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

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==See Also==
==See Also==
{{AMC12 box|year=2002|ab=A|num-b=22|after=Last<br>Problem}}
{{AMC12 box|year=21|ab=A|num-b=22|after=23}}

Revision as of 15:26, 14 December 2010


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]

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


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

21 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
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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