MathMate Help!!!

This was the question before that you answered with more information.
"Two altitudes of a triangle have lengths 12 and 14. What is the longest possible length of the third altitude, if it is a positive integer?"
I don't get it. Could you at least help me for the first couple of steps???

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  1. We know that given sides A,B,C, with A>=B,

    A-B < C < A+B

    With a little algebra (as seen here):

    http://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.577996.html

    we know that if the altitudes are a,b,c then

    1/a - 1/b < c < 1/a + 1/b

    So, for your triangle, the 3rd altitude must satisfy

    1/12 - 1/14 < c < 1/12 + 1/14

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    posted by Steve
  2. oops - make that

    1/a + 1/b < 1/c < 1/a - 1/b

    1/84 < 1/c < 13/84
    84/13 < c < 84

    So, 83 is the maximum integer side

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    posted by Steve
  3. Thank you for the brilliant and elegant proof, Steve!

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  4. Google is your friend. It gave several hits, and this one was indeed elegant and simple.

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    posted by Steve
  5. This is a question from the Intermediate Algebra homework. People shouldn't be asking for the answer on different websites. Thanks!

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  6. You're right. We're currently tracking down the IP addresses of the users who post for help here, so that we can contact and deactivate their AoPS accounts.

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    posted by AoPS

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