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

(Solution 3)
(Solution 3)
Line 63: Line 63:
 
defaultpen(fontsize(10));
 
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;
 
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(A--B--C--D--cycle);draw(C--E);draw(
 
draw(Arc((1,0),1,0,180));draw((A+B)/2--F);
 
draw(Arc((1,0),1,0,180));draw((A+B)/2--F);
 
label("$A$",A,(-1,-1));
 
label("$A$",A,(-1,-1));
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label("$x$",(A+E)/2,(-1, 0));
 
label("$x$",(A+E)/2,(-1, 0));
 
label("$x$",(E+F)/2,( 0, 1));
 
label("$x$",(E+F)/2,( 0, 1));
label(
+
label("$2$",(F+C)/2,( 0, 1));
 +
label("$2$",(D+C)/2,( 0, 1));
 +
label("$2$",(B+C)/2,( 1, 0));
 +
label("$O$, (1,0), S);
 
</asy>
 
</asy>
  

Revision as of 14:24, 25 July 2015

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

2004 AMC12A-18.png

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

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(
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("$O$, (1,0), S);
 (Error compiling LaTeX. f1ac31ee1432b29fdca53b4f270214bb276b48e9.asy: 7.25: syntax error
error: could not load module 'f1ac31ee1432b29fdca53b4f270214bb276b48e9.asy')

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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