1982 AMC 12 第 10 题

Because $O$ is the incenter, its distance from $BC$ is the inradius

Compare the altitude of $\triangle AMN$ with the altitude of $\triangle ABC$

The sides $12,$ $18,$ $24$ have semiperimeter $27$ and area $$\sqrt{27\cdot15\cdot9\cdot3}=27\sqrt{15},$$ so the inradius is $r=\sqrt{15}.$ Taking $BC$ as base, the altitude is $h=\frac{2(27\sqrt{15})}{24}=\frac{9\sqrt{15}}{4}.$ Thus the similarity scale from $\triangle ABC$ to $\triangle AMN$ is $1-\frac{r}{h}=1-\frac{4}{9}=\frac{5}{9}.$ Its perimeter is $(\frac{5}{9})(12+18+24)=30.$

Therefore, the correct answer is A.