What is the standard deviation of a, b and c?

Very nice problem, congrats Max!

\(? = \sigma \left( {a,b,c} \right)\)

\(\left( 1 \right)\,\,{a^2} + {b^2} + {c^2} = 77\,\,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {a;b;c} \right) = \left( {\sqrt {\frac{{77}}{3}} \,;\sqrt {\frac{{77}}{3}} \,;\sqrt {\frac{{77}}{3}} \,} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{0}}\,\, \hfill \\\\
\,\,{\text{Take}}\,\,\left( {a;b;c} \right) = \left( {\sqrt {77} \,;0\,;0\,} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\text{?}}\,\, \ne \,\,{\text{0}}\,\,\,\, \hfill \\ \\
\end{gathered} \right.\)


\(\left( 2 \right)\,\,a + b + c = 15\,\,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {a;b;c} \right) = \left( {5\,;5\,;5\,} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{0}}\,\, \hfill \\\\
\,\,{\text{Take}}\,\,\left( {a;b;c} \right) = \left( {15\,;0\,;0\,} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\text{?}}\,\, \ne \,\,{\text{0}}\,\,\,\, \hfill \\ \\
\end{gathered} \right.\)


\(\left( {1 + 2} \right)\,\,\,\,\mu = \frac{{a + b + c}}{3} = 5\)

\(? = \sqrt {\frac{{{{\left( {a - \mu } \right)}^2} + {{\left( {b - \mu } \right)}^2} + {{\left( {c - \mu } \right)}^2}}}{3}} \,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{\,\,\,? = {{\left( {a - 5} \right)}^2} + {{\left( {b - 5} \right)}^2} + {{\left( {c - 5} \right)}^2}\,\,\,}\)

\(?\,\,\, = \,\,\,{\left( {a - 5} \right)^2} + {\left( {b - 5} \right)^2} + {\left( {c - 5} \right)^2}\,\,\, = \,\,\,\,\underbrace {{a^2} + {b^2} + {c^2}}_{77} - 10\underbrace {\left( {a + b + c} \right)}_{15} + 3 \cdot 25\,\,\,\,\,{\text{unique}}\)

The correct answer is (C), indeed.


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.