Let $k$ $(\geq 2)$ be an integer, to find $k$ points on the sphere or in the ball of a Euclidean $n$-space $E^{n}$ such that their pairwise distances are as large as possible is a long-standing problem in geometry. Let $\mathcal{C}$ be a plane convex body. Many authors considered this problem in the sense of the following notion of $\mathcal{C}$-distance of points in a plane convex body [3]. Some results concerning this kind of distance appeared in [1, 2, 4], and [6-10].

We recall the following definitions. For arbitrary different points $A, B\in E^{2}$, denote by $AB$ the line-segment connecting the points $A$ and $B$, by $|AB|$ the Euclidean length of the line-segment $AB$, by $\overrightarrow{AB}$ the ray starting at the point $A$ and passing through the point $B$, and by $\overline{AB}$ the straight line passing through the points $A$ and $B$. Let $\mathcal{C}$ be a plane convex body and let $A_{1}B_{1}$ be a longest chord of $\mathcal{C}$ parallel to $AB$. The $\mathcal{C}$-distance $d_{\mathcal{C}}(A, B)$ between the points $A$ and $B$ is defined by the ratio of $|AB|$ to $\frac{1}{2}|A_{1}B_{1}|$. If there is no confusion about $\mathcal{C}$, we may use the term relative distance between $A$ and $B$. Observe that for arbitrary points $A, B\in E^{2}$ the $\mathcal{C}$-distance between $A$ and $B$ is equal to their $[\frac{1}{2}(\mathcal{C}+(-\mathcal{C}))]$-distance. Thus the metric $d_{\mathcal{C}}(A, B)$ is the metric of $E^{2}$ whose unit ball is $\frac{1}{2}(\mathcal{C}+(-\mathcal{C}))$.

In this paper we consider the related problem. That is, let $A$, $B$, $C$ be three fixed points in the plane and let $\mathscr{T}:=ABC$ be the triangle formed by the points $A$, $B$, and $C$. We prove that, for any given real number $\tau>4$, the locus of the points $P$ satisfying the condition $d_{\mathscr{T}}(P, A)+d_{\mathscr{T}}(P, B)+d_{\mathscr{T}}(P, C)=\tau$, is a convex dodecagon (or nonagon) (That is, Theorem 2.4).

For simplicity, if two lines $\overline{PQ}$ and $\overline{RS}$ are parallel, we write $\overline{PQ}\|\overline{RS}$. Denote by $A(\mathcal{P})$ the area of the polygon $\mathcal{P}$. For a plane convex body $\mathcal{C}$, a chord $PQ$ of $\mathcal{C}$ is called an affine diameter if there is no longer chord parallel to $PQ$ in $\mathcal{C}$.

We first apply mass point geometry [5] to prove the following lemma. A mass point is a pair $(\alpha, P)$, where $\alpha$ is a positive number (the mass) and $P$ is a point in the plane. By the Archimedes principle of the lever, one can have the following addition rule for mass points.

Addition Rule: $(\varphi, A)+(\mu, B)=(\varphi+\mu, C)$, where point $C$ is on $AB$ with $|AC|:|CB|=\mu:\varphi$.

Lemma2.1   Let $T:=PAB$ be a triangle. Suppose that $X\in \overrightarrow{PA}$, $Y\in \overrightarrow{PB}$, and $Z\in XY$ with $Z=\lambda\cdot X+(1-\lambda)\cdot Y$, $0\leq\lambda\leq 1$. Then $d_{T}(P, Z)=\lambda\cdot d_{T}(P, X)+(1-\lambda)\cdot d_{T}(P, Y)$.

Proof   Denote by $C$ the intersection point of the lines $\overline{PZ}$ and $\overline{AB}$. If $X\notin PA$ or $Y\notin PB$, one may take $X'\in PA$, $Y'\in PB$, and $Z'\in PC$ with $\frac{|PX'|}{|PX|}=\frac{|PY'|}{|PY|}=\frac{|PZ'|}{|PZ|}$ (see the right picture in Figure 1), thus $\overline{XY}\|\overline{X'Y'}$ and $Z'=\lambda\cdot X'+(1-\lambda)\cdot Y'$, which implies that $d_{T}(P, Z)=\lambda\cdot d_{T}(P, X)+(1-\lambda)\cdot d_{T}(P, Y)$ if and only if $d_{T}(P, Z')=\lambda\cdot d_{T}(P, X')+(1-\lambda)\cdot d_{T}(P, Y')$. So without loss of generality, we may assume that $X\in PA$ and $Y\in PB$. Let $|PX|=\alpha_{1}$, $|PY|=\alpha_{2}$, $|PA|=\beta_{1}$, and $|PB|=\beta_{2}$. We assign masses $\alpha_{1}\varphi$, $\alpha_{2}\mu$, and $(\beta_{1}-\alpha_{1})\varphi+(\beta_{2}-\alpha_{2})\mu$ to points $A$, $B$, and $P$, respectively, where $\varphi$ and $\mu$ satisfy $\beta_{1}\varphi:\beta_{2}\mu=\lambda:(1-\lambda)$. Now we apply the addition rule in mass point geometry and prove that $Z$ is the center of the mass system $PAB$:

$ \begin{eqnarray*} &&(\alpha_{1}\varphi, A)+((\beta_{1}-\alpha_{1})\varphi+(\beta_{2}-\alpha_{2})\mu, P)+(\alpha_{2}\mu, B)\\&=&[(\alpha_{1}\varphi, A)+((\beta_{1}-\alpha_{1})\varphi, P)]+[(\beta_{2}-\alpha_{2})\mu, P)+(\alpha_{2}\mu, B)]\\&=&(\beta_{1}\varphi, X)+(\beta_{2}\mu, Y)=(\beta_{1}\varphi+\beta_{2}\mu, Z). \end{eqnarray*} $

Since $Z$ is the center of the mass system, the mass at $C$ can be obtained by adding the mass points of $A$ and $B$:

$ \begin{eqnarray*} &&(\alpha_{1}\varphi, A)+(\alpha_{2}\mu, B)=(\alpha_{1}\varphi+\alpha_{2}\mu, C), \\ &&|PZ|:|ZC|=(\alpha_{1}\varphi+\alpha_{2}\mu):[(\beta_{1}-\alpha_{1})\varphi+(\beta_{2}-\alpha_{2})\mu]. \end{eqnarray*} $

Thus

$ \begin{eqnarray*} d_{T}(P, Z)&=&\frac{2|PZ|}{|PC|}=\frac{2|PZ|}{|PZ|+|ZC|}=2\cdot\frac{\alpha_{1}\varphi+\alpha_{2}\mu}{\beta_{1}\varphi+\beta_{2}\mu}\\ &=&\lambda\cdot\frac{2\alpha_{1}}{\beta_{1}}+(1-\lambda)\cdot\frac{2\alpha_{2}}{\beta_{2}}=\lambda\cdot d_{T}(P, X)+(1-\lambda)\cdot d_{T}(P, Y), \end{eqnarray*} $

where the second last equality holds since $\lambda\beta_{2}\mu=(1-\lambda)\beta_{1}\varphi$.

Lemma2.2   Let $ABDC$ be a parallelogram and let $\mathscr{T}:=ABC$ be the triangle formed by the points $A$, $B$, and $C$. Suppose $U\in BD$ and $V\in \overrightarrow{AB}$ with $d_{\mathscr{T}}(U, A)+d_{\mathscr{T}}(U, B)+d_{\mathscr{T}}(U, C)=d_{\mathscr{T}}(V, A)+d_{\mathscr{T}}(V, B)+d_{\mathscr{T}}(V, C)=\tau$ (see Figure 2), then $d_{\mathscr{T}}(W, A)+d_{\mathscr{T}}(W, B)+d_{\mathscr{T}}(W, C)=\tau$ for any point $W\in UV$ with $W=\lambda\cdot U+(1-\lambda)\cdot V$, $0\leq\lambda\leq 1$.

Proof   Since this lemma satisfies the conditions of Lemma 2.1 (translate some triangle if necessary), we obtain that

$ \begin{eqnarray*} &&d_{\mathscr{T}}(A, W)=\lambda\cdot d_{\mathscr{T}}(A, U)+(1-\lambda)\cdot d_{\mathscr{T}}(A, V), \\ &&d_{\mathscr{T}}(B, W)=\lambda\cdot d_{\mathscr{T}}(B, U)+(1-\lambda)\cdot d_{\mathcal{\mathscr{T}}}(B, V), \\ &&d_{\mathscr{T}}(C, W)=\lambda\cdot d_{\mathscr{T}}(C, U)+(1-\lambda)\cdot d_{\mathscr{T}}(C, V). \end{eqnarray*} $

Thus

$ \begin{eqnarray*} &&d_{\mathscr{T}}(W, A)+d_{\mathscr{T}}(W, B)+d_{\mathscr{T}}(W, C)\\&=&\lambda\cdot(d_{\mathscr{T}}(A, U)+d_{\mathscr{T}}(B, U)+d_{\mathscr{T}}(C, U))+(1-\lambda)\cdot(d_{\mathscr{T}}(A, V)+d_{T}(B, V)+\\ &&d_{\mathscr{T}}(C, V))\\&=&\lambda\cdot\tau+(1-\lambda)\cdot\tau=\tau.\end{eqnarray*} $

Theorem2.3   Let $\mathscr{T}:=ABC$ be a triangle. If a point $P$ lies in the interior of $\mathscr{T}$, then $d_{\mathscr{T}}(P, A)+d_{\mathscr{T}}(P, B)+d_{\mathscr{T}}(P, C)=4$.

Proof   Denote by $D$ the intersection point of the lines $\overline{AP}$ and $\overline{BC}$, and denote by $\theta$ the angle formed by the lines $\overline{AD}$ and $\overline{BC}$, as shown in Figure 3. Then we have

$ \begin{eqnarray*} &d_{\mathscr{T}}(P, A)=\frac{2|PA|}{|AD|}=2\cdot(1-\frac{|PD|}{|AD|})= 2\cdot(1-\frac{|PD|\cdot|BC|\cdot\sin(\theta)/2}{|AD|\cdot|BC|\cdot\sin(\theta)/2})=\\ &2\cdot(1-\frac{A(PBC)}{A(ABC)}).\end{eqnarray*} $

Similarly, we get $d_{\mathscr{T}}(P, B)=2\cdot(1-\frac{A(PAC)}{A(ABC)})$, and $d_{\mathscr{T}}(P, C)=2\cdot(1-\frac{A(PAB)}{A(ABC)})$. Hence

$ \begin{eqnarray*}d_{\mathscr{T}}(P, A)+d_{\mathscr{T}}(P, B)+d_{\mathscr{T}}(P, C)&=&2\cdot(3-\frac{A(PBC)+A(PAC)+A(PAB)}{A(ABC)})\\&=&2\cdot(3-1)=4.\end{eqnarray*} $

Theorem2.4   Let $\mathscr{T}:=ABC$ be a triangle. Then for any real number $\tau>4$, the locus of the points $P$ satisfying the condition $d_{\mathscr{T}}(P, A)+d_{\mathscr{T}}(P, B)+d_{\mathscr{T}}(P, C)=\tau$, is either a convex dodecagon (when $\tau\neq 8$) or a nonagon (when $\tau=8$).

Proof   The proof follows from the following steps.

Step1   Draw a line-segment $CA_{1}$ with $CA_{1}\|AB$, and draw a line-segment $BA_{1}$ with $BA_{1}\|AC$ (see the left in Figure 4). Let $P$ be an arbitrary point in the triangle $A_{1}BC$, and let $D$ the be the intersection point of $AP$ and $BC$. Then $d_{\mathscr{T}}(P, A)=\frac{2|PA|}{|AD|}=2\cdot\frac{A(ABC)+A(PBC)}{A(ABC)}$. Similarly, one can have

$ \begin{eqnarray*} &&d_{\mathscr{T}}(P, B)=2\cdot\frac{A(PAB)}{A(ABC)}, \\ &&d_{\mathscr{T}}(P, C)=2\cdot\frac{A(PAC)}{A(ABC)}.\end{eqnarray*} $

Thus

$ d_{\mathscr{T}}(P, A)+d_{\mathscr{T}}(P, B)+d_{\mathscr{T}}(P, C)=4\cdot\frac{A(ABC)+A(PBC)}{A(ABC)}=4+4\cdot\frac{A(PBC)}{A(ABC)}. $

So in this case the locus of points $P$ must be a line-segment $XY$ parallel to $BC$.

Step2   Let $P$ belong to the unbounded angular region $MCN$ bounded by the lines $\overline{AC}$ and $\overline{BC}$, see the right in Figure 4. Then we get

$ \begin{eqnarray*} &&d_{\mathscr{T}}(P, A)=2\cdot\frac{A(PAB)}{A(ABC)}, \\ &&d_{\mathscr{T}}(P, B)=2\cdot\frac{A(PBA)}{A(ABC)}, \\ &&d_{\mathscr{T}}(P, C)=2\cdot\frac{A(PAB)-A(ABC)}{A(ABC)}. \end{eqnarray*} $

And thus

$ d_{\mathscr{T}}(P, A)+d_{\mathscr{T}}(P, B)+d_{\mathscr{T}}(P, C)=6\cdot\frac{A(PAB)}{A(ABC)}-2. $

So in this case the locus of points $P$ must be a line-segment $UV$ parallel to $AB$.

Step3   Take lines $\overline{BE}$ and $\overline{CF}$ such that $\overline{BE}\|\overline{AC}$, $\overline{CF}\|\overline{AB}$, respectively. Denote by $A_{1}$ the intersection point of the lines $\overline{BE}$ and $\overline{CF}$ (See the left in Figure 5). Let $P$ lie in the angular region $EA_{1}F$. Then

$ \begin{eqnarray*} &&d_{\mathscr{T}}(P, A)=2\cdot\frac{A(ABC)+A(PBC)}{A(ABC)}, \\ &&d_{\mathscr{T}}(P, B)=2\cdot\frac{A(PBC)}{A(ABC)}, \\ &&d_{\mathscr{T}}(P, C)=2\cdot\frac{A(PCB)}{A(ABC)}. \end{eqnarray*} $

And thus

$ d_{\mathscr{T}}(P, A)+d_{\mathscr{T}}(P, B)+d_{\mathscr{T}}(P, C)=2+6\cdot\frac{A(PBC)}{A(ABC)}. $

So in this case the locus of points $P$ must be a line-segment $ST$ parallel to $BC$.

By symmetry and by Lemma 2.2, from the three steps above we conclude that the locus of the points $P$ satisfying the condition, $d_{\mathscr{T}}(P, A)+d_{\mathscr{T}}(P, B)+d_{\mathscr{T}}(P, C)=\tau$, is a convex dodecagon (when $\tau\neq 8$), (see the right in Figure 5) or a nonagon (when $\tau=8$). The proof is complete.

We now generalize the result of Theorem 2.3 as follows.

Theorem2.5   Let $\mathcal{Q}:=ABCD$ be any convex quadrangle. If a point $P$ lies in the interior of $\mathcal{Q}$, then

$ 4\leq d_{\mathcal{Q}}(P, A)+d_{\mathcal{Q}}(P, B)+d_{\mathcal{Q}}(P, C)+d_{\mathcal{Q}}(P, D)\leq 6. $

Proof   Denote by $O$ the intersection point of the line-segments $AC$ and $BD$. The point $P$ must be in at least one of the four triangles $OAB$, $OBC$, $OCD$, and $ODA$. We suppose without loss of generality that $P$ lies in the interior of $OAB$ (see Figure 6). Since $d_{\mathcal{Q}}(A, C)=d_{\mathcal{Q}}(B, D)=2$, by the triangle inequality, we have

$ \begin{eqnarray*} &&d_{\mathcal{Q}}(P, A)+d_{\mathcal{Q}}(P, C)\geq d_{\mathcal{Q}}(A, C)=2, \\ &&d_{\mathcal{Q}}(P, B)+d_{\mathcal{Q}}(P, D)\geq d_{\mathcal{Q}}(B, D)=2.\end{eqnarray*} $

So

$ d_{\mathcal{Q}}(P, A)+d_{\mathcal{Q}}(P, B)+d_{\mathcal{Q}}(P, C)+d_{\mathcal{Q}}(P, D)\geq 4. $

Let $\mathscr{T}:=ABC$. Since $P$ lies in the interior of $T$, by Theorem 2.3 we have $d_{\mathscr{T}}(P, A)+d_{\mathscr{T}}(P, B)+d_{\mathscr{T}}(P, C)=4$. Since $\mathscr{T}\subset \mathcal{Q}$, we get $d_{\mathcal{Q}}(P, A)+d_{\mathcal{Q}}(P, B)+d_{\mathcal{Q}}(P, C)\leq 4$. From $d_{\mathcal{Q}}(P, D)\leq 2$, we conclude that

$ d_{\mathcal{Q}}(P, A)+d_{\mathcal{Q}}(P, B)+d_{\mathcal{Q}}(P, C)+d_{Q}(P, D)\leq 6. $

We also have the following proposition.

Corollary2.6   Let $\mathcal{S}:=ABCD$ be a unit square. Then the locus of the points $P$ satisfying the condition,

$ d_{\mathcal{S}}(P, A)+d_{\mathcal{S}}(P, B)+d_{\mathcal{S}}(P, C)+d_{\mathcal{S}}(P, D)=\tau, 4\leq\tau\leq6 $

is also a square.

Proof   We take a Cartesian coordinate system such that the coordinates of the points $A$, $B$, $C$, and $D$ are $(0, 0)$, $(1, 0)$, $(1, 1)$, $(0, 1)$, respectively. Denote by $E$ the intersection point of the line-segments $AC$ and $BD$. Let $P=(x, y)$ and let $P$ lie in the triangle $EBC$ (see the left in Figure 7). Then

$ \begin{eqnarray*} &&d_{\mathcal{S}}(P, A)=\frac{x}{\frac{1}{2}\cdot1}=2x, d_{\mathcal{S}}(P, B)=\frac{y}{\frac{1}{2}\cdot1}=2y, \\ &&d_{\mathcal{S}}(P, C)=\frac{1-y}{\frac{1}{2}\cdot1}=2-2y, d_{\mathcal{S}}(P, D)=\frac{x}{\frac{1}{2}\cdot1}=2x. \end{eqnarray*} $

So

$ d_{\mathcal{S}}(P, A)+d_{\mathcal{S}}(P, B)+d_{\mathcal{S}}(P, C)+d_{\mathcal{S}}(P, D)=4x+2. $

Thus we obtain that $4x+2=\tau$, that is, $x=\frac{\tau-2}{4}$.

Similarly, when $P\in ECD$, we have $y=\frac{\tau-2}{4}$. When $P\in EAB$, we get

$ \begin{eqnarray*} &&d_{\mathcal{S}}(P, A)=\frac{x}{\frac{1}{2}\cdot1}=2x, d_{\mathcal{S}}(P, B)=\frac{1-x}{\frac{1}{2}\cdot1}=2-2x, \\ &&d_{\mathcal{S}}(P, C)=\frac{1-y}{\frac{1}{2}\cdot1}=2-2y, d_{\mathcal{S}}(P, D)=\frac{1-y}{\frac{1}{2}\cdot1}=2-2y. \end{eqnarray*} $

So

$ d_{\mathcal{S}}(P, A)+d_{S}(P, B)+d_{\mathcal{S}}(P, C)+d_{\mathcal{S}}(P, D)=6-4y. $

Thus $6-4y=\tau$, that is, $y=\frac{6-\tau}{4}$. Similarly, when $P\in EAD$, we have $x=\frac{6-\tau}{4}$. It is clear that if $P$ lies in the boundary of $\mathcal{S}$, then

$ d_{\mathcal{S}}(P, A)+d_{\mathcal{S}}(P, B)+d_{\mathcal{S}}(P, C)+d_{\mathcal{S}}(P, D)=6, $

and if $P=E$, then

$ d_{\mathcal{S}}(P, A)+d_{\mathcal{S}}(P, B)+d_{\mathcal{S}}(P, C)+d_{\mathcal{S}}(P, D)=4. $

Then from the discussions above we conclude that the locus of the points $P$ satisfying the condition,

$ d_{\mathcal{S}}(P, A)+d_{\mathcal{S}}(P, B)+d_{\mathcal{S}}(P, C)+d_{\mathcal{S}}(P, D)=\tau, 4<\tau\leq6 $

is a square, whose center is the same as that of $\mathcal{S}$ (see the right in Figure 7). The proof is complete.