# Which points in the interior of a parallelogram are as far as possible from the corners?

Question 1: Given a parallelogram $P=ABCD$, how does one construct/determine the points $X \in P$ which are as far as possible from the corners? That is, the points $X$ for which $$\min(|X-A|,|X-B|,|X-C|,|X-D|)$$ is maximized?

If $P$ is a rectangle, then there is only one such point $X$, the center of mass of the rectangle. Otherwise, it looks as though there are two such points.

In the picture above, it looks as though the two points in question are also determined by being equidistant from three of the corners of the parallelogram. So, maybe another relevant question here could be:

Question 2: Given three points in the plane, how can one construct/determine the (unique?) point which is equidistant from the three points (assuming it exists)?

Added: It looks as though there are some other cases which need to be considered in order to answer Question 1. For example, we could have something like the situation pictured below. In this case the desired points seem to occur on one of the quadrilateral's sides, and equidistant from two diagonally opposite corners of the parallelogram.

By the way, the reason I asked this question in the first place was as a means to answering another question, which I now post below.

Question 3: Let $P$ be a parallelogram. What is the smallest positive number $D$ such that every point in $P$ is within distance $D$ of one of the corners? Ideally, $D$ should be expressed in terms of things like the side lengths and angles of $P$.