A curve is a circle or a line

Question

Let $\gamma(t):\mathbb{R}\to\mathbb{R}^2$ be a continuous curve in the plane such that for every $t_1,t_2\in\mathbb{R}$ the euclidean distance $d(\gamma(t_1),\gamma(t_2))$ depends only on $|t_1-t_2|$. Must the curve be a circle or a line?

I believe the answer is affirmative and tried to find a Möbius transformation that maps the curve to a generalized circle, or to show that the cross-ratio $[\gamma(0),\gamma(1),\gamma(2),\gamma(t)]$ (for arbitrary $t$) is a real number, but couldn't figure it out. Appreciate any advice!

Answer 1

The core of this problem is a simple geometry problem. Let $A=\gamma (t_0), B=\gamma (t_0+t), C=\gamma (t_0+2t), D=\gamma (t_0+3t)$. We have the following distance equalities: $|A-B|=|B-C|=|C-D|$ and $|A-C|=|B-D|$. Triangle $A-B-C$ is thus congruent to triangle $B-C-D$. If points $A, B, C etc.$ are collinear then the curve will turn out to be a line, because $t_0$, and $t$ are arbitrary. Otherwise Let $L_1$ be the line through point $B$ which bisects angle $A-B-C$ let $L_2$ be the line through point $C$ which bisects angle $B-C-D$, and point $G$ be the intersection of $L_1$, and $L_2$. Triangle $B-G-C$ is an isosceles triangle, because angle $G-B-C$ equals angle $G-C-B$. At this point distance $|B-G|=|C-G|$. Wrapping things up, this construction works for any $t_0$, and the triangles involved are always congruent, so point $G$ is distance $|B-G|$ from every point on the curve.

Answer 2

Suppose that the image of $\gamma$ contains three noncollinear points $A_k=\gamma(a_k) (k=1,2,3)$.

Let $s\in{\mathbb R}$. If we put $\eta_s(t)=\gamma(s+t)$, then we have $d(\eta_s(t_1),\eta_s(t_2))=d(\gamma(t_1),\gamma(t_2))$ for any $t_1,t_2$. By a well-known exercise, there is a isometry of $\mathbb R^2$ sending each $\gamma(t)$ to the corresponding $\eta_s(t)$. This isometry is in fact unique because an isometry is uniquely determined by its values on the three points $A_1,A_2,A_3$.

We shall denote this unique isometry by $R_s$ ; the fundamental equation is then

$$ \gamma(x+t)=R_t(\gamma(x)) \tag{1} $$

Evaulating $\gamma(x+t_1+t_2)$ in two ways using (1), we deduce

$$ R_{t_1+t_2}=R_{t_1} \circ R_{t_2} \tag{2} $$

If $R_1$ is a rotation with center $W$ and angle $\theta$, it is easy to see that $R_t$ is the rotation with center $W$ and angle $t\theta$ (do it first for rational $t$ by algebra, then extend to all of $\mathbb R$ using the continuity of $\gamma$). It follows then from (1) that the image of $\gamma$ is a circle centered at $W$.

If $R_1$ is a translation with vector $v$, it is easy to see that $R_t$ is the translation with vector $tv$ (do it first for rational $t$ by algebra, then extend to all of $\mathbb R$ using the continuity of $\gamma$). It follows then from (1) that the image of $\gamma$ is a line directed by $v$.

APPENDIX: The well-known exercise.

Theorem. Let $X$ be a set, and let $\eta,\gamma$ be two maps $\eta : X \to {\mathbb R}^a, \gamma : X \to {\mathbb R}^b$ be two that are "isometric", i.e. we have $d(\eta(x),\eta(x'))=d(\gamma(x),\gamma(x'))$ for any $x,x'\in X$. Then there is an isometry $i$ (from an affine subspace of ${\mathbb R}^a$ to an affine subspace of ${\mathbb R}^b$ sending $\gamma$ to $\eta$, i.e. $i(\gamma(x))=\eta(x)$ for any $x\in X$.

Proof : We may assume that $X$ is nonempty. Let $x_0 \in X$. Composing $\gamma$ and $\eta$ on the left with translations, we can assume that $\gamma(x_0)=\eta(x_0)=0$, which reduces the initial "affine" problem to a purely "vectorial" one.

The advantage of this reduction is that all this distance stuff may be reformulated in terms of the scalar product (using the formula $\langle u,v\rangle=\frac{d(u,0)^2+d(v,0)^2-d(u,v)^2}{2}$) : we now have $(*) : \langle\eta(x),\eta(x')\rangle=\langle\gamma(x),\gamma(x')\rangle$ for any $x,x'\in X$.

Decreasing $a$ and $b$ if necessary, we may assume that $\gamma(X)$ spans all of ${\mathbb R}^a$ and $\eta(X)$ spans all of ${\mathbb R}^b$.

There is a $X_0 \subseteq X$ with cardinality $a$, such that $\gamma(X_0)$ is a basis of ${\mathbb R}^a$. Write $X_0=\lbrace x_1,x_2,\ldots,x_a \rbrace$, and put $g_k=\gamma(x_k) (0\leq k \leq a)$, so that for example $g_0=0$.

Because of (*), the unique linear map $i$ sending $g_k$ to $\eta(x_k)$ is in fact an isometry. Finally, for any $x\in X$ the two vectors $i(\gamma(x))$ and $\eta(x)$ are the same, because they have the same coordinates in the system defined by $(\eta(x_1),\ldots,\eta(x_k))$. This finishes the proof.

Answer 3

(Another) geometric approach to solve it:

Let $P_n = \gamma(t_0 + n \cdot t)$. Then, it can shown (*) that $P_{-2}P_{2}P_{1}P_{-1}$ is an isosceles trapezoid or that they are collinear. Therefore, if 3 points are collinear, all of them are so. Otherwise, all of them belong to the circumference defined by three points, since the fourth point always completes an isosceles trapezoid, which is a cyclic quadrilateral.

The proof of (*) goes as follows: the two legs of $P_{-2}P_{2}P_{1}P_{-1}$ are equal, and the two diagonals too. If we fix the base $P_{-2}P_{2}$, there are only two alternatives for the other two vertices. They are either on the line $P_{-2}P_{2}$ (getting collinear points), on the same half-plane (thus getting an isosceles trapezoid), or on a different half-plane. The latter is actually impossible: it implies that $P_0$ is in the intersection of $P_{-2}P_{2}$ and $P_{-1}P_{1}$, and we can use the triangular inequality on $\triangle P_0P_1P_2$ to derive the contradiction $f(|2t|) < f(|2t|)$:

$f(|2t|) = d(P_0, P_2) < d(P_0, P_1) + d(P_1, P_2) = d(P_{-1}, P_0) + d(P_0, P_1) = d(P_{-1}, P_1) = f(|2t|)$.

Answer 4

Assume that $\overline{x}=\overline{x}(t)=(f(t),g(t))$, $t\in\textbf{R}$ is a plane curve, such that for every $t_1,t\in \textbf{R}$ we have
$$ |\overline{x}(t_1)-\overline{x}(t)|=h\left(t_1-t\right).\tag 1 $$

If we assume that $h(t)$ is such that $$ \lim_{t\rightarrow 0}\frac{h(t)}{t}=l<\infty, $$ then $\overline{x}(t)$ is differentiatable and $\overline{x}(t)$ is circle or line.

PROOF. We can write $$ |\overline{x}(t)-\overline{x}(t_1)|=h(t-t_1)\Rightarrow \left|\frac{\overline{x}(t)-\overline{x}(t_1)}{t-t_1}\right|= $$ $$ =\sqrt{\left(\frac{f(t)-f(t_1)}{t-t_1}\right)^2+\left(\frac{g(t)-g(t_1)}{t-t_1}\right)^2}=\frac{h(t-t_1)}{t-t_1}.\tag 2 $$ Hence taking the limit $t \rightarrow t_1$ in both sides of $(2)$ we get $$ \sqrt{f'(t_1)^2+g'(t_1)^2}=l=constant\textrm{, for all }t_1\in D\subseteq \textbf{R}. $$ Hence for the lenght $s=s(t)$ of the curve $\overline{x}(t)$, we have $$ s=lt+c_1. $$ Assumming the polar coordinates $r=r(\theta)$, we get (the length of a curve is an invriant) $$ \dot{s}^2=\left(\frac{ds}{d\theta}\right)^2=c^2=r(\theta)^2+\left(\frac{dr}{d\theta}\right)^2\tag 3 $$ Hence solving the DE $(3)$, we get $$ r(\theta)=c\frac{\tan(\theta\pm C_1)}{\sqrt{1+\tan(\theta\pm C_1)^2}}, $$ which are circles and lines, since $$ \frac{\tan(\theta\pm C_1)}{\sqrt{1+\tan(\theta\pm C_1)^2}}=\frac{\left|\cos(\theta\pm C_1)\right|\tan(\theta\pm C_1)}{\sqrt{\cos(\theta\pm C_1)^2+\sin(\theta\pm C_1)^2}} $$ and thus $$ r(\theta)=c\left|\cos(\theta\pm C_1)\right|\tan(\theta\pm C_1) $$

NOTE. In case $y=f(t)$ i.e. the curve $\overline{x}(t)=(t,f(t))$ is a function, then condition $(2)$ becomes $$ \sqrt{1+f'(t)^2}=l $$ and $y=f(t)$ is a line.