In a plane, points P and Q are 20 inches apart. If point R is randomly chosen from all the points in the plane that are 20 inches from P, what is the probability that R is closer to P than it is to Q?

A. \(0\)

B. \(\frac{1}{4}\)

C. \(\frac{1}{3}\)

D. \(\frac{1}{2}\)

E. \(\frac{2}{3}\)


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IMO OA is 0
All the points of R form a circle with P as the center of the circle. Point Q and any point R always lie on the circle. so the Point P is always at equal distance from R & Q and there is no chance where R can be closer to P than Q. So the combinations are 0 hence the probability is 0

rajatchopra1994
I read the solution from nick1816

I have a few questions,

1. If point R is on the circle, with P as the center, Isn't the distance between them is now the radius of the circle =20
and Q is also 20 inches away from P?

2. From the diagram, Let us consider the other two points be A & B. I can see that PA= 20 & PB= 20 as they are radius,
but how do we get the other side (QA & QB length as 20? why did we assume the angle BPA as 120 and line segment PQ bisects the angle BPA? Why can't the angles be 180 or 60 only?

Please explain in detail,
Thank you.

In the non-highlighted potion, R is closer to Q (less than radius of circle) than it is to P. We are only looking for the portion where R is closer to P than it is to Q.