IMC 2004 · Day 1 · P4

Define $f : M \to \{-1, 1\}$, $f(X) = \begin{cases} -1, & \text{if $X$ is white} \\ 1, & \text{if $X$ is black} \end{cases}$. The given condition becomes $\sum\limits_{X \in S} f(X) = 0$ for any sphere $S$ which passes through at least 4 points of $M$. For any 3 given points $A$, $B$, $C$ in $M$, denote by $S(A,B,C)$ the set of all spheres which pass through $A$, $B$, $C$ and at least one other point of $M$ and by $|S(A,B,C)|$ the number of these spheres. Also, denote by $\sum$ the sum $\sum\limits_{X \in M} f(X)$.

We have $$\tag{1} 0 = \sum_{S \in S(A,B,C)} \sum_{X \in S} f(X) = (|S(A,B,C)| - 1)(f(A) + f(B) + f(C)) + \sum$$ since the values of $A$, $B$, $C$ appear $|S(A,B,C)|$ times each and the other values appear only once.

If there are 3 points $A$, $B$, $C$ such that $|S(A,B,C)| = 1$, the proof is finished.

If $|S(A,B,C)| > 1$ for any distinct points $A$, $B$, $C$ in $M$, we will prove at first that $\sum = 0$.

Assume that $\sum > 0$. From (1) it follows that $f(A) + f(B) + f(C) < 0$ and summing by all $\binom{n}{3}$ possible choices of $(A, B, C)$ we obtain that $\binom{n}{3} \sum < 0$, which means $\sum < 0$ (contradicts the starting assumption). The same reasoning is applied when assuming $\sum < 0$.

Now, from $\sum = 0$ and (1), it follows that $f(A) + f(B) + f(C) = 0$ for any distinct points $A$, $B$, $C$ in $M$. Taking another point $D \in M$, the following equalities take place $$\begin{align*} f(A) + f(B) + f(C) &= 0 \\ f(A) + f(B) + f(D) &= 0 \\ f(A) + f(C) + f(D) &= 0 \\ f(B) + f(C) + f(D) &= 0 \end{align*}$$ which easily leads to $f(A) = f(B) = f(C) = f(D) = 0$, which contradicts the definition of $f$.