2004 AMC 10A Problem 22

Let $F$ be the point where $\overline{CE}$ touches the semicircle. Because $\overline{BC}$ and $\overline{AD}$ are perpendicular to the diameter $\overline{AB}$ at its endpoints, they are tangent to the semicircle at $B$ and $A$.

Equal tangents from an external point:

• From $C$: $CF = CB = 2$.
• From $E$: $EF = EA$. Call this length $x$.

Then $CE = CF + FE = 2 + x$, while along the side $DE = DA - EA = 2 - x$.

Apply the Pythagorean theorem to right triangle $CDE$ (right angle at $D$):
$$(2 - x)^2 + 2^2 = (2 + x)^2.$$
Expanding, $4 - 4x + x^2 + 4 = 4 + 4x + x^2$, so $8x = 4$ and $x = \tfrac12$.

Therefore $CE = 2 + \tfrac12 = \tfrac52$. (Check: $DE = \tfrac32$, $DC = 2$, $CE = \tfrac52$ is a scaled $3$-$4$-$5$ triangle.)

The answer is $\boxed{\textbf{(D)}\ \frac{5}{2}}$.

Whenever a tangent line meets other tangent lines, mark the tangency points and label equal tangent segments; that converts the circle into pure length bookkeeping. Here the square's sides are hidden tangents, which is why the unknown $x$ appears both along $\overline{CE}$ and along $\overline{AD}$ and a single right triangle finishes the problem.

Coordinates: $A = (0,0)$, $B = (2,0)$, $C = (2,2)$, center $(1,0)$, radius $1$. A line through $C$ with slope $m$ is $mx - y + 2 - 2m = 0$; its distance from $(1,0)$ is $\frac{|2 - m|}{\sqrt{m^2+1}} = 1$, so $(2-m)^2 = m^2 + 1$ and $m = \tfrac34$. At $x = 0$ the line has $y = \tfrac12$, so $E = (0, \tfrac12)$ and $CE = \sqrt{4 + \tfrac94} = \tfrac52$.

• Not recognizing that B and A are tangency points, so the equal-tangent lengths CB = CF and AE = EF go unused.
• Writing $DE = x$ instead of $2 - x$ in the Pythagorean equation.