2004 AMC 12B Problems/Problem 10

The following problem is from both the 2004 AMC 12B #10 and 2004 AMC 10B #12, so both problems redirect to this page.

Problem

An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$, with $b>c$. Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$, and let $OY$ be the radius of the larger circle that contains $Z$. Let $a=XZ$, $d=YZ$, and $e=XY$. What is the area of the annulus?

$[asy] unitsize(1.5cm); defaultpen(0.8); real r1=1.5, r2=2.5; pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2); pair Y=(0,r2), Z=(0,r1), X=intersectionpoint( Z--(Z+(10,0)), outer ); filldraw(outer,lightgray,black); filldraw(inner,white,black); draw(X--O--Y); draw(Y--X--Z); label("O",O,SW); label("X",X,E); label("Y",Y,N); label("Z",Z,SW); label("a",X--Z,N); label("b",0.25*X,SE); label("c",O--Z,E); label("d",Y--Z,W); label("e",Y*0.65 + X*0.35,SW); defaultpen(0.5); dot(O); dot(X); dot(Z); dot(Y); [/asy]$

$\mathrm{(A) \ } \pi a^2 \qquad \mathrm{(B) \ } \pi b^2 \qquad \mathrm{(C) \ } \pi c^2 \qquad \mathrm{(D) \ } \pi d^2 \qquad \mathrm{(E) \ } \pi e^2$

Solution

The area of the large circle is $\pi b^2$, the area of the small one is $\pi c^2$, hence the shaded area is $\pi(b^2-c^2)$.

From the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$, hence $b^2-c^2=a^2$ and thus the shaded area is $\boxed{\mathrm{(A)\ }\pi a^2}$.

Solution 2

Set $c=0,$ then the shaded area is just the area of a circle with radius $a,$ which is $\boxed{\mathrm{(A)\ }\pi a^2}$.