An isosceles triangle has angles of $50^\circ,x^\circ,$ and $y^\circ$. Find the maximum possible value of $x-y$.

[i]Proposed by Nathan Ramesh
1. **Identify the properties of the isosceles triangle:**
   - An isosceles triangle has two equal angles.
   - The sum of the angles in any triangle is $180^\circ$.

2. **Set up the equations based on the given angles:**
   - Let the angles of the isosceles triangle be $50^\circ$, $x^\circ$, and $y^\circ$.
   - Since it is an isosceles triangle, two of these angles must be equal.

3. **Consider the possible cases for the angles:**
   - Case 1: $50^\circ$ is one of the equal angles.
     - If $50^\circ$ is one of the equal angles, then the other equal angle must also be $50^\circ$.
     - Therefore, the third angle must be $180^\circ - 50^\circ - 50^\circ = 80^\circ$.
     - This gives us the angle sets $(50^\circ, 50^\circ, 80^\circ)$.

   - Case 2: $50^\circ$ is the unique angle.
     - If $50^\circ$ is the unique angle, then the other two angles must be equal.
     - Let these equal angles be $x^\circ$.
     - Therefore, $50^\circ + x^\circ + x^\circ = 180^\circ$.
     - Solving for $x$, we get $2x = 130^\circ \implies x = 65^\circ$.
     - This gives us the angle sets $(50^\circ, 65^\circ, 65^\circ)$.

4. **Calculate $x - y$ for each case:**
   - For the angle set $(50^\circ, 50^\circ, 80^\circ)$:
     - If $x = 80^\circ$ and $y = 50^\circ$, then $x - y = 80^\circ - 50^\circ = 30^\circ$.
     - If $x = 50^\circ$ and $y = 80^\circ$, then $x - y = 50^\circ - 80^\circ = -30^\circ$.

   - For the angle set $(50^\circ, 65^\circ, 65^\circ)$:
     - If $x = 65^\circ$ and $y = 65^\circ$, then $x - y = 65^\circ - 65^\circ = 0^\circ$.

5. **Determine the maximum possible value of $x - y$:**
   - From the calculations above, the maximum possible value of $x - y$ is $30^\circ$.

The final answer is $\boxed{30^\circ}$