2024 AMC 10B Problem 21

A lemma. Let two circles of radii $r$ and $s$ rest on the floor and touch each other. Their centers are at heights $r$ and $s$, and the distance between centers is $r + s$. Drop a perpendicular from the higher center to the horizontal line through the lower one: the horizontal leg $d$ (the distance between the two floor contact points) satisfies
$$d^2 = (r + s)^2 - (r - s)^2 = 4rs, \qquad d = 2\sqrt{rs}.$$

For the given pipes, the contact points are $2\sqrt{1 \cdot \tfrac14} = 1$ apart.

The third pipe, radius $x$, has a contact point at horizontal distance $2\sqrt{x}$ from the big pipe's and $2\sqrt{x/4} = \sqrt{x}$ from the small pipe's. Its contact point is either between the other two or outside them.

• Between: $2\sqrt{x} + \sqrt{x} = 1$, so $\sqrt{x} = \tfrac13$ and $x = \tfrac19$.
• Beyond the small pipe: $2\sqrt{x} = 1 + \sqrt{x}$, so $\sqrt{x} = 1$ and $x = 1$.
• Beyond the big pipe: $\sqrt{x} = 1 + 2\sqrt{x}$ has no positive solution.

The possible radii are $\tfrac19$ and $1$, with sum $\tfrac{10}{9}$.

The answer is $\boxed{\textbf{(C)}\ \frac{10}{9}}$.

Circles resting on a common line and touching each other are governed by one clean relation: contact points $2\sqrt{rs}$ apart. That converts a tangency problem into additions and subtractions of square roots of radii along the floor, so the whole configuration lives on one number line. The question asks for a sum, a hint that more than one pipe works; check every side.

Descartes' circle theorem with the floor as a circle of curvature $0$: curvatures $k$ satisfy $(1 + 4 + k)^2 = 2(1 + 16 + k^2)$, i.e. $k^2 - 10k + 9 = 0$, so $k = 1$ or $9$ and the radii are $1$ and $\frac19$.

• Finding only the small pipe wedged between the two and answering 1/9, forgetting that a pipe on the far side of the small one is also in contact with both.
• Setting the center-to-center distance equal to the horizontal gap (ignoring the vertical offset $r - s$), which breaks the $2\sqrt{rs}$ relation.