Напомню что мы рассматриваем восьмиугольник

:
Требуется предъявить ограничения на

и

при которых многоугольник является простым и невыпуклым.
 alt="$$
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$$ $$
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$$](https://dxdy-03.korotkov.co.uk/f/2/f/b/2fb1733fb8bc1eafac4c05b123bbabc782.png)
Рассмотрим две ломаные:

и

. Я буду доказывать что первая ломаная "правее" второй в следующем смысле: любая горизонтальная прямая проходящая выше

и ниже

пересекает первую ломаную в точке, абсцисса которой больше абсциссы точки пересечения со второй ломаной.
Координаты вершин:

Нужный порядок вершин по вертикали:

.
Приводит к неравенствам:

.
Откуда

.
Чтобы одна ломаная оказалась "правее" другой требуется чтобы выполнялись условия для вершин. В нашем случае такие:
1. Вершина

должна быть "правее" ребра

.
2. Ребро

должнo быть "правее" вершины

.
3. Вершина

должна быть "правее" вершины

.
4. Ребро

должнo быть "правее" вершины

.
5. Вершина

должна быть "правее" ребра

.
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.

2.

4.

5.

Все пять условий:
1.

2.

3.

4.

5.

Убираем дупликаты, учитываем что

:
1.

2.

3.

Зaметим что если

то выполнены все три условия выше.
Простота получена. Для невыпуклости достаточно проверить что вершина

"правее" отрезка

. Определитель не понадобится так как

.
Получается условие:

. Откуда

. Откуда

. Это условие выполнено так как

и

.
Доказано, что при

и

восьмиугольник

является простым и невыпуклым.