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?
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To find the longest possible length of the third altitude of the triangle, we need to consider the properties of an altitude of a triangle.
An altitude of a triangle is a line segment drawn from a vertex of the triangle to the opposite side and is perpendicular to that side. The length of an altitude can be calculated using the formula:
Altitude = (2 * Area) / Base
where Area is the area of the triangle and Base is the length of the corresponding base.
Since we are given the lengths of two altitudes (12 and 14), we can calculate the areas of two triangles using these altitudes as bases. Let's refer to these as Triangle A and Triangle B.
Area of Triangle A = (12 * Base A) / 2 = 6 * Base A
Area of Triangle B = (14 * Base B) / 2 = 7 * Base B
To find the upper bound for the length of the third altitude, we need to maximize the sum of Base A and Base B. In a triangle, the sum of any two sides must be greater than the third side. Therefore, we can conclude that:
Base A + Base B > Largest Side
Since we know that the longest possible length of the third altitude is a positive integer, we can assume that the longest side of the triangle is 1 less than the sum of the lengths of the two given altitudes (12 + 14 - 1 = 25).
Now, to maximize the sum of Base A and Base B, we need to distribute the length of the longest side (25) between these two sides as evenly as possible. Therefore, the longest possible length of the third altitude is the largest integer less than the average of the two given altitudes, which is:
(12 + 14 - 1) / 2 = 25 / 2 = 12.5
Since the length of the third altitude must be a positive integer, the longest possible length is 12.
So, the longest possible length of the third altitude, if it is a positive integer, is 12.