HSC 2013 MX2 Marathon (archive) (10 Viewers)

Trebla · Feb 16, 2013

Re: HSC 2013 4U Marathon

Prove that

$\displaystyle\sum_{k=1}^n x_k^2 \geq \dfrac{1}{n}\left(\displaystyle\sum_{k=1}^n x_k\right)^2$

Sy123 · Feb 17, 2013

Re: HSC 2013 4U Marathon

Kipling said:
$P(x) is a polynomial of degree at least 2 such that P'(a) = 0. Show that when P(x) is divided by$ (x-a)^2 $the remainder is P(a)$

$P(x)=(x-a)^2 Q(x) + R(x)$

$P'(x)=2(x-a)Q(x) + (x-a)^2Q'(x)+R'(x)$

In order for P'(a) = 0

R'(a) = 0

However R(x) does not contain any factors of (x-a)^2 since its the remainder, therefore the only consequence is that if R(x) = Ax+b or R(x) = C
In order for R'(a) = 0
Therefore R(x) = C where C is a constant

$P(x)=(x-a)^2 Q(x) + C$

x=a

$P(a)=0 + C \ \ C=P(a)$

Therefore the remainder is P(a)

hayabusaboston said:
and how come you havent replied to my messaage?
lol
I am curious

I don't feel like revealing it to you

Trebla said:
Prove that

$\displaystyle\sum_{k=1}^n x_k^2 \geq \dfrac{1}{n}\left(\displaystyle\sum_{k=1}^n x_k\right)^2$

Excellent question, I may have done it a long way but I will wait for a bit before posting my solution

Trebla · Feb 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Excellent question, I may have done it a long way but I will wait for a bit before posting my solution

Try finding a second method.

Edit: Actually nvm both the methods I know of are closely related (though if you can find a different one then that would be awesome)

Lieutenant_21 · Feb 17, 2013

Re: HSC 2013 4U Marathon

Trebla said:
Prove that

$\displaystyle\sum_{k=1}^n x_k^2 \geq \dfrac{1}{n}\left(\displaystyle\sum_{k=1}^n x_k\right)^2$

$With some simple algebraic manipulations, we get:$
$\sum_{k=1}^n x_k \leq \sqrt{n}\sqrt{\sum_{k=1}^n x_k^2}$
$I will leave it to someone else to complete (Hint: this is very similar to the Cauchy–Schwarz inequality).$

Lieutenant_21 · Feb 17, 2013

Re: HSC 2013 4U Marathon

asianese said:
Good observation on the cos3x expansion Sy!

I would definitely like to see it 'solved in 1-2 lines'...but the 'long way' would be expanding it all and factorising l0l

$Sub x=cosa, \therefore cos^{2}a+cos^{2}3a=1\Rightarrow cos2a+cos6a=0\therefore 2cos2acos4a=0$
$\therefore 2a=\frac{\pi }{2}, \frac{3\pi }{2} or 4a=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2}$ $we get$ $x=\pm \frac{\sqrt{2}}{2}and\pm \sqrt{\frac{2\pm \sqrt{2}}{2}}$

Lieutenant_21 · Feb 17, 2013

Re: HSC 2013 4U Marathon

$Let$ $a_{1},a_{2},\cdot\cdot\cdot,a_{n}$ $be non-negative numbers such that$ $a_{1}.a_{2}\cdot\cdot\cdot a_{k}\geq \frac{1}{\left ( 2k \right )!}$ $for all k.$
$$Prove that:$ $a_{1}+a_{2}+\cdot\cdot\cdot+a_{n}\geq \frac{1}{n+1}+\frac{1}{n+2}+\cdot\cdot\cdot+ \frac{1}{2n}$

Sy123 · Feb 17, 2013

Re: HSC 2013 4U Marathon

Trebla said:
Prove that

$\displaystyle\sum_{k=1}^n x_k^2 \geq \dfrac{1}{n}\left(\displaystyle\sum_{k=1}^n x_k\right)^2$

Alternatively,

$a^2+b^2 \geq 2ab$

$$We need to sub into a and b, all the combinations of 2 numbers from the set$ \ \ x_1, \ x_2, \ x_2 \ \dots, \ x_n$

$x_1^2+x_1^2 \geq 2x_1 x_2$

$x_1^2+x_3^2 \geq 2x_1x_3$

.
.
.

$x_1^2+x_n^2 \geq 2x_1x_n$

$x_2^2+x_3^2 \geq 2x_2 x_3$

And so on, then we add these inequalities side by side.

$(n-1)(x_1^2+x_2^2 \dots + x_n^2) \geq 2(x_1x_2+x_1x_3+ \dots)$

$n(x_1^2+x_2^2+\dots+x_n^2) \geq x_1^2+x_2^2+\dots + x_n^2 + 2(x_1x_2+\dots$

The RHS is the expanded form of:

$n\sum_{k=1}^{n} x_k^2 \geq \left(\sum_{k=1}^{n}x_k \right)^2$

$\sum_{k=1}^{n} x_k^2 \geq \frac{1}{n}\left(\sum_{k=1}^{n}x_k \right)^2$

hayabusaboston · Feb 17, 2013

Re: HSC 2013 4U Marathon

Wow william sidis.... dat sig.....496/500 aggreggate at least if you manage that. Good luck

Trebla · Feb 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Alternatively,

$a^2+b^2 \geq 2ab$

$$We need to sub into a and b, all the combinations of 2 numbers from the set$ \ \ x_1, \ x_2, \ x_2 \ \dots, \ x_n$

$x_1^2+x_1^2 \geq 2x_1 x_2$

$x_1^2+x_3^2 \geq 2x_1x_3$

.
.
.

$x_1^2+x_n^2 \geq 2x_1x_n$

$x_2^2+x_3^2 \geq 2x_2 x_3$

And so on, then we add these inequalities side by side.

$(n-1)(x_1^2+x_2^2 \dots + x_n^2) \geq 2(x_1x_2+x_1x_3+ \dots)$

$n(x_1^2+x_2^2+\dots+x_n^2) \geq x_1^2+x_2^2+\dots + x_n^2 + 2(x_1x_2+\dots$

The RHS is the expanded form of:

$n\sum_{k=1}^{n} x_k^2 \geq \left(\sum_{k=1}^{n}x_k \right)^2$

$\sum_{k=1}^{n} x_k^2 \geq \frac{1}{n}\left(\sum_{k=1}^{n}x_k \right)^2$

That is actually quite neat. The methods I had in mind relate to the derivation of the CS inequality.
i.e. Note that the discriminant of the following quadratic in z is non-negative

$\displaystyle\sum_{k=1}^n (x_k z + 1)^2$

or alternatively manipulate

$\displaystyle\sum_{k=1}^n \left(x_k - \dfrac{1}{n}\displaystyle\sum_{j=1}^n x_j\right)^2 \geq 0$

Side note: Anyone who has done stats would recognise the above inequality as a consequence of the sample variance (or standard deviation) being always non-negative.

Lieutenant_21 · Feb 17, 2013

Re: HSC 2013 4U Marathon

William Sidis said:
$Let$ $a_{1},a_{2},\cdot\cdot\cdot,a_{n}$ $be non-negative numbers such that$ $a_{1}.a_{2}\cdot\cdot\cdot a_{k}\geq \frac{1}{\left ( 2k \right )!}$ $for all k.$
$$Prove that:$ $a_{1}+a_{2}+\cdot\cdot\cdot+a_{n}\geq \frac{1}{n+1}+\frac{1}{n+2}+\cdot\cdot\cdot+ \frac{1}{2n}$

If no one does this question I will post the solution soon. It is very tricky.

Lieutenant_21 · Feb 18, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
Wow william sidis.... dat sig.....496/500 aggreggate at least if you manage that. Good luck

If I don't manage to achieve it I will still be perfectly happy knowing that the HSC is over because all I want is to go to uni

asianese · Feb 18, 2013

Re: HSC 2013 4U Marathon

Is it by induction?

Lieutenant_21 · Feb 18, 2013

Re: HSC 2013 4U Marathon

$No.$
$You start with this:$ $a_{1}+a_{2}+\cdot\cdot\cdot+a_{n}=$
$\left ( 1-\frac{1}{2} \right )(1.2a_{1})+\left ( \frac{1}{3}-\frac{1}{4} \right )(3.4a_{2})+\cdot\cdot\cdot+\left ( \frac{1}{2n-1}-\frac{1}{2n} \right )((2n-1).2na_{n})$
$Think about making use of the AM-GM inequality$

asianese · Feb 18, 2013

Re: HSC 2013 4U Marathon

I don't see why induction wouldn't work but alright.

Sy123 · Feb 18, 2013

Re: HSC 2013 4U Marathon

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

$$Has 2 points$ \ \ P(a\cos \theta, b\sin \theta) \ \ $and$ \ \ Q(a\cos \alpha, b\sin \alpha)$

$The tangents at the points P and Q intersect at R at a right angle$

$$i) Prove that the locus of R is a circle with constant radius$ \ \ \ \fbox{1}$

$$ii) Hence find the equation of the locus of R as P and Q vary$ \ \ \ \fbox{2}$

Nws m8 · Feb 18, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$The ellipse$ \ \ \ \frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$

$$Has 2 points$ \ \ P(a\cos \theta, b\sin \theta) \ \ $and$ \ \ Q(a\cos \alpha, b\sin \alpha)$

$The tangents at the points P and Q intersect at R at a right angle$

$$i) Prove that the locus of R is a circle with constant radius$ \ \ \ \fbox{1}$

$$ii) Hence find the equation of the locus of R as P and Q vary$ \ \ \ \fbox{2}$

This is from a past hsc/trial paper iirc

Sy123 · Feb 18, 2013

Re: HSC 2013 4U Marathon

Nws m8 said:
This is from a past hsc/trial paper iirc

Really? I derived it myself.

Nws m8 · Feb 18, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Really? I derived it myself.

Did ya? Lol
I'm pretty sure it's from past paper or maybe from Patel , yeah I think it's from Patel, or maybe a similar question ....

twinklegal19 · Feb 18, 2013

Re: HSC 2013 4U Marathon

Just thought of this interesting curve sketching question =)

$y=\log_{x}e$

Just draw on paint/scan working/type up the general outline of the steps/whatever

Sy123 · Feb 18, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
Just thought of this interesting curve sketching question =)

$y=\log_{x}e$

Just draw on paint/scan working/type up the general outline of the steps/whatever

By change of base law:

$\log_x e = \frac{\log_e e}{\log_e x} = \frac{1}{\ln x}$

And its straightforward graph manipulation from there.

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