Ars Mathematica: Solución al problema 136

s = \frac{a + b + c}{2}

\begin{matrix} ▵^{2} & = & \frac{a + b + c}{2} (\frac{a + b + c}{2} - a) (\frac{a + b + c}{2} - b) (\frac{a + b + c}{2} - c) = \end{matrix}

\frac{(a + b + c) (b + c - a) (a + c - b) (a + b - c)}{16} \Rightarrow

\Rightarrow 16 ▵^{2} = (a + b + c) (b + c - a) (a + c - b) (a + b - c)

(a + b + c) (b + c - a) (a + c - b) (a + b - c) = - | \begin{matrix} 0 & a & b & c \\ a & 0 & c & b \\ b & c & 0 & a \\ c & b & a & 0 \end{matrix} |

| \begin{matrix} 0 & a & b & c \\ a & 0 & c & b \\ b & c & 0 & a \\ c & b & a & 0 \end{matrix} | \underset{⟶}{F_{4} + F_{3} + F_{2} + F_{1}} | \begin{matrix} 0 & a & b & c \\ a & 0 & c & b \\ b & c & 0 & a \\ a + b + c & a + b + c & a + b + c & a + b + c \end{matrix} | =

= (a + b + c) | \begin{matrix} 0 & a & b & c \\ a & 0 & c & b \\ b & c & 0 & a \\ 1 & 1 & 1 & 1 \end{matrix} | \underset{⟶}{\begin{matrix} C_{4} - C_{1} \\ C_{3} - C_{1} \\ C_{2} - C_{1} \end{matrix}} | \begin{matrix} 0 & a & b & c \\ a & - a & c - a & b - a \\ b & c - b & - b & a - b \\ 1 & 0 & 0 & 0 \end{matrix} | =

= - (a + b + c) | \begin{matrix} a & b & c \\ - a & c - a & b - a \\ c - b & - b & a - b \end{matrix} | \underset{⟶}{C_{1} + C_{2} - C_{3}} =

= - (a + b + c) | \begin{matrix} a + b - c & b & c \\ - a + c - a - b + a & c - a & b - a \\ c - b - b - a + b & - b & a - b \end{matrix} | =

= - (a + b + c) | \begin{matrix} a + b - c & b & c \\ a (a + b - c) & c - a & b - a \\ - (a + b - c) & - b & a - b \end{matrix} | =

- (a + b + c) (a + b - c) | \begin{matrix} 1 & b & c \\ - 1 & c - a & b - a \\ - 1 & - b & a - b \end{matrix} | =

\underset{⟶}{\begin{matrix} F_{2} + F_{1} \\ F_{3} + F_{1} \end{matrix}} - (a + b + c) (a + b - c) | \begin{matrix} 1 & b & c \\ 0 & c - a + b & b - a + c \\ 0 & 0 & a - b + c \end{matrix} | =

= - (a + b + c) (a + b - c) | \begin{matrix} b + c - a & b + c - a \\ 0 & a + c - b \end{matrix} | =

= - (a + b + c) (a + b - c) (b + c - a) (a + c - b)

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