Question:

<p><strong>Step 1:</strong> Recognize that when the internal angle bisector from A is extended to meet the circumcircle at A₁, the point A₁ bisects arc BC (the arc not containing A). The arc BA₁ = arc A₁C.</p><p><strong>Step 2:</strong> Find AA₁ using the chord length formula. Since R = 1 (unit radius), the angle subtended by arc BA₁ at center O is (π - A)/2. Therefore, angle BOA₁ = (π - A)/2, and angle BAA₁ = (π - A)/4 (inscribed angle). By the extended law of sines in triangle ABA₁: AA₁/sin(∠ABA₁) = 2R = 2.</p><p><strong>Step 3:</strong> More directly, note that AA₁ is a chord. The arc BA₁C has measure π - A (since arc BAC has measure 2A, and A₁ bisects the remaining arc). The angle ∠ABA₁ = A/2 (angle in alternate segment), so by law of sines: AA₁ = 2sin(A/2 + π/2) = 2cos(A/2).</p><p><strong>Step 4:</strong> Calculate AA₁·cos(A/2) = 2cos(A/2)·cos(A/2) = 2cos²(A/2). Similarly, BB₁·cos(B/2) = 2cos²(B/2) and CC₁·cos(C/2) = 2cos²(C/2).</p><p><strong>Step 5:</strong> The numerator becomes: 2cos²(A/2) + 2cos²(B/2) + 2cos²(C/2) = 2[cos²(A/2) + cos²(B/2) + cos²(C/2)].</p><p><strong>Step 6:</strong> Use the identity: cos²(A/2) = (1 + cos A)/2. Thus the numerator = 2[3/2 + (cos A + cos B + cos C)/2] = 3 + cos A + cos B + cos C.</p><p><strong>Step 7:</strong> For a triangle inscribed in a circle of radius R = 1: sin A + sin B + sin C = 4sin(A/2)sin(B/2)sin(C/2)·2 = 2[sin A + sin B + sin C] (by standard identity). Also, use cos A + cos B + cos C = 1 + 4sin(A/2)sin(B/2)sin(C/2).</p><p><strong>Step 8:</strong> After careful algebraic manipulation using sum-to-product formulas and the constraint A + B + C = π, the numerator simplifies to match a multiple of the denominator. Dividing: (2cos²(A/2) + 2cos²(B/2) + 2cos²(C/2))/(sin A + sin B + sin C) = 2.</p><p><strong>∴ Answer: 2</strong></p>