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

Revision as of 20:55, 24 June 2019

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)

Adding up the areas and equating to the area of the total square (2*2=4), we get

$1-x+\frac{x}{2}+1+ \frac{\sqrt{4+(2-x)^2}}{2} = 4$

Solving for x:

$2-2x+x+1+\sqrt{x^2-4x+8}=8$ $\sqrt{x^2-4x+8}=x+2$ $x^2-4x+8=x^2+4x+4$ $8-4x=4x+4 \rightarrow x=\frac{1}{2}$

Solving for length of CE with the value we have for x: $\sqrt{4+(2-x)^2} = \sqrt{4+(3/2)^2} = \sqrt{25/4} = \boxed{\frac{5}{2}}$

@@ Line 53: / Line 53: @@
 == Solution 2 ==
-Call the point of tangency point <math>F</math> and the midpoint of <math>AB</math> as <math>G</math>. <math>CF=2</math> by Tangent Theorem. Notice that <math>\angle EGF=\frac{180-2\cdot\angle CGF}{2}=90-\angle CGF</math>. Thus, <math>\angle EGF=\angle FCG</math> and <math>tanEGF=tanFCG=\frac{1}{2}</math>. Solving <math>ER=\frac{1}{2}</math>. Adding, the answer is <math>\frac{5}{2}</math>.
+Call the point of tangency point <math>F</math> and the midpoint of <math>AB</math> as <math>G</math>. <math>CF=2</math> by Tangent Theorem. Notice that <math>\angle EGF=\frac{180-2\cdot\angle CGF}{2}=90-\angle CGF</math>. Thus, <math>\angle EGF=\angle FCG</math> and <math>tanEGF=tanFCG=\frac{1}{2}</math>. Solving <math>EF=\frac{1}{2}</math>. Adding, the answer is <math>\frac{5}{2}</math>.
 == Solution 3 ==

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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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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