2004 AMC 10A Problem 23

Circle $A$ is internally tangent to $D$ and passes through $D$'s center, so a diameter of $A$ lies along a radius of $D$: circle $D$ has radius $2$.

Place $D$'s center at the origin $O$ and $A$'s center at $P = (-1, 0)$. By the symmetry of $B$ and $C$, their centers are $Q = (x, r)$ and $(x, -r)$, where $r$ is the unknown radius; the two circles then touch each other automatically, since their centers are $2r$ apart.

Two tangency conditions remain.

Circle $B$ inside $D$: $OQ = 2 - r$, so
$$x^2 + r^2 = (2 - r)^2 \quad\Longrightarrow\quad x^2 = 4 - 4r.$$

Circle $B$ touching $A$ externally: $PQ = 1 + r$, so
$$(x + 1)^2 + r^2 = (1 + r)^2 \quad\Longrightarrow\quad x^2 + 2x = 2r.$$

Substitute the first into the second: $4 - 4r + 2x = 2r$, giving $x = 3r - 2$. Then
$$(3r - 2)^2 = 4 - 4r \quad\Longrightarrow\quad 9r^2 - 12r + 4 = 4 - 4r \quad\Longrightarrow\quad 9r^2 = 8r.$$
Since $r \neq 0$, $r = \tfrac89$. (Then $x = \tfrac23$, consistent with $x^2 = 4 - \tfrac{32}{9} = \tfrac49$.)

The answer is $\boxed{\textbf{(D)}\ \frac{8}{9}}$.

Tangent circles are entirely described by their centers and radii: external tangency means center distance equals the sum of the radii, internal tangency means the difference. Choosing coordinates that respect the symmetry (the $x$-axis through the centers of $A$ and $D$) leaves two unknowns and two distance equations, and subtracting them removes the squares.

Descartes' circle theorem with curvatures $k_D = -\tfrac12$ (enclosing), $k_A = 1$, $k_B = k_C = \tfrac1r$: $\left(-\tfrac12 + 1 + \tfrac2r\right)^2 = 2\left(\tfrac14 + 1 + \tfrac{2}{r^2}\right)$. Expanding, $\tfrac14 + \tfrac2r + \tfrac{4}{r^2} = \tfrac52 + \tfrac{4}{r^2}$, so $\tfrac2r = \tfrac94$ and $r = \tfrac89$.

• Forgetting that internal tangency means the center distance is 2 - r (not 2 + r), or taking the radius of D to be 1.
• Squaring an equation like $2\sqrt{1 - r} = 3r - 2$ without discarding the extraneous root, or an arithmetic slip in expanding $(3r-2)^2$.