Difference between revisions of "2003 AMC 12B Problems/Problem 21"
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<math>WLOG</math>, let the object turn clockwise. | <math>WLOG</math>, let the object turn clockwise. | ||
− | 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> center <math>A</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></math>\frac{\angle ABO}{180 ^\circ}<math>. |
− | Let <math>B = (0, 0)< | + | Let </math>B = (0, 0)<math>. The function of </math>\odot B = x^2 + y^2 = 25<math>, the function of </math>\odot A = x^2 + (y+8)^2 = 49<math>. |
− | <math>O< | + | </math>O<math> is the point that satisfies both functions. |
− | <math>x^2 + (y+8)^2 - x^2 - y^2 = 49 - 25< | + | </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< | + | 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< | + | Therefore the probability that </math>AC < 7<math> is </math>\frac{\angle ABO}{180 ^\circ} = \frac{60 ^\circ}{180 ^\circ} = \boxed{\textbf{(D) } \frac13 }$ |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Revision as of 11:53, 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
, let the object turn clockwise.
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 $$ (Error compiling LaTeX. Unknown error_msg)\frac{\angle ABO}{180 ^\circ}$.
Let$ (Error compiling LaTeX. Unknown error_msg)B = (0, 0)\odot B = x^2 + y^2 = 25
\odot A = x^2 + (y+8)^2 = 49
O
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)
\triangle BDO
30-60-90
BO = 5
BD = \frac{5 \sqrt{3}}{2}
DO = \frac52
\angle CBO = 30 ^\circ
\angle ABO = 60 ^\circ$.
Therefore the probability that$ (Error compiling LaTeX. Unknown error_msg)AC < 7\frac{\angle ABO}{180 ^\circ} = \frac{60 ^\circ}{180 ^\circ} = \boxed{\textbf{(D) } \frac13 }$
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.