In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$.  The line through $I$ parallel to $BC$ intersects $AB$ and $AC$ at $M$ and $N$, respectively.  If $AB = 17$, $AC = 24$, and $BC = 33$, then find the perimeter of triangle $AMN$.
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 $NI = NC$.

[asy]
import geometry;

unitsize(1 cm);

pair A, B, C, I, M, N;

A = (1,3);
B = (0,0);
C = (4,0);
I = incenter(A,B,C);
M = extension(I, I + B - C, A, B);
N = extension(I, I + B - C, A, C);

draw(A--B--C--cycle);
draw(B--I--C);
draw(M--N);

label("$A$", A, dir(90));
label("$B$", B, SW);
label("$C$", C, SE);
label("$I$", I, dir(90));
label("$M$", M, NW);
label("$N$", N, NE);
[/asy]

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