probability
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?
Well although I doesn't make sense to put the words prove and probability in the same sentence, I'll answer this anyway. The question should be rephrased as find the probability and not prove the probability.
Anyway, the simplest definition of probability is No. of favorable outcomes divided by Total NO. of outcomes. Since, we are selecting a point from the square, the area of the square is the total no of outcomes. And, the favorable is the area of the circle, since, that is where we want the point. So, answer is area of circle divided by area of square.
Your thought process looks good, but the integration is actually a bit tricky. Here is your dartboard:
We'd like to calculate the area of the center shape, and as you say, we can split this into eight slices of equal area. Here we see the two slices in the first quadrant:
Let's compute the area of the slice from $45^{\circ}$ to $90^{\circ}$:
$$\int_{0}^{\sqrt{2}-1}(\frac{1-x^2}{2}-x)dx=\frac{1}{6}(4\sqrt{2}-5)$$
Multiplying this by $8$ gives us the area of the center shape. Then, we divide by $4$, the area of the dartboard. The final probability is then:
$$\frac{1}{3}(4\sqrt{2}-5)\approx.2190$$
Judging by our picture above, this seems right. The center shape is about $1/5$ the area of the dartboard.
Best Answer
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 $C$.