Let $ABC$ be a triangle and $O$ be its circumcircle. Let $A', B', C'$ be the midpoints of minor arcs $AB$, $BC$ and $CA$ respectively. Let $I$ be the center of incircle of $ABC$. If $AB = 13$, $BC = 14$ and $AC = 15$, what is the area of the hexagon $AA'BB'CC'$?
Suppose $m \angle BAC = \alpha$ , $m \angle CBA = \beta$, and $m \angle ACB = \gamma$.


[b]p10.[/b] Let the incircle of $ABC$ be tangent to $AB, BC$, and $AC$ at $J, K, L$, respectively. Compute the angles of triangles $JKL$ and $A'B'C'$ in terms of $\alpha$, $\beta$, and $\gamma$, and conclude that these two triangles are similar. 


[b]p11.[/b] Show that triangle $AA'C'$ is congruent to triangle $IA'C'$. Show that $AA'BB'CC'$ has twice the area of $A'B'C'$. 

[b]p12.[/b] Let $r = JL/A'C'$ and the area of triangle $JKL$ be $S$. Using the previous parts, determine the area of hexagon $AA'BB'CC'$ in terms of $ r$ and $S$. 

[b]p13.[/b] Given that the circumradius of triangle $ABC$ is $65/8$ and that $S = 1344/65$, compute $ r$ and the exact value of the area of hexagon $AA'BB'CC'$.

PS. You had better use hide for answers.
1. **Determine the angles of triangles \(JKL\) and \(A'B'C'\) in terms of \(\alpha\), \(\beta\), and \(\gamma\):**

   - For triangle \(A'B'C'\):
     \[
     \angle A'B'C' = \frac{\angle B + \angle C}{2} = \frac{\beta + \gamma}{2}
     \]
     \[
     \angle B'A'C' = \frac{\angle A + \angle B}{2} = \frac{\alpha + \beta}{2}
     \]
     \[
     \angle B'C'A' = \frac{\angle A + \angle C}{2} = \frac{\alpha + \gamma}{2}
     \]

   - For triangle \(JKL\):
     \[
     \angle JKL = \frac{\beta + \gamma}{2}
     \]
     \[
     \angle JLK = \frac{\alpha + \gamma}{2}
     \]
     \[
     \angle KJL = \frac{\alpha + \beta}{2}
     \]

   Since the angles of triangles \(A'B'C'\) and \(JKL\) are the same, the two triangles are similar by the AA (Angle-Angle) similarity criterion.

2. **Show that triangle \(AA'C'\) is congruent to triangle \(IA'C'\):**

   - Since \(A'\) and \(C'\) are midpoints of the arcs \(AB\) and \(CA\) respectively, they lie on the perpendicular bisectors of \(AB\) and \(CA\). Therefore, \(A'\) and \(C'\) are equidistant from \(A\) and \(B\), and from \(C\) and \(A\) respectively.
   - The line segment \(A'I\) is the angle bisector of \(\angle BAC\), and similarly, \(C'I\) is the angle bisector of \(\angle ACB\).
   - Since \(A'\) and \(C'\) are midpoints of the arcs, \(A'\) and \(C'\) are also equidistant from \(I\), the incenter of \(\triangle ABC\).
   - Therefore, \(\triangle AA'C'\) is congruent to \(\triangle IA'C'\) by the SAS (Side-Angle-Side) congruence criterion.

3. **Show that hexagon \(AA'BB'CC'\) has twice the area of \(\triangle A'B'C'\):**

   - Since \(\triangle AA'C'\) is congruent to \(\triangle IA'C'\), the area of \(\triangle AA'C'\) is equal to the area of \(\triangle IA'C'\).
   - Similarly, \(\triangle BB'A'\) is congruent to \(\triangle IB'A'\), and \(\triangle CC'B'\) is congruent to \(\triangle IC'B'\).
   - Therefore, the area of hexagon \(AA'BB'CC'\) is twice the area of \(\triangle A'B'C'\).

4. **Determine the area of hexagon \(AA'BB'CC'\) in terms of \(r\) and \(S\):**

   - Let \(r = \frac{JL}{A'C'}\) and the area of \(\triangle JKL\) be \(S\).
   - Since \(\triangle JKL\) is similar to \(\triangle A'B'C'\), the ratio of their areas is the square of the ratio of their corresponding sides:
     \[
     \frac{\text{Area of } \triangle JKL}{\text{Area of } \triangle A'B'C'} = r^2
     \]
     \[
     S = r^2 \times \text{Area of } \triangle A'B'C'
     \]
     \[
     \text{Area of } \triangle A'B'C' = \frac{S}{r^2}
     \]
   - The area of hexagon \(AA'BB'CC'\) is twice the area of \(\triangle A'B'C'\):
     \[
     \text{Area of hexagon } AA'BB'CC' = 2 \times \frac{S}{r^2}
     \]

5. **Given that the circumradius of \(\triangle ABC\) is \(\frac{65}{8}\) and that \(S = \frac{1344}{65}\), compute \(r\) and the exact value of the area of hexagon \(AA'BB'CC'\):**

   - Using the given circumradius \(R = \frac{65}{8}\), we can find the side length \(A'C'\) of \(\triangle A'B'C'\) using the formula for the circumradius of a triangle:
     \[
     A'C' = 2R \sin\left(\frac{\alpha}{2}\right)
     \]
   - Given \(S = \frac{1344}{65}\), we can solve for \(r\):
     \[
     r = \frac{JL}{A'C'}
     \]
   - The exact value of the area of hexagon \(AA'BB'CC'\) is:
     \[
     \text{Area of hexagon } AA'BB'CC' = 2 \times \frac{S}{r^2}
     \]

The final answer is \( \boxed{ 2 \times \frac{S}{r^2} } \)