I was doing random things when I noticed something which seemed strange to me.

What I did was the following.

  • Take two points $A$ and $C$ of the plane. We denote $\ell$ the distance $AC$.

  • Draw a circle $\mathscr C_0$ of center $A$ and radius $r_0$.

  • Then define $M_0$ to be a point of $\mathscr C_0$, and $M_1$ the middle of $[M_0C]$.

  • Finally, define $M_2$ and $M_3$ such that $M_0M_1M_2M_3$ is square.

It should look like something like this:

enter image description here

Then we are interested in the trajectory of $M_1$, $M_2$ and $M_3$ when $M_0$ move along the circle $\mathscr C_0$.

enter image description here

We notice that every $M_i$ seems to move on a circle $\mathscr C_i$ of a unique radius $r_i$.

enter image description here

But I don't get why this would be true, which leads us to the first questions:

Question 1. Are all trajectories $\mathscr C_i$ circles?

Question 2. What are the radius $r_i$ in terms of $\ell$ and $r_0$?

Question 3. Where are located the centers of those circles?

What I noticed is that if the radius $r_0$ changes, we still get three other circles, and they are all concentric:

enter image description here

And this is the case even when $r_0\geqslant \ell$:

enter image description here

It looks like this if $r_0$ varies continuously:

enter image description here

What I did to try to find the centers (since all three circles seems to have the same three centers for different radius $r_0$) is to see what it would look like with $r_0$ really small:

enter image description here

So the centers seems to be:

  • the middle point $A_1$ of $[AC]$,

  • the two points $A_2$, $A_3$ such that $AA_1A_2A_3$ is a square.


2 Answers 2



Let $A$ be the origin in the complex plane, and let $C$ be the point $c\in \mathbb{R}$

Let $M_0=re^{i \theta}$ so that $$M_1=\frac 12(c+M_0)$$

Then $$M_2=M_1+i(c-M_1)$$ and $$M_3=M_0+i(M_1-M_0)$$

Now you can obtain the parametric equations of the loci of $M_{1,2,3}$


Consider that vector ${CM_0} = {CA}+{AM_0} $, a combination of a fixed vector and a rotating constant-length vector.
${CM_1} = \frac 12{CA}+\frac 12{AM_0} $
${CM_2} = {CM_1} +{CM_1}^\perp = \frac 12({CA}+ {CA}^\perp) +\frac 12({AM_0} +{AM_0}^\perp)$
${CM_3} = {CM_0} +{CM_1}^\perp = ({CA} + \frac 12 {CA}^\perp) + ({AM_0} +\frac 12{AM_0}^\perp)$

So at the other points of the square we have in each case a fixed vector to the centre of the circle and a rotating constant-length vector to a point on the circle. Due to the combination of perpendicular vectors we should see $M_2$ and $M_3$ with shifted phase by $45°$ and about $26°$ respectively, which is borne out by your graphics.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.