What is the probability of center of rectangle being inside the circle which is formed by taking 2 random points inside the rectangle as the diameter?
Suppose that the probability is x/y then x and y are natural numbers with no common factors.
That is all the information I was provided with and I was not able to solve it.
I divided the rectangle into octets and I noticed that the octets which are adjacent and lie on the shorter side and adjacent octets which divide the cartesian quadrant do not allow for the formation of circles with center in the interior this gave me an upper bound of probability at $7/8$ but that's not the answer.
Tried searching online could not find any reliable solutions there as well.
Let $O$ be the centre of the rectangle. Notice that, whenever $\angle AOB\lt 90^\circ$, the angle $A'OB\gt 90^\circ$, and vice versa, where $A'$ is the result of central reflection of the point $A$ with the centre $O$.
This tells us that the map $(A, B)\mapsto (A', B)$, which is a(n) ($4$-dimensional) isometry, maps the locus of all the pairs $(A, B)$ making an obtuse angle $\angle AOB$ (and therefore $O$ being in the circle with diameter $AB$) onto the locus of all pairs $(A', B)$ making an acute angle (and therefore $O$ not being in the circle) and vice versa. Thus the ($4$-dimensional) volumes of those loci must be equal to each other. With a reasonable assumption that the volume of the locus where the angle is exactly $90^\circ$ is zero, it follows that each volume is exactly one half of the total volume of the space of all pairs $(A, B)$ of points in the rectangle.
Thus, the solution is $\frac{1}{2}$.