Isoperimetric Enclosures

Abstract

Given a number $P$, we study the following three isoperimetric problems introduced by Besicovitch in 1952: (1) Let $S$ be a set of $n$ points in the plane. Among all the curves with perimeter $P$ that enclose $S$, what is the curve that encloses the maximum area? (2) Let $Q$ be a convex polygon with $n$ vertices. Among all the curves with perimeter $P$ contained in $Q$, what is the curve that encloses the maximum area? (3) Let $r_{\circ }$ be a positive number. Among all the curves with perimeter $P$ and circumradius $r_{\circ }$, what is the curve that encloses the maximum area? In this paper, we provide a complete characterization for the solutions to Problems 1, 2 and 3. We show that there are cases where the solution to Problem 1 cannot be computed exactly. However, it is possible to compute in $O(n \log n)$ time the exact combinatorial structure of the solution. In addition, we show how to compute an approximation of this solution with arbitrary precision. For Problem 2, we provide an $O(n\log n)$-time algorithm to compute its solution exactly. In the case of Problem 3, we show that the problem can be solved in constant time. As a side note, we show that if $S$ is a set of $n$ points in the plane, then finding the area of the curve of perimeter $P$ that encloses $S$ and has minimum area is NP-hard.

Appendix: Missing Proofs of Section 2.1

In this appendix, we provide the proofs of Corollaries 1 and 2. The proofs are split into four lemmas: Lemmas 21 and 22 deal with perimeter approximations, while lemmas 23 and 24 deal with area approximations.

Recall that $\fancyscript{C}= (v_0, v_1, \ldots , v_k, v_0)$ denotes the sequence of $k$ vertices of $Q$ traversed by the $(P,Q)$-curve of maximum area, where $v_i$ and $v_{i+1}$ are connected by a circular arc $a_i$ of radius $r$ centered on $c_i$. Let $s_i = v_iv_{i+1}$, i.e., $s_i$ is the edge connecting the endpoints of the arc $a_i$. Recall that $Y = CH(v_0, v_1, \ldots , v_k)$ is the convex hull of the vertices of $\fancyscript{C}$. For each arc $a_i$ with endpoints $v_i$ and $v_{i+1}$, let $\alpha _i$ be the angle $\angle v_ic_ic_{i+1}$. Let $A_i$ denote the area of $CH(a_i)$. Therefore, the perimeter and area of $\fancyscript{C}$ can be expressed as $P = \sum _{i=0}^k r\alpha _i$ and $A = \sum _{i=0}^k A_i + \textsc {area}(Y)$, respectively.

By Lemma 12, we can compute an interval $(x,y)$ such that (1) $r\in (x,y)$ and (2) for any $r'\in (x,y)$, the boundary of $\varphi _{r'}$ and $\varphi _r = \fancyscript{C}$ have the same combinatorial structure.

Let $r'\in (x,y)$, and recall that $y$ may be equal to $+\infty$. Let $P' = \textsc {peri}(\varphi _{r'})$ and $A' = \textsc {area}(\varphi _{r'})$ denote the perimeter and area of $\varphi _{r'}$, respectively. In this appendix, we provide sufficient conditions to bound the difference between $P, P'$ and $A,A'$ in terms of the length of $(x,y)$, i.e., in terms of the difference between $r$ and $r'$.

Since the combinatorial structure of $\varphi _{r'}$ is equal to that of $\fancyscript{C}$ we can define $a'_i$, $c'_i$, $\alpha '_i$ and $A'_i$ analogously. Thus, the perimeter and area of $\varphi _{r'}$ can be expressed as $P' = \sum _{i=0}^k r' \alpha '_i$ and $A' = \sum _{i=0}^k A'_i + \textsc {area}(Y)$, respectively.

Lemma 21

Let $s_{*}$ be the largest $s_i$ and suppose that $r' < r$. If

$$\begin{aligned} r\left( 1 - \frac{1}{2}\left( \frac{\epsilon }{2s_{*}k}\right) ^2\right) < r', \end{aligned}$$

then $P' < P + \epsilon$.

Proof

We have $\frac{1}{2}s_{*} \le r' < r$. Moreover, $\alpha _i' > \alpha _i$ and $|a_i'| > |a_i|$. Suppose that we can find an $r'$ such that for all $1\le i \le k$,

$$\begin{aligned} |a_i'| - |a_i|&= r'\alpha _i' - r\alpha _i \nonumber \\&= r'\alpha _i' - r'\alpha _i + r'\alpha _i - r\alpha _i \nonumber \\&= r'(\alpha _i' - \alpha _i) - \alpha _i(r-r') \nonumber \\&< r'(\alpha _i' - \alpha _i) \qquad \text {since~} r > r', \\&< \frac{1}{k}\epsilon .\nonumber \end{aligned}$$

(6)

Then we get

$$\begin{aligned} P' = \sum _{i=0}^k |a_i'| < \sum _{i=0}^k \left( |a_i| + \frac{1}{k}\epsilon \right) = P + \epsilon . \end{aligned}$$

From (6), we see that it is sufficient to find an $r'$ such that for all $1\le i \le k$, $\alpha _i' - \alpha _i < \frac{\epsilon }{r'k}$. We have $\alpha _i=2\arcsin (\frac{s_i}{2r})$ and $\alpha _i'=2\arcsin (\frac{s_i}{2r'})$. The largest difference $\alpha _i' - \alpha _i$ occurs when $s_i$ is largest.³ Hence, if we find an $r'$ such that

$$\begin{aligned} 2\arcsin \left( \frac{s_{*}}{2r'}\right) - 2\arcsin \left( \frac{s_{*}}{2r}\right) < \frac{\epsilon }{r'k}, \end{aligned}$$

then it satisfies $\alpha _i' - \alpha _i < \frac{\epsilon }{r'k}$ for all $1\le i \le k$.

Since, by elementary calculus, $\arcsin (x) \le 2x$, it is sufficient to find an $r'$ such that

$$\begin{aligned} \arcsin \left( \frac{s_{*}}{2r'}\right) -\arcsin \left( \frac{s_{*}}{2r}\right) < \arcsin \left( \frac{\epsilon }{4r'k}\right) . \end{aligned}$$

We have

$$\begin{aligned} \arcsin \left( \frac{s_{*}}{2r'}\right) -\arcsin \left( \frac{s_{*}}{2r}\right)&< \arcsin \left( \frac{\epsilon }{4r'k}\right) \nonumber \\ \frac{s_{*}}{2r'}\sqrt{1-\left( \frac{s_{*}}{2r}\right) ^2} - \frac{s_{*}}{2r}\sqrt{1-\left( \frac{s_{*}}{2r'}\right) ^2}&< \frac{\epsilon }{4r'k} \qquad \text {by taking} \sin (\cdot ) \text {~on both sides,}\nonumber \\ \frac{s_{*}}{2rr'}\left( \sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2} - \sqrt{r'^2 - \left( \frac{s_{*}}{2}\right) ^2}\right)&< \frac{\epsilon }{4r'k} \nonumber \\ \sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2} - \sqrt{r'^2 - \left( \frac{s_{*}}{2}\right) ^2}&< \frac{\epsilon r}{2s_{*}k} \nonumber \\ \sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2} - \frac{\epsilon r}{2s_{*}k}&< \sqrt{r'^2 - \left( \frac{s_{*}}{2}\right) ^2}. \end{aligned}$$

(7)

If $\sqrt{r^2 - (\frac{s_{*}}{2})^2} - \frac{\epsilon r}{2s_{*}k} < 0$, then (7) is true for any $\frac{1}{2}s_{*} < r' < r$. Otherwise, we have $\sqrt{r^2 - (\frac{s_{*}}{2})^2} - \frac{\epsilon r}{2s_{*}k} \ge 0$. Equivalently,

$$\begin{aligned} \epsilon \le \frac{2s_{*}k\sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2}}{r}. \end{aligned}$$

(8)

Then, (7) becomes

$$\begin{aligned} \left( \sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2} - \frac{\epsilon r}{2s_{*}k}\right) ^2&< r'^2 - \left( \frac{s_{*}}{2}\right) ^2 \\ r\sqrt{1 + \frac{\epsilon ^2}{4s_{*}^2k^2} - \frac{\epsilon \sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2}}{s_{*}kr}}&< r'. \end{aligned}$$

Therefore, from (8), it is sufficient to find an $r'$ such that

$$\begin{aligned} r\sqrt{1 + \frac{\epsilon ^2}{4s_{*}^2k^2} - \frac{\epsilon ^2}{2s_{*}^2k^2} }&< r' \\ r\sqrt{1 - \left( \frac{\epsilon }{2s_{*}k}\right) ^2 }&< r'. \end{aligned}$$

And thus, from elementary calculus, it is sufficient to find an $r'$ such that

$$\begin{aligned} r\left( 1 - \frac{1}{2}\left( \frac{\epsilon }{2s_{*}k}\right) ^2\right) < r'. \end{aligned}$$

$\square$

Lemma 22

Let $s_{*}$ be the largest $s_i$. If $P > \sum _{i = 1}^k s_i + \epsilon$, then $r < \frac{\epsilon ^2 + s_{*}^2 k^2}{4 \epsilon k}$.

Proof

We have

$$\begin{aligned}&P > \sum _{i = 1}^k s_i + \epsilon \\&P - \sum _{i = 1}^k s_i > \epsilon \\&\sum _{i = 1}^k (\alpha _i r - s_i) > \epsilon \\&\sum _{i = 1}^k \left( 2\arcsin \left( \frac{s_i}{2r}\right) r - s_i\right) > \epsilon . \end{aligned}$$

Let $s_{*}$ be the largest $s_i$. We have⁴

$$\begin{aligned} \epsilon&< \sum _{i = 1}^k \left( 2\arcsin \left( \frac{s_i}{2r}\right) r - s_i\right) \\&\le \sum _{i = 1}^k \left( 2\arcsin \left( \frac{s_{*}}{2r}\right) r - s_{*} \right) \\&= k\left( 2\arcsin \left( \frac{s_{*}}{2r}\right) r - s_{*}\right) \\&< 2k\left( r-\sqrt{r^2-\left( \frac{s_{*}}{2}\right) ^2}\right) \qquad \text {by elementary calculus.} \end{aligned}$$

Consequently,

$$\begin{aligned} r < \frac{\epsilon ^2 + s_{*}^2 k^2}{4 \epsilon k}. \end{aligned}$$

$\square$

Lemma 23

Let $s_{*}$ be the largest $s_i$ and suppose that $r' < r$. If

$$\begin{aligned} r\left( 1 - \frac{1}{2}\left( \frac{\epsilon }{2s_{*}kr}\right) ^2\right) < r', \end{aligned}$$

then $A' < A + \epsilon$.

Proof

The proof is similar to that of Lemma 21. We have $\frac{1}{2}s_{*} \le r' < r$. Moreover, $\alpha _i' > \alpha _i$ and $A_i' > A_i$. Suppose that we can find an $r'$ such that for all $1\le i \le k$,

$$\begin{aligned} A_i' - A_i= & {} \left( \frac{1}{2}r'^2\alpha _i' - \frac{s_i}{2}\sqrt{r'^2-\left( \frac{s_i}{2}\right) ^2}\right) - \left( \frac{1}{2}r^2\alpha _i - \frac{s_i}{2}\sqrt{r^2-\left( \frac{s_i}{2}\right) ^2}\right) \nonumber \\= & {} \frac{1}{2}\left( r'^2\left( \alpha _i' - \alpha _i\right) \!-\! \alpha _i\left( r^2-r'^2\right) \right) \!+\! \frac{s_i}{2}\left( \sqrt{r^2-\left( \frac{s_i}{2}\right) ^2}-\sqrt{r'^2-\left( \frac{s_i}{2}\right) ^2}\right) \nonumber \\< & {} \frac{1}{2}r'^2\left( \alpha _i' - \alpha _i\right) + \frac{s_i}{2}\left( \sqrt{r^2-\left( \frac{s_i}{2}\right) ^2}-\sqrt{r'^2-\left( \frac{s_i}{2}\right) ^2}\right) \qquad \text {since~} r > r',\nonumber \\< & {} \frac{1}{k}\epsilon . \end{aligned}$$

(9)

Then we get

$$\begin{aligned} A' = \textsc {area}(Y) + \sum _{i=0}^k A_i' < \textsc {area}(Y) + \sum _{i=0}^k \left( A_i + \frac{1}{k}\epsilon \right) = A + \epsilon . \end{aligned}$$

From (9), it is sufficient to find an $r'$ such that for all $1\le i \le k$,

$$\begin{aligned} \alpha _i' - \alpha _i < \frac{1}{r'^2k}\epsilon \end{aligned}$$

(10)

and

$$\begin{aligned} s_i\left( \sqrt{r^2-\left( \frac{s_i}{2}\right) ^2}-\sqrt{r'^2-\left( \frac{s_i}{2}\right) ^2}\right) < \frac{1}{k}\epsilon . \end{aligned}$$

(11)

We first look at (11). The largest difference $s_i(\sqrt{r^2-(\frac{s_i}{2})^2}-\sqrt{r'^2-(\frac{s_i}{2})^2})$ occurs when $s_i$ is largest.⁵ Hence, if we find an $r'$ such that

$$\begin{aligned} s_{*} \left( \sqrt{r^2-\left( \frac{s_{*}}{2}\right) ^2}-\sqrt{r'^2-\left( \frac{s_{*}}{2}\right) ^2}\right) < \frac{1}{k}\epsilon , \end{aligned}$$

(12)

then it satisfies $s_i(\sqrt{r^2-(\frac{s_i}{2})^2}-\sqrt{r'^2-(\frac{s_i}{2})^2}) < \frac{1}{k}\epsilon$ for all $1\le i \le k$. Therefore, we explain how to find an $r'$ that satisfies (10) and (12). Notice that if $r'$ satisfies

$$\begin{aligned} \sqrt{r^2-\left( \frac{s_{*}}{2}\right) ^2}-\sqrt{r'^2-\left( \frac{s_{*}}{2}\right) ^2} < \frac{1}{2s_{*} k}\epsilon , \end{aligned}$$

(13)

then it satisfies (12). Next we show that if $r'$ satisfies (13), then it satisfies (10), which completes the proof.

The largest difference $\alpha _i' - \alpha _i$ occurs when $s_i$ is largest. Hence, if we find an $r'$ such that

$$\begin{aligned} 2\arcsin \left( \frac{s_{*}}{2r'}\right) - 2\arcsin \left( \frac{s_{*}}{2r}\right) < \frac{\epsilon }{r'^2k}, \end{aligned}$$

then it satisfies (10). Therefore, it is sufficient to find an $r'$ such that

$$\begin{aligned} \arcsin \left( \frac{s_{*}}{2r'}\right) -\arcsin \left( \frac{s_{*}}{2r}\right) < \arcsin \left( \frac{\epsilon }{4r'^2k}\right) . \end{aligned}$$

We have

$$\begin{aligned} \arcsin \left( \frac{s_{*}}{2r'}\right) -\arcsin \left( \frac{s_{*}}{2r}\right)&< \arcsin \left( \frac{\epsilon }{4r'^2k}\right) \\ \frac{s_{*}}{2rr'}\left( \sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2} - \sqrt{r'^2 - \left( \frac{s_{*}}{2}\right) ^2}\right)&< \frac{\epsilon }{4r'^2k} \qquad \text {by taking} \sin (\cdot ) \text {on both sides,} \\ \sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2} - \sqrt{r'^2 - \left( \frac{s_{*}}{2}\right) ^2}&< \frac{\epsilon r}{2r's_{*}k}. \end{aligned}$$

Since $r' < r$, it is sufficient to find an $r'$ such that

$$\begin{aligned} \sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2} - \sqrt{r'^2 - \left( \frac{s_{*}}{2}\right) ^2} < \frac{\epsilon }{2s_{*}k}, \end{aligned}$$

Which is exactly (13). Equivalently,

$$\begin{aligned} \sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2} - \frac{\epsilon }{2s_{*}k} < \sqrt{r'^2 - \left( \frac{s_{*}}{2}\right) ^2}. \end{aligned}$$

(14)

If $\sqrt{r^2 - (\frac{s_{*}}{2})^2} - \frac{\epsilon }{2s_{*}k} < 0$, then (14) is true for any $\frac{1}{2}s_{*} < r' < r$. Otherwise, we have $\sqrt{r^2 - (\frac{s_{*}}{2})^2} - \frac{\epsilon }{2s_{*}k} \ge 0$. Equivalently,

$$\begin{aligned} \epsilon \le 2s_{*}k\sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2}. \end{aligned}$$

(15)

Then, (14) becomes

$$\begin{aligned} \left( \sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2} - \frac{\epsilon }{2s_{*}k}\right) ^2&< r'^2 - \left( \frac{s_{*}}{2}\right) ^2 \\ r\sqrt{1 + \frac{\epsilon ^2}{4s_{*}^2k^2r^2} - \frac{\epsilon \sqrt{r^2 - \left( \frac{s_{*}}{2}\right) ^2}}{s_{*}kr^2}}&< r'. \end{aligned}$$

Therefore, from (15), it is sufficient to find an $r'$ such that

$$\begin{aligned} r\sqrt{1 + \frac{\epsilon ^2}{4s_{*}^2k^2r^2} - \frac{\epsilon ^2}{2s_{*}^2k^2r^2} }&< r' \\ r\sqrt{1 - \left( \frac{\epsilon }{2s_{*}kr}\right) ^2 }&< r'. \end{aligned}$$

And thus, from elementary calculus, it is sufficient to find an $r'$ such that

$$\begin{aligned} r\left( 1 - \frac{1}{2}\left( \frac{\epsilon }{2s_{*}kr}\right) ^2\right) < r'. \end{aligned}$$

$\square$

Lemma 24

Let $s_{*}$ be the largest $s_i$. If $A > \textsc {area}(Y) + \epsilon$, then $r < \frac{4 \epsilon ^2 + s_{*}^4 k^2}{8 s_{*} \epsilon k}$.

Proof

We have

$$\begin{aligned}&A > \textsc {area}(Y) + \epsilon \\&A - \textsc {area}(Y) > \epsilon \\&\sum _{i = 1}^k A_i > \epsilon \\&\sum _{i = 1}^k \left( \frac{1}{2}r^2\alpha _i - \frac{s_i}{2}\sqrt{r^2-\left( \frac{s_i}{2}\right) ^2}\right) > \epsilon \\&\sum _{i = 1}^k \left( r^2\arcsin \left( \frac{s_i}{2r}\right) - \frac{s_i}{2}\sqrt{r^2-\left( \frac{s_i}{2}\right) ^2}\right) > \epsilon . \end{aligned}$$

Let $s_{*}$ be the largest $s_i$. We have⁶

$$\begin{aligned} \epsilon&< \sum _{i = 1}^k \left( r^2\arcsin \left( \frac{s_i}{2r}\right) - \frac{s_i}{2}\sqrt{r^2-\left( \frac{s_i}{2}\right) ^2}\right) \\&\le \sum _{i = 1}^k \left( r^2\arcsin \left( \frac{s_{*}}{2r}\right) - \frac{s_{*}}{2}\sqrt{r^2-\left( \frac{s_{*}}{2}\right) ^2}\right) \\&= k\left( r^2\arcsin \left( \frac{s_{*}}{2r}\right) - \frac{s_{*}}{2}\sqrt{r^2-\left( \frac{s_{*}}{2}\right) ^2}\right) \\&< k s_{*}\left( r-\sqrt{r^2-\left( \frac{s_{*}}{2}\right) ^2}\right) \qquad \text {by elementary calculus.} \end{aligned}$$

Consequently,

$$\begin{aligned} r < \frac{4 \epsilon ^2 + s_{*}^4 k^2}{8 s_{*} \epsilon k}. \end{aligned}$$

$\square$

Corollary 1. Let $s_{*}$ be the largest $s_i$, and suppose that $s_{*} = 2$ and $r' < r$. If $P > \sum _{i = 1}^k s_i + \epsilon$ or $A > \textsc {area}(Y) + \epsilon$, then $r < \frac{\epsilon }{4k}+\frac{k}{\epsilon }$.

Proof

If $P > \sum _{i = 1}^k s_i + \epsilon$, then

$$\begin{aligned} r < \frac{\epsilon ^2 + s_{*}^2 k^2}{4 \epsilon k} = \frac{\epsilon ^2 + 2^2 k^2}{4 \epsilon k} = \frac{\epsilon }{4k}+\frac{k}{\epsilon } \end{aligned}$$

by Lemma 22. If $A > \textsc {area}(Y) + \epsilon$, then

$$\begin{aligned} r < \frac{4 \epsilon ^2 + s_{*}^4 k^2}{8 s_{*} \epsilon k} = \frac{4 \epsilon ^2 + 2^4 k^2}{8 \cdot 2 \epsilon k} = \frac{\epsilon }{4k}+\frac{k}{\epsilon } \end{aligned}$$

by Lemma 24. $\square$

Corollary 2. Let $s_{*}$ be the largest $s_i$, and suppose that $s_{*} = 2$ and $r' < r$. Moreover, suppose that $P > \sum _{i = 1}^k s_i + \epsilon$ or $A > \textsc {area}(Y) + \epsilon$. If

$$\begin{aligned} r - r' < \frac{1}{2}\left( \frac{\epsilon ^2}{\epsilon ^2 + 4 k^2}\right) ^2, \end{aligned}$$

then $|P' - P|< \epsilon$ and $|A' - A| < \epsilon$.

Proof

Let $\delta = \frac{1}{2}(\frac{\epsilon ^2}{\epsilon ^2 + 4 k^2})^2$. We have that $r -\delta < r'$ and $r\ge 1$ since we assumed that $s_* = 2$. Therefore, $r'> r-\delta \ge r-\delta r = r(1-\delta )$.

We have that

$$\begin{aligned} r\left( 1 - \frac{1}{2}\left( \frac{\epsilon }{2s_{*}k}\right) ^2\right)&\le r\left( 1 - \frac{1}{2}\left( \frac{\epsilon }{2\cdot 2kr}\right) ^2\right) \qquad \text {since~} r \ge \frac{1}{2}s_{*}, \\&\le r\left( 1 - \frac{1}{2}\left( \frac{\epsilon }{2\cdot 2k\left( \frac{\epsilon }{4k}+\frac{k}{\epsilon }\right) }\right) ^2\right) \qquad \text {by Corollary 1} \\&= r\left( 1 - \frac{1}{2}\left( \frac{\epsilon ^2}{\epsilon ^2 + 4 k^2}\right) ^2\right) . \end{aligned}$$

By Lemmas 21 and 23 and from the fact that

$$\begin{aligned} r(1-\delta ) = r\left( 1 - \frac{1}{2}\left( \frac{\epsilon ^2}{\epsilon ^2 + 4 k^2}\right) ^2\right) < r', \end{aligned}$$

we conclude that both $P' < P + \epsilon$ and $A' < A + \epsilon$. Moreover, since $r' < r$, we know that $P' > P$ and $A' >A$ which competes the proof. $\square$