Question

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?

I have no idea what to do. Someone please help!

Answer 1

Let the sides of the triangle be a, b, and c, and let the altitudes to these sides be h_A = 12, h_B = 14, and h_C, respectively. Let K denote the area of the triangle. Then

1/2 ah_A = 1/2 bh_B = 1/2 ch_C = K

so

a = 2K/h_A, b = 2K/h_B, c = 2K/h_C

By the triangle inequality,

a + b > c
a + c > b
b + c > a

Substituting our formulas above, 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

which simplify to

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

Substituting 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

From the first inequality,

1/h_C < 1/12+1/14=13/84
so h_C > 84/13.

Since 1/12 > 1/14, the second inequality is always satisfied.

And from the third inequality,

1/h_C >1/12 - 1/14 = 1/84
so h_C < 84.

Therefore, the largest possible value of h_C is 83.

Answer 2

Thanks!