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

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Revision as of 14:27, 14 December 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 21 		Followed by
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 12 Problems and Solutions
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