[Geometry]
Problem: Given quadrilateral $A R M L$ with $A R=20, R M=23, M L=25$, and $A M=32$, compute the number of different integers that could be the perimeter of $A R M L$.

Solution: Notice that $\triangle A R M$ is fixed, so the number of integers that could be the perimeter of $A R M L$ is the same as the number of integers that could be the length $A L$ in $\triangle A L M$. By the Triangle Inequality, $32-25<A L<32+25$, so $A L$ is at least 8 and no greater than 56 . The number of possible integer values for $A L$ is $56-8+1=49$.

Final answer: 49