+0  
 
0
97
2
avatar

Two points on a circle of radius 1 are chosen at random. Find the probability that the distance between the two points is at most 1.5. 

 Mar 10, 2023
 #1
avatar
0

Let's denote the two points on the circle as A and B, and let's assume that A is fixed at the top of the circle (i.e., at the point (0,1)). We can then use polar coordinates to describe the position of point B on the circle. Let θ be the angle that the line segment AB makes with the positive x-axis, measured in radians.

We can then write the position of point B as (cos θ, sin θ), since it lies on the circle of radius 1 centered at the origin. Since we are choosing point B at random, the angle θ is uniformly distributed on the interval [0, 2π).

To find the probability that the distance between A and B is at most 1.5, we need to find the set of values of θ that satisfy this condition. The distance between A and B is given by the formula:

distance AB = sqrt((cos θ - 0)^2 + (sin θ - 1)^2) = sqrt(cos^2 θ + (sin θ - 1)^2)

So, the condition that the distance between A and B is at most 1.5 is equivalent to the inequality:

sqrt(cos^2 θ + (sin θ - 1)^2) ≤ 1.5

Squaring both sides and simplifying, we get:

cos^2 θ + (sin θ - 1)^2 ≤ 2.25

Expanding the second term and simplifying, we get:

sin^2 θ - 2sin θ + 2 ≤ 0

Using the quadratic formula to solve for sin θ, we get:

sin θ ≤ 1 ± sqrt(3) / 2

Since θ is uniformly distributed on the interval [0, 2π), the probability that sin θ satisfies this inequality is equal to the ratio of the length of the interval [0, 2π) for which sin θ satisfies this inequality to the length of the entire interval [0, 2π). The length of the interval [0, 2π) for which sin θ satisfies this inequality can be found by considering the two cases:

sin θ ≤ 1 + sqrt(3) / 2: In this case, we have:

0 ≤ θ ≤ pi/3 or 5pi/3 ≤ θ ≤ 2pi

The length of this interval is pi/3 + (2pi - 5pi/3) = 4pi/3.

sin θ ≤ 1 - sqrt(3) / 2: In this case, we have:

pi/3 ≤ θ ≤ 2pi/3 or 4pi/3 ≤ θ ≤ 5pi/3

The length of this interval is (2pi/3 - pi/3) + (5pi/3 - 4pi/3) = 2pi/3.

Therefore, the probability that the distance between A and B is at most 1.5 is:

 

(4pi/3 + 2pi/3) / 2pi = 1

So the probability that the distance between two randomly chosen points on the circle of radius 1 is at most 1.5 is 1 or 100%

 Mar 11, 2023

6 Online Users

avatar
avatar