Question

A circle of radius $1$ is tangent to a circle of radius $2.$ The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $AB$ and $AC$ are congruent. What is the area of $\triangle ABC?$

$\dfrac{35}{2}$

$15\sqrt{2}$

$\dfrac{64}{3}$

$16\sqrt{2}$

$24$

Solution:

Let $O, O'$ be the centers of the small and large circles, and let $D$ be the point where the small circle touches $AC.$ The right triangles cut off along $AC$ are similar, so $\frac{AO}{1} = \frac{AO + 3}{2},$ giving $AO = 3$ and $AO' = 6.$

The tangent length is $AD = \sqrt{AO^2 - 1^2} = \sqrt{3^2 - 1^2} = 2\sqrt2.$ Let $F$ be the midpoint of $BC$ ; then $AF = AO' + 2 = 8.$

Since $\triangle ADO \sim \triangle AFC,$ we get $\frac{FC}{1} = \frac{AF}{2\sqrt2} = \frac{8}{2\sqrt2} = 2\sqrt2.$ Thus $BC = 4\sqrt2,$ and the area is $\frac12 \cdot BC \cdot AF = \frac12 \cdot 4\sqrt2 \cdot 8 = 16\sqrt2.$

Thus, the correct answer is D.

2006 AMC 10A Problem 16

Solution:

Problem 16 in Other Years