Given a rhombus. What is the geometric place of points in the plane of the rhombus for which the square of the distance from one of the vertices is equal to the sum of the squares of the distances from the other three vertices?
Let the lengths of the diagonals of the rhombus be denoted by $2e$ and $2f$, and choose the lines of the diagonals as the axes of our coordinate system; thus, the coordinates of the vertices of the rhombus are: $A(e, 0), B(0, f), C(-e, 0), D(0, -f)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e9c598a4e06cf4e46fe7g-1.jpg?height=716&width=1412&top_left_y=260&top_left_x=346)

First, assign the highlighted role to vertex $A$. For this case, the definition of the geometric locus as a function of $e$ and $f$ is given by the following equation:

$$
PA^2 = PB^2 + PC^2 + PD^2
$$

which is

$$
(x-e)^2 + y^2 = \left[x^2 + (y-f)^2\right] + \left[(x+e)^2 + y^2\right] + \left[x^2 + (y+f)^2\right]
$$

where $P(x, y)$ is any point in the point set. By rearranging and completing the square immediately,

$$
(x+e)^2 + y^2 = e^2 - f^2
$$

This is the equation of a circle with center $C$ (the vertex opposite to $A$) and radius $r = \sqrt{e^2 - f^2}$ if $e^2 - f^2 > 0$, i.e., $e > f (> 0)$. If $e = f$, then (1) is satisfied only by the point $C$. Finally, if $0 < e < f$, then (1) has no solution.

For the case $e > f$, by assigning the role of "some" vertex to $A$, $B$, $C$, and $D$ in turn, we get one circle each centered at $C$ and $A$ with the above radius $r$, while the 2nd and 4th cases do not add to the set, so the geometric locus is the union of the circles centered at $A$ and $C$ (the acute vertices of the rhombus) with radius $r$.

If, however, $e = f$, then the requirement is satisfied by the four vertices of the specialized rhombus, no other point.

Our steps are reversible, so the points of the two circles indeed satisfy the requirement.

A simple construction method can be given for the circles: let the common point of the Thales circle above the $OC$ semi-diagonal and the circle centered at $O$ passing through $D$ be $E$, then the circle centered at $C$ passes through $E$ (if $e < f$, $E$ does not exist).

Remark. Some chose one vertex of the rhombus as the origin and one of the sides starting from it as one of the coordinate axes. The calculation is somewhat longer this way, but the interesting result is that the radius of the two circles is $AB \sqrt{\cos \angle DAB}$, and the angle was naturally necessary as a parameter for the other two vertices.

This result can of course be derived from the above result as well:

$$
e = AB \cos \frac{\varphi}{2}, \quad f = AB \sin \frac{\varphi}{2}
$$

and

$$
e^2 - f^2 = AB^2 \left(\cos^2 \frac{\varphi}{2} - \sin^2 \frac{\varphi}{2}\right) = AB^2 \cos \varphi
$$

And another construction can be read from it: if the projection of $D$ onto the side $AB$ is $F$ (and $AF < AB$), then $r = AG$ is the geometric mean of the segment pair $AB, AF$, and $G$ is also a point of the geometric locus.