometric series sum and got all imaginary bits:

$\frac{\sin(\frac{\pi}{2015})}{1-cos(\frac{\pi}{2015})} = 1282.7885...$

$$ However, $ cot(\frac{\pi}{3030})=964.478...$

But if I was to sub in 4030 instead of 3030, it would get the same answer...

FrankXie · Feb 12, 2015

Re: HSC 2015 4U Marathon

There are seven different coloured jars. Each jar contains ten identical marbles, each being the
same colour as the jar it is in. A total of ten marbles is selected from these seven jars.
(a) How many different selections are possible if at least one marble is selected from each jar?
(b) How many different selections are possible if no restrictions?

FrankXie · Feb 12, 2015

Re: HSC 2015 4U Marathon

Ekman said:
I did something a little different. I made a geometric sum of this:

$cisx+ cis2x+ cis3x+cis4x+...+cisnx = cisx+(cisx)^2+(cisx)^3+...+(cisx)^n$

Then made the whole geometric series sum and got all imaginary bits:

$\frac{\sin(\frac{\pi}{2015})}{1-cos(\frac{\pi}{2015})} = 1282.7885...$

$$ However, $ cot(\frac{\pi}{3030})=964.478...$

But if I was to sub in 4030 instead of 3030, it would get the same answer...

sorry 3030 was a typo, i actually corrected into 4030

braintic · Feb 12, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
There are seven different coloured jars. Each jar contains ten identical marbles, each being the
same colour as the jar it is in. A total of ten marbles is selected from these seven jars.
(a) How many different selections are possible if at least one marble is selected from each jar?
(b) How many different selections are possible if no restrictions?

(a) Not sure ... is it 343 ?

(EDIT : no, I'm double counting, aren't I)

seanieg89 · Feb 12, 2015

Re: HSC 2015 4U Marathon

Ekman said:
I did something a little different. I made a geometric sum of this:

$cisx+ cis2x+ cis3x+cis4x+...+cisnx = cisx+(cisx)^2+(cisx)^3+...+(cisx)^n$

Then made the whole geometric series sum and got all imaginary bits:

$\frac{\sin(\frac{\pi}{2015})}{1-cos(\frac{\pi}{2015})} = 1282.7885...$

$$ However, $ cot(\frac{\pi}{3030})=964.478...$

But if I was to sub in 4030 instead of 3030, it would get the same answer...

The common ration has modulus 1, so this is not a convergent series. Using the infinite geometric series formula here is meaningless.

seanieg89 · Feb 12, 2015

Re: HSC 2015 4U Marathon

braintic said:
(a) Not sure ... is it 343 ?

(EDIT : no, I'm double counting, aren't I)

I got 84, form a string of 6 slashes 3 dots etc.

Drsoccerball · Feb 13, 2015

Re: HSC 2015 4U Marathon

Using cartesian coordinates, show that the sum of the distance from any point P on the ellipse
$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1 $ to its two foci is equal to 2a. $$
Ekman dont post your solution let someone else do it i want to see their method

VBN2470 · Feb 13, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Using cartesian coordinates, show that the sum of the distance from any point P on the ellipse
$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1 $ to its two foci is equal to 2a. $$
Ekman dont post your solution let someone else do it i want to see their method

$Using the fact that $PS=ePM$ ($0<\,$e$\,<1$), we can use this to show that $PS+PS'=ePM+ePM'=e(\frac{2a}{e})=2a$ i.e. the set of points such that the sum of the distances to two fixed points (the foci) is constant.$

Drsoccerball · Feb 13, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$$PS+PS'=ePM+ePM'=e(\frac{2a}{e})=2a$$

how does the first = second

Drsoccerball · Feb 13, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
how does the first = second

$PS=\sqrt{(x-ae)^2 +y^2}$

$PM=\sqrt{(x-ae)^2}$ ?

VBN2470 · Feb 13, 2015

Re: HSC 2015 4U Marathon

It is the standard definition of the ellipse you learn when you start Conics i.e. the ellipse is defined as the set of points such that the distance from any point in that set to a given point in the plane (a focus) is a constant positive fraction less than 1 (the eccentricity) of the perpendicular distance of the point in the set to a given line (called the directrix) or mathematically, $$PS=ePM$ $(0<\,$e$\,<1)$$

Drsoccerball · Feb 13, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
It is the standard definition of the ellipse you learn when you start Conics i.e. the ellipse is defined as the set of points such that the distance from any point in that set to a given point in the plane (a focus) is a constant positive fraction less than 1 (the eccentricity) of the perpendicular distance of the point in the set to a given line (called the directrix) or mathematically, $$PS=ePM$ $(0<\,$e$\,<1)$$

Weve started conics we did the ellipse but how does that make PS=ePM

Drsoccerball · Feb 13, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Weve started conics we did the ellipse but how does that make PS=ePM

In fact wouldn't $\frac{PS}{e^2}=PM?$

InteGrand · Feb 13, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
In fact wouldn't $\frac{PS}{e^2}=PM?$

The eccentricity e is defined by $e\equiv\frac{PS}{PM}$ .

VBN2470 · Feb 13, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
In fact wouldn't $\frac{PS}{e^2}=PM?$

Why $$e^2$$ ?

Drsoccerball · Feb 13, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
The eccentricity e is defined by $e\equiv\frac{PS}{PM}$ .

Forgot that

sorry but yeah now makes sense

Drsoccerball · Feb 13, 2015

Re: HSC 2015 4U Marathon

I think we should start posting conics and harder extension 1 questions

VBN2470 · Feb 13, 2015

Re: HSC 2015 4U Marathon

NEW QUESTION (HARDER):

$Prove that the tangent at a point $P$ on the ellipse is equally inclined to the focal chords through $P$.$

Drsoccerball · Feb 13, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEW QUESTION:

$Prove that the tangent at a point $P$ on the ellipse is equally inclined to the focal chords through $P$.$

$\frac{x^2}{a^2} +\frac{y^2}{b^2}=1$

$\frac{2x}{a^2} + \frac{2y \times y'}{b^2}=0$

$\therefore y'= \frac{-2b^2x_1}{2y_1a^2}$

$$ Focal chord gradient = $ \frac{y_1}{x_1-ae}$

$Use equations $ b^2=a^2(1-e^2) $ to equate$

Drsoccerball · Feb 13, 2015

Re: HSC 2015 4U Marathon

at school cant really complete now

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