# HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

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#### Sy123

##### This too shall pass
Re: HSC 2014 4U Marathon - Advanced Level

$\bg_white Let a,b,c\geq 0.\\ \\Prove that: \\ \\ abc\geq (a+b-c)(a+c-b)(b+c-a).$
$\bg_white \\ Let \ x = a+b-c \ , \ y = b + c - a \ , \ z = a+c - b \\ \\ \Rightarrow \ Prove \ \frac{x+y}{2} \cdot \frac{x+z}{2} \cdot \frac{y+z}{2} \geq xyz \\ \\ If \ x,y, z \ are positive, (i.e. sides of a triangle) \\ \frac{x+y}{2} \geq \sqrt{xy} \ \ (1) \ , \frac{y+z}{2} \geq \sqrt{yz} \ \ (2) \ , \ \frac{x+z}{2} \geq \sqrt{xz} \ \ (3) \\ \\ (1) \cdot (2) \cdot (3) \\ \\ \Rightarrow \ (x+y)(x+z)(y+z) \geq 8xyz \ \square$

$\bg_white \\ If \ a,b,c \ are not sides of a triangle, and they are positive, then, at least 1 of \ x,y,z \ is negative \\ \\ If \ x,y,z \ at least 2 are negative \\ \\ a+b \leq c \ , \ (1) \\ a+c \leq b \Rightarrow \ -b \leq -a-c \ (2) \\ \\ (1) + (2) \Rightarrow \ \therefore \ a+b-b \leq -a-c+c \Rightarrow \ a\leq -a \Rightarrow \ a\leq 0 \\ \\ Similary we do the same to get \ b \leq 0 \ and \ c \leq 0 \\ \\ But since \ a,b,c \geq 0 \Rightarrow \ a=b=c=0 \\ \\ Meaning \ 0 \geq 0 \ which is true \ \square$

$\bg_white \\ If only one of \ x,y,z \ is negative, then \ abc \geq 0 \geq xyz \ \square$

EDIT: There is a small error with my solution when looking to the case that at least 2 of x,y,z are negative, it has been amended

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#### Sy123

##### This too shall pass
Re: HSC 2014 4U Marathon - Advanced Level

#### seanieg89

##### Well-Known Member
Re: HSC 2014 4U Marathon - Advanced Level

$\bg_white \\ Let \ x = a+b-c \ , \ y = b + c - a \ , \ z = a+c - b \\ \\ \Rightarrow \ Prove \ \frac{x+y}{2} \cdot \frac{x+z}{2} \cdot \frac{y+z}{2} \geq xyz \\ \\ If \ x,y, z \ are positive, (i.e. sides of a triangle) \\ \frac{x+y}{2} \geq \sqrt{xy} \ \ (1) \ , \frac{y+z}{2} \geq \sqrt{yz} \ \ (2) \ , \ \frac{x+z}{2} \geq \sqrt{xz} \ \ (3) \\ \\ (1) \cdot (2) \cdot (3) \\ \\ \Rightarrow \ (x+y)(x+z)(y+z) \geq 8xyz \ \square$

$\bg_white \\ If \ a,b,c \ are not sides of a triangle, and they are positive, then, at least 1 of \ x,y,z \ is negative \\ \\ If \ x,y,z \ at least 2 are negative \\ \\ a+b \leq c \ , \ (1) \\ a+c \leq b \Rightarrow \ -b \leq -a-c \ (2) \\ \\ (1) + (2) \Rightarrow \ \therefore \ a+b-b \leq -a-c+c \Rightarrow \ a\leq -a \Rightarrow \ a=0 \\ \\ Similary we do the same to get \ b=0 \ and \ c=0 \\ \\ Meaning \ 0 \geq 0 \ which is true \ \square$

$\bg_white \\ If only one of \ x,y,z \ is negative, then \ abc \geq 0 \geq xyz \ \square$
Yep, good. It should be noted that this question is just a rearrangement of Schur's inequality, which is why the method is somewhat similar to how Schur's is proven.

#### TL1998

##### Member
Re: HSC 2014 4U Marathon - Advanced Level

Someone try this

#### seanieg89

##### Well-Known Member
Re: HSC 2014 4U Marathon - Advanced Level

$\bg_white For brevity of notation, we set:\\y:=\tan(x),u:=\tan(\alpha),v:=\tan(\beta),s:=u+v,t:=uv.\\ \\ Then the assumption translates to:\\ \\ y^2=\frac{(y-u)(y-v)}{(1+uy)(1+vy)}\\ \Rightarrow t(y^4-1)+s(y^3+y)=0\\ \Rightarrow \frac{t}{s}=\frac{y}{1-y^2}. \\ \\ So: \\ \\\frac{2\sin(\alpha)\sin(\beta)}{\sin(\alpha + \beta)}=\frac{2t}{s}=\frac{2y}{1-y^2}=\tan(2x).$

#### TL1998

##### Member
Re: HSC 2014 4U Marathon - Advanced Level

$\bg_white For brevity of notation, we set:\\y:=\tan(x),u:=\tan(\alpha),v:=\tan(\beta),s:=u+v,t:=uv.\\ \\ Then the assumption translates to:\\ \\ y^2=\frac{(y-u)(y-v)}{(1+uy)(1+vy)}\\ \Rightarrow t(y^4-1)+s(y^3+y)=0\\ \Rightarrow \frac{t}{s}=\frac{y}{1-y^2}. \\ \\ So: \\ \\\frac{2\sin(\alpha)\sin(\beta)}{\sin(\alpha + \beta)}=\frac{2t}{s}=\frac{2y}{1-y^2}=\tan(2x).$
Nice! However, is there another way around this problem?

#### seanieg89

##### Well-Known Member
Re: HSC 2014 4U Marathon - Advanced Level

Nice! However, is there another way around this problem?
I am sure there are lots of ways, its just an algebraic manipulation so you always have lots of choices with these things.

#### Sy123

##### This too shall pass
Re: HSC 2014 4U Marathon - Advanced Level

$\bg_white \\ If \ a,b,c \ are integers such that \ \ ab + bc +ac = 1 \ \ show that \\ (1+a^2)(1+b^2)(1+c^2) \ is a perfect square$

#### RealiseNothing

##### what is that?It is Cowpea
Re: HSC 2014 4U Marathon - Advanced Level

$\bg_white \\ If \ a,b,c \ are integers such that \ \ ab + bc +ac = 1 \ \ show that \\ (1+a^2)(1+b^2)(1+c^2) \ is a perfect square$
$\bg_white (1+a^2)(1+b^2)(1+c^2) = (a+b+c-abc)^2$

#### dunlop1234

##### New Member
Re: HSC 2014 4U Marathon - Advanced Level

Hey guys!

Guess what, i forgot how to solve 4e^(-x)-6e^(-2x)=0. Can someone how to remind me how to find x when its in such form? (came up in some graph question - you need to find point of intersection with the axis blah blah)

#### TL1998

##### Member
Re: HSC 2014 4U Marathon - Advanced Level

#### Sy123

##### This too shall pass
Re: HSC 2014 4U Marathon - Advanced Level

$\bg_white \\ Find \ \ \sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$

#### Sy123

##### This too shall pass
Re: HSC 2014 4U Marathon - Advanced Level

$\bg_white \\ Let the sides of the square be \ x \\ In the triangle \ \triangle ADP \ by the cosine rule: \\ \\ \cos \theta = \frac{5-x^2}{4} \\ \\ By the sine rule: \ \sin \theta = x \sin \alpha \ (*) \ \ \ \ (\alpha = \angle DAP) \\ \\ By the cosine rule in \ \triangle PAB \\ \\ 4x\sin \alpha = x^2-5 \\ \\ Using \ (*) \\ \\ \Rightarrow \ \sin \theta = \frac{x^2-5}{4} \\ \\ \Rightarrow \ \tan \theta = - 1 \Rightarrow \ \theta = \frac{3\pi}{4} \\ (This solution does not yield \ x = \sqrt{5})$

$\bg_white \\ For the second part, it is clear that if \ P \ was outside the square, then at least one triangle will be obtuse angled, this will change the subsequent signs of \ \sin \theta \ or \cos \theta$

$\bg_white \\ For the third part, simply follow the procedure in the first part, we see that it is indeed possible$

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#### RealiseNothing

##### what is that?It is Cowpea
Re: HSC 2014 4U Marathon - Advanced Level

$\bg_white \\ Find \ \ \sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$
Let:

$\bg_white S(n)=\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$

Using partial fractions we get:

$\bg_white S(n) = \frac{1}{2}(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1})$

Now for $\bg_white n^2-n+1$, we will let $\bg_white n=m+1$

$\bg_white (m+1)^2-(m+1)+1 = m^2+m+1$

Now this is just $\bg_white n^2+n+1$ but for the next term. Hence what we have is a telescoping sum:

$\bg_white S(n) = (1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{7}) + (\frac{1}{7}-\frac{1}{13}) + (\frac{1}{13}-\frac{1}{21})+...$

And we are left with just:

$\bg_white S(n) = 1$

##### i'm the cook
Re: HSC 2014 4U Marathon - Advanced Level

$\bg_white \\ Let the sides of the square be \ x \\ In the triangle \ \triangle ADP \ by the cosine rule: \\ \\ \cos \theta = \frac{5-x^2}{4} \\ \\ By the sine rule: \ \sin \theta = x \sin \alpha \ (*) \ \ \ \ (\alpha = \angle ADP) \\ \\ By the cosine rule in \ \triangle DPC \\ \\ 4x\sin \alpha = x^2-5 \\ \\ Using \ (*) \\ \\ \Rightarrow \ \sin \theta = \frac{x^2-5}{4} \\ \\ \Rightarrow \ \tan \theta = - 1 \Rightarrow \ \theta = \frac{3\pi}{4} \\ (This solution does not yield \ x = \sqrt{5})$

$\bg_white \\ For the second part, it is clear that if \ P \ was outside the square, then at least one triangle will be obtuse angled, this will change the subsequent signs of \ \sin \theta \ or \cos \theta$

$\bg_white \\ For the third part, simply follow the procedure in the first part, we see that it is indeed possible$
I'm just a little confused, by using the sine rule with alpha as angle ADP shouldn't we get sin(theta) = xsin(alpha)/2? unless that's meant to be angle DAP?

#### Sy123

##### This too shall pass
Re: HSC 2014 4U Marathon - Advanced Level

Thanks Dude.
If I may ask, did you state rank for any of the mathematics courses?
Yes I did for 4U

I don't get how you used the cosine rule to get the value of sin alpha. Did you a certain value for PC?
I meant triangle PAB, not sure why I wrote DPC!

Basically, the angle PAB = 90 - alpha, when you do cosine rule, it becomes a sin(alpha)

I'm just a little confused, by using the sine rule with alpha as angle ADP shouldn't we get sin(theta) = xsin(alpha)/2? unless that's meant to be angle DAP?
Yes my bad that is supposed to be angle DAP

#### Sy123

##### This too shall pass
Re: HSC 2014 4U Marathon - Advanced Level

Let:

$\bg_white S(n)=\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$

Using partial fractions we get:

$\bg_white S(n) = \frac{1}{2}(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1})$

Now for $\bg_white n^2-n+1$, we will let $\bg_white n=m+1$

$\bg_white (m+1)^2-(m+1)+1 = m^2+m+1$

Now this is just $\bg_white n^2+n+1$ but for the next term. Hence what we have is a telescoping sum:

$\bg_white S(n) = (1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{7}) + (\frac{1}{7}-\frac{1}{13}) + (\frac{1}{13}-\frac{1}{21})+...$

And we are left with just:

$\bg_white S(n) = 1$
Don't forget to divide by half!

But yes that was my method as well

#### RealiseNothing

##### what is that?It is Cowpea
Re: HSC 2014 4U Marathon - Advanced Level

Don't forget to divide by half!

But yes that was my method as well
Ahh shit, always doing dem sillies.

#### RealiseNothing

##### what is that?It is Cowpea
Re: HSC 2014 4U Marathon - Advanced Level

Prove that, for $\bg_white m \geq n$

$\bg_white \frac{(2m)!(m-n)!}{(m)!(2m-n)!} > 2^n$

##### Active Member
Re: HSC 2014 4U Marathon - Advanced Level

$\bg_white \\ Let \ x = a+b-c \ , \ y = b + c - a \ , \ z = a+c - b \\ \\ \Rightarrow \ Prove \ \frac{x+y}{2} \cdot \frac{x+z}{2} \cdot \frac{y+z}{2} \geq xyz \\ \\ If \ x,y, z \ are positive, (i.e. sides of a triangle) \\ \frac{x+y}{2} \geq \sqrt{xy} \ \ (1) \ , \frac{y+z}{2} \geq \sqrt{yz} \ \ (2) \ , \ \frac{x+z}{2} \geq \sqrt{xz} \ \ (3) \\ \\ (1) \cdot (2) \cdot (3) \\ \\ \Rightarrow \ (x+y)(x+z)(y+z) \geq 8xyz \ \square$

$\bg_white \\ If \ a,b,c \ are not sides of a triangle, and they are positive, then, at least 1 of \ x,y,z \ is negative \\ \\ If \ x,y,z \ at least 2 are negative \\ \\ a+b \leq c \ , \ (1) \\ a+c \leq b \Rightarrow \ -b \leq -a-c \ (2) \\ \\ (1) + (2) \Rightarrow \ \therefore \ a+b-b \leq -a-c+c \Rightarrow \ a\leq -a \Rightarrow \ a\leq 0 \\ \\ Similary we do the same to get \ b \leq 0 \ and \ c \leq 0 \\ \\ But since \ a,b,c \geq 0 \Rightarrow \ a=b=c=0 \\ \\ Meaning \ 0 \geq 0 \ which is true \ \square$

$\bg_white \\ If only one of \ x,y,z \ is negative, then \ abc \geq 0 \geq xyz \ \square$

EDIT: There is a small error with my solution when looking to the case that at least 2 of x,y,z are negative, it has been amended
A hsc question? Remember doing this.

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