\[\boxed{\mathbf{982.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[B \in AC;\]
\[AB = BC;\]
\[AC = 2;\]
\[\textbf{а)}\ AM^{2} + BM^{2} + CM^{2} = 50;\]
\[\textbf{б)}\ AM^{2} + 2BM^{2} + 3CM^{2} = 4.\]
\[\mathbf{Найти:}\]
\[множество\ точек\ \text{M.}\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ Введем\ систему\ \]
\[координат:\]
\[A( - 1;0);C(1;0);M(x;y);\]
\[B(0;0);\]
\[\left\{ \begin{matrix} AM^{2} = (x + 1)^{2} + y^{2} \\ BM^{2} = \left( x^{2} \right) + \left( y^{2} \right)\text{\ \ \ \ } \\ CM^{2} = (x - 1)^{2} + y^{2} \\ \end{matrix} \right.\ \]
\[3x^{2} + 3y^{2} = 48\]
\[x^{2} + y^{2} = 16.\]
\[Множество\ точек\ M:\ \]
\[окружность\ с\ центром\ в\ \]
\[точке\ B(0;0)\ и\ R = 4.\]
\[\textbf{б)}\ 1)\ Введем\ систему\ \]
\[координат:\]
\[A( - 1;0);C(1;0);M(x;y);\]
\[B(0;0);\]
\[\left\{ \begin{matrix} AM^{2} = (x + 1)^{2} + y^{2} \\ BM^{2} = \left( x^{2} \right) + \left( y^{2} \right)\text{\ \ \ \ } \\ CM^{2} = (x - 1)^{2} + y^{2} \\ \end{matrix} \right.\ \]
\[6x^{2} - 4x + 6y^{2} = 0\]
\[3x^{2} - 2x + 3y^{2} = 0\]
\[3\left( x^{2} - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9} \right) + 3y^{2} = 0\]
\[3\left( x - \frac{1}{3} \right)^{2} - \frac{1}{3} + 3y^{2} = 0\]
\[\left( x - \frac{1}{3} \right)^{2} + y^{2} = \frac{1}{9}.\]
\[Множество\ точек\ M:\]
\[окружность\ с\ центром\ в\ \]
\[точке\ \left( \frac{1}{3};0 \right)\ и\ R = \frac{1}{3}.\]
\[\boxed{\mathbf{982}\mathbf{.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ \mathbf{задачи:}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\ \]
\[AB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B\]
\[BC \parallel B_{3}C_{3} \parallel B_{2}C_{2} \parallel B_{1}C_{1}\]
\[B_{1}C_{1} = 3,4\ см.\]
\[\mathbf{Найти:}\]
\[B_{2}C_{2};\ \ B_{3}C_{3}.\]
\[\mathbf{Решение.}\]
\[1)\ По\ условию:\]
\[AB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B;\]
\[BC \parallel B_{3}C_{3} \parallel B_{2}C_{2} \parallel B_{1}C_{1}.\]
\[По\ теореме\ Фалеса:\]
\[AC_{1} = C_{1}C_{2} = C_{2}C_{3} = C_{3}\text{C.}\]
\[2)\ Рассмотрим\ \mathrm{\Delta}AB_{2}C_{2}:\ \]
\[B_{1}C_{1} - средняя\ линия:\]
\[AB_{1} = B_{1}B_{2};\ \ \ \]
\[AC_{1} = C_{1}C_{2}.\]
\[Следовательно:\]
\[B_{1}C_{1} = \frac{B_{2}C_{2}}{2}.\]
\[\text{\ \ }B_{2}C_{2} = 2 \bullet 3,4 = 6,8\ см.\]
\[3)\ Рассмотрим\ трапецию\ \]
\[C_{2}B_{2}\text{BC}:\ \]
\[B_{3}C_{3} - средняя\ линия:\]
\[B_{2}B_{3} = B_{3}B;\ \ \ \]
\[C_{2}C_{3} = C_{3}\text{C.}\]
\[C_{2}B_{2} - средняя\ линия\ \mathrm{\Delta}\ ABC:\]
\[BC = 2 \bullet 6,8 = 13,6\ см.\]
\[Следовательно:\]
\[B_{3}C_{3} = \frac{B_{2}C_{2} + BC}{2} =\]
\[= \frac{6,8 + 13,6}{2} = 10,2\ см.\]
\[Ответ:B_{2}C_{2} = 6,8\ см;\ \ \]
\[B_{3}C_{3} = 10,2\ см.\]