$32 \cdot 60$ In a scalene $\triangle ABC$, two of the altitudes have lengths 4 and 12. If the third altitude also has an integer length, then the largest it can be is
(A) 4.
(B) 5.
(C) 6.
(D) 7.
(E) None of the above.
(37th American High School Mathematics Examination, 1986)
［Solution 1］Let $AB = c$. By the triangle area formula, we have
$12c = 4 \cdot AC$, which means $AC = 3c$.
$$
\because \quad AC - AB \frac{1}{h_{c}} \quad (here \, h_{a} \geqslant h_{b} \geqslant h_{c} > 0).
$$
Thus, we have
$$
\left\{\begin{array}{l}
\frac{1}{4} < \frac{1}{12} + \frac{1}{h} \\
\frac{1}{h} < \frac{1}{4} + \frac{1}{12}
\end{array}\right.
$$

Solving this, we get $3 < h < 6$. Since $h$ is an integer, then $h_{\max} = 5$.
Therefore, the answer is $(B)$.
[Solution 3] Let the length of $BC$ be $a$, and the height on this side be $h$, with the area of $\triangle ABC$ being $s$. From the given conditions,
we have $AB = \frac{4}{\sin A}, \quad AC = \frac{12}{\sin A}$,
and by the cosine rule,
$$
\begin{array}{l}
a^{2} = \frac{16}{\sin^{2} A} + \frac{144}{\sin^{2} A} - \frac{96 \cos A}{\sin^{2} A} = \frac{160 - 96 \cos A}{\sin^{2} A}. \\
\because \quad 2s = \frac{48}{\sin A} = a h. \\
\therefore \quad h^{2} = \frac{48^{2}}{a^{2} \sin^{2} A} = \frac{48^{2}}{160 - 96 \cos A} = \frac{72}{5 - 3 \cos A} < \frac{72}{5 - 3} = 36, \\
\end{array}
$$

Thus, $h < 6$, and $h \neq 4$ (the triangle is not isosceles),
$$
\therefore \quad h = 5.
$$

Therefore, the answer is $(B)$.