HSC 2015 MX2 Marathon ADVANCED (archive) (3 Viewers)

Kaido · Jan 16, 2015

Re: HSC 2015 4U Marathon - Advanced Level

^ Yeah the inequality shifts as well...

pretty sure there's some kind of 'inequality result' concerning this problem set.
any other hints?

Chlee1998 · Jan 16, 2015

Re: HSC 2015 4U Marathon - Advanced Level

yes indeed i forgot to mention that A,B,C,D >0.

The method I used was AM -Gm, although the problem is still very difficult...

dan964 · Jan 21, 2015

Re: HSC 2015 4U Marathon - Advanced Level

you just have to prove that A^2*C*D + A*B^2*D + A*B*C^2 + B*C*D^2 <= A+ B+C+D

dan964 · Jan 21, 2015

Re: HSC 2015 4U Marathon - Advanced Level

not really that hard...

(A+B) ≥2√AB
a+b≥2√ab
c+d≥2√cd
(a+b+c+d)≥2√ab+2√cd
Substitute RHS into LINE 1
2(√ab+2√cd)≥4×√(√ab×√cd)
(a+b+c+d)/4≥∜abcd, a,b,c,d≥0
∴∜ABCD≤1
∴ABCD≤1

Let a=A^2 CD, b=AB^2 D, c=ABC^2, d= BCD^2
A^2 CD+AB^2 D+ABC^2+BCD^2
≥4∜(A^4 B^4 C^4 D^4 )
≥ 4ABCD
A^2 CD+AB^2 D+ABC^2+BCD^2 ≤4
using 0< ABCD≤1
Divide both sides by ABCD gives the required result

Chlee1998 · Jan 21, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
not really that hard...

(A+B) ≥2√AB
a+b≥2√ab
c+d≥2√cd
(a+b+c+d)≥2√ab+2√cd
Substitute RHS into LINE 1
2(√ab+2√cd)≥4×√(√ab×√cd)
(a+b+c+d)/4≥∜abcd, a,b,c,d≥0
∴∜ABCD≤1
∴ABCD≤1

Let a=A^2 CD, b=AB^2 D, c=ABC^2, d= BCD^2
A^2 CD+AB^2 D+ABC^2+BCD^2
≥4∜(A^4 B^4 C^4 D^4 )
≥ 4ABCD
A^2 CD+AB^2 D+ABC^2+BCD^2 ≤4
using 0< ABCD≤1
Divide both sides by ABCD gives the required result

Not that hard? Oh please, your solution has logical flaws.
Let's look at one. Look at the bolded bit. You can't just switch the inequality sign like that, division by a factor whcih is less than 1 does not mean that.
Think again, when u give up I'll give time for someone else to have a go. if no one answers in like a week or so, I will post my solution.

Chlee1998 · Jan 21, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Just to clarify in case you don't understand, ur previous line said
'Let a=A^2 CD, b=AB^2 D, c=ABC^2, d= BCD^2
A^2 CD+AB^2 D+ABC^2+BCD^2
≥4∜(A^4 B^4 C^4 D^4 )
≥ 4ABCD'

But in the line which follows, u swapped the inequality sign, which u cannot do. Hence you have not acquired the desired result.

FrankXie · Jan 21, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
not really that hard...

(A+B) ≥2√AB
a+b≥2√ab
c+d≥2√cd
(a+b+c+d)≥2√ab+2√cd
Substitute RHS into LINE 1
2(√ab+2√cd)≥4×√(√ab×√cd)
(a+b+c+d)/4≥∜abcd, a,b,c,d≥0
∴∜ABCD≤1
∴ABCD≤1

Let a=A^2 CD, b=AB^2 D, c=ABC^2, d= BCD^2
A^2 CD+AB^2 D+ABC^2+BCD^2
≥4∜(A^4 B^4 C^4 D^4 )
≥ 4ABCD
A^2 CD+AB^2 D+ABC^2+BCD^2 ≤4
using 0< ABCD≤1
Divide both sides by ABCD gives the required result

This solution is invalid.

FrankXie · Jan 21, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Chlee1998 said:
Since no one has psoted a new question, here's one:

given A+B+C+D = 4, prove that 4/(ABCD) >= A/B +B/C + C/D + D/A

$$ The inequality is equivalent to $ LHS=cda^2+dab^2+abc^2+bcd^2\leq4 $ by using common denominator for the desired result.$
$$ Now $ LHS=(cda^2+abc^2)+(dab^2+bcd^2)=ac(ad+bc)+bd(ab+cd),$
$$ in case (i) if $ ad+bc\leq ab+cd, $ then $ LHS\leq ac(ab+cd)+bd(ab+cd)=(ac+bd)(ab+cd)\leq\left(\frac{ac+bd+ab+cd}{2}\right)^2=\left(\frac{(a+d)(b+c)}{2}\right)^2=\frac{1}{4}[(a+d)(b+c)]^2\leq\frac{1}{4}\left[\left(\frac{a+d+b+c}{2}\right)^2\right]^2=4$
$in case (ii) if $ ab+cd\leq ad+bc, $ the proof is similar.$
Q.E.D.

dan964 · Jan 21, 2015

Re: HSC 2015 4U Marathon - Advanced Level

FrankXie said:
This solution is invalid.

thought it might have been
I was thinking that
but not sure why.

dan964 · Jan 21, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Chlee1998 said:
Not that hard? Oh please, your solution has logical flaws.
Let's look at one. Look at the bolded bit. You can't just switch the inequality sign like that, division by a factor whcih is less than 1 does not mean that.
Think again, when u give up I'll give time for someone else to have a go. if no one answers in like a week or so, I will post my solution.

correction: multiplication but anyway same problem I guess.

it is invalid not because of division (which I didn't do) but because assuming that 4*(a number less than 1) would account for the difference between the LHS and RHS, and result in the LHS <4 , and hence invalid. Yes I shouldn't have swapped the inequality.

It is also invalid, because I didn't prove anything useful, ended up proving it with the inequality reversed, if my working had been correct. It is actually the 5th line in my working that makes that incorrect assumption.

Drsoccerball · Jan 23, 2015

Re: HSC 2015 4U Marathon - Advanced Level

A question from 2014 maths marathon was posted. z^13=w and w^11=z The imaginary part of z is sin (Mpi/n) where m,n are positive integers with no common factors. Find n. Is the answer 143 ? Thanks

dan964 · Jan 23, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Raise the first one to the power of 11.
z^143-z=0
z=0
or z^142 = 1
The general solution form is cos (M*2pi/n)+isin(M*2*pi/n)
So it would be n=71

Drsoccerball · Jan 23, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Z^1/11 =z^13
Z=z^143
Cistheta=cis143theta
Therefore n must= 143
Because of the rule how if z^5=1 or whatever

FrankXie · Jan 23, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Drsoccerball said:
Z^1/11 =z^13
Z=z^143
Cistheta=cis143theta
Therefore n must= 143
Because of the rule how if z^5=1 or whatever

whatsoever, z=z^5 is totally different from z^5=1, coz z=z^5 is kind of z^4=1

Ekman · Jan 24, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Drsoccerball said:
A question from 2014 maths marathon was posted. z^13=w and w^11=z The imaginary part of z is sin (Mpi/n) where m,n are positive integers with no common factors. Find n. Is the answer 143 ? Thanks

x=Mpi/N
z^143 = z
z(z^142 - 1) = 0
z=0 has no argument that satisfies this
z^142 = 1
cis142x = cis0
142x = 0 + 2*pi*k
x = (2*pi*k) / 142
x = pi*k/71
thus Mpi/N = pi*k/71
N = 71

seanieg89 · Jan 24, 2015

Re: HSC 2015 4U Marathon - Advanced Level

For positive real x,y,z, show that:

$\frac{xyz}{x^3+y^3+xyz}+\frac{xyz}{z^3+y^3+xyz}+ \frac{xyz}{x^3+z^3+xyz}\leq 1$

seanieg89 · Jan 24, 2015

Re: HSC 2015 4U Marathon - Advanced Level

And an easier one:

Prove that

$\sum_{k=1}^n x_ky_{\sigma(k)}\leq \sum_{k=1}^n x_ky_k$

for increasing sequences $(x_k),(y_k)$ and an arbitrary permutation $\sigma$ of the first n positive integers.)

Drsoccerball · Jan 24, 2015

Re: HSC 2015 4U Marathon - Advanced Level

take out xyz and make the denominator such that we are able to do x^3 +y^3 +z^3 therfore bottom becomes x^3 +y^3 +z^3 +xyz - extra value. Then make an equation ax^3 + bx^2 +cx + d where x+y+z=-B/a where -B/a is a negative number. xy +yz +xz= c/a positive and -d/a = xyz negative number. sub in x into equation to find sum of roots cubed. Therfore at you get a^3/ -(b^3 +2a^2d +3abc -x^3a^3) for all the rest aswell but then you're going to have to find a common denominator at this point i gave up ahaha what did i do wrong

Sy123 · Jan 25, 2015

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
For positive real x,y,z, show that:

$\frac{xyz}{x^3+y^3+xyz}+\frac{xyz}{z^3+y^3+xyz}+ \frac{xyz}{x^3+z^3+xyz}\leq 1$

$\\ x^2 + y^2 \geq 2xy \\ \\ x^2 - xy + y^2 \geq xy \\ \\ \therefore \ x^3 + y^3 = (x+y)(x^2 -xy +y^2) \geq xy(x+y) \ $for$ \ x,y \ $positive$ \\\\ \therefore \ x^3 + y^3 + xyz \geq xy(x+y+z) \\ \\ \therefore \ \frac{xyz}{x^3 + y^3 + xyz} \leq \frac{xyz}{xy(x+y+z)} = \frac{z}{x+y+z}$

$\\ \therefore \ \frac{xyz}{x^3+y^3+xyz}+\frac{xyz}{z^3+y^3+xyz}+ \frac{xyz}{x^3+z^3+xyz} \\ \leq \frac{x}{x+y+z} + \frac{y}{x+y+z} + \frac{z}{x+y+z} \\ = 1 \\ \square$

Sy123 · Jan 25, 2015

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
And an easier one:

Prove that

$\sum_{k=1}^n x_ky_{\sigma(k)}\leq \sum_{k=1}^n x_ky_k$

for increasing sequences $(x_k),(y_k)$ and an arbitrary permutation $\sigma$ of the first n positive integers.)

I found this harder than the first inequality, probably because the answer doesn't require much algebra!

$\\ $There are two cases to consider$ \\ $Case 1:$ \ \sigma (k) \leq k \ \Rightarrow \ x_k y_{\sigma (k)} \leq x_k y_k \\ \\ $Case 2:$ \ \sigma (k) \geq k \ \Rightarrow \ x_k y_{\sigma (k)} \leq x_{\sigma (k)} y_{\sigma (k)}$

$\\ $So, separate the sum$ \ \sum_{k=1}^{n} x_k y_{\sigma (k)} \ $into two sums, one group in which$ \ \sigma (k) < k \ $and another group in which$ \ \sigma (k) \geq k \\ \\ $Then apply the inequality cases above to gain two sums, that when added together yield$ \ \sum_{k=1}^n x_ky_k$

I don't exactly know how to symbolize the reasoning but it should be clear

HSC 2015 MX2 Marathon ADVANCED (archive) (3 Viewers)

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