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2002 AMC 12A Problems/Problem 22: Difference between revisions

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==Problem==
==Problem==


{{Empty}}
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==


{{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 01:44, 31 January 2010

Problem

Triangle $ABC$ is a right triangle with $\angle ACB$ as its right angle, $m\angle ABC = 60^\circ$ , and $AB = 10$. Let $P$ be randomly chosen inside $ABC$ , and extend $\overline{BP}$ to meet $\overline{AC}$ at $D$. What is the probability that $BD > 5\sqrt2$?

[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]


$\textbf{(A)}\ \frac{2-\sqrt2}{2}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{3-\sqrt3}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{5-\sqrt5}{5}$

Solution

Clearly $BC=5$ and $AC=5\sqrt{3}$. Choose a $P'$ and get a corresponding $D'$ such that $BD'= 5\sqrt{2}$ and $AD'=5$. For $BD > 5\sqrt2$ we need $AD>5$. Thus the point $P$ may only lie in the triangle $ABD'$. The probability of it doing so is the ratio of areas of $ABD'$ to $ABC$, or equivalently, the ratio of $AD'$ to $AC$ because the triangles have identical altitudes when taking $AD'$ and $AC$ as bases. This ratio is equal to $\frac{AC-BD'}{AC}=1-\frac{BD'}{AC}=1-\frac{5}{5\sqrt{3}}=1-\frac{\sqrt{3}}{3}= \frac{3-\sqrt{3}}{3}$. Thus the answer is $C$.

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
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