Cho tam giác đều $ABC$ cạnh $a$ nội tiếp đường tròn $\left( {O,R} \right)$. Tìm tập hợp các điểm $M$ sao cho: $\overrightarrow {MB} .\overrightarrow {MC}  + \overrightarrow {MC} .\overrightarrow {MA}  + \overrightarrow {MA} .\overrightarrow {MB}  = 3{a^2}$
Cách 1:    $\overrightarrow {MB} .\overrightarrow {MC}  + \overrightarrow {MC} .\overrightarrow {MA}  + \overrightarrow {MA} .\overrightarrow {MB}  = $
        $\begin{array}{l}
 = \frac{1}{2}\left( {M{B^2} + M{C^2} - B{C^2}} \right) + \frac{1}{2}\left( {M{C^2} + M{A^2} - C{A^2}} \right) + \frac{1}{2}\left( {M{A^2} + M{B^2} - A{B^2}} \right)\\
 = \left( {M{A^2} + M{B^2} + M{C^2}} \right) - \frac{1}{2}\left( {B{C^2} + C{A^2} + A{B^2}} \right)\\
 = 3M{O^2} + \left( {O{A^2} + O{B^2} + O{C^2}} \right) - \frac{3}{2}{a^2}\\
 = 3M{O^2} + 3{R^2} - \frac{3}{2}{a^2} = 3M{O^2} - \frac{{{a^2}}}{2}    (R = \frac{a}{\sqrt 3} )
\end{array}$
Do đó $\overrightarrow {MB} .\overrightarrow {MC}  + \overrightarrow {MC} .\overrightarrow {MA}  + \overrightarrow {MA} .\overrightarrow {MB}  = 3{a^2}$
    $ \Leftrightarrow 3O{M^2} = 3{a^2} + \frac{{{a^2}}}{2} \Leftrightarrow O{M^2} = \frac{{7{a^2}}}{6} \Leftrightarrow OM = a\sqrt {\frac{7}{6}} $
    $ \Leftrightarrow M  $ thuộc đường tròn $\left( {O;a\sqrt {\frac{7}{6}} } \right)$

Cách 2:     $\overrightarrow {MB} .\overrightarrow {MC}  + \overrightarrow {MC} .\overrightarrow {MA}  + \overrightarrow {MA} .\overrightarrow {MB} $
    $\begin{array}{l}
 = \frac{1}{2}\left[ {{{\left( {\overrightarrow {MA}  + \overrightarrow {MB}  + \overrightarrow {MC} } \right)}^2} - \left( {M{A^2} + M{B^2} + M{C^2}} \right)} \right]\\
 = \frac{1}{2}\left[ {{{\left( {3\overrightarrow {MO} } \right)}^2} - \left( {3M{O^2} + O{A^2} + O{B^2} + O{C^2}} \right)} \right]\\
 = \frac{1}{2}\left( {6M{O^2} - 3{R^2}} \right)\\
 = \frac{1}{2}\left( {6M{O^2} - {a^2}} \right)
\end{array}$
Đến đây, ta biến đổi tiếp tục như cách $1$.

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