Question

Circles $A,$ $B,$ and $C$ are externally tangent to each other and internally tangent to circle $D.$ Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D.$ What is the radius of circle $B?$

$\dfrac{2}{3}$

$\dfrac{\sqrt{3}}{2}$

$\dfrac{7}{8}$

$\dfrac{8}{9}$

$\dfrac{1 + \sqrt{3}}{3}$

Solution:

Because circle $A$ passes through $D$ 's center and is internally tangent to $D,$ circle $D$ has radius $2.$ Place $D$ 's center at the origin and $A$ 's center at $(-1, 0).$

Let circle $B$ have radius $r$ and center $(x, r),$ using the symmetry of $B$ and $C$ about the horizontal axis. Tangency gives $(x + 1)^2 + r^2 = (1 + r)^2, \qquad x^2 + r^2 = (2 - r)^2.$

Subtracting yields $x = 3r - 2.$ Substituting into the second equation gives $9r^2 - 8r = 0,$ so $r = \dfrac{8}{9}.$

Thus, the correct answer is D.

2004 AMC 10A Problem 23

Solution:

Problem 23 in Other Years