This was the question before that you answered with more information.

Question

"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???

Answer 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

Answer 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

Answer 3

Thank you for the brilliant and elegant proof, Steve!

Answer 4

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

Answer 5

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

Answer 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.