Question

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 $\mathcal{C}$ with radius $30.$ Let $K$ be the area of the region inside $\mathcal{C}$ and outside all of the six circles in the ring. Find $\lfloor K\rfloor.$ (The notation $\lfloor K\rfloor$ denotes the greatest integer that is less than or equal to $K.$ )

Answer

Solution:

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

Therefore $K = \pi\left(30^2 - 6 \cdot 10^2\right) = 300\pi \approx 942.48,$ and $\lfloor K\rfloor = 942.$

2005 AIME I Problem 1

Solution:

Problem 1 in Other Years