Problem:
Alex divides a disk into four quadrants with two perpendicular diameters intersecting at the center of the disk. He draws more line segments through the disk, drawing each segment by selecting two points at random on the perimeter of the disk in different quadrants and connecting those two points. Find the expected number of regions into which these line segments divide the disk.
Solution:
Assuming that no three lines pass through the same point, we can count the number of regions the disk is split up into as (the number of intersections the number of lines ). This can be seen by taking an existing disk with lines drawn through it and drawing a new one. In addition to the first region, every time a new line intersects another line in the interior of the circle, a new region is formed.
Note that since the probability of three lines intersecting through the same point is , for the purposes of expected value, we may ignore this case.
By Linearity of Expectation:
We can further break down the expected number of intersections into three cases - the intersection between the two diameters, intersections between the lines and diameters, and intersections between the lines themselves.
To compute the probability that two of the lines intersect, we have three cases:
Case 1: Both lines connect adjacent quadrants:
There is a probability a line connects adjacent quadrants, and therefore a probability both lines are as such. If the two lines share a quadrant there is a probability they intersect. To see this, first fix the points used to connect the line segments. There are two ways to connect them one of which results in them intersecting. There is a probability they share a quadrant. In the other case, the lines never intersect, so the probability of this case occuring and resulting in an intersection is:
Case 2: One line connects opposite quadrants and the other connects adjacent quadrants There is a chance of this case occuring and regardless of the orientation there is probability the lines intersect. The probability for this case is
Case 3: Both lines connect opposite quadrants There is a chance of this case occuring. If both lines are joined from the same two quadrants there is a probability they intersect and if they join from different quadrants they always intersect. The probability for this case is
Computing the contribution to the expected value:
To compute the expected number of intersections between one of the lines and the diameters, note that if the line is drawn between opposite quadrants, there are intersections and if the line is drawn between adjacent quadrants, there is . This contributes
to the expected value. Don't forget about the intersection between the two diameters. This gives a final answer of
The problems on this page are the property of the MAA's American Mathematics Competitions