Minimize the area of $n$ intersecting circles

Question

Let $0 < c < 2/(n-1)$. For $p_1,...,p_n \in \mathbb{R}^2$, consider the property $$ |p_{k+1}-p_k| = c, \ k=1,...,n-1 \tag 1 $$ where $| \cdot |$ is the Euclidean distance. Denote $C_k$ for the unit circle centered at $p_k, \ k=1,...,n$.

$\textbf{Question}$: For what geometric configuration of $p_1,...,p_n \in \mathbb{R}^2$ satisfying $(1)$ is the area of $C_1 \cap \cdots \cap C_n$ minimized? (note $c$ is fixed)

Comment: I've been trying to prove the solution is a straight line, as it appears to lead to a solution for the following: if you are placed at a random position in a circle, which path minimizes the average time it takes for you to escape the circle? (something I really want to solve)

$\textbf{Update} \ (2/19)$: I solved the alternate question, that the area of $C_1 \cup \cdots \cup C_n$ is maximized when $p_k$ are in line. The proof is below.

Write $\mu(A)$ for the area of a set $A$. Write $B_1 = C_1 \cup \cdots \cup C_{n-1}$ and $B_2 = C_2 \cup \cdots \cup C_n$. Then

\begin{align} \mu(C_1 \cup \cdots \cup C_n) &= \mu(B_1 \cup B_2) \\ &= \mu(B_1) + \mu(B_2) - \mu(B_1 \cap B_2) \\ &= \mu(B_1)+\mu(B_2) - \mu\big((C_1 \cap C_n) \cup (C_2 \cup \cdots \cup C_{n-1})\big) \\ &\leq \mu(B_1) + \mu(B_2) - \mu(C_2 \cup \cdots \cup C_{n-1}) \tag 2 \\ &= \mu(B_1 \backslash (C_2 \cup \cdots \cup C_{n-1})) + \mu (B_2) \\ &= \mu(C_1 \backslash (C_2 \cup \cdots \cup C_{n-1})) + \mu(B_2) \\ &\leq \mu(C_1 \backslash C_2) + \mu(B_2) \tag 3 \end{align}
We will now prove by induction $\mu(C_1 \cup \cdots \cup C_n)$ is maximized when $p_k$ are in a sequential line. For case $n=3$, notice $\mu(B_1)$ and $\mu(B_2)$ only depend on $c$, therefore are constant. Let $p_1, p_2, p_3$ be in a line. One can prove with law of consines that $C_1 \cap C_3 \subset C_2$. So equality holds at $(2)$, which shows the result.

For case $n > 3$, assume case $n-1$ holds. Let $p_1,...,p_n$ be in a line. By the induction hypothesis, $\mu(B_2)$ is maximized. $\mu(C_1 \backslash C_2)$ only depends on $c$, so is constant. So $(3)$ is maximized. Also, by law of cosines one can show $C_1 \cap C_n \subset C_2 \cup \cdots \cup C_{n-1}$ and $C_1 \backslash C_2 = C_1 \backslash (C_2 \cup \cdots \cup C_n)$. So equality holds at $(2)$ and $(3)$. This proves case $n$.

Answer 1

$\forall x \in \mathbb{R}^2$, let $B_1(x)$ be the closed unit ball in $\mathbb{R}^2$ with center $x$. Let $\mu$ be the Lebesgue measure.

We will outline a proof of the stronger statement (which hopefully can make you a believer, because many details are needed to make the proof rigorous):

For all paths $\gamma: [0,L] \rightarrow \mathbb{R}^2$ parameterized by arclength with $\gamma(0) = 0$ and length $L < 2$, $$ \mu\Big(\bigcap_{0 \leq s \leq L} B_1(\gamma(s))\Big) \geq \mu\Big(B_1(0) \cap B_1((L,0)) \Big). \tag 4 $$

It is easy to show with law of cosines that equality holds when $\gamma$ is a straight line.

$(4)$ solves the question asked, since one can take $L = c(n-1)$ and $\gamma$ to be a polygonal path with vertices $p_1,...,p_n$, making the LHS of $(4) \leq \mu(C_1 \cap \cdots \cap C_n)$, whereas if $p_1,...,p_n$ are put in a sequential line one has $\mu(C_1 \cap \cdots \cap C_n) = $ RHS of $(4)$.

Fix $L$ and a $\gamma$ satisfying the conditions of $(4)$. Denote $$ I(s) = \bigcap_{0 \leq s' \leq s} B_1(\gamma(s')), \ \forall s \in [0,L]. \\ $$ Note $\forall s' \in [0,s], \ B_1(\gamma(s')) \cap \partial B_1(\gamma(s))$ is an arc of $\partial B_1(\gamma(s))$. So $I(s) \cap \partial B_1(\gamma(s))$ is an intersection of arcs, hence also an arc of $\partial B_1(\gamma(s))$.

There are two main ideas:

(i): $\frac{d}{ds} \mu(I(s))$ is directly limited by the length of the arc $I(s) \cap \partial B_1(\gamma(s))$. This is explained as follows: $x \in I(s)$ gets removed from $I(s)$ as $s$ increases iff the arc $I(s) \cap \partial B_1(\gamma(s))$ is "dragged over" $x$ as $s$ increases. So a shorter arc leads to a slower decrease of area.

Similarly $\frac{d}{ds} \mu\big(B_1(0) \cap B_1((s,0))\big)$ is directly limited by the length of the arc $B_1(0) \cap \partial B_1((s,0))$.

(ii): Let $A$ be any arc of any unit circle. If $I(s) \supset A$ then (due to $I(s)$ being an intersection of closed unit balls) the smallest $I(s)$ can be is the intersection of all closed unit balls which contain $A$. i.e. $$ I(s) \supset A \implies I(s) \supset \bigcap_{\substack{x \in \mathbb{R}^2 \\ B_1(x) \supset A}} B_1(x). \tag 5 $$ It can be shown $$ \bigcap_{\substack{x \in \mathbb{R}^2 \\ B_1(x) \supset A}} B_1(x) = B_1(x_1) \cap B_1(x_2) \tag 6 $$
where $x_1$ is such that $A \subset \partial B_1(x_1)$ (recall $A$ is an arc of a unit circle) and $x_2$ is such that $\partial B_1(x_2)$ intersects $A$ at precisely the two endpoints of $A$.

Now, if for some $s \in [0,L]$, $$ \frac{d}{ds} \mu(I(s)) < \frac{d}{ds} \mu(B_1(0) \cap B_1((s,0)), \tag 7 $$ then by (i) the arc $I(s) \cap \partial B_1(\gamma(s))$ must be longer than the arc $B_1(0) \cap \partial B_1((s,0))$.

The intersection of all closed unit balls containing $B_1(0) \cap \partial B_1((s,0))$ is precisely $B_1(0) \cap B_1((s,0))$. Since $I(s) \cap \partial B_1(\gamma(s))$ is longer than $B_1(0) \cap \partial B_1((s,0))$, by $(5)$ and $(6)$ this implies $I(s)$ is a larger area than $B_1 \cap B_1((s,0))$.

In summary, $(7) \implies \mu(I(s)) > \mu\big(B_1(0) \cap B_1((s,0))\big)$. Denote $$ s_0 = \sup \big\{s \in [0,L] : \mu(I(s)) > \mu\big(B_1(0) \cap B_1((s,0))\big) \big\}. $$ So $s > s_0$ implies $(7)$ does not hold. It follows \begin{align} \mu(I(L)) & = \mu(I(s_0)) + \int_{s_0}^L \frac{d}{ds} \mu(I(s)) \ ds \\ & \geq \mu\big(B_1(0) \cap B_1((s_0,0))\big) + \int_{s_0}^L \frac{d}{ds} \mu\big(B_1(0) \cap B_1((s,0))\big) \ ds \\ &= \mu\big(B_1(0) \cap B_1((L,0)) \big) \end{align} which is $(4)$.