During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

Difference between revisions of "2002 AMC 12A Problems/Problem 22"

(need solution)
 
Line 1: Line 1:
 
==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 (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS