In the diagram, $\angle AOB = \angle  BOC$ and$\angle COD = \angle DOE = \angle EOF$. Given that $\angle AOD = 82^o$ and $\angle BOE = 68^o$. Find $\angle AOF$.
[img]https://cdn.artofproblemsolving.com/attachments/b/2/deba6cd740adbf033ad884fff8e13cd21d9c5a.png[/img]
1. Let $\angle AOB = x$ and $\angle COD = y$. Given that $\angle AOB = \angle BOC$ and $\angle COD = \angle DOE = \angle EOF$, we can set up the following equations based on the given angles:
   \[
   2x + y = 82^\circ \quad \text{(since } \angle AOD = \angle AOB + \angle BOC + \angle COD = 2x + y)
   \]
   \[
   x + 2y = 68^\circ \quad \text{(since } \angle BOE = \angle BOC + \angle COD + \angle DOE = x + 2y)
   \]

2. We now have a system of linear equations:
   \[
   \begin{cases}
   2x + y = 82 \\
   x + 2y = 68
   \end{cases}
   \]

3. To solve this system, we can use the method of substitution or elimination. Here, we will use elimination. First, multiply the second equation by 2 to align the coefficients of $x$:
   \[
   2(x + 2y) = 2 \cdot 68 \implies 2x + 4y = 136
   \]

4. Now subtract the first equation from this result:
   \[
   (2x + 4y) - (2x + y) = 136 - 82
   \]
   \[
   3y = 54 \implies y = \frac{54}{3} = 18
   \]

5. Substitute $y = 18$ back into the first equation to find $x$:
   \[
   2x + 18 = 82 \implies 2x = 82 - 18 \implies 2x = 64 \implies x = \frac{64}{2} = 32
   \]

6. Now, we need to find $\angle AOF$. Since $\angle AOF = \angle AOB + \angle BOC + \angle COD + \angle DOE + \angle EOF$, and we know:
   \[
   \angle AOB = x = 32^\circ
   \]
   \[
   \angle BOC = x = 32^\circ
   \]
   \[
   \angle COD = y = 18^\circ
   \]
   \[
   \angle DOE = y = 18^\circ
   \]
   \[
   \angle EOF = y = 18^\circ
   \]

7. Therefore:
   \[
   \angle AOF = \angle AOB + \angle BOC + \angle COD + \angle DOE + \angle EOF = 32^\circ + 32^\circ + 18^\circ + 18^\circ + 18^\circ
   \]
   \[
   \angle AOF = 32 + 32 + 18 + 18 + 18 = 118^\circ
   \]

The final answer is $\boxed{118^\circ}$