Is is possible to have a third point C on this circle, so that A,B,C all fairly placed?

Question

Consider a circle with radius $3$ and centre at the origin. Points $A$ and $B$ have coordinates $(2,0)$ and $(-2,0)$, respectively.

If a dart is thrown at the circle, then assuming a uniform distribution, it is clear that the probabilities of the following two events are equal:

• The dart is closer to A than to B

• The dart is closer to B than to A

Is it possible to have a third point C on this circle, so that all $A, B, C$ are fairly placed?

What's the key to solving this question? I have tried to express the areas in terms of the coordinates of C but failed.

Answer 1

Let $p_A$ be the probability that $A$ is the nearest point, and so on.

These probabilities are proportional to the Voronoi areas.

Assume $C$ is at $(0,y)$, then by symmetry $p_B=p_A$

If $C$ is at $(0,2)$ then clearly $p_C < p_A$

If $C$ is at $(0,0)$ then clearly $p_C > p_A$

Because the probabilities (areas) are continuous on $y$, then there must be some $0<y<2$ such that $p_C=p_A=p_B$