8. For $\triangle A B C$ with side lengths $a, b, c(a \leqslant b \leqslant c)$, the following conditions are satisfied: (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. The number of all possible triples $(a, b, c)$ is $\qquad$
8. 10 Detailed Explanation: According to the problem, positive integers $a, b, c$ satisfy $a \leqslant b \leqslant c, a+b>c, b^{2}=a c$, and at least one of $a, c$ is 100. (i) If $a=100$, then $b^{2}=100 c$, so $10 \mid b$. We can take $100+b>c=\frac{b^{2}}{100}, b^{2}-100 b-100^{2}100, b^{2}+100 b-100^{2}>0$, hence $50(\sqrt{5}-1)<b \leqslant$ 100. Noting that $10 \mid b$, then $b$ can be $70,80,90,100$, and the corresponding $a=49,64,81,100$. In summary, the triplet $(a, b, c)$ has 10 possible solutions: $(49,70,100),(64,80,100),(81,90,100),(100,100,100),(100,110,121),(100,120$, $144),(100,130,169),(100,140,196),(100,150,225),(100,160,256)$.