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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.

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2 Answers 2

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Hint

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 $M_{1,2,3}$

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Consider that vector ${CM_0} = {CA}+{AM_0} $, a combination of a fixed vector and a rotating constant-length vector.
Then:
${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.

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