# Difference between revisions of "2004 AMC 12A Problems/Problem 18"

The following problem is from both the 2004 AMC 12A #18 and 2004 AMC 10A #22, so both problems redirect to this page.

## Problem

Square $ABCD$ has side length $2$. A semicircle with diameter $\overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $\overline{AD}$ at $E$. What is the length of $\overline{CE}$?

$[asy] size(100); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180)); label("A",A,(-1,-1)); label("B",B,( 1,-1)); label("C",C,( 1, 1)); label("D",D,(-1, 1)); label("E",E,(-1, 0)); [/asy]$

$\mathrm{(A) \ } \frac{2+\sqrt{5}}{2} \qquad \mathrm{(B) \ } \sqrt{5} \qquad \mathrm{(C) \ } \sqrt{6} \qquad \mathrm{(D) \ } \frac{5}{2} \qquad \mathrm{(E) \ } 5-\sqrt{5}$

## Solution 1

$[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2; draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label("A",A,(-1,-1)); label("B",B,( 1,-1)); label("C",C,( 1, 1)); label("D",D,(-1, 1)); label("E",E,(-1, 0)); label("F",F,( 0, 1)); label("x",(A+E)/2,(-1, 0)); label("x",(E+F)/2,( 0, 1)); label("2",(F+C)/2,( 0, 1)); label("2",(D+C)/2,( 0, 1)); label("2",(B+C)/2,( 1, 0)); label("2-x",(D+E)/2,(-1, 0)); [/asy]$ Let the point of tangency be $F$. By the Two Tangent Theorem $BC = FC = 2$ and $AE = EF = x$. Thus $DE = 2-x$. The Pythagorean Theorem on $\triangle CDE$ yields

\begin{align*} DE^2 + CD^2 &= CE^2\\ (2-x)^2 + 2^2 &= (2+x)^2\\ x^2 - 4x + 8 &= x^2 + 4x + 4\\ x &= \frac{1}{2}\end{align*}

Hence $CE = FC + x = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.

## Solution 2

Clearly, $EA = EF = BG$. Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = 5/2$.

== Solution 3 ==


TBE