Is it possible that $\frac{|A \cap B|}{|A \cup B|} > \frac 12$ and $\frac{|A \cap C|}{|A \cup C|} > \frac 12$ given that $|B \cap C| = 0$? Given three sets, $A$, $B$, $C$, I have that  $B$ and $C$ are disjoint, and $|\cdot|$ represents the number of elements in the set:
$$|B \cap C| = 0$$ 
Is it possible that both:
$$\frac{|A \cap B|}{|A \cup B|} > \frac 12 $$
and
$$\frac{|A \cap C|}{|A \cup C|} > \frac 12 $$
are satisfied at the same time?
 A: The answer is no, an easy way to see this
Suppose 
$$\frac{|A \cap B|}{|A \cup B|} > \frac{1}{2} \Rightarrow |A \cap B| > \frac{1}{2} |A \cup B| \ge \frac{1}{2}|A| $$ 
hence $|A \cap B| > \frac{1}{2}|A|$, same argument shows that $|A \cap C| > \frac{1}{2}|A|$
since $B$ and $C$ are disjoint then $$|A| \ge |A \cap B| + |A \cap C| > \frac{1}{2}|A| + \frac{1}{2}|A| = |A|$$
Clearly impossible.
A: Disclaimer: I assume $A,B,C$ are finite, otherwise the division makes little sense.
No, this is not possible.
First of all, you can see that if $B$ and $C$ are not subsets of $A$, you can repace them with $B'=B\cap A$ and $C'=C\cap A$, and the quantities you want to be bigger than $\frac12$ will merely increase. In other words, without loss of generality, you can say that $B,C\subseteq A$.
But, if $B$ and $C$ are indeed subsets of $A$, then $A\cap B = B, A\cap C=C, A\cup B = A\cup C = A$.
Therefore, the question then becomes:
Does there exist a pair of disjoint sets $A,B\subseteq A$ such that $$\frac{|B|}{|A|}>\frac12 \text{ and } \frac{|C|}{|A|} > \frac12$$
This question can more easiliy be answered: no
A: $\frac {|A \cap B|}{|A|} \ge  \frac {|A \cap B|}{|A \cup B|} > 1/2$ so more than half of the elements of A are in B.  Likewise more than half the elements of A are in C.  B and C can't both have more than 1/2 the elements and be distinct.  That'd means there'd be more then 100% of the elements accounted for.
