2004 AMC 10A Problem 22

1. Let $F$ be the point where segment $CE$ touches the semicircle.
2. Sides $BC$ and $AD$ are perpendicular to the diameter $AB$ at its ends, so they also just touch the semicircle, at $B$ and at $A$.
3. From a point outside a circle, the two segments that just touch the circle have equal length. Use this twice.
- From $C$: $CF = CB = 2$.
- From $E$: $EF = EA$. Call this length $x$.
4. Now write two lengths using $x$:
$$CE = CF + FE = 2 + x, \qquad DE = DA - EA = 2 - x.$$
5. Triangle $CDE$ has a right angle at $D$, with legs $DE$ and $DC = 2$. By the Pythagorean theorem,
$$(2 - x)^2 + 2^2 = (2 + x)^2.$$
6. Multiply out: $4 - 4x + x^2 + 4 = 4 + 4x + x^2$. The $x^2$ terms cancel, leaving $4 = 8x$, so $x = \tfrac12$.
7. So $CE = 2 + \tfrac12 = \tfrac52$. Check: the triangle has sides $\tfrac32$, $2$, $\tfrac52$, which is a $3$-$4$-$5$ triangle cut in half.

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

When lines touch a circle, mark every touching point and label the equal lengths from each outside point. That turns the circle problem into adding and subtracting lengths. Here the square's own sides are hidden touching lines, which is why the unknown $x$ shows up both along $CE$ and along $AD$, and one right triangle finishes the job.

Use coordinates. Put $A = (0,0)$, $B = (2,0)$, $C = (2,2)$, $D = (0,2)$. The semicircle has center $(1,0)$ and radius $1$.

1. Point $E$ is $(0, e)$ for some $e$. The line through $C$ and $E$ must be exactly $1$ unit from the center $(1,0)$.
2. The line through $(2,2)$ with slope $m$ has distance $\dfrac{|2 - m|}{\sqrt{m^2 + 1}}$ from $(1,0)$. Setting this equal to $1$ gives $(2 - m)^2 = m^2 + 1$, so $4 - 4m = 1$ and $m = \tfrac34$.
3. Going left $2$ units from $C$ drops the line by $\tfrac32$, so $E = (0, \tfrac12)$.
4. Then $CE = \sqrt{2^2 + \left(\tfrac32\right)^2} = \sqrt{\tfrac{25}{4}} = \tfrac52$.