Difference between revisions of "2003 AMC 12B Problems/Problem 21"

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==Solution 2==
 
==Solution 2==
  
[[File:2003AMC12BP21.png|center|700px]]
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[[File:2003AMC12BP21.png|center|500px]]
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Note that the possible points of <math>C</math> create a semi-circle of radius <math>5</math> and center <math>B</math>. The
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Let <math>B = (0, 0)</math>. Function of <math>\odot B = x^2 + y^2 = 25</math>, function of <math>\odot A = x^2 + (y-8)^2 = 49</math>.
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<math>O</math> is the point that satisfies both functions.
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<math>x^2 + (y-8)^2 - x^2 - y^2 = 49 - 25</math>, <math>64 - 16y =24</math>, <math>y = \frac52</math>, <math>x = \frac{5 \sqrt{3}}{2}</math>, <math>O = (\frac{5 \sqrt{3}}{2}, \frac52)</math>
  
  

Revision as of 11:22, 31 August 2022

Problem

An object moves $8$ cm in a straight line from $A$ to $B$, turns at an angle $\alpha$, measured in radians and chosen at random from the interval $(0,\pi)$, and moves $5$ cm in a straight line to $C$. What is the probability that $AC < 7$?

$\mathrm{(A)}\ \frac{1}{6} \qquad\mathrm{(B)}\ \frac{1}{5} \qquad\mathrm{(C)}\ \frac{1}{4} \qquad\mathrm{(D)}\ \frac{1}{3} \qquad\mathrm{(E)}\ \frac{1}{2}$

Solution 1 (Trigonometry)

By the Law of Cosines, \begin{align*} AB^2 + BC^2 - 2 AB \cdot BC \cos \alpha = 89 - 80 \cos \alpha = AC^2 &< 49\\ \cos \alpha &> \frac 12\\ \end{align*}

It follows that $0 < \alpha < \frac {\pi}3$, and the probability is $\frac{\pi/3}{\pi} = \frac 13 \Rightarrow \mathrm{(D)}$.

Solution 2

2003AMC12BP21.png

Note that the possible points of $C$ create a semi-circle of radius $5$ and center $B$. The

Let $B = (0, 0)$. Function of $\odot B = x^2 + y^2 = 25$, function of $\odot A = x^2 + (y-8)^2 = 49$.

$O$ is the point that satisfies both functions.

$x^2 + (y-8)^2 - x^2 - y^2 = 49 - 25$, $64 - 16y =24$, $y = \frac52$, $x = \frac{5 \sqrt{3}}{2}$, $O = (\frac{5 \sqrt{3}}{2}, \frac52)$


~isabelchen

See also

2003 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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