Đáp án:
\[c = 2b\]
Giải thích các bước giải:
Đặt \(\overrightarrow {AB} = \overrightarrow x ;\,\,\,\overrightarrow {AC} = \overrightarrow y \Rightarrow \left\{ \begin{array}{l}
\left( {\overrightarrow x ;\overrightarrow y } \right) = 60^\circ \\
\left| {\overrightarrow x } \right| = c;\,\,\,\left| {\overrightarrow y } \right| = b
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\overrightarrow {AM} = \overrightarrow {AB} + \overrightarrow {BM} = \overrightarrow {AB} + \frac{1}{3}\overrightarrow {BC} = \overrightarrow {AB} + \frac{1}{3}.\left( {\overrightarrow {BA} + \overrightarrow {AC} } \right)\\
= \overrightarrow {AB} - \frac{1}{3}\overrightarrow {AB} + \frac{1}{3}\overrightarrow {AC} = \frac{2}{3}\overrightarrow {AB} + \frac{1}{3}\overrightarrow {AC} = \frac{2}{3}\overrightarrow x + \frac{1}{3}\overrightarrow y \\
\overrightarrow {CN} = \overrightarrow {CA} + \overrightarrow {AN} = - \overrightarrow {AC} + \frac{1}{3}\overrightarrow {AB} = \frac{1}{3}\overrightarrow x - \overrightarrow y \\
\overrightarrow x .\overrightarrow y = \left| {\overrightarrow x } \right|.\left| {\overrightarrow y } \right|.\cos \left( {\overrightarrow x ;\overrightarrow y } \right) = b.c.\cos 60^\circ = \frac{1}{2}bc\\
AM \bot CN \Leftrightarrow \overrightarrow {AM} .\overrightarrow {CN} = 0\\
\Leftrightarrow \left( {\frac{2}{3}\overrightarrow x + \frac{1}{3}\overrightarrow y } \right).\left( {\frac{1}{3}\overrightarrow x - \overrightarrow y } \right) = 0\\
\Leftrightarrow \frac{2}{9}{\overrightarrow x ^2} - \frac{5}{9}\overrightarrow x .\overrightarrow y - \frac{1}{3}{\overrightarrow y ^2} = 0\\
\Leftrightarrow \frac{2}{9}{c^2} - \frac{5}{9}.\frac{1}{2}bc - \frac{1}{3}{b^2} = 0\\
\Leftrightarrow 4{c^2} - 5bc - 6{b^2} = 0\\
\Leftrightarrow \left( {c - 2b} \right)\left( {4c + 3b} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
c = 2b\\
4c = - 3b\,\,\,\,\left( L \right)
\end{array} \right.\\
\Leftrightarrow c = 2b
\end{array}\)
