Difference between revisions of "2003 AMC 12B Problems/Problem 25"
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==Solution 3== | ==Solution 3== | ||
− | The probability that an angle between <math>2</math> points is smaller than <math>60^\circ</math> is <math>\frac{60^\circ}{360^\circ} = \frac16</math> | + | There are <math>\binom32 = 3</math> different ways to choose <math>2</math> points out of <math>3</math>. The probability that an angle between <math>2</math> points is smaller than <math>60^\circ</math> is <math>\frac{60^\circ}{360^\circ} = \frac16</math>. The probability that the third point is inside the <math>60^\circ</math> angle is <math>\frac{60^\circ}{360^\circ} = \frac16</math> |
− | <math>\frac16 \cdot \frac16 | + | <math>3 \cdot \frac16 \cdot \frac16 = \boxed{\text{(D)} \frac{1}{12} }</math> |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Latest revision as of 11:46, 1 September 2022
Problem
Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distance between the points are less than the radius of the circle?
Solution 1
The first point is placed anywhere on the circle, because it doesn't matter where it is chosen.
The next point must lie within degrees of arc on either side, a total of
degrees possible, giving a total
chance. The last point must lie within
degrees of both points.
The minimum area of freedom we have to place the third point is a degrees arc(if the first two are
degrees apart), with a
probability.
The maximum amount of freedom we have to place the third point is a
degree arc(if the first two are the same point), with a
probability.
As the second point moves farther away from the first point, up to a maximum of degrees, the probability changes linearly (every degree it moves, adds one degree to where the third could be).
Therefore, we can average probabilities at each end to find , the average probability we can place the third point based on a varying second point.
Therefore the total probability is or
Solution 2
We will use geometric probability.
The first point can be anywhere. Each point must be or less away from each other.
Define be the amount of radians away the second point is from the first. We limit
to be in the interval
. Define
be the amount of radians away the third point is from the first. We limit
to be in the interval
. Now, we can deduct that:
and
We now begin plotting these on the coordinate grid.
First note that the area of the points that and
can be (ignoring the conditions) is
(remember what we restricted
and
to).
Now, we can graph the equations we deduced on the coordinate grid. That should look like this:
The area of the shaded region can be calculated in many ways. Eventually, you will find that the area is
.
Thus, the probability is , or
.
~superagh
Solution 3
There are different ways to choose
points out of
. The probability that an angle between
points is smaller than
is
. The probability that the third point is inside the
angle is
See Also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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.