2007 AMC 10B Problem 18

When circles touch, the distance between their centers is the sum of their radii. So join centers and look for a triangle.

1. Call the middle circle's center $O$. Pick two neighboring outer circles, with centers $P$ and $Q$.
2. $P$'s circle touches the middle circle, so $OP = 1 + r$. Likewise $OQ = 1 + r$.
3. The two outer circles touch each other, so $PQ = r + r = 2r$.
4. By symmetry, the four outer centers are spread evenly around $O$, a quarter turn apart. So the angle at $O$ between $OP$ and $OQ$ is $90^\circ$.
5. Triangle $OPQ$ is a right triangle with two equal legs of $1 + r$. Its long side is a leg times $\sqrt2$:
$$2r = (1 + r)\sqrt2.$$
6. Solve. Multiply out: $2r = \sqrt2 + \sqrt2\,r$. Collect the $r$ terms: $r(2 - \sqrt2) = \sqrt2$. So
$$r = \frac{\sqrt2}{2 - \sqrt2}.$$
Multiply top and bottom by $2 + \sqrt2$:
$$r = \frac{\sqrt2\,(2 + \sqrt2)}{4 - 2} = \frac{2\sqrt2 + 2}{2} = 1 + \sqrt2.$$
7. Check: $r \approx 2.414$. Then $1 + r \approx 3.414$, and $3.414 \times 1.414 \approx 4.83 \approx 2r$.

The answer is $\boxed{\textbf{(B)}\ 1+\sqrt{2}}$.

Touching circles become a problem about triangles once you join their centers: every point of contact gives a segment whose length is the sum of two radii. Here the symmetry of four circles around one makes a right angle at the middle, and the Pythagorean theorem does the rest.

Join all four outer centers. They form a square with side $2r$, centered at $O$. The distance from the middle of a square to a corner is half the diagonal, $\frac{2r\sqrt2}{2} = r\sqrt2$. That same distance is $1 + r$, so $r\sqrt2 = 1 + r$, giving $r(\sqrt2 - 1) = 1$ and $r = \frac{1}{\sqrt2 - 1} = \sqrt2 + 1$.