2011 AMC 12A Problem 17

Question

Circles with radii

1,

2,

and

3

are mutually externally tangent. What is the area of the triangle determined by the points of tangency?

\dfrac{3}{5}

\dfrac{4}{5}

1

\dfrac{6}{5}

\dfrac{4}{3}

The centers form a triangle with sides

1+2,

2+3,

1+3,

i.e. a

3

-

4

-

5

right triangle

Put the right angle at the center of the radius-

1

circle and locate each tangency point on the connecting segment

The centers are separated by the sums of radii:

3,

4,

and

5,

a right triangle with the right angle at the radius-

1

center. Place that center at

(0,0),

the radius-

2

center at

(3,0),

and the radius-

3

center at

(0,4).

The tangency points lie on the segments at distances equal to the radii:

(1, 0),

(0, 1),

and on the hypotenuse at

(3,0) + 2 \cdot \tfrac{(-3,4)}{5} = \left(\tfrac95, \tfrac85\right).

By the shoelace formula the area is

\begin{gathered} \tfrac12\left| 1\left(1 - \tfrac85\right) + 0 + \tfrac95(0 - 1) \right| \\ = \tfrac12 \cdot \tfrac{12}{5} = \tfrac65. \end{gathered}

Thus, the correct answer is D.

Problem 17 in other years