If `a + b + c = 2s`, then `((s-a)^2 + (s-b)^2 + (s-c)^2 + s^2)/(a^2 + b^2 + c^2)` is equal to

Options

A. `a^2 + b^2 + c^2`
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to evaluate the expression given the condition \( a + b + c = 2s \). We will follow these steps:

### Step 1: Understand the given condition
We have the equation:
\[ a + b + c = 2s \]
This means that the sum of \( a \), \( b \), and \( c \) is twice the value of \( s \).

### Step 2: Assume values for \( a \), \( b \), and \( c \)
Since we have one equation with three variables, we can assume values for \( a \), \( b \), and \( c \) to simplify our calculations. Let's assume:
- \( a = 0 \)
- \( b = 1 \)
- \( c = 1 \)

Now we can check if these values satisfy the equation:
\[ 0 + 1 + 1 = 2 \quad \text{which implies} \quad 2s = 2 \quad \Rightarrow \quad s = 1 \]

### Step 3: Substitute the values into the expression
We need to evaluate the expression:
\[ \frac{(s-a)^2 + (s-b)^2 + (s-c)^2 + s^2}{a^2 + b^2 + c^2} \]

Substituting \( s = 1 \), \( a = 0 \), \( b = 1 \), and \( c = 1 \):
- \( s - a = 1 - 0 = 1 \)
- \( s - b = 1 - 1 = 0 \)
- \( s - c = 1 - 1 = 0 \)

Now we can calculate:
\[ (s-a)^2 = 1^2 = 1 \]
\[ (s-b)^2 = 0^2 = 0 \]
\[ (s-c)^2 = 0^2 = 0 \]
\[ s^2 = 1^2 = 1 \]

Now, substituting these values into the expression:
\[ \frac{1 + 0 + 0 + 1}{a^2 + b^2 + c^2} \]

### Step 4: Calculate the denominator
Now we calculate \( a^2 + b^2 + c^2 \):
\[ a^2 = 0^2 = 0 \]
\[ b^2 = 1^2 = 1 \]
\[ c^2 = 1^2 = 1 \]
Thus,
\[ a^2 + b^2 + c^2 = 0 + 1 + 1 = 2 \]

### Step 5: Final calculation
Now we can substitute back into the expression:
\[ \frac{1 + 0 + 0 + 1}{0 + 1 + 1} = \frac{2}{2} = 1 \]

### Conclusion
The value of the expression is:
\[ \boxed{1} \]