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HSC 2015 MX2 Marathon ADVANCED (archive) (4 Viewers)

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Sy123

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Re: HSC 2015 4U Marathon - Advanced Level

 

glittergal96

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Re: HSC 2015 4U Marathon - Advanced Level

Let x=a+b, y=a+c, z=b+c.

Then LHS=[(x+y-z)/z+(x+z-y)/y+(y+z-x)/x]/2=[(x/y+y/x)+(x/z+z/x)+(y/z+z/y)-3]/2 >= 3/2.

(We used AM-GM thrice in the last inequality to bound the (t+1/t) terms below by 2.)
 

Sy123

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Re: HSC 2015 4U Marathon - Advanced Level

Let x=a+b, y=a+c, z=b+c.

Then LHS=[(x+y-z)/z+(x+z-y)/y+(y+z-x)/x]/2=[(x/y+y/x)+(x/z+z/x)+(y/z+z/y)-3]/2 >= 3/2.

(We used AM-GM thrice in the last inequality to bound the (t+1/t) terms below by 2.)
nice method, mine was a little longer:







 
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Chlee1998

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Re: HSC 2015 4U Marathon - Advanced Level

suppose that a polynomial with integer coefficients satisfies P(100)=100. Find the maximum of integer solutions to P(k)=k^3 such a polynomial can have.
 

kawaiipotato

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Re: HSC 2015 4U Marathon - Advanced Level

suppose that a polynomial with integer coefficients satisfies P(100)=100. Find the maximum of integer solutions to P(k)=k^3 such a polynomial can have.
that seems familiar to a question in the maths competition by AMT last year. couldnt do it lol
 

glittergal96

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Re: HSC 2015 4U Marathon - Advanced Level

suppose that a polynomial with integer coefficients satisfies P(100)=100. Find the maximum of integer solutions to P(k)=k^3 such a polynomial can have.
Can you please clarify the exact question statement?

Are these polynomials definitely required to have integer coefficients?

Is it definitely P(100)=100?

Are we meant to find the maximum number of solutions to f(k) = k^3 possible, or the largest possible k such that f(k)=k^3?

(I just want to be clear the problem is correctly written before spending time on it.)
 

Chlee1998

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Re: HSC 2015 4U Marathon - Advanced Level

Can you please clarify the exact question statement?

Are these polynomials definitely required to have integer coefficients?

Is it definitely P(100)=100?

Are we meant to find the maximum number of solutions to f(k) = k^3 possible, or the largest possible k such that f(k)=k^3?

(I just want to be clear the problem is correctly written before spending time on it.)
a polynomial is called self centered if it has integer coefficients and p (100) = 100. if p (x) is a self centred polynomial, what is the maximum number of integer solutions to p (k) =k^3?
 

Drsoccerball

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Re: HSC 2015 4U Marathon - Advanced Level

a polynomial is called self centered if it has integer coefficients and p (100) = 100. if p (x) is a self centred polynomial, what is the maximum number of integer solutions to p (k) =k^3?
is the answer
?
EDIT: n-2
 
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glittergal96

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Re: HSC 2015 4U Marathon - Advanced Level

Okay cool, seems hard! Will spend more time on it tomorrow morning, but here is some first progress:

Since P has integer coefficients, we must have k - 100 | P(k) - P(100) = k^3 - 100.

But k^3 - 100 = (k - 100)(k^2 -100k + 10000) + 999900.

So the only k for which the identity P(k) = k^3 can possibly hold true are k such that k - 100 | 999900.

We can prime factorise this large number to obtain (2^2)(3^2)(5^2)(11)(101), which has: 2.3.3.3.2.2=216 integer factors.

So there are at most 216 integers k satisfying the property P(k) = k^3.

It is not clear however, that we can find an INTEGER polynomial passing through the points (k,k^3) for k ranging over the set {100 + d : d | 999900}, so we have only shown that 216 is an upper bound, rather than a maximum.

Will return to this tomorrow.
 

glittergal96

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Re: HSC 2015 4U Marathon - Advanced Level

Lol, woke up with an idea that I think does it.

Last post we showed that the subset S of k satisfying P(k) = k^3 consists of integers m with 100 - m a divisor of 999900.

We know that P(x) - x^3 vanishes at k in S.

So by the factor theorem we have



for some integer polynomial Q.

Evaluating at 100, we get:



So we need to find a maximal set of distinct divisors of 999900 with product a factor of 999900.

By inspection, such a set is 1,-1,2,-2,3,-3,5,-5,11,101.

So: Taking Q(x)=-1 as constant and taking S={99,101,98,102,97,103,95,105,89,-1} gives us a concrete expression for an optimal P, and the answer to the question is 10.

(will double-check this in the morning, hopefully no sillies!)
 
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simpleetal

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Re: HSC 2015 4U Marathon - Advanced Level

here is a great question I was doing a couple of weeks back. show that in every set of 9 positive integers, a subset of 5 positive integers exists such that the sum of the numbers is divisble by 5.
 

glittergal96

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Re: HSC 2015 4U Marathon - Advanced Level

Nesbitt's inequality is nice.






















Cauchy-Schwartz states for real variables we have:

(ax+by+cz)^2 <= (a^2+b^2+c^2)(x^2+y^2+z^2)

with equality iff (x,y,z)=(ta,tb,tc) for some real c.

Lots of proofs of Cauchy-Schwartz are on this site, so I will assume it. (Can provide a short proof if you like, such as the one based on non-negative polynomials having non-positive discriminant).

Then C-S implies

LHS^2 =< ((6x+1)+(6y+1)+(6z+1))(1+1+1)=27.

And taking square roots of both sides finishes the question. The C-S equality condition also implies x=y=z=1/3 if we have equality.
 

simpleetal

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Re: HSC 2015 4U Marathon - Advanced Level

here is a great question I was doing a couple of weeks back. show that in every set of 9 positive integers, a subset of 5 positive integers exists such that the sum of the numbers is divisble by 5.
Should I post a hint?
 

SpiralFlex

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Re: HSC 2015 4U Marathon - Advanced Level

Cauchy-Schwartz states for real variables we have:

(ax+by+cz)^2 (a^2+b^2+c^2)(x^2+y^2+z^2)

with equality iff (x,y,z)=(ta,tb,tc) for some real c.

Lots of proofs of Cauchy-Schwartz are on this site, so I will assume it. (Can provide a short proof if you like, such as the one based on non-negative polynomials having non-positive discriminant).

Then C-S implies

LHS^2 =< ((6x+1)+(6y+1)+(6z+1))(1+1+1)=27.

And taking square roots of both sides finishes the question. The C-S equality condition also implies x=y=z=1/3 if we have equality.




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