2022 AIME II Problems/Problem 9
Problem
Let and be two distinct parallel lines. For positive integers and , distinct points lie on , and distinct points lie on . Additionally, when segments are drawn for all and , no point strictly between and lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when and . The figure shows that there are 8.0 regions when and
Solution 1
We can use recursion to solve this problem:
1. Fix 7 points on , then put one point on . Now, introduce a function that indicates the number of regions created, where x is the number of points on . For example, because there are 6 regions.
2. Now, put the second point on . Join and will create new regions (and we are not going to count them again), and split the existing regions. Let's focus on the spliting process: line segment formed between and intersect lines , , ..., at points creating regions (we already count one region at first), then points creating regions (we already count one region at first), 4 points, etc. So, we have:
3. If you still need one step to understand this: and will still create new regions. Intersecting at points, creating regions, etc. Thus, we have:
Yes, you might already notice that:
5. (Finally) we have , and . Therefore, the answer is .
Note: we could deduce a general formula of this recursion: , where is the number of points on .
~DSAERF-CALMIT (https://binaryphi.site)
Solution 2
We want to derive a general function that indicates the number of bounded regions. Observing symmetry, we know this is a symmetric function about and . Now let's focus on , which is the difference caused by adding one point to the existing points of line . This new point, call it #m, when connected to point #1 on , crosses lines, thus making additional bounded regions; when connected to point #2 on , it crosses lines, thus making additional bounded regions; etc. By simple algebra/recursion methods, we see
Notice . Not very difficult to figure out:
The fact that makes us more confident about the formula. Now plug in , we get the final answer of .
Solution 3
Let some number of segments be constructed. We construct a new segment. We start from the straight line WLOG from point Segment will cross several existing segments (points ) and enter one of the points of the line
Each of these points adds exactly 1 new bounded region (yellow bounded regions).
The exception is the only first segment which does not create any bounded region. Thus, the number of bounded regions is less than the number of points of intersection of the segments plus the number of points of arrival of the segments to
Each point of intersection of two segments is determined uniquely by the choice of pairs of points on each line.
The number of such pairs is
Exactly one segment comes to each of the points of the line from each of the points of the line The total number of arrivals is equal to Hence, the total number of bounded regions is
We plug in , we get the final answer of .
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
~MathProblemSolvingSkills.com
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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