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.

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Notes

  1. 1.

    Throughout this paper, all index manipulations are modulo \(k+1\).

  2. 2.

    In this case, \(r' \approx 0.94683\), from which \(r\approx 1.23219\).

  3. 3.

    Consider the function \(\iota (s_i) = 2\arcsin (\frac{s_i}{2r'}) - 2\arcsin (\frac{s_i}{2r})\). We can show, by elementary calculus, that \(\iota \) increases when \(s_i\) increases.

  4. 4.

    Consider the function \(\zeta (s_i) = 2\arcsin (\frac{s_i}{2r})r - s_i\). We can show, by elementary calculus, that \(\zeta \) increases when \(s_i\) increases.

  5. 5.

    Consider the function \(\xi (s_i) = s_i(\sqrt{r^2-(\frac{s_i}{2})^2}-\sqrt{r'^2-(\frac{s_i}{2})^2})\). We can show, by elementary calculus, that \(\xi \) increases when \(s_i\) increases.

  6. 6.

    Consider the function \(\chi (s_i) = r^2\arcsin (\frac{s_i}{2r}) - \frac{s_i}{2}\sqrt{r^2-(\frac{s_i}{2})^2}\). We can show, by elementary calculus, that \(\chi \) increases when \(s_i\) increases.

References

  1. 1.

    Aggarwal, A., Guibas, L., Saxe, J., Shor, P.: A linear time algorithm for computing the Voronoi diagram of a convex polygon. In: Proceedings of STOC, pp. 39–45. ACM, New York (1987)

  2. 2.

    Aggarwal, A., Guibas, L.J., Saxe, J., Shor, P.W.: A linear-time algorithm for computing the voronoi diagram of a convex polygon. Discret. Comput. Geom. 4(1), 591–604 (1989)

    Article  MATH  MathSciNet  Google Scholar 

  3. 3.

    Baker, A.: Transcendental Number Theory. Cambridge University Press, London (1990)

  4. 4.

    Banchoff, T., Giblin, P.: On the geometry of piecewise circular curves. Am. Math. Mon. 101(5), 403–416 (1994)

  5. 5.

    Barba, L.: Disk constrained 1-center queries. In: Proceedings of the Canadian Conference on Computational Geometry, CCCG, vol. 12, pp. 15–19 (2012)

  6. 6.

    Besicovitch, A.S.: Variants of a classical isoperimetric problem. Q. J. Math. Oxf. Ser. (2) 3, 42–49 (1952)

  7. 7.

    Blåsjö, V.: The isoperimetric problem. Am. Math. Mon. 112(6), 526–566 (2005)

    Article  MATH  Google Scholar 

  8. 8.

    Borwein, J.M., Crandall, R.E.: Closed forms: what they are and why we care. Not. Am. Math. Soc. 60(1), 50–65 (2013)

    Article  MATH  MathSciNet  Google Scholar 

  9. 9.

    Bose, P., De Carufel, J.-L.: Isoperimetric triangular enclosures with a fixed angle. J. Geom. 104(2), 229–255 (2013)

    Article  MATH  MathSciNet  Google Scholar 

  10. 10.

    Chandrasekaran, K., Dadush, D., Vempala, S.: Thin partitions: isoperimetric inequalities and a sampling algorithm for star shaped bodies. In: SODA, pp. 1630–1645 (2010)

  11. 11.

    Chow, T.Y.: What is a closed-form number? Am. Math. Mon. 106(5), 440–448 (1999)

    Article  MATH  Google Scholar 

  12. 12.

    de Berg, M., van Kreveld, M., Overmars, M., Schwarzkopf, O.: Computational Geometry: Algorithms and Applications, 3rd edn. Springer, Berlin (2008)

  13. 13.

    Garey, M.R., Graham, R.L., Johnson, D.S.: The complexity of computing Steiner minimal trees. SIAM J. Appl. Math. 32(4), 835–859 (1977)

    Article  MATH  MathSciNet  Google Scholar 

  14. 14.

    Hwang, F.K., Richards, D.S., Winter, P.: The Steiner Tree Problem. Elsevier, Amsterdam (1992)

  15. 15.

    Koutsoupias, E., Papadimitriou, C.H., Sideri, M.: On the optimal bisection of a polygon. INFORMS J. Comput. 4(4), 435–438 (1992)

    Article  MATH  MathSciNet  Google Scholar 

  16. 16.

    Megiddo, N.: Linear-time algorithms for linear programming in \(\mathbb{R}^3\) and related problems. SIAM J. Comput. 12(4), 759–776 (1983)

    Article  MATH  MathSciNet  Google Scholar 

  17. 17.

    Morgan, F., Hutchings, M., Howards, H.: The isoperimetric problem on surfaces of revolution of decreasing gauss curvature. Trans. Am. Math. Soc. 352(11), 4889–4909 (2000)

    Article  MATH  MathSciNet  Google Scholar 

  18. 18.

    Polya, G.: Mathematics and Plausible Reasoning: Induction and Analogy in Mathematics, vol. 1. Princeton University Press, New Jersey (1990)

  19. 19.

    Preparata, F., Shamos, M.: Computational Geometry—An Introduction. Springer, Berlin (1985)

  20. 20.

    Preparata, F.P.: The medial axis of a simple polygon. In: Mathematical Foundations of Computer Science 1977, pp. 443–450. Springer, Berlin (1977)

  21. 21.

    Singer, D.A.: Geometry: Plane and Fancy. Springer, Berlin (1998)

  22. 22.

    Welzl, E.: Smallest enclosing disks (balls and ellipsoids). In: Results and New Trends in Computer Science, pp. 359–370. Springer, Berlin (1991)

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Correspondence to Luis Barba.

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J.-L. De Carufel: Research supported in part by NSERC and FQRNT. S. Langerman: Directeur de recherches du F.R.S.-FNRS. D. L. Souvaine NSF Grant #CCF-0830734.

Appendix: Missing Proofs of Section 2.1

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.Footnote 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 haveFootnote 4

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

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.Footnote 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 haveFootnote 6

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

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. \(\square \)

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Aloupis, G., Barba, L., De Carufel, J. et al. Isoperimetric Enclosures. Graphs and Combinatorics 31, 361–392 (2015). https://doi.org/10.1007/s00373-015-1553-2

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Keywords

  • Isoperimetric problems
  • Voronoi diagram
  • Enclosing curves
  • Optimization problems