Difference between revisions of "2003 AMC 12B Problems/Problem 21"
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==Solution 2== | ==Solution 2== | ||
| − | [[File:2003AMC12BP21.png|center| | + | [[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 10:22, 31 August 2022
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
Note that the possible points of create a semi-circle of radius and center . The
Let 
. Function of 
, function of 
.
is the point that satisfies both functions.
, 
, 
, 
, 
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. 
