Difference between revisions of "2022 AIME II Problems/Problem 9"
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==Problem== | ==Problem== | ||
− | Let <math>\ell_A</math> and <math>\ell_B</math> be two distinct parallel lines. For positive integers <math>m</math> and <math>n</math>, distinct points <math>A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m</math> lie on <math>\ell_A</math>, and distinct points <math>B_1, B_2, B_3, \ldots, B_n</math> lie on <math>\ell_B</math>. Additionally, when segments <math>\overline{A_iB_j}</math> are drawn for all <math>i=1,2,3,\ldots, m</math> and <math>j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n</math>, no point strictly between <math>\ell_A</math> and <math>\ell_B</math> lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when <math>m=7</math> and <math>n=5</math>. The figure shows that there are 8 regions when <math>m=3</math> and <math>n=2</math> | + | Let <math>\ell_A</math> and <math>\ell_B</math> be two distinct parallel lines. For positive integers <math>m</math> and <math>n</math>, distinct points <math>A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m</math> lie on <math>\ell_A</math>, and distinct points <math>B_1, B_2, B_3, \ldots, B_n</math> lie on <math>\ell_B</math>. Additionally, when segments <math>\overline{A_iB_j}</math> are drawn for all <math>i=1,2,3,\ldots, m</math> and <math>j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n</math>, no point strictly between <math>\ell_A</math> and <math>\ell_B</math> lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when <math>m=7</math> and <math>n=5</math>. The figure shows that there are 8 regions when <math>m=3</math> and <math>n=2</math>. |
<asy> | <asy> | ||
import geometry; | import geometry; | ||
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Yes, you might already notice that: <cmath>f(n+1) = f(n)+7+(6+5+...+1)\cdot n = f(n) + 7 + 21n.</cmath> | Yes, you might already notice that: <cmath>f(n+1) = f(n)+7+(6+5+...+1)\cdot n = f(n) + 7 + 21n.</cmath> | ||
− | <b>5. | + | <b>5.</b> Finally, we have <math>f(4) = 153</math>, and <math>f(5)=244</math>. Therefore, the answer is <math>\boxed{244}</math>. |
Note: we could deduce a general formula of this recursion: <math>f(n+1)=f(n)+N_a+\frac{n\cdot (N_a) \cdot (N_a-1)}{2}</math>, where <math>N_a</math> is the number of points on <math>\ell_A</math> | Note: we could deduce a general formula of this recursion: <math>f(n+1)=f(n)+N_a+\frac{n\cdot (N_a) \cdot (N_a-1)}{2}</math>, where <math>N_a</math> is the number of points on <math>\ell_A</math> | ||
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
− | ==Solution 4 (Recursion | + | == Solution 4 (Recursion and Complementary Counting) == |
− | + | When a new point is added to a line, the number of newly bounded regions it creates with each line segment will be one more than the number of intersection points the line makes with other lines. | |
− | Case <math> | + | Case 1: If a new point <math>P</math> is added to the right on a line when both lines have an equal amount of points. |
− | + | WLOG, let the point be on line <math>\ell_A</math>. We consider the complement, where new lines don't intersect other line segments. Simply observing, we see that the only line segments that don't intersect with the new lines are lines attached to some point that a new line does not pass through. If we look at a series of points on line <math>\ell_B</math> from left to right and a line connects <math>P</math> to an arbitrary point, then the lines formed with that point and with remaining points on the left of that point never intersect with the line with <math>P</math>. Let there be <math>s</math> points on lines <math>\ell_A</math> and <math>\ell_B</math> before <math>P</math> was added. For each of the <math>s</math> points on <math>\ell_B</math>, we subtract the total number of lines formed, which is <math>s^2</math>, not counting <math>P</math>. Considering all possible points on <math>\ell_B</math>, we get <math>(s^2-s)+(s^2-2s)\cdots(s^2-s^2)</math> total intersections. However, for each of the lines, there is one more bounded region than number of intersections, so we add <math>s</math>. Simplifying, we get <math>s^3-s\sum_{i=1}^{s}{i}+s\Longrightarrow s(s^2-\sum_{i=1}^{s}{i}+1)</math>. Note that this is only a recursion formula to find the number of new regions added for a new point <math>P</math> added to <math>\ell_A</math>. | |
− | |||
− | Case <math> | + | Case 2: If a new point <math>P</math> is added to the right of a line that has one less point than the other line. |
− | + | Continuing on case one, let this point <math>P</math> be on line <math>\ell_B</math>. With similar reasoning, we see that the idea remains the same, except <math>s+1</math> lines are formed with <math>P</math> instead of just <math>s</math> lines. Once again, each line from <math>P</math> to a point on line <math>\ell_A</math> creates <math>s</math> non-intersecting lines for that point and each point to its left. Subtracting from <math>s(s+1)</math> lines and considering all possible lines created by <math>P</math>, we get <math>(s(s+1)-s)+(s(s+1)-2s)\cdots(s(s+1)-s(s+1)</math> intersections. However, the number of newly bounded regions is the number of intersections plus the number of points on line <math>\ell_A</math>. Simplying, we get <math>s(s+1)^2-s\sum_{i=1}^{s+1}{i}+(s+1)</math> newly bounded regions. | |
− | For the base case <math> | + | For the base case <math>s=2</math> for both lines, there are <math>4</math> bounded regions. Next, we plug in <math>s=2,3,4</math> for both formulas and plug <math>s=5</math> for the first formula to find the number of regions when <math>m=6</math> and <math>n=5</math>. Notice that adding a final point on <math>\ell_A</math> is a variation of our Case 1. The only difference is for each of the <math>s</math> lines formed by <math>P</math>, there are <math>s+1</math> points that can form a non-intersecting line. Therefore, we are subtracting a factor of <math>s+1</math> lines instead of <math>s</math> lines from a total of <math>s(s+1)</math> lines. However, the number of lines formed by <math>P</math> remains the same so we still add <math>s</math> at the end when considering intersection points. Thus, the recursive equation becomes <math>(s(s+1)-(s+1))+(s(s+1)-2(s+1))\cdots(s(s+1)-s(s+1))+s\Longrightarrow s^2(s+1)-(s+1)\sum_{i=1}^{s}{i}+s</math>. Plugging <math>s=5</math> into this formula and adding the values we obtained from the other formulas, the final answer is <math>4+4+9+12+22+28+45+55+65=\boxed{244}</math>. |
~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja] | ~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja] | ||
+ | |||
+ | == Solution 5 (Euler's Graph Formula, similar to solution 3) == | ||
+ | We know the by Euler's Formula for planar graphs that <math>F-E+V=2</math>, where <math>F</math> is the number of bounded faces, plus the outer region, <math>E</math> is the number of edges, and <math>V</math> is the number of vertices. Temporarily disregarding the intersections between the lines, we can easily calculate that: | ||
+ | |||
+ | <math>V_{i}=7+5=12</math> | ||
+ | |||
+ | <math>E_{i}=6+4+7\cdot5=45</math> | ||
+ | |||
+ | However, the resulting graph is not planar, as the edges clearly intersect. To account for this, we must turn all intersection points into vertices, and update our values accordingly. | ||
+ | |||
+ | Observe that each intersection point can be mapped to two points on either line, and analogously, two points on either line can be mapped to one intersection point, uniquely. Thus, to count intersection points, we simply calculate: | ||
+ | |||
+ | <math>{7 \choose 2}{5 \choose 2} = 210</math> | ||
+ | |||
+ | And thus, | ||
+ | |||
+ | <math>V=V_{i}+210=222</math> | ||
+ | |||
+ | We must also account for the edges. Observe that each intersection point turns the two edges that make it into four, that is, each intersection point adds <math>2</math> to the number of edges. Therefore, | ||
+ | |||
+ | <math>E=E_{i}+2\cdot210=465</math> | ||
+ | |||
+ | Plugging these into Euler's Formula we get: | ||
+ | <math>F-E+V=2</math> | ||
+ | |||
+ | <math>F-465+222=2</math> | ||
+ | |||
+ | <math>F=245</math> | ||
+ | |||
+ | Disregarding the outer region, we conclude that our answer is <math>F-1=245-1=\boxed {244}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Shadowleafy Shadowleafy] | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 14:56, 25 January 2024
Contents
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 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
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
Solution 4 (Recursion and Complementary Counting)
When a new point is added to a line, the number of newly bounded regions it creates with each line segment will be one more than the number of intersection points the line makes with other lines.
Case 1: If a new point is added to the right on a line when both lines have an equal amount of points.
WLOG, let the point be on line . We consider the complement, where new lines don't intersect other line segments. Simply observing, we see that the only line segments that don't intersect with the new lines are lines attached to some point that a new line does not pass through. If we look at a series of points on line
from left to right and a line connects
to an arbitrary point, then the lines formed with that point and with remaining points on the left of that point never intersect with the line with
. Let there be
points on lines
and
before
was added. For each of the
points on
, we subtract the total number of lines formed, which is
, not counting
. Considering all possible points on
, we get
total intersections. However, for each of the lines, there is one more bounded region than number of intersections, so we add
. Simplifying, we get
. Note that this is only a recursion formula to find the number of new regions added for a new point
added to
.
Case 2: If a new point is added to the right of a line that has one less point than the other line.
Continuing on case one, let this point be on line
. With similar reasoning, we see that the idea remains the same, except
lines are formed with
instead of just
lines. Once again, each line from
to a point on line
creates
non-intersecting lines for that point and each point to its left. Subtracting from
lines and considering all possible lines created by
, we get
intersections. However, the number of newly bounded regions is the number of intersections plus the number of points on line
. Simplying, we get
newly bounded regions.
For the base case for both lines, there are
bounded regions. Next, we plug in
for both formulas and plug
for the first formula to find the number of regions when
and
. Notice that adding a final point on
is a variation of our Case 1. The only difference is for each of the
lines formed by
, there are
points that can form a non-intersecting line. Therefore, we are subtracting a factor of
lines instead of
lines from a total of
lines. However, the number of lines formed by
remains the same so we still add
at the end when considering intersection points. Thus, the recursive equation becomes
. Plugging
into this formula and adding the values we obtained from the other formulas, the final answer is
.
Solution 5 (Euler's Graph Formula, similar to solution 3)
We know the by Euler's Formula for planar graphs that , where
is the number of bounded faces, plus the outer region,
is the number of edges, and
is the number of vertices. Temporarily disregarding the intersections between the lines, we can easily calculate that:
However, the resulting graph is not planar, as the edges clearly intersect. To account for this, we must turn all intersection points into vertices, and update our values accordingly.
Observe that each intersection point can be mapped to two points on either line, and analogously, two points on either line can be mapped to one intersection point, uniquely. Thus, to count intersection points, we simply calculate:
And thus,
We must also account for the edges. Observe that each intersection point turns the two edges that make it into four, that is, each intersection point adds to the number of edges. Therefore,
Plugging these into Euler's Formula we get:
Disregarding the outer region, we conclude that our answer is
Video Solution
~MathProblemSolvingSkills.com
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.