have been looking at your post ...
http://www.jiskha.com/display.cgi?id=1357047578
interesting question...
let the altitude of 12 hit base a
let the altitude of 14 hit base b
then (1/2)(12)a = (1/2)(14)b
6a = 7b
a = 7b/6
let the third side be c and its altitude be x
from the properties of the sides of a triangle
c < a+b
c < 7b/6 + b
c < 13b/6
but by area property
cx = 7b or cx = 6a
take cx = 7b
the absolute maximum value that c could be is a number just a "smidgeon" less than 13b/6 , which is positive , so divide both sides of
cx = 7b by that
cx/c < 7b / (13b/6)
c < 42/13
since you want the largest integer, that would be 3
( I am a little uneasy about my last division of the inequality)
me too. You are dividing by c, which is less than 13b/6. The result will be larger than 42/13. What do you think of my attempt on the question as posted earlier?
thanks a lot guys for your interest. Yes, it is somewhat of a confusing problem. @Steve: I appreciated it a lot and you did a very good job at explaining it.
Thanks a lot guys and Happy New year. take care!
In a right triangle, if the legs (altitudes are 12 and 14), then
if we pick an angle θ and let h be the altitude to the hypotenuse,
h/14 = sinθ
h/12 = cosθ
so h^2/144 + h^2/196 = 1, and h =~ 9.1
So, Reiny's estimate is way low. Now the question is, does a right triangle provide maximum 3rd altitude h?
More thought needed.
Take a look at
http://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.577996.html
where this very question is discussed.
Just a final note. Since the 3rd altitude h is bounded by
1/a + 1/b <= 1/h <= 1/a - 1/b
where a < b,
the limiting case where a=b (an isosceles triangle) shows that
1/a+1/a <= 1/h <= 1/a - 1/a
2/a <= 1/h <= 0
shows that h >= a/2, and may grow without bound.
In the given problem, we are dealing with a triangle with altitudes and bases. To find the largest possible integer value of c, we need to understand the relationship between the altitudes and the bases.
First, let's assign variables to the altitudes and bases of the triangle. Let the altitude of the side with length 12 be labeled as a, and the altitude of the side with length 14 be labeled as b.
According to the properties of triangles, the area of a triangle is given by the formula: (1/2)(base)(altitude). Since the triangle has two different bases, the areas on both sides of the triangle must be equal.
So, we have (1/2)(12)a = (1/2)(14)b. Simplifying this equation, we get 6a = 7b.
To find the value of a in terms of b, we divide both sides of the equation by 6, resulting in a = 7b/6.
Now, let's introduce the third side of the triangle with length c and its altitude x.
From the properties of triangles, we know that the sum of the lengths of any two sides of a triangle must always be greater than the length of the third side. Therefore, we have c < a + b.
Substituting the value of a we found earlier, we get c < 7b/6 + b. Simplifying this inequality, we get c < 13b/6.
However, we also know that the area of the triangle formed by the third side and its altitude is equal to 7b (from the given problem). So another relationship can be written as cx = 7b or cx = 6a.
If we consider cx = 7b, we can find an upper bound for the value of c. Dividing both sides of the inequality cx < 7b by 13b/6, we get c < 42/13.
Since we are looking for the largest possible integer value for c, we take the integer part of 42/13, which is 3.
Therefore, our conclusion is that the largest possible integer value for c is 3.