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???

6 answers

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
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
Thank you for the brilliant and elegant proof, Steve!
Google is your friend. It gave several hits, and this one was indeed elegant and simple.
This is a question from the Intermediate Algebra homework. People shouldn't be asking for the answer on different websites. Thanks!
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.