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?

$\dfrac{3}{5}$

$\dfrac{4}{5}$

$1$

$\dfrac{6}{5}$

$\dfrac{4}{3}$

Solution:

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

Thus, the correct answer is D.

2011 AMC 12A Problem 17

Solution:

Problem 17 in Other Years