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Difference between revisions of "2022 AIME II Problems/Problem 9"
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==Solution==
 
==Solution==
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We could use recursion:
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<b>1.</b> Fix 7 points on <math>\ell_A</math>, then put one point <math>B_1</math> on <math>\ell_B</math>. Now, introduce a function <math>f(x)</math> that indicates the number of regions created, where x is the number of points on <math>\ell_B</math>. For example, <math>f(1) = 6</math> because there are 6 regions.
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<b>2.</b> Now, put the second point <math>B_2</math> on <math>\ell_B</math>. Join <math>A_1~A_7</math> and <math>B_2</math> will create <math>7</math> 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 <math>B_2</math> and <math>A_1</math> intersect lines <math>\overline{B_1A_2}</math>, <math>\overline{B_1A_3}</math>, ..., <math>\overline{B_1A_7}</math> at <math>6</math> points <math>\Longrightarrow</math> creating <math>6</math> regions (we already count one region at first), then <math>5</math> points <math>\Longrightarrow</math> creating <math>5</math> regions (we already count one region at first), 4 points, etc. So, we have: <cmath>f(2) = f(1) + 7 + (6+5+...+1) = 34.</cmath>
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<b>3.</b> If you still need one step to understand this: <math>A_1~A_7</math> and <math>B_3</math> will still create <math>7</math> new regions. Intersecting <cmath>\overline{A_2B_1}, \overline{A_2B_2};</cmath> <cmath>\overline{A_3B_1}, \overline{A_3B_2};</cmath> <cmath> ...</cmath> <cmath>\overline{A_7B_1}, \overline{A_7B_2}</cmath> at <math>12</math> points, creating <math>12</math> regions, etc. Thus, we have: <cmath>f(3) = f(2)+7+(12+10+8+...+2)=34+7+6\cdot 7=83.</cmath>
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Yes, you might already notice that <math>f(n+1) = f(n)+7+(6+5+...+1)\cdot n = f(n) + 7 + 21n.</math>
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<b>Finally, </b> we have <math>f(4) = 153</math>, and <math>f(5)=244</math>. Therefore, the answer is <math>\boxed{244}</math>.
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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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~DSAERF-CALMIT (https://binaryphi.site)
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=II|num-b=8|num-a=10}}
 
{{AIME box|year=2022|n=II|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 08:37, 18 February 2022

Problem

Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$, distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$, and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$. Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$, no point strictly between $\ell_A$ and $\ell_B$ lies on more than two of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$. The figure shows that there are 8 regions when $m=3$ and $n=2$. [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy]

Solution

We could use recursion:

1. Fix 7 points on $\ell_A$, then put one point $B_1$ on $\ell_B$. Now, introduce a function $f(x)$ that indicates the number of regions created, where x is the number of points on $\ell_B$. For example, $f(1) = 6$ because there are 6 regions.

2. Now, put the second point $B_2$ on $\ell_B$. Join $A_1~A_7$ and $B_2$ will create $7$ 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 $B_2$ and $A_1$ intersect lines $\overline{B_1A_2}$, $\overline{B_1A_3}$, ..., $\overline{B_1A_7}$ at $6$ points $\Longrightarrow$ creating $6$ regions (we already count one region at first), then $5$ points $\Longrightarrow$ creating $5$ regions (we already count one region at first), 4 points, etc. So, we have: \[f(2) = f(1) + 7 + (6+5+...+1) = 34.\]

3. If you still need one step to understand this: $A_1~A_7$ and $B_3$ will still create $7$ new regions. Intersecting \[\overline{A_2B_1}, \overline{A_2B_2};\] \[\overline{A_3B_1}, \overline{A_3B_2};\] \[...\] \[\overline{A_7B_1}, \overline{A_7B_2}\] at $12$ points, creating $12$ regions, etc. Thus, we have: \[f(3) = f(2)+7+(12+10+8+...+2)=34+7+6\cdot 7=83.\]

Yes, you might already notice that $f(n+1) = f(n)+7+(6+5+...+1)\cdot n = f(n) + 7 + 21n.$

Finally, we have $f(4) = 153$, and $f(5)=244$. Therefore, the answer is $\boxed{244}$.

Note: we could deduce a general formula of this recursion: $f(n+1)=f(n)+N_a+\frac{n\cdot (N_a) \cdot (N_a-1)}{2}$, where $N_a$ is the number of points on $\ell_A$.

~DSAERF-CALMIT (https://binaryphi.site)

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

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