직각삼각형 곱셈공식

$\displaystyle {\overline{\mathrm{MH}}} = a$, $\displaystyle {\overline{\mathrm{NH}}} = b$라 하자. 점 ${\mathrm{M}}$, ${\mathrm{N}}$은 각각 직각삼각형 ${\mathrm{ABH}}$, ${\mathrm{BCH}}$의 외심이므로 $\displaystyle {\overline{\mathrm{MH}}} = {\overline{\mathrm{MA}}} = {\overline{{\mathrm{MB}}}}$, $\displaystyle {\overline{\mathrm{NH}}} = {\overline{\mathrm{NB}}} = {\overline{\mathrm{NC}}}$ $\displaystyle {\overline{\mathrm{AB}}} = 2 \times {\overline{\mathrm{MH}}} = 2 a$ $\displaystyle {\overline{\mathrm{BC}}} = 2 \times {\overline{\mathrm{NH}}} = 2 b$ $\displaystyle {\overline{\mathrm{AC}}} ^{2} = \left( 2 a \right) ^{2} + \left( 2 b \right) ^{2} = 56$, $a ^{2} + b ^{2} = 14$ $\left( \right.$삼각형 ${\mathrm{ABC}}$의 넓이$\displaystyle \left. \right) = \frac{1}{2} \times 2 a \times 2 b = 2 ab = 10$ $ab = 5$ $\left( a + b \right) ^{2} = a ^{2} + b ^{2} + 2 ab = 14 + 10 = 24$, $\displaystyle a + b = 2 \sqrt{6}$ 따라서 $\displaystyle a ^{3} + b ^{3} = \left( a + b \right) ^{3} - 3 ab \left( a + b \right) = 18 \sqrt{6}$