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Putnam and beyond prob 113

  1. May 15, 2008 #1
    [SOLVED] putnam and beyond prob 113

    1. The problem statement, all variables and given/known data
    Let a,b,c be side lengths of a triangle with the property that for any positive integer n, the numbers a^n, b^n, and c^n can also be the side lengths of a triangle. Prove that the triangle is necessarily isosceles.


    2. Relevant equations



    3. The attempt at a solution
    I can do the case where a,b>1 and c<1 using the triangle inequalities a^n+c^n>b^n>a^n-c^n and squeezing but I cannot do the case where a,b,c>1 or a,b,c<1 unfortunately. Can someone give me a hint?
     
  2. jcsd
  3. May 16, 2008 #2

    tiny-tim

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    Hi ehrenfest! :smile:

    Hint: if it's not isoceles, then name the sides so that a > b > c > 1.

    You need to find an n such that a^n > b^n + c^n. :smile:
     
  4. May 16, 2008 #3
    I got it. When n is large enough

    [tex]\frac{c}{a-b}\left( \frac{c}{a} \right)^{n-1} < 1[/tex]

    or

    [tex](a-b) a^{n-1} > c^n[/tex]

    Then we have

    [tex]a^n -b^n = (a-b)(a^{n-1}+ \cdots b^{n-1}) > (a-b)a^{n-1} >c^n[/tex]

    which is what we want. Thanks. Please confirm that this is correct.
     
    Last edited: May 16, 2008
  5. May 16, 2008 #4

    tiny-tim

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    hmm … a bit complicated …

    For a simpler proof …

    Hint: forget c, and just find an n with a^n > 2(b^n). :smile:
     
  6. May 16, 2008 #5
    That works too.

    EDIT: and wow that is much simpler since you avoid dividing into cases
     
    Last edited: May 16, 2008
  7. May 16, 2008 #6

    tiny-tim

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    Yes … very simple questions often have very simple answers if you look long enough! :rolleyes:

    Moral … simplify the problem as much as possible … then solve it! :smile:
     
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