2004 AMC 10B Problem 12

1. The ring (annulus) is the big disk with the small disk cut out. Its area is
$$\pi b^2 - \pi c^2 = \pi(b^2 - c^2).$$
So we need to connect $b^2 - c^2$ to one of the lengths $a, d, e$.
2. Look for a right triangle, since squares of lengths suggest the Pythagorean theorem. $XZ$ touches the small circle at $Z$ (it is tangent there), and $OZ$ is a radius of the small circle. A radius always meets a tangent line at a right angle, so the angle at $Z$ in triangle $OZX$ is $90^\circ$.
3. In that right triangle, the legs are $OZ = c$ and $ZX = a$, and the longest side is $OX = b$ (a radius of the big circle). So
$$a^2 + c^2 = b^2, \qquad \text{which means} \qquad b^2 - c^2 = a^2.$$
4. The area of the ring is $\pi a^2$.

The answer is $\boxed{\textbf{(A)}\ \pi a^2}$.

"Radius meets tangent at a right angle" is the standard way to find a right triangle inside a circle problem. The area of a ring is $\pi$ times a difference of two squares, and a right triangle is exactly what turns a difference of squares into one square. The lengths $d$ and $e$ are in the picture only to tempt you.

Test an extreme case. Let the small circle shrink to a point, so $c = 0$. Then the ring is the whole big disk, with area $\pi b^2$. Also $Z$ lands on $O$, so $a = XZ = XO = b$. Choice (A) gives $\pi a^2 = \pi b^2$, which matches. This is a quick check; the right-triangle argument is what proves it for every size.

The annulus is the big disk minus the small disk, so its area is
$$\pi b^2 - \pi c^2 = \pi\,(b^2 - c^2).$$

Now relate $b^2 - c^2$ to the labeled segments. $XZ$ is tangent to the small circle at $Z$, and $OZ$ is a radius of that circle, so $OZ \perp XZ$. Triangle $OZX$ is therefore right-angled at $Z$ with legs $OZ = c$ and $ZX = a$ and hypotenuse $OX = b$. The Pythagorean theorem gives
$$a^2 + c^2 = b^2 \quad\Longrightarrow\quad b^2 - c^2 = a^2.$$

Hence the annulus has area $\pi a^2$.

The answer is $\boxed{\textbf{(A)}\ \pi a^2}$.