Difference between revisions of "2003 AMC 12B Problems/Problem 21"
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==Solution 3 (Geometric Probability)== | ==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>. | + | 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 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.