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

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Sy123

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Re: HSC 2014 4U Marathon - Advanced Level

 

Sy123

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Re: HSC 2014 4U Marathon - Advanced Level

the inequality shows symmetry in a, b and c. nature does not favor one or the other.
minimum value when a=b=c=3/4
Wait is this an actual theorem? What is it called?
 

seanieg89

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Re: HSC 2014 4U Marathon - Advanced Level

the inequality shows symmetry in a, b and c. nature does not favor one or the other.
minimum value when a=b=c=3/4
Symmetry in variables does not mean that minima occur when all variables are equal...
 

seanieg89

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Re: HSC 2014 4U Marathon - Advanced Level

in this case when does minimum occur?
In this question it happens to occur when x,y and z are equal. But this isn't because of symmetry, this is because Davo has shown it above (the only steps where inequality was introduced was the applications of AM-GM, whose equality conditions we understand well.) Writing what you did is not valid because there is no reason (that you have provided) why such an expression HAS to be minimised at x=y=z.
 

jyu

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Re: HSC 2014 4U Marathon - Advanced Level

A triangle with side lengths a, b, c is inscribed in a circle of radius 1. Prove that a+b+c>=abc.
 

jyu

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Re: HSC 2014 4U Marathon - Advanced Level

In this question it happens to occur when x,y and z are equal. But this isn't because of symmetry, this is because Davo has shown it above (the only steps where inequality was introduced was the applications of AM-GM, whose equality conditions we understand well.) Writing what you did is not valid because there is no reason (that you have provided) why such an expression HAS to be minimised at x=y=z.
In this case the minimum 3/4 is given. The expression equals the min value when a=b=c, this is chosen due to symmetry, or should I say the cyclic nature of the expression. If a, b and c are different then the expression will be >= 3/4.
 

seanieg89

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Re: HSC 2014 4U Marathon - Advanced Level

In this case the minimum 3/4 is given. The expression equals the min value when a=b=c, this is chosen due to symmetry, or should I say the cyclic nature of the expression. If a, b and c are different then the expression will be >= 3/4.
Yes, it does equal 3/4 when a=b=c. But this does not prove that this is a minimum, and you cannot assume 3/4 is a minimum because that is what the question is asking you to prove.

Symmetry means that a=b=c is often a good candidate for a minimum, but doesn't prove anything without additional information.
 

seanieg89

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Re: HSC 2014 4U Marathon - Advanced Level

Yes, it does equal 3/4 when a=b=c. But this does not prove that this is a minimum, and you cannot assume 3/4 is a minimum because that is what the question is asking you to prove.

Symmetry means that a=b=c is often a good candidate for a minimum, but doesn't prove anything without additional information.
Picture it this way in two variables x and y:

A real function f can be viewed as a graph z=f(x,y).

Saying such a function is symmetric, is just saying that it's graph has reflective symmetry about the plane x=y.

This just means that any extrema will have a "mirrored" extrema on the other side of this plane. It definitely does not imply that every extrema must lie on x=y.
 

jyu

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Re: HSC 2014 4U Marathon - Advanced Level

Look at some easier examples to picture. Area of a rectangle of fixed perimeter = xy, symmetry in x and y; volume of a cuboid of fixed surface area = xyz, symmetry in x, y and z.
 

seanieg89

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Re: HSC 2014 4U Marathon - Advanced Level

Look at some easier examples to picture. Area of a rectangle of fixed perimeter = xy, symmetry in x and y; volume of a cuboid of fixed surface area = xyz, symmetry in x, y and z.
So what if some symmetric expressions have extrema when all variables are equal? This does not mean that all do, so we can't use it in proofs.

Here is a symmetric expression in two variables, (switching x and y just switches the roles of the two bracketed terms).



But P only attains it's minimum of 0 at (1,-1) and (-1,1). When x=y, it is positive.

Why does your "rule" apply for the question Sy asked, but not for the expression in this post?
 
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jyu

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Re: HSC 2014 4U Marathon - Advanced Level

So what if some symmetric expressions have extrema when all variables are equal? This does not mean that all do, so we can't use it in proofs.

Here is a symmetric expression in two variables, (switching x and y just switches the roles of the two bracketed terms).



But P only attains it's minimum of 0 at (1,-1) and (-1,1). When x=y, it is positive.

Why does your "rule" apply for the question Sy asked, but not for the expression in this post?
Because of the constants.
Change x+1 to X and y-1 to Y,
 

seanieg89

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Re: HSC 2014 4U Marathon - Advanced Level

The constants? Which constants and what about them?

What is the purpose of this coordinate change? P is not symmetric in X and Y (and we should not expect it to be, because the relationship between (x,y) and (X,Y) breaks symmetry.

Your claim was:

Symmetry in variables => Any minimum must happen when all variables are equal.

I provided an example of an expression that is (A) symmetric, and (B) does not have a minimal point with x=y, but has a minimum elsewhere.

If you think this is not a counterexample to your claim then you must contest one of these two assertions. Which one, and why? Please be precise, as I am making an effort to.
 
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