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

Call the point of tangency point $F$ and the midpoint of $AB$ as $G$ . $CF=2$ by Tangent Theorem. Notice that $\angle EGF=\frac{180-2\cdot\angle CGF}{2}=90-\angle CGF$ . Thus, $\angle EGF=\angle FCG$ and $tanEGF=tanFCG=\frac{1}{2}$ . Solving $EF=\frac{1}{2}$ . Adding, the answer is $\frac{5}{2}$ .

Solution 3

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 4

$[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label("$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)); label("$G$",G,(0,-1)); dot(G); draw(G--C); label("$\sqrt{5}$",(G+C)/2,(-1,0)); [/asy]$

Let us call the midpoint of side $AB$ , point $G$ . Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$ . We get $GC=\sqrt{5}$ . We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$ , we get similar triangles $EFG$ and $GFC$ . Solving the ratios, we get $x=\frac{1}{2}$ , so the answer is $\frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$ .

Solution 5

Using the diagram as drawn in Solution 5, let the total area of square $ABCD$ be divided into the triangles $DCE$ , $EAG$ , $CGB$ , and $EGC$ . Let x be the length of AE. Thus, the area of each triangle can be determined as follows:

$DCE = \frac{DC\cdot{DE}}{2} = \frac{2\cdot(2-x)}{2} = 1-x$

$EAG= \frac{AE\cdot{AG}}{2} = \frac{1\cdot{x}}{2} = \frac{x}{2}$

$CGB = \frac{GB\cdot{CB}}{2} = \frac{1\cdot(2)}{2} = 1$

$EGC= \frac{EG\cdot{GC}}{2} = \frac{\sqrt{4+(2-x)^2}}{2}$ (the length of CE is calculated with the Pythagorean Theorem, lines GE and CE are perpendicular by definition of tangent)