2007 AMC 10B Problem 18

Connect the centers of the four outer circles. Adjacent outer circles are tangent, so neighboring centers are $2r$ apart, and by symmetry the four centers form a square of side $2r$ centered at the small circle's center $O$.

The distance from $O$ to a corner of that square is half the diagonal:
$$\frac{2r\sqrt2}{2} = r\sqrt2.$$
But each outer circle is also tangent to the central circle, so that same distance equals the sum of the radii, $1 + r$. Therefore
$$r\sqrt2 = 1 + r \quad\Longrightarrow\quad r(\sqrt2 - 1) = 1 \quad\Longrightarrow\quad r = \frac{1}{\sqrt2 - 1} = \sqrt2 + 1.$$

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

Tangent-circle problems become polygon problems once you join the centers: every tangency is a segment of known length (sum of radii). Here the square of centers gives two expressions for the same distance, one from geometry ($r\sqrt2$) and one from tangency ($1 + r$).

Look at the right isosceles triangle formed by $O$ and two adjacent outer centers: its legs are $1 + r$ and its hypotenuse is $2r$, so $(1 + r)\sqrt2 = 2r$, again giving $r = \sqrt2 + 1$. Numerically $r \approx 2.41$, and only choice (B) is close.

• Using the square's side or a leg of the wrong triangle as 1 + r, producing r = 1 or r = sqrt(2).
• Stopping at $r = \frac{1}{\sqrt2 - 1}$ without rationalizing and failing to match it to $1 + \sqrt2$.