HSC 2013 MX2 Marathon (archive) (4 Viewers)

Sy123 · Nov 18, 2012

Re: HSC 2013 4U Marathon

$$When a polynomial $ P(x) $ is divided by $ x-3 $ the remainder is 5 and when it is divided by $ x-4 $ the remainder is 9. Find the remainder when $ P(x) $ is divided by $ (x-3)(x-4)$

seanieg89 · Nov 18, 2012

Re: HSC 2013 4U Marathon

As no-one is answering spirals question I will give a solution for b). The rest is easy.

$(z+1)^n+(z-1)^n=0\Rightarrow (z+1)^n=-(z-1)^n\Rightarrow |z+1|^n=|z-1|^n\Rightarrow |z+1|=|z-1|.$

So any roots of this polynomial must lie on the perpendicular bisector of -1 and 1, that is the imaginary axis.

seanieg89 · Nov 18, 2012

Re: HSC 2013 4U Marathon

Another question on polys/complex numbers:

$Let $a_k>0$ for $k=0,1,\ldots,n$.\\ Prove that the polynomial\\ \\ $P(z):=\sum_{k=0}^n a_kz^k$ \\ \\has no complex roots $\alpha$ with $|\arg(\alpha)|\leq\pi/n.$$

Sy123 · Nov 18, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Another question on polys/complex numbers:

$Let $a_k>0$ for $k=0,1,\ldots,n$.\\ Prove that the polynomial\\ \\ $P(z):=\sum_{k=0}^n a_kz^k$ \\ \\has no complex roots $\alpha$ with $|\arg(\alpha)|\leq\pi/n.$$

Is a_k real?

seanieg89 · Nov 18, 2012

Re: HSC 2013 4U Marathon

Yep, implied by the use of inequality signs.

twinklegal19 · Nov 18, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$When a polynomial $ P(x) $ is divided by $ x-3 $ the remainder is 5 and when it is divided by $ x-4 $ the remainder is 9. Find the remainder when $ P(x) $ is divided by $ (x-3)(x-4)$

$\\P(3)=5\\P(4)=9\\\therefore P(x)=(x-3)(x-4)Q(x)+Ax+B\\\\P(3)=3A+B=5\Rightarrow (1)\\P(4)=4A+B=9\Rightarrow(2)\\(2)-(1)\Rightarrow A=4\\$from $(1)\\3(4)+B=5\\B=-7\\\\\therefore $ the remainder is $4x-7$

RealiseNothing · Nov 18, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
$\\P(3)=5\\P(4)=9\\\therefore P(x)=(x-3)(x-4)Q(x)+Ax+B\\\\P(3)=3A+B=5\Rightarrow (1)\\P(4)=4A+B=9\Rightarrow(2)\\(2)-(1)\Rightarrow A=4\\$from $(1)\\3(4)+B=5\\B=-7\\\\\therefore $ the remainder is $4x-7$

Did you angle chase this too?

twinklegal19 · Nov 18, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Did you angle chase this too?

yep totes

twinklegal19 · Nov 18, 2012

Re: HSC 2013 4U Marathon

$$Prove that the diagonals of a rhombus intersect at right angles$\\$(Hint: $\left | z_1 \right |=\left | z_2 \right |)$

RealiseNothing · Nov 18, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
$$Prove that the diagonals of a rhombus intersect at right angles$\\$(Hint: $\left | z_1 \right |=\left | z_2 \right |)$

Let:

$z_1 = a + ib$

$z_2 = c + id$

Then the diagonals of the quadrilateral formed are:

$z_1 + z_2 = (a+c) + (b+d)i$

$z_1 - z_2 = (a-c) + (b-d)i$

Now multiply the gradients of the diagonals:

$\frac{d-b}{c-a} \times \frac{b+d}{a+c} = \frac{d^2 - b^2}{c^2 - a^2}$

Suppose these diagonals intersect at right angles, hence:

$\frac{d^2 - b^2}{c^2 - a^2} = -1$

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

$a^2 + b^2 = c^2 + d^2$

But:

$a^2 + b^2 = |z_1|^2$

$c^2 + d^2 = |z_2|^2$

Hence:

$|z_1| = |z_2|$

Thus the quadrilateral formed is a rhombus if the diagonals intersect at right angles. Therefore the converse is true.

Sy123 · Nov 18, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
$$Prove that the diagonals of a rhombus intersect at right angles$\\$(Hint: $\left | z_1 \right |=\left | z_2 \right |)$

Trying my best using complex numbers (otherwise its quite simple to prove that property, just prove that the 4 triangles that are created with diagonals of rhombus are all congruent)

$z_1=x+iy$
$z_2=u+iv$

Now when we form the rhombus fixing one end at the origin, the diagonal coming from the origin will be z_1+z_2, and the other diagonal will be z_1-z_2.

Now lets see what we get when we take the argument of z_1+z_2 and z_1-z_2.

$arg(z_1+z_2)-arg(z_1-z_2) = k \\ \\ \tan^{-1} \frac{y+v}{x+u}-\tan^{-1} \frac{y-v}{x-u} = k \\ \\ \tan k = \frac{2vx-2uy}{x^2+y^2-u^2-v^2}$

That is after simplification, but notice

$| z_1|=|z_2|=x^2+y^2=u^2+v^2$

$\tan k = \frac{2vx-2uy}{|z_1|-|z_2|}$

But this is unattainable, hence we can then deduce that k = pi/2, 3pi/2 etc.
Either way we have proven perpendicular property.

Is it mathematically legal for me to use this argument?

(Not a very elegant proof, I know, but is it possible for me to do this?)

seanieg89 · Nov 18, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Another question on polys/complex numbers:

$Let $a_k>0$ for $k=0,1,\ldots,n$.\\ Prove that the polynomial\\ \\ $P(z):=\sum_{k=0}^n a_kz^k$ \\ \\has no complex roots $\alpha$ with $|\arg(\alpha)|\leq\pi/n.$$

At Sy's request a hint:

Suppose to the contrary that a root alpha exists in this sector. Evaluating the polynomial P at this alpha we get a sum of (n+1) complex numbers which is supposedly equal to zero. Think about this sum geometrically and a problem should become pretty obvious. Its one of those things that may take a little while to spot, but when you see it you will slap yourself for not seeing it sooner

.

If the above is not enough of a hint then try to do this problem for n=2 before attempting the general case.

RealiseNothing · Nov 18, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Another question on polys/complex numbers:

$Let $a_k>0$ for $k=0,1,\ldots,n$.\\ Prove that the polynomial\\ \\ $P(z):=\sum_{k=0}^n a_kz^k$ \\ \\has no complex roots $\alpha$ with $|\arg(\alpha)|\leq\pi/n.$$

Suppose $\alpha$ is a root of the polynomial.

Let $\alpha = Rcis\theta$ such that $\theta \leq \frac{\pi}{n}$

Then by substituting this into our polynomial and letting it equal to 0 (since we are assuming it's a root) and by using De Moivre's Theorem, we obtain:

$a_nR^ncis(n)\theta + a_{n-1}R^{n-1}cis(n-1)\theta + ... + a_1Rcis\theta + a_0 = 0$

If we plotted these points on an Argand Diagram, $a_0$ would be on the positive real axis given that $a_k > 0$ .

The leading term would lay in the first two quadrants, unless $\theta = \frac{\pi}{n}$ , in which case it would lay on the negative real axis.

All terms in between would lay in the first two quadrants such that their imaginary components are positive.

By vector addition, since all points have $Im(z) > 0$ , we can never arrive at 0, and hence $P(z) \neq 0$

Therefore we have a contradiction - our assumption is wrong. Hence we can deduce that $P(z)$ has no roots $\alpha$ where $|arg(\alpha)| \leq \frac{\pi}{n}$

Note by a symmetrical argument we can justify it holds true when all the points obey the condition $Im(z) < 0$ .

seanieg89 · Nov 18, 2012

Re: HSC 2013 4U Marathon

Bingo. Nice result right? I will try to think of harder questions along these lines later.

RealiseNothing · Nov 18, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Bingo. Nice result right? I will try to think of harder questions along these lines later.

The Cauchy Schwarz Inequality can be proven like that as well yes? I remember carrot giving me that as a question at USYD when we went.

seanieg89 · Nov 18, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
The Cauchy Schwarz Inequality can be proven like that as well yes? I remember carrot giving me that as a question at USYD when we went.

I can't think of how off the top of my head, although there are pretty simple ways to visualise C-S geometrically...in many ways it is more fundamental than the triangle inequality. Was Carrot's question proving C-S for z,w in C^n? Or just in R^n?

RealiseNothing · Nov 18, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
I can't think of how off the top of my head, although there are pretty simple ways to visualise C-S geometrically...in many ways it is more fundamental than the triangle inequality. Was Carrot's question proving C-S for z,w in C^n? Or just in R^n?

It was showing that $\sum_{k=0}^{n} (x_kz + y_k)^2 \geq 0$

Which you can show by adding all the terms graphically.

Then use the discriminant and the inequality falls straight out.

(something like this, might not be the exact question).

seanieg89 · Nov 18, 2012

Re: HSC 2013 4U Marathon

Well pretty different theme, everything is real here, and the inequality comes directly from squares being positive. But yeah, that is one of the common proofs for the real version of C-S.

Sy123 · Nov 19, 2012

Re: HSC 2013 4U Marathon

Basic but because I cant think of any original questions at the moment and to keep the marathon going:

$If the complex numbers $ z_1, z_2, z_3, z_4 $ lie on the Argand plane. Show that if these points form a cyclic quadrilateral then, the complex number represented by $ \frac{(z_4-z_1)(z_2-z_3)}{(z_2-z_1)(z_4-z_3)} $ is a negative real number$

twinklegal19 · Nov 19, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Basic but because I cant think of any original questions at the moment and to keep the marathon going:

$If the complex numbers $ z_1, z_2, z_3, z_4 $ lie on the Argand plane. Show that if these points form a cyclic quadrilateral then, the complex number represented by $ \frac{(z_4-z_1)(z_2-z_3)}{(z_2-z_1)(z_4-z_3)} $ is a negative real number$

$let the points A,B,C and D represent the complex numbers $z_1, z_2, z_3, z_4\\$Since the points are concyclic$, \angle BAD+ \angle ADC=\pi$ (opposite angles in cyclic quadrilaterals are supplementary)$\\\therefore \arg\left ( \frac{{\overrightarrow{AB}}}{{\overrightarrow{AD}}} \right )+\arg\left ( \frac{{\overrightarrow{CB}}}{{\overrightarrow{CD}}} \right )=\pi \\\arg\left ( \frac{z_4-z_1}{z_2-z_1} \right )+\arg\left ( \frac{z_2-z_3}{z_4-z_3} \right )=\pi\\\arg\left ( \frac{(z_4-z_1)(z_2-z_3)}{(z_2-z_1)(z_4-z_3)} \right )=\pi\\ \frac{(z_4-z_1)(z_2-z_3)}{(z_2-z_1)(z_4-z_3)} \right )=-k, (k> 0)\\\therefore$ it is a negative real number$

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HSC 2013 MX2 Marathon (archive) (4 Viewers)

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