Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AIME II Problems/Problem 6"

(→‎Solution 3 (Geometry))
(→‎Solution 4 (Coordinates, similar to Solution 1))
 
(2 intermediate revisions by 2 users not shown)
Line 25: Line 25:
  
 
By symmetry this has probability <math>\frac{1}{2}</math>. Also by symmetry, the probability the y-coordinate works as well is <math>\frac{1}{2}</math>. Thus the probability that the midpoint is outside the L-shape is:  
 
By symmetry this has probability <math>\frac{1}{2}</math>. Also by symmetry, the probability the y-coordinate works as well is <math>\frac{1}{2}</math>. Thus the probability that the midpoint is outside the L-shape is:  
−
<cmath>\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{2}</cmath>
+
<cmath>\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{18}</cmath>
−
<cmath>\frac{1}{18}</cmath>
 
  
 
We want the probability that the point is inside the L-shape however, which is <math>1-\frac{1}{18}=\frac{17}{18}</math>, yielding the answer of <math>17+18=\boxed{35}</math>
 
We want the probability that the point is inside the L-shape however, which is <math>1-\frac{1}{18}=\frac{17}{18}</math>, yielding the answer of <math>17+18=\boxed{35}</math>
Line 114: Line 113:
 
==Solution 4 (Coordinates, similar to Solution 1)==
 
==Solution 4 (Coordinates, similar to Solution 1)==
 
Consider this diagram:
 
Consider this diagram:
 +
 
[[File:Screenshot_2024-01-10_101738.png|500px]]
 
[[File:Screenshot_2024-01-10_101738.png|500px]]
 +
 
First, the one of points must be in the uppermost box and the other in the rightmost box. This happens with probability 2/3*1/3=2/9.
 
First, the one of points must be in the uppermost box and the other in the rightmost box. This happens with probability 2/3*1/3=2/9.
  
Line 128: Line 129:
  
 
[[File:Screenshot_2024-01-10_101905.png|500px]]
 
[[File:Screenshot_2024-01-10_101905.png|500px]]
 +
 
Thus, the desired is 2/9*1/4=1/18 => 1-1/18=17/18.
 
Thus, the desired is 2/9*1/4=1/18 => 1-1/18=17/18.
  
 
mathboy282
 
mathboy282
 +
 +
==Solution 5==
 +
 +
We casework on the squares the two points are in. Unless the two squares are on opposite corners, which happens with probability <math>\frac{2}{9}</math>. If the points are on opposite corners, the points must be on the upper half of the splitting diagonal for the midpoint to not be in the square. This occurs with probability <math>\frac{1}{4}</math>. We take the complement and get final answer <math>(\frac{2}{9})(\frac{3}{4}) + (\frac{7}{9}) = \frac{17}{18}</math>.
  
 
==Video Solution by The Power of Logic==
 
==Video Solution by The Power of Logic==

Latest revision as of 14:42, 13 November 2024

Problem

Consider the L-shaped region formed by three unit squares joined at their sides, as shown below. Two points $A$ and $B$ are chosen independently and uniformly at random from inside the region. The probability that the midpoint of $\overline{AB}$ also lies inside this L-shaped region can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ [asy] unitsize(2cm); draw((0,0)--(2,0)--(2,1)--(1,1)--(1,2)--(0,2)--cycle); draw((0,1)--(1,1)--(1,0),dashed); [/asy]

Video Solution by MOP 2024

https://youtube.com/watch?v=0UViBt-LYTo&t=0s

Solution 1

We proceed by calculating the complement.

Note that the only configuration of the 2 points that makes the midpoint outside of the L shape is one point in the top square, and one in the right square. This occurs with $\frac{2}{3} \cdot \frac{1}{3}$ probability.

Let the topmost coordinate have value of: $(x_1,y_1+1)$, and rightmost value of: $(x_2+1,y_2)$.

The midpoint of them is thus: \[\left(\frac{x_1+x_2+1}{2}, \frac{y_1+y_2+1}{2} \right)\]

It is clear that $x_1, x_2, y_1, y_2$ are all between 0 and 1. For the midpoint to be outside the L-shape, both coordinates must be greater than 1, thus: \[\frac{x_1+x_2+1}{2}>1\] \[x_1+x_2>1\]

By symmetry this has probability $\frac{1}{2}$. Also by symmetry, the probability the y-coordinate works as well is $\frac{1}{2}$. Thus the probability that the midpoint is outside the L-shape is: \[\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{18}\]

We want the probability that the point is inside the L-shape however, which is $1-\frac{1}{18}=\frac{17}{18}$, yielding the answer of $17+18=\boxed{35}$ ~SAHANWIJETUNGA

Solution 2（Calculus)

We assume each box has side length 1. We index the upper left box, the bottom left box, the bottom right box as II, III, IV, respectively. We index the missing upper right box as I. We put the graph to a coordinate system by putting the intersecting point of four foxes at the origin, the positive direction of the $x$axis at the intersecting line of boxes I and IV, and the positive direction of the $y$-axis at the intersecting line of boxes I and II. We denote by $M$ the midpoint of $AB$.

Therefore, \begin{align*} \Bbb P \left( M \in II \cup III \cup IV \right) & = 1 - \Bbb P \left( M \in I \right) . \end{align*}

We observe that a necessary for $M \in I$ is either $A \in II$ and $B \in IV$, or $A \in IV$ and $B \in II$. In addition, by symmetry, \[ \Bbb P \left( M \in I | A \in II, B \in IV \right) = \Bbb P \left( M \in I | A \in IV, B \in II \right) . \]

Thus, \begin{align*} \Bbb P \left( M \in I \right) & = 2 \Bbb P \left( M \in I | A \in II, B \in IV \right) \Bbb P \left( A \in II, B \in IV \right) \\ & = 2 \Bbb P \left( M \in I | A \in II, B \in IV \right) \Bbb P \left( A \in II \right) \Bbb P \left( B \in IV \right) \\ & = 2 \Bbb P \left( M \in I | A \in II, B \in IV \right) \cdot \frac{1}{3} \cdot \frac{1}{3} \\ & = \frac{2}{9} \Bbb P \left( M \in I | A \in II, B \in IV \right) \\ & = \frac{2}{9} \Bbb P \left( x_M > 0, y_M > 0 | x_A < 0, y_A > 0, x_B > 0, y_B < 0 \right) \\ & = \frac{2}{9} \Bbb P \left( x_M > 0 | x_A < 0, x_B > 0 \right) \cdot \Bbb P \left( y_M > 0 | y_A > 0, y_B < 0 \right) \\ & = \frac{2}{9} \Bbb P \left( x_M > 0 | x_A < 0, x_B > 0 \right)^2 \\ & = \frac{2}{9} \Bbb P \left( x_A + x_B > 0 | x_A < 0, x_B > 0 \right)^2 \\ & = \frac{2}{9} \left( \int_{x_A = -1}^0 \int_{x_B = 0}^1 \mathbf 1 \left\{ x_A + x_B > 0 \right\} dx_B dx_A \right)^2 \\ & = \frac{2}{9} \left( \int_{x_A = -1}^0 \int_{x_B = - x_A}^1 dx_B dx_A \right)^2 \\ & = \frac{2}{9} \left( \int_{x_A = -1}^0 \left( 1 + x_A \right) dx_A \right)^2 \\ & = \frac{2}{9} \left( \frac{1}{2} \right)^2 \\ & = \frac{1}{18} . \end{align*} The second equality follows from the condition that the positions of $A$ and $B$ are independent. The sixth equality follows from the condition that for each point of $A$ and $B$, the $x$ and $y$ coordinate are independent.

Therefore, \begin{align*} \Bbb P \left( M \in II \cup III \cup IV \right) & = 1 - \frac{1}{18} \\ & = \frac{17}{18} . \end{align*}

Therefore, the answer is $17 + 18 = \boxed{\textbf{(035) }}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3 (Geometry)

2023 AIME I 6.png

The only configuration of two points that makes the midpoint outside of the $L-$shape is one point in the top square, and one in the right square with probability $\frac{2}{9}$ (see Solution $1.)$

We use the coordinate system shown in diagram. Let the arbitrary point $A(x, -y), x \in [0,1], y \in [0,1]$ be in right square. It is clear that iff point $B$ lies in yellow rectangle with sides $x$ and $1 – y,$ then the midpoint $AB$ lies outside of the $L.$

Probability of this is $p = x(1 – y).$

Consider the points $A_1(1-y,-x), A_2(1-x,y-1), A_3(y,x-1).$

Similarly we find $p_1 = (1-y)(1-x), p_2= (1-x)y, p_3 = xy.$

The probability that the midpoint of one of the segments $A_1B, A_2B, A_3B$ and $AB$ is outside of the $L-$shape is \[\frac {p_1+p_2+p_3+p}{4} = \frac {1}{4}.\] If point $A$ is in the right square, point $B$ in the top square, the probability that the midpoint of $\overline{AB}$ lies outside $L-$shape is $\frac {1}{4} \implies$

If points $A$ and $B$ are chosen independently and uniformly at random in $L-$shape, then the probability that the midpoint of $\overline{AB}$ lies outside $L-$shape is $\frac {1}{4} \cdot \frac {2}{9} = \frac {1}{18}.$ Therefore the probability that the point is inside the $L-$shape is $1-\frac{1}{18}=\frac{17}{18} \implies \boxed{35}.$

vladimir.shelomovskii@gmail.com, vvsss

Solution 4 (Coordinates, similar to Solution 1)

Consider this diagram:

Screenshot 2024-01-10 101738.png

First, the one of points must be in the uppermost box and the other in the rightmost box. This happens with probability 2/3*1/3=2/9.

We need the midpoints of the $x$ coordinates to be greater than $1$ but less than $2.$

We need the midpoints of the $y$ coordinates to be greater than $1$ but less than $2.$

Thus, we set up: \[1<\frac{x_1+x_2}{2}<2\] \[1<\frac{y_1+y_2}{2}<2\]

Using geometric probability, as shown below, we get that the probability of happening is 1/2. This is for both. So we have probability 1/2*1/2=1/4.

Screenshot 2024-01-10 101905.png

Thus, the desired is 2/9*1/4=1/18 => 1-1/18=17/18.

mathboy282

Solution 5

We casework on the squares the two points are in. Unless the two squares are on opposite corners, which happens with probability $\frac{2}{9}$. If the points are on opposite corners, the points must be on the upper half of the splitting diagonal for the midpoint to not be in the square. This occurs with probability $\frac{1}{4}$. We take the complement and get final answer $(\frac{2}{9})(\frac{3}{4}) + (\frac{7}{9}) = \frac{17}{18}$.

Video Solution by The Power of Logic

https://youtu.be/KngzOXcmM-E

See also

2023 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_II_Problems/Problem_6&oldid=233286"