Difference between revisions of "2003 AMC 12B Problems/Problem 21"
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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>. | 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 | + | <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> | <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> | ||
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+ | ==Solution 3 (Geometric Probability)== | ||
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+ | 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). | ||
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+ | ~PeterDoesPhysics | ||
== See also == | == See also == |
Revision as of 22:29, 2 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.