2000 AMC 12 Problem 24

Attempt Problem 24 of the 2000 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 12 solutions, or check the answer key.

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24.

If circular arcs ACAC and BCBC have centers at BB and A,A, respectively, then there exists a circle tangent to both arc ACAC and arc BC,BC, and to AB.\overline{AB}. If the length of arc BCBC is 12,12, then what is the circumference of the circle?

2424

2525

2626

2727

2828

Answer: D
Concepts:power of a pointtangent circlesarc
Difficulty rating: 2390
Solution:

Each arc has radius AB,AB, and CC is at distance ABAB from both AA and B,B, so ABC\triangle ABC is equilateral. Thus arc BCBC subtends 6060^\circ of a circle of radius AB,AB, whose full circumference is 612=72.6 \cdot 12 = 72.

Let the small circle have radius rr and touch AB\overline{AB} at its midpoint D,D, where AD=12AB.AD = \tfrac12 AB. By Power of a Point, AD2=AB(AB2r),AD^2 = AB(AB - 2r), so AB24=AB22rAB, \frac{AB^2}{4} = AB^2 - 2r\,AB, giving 2r=34AB,2r = \tfrac34 AB, hence r=38AB.r = \tfrac38 AB.

The circumferences are in the ratio of the radii, so the small circle's circumference is 3872=27.\tfrac38 \cdot 72 = 27.

Thus, the correct answer is D.

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