HSC 2013 MX2 Marathon (archive) (5 Viewers)

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gustavo28

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Re: HSC 2013 4U Marathon








edit: my bad
Assume that x>y since the problem is symmetric in x and y.
We can manipulate the inequality into wanting to prove that

x^y(x^(x-y)-1)/y^y >y^(x-y)-1. Now, x^a>y^a if a,x,y are positive since x^a is increasing for positive x (f'(x)=ax^(a-1)). So, x^(x-y)-1>y^(x-y)-1. Moreover x^y/y^y>1. So multiplying these two inequalities together we get the desired result. (Note y^(x-y)-1 is positive since x-y>0)
 

gustavo28

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Re: HSC 2013 4U Marathon

So, first of all suppose there is a monic rational quadratic with the root X+1/X with X as the cube root of a non-cube rational. From now on this root will be known as T.
Let this quadratic be x^2+ax+b. So, (x-a)(x^2+ax+b)=x^3+ax^2+bx-ax^2-a^2x-ab=x^3+x(b-a^2)-ab=Q.

So, using the P that has been referred to previously, we get P-Q=some linear rational polynomial which also must have T as a root. But this cant have the irrational root T.

Now, rational polynomials of degree 1 and 2 cant have T as a root then. Suppose some rational polynomial S(x) has T as a root. Note it must be of degree greater than or equal to 2. So, S(x)=M(x)P(x)+R(x) where deg R is 2 or less. Now, R has T as a root. But R can't be a constant, linear or quadratic rational polynomial. So R=0 and P has to divide S.
I think there's a fairly subtantial problem with my proof in that I assume that X+1/X is irrational where X is the cube root of a non-cube rational. However, I'm not entirely sure whether or not this is provable within the syllabus.
 

psychotropic

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Re: HSC 2013 4U Marathon

I think there's a fairly subtantial problem with my proof in that I assume that X+1/X is irrational where X is the cube root of a non-cube rational. However, I'm not entirely sure whether or not this is provable within the syllabus.
I think there is a much simpler way to do this- btw who are you?
A proof would be along these lines
x^(1/3)+x^(-1/3) is rational =>
(x^(1/3)+x^(-1/3))^2=x^(2/3)+x^(-2/3)+2 is rational=>
x^(2/3)+x^(-2/3) is rational=>
x(x^(1/3)+x^(-1/3))=x^(2/3)+x^(4/3) is rational=>
(x^(2/3)+x^(-2/3))-(x^(2/3)+x^(4/3))=x^(-2/3)+x^(4/3)=(x+x^(-1))x^(1/3) is rational=>
x^(1/3) is rational.
therefore x has to be a rational cube?
 

psychotropic

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Re: HSC 2013 4U Marathon

Given four concyclic points. For each subset of three points take the incenter. Show that the four incenters from a rectangle.
 

bottleofyarn

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Re: HSC 2013 4U Marathon

Here's a doable one, four lines of working and one of my favourites.



 

Sy123

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Re: HSC 2013 4U Marathon

View attachment 28799

lol ok here it is again




We are being asked for the roots of the polynomial



So, basically, for what values of x will that equation hold, well it is given from the above property, if:



then the equation holds hence the roots of the polynomial (x+1) P(x) - x









Then just re-arrange
 

seanieg89

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Re: HSC 2013 4U Marathon

I think there's a fairly subtantial problem with my proof in that I assume that X+1/X is irrational where X is the cube root of a non-cube rational. However, I'm not entirely sure whether or not this is provable within the syllabus.
This is easy enough:

If X+1/X=r rational, then X is a nonzero irrational root of a rational quadratic. So we can write X=a+b*sqrt(c), with a,b,c rational, c not a perfect square (from the definition of X). By cubing this implies that we must have 3a^2b+cb^3=0. As c is irrational, this forces a=b=0, contradiction.
 
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