In triangle $ABC$, let $I$ be the incenter of triangle $ABC$. The line through $I$ parallel to $BC$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. If $AB = 5$, $AC = 5$, and $BC = 8$, then find the perimeter of triangle $AMN$.

Answer 1

Let $s$ be the semiperimeter of triangle $ABC$, so $s = \frac{AB + AC + BC}{2} = 9$. As shown below, let $D$ and $E$ be the feet of the perpendiculars from $I$ to $AB$ and $AC$, respectively, and let $F$ be the foot of the perpendicular from $D$ to $AC$. Similarly, let $G$ be the foot of the perpendicular from $E$ to $AB$. [asy]
unitsize(1 cm);

pair A, B, C, D, E, F, G, I, M, N;

A = (0,0);
B = (3,4);
C = (10,0);
I = incenter(A,B,C);
D = (A + I)/2;
E = (C + I)/2;
F = (2*D + I)/3;
G = (2*E + I)/3;
M = extension(A,B,I,I + (10,8)/3);
N = extension(A,C,I,I + (10,8)/3);

draw(A--B--C--cycle);
draw(A--I--C);
draw(D--F);
draw(E--G);

label("$A$", A, S);
label("$B$", B, N);
label("$C$", C, NE);
label("$D$", D, NW);
label("$E$", E, NE);
label("$F$", F, S);
label("$G$", G, SW);
label("$I$", I, NW);
label("$M$", M, N);
label("$N$", N, NE);
[/asy]

Since $\angle BID$ and $\angle CIE$ are right angles, quadrilaterals $DIIM$ and $EIIN$ are cyclic. We can find the radius of the circumcircle of triangle $DIIM$ by drawing the perpendicular bisectors of $IM$ and $ID$, as shown below. We can find the radius of the circumcircle of cyclic quadrilateral $DIIM$ by drawing the perpendicular from the center of this circumcircle to $IM$, as shown below. [asy]
unitsize(1 cm);

pair D, I, M, O, P;

D = (0,0);
M = (1,0);
I = (1,0.8);
O = (1/2,0);
P = (1,0);

draw(Circle(O,1/2));
draw(M--P);

label("$D$", D, SW);
label("$M$", M, S);
label("$I$", I, N);
label("$O$", O, SW);
[/asy]

Since angle $MDP$ is right, we have $MP = \frac{DI}{2} = \frac{IN}{2}$, so the radius of the circumcircle of triangle $DIIM$ is $\frac{DN}{2}$. Similarly, the radius of the circumcircle of triangle $EIIN$ is $\frac{DM}{2}$. It follows that triangles $DEM$ and $DEN$ are 8-15-17 triangles, so triangles $DMN$ and $DEN$ are 3-4-5 triangles, as shown below. [asy]
unitsize(1 cm);

pair A, B, C, D, E, F, N, L, M;

A = (0,0);
B = (3,4);
C = (8,0);
D = (A + B)/2;
E = (A + C)/2;
F = (B + C)/2;
N = extension(B,C,D,D + 5*dir(F - D));
L = extension(B,C,E,E + 4*dir(F - E));
M = extension(A,L,B,C);

draw(A--B--C--cycle);
draw(A--D);
draw(A--E);
draw(D--M);
draw(E--N);

label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, NW);
label("$E$", E, S);
label("$F$", F, SE);
label("$N$", N, NE);
label("$L$", L, N);
label("$M$", M, S);
label("$s - b = \frac{9}{2} - \frac{5}{2} = 2$", (B + D)/2, NW);
label("$s - c = \frac{9}{2} - \frac{4}{2} = 2.5$", (C + E)/2, SW);
label("$2.5$", (E + L)/2, SW);
label("$2 + 2.5 = 4.5$", (M + N)/2, NE);
[/asy] Hence, the perimeter of triangle $AMN$ is $AM + MN + NA = \boxed{4.5}$.

Notice that $4.5$ is small compared to $18$, which is the perimeter of triangle $ABC$. This is because $MN$ is parallel to $BC$, so triangle $AMN$ is similar to triangle $ABC$.