2005 AIME I Problem 1

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

Answer: 942

Concepts:tangent circles regular polygon circle area

Difficulty rating: 2010

Small hint:

The six centers form a regular hexagon of side

2r,

and a regular hexagon’s circumradius equals its side

Big hint:

Internal tangency to

\mathcal{C}

gives

2r + r = 30;

then subtract the six small disks from the big one

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

\begin{aligned} K &= \pi\left(30^2 - 6 \cdot 10^2\right) \\ &= 300\pi \approx 942.48, \end{aligned}

and

\lfloor K\rfloor = 942.