In a triangle, the two heights are 12 and 20. Find the maximum possible integer length of the third height.

The solution must begin with the argument that the area of ​​a triangle is equal to the half-product of its side by the height lowered to this side.
Let the sides of the triangle be equal to a, b, c. And the height lowered to the third side will be denoted by x. Then the equality is true: 12 * a = 20 * b = x * c.
The maximum height x will be provided that side c is minimal, but side c is connected to the other sides by inequality: c <a + b. 12 * a = 20 * b, b = 3 \ 5 * a, Next, we write other triangle inequalities: a <b + c, a <3 \ 5 * a + c,
2 \ 5 * a <c. Hence, the minimum c = 2 \ 5 * a Then we find the maximum x.
12 * a = 2 \ 5 * a * x.
x max = 6 * 5 = 30