직육면체 곱셈공식

[출제의도] 다항식의 곱셈공식을 활용하여 문제해결하기 $\displaystyle \overline{{\mathrm{AB}}} = x$, $\displaystyle \overline{{\mathrm{AD}}} = y$, $\displaystyle \overline{{\mathrm{AE}}} = z$라 하면 $\displaystyle l _{1} = 3 x + 3 y + 3 z + \overline{{\mathrm{AC}}} + \overline{{\mathrm{CF}}} + \overline{{\mathrm{FA}}}$ $\displaystyle l _{2} = x + y + z + \overline{{\mathrm{AC}}} + \overline{{\mathrm{CF}}} + \overline{{\mathrm{FA}}}$ 이므로 $l _{1} - l _{2} = 2 x + 2 y + 2 z = 28$에서 $x + y + z = 14$ $\displaystyle S _{1} = x y + y z + z x + \frac{1}{2} x y + \frac{1}{2} y z + \frac{1}{2} z x$ $+$ (삼각형 ${\mathrm{AFC}}$의 넓이) $\displaystyle S _{2} = \frac{1}{2} x y + \frac{1}{2} y z + \frac{1}{2} z x + ($삼각형 ${\mathrm{AFC}}$의 넓이$)$ 이므로 $S _{1} - S _{2} = x y + y z + z x = 61$ $\displaystyle {\overline{\mathrm{AC}}} ^{2} + {\overline{\mathrm{CF}}} ^{2} + {\overline{\mathrm{FA}}} ^{2}$$= \left( x ^{2} + y ^{2} \right) + \left( y ^{2} + z ^{2} \right) + \left( z ^{2} + x ^{2} \right)$ $= 2 \left( x ^{2} + y ^{2} + z ^{2} \right)$$= 2 \left\{ ( x + y + z ) ^{2} - 2 ( x y + y z + z x ) \right.$ $= 2 \times \left( 14 ^{2} - 2 \times 61 \right) = 148$