Невыпуклый многоугольник

[waxtep]

slavav в сообщении #1439749 писал(а):

Восьмиугольник удобно записать в радикалах:
$(8, \sqrt 2 + (8\sqrt 2 - 2)i, -10 + 16i, -\sqrt 2 + (8\sqrt 2 + 2)i, 12, -\sqrt 2 - (8\sqrt 2 + 2)i, -10 - 16i, \sqrt 2 - (8\sqrt 2 - 2)i)$

ух ты, даже 8-угольник! И, раз там только целые и корни из двух, любую его итерацию же можно выписать в явном виде, по крайней мере в принципе. Красота!

[staric]

Гм. Пока полностью согласен только с ответом на п. а).
По б). Действительно, нетрудно проверить домножением указанного slavav вектора для восьмиугольника на матрицу

$$\begin{bmatrix} 1 & \frac {2+\sqrt{2}}{4} & \frac12 & \frac {2-\sqrt{2}}{4} & 0 & \frac {2-\sqrt{2}}{4} & \frac12 & \frac {2+\sqrt{2}}{4}\\ \frac {2+\sqrt{2}}{4} & 1 & \frac {2+\sqrt{2}}{4} & \frac12 & \frac {2-\sqrt{2}}{4} & 0 & \frac {2-\sqrt{2}}{4} & \frac12 \\ \frac12 & \frac {2+\sqrt{2}}{4} & 1 & \frac {2+\sqrt{2}}{4} & \frac12 & \frac {2-\sqrt{2}}{4} & 0 & \frac {2-\sqrt{2}}{4}\\ \frac {2-\sqrt{2}}{4} & \frac12 & \frac {2+\sqrt{2}}{4} & 1 & \frac {2+\sqrt{2}}{4} & \frac12 & \frac {2-\sqrt{2}}{4} & 0 \\ 0 & \frac {2-\sqrt{2}}{4} & \frac12 & \frac {2+\sqrt{2}}{4} & 1 & \frac {2+\sqrt{2}}{4} & \frac12 & \frac {2-\sqrt{2}}{4} \\ \frac {2-\sqrt{2}}{4} & 0 & \frac {2-\sqrt{2}}{4} & \frac12 & \frac {2+\sqrt{2}}{4} & 1 & \frac {2+\sqrt{2}}{4} & \frac12 \\ \frac12 & \frac {2-\sqrt{2}}{4} & 0 & \frac {2-\sqrt{2}}{4} & \frac12 & \frac {2+\sqrt{2}}{4} & 1 & \frac {2+\sqrt{2}}{4} \\ \frac {2+\sqrt{2}}{4} & \frac12 & \frac {2-\sqrt{2}}{4} & 0 & \frac {2-\sqrt{2}}{4} & \frac12 & \frac {2+\sqrt{2}}{4} & 1 \\ \end{bmatrix}$$

,что представленный восьмиугольник вытягивается в пределе итераций в вертикальную струнку, однако остается пара вопросов:
- не потеряет ли он невыпуклость "по дороге"?
- как насчет n=6 и n=7?
Хотя в целом красиво сработано, что да - то да.

[slavav]

Первая геометрическая лемма
Образ круга - эллипс:
Рассмотрим $M: \mathbb{C} \to \mathbb{C}, M(x) = ax + b\overline{x}$, где $\overline{x}$ - операция сопряжения.
Можно доказать что образ единичной окружности $M({\left\lvert x\right\rvert = 1})$ есть эллипс с полуосями $\left\lvert a \right\rvert + \left\lvert b \right\rvert$ и $\left\lvert \left\lvert a \right\rvert - \left\lvert b \right\rvert \right\rvert$.
Если $\left\lvert a \right\rvert = \left\lvert b \right\rvert$, то эллипс вырождается в отрезок длины $4\left\lvert a \right\rvert$.

Вторая геометрическая лемма
Малые возмущения не портят выпуклость:
Дан выпуклый многоугольник $u$. $w_m$ последовательность ломаных (возможно самопересекающихся).
Можно доказать что если $\lim\limits_{m \to \infty}w_m = 0$, то найдётся $m_0$ такое что $u + w_{m_0}$ - выпуклый.

Обозначения
Первообразный корень из единицы порядка $n$: $\varepsilon_n = e^{\frac{2\pi i}{n}}$.
Отображение многоугольников задаёт оператор $A$ с собственными векторами и значениями:
$v_k=(1,\varepsilon_n^k, \varepsilon_n^{2k}, \dots, \varepsilon_n^{(n-2)k}, \varepsilon_n^{(n-1)k}), k = 0 \dots n-1$,
$\lambda_k=\frac{1 + \varepsilon_n^k}2, k = 0 \dots n-1$.

Пары собственных чисел и векторов сопряжены: $\lambda_k = \overline{\lambda_{n-k}}, v_k = \overline{v_{n-k}}, k = 1 \dots n - 1$.

Модули собственных чисел убывают к середине: $\left\lvert \lambda_k \right\rvert > \left\lvert \lambda_{k+1} \right\rvert, k = 0, \dots, \left\lfloor\frac{n-1}{2}\right\rfloor$.

Произвольный многоугольник в базисе из собственных векторов: $\sum\limits_{k=0}^{n-1} a_k v_k$.

Отобразим его $m$ раз: $A^m(\sum\limits_{k=0}^{n-1} a_k v_k) = \sum\limits_{k=0}^{n-1} a_k \lambda_k^m v_k$.

"Предел" итераций по $m$

Положим $a_0 = 0$. Он отвечает только за трансляцию всего многоугольника и не влияет на выпуклость так как $v_0 = (1, \dots, 1)$.
Положим $\left\lvert a_1 \right\rvert \ne \left\lvert a_{n - 1} \right\rvert$.

Все вершины многоугольника можно умножать или делить на один и тот же коэффициент. На выпуклость не влияет:
$\frac{A^m(\sum\limits_{k=1}^{n - 1} a_k v_k)}{\left\lvert\lambda_1^m\right\rvert} = \sum\limits_{k=1}^{n - 1} a_k \frac{\lambda_k^m}{\left\lvert\lambda_1^m\right\rvert} v_k =$

$= [a_1 \frac{\lambda_1^m}{\left\lvert\lambda_1^m\right\rvert} v_1 + a_{n - 1} \frac{\lambda_{n - 1}^m}{\left\lvert\lambda_1^m\right\rvert} v_{n - 1}] + \sum\limits_{k=2}^{n - 2} a_k \frac{\lambda_k^m}{\left\lvert\lambda_1^m\right\rvert} v_k =$

$= [a_1 \frac{\lambda_1^m}{\left\lvert\lambda_1^m\right\rvert} v_1 + a_{n - 1} \frac{\overline{\lambda_1}^m}{\left\lvert\lambda_1^m\right\rvert} v_{n - 1}] + \sum\limits_{k=2}^{n - 2} a_k \frac{\lambda_k^m}{\left\lvert\lambda_1^m\right\rvert} v_k$.

$= [a_1 (\frac{\lambda_1}{\left\lvert\lambda_1\right\rvert})^m v_1 + a_{n - 1} (\frac{\overline{\lambda_1}}{\left\lvert\lambda_1\right\rvert})^m \overline{v_1}] + \sum\limits_{k=2}^{n - 2} a_k \frac{\lambda_k^m}{\left\lvert\lambda_1^m\right\rvert} v_k$.

$= \left\lbrace a_1 (\frac{\lambda_1}{\left\lvert\lambda_1\right\rvert})^m v_1 + a_{n - 1} \overline{(\frac{\lambda_1}{\left\lvert\lambda_1\right\rvert})^m v_1}\right\rbrace + \sum\limits_{k=2}^{n - 2} a_k \frac{\lambda_k^m}{\left\lvert\lambda_1^m\right\rvert} v_k$.

По первой геометрической лемме первое слагаемое отображает $v_1$ в выпуклый многоугольник вписанный в эллипс с полуосями $\left\lvert a_1 \right\rvert + \left\lvert a_{n - 1} \right\rvert$ и $\left\lvert \left\lvert a_1 \right\rvert - \left\lvert a_{n - 1} \right\rvert \right\rvert$.
Второе слагаемое стремится к нулю так как $\lim\limits_{m \to \infty}(\frac{\lambda_k}{\left\lvert\lambda_1\right\rvert})^m = 0, k = 2, \dots, n - 2$.

По второй геометрической лемме в последовательности отображений найдётся выпуклый многоугольник.

Вывод
Так как мы ищем многоугольник который останется невыпуклым после любого $A^m$, то надо рассматривать только случай $\left\lvert a_1 \right\rvert = \left\lvert a_{n - 1} \right\rvert$.

[slavav]

Почему шестиугольник не годится?
Базис выглядит так:

$$\begin{tikzpicture}\draw [black, thin](1.48,0.0000) -- (1.5000,0.0000);\draw [black, thin](2.6000,0.0000) -- (2.3500,0.4330) -- (1.8500,0.4330) -- (1.6000,0.0000) -- (1.8500,-0.4330) -- (2.3500,-0.4330) -- (2.6000,0.0000);\draw [black, thin](3.7000,0.0000) -- (2.9500,0.4330) -- (2.9500,-0.4330) -- (3.7000,-0.0000) -- (2.9500,0.4330) -- (2.9500,-0.4330) -- (3.7000,0.0000);\draw [black, thin](4.8000,0.0000) -- (3.8000,0.0000) -- (4.8000,-0.0000) -- (3.8000,0.0000) -- (4.8000,-0.0000) -- (3.8000,0.0000) -- (4.8000,0.0000);\draw [black, thin](5.9000,0.0000) -- (5.1500,-0.4330) -- (5.1500,0.4330) -- (5.9000,-0.0000) -- (5.1500,-0.4330) -- (5.1500,0.4330) -- (5.9000,0.0000);\draw [black, thin](7.0000,0.0000) -- (6.7500,-0.4330) -- (6.2500,-0.4330) -- (6.0000,0.0000) -- (6.2500,0.4330) -- (6.7500,0.4330) -- (7.0000,0.0000);\end{tikzpicture}$$

Предположим что нам удалось построить шестиугольник, который остаётся не выпуклым без самопересечений после применения $A^m, m \geqslant 0$.
Так как нулевой вектор не меняет форму многоугольника то $a_0 = 0$.
Так как $\lambda_3 = 0$, то не умоляя общности $a_3 = 0$.
Из предыдущего сообщения $\left\lvert a_1 \right\rvert = \left\lvert a_5 \right\rvert$.
Последнее условие означает что шестиугольник проектируется на прямую (временно примем $a_2 = a_4 = 0$). Обозначим проекции $p_k$. В каком порядке они могут оказаться на прямой? Вариантов конечное количество. Рассмотрим основные варианты:
1. $p_0 < p_1 = p_5 < p_2 = p_4 < p_3$.
2. $p_0 < p_1 < p_5 < p_2 < p_4 < p_3$.
3. $p_0 = p_1 < p_5 = p_2 < p_4 = p_3$.

Оператор $A$ переводит вариант 1 в 3 и наоборот. Вариант 2 переходит сам в себя с циклической перестановкой индексов.

Вариант 3 не подходит: после проекции пятая и вторая вершины совпадают. Они совпадают и в векторах $\varepsilon^2_6$ и $\varepsilon^4_6$ (в базисе это треугольники). Тогда они совпадут и в любой линейной комбинации. Самопересечение.
Вариант 1 перейдёт в вариант 3 и даст самопересечение на следующем шаге.

Вариант 2
Применим оператор $A$ много раз. Модули $\lambda_2^m a_2$ и \lambda_4^m $a_4$ есть o-малые от модуля $\lambda_1^m a_1$. Тогда начиная с некоторой итерации добавление $\lambda_2^m a_2 \varepsilon^2_6$ + $\lambda_4^m a_4 \varepsilon^4_6$ к точкам $p_i$ не будет влиять на порядок их проекций на прямую.

Повернём прямую чтобы она стала горизонтальной. Нулевая и третья вершины на одной высоте. То же верно для первой и четвёртой. И для второй и пятой. Среди этих трёх высот есть минимальная и максимальная. Выберем любой экстремум, но не пару ноль-три. Пусть для определённости это пара один-четыре, максимум. Первая вершина принадлежит цепи $0-1-2-3$, четвёртая - цепи $3-4-5-0$. Из этих двух цепей одна выше другой. Следовательно у них должны быть разные максимумы. Противоречие, которое разрешается только если цепи пересекаются.

То есть, шестиугольник не может решить задачу.

[slavav]

Про восьмиугольник.
Введём функцию, отображающую пару вещественных чисел в восьмиугольник: $v\left(c, d\right) = \left(v_1 - v_7\right) + c v_2 + c d \left(v_3 + v_5\right)$.
Первое слагаемое $v\left(0, 0\right) = v_1 - v_7$ выглядит так:

$$ \begin{tikzpicture} \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,4.2426) -- (0.0354,4.2780) -- (0.0000,4.2926) -- (-0.0354,4.2780) -- (-0.0500,4.2426) -- (-0.0354,4.2073) -- (0.0000,4.1926) -- (0.0354,4.2073) -- cycle; \draw [black, fill](0.0500,6.0000) -- (0.0354,6.0354) -- (0.0000,6.0500) -- (-0.0354,6.0354) -- (-0.0500,6.0000) -- (-0.0354,5.9646) -- (0.0000,5.9500) -- (0.0354,5.9646) -- cycle; \draw [black, fill](0.0500,4.2426) -- (0.0354,4.2780) -- (-0.0000,4.2926) -- (-0.0354,4.2780) -- (-0.0500,4.2426) -- (-0.0354,4.2073) -- (-0.0000,4.1926) -- (0.0354,4.2073) -- cycle; \draw [black, fill](0.0500,-0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,-0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,-4.2426) -- (0.0354,-4.2073) -- (-0.0000,-4.1926) -- (-0.0354,-4.2073) -- (-0.0500,-4.2426) -- (-0.0354,-4.2780) -- (-0.0000,-4.2926) -- (0.0354,-4.2780) -- cycle; \draw [black, fill](0.0500,-6.0000) -- (0.0354,-5.9646) -- (0.0000,-5.9500) -- (-0.0354,-5.9646) -- (-0.0500,-6.0000) -- (-0.0354,-6.0354) -- (0.0000,-6.0500) -- (0.0354,-6.0354) -- cycle; \draw [black, fill](0.0500,-4.2426) -- (0.0354,-4.2073) -- (-0.0000,-4.1926) -- (-0.0354,-4.2073) -- (-0.0500,-4.2426) -- (-0.0354,-4.2780) -- (-0.0000,-4.2926) -- (0.0354,-4.2780) -- cycle; \node [right, black] at (0.0000,0.0000) {0}; \node [right, black] at (0.0000,4.2426) {1}; \node [above, black] at (0.0000,6.0000) {2}; \node [left, black] at (-0.0000,4.2426) {3}; \node [left, black] at (0.0000,-0.0000) {4}; \node [left, black] at (-0.0000,-4.2426) {5}; \node [below, black] at (0.0000,-6.0000) {6}; \node [right, black] at (-0.0000,-4.2426) {7}; \draw [black, thick](0.0000,0.0000) -- (0.0000,4.2426) -- (0.0000,6.0000) -- (-0.0000,4.2426) -- (0.0000,-0.0000) -- (-0.0000,-4.2426) -- (0.0000,-6.0000) -- (-0.0000,-4.2426) -- cycle; \end{tikzpicture} $$

[slavav]

Добавим второе слагаемое.
На следующем рисунке в верхнем ряду разные варианты второго слагаемого $0, \frac19 v_2, \frac29 v_2, \frac13 v_2$
Под ними сумма двуx первых слагаемых $v\left(0, 0\right), v\left(\frac19, 0\right), v\left(\frac29, 0\right), v\left(\frac13, 0\right)$.
Серым нарисован $v\left(0, 0\right)$.

$$ \begin{tikzpicture} \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \node [right, black] at (0.0000,0.0000) {0}; \node [above, black] at (0.0000,0.0000) {1}; \node [left, black] at (0.0000,0.0000) {2}; \node [below, black] at (0.0000,0.0000) {3}; \node [left, black] at (0.0000,0.0000) {4}; \node [below, black] at (0.0000,0.0000) {5}; \node [right, black] at (0.0000,0.0000) {6}; \node [above, black] at (0.0000,0.0000) {7}; \draw [black, thick](0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- cycle; \draw [black, fill](4.3833,0.0000) -- (4.3687,0.0354) -- (4.3333,0.0500) -- (4.2980,0.0354) -- (4.2833,0.0000) -- (4.2980,-0.0354) -- (4.3333,-0.0500) -- (4.3687,-0.0354) -- cycle; 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\draw [black, fill](4.0500,-0.3333) -- (4.0354,-0.2980) -- (4.0000,-0.2833) -- (3.9646,-0.2980) -- (3.9500,-0.3333) -- (3.9646,-0.3687) -- (4.0000,-0.3833) -- (4.0354,-0.3687) -- cycle; \node [right, black] at (4.3333,0.0000) {0}; \node [above, black] at (4.0000,0.3333) {1}; \node [left, black] at (3.6667,0.0000) {2}; \node [below, black] at (4.0000,-0.3333) {3}; \node [left, black] at (4.3333,-0.0000) {4}; \node [below, black] at (4.0000,0.3333) {5}; \node [right, black] at (3.6667,0.0000) {6}; \node [above, black] at (4.0000,-0.3333) {7}; \draw [black, thick](4.3333,0.0000) -- (4.0000,0.3333) -- (3.6667,0.0000) -- (4.0000,-0.3333) -- (4.3333,-0.0000) -- (4.0000,0.3333) -- (3.6667,0.0000) -- (4.0000,-0.3333) -- cycle; \draw [black, fill](8.7167,0.0000) -- (8.7020,0.0354) -- (8.6667,0.0500) -- (8.6313,0.0354) -- (8.6167,0.0000) -- (8.6313,-0.0354) -- (8.6667,-0.0500) -- (8.7020,-0.0354) -- cycle; \draw [black, fill](8.0500,0.6667) -- (8.0354,0.7020) -- (8.0000,0.7167) -- (7.9646,0.7020) -- (7.9500,0.6667) -- (7.9646,0.6313) -- (8.0000,0.6167) -- (8.0354,0.6313) -- cycle; 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\node [right, black] at (0.0000,-3.7574) {1}; \node [above, black] at (0.0000,-2.0000) {2}; \node [left, black] at (-0.0000,-3.7574) {3}; \node [left, black] at (0.0000,-8.0000) {4}; \node [left, black] at (-0.0000,-12.2426) {5}; \node [below, black] at (0.0000,-14.0000) {6}; \node [right, black] at (-0.0000,-12.2426) {7}; \draw [black, thick](0.0000,-8.0000) -- (0.0000,-3.7574) -- (0.0000,-2.0000) -- (-0.0000,-3.7574) -- (0.0000,-8.0000) -- (-0.0000,-12.2426) -- (0.0000,-14.0000) -- (-0.0000,-12.2426) -- cycle; \draw [black, fill](4.3833,-8.0000) -- (4.3687,-7.9646) -- (4.3333,-7.9500) -- (4.2980,-7.9646) -- (4.2833,-8.0000) -- (4.2980,-8.0354) -- (4.3333,-8.0500) -- (4.3687,-8.0354) -- cycle; \draw [black, fill](4.0500,-3.4240) -- (4.0354,-3.3887) -- (4.0000,-3.3740) -- (3.9646,-3.3887) -- (3.9500,-3.4240) -- (3.9646,-3.4594) -- (4.0000,-3.4740) -- (4.0354,-3.4594) -- cycle; \draw [black, fill](3.7167,-2.0000) -- (3.7020,-1.9646) -- (3.6667,-1.9500) -- (3.6313,-1.9646) -- (3.6167,-2.0000) -- (3.6313,-2.0354) -- (3.6667,-2.0500) -- (3.7020,-2.0354) -- cycle; \draw [black, fill](4.0500,-4.0907) -- (4.0354,-4.0553) -- (4.0000,-4.0407) -- (3.9646,-4.0553) -- (3.9500,-4.0907) -- (3.9646,-4.1260) -- (4.0000,-4.1407) -- (4.0354,-4.1260) -- cycle; \draw [black, fill](4.3833,-8.0000) -- (4.3687,-7.9646) -- (4.3333,-7.9500) -- (4.2980,-7.9646) -- (4.2833,-8.0000) -- (4.2980,-8.0354) -- (4.3333,-8.0500) -- (4.3687,-8.0354) -- cycle; \draw [black, fill](4.0500,-11.9093) -- (4.0354,-11.8740) -- (4.0000,-11.8593) -- (3.9646,-11.8740) -- (3.9500,-11.9093) -- (3.9646,-11.9447) -- (4.0000,-11.9593) -- (4.0354,-11.9447) -- cycle; \draw [black, fill](3.7167,-14.0000) -- (3.7020,-13.9646) -- (3.6667,-13.9500) -- (3.6313,-13.9646) -- (3.6167,-14.0000) -- (3.6313,-14.0354) -- (3.6667,-14.0500) -- (3.7020,-14.0354) -- cycle; \draw [black, fill](4.0500,-12.5760) -- (4.0354,-12.5406) -- (4.0000,-12.5260) -- (3.9646,-12.5406) -- (3.9500,-12.5760) -- (3.9646,-12.6113) -- (4.0000,-12.6260) -- (4.0354,-12.6113) -- cycle; \node [right, black] at (4.3333,-8.0000) {0}; \node [right, black] at (4.0000,-3.4240) {1}; \node [above, black] at (3.6667,-2.0000) {2}; \node [left, black] at (4.0000,-4.0907) {3}; \node [left, black] at (4.3333,-8.0000) {4}; \node [left, black] at (4.0000,-11.9093) {5}; \node [below, black] at (3.6667,-14.0000) {6}; \node [right, black] at (4.0000,-12.5760) {7}; \draw [black, thick](4.3333,-8.0000) -- (4.0000,-3.4240) -- (3.6667,-2.0000) -- (4.0000,-4.0907) -- (4.3333,-8.0000) -- (4.0000,-11.9093) -- (3.6667,-14.0000) -- (4.0000,-12.5760) -- cycle; \draw [black, fill](8.7167,-8.0000) -- (8.7020,-7.9646) -- (8.6667,-7.9500) -- (8.6313,-7.9646) -- (8.6167,-8.0000) -- (8.6313,-8.0354) -- (8.6667,-8.0500) -- (8.7020,-8.0354) -- cycle; \draw [black, fill](8.0500,-3.0907) -- (8.0354,-3.0553) -- (8.0000,-3.0407) -- (7.9646,-3.0553) -- (7.9500,-3.0907) -- (7.9646,-3.1260) -- (8.0000,-3.1407) -- (8.0354,-3.1260) -- cycle; \draw [black, fill](7.3833,-2.0000) -- (7.3687,-1.9646) -- (7.3333,-1.9500) -- (7.2980,-1.9646) -- (7.2833,-2.0000) -- (7.2980,-2.0354) -- (7.3333,-2.0500) -- (7.3687,-2.0354) -- cycle; \draw [black, fill](8.0500,-4.4240) -- (8.0354,-4.3887) -- (8.0000,-4.3740) -- (7.9646,-4.3887) -- (7.9500,-4.4240) -- (7.9646,-4.4594) -- (8.0000,-4.4740) -- (8.0354,-4.4594) -- cycle; \draw [black, fill](8.7167,-8.0000) -- (8.7020,-7.9646) -- (8.6667,-7.9500) -- (8.6313,-7.9646) -- (8.6167,-8.0000) -- (8.6313,-8.0354) -- (8.6667,-8.0500) -- (8.7020,-8.0354) -- cycle; \draw [black, fill](8.0500,-11.5760) -- (8.0354,-11.5406) -- (8.0000,-11.5260) -- (7.9646,-11.5406) -- (7.9500,-11.5760) -- (7.9646,-11.6113) -- (8.0000,-11.6260) -- (8.0354,-11.6113) -- cycle; \draw [black, fill](7.3833,-14.0000) -- (7.3687,-13.9646) -- (7.3333,-13.9500) -- (7.2980,-13.9646) -- (7.2833,-14.0000) -- (7.2980,-14.0354) -- (7.3333,-14.0500) -- (7.3687,-14.0354) -- cycle; \draw [black, fill](8.0500,-12.9093) -- (8.0354,-12.8740) -- (8.0000,-12.8593) -- (7.9646,-12.8740) -- (7.9500,-12.9093) -- (7.9646,-12.9447) -- (8.0000,-12.9593) -- (8.0354,-12.9447) -- cycle; \node [right, black] at (8.6667,-8.0000) {0}; \node [right, black] at (8.0000,-3.0907) {1}; \node [above, black] at (7.3333,-2.0000) {2}; \node [left, black] at (8.0000,-4.4240) {3}; \node [left, black] at (8.6667,-8.0000) {4}; \node [left, black] at (8.0000,-11.5760) {5}; \node [below, black] at (7.3333,-14.0000) {6}; \node [right, black] at (8.0000,-12.9093) {7}; \draw [black, thick](8.6667,-8.0000) -- (8.0000,-3.0907) -- (7.3333,-2.0000) -- (8.0000,-4.4240) -- (8.6667,-8.0000) -- (8.0000,-11.5760) -- (7.3333,-14.0000) -- (8.0000,-12.9093) -- cycle; \draw [black, fill](13.0500,-8.0000) -- (13.0354,-7.9646) -- (13.0000,-7.9500) -- (12.9646,-7.9646) -- (12.9500,-8.0000) -- (12.9646,-8.0354) -- (13.0000,-8.0500) -- (13.0354,-8.0354) -- cycle; \draw [black, fill](12.0500,-2.7574) -- (12.0354,-2.7220) -- (12.0000,-2.7074) -- (11.9646,-2.7220) -- (11.9500,-2.7574) -- (11.9646,-2.7927) -- (12.0000,-2.8074) -- (12.0354,-2.7927) -- cycle; \draw [black, fill](11.0500,-2.0000) -- (11.0354,-1.9646) -- (11.0000,-1.9500) -- (10.9646,-1.9646) -- (10.9500,-2.0000) -- (10.9646,-2.0354) -- (11.0000,-2.0500) -- (11.0354,-2.0354) -- cycle; \draw [black, fill](12.0500,-4.7574) -- (12.0354,-4.7220) -- (12.0000,-4.7074) -- (11.9646,-4.7220) -- (11.9500,-4.7574) -- (11.9646,-4.7927) -- (12.0000,-4.8074) -- (12.0354,-4.7927) -- cycle; \draw [black, fill](13.0500,-8.0000) -- (13.0354,-7.9646) -- (13.0000,-7.9500) -- (12.9646,-7.9646) -- (12.9500,-8.0000) -- (12.9646,-8.0354) -- (13.0000,-8.0500) -- (13.0354,-8.0354) -- cycle; \draw [black, fill](12.0500,-11.2426) -- (12.0354,-11.2073) -- (12.0000,-11.1926) -- (11.9646,-11.2073) -- (11.9500,-11.2426) -- (11.9646,-11.2780) -- (12.0000,-11.2926) -- (12.0354,-11.2780) -- cycle; \draw [black, fill](11.0500,-14.0000) -- (11.0354,-13.9646) -- (11.0000,-13.9500) -- (10.9646,-13.9646) -- (10.9500,-14.0000) -- (10.9646,-14.0354) -- (11.0000,-14.0500) -- (11.0354,-14.0354) -- cycle; \draw [black, fill](12.0500,-13.2426) -- (12.0354,-13.2073) -- (12.0000,-13.1926) -- (11.9646,-13.2073) -- (11.9500,-13.2426) -- (11.9646,-13.2780) -- (12.0000,-13.2926) -- (12.0354,-13.2780) -- cycle; \node [right, black] at (13.0000,-8.0000) {0}; \node [right, black] at (12.0000,-2.7574) {1}; \node [above, black] at (11.0000,-2.0000) {2}; \node [left, black] at (12.0000,-4.7574) {3}; \node [left, black] at (13.0000,-8.0000) {4}; \node [left, black] at (12.0000,-11.2426) {5}; \node [below, black] at (11.0000,-14.0000) {6}; \node [right, black] at (12.0000,-13.2426) {7}; \draw [black, thick](13.0000,-8.0000) -- (12.0000,-2.7574) -- (11.0000,-2.0000) -- (12.0000,-4.7574) -- (13.0000,-8.0000) -- (12.0000,-11.2426) -- (11.0000,-14.0000) -- (12.0000,-13.2426) -- cycle; \end{tikzpicture} $$

[slavav]

Добавим третье слагаемое.
В верхнем ряду разные варианты третьего слагаемого $0, \frac1{63}\left(v_3 + v_5\right), \frac2{63}\left(v_3 + v_5\right), \frac1{21}\left(v_3 + v_5\right)$.
Под ними сумма трёх слагаемых $v\left(\frac13, 0\right), v\left(\frac13, \frac1{21}\right), v\left(\frac13, \frac2{21}\right), v\left(\frac13, \frac17\right)$.
Серым нарисован $v\left(\frac13, 0\right)$.

$$ \begin{tikzpicture} \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \draw [black, fill](0.0500,0.0000) -- (0.0354,0.0354) -- (0.0000,0.0500) -- (-0.0354,0.0354) -- (-0.0500,0.0000) -- (-0.0354,-0.0354) -- (-0.0000,-0.0500) -- (0.0354,-0.0354) -- cycle; \node [right, black] at (0.0000,0.0000) {0}; \node [above, black] at (0.0000,0.0000) {1}; \node [below, black] at (0.0000,0.0000) {2}; \node [above, black] at (0.0000,0.0000) {3}; \node [left, black] at (0.0000,0.0000) {4}; \node [below, black] at (0.0000,0.0000) {5}; \node [above, black] at (0.0000,0.0000) {6}; \node [below, black] at (0.0000,0.0000) {7}; \draw [black, thick](0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- (0.0000,0.0000) -- cycle; \draw [black, fill](4.1452,0.0000) -- (4.1306,0.0354) -- (4.0952,0.0500) -- (4.0599,0.0354) -- (4.0452,0.0000) -- (4.0599,-0.0354) -- (4.0952,-0.0500) -- (4.1306,-0.0354) -- cycle; \draw [black, fill](3.9827,0.0000) -- (3.9680,0.0354) -- (3.9327,0.0500) -- (3.8973,0.0354) -- (3.8827,0.0000) -- (3.8973,-0.0354) -- (3.9327,-0.0500) -- (3.9680,-0.0354) -- cycle; \draw [black, fill](4.0500,0.0000) -- (4.0354,0.0354) -- (4.0000,0.0500) -- (3.9646,0.0354) -- (3.9500,0.0000) -- (3.9646,-0.0354) -- (4.0000,-0.0500) -- (4.0354,-0.0354) -- cycle; \draw [black, fill](4.1173,-0.0000) -- (4.1027,0.0354) -- (4.0673,0.0500) -- (4.0320,0.0354) -- (4.0173,-0.0000) -- (4.0320,-0.0354) -- (4.0673,-0.0500) -- (4.1027,-0.0354) -- cycle; \draw [black, fill](3.9548,0.0000) -- (3.9401,0.0354) -- (3.9048,0.0500) -- (3.8694,0.0354) -- (3.8548,0.0000) -- (3.8694,-0.0354) -- (3.9048,-0.0500) -- (3.9401,-0.0354) -- cycle; \draw [black, fill](4.1173,-0.0000) -- (4.1027,0.0354) -- (4.0673,0.0500) -- (4.0320,0.0354) -- (4.0173,-0.0000) -- (4.0320,-0.0354) -- (4.0673,-0.0500) -- (4.1027,-0.0354) -- cycle; \draw [black, fill](4.0500,0.0000) -- (4.0354,0.0354) -- (4.0000,0.0500) -- (3.9646,0.0354) -- (3.9500,0.0000) -- (3.9646,-0.0354) -- (4.0000,-0.0500) -- (4.0354,-0.0354) -- cycle; \draw [black, fill](3.9827,0.0000) -- (3.9680,0.0354) -- (3.9327,0.0500) -- (3.8973,0.0354) -- (3.8827,0.0000) -- (3.8973,-0.0354) -- (3.9327,-0.0500) -- (3.9680,-0.0354) -- cycle; \node [right, black] at (4.0952,0.0000) {0}; \node [above, black] at (3.9327,0.0000) {1}; \node [below, black] at (4.0000,0.0000) {2}; \node [above, black] at (4.0673,-0.0000) {3}; \node [left, black] at (3.9048,0.0000) {4}; \node [below, black] at (4.0673,-0.0000) {5}; \node [above, black] at (4.0000,0.0000) {6}; \node [below, black] at (3.9327,0.0000) {7}; \draw [black, thick](4.0952,0.0000) -- (3.9327,0.0000) -- (4.0000,0.0000) -- (4.0673,-0.0000) -- (3.9048,0.0000) -- (4.0673,-0.0000) -- (4.0000,0.0000) -- (3.9327,0.0000) -- cycle; \draw [black, fill](8.2405,0.0000) -- (8.2258,0.0354) -- (8.1905,0.0500) -- (8.1551,0.0354) -- (8.1405,0.0000) -- (8.1551,-0.0354) -- (8.1905,-0.0500) -- (8.2258,-0.0354) -- cycle; \draw [black, fill](7.9153,0.0000) -- (7.9007,0.0354) -- (7.8653,0.0500) -- (7.8300,0.0354) -- (7.8153,0.0000) -- (7.8300,-0.0354) -- (7.8653,-0.0500) -- (7.9007,-0.0354) -- cycle; 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\node [right, black] at (8.1905,0.0000) {0}; \node [above, black] at (7.8653,0.0000) {1}; \node [below, black] at (8.0000,0.0000) {2}; \node [above, black] at (8.1347,-0.0000) {3}; \node [left, black] at (7.8095,0.0000) {4}; \node [below, black] at (8.1347,-0.0000) {5}; \node [above, black] at (8.0000,0.0000) {6}; \node [below, black] at (7.8653,0.0000) {7}; \draw [black, thick](8.1905,0.0000) -- (7.8653,0.0000) -- (8.0000,0.0000) -- (8.1347,-0.0000) -- (7.8095,0.0000) -- (8.1347,-0.0000) -- (8.0000,0.0000) -- (7.8653,0.0000) -- cycle; \draw [black, fill](12.3357,0.0000) -- (12.3211,0.0354) -- (12.2857,0.0500) -- (12.2504,0.0354) -- (12.2357,0.0000) -- (12.2504,-0.0354) -- (12.2857,-0.0500) -- (12.3211,-0.0354) -- cycle; \draw [black, fill](11.8480,0.0000) -- (11.8333,0.0354) -- (11.7980,0.0500) -- (11.7626,0.0354) -- (11.7480,0.0000) -- (11.7626,-0.0354) -- (11.7980,-0.0500) -- (11.8333,-0.0354) -- cycle; \draw [black, fill](12.0500,0.0000) -- (12.0354,0.0354) -- (12.0000,0.0500) -- (11.9646,0.0354) -- (11.9500,0.0000) -- (11.9646,-0.0354) -- (12.0000,-0.0500) -- (12.0354,-0.0354) -- cycle; 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[slavav]

Все эти картинки имеют цель убедить вас что $v\left(c, d\right) = \left(v_1 - v_7\right) + c v_2 + c d \left(v_3 + v_5\right), 0 < c \le \frac13, 0 < d \le \frac{c}3$ порождает простой невыпуклый многоугольник.
У меня нет точного доказательства этого факта, но есть идеи как это сделать. Останавливает только объём вычислений, сами вычисления кажутся элементарными.
Неравенство $d \le \frac{c}3$ появилось в последний момент. До того предполагалось что $\left(c, d\right)$ может быть любой парой из небольшого прямоугольника, но, увы, не может. Получился треугольник, для дальнейших рассуждений его хватает.

Введём оператор $\overline{B}$. Он переставляет вершины многоугольника циклически: $n-1 \to 0, 0 \to 1, 1 \to 2, \dots, n-2 \to n-1$.
Собственные вектора $\overline{B}$ и $A$ общие: $v_k=\left(1,\varepsilon_n^k, \varepsilon_n^{2k}, \dots, \varepsilon_n^{\left(n-2\right)k}, \varepsilon_n^{\left(n-1\right)k}\right), k = 0 \dots n-1$,
Собственные значения оператора $\overline{B}$: $\beta_k=\varepsilon_n^{-k}, k = 0 \dots n-1$.

Оператор $\overline{B}$ сохраняет все свойства многоугольника как множества: (не)выпуклость, (не)простоту.

Рассмотрим $\overline{B}A^2\left(v\left(c, d\right)\right)$:

$$\overline{B}A^2\left(v\left(c, d\right)\right) = \overline{B}A^2\left(\left(v_1 - v_7\right) + c v_2 + c d \left(v_3 + v_5\right)\right) = \overline{B}A^2\left(v_1 - v_7\right) + c \overline{B}A^2\left(v_2\right) + c d \overline{B}A^2\left(v_3 + v_5\right) $$

Слагаемые по отдельности:

$$\overline{B}A^2\left(v_1 - v_7\right) = \overline{B}A^2\left(v_1\right) - \overline{B}A^2\left(v_7\right) = \beta_1\lambda_1^2v_1 - \beta_7\lambda_7^2v_7 = \varepsilon_8^{-1}{\left(\frac{1 + \varepsilon_8}2\right)}^2 v_1 - \varepsilon_8^{-7}{\left(\frac{1 + \varepsilon_8^7}2\right)}^2 v_7 =$$

$$= \frac{2 + \sqrt{2}}4 \left(v_1 - v_7\right) $$

$$\overline{B}A^2\left(v_2\right) = \beta_2\lambda_2^2v_2 = \varepsilon_8^{-2}{\left(\frac{1 + \varepsilon_8^2}2\right)}^2 v_2 = \frac{v_2}2$$

$$\overline{B}A^2\left(v_3 + v_5\right) = \overline{B}A^2\left(v_3\right) + \overline{B}A^2\left(v_5\right) = \beta_3\lambda_3^2v_3 + \beta_5\lambda_5^2v_5 = \varepsilon_8^{-3}{\left(\frac{1 + \varepsilon_8^3}2\right)}^2 v_3 + \varepsilon_8^{-5}{\left(\frac{1 + \varepsilon_8^5}2\right)}^2 v_5 =$$

$$= \frac{2 - \sqrt{2}}4 \left(v_3 + v_5\right) $$

Вся сумма вместе:

$$\overline{B}A^2\left(v\left(c, d\right)\right) = \frac{2 + \sqrt{2}}4 \left(v_1 - v_7\right) + c \frac{v_2}2 + c d \frac{2 - \sqrt{2}}4 \left(v_3 + v_5\right) =$$

$$= \frac{2 + \sqrt{2}}4 \left[ \left(v_1 - v_7\right) + \left(2 - \sqrt2\right) c v_2 + \left(3 - 2\sqrt2\right) c d \left(v_3 + v_5\right) \right] =$$

$$= \frac{2 + \sqrt{2}}4 \left[ \left(v_1 - v_7\right) + \left(2 - \sqrt2\right) c v_2 + \left(2 - \sqrt2\right) c \frac{2 - \sqrt2}2 d \left(v_3 + v_5\right) \right] =$$

$$= \frac{2 + \sqrt{2}}4 v\left(\left(2 - \sqrt2\right) c, \frac{2 - \sqrt2}2 d\right) $$

Оператор $\overline{B}A^2$ отображает пару $\left(c, d\right)$ в $\left(\left(2 - \sqrt2\right) c, \frac{2 - \sqrt2}2 d\right)$. Если исходная пара принадлежала треугольнику ограничений, то и новая тоже попадёт в этот треугольник.

Вывод: предъявлен невыпуклый простой восьмиугольник, который сохраняет эти свойства при отображениях вида $A^{2m}$.
Оставшиеся дыры:
точное доказательство свойств восьмиугольников $v\left(c, d\right), 0 < c \le \frac13, 0 < d \le \frac{c}3$;
доказательство свойств восьмиугольников $A\left(v\left(c, d\right)\right), 0 < c \le \frac13, 0 < d \le \frac{c}3$.

[slavav]

Напомню что мы рассматриваем восьмиугольник $v\left(c, d\right) = \left(v_1 - v_7\right) + c v_2 + c d \left(v_3 + v_5\right)$:
Требуется предъявить ограничения на $c$ и $d$ при которых многоугольник является простым и невыпуклым.

$$ \begin{tikzpicture} \draw [black, fill](0.6611,0.0000) -- (0.6465,0.0354) -- (0.6111,0.0500) -- (0.5758,0.0354) -- (0.5611,0.0000) -- (0.5758,-0.0354) -- (0.6111,-0.0500) -- (0.6465,-0.0354) -- cycle; \draw [black, fill](-0.0286,2.6213) -- (-0.0432,2.6567) -- (-0.0786,2.6713) -- (-0.1139,2.6567) -- (-0.1286,2.6213) -- (-0.1139,2.5860) -- (-0.0786,2.5713) -- (-0.0432,2.5860) -- cycle; \draw [black, fill](-0.4500,3.0000) -- (-0.4646,3.0354) -- (-0.5000,3.0500) -- (-0.5354,3.0354) -- (-0.5500,3.0000) -- (-0.5354,2.9646) -- (-0.5000,2.9500) -- (-0.4646,2.9646) -- cycle; \draw [black, fill](0.1286,1.6213) -- (0.1139,1.6567) -- (0.0786,1.6713) -- (0.0432,1.6567) -- (0.0286,1.6213) -- (0.0432,1.5860) -- (0.0786,1.5713) -- (0.1139,1.5860) -- cycle; \draw [black, fill](0.4389,-0.0000) -- (0.4242,0.0354) -- (0.3889,0.0500) -- (0.3535,0.0354) -- (0.3389,-0.0000) -- (0.3535,-0.0354) -- (0.3889,-0.0500) -- (0.4242,-0.0354) -- cycle; \draw [black, fill](0.1286,-1.6213) -- (0.1139,-1.5860) -- (0.0786,-1.5713) -- (0.0432,-1.5860) -- (0.0286,-1.6213) -- (0.0432,-1.6567) -- (0.0786,-1.6713) -- (0.1139,-1.6567) -- cycle; \draw [black, fill](-0.4500,-3.0000) -- (-0.4646,-2.9646) -- (-0.5000,-2.9500) -- (-0.5354,-2.9646) -- (-0.5500,-3.0000) -- (-0.5354,-3.0354) -- (-0.5000,-3.0500) -- (-0.4646,-3.0354) -- cycle; \draw [black, fill](-0.0286,-2.6213) -- (-0.0432,-2.5860) -- (-0.0786,-2.5713) -- (-0.1139,-2.5860) -- (-0.1286,-2.6213) -- (-0.1139,-2.6567) -- (-0.0786,-2.6713) -- (-0.0432,-2.6567) -- cycle; \node [right, black] at (0.6111,0.0000) {0}; \node [right, black] at (-0.0786,2.6213) {1}; \node [above, black] at (-0.5000,3.0000) {2}; \node [left, black] at (0.0786,1.6213) {3}; \node [left, black] at (0.3889,-0.0000) {4}; \node [left, black] at (0.0786,-1.6213) {5}; \node [below, black] at (-0.5000,-3.0000) {6}; \node [right, black] at (-0.0786,-2.6213) {7}; \draw [black, thick](0.6111,0.0000) -- (-0.0786,2.6213) -- (-0.5000,3.0000) -- (0.0786,1.6213) -- (0.3889,-0.0000) -- (0.0786,-1.6213) -- (-0.5000,-3.0000) -- (-0.0786,-2.6213) -- cycle; \end{tikzpicture} $$

Рассмотрим две ломаные: $(p_6, p_7, p_0, p_1, p_2)$ и $(p_2, p_3, p_4, p_5, p_6)$. Я буду доказывать что первая ломаная "правее" второй в следующем смысле: любая горизонтальная прямая проходящая выше $p_6$ и ниже $p_2$ пересекает первую ломаную в точке, абсцисса которой больше абсциссы точки пересечения со второй ломаной.

Координаты вершин:

$$ \begin{tabular}{l|c|c} i & x_i & y_i \\ \hline 0 & 2cd + c & 0 \\ 1 & -\sqrt{2}cd & c + \sqrt{2} \\ 2 & -c & 2 \\ 3 & \sqrt{2}cd & -c + \sqrt{2} \\ 4 & -2cd + c & 0 \\ 5 & \sqrt{2} c d & c - \sqrt{2} \\ 6 & -c & -2 \\ 7 & -\sqrt{2}cd & -c - \sqrt{2} \end{tabular} $$

Нужный порядок вершин по вертикали: $y_6 < y_7 < y_5 < y_4 = y_0 < y_3 < y_1 < y_2$.
Приводит к неравенствам: $-2 < -c - \sqrt{2} < c - \sqrt{2} < 0 = 0 < -c + \sqrt{2} < c + \sqrt{2} < 2$.
Откуда $0 < c < 2 - \sqrt{2}$.

Чтобы одна ломаная оказалась "правее" другой требуется чтобы выполнялись условия для вершин. В нашем случае такие:
1. Вершина $p_1$ должна быть "правее" ребра $(p_2, p_3)$.
2. Ребро $(p_0, p_1)$ должнo быть "правее" вершины $p_3$.
3. Вершина $p_0$ должна быть "правее" вершины $p_4$.
4. Ребро $(p_7, p_0)$ должнo быть "правее" вершины $p_5$.
5. Вершина $p_7$ должна быть "правее" ребра $(p_5, p_6)$.

Условия типа ребро-вершина выражается через проверку площади со знаком треугольника натянутого на ребро и вершину. Площадь со знаком считается как определитель матрицы три на три. Определители для условий 1, 2, 4, 5:
1. $\begin{vmatrix}x_{2} & y_{2} & 1\\x_{3} & y_{3} & 1\\x_{1} & y_{1} & 1\end{vmatrix} = \begin{vmatrix}- c & 2 & 1\\\sqrt{2} c d & - c + \sqrt{2} & 1\\- \sqrt{2} c d & c + \sqrt{2} & 1\end{vmatrix} = 2 c \left(c - 2d \left(\sqrt{2} - 1\right)\right)$

2. $\begin{vmatrix}x_{0} & y_{0} & 1\\x_{1} & y_{1} & 1\\x_{3} & y_{3} & 1\end{vmatrix} = \begin{vmatrix}2 c d + c & 0 & 1\\- \sqrt{2} c d & c + \sqrt{2} & 1\\\sqrt{2} c d & - c + \sqrt{2} & 1\end{vmatrix} = 2 c \left(c - 2d \left(1 - c\right)\right)$

4. $\begin{vmatrix}x_{7} & y_{7} & 1\\x_{0} & y_{0} & 1\\x_{5} & y_{5} & 1\end{vmatrix} = \begin{vmatrix}- \sqrt{2} c d & - c - \sqrt{2} & 1\\2 c d + c & 0 & 1\\\sqrt{2} c d & c - \sqrt{2} & 1\end{vmatrix} = 2 c \left(c - 2d \left(1 - c\right)\right)$

5. $\begin{vmatrix}x_{5} & y_{5} & 1\\x_{6} & y_{6} & 1\\x_{7} & y_{7} & 1\end{vmatrix} = \begin{vmatrix}\sqrt{2} c d & c - \sqrt{2} & 1\\- c & -2 & 1\\- \sqrt{2} c d & - c - \sqrt{2} & 1\end{vmatrix} = 2 c \left(c - 2d \left(\sqrt{2} - 1\right)\right)$

Все пять условий:
1. $2c \left(c - 2d \left(\sqrt{2} - 1\right)\right) > 0$
2. $2c \left(c - 2d \left(1 - c\right)\right) > 0$
3. $2cd + c > -2cd + c$
4. $2c \left(c - 2d \left(1 - c\right)\right) > 0$
5. $2c \left(c - 2d \left(\sqrt{2} - 1\right)\right) > 0$

Убираем дупликаты, учитываем что $0 < c < 2 - \sqrt{2}$:
1. $d < \frac{c}{2\left(\sqrt{2} - 1\right)}$
2. $d < \frac{c}{2\left(1 - c\right)}$
3. $d > 0$

Зaметим что если $0 < d < \frac{c}{2}$ то выполнены все три условия выше.

Простота получена. Для невыпуклости достаточно проверить что вершина $p_4$ "правее" отрезка $(p_2, p_6)$. Определитель не понадобится так как $x_2 = x_6$.
Получается условие: $x_2 < x_4$. Откуда $-c < -2cd + c$. Откуда $d < 1$. Это условие выполнено так как $d < \frac{c}{2}$ и $c < 2 - \sqrt{2}$.

Доказано, что при $0 < c < 2 - \sqrt{2}$ и $0 < d < \frac{c}{2}$ восьмиугольник $v\left(c, d\right)$ является простым и невыпуклым.

[slavav]

Рассмотрим $A\left(v\left(c, d\right)\right)$. Чтобы уменьшить количество дробей умножим его на два.

$2A\left(v\left(c, d\right)\right)$ в координатах:

$$ \begin{tabular}{l|c|c} i & x_i & y_i \\ \hline 0 & -\sqrt{2}cd + 2cd + c & c + \sqrt{2} \\ 1 & -\sqrt{2}cd - c & c + \sqrt{2} + 2 \\ 2 & \sqrt{2}cd - c & -c + \sqrt{2} + 2 \\ 3 & -2cd + \sqrt{2}cd + c & -c + \sqrt{2} \\ 4 & -2cd + \sqrt{2}cd + c & c - \sqrt{2} \\ 5 & \sqrt{2}cd - c & c - 2 - \sqrt{2} \\ 6 & -\sqrt{2}cd - c & -c - 2 - \sqrt{2} \\ 7 & -\sqrt{2}cd + 2cd + c & -c - \sqrt{2} \end{tabular} $$

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Рассмотрим две ломаные: $(p_6, p_7, p_0, p_1)$ и $(p_1, p_2, p_3, p_4, p_5, p_6)$. Первая должна быть "правее" второй.

Нужный порядок вершин по вертикали: $y_6 < y_5 < y_7 < y_4 < y_3 < y_0 < y_2 < y_1$.
Что приводит к неравенствам:

$$-c - 2 - \sqrt{2} < c - 2 - \sqrt{2} < -c - \sqrt{2} < c - \sqrt{2} < -c + \sqrt{2} < c + \sqrt{2} < -c + \sqrt{2} + 2 < c + \sqrt{2} + 2$$

.
После упрощения: $0 < c < 1$. Это неравенство выполнено, так как выполнено более строгое $0 < c < 2 - \sqrt{2}$.

Все условия для вершин имеют вид определителей:

$$\begin{vmatrix}x_{6} & y_{6} & 1\\x_{7} & y_{7} & 1\\x_{5} & y_{5} & 1\end{vmatrix} = \begin{vmatrix}- \sqrt{2} c d - c & - c - 2 - \sqrt{2} & 1\\- \sqrt{2} c d + 2 c d + c & - c - \sqrt{2} & 1\\\sqrt{2} c d - c & c - 2 - \sqrt{2} & 1\end{vmatrix} = 4 c \left(c + d \left(c - \sqrt{2}\right)\right) > 0$$

$$\begin{vmatrix}x_{4} & y_{4} & 1\\x_{5} & y_{5} & 1\\x_{7} & y_{7} & 1\end{vmatrix} = \begin{vmatrix}- 2 c d + \sqrt{2} c d + c & c - \sqrt{2} & 1\\\sqrt{2} c d - c & c - 2 - \sqrt{2} & 1\\- \sqrt{2} c d + 2 c d + c & - c - \sqrt{2} & 1\end{vmatrix} = 4 c \left(c + d \left(- c - \sqrt{2} + 2\right)\right) > 0$$

$$\begin{vmatrix}x_{7} & y_{7} & 1\\x_{0} & y_{0} & 1\\x_{4} & y_{4} & 1\end{vmatrix} = \begin{vmatrix}- \sqrt{2} c d + 2 c d + c & - c - \sqrt{2} & 1\\- \sqrt{2} c d + 2 c d + c & c + \sqrt{2} & 1\\- 2 c d + \sqrt{2} c d + c & c - \sqrt{2} & 1\end{vmatrix} = 4 c d \left(2 - \sqrt{2}\right) \left(c + \sqrt{2}\right) > 0$$

$$\begin{vmatrix}x_{7} & y_{7} & 1\\x_{0} & y_{0} & 1\\x_{3} & y_{3} & 1\end{vmatrix} = \begin{vmatrix}- \sqrt{2} c d + 2 c d + c & - c - \sqrt{2} & 1\\- \sqrt{2} c d + 2 c d + c & c + \sqrt{2} & 1\\- 2 c d + \sqrt{2} c d + c & - c + \sqrt{2} & 1\end{vmatrix} = 4 c d \left(2 - \sqrt{2}\right) \left(c + \sqrt{2}\right) > 0$$

$$\begin{vmatrix}x_{2} & y_{2} & 1\\x_{3} & y_{3} & 1\\x_{0} & y_{0} & 1\end{vmatrix} = \begin{vmatrix}\sqrt{2} c d - c & - c + \sqrt{2} + 2 & 1\\- 2 c d + \sqrt{2} c d + c & - c + \sqrt{2} & 1\\- \sqrt{2} c d + 2 c d + c & c + \sqrt{2} & 1\end{vmatrix} = 4 c \left(c + d \left(- c - \sqrt{2} + 2\right)\right) > 0$$

$$\begin{vmatrix}x_{0} & y_{0} & 1\\x_{1} & y_{1} & 1\\x_{2} & y_{2} & 1\end{vmatrix} = \begin{vmatrix}- \sqrt{2} c d + 2 c d + c & c + \sqrt{2} & 1\\- \sqrt{2} c d - c & c + \sqrt{2} + 2 & 1\\\sqrt{2} c d - c & - c + \sqrt{2} + 2 & 1\end{vmatrix} = 4 c \left(c + d \left(c - \sqrt{2}\right)\right) > 0$$

Убираем дупликаты, учитываем что $0 < c < 2 - \sqrt{2}$, получаем $0 < d < \frac{c}{\left(\sqrt{2} - c\right)}$.
Это условие выполняется так как $0 < d < \frac{c}{2}$.

Простота восьмиугольника проверена, осталась невыпуклость. Для неё достаточно убедится что $x_2 > x_1$.
Подстановка показывает что это правда: $\sqrt{2}cd - c > -\sqrt{2}cd - c$.

Доказано, что при $0 < c < 2 - \sqrt{2}$ и $0 < d < \frac{c}{2}$ восьмиугольник $A\left(v\left(c, d\right)\right)$ является простым и невыпуклым.

[Утундрий]

Ну, Ок. С какого конца это начинать есть?

[slavav]

Утундрий в сообщении #1446639 писал(а):

Ну, Ок. С какого конца это начинать есть?

Оглавление
Глава первая, в которой представлено двухпараметрическое семейство восьмиугольников
Глава вторая, в которой показано как действует первый параметр
Глава третья, в которой показано как действует второй параметр
Глава четвёртая, в которой доказывается, что $A^2$ порождает восьмиугольник из того же семейства
Глава пятая, в которой доказывается, что восьмиугольники семейства простые и невыпуклые
Глава шестая, в которой доказывается, что $A$ порождает простые и невыпуклые многоугольники

Из последних трёх глав по индукции следует, что любой восьмиугольник семейства решает пункт б) исходной задачи.

[staric]

Да, все так, сделано даже больше, чем ожидалось. Хорошая работа.

[slavav]

staric, спасибо. Откуда возникла такая задача?

Оставшиеся вопросы:
Может ли семиугольник решить пункт б)? А девятиугольник?
Доказательства тяжёлые и неуклюжие - всё какие-то частные случаи. Есть ли общая теория связывающая линейные операторы и "простоту" многоугольника?

Благодарности:
staric, ещё раз спасибо. Хорошая задача!
Спасибо DeBill за спектр оператора $A$.
Спасибо Jim Fowler (kisonecat) за библиотеку tikzjax, с помощью которой я строил картинки.
Спасибо авторам библиотеки SymPy, которая составляла, решала и печатала неравенства.