geometry

Let's apply the formula for the area of a triangle using its altitudes. Let A be the area of triangle ABC, and let h be the length of altitude CF. Then we have:

A = (1/2) * BC * AD = (1/2) * AC * BE = (1/2) * AB * h

Equating these expressions for A, we get:

BC * AD = AC * BE = AB * h

Substituting AD = 12 and BE = 18, and using the fact that BC = AB - AC, we can write:

(AB - AC) * 12 = AC * 18

Simplifying this expression, we get:

6AB = 7AC

Since AB and AC are positive integers, this implies that AC is a multiple of 6. Let AC = 6k for some positive integer k. Then we have:

AB = (7/6) * AC = (7/6) * 6k = 7k

Now let's consider the altitude CF. Using the formula we derived earlier, we have:

h = (2A) / AB = (AB * BC) / AB = BC

Since BC is also an altitude of triangle ABC, we have:

h <= min{AD, BE, CF} = min{12, 18, CF} = 12

Therefore, the largest possible value of CF is 12 12. We can achieve this value by constructing triangle ABC to be a right triangle with legs of length 7 and 24, so that CF (the hypotenuse) has length 25, which is a multiple of 5 and therefore meets the condition of being a multiple of 15.

Thus, the largest possible value of CF is 12.

Since the greatest altitude = 18, therefore:

 

18  x  2  -  1 = 35 - the largest possible value of CF

The altitude of a triangle is the perpendicular line segment from a vertex to the opposite side. The three altitudes of a triangle always intersect at a point called the orthocenter.

The orthocenter of a triangle is always inside the triangle if and only if the triangle is acute. In an acute triangle, the three altitudes are shorter than the sides of the triangle.

In this case, we are given that AD = 12 and BE = 18. Since AD and BE are the altitudes of an acute triangle, we know that CF must be less than or equal to 12 + 18 = 30.

The largest possible value of CF is 30, which occurs when the triangle is equilateral. In an equilateral triangle, all three sides are congruent and all three angles are 60 degrees. The altitudes of an equilateral triangle are also congruent, so CF = 30.

However, it is also possible for CF to be less than 30. For example, if the triangle is right isosceles, then CF = 15.

Therefore, the largest possible value of CF is 20​.

(1/12  +  1/18  +  1/CF) / 2 > 1/12, solve for CF

 

CF = 0 < CF < 36

 

Therefore CF(max) = 35

 

With CF =35, you have the maximum area of the triangle ABC =1,100.179 and sides:62.867,  122.242,  183.363.

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With CF=20, the maximum area of the triangle ABC=185.631 and sides: 18.563,  20.626,  30.939.