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

(sol, {{dup}})
Line 4: Line 4:
 
[[Square]] <math>ABCD</math> has side length <math>2</math>. A [[semicircle]] with [[diameter]] <math>\overline{AB}</math> is constructed inside the square, and the [[tangent (geometry)|tangent]] to the semicircle from <math>C</math> intersects side <math>\overline{AD}</math> at <math>E</math>. What is the length of <math>\overline{CE}</math>?
 
[[Square]] <math>ABCD</math> has side length <math>2</math>. A [[semicircle]] with [[diameter]] <math>\overline{AB}</math> is constructed inside the square, and the [[tangent (geometry)|tangent]] to the semicircle from <math>C</math> intersects side <math>\overline{AD}</math> at <math>E</math>. What is the length of <math>\overline{CE}</math>?
  
−
<center>[[Image:AMC10_2004A_22.png]]</center>
+
<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>
  
 
<math> \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} </math>
 
<math> \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} </math>
Line 11: Line 22:
 
==Solution==
 
==Solution==
 
=== Solution 1 ===
 
=== 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 <math>F</math>. By the [[Two Tangent Theorem]] <math>BC = FC = 2</math> and <math>AE = EF = x</math>. Thus <math>DE = 2-x</math>. The [[Pythagorean Theorem]] on <math>\triangle CDE</math> yields
 
Let the point of tangency be <math>F</math>. By the [[Two Tangent Theorem]] <math>BC = FC = 2</math> and <math>AE = EF = x</math>. Thus <math>DE = 2-x</math>. The [[Pythagorean Theorem]] on <math>\triangle CDE</math> yields
  
−
<cmath>\begin{eqnarray*}(2-x)^2 + 2^2 &=& (2+x)^2\\
+
<cmath>\begin{align*}
−
x^2 - 4x + 8 &=& x^2 + 4x + 4\\
+
DE^2 + CD^2 &= CE^2\\
−
x &=& \frac{1}{2}\end{eqnarray*}</cmath>
+
(2-x)^2 + 2^2 &= (2+x)^2\\
 +
x^2 - 4x + 8 &= x^2 + 4x + 4\\
 +
x &= \frac{1}{2}\end{align*}</cmath>
  
 
Hence <math>CE = FC + x = \frac{5}{2} \Rightarrow \mathrm{(D)}</math>.
 
Hence <math>CE = FC + x = \frac{5}{2} \Rightarrow \mathrm{(D)}</math>.

Revision as of 12:07, 13 January 2012

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

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

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 $DC = 2$, $CE = 5/2$.

See also

2004 AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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
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