2015 USAMO Problems/Problem 3

Problem

Let $S = \{1, 2, ..., n\}$ , where $n \ge 1$ . Each of the $2^n$ subsets of $S$ is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set $T \subseteq S$ , we then write $f(T)$ for the number of subsets of T that are blue.

Determine the number of colorings that satisfy the following condition: for any subsets $T_1$ and $T_2$ of $S$ , $f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2).$

Solution

Define function: $C(T)=1$ if the set T is colored blue, and $C(T)=0$ if $T$ is colored red. Define the $\text{Core} =\text{intersection of all } T \text{ where } C(T)=1$ .

The empty set is denoted as $\varnothing$ , $\cap$ denotes intersection, and $\cup$ denotes union. T1<T2 means T1 is a subset of T2 but not =T2. Let $Sn={n}$ are one-element subsets.

Let mCk denote m choose k = $\frac{m!}{k!(m-k)!}$

(Case I) $f(\null)=1$ . Then for distinct m and k, $f(Sm+Sk)=f(Sm)f(Sk)$ , meaning only if Sm and Sk are both blue is their union blue. Namely $C(Sm+Sk)=C(Sm)C(Sk).$

Similarly, for distinct m,n,k, f(Sm \cup Sk \cup Sn)=f(Sm \cup Sk)f(Sn), C(Sm \cup Sk \cup Sn)=C(Sm)C(Sk)C(Sn). This procedure of determination continues to S. Therefore, if $T=\{a_1,a_2, \cdots a_k\}$ , then $C(T)=C(Sa1)C(Sa2)...C(Sak)$ . All colorings thus determined by the free colors chosen for subsets of one single elements satisfy the condition. There are 2^n colorings in this case.

(Case II.) $f(\varnothing)=0$ .

(Case II.1) $\text{Core}=\varnothing$ . Then either (II.1.1) there exist two nonintersecting subsets A and B, $C(A)=C(B)=1$ , but f $(A)f(B)=0$ which is a contradiction, or (II.1.2) all subsets has $C(T)=0$ , which is easily confirmed to satisfy the condition $f(T1)f(T2)=f(T1 \cap T2)f(T1 \cup T2)$ . There is one coloring in this case.

(Case II.2) Core = a subset of 1 element. WLOG, C(S1)=1. Then f(S1)=1, and subsets containing element 1 may be colored Blue.

, namely . Now S1 functions as the  in case I, with n-1 elements to combine into a base of n-1 2-element sets, and all the other subsets are determined. There are 2^(n-1) legit colorings for each choice of core. But there are nC1 (i.e. n choose 1) = n such cores. Hence altogether there are  colorings in this case.

(Case II.3) Core = a subset of 2 elements. WLOG, C(S1+S2)=1. Only subsets containing the core may be colored Blue. With the same reasoning as in the preceding case, there are $(nC2)2^(n-2)$ colorings.

... (Case II.n+1) Core = S. Then C(S)=1, with all other subsets $C(T)=0$ , there is $1=(nCn)2^0$

Combining all the cases, we have $1+[1+(\dbinom{n}{1})2^{n-1}+(\dbinom{n}{2})2^{n-2}+\cdot \cdot \cdot +(\dbinom{n}{n})2^0]=\boxed{1+3^n}$ is the total number of colorings satisfying the given condition.

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2015 USAMO Problems/Problem 3

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