2005 AIME I Problem 1

Attempt Problem 1 of the 2005 AIME I 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 2005 AIME I solutions, or check the answer key.

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1.

Six congruent circles form a ring with each circle externally tangent to the two circles adjacent to it. All six circles are internally tangent to a circle C\mathcal{C} with radius 30.30. Let KK be the area of the region inside C\mathcal{C} and outside all of the six circles in the ring. Find K.\lfloor K\rfloor. (The notation K\lfloor K\rfloor denotes the greatest integer that is less than or equal to K.K.)

Answer: 942
Concepts:tangent circlesregular polygoncircle area
Difficulty rating: 2010
Solution:

Let rr be the common radius of the six circles. Adjacent circles are externally tangent, so their centers are 2r2r apart, and the six centers form a regular hexagon with side 2r.2r. Since a regular hexagon's circumradius equals its side length, each center is at distance 2r2r from the center OO of C.\mathcal{C}. Internal tangency to C\mathcal{C} means the distance from OO to each small center plus rr equals 30,30, so 3r=303r = 30 and r=10.r = 10.

Therefore K=π(3026102)=300π942.48, \begin{aligned} K &= \pi\left(30^2 - 6 \cdot 10^2\right) \\ &= 300\pi \approx 942.48, \end{aligned} and K=942.\lfloor K\rfloor = 942.

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