6
$\begingroup$

In the argand plane $$C:|z-1|+|z-2|+|z-3|=6 ~~~(1)$$ represents a bounded curve which is a rounded blob mimicking an "ellipse".

Using the the inequality $$|Z_1+Z_2+Z_3| \le |Z_1|+|Z_2|+|Z_3|, ~~~~(2)$$ we can see that $$6=|z-1|+|z-2|+|z-3|\ge |3z-6| \implies |z-2| \le 2. ~~~(3)$$ The circle $C_1: |z-2|=2$ touches and remains outside $C$. Next, it will be interesting to use some other inequality or something to find the equation of a bounded curve $C_2$ which touches and remains inside $C$. Eventually, $C$ will touch $C_1$ and $C_2$ and it will be enclosed between them.

The question is to find the curve $C_2.$

Edit:
For $C_2$, the RMS-AM inequality gives: $$\sqrt{\frac{|z-1|^2+|z-2|^2+|z-3|^2}{3}} \ge \frac{|z-1|+|z-2|+|z-3|}{3}=2~~~~(4)$$
The equality in above doesn't holds as $|z-1|=|z-2|=|z-3|$ cannot be met $$\implies(x-2)^2+y^2 >\sqrt{\frac{10}{3}}=1.82574~~~~(5)$$
This circle $|z-2|=\sqrt{\frac{10}{3}}$ cannot be the required $C_2$ as it cannot touch $C$ in (1).

$\endgroup$
6
  • 3
    $\begingroup$ What is your question? $\endgroup$ Sep 1, 2020 at 7:06
  • 2
    $\begingroup$ Indeed the shape is called a "$3$-ellipse": math.stackexchange.com/questions/124333/… $\endgroup$ Sep 1, 2020 at 7:07
  • $\begingroup$ Maybe he wants to plot the curve $\endgroup$ Sep 1, 2020 at 7:07
  • 2
    $\begingroup$ To find an equation for $C_2$. $\endgroup$
    – Z Ahmed
    Sep 1, 2020 at 7:09
  • $\begingroup$ $|1-z|+|z-2|+|z-3|=6$ thus using your inequality we get$ |z-4|\le 6$ $\endgroup$ Sep 1, 2020 at 7:22

2 Answers 2

2
$\begingroup$

Our claim is that the circle $$C_2:|z-2|=\sqrt\frac{44}{3}-2$$ lies entirely inside $C$. It is not hard to show that this value is the unique real $r$ for which $2\pm ir$ are on $C$.

We can calculate that $$|z-2|^2=\frac{|z-1|^2+|z-3|^2}{2}-1.$$ We have that $$\frac{|z-1|^2+|z-3|^2}{2}\geq\left(\frac{|z-1|+|z-3|}{2}\right)^2,$$ with equality if and only if $|z-1|=|z-3|$, so $$|z-1|+|z-3|\leq 2\sqrt{|z-2|^2+1},$$ again with equality if and only if $|z-1|=|z-3|$. So, for a point $z$ on $C$, $$6=|z-1|+|z-2|+|z-3|\leq |z-2|+2\sqrt{|z-2|^2+1};$$ since the right side is increasing in $|z-2|$, this implies that $|z-2|$ is at least the real $r$ for which $6=r+2\sqrt{r^2+1}$, with equality reached if and only if $z=2\pm ir$. It can be calculated that $r=\sqrt\frac{44}3-2$ is the positive value of $r$ that satisfies $6=r+2\sqrt{r^2+1}$, so we are done.

$\endgroup$
1
  • $\begingroup$ Very well done, you may note that $\sqrt{\frac{44}{3}} -2=1.82971$. You may see my edit in this regard. $\endgroup$
    – Z Ahmed
    Sep 1, 2020 at 8:01
2
$\begingroup$

Let us consider the equivalent problem, with a $(-2,0)$ translation:

$$|z+1|+|z|+|z-1|=6\tag{1}$$

Let (C) be the corresponding curve (generalized elllipse). It is possible to give an intuitive graphical construction of (C) (see Fig. 2) that will naturaly involve the maximal circle. Morover, in this way, we will be able to provide parametric equations of (C) (see (4)).

Due to the invariance of (1) with respect to transformations $z \to -z$ and $z \to $\overline{z}$, it is enough to study (C) in the first quadrant.

enter image description here

Fig. 1: The thick blue curve is the part of (C) belonging to the first quadrant (passing in particular through point $(2,0))$. It has been plotted using parametric equations (4). The other curves are level curves with equations $|z+1|+|z|+|z-1|=k$ for different values of $k$.

Let us introduce a parameter $m$ transforming (1) into the equivalent issue:

$$\begin{cases}\text{Circle }(C_m):&|z|&=&2m\\\text{Ellipse } (E_m):&|z+1|+|z-1|&=&6-2m\end{cases}\tag{2}$$

This is represented on Fig. 2:

enter image description here

Fig. 2: Circles $(C_m)$ and ellipses $(E_m)$ for values of $m \in [0,1]$ from red (small values of) $m$ to blue (values of $m$ close to $1$). Curve (C) in black. In fact $m \in [m_0/2,1]$ where $m_0=\sqrt{\tfrac{44}{3}}-2$... We have found back this critical value! The corresponding circle is thickened : it is the biggest circle that can be included into (C), and tangent to it (a single intersection solution).

Clearly, $m \in [0,3]$. In fact, the domain of variation of $m$ is much more narrow: this is easy to understand on Fig. 2: $m$ cannot be too small, because in this case the circle is itself too small to have an intersection with the ellipse.

The foci of $(E_m)$ are $z=-1$ and $z=1$ ; its semiaxes are:

$$a=3-m \ \ \text{and} \ \ b=\sqrt{a^2-1}.$$

A classical parametrization for the ellipse is:

$$\begin{cases}x&=&a \cos \theta\\y&=&b \sin \theta \end{cases} \ \iff \ \begin{cases}x^2&=&(3-m)^2 \cos^2 \theta\\y^2&=&((3-m)^2-1) \sin^2 \theta \end{cases}\tag{3}$$

If $(x,y)\in(C_m)\cap(E_m)$, we add the two right hand equations:

$$(2m)^2=(3-m)^2-\sin^2 \theta$$

For the first quadrant, where $\sin \theta \ge 0$ and $\cos \theta \ge 0$, we have a unique solution:

$$\sin \theta=\sqrt{(3-m)^2-4m^2}=\sqrt{-3(m-1)(m+3)}\tag{4}$$

meaning that the domain of variation of $m$ is included into $[0,1]$ (in fact, smaller).

We deduce from (3) and (4) the following parametric equations for the locus :

$$\begin{cases}x&=&(3-m)\sqrt{3m^2+6m-8}\\ y&=&\sqrt{-3(4-m)(2-m)(m-1)(m+3)}\end{cases}\tag{5}$$

(valid for the first quadrant as said upwards).

Remark: the critical value of the radius can be obtained (see legend of Fig. 2) by setting $x=0$ in (4), i.e. by solving $3m^2+6m-8=0$.

$\endgroup$
1
  • $\begingroup$ About generalized ellipses, see this article whose author is... the great Maxwell whose fields of interest were large... $\endgroup$
    – Jean Marie
    Apr 13, 2023 at 18:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .