In $\triangle ABC$ with integer side lengths, $\cos A=\frac34,\cos B=\frac9{16},\cos C=\frac18$. What is the least possible perimeter?

1. The angles determine the triangle up to similarity, and by the law of sines the sides are proportional to $\sin A,\sin B,\sin C$.
2. A $4$-$5$-$6$ triangle has exactly these cosines: $\cos A=\frac{5^2+6^2-4^2}{2\cdot5\cdot6}=\frac34$, $\cos B=\frac{4^2+6^2-5^2}{2\cdot4\cdot6}=\frac9{16}$, $\cos C=\frac{4^2+5^2-6^2}{2\cdot4\cdot5}=\frac18$. So the sides are in the ratio $4:5:6$.
3. The least perimeter (already in lowest terms) is $4+5+6=\boxed{15}$.