Example 5 (1999 Shanghai Mathematics Competition) The side lengths $a, b, c (a \leqslant b \leqslant c)$ of $\triangle A B C$ simultaneously satisfy the following three conditions:
( i ) $a, b, c$ are all integers;
(ii) $a, b, c$ form a geometric sequence;
(iii) at least one of $a$ and $c$ equals 100.
Find all possible solutions for the triplet $(a, b, c)$.
Solution: According to the problem, the positive integers $a, b, c$ satisfy $\left\{\begin{array}{l}a \leqslant b \leqslant c \\ a+b>c \\ b^{2}=a c \\ a \text { or } c \text { is at least one } 100\end{array}\right.$

(1) If $a=100$, then $b^{2}=100 c$, so $10 \mid b$.

Since $100+b>c=\frac{b^{2}}{100}, b^{2}-100 b-100^{2}100, b^{2}+100 b-100^{2}>0$.
Therefore, $50(\sqrt{5}-1)<b \leqslant 100$.
Since it has been proven that $10 \mid b$, $b$ can take $70,80,90,100$, and accordingly $a=49,64,81,100$.
In summary, the triplet $(a, b, c)$ has 10 possible solutions:
$(49,70,100),(64,80,100),(81,90,100),(100,100,100),(100,110,121)$, $(100,120,144),(100,130,169),(100,140,196),(100,150,225),(100,160,256)$.

Analysis: The discussion here is not complicated. If it were changed to a fill-in-the-blank question asking for the number of solutions, the difficulty would be slightly higher, as inattention could lead to missing solutions.