HSC 2013 MX2 Marathon (archive) (7 Viewers)

shongaponga · May 11, 2013

Re: HSC 2013 4U Marathon

bleakarcher said:
Did it. ^ That happened.

You copied the wrong code, swing me a PM so we aren't spamming this thread and i'll teach you how to do it.

Registered User · May 11, 2013

Re: HSC 2013 4U Marathon

$Prove the inequality:$

${a^{2}}+b^2+c^2+\frac{2}{5}abc<50$

$where:$

$a,b,c are the lengths of triangle's sides$

$and the perimeter of the triangle is 10.$

Registered User · May 12, 2013

Re: HSC 2013 4U Marathon

$Prove that$

$\prod_{k=1}^{n}\left ( 1+2cos\frac{2\pi .3^{k}}{3^{n}+1} \right )=1$

Registered User · May 12, 2013

Re: HSC 2013 4U Marathon

$Prove that$

$\left ( \frac{1}{2}+cos\frac{\pi }{20} \right )\left ( \frac{1}{2}+cos\frac{3\pi }{20} \right )\left ( \frac{1}{2}+cos\frac{9\pi }{20} \right )\left ( \frac{1}{2}+cos\frac{27\pi }{20} \right )=\frac{1}{16}$

Registered User · May 12, 2013

Re: HSC 2013 4U Marathon

bleakarcher said:
1) [(1/a)-(1/b)]^2 + [(1/a)-(1/c)]^2 + [(1/b)-(1/c)]^2 >=0 (equality occurs iff a=b=c)
2/a^2 + 2/b^2 + 2/c^2 -2[(1/ab)+(1/bc)+(1/ac)] >=0
=> (1/a)^2 + (1/b)^2 + (1/c)^2 >= 1/bc + 1/ac + 1/ab for all real a,b,c=/=0 (*)
Hence, bc/a + ac/b + ab/c = abc[(1/a^2)+(1/b^2) +(1/c^2)]>=abc[(1/ab)+(1/bc)+(1/ac)]=a+b+c as req.
2) Refer back to (*), replace a with sqrt(a), b with sqrt(b) and c with sqrt(c) and the second inequality immediately falls out for a,b,c>0

Sorry if it's hard to read, still haven't learnt latex.

$1)\ (\frac{1}{a}-\frac{1}{b})^{2} + (\frac{1}{a}-\frac{1}{c})^{2} + (\frac{1}{b}-\frac{1}{c})^{2} > 0$ $(equality occurs iff a=b=c)$

$\frac{2}{a^{2}} + \frac{2}{b^{2}} + \frac{2}{c^{2}} -2\left ( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} \right ) \geq 0$

$\Rightarrow \left ( \frac{1}{a} \right )^{2} + \left ( \frac{1}{b} \right )^{2} + \left ( \frac{1}{c} \right )^{2} \geq \frac{1}{bc} + \frac{1}{ac} + \frac{1}{ab}$ $for all real$ $a,b,c\neq 0 (*)$

$Hence,$ $\frac{bc}{a} + \frac{ac}{b} + \frac{ab}{c} = abc\left ( \frac{1}{a^{2}}+ \frac{1}{b^{2}} +\frac{1}{c^{2}} \right )\geq \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geqslant (a+b+c)$ $as required.$

$2) Refer back to (*), replace a with$ $\sqrt{a},$ $b with$ $\sqrt{b},$ $and c with$ $\sqrt{c}$ $and the second inequality immediately falls out for$ $a,b,c > 0$

http://www.codecogs.com/latex/eqneditor.php

Make the latex there then put $[/ tex](*) For writing put [tex]$ writing $[/ tex] (*) Without the space$

Registered User · May 12, 2013

Re: HSC 2013 4U Marathon

$For any positive real numbers$ $a,b,c$ $show that the following inequality holds$ $\left ( \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right )\geq \left ( \frac{c+a}{c+b}+\frac{a+b}{a+c}+\frac{b+c}{b+a} \right )$

Sy123 · May 12, 2013

Re: HSC 2013 4U Marathon

Registered User said:
$Prove that$

$\prod_{k=1}^{n}\left ( 1+2cos\frac{2\pi .3^{k}}{3^{n}+1} \right )=1$

$\sin(3\theta) = \sin \theta \cos(2\theta) + \cos \theta \sin(2\theta)$

$\frac{\sin 3\theta}{\sin \theta} = \cos 2\theta + 2\cos^2 \theta$

$\therefore \ \ \frac{\sin 3\theta}{\sin \theta} = 1+ 2\cos 2\theta \ \ \dots \ \ (1)$

$$Let$ \ \ \theta_k = \frac{\pi 3^k}{3^n+1}$

$$Due to$ \ \ (1)$

$\therefore \ \ P = \prod_{k=1}^{n} \left (1+ 2\cos 2\theta_k \right ) = \frac{\sin 3\theta_1}{\sin \theta_1} \cdot \frac{\sin 3\theta_2}{\sin \theta_2} \dots \frac{\sin 3\theta_n}{\sin \theta_n}$

$\theta_k = 3\theta_{k-1}$

$A 'telescoping product' is formed$

$P = \frac{\sin \theta_{n+1}}{\sin \theta_1}$

$\therefore P = \prod_{k=1}^{n} \left (1+ 2\cos 2\theta_k \right ) = \frac{\sin \left (3\pi - \theta_1 \right)}{\sin (\theta_1)} = 1$

$\square$

Registered User · May 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$\sin(3\theta) = \sin \theta \cos(2\theta) + \cos \theta \sin(2\theta)$

$\frac{\sin 3\theta}{\sin \theta} = \cos 2\theta + 2\cos^2 \theta$

$\therefore \ \ \frac{\sin 3\theta}{\sin \theta} = 1+ 2\cos 2\theta \ \ \dots \ \ (1)$

$$Let$ \ \ \theta_k = \frac{\pi 3^k}{3^n+1}$

$$Due to$ \ \ (1)$

$\therefore \ \ P = \prod_{k=1}^{n} \left (1+ 2\cos 2\theta_k \right ) = \frac{\sin 3\theta_1}{\sin \theta_1} \cdot \frac{\sin 3\theta_2}{\sin \theta_2} \dots \frac{\sin 3\theta_n}{\sin \theta_n}$

$\theta_k = 3\theta_{k-1}$

$A 'telescoping product' is formed$

$P = \frac{\sin \theta_{n+1}}{\sin \theta_1}$

$\therefore P = \prod_{k=1}^{n} \left (1+ 2\cos 2\theta_k \right ) = \frac{\sin \left (3\pi - \theta_1 \right)}{\sin (\theta_1)} = 1$

$\square$

Nice work.

Registered User · May 14, 2013

Re: HSC 2013 4U Marathon

$Prove that$

$\frac{ \sin x}{ \cos x}+\frac{\sin2x}{\cos^{2}x}+\frac{\sin3x}{\cos^{3}x}+\cdots+\frac{\sin nx}{\cos^{n}x}=\cot x-\frac{\cos(n+1)x}{\sin x \cos^{n}x}$

Do not use induction.

Sy123 · May 14, 2013

Re: HSC 2013 4U Marathon

Registered User said:
$Prove that$

$\frac{ \sin x}{ \cos x}+\frac{\sin2x}{\cos^{2}x}+\frac{\sin3x}{\cos^{3}x}+\cdots+\frac{\sin nx}{\cos^{n}x}=\cot x-\frac{\cos(n+1)x}{\sin x \cos^{n}x}$

Do not use induction.

$By DeMoivere's theorem$

$(\cos x + i\sin x)^k = \cos(kx) + i\sin(kx)$

$(1+ i\tan x)^k = \frac{\cos(kx)}{\cos^kx} + i \frac{\sin(kx)}{\cos^k x}$

$$Im$ \left(\sum_{k=1}^{n} (1+i\tan x)^k \right) = \sum_{k=1}^{n} \frac{\sin kx}{\cos^kx} \ \ \dots \ \ (1)$

$Consider the geometric series$

$(1+i\tan x) + (1+i\tan x)^2 + \dots + (1+i\tan x)^n = \frac{(1+i\tan x)^{n+1} - (1+i\tan x)}{1+i\tan x - 1}$

$\Rightarrow \ \ $Im$ \frac{\frac{\cos(n+1)x}{\cos^{n+1}x} + \frac{i\sin(n+1)x}{\cos^{n+1}x} - 1 - i\tan x}{i\tan x} = \cot x-\frac{\cos(n+1)x}{\sin x \cos^{n}x}$

$\therefore \ \frac{ \sin x}{ \cos x}+\frac{\sin2x}{\cos^{2}x}+\frac{\sin3x}{\cos^{3}x}+\cdots+\frac{\sin nx}{\cos^{n}x}=\cot x-\frac{\cos(n+1)x}{\sin x \cos^{n}x}$

bleakarcher · May 14, 2013

Re: HSC 2013 4U Marathon

Hey guys, here's a question from last year's UNSW maths competition. Let f(x) denote a strictly positive continuous function defined on all real numbers with the property that f(2012)=2012 and f(x)=f(x+f(x)) for all x. Prove that f(x)=2012 for all x.

Sy123 · May 14, 2013

Re: HSC 2013 4U Marathon

bleakarcher said:
Hey guys, here's a question from last year's UNSW maths competition. Let f(x) denote a strictly positive continuous function defined on all real numbers with the property that f(2012)=2012 and f(x)=f(x+f(x)) for all x. Prove that f(x)=2012 for all x.

$$For some number$ \ \ a$

$f(a) = f(a+f(a+f(a+f(a+ \dots +f(a)) \dots ) = f(k)$ for some number k, f is continuous and defined on all real therefore a and k can be anything.
If that is the case then a and k can be such that they represent any pair of real numbers. The function of these numbers are all equal, therefore

$f(x) = C$ for some constant C since f(a) = f(k)

$f(2012) = 2012$

$\therefore f(x) = 2012$

Is that good enough?

============

$Determine all solutions to the simultaneous equations$

$x+y+z = 3$

$x^2+y^2+z^2 = 3$

$x^3+y^3+z^3 = 3$

RealiseNothing · May 14, 2013

Re: HSC 2013 4U Marathon

edit: I get it.

Makematics · May 14, 2013

Re: HSC 2013 4U Marathon

bleakarcher said:
Hey guys, here's a question from last year's UNSW maths competition. Let f(x) denote a strictly positive continuous function defined on all real numbers with the property that f(2012)=2012 and f(x)=f(x+f(x)) for all x. Prove that f(x)=2012 for all x.

omg thank you so much for asking that. really wanted to know the solution

Registered User · May 14, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$By DeMoivere's theorem$

$(\cos x + i\sin x)^k = \cos(kx) + i\sin(kx)$

$(1+ i\tan x)^k = \frac{\cos(kx)}{\cos^kx} + i \frac{\sin(kx)}{\cos^k x}$

$$Im$ \left(\sum_{k=1}^{n} (1+i\tan x)^k \right) = \sum_{k=1}^{n} \frac{\sin kx}{\cos^kx} \ \ \dots \ \ (1)$

$Consider the geometric series$

$(1+i\tan x) + (1+i\tan x)^2 + \dots + (1+i\tan x)^n = \frac{(1+i\tan x)^{n+1} - (1+i\tan x)}{1+i\tan x - 1}$

$\Rightarrow \ \ $Im$ \frac{\frac{\cos(n+1)x}{\cos^{n+1}x} + \frac{i\sin(n+1)x}{\cos^{n+1}x} - 1 - i\tan x}{i\tan x} = \cot x-\frac{\cos(n+1)x}{\sin x \cos^{n}x}$

$\therefore \ \frac{ \sin x}{ \cos x}+\frac{\sin2x}{\cos^{2}x}+\frac{\sin3x}{\cos^{3}x}+\cdots+\frac{\sin nx}{\cos^{n}x}=\cot x-\frac{\cos(n+1)x}{\sin x \cos^{n}x}$

Can also be done by expanding $cos(nx+x)$ but your method is nice too.

==============================

$What does$ $\sum_{k=0}^{n}k\times k!$ $converge to?$

RealiseNothing · May 14, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Determine all solutions to the simultaneous equations$

$x+y+z = 3$

$x^2+y^2+z^2 = 3$

$x^3+y^3+z^3 = 3$

Is there any other solutions besides $x=y=z=1$ because my method yields that as the only solution.

Sy123 · May 14, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Is there any other solutions besides $x=y=z=1$ because my method yields that as the only solution.

Yep those are the only solutions, what was your method?

RealiseNothing · May 14, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep those are the only solutions, what was your method?

My aim was to construct a cubic polynomial which has roots x, y, z, using sum of roots/product of roots/two at a time. Then by finding the roots of this polynomial, we find the values of x, y, z, that satisfy the system of equations. The cubic is in the form:

$X^3 - aX^2 + bX - c$

The sum of roots is easy as it's already given:

$x+y+z=~$sum of roots$~ = 3 = a$

So we have:

$X^3 - 3X^2 + bX - c$

Now we get the value of 'b' by finding the sum of the pair roots (xy+xz+yz). We know that:

$x^2+y^2+z^2=(x+y+z)^2 - 2(xy+xz+yz)$

$3=3^2-2(xy+xz+yz)$

$xy+xz+yz = 3 = b$

So we now have:

$X^3 - 3X^2 + 3X - c$

Now we know that x, y, z and the roots of this equation, so by substitution we arrive at:

$x^3 - 3x^2 + 3x - c = 0$

$y^3 - 3y^2 + 3y - c = 0$

$z^3 - 3z^2 + 3z - c = 0$

If we add all these together we get:

$(x^3+y^3+z^3) - 3(x^2+y^2+z^2) + 3(x+y+z) = 3c$

$(3)-3(3)+3(3) = 3c$

$c=1$

So our polynomial is:

$X^3-3X^2+3X-1$

Which simplifies to:

$(X-1)^3 = 0$

Which has all roots equal to 1. So x, y, z are all equal to 1. And the polynomial can not have other roots and so this is the only possible solution set to the given system of equations.

bleakarcher · May 14, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$For some number$ \ \ a$

$f(a) = f(a+f(a+f(a+f(a+ \dots +f(a)) \dots ) = f(k)$ for some number k, f is continuous and defined on all real therefore a and k can be anything.
If that is the case then a and k can be such that they represent any pair of real numbers. The function of these numbers are all equal, therefore

$f(x) = C$ for some constant C since f(a) = f(k)

$f(2012) = 2012$

$\therefore f(x) = 2012$

Is that good enough?

============

$Determine all solutions to the simultaneous equations$

$x+y+z = 3$

$x^2+y^2+z^2 = 3$

$x^3+y^3+z^3 = 3$

Think it works dude, nice. I'm not so sure about my method though:

f(x)=f(x+f(x))
f '(x)=[1+f '(x)]*f '(x+f(x))=[1+f '(x)]*f '(x) => f '(x)=0 for all x. Since f(x) is continuous and f(2012)=2012, f(x)=2012 for all x.

RealiseNothing · May 14, 2013

Re: HSC 2013 4U Marathon

bleakarcher said:
Think it works dude, nice. I'm not so sure about my method though:

f(x)=f(x+f(x))
f '(x)=[1+f '(x)]*f '(x+f(x))=[1+f '(x)]*f '(x) => f '(x)=0 for all x. Since f(x) is continuous, f(x)=2012 for all x.

How did you get that $f'(x+f(x)) = f'(x)$ ?

HSC 2013 MX2 Marathon (archive) (7 Viewers)

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