17-114 Among the following sets of numbers, which set cannot be the lengths of the altitudes of a triangle
(A) $1, \sqrt{3}, 2$.
(B) $3,4,5$.
(C) $5,12,13$.
(D) $7,8, \sqrt{113}$.
(E) $8,15,17$.
(41st American High School Mathematics Examination, 1990)
[Solution 1] If $h_{a}, h_{b}, h_{c}$ are the altitudes of $\triangle A B C$ on sides $a, b, c$ respectively, then the area of the triangle is
$$
S_{\triangle A B C}=\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=\frac{1}{2} c h_{c},
$$

which means
$$
a h_{a}=b h_{b}=c h_{c}.
$$

Thus, $a: b: c=h_{b} h_{c}: h_{a} h_{c}: h_{a} h_{b}$. 
For option (C), $a: b: c=12 \cdot 13: 5 \cdot 13: 5 \cdot 12$.
Noting that $12 \cdot 13 > 5 \cdot 13 + 5 \cdot 12$, the ratio of $a, b, c$ cannot form a triangle. This means that $5, 12, 13$ cannot be the three altitudes of a triangle. (It is easy to verify that the other four groups can)
Therefore, the answer is (C).
[Solution 2] Similarly, we can have $a: b: c=\frac{1}{h_{a}}: \frac{1}{h_{b}}: \frac{1}{h_{c}}$, then for (C), we have $a: b: c=\frac{1}{5}: \frac{1}{12}: \frac{1}{13}$, and $\frac{1}{5}=\frac{1}{10}+\frac{1}{10}>\frac{1}{12}+\frac{1}{13}$,
$\therefore$ this set of numbers $(5,12,13)$ cannot be the three altitudes of a triangle. Hence, the answer is (C).