AoPS - Problem

Let $b_1, \ldots , b_{|B|} \in B$. For any $b_i \in B$, let $M(b_i)$ be a subset of $A$ consisting of all $a \in M(b_i)$ then $\delta(a,b_i) \le t$. So we easily see that $|M(b_i)| = \sum_{k=0}^{t}\binom{n}{k} (q-1)^k $. Now let $i \ne j$, and $b_i, b_j \in B$, then we are given that $\delta(b_i, b_j) \ge 2t+1$ so we have that $M(b_i) \cap M(b_j) = \emptyset$. It is given that for any $a \in A$, there exists a $b \in B$, such that $\delta(a,b) \le t$;and it follows that for any $a$ there exists only one such $b$. So $\sum_{i=1}^{|B|}M(b_i) = |B| \sum_{k=0}^{t}\binom{n}{k} (q-1)^k = |A| = q^n$; giving that (a) implies (b). To show that (b) implies (a), we note that $\bigcup_{i=1}^{|B|} M(b_i) $ is a subset of $A$. Since $M(b_i)$ and $M(b_j)$ are pairwise disjoint, $\bigcup_{i=1}^{|B|} M(b_i) = \sum_{k=0}^{t}\binom{n}{k} (q-1)^k = q^n = |A|$ according to (b).So $\bigcup_{i=1}^{|B|} M(b_i) = A$; which implies (a).