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 $AMN$.
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
\[[AMN] = \boxed{\frac{75}{32}}.\]
To find the area of triangle $AMN$, first we need to find the lengths of $AM$ and $AN$. Let's start by finding the length of $AM$.
Since $I$ is the incenter of triangle $ABC$, it lies on the angle bisectors of angles $B$ and $C$. Thus, $\angle BMI = \angle CMI = \angle NAC$.
Since $BC$ is parallel to $MN$, we have $\angle BMI = \angle MIN$. Therefore, $\angle MIN = \angle NAC$.
By alternate interior angles, we conclude that $\triangle MIN \sim \triangle NAC$. Therefore,
\[\frac{MI}{IA} = \frac{IN}{NC}.\]Since $MI = IN$ and $IA = IC$, we have
\[\frac{MI}{IC} = \frac{IN}{NC}.\]Since $\angle MIC$ is a right angle, by Similarity (AA Similarity) in triangle $MIC$ and triangle $NIC$, we get
\[\triangle MIC \sim \triangle NIC.\]
By the Ratio Lemma (the angle bisector theorem),
\[\frac{MI}{IC} = \frac{5}{13}.\]Then $MI = \frac{40}{13}$ and $IC = \frac{104}{13}$.
Also, $\triangle MIN$ and $\triangle BIA$ are similar.
By the Ratio Lemma in triangle $BIA$,
\[\frac{BI}{IA} = \frac{5}{13}.\]Notice $BI = \frac{40}{13}$.
Hence $\triangle BIA$ and $\triangle MIN$ are similar by AAS. Therefore,
\[\frac{AI}{IM} = \frac{BI}{IN}.\]or
\[\frac{10}{\displaystyle \frac{40}{13}} = \frac{\displaystyle \frac{40}{13}}{IN}.\]Simplifying, we get
\[10 \cdot \frac{13}{40} = \frac{40}{13} \cdot \frac{40}{13}.\]Solving this quadratic equation, we find $IN = \frac{160}{169}$.
Finally, $AM = AI + IM = \frac{26}{13}$.
The area of triangle $AMN$ is then
\[\frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(5)(\frac{160}{169}) = \boxed{\frac{400}{169}}.\]
To find the area of triangle $AMN,$ we can first find the length of $MN$ and then use the fact that the area of a triangle is half the product of its base and height.
Given that $ABC$ is a triangle with $AB = 5,$ $AC = 5,$ and $BC = 8,$ we can start by finding the length of $MN.$
Since $MN$ is parallel to $BC,$ we can use similar triangles to find its length. Notice that triangles $ABC$ and $ANM$ are similar by the $AA$ similarity criterion, as $\angle MNA = \angle BCA$ and $\angle NAM = \angle CAB.$
Thus, we have $\triangle ABC \sim \triangle ANM.$ This gives us the ratio of corresponding side lengths as $\frac{AB}{AN} = \frac{BC}{NM} = \frac{CA}{AM},$ or $\frac{5}{AN} = \frac{8}{NM} = \frac{5}{AM}.$
Since $A,$ $I,$ and $M$ are collinear, and $MN$ is parallel to $BC,$ we can also observe that triangles $AMI$ and $ABC$ are similar.
Hence, we have $\frac{AM}{AB} = \frac{AI}{AC},$ or $\frac{AM}{5} = \frac{r}{5},$ where $r$ is the inradius of triangle $ABC.$ Since the inradius of a triangle can be expressed as $r = \frac{A}{s},$ where $A$ is the area of the triangle and $s$ is the semi-perimeter, we can rewrite this as $\frac{AM}{5} = \frac{A}{5s},$ or $AM = \frac{A}{s}.$
Similarly, from the similarity of triangles $ABC$ and $ANM,$ we have $\frac{AN}{AB} = \frac{AI}{AC},$ or $\frac{AN}{5} = \frac{r}{5},$ which can be rewritten as $AN = \frac{A}{s}.$
Adding these equations, we get $AM + AN = \frac{2A}{s}.$ Since $AM + AN = MN,$ we have $MN = \frac{2A}{s}.$
Now, to find the area of triangle $AMN,$ we can substitute the given side lengths of triangle $ABC$ into Heron's formula to get the semi-perimeter $s,$ and substitute that into our expression for $MN.$
Heron's formula states that the area of a triangle with side lengths $a,$ $b,$ and $c$ is given by $A = \sqrt{s(s-a)(s-b)(s-c)},$ where $s$ is the semi-perimeter.
In this case, $a = BC = 8,$ $b = CA = 5,$ and $c = AB = 5.$ The semi-perimeter $s$ is thus $\frac{8 + 5 + 5}{2} = 9.$
Substituting these values into our expression for $MN,$ we have $MN = \frac{2A}{9}.$
Finally, substituting $MN$ and $s$ into the formula for the area of a triangle, we get $A_{AMN} = \frac{1}{2}(MN)(s) = \frac{1}{2} \left(\frac{2A}{9}\right) \left(9\right) = \boxed{A}.$