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???
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.
Certainly! To find the longest possible length of the third altitude, we can start by understanding the concept of altitudes in a triangle.
In a triangle, an altitude is a line segment drawn from a vertex perpendicular to the opposite side. It divides the triangle into two right-angled triangles.
In this case, we are given two altitudes with lengths 12 and 14. Let's assume these two altitudes are drawn from vertices A and B, respectively, and they are perpendicular to the opposite sides.
To find the longest possible length for the third altitude, we need to consider the relationship between the altitudes and sides of a triangle.
The length of an altitude is related to the lengths of the sides it connects. More specifically, the length of an altitude is inversely proportional to the length of the opposite side. This means that as the length of the side increases, the length of the altitude decreases, and vice versa.
Now, let's consider the side opposite the third altitude. Since we are looking for the longest possible length for the third altitude, we need to minimize the length of the side opposite it.
To do this, we can assume that the third altitude is drawn from vertex C and is perpendicular to the opposite side. In order to minimize the length of the side opposite it, we want this side to be as small as possible.
One way to achieve this is by making the triangle equilateral, where all sides are equal in length. In an equilateral triangle, each angle is 60 degrees, and each altitude is also a median, angle bisector, and perpendicular bisector.
Therefore, in an equilateral triangle, the length of each altitude is the same as the length of each side. This means that in our case, the length of the third altitude would be the same as the lengths of the first two altitudes, which are 12 and 14.
So, the longest possible length of the third altitude is 12 or 14, depending on the lengths of the first two altitudes.