HSC 2015 MX2 Marathon (archive) (6 Viewers)

RealiseNothing · Feb 15, 2015

Re: HSC 2015 4U Marathon

braintic said:
I can think of good reasons. But if those reasons are correct, I sure hope he goes to a co-ed school.

But maths over girls anyday.

Drsoccerball · Feb 16, 2015

Re: HSC 2015 4U Marathon

braintic said:
I can think of good reasons. But if those reasons are correct, I sure hope he goes to a co-ed school.

sadly i go to an all boys school... Thats not why I was distracted though ahahaha

Ekman · Feb 16, 2015

Re: HSC 2015 4U Marathon

RealiseNothing said:
But maths over girls anyday.

Maths is my lover

braintic · Feb 16, 2015

Re: HSC 2015 4U Marathon

RealiseNothing said:
But maths over girls anyday.

Ekman said:
Maths is my lover

Imaginary roots can provide only so much enjoyment.

Drsoccerball · Feb 16, 2015

Re: HSC 2015 4U Marathon

braintic said:
imaginary roots can provide only so much enjoyment.

ahahahahahahah

Drsoccerball · Feb 16, 2015

Re: HSC 2015 4U Marathon

Made me day :"))

Ekman · Feb 16, 2015

Re: HSC 2015 4U Marathon

braintic said:
Imaginary roots can provide only so much enjoyment.

The innuendoes are real!

Anyways back to maths:

Ekman said:
$$The sides of a rhombus are tangents to the ellipse: $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ and the rhombus diagonals are along the coordinate axes. The angles whose vertices are on the x-axis are of magnitude $ 2\beta $. Show that the area of the rhombus is: $ 2(a^2\tan(\beta) + b^2\cot(\beta))$

FrankXie · Feb 18, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$The sides of a rhombus are tangents to the ellipse: $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ and the rhombus diagonals are along the coordinate axes. The angles whose vertices are on the x-axis are of magnitude $ 2\beta $. Show that the area of the rhombus is: $ 2(a^2\tan(\beta) + b^2\cot(\beta))$

Probably people are looking forward to your method, I can't see other reasons why this one has been outstanding so long

Sy123 · Feb 18, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$The sides of a rhombus are tangents to the ellipse: $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ and the rhombus diagonals are along the coordinate axes. The angles whose vertices are on the x-axis are of magnitude $ 2\beta $. Show that the area of the rhombus is: $ 2(a^2\tan(\beta) + b^2\cot(\beta))$

$\\ $Finding the area of the rhombus is the same as finding the four times the area of the triangle of the first quadrant$ \\\\ $The triangle is represented by the equation$ \ y = k - x\tan \beta \\ \\ $Where$ \ k \ $is a constant such that the line is tangent to the ellipse, and the gradient of the line is such that the angle made with the$ \ x \ $axis clockwise is$ \ \pi - \beta \\ \\ $So, the equations$ \ y = k - x\tan \beta \ $and$ \ b^2x^2 + a^2y^2 = a^2b^2 \ $has one solution$ \\ \\ $So the equation$ \ b^2x^2 + a^2(k^2 - 2k\tan \beta x + x^2\tan^2 \beta) = a^2b^2 \ $has one solution, meaning discriminant is zero$ \\ \\ $So,$ \ \triangle = 0 \\ \\ \Rightarrow \ 4k^2a^4\tan^2\beta = 4(b^2+a^2\tan^2\beta)(a^2k^2 - a^2b^2) \\ \\ \Rightarrow \ k^2 = \dots \ \ (*)$

$\\ $The area of the triangle is given by$ \ \frac{1}{2} \times (x \ $intercept of line$) \times (y \ $intercept of line$) \\ \\ \therefore \ $Area of ellipse is$ \ A = 2\frac{k^2}{\tan \beta} \\ \\ $Using$ \ (*) \ $then$ \ A = 2(a^2 \tan \beta + b^2\cot \beta) \\ \\ \square$

Sy123 · Feb 18, 2015

Re: HSC 2015 4U Marathon

$\\ $i) Expand$ \ (4+i)(5+3i) \\ $ii) Hence show that$ \ \tan^{-1} \left( \frac{1}{4} \right) + \tan^{-1} \left(\frac{3}{5} \right) = \frac{\pi}{4}$

Drsoccerball · Feb 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $i) Expand$ \ (4+i)(5+3i) \\ $ii) Hence show that$ \ \tan^{-1} \left( \frac{1}{4} \right) + \tan^{-1} \left(\frac{3}{5} \right) = \frac{\pi}{4}$

$i) =17+17i=(\sqrt{17}cis(tan^{-1}\frac{1}{4})...$

$ii) $ use $ arg(ab)=arg(a) +arg(b) $ rule $$

FrankXie · Feb 18, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$i) =17+17i=(\sqrt{17}cis(tan^{-1}\frac{1}{4})...$

$ii) $ use $ arg(ab)=arg(a) +arg(b) $ rule $$

\sqrt{17} ?

Drsoccerball · Feb 18, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
\sqrt{17} ?

modulus is root 17 and root 34

VBN2470 · Feb 18, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
modulus is root 17

Modulus is $17\sqrt2$ , doesn't matter though since the question at hand is only asking about the properties of arguments.

Drsoccerball · Feb 18, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
Modulus is $17\sqrt2$ , doesn't matter though since the question at hand is only asking about the properties of arguments.

which is root 17 xroot34...

VBN2470 · Feb 18, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION

$Solve $3x^2-2x-2\le|3x|$.$

Ekman · Feb 18, 2015

Re: HSC 2015 4U Marathon

For those who wanted to know what I did for:

Ekman said:
$$The sides of a rhombus are tangents to the ellipse: $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ and the rhombus diagonals are along the coordinate axes. The angles whose vertices are on the x-axis are of magnitude $ 2\beta $. Show that the area of the rhombus is: $ 2(a^2\tan(\beta) + b^2\cot(\beta))$

Its quite simply:
$$Let P be $(x_1 , y_1)$

$$ X-intercept of the tangent at P $= \frac{a^2}{x_1} $ and the y-intercept at the tangent at P $ = \frac{b^2}{y_1}$

$\therefore Area = \frac{2 a^2 b^2}{x_1 y_1}$

$$ Equation of ellipse (with point P subbed in): $ b^2(x_1)^2+a^2(y_1)^2 =a^2b^2$

$\therefore \frac{2(b^2(x_1)^2+a^2(y_1)^2)}{x_1 y_1}$

$\tan\beta = \frac{b^2 x_1}{a^2 y_1} $ and $ \cot\beta = \frac{a^2 y_1}{b^2 x_1} $ (This is the gradient of the tangent)$$

$\therefore 2(a^2\tan(\beta) + b^2\cot(\beta)) $ (through substitution)$$

Ekman · Feb 18, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEXT QUESTION

$Solve $3x^2-2x-2\le|3x|$.$

$-1\leq x \leq 2$

FrankXie · Feb 18, 2015

Re: HSC 2015 4U Marathon

new question

$$ Find the number of real roots of the equation $ x-2015\sin x=0.$

Sy123 · Feb 19, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
new question

$$ Find the number of real roots of the equation $ x-2015\sin x=0.$

$\\ $The number of real roots is equal to the number of non-negative roots multiplied by 2 and take away 1 (to avoid double counting zero)$$

$\\ 641 \pi < 2015 < 642\pi$

$\\ $Therefore, there are no roots to the equation$ \ \frac{x}{2015} = \sin x \ $for$ \ x \geq \frac{642}{\pi}$

$\\ $For$ \ 2(n-1)\pi \leq x \leq (2n-1)\pi \ $where$ \ 1 \leq n \leq 321 \ $for integers$ \ n \\ \\ $Each interval has contains 2 real roots, this is because for the function$ \\ \\ f(x) = \sin x \Rightarrow \ f(2(n-1)\pi) = 0 \ , \ f((2n-1)\pi) = 0 \ , \ f\left(\left((2n - \frac{3}{2}\right) \pi \right) = 1 \ $and$ \ f(x) \ $is continuous$ \\ \\ $Moreover,$ \ 0 < \frac{x}{2015} < 1 \ $for each interval, therefore there are 2 roots in each interval$ \\ \\ $Meaning, total non-negative roots are$ \ 642 \\ \\ $Therefore, total roots are$ \ 1283$

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