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Difference between revisions of "2003 AMC 12B Problems/Problem 25"

m (→‎Solution 3)
m (fixed typo, "all three pairwise distance" should be "all three pairwise distances")
 
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== Problem ==
 
== Problem ==
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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?
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Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?
  
 
<math>\mathrm{(A)}\ \dfrac{1}{36}
 
<math>\mathrm{(A)}\ \dfrac{1}{36}
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==Solution 3==
 
==Solution 3==
  
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The first point can be placed anywhere on the circle. The second point must lie <math>60^\circ</math> within the first point, so as the third point. The three points can be switched around in <math>\binom32 = 3</math> different ways.
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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>
  
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<math>\frac16 \cdot \frac16 \cdot 3 = \boxed{\text{(D)} \frac{1}{12} }</math>
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<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 00:42, 14 October 2024

Problem

Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?

$\mathrm{(A)}\ \dfrac{1}{36} \qquad\mathrm{(B)}\ \dfrac{1}{24} \qquad\mathrm{(C)}\ \dfrac{1}{18} \qquad\mathrm{(D)}\ \dfrac{1}{12} \qquad\mathrm{(E)}\ \dfrac{1}{9}$

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 $60$ degrees of arc on either side, a total of $120$ degrees possible, giving a total $\frac{1}{3}$ chance. The last point must lie within $60$ degrees of both points.

The minimum area of freedom we have to place the third point is a $60$ degrees arc(if the first two are $60$ degrees apart), with a $\frac{1}{6}$ probability. The maximum amount of freedom we have to place the third point is a $120$ degree arc(if the first two are the same point), with a $\frac{1}{3}$ probability.

As the second point moves farther away from the first point, up to a maximum of $60$ 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 $\frac{1}{4}$, the average probability we can place the third point based on a varying second point.

Therefore the total probability is $1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$ or $\boxed{\text{(D)}}$

Solution 2

We will use geometric probability.

The first point can be anywhere. Each point must be $\frac{\pi}{3}$ or less away from each other.

Define $x$ be the amount of radians away the second point is from the first. We limit $x$ to be in the interval $[-\pi, \pi]$. Define $y$ be the amount of radians away the third point is from the first. We limit $y$ to be in the interval $[-\pi, \pi]$. Now, we can deduct that: \[|x| \le \frac{\pi}{3},\] \[|y| \le \frac{\pi}{3},\] and \[|x-y| \le \frac{\pi}{3}.\] We now begin plotting these on the coordinate grid.

First note that the area of the points that $x$ and $y$ can be (ignoring the conditions) is $(2\pi)^2$ (remember what we restricted $x$ and $y$ to).

Now, we can graph the equations we deduced on the coordinate grid. That should look like this: [asy] fill((40, 0)--(40, 40)--(0, 40)--(-40, 0)--(-40, -40)--(0, -40)--cycle, green); draw((-50, 0)--(50, 0)); draw((0, -50)--(0, 50)); draw((-50, -10)--(10, 50),red); draw((-10, -50)--(50, 10),red); draw((-40, 40)--(40, 40),red); draw((-40, -40)--(40, -40),red); draw((40, 40)--(40, -40),red); draw((-40, 40)--(-40, -40),red); label("$\frac{\pi}{3}$", (40, 0), SE); label("$\frac{\pi}{3}$", (-40, 0), SW); label("$\frac{\pi}{3}$", (0, 40), NE); label("$\frac{\pi}{3}$", (0, -40), NW); [/asy] The area of the shaded region can be calculated in many ways. Eventually, you will find that the area is $\frac{\pi^2}{3}$.

Thus, the probability is $\frac{\frac{\pi^2}{3}}{(2\pi)^2} = \frac{1}{12}$, or $\boxed{\text{(D)}}$.

~superagh

Solution 3

There are $\binom32 = 3$ different ways to choose $2$ points out of $3$. The probability that an angle between $2$ points is smaller than $60^\circ$ is $\frac{60^\circ}{360^\circ} = \frac16$. The probability that the third point is inside the $60^\circ$ angle is $\frac{60^\circ}{360^\circ} = \frac16$

$3 \cdot \frac16 \cdot \frac16 = \boxed{\text{(D)} \frac{1}{12} }$

~isabelchen

See Also

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AMC 12 Problems and Solutions

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