Since the angles $MIA$ and $INA$ are right angles, quadrilateral $A\! M\! I\! N$ is cyclic.
[asy]
unitsize(2 cm);
pair A, B, C, I, M, N;
B = (0,0);
A = (8,0);
C = intersectionpoint(arc(B,5,0,180),arc(A,5,0,180));
I = incenter(A,B,C);
M = extension(I,I + (B - C),A,B);
N = extension(I,I + (B - C),A,C);
draw(M--I--N);
draw(A--B--C--cycle);
draw(arc(I,I + (B - C),I + (B - A)));
draw(arc(I,I + (C - B),I + (C - A)));
label("$A$", A, S);
label("$B$", B, SW);
label("$C$", C, SE);
label("$I$", I, NE);
label("$M$", M, NW);
label("$N$", N, NE);
[/asy]
As in Solution 1, let $r$ be the inradius. Then
\[[AMN] = \frac{1}{2} AM \cdot AN \cdot \sin \angle MAN = \frac{1}{2} \left( (5 - r) \cdot (5 - r) \cdot \frac{4r}{rn} \right),\]where $n$ is the angle bisector of $\angle A$.
By the Angle Bisector Theorem, $\frac{AN}{NC} = 5:3$, so $AN = \frac{25}{8}$ and $NC = \frac{15}{8}$.
By the Law of Sines on triangle $AMN,$
\[\frac{AN}{\sin \angle MAN} = \frac{AM}{\sin \angle NAM}.\]Substituting $\frac{SN}{8} = \sin \angle MAN$ and $\frac{SM}{8} = \sin \angle NAM,$ we get
\[\frac{25}{8 \cdot \frac{5r}{6}} = \frac{5 - r}{8 \cdot \frac{r}{6}}.\]Therefore, $19 - 12r + r^2 = 0$, which factors as $(r - 3)(r - 16) = 0$. Since $r$ is positive, $r = 3$.
Note: The quadratic is the result of $a (s - a) + a (t - a) = nt$, where $s + t = 8$ and $a$ is the inradius; see Solution 1 for more detail. Similarly, $a (s - a) = bs$.
Thus, $AM = 2$, $AN = \frac{25}{8}$ and
\[[AMN] = \boxed{\frac{75}{32}}.\]
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 area of triangle $AMN$.
1 answer