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Difference between revisions of "2011 AMC 10A Problems/Problem 20"

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== Problem 20 ==
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Two points on the circumference of a circle of radius <math>r</math> are selected independently and at random. From each point a chord of length <math>r</math> is drawn in a clockwise direction. What is the probability that the two chords intersect?
  
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<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math>
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[[Category: Introductory Geometry Problems]]
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==Solution 1==
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Fix a point <math>A</math> from which we draw a clockwise chord. In order for the clockwise chord from another point <math>B</math> to intersect that of point <math>A</math>, <math>A</math> and <math>B</math> must be no more than <math>r</math> units apart. By drawing the circle, we quickly see that <math>B</math> can be on <math>\frac{120}{360}=\boxed{\frac{1}{3}  \ \textbf{(D)}}</math> of the perimeter of the circle. (Imagine a regular hexagon inscribed in the circle)
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==Solution 2==
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Do what Solution 1 did until the guessing part. We then realize that the chords and radii make an equilateral triangle of length <math>r</math>. Therefore the arc degree is <math>60.</math> The other arc degree is also <math>60.</math> Therefore the sum is <math>120.</math> Continue as follows.
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== See Also ==
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{{AMC10 box|year=2011|ab=A|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 19:42, 16 January 2021

Problem 20

Two points on the circumference of a circle of radius $r$ are selected independently and at random. From each point a chord of length $r$ is drawn in a clockwise direction. What is the probability that the two chords intersect?

$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}$

Solution 1

Fix a point $A$ from which we draw a clockwise chord. In order for the clockwise chord from another point $B$ to intersect that of point $A$, $A$ and $B$ must be no more than $r$ units apart. By drawing the circle, we quickly see that $B$ can be on $\frac{120}{360}=\boxed{\frac{1}{3} \ \textbf{(D)}}$ of the perimeter of the circle. (Imagine a regular hexagon inscribed in the circle)

Solution 2

Do what Solution 1 did until the guessing part. We then realize that the chords and radii make an equilateral triangle of length $r$. Therefore the arc degree is $60.$ The other arc degree is also $60.$ Therefore the sum is $120.$ Continue as follows.

See Also

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 10 Problems and Solutions

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