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

2021 AIME II Problems/Problem 6

Revision as of 11:44, 24 March 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 3 (Casework Bash))

Problem

For any finite set $S$, let $|S|$ denote the number of elements in $S$. FInd the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]

Solution 1

By PIE, $|A|+|B|-|A \cap B| = |A \cup B|$, and after some algebra you see that we need $A \subseteq B$ or $B \subseteq A$. WLOG $A\subseteq B$, then for each element there are $3$ possibilities, either it is in both $A$ and $B$, it is in $B$ but not $A$, or it is in neither $A$ nor $B$. This gives us $3^{5}$ possibilities, and we multiply by $2$ since it could of also been the other way around. Now we need to subtract the overlaps where $A=B$, and this case has $2^{5}=32$ ways that could happen. It is $32$ because each number could be in the subset or it could not be in the subset. So the final answer is $2\cdot 3^5 - 2^5 = \boxed{454}$.

~ math31415926535

Solution 2

We denote $\Omega = \left\{ 1 , 2 , 3 , 4 , 5 \right\}$. We denote $X = A \cap B$, $Y = A \backslash \left( A \cap B \right)$, $Z = B \backslash \left( A \cap B \right)$, $W = \Omega \backslash \left( A \cup B \right)$.

Therefore, $X \cup Y \cup Z \cup W = \Omega$ and the intersection of any two out of sets $X$, $Y$, $Z$, $W$ is an empty set. Therefore, $\left( X , Y , Z , W \right)$ is a partition of $\Omega$.

Following from our definition of $X$, $Y$, $Z$, we have $A \cup B = X \cup Y \cup Z$.

Therefore, the equation

\[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]

can be equivalently written as

\[\left( | X | + | Y | \right) \left( | X | + | Z | \right) = | X | \left( | X | + | Y | + | Z | \right) .\]

This equality can be simplified as

\[| Y | \cdot | Z | = 0 .\]

Therefore, we have the following three cases: (1) $|Y| = 0$ and $|Z| \neq 0$, (2) $|Z| = 0$ and $|Y| \neq 0$, (3) $|Y| = |Z| = 0$. Next, we analyze each of these cases, separately.

Case 1: $|Y| = 0$ and $|Z| \neq 0$.

In this case, to count the number of solutions, we do the complementary counting.

First, we count the number of solutions that satisfy $|Y| = 0$.

Hence, each number in $\Omega$ falls into exactly one out of these three sets: $X$, $Z$, $W$. Following from the rule of product, the number of solutions is $3^5$.

Second, we count the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$.

Hence, each number in $\Omega$ falls into exactly one out of these two sets: $X$, $W$. Following from the rule of product, the number of solutions is $2^5$.

Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy $|Y| = 0$ minus the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$, i.e., $3^5 - 2^5$.

Case 2: $|Z| = 0$ and $|Y| \neq 0$.

This case is symmetric to Case 1. Therefore, the number of solutions in this case is the same as the number of solutions in Case 1, i.e., $3^5 - 2^5$.

Case 3: $|Y| = 0$ and $|Z| = 0$.

Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is $2^5$.

By putting all cases together, following from the rule of sum, the total number of solutions is equal to

\begin{align*} \left( 3^5 - 2^5 \right) + \left( 3^5 - 2^5 \right) + 2^5 & = 2 \cdot 3^5 - 2^5 \\ & = \boxed{454} . \end{align*}

~ Steven Chen (www.professorchenedu.com)

Solution 3 (Casework Bash)

By the Principle of Inclusion-Exclusion (abbreviated PIE), we have $|A \cup B|=|A|+|B|-|A \cap B|,$ from which we rewrite the given equation as \[|A| \cdot |B| = |A \cap B| \cdot \left(|A|+|B|-|A \cap B|\right).\] Rearranging and applying Simon's Favorite Factoring Trick give \begin{align*} |A| \cdot |B| &= |A \cap B|\cdot|A| + |A \cap B|\cdot|B| - |A \cap B|^2 \\ |A| \cdot |B| - |A \cap B|\cdot|A| - |A \cap B|\cdot|B| &= - |A \cap B|^2 \\ \left(|A| - |A \cap B|\right)\cdot\left(|B| - |A \cap B|\right) &=0. \end{align*} We will use casework on the value of $|A \cap B|:$ \[\begin{array}{c|c|l} & & \\ [-2.25ex] \boldsymbol{|A \cap B|} & \textbf{Equation} & \textbf{Condition} \\ [0.5ex] \hline & & \\ [-2ex] 1 & 1 & 1 \\ [0.5ex] \end{array}\]

SOLUTION IN PROGRESS. NO EDIT PLEASE--A MILLION THANKS!

~MRENTHUSIASM

See Also

2021 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=2021_AIME_II_Problems/Problem_6&oldid=150099"