2023 AMC 10A Problem 22

Two rules about touching circles turn everything into distances between centers:

• If a small circle touches a big one from the inside, the centers are (big radius) − (small radius) apart.
• If two circles touch from the outside, the centers are (one radius) + (other radius) apart.

1. Set up. Call the centers of $C_1$ and $C_2$ $O_1$ and $O_2$. They are $\frac12$ apart. Let $M$ be the point halfway between them, so $M$ is $\frac14$ from each.
2. Find $C_3$. The largest circle inside both big circles sits right in the middle of their overlap, centered at $M$. It touches $C_1$ from the inside, so its radius is $1 - \frac14 = \frac34$.
3. Where is $C_4$'s center? $C_4$ touches $C_1$ and $C_2$ in the same way, so its center $O_4$ is the same distance from $O_1$ as from $O_2$. That puts it on the line through $M$ at a right angle to $O_1O_2$ (straight "above" $M$ in the picture).
4. Write the distances. Call the radius of $C_4$ $r$.
- $C_4$ touches $C_3$ from the outside, so $MO_4 = \frac34 + r$.
- $C_4$ touches $C_1$ from the inside, so $O_1O_4 = 1 - r$.
- And $O_1M = \frac14$.
5. Right triangle. Triangle $O_1 M O_4$ has a right angle at $M$. By the Pythagorean theorem,
$$\left(\tfrac14\right)^2 + \left(\tfrac34 + r\right)^2 = (1 - r)^2.$$
6. Solve. Expand both sides:
$$\tfrac{1}{16} + \tfrac{9}{16} + \tfrac32 r + r^2 = 1 - 2r + r^2.$$
The $r^2$ cancels. Collect terms: $\tfrac32 r + 2r = 1 - \tfrac{10}{16}$, so $\tfrac72 r = \tfrac38$ and
$$r = \frac38 \times \frac27 = \frac{3}{28}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{3}{28}}$.

Touching circles are really about distances between centers: subtract the radii for touching inside, add them for touching outside. Once you place the centers (here the picture's line of symmetry does most of the work), the whole problem is one right triangle. Always look for the symmetry line first in a problem with several touching circles.