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

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\qquad\mathrm{(E)}\ \dfrac{1}{9}</math>
 
\qquad\mathrm{(E)}\ \dfrac{1}{9}</math>
  
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==Solution==
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==Solution 1==
  
 
The first point anywhere on the circle, because it doesn't matter where it is chosen.
 
The first point anywhere on the circle, because it doesn't matter where it is chosen.
Line 22: Line 22:
  
 
Therefore the total probability is <math>1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}</math> or <math>\boxed{\text{(D)}}</math>
 
Therefore the total probability is <math>1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}</math> or <math>\boxed{\text{(D)}}</math>
 +
 +
==Solution 2==
 +
 +
We will use geometric probability.
 +
 +
The first point can be anywhere. Each point must be <math>\frac{\pi}{3}</math> or less away from each other.
 +
 +
Define <math>x</math> be the amount of radians away the second point is from the first. We limit <math>x</math> to be in the interval <math>[-\pi, \pi]</math>. Define <math>y</math> be the amount of radians away the third point is from the first. We limit <math>y</math> to be in the interval <math>[-\pi, \pi]</math>. Now, we can deduct that: <cmath>|x| \le \frac{\pi}{3},</cmath> <cmath>|y| \le \frac{\pi}{3},</cmath> and <cmath>|x-y| \le \frac{\pi}{3}.</cmath> We now begin plotting these on the coordinate grid.
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 +
First note that the area of the points that <math>x</math> and <math>y</math> can be (ignoring the conditions) is <math>(2\pi)^2</math> (remember what we restricted <math>x</math> and <math>y</math> to).
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 +
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));
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draw((0, -50)--(0, 50));
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draw((-50, -10)--(10, 50),red);
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draw((-10, -50)--(50, 10),red);
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draw((-40, 40)--(40, 40),red);
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draw((-40, -40)--(40, -40),red);
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draw((40, 40)--(40, -40),red);
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draw((-40, 40)--(-40, -40),red);
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label("$\frac{\pi}{3}$", (40, 0), SE);
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label("$\frac{\pi}{3}$", (-40, 0), SW);
 +
label("$\frac{\pi}{3}$", (0, 40), NE);
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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 <math>\frac{\pi^2}{3}</math>.
 +
 +
Thus, the probability is <math>\frac{\frac{\pi^2}{3}}{(2\pi)^2} = \frac{1}{12}</math>, or <math>\boxed{\text{(D)}}</math>.
 +
 +
~superagh
  
 
==See Also==
 
==See Also==

Revision as of 19:05, 27 August 2020

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?

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

Solution 1

The first point 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.

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 $\frac{1}{4}$ to find 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

See Also

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

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