2006 AIME I Problem 10

Question

Eight circles of diameter

1

are packed in the first quadrant of the coordinate plane as shown. Let region

\mathcal{R}

be the union of the eight circular regions. Line

\ell,

with slope

3,

divides

\mathcal{R}

into two regions of equal area. Line

\ell

’s equation can be expressed in the form

ax = by + c,

where

a,

b,

and

c

are positive integers whose greatest common divisor is

1.

Find

a^2 + b^2 + c^2.

Answer

Any line through the point where two congruent circles touch splits that pair’s area evenly

The line through

\left(1, \frac{1}{2}\right)

and

\left(\frac{3}{2}, 2\right)

has slope

3

and leaves two whole circles on each side

The circles have radius

\frac{1}{2}

and centers at

\left(\frac{1}{2}, \frac{1}{2}\right),

\left(\frac{3}{2}, \frac{1}{2}\right),

\left(\frac{5}{2}, \frac{1}{2}\right),

\left(\frac{1}{2}, \frac{3}{2}\right),

\left(\frac{3}{2}, \frac{3}{2}\right),

\left(\frac{5}{2}, \frac{3}{2}\right),

\left(\frac{1}{2}, \frac{5}{2}\right),

and

\left(\frac{3}{2}, \frac{5}{2}\right).

The pair of circles tangent at

A = \left(1, \frac{1}{2}\right)

is symmetric about

A,

so any line through

A

bisects that pair’s area; similarly for the pair tangent at

B = \left(\frac{3}{2}, 2\right).

The line

AB

has slope

\frac{2 - \frac{1}{2}}{\frac{3}{2} - 1} = 3.

Line

AB

misses the remaining four circles entirely, and exactly two of them lie on each side of it, so it divides

\mathcal{R}

into two regions of equal area. Sliding a slope-

3

line strictly shifts area from one side to the other, so

\ell

must be this line.

Its equation is

y - \frac{1}{2} = 3(x - 1),

that is,

6x = 2y + 5.

With

\gcd(6, 2, 5) = 1,

the answer is

a^2 + b^2 + c^2 = 36 + 4 + 25 = 65.

Problem 10 in other years