Difference between revisions of "2003 AMC 12B Problems/Problem 21"
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\qquad\mathrm{(E)}\ \frac{1}{2}</math> | \qquad\mathrm{(E)}\ \frac{1}{2}</math> | ||
− | == Solution 1 == | + | == Solution 1 (Trigonometry) == |
By the [[Law of Cosines]], | By the [[Law of Cosines]], | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
Line 15: | Line 15: | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | It follows that <math>0 < \alpha < \frac {\pi}3</math>, and the probability is <math>\frac{\pi/3}{\pi} = \ | + | It follows that <math>0 < \alpha < \frac {\pi}3</math>, and the probability is <math>\frac{\pi/3}{\pi} = \boxed{\textbf{(D) } \frac13 }</math>. |
− | ==Solution 2== | + | ==Solution 2 (Analytic Geometry)== |
− | [[File:2003AMC12BP21.png|500px]] | + | [[File:2003AMC12BP21.png|center|500px]] |
+ | <math>WLOG</math>, let the object turn clockwise. | ||
+ | |||
+ | Let <math>B = (0, 0)</math>, <math>A = (0, -8)</math>. | ||
+ | |||
+ | Note that the possible points of <math>C</math> create a semi-circle of radius <math>5</math> and center <math>B</math>. The area where <math>AC < 7</math> is enclosed by a circle of radius <math>7</math> and center <math>A</math>. The probability that <math>AC < 7</math> is <math>\frac{\angle ABO}{180 ^\circ}</math>. | ||
+ | |||
+ | The function of <math>\odot B</math> is <math>x^2 + y^2 = 25</math>, the function of <math>\odot A</math> is <math>x^2 + (y+8)^2 = 49</math>. | ||
+ | |||
+ | <math>O</math> is the point that satisfies the system of equations: <math>\begin{cases} x^2 + y^2 = 25 \\ x^2 + (y+8)^2 = 49 \end{cases}</math> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | Note that <math>\triangle BDO</math> is a <math>30-60-90</math> triangle, as <math>BO = 5</math>, <math>BD = \frac{5 \sqrt{3}}{2}</math>, <math>DO = \frac52</math>. As a result <math>\angle CBO = 30 ^\circ</math>, <math>\angle ABO = 60 ^\circ</math>. | ||
+ | |||
+ | Therefore the probability that <math>AC < 7</math> is <math>\frac{\angle ABO}{180 ^\circ} = \frac{60 ^\circ}{180 ^\circ} = \boxed{\textbf{(D) } \frac13 }</math> | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 3 (Geometric Probability)== | ||
+ | |||
+ | Setting <math>A = (0,0)</math> we get that <math>B = (8,0)</math>, after assuming segment AB to be straight in the x-direction relative to our coordinate system (in other words, due to symmetrically we can set <math>x = 8</math> for point B). This gives <math>C = (8 + 5cos(\alpha), 5sin(\alpha))</math>. Using the distance formula we get <math>sqrt((8 + 5cos(\alpha))^2 + (5sin(\alpha))^2) < 7</math>. After algebra, this simplifies to <math>cos(\alpha) < -\frac{1}{2}</math>. After evaluating the constraints of the problem, we land on option (D). | ||
+ | |||
+ | ~PeterDoesPhysics | ||
== See also == | == See also == |
Latest revision as of 09:23, 20 September 2024
Contents
Problem
An object moves cm in a straight line from to , turns at an angle , measured in radians and chosen at random from the interval , and moves cm in a straight line to . What is the probability that ?
Solution 1 (Trigonometry)
By the Law of Cosines,
It follows that , and the probability is .
Solution 2 (Analytic Geometry)
, let the object turn clockwise.
Let , .
Note that the possible points of create a semi-circle of radius and center . The area where is enclosed by a circle of radius and center . The probability that is .
The function of is , the function of is .
is the point that satisfies the system of equations:
, , , ,
Note that is a triangle, as , , . As a result , .
Therefore the probability that is
Solution 3 (Geometric Probability)
Setting we get that , after assuming segment AB to be straight in the x-direction relative to our coordinate system (in other words, due to symmetrically we can set for point B). This gives . Using the distance formula we get . After algebra, this simplifies to . After evaluating the constraints of the problem, we land on option (D).
~PeterDoesPhysics
See also
2003 AMC 12B (Problems • Answer Key • Resources) | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.