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?

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.

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step-by-step explanation:

answer: approximately 206 bones

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