HSC 2013 MX2 Marathon (archive) (1 Viewer)

SpiralFlex · Nov 15, 2012

Re: HSC 2013 4U Marathon

Nice question Sy.

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
~~still editing~~

$\\$i) by inspection, $(x+1)$ is a factor$\\S(x)=(x+1)(ax^2+cx+a)\\$equating coefficients of $x\\b=a+c\\c=b-a\\S(x)=(x+1)[ax^2+(b-a)x+a]$
$\\ax^2+(b-a)x+a=0\\\bigtriangleup =(b-a)^2-4a^2\\$For real roots, \bigtriangleup \geq 0$\\(b-a)^2-4a^2\geq 0\\(b-a-2a)(b-a+2a)\geq 0\\(b-3a)(b+a)\geq 0$

Edit: sorry, got to study for English now, someone feel free to take over lol.

Very good, a different approach to what I had in mind for the second one however. The third one is straightforward when b=3a then there is a horizontal point of inflection at x=1.

Can someone else post a question now? (preferably Complex and Polynomials, (or also anything that can be done with 2U))

SpiralFlex · Nov 15, 2012

Re: HSC 2013 4U Marathon

^2U questions should be placed in the 2U thread. I'll post a question in a few minutes.

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
2U questions should be placed in the 2U thread. I'll post a question in a few minutes.

Well the really hard ones should probably not be. For example my triangular numbers question should be in this thread imo (or 3U one)

SpiralFlex · Nov 15, 2012

Re: HSC 2013 4U Marathon

Would you like circle geometry problems?

4U circle isn't any real extra theorems, it's just harder problems to 3U.

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
Would you like circle geometry problems?

4U circle isn't any real extra theorems, it's just harder problems to 3U.

Err yeah sure I guess, it isnt my strength but practise makes perfect.

twinklegal19 · Nov 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Very good, a different approach to what I had in mind for the second one however. The third one is straightforward when b=3a then there is a horizontal point of inflection at x=1.

Can someone else post a question now? (preferably Complex and Polynomials, (or also anything that can be done with 2U))

urgh so tired of belonging

How come $b+a\ngeqslant0$ though?

Anyway here's a difficult complex/circle geo question that keeps propping up everywhere lol

$\\$The points representing the complex numbers $z_{1}, z_{2}$ and $z_{3}$ lie on a circle which passes through the origin. Prove that the points representing $\frac{1}{z_{1}},\frac{1}{z_{2}}$ and $\frac{1}{z_{3}}$ are collinear.$$

RealiseNothing · Nov 15, 2012

Re: HSC 2013 4U Marathon

i) Show that $\frac{a+b}{2} \geq \sqrt{ab}$

ii) Consider the quadratic $ax^2 + bx + c$ where the co-efficients are all positive.

If the roots are both complex, show that $b < a + c$

iii) If one of the roots is $x + iy$ , deduce that:

$(x-1)^2 + y^2 > 2$

RealiseNothing · Nov 15, 2012

Re: HSC 2013 4U Marathon

Actually I just remembered an awesome question (aye aye spiral

)

If $a ~ b ~ c ~ d$ are all distinct positive integers, find what they are so that:

$a^2 - b^2 = c^2 - d^2 = 957$

SpiralFlex · Nov 15, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Actually I just remembered an awesome question (aye aye spiral )

If $a ~ b ~ c ~ d$ are all distinct positive integers, find what they are so that:

$a^2 - b^2 = c^2 - d^2 = 957$

M8 this question cost you a generation of babies + you know.

RealiseNothing · Nov 15, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
M8 this question cost you a generation of babies + you know.

No, YOU cost me.

You were the one who got it wrong.

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
urgh so tired of belonging How come $b+a\ngeqslant0$ though?

Anyway here's a difficult complex/circle geo question that keeps propping up everywhere lol

$\\$The points representing the complex numbers $z_{1}, z_{2}$ and $z_{3}$ lie on a circle which passes through the origin. Prove that the points representing $\frac{1}{z_{1}},\frac{1}{z_{2}}$ and $\frac{1}{z_{3}}$ are collinear.$$

I just found a mistake in my question, when I did it I divided the an inequality by b, which is not allowed since that is assuming b>0 there is indeed two domains for b such that there are 3 real roots.

Will do the question now

SpiralFlex · Nov 15, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
No, YOU cost me.

You were the one who got it wrong.

But! The mistake was critical if I had gotten is right you wouldn't be how you say "IN" at the moment.

RealiseNothing · Nov 15, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
But! The mistake was critical if I had gotten is right you wouldn't be how you say "IN" at the moment.

wait wat.

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
urgh so tired of belonging How come $b+a\ngeqslant0$ though?

Anyway here's a difficult complex/circle geo question that keeps propping up everywhere lol

$\\$The points representing the complex numbers $z_{1}, z_{2}$ and $z_{3}$ lie on a circle which passes through the origin. Prove that the points representing $\frac{1}{z_{1}},\frac{1}{z_{2}}$ and $\frac{1}{z_{3}}$ are collinear.$$

Solution

bobmcbob365 · Nov 15, 2012

Re: HSC 2013 4U Marathon

Lol, is the answer a = 479, b=478; c=31, d=2; a=45, b = 42; c=161, d=158

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
i) Show that $\frac{a+b}{2} \geq \sqrt{ab}$

ii) Consider the quadratic $ax^2 + bx + c$ where the co-efficients are all positive.

If the roots are both complex, show that $b < a + c$

iii) If one of the roots is $x + iy$ , deduce that:

$(x-1)^2 + y^2 > 2$

Assuming a, and b real:

Construct:

$(\sqrt{a}-\sqrt{b})^2 \geq 0$

$a-2\sqrt{ab}+b \geq 0 \\ \therefore \frac{a+b}{2} \geq \sqrt{ab}$

$ii) \ \ \ \ $If both roots are complex discriminant less than zero$ \\ \\ b^2-4ac < 0 \\ \\ b^2 < 4ac \\ b<2\sqrt{ac} \ \ \ \ \ \ $(Disregard other one since b>0)$$

$$By part i using the inequality we arrive at: $ 2\sqrt{ac} \leq a+c \\ \therefore b < 2\sqrt{ac}\leq a+c \\ \\ b<a+c$

$iii) x+iy, x-iy $ roots, by conjugate root theorem (assuming a b and c are real from start of question)$ \\ \\ b<a+c \\ \frac{-b}{a}=x+iy+x-iy \\ \\ -2x=\frac{b}{a} \\ \\ (x-iy)(x+iy)=\frac{c}{a} \\ x^2+y^2 = \frac{c}{a} \\ \\ \therefore -2ax < a + a(x^2+y^2) \ \ \ \ a>0 \\ \\ -2x<1+x^2+y^2 \\ (x+1)^2+y^2>0 ???$

Are you sure the last result is correct?

cutemouse · Nov 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Solution

There's another way to do this problem using mappings in complex analysis. It's WAY more elegant.

EDIT: Here is my sketch solution:

Let 'c' be the complex number representing centre of the circle. Then the equation of the circle is |z-c| = |c| (since the circle passes through O).

Define the mapping w=1/z

Then |1/w - c| = |c|

This leads to |w-1/c| = |w|

This represents a line (in the w plane that is the right bisector of the join from O to 1/c).

As z_1, z_2, z_3 lie on |z-c|=|c| therefore 1/z_1, 1/z_2 and 1/z_3 lie on |w-1/c| = |w| and therefore are collinear.

twinklegal19 · Nov 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Solution

nice work! Another way is to use exterior angle of a cyclic quad or angles in the same segment to prove arg(1/z3-1/z1)=arg(1/z2-1/z1)

RealiseNothing · Nov 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Are you sure the last result is correct?

Woops, just realised in my working out I forgot a minus sign, which lead to my result.

Yours is correct though.

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