A unit square is drawn in the Cartesian plane with vertices at (0,0),(0,1),(1,0),(1,1). Two points P,Q are chosen uniformly at random, P from the boundary of the square and Q from the interior of the square. The line L1 through P and Q is drawn. The probability that the points (0,0) and (1,1) are both on the same side of L1 can be expressed as a/b where a and b are coprime positive integers. What is the value of a+b?
To solve this problem, we need to consider the different possibilities for the position of the line L1 with respect to the unit square.
First, let's consider the position of point P. It can be on any of the four sides of the square. Let's analyze each case:
Case 1: P is on the side connecting (0,0) and (0,1).
In this case, the line L1 can pass through the square in two ways:
1. It can pass through the interior of the square, which will result in points (0,0) and (1,1) being on the same side of L1.
2. It can pass through the square such that (0,0) and (1,1) are on opposite sides of L1.
Case 2: P is on the side connecting (0,0) and (1,0).
In this case, similarly, the line L1 can pass through the square in two ways:
1. It can pass through the interior of the square, resulting in (0,0) and (1,1) being on opposite sides of L1.
2. It can pass through the square such that (0,0) and (1,1) are on the same side of L1.
Case 3: P is on the side connecting (0,1) and (1,1).
Again, the line L1 can pass through the square in two ways:
1. It can pass through the interior of the square, such that (0,0) and (1,1) are on the same side of L1.
2. It can pass through the square such that (0,0) and (1,1) are on opposite sides of L1.
Case 4: P is on the side connecting (1,0) and (1,1).
Similarly, the line L1 can pass through the square in two ways:
1. It can pass through the interior of the square, resulting in (0,0) and (1,1) being on opposite sides of L1.
2. It can pass through the square, and (0,0) and (1,1) will be on the same side of L1.
Overall, there are 2 * 4 = 8 possible ways for the line L1 to intersect the square.
Out of these 8 possibilities, in 4 cases, (0,0) and (1,1) are on the same side of L1, and in the remaining 4 cases, they are on opposite sides.
Since P is chosen uniformly at random from the boundary of the square, the probability that any specific case occurs is 1/4.
Therefore, the probability that (0,0) and (1,1) are on the same side of L1 is 4/8 = 1/2.
Finally, a + b = 1 + 2 = 3.