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.
I don't understand this problem. Can someone help?
Let's denote the two points on the circle as A and B.
To find the probability that the distance between A and B is at most 1, we can start by drawing a diagram of the circle and marking a region where the distance between A and B is at most 1.
This region consists of all the points on the circle that are within a distance of 1 from each other. We can visualize this as two semicircles with radius 1 that intersect each other, as shown in the diagram below:
Let A and B be the two points that we choose at random. We want to find the probability that the distance between A and B is at most 1.
One way to approach this problem is to use geometric probability. The total area of the circle is πr^2, where r = 1 is the radius of the circle.
The area of the region where the distance between A and B is at most 1 is the area of the two intersecting semicircles.
To find this area, we can subtract the area of the two triangles ABA' and ABB' from the area of the two semicircles.
The area of each semicircle is 1/2 πr^2 = 1/2 π. The area of each triangle is 1/2 base x height, where the base is the chord AB and the height is the distance from the midpoint of AB to the center of the circle.
The chord AB has length 2 sinθ, where θ is the central angle between A and B. The midpoint of AB is the midpoint of the chord, which has a distance of sinθ from the center of the circle.
Therefore, the height of each triangle is sinθ, and the area of each triangle is 1/2 (2 sinθ)(sinθ) = sin^2 θ.
The area of the region where the distance between A and B is at most 1 is then:
2(1/2 π) - 2(sin^2 θ) = π - 2sin^2 θ
Now, we need to find the probability that the distance between A and B is at most 1. This is the ratio of the area of the region where the distance between A and B is at most 1 to the total area of the circle:
P(distance ≤ 1) = (π - 2sin^2 θ)/(π) = 1 - 2sin^2 θ/π
To find the probability, we need to integrate this expression over all possible values of θ from 0 to π.
∫[0, π] (1 - 2sin^2 θ/π) dθ
Using the identity ∫ sin^2 x dx = (1/2) (x - sin x cos x) we get
= 1 - 2/π ∫[0, π] sin^2 θ dθ = 1 - 2/π [π/2 - 1/2 sin(2π)] = 1 - 1/π
Therefore, the probability that the distance between two randomly chosen points on a circle of radius 1 is 1 - 1/pi.
