If `a,b,c gt 0,` then `(a )/(2a + b +c ) = (b)/(a + 2b +c)= (c )/(a +b + 2c) = ?`

Options

A. `2`
B. 1/4
C. `-1`
D. 0

Answer

To solve the equation

\[ \frac{a}{2a + b + c} = \frac{b}{a + 2b + c} = \frac{c}{a + b + 2c} = k \]

where \( k \) is a constant, we can proceed as follows:

### Step 1: Set up the equations
From the equation, we can express each variable in terms of \( k \):

1. From \(\frac{a}{2a + b + c} = k\), we can write:
\[ a = k(2a + b + c) \]
Rearranging gives:
\[ a - 2ka - kb - kc = 0 \quad \Rightarrow \quad a(1 - 2k) = kb + kc \]

2. From \(\frac{b}{a + 2b + c} = k\), we can write:
\[ b = k(a + 2b + c) \]
Rearranging gives:
\[ b - ka - 2kb - kc = 0 \quad \Rightarrow \quad b(1 - 2k) = ka + kc \]

3. From \(\frac{c}{a + b + 2c} = k\), we can write:
\[ c = k(a + b + 2c) \]
Rearranging gives:
\[ c - ka - kb - 2kc = 0 \quad \Rightarrow \quad c(1 - 2k) = ka + kb \]

### Step 2: Analyze the equations
We have three equations:

1. \( a(1 - 2k) = kb + kc \)
2. \( b(1 - 2k) = ka + kc \)
3. \( c(1 - 2k) = ka + kb \)

### Step 3: Assume \( a = b = c \)
To simplify the problem, let's assume \( a = b = c \). Let’s denote \( a = b = c = x \). Then:

1. The first equation becomes:
\[ x(1 - 2k) = kx + kx \quad \Rightarrow \quad x(1 - 2k) = 2kx \]
Dividing by \( x \) (since \( x > 0 \)):
\[ 1 - 2k = 2k \quad \Rightarrow \quad 1 = 4k \quad \Rightarrow \quad k = \frac{1}{4} \]

### Step 4: Conclusion
Thus, we find that:

\[ \frac{a}{2a + b + c} = \frac{b}{a + 2b + c} = \frac{c}{a + b + 2c} = \frac{1}{4} \]

### Final Answer
The value of \( k \) is \( \frac{1}{4} \).

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