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Difficulty: HardCircles, Arc Lengths, and Sector Areas

Two concentric circles centered at point OO have radii rr and RR, where r<Rr < R. A sector bounded by radii OAOA and OBOB of the outer circle has central angle θ∘\theta^\circ. Region SS is the region lying inside sector AOBAOB but outside the inner circle. The area of region SS is equal to 33 times the area of the sector of the inner circle bounded by central angle θ∘\theta^\circ. If the perimeter of region SS is equal to 116\frac{11}{6} times the length of arc ABAB, what is the value of θ\theta?

  1. A
    60π\frac{60}{\pi}
  2. B
    270π\frac{270}{\pi}
  3. C
    480π\frac{480}{\pi}
  4. 540π\frac{540}{\pi}Answer
  5. E
    1080π\frac{1080}{\pi}

Answer

540π\frac{540}{\pi}
The area of region SS is θ360π(R2−r2)\frac{\theta}{360}\pi(R^2 - r^2), which is equal to 33 times the area of the inner sector θ360πr2\frac{\theta}{360}\pi r^2. This simplifies to R2−r2=3r2  ⟹  R=2rR^2 - r^2 = 3r^2 \implies R = 2r. The perimeter of region SS consists of outer arc length LouterL_{\text{outer}}, inner arc length Linner=12LouterL_{\text{inner}} = \frac{1}{2} L_{\text{outer}}, and two straight line segments each of length R−r=rR - r = r. Setting the total perimeter 32Louter+2r\frac{3}{2} L_{\text{outer}} + 2r equal to 116Louter\frac{11}{6} L_{\text{outer}} yields 2r=13Louter2r = \frac{1}{3} L_{\text{outer}}, or Louter=6rL_{\text{outer}} = 6r. Substituting Louter=θ360(2πR)=4πrθ360L_{\text{outer}} = \frac{\theta}{360}(2\pi R) = \frac{4\pi r \theta}{360} gives 4πrθ360=6r\frac{4\pi r \theta}{360} = 6r, which solves to θ=540π\theta = \frac{540}{\pi}.

Step-by-Step Solution

1
Relate outer radius RR to inner radius rr using sector areas.
R=2rR = 2r
The area of region SS is θ360π(R2−r2)\frac{\theta}{360}\pi(R^2 - r^2) and the inner sector area is θ360πr2\frac{\theta}{360}\pi r^2. Setting θ360π(R2−r2)=3⋅θ360πr2\frac{\theta}{360}\pi(R^2 - r^2) = 3 \cdot \frac{\theta}{360}\pi r^2 gives R2−r2=3r2R^2 - r^2 = 3r^2, so R2=4r2R^2 = 4r^2 and R=2rR = 2r.
2
Express the perimeter of region SS in terms of outer arc length LouterL_{\text{outer}} and radius rr.
Perimeter(S)=32Louter+2r\text{Perimeter}(S) = \frac{3}{2} L_{\text{outer}} + 2r
Region SS is bounded by outer arc ABAB (LouterL_{\text{outer}}), inner arc CDCD (LinnerL_{\text{inner}}), and two straight segments ACAC and BDBD of length R−r=2r−r=rR - r = 2r - r = r. Since R=2rR = 2r, Linner=12LouterL_{\text{inner}} = \frac{1}{2} L_{\text{outer}}, making total perimeter Louter+12Louter+2(r)=32Louter+2rL_{\text{outer}} + \frac{1}{2} L_{\text{outer}} + 2(r) = \frac{3}{2} L_{\text{outer}} + 2r.
3
Use the given perimeter relationship to express LouterL_{\text{outer}} in terms of rr.
Louter=6rL_{\text{outer}} = 6r
Setting 32Louter+2r=116Louter\frac{3}{2} L_{\text{outer}} + 2r = \frac{11}{6} L_{\text{outer}} yields 2r=(116−96)Louter=13Louter2r = \left(\frac{11}{6} - \frac{9}{6}\right) L_{\text{outer}} = \frac{1}{3} L_{\text{outer}}, so Louter=6rL_{\text{outer}} = 6r.
4
Solve for θ\theta using the definition of outer arc length.
θ=540π\theta = \frac{540}{\pi}
Louter=θ360(2πR)=θ360(4πr)L_{\text{outer}} = \frac{\theta}{360}(2\pi R) = \frac{\theta}{360}(4\pi r). Setting 4πrθ360=6r\frac{4\pi r \theta}{360} = 6r simplifies to πθ90=6\frac{\pi \theta}{90} = 6, giving θ=540π\theta = \frac{540}{\pi}.

Key Concept

Annular sector area and perimeter relations combining arc length formulas and concentric circle geometry.
Estimated Time:2m 30s
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Topics

Circles, Arc Lengths, and Sector Areas

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