Problem:
Which of these triples could not be the lengths of the three altitudes of a triangle?
Answer Choices:
A. 1,3,2
B. 3,4,5
C. 5,12,13
D. 7,8,113
E. 8,15,17
Solution:
Let x,y and z denote the sides of a triangle, hx,hy and hz the corresponding altitudes, and A the area. Since xhx=yhy=zhz=2A, the sides are inversely proportional to the altitudes. If x,y and z form a triangle with largest side x, then x<y+z. Thus
hx2A<hy2A+hz2A or hx1<hy1+hz1(*)
Only triple (C) fails to satisfy (∗). To show that the other four choices (a,b,c) do correspond to possible triangles, just build a triangle T with sides a1,b1 and c1. The altitudes of T are in the ratio a:b:c, so some triangle similar to Tb has altitudes a,b and c.