2024 AMC 10B Problem 21

Look at the pipes head-on: they are circles sitting on a line (the floor). The trick is to measure everything along the floor.

A useful rule: where two touching circles meet the floor.

1. Take circles of radius $r$ and $s$ on the floor, touching each other. Their centers are at heights $r$ and $s$, and the distance between the centers is $r + s$.
2. Draw a right triangle: the slanted side joins the centers (length $r + s$), the vertical side is the difference in heights ($r - s$), and the horizontal side $d$ is the distance between the two floor points.
3. By the Pythagorean theorem:
$$d^2 = (r + s)^2 - (r - s)^2 = 4rs, \qquad \text{so } d = 2\sqrt{rs}.$$

Use the rule.

4. For the given pipes ($1$ and $\tfrac14$): the floor points are $2\sqrt{1 \cdot \tfrac14} = 2 \cdot \tfrac12 = 1$ apart.
5. Let the third pipe have radius $x$. Its floor point is $2\sqrt{x}$ from the big pipe's floor point and $2\sqrt{x \cdot \tfrac14} = \sqrt{x}$ from the small pipe's.
6. The third pipe can sit in three places. Check each:
- Between the two pipes. The two distances add to $1$: $2\sqrt{x} + \sqrt{x} = 1$, so $\sqrt{x} = \tfrac13$ and $x = \tfrac19$.
- Beyond the small pipe. The distance to the big pipe is $1$ more than the distance to the small pipe: $2\sqrt{x} = 1 + \sqrt{x}$, so $\sqrt{x} = 1$ and $x = 1$.
- Beyond the big pipe. Then $\sqrt{x} = 1 + 2\sqrt{x}$, which would make $\sqrt{x}$ negative. Impossible.
7. The possible radii are $\tfrac19$ and $1$. Their sum is $\tfrac19 + 1 = \tfrac{10}{9}$.

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

Circles that sit on the same line and touch each other follow one simple rule: their floor points are $2\sqrt{rs}$ apart. That turns a circle problem into adding and subtracting distances along one line. The question asks for a sum of radii, which is a hint that more than one position works, so check every place the third pipe could go.