Singapur, 2015

Let $x < y$ be two numbers in $A$. For them to determine a unique isosceles triangle, we must have $2x \le y$. If $A = \{2^0, 2^1, 2^2, \ldots, 2^{10}\}$, then any two of the numbers $x < y$ satisfy $2x \le y$. So the maximum size is $\ge 11$.

Now suppose that there is a set $A$ with $|A| = 12$ that has the property. Let $a_1, a_2, \ldots, a_{12}$ be the elements of $A$ in increasing order. Then $a_2 \ge 2a_1$, $a_3 \ge 2a_2 \ge 2^2a_1$, $\ldots$, $a_{12} \ge 2a_{11} \ge 2^{11}a_1 \ge 2^{11} = 2048$, a contradiction (since $2015 < 2048$). Thus the maximum size is $11$.