Difference between revisions of "2011 AMC 10A Problems/Problem 20"
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[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
| − | ==Solution== | + | ==Solution 1== |
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) | 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) | ||
| + | |||
| + | ==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 <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. | ||
== See Also == | == See Also == | ||
Latest revision as of 19:42, 16 January 2021
Contents
Problem 20
Two points on the circumference of a circle of radius 
are selected independently and at random. From each point a chord of length is drawn in a clockwise direction. What is the probability that the two chords intersect?
Solution 1
Fix a point 
from which we draw a clockwise chord. In order for the clockwise chord from another point 
to intersect that of point , and must be no more than units apart. By drawing the circle, we quickly see that can be on 
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 . Therefore the arc degree is 
The other arc degree is also Therefore the sum is 
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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