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2018 AMC 10B Problem 12

Problem 12 of 25IntermediateGeometry

Line segment ABAB is a diameter of a circle with AB=24.AB = 24. Point C,C, not equal to AA or B,B, lies on the circle. As point CC moves around the circle, the centroid (center of mass) of ABC\triangle ABC traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?

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Solution

Put the center OO at the origin, so A=(12,0)A = (-12, 0) and B=(12,0),B = (12, 0), while CC runs over the circle of radius 12.12. Then A+B=0,A + B = 0, so the centroid is 13(A+B+C)=13C.\tfrac13(A + B + C) = \tfrac13 C. As CC circles, 13C\tfrac13 C traces a circle of radius 123=4\tfrac{12}{3} = 4 (minus the two points where C=AC = A or BB). Its area is π42=16π50.\pi \cdot 4^2 = 16\pi \approx 50. Therefore, the answer is C.

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Concepts: centroid · homothety · circle area

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.