31 In a triangle $A B C$, the length of the altitudes $A D$ and $B E$ are 4 and 12 respectively. Find the largest possible integer value for the length of the third altitude $C F$.
31 Answer: (5)
Area of the triangle is $\frac{1}{2} \times A D \times B C=\frac{1}{2} \times A C \times B E=\frac{1}{2} \times A B \times C F$.
Since $A D=4$ and $B E=12, B C: A C=3: 1$. Let $A C=x$, then $B C=3 x$.
Using Triangle Inequality, $A B<A C+B C$ and $B C<A B+A C \Rightarrow 2 x<A B<4 x$.
From $C F=\frac{12 x}{A B}, 3<C F<6 \Rightarrow$ The largest integer value for $C F$ is 5 .