# 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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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.

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## Author information

Authors

### Corresponding author

Correspondence to Luis Barba.

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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