parmenides51 wrote: 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 $A$? By the triangle inequality, for any two elements $x$ and $y$ in $A$, either $x > 2y$ or $y > 2x$, and only one of them can hold, otherwise the triangle won’t be unique. Now let the elements of $A$ be $s_1, s_2, s_3 \cdots s_n$ where $s_1< s_2 < s_3 \cdots < s_n $ where $n= |A|$. Notice that $s_2 \geq 2s_1+ 1$, $s_3 \geq 2s_2+1$ and so on. Substituting this recursive relation repeatedly, we get that $s_k \geq (2^k -1)s_1$ where $1 \leq k \leq n $. So $2015 \geq s_n \geq (2^n-1)s_1 $ Since we need to maximize $n$, we should let $s_1$ be as small as possible, i.e. 1. Therefore , $2015 \geq 2^n-1 $ so the maximum value of $n$ is $10$ . Checking, we see that $A= \{1, 3, 7, 15, 31, 63, 127, 255, 511, 1023\}$ works.