If a point is selected inside a rectangle what's the probability that the point is closer to center than vertex? 
If a point is selected inside a rectangle what's the probability that the point is closer to center than vertex?

I thought of working on this problem using the concept of area.
If I draw two concentric rectangles where length and width of inner rectangle are half of the outer one, then the probability should be the ratio of areas of both rectangles.
Therefore $P(E) = \dfrac{l\times B}{2l\times 2b} = 1/4$
But the answer given in my book is $1/2$. What's the problem here?
 A: The region of points that are closer to the center than to any of the vertices doesn't look like a smaller rectangle. In a square that region is a $45^\circ$ diagonal square whose corners are the midpoints of the sides of the large square. For an oblong, the shape is hexagonal, with two edges aligning with the long sides of the oblong.
As for how to solve it, I would divide the rectangle into four quarters (top and bottom, right and left). Then look at each quarter of the rectangle separately. That way we already know which corner is the closest corner, and it is easier to see that the probability is $\frac12$ in each of them.
A: Perpendicular bisector of segment $P_{1}P_{2}$ divide the area into region in which all points are closer to $P_{1}$ and region in which all points are closer to $P_{2}$. Therefore, the perpendicular bisectors of segments connecting the centre and the vertices create a region inside of which all points are closer to the centre than any vertex. For visual cue, all points inside the red hexagon are closer to the centre than any vertex.

I still like Arthur's solution more though
We divide the rectangle into four quadrants. For example if the point lies on the top right quadrant then proceed as before, draw a perpendicular bisector of the segment connecting rectangle's centre and the vertex. This bisector clearly divides the quadrant into two parts with equal area, making the probability $\frac{1}{2}$ more pronounced. Here all points in red polygon are closer to rectangle's centre while all points in blue polygon are closer to vertex $C$.

A: I think you've answered a different question, namely the probability that the random point is closer to the center than to the perimeter of the rectangle. But the stated question asks about the rectangle's vertices, not its perimeter.
By the way, anytime we select something at random, we need to explicitly say what the probability distribution is. Presumably we are selecting a point uniformly at random from the rectangle—but we shouldn't let the "uniform" go unsaid, for then the probability distribution is not well defined.
