Difference between revisions of "2004 AMC 12A Problems/Problem 18"
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[[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>? | ||
− | < | + | <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> | ||
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==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{ | + | <cmath>\begin{align*} |
− | x^2 - 4x + 8 &= | + | DE^2 + CD^2 &= CE^2\ |
− | x &= | + | (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 has side length . A semicircle with diameter is constructed inside the square, and the tangent to the semicircle from intersects side at . What is the length of ?
Contents
[hide]Solution
Solution 1
Let the point of tangency be . By the Two Tangent Theorem and . Thus . The Pythagorean Theorem on yields
Hence .
Solution 2
Clearly, . Thus, the sides of right triangle are in arithmetic progression. Thus it is similar to the triangle and since , .
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 |