P , Q and R are chosen uniformly at random from the circumference of a circle with centre O . Let x be the probability of triangle P Q R contains the point O . Find the value of ⌊ 1 0 0 0 x ⌋ , where ⌊ ⋅ ⌋ denotes the floor function .
Three points, says
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A very similar problem was posts about one month ago . I'll copy the relevant portion of my solution below.
Given two points A , B on a circle, the third point C must lie in the red region below if △ A B C is to contain the circle's center:
Now, we find the probability of this happening. Fix A and B on the circle, and let θ be the angle between them ( ∠ A O B ). Then 0 < θ ≤ π . Note that the angle intercepting the red region in the diagram is precisely θ . Thus the probability of a random point C being placed in the red area is 2 π θ . We integrate to compute the average probability over all possible θ : π − 0 1 ∫ 0 π 2 π θ d θ = π 1 [ 4 π θ 2 ] θ = 0 π = 4 π 2 π 2 = 4 1
⌊ 1 0 0 0 × 4 1 ⌋ = ⌊ 2 5 0 ⌋ = 2 5 0
Great answer! This is much easier to understand than Parth Sankhe's answer.
Probability is simply the probability of an acute triangle from 3 random points on a circle.
= π − 0 1 ∫ 0 π 2 π θ d θ = 0 . 2 5
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Suppose P and Q placed randomly. Join the points P and O and extend it to the point A on the circumference. Do the similar to the segment Q O B
If the triangle P Q R contains the point O , then the final point R must be placed within the minor arc A B colored red (as shown). So the probability, x is the length of minor arc A B divided by the circumference. Suppose the circumference is of length 1 unit. Then x is the length of minor arc A B , which is the same as the length of minor arc P Q .
Now fix the point P and draw the segment P O A . By symmetry, the point Q is either above P A or below P A . Without loss of generality, let the point Q be above P A . As Q is chosen uniformly at random on the arc P A , averagely the point Q will fall on the center of the arc P A , which means that the length of minor arc P Q is 4 1 , and so the length of minor arc A B . This implies that x = 4 1 and hence ⌊ 1 0 0 0 x ⌋ = 2 5 0 .