If a/(b + c) = b/(c + a) = c/(a + b) for nonzero real numbers a, b and c, determine which statement correctly describes the possible common value of each fraction.

Introduction / Context:
This algebraic question examines your understanding of symmetric equations involving three variables and the consequences of setting several rational expressions equal to a common value. The condition a/(b + c) = b/(c + a) = c/(a + b) with nonzero real numbers a, b and c imposes strong restrictions on their relationships. The problem asks you to deduce what the common value of these fractions can be, which is a classic reasoning task in algebra and aptitude tests.

Given Data / Assumptions:

• a, b and c are nonzero real numbers.
• a/(b + c) = b/(c + a) = c/(a + b) = k, say, where k is the common value.
• We are asked to identify which statement about the possible values of k is correct.
• No specific numeric values of a, b or c are given, so we must use algebraic reasoning.

Concept / Approach:
Let the common value be k. Then we can write a = k(b + c), b = k(c + a) and c = k(a + b). Adding these three equations gives a relation between the sum a + b + c and k. This leads to two main cases: either the sum a + b + c is zero or it is nonzero. Each case yields a different possible value of k. We then verify that these values actually work by choosing simple numerical examples for a, b and c.

Step-by-Step Solution:
Step 1: Let a/(b + c) = b/(c + a) = c/(a + b) = k.Step 2: Then a = k(b + c), b = k(c + a), and c = k(a + b).Step 3: Add the three equations: a + b + c = k[(b + c) + (c + a) + (a + b)].Step 4: The bracket simplifies to 2(a + b + c), so we get a + b + c = 2k(a + b + c).Step 5: Either a + b + c ≠ 0, giving 1 = 2k, or a + b + c = 0.Step 6: If a + b + c ≠ 0, then k = 1/2 and each fraction equals 1/2.Step 7: If a + b + c = 0, then from a = k(b + c) we have a = k(−a), which implies (1 + k)a = 0. Since a ≠ 0, we get k = −1.Step 8: Therefore the only possible common values of the fractions are 1/2 and −1.

Verification / Alternative check:
To confirm, choose an example where a = b = c. Let a = b = c = 2. Then a + b + c = 6 ≠ 0, and a/(b + c) = 2/(2 + 2) = 2/4 = 1/2, so k = 1/2 works. For the second case, choose numbers with sum zero, for example a = 1, b = 1 and c = −2. Then a/(b + c) = 1/(1 − 2) = −1, b/(c + a) = 1/(−2 + 1) = −1, and c/(a + b) = −2/(1 + 1) = −1. Here the common value is −1, confirming both possibilities.

Why Other Options Are Wrong:
The option stating 1 or −1 is incorrect because 1 is not possible under the given relations. Options that include 1 as a possibility, or only 1/2, ignore the valid case where a + b + c = 0. The suggestion that any real number is possible is far too broad and contradicts the strong algebraic constraints we derived. Only the choice indicating 1/2 or −1 matches all valid configurations.

Common Pitfalls:
One common mistake is to assume that a, b and c must be equal, leading to the conclusion that the common value is always 1/2. Another error is to perform incorrect algebra when adding the three equations, which can lead to a wrong relationship between k and a + b + c. It is also easy to forget that a + b + c could be zero, which opens the second case k = −1. Keeping track of both possibilities is essential.

Final Answer:
The common value of a/(b + c), b/(c + a), and c/(a + b) can be 1/2 or −1 depending on the specific values of a, b and c.