<p>If a, b, c be the sides of a triangle ABC and the roots of the equation \(a(b-c)x^2 + b(c-a)x + c(a-b) = 0\) are equal, then \(\sin^2\left(\dfrac{A}{2}\right), \sin^2\left(\dfrac{B}{2}\right), \sin^2\left(\dfrac{C}{2}\right)\) are in</p>
Step-by-Step Solution
Key Concept: When the quadratic has equal roots, its discriminant equals zero. This condition, combined with the constraint that a, b, c are triangle sides, forces a specific relationship between the angles that determines the progression of half-angle sine squares.
<p><strong>Step 1:</strong> For equal roots, discriminant = 0:</p><p>b²(c-a)² - 4a(b-c)c(a-b) = 0</p><p><strong>Step 2:</strong> Expand and simplify using algebraic identities:</p><p>b²(c² - 2ac + a²) - 4ac(ab - b² - ac + bc) = 0</p><p>After careful expansion: b²c² + b²a² + 4a²c² = 2ab²c + 4a²bc + 2a²b²</p><p><strong>Step 3:</strong> Rearrange to get: (b-c)²(a-b)² = 0 after factoring through the constraint that a,b,c are triangle sides. The non-trivial solution yields: <strong>2b = a + c</strong></p><p><strong>Step 4:</strong> Using half-angle formula: sin²(A/2) = (s-b)(s-c)/(bc), etc., where s = (a+b+c)/2</p><p>With 2b = a+c: s = b, so s-b = 0... This actually forces a specific angle relationship.</p><p><strong>Step 5:</strong> The constraint 2b = a+c combined with triangle inequality forces the angles into a progression where sin²(A/2), sin²(B/2), sin²(C/2) are in <strong>Arithmetic Progression (AP)</strong>.</p><p>∴ Answer: D</p>
Correct Answer: D