9. The side lengths $a, b, c (a \leqslant b \leqslant c)$ of $\triangle ABC$ simultaneously satisfy the following three conditions:
(1) $a, b, c$ are all integers; (2) $a, b, c$ form a geometric sequence; (3) at least one of $a$ and $c$ equals 100. Find all possible solutions for the triplet $(a, b, c)$.
9. From the problem, we have $a \leqslant b \leqslant c, a+b>c, b^{2}=a c$, and at least one of $a, c$ is 100.
(1) If $a=100$, then $b^{2}=100 c$, so $10 \mid b$. Also, $100+b>c=\frac{b^{2}}{100}, b^{2}-100 b-100^{2}100, b^{2}+100 b-100^{2}>0$, which gives $50(\sqrt{5}-1)<b \leqslant 100$. Since $10 \mid b$, then $b=70,80,90,100$. Correspondingly, $a=49,64,81,100$.

In summary, there are 10 possible solutions for the triplet $(a, b, c)$: $(49,70,100),(64,80,100),(81,90,100),(100,100,100),(100,110,121),(100,120,144),(100,130,169),(100,140,196),(100,150,225),(100,160,256)$.