Difference between revisions of "2003 AMC 12B Problems/Problem 21"
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It follows that <math>0 < \alpha < \frac {\pi}3</math>, and the probability is <math>\frac{\pi/3}{\pi} = \frac 13 \Rightarrow \mathrm{(D)}</math>. | It follows that <math>0 < \alpha < \frac {\pi}3</math>, and the probability is <math>\frac{\pi/3}{\pi} = \frac 13 \Rightarrow \mathrm{(D)}</math>. | ||
− | ==Solution 2== | + | ==Solution 2 (Analytic Geometry)== |
[[File:2003AMC12BP21.png|center|500px]] | [[File:2003AMC12BP21.png|center|500px]] | ||
<math>WLOG</math>, let the object turn clockwise. | <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>. | 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 = x^2 + y^2 = 25</math>, the function of <math>\odot A = x^2 + (y+8)^2 = 49</math>. | |
<math>O</math> is the point that satisfies both functions. | <math>O</math> is the point that satisfies both functions. |
Revision as of 11:57, 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 (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 , the function of
.
is the point that satisfies both functions.
,
,
,
,
Note that is a
triangle, as
,
,
. As a result
,
.
Therefore the probability that is
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.