\[\boxed{\mathbf{1133.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[A_{1}A_{2}\ldots A_{12} - правильный\ \]
\[двенадцатиугольник;\]
\[A_{1}A_{6} \cap A_{2}A_{9} = B.\]
\[\mathbf{Доказать:}\]
\[\textbf{а)}\ \mathrm{\Delta}A_{1}A_{2}\text{B\ }и\ \mathrm{\Delta}A_{6}A_{9}B -\]
\[правильные;\]
\[\textbf{б)}\ A_{1}A_{6} = 2r.\]
\[\mathbf{Доказательство.}\]
\[\textbf{а)}\ Правильный\ многоугольник\ \]
\[вписан\ в\ окружность,\ дуга\ \]
\[равна:\]
\[A_{1}A_{2} = A_{2}A_{3} = \ldots = A_{11}A_{12} =\]
\[= 360{^\circ}:12 = 30{^\circ};\]
\[\angle A_{2}A_{1}B = \frac{1}{2} \cap A_{2}A_{4}A_{6} =\]
\[= \frac{1}{2} \bullet 120{^\circ} = 60{^\circ};\]
\[\angle A_{1}A_{2}B = \frac{1}{2} \cap A_{1}A_{11}A_{9} =\]
\[= \frac{1}{2} \bullet 120{^\circ} = 60{^\circ};\]
\[\angle A_{9}A_{6}B = \frac{1}{2} \cap A_{1}A_{11}A_{9} =\]
\[= \frac{1}{2} \bullet 120{^\circ} = 60{^\circ};\]
\[\angle A_{9}A_{6}B = \frac{1}{2} \cap A_{2}A_{4}A_{6} =\]
\[= \frac{1}{2} \bullet 120{^\circ} = 60{^\circ}.\]
\[Сумма\ углов\ \]
\[в\ треугольнике\ 180{^\circ}:\]
\[\angle A_{1}BA_{2} = \angle A_{6}BA_{9} = 60{^\circ};\]
\[\mathrm{\Delta}A_{1}A_{2}\text{B\ }и\ \mathrm{\Delta}A_{6}A_{9}B -\]
\[правильные.\]
\[Что\ и\ требовалось\ доказать.\]
\[\textbf{б)}\ \angle A_{1}A_{6}A_{7} - вписанный;\]
\[\angle A_{1}A_{6}A_{7} = \frac{1}{2} \cap A_{1}A_{10}A_{7} = 90{^\circ}:\]
\[\ A_{6}A_{1}\bot A_{1}A_{12}\ и\ OH_{2}\bot A_{1}A_{12};\]
\[OH_{2} \parallel A_{1}A_{6}.\]
\[Так\ же:\]
\[\ \angle A_{12}A_{1}A_{6} = 90{^\circ};\ \]
\[A_{1}A_{6}\bot A_{6}A_{7}\ и\ OH_{1}\bot A_{6}A_{7};\]
\[OH_{1} \parallel A_{1}A_{6}.\]
\[Получаем:\]
\[A_{1}A_{6}H_{1}H_{2} - прямоугольник.\]
\[Отсюда:\]
\[A_{1}A_{6} = H_{1}H_{2} = 2r.\]
\[Что\ и\ требовалось\ доказать.\]
\[\boxed{\mathbf{1133.еуроки - ответы\ на\ пятёрку}}\]
\[Решение.\]
\[\textbf{а)}\ \overrightarrow{a}\left\{ \frac{1}{4}; - 1 \right\};\ \overrightarrow{b}\left\{ 2;3 \right\}:\]
\[\overrightarrow{a} \bullet \overrightarrow{b} = \frac{1}{4} \bullet 2 + ( - 1) \bullet 3 =\]
\[= \frac{1}{2} - 3 = - 2,5.\]
\[\textbf{б)}\ \overrightarrow{a}\left\{ - 5;6 \right\};\ \overrightarrow{b}\left\{ 6;5 \right\}:\]
\[\overrightarrow{a} \bullet \overrightarrow{b} = - 5 \bullet 6 + 6 \bullet 5 =\]
\[= - 30 + 30 = 0.\]
\[\textbf{в)}\ \overrightarrow{a}\left\{ 1,5;2 \right\};\ \overrightarrow{b}\left\{ 4; - 0,5 \right\}:\]
\[\overrightarrow{a} \bullet \overrightarrow{b} = 1,5 \bullet 4 + 2 \bullet ( - 0,5) =\]
\[= 6 - 1 = 5.\]