<p>In triangle ABC, if \(AB = x\), \(BC = x + 1\), and \(\angle C = \frac{\pi}{3}\), then the least integer value of \(x\) is</p>
Step-by-Step Solution
Key Concept: Apply the Law of Cosines to relate the three sides and the given angle, then use the triangle inequality constraints to find when a valid triangle exists. The expression will yield an inequality in x that determines the minimum integer value.
<p><strong>Step 1: Apply Law of Cosines</strong></p><p>In triangle ABC with sides AB = c = x, BC = a = x+1, AC = b, and ∠C = π/3:</p><p>By the Law of Cosines: c² = a² + b² - 2ab·cos(C)</p><p>x² = (x+1)² + b² - 2(x+1)b·cos(π/3)</p><p>x² = (x+1)² + b² - 2(x+1)b·(1/2)</p><p>x² = (x+1)² + b² - (x+1)b</p><p><strong>Step 2: Rearrange as quadratic in b</strong></p><p>b² - (x+1)b + (x+1)² - x² = 0</p><p>b² - (x+1)b + (x² + 2x + 1 - x²) = 0</p><p>b² - (x+1)b + (2x+1) = 0</p><p><strong>Step 3: Apply discriminant condition</strong></p><p>For b to be real: Δ ≥ 0</p><p>(x+1)² - 4(2x+1) ≥ 0</p><p>x² + 2x + 1 - 8x - 4 ≥ 0</p><p>x² - 6x - 3 ≥ 0</p><p><strong>Step 4: Solve the quadratic inequality</strong></p><p>Using the quadratic formula: x = (6 ± √(36+12))/2 = (6 ± √48)/2 = (6 ± 4√3)/2 = 3 ± 2√3</p><p>Since √3 ≈ 1.732, we have 2√3 ≈ 3.464</p><p>x = 3 + 2√3 ≈ 6.464 or x = 3 - 2√3 ≈ -0.464</p><p>Since x must be positive (it's a side length), we need: x ≥ 3 + 2√3 ≈ 6.464</p><p><strong>Step 5: Apply triangle inequality</strong></p><p>We also need AB + AC > BC, i.e., x + b > x+1, which gives b > 1 (automatically satisfied for valid triangles in this range).</p><p>The critical constraint is from the discriminant condition.</p><p><strong>Step 6: Find the least integer value</strong></p><p>Since x ≥ 6.464, the least integer value is x = 7</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B