The solution would be like this for this specific problem:
h_A = 12, h_B = 14, and h_C
1/2 ah_A = 1/2 bh_B = 1/2 ch_C = K
a= 2K/h_A, b = 2K/h_B, c = 2K/h_C
a + b > c a + c > b b + c > a
After substituting, we get:
2K/h_A + 2K/h_B > 2K/h_C 2K/h_A + 2K/h_C > 2K/h_B 2K/h_B + 2K/h_C > 2K/h_A
and after simplifying we get:
1/h_A + 1/h_B > 1/h_C 1/h_A + 1/h_C > 1/h_B 1/h_B + 1/h_C > 1/h_A
h_A = 12 and h_B = 14, we get 1/12 + 1/14 > 1/h_C 1/12 + 1/h_C > /14 1/14 + 1/h_C > 1/12
1st inequality: 1/h_C < 1/12+1/14=13/84 so h_C > 84/13. Since 1/12 > 1/14, 2nd inequality is always satisfied.
3rd inequality:1/h_C >1/12 - 1/14 = 1/84 so h_C < 84
Therefore, the longest possible length of the third altitude, if it is a positive integer is h_C is 83.