Finding properties of parallelogram given three of its vertices passing through a circle 
Given the above construction of parallelogram ABCD, with point E between B and C lying on the circle, the sine of angle A must be greater than √(x)/y where x and y are positive integers. What is the greatest possible value of √(x)/y? Answer: √(5)/3
At first I considered using the law of cosines and took the sine of the arccosine of (12^2 + 16^2 - BD^2)/(2*12*16) on the domain [4, 28] because of the triangle inequality with 4 ≤ BD ≤ 28. I got a maximum value of sin(A) = 1 at BD = 20. However, it's obvious this is not fully correct since I imagine the parallel sets of lines and circle itself would offer some other upper constraint that I can't figure out. Even once I figured out the maximum of sin(A), how would I find the maximum value of √(x)/y with x and y as integers, or what might √(x)/y represent in terms of the problem given?
 A: 
the sine of angle A must be greater than $\sqrt{x}/y$ where x and y are positive integers. What is the greatest possible value of $\sqrt{x}/y$?

First note that this is a roundabout way to ask "what is the minimum value of $\sin A$". As it turns out, $A$ must be an acute angle, so this is equivalent to finding the minimum value of $A\,$.

I got a maximum value of $\sin(A) = 1$ at $BD = 20$

This is correct, but it doesn't help with finding the minimum value, which the problem asks for.
Hint for the minimum $A\,$: the driving constraint of the problem is (as noted in a comment already) that point $E$ must lie between $B$ and $C\,$. It is easier to determine the range of eligible angles $A$ by focusing on the limit points where $E$ crosses out of the segment $BC\,$.


*

*At one end of the range, this happens when $E \equiv C\,$, which is the case where $ABCD$ is a rectangle and $A=\pi/2\,$. This is the case that gives $BD=20$ and the maximum $\sin(A)= 1$. This also shows that angle $A$ must be acute.

*At the other end of the range, the limit case is $E \equiv B$ which means that $BC$ is tangent to the circle, so $B$ lies on the perpendicular bisector of $AD\,$. This is the case that gives the minimum angle $A\,$ and also, by the previous point about $A$ being acute, the minimum $\sin(A)\,$.
Therefore, the minimum $A$ is the base angle of the isosceles triangle $\triangle BAD$ with equal sides $BA=BD=12$ and base $AD=16\,$. Solving the triangle gives in the end $\,\sin(A)=\sqrt{5}/3\,$.
