2004 AMC 10A Problem 23

Two circles that touch from the outside have centers (sum of radii) apart. If one circle touches another from the inside, the centers are (difference of radii) apart. Everything below uses just these two facts.

1. Radius of $D$. Circle $A$ touches $D$ from the inside and also passes through $D$'s center. So a full diameter of $A$, which is $2$, stretches from $D$'s center to $D$'s edge. Circle $D$ has radius $2$.
2. Coordinates. Put $D$'s center at $O = (0, 0)$ and $A$'s center at $P = (-1, 0)$. Circles $B$ and $C$ are the same size and sit symmetrically above and below the horizontal line, so call $B$'s center $Q = (x, r)$, where $r$ is the radius we want. ($C$'s center is then $(x, -r)$, exactly $2r$ from $Q$, so $B$ and $C$ touch automatically.)
3. $B$ touches $D$ from the inside, so $OQ = 2 - r$:
$$x^2 + r^2 = (2 - r)^2 = 4 - 4r + r^2 \quad\Longrightarrow\quad x^2 = 4 - 4r.$$
4. $B$ touches $A$ from the outside, so $PQ = 1 + r$:
$$(x + 1)^2 + r^2 = (1 + r)^2 \quad\Longrightarrow\quad x^2 + 2x + 1 = 1 + 2r \quad\Longrightarrow\quad x^2 + 2x = 2r.$$
5. Put step 3 into step 4: $4 - 4r + 2x = 2r$, so $2x = 6r - 4$ and $x = 3r - 2$.
6. Put this back into step 3:
$$(3r - 2)^2 = 4 - 4r \quad\Longrightarrow\quad 9r^2 - 12r + 4 = 4 - 4r \quad\Longrightarrow\quad 9r^2 = 8r.$$
7. The radius is not $0$, so divide by $r$: $9r = 8$ and $r = \tfrac89$.
8. Check: $x = 3 \cdot \tfrac89 - 2 = \tfrac23$, and $x^2 = \tfrac49 = 4 - \tfrac{32}{9}$. It fits.

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

Touching circles are completely described by their centers and radii: from the outside, the centers are the sum of the radii apart; from the inside, the difference. Putting the coordinates along the line of symmetry leaves two unknowns and two distance equations, and subtracting one from the other removes the squared terms.

There is a known formula, Descartes' circle theorem, for four circles that all touch each other. Using "bending" $= \tfrac{1}{\text{radius}}$, with the outer circle $D$ counted as negative: $-\tfrac12$ for $D$, $1$ for $A$, and $\tfrac1r$ for each of $B$ and $C$. The theorem says (sum of bendings)$^2$ $= 2 \times$ (sum of squared bendings):
$$\left(-\tfrac12 + 1 + \tfrac2r\right)^2 = 2\left(\tfrac14 + 1 + \tfrac{2}{r^2}\right).$$
Multiplying out gives $\tfrac14 + \tfrac2r + \tfrac{4}{r^2} = \tfrac52 + \tfrac{4}{r^2}$, so $\tfrac2r = \tfrac94$ and $r = \tfrac89$. This is quick if you know the formula, but it is outside the usual school toolkit.

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