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.

enter image description here

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.

enter image description here

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 $P$.


Given three (noncollinear) points in the plane, the point equidistant from the three points always exists. It defines the center of a circle, known as the circumcircle, that passes through all three points. The center point is the circumcenter. You construct the circumcenter by dropping perpendicular bisectors from any two sides of the triangle formed by the three points. See http://www.mathopenref.com/constcircumcenter.html

Here's an illustration of this construction, as it applies to to the given situation with the parallelogram. One draws the perpendicular bisector of each of the parallelogram's sides, and then finds the intersection points of the bisecting lines just drawn. Two of those intersection points occur inside the quadrilateral, and those are the points of interest.

enter image description here

  • $\begingroup$ Thanks for the reply. I added an illustration showing how this applies to the quadrilateral case. $\endgroup$ – Mike F May 19 '15 at 19:31
  • $\begingroup$ @MikeF : Nice illustration. And the construction produces the right result when the parallelogram is a rectangle. $\endgroup$ – grand_chat May 19 '15 at 20:36
  • $\begingroup$ Thanks! While the construction you described certainly solves my Question 2, it is not quite enough to solve Question 1. See my recent edit. $\endgroup$ – Mike F May 19 '15 at 20:38

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