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Thread: Tough Problem from an olympiad

  1. #1
    Senior Member pankaj's Avatar
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    Tough Problem from an olympiad

    Let $\displaystyle a<b<c$ be real numbers such that $\displaystyle a+b+c=6$ and $\displaystyle ab+bc+ca=9$ then prove that $\displaystyle 0<a<1<b<3<c<4$.
    Also,find the range of $\displaystyle a^3+b^3+c^3$.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Hello Sir,
    I haven't solved the problem

    but I think this can help a bit for the second part
    a+b+c=6
    b+bc+ca=9
    (a+b+c)^2 = 36
    a^2+b^2+c^2 +2(ab+bc+ca)=36
    a^2+b^2+c^2=18
    now
    a^3+b^3+c^3 =(a+b+c)(a^2+b^2+c^2-ab-bc-ca) +3abc
    =6(18-9)+3abc
    =54+3abc

    I stopped here

    But I think
    using a cubic equation like
    x^3+6x^2+9x+m=0
    can help
    where m=abc
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  3. #3
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    Quote Originally Posted by pankaj View Post

    Let $\displaystyle a<b<c$ be real numbers such that $\displaystyle a+b+c=6$ and $\displaystyle ab+bc+ca=9$ then prove that $\displaystyle 0<a<1<b<3<c<4.$
    so $\displaystyle a,b,c$ are roots of: $\displaystyle x^3-6x^2 + 9x -abc=0,$ which gives us: $\displaystyle (a-3)^2=bc, \ (b-3)^2=ac, \ (c-3)^2=ab.$ call this (1). note that all $\displaystyle a,b,c$ are non-zero, because otherwise we would have

    $\displaystyle abc=0$ and thus $\displaystyle 0=x^3-6x^2+9x=x(x-3)^2,$ which contradicts that $\displaystyle a,b,c$ are pairwise distinct. now by (1) either all $\displaystyle a,b,c$ are positive or all of them are negative. but since $\displaystyle a+b+c >0,$

    we must have that all $\displaystyle a,b,c$ are positive. also since $\displaystyle a+b+c=6$ and $\displaystyle a<b<c,$ we have $\displaystyle a < 3.$ hence by (1): $\displaystyle a=3-\sqrt{bc}$ and thus: $\displaystyle 6=a+b+c=b-\sqrt{bc}+c+3.$ that will give us

    $\displaystyle b-\sqrt{bc} + c - 3=0.$ call this (2). the discriminant of $\displaystyle b -\sqrt{bc} + c - 3,$ as a quadratic function in $\displaystyle \sqrt{b},$ is $\displaystyle 12 - 3c,$ which must be positive. therefore $\displaystyle \boxed{c < 4}.$ also since $\displaystyle b < c$ and by (2) we have

    $\displaystyle \sqrt{b}(\sqrt{b}-\sqrt{c})=3-c,$ and $\displaystyle \sqrt{c}(\sqrt{c}-\sqrt{b})=3-b,$ we must have $\displaystyle \boxed{c > 3}$ and $\displaystyle \boxed{b < 3}.$ finally since c > 3, we'll get from (1) that $\displaystyle c=3+\sqrt{ab}.$ thus: $\displaystyle 6=a+b+c=a+b+\sqrt{ab}+3.$ hence

    $\displaystyle a+b+\sqrt{ab}=3.$ but: $\displaystyle 3a < a+b+\sqrt{ab} < 3b,$ because $\displaystyle a<b.$ thus $\displaystyle \boxed{a< 1}$ and $\displaystyle \boxed{b>1}$ and we are done.


    Also,find the range of $\displaystyle a^3+b^3+c^3.$
    since $\displaystyle abc=a(a-3)^2$ and $\displaystyle 0 < a < 1,$ it's clear that $\displaystyle 0 < abc < 4.$ but this is not enough to claim that the range of $\displaystyle abc$ is $\displaystyle (0,4)$ because it might be smaller. in fact we'll prove that the range is

    indeed the interval $\displaystyle (0,4)$: let $\displaystyle abc=k.$ the cubic $\displaystyle x^3 -6x^2+9x - k=0$ must have 3 distinct real roots $\displaystyle a,b,c,$ which happens iff $\displaystyle \Delta,$ the discriminant of the cubic, is positive. a simple calculation

    shows that $\displaystyle \Delta=27k(4-k). \ \Box$
    Last edited by NonCommAlg; Feb 15th 2009 at 05:43 PM. Reason: typo!
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  4. #4
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    ok, the second part of the question was to find the range of $\displaystyle a^3 + b^3 + c^3$ and, for some strange reasons, i thought it was to find the range of $\displaystyle abc.$ however, since $\displaystyle a^3+b^3+c^3=54+3abc,$

    as ADARSH showed, and the range of $\displaystyle abc,$ as i showed, is (0,4), the range of $\displaystyle a^3+b^3+c^3$ will be $\displaystyle (54, 66).$
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  5. #5
    Senior Member pankaj's Avatar
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    Thanks NCA

    The value of $\displaystyle abc$ can also be found by the fact that if $\displaystyle f(x)=0$ will have $\displaystyle 3$ roots then $\displaystyle f(1)f(3)<0$ where $\displaystyle x=1$ and $\displaystyle x=3$ are roots of $\displaystyle f'(x)=0$
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  6. #6
    Senior Member pankaj's Avatar
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    Hello Adarsh and NCA there is also a Caculus approach to this problem.

    You see there are two critical points of $\displaystyle f(x)=0$,$\displaystyle viz.x=1,3.$
    Since $\displaystyle f(x)$ is a cubic($\displaystyle f(x)=(x-a)(x-b)(x-c)=x^3-6x^2+9x-abc)$ with coefficient of $\displaystyle x^3$ positve it is obvious that $\displaystyle x=1$ is a point of maxima and $\displaystyle x=3$ is a point of minima(can also be ascertained by the double derivative test).
    $\displaystyle f(0)=-abc$
    $\displaystyle f(1)=4-abc$
    $\displaystyle f(3)=-abc$
    $\displaystyle f(4)=4-abc$
    As I said before that for $\displaystyle f(x)=0$ to have $\displaystyle 3$ roots $\displaystyle f(1)f(3)<0$ which gives $\displaystyle 0<abc<4$.
    Now it is obvious that there is one root between $\displaystyle x=0$ and $\displaystyle x=1$,another root between $\displaystyle x=1$ and $\displaystyle x=3$ and the third root between $\displaystyle x=3$ and $\displaystyle x=4$
    Last edited by pankaj; Feb 15th 2009 at 05:15 PM.
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  7. #7
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    Quote Originally Posted by pankaj View Post
    Hello Adarsh and NCA there is also a Caculus approach to this problem.

    You see there are two critical points of $\displaystyle f(x)=0$,$\displaystyle viz.x=1,3.$
    Since $\displaystyle f(x)$ is a cubic with coefficient of $\displaystyle x^3$ positve it is obvious that $\displaystyle x=1$ is a point of maxima and $\displaystyle x=3$ is a point of minima(can also be ascertained by the double derivative test).
    $\displaystyle f(0)=-abc$
    $\displaystyle f(1)=4-abc$
    $\displaystyle f(3)=-abc$
    $\displaystyle f(4)=4-abc$
    As I said before that for $\displaystyle f(x)=0$ to have $\displaystyle 3$ roots $\displaystyle f(1)f(3)<0$ which gives $\displaystyle 0<abc<4$.
    Now it is obvious that there is one root between $\displaystyle x=0$ and $\displaystyle x=1$,another root between $\displaystyle x=1$ and $\displaystyle x=3$ and the third root between $\displaystyle x=3$ and $\displaystyle x=4$
    this solution, although not very accurate in this form, will work. the reason that i didn't use calculus was because you posted the question in the pre-algebra subforum! so i thought you were

    looking for an elementary solution.
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  8. #8
    Senior Member pankaj's Avatar
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    Question was posted a long time ago.How ADARSH retrieved it,well that goes to his credit.
    In fact I was looking for an algebraical solution which you have provided.
    The method that I have pointed out is not genral one and may work for this specific problem.
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