Difficulty: Hard
Correct Answer: 2
Explanation:
Introduction / Context:
This question is a classic symmetric-algebra simplification problem. You are told that three different expressions are equal to the same constant k. Each expression compares a sum of two squares to the remaining square, but in a cyclic way. Instead of trying to solve for a, b, and c individually, the smart approach is to rewrite each equality in a comparable form and then add them. The symmetry makes many terms combine neatly, allowing k to be determined uniquely under the non-zero assumption.
Given Data / Assumptions:
Concept / Approach:
Convert each ratio equation into a linear relation in squares by multiplying both sides by the denominator. Then add the three resulting equations. The left side becomes 2(a^2 + b^2 + c^2), while the right side becomes k(a^2 + b^2 + c^2). Since a, b, c are non-zero, the sum a^2 + b^2 + c^2 is strictly positive, so we can divide by it safely to get k.
Step-by-Step Solution:
1) From (a^2 + b^2)/c^2 = k, multiply both sides by c^2:
a^2 + b^2 = k * c^2
2) From (b^2 + c^2)/a^2 = k, multiply both sides by a^2:
b^2 + c^2 = k * a^2
3) From (c^2 + a^2)/b^2 = k, multiply both sides by b^2:
c^2 + a^2 = k * b^2
4) Add all three equations (left sides together, right sides together):
(a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) = k(c^2 + a^2 + b^2)
5) Combine like terms on the left:
2a^2 + 2b^2 + 2c^2 = k(a^2 + b^2 + c^2)
6) Factor out 2 on the left:
2(a^2 + b^2 + c^2) = k(a^2 + b^2 + c^2)
7) Since a, b, c are non-zero, a^2 + b^2 + c^2 > 0, so divide both sides by it:
k = 2
Verification / Alternative check:
Pick any convenient non-zero values with symmetry, for example a = b = c = 1:
(a^2 + b^2)/c^2 = (1 + 1)/1 = 2, and similarly the other two ratios are also 2. This confirms that k = 2 is consistent and matches the derived unique value.
Why Other Options Are Wrong:
• 1 or 1/2: would contradict the summed relation 2(a^2 + b^2 + c^2) = k(a^2 + b^2 + c^2).
• 3 or 4: would require the left side to be 3 or 4 times the sum of squares, which is impossible because the left side is exactly 2 times that sum.
Common Pitfalls:
• Trying to solve for a, b, c individually instead of using symmetry.
• Forgetting that a^2 + b^2 + c^2 is positive when a, b, c are real and non-zero.
Final Answer:
2
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