부채꼴 속 삼각형 넓이

[출제의도] 삼각비를 이용하여 삼각형의 넓이를 구하는 문제를 해결한다.

점 $\mathrm{F}$에서 선분 $\mathrm{OD}$에 내린 수선의 발을 $\mathrm{H}$라 하자. $\displaystyle \mathrm{\overline{OF}} = \mathit{x}$라 하면 직각삼각형 $\mathrm{OFH}$에서 $\displaystyle \cos 30 ^\circ = \frac{\mathrm{\overline{OH}}}{\mathrm{\overline{OF}}}$이므로 $\displaystyle \frac{\sqrt{3}}{2} = \frac{\mathrm{\overline{OH}}}{x}$ $\displaystyle \mathrm{\overline{OH}} = \frac{\mathit{\sqrt{3}}}{2} x$ 부채꼴 $\mathrm{OAB}$의 반지름의 길이를 $r$라 하면 $r$$\displaystyle = \mathrm{2} {\overline{OH}}$$\displaystyle = \sqrt{3} x$이므로 $r$$\displaystyle = \mathrm{\overline{OF}} + \mathrm{\overline{BF}}$$\displaystyle = x + \frac{2 \sqrt{3}}{3}$$\displaystyle = \sqrt{3} x$ $\displaystyle \left( \sqrt{3} - 1 \right) x = \frac{2 \sqrt{3}}{3}$ $x$$\displaystyle = \frac{2 \sqrt{3}}{3 ( \sqrt{3} - 1 )}$$\displaystyle = \frac{2 \sqrt{3} \left( \sqrt{3} + 1 \right)}{3 \left( \sqrt{3} - 1 \right) \left( \sqrt{3} + 1 \right)}$$\displaystyle = \frac{2 \times 3 + 2 \sqrt{3}}{3 \times 2}$$\displaystyle = \frac{3 + \sqrt{3}}{3}$ $r$$\displaystyle \mathrm{=} \mathit{\sqrt{3}} x$$\displaystyle \mathrm{=} \mathit{\sqrt{3}} \times \frac{3 + \sqrt{3}}{3}$$\displaystyle \mathrm{=} \frac{3 \sqrt{3} + 3}{3}$$\displaystyle = \sqrt{3} + 1$ 점 $\mathrm{E}$에서 선분 $\mathrm{OD}$에 내린 수선의 발을 $\mathrm{I}$라 하고 $\displaystyle \mathrm{\overline{OI}} = \mathit{y}$라 하면 $\mathrm{\angle} EOI$$= \mathrm{\angle} AOB - \mathrm{\angle} AOC - \mathrm{\angle} DOB$ $= \mathrm{120} ^\circ - \mathrm{30} ^\circ - 30 ^\circ$$= \mathrm{60} ^\circ$ 직각삼각형 $\mathrm{EOI}$에서 $\tan \left( \angle \mathrm{EOI} \right)$$= \tan 60 ^\circ$$\displaystyle = \frac{\mathrm{\overline{EI}}}{\mathrm{\overline{OI}}}$$\displaystyle = \frac{\mathrm{\overline{EI}}}{y}$ $\displaystyle \mathrm{\overline{EI}}$$= y \times \mathrm{\tan} 60 ^\circ$$\displaystyle = \sqrt{3} y$ $\displaystyle \mathrm{\overline{OA}} = \mathrm{\overline{OD}}$인 이등변삼각형 $\mathrm{AOD}$에서 $\mathrm{\angle} AOD$$= \mathrm{\angle} AOC + \mathrm{\angle} COD$$= \mathrm{\angle} AOC + \mathrm{\angle} EOI$$= \mathrm{30} ^\circ + \mathrm{60} ^\circ$$= \mathrm{90} ^\circ$ 이므로 삼각형 $\mathrm{AOD}$가 직각삼각형이다. 그러므로 $\mathrm{\angle} EDI = \angle ADO = 45 ^\circ$ $\tan \left( \angle \mathrm{EDI} \right)$$= \tan 45 ^\circ$$\displaystyle = \frac{\mathrm{\overline{EI}}}{\mathrm{\overline{DI}}}$$\displaystyle = \frac{\mathit{\sqrt{3}} y}{\mathrm{\overline{DI}}}$ $\displaystyle \mathrm{\overline{DI}}$$\displaystyle = \sqrt{3} y \times \frac{\mathrm{1}}{\tan 45 ^\circ}$$\displaystyle = \sqrt{3} y$ $\displaystyle \mathrm{\overline{OD}}$$\displaystyle \mathrm{=} {\overline{OI}} + {\overline{DI}}$$\displaystyle = y + \sqrt{3} y$$\displaystyle = \left( \sqrt{3} + 1 \right) y$ $\displaystyle \sqrt{3} + 1 = \left( \sqrt{3} + 1 \right) y$ $y = 1$ 따라서 $\triangle \mathrm{ODE}$$\displaystyle = \frac{1}{2} \times \mathrm{\overline{OD}} \times \overline{EI}$ $\displaystyle = \frac{1}{2} \times r \times \sqrt{3} y$ $\displaystyle = \frac{1}{2} \times \left( \sqrt{3} + 1 \right) \times \sqrt{3}$ $\displaystyle = \frac{3 + \sqrt{3}}{2}$