HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)

Sy123 · Jul 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
$$Let $a,b,c\geq 0.\\ \\$Prove that$: \\ \\ abc\geq (a+b-c)(a+c-b)(b+c-a).$

$\\ $Let$ \ 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$

$\\ $If$ \ 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$

$\\ $If only one of$ \ 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

Sy123 · Jul 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 · Jul 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let$ \ 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$

$\\ $If$ \ 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$

$\\ $If only one of$ \ 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 · Jul 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Someone try this

seanieg89 · Jul 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
View attachment 30600

Someone try this

$For brevity of notation, we set:\\$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 · Jul 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
$For brevity of notation, we set:\\$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 · Jul 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
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 · Jul 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $If$ \ 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 · Jul 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $If$ \ 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$$

$(1+a^2)(1+b^2)(1+c^2) = (a+b+c-abc)^2$

dunlop1234 · Jul 3, 2014

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 · Jul 7, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 · Jul 7, 2014

Re: HSC 2014 4U Marathon - Advanced Level

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

Sy123 · Jul 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
View attachment 30621

$\\ $Let the sides of the square be$ \ 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}$)$$

$\\ $For the second part, it is clear that if$ \ 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$

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

RealiseNothing · Jul 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find$ \ \ \sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$

Let:

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

Using partial fractions we get:

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

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

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

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

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

And we are left with just:

$S(n) = 1$

aDimitri · Jul 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let the sides of the square be$ \ 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}$)$$

$\\ $For the second part, it is clear that if$ \ 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$

$\\ $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 · Jul 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
Thanks Dude.
If I may ask, did you state rank for any of the mathematics courses?

Yes I did for 4U

TL1998 said:
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)

aDimitri said:
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 · Jul 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
Let:

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

Using partial fractions we get:

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

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

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

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

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

And we are left with just:

$S(n) = 1$

Don't forget to divide by half!

But yes that was my method as well

RealiseNothing · Jul 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Don't forget to divide by half!

But yes that was my method as well

Ahh shit, always doing dem sillies.

RealiseNothing · Jul 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Prove that, for $m \geq n$

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

Fade1233 · Jul 18, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let$ \ 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$

$\\ $If$ \ 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$

$\\ $If only one of$ \ 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.

HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)

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