# What is the form of the $(v_0,v_1)$-pizza curve?

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.

• 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, $y=\pm \sqrt{(1-2x)v_0^2 /(v_1^2 -v_0^2 ) -x^2 }$. – Michael Engelhardt Oct 3 '19 at 5:58
• @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 $U_0,U_1$. – Taras Banakh Oct 3 '19 at 6:25
• Ah! I didn't catch that essential detail. Now I appreciate better what you're asking, thank you. – Michael Engelhardt Oct 3 '19 at 6:27
• i think of $v_0$ as delivery by bike and $v_1$ as delivery by motorcycle – Matt F. Oct 3 '19 at 11:45

## 1 Answer

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 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.

• Will you (or will anyone) include a graph for $c=2$? – Matt F. Oct 3 '19 at 11:56
• @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. – Taras Banakh Oct 3 '19 at 12:24
• 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 $c$. – LSpice May 12 at 16:49