Let $A$ be a set of numbers chosen from $1,2,..., 2015$ with the property that any two distinct numbers, say $x$ and $y$, in $A$ determine a unique isosceles triangle (which is non equilateral) whose sides are of length $x$ or $y$. What is the largest possible size of $A$?
1. **Understanding the Problem:**
   We need to find the largest possible size of a set \( A \) of numbers chosen from \( \{1, 2, \ldots, 2015\} \) such that any two distinct numbers \( x \) and \( y \) in \( A \) determine a unique isosceles triangle (which is non-equilateral) with sides of length \( x \) or \( y \).

2. **Triangle Inequality Condition:**
   For any two elements \( x \) and \( y \) in \( A \), the triangle inequality must hold:
   \[
   x + x > y, \quad x + y > x, \quad y + y > x
   \]
   Simplifying these, we get:
   \[
   2x > y \quad \text{and} \quad 2y > x
   \]
   However, for the triangle to be unique and non-equilateral, we need either \( x > 2y \) or \( y > 2x \), but not both.

3. **Ordering the Elements:**
   Let the elements of \( A \) be \( s_1, s_2, s_3, \ldots, s_n \) where \( s_1 < s_2 < s_3 < \cdots < s_n \). We need to ensure that:
   \[
   s_2 \geq 2s_1 + 1, \quad s_3 \geq 2s_2 + 1, \quad \ldots, \quad s_k \geq 2s_{k-1} + 1
   \]

4. **Recursive Relation:**
   Using the recursive relation, we can express \( s_k \) in terms of \( s_1 \):
   \[
   s_2 \geq 2s_1 + 1
   \]
   \[
   s_3 \geq 2(2s_1 + 1) + 1 = 4s_1 + 3
   \]
   \[
   s_4 \geq 2(4s_1 + 3) + 1 = 8s_1 + 7
   \]
   Continuing this pattern, we get:
   \[
   s_k \geq (2^k - 1)s_1
   \]

5. **Maximizing \( n \):**
   Since \( s_n \leq 2015 \), we have:
   \[
   2015 \geq (2^n - 1)s_1
   \]
   To maximize \( n \), we should let \( s_1 \) be as small as possible, ideally \( s_1 = 1 \):
   \[
   2015 \geq 2^n - 1
   \]
   \[
   2016 \geq 2^n
   \]
   \[
   n \leq \log_2 2016
   \]
   Calculating the logarithm:
   \[
   \log_2 2016 \approx 11.0
   \]
   Therefore, the largest integer \( n \) is \( 10 \).

6. **Verification:**
   We check if the set \( A = \{1, 3, 7, 15, 31, 63, 127, 255, 511, 1023\} \) works:
   \[
   3 \geq 2 \cdot 1 + 1, \quad 7 \geq 2 \cdot 3 + 1, \quad 15 \geq 2 \cdot 7 + 1, \quad \ldots, \quad 1023 \geq 2 \cdot 511 + 1
   \]
   All conditions are satisfied, confirming that the set works.

The final answer is \(\boxed{10}\).