\[\boxed{\mathbf{775.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[\text{ABCD}A_{1}B_{1}C_{1}D_{1} - куб;\]
\[b - прямая;\]
\[O \in B\]
\[Доказать:\]
\[сумма\ квадратов\ расстояний\ от\]
\[вершин\ куба\ до\ b - const.\]
\[Доказательство.\]
\[1)\ Пусть\ a - сторона\ куба,\ \]
\[O - центр\ куба\text{.\ }\]
\[2)\ Построим\ AA_{2}\bot b\ \ и\ \ CC_{2}\bot b.\]
\[Пусть\ \angle AOA_{2} = \angle C_{1}OC = \varphi_{1};\ \]
\[\angle COA_{2} = \angle A_{1}OC = \varphi_{2};\]
\[\angle A_{1}OA_{2} = \angle COC_{2} = \varphi_{3};\ \ \]
\[\angle A_{2}OD = \varphi_{4}.\]
\[3)\ OA - половина\ диагонали\ \]
\[куба:\ \ \]
\[OA = \frac{a\sqrt{3}}{2};\]
\[AA_{2} = OA \bullet \sin\varphi_{1};\]
\[S = \frac{3}{2}a^{2} \bullet \sum_{1}^{4}{\sin^{2}\varphi_{n}}.\]
\[4)\ b \cap ABCD = A_{3}.\]
\[5)\ Пусть\ OY - AB;\text{\ \ }OX - AD:\]
\[A_{3}A^{2} = x^{2} + y^{2};\ \ \]
\[A_{3}B^{2} = (a - x) + y^{2};\]
\[A_{3}C^{2} = (a - x)^{2} + (a - y)^{2};\ \ \]
\[A_{3}D^{2} = x^{2} + (a - y)^{2}.\]
\[6)\ В\ \mathrm{\Delta}\ OA_{3}A:\]
\[A_{3}O^{2} =\]
\[= \left( x - \frac{a}{2} \right)^{2} + \left( y - \frac{a}{2} \right)^{2} + \frac{a^{2}}{4}.\]
\[7)\ 2OP \bullet A_{3}O \bullet \cos{\angle AOA_{3}} =\]
\[= \frac{3a^{2}}{2} - a(x + y)\]
\[2OP \bullet A_{3}O \bullet \cos{\angle A_{3}\text{OB}} =\]
\[= \frac{a^{2}}{2} + a(x - y)\]
\[2OP \bullet A_{3}O \bullet \cos{\angle A_{3}\text{OC}} =\]
\[= \frac{a^{2}}{2} + a(x + y)\]
\[2OP \bullet A_{3}O \bullet \cos{\angle A_{3}\text{OD}} =\]
\[= \frac{a^{2}}{2} - a(x - y)\]
\[4a^{2}OP^{2} =\]
\[= 4OP^{2} \bullet A_{3}O^{2} \bullet \sum_{1}^{4}{\cos^{2}\varphi_{n}}\]
\[\sum_{1}^{4}{\cos^{2}\varphi_{n}} = \frac{a^{2}}{A_{3}O^{2}} = \frac{a^{2} \bullet 4}{3a^{2}} = \frac{4}{3}\]
\[\sum_{1}^{4}{\sin^{2}\varphi_{n}} = 4 - \sum_{1}^{4}{\cos^{2}\varphi_{n}} =\]
\[= \frac{8}{3}.\]
\[8)\ Сумма\ кважратов\ \]
\[расстояний\ от\ вершин\ куба\ до\ \]
\[прямой:\]
\[s = \frac{3}{2}a^{2} \bullet \frac{8}{3} = 4a^{2}.\]
\[Что\ и\ требовалось\ доказать.\]