Question:

<p><strong>Step 1: Set up the geometry.</strong> Original sheet has radius R = 1, so A₂ = π. Let θ be the angle of the sector removed. The remaining sector has angle (2π - θ).</p><p><strong>Step 2: Form the cone.</strong> The remaining sector (with radius 1) is rolled into a cone. The arc length of remaining sector = 1·(2π - θ) becomes the base circumference of cone. So 2πr = 2π - θ, where r is cone's base radius. Thus r = (2π - θ)/(2π).</p><p><strong>Step 3: Find cone's height.</strong> The slant height of cone = 1 (original radius). Using l² = h² + r²: h² = 1 - r² = 1 - [(2π - θ)/(2π)]². So h = √[1 - (2π - θ)²/(4π²)].</p><p><strong>Step 4: Express volume.</strong> V = (1/3)πr²h = (1/3)π · [(2π - θ)/(2π)]² · √[1 - (2π - θ)²/(4π²)]. Let u = (2π - θ)/(2π), so u ∈ (0,1). Then V = (1/3)π u² √(1 - u²).</p><p><strong>Step 5: Maximize volume.</strong> V = (1/3)π u² √(1 - u²). Taking dV/du: dV/du = (1/3)π[2u√(1 - u²) + u² · (-u)/√(1 - u²)] = (1/3)π · [2u(1 - u²) - u³]/√(1 - u²) = (1/3)π · u(2 - 3u²)/√(1 - u²). Setting dV/du = 0: 2 - 3u² = 0, so u² = 2/3, giving u = √(2/3).</p><p><strong>Step 6: Find removed angle.</strong> From u = (2π - θ)/(2π) = √(2/3): 2π - θ = 2π√(2/3). Thus θ = 2π[1 - √(2/3)] = 2π[(√3 - √2)/√3].</p><p><strong>Step 7: Calculate A₁.</strong> A₁ = (θ/2π) · π · 1² = θ/2 = π[1 - √(2/3)] = π[(√3 - √2)/√3].</p><p><strong>Step 8: Simplify A₁.</strong> A₁ = π - π√(2/3) = π - π√2/√3 = π[1 - √(2/3)].</p><p><strong>Step 9: Calculate ratio.</strong> A₂/A₁ = π / {π[1 - √(2/3)]} = 1/[1 - √(2/3)]. Rationalizing: multiply by [1 + √(2/3)]/[1 + √(2/3)] = [1 + √(2/3)]/[1 - 2/3] = [1 + √(2/3)]/(1/3) = 3[1 + √(2/3)] = 3 + 3√(2/3) = 3 + √(18/3) = 3 + √6.</p><p><strong>Step 10: Identify m and n.</strong> A₂/A₁ = 3 + √6, so m = 3, n = 6. Therefore m + n = 9.</p><p><strong>∴ Answer: 9</strong></p>