Is it true that $4$ distinct points in the plane form a rhombus if and only if the set of their distances contains exactly $2$ or $3$ elements?
The problem can be stated formally as:
Let $I=\{1,2,3,4\}$ be a set of indices,
Let $\vec{x}_1,\vec{x}_2,\vec{x}_3,\vec{x}_4\in\mathbb{R}^2$ be some distinct points in the plane, I.e. they satisfy:
$\forall i,j\in I,i\neq j \longrightarrow \vec{x}_i\neq\vec{x}_j$
And let $D=\{d(\vec{x}_i,\vec{x}_j)|i,j\in I, i\neq j\}$ be the set of all distances between those distinct points.
Prove that the set $\{\vec{x}_1,\vec{x}_2,\vec{x}_3,\vec{x}_4\}$ forms a rhombus (with respect to $d$) in $\mathbb{R}^2$ if and only if $|D|\in\{2,3\}$.
(Note: Here $d(\vec{x},\vec{y})$ is the standard Euclidean distance in $\mathbb{R}^2$, I.e. $\forall \vec{x},\vec{y}\in\mathbb{R}^2,d(\vec{x},\vec{y})=||\vec{x}-\vec{y}||_2$, But I think that it is true even for other general norms)
Trigonometric Proof of $\Longrightarrow$:
Suppose that the set $\{\vec{x}_1,\vec{x}_2,\vec{x}_3,\vec{x}_4\}$ forms a rhombus in $\mathbb{R}^2$,
Then without loss of generality we may assume that the sides of the rhombus are $\{\vec{x}_1,\vec{x}_2\}$, $\{\vec{x}_1,\vec{x}_4\}$, $\{\vec{x}_2,\vec{x}_3\}$ and $\{\vec{x}_3,\vec{x}_4\}$ and that the diagonals are $\{\vec{x}_1,\vec{x}_3\}$ and $\{\vec{x}_2,\vec{x}_4\}$,
Let’s denote by $a\in (0,\infty)$ the length of each side of the rhombus, I.e. $a:=d(\vec{x}_1,\vec{x}_2)=d(\vec{x}_1,\vec{x}_4)=d(\vec{x}_2,\vec{x}_3)=d(\vec{x}_3,\vec{x}_4)$,
By $d_1\in (0,\infty)$ the length of the diagonal $\{\vec{x}_2,\vec{x}_4\}$, I.e. $d_1:=d(\vec{x}_2,\vec{x}_4)$,
By $d_2\in (0,\infty)$ the length of the diagonal $\{\vec{x}_1,\vec{x}_3\}$, I.e. $d_2:=d(\vec{x}_1,\vec{x}_3)$,
And by $\theta\in (0,\pi)$ the angle between the side $\{\vec{x}_1,\vec{x}_2\}$ and the side $\{\vec{x}_1,\vec{x}_4\}$.
(See picture).
(Note: Because the points are all distinct, We clearly have that indeed $a,d_1,d_2\in (0,\infty)$, The fact that $\theta\in (0,\pi)$ may not be immediately clear and here is an explanation: Clearly $\theta\in[0,\pi]$, We’ll show that it is impossible for $\theta$ to equal either $0$ or $\pi$, Because if it were, We would have got that either the points $\vec{x}_2$ and $\vec{x}_4$ would merge into one single point and that the diagonal $d_1$ would vanish to $0$ or that the points $\vec{x}_1$ and $\vec{x}_3$ would merge into one single point and that the diagonal $d_2$ would vanish to $0$ respectively, But this contradicts the fact that all points are distinct, And thus it must be the case that $\theta\in(0,\pi)$. Intuitively this says that the rhombus cannot be a degenerate rhombus)
Therefore we get by definition of $D$, That $D=\{a,d_1,d_2\}$.
We will prove by using trigonometry that $D$ contains exactly $2$ or $3$ elements:
By applying the law of sines to the triangle $\bigtriangleup\vec{x}_1\vec{x}_2\vec{x}_4$ we get $\frac{d_1}{\sin\theta}=\frac{a}{\sin\frac{\pi-\theta}{2}}$, And so:
$$d_1=a\cdot \frac{\sin\theta}{\sin\frac{\pi-\theta}{2}}=a\cdot \frac{\sin\theta}{\sin(\frac{\pi}{2}-\frac{\theta}{2})}=a\cdot \frac{\sin\theta}{\cos\frac{\theta}{2}}=\\a\cdot \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}}=2a\sin\frac{\theta}{2}$$
And by applying the law of sines to the triangle $\bigtriangleup\vec{x}_1\vec{x}_2\vec{x}_3$ we get $\frac{d_2}{\sin(\pi-\theta)}=\frac{a}{\sin\frac{\theta}{2}}$, And so:
$$d_2=a\cdot \frac{\sin(\pi-\theta)}{\sin\frac{\theta}{2}}=a\cdot \frac{\sin\theta}{\sin\frac{\theta}{2}}=\\a\cdot \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}=2a\cos\frac{\theta}{2}$$
Now let’s find for which $\theta\in(0,\pi)$ we have $d_1=d_2$ and for which $\theta\in(0,\pi)$ we have $d_1\neq d_2$:
$\begin{gather} d_1 = d_2 \\ \Updownarrow \\ 2a\sin\frac{\theta}{2}=2a\cos\frac{\theta}{2} \\ \Updownarrow \\ \sin\frac{\theta}{2}=\cos\frac{\theta}{2} \\ \Updownarrow \\ \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}=0 \\ \Updownarrow \\ \tan\frac{\theta}{2}=0 \\ \Updownarrow \\ \frac{\theta}{2}=\frac{\pi}{4} \\ \Updownarrow \\ \theta = \frac{\pi}{2} \end{gather}$
(Note: We have here $\Updownarrow$/iff because $\frac{\theta}{2}\in (0,\frac{\pi}{2})$ is an acute angle)
Thus we see that for $\theta =\frac{\pi}{2}$, We have $d_1=d_2$, And for $\theta\neq\frac{\pi}{2}$, We have $d_1\neq d_2$.
Now we will separate for cases:
If $\theta=\frac{\pi}{2}$, Then we get that $d_1=d_2$, Let’s denote their common value by $d:=d_1=d_2$, And so $D=\{a,d\}$, Now since $d=2a\sin\frac{\theta}{2}=2a\sin\frac{\pi}{4}=2a\cdot \frac{\sqrt{2}}{2}=\sqrt{2}a$, And since $1\lt \sqrt{2}$, Which implies that $a\lt \sqrt{2} a$ (because $a\gt 0$), We conclude that $a\lt d$, And in particular $a\neq d$, And we conclude that $|D|=|\{a,d\}|=2\in\{2,3\}$.
If $\theta\neq\frac{\pi}{2}$, Then we get that $d_1\neq d_2$, And we conclude that $2=|\{d_1,d_2\}|\leq |\{a,d_1,d_2\}|=|D|$, Also since it is clear that $|D|=|\{a,d_1,d_2\}|\leq 3$, We conclude that $2\leq |D|\leq 3$, And so $|D|\in \{2,3\}$.
Thus we see that it is always the case that $|D|\in\{2,3\}$ as was to be shown.
So, some thoughts:
I do not think that there is a need to use Trigonometry to prove the first direction, Maybe some clever Triangle Inequality tricks can do it, But I wasn’t able to prove it in this way.
And about the second direction, I think Trigonometry will not even work here, I think there is some need to use some Graph Theoretic argument with some clever use of the Triangle Inequality.
Also, I think the proposition is true even with general norms defined on $\mathbb{R}^2$ or any other normed vector space of dimension $2$, But I guess we need for this the clever argument.
Thanks for any help for proving this in general...
(Clearly in $\mathbb{R}^3$, The proposition is false, Just take a tetrahedron as a counter-example).
There are six arrangements of points where $D=2$, only two of which are rhombuses. Therefore, your test does not work, as it will falsely conclude all of the below arrangements are rhombuses, though not all of them are.
However, this test does. Four points form a rhombus if
So compute all $6$ distances, then do at most $9$ equality checks.