Difference between revisions of "2002 AMC 12A Problems/Problem 22"
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==Problem== | ==Problem== | ||
− | {{ | + | Triangle <math> ABC </math> is a right triangle with <math> \angle ACB </math> as its right angle, <math> m\angle ABC = 60^\circ </math> , and <math> AB = 10 </math>. Let <math>P</math> be randomly chosen inside <math>ABC</math> , and extend <math> \overline{BP} </math> to meet <math> \overline{AC} </math> at <math>D</math>. What is the probability that <math> BD > 5\sqrt2 </math>? |
+ | |||
+ | <asy> | ||
+ | import math; | ||
+ | unitsize(4mm); | ||
+ | defaultpen(fontsize(8pt)+linewidth(0.7)); | ||
+ | dotfactor=4; | ||
+ | |||
+ | pair A=(10,0); | ||
+ | pair C=(0,0); | ||
+ | pair B=(0,10.0/sqrt(3)); | ||
+ | pair P=(2,2); | ||
+ | pair D=extension(A,C,B,P); | ||
+ | |||
+ | draw(A--C--B--cycle); | ||
+ | draw(B--D); | ||
+ | dot(P); | ||
+ | label("A",A,S); | ||
+ | label("D",D,S); | ||
+ | label("C",C,S); | ||
+ | label("P",P,NE); | ||
+ | label("B",B,N);</asy> | ||
+ | |||
+ | |||
+ | <math> \textbf{(A)}\ \frac{2-\sqrt2}{2}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{3-\sqrt3}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{5-\sqrt5}{5} </math> | ||
==Solution== | ==Solution== | ||
− | {{ | + | Clearly <math>BC=5</math> and <math>AC=5\sqrt{3}</math>. Choose a <math>P'</math> and get a corresponding <math>D'</math> such that <math>BD'= 5\sqrt{2}</math> and <math>AD'=5</math>. For <math> BD > 5\sqrt2 </math> we need <math>AD>5</math>. Thus the point <math>P</math> may only lie in the triangle <math>ABD'</math>. The probability of it doing so is the ratio of areas of <math>ABD'</math> to <math>ABC</math>, or equivalently, the ratio of <math>AD'</math> to <math>AC</math> because the triangles have identical altitudes when taking <math>AD'</math> and <math>AC</math> as bases. This ratio is equal to <math>\frac{AC-BD'}{AC}=1-\frac{BD'}{AC}=1-\frac{5}{5\sqrt{3}}=1-\frac{\sqrt{3}}{3}= \frac{3-\sqrt{3}}{3}</math>. Thus the answer is <math>C</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2002|ab=A|num-b=24|after=Last<br>Problem}} | {{AMC12 box|year=2002|ab=A|num-b=24|after=Last<br>Problem}} |
Revision as of 02:44, 31 January 2010
Problem
Triangle is a right triangle with as its right angle, , and . Let be randomly chosen inside , and extend to meet at . What is the probability that ?
Solution
Clearly and . Choose a and get a corresponding such that and . For we need . Thus the point may only lie in the triangle . The probability of it doing so is the ratio of areas of to , or equivalently, the ratio of to because the triangles have identical altitudes when taking and as bases. This ratio is equal to . Thus the answer is .
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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 |