Consider a simple closed curve $C$ in $\mathbb{R}^2$. For any points $a$ and $b$ on this curve we associate a point $c_1$ on the left and $c_2$ on the right side to the chord ab, such that $ac_1bc_2$ is a square. Continuously moving (sliding) the chord ab in such a way that it goes to ba (and $a(t)$ never equals $b(t)$, i.e. $\|a(t)−b(t)\|>0 ~\forall t$), the points $c_1$ and $c_2$ will draw the curves $L_1$ and $L_2$. These curves will form a closed curve $L = L_1 \cup L_2$. My conjecture is that for any such trajectory and for any $C$, $L$ will intersect $C$, i.e. the curve $L$ can never be completely inside or outside $C$. Is it true?

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P.S. Curve $C$ can be non smooth.


1 Answer 1


This is yet another idea to to resolve this resistant problem. Let us parametrize everything. Identify the plane containing the moving square with the complex plane $\mathbb C$ and let $\mathbb T=\{z\in\mathbb C:|z|=1\}$ be the unit circle.

Let $a(t),b(t)$, $t\in\mathbb T$, be parametrizations of the motion of the points $a,b$ such that $a(-t)=b(t)$ for every $t\in\mathbb T$. Let $z(t)=\frac12(a(t)+b(t))$ be the parametrization of the motion of the center of the square. It follows that $z(-t)=z(t)$ is an even function.

By the assumption, the value $r(t)=|a(t)-z(t)|$ is strictly positive. So, the function $\varphi:\mathbb T\to\mathbb T$, $\varphi:t\mapsto \frac{a(t)-z(t)}{|a(t)-z(t)|}$, is well-defined and odd.

It follows that $a(t)=z(t)+r(t)\varphi(t)$ for $t\in\mathbb T$. Then $c(t)=z(t)\pm r(t)i\varphi(t)$, $t\in\mathbb T$ is the trajectory of the motion of the vertices $c_1$ and $c_2$.

Our aim is to prove that the curves $C=a(\mathbb T)$ and $L=c(\mathbb T)$ intersect each other. To derive a contradiction, assume that they are disjoint. Replacing the functions $z(t),r(t),\varphi(t)$ by their smooth approximations, we can assume that these functions are smooth and the curve $C$ has finite intersection with each line on $\mathbb C$.

For each point $t\in\mathbb T$ consider the map $f_t:\mathbb R\to\mathbb C$ defined by $f_t(x)=z(t)+i\varphi(t)x$ and let $\Phi(t)=f_x^{-1}(C)$. Our assumption on $C$ guarantees that the set $\Phi(t)$ is finite. Moreover, since the curve connects the points $a(t)=z(t)+r(t)\varphi(t)$ and $b(t)=z(t)=r(t)\varphi(t)$ that lie on opposite sides of the line $f_t(\mathbb R)$, the set $\Phi(t)$ is non-empty. So, we get the multi-valued map $\Phi:\mathbb T\multimap\mathbb R$ with finite non-empty values. This map is odd in the sense that $\Phi(-t)=-\Phi(t)$ for every $t\in\mathbb T$.

To finish the proof it suffices to find a point $t\in\mathbb T$ such that $r(t)\in\Phi(t)$. Further I have no idea how to do that. Maybe some connectedness of mean value argument (taking into account that the function $r(t)$ is even, $\Phi(t)$ is odd, and each value $r(t)$ is contained in some $\Phi(\tau)$)?

  • $\begingroup$ I thought about it, but the problem is that you don't take in acount the movement of the center of the chord. It's not trivial to show, that even with moving center, curves will intersect (at least I could not) $\endgroup$
    – makkostya
    Aug 18, 2017 at 20:18
  • $\begingroup$ Correct me if I miss understood your arguments. You say: 1. $L$ intersects $C$ $\iff$ $C'$ intersects $C''$ 2. Then you show in details that $C'$ intersects $C''$. But I had problems with the first argument. For the second one I had an idea of proof : shapes bounded by $C'$ and $C''$ have the same area and at least one point in common (origin). I imagine that it is enough to say that $C'$ intersects $C''$. $\endgroup$
    – makkostya
    Aug 19, 2017 at 9:05
  • $\begingroup$ But the first argument is not obvious for me. I mean why changing coordinate system in such a complex way preserves the fact of intersection? Is it some fundamental property? (I'm not at all specialist in this domain) $\endgroup$
    – makkostya
    Aug 19, 2017 at 9:05
  • $\begingroup$ @makkostya You are right there is a problem with this equivalence of intersections. Let me think a bit how to fix it. $\endgroup$ Aug 19, 2017 at 9:49
  • 2
    $\begingroup$ It's not true that $n_1$ and $n_2$ are both odd. On a more fundamental level, can you show where does your proof fail if you replace $c=z+i r \phi$ with $c=z+1024 i r\phi$? This is not a purely topological problem, it is also metric... $\endgroup$ Aug 19, 2017 at 17:51

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