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

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=== Solution 5 ===
 
=== Solution 5 ===
Using the diagram as drawn in Solution 5, let the total area of square <math>ABCD</math> be divided into the triangles <math>DCE</math>, <math>EAG</math>, <math>CGB</math>, and <math>EGC</math>. Let x be the length of AE. Thus, the area of each triangle can be determined as follows:  
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Using the diagram as drawn in Solution 4, let the total area of square <math>ABCD</math> be divided into the triangles <math>DCE</math>, <math>EAG</math>, <math>CGB</math>, and <math>EGC</math>. Let x be the length of AE. Thus, the area of each triangle can be determined as follows:  
  
 
<cmath>DCE = \frac{DC\cdot{DE}}{2} = \frac{2\cdot(2-x)}{2} = 1-x</cmath>
 
<cmath>DCE = \frac{DC\cdot{DE}}{2} = \frac{2\cdot(2-x)}{2} = 1-x</cmath>
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Solving for length of CE with the value we have for x:
 
Solving for length of CE with the value we have for x:
 
<cmath> \sqrt{4+(2-x)^2} = \sqrt{4+(3/2)^2} = \sqrt{25/4} = \boxed{\frac{5}{2}} </cmath>
 
<cmath> \sqrt{4+(2-x)^2} = \sqrt{4+(3/2)^2} = \sqrt{25/4} = \boxed{\frac{5}{2}} </cmath>
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=== Video Solution ===
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https://youtu.be/pM0zICtH6Lg
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Education, the Study of Everything
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2004|ab=A|num-b=17|num-a=19}}
 
{{AMC12 box|year=2004|ab=A|num-b=17|num-a=19}}
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{{AMC10 box|year=2004|ab=A|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:21, 21 April 2021

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]

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

Solutions

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

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 $DC = 2$, $CE = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.

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 4, 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: \[\sqrt{4+(2-x)^2} = \sqrt{4+(3/2)^2} = \sqrt{25/4} = \boxed{\frac{5}{2}}\]

Video Solution

https://youtu.be/pM0zICtH6Lg

Education, the Study of Everything

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

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