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Difference between revisions of "2023 AIME II Problems/Problem 11"

(Created page with "==Solution== Denote by <math>\mathcal C</math> a collection of 16 distinct subsets of <math>\left\{ 1, 2, 3, 4, 5 \right\}</math>. Denote <math>N = \min \left\{ |S|: S \in \m...")
 
(→‎Solution)
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Because <math>mathcal C</math> contains 16 subsets.
 
Because <math>mathcal C</math> contains 16 subsets.
 
We must have <math>\mathcal C = \left\{ \{a_1\} \cup A : \forall \ A \subseteq \left\{ 1, 2, 3, 4, 5 \right\} \backslash \left\{a_1 \right\} \right\}</math>.
 
We must have <math>\mathcal C = \left\{ \{a_1\} \cup A : \forall \ A \subseteq \left\{ 1, 2, 3, 4, 5 \right\} \backslash \left\{a_1 \right\} \right\}</math>.
−
Therefore, for any <math>X, Y \in \mathcal C</math>, we must have <math>X \cap Y \supseteq \{1\}</math>.
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Therefore, for any <math>X, Y \in \mathcal C</math>, we must have <math>X \cap Y \supseteq \{a_1\}</math>.
 
So this is feasible.
 
So this is feasible.
  

Revision as of 16:51, 16 February 2023

Solution

Denote by $\mathcal C$ a collection of 16 distinct subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$. Denote $N = \min \left\{ |S|: S \in \mathcal C \right\}$.

Case 1: $N = 0$.

This entails $\emptyset \in \mathcal C$. Hence, for any other set $A \in \mathcal C$, we have $\emptyset \cap A = \emptyset$. This is infeasible.

Case 2: $N = 1$.

Let $\{a_1\} \in \mathcal C$. To get $\{a_1\} \cap A \neq \emptyset$ for all $A \in \mathcal C$. We must have $a_1 \in \mathcal A$.

The total number of subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain $a_1$ is $2^4 = 16$. Because $mathcal C$ contains 16 subsets. We must have $\mathcal C = \left\{ \{a_1\} \cup A : \forall \ A \subseteq \left\{ 1, 2, 3, 4, 5 \right\} \backslash \left\{a_1 \right\} \right\}$. Therefore, for any $X, Y \in \mathcal C$, we must have $X \cap Y \supseteq \{a_1\}$. So this is feasible.

Now, we count the number of $\mathcal C$ in this case. We only need to determine $a_1$. Therefore, the number of solutions is 5.

Case 3: $N = 2$.

Case 3.1: There is exactly one subset in $\mathcal C$ that contains 2 elements.

Denote this subset as $\left\{ a_1, a_2 \right\}$. We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$, except $\left\{ a_3, a_4, a_5 \right\}$. This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$.

Now, we count the number of $\mathcal C$ in this case. We only need to determine $\left\{ a_1, a_2 \right\}$. Therefore, the number of solutions is $\binom{5}{2} = 10$.

Case 3.2: There are exactly two subsets in $\mathcal C$ that contain 2 elements.

They must take the form $\left\{ a_1, a_2 \right\}$ and $\left\{ a_1, a_3 \right\}$.

We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$, except $\left\{ a_3, a_4, a_5 \right\}$ and $\left\{ a_2, a_4, a_5 \right\}$. This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$.

Now, we count the number of $\mathcal C$ in this case. We only need to determine $\left\{ a_1, a_2 \right\}$ and $\left\{ a_1, a_3 \right\}$. Therefore, the number of solutions is $5 \cdot \binom{4}{2} = 30$.

Case 3.3: There are exactly three subsets in $\mathcal C$ that contain 2 elements. They take the form $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_1, a_4 \right\}$.

We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$, except $\left\{ a_3, a_4, a_5 \right\}$, $\left\{ a_2, a_4, a_5 \right\}$, $\left\{ a_2, a_3, a_5 \right\}$. This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$.

Now, we count the number of $\mathcal C$ in this case. We only need to determine $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_1, a_4 \right\}$. Therefore, the number of solutions is $5 \cdot \binom{4}{3} = 20$.

Case 3.4: There are exactly three subsets in $\mathcal C$ that contain 2 elements. They take the form $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_2, a_3 \right\}$.

We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$, except $\left\{ a_3, a_4, a_5 \right\}$, $\left\{ a_2, a_4, a_5 \right\}$, $\left\{ a_1, a_4, a_5 \right\}$. This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$.

Now, we count the number of $\mathcal C$ in this case. We only need to determine $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_2, a_3 \right\}$. Therefore, the number of solutions is $\binom{5}{3} = 10$.

Case 3.5: There are exactly four subsets in $\mathcal C$ that contain 2 elements. They take the form $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_1, a_4 \right\}$, $\left\{ a_1, a_5 \right\}$.

We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$, except $\left\{ a_3, a_4, a_5 \right\}$, $\left\{ a_2, a_4, a_5 \right\}$, $\left\{ a_1, a_4, a_5 \right\}$, $\left\{ a_2, a_3, a_4 \right\}$. This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$.

Now, we count the number of $\mathcal C$ in this case. We only need to determine $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_1, a_4 \right\}$, $\left\{ a_1, a_5 \right\}$. Therefore, the number of solutions is 5.

Putting all subcases together, the number of solutions is this case is $10 + 30 + 20 + 10 + 5 = 75$.

Case 4: $N \geq 3$.

The number of subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements is $\sum_{i=3}^5 \binom{5}{3} = 16$. Because $\mathcal C$ has 16 elements, we must select all such subsets into $\mathcal C$. Therefore, the number of solutions in this case is 1.

Putting all cases together, the total number of $\mathcal C$ is $5 + 75 + 1 = \boxed{\textbf{(081) }}$.

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

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