Puzzle #6: Probability of centre of square inside circle
Select two points uniformly randomly inside a square. What is the probability that the center of the square will lie inside the circle drawn with these two points as ends of diameter?
Let the points be A and B. Centre of square (say O) will lie inside the circle with AB as diameter iff angle AOB is in [90,270]. After selecting A randomly AOB is in [90,270] if B lies on the otherside of line perpendicular to AO. This summed over all possible selection of A is 0.5
I agree...I'll try to explain why it is 0.5. The argument is based on symmetry. Given the randomly selected point A, the eligible candidates for B are points which lie on the otherside of line perpendicular to AO. Now consider reflection of A through line perpendicular to AO (call this A'). For A' as the first random point, eligible points for B are points which lie on the same side as A. Thus, A and A' combined together, eligible candidates for B are all the points inside the square. This summed over all the choices of A and A' is 0.5 because we have just seen the P([A,B]|A)+P([A',B]|A') = 1.
Let the points be A and B. Centre of square (say O) will lie inside the circle with AB as diameter iff angle AOB is in [90,270]. After selecting A randomly AOB is in [90,270] if B lies on the otherside of line perpendicular to AO. This summed over all possible selection of A is 0.5
ReplyDelete@Piyush Nice problem and solution. "This summed over all possible selection of A is 0.5" The statement is a bit too brief :D.
ReplyDeleteI agree...I'll try to explain why it is 0.5. The argument is based on symmetry. Given the randomly selected point A, the eligible candidates for B are points which lie on the otherside of line perpendicular to AO. Now consider reflection of A through line perpendicular to AO (call this A'). For A' as the first random point, eligible points for B are points which lie on the same side as A. Thus, A and A' combined together, eligible candidates for B are all the points inside the square. This summed over all the choices of A and A' is 0.5 because we have just seen the P([A,B]|A)+P([A',B]|A') = 1.
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