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Assume that there are two (competing) pizza houses situated at the points $0$ and $1$ on the complex plane. These pizza houses can deliver pizza to points of the plane with the largest velocities $v_0$ and $v_1$, respectively.

Definition. A closed subset $P$ of the complex plane is called a $(v_0,v_1)$-pizza curve if $P$ is a common boundary of two open connected sets $U_0,U_1$ in $\mathbb C$ such that

  • $0\in U_0$ and $1\in U_1$;

  • $U_0\cap U_1=\emptyset$ and $U_0\cup P\cup U_1=\mathbb C$.

  • for any point $z\in P$ there exists a positive real number $t_z$ such that

    (i) for every $\varepsilon>0$ and every $k\in\{0,1\}$ there exists a smooth curve $\gamma_k:[0,t_z+\varepsilon)\to U_k$ such that $\gamma_k(0)=k$, $\lim_{t\to t_z+\varepsilon}\gamma_k(t)=z$ and $\lvert\gamma_k'(t)\rvert\le v_k$ for every $t\in[0,t_z+\varepsilon)$;

    (ii) for every $\varepsilon>0$ and every $k\in\{0,1\}$ there exists no smooth curve $\gamma_0:[0,t_z-\varepsilon)\to U_k$ such that $\gamma_k(0)=k$, $\lim_{t\to t_z-\varepsilon}\gamma_k(t)=z$ and $\lvert\gamma_k'(t)\rvert\le v_k$ for every $t\in[0,t_z-\varepsilon)$.

Problem. What is the form of a $(v_0,v_1)$-pizza curve? Is it unique?

Remark 1. If $v_0=v_1$, then the answer to this problem is well-known: the $(v_0,v_1)$-pizza curve is unique and coincides with the line $\{z\in\mathbb C:\Re(z)=\frac12\}$. So the problem essentially concerns the case $v_0\ne v_1$.

Remark 2. It can be shown that each $(v_0,v_1)$-pizza curve locally coincides with the graph of some Lipschitz function.

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  • $\begingroup$ Maybe this is a bit too heuristic for your taste, but my understanding is that the pizza curve consists of those points $(x,y)$ that both houses reach simultaneously if they go as quick as they can. Well, this implies that they travel in straight lines at maximal speed, so the condition is simply $(x^2 +y^2 )/v_0^2 = ((x-1)^2 + y^2 )/v_1^2 $, or, $y=\pm \sqrt{(1-2x)v_0^2 /(v_1^2 -v_0^2 ) -x^2 }$. $\endgroup$ – Michael Engelhardt Oct 3 '19 at 5:58
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    $\begingroup$ @MichaelEngelhardt The problem is that they are allowed to run only by their own territory, i.e. their trajectories should be contained in the closures of the sets $U_0,U_1$. $\endgroup$ – Taras Banakh Oct 3 '19 at 6:25
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    $\begingroup$ Ah! I didn't catch that essential detail. Now I appreciate better what you're asking, thank you. $\endgroup$ – Michael Engelhardt Oct 3 '19 at 6:27
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    $\begingroup$ i think of $v_0$ as delivery by bike and $v_1$ as delivery by motorcycle $\endgroup$ – Matt F. Oct 3 '19 at 11:45
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I can present an example of a pizza curve. We lose no generality assuming that $v_0=1$ and $v_1=c>1$.

A $(1,c)$-pizza curve $P$ is the union $A\cup B\cup \tilde B$ of three pieces:

  • the (shorter) arc $A$ of the circle $C$ (with equation $c^2(x^2+y^2)=(1-x)^2+y^2$), connecting the (conjugated) points $z_0$ and $\bar z_0$ such that the segments $[z_0,1]$, $[\bar z_0,1]$ lie in the tangent lines from $1$ to the circle $C$.

  • the curve $B=\{z_0e^{t(i+1/\sqrt{c^2-1})}:t\in [0,\pi-\arg(z_0)]\}$.

  • the curve $\tilde B$, symmetric to $B$ with respect to the real axis.

The (1, 2)-pizza curve described above

The curve $B$ has the property that for any $t\in[0,\pi-\arg(z_0)]$ the length $L_t$ of the curve from the initial point $z_0$ to the point $z_t=z_0e^{t(i+1/\sqrt{c^2-1})}$ satisfies the equation $L+L_t=c\lvert z_t\rvert$ where $L$ is the length of the segment $[1,z_0]$. Observe that $\lvert z_t\rvert$ and $\frac1c(L+L_t)$ are equal to the times spent by the pizza runners to get the point $z_t$ running on its own territory (including the boundary).

So, the problem is whether each $(1,c)$-pizza curve coincides with the pizza curve $P$ described above.

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  • $\begingroup$ Will you (or will anyone) include a graph for $c=2$? $\endgroup$ – Matt F. Oct 3 '19 at 11:56
  • $\begingroup$ @MattF. Sorry, I have no experience in plotting such thing (and have no time trying to learn how to do that using online graphics tools). I hope somebody with more skills can draw it fast. I just imagine that such a curve resembles a (rotated) drop of rain. $\endgroup$ – Taras Banakh Oct 3 '19 at 12:24
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    $\begingroup$ I believe that $(c^2 - 1)z_0^2 = -1$; in particular, $z_0$ is on the imaginary axis. I have edited in the image requested by @MattF.; I generated it using a Desmos calculator with a slider for $c$. $\endgroup$ – LSpice 2 days ago

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