1. ## Geometry Probability

A point is chosen at random from the interior of square ABCD. What is the probability that the point is closer to either A or C than it is to the midpoint of the diagonal connecting A and C.

My attempt at solving the problem follows.

Let $x$ be the length of a side of the square. Then, there are $x^2$ points inside of the square. Furthermore, the midpoint of the diagonal connecting A and C is at the point $(\dfrac{x}{2},\dfrac{x}{2})$.

For my square, point A is at point $(0,x)$ and point C is at point $(x,0)$. I found the distances between points A and C, point A and the midpoint, and point C and the midpoint. However, I don't think I am approaching the solution.

2. Notice the diagram. If a point lies in the YELLOW region, it will be closer to the midpoint of the diagonal than A or C; and if the point lies in the GREEN region, it will be closer to A or C than the midpoint of the diagonal. Let x be the length of each side of the square. So the required probability is total area of the GREEN region / total area of the square = [{(1/2)*(x/2)*(x/2)}*2] / x*x = 1/8

3. I appreciate your help, particularly your diagram. However, I do not understand how you knew points in the yellow region are closer to the midpoint than to either point A or point B.

4. The line EF divides AO in two equal parts, that is, AI=AO. So any point lying in the part AI will be closer to A than to O. Similarly, any point lying in the part IO will be closer to O than to A. In consequence, any point in the triangle AEF will be closer to A and any point in the triangle OEF will be closer to O.