2025 AIME I Problems/Problem 13
Problem
Alex divides a disk into four quadrants with two perpendicular diameters intersecting at the center of the disk. He draws 
more lines segments through the disk, drawing each segment by selecting two points at random on the perimeter of the disk in different quadrants and connecting these two points. Find the expected number of regions into which these 
line segments divide the disk.
Solution 1
First, we calculate the probability that two segments intersect each other. Let the quadrants be numbered 
through 
in the normal labeling of quadrants, let the two perpendicular diameters be labeled the 
-axis and 
-axis, and let the two segments be 
and 
[b]Case 1:[/b] Segment has endpoints in two opposite quadrants. This happens with probability 
WLOG let the two quadrants be and 
We do cases in which quadrants segment 
lies in.
- Quadrants
and 
and 
and 
and 
and : These share one quadrant with 
and it is clear that for any of them to intersect it must be on a certain side of 
For example, if it was quadrants and then the point in quadrant must be closer to the 
-axis than the endpoint of 
in quadrant 
This happens with probability 
Additionally, segment 
has a 
to have endpoints in any set of two quadrants, so this case contributes to the total probability
and 
and 
and 
and it is clear that for any of them to intersect 
For example, if it was quadrants 
This happens with probability 
Additionally, segment 
to have endpoints in any set of two quadrants, so this case contributes to the total probability![\[\dfrac{1}{3}\left(\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}\right)=\dfrac{1}{9}\]](http://latex.artofproblemsolving.com/1/4/2/14227a8403138b996a6c4d078e922def5b9f4ec9.png)
- Quadrants and
This always intersects segment so this case contributes to the total probability
This always intersects segment ![\[\dfrac{1}{3}\cdot\dfrac{1}{6}=\dfrac{1}{18}\]](http://latex.artofproblemsolving.com/f/3/d/f3d09d875422accb6e2911c24ead009e16b84a6b.png)
- Quadrants and
We will first choose the endpoints, and then choose the segments from the endpoints. Let the endpoints of the segments in quadrant be 
and 
and the endpoints of the segments in quadrant be 
and 
such that 
and are in clockwise order. Note that the probability that and intersect is the probability that 
is paired with 
which is 
Thus, this case contributes to the total probability
and 
and the endpoints of the segments in quadrant 
and 
such that 
and 
is paired with 
which is 
Thus, this case contributes to the total probability![\[\dfrac{1}{3}\cdot\dfrac{1}{6}\cdot\dfrac{1}{2}=\dfrac{1}{36}.\]](http://latex.artofproblemsolving.com/c/0/8/c089cfd1672bac71787e7ebe317e2593ec156797.png)
In total, case contributes
![\[\dfrac{1}{9}+\dfrac{1}{18}+\dfrac{1}{36}=\dfrac{7}{36}\]](http://latex.artofproblemsolving.com/3/1/8/318a6bb712fd1d3bf104e7d8bdcb7643dbdac919.png)
to the total probability that two segments intersect.
(Work in Progress)
See also
| 2025 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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