jmc

Since $MN$ is parallel to $BC$, $\angle MIB=\angle IBC$. But $BI$ is an angle bisector, so $\angle IBC=\angle IBM$. Hence, triangle $MIB$ is isosceles with $MI=MB$. By the same argument, triangle $NIC$ is isosceles, with $NI=NC$.



Therefore, the perimeter of triangle $AMN$ is simply $$\begin{array}{rl}AM+AN+MN& =AM+AN+MI+NI\\ & =AM+AN+MB+NC\\ & =(AM+MB)+(AN+NC)\\ & =AB+AC\\ & =17+24\\ & =\boxed{41}.\end{array}$$