Distance Between Two Points

If two points are selected at random from the interval \left [0,1\right ], what is the probability that the distance between them is less than one fourth?
Source: NCTM Mathematics Teacher, February 2006

SOLUTION
image
Consider two points A(0,0) and B(1/4,0) on the \left [0,1\right ] interval. Create an isosceles right triangle ABC by drawing a perpendicular \overline{BC} of length 1/4 to \overline{AB}.

Keeping the length of the segment \overline{AB} at a fixed value of 1/4, move the segment \overline{AB} from left to right until A is at (3/4,0) and B at (1,0). As the segment \overline{AB} moves from left to right the isosceles right triangle ABC moves within the confines of the trapezoid ABEF from the lower left corner to the upper right hand corner ending at the location of triangle DEF.

What we have done is convert the problem of two points A and B on the interval \left [0,1\right ] to the problem of an isosceles right triangle ABC moving inside the trapezoid ABEF. The probability that the distance between A and B is less than one fourth equals the probability that the isosceles right triangle ABC stays within the confines of the trapezoid ABEF.

The sample space is the isosceles right triangle of side length 1. Its area equals (1/2)1^2=1/2.
image
The figure shows that the trapezoid is made up of 7 isosceles triangles ABC.
Area of trapezoid = 7(1/2)(1/4)^2 = 7/32
Probability that the distance between A and B is less than 1/4 equals
7/32\div 1/2=7/16

Answer: 7/16

Alternative solution
Let A(x_A,0) and B(x_B,0) two points on the interval \left [0,1\right]. We want |x_A-x_B|<1/4.
image
The diagonal from (0,0) to (1,1) in the above figure is the set of points A and B such that x_A=x_B. The line above the diagonal is the set of points x_B-x_A=1/4 and the line under it is the set of points x_B-x_A=\textrm{-}1/4. As long as the coordinates (x_A,x_B) stay inside the region delineated by these two lines, |x_A-x_B|<1/4.
The square of side 1 represents the sample space; its area equals 1^2=1.
Area of the region = area of square – area of two isosceles right triangles of side length 3/4
=1-2(1/2)(3/4)^2= 7/16
Probability that the distance between A and B is less than 1/4 equals
7/16\div 1=7/16

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About mvtrinh

Retired high school math teacher.
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