If a, b, c are positive integers such that `a^2+ 2b^2-2ab = 169` and `2bc - c^2= 169` then a + b + c is:

Answer

To solve the problem step by step, we start with the given equations:

1. **Equations**:
\[ a^2 + 2b^2 - 2ab = 169 \quad \text{(1)} \]
\[ 2bc - c^2 = 169 \quad \text{(2)} \]

2. **Equate the Left-Hand Sides**:
Since both equations equal 169, we can set their left-hand sides equal to each other:
\[ a^2 + 2b^2 - 2ab = 2bc - c^2 \]

3. **Rearranging the Equation**:
Rearranging gives:
\[ a^2 - 2ab + 2b^2 + c^2 - 2bc = 0 \]

4. **Grouping Terms**:
We can rewrite the equation as:
\[ (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) = 0 \]
This simplifies to:
\[ (a - b)^2 + (b - c)^2 = 0 \]

5. **Understanding the Squares**:
Since squares of real numbers are non-negative, the only way their sum can be zero is if both squares are zero:
\[ (a - b)^2 = 0 \quad \text{and} \quad (b - c)^2 = 0 \]
This implies:
\[ a - b = 0 \quad \Rightarrow \quad a = b \]
\[ b - c = 0 \quad \Rightarrow \quad b = c \]
Therefore, we have:
\[ a = b = c \]

6. **Substituting Back**:
Now we substitute \(b = a\) and \(c = a\) back into equation (1):
\[ a^2 + 2a^2 - 2a^2 = 169 \]
This simplifies to:
\[ a^2 = 169 \]

7. **Solving for a**:
Taking the square root gives:
\[ a = 13 \quad (\text{since } a, b, c \text{ are positive integers}) \]

8. **Finding b and c**:
Since \(b = a\) and \(c = a\), we have:
\[ b = 13, \quad c = 13 \]

9. **Calculating a + b + c**:
Finally, we calculate:
\[ a + b + c = 13 + 13 + 13 = 39 \]

Thus, the final answer is:
\[ \boxed{39} \]