Quadratic Equations
Discriminant and inequalities
Grade 11
Question:
<p>If \(a\), \(b\) and \(c\) are side lengths of a triangle \(ABC\) such that \(x^2 - 2(a+b+c)x + 3k(ab+bc+ca) = 0\), where \(k < \dfrac{p}{q}\) (where \(p\) and \(q\) are relatively prime), has real roots, find \((p+q)\).</p>
Step-by-Step Solution
Key Concept: For a quadratic equation to have real roots, the discriminant must be non-negative. Combined with the triangle inequality constraints on sides a, b, c, we can determine the range of k. The critical insight is that the product of roots relates to the triangle's semiperimeter through the constraint that real roots exist.
<p><strong>Step 1:</strong> For the quadratic x² - 2(a+b+c)x + 3k(ab+bc+ca) = 0 to have real roots:</p><p>Δ = 4(a+b+c)² - 12k(ab+bc+ca) ≥ 0</p><p><strong>Step 2:</strong> Expand: 4(a² + b² + c² + 2ab + 2bc + 2ca) - 12k(ab+bc+ca) ≥ 0</p><p>4(a² + b² + c²) + 8(ab+bc+ca) - 12k(ab+bc+ca) ≥ 0</p><p>4(a² + b² + c²) + (8 - 12k)(ab+bc+ca) ≥ 0</p><p><strong>Step 3:</strong> For a triangle, a² + b² + c² < 2(ab+bc+ca) (from triangle inequalities).</p><p><strong>Step 4:</strong> Let s = ab+bc+ca. Then: 4(a² + b² + c²) ≥ (12k - 8)s</p><p>Since a² + b² + c² < 2s, we need: 4(2s) > (12k - 8)s (at maximum)</p><p>8s ≥ (12k - 8)s</p><p>8 ≥ 12k - 8</p><p>16 ≥ 12k</p><p>k ≤ 4/3</p><p><strong>Step 5:</strong> For positive triangle sides, we also need k > 0 (from the structure of the problem).</p><p>∴ Answer: <strong>0 < k ≤ 4/3</strong></p>
Correct Answer: 0