If $\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} = K$, then the value of $K$ is

### Concept & Formula This problem requires evaluating a continuous proportion under different conditions. Like the previous problem, it utilizes the property of equal fractions. However, you must account for the mathematical condition where the sum of the variables could be zero, which splits the problem into two distinct cases. $$ \text{If } \frac{p}{q} = \frac{r}{s} = \frac{t}{u} = K, \text{ then } K = \frac{p+r+t}{q+s+u} \text{ (provided } q+s+u \neq 0) $$ ### Step-by-Step Solution * **Given:** * $\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} = K$ * **Calculation:** 1. **Case 1: Assume $(a + b + c) \neq 0$** Apply the sum property of equal fractions. $K = \frac{\text{Sum of numerators}}{\text{Sum of denominators}}$ $K = \frac{a + b + c}{(b + c) + (c + a) + (a + b)}$ $K = \frac{a + b + c}{2a + 2b + 2c}$ Factor the denominator: $K = \frac{a + b + c}{2(a + b + c)}$. Cancel the common term: $K = \frac{1}{2}$. 2. **Case 2: Assume $(a + b + c) = 0$** If the sum is zero, we cannot use the theorem above as it would cause division by zero. Instead, rearrange the equation: $b + c = -a$. Substitute this into the very first given fraction: $K = \frac{a}{b + c}$. $K = \frac{a}{-a}$. $K = -1$. 3. Since both cases are mathematically valid depending on the values of $a, b, c$, $K$ can be either $\frac{1}{2}$ or $-1$. ### Exam Strategy & Shortcut Test the extreme cases using assumed numbers. Case A (Symmetry): Let $a = b = c = 1$. $K = \frac{1}{1+1} = \frac{1}{2}$. This eliminates options (c). Case B (Zero Sum): Let $a = 2$, $b = -1$, $c = -1$. (Notice $2 - 1 - 1 = 0$). Check the fractions: $\frac{2}{-1-1} = \frac{2}{-2} = -1$ $\frac{-1}{-1+2} = \frac{-1}{1} = -1$ $\frac{-1}{2-1} = \frac{-1}{1} = -1$ So $K$ also equals $-1$. Therefore, the answer must contain both values. ### Common Pitfall The most common trap is only calculating Case 1 using the sum theorem and immediately choosing $\frac{1}{2}$ (Option D). In higher-level exams, always check if the denominator sum can equal zero. If no constraint (like "$a,b,c$ are positive integers") is given, you must evaluate the zero-sum scenario. ### Final Answer Therefore, the correct answer is $\frac{1}{2}$ or $-1$.