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Math Help - Tough Problem from an olympiad

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

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

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

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

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

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

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


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

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

    shows that \Delta=27k(4-k). \ \Box
    Last edited by NonCommAlg; February 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 a^3 + b^3 + c^3 and, for some strange reasons, i thought it was to find the range of abc. however, since a^3+b^3+c^3=54+3abc,

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

    The value of abc can also be found by the fact that if f(x)=0 will have 3 roots then f(1)f(3)<0 where x=1 and x=3 are roots of 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 f(x)=0, viz.x=1,3.
    Since f(x) is a cubic( f(x)=(x-a)(x-b)(x-c)=x^3-6x^2+9x-abc) with coefficient of x^3 positve it is obvious that x=1 is a point of maxima and x=3 is a point of minima(can also be ascertained by the double derivative test).
    f(0)=-abc
    f(1)=4-abc
    f(3)=-abc
    f(4)=4-abc
    As I said before that for f(x)=0 to have 3 roots f(1)f(3)<0 which gives 0<abc<4.
    Now it is obvious that there is one root between x=0 and x=1,another root between x=1 and x=3 and the third root between x=3 and x=4
    Last edited by pankaj; February 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 f(x)=0, viz.x=1,3.
    Since f(x) is a cubic with coefficient of x^3 positve it is obvious that x=1 is a point of maxima and x=3 is a point of minima(can also be ascertained by the double derivative test).
    f(0)=-abc
    f(1)=4-abc
    f(3)=-abc
    f(4)=4-abc
    As I said before that for f(x)=0 to have 3 roots f(1)f(3)<0 which gives 0<abc<4.
    Now it is obvious that there is one root between x=0 and x=1,another root between x=1 and x=3 and the third root between x=3 and 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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