HSC 2015 MX2 Marathon ADVANCED (archive) (4 Viewers)
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dan964
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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
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
what
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
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
Last edited:
Re: HSC 2015 4U Marathon - Advanced Level
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.
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.
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.
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
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FrankXie
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dan964
what
Re: HSC 2015 4U Marathon - Advanced Level
I was thinking that
but not sure why.
thought it might have been
I was thinking that
but not sure why.
dan964
what
Re: HSC 2015 4U Marathon - Advanced Level
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.
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.
Last edited:
Drsoccerball
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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
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
what
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
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
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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
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
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Re: HSC 2015 4U Marathon - Advanced Level
whatsoever, z=z^5 is totally different from z^5=1, coz z=z^5 is kind of z^4=1
Ekman
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Re: HSC 2015 4U Marathon - Advanced Level
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
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
Drsoccerball
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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
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
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Last edited:
Sy123
This too shall pass
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Re: HSC 2015 4U Marathon - Advanced Level
 \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)} )
} \ $into two sums, one group in which$ \ \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
I found this harder than the first inequality, probably because the answer doesn't require much algebra!And an easier one:
Prove that
for increasing sequencesand an arbitrary permutation
of the first n positive integers.)
I don't exactly know how to symbolize the reasoning but it should be clear
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